# Fundamental Techniques

## Presentation on theme: "Fundamental Techniques"— Presentation transcript:

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

Complexity of Algorithms
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

Complexity of Algorithms
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

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 bi and a positive weight wi, 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

Fractional Knapsack Problem
I.e., we can take an amount xi of each item i such that The total benefit of the items taken is determined by the objective function week 4 Complexity of Algorithms

Fractional Knapsack Problem
week 4 Complexity of Algorithms

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). week 4 Complexity of Algorithms

Fractional Knapsack Problem
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

Complexity of Algorithms
Task Scheduling Suppose we are given a set T of n tasks, s.t., each task i has a start time si and a completion time fi. 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

Complexity of Algorithms
Task Scheduling The task scheduling problem is to schedule all the tasks in T on the fewest machines possible in a non-conflicting way week 4 Complexity of Algorithms

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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

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. week 4 Complexity of Algorithms

Complexity of Algorithms
Divide and Conquer 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 sub-problems and merge them into a solution to the original problem week 4 Complexity of Algorithms

Complexity of Algorithms
Divide and Conquer To analyse the running time of a divide-and-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., week 4 Complexity of Algorithms

Complexity of Algorithms
Substitution Method 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

Recursion Tree (visual approach)
In recursion tree method, some overhead (forming a part of a recurrence equation) is associated with every node of the tree. E.g., having 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

Complexity of Algorithms
Guess-and-Prove In guess-and-prove method the solution to a recurrence equation is guessed and then proved by mathematical induction 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

Complexity of Algorithms
The Master Method week 4 Complexity of Algorithms

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: Which naturally leads to a simple O(n3) time algorithm. week 4 Complexity of Algorithms

Matrix Multiplication
Another way of viewing this product is in terms of sub-matrices: where However this gives a divide-and-conquer algorithm with running time T(n), s.t., T(n) =8T(n/2) +bn2 = O(n3) week 4 Complexity of Algorithms

Complexity of Algorithms
Strassen’s Algorithm Define seven matrix products: week 4 Complexity of Algorithms

Complexity of Algorithms
Strassen’s Algorithm Having Sis we can represent I, J, K, L: week 4 Complexity of Algorithms

Complexity of Algorithms
Strassen’s Algorithm 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(nlog 7) = O(n2.808) time. week 4 Complexity of Algorithms

Complexity of Algorithms
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. week 4 Complexity of Algorithms

Complexity of Algorithms
Dynamic Programming 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

Complexity of Algorithms
Dynamic Programming 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 sub-problems in common week 4 Complexity of Algorithms

Complexity of Algorithms
0-1 Knapsack Problem In 0-1 knapsack problem, is the knapsack problem where taking fractions of items is not allowed, i.e., each item si  S, for 1  i  n, must be entirely accepted or rejected Item si has a benefit bi (s.t., b1  b2  …  bn) and an integer weight wi We have the following objective: where T S week 4 Complexity of Algorithms

Complexity of Algorithms
0-1 Knapsack Problem Exponential solution: we can easily solve 0-1 knapsack problem in O(2n) 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 week 4 Complexity of Algorithms

Complexity of Algorithms
0-1 Knapsack Problem Let Sk = {si: i= 1,2,…,k} Let B[k,w] be the maximum total benefit of a subset of Sk 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

Complexity of Algorithms
0-1 Knapsack Problem week 4 Complexity of Algorithms

Complexity of Algorithms
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