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Algorithms Midterm 2011

Algorithms Midterm 2011. Karl Lieberherr. Question 1. The textbook answer: HSR( n,k ) <= 2*k*n^(1/k) is not precise enough to answer the questions. Need precise answer by modified Pascal Triangle. HSR(99,4) <= 7 is true HSR(99,4) <= 6 is false HSR(56,3) <= 7 is true

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Algorithms Midterm 2011

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  1. Algorithms Midterm 2011 Karl Lieberherr

  2. Question 1 • The textbook answer: HSR(n,k) <= 2*k*n^(1/k) is not precise enough to answer the questions. • Need precise answer by modified Pascal Triangle. • HSR(99,4) <= 7 is true • HSR(99,4) <= 6 is false • HSR(56,3) <= 7 is true • Decision tree construction based on MR recurrence, MR(k,q) = MR(k-1,q-1) + MR(k,q-1); MR(q,q) = 2^q;MR(0,q)=1.

  3. Question 2a • Alice defends 1000*n + n^2 in O(n^2). • Alice’ claim is : Exists C,n0>0 ForAll n>n0: 1000*n + n^2 <= C*n^2. • The refutation protocol is: Alice claims, Bob tries to refute, Alice provides C,n0 and Bob provides n so that 1000*n + n^2 > C*n^2. • Bing/Wolfram|Alpha: If Alice provides C=2 and n0=1000, Bob will not be able to refute. Next slide.

  4. 1000*n + n^2 = 2 * n^2 n =0 and n = 1000 Only highest degree counts.

  5. Question 2b • Alice’ Claim: 1000*log(n) in O(n) • Bing/Wolfram|Alpha: Alice chooses C = 1 and n = 10000 and Bob will not be able to refute. • Refutation protocol is the same as in 2a.

  6. Decimal approximation 1000 * log(10000) =

  7. Question 3 • Does graph G have an independent set of size 7? • The algorithm tries all (n choose 7) subsets of the n nodes. Is Theta(n^7) because (n choose 7) in Theta(n^7). • Can use other problems/algorithms.

  8. Question 4 • Reverse edges of graph. O(n+m) • Do depth-first or breadth-first from v3. If you encounter v1 or v2, stop and return true else return false. • O(n+m)

  9. Question 5 • Improve the following Topological Ordering Algorithm: • TopologicalOrdering(G) • Find node v with in-degree 0 and order it first. • Recursively compute TopologicalOrdering(G-v) and append this order after v.

  10. Question 5 (continued) • Need second algorithm Predec: For each node compute the number of predecessors M: V-> non-negative int. Predec: G -> M. • Need third algorithm IU to incrementally update M, G when node v is deleted: IU: G,M,v -> G’,M’. • Need fourth algorithm InitZ: M -> Z where Z is the set of all nodes with zero predecessors in M. • Need fifth algorithm IUWatch to watch updates of M when v is deleted and update set Z of nodes with 0 predecessors. IUWatch: G,M,v,Z -> G’,M’,Z’.

  11. Question 5 (continued) • compose algorithms • preprocess: compose Predec: G -> M with InitZ: M -> Z. Composition can be done more efficiently than sequential composition. Not an asymptotic improvement. • Around “Find a node v with in-degree 0” : “choose element of Z”. • IUWatch: G,M,v,Z -> G’,M’,Z’. Composing IU: G,M,v -> G’,M’ and InitZ: M’ -> Z’ achieves the same result but inefficiently. Composition must be done more efficiently to achieve asymptotic improvement. • TS(G’,M’,Z’)

  12. Question 5 (continued)summary • TopologicalOrdering(G) • Preprocess: compose Predec: G -> M with InitZ: M -> Z. • TopOrder(G,M,Z) • Find node v with in-degree 0 is replaced by: choose v in Z. Order v first. • IUWatch: G,M,v,Z -> G’,M’,Z’. • Recursively compute TopOrder(G’,M’,Z’) and append this order after v.

  13. Question 5 (continued)analysis • G=(V,E), |V|=n, |E|=m. • TopologicalOrdering(G) • Preprocess: compose Predec: G -> M with InitZ: M -> Z. O(n+m). M[0..n-1] as array. • TopOrder(G,M,Z) • Find node v with in-degree 0 is replaced by: choose v in Z. Order v first. O(1). • IUWatch: G,M,v,Z -> G’,M’,Z’. overall: O(n+m) • Recursively compute TopOrder(G’,M’,Z’) and append O(1) this order after v. • O(n+m)

  14. Question 6 • Alice claims Bertrand’s Postulate: ForAll natural numbers n>1 Exists a prime p: n < p < 2*n. • Refutation protocol: Alice claims. Bob tries to refute. Bob provides n. Alice provides p. Refutation is successful iff • !(prime(p) and n<p<2*n)

  15. Question 7: TentGraph Cartesian Product

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