Restructuring
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CT.rbcomposer.jsoncomposer.lockct.bat
data/primes
nbproject
phpunit.xmlsettings.ymlsolutions
CodeChef/easy
CodeGolf/guns-and-zips
Generic
ProjectEuler
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desc.ymlsrc
template.rttests
vendor
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solutions/ProjectEuler/015/desc.yml
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solutions/ProjectEuler/015/desc.yml
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title: Starting in the top left corner in a 20 by 20 grid, how many routes are there to the bottom right corner?
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url: http://projecteuler.net/problem=15
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desc: |
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Starting in the top left corner of a 22 grid, there are 6 routes (without backtracking) to the bottom right corner.
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How many routes are there through a 2020 grid?
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solution: |
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(2n)!/n!^2 - Where n is the size of the grid -
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todo: Find a more "programmable way" instead of this cheating mathsolution
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solutions:
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solve.php:
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desc: Using BCMath to cope with large numbers
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language: php
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solve.rb:
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desc: Basic solution
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language: ruby
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solutions/ProjectEuler/015/solve.php
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solutions/ProjectEuler/015/solve.php
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<?php
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define('GRID_SIZE',20);
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function factorial($num) {
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if($num == 0)
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return 1;
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else
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return bcmul($num,factorial(bcsub($num,1)));
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}
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echo bcdiv(factorial(2* GRID_SIZE),bcpow(factorial( GRID_SIZE),2));
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solutions/ProjectEuler/015/solve.rb
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solutions/ProjectEuler/015/solve.rb
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GRID_SIZE = 20
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def factorial(num)
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if(num == 0)
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return 1
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else
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return num * factorial(num - 1)
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end
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end
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puts factorial(2 * GRID_SIZE) / factorial(GRID_SIZE) ** 2
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