Torres de Hanói

• •No hay requisitos previos especiales, puede importar y usarlo directamente

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        "content": "### Input\n- **numberOfDiscs:** 4 "
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        "content": "### Setup\n- **numberOfDiscs:** 4  \n- **stackA**\n  - **name:** A  \n  - **items:** 4, 3, 2, 1  \n- **stackB**\n  - **name:** B  \n  - **items:** -  \n- **stackC**\n  - **name:** C  \n  - **items:** -  \n"
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        "content": "### Result\n- **numberOfDiscs:** 4  \n- **stackX**\n  - **name:** A  \n  - **items:** -  \n- **stackY**\n  - **name:** B  \n  - **items:** -  \n- **stackZ**\n  - **name:** C  \n  - **items:** 4, 3, 2, 1"
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        "content": "### Solution\nMove disc 1 from A to B,\nthen A 2→ C,\nB 1→ C, A 3→ B,\nC 1→ A, C 2→ B,\nA 1→ B, A 4→ C,\nB 1→ C, B 2→ A,\nC 1→ A, B 3→ C,\nA 1→ B, A 2→ C,\nB 1→ C."
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        "content": "### Node input\n- **numberOfDiscs** -1 \n- stackX = **stackX** \n- stackY = **stackZ** \n- stackZ = **stackY** "
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        "content": "### Node steps\n- disc = **stackX**.pop()\n- **stackZ**.push(disc)"
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        "content": "### Node steps\n- **numberOfDiscs** += 1\n- YXZ to XYZ\n  - stack = **stackY**\n  - **stackY** = **stackX**\n  - **stackX** = stack"
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      "id": "9baa2d1d-6d3b-4d8f-9d95-db479479a4e2",
      "name": "Torres de Hanói",
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        "content": "## Towers of Hanoi\nThis is an **example of using sub-workflow nodes** and a proof of concept showing that it’s possible to **solve and explain recursive problems with n8n**.\n\n### Task\nMove a stack of n disks from rod A to rod C, using rod B as auxiliary. Only one disk can be moved at a time, and no disk may be placed on a smaller disk.\n\n### Example\n```\nn=4\n     |          |          |\n     =          |          |\n    ===         |          |\n   =====        |          |\n  =======       |          |\n ---------  ---------  ---------\n     A          B          C\n```\n### Algorithm\n```\nprocedure Hanoi(n, X, Y, Z):\n  if n == 1:\n    move disk from X to Z\n  else:\n    Hanoi(n-1, X, Z, Y)\n    move disk from X to Z\n    Hanoi(n-1, Y, X, Z)\n```\nAnother variant, with better readability and no redundant code, but it includes empty n8n executions (when n is 0):\n```\nprocedure Hanoi(n, X, Y, Z):\n    if n > 0:\n        Hanoi(n-1, X, Z, Y)\n        move disk from X to Z\n        Hanoi(n-1, Y, X, Z)\n```\n### Notes\n- This is a learning example. In a real scenario, you would probably use an iterative approach with only a single code node.\n- The workflow variables X, Y, and Z represent the positions of the rods. E.g., if rods A and B are swapped, A is at position Y and B is at position X.\n- As the modified number of discs and positions are returned when a sub-workflow finishes, the \"Update\" nodes reset these values.\n- When experimenting with recursion, make sure to define a termination condition first. Also, be aware of the \"Restart workspace\" link in the [n8n Dashboard](https://app.n8n.cloud/manage).\n- Check Towers of Hanoi on [GitHub](https://github.com/adibaba/n8n.workflows)\n- Learn more about [Recursion on Wikipedia](https://en.wikipedia.org/w/index.php?title=Recursion_(computer_science)&oldid=1301600240#Towers_of_Hanoi)."
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