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Arithmetic operations in binary

Arithmetic operations in binary. Addition / subtraction 01011 + 111 ? “Method” exatly the same as decimal. Arithmetic operations in binary. Addition X = x n … x i … x 0 + Y = y n … y i … y 0 __________________ d i d i = (x i + y i ) mod r + carry-in.

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Arithmetic operations in binary

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  1. Arithmetic operations in binary • Addition / subtraction 01011 + 111 ? • “Method” exatly the same as decimal

  2. Arithmetic operations in binary • Addition X = xn … xi … x0 + Y = yn … yi … y0 __________________ di di = (xi + yi ) mod r + carry-in

  3. Addition For ( i = 0…n ) do di = (xi + yi + carry-in) mod r carry-out = (xi + yi + carry-in) div r End for

  4. Arithmetic operations in binary • Subtraction • Can you write an equivalent method (algorithm) for subtraction? • Algorithm: a systematic sequence of steps/operations that describes how to solve a problem

  5. Arithmetic operations in binary • Subtraction X = xn … xi … x0  Y = yn … yi … y0 ___________________________________________________ di For ( i = 0…n ) do di = ( xi yi bi-1 ) mod r borrow-out = ( xi yi bi-1 ) div r End for

  6. Subtraction

  7. Arithmetic operations in binary • Multiplication For ( i = 0…n ) do di = ( xi yi+ ci-1 ) mod r ci = ( xi yi+ ci-1 ) div r End for Is this correct?

  8. n = number of symbols (digits, bits, etc.) r = radix radix complement of x is Negative numbers (4 traditions): Signed magnitude Radix complement Diminished radix complement Excess-b (biased) e.g. n = 4, r = 10 7216 --> 9999 - 7216 + 1 = 2784 (10s complement) n = 4, r = 2 0101 --> 1111 - 0101 + 1 = 1011 (2s complement)

  9. diminished radix complement is e.g. n = 4, r = 10 7216 --> 9999 - 7216 = 2783 (9s complement) n = 4, r = 2 0101 --> 1111 - 0101 = 1010 (1s complement) Note: operation is reversible (ignoring overflow) i.e. complement of complement returns original pattern (for both radix and diminished radix forms) for zero and its complement: 10s complement: 0000 --> 9999 + 1 = 10000 --> 0000 9s complement: 0000 --> 9999 --> 0000

  10. Arithmetic with complements n = 4 r = 2 Excess-8

  11. Signed number representations [[N]] = rn – ( rn – N ) = N ( N + [N] ) mod rn = 0 • Example: (n = 4) +5  0101 -5  1011 10000  24 • N + [N] mod 24 = 0 • N = 5 r = 10 N = 32546 • [N] = 105 – 32546 = (67454)10

  12. Two’s complement arithmetic ( an-1, an-2, … a0 ) in 2-s complement is • Example: (n = 4) 0101  023 + 122 + 02 + 11 = 4 + 1 = 5 1011  123 + 022 + 12 + 11 = 8 + 2 + 1 = 5 • Addition/subtraction (n = 5) +10 01010 +3 00011 01101  13

  13. Two’s complement arithmetic • Addition/subtraction (n = 5) +10 01010 +7 00111 Overflow10001 15 +15 01010 13 10011 Discard100010  2 When can overflow happen? Only when both operands have the same sign and the sign bit of the result is different.

  14. Cyclic representation (n = 4, r = 2) avoid discontinuity between 0111 and 1000 Add x: move x positions clockwise Subtract x: move x positions counterclockwise move (16 - x) positions clockwise (i.e. add radix complement)

  15. 5 + 6 11 0101 + 0110 1011 -5 + -6 -11 1011 + 1010 10101 overflow but, carry into most significant bit is 0 while carry out is 1 How to detect discontinuity? no overflow (in 4-bit register) but, carry into most significant bit is 1 while carry out is 0

  16. Same circuitry signed numbers add subtract (use 2s complement of subtrahend) Intel architecture OF (overflow flag) detects out-of-range result unsigned numbers same protocol but discontinuity is now between 0000 and 1111 detect with overflow for addition lack of overflow for subtraction Intel uses CF (carry flag) to detect out-of-range result

  17. Codes 4-bit codes for digits 0 through 9 (BCD : 8421 weighted) 2421 and Excess-3 are self-complementing (complement bits to get 9’s complement representation) Biquinary has two ranges 01… and 10… one-bit error produces invalid codeword

  18. Number representation inside computer • Integers – represented in 2’s complement • Single precision – 32 bits Largest positive integer is 231-1 = 2,147,483,647 Smallest negative integer is -231 = 2,147,483,648

  19. Number representation inside computer • Floating point • Scientific notation 0.0043271 = 0.4327110-2 normalized number The digit after the decimal point is  0 Normalized notation maximizes the use of significant digits.

  20. Floating point numbers • Floating point N = (-1)S m  rE S = 0  positive S = 1  negative m  normalized fraction for radix r = 2 As MSB digit is always 1, no need to explicitly store it Called hidden bit  gives one extra bit of precision

  21. Floating point formats • IEEE format: (-1)S(1.m)2E-127 • DEC format: (-1)S(0.m)2E-128

  22. Floating point formats • Different manufacturers used different representations

  23. Mechanical encoder with sequential sector numbering At boundaries can get a code different than either adjacent sector

  24. Gray code:

  25. Gray code algorithm: input: (binary) output: (Gray code) Alternative algorithm (n bits) If n = 1 use 0->0 and 1->1 If n > 1 first half is Gray code on (n - 1) bits preceded by 0 second half is reversed Gray code on (n - 1) bits preceded by 1 e.g. n = 3: 6 -> 110 -> 101 n = 4: 10 -> 1010 -> 1111

  26. Representing non-numeric data • Code: systematic and preferably standardized way of using a set of symbols for representing information • Example: BCD (binary coded decimal) for decimal #s • It is a “weighted” code  the value of a symbol is a weighted sum • Extending number of symbols to include alphabetic characters • EBCDIC (by IBM): Extended BCD Interchange Code • ASCII: American Standard Code for Information Interchange

  27. Codes

  28. Cyclic codes • A circular shift of any code word produces a valid code word (maybe the same) • Gray code – example of a cyclic code • Code words for consecutive numbers differ by one bit only

  29. Ascii, ebcdic codes State codes, e.g.

  30. Consequences of binary versus one-hot coding:

  31. n-cubes of codewords: Hamming distance between x and y is count of positions where x-bit differs from y-bit Also equals link count in shortest path from x to y

  32. Gray code is path that visits each vertex exactly once

  33. Solution: parity Problem Error-detecting codes concept: choose code words such that corruption generates a non-code word to detect single-bit error, code words must occupy non-adjacent cube vertices

  34. Example: correct one bit errors or detect two-bit errors Two-bit error from 0001011 Error-correcting codes minimum distance between code words > 1

  35. Write minimum distance as 2c + d + 1 bits ==> corrects c-bit errors and detects (c + d)-bit errors Example: min distance = 4 = 2(1) + 1 + 1 But also, 4 = 2(0) + 3 + 1

  36. Why? suppose minimum distance is h c c h - 2c  d = h - 2c -1 pair of closest nodes maximally distant from left with entering correction zone 2c + d + 1 = 2c + (h - 2c - 1) + 1 = h

  37. Hamming codes: number bit positions 1, 2, 3, ... n from right to left bit position is a power of 2 => check bit else => information bit e.g. (n = 7) check bits in positions 1, 2, 4 information bits in positions 3, 5, 6, 7 Create group for each check bit Express check bit position in binary Group all information bits whose binary position has a one in same place e.g. (n = 7) check information 1 (001) 3 (011), 5 (101), 7 (111) 2 (010) 3 (011), 6 (110), 7 (111) 4 (100) 5 (101), 6 (110), 7 (111)

  38. Code information packets to maintain even parity in groups e.g. (n = 7) packet is 1011 => positions 7, 6, 5, 3 7 6 5 4 3 2 1 1 0 1 x 1 x x Consult group memberships to compute check bits check information 1 3, 5, 7 => bit 1 is 1 2 3, 6, 7 => bit 2 is 0 4 5, 6, 7 => bit 4 is 0 Code word is 1010101

  39. Note: Single bit error corrupts one or more parity groups => minimum distance > 1 Two-bit error in locations x, y corrupts at least one parity group => minimum distance > 2 Three-bit error (i.e. 1, 4, 5) goes undetected => minimum distance = 3 3 = 2(1) + 0 + 1 = 2(0) + 2 + 1 => can correct 1-bit errors or detect errors of size 1 or 2.

  40. Group 8 Group 4 Group 2 Group 1 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 • Pattern generalizes to longer bit strings: • Single bit error corrupts one or more parity groups => minimum distance > 1 • Consider two-bit error in positions x and y. To avoid corrupting all groups: -- avoid group 1: bit 1 (lsb) same for both x and y -- avoid group 2: bit 2 same for both -- avoid group 4: bit 3 same for both, etc. -- forces x = y • So true two-bit error corrupts at least one parity group => min distance > 2 • Three-bit error (two msb and lsb) goes undetected => minimum distance = 3 • Conclude: process always produces a Hamming code with min distance = 3

  41. Traditional to permute check bits to far right Used in memory protection schemes

  42. Add traditional parity bit (for entire word) to obtain code with minimum distance = 4

  43. For minimum distance = 4: Number of check bits grows slowly with information bits

  44. Two-dimensional codes (product codes) min distance is product of row and column distances for simple parity on each (below) min distance is 4

  45. Scheme for RAID storage CRC is extension of Hamming codes each disk is separate row column is disk block (say 512 characters) rows have CRC on row disk columns have even parity

  46. Checksum codes mod-256 sum of info bytes becomes checksum byte mod-255 sum used in IP packets m-hot codes (m out of n codes) each valid codes has m ones in a frame of n bits min distance is 2 detect all unidirectional errors bit flips are all 0 to 1 or 1 to 0

  47. Serial data transmission (simple) transmit clock and sync with data (3 lines) (complex) recover clock and/or sync from data line

  48. Serial line codes: NRZ: clock recovery except during long sequences of zeros or ones NRZ1: nonreturn to zero, invert on one zero => no change, one => change polarity RZ: clock recovery except during long sequences of zero DC balance Bipolar return to zero (aka alternate mark inversion: send one as +1 or -1) Manchester: zero => 0 to 1 transition, one => 1 to 0 transition at interval center

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