1. Floating Point in computers Comply with standards: IEEE 754 ISO/IEC 559

2. Timeline Introductionquite short Binary reviewnot so long Integer Arithmetic1/3 Floating Point1/3 Floating Point Arithmetic1/3 Other issuesextra short

3. Introduction Who does computer arithmetic? Intel’s spare money How is it done in hardware? How Integer relates to Floating point Now, we go back to “computer structure”

4. Binary numbers What is 1 0 0 1 0 1 1. 0 0 1 0 1 ? 64 8 2 1

5. Signed Binary Integers Sign-magnitude 2’s complement 1’s complement biased

6. Sign-Magnitude High order bit = Sign 0101 = 5 1101 = -5 2 zero’s

7. 2’s complement Number + Negative = 2 n 0101 = 5 1011 = -5 Easy addition (drop carry) Formula: -a n-1 2 n-1 + a n-2 2 n-2 + … +a 1 2 1 + a 0

8. 1’s Complement Negative - complement to 1 0101 = 5 1010 = -5 2 zero’s Number + Negative = 2 n -1

9. Biased Binary = Number + Bias Bias = 5: 1101 = 55+5=10 0000 = -5(-5)+5 = 0 Relative order remains

10. Integer Arithmetic

11. Adding (usigned) Integers Elementry school : 1 1 0 0 1 1 0 1 1 0 0 0 0 1 1 0 + 110 1 0 1 1010 1 1 Result has n+1 bits!

12. Adding Integers - hardware Half Adder ab C in s C out ab s Full Adder 2 logical levels

13. Ripple carry Adder a n-1 b n-1 s n-1 C out a n-2 b n-2 C in s n-2 a1a1 b1b1 s1s1 a0a0 b0b0 s0s0 Slow - 2n logical levels Small constant (CMOS) Other ways exist

14. Adding Signed Integers In 2’s complement: b + (-a)= b + (2 n -a)= 2 n + (b-a) hence - add as integers, discard carry out Example:0011 + 1100 = ? = (2 n - (b+a)) + 2 n = (2n-b)+(2n-a)(-b) + (-a)

15. Substracting Integers Add the negation Negating 2’s complement: 11010100101011000110000 = ? 00001001010110101001110

16. Integer (unsigned) Multiplication Elementry school :1 1 0 1 1 0 0 1 1 1 0 1 0 0 1 1 0 1 0 1 1 1 0 1 0 1 * Result is 2n bits !

17. Hardware Multiplier P=0 loop:(i) if A 0 =1, add B to P (ii) right-shift P & A AP B Shift n n Carry n

18. Integer (unsigned) Division Elementry school : 110111 0 00 011 1 11 000 0 00 001 0 00 01 Result: 0100, Rem 1 Dec: 13/3=4, Rem 1

19. Hardware Divider P=0 loop:(i) left-shift P & A (ii) Sub. B from P: positive: a 0 =1 negative: a 0 =0, restore P (add B) AP B Shift n n+1 0

20. Example 13 / 3 = 4 (1) n=4 A=1101B=00011P=00000

21. PAB 0 0 0 1 10 0 0 0 0 1 1 0 1

22. PAB 0 0 0 1 10 0 0 0 1 0 1 0 0 Quotient Remainder

23. Division - remarks Non-restoring Algorithm Load P only if positive Check for 0 (Total) Result is 2n bits!

24. Integer arithmetic - remarks Signed Multiply and Division –Algorithms exist –We will not use them What to do with extra bits? Faster methods

25. Floating Point

26. Non Integers - Other Methods Fixed Point –example: # # #. # –Binary point shifted –Integer arithmetic (extra shifting) –Small number magnitude Rational –a/b(a,b  Z)

27. Floating Point Exponent + Significand (= Mantisa) x = s 2 e Example: s=101 e=011 x = 101 2 11 = 40= 5 2 3 = 101000

28. Uniqueness Denormal Numbers:123.456  10 7 0.123  10 4 Normalized:#.###  10 # 1.123  10 4 What about 0 ?

29. Floating Point Standard Why Standartize? –Hardware accelerators –Software compatibility –Build Software Libraries –etc….. IEEE 754-1985ISO/IEC 559 Includes: Structure, Arithmetic results

30. Float Types 4 Precision Types: –Single –Single extended –Double –Double extended

31. Single Precision 32 bits: Exponent (e):Biased ( + 127) Significand (f):Fixed fraction: 0. # # # … Nuber:1.f 2 e-127 11111111111111111111111111111111 Sign(1)Exponent(8)Significand(23)

32. Single Precision - Example 1 10000001 01000000000000000000000 10000001 = 129 01000… = 0.01000…  129-127=2 X = - 1.25 2 2 X = - 5  1.01= 1.25

33. Single Precision - Range E max = 127(e = 254) E min = -126(e = 1) Why |E min |<|E max |? –1/2 E min does not overflow Why Biased notation? What about 0 and 255 ?

34. Floating Point Precision

35. Exmaples We shall use base 10 sometimes: f will have 3 digits E max will be 98 E min will be -97 Ex:5.34  10 70

36. NaN Not a Number Result of ilegal computation: – –Any computation involving a NaN e = E max + 1&f  0 # 11111111 ####################### Many NaN’s (different f’s)

37. NaN’s in use Zero finder outside domain –f(x) = sqrt(x) - 1 Works since all computations NaN No exception caused !

38. Zero’s 0 00000000 00000000000000000000000 ? this is NOT 1.0  2 E min 1 00000000 00000000000000000000000 ?  0 is signed!  0 both exits! What is the difference?

39. Signed 0’os +0 = -0 BUT: Multiply/Divide keep sign rules: Monivation: –Using inf correctly (describe later) –log(x): log(0)=-inflog(negative)=Nan log(x) if x  (-0) ?

40. ± inf More logic: e = E max + 1&f = 0 # 11111111 00000000000000000000000

41. Inf usage Example (If tan -1 is defined properly)

42. More on 0’os and inf’s General Rule for 0/inf arithmetic: –Take appropriate limit: 1/(1/x) where x=0 or inf Why not Max # instead?

43. Zero’s and inf’s - yet again X/(x 2 +1) is bad!Why? 1/(x+x -1 ) is better Do we need to check for x=0? Using 2 zero’s and inf’s saves some special cases checks.

44. Denormalized numbers Example: –x=1.2310 - 98 y=1.1110 - 98 –x-y = 1.2010 - 99 = 0 –so: x-y=0 but: x  y –think of:if(x  y) then z=1/(x-y) Soluition: –use denormalized numbers!

45. Denormal Numbers Smallest normal: 1.0 2 E min Below, use denormal: 0.f 2 E min e = E min - 1&f  0 # 00000000 ####################### Gradual underflow: 1.23 10 -4 ( /10 ) 0.12 10 -4 ( /10 ) 0.01 10 -4 ( /10 ) 0

46. Denormal Numbers Back to our Example: –x=1.2310 - 98 y=1.1110 - 98 –x-y = 0.1210 - 98 –and this is not 0 !

47. Flush to 0 Vs Gradual Underflow 02 -4 2 -3 2 -1 2 -2 02 -4 2 -3 2 -1 2 -2

48. Special Values - Summary ExponentFractionRepresents E min -1 f=0  0 E min -1 f  0 0.f  2 E min E min  e  E max ---- 1.f  2 e E max +1 f=0  0 E max +1 f  0 0.f  2 E min

49. Rounding Why is rounding needed? Infinit numbers  Finit representation Integers only overflow Almost all operations need rounding IEEE - specifies algorithms for arithmetic

50. Numbers need rounding Out of range: –x>2  2 E max x<1  2 E min Between 2 floats: –0.1 10 = 0.00011001100…. 2 = 1.1001100….  2 -4 –1.1001  2 -4

51. Measuring Error ULPS(units in last place) –1.12  10 -1 Vs 0.124: 0.4 ulps –1.12  10 -1 Vs 0.118: 0.2 ulps Relative Error –Difference/Original –1.12  10 -1 Vs 0.124: Err=0.004/0.124=0.032

52. Calculate Using Rounding Benign cancellation –Calculate 10.1-9.93 (= 0.17) 1.01  10 1 0.99  10 1 0.02  10 1 = 2.00  10 -1 –30 upls!

53. Rounding problems Catastrophic cancellation –b 2 -4ac –both b 2 and 4ac are rounded –the (-) exposes the error –b=3.34 a=1.22 c=2.28 b 2 =11.2 4ac=11.1 b 2 -4ac=0.10 correct=0.0292(70.08 upls)

54. IEEE Arithmetic Requirement: + -    shold be EXACTLY rounded remaindershold be EXACTLY rounded Integer conv.shold be EXACTLY rounded Not all (transcendental, binary to decimal) “Tie break” - Round to Even

55. Round to Even How will 1.005 be rounded ? –Round Up:1.01 –Round Even:1.00 Why? Example: –x i =x i-1 +y-yx0=1.00 y=0.125 –Round up:1.00, 1.01, 1.02, …. –Round even:1.00, 1.00, 1.00, ….

56. Float Multiplication Integer multiply Biased additio n “ Biased addition ” : ­detect Overflow: Use n+1 bit adder ­detect Underflow:Harder (Denormals)

57. Rounding Multiplication 1.23 6.78 8.3394 X Round to 8.34 2.83 4.47 12.6501 X Round to 1.27 1.28 7.81 09.9968 X Round to 1.00 1.0001 1 1.0010 0 1.0010 1 0.1101 0 Round bit 0 Round bit 1 All rest 0 Round bit 1 All rest 0 Shift needed

58. Round, Guard, Sticky 0. 1 1 0 1 0 0 0 1 0 numberguardroundsticky 1. 0 0 1 0 0 0 1 0 0 numberroundsticky

59. Rounding Multiplication AP B Shift n n Carry n x 0 x 1.x 2 x 3 x 4 x 5 g r s s s s x 1.x 2 x 3 x 4 x 5 g X 0. x 1 x 2 x 3 x 4 x 5 Case 1: x 0 =0, shift Case 2: x 0 =1, inc. exp Product Results: Roun d digit Sticky bit

60. Rounding rules r=0  rounded OK r=1, s=1  add 1 to LSB r=1, s=0  add 1 if LSB=1 Denormals  Extra shifting

61. Float addition Compute all digits and round? –1.00  2 20 + 1.00  2 -20 = 10000000….0000001 –too long! Use Round and Sticky bits: –shift to same exponent –r = first discarded digit –s = OR of rest discarded

62. Float addition - example 1.10011.00001 1.10100 + r=1, s=1 Round needed!  1.10101 Calculate:1.10011  2 0 + 1.10001  2 -5 Shift exponents:1.10011  2 0 + 0.0000110001  2 0 r=1 s=0|0|0|1= 1

63. Signed Addition/Substraction Simplest way- convert to 2’s cmpl. Cancellation of high order bit - shift more bits cancel - How many guard digits? 1.00000 1.11111 0.11111 + 1.00000 0.00000101111 - 1.1111101000 1 cmpl

64. Float Division Integer division Biased substractio n Very similar to Multiplication Dividing using integer divide Compute 2 more bits (round, guard) Use remainder as sticky bit (Why?) Sign bit: XOR

65. More on floats

66. Rounding modes IEEE specifies 4 modes: –Nearest(default) –towards 0 –towards +inf –towards -inf affects overflow (How?)

67. Exceptions Set a flag at: –Underflow1.0  2 E min x 1.0  2 E min –Overflow1.0  2 E max x 1.0  2 E max –divide by 01/0 –inexactRounded was needed –invalidNaN return operations flags are sticky

68. Speeding up Different algorithms may be used Result should be exact divide SRT algorithm in pentium –5/2048 entries in a table –1/9,000,000 chance –check:

69. Precision Why extended precisions? –Return higher accuracy (D*D  ext. D) –use for computations: