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88 lines
2.6 KiB
Matlab
88 lines
2.6 KiB
Matlab
function [] = Problem24()
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%ProjectEuler/Octave/Problem24.m
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%Matthew Ellison
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% Created: 03-25-19
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%Modified: 03-28-19
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%What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
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%{
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Copyright (C) 2019 Matthew Ellison
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This program is free software: you can redistribute it and/or modify
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it under the terms of the GNU Lesser General Public License as published by
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the Free Software Foundation, either version 3 of the License, or
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(at your option) any later version.
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This program is distributed in the hope that it will be useful,
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but WITHOUT ANY WARRANTY; without even the implied warranty of
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MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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GNU Lesser General Public License for more details.
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You should have received a copy of the GNU Lesser General Public License
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along with this program. If not, see <https://www.gnu.org/licenses/>.
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%}
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NEEDED_PERM = 1000000; %The number of the permutation that you need
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%Setup the variables
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nums = "0123456789";
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%Start the time
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startTime = clock();
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%Get all permutations of the string
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permutations = getPermutations(nums);
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%Stop the timer
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endTime = clock();
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%Print the results
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printf("The 1 millionth permutation is %s\n", permutations{NEEDED_PERM})
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printf("It took %f second to run this algorithm\n", etime(endTime, startTime))
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end %Problem24()
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function [perms] = getPermutations(master, num = 1)
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perms = {};
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%Check if the number is out of bounds
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if((num > size(master)(2)) || (num <= 0))
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%Do nothing and return and empty list
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perms;
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%If this is the last possible recurse just return the current string
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elseif(num == size(master)(2))
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perms(end + 1) = master;
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%If there are more possible recurses, recurse with the current permutations
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else
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temp = getPermutations(master, num + 1);
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perms = {perms{:}, temp{:}};
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%You need to swap the current letter with every possible letter after it
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%The ones needed to swap before will happen automatically when the function recurses
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for cnt = 1 : (size(master)(2) - num)
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%Swap two elements
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temp = master(num);
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master(num) = master(num + cnt);
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master(num + cnt) = temp;
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%Get the permutations after swapping two elements
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temp = getPermutations(master, num + 1);
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perms = {perms{:}, temp{:}};
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%Swap the elements back
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temp = master(num);
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master(num) = master(num + cnt);
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master(num + cnt) = temp;
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end
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end
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%The array is not necessarily in alpha-numeric order. So if this is the full array sort it before returning
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if(num == 1)
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perms = sort(perms);
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end
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end %getPermutations()
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%{
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Results:
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The 1 millionth permutation is 2783915460
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It took 433.922920 second to run this algorithm
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%}
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