Algorithm

Getting Started with AlgorithmWhat is an Algorithm?

Characteristics of Algorithm1 Topic

Analysis Framework

Performance Analysis3 Topics

Mathematical Analysis2 Topics

Sorting AlgorithmSorting Algorithm10 Topics

Searching Algorithm6 Topics

Fundamental of Data StructuresStacks

Queues

Graphs

Trees

Sets

Dictionaries

Divide and ConquerGeneral Method

Binary Search

Recurrence Equation for Divide and Conquer

Finding the Maximum and Minimum

Merge Sort

Quick Sort

Stassen’s Matrix Multiplication

Advantages and Disadvantages of Divide and Conquer

Decrease and ConquerInsertion Sort

Topological Sort

Greedy MethodGeneral Method

Coin Change Problem

Knapsack Problem

Job Sequencing with Deadlines

Minimum Cost Spanning Trees2 Topics

Single Source Shortest Paths1 Topic

Optimal Tree Problem1 Topic

Transform and Conquer Approach1 Topic

Dynamic ProgrammingGeneral Method with Examples

Multistage Graphs

Transitive Closure1 Topic

All Pairs Shortest Paths6 Topics

BacktrackingGeneral Method

NQueens Problem

Sum of Subsets problem

Graph Coloring

Hamiltonian Cycles

Branch and Bound2 Topics

0/1 Knapsack problem2 Topics

NPComplete and NPHard Problems1 Topic
Stassen’s Matrix Multiplication
Let A and B be two n x n matrices. The product matrix C= AB is also an n x n matrix whose i,jth element is formed by taking the elements in the ith row of A and the jth column of Band multiplying them to get for all i and j between 1 and n. To compute C(i,j) using this formula, we need n multiplications.As the matrix C has n^2 elements,the time for the resulting matrix multiplication algorithm,which we refer to as the conventional method is theta(n^3).
The divideandconquer strategy suggests another way to compute the product of two n x n matrices. For simplicity we assume that n is a power of 2, that is,that there exists a non negative integer k such that n = 2^k. In case? i is not a power of two, then enough rows and columns of zero scan be added to both A and B so that the resulting dimensions area power of two (see the exercises for more on this subject). Imagine that A and B are each partitioned into four square sub matrices, each sub matrix having dimensions n/2*n/2. Then the product AB can be computed by using the above formula for the product of 2 x 2 matrices: if AB is
If n = 2, then formulas (a) and (b) are computed using a multiplication operation for the elements of A and B. These elements are typically floating point numbers. For n > 2, the elements of C can be computed using matrix multiplication and addition operations applied to matrices of size n/2 * n/2. Since n is a power of 2, these matrix products can be recursively computed by the same algorithm we are using for the n*n case. This algorithm will continue applying itself to smallersized sub matrices until n becomes suitably small(n = 2) so that the product is computed directly.
To compute AB using (3.12), we need to perform eight multiplications of n/2*n/2 matrices and four additions of n/2*n/2 matrices. Since two n/2*n/2 matrices can be added in time cn^2 for some constant c,the overall computing time T(n) of the resulting divideandconquer algorithm is given by the recurrence
where b and c are constants.
This recurrence can be solved in the same way as earlier recurrences to obtain T(n)= 0(n^3). Hence no improvement over the conventional method has been made. Since matrix multiplications are more expensive than matrix additions (0(n^3) versus 0(n^2)),we can attempt to reformulate the equations for Cij so as to have fewer multiplications and possibly more additions. Volker Strassen has discovered a way to compute the Cij’s of (formula b) using only 7 multiplications and 18 additions or subtractions. His method involves first computing the seven n/2*n/2 matrices P, Q, R, S, T, U, and V as in (formula c).Then the Cij’s are computed using the formulas in (formula d). As can be seen P, Q, R, S, T, U, and V can be computed using 7 matrix multiplications and 10 matrix additions or subtractions. The Cij’s require an additional 8 additions or subtraction.