7052ec0c6e1d
PMCH
author | Steve Losh <steve@stevelosh.com> |
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date | Fri, 22 Feb 2019 19:23:17 -0500 |
parents | e0c558fa549b |
children | ca95211a2d98 |
branches/tags | (none) |
files | src/problems/pmch.lisp |
Changes
--- /dev/null Thu Jan 01 00:00:00 1970 +0000 +++ b/src/problems/pmch.lisp Fri Feb 22 19:23:17 2019 -0500 @@ -0,0 +1,32 @@ +(in-package :rosalind) + +(defparameter *input-pmch* ">Rosalind_23 +AGCUAGUCAU") + +(defparameter *output-pmch* "12") + +;; We can make a few observations to make things easier (well, trivial). +;; +;; First: the adjacency edges are a red herring. Ignore them. +;; +;; Next: because adenine only interacts with uracil and guanine only interacts +;; with cytosine, we can split the problem apart into two separate graphs. We +;; can compute the number of perfect matchings of each graph separately and then +;; multiply them together at the end to find the total. +;; +;; For each sub graph, every node has edges to all nodes of the complementary +;; base. The problem description also guarantees that the number of +;; complementary bases are equal. +;; +;; Say we're looking at the A/U graph, and there are N adenine bases and +;; N uracil bases. For each adenine, we need to pick a uracil to match it with. +;; For the first adenine we have N uracil's to choose from. For the second +;; adenine we have N-1 uracils. And so on down to the final pair. So the total +;; number of choices we have for each graph is N(N-1)(N-2)…(1) = N! + +(define-problem pmch (data stream) + *input-pmch* + *output-pmch* + (let ((bases (nth-value 1 (read-fasta data)))) + (* (factorial (count #\A bases)) + (factorial (count #\G bases)))))