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Finding the most profitable way to cut a rod, using bottom-up dynamic programming (CLRS Section 15.1)
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var assert = require("assert"); | |
class Result { | |
constructor(n, seq, score) { | |
this.n = n; | |
this.seq = seq; | |
this.score = score; | |
} | |
betterThan(other) { | |
assert(this.n == other.n, "Unable to compare rods of different lengths"); | |
return this.score > other.score; | |
} | |
} | |
function findOptimalSplit(n, costs) { | |
console.log("Finding optimal split for " + n + ":"); | |
var cache = {}; | |
cache[0] = new Result(0, [], 0); | |
for (var i = 1; i <= n; i++) { | |
console.log(" - subproblem " + i); | |
var best = null; | |
for (var j = 1; j <= i; j++) { | |
assert(costs[j] >= 0, "No cost defined for length " + j); | |
var left = new Result(j, [j], costs[j]); | |
var right = cache[i - j]; | |
var combined = new Result(left.n + right.n, | |
left.seq.concat(right.seq), | |
left.score + right.score); | |
console.log(" - " + combined.score + " for " + combined.seq.join(" + ")); | |
if (!best || combined.betterThan(best)) { | |
best = combined; | |
} | |
} | |
assert(best); | |
cache[i] = best; | |
} | |
return cache[n].seq; | |
} | |
console.log(findOptimalSplit(10, [0, 1, 5, 8, 9, 10, 17, 17, 20, 24, 30])); |
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