By Toshihide Ibaraki (auth.), Hon Wai Leong, Hiroshi Imai, Sanjay Jain (eds.)

ISBN-10: 3540638903

ISBN-13: 9783540638902

This publication constitutes the refereed complaints of the eighth overseas Symposium on Algorithms and Computation, ISAAC'97, held in Singapore in December 1997. The forty two revised complete papers provided have been chosen from a complete of ninety eight submissions. The scope of the amount spans the complete sector of algorithms from discrete arithmetic and complexity idea to algorithms layout and assessment in quite a few applicational components. one of the issues addressed are scheduling and logistics, networking and routing, combinatorial optimization, graph-computations, algorithmic studying, computational geometry, etc.

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**Extra info for Algorithms and Computation: 8th International Symposium, ISAAC '97 Singapore, December 17–19, 1997 Proceedings**

**Sample text**

One can not require A < β m B at input, since this condition may not be satisfied in the recursive calls. Consider for example A = 5517, B = 56 with β = 10: the first recursive call will divide 55 by 5, which yields a two-digit quotient 11. Even A ≤ β m B is not recursively fulfilled, as this example shows. , A < β m (B + 1). 19). 22. 2 Algorithm RecursiveDivRem is correct, and uses D(n + m, n) operations, where D(n + m, n) = 2D(n, n − m/2) + 2M (m/2) + O(n). In particular D(n) := D(2n, n) satisfies D(n) = 2D(n/2)+2M (n/2)+ O(n), which gives D(n) ∼ M (n)/(2α−1 − 1) for M (n) ∼ nα , α > 1.

This proves that 2j1 a′ , 2j1 b′ are two consecutive terms of the remainder sequence of a, b. Since a and a1 differ by a multiple of 22k1 +1 , a′ and a′1 differ by a multiple of 22k1 +1−2j1 ≥ 2 since j1 ≤ k1 by induction. It follows that ν(a′ ) = 0. Similarly, b′ and b′1 differ by a multiple of 2, thus j0 = ν(b′ ) > 0. The second recursive call uses k2 < k, since by induction j1 ≥ 0 and we just showed j0 > 0. It easily follows that j1 + j0 + j2 > 0, and thus j ≥ 0. If we exit at step 9, we have j = j1 ≤ k1 < k.

If the while-loop is entered, the same reasoning as above holds. We conclude that when the for-loop ends, 0 ≤ A < B holds, and since m ( j qj β j )B + A is invariant throughout the algorithm, the quotient Q and remainder R are correct. The most expensive part is step 5, which costs O(n) operations for qj B (the multiplication by β j is simply a word-shift); the total cost is O(n(m+1)). ) Here is an example of algorithm BasecaseDivRem for the inputs A = 766 970 544 842 443 844 and B = 862 664 913, with β = 1000, which gives quotient Q = 889 071 217 and remainder R = 778 334 723.

### Algorithms and Computation: 8th International Symposium, ISAAC '97 Singapore, December 17–19, 1997 Proceedings by Toshihide Ibaraki (auth.), Hon Wai Leong, Hiroshi Imai, Sanjay Jain (eds.)

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