CSE101: Design and Analysis of Algorithms. Ragesh Jaiswal, CSE, UCSD
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2 Greedy s
3 Greedy s Shortest path Claim 2: Let S be a subset of vertices containing s such that we know the shortest path length l(s, u) from s to any vertex in u S. Let e = (u, v) be an edge such that 1 u S, v V \ S, 2 (l(s, u) + W e ) is the least among all such cut edges. Then l(s, v) = l(s, u) + W e. Dijkstra s (G, s) - S {s} - d(s) 0 - While S does not contain all vertices in G - Let e = (u, v) be a cut edge across (S, V \ S) with minimum value of d(u) + W e - d(v) d(u) + W e - S S {v} What is the running time of the above algorithm? Same as that of the Prim s algorithm. O( E log V ).
4 Greedy s Shortest path Dijkstra s (G, s) - S {s} - d(s) 0 - While S does not contain all vertices in G - Let e = (u, v) be a cut edge across (S, V \ S) with minimum value of d(u) + W e - d(v) d(u) + W e - S S {v} Figure: d(s) = 0
5 Greedy s Shortest path Dijkstra s (G, s) - S {s} - d(s) 0 - While S does not contain all vertices in G - Let e = (u, v) be a cut edge across (S, V \ S) with minimum value of d(u) + W e - d(v) d(u) + W e - S S {v} Figure: d(s) = 0; d(a) = 1
6 Greedy s Shortest path Dijkstra s (G, s) - S {s} - d(s) 0 - While S does not contain all vertices in G - Let e = (u, v) be a cut edge across (S, V \ S) with minimum value of d(u) + W e - d(v) d(u) + W e - S S {v} Figure: d(s) = 0; d(a) = 1; d(e) = 9
7 Greedy s Shortest path Dijkstra s (G, s) - S {s} - d(s) 0 - While S does not contain all vertices in G - Let e = (u, v) be a cut edge across (S, V \ S) with minimum value of d(u) + W e - d(v) d(u) + W e - S S {v} Figure: d(s) = 0; d(a) = 1; d(e) = 9; d(d) = 15
8 Greedy s Shortest path Dijkstra s (G, s) - S {s} - d(s) 0 - While S does not contain all vertices in G - Let e = (u, v) be a cut edge across (S, V \ S) with minimum value of d(u) + W e - d(v) d(u) + W e - S S {v} Figure: d(s) = 0; d(a) = 1; d(e) = 9; d(d) = 15; d(c) = 17
9 Greedy s Shortest path Dijkstra s (G, s) - S {s} - d(s) 0 - While S does not contain all vertices in G - Let e = (u, v) be a cut edge across (S, V \ S) with minimum value of d(u) + W e - d(v) d(u) + W e - S S {v} Figure: d(s) = 0; d(a) = 1; d(e) = 9; d(d) = 15; d(c) = 17; d(b) = 21
10 Greedy s Shortest path Dijkstra s (G, s) - S {s} - d(s) 0 - While S does not contain all vertices in G - Let e = (u, v) be a cut edge across (S, V \ S) with minimum value of d(u) + W e - d(v) d(u) + W e - S S {v} Figure: The algorithm also implicitly produces a shortest path tree that gives the shortest paths from s to all vertices.
11 Course Overview Basic graph algorithms Design Techniques: Greedy s Divide and Conquer Dynamic Programming Network Flows Computational Intractability
12
13 Introduction You have already seen multiple examples of Divide and Conquer algorithms: Binary Search Quick Sort Multiplying two n-bit numbers in O ( n log 2 3) time.
14 Main Idea Main Idea: Divide the input into smaller parts. Solve the smaller parts and combine their solution.
15 Problem Given an array of unsorted integers, output a sorted array. MergeSort(A) - If ( A = 1) return(a) - Divide A into two equal parts A L and A R - B L MergeSort(A L ) - B R MergeSort(A R ) - B Merge(B L, B R ) - return(b)
16 MergeSort(A) - If ( A = 1) return(a) - Divide A into two equal parts A L and A R - B L MergeSort(A L ) - B R MergeSort(A R ) - B Merge(B L, B R ) - return(b)
17 MergeSort(A) - If ( A = 1) return(a) - Divide A into two equal parts A L and A R - B L MergeSort(A L ) - B R MergeSort(A R ) - B Merge(B L, B R ) - return(b)
18 MergeSort(A) - If ( A = 1) return(a) - Divide A into two equal parts A L and A R - B L MergeSort(A L ) - B R MergeSort(A R ) - B Merge(B L, B R ) - return(b)
19 MergeSort(A) - If ( A = 1) return(a) - Divide A into two equal parts A L and A R - B L MergeSort(A L ) - B R MergeSort(A R ) - B Merge(B L, B R ) - return(b)
20 MergeSort(A) - If ( A = 1) return(a) - Divide A into two equal parts A L and A R - B L MergeSort(A L ) - B R MergeSort(A R ) - B Merge(B L, B R ) - return(b)
21 MergeSort(A) - If ( A = 1) return(a) - Divide A into two equal parts A L and A R - B L MergeSort(A L ) - B R MergeSort(A R ) - B Merge(B L, B R ) - return(b)
22 MergeSort(A) - If ( A = 1) return(a) - Divide A into two equal parts A L and A R - B L MergeSort(A L ) - B R MergeSort(A R ) - B Merge(B L, B R ) - return(b)
23 MergeSort(A) - If ( A = 1) return(a) - Divide A into two equal parts A L and A R - B L MergeSort(A L ) - B R MergeSort(A R ) - B Merge(B L, B R ) - return(b)
24 MergeSort(A) - If ( A = 1) return(a) - Divide A into two equal parts A L and A R - B L MergeSort(A L ) - B R MergeSort(A R ) - B Merge(B L, B R ) - return(b)
25 MergeSort(A) - If ( A = 1) return(a) - Divide A into two equal parts A L and A R - B L MergeSort(A L ) - B R MergeSort(A R ) - B Merge(B L, B R ) - return(b) How do we argue correctness?
26 MergeSort(A) - If ( A = 1) return(a) - Divide A into two equal parts A L and A R - B L MergeSort(A L ) - B R MergeSort(A R ) - B Merge(B L, B R ) - return(b) How do we argue correctness? Proof of correctness of Divide and Conquer algorithms are usually by induction. Base case: This corresponds to the base cases of the algorithm. For the MergeSort, the base case is that the algorithm correctly sorts arrays of size 1. Inductive step: In general, this corresponds to correctly combining the solutions of smaller subproblems. For MergeSort, this is just proving that the Merge routine works correctly. This may again be done using induction and is left as an exercise.
27 MergeSort(A) - If ( A = 1) return(a) - Divide A into two equal parts A L and A R - B L MergeSort(A L ) - B R MergeSort(A R ) - B Merge(B L, B R ) - return(b) Let n be a power of 2 (e.g., n = 256) Let T (n) denote the worst case running time for the algorithm. Claim 1: T (1) c for some constant c.
28 MergeSort(A) - If ( A = 1) return(a) - Divide A into two equal parts A L and A R - B L MergeSort(A L ) - B R MergeSort(A R ) - B Merge(B L, B R ) - return(b) Let n be a power of 2 (e.g., n = 256) Let T (n) denote the worst case running time for the algorithm. Claim 1: T (1) c for some constant c. Claim 2: T (n) 2 T (n/2) + cn for all n 2.
29 MergeSort(A) - If ( A = 1) return(a) - Divide A into two equal parts A L and A R - B L MergeSort(A L ) - B R MergeSort(A R ) - B Merge(B L, B R ) - return(b) Let n be a power of 2 (e.g., n = 256) Let T (n) denote the worst case running time for the algorithm. Claim 1: T (1) c for some constant c. Claim 2: T (n) 2 T (n/2) + cn for all n 2. T (n) 2 T (n/2) + cn for n 2 and T (1) c is called a recurrence relation for the running time T (n). How do we solve such recurrence relation to obtain the value of T (n) as a function of n?
30 Let n be a power of 2 (e.g., n = 256) Let T (n) denote the worst case running time for the algorithm. Claim 1: T (1) c for some constant c. Claim 2: T (n) 2 T (n/2) + cn for all n 2. T (n) 2 T (n/2) + cn for n 2 and T (1) c is called a recurrence relation for the running time T (n). How do we solve such recurrence relation to obtain the value of T (n) as a function of n? Unrolling the recursion: Rewrite T (n/2) in terms of T (n/4) and so on until a pattern for the running time with respect to all levels of the recursion is observed. Then, combine these and get the value of T (n).
31 Recurrence relation for : T (n) 2 T (n/2) + cn for n 2 and T (1) c. How do we solve such recurrence relation to obtain the value of T (n) as a function of n? Unrolling the recursion: Rewrite T (n/2) in terms of T (n/4) and so on until a pattern for the running time with respect to all levels of the recursion is observed. Then, combine these and get the value of T (n).
32 Recurrence relation for : T (n) 2 T (n/2) + cn for n 2 and T (1) c. How do we solve such recurrence relation to obtain the value of T (n) as a function of n? So, the running time T (n) cn log n = O(n log n).
33 Solving recurrence relations Question: Suppose there is a divide and conquer algorithm where the recurrence relation for running time T (n) is the following: T (n) 2T (n/2) + cn 2 for n 2, and T (1) c. What is the solution of this recurrence relation in big-oh notation?
34 Master theorem Theorem Master Theorem: Let T (n) a T ( n b ) + c n k and T (1) c, Then? if a < b k T (n) =? if a = b k? if a > b k
35 End
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