Computer Arithmeticand the Arithmetic Unit

Lesson 2 - Ioan Despi

Digital Number Systems

• Digital number systems have a base or radix b

• Using positional notation, an m-digit base b number is written

x = xm-1 xm-2 ... x1 x0

0 xi b-1, 0 i < m

• The value of this unsigned integer is

i=0

m-1xibivalue(x) =

Range of Unsigned m Digit Base b Numbers

• The largest number has all of its digits equal to b-1, the largest possible base b digit

• Its value can be calculated in closed form

xmax = i=0

m-1(b-1)bi = (b-1)

i=0

m-1bi = bm - 1

• An important summation—geometric series

i=0

m-1bi =

bm - 1b - 1

Number Systems

• base 10 (decimal)10 digits

0, 1, 2, 3, 4, 5, 6, 7, 8, 9

positional notation

1011ten = 1 x 103 + 0 x 102 + 1 x 101 + 1 x 100

• base 2 (binary)2 digits

0,1

1011two = 1 x 23 + 0 x 22 + 1 x 21 + 1 x 20

1011two = 11ten

Number Systems

• base 8 (octal)8 digits

0, 1, 2, 3, 4, 5, 6, 7

1001eight = 1 x 83 + 0 x 82 + 1 x 81 + 1 x 80

1001eight = 521ten

• base 16 (hexadecimal)16 digits

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F

1001sixteen = 1 x 163 + 0 x 162 + 1 x 161 + 1 x 160

1001sixteen = 4113ten

Number Systems: 2

• The digital logic of computers: simple, reliable circuits (AND, OR and NOT gates) that

“lock” one of two states (On / Off, True / False, 0 / 1)

• To represent numbers, a single binary digit (bit) can represent 2 possible numbers (0 or 1).

• We can add more bits to the number giving the additional bits a place value, just as we do

for decimal numbers (base 10).

• Each additional bit doubles the number of possible combinations:

Number of bits Possiblecombinations

Range

2 4 0 .. 3

3 8 0 .. 7

4 16 0.. 15

8 256 0 .. 255

To convert a binary number to decimal, one just adds up the place values of the 1-bits, as follows:

Bit position

Binary number

7 6 5 4 3 2 1 0

1 1 0 1 1 1 0 1

14

8

16

64

128-------

221You can try with 10100111

The number has 8 bits

I numbered starting with 0 on the right-->

the place value of each bit is 2 to the power of the bit number

EX:

2 to the 7th = 128

A story:

Seven computer scientists went for a walk. After a while they counted themselves: 0, 1, 2, 3, 4, 5, 6. One must be missing!

So they spent the rest of the day looking, but never figured out where the 7th one could have gone.

This is known as an “off by one” error.

It is all too common.

To convert from decimal to binary:

divide by 2 repeatedly until 0 is reached

then the remainders, reading up, are the binary number

221 /2 = 110 r = 1

110 / 2 = 55 r = 0

55 / 2 = 27 r = 1

27 / 2 = 13 r = 1

13 / 2 = 6 r = 1

6 / 2 = 3 r = 0

3 / 2 = 1 r = 1

1 / 2 = 0 r = 1

Answer: 11011101

Radix Conversion: General Matters• Converting from one number system to another involves

computation

• We call the base in which calculation is done c and the other base b

• Calculation is based on the division algorithm

— For integers a and b, there exist integers q and r such that a = qb + r, with 0 r b-1

• Notation:

q = a/b r = a mod b

Digit Symbol Correspondence Between Bases

• Each base has b (or c) different symbols to represent the digits• If b < c, there is a table of b + 1 entries giving base c symbols for

each base b symbol and b

• If the same symbol is used for the first b base c digits as for the base b digits, the table is implicit

• If c < b, there is a table of b + 1 entries giving a base c number for each base b symbol and b

• For base b digits c, the base c numbers have more than one digit

Base 12: 0 1 2 3 4 5 6 7 8 9 A B 10

Base 3: 0 1 2 10 11 12 20 21 22 100 101 102 110

Convert Base b Integer to Calculator’s Base, c

1) Start with base b x = xm-1 xm-2 ... x1 x0

2) Set x = 0 in base c

3) Left to right, get next symbol xi

4) Lookup base c number Di for symbol xi

5) Calculate in base c: x = xb + Di

6) If there are more digits, repeat from step 3

• Example: convert 3AF16 to base 10x = 0x = 16x + 3 = 3x = 163 + 10(= A) = 58x1658 + 15(= F) = 943

Convert Calculator’s Base Integer to Base b

1) Let x be the base c integer

2) Initialize i = 0 and v = x & get digits right to left

3) Set Di = v mod b & v = v/b. Lookup Di to get xi

4) i = i + 1; If v 0, repeat from step 3

• Example: convert 356710 to base 12

3587 12 = 298 (rem = 11) x0 = B

298 12 = 24 (rem = 10) x1 = A

24 12 = 2 (rem = 0) x2 = 0

2 12 = 0 (rem = 2) x3 = 2

Thus 358710 = 20AB12

Negative Numbers

• need a representation for negative numbers

• use two’s complement

• binary number negated as follows starting from rightmost bit and working left, copy up to and

including the first 1-bit; thereafter, write 1-bits for 0-bits and 0-bits for 1-bits.

• example in 16-bit word 42 0000 0000 0010 1010

-42 1111 1111 1101 0110

Complement Number Systems

• Complement number systems use unsigned numbers to represent both positive and negative numbers

• Recall that the range of an m digit base b unsigned number is

0 x bm-1

• The first half of the range is used for positive, and the second half for negative, numbers

• Positive numbers are simply represented by the unsigned number corresponding to their absolute value

Complement Operationsfor m-Digit Base b Numbers

• Radix complement of m-digit base b number x

xc = (bm - x) mod bm

• Diminished radix complement of x

xc = bm - 1 - x

• The complement of a number in the range 0xbm-1 is in the same range

• The mod bm in the radix complement definition makes this true for x = 0; it has no effect for any other value of x

• Specifically, the radix complement of 0 is 0

Complement Representations of Negative Numbers

• For even b, radix complement system represents one more negative than positive value

• While diminished radix complement system has 2 zeros but represents same number of positive & negative values

Radix Complement Diminished Radix Complement

Number NumberRepresentation Representation

0 0 0 0 or bm-1

0<x<bm/2 x 0<x<bm/2 x

-bm/2x<0 |x|c = bm - |x| |x|c = bm - 1 - |x|-bm/2<x<0

Base 2 Complement Representations

• In 1’s complement, 255 = 111111112 is often called -0

• In 2’s complement, -128 = 100000002 is a legal value, but trying to negate it gives overflow

8 Bit 2’s Complement 8 Bit 1’s Complement

Number NumberRepresentation Representation

0 0 0 0 or 255

0<x<128 x 0<x<128 x

-128x<0 256 - |x| 255 - |x|-127x<0

2562 1282 87

Digitwise Computation of the Diminished Radix Complement

• Using the geometric series formula, the b-1’s complement of x can be written

i=0

m-1(b-1)bi -xc = bm-1-x =

i=0

m-1xibi

i=0

m-1(b-1-xi)bi=

• If 0xib-1, then 0(b-1-xi)b-1, so last formula is just an m-digit base b number with each digit obtained from the corresponding digit of x

Conversion Between Related Bases by Digit Grouping

• Let base b = ck; for example b = c2

• Then base b number x1x0 is base c number y3y2y1y0,

where x1 base b = y3y2 base c and x0 base b = y1y0 base c

• Examples: 1021304 = 10 21 304 = 49C16

49C16 = 0100 1001 11002

1021304 = 01 00 10 01 11 002

0100100111002 = 010 010 011 1002 = 22348

Addition and Subtraction

• additionbits are added and carry propagated leftwards

0000 0111 7ten

+ 0000 0110 6ten

= 0000 1101 13ten

• subtractionuse negation and addition i.e. x - y = x + (-y)

0000 0111 7ten

- 1111 1010 - 6ten (2’s complement)

= (1) 0000 0001 1ten

Addition and Subtraction

• Addition (and subtraction) of binary numbers is similar to ordinary addition.

• If the sum won’t fit in one bit, you need to carry one to the next place on the left. The rules are:

0 + 0 = 0 1 + 0 = 1 0 + 1 = 1

1 + 1 = 0 and a carry

1+ 1 + 1 (from a previous carry) = 1 and a carry

Examples:

(the second illustrates a “ripple effect”)

10 10 11 00 01 11 11 11

+00 00 11 11 00 00 00 10------------------------------------------- ------------------------------------------

10 11 10 11 10 00 00 01

0 0 0 1 1 0 0 <-- carries --> 1 1 1 1 1 1 0

Long binary numbers get hard for humans to read:

1100 1010 1110 1100 1011 0101 1100

1000 1111 1010 0111 0110 0001 0010 0011 0100 0110 0101, etc…

and it takes a long time to convert them into decimal ===>

we need a more readable representation: Bases 8, 16we need a more readable representation: Bases 8, 16

Carry and Overflow

• need to take account of restricted word size

• carry : when unsigned numbers are added, a carry occurs when a carry is propagated from the leftmost bit

1100 0000 192ten

+ 0100 0000 64ten

= (1) 0000 0000 0ten (256)

• overflow : when signed numbers are added, a overflow occurs when the carry into the sign bit differs from the carry out of it

0100 0000 64ten

+ 0100 0000 64ten

1000 0000 -128ten

Multiplication

• algorithm operates on two unsigned n-bit numbers, a and b

• initialisationload a into multiplier register A

load b into multiplier register B

clear register P

• repeat n times1. if rightmost bit of A is 1-bit, then add B to P

2. shift P and A right, with rightmost bit of P shifting into leftmost bit of A

• product is in P and A

Multiplication

Example a =5, b = 3

P: 0000 A: 0101

B: 0011

Iteration 1

P: 0001 A: 1010

B: 0011

Iteration 2

P: 0000 A: 1101

B: 0011

Iteration 3

P: 0001 A: 1110

B: 0011

Iteration 4

P: 0000 A: 1111

B: 0011

product in PA : 0000 1111

answer = 15

Division

• algorithm operates on two unsigned n-bit numbers, a and b

• initialisationload a into divider register A, load b into divider register B

clear register P

• repeat n times1. shift P and A left, with leftmost bit of A shifting into rightmost bit of P

2. subtract B from P

3. if result of subtraction is negative, set rightmost bit of A to 0-bit, otherwise set to 1-bit and add B to P restoring its value

• quotient in A, remainder in P

Division

example, a = 14, b = 3

P: 0000 A: 1110

B: 0011

Iteration 1

P: 0001 A: 1100

B: 0011

Iteration 2

P: 0000 A: 1001

B: 0011

Iteration 3

P: 0001 A: 0010

B: 0011

Iteration 4

P: 0010 A: 0100

B: 0011

quotient in A = 4

remainder in P = 2