95 lines
2.2 KiB
Plaintext
95 lines
2.2 KiB
Plaintext
{
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"cells": [
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{
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"cell_type": "markdown",
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"id": "35b56ed4",
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"metadata": {},
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"source": [
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"# [Maximum Path Sum II](https://projecteuler.net/problem=67)\n",
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"\n",
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"Since we took the time to find an efficient method in [problem 18](https://projecteuler.net/problem=18), we can just reuse that method here.\n",
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"\n",
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"So just read the triangle into a list-of-lists."
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]
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},
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{
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"cell_type": "code",
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"execution_count": 1,
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"id": "194eef78",
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"metadata": {},
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"outputs": [],
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"source": [
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"with open(\"txt/0067_triangle.txt\") as f:\n",
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" triangle = [[int(n) for n in line.split(' ')] for line in f]"
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]
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},
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{
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"cell_type": "markdown",
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"id": "b39b979e",
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"metadata": {},
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"source": [
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"Then we'll use the bottom-up method from before (you could also use the top-down method)."
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]
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},
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{
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"cell_type": "code",
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"execution_count": 2,
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"id": "fa8a353f",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"7273"
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]
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},
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"execution_count": 2,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"def max_path_sum(tri):\n",
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" for i in reversed(range(0, len(tri) - 1)):\n",
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" for j in range(0, len(tri[i])):\n",
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" tri[i][j] += max(tri[i+1][j], tri[i+1][j+1])\n",
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" \n",
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" return tri[0][0]\n",
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"\n",
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"max_path_sum(triangle)"
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]
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},
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{
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"cell_type": "markdown",
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"id": "f06b0164",
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"metadata": {},
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"source": [
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"#### Copyright (C) 2025 filifa\n",
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"\n",
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"This work is licensed under the [Creative Commons Attribution-ShareAlike 4.0 International license](https://creativecommons.org/licenses/by-sa/4.0/) and the [BSD Zero Clause license](https://spdx.org/licenses/0BSD.html)."
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "SageMath 9.5",
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"language": "sage",
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"name": "sagemath"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.11.2"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 5
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}
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