1. Exact Computation Theory and Techniques Serge Adamowsky & Lina Ourima Seminar Algorithmen WS 03 / 04

2. Contents Evaluating One Determinant Expression Error-bounds Versus Precision-bounds Error-Bounds Precision-bounds Theory of Exact Computation

3. Evaluating One Determinant Expression View Expression as labled dag (=directed acyclic graph) Each sink node is a number Each internal node is an operator

4. Example

5. Error-bounds Versus Precision- bounds Error-bounds Bottom-up Deterministic Depends on error of input Precision-bounds Top down Not deterministic User-specified quantity

6. Example: Error-bounds

7. Example: Precision-bounds

8. Error Bounds Number Representation (Notation) Normalizing Error Digits Examples for Normalization Algorithms for Maintaining bigFloat Ranges Addition Multiplication Square Root

9. Number Representation (Notation) bigFloats used as aproximation b = 1212/343434  0.35 * 10 -2 bigFloats used with error-bounds: b‘ = (0.35  0.01) * 10-2 = (35, -2, 1) (f, e, d) :=  f  d  *B e The number is exact when d = 0

10. Normalizing Error Digits (35, -2, 1) 2 = (1225, -4, 71) Storing many insignificant digits is a waste of space and time Number is normalized when f doesnt start with 0 and 0  d < B Delayed normalization improves accuracy

11. Examples for Normalization

12. Algorithms for Maintaining bigFloat Ranges Addition Multiplication Square root With a global precision bound (gpb) c‘ = a‘  b‘: the system produces the most accurate range without the exceeding global p. bound

13. Addition A safe Algorithm: Align the least significant bits Add the digits and error values Normalize (123, 0, 1) + (246, -2, 3) (12300, 0, 100) + (246, -2, 3) = (12546, 0, 103) Normalized: (125, 0, 2)

14. Addition Problem: far more digits computed then nessesary Solution: Find the least significant bit (gpb, magnitude of both summands) and don‘t use smaller bits (123, 0, 1) + (2, -2, 1) = (125, 0, 2)

15. Multiplication (f1, e1, d1) * (f2, e2, d2) = (f1 f2, e1+ e2, f1 d2+ f2 d1+ e1 e2) If f1, f2 positive f1 and f2 (both n digits) not exact => result doesnt have 2n but n significant digits Don‘t compute more digits then the significant ones (like in Addition)

16. Square Root  Newton iteration:  i+1  ½  i  i   1,  2,... converges toward   i would be exact but  Might be not exact and operations can‘t be done accurate still  lies between  i and  i+1

17. Composite Precision Bounds X approximates x absolute precision(AP): |X-x|  2 -a relative precision(RP): |X-x|  |x| 2 -r composite precision(CP): |X-x|  max(2 -a, |x| 2 -r ) written: X = x[a,r] CP  AP r =  CP  RP a = 

18. Algorithms Most significant bit (MSB) mx :=  log|x|  mx[a,r] = MX  2Mx  |x| < 21+Mx + max{2e-r, 2a} Computing Mx is expensive Often a lower bound for Mx is sufficient

19. Theory of Exact Computation Computation of semi-algebraic problems is theoretically possible. An algebraic number(AN) is any complex root of a polynomial (with integer coefficients)  2 is AN, since it is root of x 2 -2 Isolating interval representation:  2 = (x2-2, [1,2])

20. Semi-algebraic Predicates  (x1,..., xn) is a Boolean combination of P(x1,..., x) p 0 p  {=,, ,  } Example:  0(x, y) : (x2 + y2 -1 < 0)  (x + y -5 = 0) Is the union of an open disc and a straight line

21. Tarski Formula A first-order formula based on semi- algebraic predicates is a true Tarski sentence. Example: (  x)(  y)  0(x,y) Sets defined by Tarski formulas are semi- algebraic.

22. Practical Use? Theorem: A semi-algebraic problem can be solved in double-exponential time on a Turing-machine. Mainly of theoretical interest Practical use if some instances can be computed quickly (double exponential time is worst case)

23. Theory of Root Bounds Comparing values is a basic operation It is sufficient to compare to zero  determine the sign of an expression How can we know a number is 0 without looking at infinite signs and without using „some heuristic cut-off epsilon value“?

24. Detecting Zero h is a height bound on  The height of an algebraic number is the maximum value of the coefficients in its minimal polynomial Cauchy‘s bound says:  is non-zero iff |  | > 1 / (1+h) Evaluate  to an AP of 1+log(1+h) to determine the sign