src/problems/inod.lisp @ dbd94aef5f92
SETO
author |
Steve Losh <steve@stevelosh.com> |
date |
Fri, 20 Dec 2019 17:12:23 -0500 |
parents |
049e0d632763 |
children |
2735aa6aab79 |
(in-package :rosalind)
;; This one is trivial once you know the closed-form solution of N-2. The
;; intuition for that can come in two parts.
;;
;; First, a rooted binary tree has N-1 internal nodes. This is because at any
;; given point as you're building the tree, you select 2 of the remaining nodes
;; and join them together with an internal node, which reduces the total
;; remaining by 1. You end when there is only one remaining node (the root) and
;; so you did N-1 subtractions.
;;
;; To convert this to an unrooted tree, you replace the root node with an edge,
;; which subtracts one more internal node from the graph. So you're left with
;; N-2 internal nodes.
(define-problem inod (data stream)
"4"
"2"
(- (read data) 2))
#; Scratch --------------------------------------------------------------------
(problem-inod)