How

Computers

Work

Processor and

Main Memory

Roger Young

© Copyright 2001, Roger Stephen Young

All rights reserved.

No part of this book may be reproduced, stored in a retrieval system, or transmitted by any means, electronic, mechanical, photocopying, recording, or otherwise, without written permission from the author.

1

FREE

An identical Internet version of this book is available for free (for personal use and, possibly, for a limited time) at http://howcomputers.com

and/or at

http://www.geocities.com/thinkorrr/howcomputerswork/p1.html I’m at

thinkorrr@yahoo.com

and I’ll (possibly for a limited time) send you a free Microsoft Word 2000 .doc file version (for personal use) as an attachment to an email if you email me and want one.

March 20, 2002

2

Introduction

Computers are the most complex machines that have ever been created. Very few people really know how they work. This book will tell you how they work and no technical knowledge is required.

It explains the operation of a simple, but fully functional, computer in complete detail.

The simple computer described consists mainly of a processor and main memory. Relays, which are explained, are used in the circuitry instead of transistors for simplicity. This book does not cover peripherals like modems, mice, disk drives, or monitors.

Did you ever wonder what a bit, a pixel, a latch, a word (of memory), a data bus, an address bus, a memory, a register, a processor, a timing diagram, a clock (of a processor), an instruction, or machine code is? Though most explanations of how computers work are a lot of analogies or require a background in electrical engineering, this book will tell you precisely what each of them is and how each of them works without requiring any previous knowledge of computers or electronics. However, this book starts out very easy and gets harder as it goes along. You must read the book starting at the first page and not skip around because later topics depend on understanding earlier topics. How far you can get may depend on your background. A junior high school science background should be enough. There is no mathematics required other than simple addition and multiplication.

This is a short book, but it must be studied carefully. This means that you will have to read some parts more than once to understand them. Get as far as you can. You will be much more knowledgeable about how computers work when you are done than when you started, even if you are not able to get through the whole text. This is a technical book though it is aimed at a non-technical audience. Though this book takes considerable effort to understand, it is very easy for what it explains. After you have studied this book, if you go back and read it, it will seem simple. Good Luck!

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4

CONTENTS

BASICS …………………………………………… 7

MEMORY ………………………………………. 43

INSTRUCTIONS ……………………………….. 81

PROCESSOR ………………………………….. 101

PROGRAMMING …………………………….. 132

MISCELLANEOUS ……………………………156

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6

BASICS

7

Simple Circuit

The picture above shows a ‘battery’ connected to a ‘light bulb’ by a ‘power wire’ and a

‘ground wire.’ A power wire is a wire connected directly to the top of the battery. A ground wire is a wire connected directly to the bottom of the battery. Any electrical machine is called a circuit.

8

Simple Diagram

The diagram above also shows a ‘battery’ connected to a ‘light bulb’ by a ‘power wire’

and a ‘ground wire.’ This diagram means the same as the picture on page 2. The ground wire is not shown because it is assumed that one connection of every light is always connected to the bottom of the battery by a ground wire in diagrams. Diagrams are simpler to draw than pictures that mean the same thing.

9

Key Circuit

The picture above shows the ‘top of’ a ‘battery’ connected by a ‘power wire’ to a ‘key’

that is connected by a ‘light wire’ to a ‘light bulb.’

A key is a flat piece of springy steel that is bent up so that the key only touches the wire to the key’s right when the key is pressed down by someone’s finger.

When someone pushes the key down, the right end of the key touches the light wire and electricity flows from the top of the battery, through the power wire, the key, and the light wire, to the light bulb, turning the light bulb on.

When the key is released, the key springs back up. Now the key does not touch the light wire and electricity can not get from the key to the light wire to the light bulb so that the light bulb goes off.

10

Key Diagram

The diagram above shows the same circuit as the preceding picture.

Again, there is also a wire from the other connection of the light bulb back to the bottom of the battery, but that wire does not need to be shown because the other connection of every light is connected to the bottom of the battery and you know the ground wire is there without drawing it.

11

Electromagnet

The picture above shows the top of a battery connected by a wire to an electromagnet.

An electromagnet is a coil of (plastic coated) wire. An electromagnet becomes magnetic when electricity goes through it, just as a light bulb glows when electricity goes through the light bulb.

The wire that makes up the coil of wire that is the electromagnet has two ends (connections).

There is also a ‘ground wire’ from the other connection of the electromagnet back to the bottom of the battery.

12

Electromagnet Diagram

The diagram above shows the same circuit as the preceding picture.

The wire that makes up the coil of wire that is the electromagnet has two ends (connections).

There is also a ground wire from the other connection of the electromagnet back to the bottom of the battery, as in the picture, but that wire does not need to be shown because the other connection of every electromagnet is connected to the bottom of the battery.

13

Relay

The picture above shows a ‘bottom key’ that controls an electromagnet.

The electromagnet, in turn, controls the top key. A key and the electromagnet that controls it are, together, called a relay. The relay is in the dashed box.

When the bottom key is pressed, the electromagnet is powered and the electromagnet becomes magnetic. That makes the electromagnet attract the top key and pull the top key down just like a finger can push a key down. A magnet (or a powered electromagnet) attracts the top key because the top key is made of steel. A magnet (or a powered electromagnet) does not attract the wires because the wires are made of copper.

Important: The electromagnet does not ever touch the top key. No electricity can go from the electromagnet to the wires attached to the top key.

14

A computer is almost entirely made up of a lot of relays (today, transistors) connected by wires. Just how the relays are connected and just what they do is the main subject of this book. Other concepts, especially programming, will also be explained.

(Today, transistors are used instead of relays for lower cost and greater speed. The design remains practically the same, however. Relays are easier to understand and, so, will be used in this explanation.)

Relay Diagram

The diagram above shows the same circuit as the previous picture in a different way.

15

One Battery and Touching Wires

In this picture, only one battery powers all the circuitry in the previous picture. Note the symbol for wires that touch.

16

One Battery and Connected Wires Diagram

This diagram shows the same circuit as the previous picture in a different way. Touching wires are connected wires.

17

Loop

Loop Diagram

18

The picture and diagram at left show a relay that controls its own electromagnet! The square of wire that takes electricity from the key of the relay to the electromagnet of the same relay is called a ‘loop.’

No electricity can get from the top of the battery to the electromagnet because the key is up. However, if someone presses the key, then electricity can get to the electromagnet.

Then, the electromagnet will hold the key down - even if the person lets go of the key! So we say that the loop remembers that the key was pressed. Remember that the key normally springs up because it is springy and bent upward.

Similarly, if someone then lifts up the key (A person is much stronger than a little electromagnet.), then no electricity will reach the electromagnet and the key will remain up even after the person releases the key. So we say that the loop remembers that the key was lifted up.

Most relays in a computer are used to make loops, or connect the loops together.

19

Pixel

Pixel Diagram

20

The picture and diagram above show a loop that controls a light bulb. A light bulb that is controlled by a loop is called a ‘pixel.’

In a diagram, where a horizontal wire and a vertical wire meet, without crossing, there is a connection of the two wires.

Therefore, when the key is pressed, electricity can flow from the top of the battery, through the key, to both the light and the electromagnet. When the key is down and the light bulb is glowing, one says that the loop has value ‘1’ and the pixel is ‘on.’ The loop has value ‘1’ even if there is not a light bulb, just so the loop wire has electricity going through it, to the electromagnet, because the key is down.

When the key is up and the light bulb is not glowing, one says that the loop has value ‘0’

and the pixel is ‘off.’ The loop has value ‘0’ even if there is not a light bulb - just so the loop wire does not have electricity going through it (because the key is up).

21

Normally Closed Key

Normally Closed Key Diagram

22

The picture and diagram at left show the top of a battery connected by a wire to a normally closed key, that is connected by another wire to a light bulb.

A diagram of an electrical machine is called a circuit diagram, a diagram, a schematic (pronounced ske-ma’-tic) diagram, or just a schematic.

The normally closed key is different from the keys described previously. The normally closed key is also a springy piece of steel, but is bent so that it normally is connected to the right wire. Therefore, the light bulb in the circuit above is normally on. However, if you push down on the normally closed key, the light bulb becomes disconnected from the

‘ power wire’ and the light goes out.

A key is called ‘closed’ when electricity can flow through it from a wire on its left to a wire on its right.

A key is called ‘open’ when electricity can’t flow through it from a wire on the left to a wire on the right.

A normally closed key is normally closed, but is open when you push it down.

A normally open key is normally open, but is closed when you push it down.

A relay is called closed if its key is closed.

A relay is called open if its key is open.

An electromagnet is called ‘powered’ if the electromagnet is connected to the top of a battery, even if that electromagnet is connected to the top of the battery through a series of closed keys. In fact, any piece of wire is called ‘powered’ if that piece of wire is connected to the top of a battery, even if that piece of wire is connected to the top of the battery through a series of closed keys.

Any piece of wire that is powered is said to have value ‘1.’

Any piece of wire that is not powered is said to have value ‘0.’

The values of the wire in a loop as described previously are a special case of these rules for assigning values to wires.

23

Normally Closed Relay

24

Normally Closed Relay Diagram

The preceding picture and diagram show a bottom key that controls an electromagnet.

The electromagnet, in turn, controls the top, normally closed key. A normally closed key and the electromagnet that controls it are, together, called a normally closed relay.

When the bottom key is pressed, the electromagnet is powered and the electromagnet becomes magnetic. That makes the electromagnet attract the top, normally closed key and pull the top, normally closed key down, just like a finger can push a normally closed key down. A magnet (or a powered electromagnet) attracts the normally closed key because the normally closed key is made of steel. When the bottom key is pressed, the light turns off.

In other words, when the bottom key is pressed, the electromagnet energizes, disconnecting the top key.

25

Clear Key

Clear Key Diagram

The picture and diagram above show a loop as before, but a normally closed key has been added. As long as the normally closed key is closed, the loop works as before.

However, if the normally closed key is pressed, then the normally closed key will be open and electricity will not reach the electromagnet, so the electromagnet will not be magnetic, and the normally open key will pop up if it was down. If the normally open key already was up, it will stay up.

Therefore, pressing the normally closed key will clear the value of the loop to ‘0.’

Therefore, this normally closed key is called the ‘clear key’ for the loop.

26

Loop to Loop Data Transfer

In the circuit above, the ‘connecting key’ connects loop A and loop B. Both loops have value 0. Temporarily pressing ‘loop key A’ gives the value 1 to loop A. Now, temporarily pressing the ‘connecting key’ will make loop B have value 1. That is because when loop A has value 1, loop key A is closed, loop wire A has value 1, and when the connecting key is closed, electricity can reach the electromagnet of loop B, giving loop B value 1.

However, if loop A has value 0, and loop B has value 0, and the connecting key is pressed, then both loops keep their values of 0.

Therefore, if one temporarily presses ‘clear key B’ to clear loop B to value 0, and then temporarily presses the connecting key, whatever value is in loop A will be copied to loop B. Then loop A and loop B will have the same value.

27

Oscillator

Oscillator Diagram

28

The picture and diagram at left show a normally closed relay that controls its own electromagnet. The square of wire that takes electricity from the normally closed key of the relay to the electromagnet of the same normally closed relay is called a feedback wire. (Notice that this circuit is different from a loop circuit, which uses a normally open relay.) This circuit is called an oscillator because the relay oscillates (changes back and forth) between open and closed.

Electricity can get from the top of the battery, through the closed, normally closed relay key to the electromagnet. The electromagnet then pulls the normally closed key down and opens the normally closed key. Because the normally closed key is now open, no electricity can get to the electromagnet. The electromagnet now no longer attracts the normally closed key and the normally closed key closes.

Thus, the normally closed key repeatedly opens and closes without anyone touching the key. The feedback wire gets value 1, then value 0, then value 1, etc. It takes a relay about a hundredth of a second to change values.

Just as a normal loop is the basis of a computer memory, this feedback circuit is a key part of a computer’s clock. A computer’s clock is a circuit that repeatedly generates signals (1 and 0 values).

29

Keys in Series

Keys in Series Diagram

In the picture and diagram above, one must press both ‘key D’ AND ‘key E’ to turn the light on.

30

AND Gate Circuit

In the circuit above, the three triangles are all the top of the same battery. When ‘key D’

AND ‘key E’ close, then the light comes on. When ‘key A’ is pressed, then ‘key D’

closes. When ‘key B’ is pressed, then ‘key E’ closes. Therefore, when ‘key A’ and ‘key B’ are pressed, the light turns on. Another way of describing the operation of the circuit is to say that ‘output wire C’ gets value 1 only when ‘input wire A’ gets value 1 AND

‘input wire B’ gets value 1.

The following table also shows that ‘output wire C’ has value 1 only when both ‘input wire A’ has value 1 AND ‘input wire B’ has value 1.

AND gate truth table

A

B

C

0

0

0

0

1

0

1

0

0

1

1

1

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AND Gate Circuit with Symbol

The diagram above shows a circuit with the symbol for an ‘AND gate’ which is shown, alone, below.

AND Gate Symbol

The light in the circuit below only comes on whey key D, key E, AND key F are all pressed.

Three Keys in Series

32

Keys in Parallel

Keys in Parallel Diagram

In the picture and diagram above, one need only press either ‘key D’ OR ‘key E’ (or both) to turn the light on.

33

OR Gate Circuit

In the circuit above, as always, the crossing wires do not touch and are not connected to each other. When ‘key D’ OR ‘key E’ (or both) closes, the light comes on. When ‘key A’ is pressed, then ‘key D’ closes. When ‘key B’ is pressed, then ‘key E’ closes.

Therefore, when ‘key A’ OR ‘key B’ is pressed, the light turns on. Another way of describing the operation of this circuit is to say that ‘output wire C’ gets value 1 only when ‘input wire A’ has value 1 OR ‘input wire B’ has value 1.

The following table also shows that ‘output wire C’ gets value 1 only when either ‘input value A’ has value 1 OR ‘input wire B’ has value 1.

OR gate truth table

A

B

C

0

0

0

0

1

1

1

0

1

1

1

1

34

OR Gate Circuit with Symbol

The diagram above shows a circuit with the symbol for an ‘OR gate’ which is shown alone, below.

OR Gate Symbol

35

Three Keys in Parallel

The light in the circuit above turns on when key D, key E, OR key F is pressed.

36

Normally Closed Key

Normally Closed Key Diagram

In the picture and diagram above, the light is on, as we have seen before. One must press the normally closed key D down to turn the light off.

37

NOT Gate Circuit

In the circuit above, the triangles are both the top of the same battery. When ‘key A’ is pressed, ‘key D’ is pulled down and the light goes off. That is, when ‘key A’ is pressed, normally closed ‘key D’ opens. Therefore, when ‘key A’ is pressed, the light goes off.

Another way of describing the operation of the circuit is to say that ‘output wire C’ gets value 0 when ‘input wire A’ gets value 1. ‘Output wire C’ gets value 1 when ‘input wire A’ gets value 0.

The following table also shows that ‘output wire C’ gets value 0 only when ‘input wire A’ gets value 1.

NOT gate truth table

A

C

0

1

1

0

38

NOT Gate Circuit with Symbol

The diagram above shows a circuit with the symbol for a ‘NOT gate’ which is shown alone, below.

NOT Gate Symbol

39

Interconnected Gates

The diagram above shows that the output of an AND gate can be the input for a NOT

gate. The circuit above can also be represented with gate symbols as below.

Interconnected Gates with Symbols

40

A ‘NAND gate’ can be constructed from an AND gate followed by a NOT gate as indicated below.

Constructed NAND Gate

A NAND gate can be represented by the single symbol in the circuit below.

NAND Gate Circuit

41

A lone NAND gate is pictured below.

NAND Gate

The truth table for the NAND gate is shown below.

NAND gate truth table

A

B

C

0

0

1

0

1

1

1

0

1

1

1

0

42

MEMORY

43

(Address) Decoder

44

The diagram at left shows a ‘decoder.’ A and B are the inputs to the decoder and I, J, K, and L are the outputs. The truth table for this circuit is shown below.

A B I J K L

0 0 1 0 0 0

0 1 0 1 0 0

1 0 0 0 1 0

1 1 0 0 0 1

Normally closed relay AA is closed.

Normally closed relay AB is also closed.

Therefore, electricity can travel from the top of the battery, through AA and AB, to light I.

If keys A and B are both pressed, then normally open relays DA and DB are closed (because their electromagnets are powered) and electricity can reach light L.

Similarly, if key A is pressed and key B is not pressed, then normally open relay CA is closed and normally closed relay CB is closed and light K is on.

Finally, if key A is not pressed and key B is pressed, then light J is on.

Wire PO is power. A1 and A0 are address wire 1 and address wire 0. PO has value 1.

A1 can have value 1 or 0, and A0 can have value 1 or 0.

45

Truth Table Generator

46

In the preceding circuit, keys A and B are the inputs and lights G and H are the outputs.

The truth table for the circuit above is shown below.

A B G H

0 0 1 0

0 1 0 1

1 0 0 0

1 1 1 1

For example, if neither A nor B is pressed, then S00 is powered (has value 1) because the normally closed relays AA and AB are then closed. BB is open so S01 is 0, CA is open so S10 is 0, and both DA and DB are open so S11 is 0. Because S00 is powered, AG is closed and electricity can go from the top of the battery (indicated by a triangle), through relay AG, to wire D1 to light G, so G is on. Relay AI is also closed but relay AI’s key is not connected to the top of the battery so no electricity gets to light H.

For another example, if both keys A and B are pressed, then A=1 and B=1 and relays DA and DB are closed. That makes S11=1 and closes relays DG and DI. Electricity can go from the top of the battery through DG and D1 to light G and through DI and D2 to light H. Therefore, A=1 and B=1 results in G=1 and H=1 as in the truth table.

D1 and D0 are data wire 1 and data wire 0. D1 can have value 1 or 0 and D0 can be 1 or 0.

47

ROM (Read-Only Memory) With Enable (EN) Key (D)

48

The circuit above has the following truth table:

EN A1 A0 D1 D0

0

0

0

0

0

0

0

1

0

0

0

1

0

0

0

0

1

1

0

0

1

0

0

1

0

1

0

1

0

1

1

1

0

0

0

1

1

1

1

1

If key D (EN) is not pressed (‘EN’ stands for ‘enable.’), then EN is 0, so no electricity gets to the electromagnets of AG and AI. Similarly, BG, BI, CG, CI, DG, and DI are open if D (EN) is not pressed. Therefore, if D (EN) is not pressed, then no electricity can get to lights G and H as indicated in the truth table.

If A and B are not pressed (A1=0 and A0=0), then electricity gets to the electromagnet of AE and closes relay AE. If D is then pressed (EN=1), then electricity can go from the top of the battery, through D and through AE to the electromagnets of AG and AI. AG and AI then close and electricity can go from the top of the battery, through AG, to wire D1

and light G.

The truth table above can also be represented as below.

EN A1 A0 D1 D0

0

X

X

0

0

1

0

0

1

0

1

0

1

0

1

1

1

0

0

0

1

1

1

1

1

The X’s mean 0 or 1. That is, the row with X’s means that if EN is 0, then D1=0 and D0=0 no matter what values A1 and A0 have.

49

Loops Added

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In the circuit above, eight loops have been added to the previous circuit. The loops are labeled AF, AH, BF, BH, CF, CH, DF, and DH. Each loop can have value 0 or 1. The truth table fore this circuit is shown below.

EN A1 A0 D1 D0

0

X

X

0

0

1

0

0

AF AH

1

0

1

BF BH

1

1

0

CF CH

1

1

1

DF DH

To make loop AF have value 1, just press key AF down. Key AF will stay down because it is part of a loop. To make AF have value 0 again, just lift key AF up. It will stay up on its own. In the truth table, ‘AF’ means the value of loop AF. The other loops, AH, BF, BH, CF, CH, DF, and DH, operate similarly.

51

Input Keys Added

52

In the diagram at left, key E and key F (bottom right) have been added to the circuit.

Keys E and F allow one to set a loop to value 1 without touching the loop’s key.

For example, to set loop AF to 1 without touching key AF, one must not push key A or key B, which closes relay AE. Then you hold down key E to put value 1 on wire D1.

Finally, temporarily pushing key D makes EN temporarily 1. Because AE is closed, EN=1 powers relay AG’s electromagnet and closes AG. D1’s value of 1 can now go through key AG to loop AF, thereby making loop AF have value 1.

53

Memory (Clear Key Added)

54

In this circuit, key C, wire CL (for CLear), relays AC, BC, CC, and DC, normally closed relays AD, BD, CD, and DD, and wires H00, H01, H10 and H11 (H for Hold, or remember) have been added to the circuit. These additions allow loops to be ‘cleared’ to value 0 by manipulating keys outside the dashed box (memory) without touching the loop keys.

The diagram above shows a memory within the dashed box. The memory can be controlled by the keys outside the dashed box at the bottom of the diagram. What a memory does will be explained first. Then, how the memory works will be described.

AF, AH, BF, BH, CF, CH, DF, and DH are each relay keys of loops. You can change the value of loop AF from 0 to 1 by simply pressing key AF down. Similarly, you can change the value of loop AF from 1 to 0 by lifting key AF. To determine whether a loop has value 0 or value 1, just look at the loop’s key. If the key is down, then the loop has value 1. If the key is up, then the loop has value 0. The value of a loop stays the same until you change it.

However, suppose that the dashed box was a physical box and you could not reach inside the box. If you buy a memory chip at a store, the circuitry is enclosed in a plastic box with wires PO, A1, A0, CL (This may be called WR for ‘WRite.’), EN, D1, D0 and GRound wire, GR, sticking out. The circuitry uses transistors instead of relays for switches, so even if you broke the box open, you couldn’t change the values by hand. (A memory from a store would probably have more address lines (wires) like A2, A3, ...

A20 and data lines like D2, D3, ... D7.)

The memory is constructed so that the values in the loops can be examined and changed using only keys A, B, C, D, E, and F and light bulbs G and H which are all outside the box and are not part of the memory.

55

Where Power Reaches in a Memory

56

The bold wires in the diagram at left, show which wires are powered.

A wire is powered only if it is connected to the top of the battery (represented by a triangle in the lower left corner of the diagram, as shown below).

power

Notice the new symbol used for keys AC and AE. Keys AC and AE are normally open key s.

However, they are closed now because their electromagnets are powered.

Therefore, they are represented as:

closed, normally open key

instead of as:

open, normally open key

Electricity can flow from left to right (or right to left) through a closed key even if it’s a closed but normally open key.

57

Similarly, an open, normally closed key is represented as:

open, normally closed key

instead of as:

closed, normally closed key

58

Notice that, in the diagram of memory, all of the loops (AF, AH, BF, BH, CF, CH, DF, and DH) have value 0 because all of those normally open keys are open.

AF is ‘bit 1’ of ‘latch 00’ and has value 0. AH is bit 0 of latch 00 and also has value 0.

You should follow the power from the top of the battery (the triangle in the lower left of the diagram of memory above) and see why certain wires are bold and the rest are normal. Remember, electricity can’t go through open keys. Electricity also does not travel between crossing wires. Crossing wires are not touching (not connected). You should also understand why some electromagnets are powered and others aren’t, and how powering the electromagnet of a key closes a normally open key and, in later diagrams, opens a normally closed key.

59

Now, suppose we want to store value 01 in latch 10. This means we want to keep key CF

open for value 0 and close key CH for value 1. This is called ‘ writing’ value 01 to address 10.

To do this, you first select latch 10 by pressing key A and not pressing key B. This selects latch 10 as indicated by the bold select 10 wire, ‘S10,’ in the diagram below. Key A controls address wire 1, labeled A1 in the diagram, and key B controls address wire 0, labeled A0 in the diagram. Both address wires, A1 and A0, together, are called the address bus. A group of similar wires are, together, called a ‘bus.’ Pressing key A and not pressing key B results in power going through the circuit as indicated by bold lines in the diagram below. Notice that horizontal wire S10 has power (is bold) while S00, S01, and S11 do not have power. This selects latch 10.

60

Selecting the Address

61

The second step in writing value 01 to address (latch) 10 is to press key F and not press key E as in the following diagram. Not pressing key E gives value 0 to data wire D1 and pressing key F gives value 1 to data wire D0. Both data wires, D1 and D0, are, together, called the ‘data bus’ just as both address wires, A1 and A0, are, together, called the

‘address bus.’ The first and second steps can be done simultaneously. This results in power going through the circuit as indicated by the bold wires in the diagram below.

Wire D0 is bold and, so, has value 1.

62

Selecting the Data to be Written

63

The third step in writing 01 to address 10 is pressing the enable key, ‘D,’ which controls the enable (‘EN’) wire. This results in power going through the circuit as indicated in bold in the following diagram. Notice that loop CH now has value 1. Loop CH got power from wire D0 through CI. No power went from wire D1 through CG to loop CF

because wire D1 is not powered.

It’s important to remember that pressing the enable key, ‘D,’ makes EN=1 and connects the loops of the selected (by the address wires A1 and A0) latch to the data bus wires, D1

and D0.

64

Pressing Enable (EN)

65

In the fourth step, key D is released and the enable (EN) wire returns to value 0

(unpowered). This results in power flowing through the memory as indicated by bold wires in the following diagram. Notice that loop CH still has value 1 even though loop CH is no longer connected to data wire D0 through relay CI (because relay CI is open).

66

Releasing Enable (EN)

67

Step five: Keys A and F are released and address wire A0 and data wire D0 get value 0

(as indicated in the following diagram). Notice that loop CH still has value 1.

Therefore, to write value 01 to latch 10, you press A and not B to select latch 10; and, to choose data 01, do not push E and push F. This makes wire A1 have value 1, wire A0

have value 0, wire D1 have value 0, and wire D0 have value 1. Then, while holding A and F down, temporarily press D to make the enable wire, EN, temporarily 1. Then, release A and F. This can be described as follows.

1. Select the address and data with the address and data keys A, B, E, and F (A1, A0, D1

and D0).

2. Temporarily press D (EN).

3. Release the address and data keys.

That’s all there is to storing data in an empty latch.

68

Releasing the Address and Data Keys

69

To find out what is in a latch, do the following.

1. Select the address of the latch you want to read with keys A (wire A1) and B (wire A0).

2. Press key ‘D’ to make the ‘EN’ wire have value 1. The lights G and H will indicate the values of the data bits stored in that latch.

3. Release the enable key D.

4. Release the address keys, A and B.

For example, to read latch 10, first press key A (and not key B) to select latch 10 (as indicated in the following diagram).

70

Selecting the Address to Read

71

Second, press key D to make the enable (EN) wire have value 1. Then light H comes on indicating that bit 0 of latch 10 has value 1 and light G stays off indicating that latch 10

has value 0 in bit 1. This is shown in the following diagram.

Notice that making wire EN have value 1 connects the loops of the selected latch to the data wires D1 and D0.

72

Enabling (EN) the Output

73

Third, release key D when done reading latch 10 (as indicated in the following diagram).

That’s all there is to reading a latch in memory.

74

Releasing Enable (EN)

75

To erase a value from a latch and make all of the latch’s loops have value 0, do the following.

1. Select the latch with keys A (A1) and B (A0).

2. Temporarily press key C to make the ‘clear’ (CL) wire temporarily have value 1.

The following diagram shows latch 10 selected by pressing A and not pressing key B. It also shows key C being pressed to clear both of latch 10’s data bits to 0. Don’t press C

until after A is pressed (so that no other latch is accidentally erased).

Notice that pressing C makes the selected ‘H’ (for ‘Hold’) wire, ‘H10,’ have value 0.

76

Selecting the Address and Clearing (with CL)

77

Release key CL after clearing latch 10 as indicated in the following diagram. Don’t release key A until after CL is released so that you don’t accidentally erase another latch.

Writing to a latch will not clear any bits that were previously 1, so always clear a latch before copying (writing) data to it. Therefore, to write to a latch, do the following.

1. Press the correct address keys (A and B) and data keys (E and F).

2. Press the clear key, C, to clear the latch.

3. Release the clear key, C.

4. Press the enable key, D, to send data from the data wires (D1 and D0) to the latch.

5. Release the enable key, D.

6. Release the address keys (A and B) and the data keys (E and F).

To read data, just do the following.

1. Press the correct address keys, A and B, to select the latch to read.

2. Press the enable key, D, to send the latch’s values to the lights, G and H.

3. Release the enable key D.

4. Release the address keys, A and B.

78

Releasing CL

79

The memory in the drawings is very small. There are only two address wires and two data wires. Because there are two address wires, there are four possible addresses: 00, 01, 10, and 11, and, so, four latches. Because there are two data wires, each latch has two loops. Each loop holds one ‘bit’ of information, a 0 or a 1. Four latches with two loops each means 8 (= 4 x 2) loops total.

The table below shows all bit values in each latch when data 01 has been written to the loops of latch 10.

latch

bit

address

values

00

00

01

00

10

01

11

00

A larger memory with four address bits and four data bits with 0 in all the loops can be represented as below.

latch bit

address values

0000 0000

0001 0000

0010 0000

0011 0000

0100 0000

0101 0000

0110 0000

0111 0000

1000 0000

1001 0000

1010 0000

1011 0000

1100 0000

1101 0000

1110 0000

1111 0000

80

INSTRUCTIONS

The next most important part of a computer, after memory, is the processor. A processor changes the values in memory as instructed by instructions stored in memory. An instruction is a group of bits (loop values) in memory that tell the processor to do something. A group of instructions that instruct the processor to do some task is called a program.

The simple kind of processor described in this book has only one type of instruction, but that instruction is sufficient to do anything, as will be seen. The instruction is ‘copy’ (and

‘go to’). Each instruction simply copies some bits of data from somewhere in one latch in memory to somewhere else in another latch in memory. The instruction indicates: 1. which latch to copy data (data is bit (loop) values) from 2. which latch to copy data to

3. which bits to change in the copied-to latch

4. how much to rotate the ‘from’ data before copying some of its bits to the ‘to’ data latch

5. which latches to get the next instruction from.

In the computer considered here, the number of address bits is the same as the number of data bits.

81

The explanation of the following example will not be clear at first, but just read through it. Then reread it. It will be clear later.

Consider a four-address-bit, four-data-bit memory with the values (in loops) below: Example Program

latch bit

address values

0000 0100 address of instruction

0001 0001

value of a

001 0

0001

not 0 = 1

001 1

0000

not 1 = 0

instruction_1

0100 0001 from address

0101 1000 to address

0110 0001 ‘to’ bits to change

0111 1000

instr.addr.and rot.amount

instruction_2

1000 0010 from address

1001 0001 to address

1010 0001 ‘to’ bits to change

1011 1100

instr.addr.and rot.amount

instruction_3

1100 0000 from address

1101 0000 to address

1110 0000 ‘to’ bits to change

1111 1100

instr.addr.and rot.amount

Only the 1's and 0's are part of the program. The rest is just comments for a person. The latch addresses are just where the program is stored . The bit values are the program.

‘ Instr.addr.and rot.amount’ is short for ‘next instruction’s address and rotate amount.’

Latch 0000 holds the value 0100 so that the first instruction is in latches 0100, 0101, 0110, and 0111, and is labeled ‘instruction_1’ in the program. Latch 0000 is special and always holds the address of the next instruction to be executed.

The first word of instruction_1 is in latch 0100 and is 0001. That means that data (bit values) is copied from latch 0001 in memory.

The second word of instuction_1 is in latch 0101 and is 1000 and indicates that the data will be copied to latch 1000.

The third word of instruction_1 is in latch 0110 and is 0001 and indicates that only the rightmost bit, and not the three leftmost bits, of latch 1000 will be changed because only the rightmost bit of 0001 is 1.

The rightmost two bits of latch 0111 are 00 and indicate that the data in latch 0001 will not be rotated at all when data is copied to latch 1000.

The leftmost two bits of latch 0111 are 10 and indicate that the next instruction will be in latches 1000, 1001, 1010, and 1011. That is, instruction_2 will be executed next.

82

After instruction_1 is executed, the memory has the following bit values.

latch bit

address values

0000 1000 <-----|

address of instruction

0001 000 1 ---| |

value of a

0010 0001

| |

not 0 = 1

0011 0000

| |

not 1 = 0

instruction_1

0100 0001

| |

from address

0101 1000

| |

to address

0110 0001

| |

‘to’ bits to change

0111 1000 ---+--|

instr.addr.and rot.amount

instruction_2

1000 0011 <--|

from address

1001 0001

to address

1010 0001

‘to’ bits to change

1011 1100

instr.addr.and rot.amount

instruction_3

1100 0000

from address

1101 0000

to address

1110 0000

‘to’ bits to change

1111 1100

instr.addr.and rot.amount

The underlined loop values (bits) were copied to when instruction_1 was executed. The italics show from where data was copied. The arrows show how data was copied. The rightmost bit of latch 0001 has been copied to the rightmost bit of latch 1000. Also, all bits of latch 0111 (that is, 1000) have been copied to latch 0000 indicating that the next instruction will be in latches 1000, 1001, 1010, and 1011 (instruction_2). That is, after the first instruction, instruction_1, is executed, latch 0000 has value 1000.

The leftmost two bits of latch 0000 are 10, so the instruction executed next is instruction_2, in latches 1000, 1001, 1010, and 1011.

1. Latch 1000 holds 0011, so data is copied from latch 0011.

2. Latch 1001 holds 0001, so data is copied to latch 0001.

3. Latch 1010 holds 0001, so only the rightmost bit of the ‘to latch,’ latch 0001, is copied to.

4. Latch 1011 holds 1100. The rightmost two bits of 1100 are 00 so the data copied from is not rotated at all. The leftmost two bits of 1100 are 11, so the next instruction to be executed will be in latches 1100, 1101, 1110, and 1111 (instruction_3).

83

After instruction_2 is executed, the memory has the following bit values.

latch bit

address values

0000 1100 <-----|

address of instruction

0001 0000 <--| |

value of a

0010 0001

| |

not 0 = 1

0011 000 0 ---| |

not 1 = 0

instruction_1

0100 0001

|

from address

0101 1000

|

to address

0110 0001

|

‘to’ bits to change

0111 1000 |

instr.addr.and rot.amount

instruction_2

1000 0011

|

from address

1001 0001

|

to address

1010 0001

|

‘to’ bits to change

1011 1100 ------|

instr.addr.and rot.amount

instruction_3

1100 0000

from address

1101 0000

to address

1110 0000

‘to’ bits to change

1111 1100

instr.addr.and rot.amount

The italics show from where data was copied. The underlining shows to where data was copied. The arrows show how data was copied.

Latch 0000 now has value 1100, so that the next instruction to be executed is instruction_3 in latches 1100, 1101, 1110, and 1111.

1. Latch 1100 holds 0000, so data will be copied from latch 0000.

2. Latch 1101 holds 0000, so data is copied to latch 0000.

3. Latch 1110 holds 0000, so no data bits are copied (to latch 0000).

4. Latch 1111 holds 1100, so 1100 is copied to latch 0000.

84

This results in the following bit values in memory.

latch bit

address values

0000 1100 <-----|

address of instruction

0001 0000

|

value of a

0010 0001

|

not 0 = 1

0011 0000

|

not 1 = 0

instruction_1

0100 0001

|

from address

0101 1000

|

to address

0110 0001

|

‘to’ bits to change

0111 1000

|

instr.addr.and rot.amount

instruction_2

1000 0011

|

from address

1001 0001

|

to address

1010 0001

|

‘to’ bits to change

1011 1100

|

instr.addr.and rot.amount

instruction_3

1100 0000

|

from address

1101 0000

|

to address

1110 0000

|

‘to’ bits to change

1111 1100 ------|

instr.addr.and rot.amount

The underlined bits have been copied to from the italic bits.

Thus, instruction_3 changes nothing (because latch 0000 already held 1100) and leads to instruction_3 being executed again and again.

Ending a program with an instruction like instruction_3 ensures that nothing else will happen after the desired instructions (instruction_1 and instruction_2) are executed. It’s just something for the computer to do until we stop the processor and look in memory for the results.

85

We will now look at some two-instruction programs. The first instruction will do something and the second instruction will do nothing. These short computer programs will show what an instruction (of this simple computer) can do.

Instruction_1 of the following program copies 1111 from latch 0001 to latch 0010.

Notice that, because latch 0110 of instrucion_1 holds 1111, all ‘to data’ bits are copied to.

Instruction_2 does nothing over and over.

Before Copy 1111 to 0010 for 1111

latch bit

address values

0000 0100 address of instruction

0001 1111

from data

0010 0000

to data

0011 0000

instruction_1

0100 0001 from address

0101 0010 to address

0110 1111

‘to’ bits to copy to

0111

1000

instr.addr.and rot.amount

instruction_2

1000 0000

1001 0000

1010 0000

1011 1000

1100 0000

1101 0000

1110 0000

1111 0000

86

After instruction_1 is executed, the memory has the following values.

After Copy 1111 to 0010 for 1111

latch bit

address values

0000 1000 <------| address of instruction

0001 1111 ---|

| from data

0010 1111 <--|

| to data

0011 0000

|

instruction_1

0100 0001

| from address

0101 0010

| to address

0110 1111

| ‘to’ bits to copy to

0111 1000 -------| instr.addr.and rot.amount

instruction_2

1000 0000

1001 0000

1010 0000

1011 1000

1100 0000

1101 0000

1110 0000

1111 0000

87

Instruction_1 of the following program copies 0011 from latch 0001 to latch 0010.

Notice that, because latch 0110 of instruction_1 holds 1111, all ‘to’ bits are copied to.

Before Copy 0011 to 0010 for 0011

latch bit

address values

0000 0100 address of instruction

0001 0011

from data

0010 0000

to data

0011 0000

instruction_1

0100 0001 from address

0101 0010 to address

0110 1111

‘to’ bits to copy to

0111 1000

instr.addr.and rot.amount

instruction_2

1000 0000

1001 0000

1010 0000

1011 1000

1100 0000

1101 0000

1110 0000

1111 0000

88

After instruction_1 is executed, the memory has the following values.

After Copy 0011 to 0010 for 0011

latch bit

address values

0000 1000 <------| address of instruction

0001 0011 ---|

| from data

0010 0011 <--|

| to data

0011 0000

|

instruction_1

0100 0001

| from address

0101 0010

| to address

0110 1111

| ‘to’ bits to copy to

0111 1000 -------| instr.addr.and rot.amount

instruction_2

1000 0000

1001 0000

1010 0000

1011 1000

1100 0000

1101 0000

1110 0000

1111 0000

89

Instruction_1 of the following program copies the rightmost three bits (111) of 1111 from latch 0001 to latch 0010 for 0111. Notice that, because latch 0110 of instruction_1 holds 0111, the rightmost three ‘to’ bits are copied to.

Before Copy 111 to 0010 for 0111

latch bit

address values

0000 0100 address of instruction

0001 1111

from data

0010 0000

to data

0011 0000

instruction_1

0100 0001 from address

0101 0010 to address

0110 0111

‘to’ bits to copy to

0111 1000

instr.addr.and rot.amount

instruction_2

1000 0000

1001 0000

1010 0000

1011 1000

1100 0000

1101 0000

1110 0000

1111 0000

90

After instruction_1 is executed, the memory has the following values.

After Copy 111 to 0010 for 0111

latch bit

address values

0000 1000 <------| address of instruction

0001 1111 ---|

| from data

0010 0111 <--|

| to data

0011 0000

|

instruction_1

0100 0001

| from address

0101 0010

| to address

0110 0111

| ‘to’ bits to copy to

0111 1000 -------| instr.addr.and rot.amount

instruction_2

1000 0000

1001 0000

1010 0000

1011 1000

1100 0000

1101 0000

1110 0000

1111 0000

91

Instruction_1 of the following program copies the rightmost three bits (000) of 0000 from latch 0001 to latch 0010 for 1000. Notice that, because latch 0110 of instruction_1 holds 0111, the rightmost three ‘to’ bits are copied to.

Before Copy 000 to 0010 for 1000

latch bit

address values

0000 0100 address of instruction

0001 0000

from data

0010 1111

to data

0011 0000

instruction_1

0100 0001 from address

0101 0010 to address

0110 0111

‘to’ bits to copy to

0111 1000

instr.addr.and rot.amount

instruction_2

1000 0000

1001 0000

1010 0000

1011 1000

1100 0000

1101 0000

1110 0000

1111 0000

92

After instruction_1 is executed, the memory has the following values.

After Copy 000 to 0010 for 1000

latch bit

address values

0000 1000 <------| address of instruction

0001 0000 ---|

| from data

0010 1000 <--|

| to data

0011 0000

|

instruction_1

0100 0001

| from address

0101 0010

| to address

0110 0111

| ‘to’ bits to copy to

0111 1000 -------| instr.addr.and rot.amount

instruction_2

1000 0000

1001 0000

1010 0000

1011 1000

1100 0000

1101 0000

1110 0000

1111 0000

93

Instruction_1 of the following program rotates the bits (0010) of latch 0001 one space to the left (for 0100) and copies all four rotated bits to latch 0010. Notice that, because latch 0110 of instruction_1 holds 1111, all four bits are copied to. Also notice that, because latch 0111 has 01 in the rightmost two bits, the from data is rotated one bit to the left.

Before Rotate 0010 One Bit Left for 0100

latch bit

address values

0000 0100 address of instruction

0001 0010

from data

0010 0000

to data

0011 0000

instruction_1

0100 0001 from address

0101 0010 to address

0110 1111

‘to’ bits to copy to

0111 1001

instr.addr.and rot.amount

instruction_2

1000 0000

1001 0000

1010 0000

1011 1000

1100 0000

1101 0000

1110 0000

1111 0000

94

After instruction_1 is executed, the memory has the following values. Latch 0000 now holds 1001. The right two bits in latch 0000 do not affect what instruction is executed next. The left two bits of 1001 (in latch 0000) are 10, so the next instruction to be executed will be instruction_2, in latches 1000, 1001, 1010, and 1011.

After Rotate 0010 One Bit Left for 0100

latch bit

address values

0000 1001 <------| address of instruction

0001 0010 ---|

| from data

0010 0100 <--|

| to data

0011 0000

|

instruction_1

0100 0001

| from address

0101 0010

| to address

0110 1111

| ‘to’ bits to copy to

0111 1001 -------| instr.addr.and rot.amount

instruction_2

1000 0000

1001 0000

1010 0000

1011 1000

1100 0000

1101 0000

1110 0000

1111 0000

95

Instruction_1 of the following program rotates the bits (0010) of latch 0001 two spaces to the left (for 1000) and copies all four rotated bits to latch 0010. Notice that, because latch 0110 of instruction_1 holds 1111, all four bits are copied to. Also notice that, because latch 0111 has 10 in the rightmost two bits, the from data is rotated two bits to the left.

Before Rotate 0010 Two Bits Left for 1000

latch bit

address values

0000 0100 address of instruction

0001 0010

from data

0010 0000

to data

0011 0000

instruction_1

0100 0001 from address

0101 0010 to address

0110 1111

‘to’ bits to copy to

0111 1010

instr.addr.and rot.amount

instruction_2

1000 0000

1001 0000

1010 0000

1011 1000

1100 0000

1101 0000

1110 0000

1111 0000

96

After instruction_1 is executed, the memory has the following values. Again, the right two bits in latch 0000 do not affect what instruction is executed next. The left two bits of 1010 (in latch 0000) are 10, so the next instruction to be executed will be instruction_2, in latches 1000, 1001, 1010, and 1011

After Rotate 0010 Two Bits Left for 1000

latch bit

address values

0000 1010 <------| address of instruction

0001 0010 ---|

| from data

0010 1000 <--|

| to data

0011 0000

|

instruction_1

0100 0001

| from address

0101 0010

| to address

0110 1111

| ‘to’ bits to copy to

0111 1010 -------| instr.addr.and rot.amount

instruction_2

1000 0000

1001 0000

1010 0000

1011 1000

1100 0000

1101 0000

1110 0000

1111 0000

97

Instruction_1 of the following program rotates the bits (0010) of latch 0001 three spaces to the left (for 0001) and copies all four rotated bits to latch 0010. Notice that, because latch 0110 of instruction_1 holds 1111, all four bits are copied to. Also notice that, because latch 0111 has 11 in the rightmost two bits, the from data is rotated three bits to the left. Notice also that rotating three bits to the left is the same as rotating one bit to the right.

Before Rotate 0010 Three Bits Left for 0001

latch bit

address values

0000 0100 address of instruction

0001 0010

from data

0010 0000

to data

0011 0000

instruction_1

0100 0001 from address

0101 0010 to address

0110 1111

‘to’ bits to copy to

0111 1011

instr.addr.and rot.amount

instruction_2

1000 0000

1001 0000

1010 0000

1011 1000

1100 0000

1101 0000

1110 0000

1111 0000

98

After instruction_1 is executed, the memory has the following values.

After Rotate 0010 Three Bits Left for 0001

latch bit

address values

0000 1011 <------| address of instruction

0001 0010 ---|

| from data

0010 0001 <--|

| to data

0011 0000

|

instruction_1

0100 0001

| from address

0101 0010

| to address

0110 1111

| ‘to’ bits to copy to

0111 1011 -------| instr.addr.and rot.amount

instruction_2

1000 0000

1001 0000

1010 0000

1011 1000

1100 0000

1101 0000

1110 0000

1111 0000

99

The right two bits of the last word (four bits here) of an instruction indicate how many bits to rotate to the left according to the following table: rotate

bit

left

values amount

00

0

01

1

10

2

11

3

If the rightmost bit value is 1, then there is 1 bit of rotation left. If the left bit value is 1, then there is an additional two bits of rotation left.

The following table shows how rotation works with the four bits of a word labeled ‘A,’

‘B,’ ‘C,’ and ‘D.’

rotate

rotate

rotate

bit

four

left

right

values

bits

amount

amount

00

ABCD

0

0

01

BCDA

1

3

10

CDAB

2

2

11

DABC

3

1

Notice again that rotating 3 bits left is the same as rotating 1 bit right. Similarly, 1 bit left is 3 bits right and 2 bits left is 2 bits right.

100

PROCESSOR

Four-Bit Memory

The circuit above shows a memory with four data wires (D3, D2, D1, and D0) and four address wires (A3, A2, A1, and A0). Because there are four address wires, there are sixteen possible latch addresses: 0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, and 1111. Only two latches, 0000 and 1111, are shown. The rest are implied by the gap in the circuit diagram.

101

Two Memories Connected

102

The circuit diagram above shows a memory with four address wires on the bottom connected with a memory with three address wires on the top. Room has been left in the top memory for additional circuitry later. The two memories share data wires D3, D2, D1, and D0. The three-address-wire memory has address wires RA2, RA1, and RA0, clear wire CLR, and enable wire ENR. In the top memory, the latches are called registers and the address wires are called RA2 for Register Address 2, etc. CLR stands for CLear Register. ENR stands for ENable Register.

Because both memories share data wires D3, D2, D1, and D0, data can be copied from a latch of the bottom memory to a register of the top memory or from a register to a latch.

To copy data from a latch to a register, first select the register with register address keys RA2, RA1, and RA0. Second, temporarily press the CLR key to clear the register loops to all 0’s. Third, select the latch address with address keys A3, A2, A1, and A0 (while continuing to select the register with RA2, RA1, and RA0). Fourth, temporarily press the enable keys, ENR and EN, to connect the selected register loops and the selected latch loops to the data wires D3, D2, D1, and D0.

Similarly, to copy data from a register to a latch, first select the latch with address keys A3, A2, A1, and A0. Second, temporarily press the clear key, CL ( not CLR), to clear the latch. Third, select the register with register address keys RA2, RA1, and RA0. Fourth, temporarily press the enable keys, ENR and EN. This connects the register and latch loops to the data bus wires, D3, D2, D1, and D0, and, thereby, to each other.

103

Loops Controlling Lights

The circuit above shows four loops controlling four lights.

104

Rotate 1 Circuitry

In the circuit above, if the ‘rotate 1’ key is not pressed, then loop3 controls light 3, loop 2

controls light 2, loop 1 controls light 1, and loop 0 controls light 0.

However, if the ‘rotate 1’ key is pressed, then loop 3 controls light 0, loop 2 controls light 3, loop 1 controls light 2, and loop 0 controls light 1. One can say that when the ‘rotate 1’ key is pressed, then the loop values are rotated one bit to the left. There is no bit to the left of bit 3, so bit 3 is rotated to the right end to bit 0.

The following table indicates what pressing the ‘rotate 1’ key does.

Rotate 1 Light Values

Rotate left

3 2 1 0 amount in bits

0

A B C D

0

1

B C D A

1

105

Rotate Two Bits Circuitry

In the circuit above, if the ‘rotate 2’ key is not pressed, then loop 3 controls light 3, loop 2

controls light 2, loop 1 controls light 1, and loop 0 controls light 0.

However, if the ‘rotate 2’ key is pressed, then loop 3 controls light 1, loop 2 controls light 0, loop 1 controls light 3, and loop 0 controls light 2. One can say that when the ‘rotate 2’ key is pressed, then the loop values are rotated two bits to the left.

The following table indicates what pressing the ‘rotate 2’ key does.

Rotate 2 Light Values

Rotate left

3 2 1 0 amount in bits

0

A B C D

0

1

C D A B

2

106

Rotate Circuitry

In the circuit above, if neither the ‘rotate 1’ key nor the ‘rotate 2’ key is pressed, then loop 3 controls light 3, loop 2 controls light 2, loop 1 controls light 1, and loop 0 controls light 0. If rotate 1 is pressed and rotate 2 is not pressed, then the loop signals are rotated 1 bit to the left on the way to the lights. If rotate 2 is pressed and rotate 1 is not pressed, then the loop values are rotated 2 bits to the left on the way to the lights. Finally, if both the rotate 1 key and the rotate 2 key are pressed, then the loop values are rotated three bits to the left. For example, the value in loop 0 is routed to light 3.

The following table indicates what pressing one or both ‘rotate’ keys does.

Rotate 2 Rotate 1

Light values

Rotate left

3 2 1 0 amount in bits

0

0

A B C D

0

0

1

B C D A

1

1

0

C D A B

2

1

1

D A B C

3

107

Mask Circuitry

In the circuit above, if loop C has value 0, then light B gets the value in loop D. If loop C

has value 1, then light B gets the value in loop A.

108

Mask Four Bits

In the left circuit above, C3 controls whether the value of A3 or the value of D3 goes to B3. The other circuits behave similarly.

109

Rotate and Mask

In the circuit above, there is a ‘rotate circuit’ and a ‘mask circuit.’

As an example of the operation of this circuit, consider the case of ‘rotate 1’ being pressed, ‘rotate 2’ not pressed, C3 = 1, C2 = 1, C1 = 0, and C0 = 0. Then B3 gets A2, B2

gets A1, B1 gets D1, and B0 gets D0. Try to follow the signals in the circuit and see why.

This will be the logic unit of our simple processor. (The logic units of most processors do arithmetic too and so are called arithmetic logic units or, abbreviating, ALU’s.) 110

Delay Circuitry

In the circuit above, when key A is pressed, electromagnet B is powered and key B

closes. It takes time for B to close after A is pressed. That is, light B comes on about one hundredth of a second after light A. This is indicated by the following ‘timing diagram’

that shows when the lights come on. Time 0 is 0 seconds. Time 1 is one hundredth of a second (later).

Timing Diagram for Lights

111

Two Delays

In the circuit above, after key A is pressed it takes one hundredth of a second for key B to close. After key B closes, it takes one hundredth of a second for key C to close.

Therefore, after key A is pressed, it takes two hundredths of a second for light C to come on. In the following timing diagram for the circuit above, time 0 is 0 seconds, time 1 is one hundredth of a second, and time 2 is two hundredths of a second.

Timing Diagram for Lights

112

Delay Line

In the circuit (called a delay line) above, light B comes on ten hundredths of a second after light A. Ten hundredths of a second is one tenth of a second, so light B comes on one tenth of a second after key A is pressed (as indicated in the diagram below). (The small amount of time between the time key A is pressed and the time light A comes on is ignored.) Time 0 is 0 seconds and time 1 is one tenth of a second in the diagram below.

(When a key closes, it can bounce open and closed a few times. This possible problem will be ignored, except to say that using normally closed relays in a delay line might reduce the problem. This problem does not exist in a delay line made with transistors.) Timing Diagram for Delay Line

113

Tapped Delay Line

The circuit above is called a tapped delay line. Wires A, B, C, D, and E are called taps.

Light B comes on one tenth of a second after light A, light C comes on two tenths of a second after light A, light D comes on three tenths of a second after light A, and light E

comes on four tenths of a second after light A (as indicated in the timing diagram below).

114

Timing Diagram for Tapped Delay Line

115

Timing Circuit

Timing Circuit’s Timing Diagram

116

The timing diagram at left corresponds to the circuit above. Lights A, B, C, D, and E

come on, in order, as before. However, the behaviors of lights F, G, H, and I are more complex.

When light B comes on at time 1, relay J closes. Then electricity can go from the top of the battery (a triangle in the circuit diagram above), through closed relay J and normally closed relay K, to light F. Therefore, when light B comes on, light F also comes on.

However, when light D comes on at time 3, normally closed relay K opens and light F

goes out. That is, at time 1, F comes on and, at time 3, F goes out as indicated in the timing diagram.

Similarly, light G comes on when light B comes on, and light G goes out when light C

comes on. Similarly, light H turns on when light D comes on, and light H goes off when light E comes on.

The behavior of light I is more complex. At time 1, light B comes on and relay P closes.

Electricity can then go through keys P and Q to light I. At time 2, light C turns on and normally closed relay Q opens, turning light I off. Therefore, light I turns on at time 1, and goes off at time 2. At time 3, light D comes on, relay R closes, and electricity goes from the top of the battery, through key R and the normally closed S key, to light I. At time 4, light E turns on and normally closed relay S opens and light I goes off.

Therefore, light I turns on at time 1, off at time 2, on at time 3, and off at time 4.

With Processor Power (PP) Loop

The circuit above is the same as the previous circuit except that all but one connection to power is replaced by a connection to loop ‘PP.’ (‘PP’ stands for Processor Power.) After key ‘PP’ is pressed, key PP stays down and power goes to the circuit. Then, when key J

is pressed and held down, output signals F, G, H, and I are generated as indicated in the timing diagram, above. Notice how the right-hand side of the circuit above looks somewhat like the right-hand part of the timing diagram above.

117

With Feedback Through Normally Closed Key K

In the circuit above, key J in the upper left has been replaced by the normally closed relay K in the lower center of the circuit. The circuit above generates the timing diagram below when loop key PP in the lower left is pressed at time 0. Loop PP stays down after it is pressed.

When PP is first pressed, electricity can flow from PP, through normally closed relay K

to light A and to the electromagnet of relay L in the upper left of the diagram. Relay L

then powers relay M. As the relays turn on, one after another, lights B, C, D, and then E

turn on. When light E turns on, the normally closed relay K opens, light A goes out, and relay L opens. One hundredth of a second after relay L opens, relay M opens because electricity is no longer getting to the electromagnet of relay M. The relays in the delay line then open one after another and lights B, C, D, and E go off one after another. When light E goes off, no power gets to the electromagnet of normally closed relay K and relay K closes. When relay K closes, electricity can get to light A and then lights B, C, D, and E turn on.

Thus, A, B, C, D, and E turn on one after another. Then A, B, C, D, and E go off one after another. Then A, B, C, D, and E turn on one after another. Then A, B, C, D, and E

go off one after another. This pattern repeats as long as loop key PP stays down.

Light F is on only when light B is on and light D is off. Similarly, light G is on only when light B is on and light C is off. Also similarly, light H is on only when light D is on and light E is off.

When light B is on and light C is off, relays N and O are closed and light I is on.

Similarly, when light D is on and light E is off, relays P and Q are closed and light I is on.

Therefore, light I is only on when light B is on and light C is off and when light D is on and light E is off.

The circuit above is called a clock. It generates signals F, G, H, and I over and over again as indicated in the timing diagram below.

118

Timing Diagram with Feedback

119

The circuit below shows two latches of memory at the bottom, latch 0000 and latch 1111.

The other fourteen latches are not shown. It also shows a processor above the memory.

The processor is mainly made of latches. Latches in a processor are called registers.

Register 001 is not a latch, however, because it doesn’t have loops.

Writing to a latch will not clear any bits that were previously 1, so always clear a latch before writing data to it. Therefore, to write to a latch, do the following.

1. Press the correct address keys (A3, A2, A1, and A0) and data keys (D3, D2, D1, and D0).

2. Press the clear key, CL, to clear the latch.

3. Release the clear key, CL.

4. Press the enable key, EN, to copy data to the latch.

5. Release the enable key, EN.

6. Release the address keys, A3, A2, A1 and A0, and the data keys, D3, D2, D1 and D0.

To read data from a latch, do the following.

1. Press the correct address keys, A3, A2, A1 and A0, to select the latch to read.

2. Press the enable key, EN, to send the latch’s values to the lights, D3, D2, D1, and D0.

3. Release the enable key EN.

4. Release the address keys, A3, A2, A1 and A0.

Notice that reading a latch connects the loops of the latch to the data bus wires, D3, D2, D1, and D0.

Similarly, writing to a register will not clear any bits that were previously 1, so always clear a register before writing data to it. Therefore, to write to a register (except ‘register’

001, which has no loops to write to), do the following.

1. Press the correct register address keys (RA2, RA1, and RA0) and data keys (D3, D2, D1, and D0).

2. Press the clear register key, CLR, to clear the register.

3. Release the clear register key, CLR.

4. Press the enable register key, ENR, to copy data to the register.

5. Release the enable register key, ENR.

6. Release the register address keys RA2, RA1 and RA0, and the data keys, D3, D2, D1

and D0.

120

To read register data, just do the following.

1. Press the correct register address keys, RA2, RA1 and RA0, to select the register to read.

2. Press the enable register key, ENR, to copy the register’s values to the lights D3, D2, D1, and D0.

3. Release the enable register key, ENR.

4. Release the register address keys, RA2, RA1 and RA0.

Notice that reading a register connects the loops of the register to the data bus wires, D3, D2, D1, and D0.

Do the following to copy data from a latch to a register (except for register 001).

1. Select the register with RA2, RA1, and RA0.

2. Temporarily press CLR to clear the register.

3. Select the latch in memory with A3, A2, A1, and A0.

4. Temporarily press ENR and EN to connect the register loops and latch loops to the data bus wires, D3, D2, D1, and D0, and so to each other.

5. Release all keys.

Do the following to copy data from a register to a latch.

1. Select the latch with A3, A2, A1, and A0.

2. Temporarily press CL to clear the latch.

3. Select the register with RA2, RA1, and RA0.

4. Temporarily press ENR and EN to connect the register loops and latch loops to the data bus wires, D3, D2, D1, and D0, and so to each other.

5. Release all keys.

121

122

An instruction is executed in nine steps. Step 1 copies the address of the instruction from latch 0000 of the memory to the processor. Steps 2, 3, 4, and 5 copy the four words of the instruction to the processor. Step 6 copies the ‘from data’ to the processor. Step 7

copies the ‘to data’ to the processor. Step 8 copies the result from the processor to the memory. Step 9 copies the address of the next instruction to be executed from the processor to latch 0000 of the memory.

The timing diagram that follows, as well as the explanation that follows, tells the order in which to press the keys to execute an instruction with the circuit in the diagram above.

Look at the timing diagram below and the circuit above as you read about each step.

The following nine steps are tedious, but try to get through them or, at least, study step 1

and read through the rest.

1. The first step in executing an instruction is to copy the value in latch 0000 to register 111. Latch 0000 holds the address of the instruction to be executed. To copy the contents of latch 0000 to register 111, the following is done.

First, RA2 is pressed (gets set to 1), RA1 gets 1, and RA0 gets 1. This selects register 111, the instruction address register. SAF, SAT, and SAI are each set to 0 (not pressed), which selects latch 0000. (This will become clear later.) Then, CLR (CLear Register) is temporarily pressed to clear register 111. Then ENM (ENable Memory) and ENR

(ENable Register) are temporarily pressed. When ENM and ENR are pressed, the loops of both latch 0000 and register 111 are connected to the data bus. Therefore, the values in latch 0000 can flow to the loops of (just cleared) register 111 causing register 111 to have the same values as latch 0000. Register 111 then (also) holds the address of the instruction to be executed next.

2. The second step in executing an instruction is to copy the first word (four bits) of the instruction to register 101, the ‘from address register,’ because the first four bits (word) of an instruction are the address from which the data will be copied. RA0 and RA2 are pressed and RA1 is released to select register 101. Key SAI, for Select Address of Instruction, is pressed, routing the left two bits of register 111 to the address wires, A3

and A2. Pressing key SAI also routes the values of SA1 (Select Address bit 1) and SA0

(Select Address bit 0) to A1 and A0 of the memory. SA1 and SA0 are not pressed, so A1

and A0 get 0. Next, CLR is temporarily pressed, thereby clearing register 101. Next, ENR and ENM are temporarily pressed, connecting to data bus register 101 and the first latch (whose address ends in 00) of the instruction to be executed. Thus the first word of the instruction is copied to latch 101.

3. Third, the second word of the instruction is copied to ‘to address register 110.’ This is done by pressing RA2, pressing RA1, and not pressing RA0. Also SAI is pressed, SA1

(Select Address bit 1) is not pressed, and SA0 is pressed. Then CLR is temporarily pressed to clear register 110. Then ENM and ENR are temporarily pressed to copy the contents of the second word (4 bits) of the instruction to register 110.

123

4. Fourth, the third word of the instruction is copied to ‘mask register 010.’ This is done by not pressing RA2, pressing RA1, and not pressing RA0. Also SAI (Select Address of Instruction) is pressed, SA1 is pressed, and SA0 is not pressed. Then CLR is temporarily pressed to clear register 010. Then ENM and ENR are temporarily pressed to copy the contents of the third word (4 bits) of the instruction to register 010.

5. Fifth, the fourth word of the instruction is copied to ‘next/rotate register 100.’ This is done by pressing RA2, not pressing RA1, and not pressing RA0. Also SAI is pressed, SA1 is pressed, and SA0 is pressed. Then CLR is temporarily pressed to clear register 100. Then ENM and ENR are temporarily pressed to copy the contents of the fourth word (4 bits) of the instruction to register 100.

6. Sixth, RA2, RA1, and RA0 are not pressed to select register 000, the ‘from data register.’ SAF (Select Address of From data) is pressed to route the address in the ‘from address register 101’ to the memory’s address wires A3, A2, A1, and A0. CLR is then temporarily pressed to clear register 000. Next, ENM and ENR are temporarily pressed to copy the contents of memory pointed to by ‘from address register 101’ to ‘from data register 000.’

7. Seventh, RA2 is not pressed, RA1 is pressed, and RA0 is pressed to select ‘to data register 011.’ SAT (Select Address of To data) is pressed to route the address in ‘to address register 110’ to the memory’s address wires A3, A2, A1, and A0. CLR is then temporarily pressed to clear register 011. Next, ENM and ENR are temporarily pressed to copy the contents of memory pointed to by ‘to address register 110’ to ‘to data register 011.’

8. Eighth, RA2 is not pressed, RA1 is not pressed, and RA0 is pressed to select ‘back data register 001.’ SAT is pressed to route the address in ‘to address register 110’ to the memory’s address wires, A3, A2, A1, and A0. Next, CLM (CLear Memory, not CLR, CLear Register) is temporarily pressed to clear the latch in memory pointed to by ‘to address register 110.’ ENR and ENM are then temporarily pressed to copy some rotated bits of ‘from data register 000’ and not rotated bits of ‘to data register 011’ to the address in memory pointed to by ‘to address register 110.’ If the rightmost bit of ‘next/rotate register 100’ is 1, then the from data is rotated 1 bit left. If the second-to-rightmost bit of

‘next/rotate register 100’ is 1, then the from data is rotated an additional 2 bits left. If a bit of ‘mask register 010’ is 0, then the corresponding bit of ‘to data register 011’ is copied back to memory. However, if a bit of ‘mask register 010’ is 1, then the corresponding rotated bit of ‘from data register 000’ is copied back to memory. Notice that because CLM was pressed instead of CLR, a latch of memory was cleared and copied to instead of a register.

124

9. Ninth, RA2 is pressed, RA1 is not pressed, and RA0 is not pressed to select

‘next/rotate register 100.’ SAI, SAF, and SAT are not pressed, so no address goes to the memory, so latch 0000 is selected. Next, CLM ( not CLR) is temporarily pressed to clear latch 0000 in memory. ENR and ENM are then temporarily pressed to copy the data in

‘next/rotate register 100’ to latch 0000 in memory. This prepares for the next instruction.

Notice, again, that because CLM was pressed instead of CLR, a latch of memory was cleared and copied to instead of a register.

125

Timing Diagram for Instruction

126

Clock Circuit for Processor

127

The preceding clock circuit shows a circuit that repeatedly generates the timing diagram signals. The outputs of this circuit can be connected to the processor to make the processor repeatedly execute instructions as indicated in the diagram of the complete (though simple) computer in the diagram below.

The operation of a clock has already been explained.

It takes about 361 hundredths of a second for the timing diagram to be generated. Then all outputs of the right hand side are 0 for another about 361 hundredths of a second.

Then the timing diagram is generated again, etc. Therefore, it takes this computer about 722 hundredths of a second to execute each instruction! This is one main reason that transistors are now used. Transistors are millions of times faster. The other reason is that a relay costs as much as millions of interconnected transistors. However, a transistor-based computer works in the same way as a relay-based computer. The cheapness of transistors allows much more memory. It also allows extra things to be added to the processor like more registers and extra circuits to do certain common things, like multiply two numbers together, more quickly.

128

The whole computer is illustrated on the following two pages. The clock, processor and memory are shown and interconnected. The processor includes the rotate and mask circuitry.

To use the computer below, first enter the program and data into the memory with the keys at the bottom of the circuit: A3, A2, A1, A0, CL, EN, D3, D2, D1, and D0. Then, press PP at the bottom of the circuit to make the computer run. Wait until the program is finished and lift up key PP. Then use the keys at the bottom of the circuit to read the results from memory.

129

130

131

PROGRAMMING

We will now consider how to program a 16-bit rather than 4-bit computer.

A 16-bit computer has 16 bits in each ‘word’ and 65536 words of memory. This is because

there

are

65536

possible

16-bit

addresses:

0000000000000000,

0000000000000001, 0000000000000010, 0000000000000011, 0000000000000100, etc.

The instruction still consists of four words, but now each instruction is 16 bits long. An example instruction is:

label

address

data

comment

instr_1 0000000000000100 0000000000000001 from address 0000000000000101 0000000000000010 to address

0000000000000110 0000000011111111 ‘to’ bits to copy to 0000000000000111 000000000010 0000

instr.addr.and rot.amount

The labels and comments are not part of the program. The addresses just show where the data is stored. The data is the program. An instruction written with instructions that are just 1’s and 0’s is a machine language instruction. Machine language instructions are called ‘machine code.’

‘Instr.addr.and rot.amount’ is short for ‘next instruction’s address and rotate amount.’

Notice that there are now four ( italic) rotate bits ( 0000). They cause the copied-from data to be rotated according to the following table.

132

16-Bit Rotate Table

rotate

rotate

rotate

bit

16 left

right

values

bits

amount

amount

0000

ABCDEFGHIJKLMNOP

0

0

0001

BCDEFGHIJKLMNOPA

1

15

0010

CDEFGHIJKLMNOPAB

2

14

0011

DEFGHIJKLMNOPABC

3

13

0100

EFGHIJKLMNOPABCD

4

12

0101

FGHIJKLMNOPABCDE

5

11

0110

GHIJKLMNOPABCDEF

6

10

0111

HIJKLMNOPABCDEFG

7

9

1000

IJKLMNOPABCDEFGH

8

8

1001

JKLMNOPABCDEFGHI

9

7

1010

KLMNOPABCDEFGHIJ

10

6

1011

LMNOPABCDEFGHIJK

11

5

1100

MNOPABCDEFGHIJKL

12

4

1101

NOPABCDEFGHIJKLM

13

3

1110

OPABCDEFGHIJKLMN

14

2

1111

PABCDEFGHIJKLMNO

15

1

The bits to the left of the rotate bits in latch 0000000000000111 are 000000000010 and indicate that the next instruction will be in latches 0000000000001000, 0000000000001001, 0000000000001010, and 0000000000001011.

Latch 0000000000000100 holds 0000000000000001, so data is copied from latch 0000000000000001.

Latch 0000000000000101 holds 0000000000000010, so data is copied to latch 0000000000000010.

Latch 0000000000000110 holds 0000000011111111, so the rightmost 8 bits of the ‘to’

latch are copied to.

133

Next, we will consider how to make a program that adds 1 to any number between 0 and 9.

First we need a way to represent the numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 with only the bit values 0 and 1. The following table shows how we will do it.

Table 1

number representation

0 0000

1 0001

2 0010

3 0011

4 0100

5 0101

6 0110

7 0111

8 1000

9 1001

Next we need a table indicating what the answer is for each possible number 0 to 9.

Table 2

number answer

0

1

1

2

2

3

3

4

4

5

5

6

6

7

7

8

8

9

9 10

Next, we rewrite this table as:

Table 3

number answer

0 01

1 02

2 03

3 04

4 05

5 06

6 07

7 08

8 09

9 10

134

Next, we use table 1 to write table 3 with 1’s and 0’s. That is, we replace all 0’s in table 3 with 0000, all 1’s with 0001, and both 2’s with 0010, etc. This gives us the following table.

Table 4

number answer

0000 00000001

0001 00000010

0010 00000011

0011 00000100

0100 00000101

0101 00000110

0110 00000111

0111 00001000

1000 00001001

1001 00010000

Next, we write table 4 as data in memory.

Table 5

label

address

data

comment

(number)

(answer)

add1tabl 0000000001000000

0000000000000001

0000000001000001

0000000000000010

0000000001000010

0000000000000011

0000000001000011

0000000000000100

0000000001000100

0000000000000101

0000000001000101

0000000000000110

0000000001000110

0000000000000111

0000000001000111

0000000000001000

0000000001001000

0000000000001001

0000000001001001

0000000000010000

This kind of table is called a lookup table because you can look up the answer in it. A lookup table can be made to do any function with a limited number of possible inputs.

This function is called ‘increment’ (add 1) and has 10 possible inputs: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.

135

Next, we decide where the number to increment and the answer will be stored in memory.

label

address

data

comment

0000000000000000 0000000000100000

start0000000000001000

0000000000000001 0000000000001001

number to increment

0000000000000010 0000000000000000

answer

Next, we write the program. The program is a list of instructions that tell the processor how to manipulate data. That is, the program tells the processor from where and to where copy data.

label

address

data

comment

instr_1 0000000000001000 0000000000000001

from address

0000000000001001 0000000000001100

to address

0000000000001010 0000000000001111

‘to’ bits to copy to

0000000000001011 0000000000110000 instr.addr.and rot.

instr_2 0000000000001100 0000000001000000

from address

0000000000001101 0000000000000010

to address

0000000000001110 0000000011111111

‘to’ bits to copy to

0000000000001111 0000000001000000 instr.addr.and rot.

instr_3 0000000000010000 0000000000000000

from address

0000000000010001 0000000000000000

to address

0000000000010010 0000000000000000

‘to’ bits to copy to

0000000000010011 0000000001000000 instr.addr.and rot.

136

The whole program, including table, data, and instructions, follows.

Program to Add 1

label

address

data

comment

0000000000000000 0000000000100000

start0000000000001000

0000000000000001 0000000000001001

number to increment

0000000000000010 0000000000000000

answer

instr_1 0000000000001000 0000000000000001

from address

0000000000001001 0000000000001100

to address

0000000000001010 0000000000001111

‘to’ bits to copy to

0000000000001011 0000000000110000 instr.addr.and rot.

Instr_2 0000000000001100 0000000001000000

from address

0000000000001101 0000000000000010

to address

0000000000001110 0000000011111111

‘to’ bits to copy to

0000000000001111 0000000001000000 instr.addr.and rot.

Instr_3 0000000000010000 0000000000000000

from address

0000000000010001 0000000000000000

to address

0000000000010010 0000000000000000

‘to’ bits to copy to

0000000000010011 0000000001000000 instr.addr.and rot.

Add1tabl 0000000001000000

0000000000000001

0000000001000001

0000000000000010

0000000001000010

0000000000000011

0000000001000011

0000000000000100

0000000001000100

0000000000000101

0000000001000101

0000000000000110

0000000001000110

0000000000000111

0000000001000111

0000000000001000

0000000001001000

0000000000001001

0000000001001001

0000000000010000

137

After instruction 1 (at ‘instr_1’) is executed, the memory has the following values. Italics show to where data was copied and arrows show from where data was copied.

After Instruction 1 Has Been Executed

label

address

data

comment

0000000000000000 0000000000110000 <----|start0000000000001000

0000000000000001 0000000000001001 ---| | number to increment 0000000000000010 0000000000000000

| | answer

| |

instr_1 0000000000001000 0000000000000001 | | from address 0000000000001001 0000000000001100

| | to address

0000000000001010 0000000000001111

| | ‘to’ bits to copy to

0000000000001011 0000000000110000 ---+-| instr.addr.and rot.

instr_2 0000000000001100 000000000100 1001 <--|

from address

0000000000001101 0000000000000010

to address

0000000000001110 0000000011111111

‘to’ bits to copy to

0000000000001111 0000000001000000 instr.addr.and rot.

instr_3 0000000000010000 0000000000000000

from address

0000000000010001 0000000000000000

to address

0000000000010010 0000000000000000

‘to’ bits to copy to

0000000000010011 0000000001000000 instr.addr.and rot.

add1tabl 0000000001000000

0000000000000001

0000000001000001

0000000000000010

0000000001000010

0000000000000011

0000000001000011

0000000000000100

0000000001000100

0000000000000101

0000000001000101

0000000000000110

0000000001000110

0000000000000111

0000000001000111

0000000000001000

0000000001001000

0000000000001001

0000000001001001

0000000000010000

138

After instruction 2 is executed, the memory has the following values. Italics show to where data was copied and arrows show from where data was copied.

After Instruction 2 Has Been Executed

label

address

data

comment

0000000000000000 0000000001000000 <----|start0000000000001000

0000000000000001 0000000000001001

| number to increment

0000000000000010 00000000 00010000 <--| | answer

| |

instr_1 0000000000001000 0000000000000001

| | from address

0000000000001001 0000000000001100

| | to address

0000000000001010 0000000000001111

| | ‘to’ bits to copy to

0000000000001011 0000000000110000

| | instr.addr.and rot.

instr_2 0000000000001100 0000000001001001

| | from address

0000000000001101 0000000000000010

| | to address

0000000000001110 0000000011111111

| | ‘to’ bits to copy to

0000000000001111 0000000001000000 ---+-| instr.addr.and rot.

instr_3 0000000000010000 0000000000000000

|

from address

0000000000010001 0000000000000000

|

to address

0000000000010010 0000000000000000

|

‘to’ bits to copy to

0000000000010011 0000000001000000

| instr.addr.and rot.

|

add1tabl 0000000001000000

0000000000000001

|

0000000001000001

0000000000000010

|

0000000001000010

0000000000000011

|

0000000001000011

0000000000000100

|

0000000001000100

0000000000000101

|

0000000001000101

0000000000000110

|

0000000001000110

0000000000000111

|

0000000001000111

0000000000001000

|

0000000001001000

0000000000001001

|

0000000001001001

0000000000010000 ---|

Now, the result of 1 being added to 9 is 10 and the result, 10, written as 00010000, is stored in the last eight bits of latch 0000000000000010.

Instruction 3 (‘instr_3’) does nothing but execute over and over until the processor is stopped.

The program calculates what one more than nine is and finds that the answer is ten. Nine is represented as 1001 in latch 0000000000000001 and ten is represented as 00010000 in latch 0000000000000010.

139

Next, we will consider how to make a program that adds two numbers, from 0 to 9

together.

Again, we will represent the numbers 0 through 9 with only 1’s and 0’s as indicated in the following table.

Table 6

Number Representation

0 0000

1 0001

2 0010

3 0011

4 0100

5 0101

6 0110

7 0111

8 1000

9 1001

Next, we need a table that shows the answer for each possible pair of numbers from 0 to 9. This is the addition table we studied so hard to learn in grade school and is reproduced below.

Addition Table

+ 0

1

2

3

4

5

6

7

8

9

0 0

1

2

3

4

5

6

7

8

9

1 1

2

3

4

5

6

7

8

9 10

2 2

3

4

5

6

7

8

9 10 11

3 3

4

5

6

7

8

9 10 11 12

4 4

5

6

7

8

9 10 11 12 13

5 5

6

7

8

9 10 11 12 13 14

6 6

7

8

9 10 11 12 13 14 15

7 7

8

9 10 11 12 13 14 15 16

8 8

9 10 11 12 13 14 15 16 17

9 9 10 11 12 13 14 15 16 17 18

140

Next, we rewrite the addition table above as below.

Addition Table Listing

0 + 0 = 00

0 + 1 = 01

0 + 2 = 02

0 + 3 = 03

0 + 4 = 04

0 + 5 = 05

0 + 6 = 06

0 + 7 = 07

0 + 8 = 08

0 + 9 = 09

1 + 0 = 01

1 + 1 = 02

.

.

.

8 + 9 = 17

9 + 0 = 09

9 + 1 = 10

9 + 2 = 11

9 + 3 = 12

9 + 4 = 13

9 + 5 = 14

9 + 6 = 15

9 + 7 = 16

9 + 8 = 17

9 + 9 = 18

Only some of the table elements are listed above to save space.

141

Next, substituting according to table 6, we rewrite the table above as below.

Addition Table for Program

label

address

data

comment

addtable 0000001000000000

0000000000000000

0 + 0 = 00

0000001000000001

0000000000000001

0 + 1 = 01

0000001000000010

0000000000000010

0 + 2 = 02

0000001000000011

0000000000000011

0 + 3 = 03

0000001000000100

0000000000000100

0 + 4 = 04

0000001000000101

0000000000000101

0 + 5 = 05

0000001000000110

0000000000000110

0 + 6 = 06

0000001000000111

0000000000000111

0 + 7 = 07

0000001000001000

0000000000001000

0 + 8 = 08

0000001000001001

0000000000001001

0 + 9 = 09

0000001000010000

0000000000000001

1 + 0 = 01

0000001000010001

0000000000000010

1 + 1 = 02

.

.

.

0000001010001001

0000000000010111

8 + 9 = 17

0000001010010000

0000000000001001

9 + 0 = 09

0000001010010001

0000000000010000

9 + 1 = 10

0000001010010010

0000000000010001

9 + 2 = 11

0000001010010011

0000000000010010

9 + 3 = 12

0000001010010100

0000000000010011

9 + 4 = 13

0000001010010101

0000000000010100

9 + 5 = 14

0000001010010110

0000000000010101

9 + 6 = 15

0000001010010111

0000000000010110

9 + 7 = 16

0000001010011000

0000000000010111

9 + 8 = 17

0000001010011001

0000000000011000

9 + 9 = 18

Of course, in the actual program’s table, all one hundred table elements must be included.

The table begins at 0000001000000000. It can begin anywhere in memory, just so table data doesn’t overlap other data.

142

Next, we write the whole program, including data, instructions, and table. We want to add 9 + 7. That is, we want to calculate C=A+B where A is 9, B is 7, and C is the answer. The underlining and italics are just to highlight data for the person reading the program and do not affect the program. In instr_4, the first two 16-bit words of data don’t matter because no bits are copied.

Addition Program

label

address

data

comment

start

0000000000000000 0000000000010000 start 0000000000000100

A

0000000000000001 000000000000 1001

9 (A)

B

0000000000000010 000000000000 0111

7 (B)

C

0000000000000011 00000000 00000000

answer (C)

instr_1 0000000000000100 0000000000000001 from address of A 0000000000000101 0000000000001100 to instr_3’s from addr.

0000000000000110 0000000011110000 copy to these bits 0000000000000111 0000000000100100 go to instr_2, rot. 4

instr_2 0000000000001000 0000000000000010 from address of B

0000000000001001 0000000000001100 to instr_3’s from addr.

0000000000001010 0000000000001111 copy to these bits 0000000000001011 0000000000110000 go to instr_3, no rot.

instr_3 0000000000001100 00000010 00000000

from addtable

0000000000001101 0000000000000011 to address of C

0000000000001110 0000000011111111 copy to these bits 0000000000001111 0000000001000000 go to instr_4, no rot.

instr_4 0000000000010000 0000000000000000 doesn’t matter 0000000000010001 0000000000000000 doesn’t matter

0000000000010010 0000000000000000 copy NO bits

0000000000010011 0000000001000000 go to this instruction

addtable 0000001000000000

0000000000000000

0 + 0 = 00

0000001000000001

0000000000000001

0 + 1 = 01

0000001000000010

0000000000000010

0 + 2 = 02

.

.

.

0000001010010110

0000000000010101

9 + 6 = 15

0000001010010111

00000000 00010110

9 + 7 = 16

0000001010011000

0000000000010111

9 + 8 = 17

0000001010011001

0000000000011000

9 + 9 = 18

143

In the program above, instr_1 copies the value of A (9) to instr_3. Instr_2 copies the value of B (7) to instr_3. Instr_3 copies the result (16) from the ‘addtable’ to C. Instr_4

does nothing repeatedly. The program below shows, after the program has run, from where the data has been copied and to where the data has been copied in italics. You should try to see from where and to where the data was copied by each instruction: instr_1, instr_2, and instr_3.

After Addition Program Has Run

label

address

data

comment

start

0000000000000000 0000000001000000 start 0000000000000100

A

0000000000000001 000000000000 1001

9 (A)

B

0000000000000010 000000000000 0111

7 (B)

C

0000000000000011 00000000 00010110 answer (C) instr_1 0000000000000100 0000000000000001 from address of A 0000000000000101 0000000000001100 to instr_3’s from addr.

0000000000000110 0000000011110000 copy to these bits 0000000000000111 0000000000100100 go to instr_2, rot. 4

instr_2 0000000000001000 0000000000000010 from address of B

0000000000001001 0000000000001100 to instr_3’s from addr.

0000000000001010 0000000000001111 copy to these bits 0000000000001011 0000000000110000 go to instr_3, no rot.

instr_3 0000000000001100 00000010 10010111

from addtable

0000000000001101 0000000000000011 to address of C

0000000000001110 0000000011111111 copy to these bits 0000000000001111 0000000001000000 go to instr_4, no rot.

instr_4 0000000000010000 0000000000000000 doesn’t matter 0000000000010001 0000000000000000 doesn’t matter

0000000000010010 0000000000000000 copy NO bits

0000000000010011 0000000001000000 go to this instruction

addtable 0000001000000000

0000000000000000

0 + 0 = 00

0000001000000001

0000000000000001

0 + 1 = 01

0000001000000010

0000000000000010

0 + 2 = 02

.

.

.

0000001010010110

0000000000010101

9 + 6 = 15

0000001010010111

00000000 00010110

9 + 7 = 16

0000001010011000

0000000000010111

9 + 8 = 17

0000001010011001

0000000000011000

9 + 9 = 18

144

The following program multiplies two numbers together. It calculates C = A X B where A is 9, B is 7 and C is the answer (63). It is the same as the addition program except that it uses a multiplication table rather than an addition table.

Multiplication Program

label

address

data

comment

start

0000000000000000 0000000000010000 start 0000000000000100

A

0000000000000001 000000000000 1001

9 (A)

B

0000000000000010 000000000000 0111

7 (B)

C

0000000000000011 00000000 00000000

answer (C)

instr_1 0000000000000100 0000000000000001 from address of A 0000000000000101 0000000000001100 to instr_3’s from addr.

0000000000000110 0000000011110000 copy to these bits 0000000000000111 0000000000100100 go to instr_2, rot. 4

instr_2 0000000000001000 0000000000000010 from address of B

0000000000001001 0000000000001100 to instr_3’s from addr.

0000000000001010 0000000000001111 copy to these bits 0000000000001011 0000000000110000 go to instr_3, no rot.

instr_3 0000000000001100 00000100 00000000

from multiply table

0000000000001101 0000000000000011 to address of C

0000000000001110 0000000011111111 copy to these bits 0000000000001111 0000000001000000 go to instr_4, no rot.

instr_4 0000000000010000 0000000000000000 doesn’t matter 0000000000010001 0000000000000000 doesn’t matter

0000000000010010 0000000000000000 copy NO bits

0000000000010011 0000000001000000 go to this instruction

multiply 0000010000000000

0000000000000000

0 X 0 = 00

0000010000000001

0000000000000000

0 X 1 = 00

0000010000000010

0000000000000000

0 X 2 = 00

.

.

.

0000010010010110

0000000001010100

9 X 6 = 54

0000010010010111

00000000 01100011

9 X 7 = 63

0000010010011000

0000000001110010

9 X 8 = 72

0000010010011001

0000000010000001

9 X 9 = 81

Notice that one can save a lot of work by salvaging (copying) instructions from a program one has already written for use in a new, similar program. Tables can often be salvaged as well.

145

The following program adds two two-digit numbers, A (99) and B (87), together for a result of C (186). First, it adds the right digits together (9+7) for a result of 16. 16 is 6

with a carry. Then, the carry, 1, is added to 9 and 8 for a result of 18. That makes the entire answer 186. Adding 1+9+8 together requires an add with carry, so we need a table with carry of 1 or 0 as below. For this table, there are 200 possibilities. There are 2

values of carry (0 or 1), 10 values of one input (0-9), and 10 values of another input (0-9) for 2 X 10 X 10 = 200 possibilities. Notice the carry, ‘+0’ and ‘+1,’ in the upper left of the tables below. The two tables below are two halves of the entire table.

Add with Carry Table

+0 0

1

2

3

4

5

6

7

8

9

0 0

1

2

3

4

5

6

7

8

9

1 1

2

3

4

5

6

7

8

9 10

2 2

3

4

5

6

7

8

9 10 11

3 3

4

5

6

7

8

9 10 11 12

4 4

5

6

7

8

9 10 11 12 13

5 5

6

7

8

9 10 11 12 13 14

6 6

7

8

9 10 11 12 13 14 15

7 7

8

9 10 11 12 13 14 15 16

8 8

9 10 11 12 13 14 15 16 17

9 9 10 11 12 13 14 15 16 17 18

+1

0

1

2

3

4

5

6

7

8

9

0

1

2

3

4

5

6

7

8

9 10

1

2

3

4

5

6

7

8

9 10 11

2

3

4

5

6

7

8

9 10 11 12

3

4

5

6

7

8

9 10 11 12 13

4

5

6

7

8

9 10 11 12 13 14

5

6

7

8

9 10 11 12 13 14 15

6

7

8

9 10 11 12 13 14 15 16

7

8

9 10 11 12 13 14 15 16 17

8

9 10 11 12 13 14 15 16 17 18

9 10 11 12 13 14 15 16 17 18 19

146

The ‘add with carry’ table for the program is shown below. Notice that the carry is represented with only 1 bit because carry can only have two values, 0 or 1. A normal digit requires 4 bits for the 10 possibilities, 0-9.

Add with Carry Table Listing

label

address

data

comment

addtable 0000001000000000

0000000000000000

0 + 0 + 0 = 00

0000001000000001

0000000000000001

0 + 0 + 1 = 01

0000001000000010

0000000000000010

0 + 0 + 2 = 02

.

.

.

0000001010010110

0000000000010101

0 + 9 + 6 = 15

0000001010010111

0000000000010110

0 + 9 + 7 = 16

0000001010011000

0000000000010111

0 + 9 + 8 = 17

0000001010011001

0000000000011000

0 + 9 + 9 = 18

0000001100000000

0000000000000001

1 + 0 + 0 = 01

0000001100000001

0000000000000010

1 + 0 + 1 = 02

0000001100000010

0000000000000011

1 + 0 + 2 = 03

.

.

.

0000001110010110

0000000000010110

1 + 9 + 6 = 16

0000001110010111

0000000000010111

1 + 9 + 7 = 17

0000001110011000

0000000000011000

1 + 9 + 8 = 18

0000001110011001

0000000000011001

1 + 9 + 9 = 19

147

In the program below, instr_1, instr_2, and instr_3 are exactly as in the previous addition program. Instr_1 and instr_2 copy the right digits of A (9) and B (7) into instr_3 and instr_3 copies the result (16) from addtable to the right two digits of C.

Instr_4, instr_5, instr_6, and instr_7 add the left digits of A and B together with the carry bit in C. Instr_4 copies the left digit of A (9) into instr_7. Instr_5 copies the left digit of B (8) into instr_7. Instr_6 copies the carry bit (1) from the middle digit of C to instr_7.

Instr_7 copies the result, (1 + 9 + 8 =) 18, into the left two digits of C for a total result in all three digits of C of 186.

Add Two Digits Program

label

address

data

comment

start

0000000000000000 0000000000010000 start 0000000000000100

A

0000000000000001 00000000 10011001

99 (A)

B

0000000000000010 00000000 10000111

87 (B)

C

0000000000000011 0000 000000000000

answer (C)

instr_1 0000000000000100 0000000000000001 from address of A 0000000000000101 0000000000001100 to instr_3’s from addr.

0000000000000110 0000000011110000 copy to these bits 0000000000000111 0000000000100100 go to instr_2, rot. 4

instr_2 0000000000001000 0000000000000010 from address of B

0000000000001001 0000000000001100 to instr_3’s from addr.

0000000000001010 0000000000001111 copy to these bits 0000000000001011 0000000000110000 go to instr_3, no rot.

instr_3 0000000000001100 00000010 00000000

from addtable

0000000000001101 0000000000000011 to address of C

0000000000001110 0000000011111111 copy to these bits 0000000000001111 0000000001000000 go to instr_4, no rot.

instr_4 0000000000010000 0000000000000001 from address of A 0000000000010001 0000000000011100 to instr_7’s from addr.

0000000000010010 0000000011110000 copy to these bits 0000000000010011 0000000001010000 go to instr_5, no rot.

instr_5 0000000000010100 0000000000000010 from address of B

0000000000010101 0000000000011100 to instr_7’s from addr.

0000000000010110 0000000000001111 copy to these bits 0000000000010111 0000000001101100 to instr_6,rot.4 right instr_6 0000000000011000 0000000000000011 from addr.of C (carry) 0000000000011001 0000000000011100 to instr_7’s from addr.

0000000000011010 0000000100000000 copy to this bit

0000000000011011 0000000001110100 to instr_7,rot.4 left instr_7 0000000000011100 0000001 000000000

from addtable

0000000000011101 0000000000000011 to address of C

0000000000011110 0000111111110000 copy to these bits 0000000000011111 0000000010000100 go to instr_8, rot. 4

instr_8 0000000000100000 0000000000000000 doesn’t matter 0000000000100001 0000000000000000 doesn’t matter

0000000000100010 0000000000000000 copy NO bits

0000000000100011 0000000010000000 go to this instruction 148

addtable 0000001000000000

0000000000000000

0 + 0 + 0 = 00

0000001000000001

0000000000000001

0 + 0 + 1 = 01

0000001000000010

0000000000000010

0 + 0 + 2 = 02

.

.

.

0000001010010110

0000000000010101

0 + 9 + 6 = 15

0000001010010111 00000000 00010110

0 + 9 + 7 = 16

0000001010011000

0000000000010111

0 + 9 + 8 = 17

0000001010011001

0000000000011000

0 + 9 + 9 = 18

0000001100000000

0000000000000001

1 + 0 + 0 = 01

0000001100000001

0000000000000010

1 + 0 + 1 = 02

0000001100000010

0000000000000011

1 + 0 + 2 = 03

.

.

.

0000001110010110

0000000000010110

1 + 9 + 6 = 16

0000001110010111

0000000000010111

1 + 9 + 7 = 17

0000001110011000 00000000 00011000

1 + 9 + 8 = 18

0000001110011001

0000000000011001

1 + 9 + 9 = 19

In some high level languages, instructions 1 through 7 can be written with one instruction, ‘C = A + B.’ You type in ‘C = A + B.’ Then you run another program that is called a compiler. The compiler converts ‘C = A + B’ into all that machine language, instr_1 through instr_7.

A compiler can greatly ease writing programs.

Writing

programs in machine language (1’s and 0’s) is relatively difficult. (Most other processors have a hardware adder, so C = A + B becomes few instructions though many bits are added.)

149

The program below shows the result of running the program above. To and from where data has been copied is underlined and in italics. The answer is 000110000110, or 186, and is stored in the word (16 bits) labeled ‘C.’ The first word (the first 16 data bits) of instr_3 now holds 1001,0111 or 9,7 and the first word of instr_7 now holds 1,1001,1000

or 1,9,8, where the 1 in 1,9,8 is represented by only 1 bit (1).

After Add Two Digits Program Has Run

label

address

data

comment

start

0000000000000000 0000000010000000 start 0000000000000100

A

0000000000000001 00000000 10011001

99 (A)

B

0000000000000010 00000000 10000111

87 (B)

C

0000000000000011 0000 000110000110

answer (C)

instr_1 0000000000000100 0000000000000001 from address of A 0000000000000101 0000000000001100 to instr_3’s from addr.

0000000000000110 0000000011110000 copy to these bits 0000000000000111 0000000000100100 go to instr_2, rot. 4

instr_2 0000000000001000 0000000000000010 from address of B

0000000000001001 0000000000001100 to instr_3’s from addr.

0000000000001010 0000000000001111 copy to these bits 0000000000001011 0000000000110000 go to instr_3, no rot.

instr_3 0000000000001100 00000010 10010111

from addtable

0000000000001101 0000000000000011 to address of C

0000000000001110 0000000011111111 copy to these bits 0000000000001111 0000000001000000 go to instr_4, no rot.

instr_4 0000000000010000 0000000000000001 from address of A 0000000000010001 0000000000011100 to instr_7’s from addr.

0000000000010010 0000000011110000 copy to these bits 0000000000010011 0000000001010000 go to instr_5, no rot.

instr_5 0000000000010100 0000000000000010 from address of B

0000000000010101 0000000000011100 to instr_7’s from addr.

0000000000010110 0000000000001111 copy to these bits 0000000000010111 0000000001101100 to instr_6,rot.4 right instr_6 0000000000011000 0000000000000011 from addr.of C (carry) 0000000000011001 0000000000011100 to instr_7’s from addr.

0000000000011010 0000000100000000 copy to this bit

0000000000011011 0000000001110100 to instr_7,rot.4 left instr_7 0000000000011100 0000001 110011000

from addtable

0000000000011101 0000000000000011 to address of C

0000000000011110 0000111111110000 copy to these bits 0000000000011111 0000000010000100 go to instr_8, rot. 4

instr_8 0000000000100000 0000000000000000 doesn’t matter 0000000000100001 0000000000000000 doesn’t matter

0000000000100010 0000000000000000 copy NO bits

0000000000100011 0000000010000000 go to this instruction 150

addtable 0000001000000000

0000000000000000

0 + 0 + 0 = 00

0000001000000001

0000000000000001

0 + 0 + 1 = 01

0000001000000010

0000000000000010

0 + 0 + 2 = 02

.

.

.

0000001010010110

0000000000010101

0 + 9 + 6 = 15

0000001010010111 00000000 00010110

0 + 9 + 7 = 16

0000001010011000

0000000000010111

0 + 9 + 8 = 17

0000001010011001

0000000000011000

0 + 9 + 9 = 18

0000001100000000

0000000000000001

1 + 0 + 0 = 01

0000001100000001

0000000000000010

1 + 0 + 1 = 02

0000001100000010

0000000000000011

1 + 0 + 2 = 03

.

.

.

0000001110010110

0000000000010110

1 + 9 + 6 = 16

0000001110010111

0000000000010111

1 + 9 + 7 = 17

0000001110011000 00000000 00011000

1 + 9 + 8 = 18

0000001110011001

0000000000011001

1 + 9 + 9 = 19

151

The program below includes an example of ‘branching.’ Branching in a program means that either some instructions or some other instructions are executed depending on a value (usually a bit) in memory (or, in most processors, in a register). Branch bits are often called flags. In the program below, the rightmost bit in latch A (at address 0000000000000001) is a flag and determines whether instr_4 or instr_5 is executed.

Instr_4 copies the ‘all 1’s pattern’ (1111111111111111) in latch C to latch B. Instr_5

copies the ‘1,0 pattern’ (1010101010101010) from latch D to latch B. If latch A has value 1, then instr_5 is executed. If A has value 0, then instr_4 is executed. In the program below, A has value 1 so instr_5 is executed and B gets the ‘1,0 pattern.’

Instr_3 does nothing but go to the next instruction because it copies no bits. Instr_3 has 00000000010 0 0000 in latch (address) 0000000000001111, so, normally, instr_4, at address 0000000000010000, would be executed next. However, instr_2 copies the rightmost bit (1) from A into the fifth-from-rightmost bit of latch 0000000000001111 so that latch 0000000000001111 contains 00000000010 1 0000 and instr_5, at address 0000000000010100, is executed after instr_3. The instructions are then executed in the following order: instr_2, instr_3, instr_5, instr_6. Notice that instr_4 is skipped.

Branching Program

label

address

data

comment

start

0000000000000000 0000000000100000 start 0000000000001000

A

0000000000000001 000000000000000 1

flag(1 or 0)right bit

B

0000000000000010 0000000000000000

value to change

C

0000000000000011 1111111111111111

all 1’s pattern

D

0000000000000100 1010101010101010

1,0 pattern

instr_2 0000000000001000 0000000000000001 from A

0000000000001001 0000000000001111 to next of instr_3

0000000000001010 0000000000010000 change this bit

0000000000001011 0000000000110100 to instr_3, rot.4

instr_3 0000000000001100 0000000000000000 doesn’t matter 0000000000001101 0000000000000000 doesn’t matter

0000000000001110 0000000000000000 copy NO bits

0000000000001111 00000000010 00000 to instr_4 OR instr_5

instr_4 0000000000010000 0000000000000011 from C pattern 0000000000010001 0000000000000010 to B

0000000000010010 1111111111111111 copy all bits

0000000000010011 0000000001100000 to instr_6, no rot.

instr_5 0000000000010100 0000000000000100 from D pattern 0000000000010101 0000000000000010 to B

0000000000010110 1111111111111111

copy all bits

0000000000010111 0000000001100000 to instr_6, no rot.

instr_6 0000000000011000 0000000000000000 doesn’t matter 0000000000011001 0000000000000000 doesn’t matter

0000000000011010 0000000000000000 copy NO bits

0000000000011011 0000000001100000 go to this instruction 152

The result of running the program above is shown below. Notice that B now holds 1010101010101010 and that address 0000000000001111 now holds 00000000010 1 0000

instead of 00000000010 0 0000. Italics show to where and from where values have been copied.

After Branching Program Has Run

label

address

data

comment

start

0000000000000000 0000000001100000

start 0000000000001000

A

0000000000000001 000000000000000 1

flag(1 or 0)right bit

B

0000000000000010 1010101010101010

value to change

C

0000000000000011 1111111111111111

all 1’s pattern

D

0000000000000100 1010101010101010

1,0 pattern

instr_2 0000000000001000 0000000000000001 from A

0000000000001001 0000000000001111 to next of instr_3

0000000000001010 0000000000010000 change this bit

0000000000001011 0000000000110100 to instr_3, rot.4

instr_3 0000000000001100 0000000000000000 doesn’t matter 0000000000001101 0000000000000000 doesn’t matter

0000000000001110 0000000000000000 copy NO bits

0000000000001111 00000000010 10000 to instr_4 OR instr_5

instr_4 0000000000010000 0000000000000011 from C pattern 0000000000010001 0000000000000010 to B

0000000000010010 1111111111111111 copy all bits

0000000000010011 0000000001100000 to instr_6, no rot.

instr_5 0000000000010100 0000000000000100 from D pattern 0000000000010101 0000000000000010 to B

0000000000010110 1111111111111111 copy all bits

0000000000010111 0000000001100000 to instr_6, no rot.

instr_6 0000000000011000 0000000000000000 doesn’t matter 0000000000011001 0000000000000000 doesn’t matter

0000000000011010 0000000000000000 copy NO bits

0000000000011011 0000000001100000 go to this instruction

153

The following program is the same as the previous one (before it was run) except that A now has value 0 instead of value 1 (in the rightmost bit). This means that instr_4 will be executed instead of instr_5 and B will get 1111111111111111 from C rather than 1010101010101010 from D. The instructions are executed in the following order: instr_2, instr_3, instr_4, instr_6. Instr_5 is not executed.

Branching Program with Flag = 0

label

address

data

comment

start

0000000000000000 0000000000100000 start 0000000000001000

A

0000000000000001 000000000000000 0

flag(1 or 0)right bit

B

0000000000000010 0000000000000000

value to change

C

0000000000000011 1111111111111111

all 1’s pattern

D

0000000000000100 1010101010101010

1,0 pattern

instr_2 0000000000001000 0000000000000001 from A

0000000000001001 0000000000001111 to next of instr_3

0000000000001010 0000000000010000 change this bit

0000000000001011 0000000000110100 to instr_3, rot.4

instr_3 0000000000001100 0000000000000000 doesn’t matter 0000000000001101 0000000000000000 doesn’t matter

0000000000001110 0000000000000000 copy NO bits

0000000000001111 00000000010 00000 to instr_4 OR instr_5

instr_4 0000000000010000 0000000000000011 from C pattern 0000000000010001 0000000000000010 to B

0000000000010010 1111111111111111 copy all bits

0000000000010011 0000000001100000 to instr_6, no rot.

instr_5 0000000000010100 0000000000000100 from D pattern 0000000000010101 0000000000000010 to B

0000000000010110 1111111111111111 copy all bits

0000000000010111 0000000001100000 to instr_6, no rot.

instr_6 0000000000011000 0000000000000000 doesn’t matter 0000000000011001 0000000000000000 doesn’t matter

0000000000011010 0000000000000000 copy NO bits

0000000000011011 0000000001100000 go to this instruction 154

The following program shows the result of executing the preceding program. Notice that B now contains 1111111111111111 from C and address 0000000000001111 still contains 00000000010 0 0000. Instr_4 has been executed instead of instr_5.

After Branching Program with Flag = 0 Has Run

label

address

data

comment

start

0000000000000000 0000000001100000

start 0000000000001000

A

0000000000000001 000000000000000 0

flag(1 or 0)right bit

B

0000000000000010 1111111111111111

value to change

C

0000000000000011 1111111111111111

all 1’s pattern

D

0000000000000100 1010110010101010

1,0 pattern

instr_2 0000000000001000 0000000000000001 from A

0000000000001001 0000000000001111 to next of instr_3

0000000000001010 0000000000010000 change this bit

0000000000001011 0000000000110100 to instr_3, rot.4

instr_3 0000000000001100 0000000000000000 doesn’t matter 0000000000001101 0000000000000000 doesn’t matter

0000000000001110 0000000000000000 copy NO bits

0000000000001111 00000000010 00000 to instr_4 OR instr_5

instr_4 0000000000010000 0000000000000011 from C pattern 0000000000010001 0000000000000010 to B

0000000000010010 1111111111111111 copy all bits

0000000000010011 0000000001100000 to instr_6, no rot.

instr_5 0000000000010100 0000000000000100 from D pattern 0000000000010101 0000000000000010 to B

0000000000010110 1111111111111111 copy all bits

0000000000010111 0000000001100000 to instr_6, no rot.

instr_6 0000000000011000 0000000000000000 doesn’t matter 0000000000011001 0000000000000000 doesn’t matter

0000000000011010 0000000000000000 copy NO bits

0000000000011011 0000000001100000 go to this instruction 155

MISCELLANEOUS

Computer with Input and Output

Inputs and outputs have been added to the computer above in place of two memory latches. When data is written to (copied to) ‘output latch 1111,’ then each loop, O3, O2, O1, and O0, will turn on its light if a 1 is stored in the loop. When data is read from (copied from) input ‘latch’ 1110 (It’s not really a latch because it doesn’t have loops.), then a 1 will be copied from key I3 if key I3 is pressed. If key 3 is not pressed, then a 0 is copied from I3. I2, I1, and I0 work similarly.

156

For example, the following one-instruction program (for the four-bit computer above) copies data from address (input ‘latch’) 1110 to address (output latch) 1111 over and over again. Therefore, when the program is running, pressing key I3 turns on light O3, pressing key I2 turns on light O2, pressing key I1 turns on light O1, and pressing key I0

turns on light O0.

label address data comment

start

0000 0100 start at 0100

instr_1 0100 1110 from inputs

.

0101 1111 to outputs

.

0110 1111 copy all bits

.

0111 0100 repeat this instruction, no rotate

The keys PP, A3, A2, …D0 along the bottom of the computer diagrammed above allow you to control the computer. You can write to memory, start the processor, stop the processor, and read the results from memory. These keys are, together, called the control panel. A control panel controlled early computers. However, today a keyboard controls a computer. A keyboard is a lot of keys similar to the input keys. The computer runs a program that checks for key presses and reacts accordingly. That program is called an operating system. A joystick may control a game computer. Inside a typical joystick are keys that the joystick bumps into. Those keys and the keys under the joystick’s buttons are also like the input keys above. The outputs can control motors (like in a disk drive) rather than lights.

157

Transistors

Modern computers use two types of transistors, which correspond to the two types of relays. An N-channel transistor corresponds to a normally open relay. A P-channel transistor corresponds to a normally closed relay. However, transistors have some idiosyncrasies and you can’t simply replace relays with transistors to make a successful transistor-based design.

It takes about twice as many transistors as relays to do something. Of course, the high speed and low cost of transistors make transistors vastly superior in spite of the extra design effort required. The millions of transistors in a modern microprocessor allow for more than one type of instruction. For example, besides, or instead of, rotate and mask, the instruction set can include add, subtract, multiply, divide, etc.

The Future

This completes the explanation of how the vast majority of computers work now. One instruction is executed at a time. In modern designs, it is common for the ensuing instruction to be started before the prior instruction finishes, so a few instructions can be executed at once. There are designs that allow many instructions to be executed at once; but such computers, though very fast, are relatively hard to program and, mainly for that reason, have not become very popular. Most such designs use many (often relatively simple) computers, each of which can execute an instruction at once, and which communicate with each other through inputs and outputs. Computers with such designs are called parallel computers and are probably what will be used in the future. For example, I have an idea for a computer that would be able to execute thousands of instructions at a time and still be programmed almost the same way as a normal computer. That will have to be the subject of another book.

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