121 lines
3.9 KiB
Plaintext
121 lines
3.9 KiB
Plaintext
{
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"cells": [
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{
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"cell_type": "markdown",
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"id": "6cc372f3",
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"metadata": {},
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"source": [
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"# [Lattice Paths](https://projecteuler.net/problem=15)\n",
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"\n",
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"Let $f(m, n)$ equal the number of routes in an $m \\times n$ grid. Intuitively, if you start at the top left corner, you immediately have two choices - you can go down or right. If you go down, there are $f(m-1,n)$ routes in the $(m - 1) \\times n$ subgrid; if you go right, there are $f(m,n-1)$ routes in the $m \\times (n-1)$ subgrid. Therefore\n",
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"$$f(m,n) = f(m-1,n) + f(m,n-1)$$\n",
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"\n",
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"There are two trivial base cases: $f(m,0) = 1$ for any $m$ and $f(0,n) = 1$ for any $n$. From these facts, you can write a simple [memoized](https://en.wikipedia.org/wiki/Memoization) recursive function to solve this problem."
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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": "9878534a",
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"metadata": {},
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"outputs": [],
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"source": [
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"from functools import cache\n",
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"\n",
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"@cache\n",
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"def f(m, n):\n",
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" if m == 0:\n",
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" return 1\n",
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" if n == 0:\n",
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" return 1\n",
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" \n",
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" return f(m - 1, n) + f(m, n - 1)"
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]
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},
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{
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"cell_type": "markdown",
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"id": "5beab0ed",
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"metadata": {},
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"source": [
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"Then the answer is given by"
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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": "f21257bc",
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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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"137846528820"
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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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"f(20,20)"
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]
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},
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{
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"cell_type": "markdown",
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"id": "3561530b",
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"metadata": {},
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"source": [
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"However, we can do better and give a non-recursive formula for $f$. It turns out, for an $m \\times n$ grid, there are $\\binom{m+n}{m}$ possible routes. Here's an outline of how you can prove this [inductively](https://en.wikipedia.org/wiki/Mathematical_induction):\n",
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"\n",
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"1. Prove $f(m, 0) = \\binom{m+0}{m} = 1$ for all $m$\n",
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" 1. Prove $f(0, 0) = \\binom{0 + 0}{0} = 1$\n",
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" 2. Assuming $f(j, 0) = \\binom{j+0}{j} = 1$ for some $j$, prove $f(j+1,0) = \\binom{j+1+0}{j+1} = 1$\n",
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"2. Assuming there exists some $k$ such that for all $m$, $f(m, k) = \\binom{m + k}{m}$, prove $f(m, k+1) = \\binom{m + k + 1}{m}$ for all $m$\n",
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" 1. Prove $f(0, k+1) = \\binom{0+k+1}{0} = 1$\n",
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" 2. Assuming $f(j, k+1) = \\binom{j+k+1}{j}$ for some $j$, prove $f(j+1,k+1) = \\binom{j+k+2}{j+1}$\n",
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"\n",
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"Note that there are *three* inductive proofs here - an overall induction on $m$ and nested inductions in both the base case and the induction step (yo dawg, I heard you like induction...).\n",
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"\n",
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"One useful identity for proving this, particularly in step 2, comes from [Pascal's triangle](https://en.wikipedia.org/wiki/Pascal%27s_triangle):\n",
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"$$\\binom{n}{k} = \\binom{n-1}{k} + \\binom{n-1}{k-1}$$"
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]
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},
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{
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"cell_type": "markdown",
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"id": "2c6e7d8c",
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"metadata": {},
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"source": [
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"## Relevant sequences\n",
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"* Central binomial coefficients: [A000984](https://oeis.org/A000984)\n",
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"* General formula: [A046899](https://oeis.org/A046899)\n",
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"\n",
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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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"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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