Basic Algorithm Design Techniques Divide and conquer Dynamic Programming Greedy Common Theme To solve a large complicated problem break it into many smaller subproblems Dynamic Programming ID: 1001363
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1. Lecture 4 Dynamic Programming
2. Basic Algorithm Design TechniquesDivide and conquerDynamic ProgrammingGreedyCommon Theme: To solve a large, complicated problem, break it into many smaller sub-problems.
3. Dynamic ProgrammingIdea: Break the problem into many closely related sub-problems, memorize the result of the sub-problems to avoid repeated computation.
4. Warmup Example: Fibonacci Numbersf(n) = f(n-1) + f(n-2), f(1) = f(2) = 1
5. Naïve solutionFollow the recursion directlyf(7)f(6)f(5)f(5)f(4)f(4)f(3)f(4)f(3)… …
6. Naïve solutionFollow the recursion directlyf(7)f(6)f(5)f(5)f(4)f(4)f(3)f(4)f(3)… …
7. Memoizationf(n)IF n < 3 THEN RETURN 1. IF f(n) has been computed before THEN RETURN stored result.res = f(n-1) + f(n-2)Mark f(n) as computed, Store f(n) = resRETURN resn12345678f(n)1123581321Dynamic Programming Table
8. Filling in the table iterativelyObservation: f(n) is always computed in the order of n = 1, 2, 3, …No need to recursef(n)IF n < 3 THEN RETURN 1. a[1] = a[2] = 1FOR i = 3 TO n a[i] = a[i-1] + a[i-2]RETURN a[n]
9. Basic Steps in Designing Dynamic Programming AlgorithmsRelate the problem recursively to smaller sub-problems. (Transition function, f(n) = f(n-1)+f(n-2))Organize all sub-problems as a dynamic programming table. (A table for all values of n)Fill in values in the table in an appropriate order.(n = 1, 2, 3, …)
10. Example 1: Longest Increasing SubsequenceInput: Array of numbersa[] = {4, 2, 5, 3, 9, 7, 8, 10, 6}Subsequence: list of numbers appearing in the same order, but may not be consecutiveExample: {4, 2, 5}, {4, 3, 8}, {2, 5, 7, 8, 10}Subsequence b[] is increasing if b[i+1] > b[i] for all i.Output: Length of the longest increasing subsequence.Example: 5, the subsequence is {2, 5, 7, 8, 10}or {2, 3, 7, 8, 10} (both have length 5)
11. Example 2 Knapsack ProblemThere is a knapsack that can hold items of total weight at most W. There are now n items with weights w1,w2,…, wn. Each item also has a value v1,v2,…,vn.Goal: Select some items to put into knapsack1. Total weight is at most W.2. Total value is as large as possible.