The Greedy Method. Fundamental Techniques. The Greedy Method. Fractional Knapsack Problem

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1 Fundamental Techniques There are some algorithmic tools that are quite specialised. They are good for problems they are intended to solve, but they are not very versatile. There are also more fundamental (general) algorithmic tools that can be applied to a wide variety of different data structure and algorithm design problems. week 4 Complexity of Algorithms 1 The Greedy Method An optimisation problem (OP) is a problem that involves searching through a set of configurations to find one that minimises or maximizes an objective function defined on these configurations The greedy method solves a given OP going through a sequence of (feasible) choices The sequence starts from well-understood starting configuration, and then iteratively makes the decision that seems best from all those that are currently possible. week 4 Complexity of Algorithms 2 The Greedy Method The greedy approach does not always lead to an optimal solution. The problems that have a greedy solution are said to posses the greedy-choice property. The greedy approach is also used in the context of hard (difficult to solve) problems in order to generate an approximate solution. week 4 Complexity of Algorithms 3 Fractional Knapsack Problem In fractional knapsack problem, where we are given a set S of n items, s.t., each item I has a positive benefit b i and a positive weight w i, and we wish to find the maximum-benefit subset that doesn t exceed a given weight W. We are also allowed to to take arbitrary fractions of each item. week 4 Complexity of Algorithms 4 1

2 Fractional Knapsack Problem Fractional Knapsack Problem I.e., we can take an amount x i of each item i such that The total benefit of the items taken is determined by the objective function week 4 Complexity of Algorithms 5 week 4 Complexity of Algorithms 6 Fractional Knapsack Problem Fractional Knapsack Problem In the solution we use a heap-based PQ to store the items of S, where the key of each item is its value index With PQ, each greedy choice, which removes an item with the greatest value index, takes O(log n) time The fractional knapsack algorithm can be implemented in time O(n log n). Fractional knapsack problem satisfies the greedy-choice property, hence Thm: Given an instance of a fractional knapsack problem with set S of n items, we can construct a maximum benefit subset of S, allowing for fractional amounts, that has a total weight W in O(n log n) time. week 4 Complexity of Algorithms 7 week 4 Complexity of Algorithms 8 2

3 Task Scheduling Task Scheduling Suppose we are given a set T of n tasks, s.t., each task i has a start time s i and a completion time f i. Each task has to be performed on a machine and each machine can execute only one task at a time. Two tasks i and j are non-conflicting if fi sj or fj si. Two tasks can be executed on the same machine only if they are non-conflicting. week 4 Complexity of Algorithms 9 The task scheduling problem is to schedule all the tasks in T on the fewest machines possible in a nonconflicting way week 4 Complexity of Algorithms 10 Task Scheduling (algorithm) Task Scheduling (analysis) week 4 Complexity of Algorithms 11 In the algorithm TaskSchedule, we begin with no machines and we consider the tasks in a greedy fashion, ordered by their start times. For each task i, if we have the machine that can handle task i, then we schedule i on that machine. Otherwise, we allocate a new machine, schedule i on it, and repeat this greedy selection process until we have considered all the tasks in T. week 4 Complexity of Algorithms 12 3

4 Task Scheduling (analysis) Divide and Conquer Task scheduling problem satisfies the greedy-choice property, hence Thm: Given an instance of a task scheduling problem with set of n tasks, the algorithm TaskSchedule produces a schedule of the tasks with the minimum number of machines in O(n log n) time. Divide: if the input size is small then solve the problem directly; otherwise divide the input data into two or more disjoint subsets Recur: recursively solve the sub-problems associated with the subsets Conquer: take the solutions to the subproblems and merge them into a solution to the original problem week 4 Complexity of Algorithms 13 week 4 Complexity of Algorithms 14 Divide and Conquer Substitution Method To analyse the running time of a divideand-conquer algorithm we utilise a recurrence equation, where T(n) denotes the running time of the algorithm on an input of size n, and Characterise T(n) using an equation that relates T(n) to values of function T for problem sizes smaller than n, e.g., One way to solve a divide-and-conquer recurrence equation is to use the iterative substitution method, a.k.a., plug-and-chug method, e.g., having We get And after i-1 substitutions we have And for i = log n, we get week 4 Complexity of Algorithms 15 week 4 Complexity of Algorithms 16 4

5 Recursion Tree (visual approach) Guess-and-Prove In recursion tree method, some overhead (forming a part of a recurrence equation) is associated with every node of the tree. E.g., having In guess-and-prove method the solution to a recurrence equation is guessed and then proved by mathematical induction Where the overhead corresponds to summand +bn. We get The value of T(n) corresponds to the sum of all overheads. In this example, depth of the tree times overhead at each level, which is O(n log n) week 4 Complexity of Algorithms 17 We guess that T(n) = O(n log n). We have to prove that T(n) < C n log n for some constant C and large enough n. We use inductive assumption that T(n/2) < C n/2 log (n/2) = Cn/2 (log n 1) = (Cn log n)/2 Cn/2. T(n) = 2T(n/2) +bn < 2((Cn log n)/2 Cn/2) +bn = Cn log n + (-Cn + bn) < Cn log n, for any C > b. week 4 Complexity of Algorithms 18 The Master Method Matrix Multiplication Suppose we are given two n x n matrices X and Y, and we wish to compute their product Z=X Y, which is defined so that: week 4 Complexity of Algorithms 19 Which naturally leads to a simple O(n 3 ) time algorithm. week 4 Complexity of Algorithms 20 5

6 Matrix Multiplication Strassen s Algorithm Another way of viewing this product is in terms of sub-matrices: Define seven matrix products: where However this gives a divide-and-conquer algorithm with running time T(n), s.t., T(n) =8T(n/2) +bn 2 = O(n 3 ) week 4 Complexity of Algorithms 21 week 4 Complexity of Algorithms 22 Strassen s Algorithm Strassen s Algorithm Having S i s we can represent I, J, K, L: Thus, we can compute Z=XY using seven recursive multiplications of matrices of size (n/2) x (n/2), where One can prove, e.g., using Master Theorem, that: Thm: We can multiply two n x n matrices in O(n log 7 ) = O(n ) time. week 4 Complexity of Algorithms 23 week 4 Complexity of Algorithms 24 6

7 Dynamic Programming Dynamic Programming The dynamic programming (DP) algorithm-design technique is similar to divide-and-conquer technique. The main difference is in replacing (possibly) repetitive recursive calls by the reference to already computed values stored in a special table. DP technique is used primarily for optimisation problems We very often apply DP where the brute-force search for the best is infeasible However DP is efficient only if the problem has a certain amount of structure that we can exploit week 4 Complexity of Algorithms 25 week 4 Complexity of Algorithms 26 Dynamic Programming 0-1 Knapsack Problem Simple sub-problems: there must be a way of braking the whole optimisation problem into smaller pieces sharing a similar structure Sub-problem optimality: an optimal solution to the global problem must be a composition of optimal sub-problem solutions Sub-problem overlap: optimal solutions to unrelated sub-problems can contain subproblems in common week 4 Complexity of Algorithms 27 In 0-1 knapsack problem, is the knapsack problem where taking fractions of items is not allowed, i.e., each item s i S, for 1 i n, must be entirely accepted or rejected Item s i has a benefit b i (s.t., b 1 b 2 b n ) and an integer weight w i We have the following objective: where T S week 4 Complexity of Algorithms 28 7

8 0-1 Knapsack Problem 0-1 Knapsack Problem Exponential solution: we can easily solve 0-1 knapsack problem in O(2 n ) time by testing all possible subsets of items Unfortunately exponential complexity is not acceptable for large n and we rather have to focus on nice characterisation for sub-problems in order to use DP approach Let S k = {s i : i= 1,2,,k} Let B[k,w] be the maximum total benefit of a subset of S k from among all those subsets having total weight exactly w We have b[0,w]=0, for each w W, and week 4 Complexity of Algorithms 29 week 4 Complexity of Algorithms Knapsack Problem 0-1 Knapsack Problem The running time of the 01Knapsack algorithm is dominated by the two nested for-loops, where the outer one iterates n times and the inner one iterates at most W times Thm: 01Knapsack algorithm finds the highest benefit subset of S with total weight at most W in O(nW) time week 4 Complexity of Algorithms 31 week 4 Complexity of Algorithms 32 8