Taelin Challenge (cf. https://github.com/youqad/bit-reversal-trees): chat trace of Haskell solutions by o3: passes all tests, and does NOT use unauthorized helper functions, so it solves the challenge!

o3-019642d6-1889-7792-ada8-81be8701d7da.json

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        "content": "```haskell\n-- Type `Tree` of perfect binary trees\ndata Tree a = Leaf a | Node (Tree a) (Tree a) \n\n{- You are an expert Haskell competitive programmer. Your goal is to implement an `invert :: Tree a -> Tree a` function that performs a bit-reversal permutation on a `Tree`. Here is what we mean by that:\n\n1. Each leaf in the binary tree has a path leading to it, which can be represented as a sequence of bits: `False` (or `0`) for left, `True` (or `1`) for right.\n2. The bit-reversal permutation swaps a leaf at path `p` with the leaf at path `reverse p`. For example, a leaf at path `[False, False, True]` (left, left, right) would be swapped with the leaf at path `[True, False, False]` (right, left, left).\n\nMANDATORY SYNTACTIC REQUIREMENTS:\n1. The `invert` function must be a standalone and pure function ONLY relying on an inner function `invertHelper :: Bool -> Tree a -> Tree a` that is itself a self-contained single pure recursive function.\n2. Only use recursion (no loops).\n3. Maintain purity (no side effects or mutability).\n\nThe `Bool` parameter is an extra boolean that you can use however you want: the goal is that `invertHelper True tree` should return the bit-reversed tree.\n\nThis is a very difficult problem, so think step-by-step before implementing your solution and carefully review it to make sure it meets all the requirements. Test your implementation against the test cases to verify its correctness. I guarantee you that it is solvable within the constraints.\n-}\n\n-- Implement the `invert` function as follows:\ninvert :: Tree a -> Tree a\ninvert tree = invertHelper tree True\n  where\n    invertHelper :: Bool -> Tree a -> Tree a\n    invertHelper flag tree = undefined  -- Replace 'undefined' with your implementation\n\n-- Tests:\ntestTree1 :: Tree Int\ntestTree1 = Node (Node (Node (Leaf 0) (Leaf 1)) (Node (Leaf 2) (Leaf 3))) \n                 (Node (Node (Leaf 4) (Leaf 5)) (Node (Leaf 6) (Leaf 7)))\n\nexpectedTree1 :: Tree Int\nexpectedTree1 = Node (Node (Node (Leaf 0) (Leaf 4)) (Node (Leaf 2) (Leaf 6))) \n                     (Node (Node (Leaf 1) (Leaf 5)) (Node (Leaf 3) (Leaf 7)))\n\ntestTree2 :: Tree Int\ntestTree2 = Node (Node (Node (Leaf 1) (Leaf 3)) (Node (Leaf 5) (Leaf 7)))\n                 (Node (Node (Leaf 0) (Leaf 2)) (Node (Leaf 4) (Leaf 6)))\n\nexpectedTree2 :: Tree Int\nexpectedTree2 = Node (Node (Node (Leaf 1) (Leaf 0)) (Node (Leaf 5) (Leaf 4)))\n                     (Node (Node (Leaf 3) (Leaf 2)) (Node (Leaf 7) (Leaf 6)))\n\ntestTree3 :: Tree Int\ntestTree3 = Node (Node (Node (Node (Leaf 10) (Leaf 14)) (Node (Leaf 4) (Leaf 6)))\n                       (Node (Node (Leaf 9) (Leaf 1)) (Node (Leaf 0) (Leaf 7))))\n                 (Node (Node (Node (Leaf 11) (Leaf 15)) (Node (Leaf 2) (Leaf 5)))\n                       (Node (Node (Leaf 13) (Leaf 12)) (Node (Leaf 8) (Leaf 3))))\n\nexpectedTree3 :: Tree Int\nexpectedTree3 = Node (Node (Node (Node (Leaf 10) (Leaf 11)) (Node (Leaf 9) (Leaf 13)))\n                           (Node (Node (Leaf 4) (Leaf 2)) (Node (Leaf 0) (Leaf 8))))\n                     (Node (Node (Node (Leaf 14) (Leaf 15)) (Node (Leaf 1) (Leaf 12)))\n                           (Node (Node (Leaf 6) (Leaf 5)) (Node (Leaf 7) (Leaf 3))))\n\nmain = do\n    mapM_ (\\(input, expected) -> do\n        putStrLn \"Expected output:\"\n        print expected\n        putStrLn \"Actual output:\"\n        print (invert input)\n    ) [(testTree1, expectedTree1),\n       (testTree2, expectedTree2),\n       (testTree3, expectedTree3)]\n```"
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          "content": "```haskell\n{-# LANGUAGE DeriveFoldable #-}\n\n-- Perfect binary tree\ndata Tree a = Leaf a | Node (Tree a) (Tree a)\n              deriving (Show, Eq, Foldable)\n\n-- | Bit‑reversal permutation of a perfect binary tree\ninvert :: Tree a -> Tree a\ninvert t = invertHelper True t\n  where\n    -- The single required inner helper.\n    --   * flag == True  – perform the actual bit–reversal work\n    --   * flag == False – ‘zip’ two already‑processed sub‑trees\n    invertHelper :: Bool -> Tree a -> Tree a\n    -- main mode --------------------------------------------------------------\n    invertHelper True (Leaf x)          = Leaf x\n    invertHelper True (Node l r)        =\n        -- recursively finish both halves, then zip them\n        let l' = invertHelper True l\n            r' = invertHelper True r\n        in  invertHelper False (Node l' r')\n\n    -- zipping mode -----------------------------------------------------------\n    -- (this puts together even/odd groups that have already been reversed)\n    invertHelper False (Leaf x)         = Leaf x      -- should occur only at depth 0\n    invertHelper False (Node a b) =\n        case (a, b) of\n          (Leaf _, Leaf _) -> Node a b                -- depth‑1: nothing more to zip\n          (Node a1 a2, Node b1 b2) ->                 -- deeper: keep zipping pair‑wise\n              Node (invertHelper False (Node a1 b1))\n                   (invertHelper False (Node a2 b2))\n          _ -> error \"invert: non‑perfect tree encountered\"\n```\n\nHow it works\n‾‾‾‾‾‾‾‾‾‾‾‾‾‾\n1. `invertHelper True` recursively bit‑reverses the two children.\n2. After both halves are individually finished, they are **zipped** together\n   by calling `invertHelper False` on `Node left right`.\n3. The “zipping” phase interleaves the already‑processed halves\n   (`Node (Node a b) (Node c d)` → `Node (Node a c) (Node b d)`)\n   and recurses until the sub‑trees are only leaves.\n4. Because every internal node is visited a constant number of times,\n   the whole procedure runs in `O(n)` time and needs only the call stack\n   for its space.\n\nThe code meets all mandatory requirements:\n• Only the pure, recursive `invertHelper` is relied upon.  \n• No loops, no mutable state or side effects are used.",
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      "content": "```haskell\n{-# LANGUAGE DeriveFoldable #-}\n\n-- Perfect binary tree\ndata Tree a = Leaf a | Node (Tree a) (Tree a)\n              deriving (Show, Eq, Foldable)\n\n-- | Bit‑reversal permutation of a perfect binary tree\ninvert :: Tree a -> Tree a\ninvert t = invertHelper True t\n  where\n    -- The single required inner helper.\n    --   * flag == True  – perform the actual bit–reversal work\n    --   * flag == False – ‘zip’ two already‑processed sub‑trees\n    invertHelper :: Bool -> Tree a -> Tree a\n    -- main mode --------------------------------------------------------------\n    invertHelper True (Leaf x)          = Leaf x\n    invertHelper True (Node l r)        =\n        -- recursively finish both halves, then zip them\n        let l' = invertHelper True l\n            r' = invertHelper True r\n        in  invertHelper False (Node l' r')\n\n    -- zipping mode -----------------------------------------------------------\n    -- (this puts together even/odd groups that have already been reversed)\n    invertHelper False (Leaf x)         = Leaf x      -- should occur only at depth 0\n    invertHelper False (Node a b) =\n        case (a, b) of\n          (Leaf _, Leaf _) -> Node a b                -- depth‑1: nothing more to zip\n          (Node a1 a2, Node b1 b2) ->                 -- deeper: keep zipping pair‑wise\n              Node (invertHelper False (Node a1 b1))\n                   (invertHelper False (Node a2 b2))\n          _ -> error \"invert: non‑perfect tree encountered\"\n```\n\nHow it works\n‾‾‾‾‾‾‾‾‾‾‾‾‾‾\n1. `invertHelper True` recursively bit‑reverses the two children.\n2. After both halves are individually finished, they are **zipped** together\n   by calling `invertHelper False` on `Node left right`.\n3. The “zipping” phase interleaves the already‑processed halves\n   (`Node (Node a b) (Node c d)` → `Node (Node a c) (Node b d)`)\n   and recurses until the sub‑trees are only leaves.\n4. Because every internal node is visited a constant number of times,\n   the whole procedure runs in `O(n)` time and needs only the call stack\n   for its space.\n\nThe code meets all mandatory requirements:\n• Only the pure, recursive `invertHelper` is relied upon.  \n• No loops, no mutable state or side effects are used.",
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  "solution": "invert :: Tree a -> Tree a\ninvert t = invertHelper True t\n  where\n    invertHelper :: Bool -> Tree a -> Tree a\n    invertHelper True (Leaf x)          = Leaf x\n    invertHelper True (Node l r)        =\n        let l' = invertHelper True l\n            r' = invertHelper True r\n        in  invertHelper False (Node l' r')\n\n    invertHelper False (Leaf x)         = Leaf x\n    invertHelper False (Node a b) =\n        case (a, b) of\n          (Leaf _, Leaf _) -> Node a b\n          (Node a1 a2, Node b1 b2) ->\n              Node (invertHelper False (Node a1 b1))\n                   (invertHelper False (Node a2 b2))\n          _ -> error \"invert: non‑perfect tree encountered\""
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solution.md

✨ Conversation 1: Found a valid implementation! 🎉 ✨

Chat trace of solutions by o3: passes all tests, and does NOT use unauthorized helper functions, so it solves the challenge! (cf. https://github.com/youqad/bit-reversal-trees)

📦 Published to https://wandb.ai/ox/bit-reversal-trees/weave/objects/019642d6-1889-7792-ada8-81be8701d7da/versions/tqH67yGo2i48AUsw4fOOawXLBgrX0oMjXF72pxziT9s

Solution has been saved (on WandB weave) with Call ID: 019642d6-1889-7792-ada8-81be8701d7da 🚀

invert :: Tree a -> Tree a
invert t = invertHelper True t
  where
    invertHelper :: Bool -> Tree a -> Tree a
    invertHelper True (Leaf x)          = Leaf x
    invertHelper True (Node l r)        =
        let l' = invertHelper True l
            r' = invertHelper True r
        in  invertHelper False (Node l' r')

    invertHelper False (Leaf x)         = Leaf x
    invertHelper False (Node a b) =
        case (a, b) of
          (Leaf _, Leaf _) -> Node a b
          (Node a1 a2, Node b1 b2) ->
              Node (invertHelper False (Node a1 b1))
                   (invertHelper False (Node a2 b2))
          _ -> error "invert: non‑perfect tree encountered"