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Binary numbers and arithmetic

Binary numbers and arithmetic. addition. Addition (decimal). Addition (binary). Addition (binary). Addition (binary). So can we count in binary?. Counting in binary (4 bits). 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15. 0000 0001 …. multiplication. Multiplication (decimal).

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Binary numbers and arithmetic

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  1. Binary numbers and arithmetic

  2. addition

  3. Addition (decimal)

  4. Addition (binary)

  5. Addition (binary)

  6. Addition (binary) So can we count in binary?

  7. Counting in binary (4 bits) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0000 0001 …

  8. multiplication

  9. Multiplication (decimal)

  10. Multiplication (binary)

  11. Multiplication (binary) It’s interesting to note that binary multiplication is a sequence of shifts and adds of the first term (depending on the bits in the second term. 110100 is missing here because the corresponding bit in the second terms is 0.

  12. Representing signed (positive and negative) numbers

  13. Representing numbers (ints) • Fixed, finite number of bits. bits bytes C/C++ Intel Sun 8 1 char [s]byte byte 16 2 short [s]word half 32 4 int or long [s]dword word 64 8 long long [s]qword xword

  14. Representing numbers (ints) • Fixed, finite number of bits. bits Intel signed unsigned 8 [s]byte -27..+27-1 0..+28-1 16 [s]word -215..+215-1 0..+216-1 32 [s]dword -231..+231-1 0..+232-1 64 [s]qword -263..+263-1 0..+264-1 In general, for k bits, the unsigned range is [0..+2k-1] and the signed range is [-2k-1..+2k-1-1].

  15. Methods for representing signed ints. • signed magnitude • 1’s complement (diminished radix complement) • 2’s complement (radix complement) • excess bD-1

  16. Signed magnitude • Ex. 4-bit signed magnitude • 1 bit for sign • 3 bits for magnitude

  17. Signed magnitude • Ex. 4-bit signed magnitude • 1 bit for sign • 3 bits for magnitude

  18. 1’s complement(diminished radix complement) • Let x be a non-negative number. • Then –x is represented by bD-1+(-x) where b = base D = (total) # of bits (including the sign bit) • Ex. Let b=2 and D=4. Then -1 is represented by 24-1-1 = 1410 or 11102.

  19. 1’s complement(diminished radix complement) • Let x be a non-negative number. • Then –x is represented by bD-1+(-x) where b = base & D = (total) # of bits (including the sign bit) • Ex. What is the 9’s complement of 1238910? Given b=10 and D=5. Then the 9’s complement of 12389 = 105 – 1 – 12389 = 100000 – 1 – 12389 = 99999 – 12389 = 87610

  20. 1’s complement(diminished radix complement) • Let x be a non-negative number. • Then –x is represented by bD-1+(-x) where b = base D = (total) # of bits (including the sign bit) • Shortcut for base 2? • All combinations used, but 2 zeros!

  21. 2’s complement(radix complement) • Let x be a non-negative number. • Then –x is represented by bD+(-x). • Ex. Let b=2 and D=4. Then -1 is represented by 24-1 = 15 or 11112. • Ex. Let b=2 and D=4. Then -5 is represented by 24 – 5 = 11 or 10112. • Ex. Let b=10 and D=5. Then the 10’s complement of 12389 = 105 – 12389 = 100000 – 12389 = 87611.

  22. 2’s complement(radix complement) • Let x be a non-negative number. • Then –x is represented by bD+(-x). • Ex. Let b=2 and D=4. Then -1 is represented by 24-1 = 15 or 11112. • Ex. Let b=2 and D=4. Then -5 is represented by 24 – 5 = 11 or 10112. • Shortcut for base 2?

  23. 2’s complement(radix complement) • Shortcut for base 2? • Yes! Flip the bits and add 1.

  24. 2’s complement(radix complement) • Are all combinations of 4 bits used? • No. (Now we only have one zero.) • 1000 is missing! • What is 1000? • Is it positive or negative? • Does -8 + 1 = -7 work in 2’s complement?

  25. excess bD-1 (biased representation) • For pos, neg, and 0, x is represented by bD-1 + x • Ex. Let b=2 and D=4. Then the excess 8 (24-1) representation for 0 is 8+0 = 8 or 10002. • Ex. Let b=2 and D=4. Then excess 8 for -1 is 8 – 1 = 7 or 01112.

  26. excess bD-1 • For pos, neg, and 0, x is represented by bD-1 + x. • Ex. Let b=2 and D=4. Then the excess 8 (24-1) representation for 0 is 8+0 = 8 or 10002. • Ex. Let b=2 and D=4. Then excess 8 for -1 is 8 – 1 = 7 or 01112.

  27. 2’s complement vs. excess bD-1 • In 2’s, positives start with 0; in excess, positives start with 1. • Both have one zero (positive). • Remaining bits are the same.

  28. Summary of methods for representing signed ints. 1000=-8| 0000 unused

  29. Signed magnitude 1’s complement 2’s complement Excess K (biased) Binary arithmetic

  30. Signed magnitude Binary Arithmetic

  31. Addition w/ signed magnitude algorithm • For A - B, change the sign of B and perform addition of A + (-B) (as in the next step) • For A + B: • if (Asign==Bsign) then { R = |A| + |B|; Rsign = Asign; } • else if (|A|>|B|) then { R = |A| - |B|; Rsign = Asign; } • else if (|A|==|B|) then { R = 0; Rsign = 0; } • else { R = |B| - |A|; Rsign = Bsign; } • Complicated?

  32. 2’s complement Binary Arithmetic

  33. Representing numbers (ints) using 2’s complement • Fixed, finite number of bits. bits Intel signed 8 sbyte -27..+27-1 16 sword -215..+215-1 32 sdword -231..+231-1 64 sqword -263..+263-1 In general, for k bits, the signed range is [-2k-1..+2k-1-1]. So where does the extra negative value come from?

  34. Representing numbers (ints) • Fixed, finite number of bits. bits Intel signed 8 sbyte -27..+27-1 16 sword -215..+215-1 32 sdword -231..+231-1 64 sqword -263..+263-1 In general, for k bits, the signed range is [-2k-1..+2k-1-1]. So where does the extra negative value come from?

  35. Addition of 2’s complement binary numbers • Consider 8-bit 2’s complement binary numbers. • Then the msb (bit 7) is the sign bit. If this bit is 0, then this is a positive number; if this bit is 1, then this is a negative number. • Addition of 2 positive numbers. • Ex. 40 + 58 = 98

  36. Addition of 2’s complement binary numbers • Consider 8-bit 2’s complement binary numbers. • Addition of a negative to a positive. • What are the values of these 2 terms? • -88 and 122 • -88 + 122 = 34

  37. So how can we perform subtraction?

  38. Addition of 2’s complement binary numbers • Consider 8-bit 2’s complement binary numbers. • Subtraction is nothing but addition of the 2’s complement. • Ex. 58 – 40 = 58 + (-40) = 18 discard carry

  39. Carry vs. overflow

  40. Addition of 2’s complement binary numbers • Carry vs. overflow when adding A + B • If A and B are of opposite sign, then overflow cannot occur. • If A and B are of the same sign but the result is of the opposite sign, then overflow has occurred (and the answer is therefore incorrect). • Overflow occurs iff the carry into the sign bit differs from the carry out of the sign bit.

  41. Addition of 2’s complement binary numbers class test { public static void main ( String args[] ) { byte A = 127; byte B = 127; byte result = (byte)(A + B); System.out.println( "A + B = " + result ); } } #include <stdio.h> int main ( int argc, char* argv[] ) { char A = 127; char B = 127; char result = (char)(A + B); printf( "A + B = %d \n", result ); return 0; } Result = -2 in both Java (left) and C++ (right). Why?

  42. Addition of 2’s complement binary numbers class test { public static void main ( String args[] ) { byte A = 127; byte B = 127; byte result = (byte)(A + B); System.out.println( "A + B = " + result ); } } Result = -2 in both Java and C++. Why? What’s 127 as a 2’s complement binary number? What is 111111102? Flip the bits: 00000001. Then add 1: 00000010. This is -2.

  43. 1’s complement Binary Arithmetic

  44. Addition with 1’s complement • Note: 1’s complement has two 0’s! • 1’s complement addition is tricky (end-around-carry).

  45. 8-bit 1’s complement addition • Ex. Let X = A816 and Y = 8616. • Calculate Y - X using 1’s complement.

  46. 8-bit 1’s complement addition • Ex. Let X = A816 and Y = 8616. • Calculate Y - X using 1’s complement. Y = 1000 01102 = -12110 X = 1010 10002 = -8710 ~X = 0101 01112 (Note: C=0 out of msb.) Y - X = -121 + 87 = -34 (base 10)

  47. 8-bit 1’s complement addition • Ex. Let X = A816 and Y = 8616. • Calculate X - Y using 1’s complement.

  48. 8-bit 1’s complement addition • Ex. Let X = A816 and Y = 8616. • Calculate X - Y using 1’s complement. X = 1010 10002 = -8710 Y = 1000 01102 = -12110 ~Y = 0111 10012 (Note: C=1 out of msb.) end around carry X - Y = -87 + 121 = 34 (base 10)

  49. Excess K (biased) Binary Arithmetic

  50. Binary arithmetic and Excess K (biased) Method: Simply add and then flip the sign bit. -1 0111 +5 1101 -- ---- +4 0100 -> flip sign -> 1100 +1 1001 -5 0011 -- ---- -4 1100 -> flip sign -> 0100 +1 1001 +5 1101 -- ---- +6 0110 -> flip sign -> 1110 -1 0111 -5 0011 -- ---- -6 1010 -> toggle sign -> 0010 (Not used for integer arithmetic but employed in IEEE 754 floating point standard.)

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