Algorithm Design Techniques An Example The Problem Algorithm 1: Cubic Time Algorithm 2: Quadratic Time Algorithm 3: O(n log n) Time Algorithm 4: Linear.

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Presentation transcript:

Algorithm Design Techniques An Example The Problem Algorithm 1: Cubic Time Algorithm 2: Quadratic Time Algorithm 3: O(n log n) Time Algorithm 4: Linear Time Comparison of Algorithms Principles

The Problem Definition. Given the real vector x [n], compute the maximum sum found in any contiguous subvector. An Example. If the input vector is then the program returns the sum of x[2..6], or 187.

A Cubic Algorithm Idea. For all pairs of integers i and j satisfying 0 <= i<= j <n, check whether the sum of x [i.. j ] is greater than the maximum sum so far. Code. maxsofar = 0 for i = [0, n) for j = [i, n) sum = 0 for k = [i, j] sum += x[k] /* sum is sum of x[i..j] */ maxsofar = max(maxsofar, sum) Run Time. O(n 3 ).

A Quadratic Algorithm Idea. The sum of x [i.. j ] is close to the previous sum, x [i.. j - 1 ]. Code. maxsofar = 0 for i = [0, n) sum = 0 for j = [i, n) sum += x[j] /* sum is sum of x[i..j] */ maxsofar = max(maxsofar, sum) Run Time. O(n 2 ).

An O(n log n) Algorithm The Divide-and-Conquer Schema. To solve a problem of size n, recursively solve two sub-problems of size n / 2 and combine their solutions. The Idea. Divide into two sub-problems. ________________________ | a | b | Recursively find maximum in subvectors. _____________________________ | | ma | | mb | | Find maximum crossing sub-vector mc _____________________________ | | | mc | | | Return max of ma, mb and mc. Run Time. O(n log n).

Code for the O(N log N) Algorithm float maxsum3(l, u) if (l > u) /* zero elements */ return 0 if (l == u) /* one element */ return max(0, x[l]) m = (l + u) / 2 /* find max crossing to left */ lmax = sum = 0 for (i = m; i >= l; i--) sum += x[i] lmax = max(lmax, sum) /* find max crossing to right */ rmax = sum = 0 for i = (m, u] sum += x[i] rmax = max(rmax, sum) return max(lmax+rmax, maxsum3(l, m), maxsum3(m+1, u))

A Linear Algorithm Idea. How can we extend a solution for x [ 0..i - 1 ] into a solution for x [ 0..i]? Key variables: Code. maxsofar = 0 maxhere = 0 for i = [0, n) /* invariant: maxhere and maxsofar are accurate for x[0..i-1] */ maxhere = max(maxhere + x[i], 0) maxsofar = max(maxsofar, maxhere) Run Time. O(n). maxsofar maxhere