Representing Numbers on the Computer

Transcription

1 Representing Numbers on the Computer Let s calculate J! Program Maxnumber * * Check what the max numbers are on the computer * Integer N * * N= Do J=,2 N=N*J Print *,J,N Enddo stop end What happened? J N=J! Winter Semester 26/7 Computational Physics I Lecture 2

2 Representing Integers E.g., single precision: 4 bytes or 32 bits bit is used for the sign ( for - for +) 3 bits for value Because start from Biggest integer = = 2 s complement is standard for integer representation. 8 bit example (from Wikipedia) Sign Bit Value -28 Winter Semester 26/7 Computational Physics I Lecture

3 Representing Integers To calculate the 2 s complement value: N * =2 n -N, where n is the number of bits used to represent an integer. N * is the 2 s complement representation of the negative of N. e.g., N =5 N 2 = (n=8) N * 2 =2n -N 2 = - = 2 s complement is convenient for computer calculations There is no rounding error - only a maximum allowed range for the values. For the mathematicians: 2 n possible values of n bits form a ring of equivalence classes Winter Semester 26/7 Computational Physics I Lecture 2 3

4 Or, invert the bits and add. E.g., 5 = Representing Integers To convert to -5, flip the bits Then add The other way, to go from -5 to 5, flip the bits And add Winter Semester 26/7 Computational Physics I Lecture 2 4

5 Representing Real Numbers Representation of real numbers (scientific notation - IEEE754): Mantissa and Exponent + sign bit. E.g., single precision (4 bytes) = 32 bits (sign) (exponent) (mantissa) Double precision = 64 bits (sign) (exponent) (mantissa) x = ( ) s iai2 b E s is sign bit a is normalized so first bit is (radix point - implicit) E = /2 of (maximum exponent -), or E= in single precision Winter Semester 26/7 Computational Physics I Lecture 2 5

6 Example: 4/7 on the computer: Representing Real Numbers 4 7 = = = = = = =. =.2 s = a = (first is implicit) b-e=- or, b 2 = -, b 2 = Winter Semester 26/7 Computational Physics I Lecture 2 6

7 Representing Real Numbers If the exponent b=, the number has a special value: if a=, value is ± depending on s else, value is NaN (not a number) If b= x=±.a 2-26 Otherwise x=±.a 2 b-27 (single precision) Precision # bits a b Relative precision Max magnitude Min magnitude (normalize d) single (255-27) double (247-23) Winter Semester 26/7 Computational Physics I Lecture 2 7

8 Calculation of As an example, consider the calculation of using the following algorithm (due to Madhava of Sangamagrama, Indian Mathematician of the 4th century) = 2 () i ( 2i + )3 i i= First 6 digits of correct value I single precision double precision Error After 2 iterations, single precision good to -7 Double precision to - Winter Semester 26/7 Computational Physics I Lecture 2 8

9 Calculation of Close to -6 Winter Semester 26/7 Computational Physics I Lecture 2 9

10 Calculation of Dear folks, 2th October 25 Our latest record which was announced already at press release time of 6-th of December, 22 was as the followings; Declared record: hexadecimal digits,24,,, decimal digits Two independent hexadecimal calculation based on two different algorithms generated more than,3,775,43, hexadecimal digits of pi and comparison of two generated sequences matched completely. Computed hexadecimal digits of pi were radix converted into base, generating more than,24,77,3, decimal digits of pi and generated decimal digits of pi were radix converted again into base 6. Radix converted hexadecimal digits of pi were compared with original hexadecimal digits of pi. There were no difference up to,24,,, decimal digits. Then we are declaring,3,7,, hexadecimal digits and,24,,, decimal digits as the new world records. Details of computed results are available on the following URL's. (hexadecimal) (decimal) Winter Semester 26/7 Computational Physics I Lecture 2

11 Rounding Errors for Simple Sum In contrast to integers, there are rounding errors for real numbers.the error resulting from adding two numbers: y = x + x 2 [ ] where rd() means computer rounding y = rd rd(x ) + rd(x 2 ) y [ x ( + ) + x 2 ( + ) ]( + ) where is the typical relative error 2 t where t is the number of bits assigned to the mantissa single precision, = y x + x 2 + (x + x 2 ) + x + x 2 2 double precision, = y y y + x x + x 2 + x 2 x + x 2 2 Can get large multiplication of relative error if x -x 2 Winter Semester 26/7 Computational Physics I Lecture 2

12 Error Propagation More generally (see Lecture Notes from Scherer): Input data x = (x,, x n ) Output data y = (y,, y m ) where y = ( x) = (r) (r ) () and the are simple functions Define x = () ( x) x i = (i) ( x i ) y = (r) ( x r ) Treat all errors as small, represent with x Winter Semester 26/7 Computational Physics I Lecture 2 2

13 Error Propagation First step: x = rd( () ( x + x)) ( () ( x) + D () x )( + E ) where D () = and E = () x i x j = n () x x x x n x n x x n x n First order in errors x = x x D () x + () ( x)e Winter Semester 26/7 Computational Physics I Lecture 2 3

14 Error Propagation y ye r + D (r) () x + D (r) (2) x E ++ D (r) xr E r D = D (r) () = y x y x n y m x y m x n The first term is the inevitable rounding error The second term contains the propagation of the input errors and initial rounding errors. The other terms depend on how the algorithm is set up. Winter Semester 26/7 Computational Physics I Lecture 2 4

15 Let s look at the individual terms: Error Propagation ye r i y i The rounding error on the final answer D (r) () x i j y i x j x j Propagation of input errors The other terms depend on the specific algorithm. The goal is for the algorithm to not give errors larger than the first two (unavoidable) errors. Winter Semester 26/7 Computational Physics I Lecture 2 5

16 Error Propagation Let us look at an example in detail - the calculation of a 2 -b 2 Procedure I:. Calculate a 2 and b 2 2. Calculate their difference Unavoidable error: x = a b x = x 2 2 x 2 y = x x 2 y = a 2 b 2 y () = a 2 b 2 + 2( a + b ) y x j = (a2 b 2 ) j x j a + (a2 b 2 ) b = 2( a + b ) Winter Semester 26/7 Computational Physics I Lecture 2 6

17 Error Propagation Let us look at an example in detail - the calculation of a 2 -b 2 Procedure I:. Calculate a 2 and b 2 2. Calculate their difference x = a b Error magnitude estimation: x = x 2 2 x 2 y = x x 2 x = a( + a ) b( + b ) x = a( + a )a( + a )( + ) b( + b )b( + b )( + 2 ) a2 ( + 2 a + ) b 2 ( + 2 b + 2 ) y = a 2 ( + 2 a + ) b 2 ( + 2 b + 2 ) ( + 2 ) y a 2 b 2 + 3(a 2 + b 2 ) Winter Semester 26/7 Computational Physics I Lecture 2 7

18 Error Propagation Procedure II:. Calculate a-b and a+b 2. Calculate their product x = a b Error magnitude estimation: x = x x 2 x + x 2 y = x ix 2 x = a( + ) a ( a( + a ) b( + b ))( + ) b( + b ) x = (a b)( + ) + a b a b ( a( + a ) + b( + b ))( + 2 ) (a + b)( + 2 ) + a a + b b y = (a 2 b 2 )( ) + 2a 2 a 2b 2 b ( + 2 ) y 3 a 2 b 2 + 2(a 2 + b 2 ) Winter Semester 26/7 Computational Physics I Lecture 2 8

19 Error Propagation Single precision a b Exact value (a 2 -b 2 ) a 2 -b 2 (a-b)(a+b) Winter Semester 26/7 Computational Physics I Lecture 2 9

20 Exercises 2. Look up a different algorithm to calculate from the one presented in the lecture and code it in single and double precision. Compare the speed of convergence to the one shown in class. 2. Calculate (a 4 -b 4 ) numerically in single and double precision. Compare the resulting accuracy to the true value for test cases. Compare to the expected precision for single and double precision calculations. Winter Semester 26/7 Computational Physics I Lecture 2 2