57139SinghHritik RajITSU2011Assignment2 2
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Dec 21, 2024
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Computational MathematicsITSU2011Victorian INstitute of TechnologyAssignment 02Weight 15%Hritik Raj SinghSID: 57139Date : 26 May 2024Lecture : Dr.Osamah Albahrey
IntroductionAssignment 02 is going to discuss four key questions that encompass topics such as coveringtrees, shortcut routes, mathematical algorithms as well as combinatorial procedures. It willgive sight of graph concepts to solve problems, evaluating and constructing simplemathematical arguments using various proof methods.Therefore, for completing this assignment, we will name the tasks below as in questions,Question Number 1Tournament MatchesGiven from the question,Four teams named T1, T2, T3 and T4, where each time will be playing against each otherteam in their group.Arranging to the playing tactics,-T1 Winner plays against Team T2 IIndnumber team.-T2 Winner plays against Team T1 IIndnumber team.-T3 Winner plays against Team T4 IIndnumber team.-T4 Winner plays against Team T3 IIndnumber team.From the above winners of all these matches will get to the semi finals match.From the above,We have the combinatorial formula,(4 : 2 ) = 4!2!(4β2)!= 6 Hence, 6matches will be played in each group.Now, for the final match,-There must be 4 quarter- final matches-2 semi- finals-1 finalFinally,Total matches;24(group stage matches) + 4(Quarter final matches) + 2 (semi finals) + 1 finals= 31 matchesTherefore, we got 31 matches to be played for this tournament.
Question 2Book store : Combinatorial SolutionWe have,3 categories of books-Love, Scifi and Enigma-Also, 6 unique authors in each category with 4 novelsProbable option to be selected areFor 1 book we have4 π₯ 3 π₯ 6=72 Similarly,For 2 books choosing,(4 : 2 ) = 4!2!(4β2)!= 6 Finally,For choosing 3 books by different authors would be,(6 : 3 ) = 6!3!(6β3)!= 20Thereofere, total choices are43= 64Sum of all = 64 x 20 x 3 = 3840Again, 72 + 108 + 3840 = 4020 total choices.Question 3Graph of spinning Tree:As given from the question image,We have,-Vertices, 4 ( A, B, C, D)-Edges, (A-B) (A-C) (A-D) (B-C) (B-D) (C-D)Cayleyβsformula,The number of the spanning trees in a complete graph beXn:
Therefore,nn-2π( ππ)=For n= 4,44-2π( π4)== 16Question 4Shortest Route - Delivery SystemGiven pentagon have 5 vertices, A, B, C, D and EDistances of vertices are;-A-B = 7, A-C = 1, A-D = 4, A-E= 3,-B-C = 2, B-D = 5, B-E = 7,-C-D =9, C-E = 8 and-D-E = 10For this solution, we have circuit solving method,HamiltonianFirst of all,Routes starting with A:1.Route A-C-E-B-D-A:-1 + 8 + 2 + 5 + 4 = 20 unit-Shortest RouteAlso we have to check all the other possibilities for the shortest route,βA-B-C-D-E-A = 31βA-B-C-E-D-A = 31βA-B-D-C-E-A = 32βA-B-D-E-C-A = 31βA-B-E-C-D-A = 35βA-B-E-D-C-A = 34βA-C-B-D-E-A = 21βA-C-B-E-D-A = 24βA-C-D-B-E-A = 25βA-C-D-E-B-A = 34
βA-C-E-D-B-A = 31βA-D-B-C-E-A = 22βA-D-B-E-C-A = 25βA-D-C-B-E-A = 25βA-D-C-E-B-A = 35βA-D-E-B-C-A = 24βA-D-E-C-B-A = 31βA-E-B-C-D-A = 25βA-E-B-D-C-A = 25βA-E-C-B-D-A = 22βA-E-C-D-B-A = 32βA-E-D-B-C-A = 21βA-E-D-C-B-A = 31Now, we came to the verification and we found,Route, A-C-E-B-D-A = 20 Unit