Computer Arithmetic and 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 = x m-1 x m-2... x 1 x 0 0  x i  b-1, 0  i < m The value of this unsigned integer is

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 An important summation—geometric series

Number Systems base 10 (decimal) 10 digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 positional notation 1011 ten = 1 x x x x 10 0 base 2 (binary) 2 digits 0, two = 1 x x x x two = 11 ten

Number Systems base 8 (octal) 8 digits 0, 1, 2, 3, 4, 5, 6, eight = 1 x x x x eight = 521 ten base 16 (hexadecimal) 16 digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F 1001 sixteen = 1 x x x x sixteen = 4113 ten

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:

To convert a binary number to decimal, one just adds up the place values of the 1-bits, as follows : Bit position Binary number You can try with 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 = / 2 = 55r = 0 55 / 2 = 27r = 1 27 / 2 = 13r = 1 13 / 2 = 6r = 1 6 / 2 = 3r = 0 3 / 2 = 1 r = 1 1 / 2 = 0 r = 1 Answer:

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 = q  b + 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: A B 10 Base 3:

Convert Base b Integer to Calculator’s Base, c 1) Start with base b x = x m-1 x m-2... x 1 x 0 2) Set x = 0 in base c 3) Left to right, get next symbol x i 4) Lookup base c number D i for symbol x i 5) Calculate in base c: x = x  b + D i 6) If there are more digits, repeat from step 3 Example: convert 3AF 16 to base 10 x = 0 x = 16 x + 3 = 3 x = 16  (= A) = 58 x  16  (= 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 D i = v mod b & v =  v/b . Lookup D i to get x i 4) i = i + 1; If v  0, repeat from step 3 Example: convert to base  12 = 298 (rem = 11)  x 0 = B 298  12 = 24 (rem = 10)  x 1 = A 24  12 = 2 (rem = 0)  x 2 = 0 2  12 = 0 (rem = 2)  x 3 = 2 Thus = 20AB 12

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

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  b m -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 Operations for m-Digit Base b Numbers Radix complement of m-digit base b number x x c = (b m - x) mod b m Diminished radix complement of x x c = b m x The complement of a number in the range 0  x  b m -1 is in the same range The mod b m 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 ComplementDiminished Radix Complement Number Representation 0000 or b m -1 0<x<b m /2x x -b m /2  x<0 |x| c = b m - |x| |x| c = b m |x| -b m /2<x<0

Base 2 Complement Representations In 1’s complement, 255 = is often called -0 In 2’s complement, -128 = is a legal value, but trying to negate it gives overflow 8 Bit 2’s Complement8 Bit 1’s Complement Number Representation or 255 0<x<128x x -128  x< |x| |x| -127  x<0

Digitwise Computation of the Diminished Radix Complement Using the geometric series formula, the b-1’s complement of x can be written If 0  x i  b-1, then 0  (b-1-x i )  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 = c k ; for example b = c 2 Then base b number x 1 x 0 is base c number y 3 y 2 y 1 y 0, where x 1 base b = y 3 y 2 base c and x 0 base b = y 1 y 0 base c Examples: = = 49C 16 49C 16 = = = =

Addition and Subtraction addition bits are added and carry propagated leftwards ten ten = ten subtraction use negation and addition i.e. x - y = x + (-y) ten ten (2’s complement) = (1) ten

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

Examples: (the second illustrates a “ripple effect”) Long binary numbers get hard for humans to read: , etc… and it takes a long time to convert them into decimal ===> we 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 ten ten = (1) ten (256) overflow : when signed numbers are added, a overflow occurs when the carry into the sign bit differs from the carry out of it ten ten ten

Multiplication algorithm operates on two unsigned n-bit numbers, a and b initialisation load a into multiplier register A load b into multiplier register B clear register P repeat n times 1. 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 : answer = 15

Division algorithm operates on two unsigned n-bit numbers, a and b initialisation load a into divider register A, load b into divider register B clear register P repeat n times 1. 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