Numeric-Symbolic Exact Rational Linear System Solver

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1 Numeric-Symbolic Exact Rational Linear System Solver B David Saunders, David Wood, and Bryan Youse University of Delaware June 10, 2011

2 Motivation Ax = b Given A Z m n and b Z m, compute x Q n Core problem specifics: Square m = n matrices Matrix entries of length d bits or fewer

3 Background Competing Approaches Purely Symbolic [Dixon, 1982] Solve system modulo p Use Hensel lifting to obtain a p-adic expansion of solution Rational reconstruction from p-adic approximants

4 Background Competing Approaches Purely Symbolic [Dixon, 1982] Solve system modulo p Use Hensel lifting to obtain a p-adic expansion of solution Rational reconstruction from p-adic approximants Numeric-Symbolic [Wan, 2006] Numeric iterative refinement to obtain dyadic number solution of high accuracy Specifically, 2 lg(h), the Hadamard bound of the input system Rational reconstruction from dyadic approximants

5 Fact If two rational numbers r 1 = a/b, r 2 = c/d are given in lowest terms, and r 1 r 2, then r 1 r 2 1/bd That is, though dense in the real number line, rational numbers with bound denominators are discrete

6 Fact If two rational numbers r 1 = a/b, r 2 = c/d are given in lowest terms, and r 1 r 2, then r 1 r 2 1/bd That is, though dense in the real number line, rational numbers with bound denominators are discrete A given real number r can be represented by a continued fraction: 1 r = a 0 +, where a i Z a a 2 + a 1 3 +

7 Iterative Refinement System

8 Iterative Refinement System Numeric Solver

9 Iterative Refinement System Numeric Solver approx soln

10 Iterative Refinement System Numeric Solver n: Update dyadic solution approx soln d: 2 k s

11 Iterative Refinement System Numeric Solver n: Update dyadic solution approx soln d: 2 k, k k + s s s

12 Iterative Refinement System Numeric Solver n: Update dyadic solution approx soln d: 2 k, k k + s s s

13 Iterative Refinement System Numeric Solver n: Update dyadic solution approx soln d: 2 k, k k + s s s enough accuracy?

14 Iterative Refinement System Numeric Solver n: Update dyadic solution approx soln d: 2 k, k k + s s s enough accuracy? yes Proceed to Rational Reconstruction

15 Iterative Refinement System n: Update dyadic solution Numeric Solver approx soln Apply matrix to soln d: 2 k, k k + s s s enough accuracy? yes no Proceed to Rational Reconstruction

16 Iterative Refinement System Obtain residual n: Update dyadic solution Numeric Solver approx soln Apply matrix to soln d: 2 k, k k + s s s enough accuracy? yes no Proceed to Rational Reconstruction

17 Iterative Refinement System Obtain residual amplify by s n: Update dyadic solution Numeric Solver approx soln Apply matrix to soln d: 2 k, k k + s s s enough accuracy? yes no Proceed to Rational Reconstruction

18 Iterative Refinement System Obtain residual n: Update dyadic solution Numeric Solver approx soln Apply matrix to soln d: 2 k, k k + s s s enough accuracy? yes no Proceed to Rational Reconstruction

19 Iterative Refinement System Obtain residual n: Update dyadic solution Numeric Solver approx soln Apply matrix to soln d: 2 k, k k + s s s enough accuracy? yes no Proceed to Rational Reconstruction

20 Iterative Refinement System Obtain residual n: Update dyadic solution Numeric Solver approx soln Apply matrix to soln d: 2 k, k k + s s s enough accuracy? yes no Proceed to Rational Reconstruction

$partial solution ˆx, then split into: ˆx int = ˆx 2 s ˆx frac = ˆx 2 s$

21 Confirmed Continuation (Overlap) At each iteration, we scale our partial solution ˆx, then split into: ˆx int = ˆx 2 s ˆx frac = ˆx 2 s ˆx int

$Confirmed Continuation (Overlap) At each iteration, we scale our partial solution ˆx, then split into: ˆx int = ˆx 2 s ˆx frac = ˆx 2 s ˆx int$

22 Confirmed Continuation (Overlap) At each iteration, we scale our partial solution ˆx, then split into: ˆx int = ˆx 2 s ˆx frac = ˆx 2 s ˆx int Before updating dyadic approximants, we confirm continuation of the iteration by verifying overlap between the current ˆx and the previous ˆx frac

23 Confirmed Continuation (Overlap) At each iteration, we scale our partial solution ˆx, then split into: ˆx int = ˆx 2 s ˆx frac = ˆx 2 s ˆx int Before updating dyadic approximants, we confirm continuation of the iteration by verifying overlap between the current ˆx and the previous ˆx frac max ˆx ˆx frac 1 ensures b bits of overlap 2 b One bit is (typically) sufficient

24 Rational Reconstruction Like rational reconstruction from p-adic digits, Euclidean remainder sequence used to recover exact solution from dyadic approximants

25 Rational Reconstruction Like rational reconstruction from p-adic digits, Euclidean remainder sequence used to recover exact solution from dyadic approximants P-adically a remainder sequence entry is (r i, s i, t i ), where r i = s i x + t i p k, the remainder r i is a numerator candidate and coefficient s i of the given residue x is a denominator candidate

26 Rational Reconstruction Like rational reconstruction from p-adic digits, Euclidean remainder sequence used to recover exact solution from dyadic approximants P-adically a remainder sequence entry is (r i, s i, t i ), where r i = s i x + t i p k, the remainder r i is a numerator candidate and coefficient s i of the given residue x is a denominator candidate Dyadically a remainder sequence entry is (r i, s i, t i ), where r i = s i x + t i 2 k, the remainder r i is a measure of error and coefficient s i of the given dyadic approximation x is a denominator candidate

27 Rational Reconstruction - single element example Example Solving Ax = b, suppose: x 1 = 9/17 the Hadamard bound for det(a) is h = 2 8 = 256 In general we need 2 lg(h) = 16 bits of approximation

28 Rational Reconstruction - single element example Example Solving Ax = b, suppose: x 1 = 9/17 the Hadamard bound for det(a) is h = 2 8 = 256 In general we need 2 lg(h) = 16 bits of approximation Dyadic approximation to 16 bits: n d = Stopping condition: a b n d 1 2d a b

29 Rational Reconstruction - single element example Example Solving Ax = b, suppose: x 1 = 9/17 the Hadamard bound for det(a) is h = 2 8 = 256 In general we need 2 lg(h) = 16 bits of approximation Dyadic approximation to 16 bits: n d = a b Stopping condition: a b n d 1 2d bn ad b 2

30 Rational Reconstruction - single element example Example Solving Ax = b, suppose: x 1 = 9/17 the Hadamard bound for det(a) is h = 2 8 = 256 In general we need 2 lg(h) = 16 bits of approximation Dyadic approximation to 16 bits: n d = a b Stopping condition: a b n d 1 2d bn ad b = 0 n + 1 d = 1 n + 0 d = 1 n + 1 d 3856 = 2 n + 1 d 3848 = 15 n + 8 d 8 = 17 n + 9 d

31 Output sensitivity (early termination) Dyadic approximation to 12 bits: n d = a b 4096 = 0 n + 1 d 2168 = 1 n + 0 d 1928 = 1 n + 1 d 240 = 2 n + 1 d 8 = 17 n + 9 d No random choice of prime, therefore no probabilistic early termination But guaranteed early termination in some cases Idea explored in-depth by Steffy [2010]

32 Vector rational reconstruction Suppose solution to Ax = b is (9/17, 3/11, 7/187) and we have computed the approximation to 12 bits: 2168,1117,

33 Vector rational reconstruction Suppose solution to Ax = b is (9/17, 3/11, 7/187) and we have computed the approximation to 12 bits: 2168,1117, x 1 : As we saw, 9/17 is found with certainty 17 = lcm of denominators so far

34 Vector rational reconstruction Suppose solution to Ax = b is (9/17, 3/11, 7/187) and we have computed the approximation to 12 bits: 2168,1117, x 1 : As we saw, 9/17 is found with certainty 17 = lcm of denominators so far x 2 : Try division once- failure ( 1491 = ) Find 3/11 with single element reconstruction 187 = lcm of denominators so far

35 Vector rational reconstruction Suppose solution to Ax = b is (9/17, 3/11, 7/187) and we have computed the approximation to 12 bits: 2168,1117, x 1 : As we saw, 9/17 is found with certainty 17 = lcm of denominators so far x 2 : Try division once- failure ( 1491 = ) Find 3/11 with single element reconstruction 187 = lcm of denominators so far x 3 : Try division once- success! ( 61 = ) 7/187 found with small enough remainder ( )

36 Performance 16 Relative Running Time Overlap Method on Zero-One systems Z vs Dixon Z vs Wan Matrix Order

37 Performance Relative Running Time: Overlap Dixon Q [27 11] R [38 30] Z [36 28] M [31 18] m S Matrix Order

38 Conclusion and Ongoing Work Overlap method results in consistently faster, more robust performance over algorithm predecessor

39 Conclusion and Ongoing Work Overlap method results in consistently faster, more robust performance over algorithm predecessor Output sensitive early termination is a proven avenue for runtime savings

40 Conclusion and Ongoing Work Overlap method results in consistently faster, more robust performance over algorithm predecessor Output sensitive early termination is a proven avenue for runtime savings Specialized numeric solvers fit easily into the framework

41 Conclusion and Ongoing Work Overlap method results in consistently faster, more robust performance over algorithm predecessor Output sensitive early termination is a proven avenue for runtime savings Specialized numeric solvers fit easily into the framework Work ongoing to incorporate highly-tuned direct (SuperLU) and iterative sparse solvers

42 The End

Rational numbers as decimals and as integer fractions

$Rational numbers as decimals and as integer fractions$ Rational numbers as decimals and as integer fractions Given a rational number expressed as an integer fraction reduced to the lowest terms, the quotient of that fraction will be: an integer, if the denominator