ocaml: phase 5.1 max_product3.ml baseline (max product of 3, with negatives -> 300)
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Sort, then compare two candidates: p1 = product of three largest values p2 = product of two smallest (potentially negative) values and the largest For [-10;-10;1;3;2]: sorted = [-10;-10;1;2;3] p1 = 3 * 2 * 1 = 6 p2 = (-10) * (-10) * 3 = 300 max = 300 Tests List.sort + Array.of_list + arr.(n-i) end-walk + candidate-pick via if-then-else. 104 baseline programs total.
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@@ -51,6 +51,7 @@
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"luhn.ml": 2,
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"mat_mul.ml": 621,
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"max_path_tree.ml": 11,
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"max_product3.ml": 300,
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"max_run.ml": 5,
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"mod_inverse.ml": 27,
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"json_pretty.ml": 24,
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11
lib/ocaml/baseline/max_product3.ml
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11
lib/ocaml/baseline/max_product3.ml
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let max_prod3 xs =
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let sorted = List.sort compare xs in
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let arr = Array.of_list sorted in
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let n = Array.length arr in
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let p1 = arr.(n - 1) * arr.(n - 2) * arr.(n - 3) in
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let p2 = arr.(0) * arr.(1) * arr.(n - 1) in
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if p1 > p2 then p1 else p2
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;;
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max_prod3 [-10; -10; 1; 3; 2]
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@@ -407,6 +407,12 @@ _Newest first._
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binary search tree (`type 'a tree = Leaf | Node of 'a * 'a tree *
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'a tree`) with insert + in-order traversal. Tests parametric ADT,
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recursive match, List.append, List.fold_left.
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- 2026-05-09 Phase 5.1 — max_product3.ml baseline (max product of
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three from a list including negatives = 300). Sort, then compare
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product of three largest vs product of two smallest negatives and
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one largest. For [-10;-10;1;3;2]: 3*2*1 = 6 vs (-10)*(-10)*3 = 300.
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Tests List.sort + Array.of_list + arr.(n-i) end-walk + candidate
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compare. 104 baseline programs total.
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- 2026-05-09 Phase 5.1 — euler9.ml baseline (Project Euler #9, abc
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product for the unique Pythagorean triple with a+b+c=1000 →
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31875000). Naive triple loop times out under contention (10-min
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