# A Generic Brute Force Backtracking Solver

One of the projects I’ve had vaguely in the back of my head is a sort of generic puzzle solver. I really love puzzles, but of the pencil and paper and video game varieties. So I think it would be awesome to write out a definition of a puzzle (say how to play Sudoku), give it input, and have it give me an answer back.

Well, I finally got around to trying it!

And it turns out for what I’m working on, it’s actually not at all complicated code:

export function makeSimpleSolver({
generateNextStates,
isValid,
isSolved,
}) {
return (state) => {
let visited = 0

let recur = () => {
visited++

if (!isValid(state)) {
return false
}

if (isSolved(state)) {
return true
}

for (let {step, unstep} of generateNextStates(state)) {
step(state)

if (recur()) {
return true
} else {
unstep(state)
}
}

return false
}
recur()

return state
}
}


Essentially for any puzzle you want to solve, I want you to define three functions:

• generateNextStates: takes in the current ‘state’ (whatever that is) and returns an interable objects containing two functions {step, unstep}:
• step: Take the current state and make one move (move a piece in chess, fill in a number in a Sudoku puzzle)
• unstep: Should be the exact inverse of the previous function
• isValid: Take a state and tell us if it’s a valid state (this may be useless if the generator functions only return valid states, but I needed it for sudoku)
• isSolved: Return true if we have solved the problem

I’ll admit the generateNextStates was particularly weird. If I was working with purely functional languages / immutable data structures, I would probably just keep the state that way. But working on a mutable state (with an undo function) lets us only keep one copy of the actual state data rather than copying it.

It’s like keeping only the current state of a chess board in memory and keeping a list of moves rather than keeping a copy of each possible chess board along the way. It should be somewhat more memory efficient.

And that’s all we need. Other than that, it’s a standard recursive backtracker:

• Take the current state:
• If it’s an invalid state, ignore it
• If it’s a solved state, return it all the way back up the call stack
• Otherwise, generate a list of moves, for each in turn:
• Try to make that move and recursively solve from that point
• If any branch of that state is a solution, keep returning
• If no state was, backtrack (undoing this step) and try another branch

Now, let’s go back to my working example, a Sudoku solver.

First, what are we using as the state of this problem. In this case, a 2D array where each value is 0-9. 1-9 are numbers and 0 is a blank space.

Next, my ‘moves’ are going to be: find the first empty spaces and try every value in it. I’m not at all making sure that a particular number can be used in a space, that’s what isValid is used for.

After that, isValid and isSolved each basically require the same set of helper functions. One each to get all of the values in a specific row (getRow), column, or cell/block (the 3x3 subsections of the puzzle). The first two are straight forward, just access that row or column in the state, but the third is… weird. I mostly had to work it out, but it’s a lot of modular math that gets the blocks from top left, across, and then down.

After that, a quick functional way of determining if a list has any non-zero duplicates:

let hasDup = (vs) => vs.some(v1 => v1 != 0 && vs.filter(v2 => v1 == v2).length != 1)


I’m actually quite happy with that function. It takes a list of values vs and asks if there is some value v1 such that if you filter the list for only values v2 that are the same as v1 there’s more than one of them (a duplicate!). I still really like functional programming, but the absolutely crazy number of libraries for JavaScript available is something to appreciate. Plus, I can run it directly in the browser!

So, putting this all together:

import { makeSimpleSolver } from './solver.js'

let indexes = [0, 1, 2, 3, 4, 5, 6, 7, 8]

let getRow = (state, row) => indexes.map(i => state[row][i])
let getCol = (state, col) => indexes.map(i => state[i][col])
let getCel = (state, cel) => indexes.map(i => state[3 * Math.floor(cel / 3) + Math.floor(i / 3)][3 * (cel % 3) + (i % 3)])

let hasDup = (vs) => vs.some(v1 => v1 != 0 && vs.filter(v2 => v1 == v2).length != 1)

let solveSudoku = makeSimpleSolver({
returnMeta: true,
generateNextStates: function*(state) {
// Find the first empty cell
for (let i = 0; i < 9; i++) {
for (let j = 0; j < 9; j++) {
if (state[i][j] == 0) {
// Try each value in it
for (let v = 1; v <= 9; v++) {
yield {
step: function(state) { state[i][j] = v },
unstep: function(state) { state[i][j] = 0 }
}
}
return
}
}
}
},
isValid: function(state) {
for (let p = 0; p < 9; p++) {
if (hasDup(getRow(state, p))
|| hasDup(getCol(state, p))
|| hasDup(getCel(state, p)))
return false
}
return true
},
isSolved: function(state) {
for (let i = 0; i < 9; i++) {
for (let j = 0; j < 9; j++) {
if (state[i][j] == 0)
return false
}
}

for (let p = 0; p < 9; p++) {
if (hasDup(getRow(state, p))
|| hasDup(getCol(state, p))
|| hasDup(getCel(state, p)))
return false
}

return true
}
})


One interesting thing I didn’t actually know you could do with JavaScript until this project: generators! I’ve always liked them in Python and they’re not that much more difficult to use in JavaScript. Just use function* instead of function and then yield each answer to return. Free (more or less) iterable!

One caveat to watch for, the step and unstep functions have to use the function form, they cannot use arrow syntax, since the latter does not make a closure. The result of that is that the current i/j/v values aren’t captured, instead only one value (the last) is used. Which doesn’t work so well. That was an interesting but to track down.

And that’s it! Let’s run a few tests with some vague timing:

import { sudokuToString, solveSudoku } from './sudoku.js'

let tests = {
'easy': [
[0,4,0,0,0,0,1,7,9],
[0,0,2,0,0,8,0,5,4],
[0,0,6,0,0,5,0,0,8],
[0,8,0,0,7,0,9,1,0],
[0,5,0,0,9,0,0,3,0],
[0,1,9,0,6,0,0,4,0],
[3,0,0,4,0,0,7,0,0],
[5,7,0,1,0,0,2,0,0],
[9,2,8,0,0,0,0,6,0]
],
'hard': [
[1,0,0,0,0,7,0,9,0],
[0,3,0,0,2,0,0,0,8],
[0,0,9,6,0,0,5,0,0],
[0,0,5,3,0,0,9,0,0],
[0,1,0,0,8,0,0,0,2],
[6,0,0,0,0,4,0,0,0],
[3,0,0,0,0,0,0,1,0],
[0,4,0,0,0,0,0,0,7],
[0,0,7,0,0,0,3,0,0]
],
'hardest': [
[8,0,0,0,0,0,0,0,0],
[0,0,3,6,0,0,0,0,0],
[0,7,0,0,9,0,2,0,0],
[0,5,0,0,0,7,0,0,0],
[0,0,0,0,4,5,7,0,0],
[0,0,0,1,0,0,0,3,0],
[0,0,1,0,0,0,0,6,8],
[0,0,8,5,0,0,0,1,0],
[0,9,0,0,0,0,4,0,0]
]
}

for (let test in tests) {
let input = tests[test]

console.log(===== \${test} =====)
console.log(input:)
console.log(sudokuToString(input))

console.time(test)
let output = solveSudoku(input)
console.timeEnd(test)
console.log()

console.log(output:)
console.log(sudokuToString(input))
}


Run:

===== easy =====
input:
040000179
002008054
006005008
080070910
050090030
019060040
300400700
570100200
928000060

easy: 46.848ms

output:
845632179
732918654
196745328
683574912
457291836
219863547
361429785
574186293
928357461

===== hard =====
input:
100007090
030020008
009600500
005300900
010080002
600004000
300000010
040000007
007000300

hard: 444.153ms

output:
162857493
534129678
789643521
475312986
913586742
628794135
356478219
241935867
897261354

===== hardest =====
input:
800000000
003600000
070090200
050007000
000045700
000100030
001000068
008500010
090000400

hardest: 2.034s

output:
812753649
943682175
675491283
154237896
369845721
287169534
521974368
438526917
796318452


Fun! And quick too. At least quick enough for what I’m looking for.

Next up, I want to implement duplicate detection so that our solver doesn’t get into infinite loops. This doesn’t happen in Sudoku (since we’re always adding elements, never removing them; other than backtacking). But in Chess for example, we could get in a state where we move the King back and forth indefiniately, which I want to avoid.

And then… actual puzzle solvers!

Onward!