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Added solution for problem 35

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Mattrixwv
2021-06-06 12:19:14 -04:00
parent c80941d3fb
commit 4ee0939c58
2 changed files with 93 additions and 4 deletions
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8
Problem34.m
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@@ -1,7 +1,7 @@
function [] = Problem34()
%ProjectEuler/ProjectEulerOctave/Problem34.lua
%Matthew Ellison
% Createe: 06-01-21
% Created: 06-01-21
%Modified: 06-01-21
%Find the sum of all numbers which are equal to the sum of the factorial of their digits
%{
@@ -22,13 +22,13 @@ function [] = Problem34()
%}
%Setup the variables
%Start the timer
startTime = clock();
MAX_NUM = 1499999; %The largest num that can be the sum of its own digits
numberFactorials = []; %Holds the pre-computed factorials of the numbers 0-9
totalSum = 0; %Holds the sum of all numbers equal to the sum of their digit's factorials
%Start the timer
startTime = clock();
%Pre-compute the possible factorials from 0! to 9!
for cnt = 1 : 9
numberFactorials(cnt) = factorial(cnt);

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Problem35.m Normal file
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function [] = Problem35()
%ProjectEuler/ProjectEulerOctave/Problem35.lua
%Matthew Ellison
% Created: 06-05-21
%Modified: 06-05-21
%Find the sum of all numbers which are equal to the sum of the factorial of their digits
%{
Copyright (C) 2021 Matthew Ellison
This program is free software: you can redistribute it and/or modify
it under the terms of the GNU Lesser General Public License as published by
the Free Software Foundation, either version 3 of the License, or
(at your option) any later version.
This program is distributed in the hope that it will be useful,
but WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
GNU Lesser General Public License for more details.
You should have received a copy of the GNU Lesser General Public License
along with this program. If not, see <https://www.gnu.org/licenses/>.
%}
%Setup the variables
MAX_NUM = 999999;
%MAX_NUM = 100;
circularPrimes = [];
%Start the timer
startTime = clock();
%Get all primes under 1,000,000
primeList = primes(MAX_NUM);
%Go through all primes, get all their rotations, and check if those numbers are also prime
for cnt = 1 : size(primeList)(2)
prime = primeList(cnt)
allRotationsPrime = true;
%Get all of the rotations of the pirme and see if they are also prime
rotations = getRotations(num2str(prime));
for rotCnt = 1 : size(rotations)(2)
rotation = rotations(rotCnt);
p = rotation;
if(~isprime(p))
allRotationsPrime = false;
break;
end
end
%If all rotations are prime add it to the list of circular primes
if(allRotationsPrime)
circularPrimes(end + 1) = prime;
end
end
%Stop the timer
endTime = clock();
%Print the results
printf("The number of all circular prime numbers under %d is %d\n", MAX_NUM, size(circularPrimes)(2));
printf("It took %f seconds to run this algorithm\n", etime(endTime, startTime))
end
%Returns a list of all rotations of a string passed to it
function [rotations] = getRotations(str)
rotations = [];
rotations(end + 1) = str2num(str);
for cnt = 2 : size(str)(2)
temp = [substr(str, 2, size(str)(2) - 1) str(1)(1)];
num = str2num(temp);
rotations(end + 1) = num;
str = temp;
end
end
function [found] = isFound(array, key)
found = false;
for location = 1 : size(array)(2)
if(key == array(location))
found = true;
return;
end
end
end
%{
Results:
The number of all circular prime numbers under 999999 is 55
It took 712.060379 seconds to run this algorithm
%}