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Fast fail on out of bounds diagonals.

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Rob Rix 2015-11-12 15:58:03 -05:00
parent e1f38f64d0
commit 17b9c0a9ec
1 changed files with 1 additions and 1 deletions
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prototype/Doubt/SES.swift
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@ -32,7 +32,7 @@ public func SES<Leaf, Annotation, C: CollectionType>(a: C, _ b: C, cost: Free<Le
let down = matrix[i, j.successor()]
let diagonal = matrix[i.successor(), j.successor()]
if let right = right, down = down, diagonal = diagonal {
if let diagonal = diagonal, right = right, down = down {
let here = recur(a[i], b[j])
// If the diff at this vertex is zero-cost, we’re not going to find a cheaper one either rightwards or downwards. We can therefore short-circuit selecting the best outgoing edge and save ourselves evaluating the entire row rightwards and the entire column downwards from this point.
//