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CSCI 3160 Design and Analysis of Algorithms Tutorial 4

CSCI 3160 Design and Analysis of Algorithms Tutorial 4. Fei Chen. Outline. Dynamic programming Chain Matrix Multiplication Longest increasing subsequence (LIS) Goal: help understand the lecture materials further. Dynamic programming.

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CSCI 3160 Design and Analysis of Algorithms Tutorial 4

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  1. CSCI 3160 Design and Analysis of AlgorithmsTutorial 4 Fei Chen

  2. Outline • Dynamic programming • Chain Matrix Multiplication • Longest increasing subsequence (LIS) • Goal: help understand the lecture materials further

  3. Dynamic programming • “A method for solving complex problems by breaking them down into simpler subproblems” (from Wikipedia) • Three steps • Find the optimal substructure • Formulate the problem recursively • Compute the values bottom up

  4. Chain Matrix Multiplication • You want to multiply N matrices A1, …, AN • Suppose the cost of multiplying an m-by-n matrix with an n-by-l matrix is mnl. • The product is an m-by-l matrix • What is the minimum total cost to get the product of the N matrices P = A1…AN?

  5. Chain Matrix Multiplication • Each order of multiplication corresponds to a parenthesization ((A1(A2A3))(A4((A5A6)A7))) • Optimal substructure • If the above parenthesization is optimal, then • (A1(A2A3)) is optimal for multiplying A1, …, A3 • (A4((A5A6)A7)) is optimal for multiplying A4, …, A7 • ((A5A6)A7)is optimal for multiplying A5, …, A7 • Every “subparenthesization” of an optimal parenthesization is optimal

  6. Chain Matrix Multiplication • Recurrence • Let C(i,j) be the optimal cost of multiplying Ai, … Aj • C(i,i) = 0 for any i (No multiplication needed) • C(i,j) = mini ≤ k ≤ j-1 {C(i,k) + C(k+1,j) + mi-1mkmj } for i ≤ j-1 • Build an N-by-N matrix to store the C(i,j)’s • Our optimal value is then C(1,N) mi (Ai) mi-1

  7. Example • Let A1 be a 5-by-20 matrix, A2 be a 20-by-10 matrix, A3 be a 10-by-3 matrix, A4 be a 3-by-2 matrix • m0 = 5, m1 = 20, m2 = 10, m3 = 3, m4 = 2

  8. Example • m0 = 5, m1 = 20, m2 = 10, m3 = 3, m4 = 2 • C(1,2) = C(1,1) + C(2,2) + m0m1m2= 5 · 20 · 10 = 1000 • C(2,3) = , C(3,4) = .

  9. Example • m0 = 5, m1 = 20, m2 = 10, m3 = 3, m4 = 2 • C(1,2) = C(1,1) + C(2,2) + m0m1m2 = 5 · 20 · 10 = 1000 • C(2,3) = 600, C(3,4) = 60.

  10. Example • m0 = 5, m1 = 20, m2 = 10, m3 = 3, m4 = 2 • C(1,3) = min { C(1,1) + C(2,3)+ m0m1m3, C(1,2) + C(3,3) + m0m2m3 } = min { 0 + 600 + 300, 1000 + 0 + 150 } = 900

  11. Example • m0 = 5, m1 = 20, m2 = 10, m3 = 3, m4 = 2 • C(2,4) = min { C(2,2) + C(3,4) + m1m2m4, C(2,3) + C(4,4) + m1m3m4 } = min { 0 + 60 + 400, 600+ 0 + 30 } = 460

  12. Example • m0 = 5, m1 = 20, m2 = 10, m3 = 3, m4 = 2 • C(1,4) = min { C(1,1) + C(2,4) + m0m1m4, , } = min { 0 + 460 + 200, , } = .

  13. Example • m0 = 5, m1 = 20, m2 = 10, m3 = 3, m4 = 2 • C(1,4) = min { C(1,1) + C(2,4) + m0m1m4, C(1,2) + C(3,4) + m0m2m4, C(1,3) + C(4,4) + m0m3m4} = min { 0 + 460 + 200, 1000 + 60 + 100, 900 + 0 + 30 } = 660.

  14. Example • Hence, the optimal cost of computing P = A1…A4is 660.

  15. Analysis • Space complexity = O(N2) • Time complexity = O(N3) • A for loop (k = i .. j-1) is run every time we compute the value of a table entry

  16. Longest increasing subsequence (LIS) • Given a sequence of N integers a1, …, aN, find a subsequence ai1, …, aik (1 ≤ i1 < … < ik ≤ N) such that • Increasing: ai1 < … < aik • Longest: its length k is maximum among all increasing subsequences • Example: 3, 1, 6, 0, 8 • Can you give another LIS?

  17. Longest increasing subsequence (LIS) • We can construct a graph • Optimal substructure • If ai1, …, aik is an LIS ending at position ik, then ai1, …, aij is an LIS ending at position ij for any j≤k • In the language of longest paths: a subpath of a longest path is longest 3 1 6 0 8

  18. Longest increasing subsequence (LIS) • Add an imaginary node “-∞” at the front • Recurrence • Let L(j) be the length of the LIS ending at position j • L(0) = 0 • L(j) = 1 + maxi : ai < aj L(i), for j ≥ 1 • Our longest length is then maxj L(j) -∞ 3 1 6 0 8

  19. Example • Use a linear array to store the values -∞ 3 1 6 0 8

  20. Example -∞ 3 1 6 0 8

  21. Example -∞ 3 1 6 0 8

  22. Example -∞ 3 1 6 0 8

  23. Example -∞ 3 1 6 0 8

  24. Example -∞ 3 1 6 0 8

  25. Example • Length of LIS = maxj L(j) = 3 -∞ 3 1 6 0 8

  26. Analysis • Space complexity = O(N) • O(N2) if you store the graph, since there can be Θ(N) arrows pointing to a node • Time complexity = O(N2) • Each arrow is gone through once

  27. End • Questions

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