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Computer Arithmetic and the Arithmetic Unit

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

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Computer Arithmetic and the Arithmetic Unit

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  1. Computer Arithmeticand the Arithmetic Unit Lesson 2 - Ioan Despi

  2. 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

  3. 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

  4. 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

  5. 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

  6. 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:

  7. To convert a binary number to decimal, one just adds up the place values of the 1-bits, as follows: 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 Bit position 7 6 5 4 3 2 1 0 Binary number 1 1 0 1 1 1 0 1 1 4 8 16 64 128 ------- 221 You can try with 10100111

  8. 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.

  9. 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

  10. 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

  11. 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

  12. 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 = xb + Di 6) If there are more digits, repeat from step 3 • Example: convert 3AF16 to base 10 x = 0 x = 16x + 3 = 3 x = 163 + 10(= A) = 58 x1658 + 15(= F) = 943

  13. 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

  14. 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

  15. 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

  16. 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 0xbm-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

  17. Complement Representations of Negative Numbers Radix Complement Diminished Radix Complement Number Representation Number Representation 0 0 0 0 or bm-1 0<x<bm/2 x 0<x<bm/2 x |x|c = bm - 1 - |x| -bm/2x<0 |x|c = bm - |x| -bm/2<x<0 • 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

  18. Base 2 Complement Representations 8 Bit 2’s Complement 8 Bit 1’s Complement Number Representation Number Representation 0 0 0 or 255 0 0<x<128 x 0<x<128 x -128x<0 256 - |x| -127x<0 255 - |x| • 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

  19. Digitwise Computation of the Diminished Radix Complement • Using the geometric series formula, the b-1’s complement of x can be written • If 0xib-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

  20. 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

  21. Addition and Subtraction • addition bits are added and carry propagated leftwards 0000 0111 7ten + 0000 0110 6ten = 0000 1101 13ten • subtraction use negation and addition i.e. x - y = x + (-y) 0000 0111 7ten - 1111 1010 - 6ten (2’s complement) = (1) 0000 0001 1ten

  22. 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

  23. Examples: (the second illustrates a “ripple effect”) 0 0 0 1 1 0 0 <-- carries --> 1 1 1 1 1 1 0 10 10 11 00 01 11 11 11 +00 00 11 11 00 00 00 10 ------------------------------------------- ------------------------------------------ 10 11 10 11 10 00 00 01 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, 16

  24. 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

  25. 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

  26. 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 Multiplication

  27. 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

  28. 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 Division

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