src/problems/pmch.lisp @ e3aefcbf364c
Cache Uniprot results on the filesystem
This will make only the first `(run-tests)` on a given computer take forever,
instead of the first `(run-tests)` of a given Lisp session. It will also
hopefully make the Uniprot folks not hate me.
author |
Steve Losh <steve@stevelosh.com> |
date |
Fri, 24 Jan 2020 23:05:16 -0500 |
parents |
2735aa6aab79 |
children |
(none) |
(defpackage :rosalind/pmch (:use :cl :rosalind :losh :iterate))
(in-package :rosalind/pmch)
(defparameter *input* ">Rosalind_23
AGCUAGUCAU")
(defparameter *output* "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* *output*
(let ((bases (nth-value 1 (u:read-fasta data))))
(* (u:factorial (count #\A bases))
(u:factorial (count #\G bases)))))