{"id":208484,"date":"2025-04-27T11:31:54","date_gmt":"2025-04-27T11:31:54","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=208484"},"modified":"2025-04-27T11:31:56","modified_gmt":"2025-04-27T11:31:56","slug":"let-mm-x-m-x-m-x-m4-where-the-mj-j-1-4-are-matrices","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/04\/27\/let-mm-x-m-x-m-x-m4-where-the-mj-j-1-4-are-matrices\/","title":{"rendered":"Let M=M, X M X M X M4, where the Mj, j = 1..4, are matrices"},"content":{"rendered":"\n<p>(a) Let M=M, X M X M X M4, where the Mj, j = 1..4, are matrices. The order of each matrix is given as follows: M is 10 X 20; M2 is 20 X 15; M3 is 15 X 50 and M is 50 X 5.<\/p>\n\n\n\n<p>(1) Using the Dynamic Programming method (See Appendix C), calculate m, 4 to complete the table below which determines the least number of multiplications which can be used to compute M. [3 marks] m1 = 0 m22 = 0 m33 = 0 m44 = 0 m2 = 3000 m3 = 15000 m34 = 3750 10500 5250 m13 m24 m4=<\/p>\n\n\n\n<p>(ii) Determine the computational complexity of the algorithm referenced in part a(i) above. Show all steps in your working.<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Part (i): Dynamic Programming to Calculate Minimum Scalar Multiplications<\/h3>\n\n\n\n<p>We are given a chain of matrix multiplications ( M_1 \\times M_2 \\times M_3 \\times M_4 ), where the matrices have the following dimensions:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( M_1 ) is ( 10 \\times 20 )<\/li>\n\n\n\n<li>( M_2 ) is ( 20 \\times 15 )<\/li>\n\n\n\n<li>( M_3 ) is ( 15 \\times 50 )<\/li>\n\n\n\n<li>( M_4 ) is ( 50 \\times 5 )<\/li>\n<\/ul>\n\n\n\n<p>We are tasked with using the dynamic programming approach to find the minimum number of scalar multiplications required to multiply these matrices.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Dynamic Programming Matrix Chain Multiplication<\/h4>\n\n\n\n<p>The dynamic programming method works by breaking down the problem into smaller subproblems and then solving each one optimally. Let\u2019s define ( m[i, j] ) as the minimum number of scalar multiplications needed to compute the matrix chain from ( M_i ) to ( M_j ). The recursive relation for ( m[i, j] ) is:<\/p>\n\n\n\n<p>[<br>m[i, j] = \\min_{i \\leq k &lt; j} \\left( m[i, k] + m[k+1, j] + p_{i-1} \\cdot p_k \\cdot p_j \\right)<br>]<\/p>\n\n\n\n<p>Where ( p_i ) denotes the dimension of matrix ( M_i ), i.e., ( M_i ) has dimension ( p_{i-1} \\times p_i ).<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Step-by-Step Calculation of ( m[i, j] )<\/h4>\n\n\n\n<p><strong>Step 1: Define the dimensions<\/strong><\/p>\n\n\n\n<p>We define the dimensions for each matrix as follows:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( p_0 = 10 ), ( p_1 = 20 ), ( p_2 = 15 ), ( p_3 = 50 ), ( p_4 = 5 )<\/li>\n<\/ul>\n\n\n\n<p>The problem now becomes minimizing the number of scalar multiplications to compute the matrix chain.<\/p>\n\n\n\n<p><strong>Step 2: Initialize the ( m ) table<\/strong><\/p>\n\n\n\n<p>The diagonal elements of the matrix ( m[i, i] ) are zero because no multiplication is needed for a single matrix.<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>m&#91;1, 1] = 0\nm&#91;2, 2] = 0\nm&#91;3, 3] = 0\nm&#91;4, 4] = 0<\/code><\/pre>\n\n\n\n<p><strong>Step 3: Fill in the table for increasing chain lengths<\/strong><\/p>\n\n\n\n<p>Let&#8217;s compute for chain lengths 2, 3, and 4.<\/p>\n\n\n\n<p><strong>Chain length 2 (i.e., multiplying two matrices)<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( m[1, 2] = 10 \\times 20 \\times 15 = 3000 )<\/li>\n\n\n\n<li>( m[2, 3] = 20 \\times 15 \\times 50 = 15000 )<\/li>\n\n\n\n<li>( m[3, 4] = 15 \\times 50 \\times 5 = 3750 )<\/li>\n<\/ul>\n\n\n\n<p><strong>Chain length 3 (i.e., multiplying three matrices)<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( m[1, 3] = \\min \\left( m[1, 1] + m[2, 3] + 10 \\times 20 \\times 50, m[1, 2] + m[3, 3] + 10 \\times 15 \\times 50 \\right) ) First calculation:<br>[<br>m[1, 3] = 0 + 15000 + 10 \\times 20 \\times 50 = 0 + 15000 + 10000 = 25000<br>] Second calculation:<br>[<br>m[1, 3] = 3000 + 0 + 10 \\times 15 \\times 50 = 3000 + 0 + 7500 = 10500<br>] So, ( m[1, 3] = 10500 ).<\/li>\n\n\n\n<li>( m[2, 4] = \\min \\left( m[2, 2] + m[3, 4] + 20 \\times 15 \\times 5, m[2, 3] + m[4, 4] + 20 \\times 50 \\times 5 \\right) ) First calculation:<br>[<br>m[2, 4] = 0 + 3750 + 20 \\times 15 \\times 5 = 0 + 3750 + 1500 = 5250<br>] Second calculation:<br>[<br>m[2, 4] = 15000 + 0 + 20 \\times 50 \\times 5 = 15000 + 0 + 5000 = 20000<br>] So, ( m[2, 4] = 5250 ).<\/li>\n<\/ul>\n\n\n\n<p><strong>Chain length 4 (i.e., multiplying all four matrices)<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( m[1, 4] = \\min \\left( m[1, 1] + m[2, 4] + 10 \\times 20 \\times 5, m[1, 2] + m[3, 4] + 10 \\times 15 \\times 5, m[1, 3] + m[4, 4] + 10 \\times 50 \\times 5 \\right) ) First calculation:<br>[<br>m[1, 4] = 0 + 5250 + 10 \\times 20 \\times 5 = 0 + 5250 + 1000 = 6250<br>] Second calculation:<br>[<br>m[1, 4] = 3000 + 3750 + 10 \\times 15 \\times 5 = 3000 + 3750 + 750 = 7500<br>] Third calculation:<br>[<br>m[1, 4] = 10500 + 0 + 10 \\times 50 \\times 5 = 10500 + 0 + 2500 = 13000<br>] So, ( m[1, 4] = 6250 ).<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">Final Table for Minimum Scalar Multiplications<\/h4>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th>i, j<\/th><th>m[i,j]<\/th><\/tr><\/thead><tbody><tr><td>m[1,1]<\/td><td>0<\/td><\/tr><tr><td>m[1,2]<\/td><td>3000<\/td><\/tr><tr><td>m[1,3]<\/td><td>10500<\/td><\/tr><tr><td>m[1,4]<\/td><td>6250<\/td><\/tr><tr><td>m[2,2]<\/td><td>0<\/td><\/tr><tr><td>m[2,3]<\/td><td>15000<\/td><\/tr><tr><td>m[2,4]<\/td><td>5250<\/td><\/tr><tr><td>m[3,3]<\/td><td>0<\/td><\/tr><tr><td>m[3,4]<\/td><td>3750<\/td><\/tr><tr><td>m[4,4]<\/td><td>0<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>So the minimum number of scalar multiplications to multiply the entire chain of matrices is <strong>6250<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Part (ii): Computational Complexity of the Algorithm<\/h3>\n\n\n\n<p>The time complexity of the matrix chain multiplication dynamic programming algorithm is ( O(n^3) ), where ( n ) is the number of matrices in the chain.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Explanation:<\/h4>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Outer Loops (k, i, j):<\/strong> The algorithm iterates over all subproblems (subchains of matrices) that we need to solve. There are three nested loops:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The outer loop runs for different chain lengths, from 2 to ( n ).<\/li>\n\n\n\n<li>The second loop goes through all possible starting points of the chain (i.e., it chooses the first matrix in the chain).<\/li>\n\n\n\n<li>The innermost loop finds the optimal split of the chain at each position.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Work per Subproblem:<\/strong> Each subproblem involves a constant amount of work \u2014 finding the minimum cost split and calculating the scalar multiplications for that split, which is ( O(1) ).<\/li>\n<\/ol>\n\n\n\n<p>Thus, the overall complexity is ( O(n^3) ), where ( n ) is the number of matrices in the chain. For ( n = 4 ), the complexity is still ( O(4^3) = O(64) ), which is a manageable size for small chains like the one in this problem.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>(a) Let M=M, X M X M X M4, where the Mj, j = 1..4, are matrices. The order of each matrix is given as follows: M is 10 X 20; M2 is 20 X 15; M3 is 15 X 50 and M is 50 X 5. (1) Using the Dynamic Programming method (See Appendix [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-208484","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/208484","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=208484"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/208484\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=208484"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=208484"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=208484"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}