CS473  Algorithms I


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1 CS473  Algorithms I Lecture 4 The DivideandConquer Design Paradigm View in slideshow mode 1
2 Reminder: Merge Sort Input array A sort this half sort this half Divide Conquer merge two sorted halves Combine 2
3 The DivideandConquer Design Paradigm 1. Divide the problem (instance) into subproblems. 2. Conquer the subproblems by solving them recursively. 3. Combine subproblem solutions. CS473 Lecture 4 3
4 Example: Merge Sort 1. Divide: Trivial. 2. Conquer: Recursively sort 2 subarrays. 3. Combine: Linear time merge. T(n) = 2 T(n/2) + Θ(n) # subproblems subproblem size work dividing and combining CS473 Lecture 4 4
5 Master Theorem: Reminder T(n) = at(n/b) + f(n) Case 1: T(n) = Case 2: T(n) = Case 3: and a f (n/b) c f (n) for c < 1 T(n) = 5
6 Merge Sort: Solving the Recurrence T(n) = 2 T(n/2) + Θ(n) a = 2, b = 2, f(n) = Θ(n), n log ba = n Case 2: T(n) = holds for k = 0 T(n) = Θ (nlgn) 6
7 Binary Search Find an element in a sorted array: 1. Divide: Check middle element. 2. Conquer: Recursively search 1 subarray. 3. Combine: Trivial. Example: Find CS473 Lecture 4 7
8 Binary Search Find an element in a sorted array: 1. Divide: Check middle element. 2. Conquer: Recursively search 1 subarray. 3. Combine: Trivial. Example: Find CS473 Lecture 4 8
9 Binary Search Find an element in a sorted array: 1. Divide: Check middle element. 2. Conquer: Recursively search 1 subarray. 3. Combine: Trivial. Example: Find CS473 Lecture 4 9
10 Binary Search Find an element in a sorted array: 1. Divide: Check middle element. 2. Conquer: Recursively search 1 subarray. 3. Combine: Trivial. Example: Find CS473 Lecture 4 10
11 Binary Search Find an element in a sorted array: 1. Divide: Check middle element. 2. Conquer: Recursively search 1 subarray. 3. Combine: Trivial. Example: Find CS473 Lecture 4 11
12 Binary Search Find an element in a sorted array: 1. Divide: Check middle element. 2. Conquer: Recursively search 1 subarray. 3. Combine: Trivial. Example: Find CS473 Lecture 4 12
13 Recurrence for Binary Search T(n) = 1 T(n/2) + Ө(1) # subproblems subproblem size work dividing and combining CS473 Lecture 4 13
14 Binary Search: Solving the Recurrence T(n) = T(n/2) + Θ(1) a = 1, b = 2, f(n) = Θ(1), n log ba = n0 = 1 Case 2: T(n) = holds for k = 0 T(n) = Θ (lgn) 14
15 Powering a Number Problem: Compute a n, where n is a natural number NaivePower (a, n) powerval 1 for i 1 to n powerval powerval. a return powerval What is the complexity? T(n) = Θ (n) 15
16 Powering a Number: Divide & Conquer Basic idea: a n = a n/2. a n/2 even a (n1)/2. a (n1)/2. a if n is if n is odd 16
17 Powering a Number: Divide & Conquer POWER (a, n) if n = 0 then return 1 else if n is even then val POWER (a, n/2) return val * val else if n is odd then val POWER (a, (n1)/2) return val * val * a 17
18 Powering a Number: Solving the Recurrence T(n) = T(n/2) + Θ(1) a = 1, b = 2, f(n) = Θ(1), n log ba = n0 = 1 Case 2: T(n) = holds for k = 0 T(n) = Θ (lgn) 18
19 Matrix Multiplication Input : A = [a ij ], B = [b ij ]. Output: C = [c ij ] = A. B. i, j = 1, 2,..., n. c 11 c c 1n a 11 a a 1n b 11 b b 1n c 21 c c 2n = a 21 a a 2n. b 21 b b 2n c n1 c n2... c nn a n1 a n2... a nn b n1 b n2... b nn c ij = 1 k n a ik.b kj CS473 Lecture 4 19
20 Standard Algorithm for i 1 to n do for j 1 to n do c ij 0 for k 1 to n do c ij c ij + a ik. b kj Running time = Θ(n 3 ) CS473 Lecture 4 20
21 Matrix Multiplication: Divide & Conquer IDEA: Divide the n x n matrix into 2x2 matrix of (n/2)x(n/2) submatrices C A B c 11 =. c 12 a 11 a 12 b 11 b 12 c 21 c 22 a 21 a 22 b 21 b 22 c 11 = a 11 b 11 + a 12 b 21 21
22 Matrix Multiplication: Divide & Conquer IDEA: Divide the n x n matrix into 2x2 matrix of (n/2)x(n/2) submatrices C A B c 11 b 11 =. c 12 a 11 a 12 b 12 c 21 c 22 a 21 a 22 b 21 b 22 c 12 = a 11 b 12 + a 12 b 22 22
23 Matrix Multiplication: Divide & Conquer IDEA: Divide the n x n matrix into 2x2 matrix of (n/2)x(n/2) submatrices C A B c 11 a 11 a 12 =. c 12 b 11 b 12 c 21 c 22 a 21 a 22 b 21 b 22 c 21 = a 21 b 11 + a 22 b 21 23
24 Matrix Multiplication: Divide & Conquer IDEA: Divide the n x n matrix into 2x2 matrix of (n/2)x(n/2) submatrices C A B c 11 c 12 a 11 b 11 =. a 12 b 12 c 21 c 22 a 21 a 22 b 21 b 22 c 22 = a 21 b 12 + a 22 b 22 24
25 Matrix Multiplication: Divide & Conquer C A B c 11 c 12 a 11 a 12 b =. 11 b 12 c 21 c 22 a 21 a 22 b 21 b 22 c 11 = a 11 b 11 + a 12 b 21 c 12 = a 11 b 12 + a 12 b 22 c 21 = a 21 b 11 + a 22 b 21 8 mults of (n/2)x(n/2) submatrices 4 adds of (n/2)x(n/2) submatrices c 22 = a 21 b 12 + a 22 b 22 25
26 Matrix Multiplication: Divide & Conquer MATRIXMULTIPLY (A, B) // Assuming that both A and B are nxn matrices if n = 1 then return A * B else partition A, B, and C as shown before c 11 = MATRIXMULTIPLY (a 11, b 11 ) + MATRIXMULTIPLY (a 12, b 21 ) c 12 = MATRIXMULTIPLY (a 11, b 12 ) + MATRIXMULTIPLY (a 12, b 22 ) c 21 = MATRIXMULTIPLY (a 21, b 11 ) + MATRIXMULTIPLY (a 22, b 21 ) c 22 = MATRIXMULTIPLY (a 21, b 12 ) + MATRIXMULTIPLY (a 22, b 22 ) return C 26
27 Matrix Multiplication: Divide & Conquer Analysis T(n) = 8 T(n/2) + Θ(n 2 ) 8 recursive calls each subproblem has size n/2 submatrix addition 27
28 Matrix Multiplication: Solving the Recurrence T(n) = 8 T(n/2) + Θ(n 2 ) a = 8, b = 2, f(n) = Θ(n 2 ), n log ba = n 3 Case 1: T(n) = T(n) = Θ (n 3 ) No better than the ordinary algorithm! 28
29 Matrix Multiplication: Strassen s Idea C A B c 11 c 12 a 11 a 12 b =. 11 b 12 c 21 c 22 a 21 a 22 b 21 b 22 Compute c 11, c 12, c 21, and c 22 using 7 recursive multiplications 29
30 Matrix Multiplication: Strassen s Idea P 1 = a 11 x (b 12  b 22 ) P 2 = (a 11 + a 12 ) x b 22 P 3 = (a 21 + a 22 ) x b 11 P 4 = a 22 x (b 21  b 11 ) P 5 = (a 11 + a 22 ) x (b 11 + b 22 ) P 6 = (a 11  a 22 ) x (b 21 + b 22 ) P 7 = ( a 11  a 21 ) x (b 11 + b 12 ) Reminder: Each submatrix is of size (n/2)x(n/2) Each add/sub operation takes Θ(n 2 ) time Compute P 1..P 7 using 7 recursive calls to matrixmultiply How to compute c ij using P 1.. P 7? 30
31 Matrix Multiplication: Strassen s Idea P 1 = a 11 x (b 12  b 22 ) P 2 = (a 11 + a 12 ) x b 22 P 3 = (a 21 + a 22 ) x b 11 P 4 = a 22 x (b 21  b 11 ) P 5 = (a 11 + a 22 ) x (b 11 + b 22 ) P 6 = (a 11  a 22 ) x (b 21 + b 22 ) P 7 = ( a 11  a 21 ) x (b 11 + b 12 ) c 11 = P 5 + P 4 P 2 + P 6 c 12 = P 1 + P 2 c 21 = P 3 + P 4 c 22 = P 5 + P 1 P 3 P 7 7 recursive multiply calls 18 add/sub operations Does not rely on commutativity of multiplication 31
32 Matrix Multiplication: Strassen s Idea P 1 = a 11 x (b 12  b 22 ) P 2 = (a 11 + a 12 ) x b 22 P 3 = (a 21 + a 22 ) x b 11 P 4 = a 22 x (b 21  b 11 ) P 5 = (a 11 + a 22 ) x (b 11 + b 22 ) P 6 = (a 11  a 22 ) x (b 21 + b 22 ) P 7 = ( a 11  a 21 ) x (b 11 + b 12 ) e.g. Show that c 12 = P 1 +P 2 c 12 = P 1 + P 2 = a 11 (b 12 b 22 )+(a 11 +a 12 )b 22 = a 11 b 12 a 11 b 22 +a 11 b 22 +a 12 b 22 = a 11 b 12 +a 12 b 22 32
33 Strassen s Algorithm 1. Divide: Partition A and B into (n/2) x (n/2) submatrices. Form terms to be multiplied using + and. 2. Conquer: Perform 7 multiplications of (n/2) x (n/2) submatrices recursively. 3. Combine: Form C using + and on (n/2) x (n/2) submatrices. Recurrence: T(n) = 7 T(n/2) + Ө(n 2 ) 33
34 Strassen s Algorithm: Solving the Recurrence T(n) = 7 T(n/2) + Θ(n 2 ) a = 7, b = 2, f(n) = Θ(n 2 ), n log ba = n lg7 Case 1: T(n) = T(n) = Θ (n lg7 ) Note: lg
35 Strassen s Algorithm The number 2.81 may not seem much smaller than 3 But, it is significant because the difference is in the exponent. Strassen s algorithm beats the ordinary algorithm on today s machines for n 30 or so. Best to date: Θ(n ) (of theoretical interest only) 35
36 VLSI Layout: Binary Tree Embedding Problem: Embed a complete binary tree with n leaves into a 2D grid with minimum area. Example: 36
37 Binary Tree Embedding Use divide and conquer root LEFT SUBTREE RIGHT SUBTREE 1. Embed the root node 2. Embed the left subtree 3. Embed the right subtree What is the minarea required for n leaves? 37
38 Binary Tree Embedding root H(n/2) EMBED LEFT SUBTREE HERE EMBED RIGHT SUBTREE HERE H(n) = H(n/2) + 1 W(n/2) W(n/2) W(n) = 2W(n/2)
39 Binary Tree Embedding Solve the recurrences: W(n) = 2W(n/2) + 1 H(n) = H(n/2) + 1 W(n) = Ө(n) H(n) = Ө(lgn) Area(n) = Ө(nlgn) 39
40 Binary Tree Embedding Example: W(n) H(n) 40
41 Binary Tree Embedding: HTree Use a different divide and conquer method root left subtree 1 subtree 2 subtree 3 right subtree 4 1. Embed root, left, right nodes 2. Embed subtree 1 3. Embed subtree 2 4. Embed subtree 3 5. Embed subtree 4 What is the minarea required for n leaves? 41
42 Binary Tree Embedding: HTree H(n/4) H(n/4) SUBTREE 1 SUBTREE 2 SUBTREE 3 SUBTREE 4 W(n) = 2W(n/4) + 1 H(n) = 2H(n/4) + 1 W(n/4) W(n/4) 42
43 Binary Tree Embedding: HTree Solve the recurrences: W(n) = 2W(n/4) + 1 H(n) = 2H(n/4) + 1 W(n) = Ө( ) H(n) = Ө( ) Area(n) = Ө(n) 43
44 Binary Tree Embedding: HTree Example: W(n) H(n) 44
45 Correctness Proofs Proof by induction commonly used for D&C algorithms Base case: Show that the algorithm is correct when the recursion bottoms out (i.e., for sufficiently small n) Inductive hypothesis: Assume the alg. is correct for any recursive call on any smaller subproblem of size k (k < n) General case: Based on the inductive hypothesis, prove that the alg. is correct for any input of size n 45
46 Example Correctness Proof: Powering a Number POWER (a, n) if n = 0 then return 1 else if n is even then val POWER (a, n/2) return val * val else if n is odd then val POWER (a, (n1)/2) return val * val * a 46
47 Example Correctness Proof: Powering a Number Base case: POWER (a, 0) is correct, because it returns 1 Ind. hyp: Assume POWER (a, k) is correct for any k < n General case: In POWER (a, n) function: If n is even: val = a n/2 (due to ind. hyp.) it returns val. val = a n If n is odd: val = a (n1)/2 (due to ind. hyp.) it returns val. val. a = a n The correctness proof is complete 47
48 Maximum Subarray Problem Input: An array of values Output: The contiguous subarray that has the largest sum of elements Input array: the maximum contiguous subarray 48
49 Maximum Subarray Problem: Divide & Conquer Basic idea: Divide the input array into 2 from the middle Pick the best solution among the following: 1. The max subarray of the left half 2. The max subarray of the right half 3. The max subarray crossing the midpoint Crosses the midpoint A Entirely in the left half Entirely in the right half 49
50 Maximum Subarray Problem: Divide & Conquer Divide: Trivial (divide the array from the middle) Conquer: Recursively compute the max subarrays of the left and right halves Combine: Compute the maxsubarray crossing the midpoint (can be done in Θ(n) time). Return the max among the following: 1. the max subarray of the left subarray 2. the max subarray of the right subarray 3. the max subarray crossing the midpoint See textbook for the detailed solution. 50
51 Conclusion Divide and conquer is just one of several powerful techniques for algorithm design. Divideandconquer algorithms can be analyzed using recurrences and the master method (so practice this math). Can lead to more efficient algorithms CS473 Lecture 4 51
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