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chore: improve ac3 code
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@TN1ck
TN1ck committed May 31, 2024
1 parent e0fb5ad commit 6cf500d
Showing 1 changed file with 96 additions and 67 deletions.
163 changes: 96 additions & 67 deletions src/engine/solverAC3.ts
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Original file line number Diff line number Diff line change
Expand Up @@ -36,6 +36,95 @@ function toDomainSudoku(grid: SimpleSudoku): DomainSudoku {
});
}

function ac3(sudoku: DomainSudoku): {
sudoku: DomainSudoku;
solvable: boolean;
} {
sudoku = sudoku.map((r) => r.map((c) => c.slice()));
// Loop until no changes are made to any domain of any cell.
// The original paper did not do this, as the iteration counts do not match.
// I still leave it here, but do not use it.
while (true) {
let change = false;
// We don't keep an actual set of constraints as some AC3 algorithm explanations do it.
// Sudoku has very well defined constraints, we can use loops to check the constraints.
for (let y = 0; y < 9; y++) {
const row = sudoku[y];
for (let x = 0; x < 9; x++) {
let domain1 = row[x];

// The coordinates of the cells that have a constraint with the
// the current cell.
const coordinates: [number, number][] = [];
// Cells in the same row.
for (let xx = 0; xx < 9; xx++) {
if (xx === x) {
continue;
}
coordinates.push([y, xx]);
}

// Cells in the same column.
for (let yy = 0; yy < 9; yy++) {
if (yy === y) {
continue;
}
coordinates.push([yy, x]);
}

// Cells in the same square.
const square = SQUARE_TABLE[squareIndex(x, y)];
for (let c = 0; c < 9; c++) {
const s = square[c];
const [xx, yy] = s;
if (xx === x && yy === y) {
continue;
}
coordinates.push([yy, xx]);
}

for (const [yy, xx] of coordinates) {
const domain2 = sudoku[yy][xx];

// If domain2 consists of only one number, remove it from domain1.
//
// This is an optimization of AC3:
// AC3 checks if there is a value in domain1 that
// does not comply the constraint with at least one value in domain2.
// But because the constraint for sudoku is inequality, the case happens only
// when the domain2 is just one variable.
let changed = false;
if (domain2.length === 1) {
const index = domain1.indexOf(domain2[0]);
if (index !== -1) {
domain1.splice(index, 1);
changed = true;
}
}

change = change || changed;
sudoku[y][x] = domain1;
}

// A domain became empty (e.g. no value works for a cell), we can't solve this Sudoku,
// continue with the next one.
if (domain1.length === 0) {
return {sudoku, solvable: false};
}
}
}
// Note: For "proper" AC3, we wouldn't simply just loop, but only add the constraints to check again if a change was made.
// The result is the same, we might do a few more comparisons, but it is easier to implement.
// TODO: I initially didn't count the ac3 iterations as proposed by the paper.
// But using them now falsifies the tests.
change = false;
if (!change) {
break;
}
}
return {sudoku, solvable: true};
}

/**
* For more information see the paper
* Rating and Generating Sudoku Puzzles Based On Constraint Satisfaction Problems
Expand All @@ -48,8 +137,8 @@ export function _solveGridAC3(
sudoku: SimpleSudoku | null;
iterations: number;
} {
loop: while (stack.length > 0) {
const [grid, ...rest] = stack;
while (stack.length > 0) {
let [grid, ...rest] = stack;

iterations++;
// evil puzzles have an average of about 500, everything more than 1000 that is actually solvable
Expand All @@ -61,72 +150,12 @@ export function _solveGridAC3(
};
}

const rows = grid;

// Loop until no changes are made to any domain of any cell.
// The original paper did not do this, as the iteration counts do not match.
// I still leave it here, but do not use it.
while (true) {
let change = false;
// We don't keep an actual set of constraints as some AC3 algorithm explanations do it.
// Sudoku has very well defined constraints, we can use loops to check the constraints.
for (let y = 0; y < 9; y++) {
const row = rows[y];
for (let x = 0; x < 9; x++) {
let domainCell1 = row[x];
// Note: I once tried to be clever and tried not to compare cells twice but this is will falsify the algorithm.

// Cells in the same row
for (let xx = 0; xx < 9; xx++) {
if (xx === x) {
continue;
}
const domainCell2 = row[xx];
const result = removeValuesFromDomain(domainCell1, domainCell2);
domainCell1 = result[0];
change = change || result[1];
row[x] = domainCell1;
}

// Cells in the same column
for (let yy = 0; yy < 9; yy++) {
if (yy === y) {
continue;
}
const domainCell2 = rows[yy][x];
const result = removeValuesFromDomain(domainCell1, domainCell2);
domainCell1 = result[0];
change = change || result[1];
row[x] = domainCell1;
}

// Cells in the same square
const square = SQUARE_TABLE[squareIndex(x, y)];
for (let c = 0; c < 9; c++) {
const s = square[c];
const [xx, yy] = s;
if (xx === x && yy === y) {
continue;
}
const domainCell2 = rows[yy][xx];
const result = removeValuesFromDomain(domainCell1, domainCell2);
domainCell1 = result[0];
change = change || result[1];
row[x] = domainCell1;
}

// A domain became empty (e.g. no value works for a cell), we can't solve this Sudoku, continue with the next one.
if (domainCell1.length === 0) {
stack = rest;
continue loop;
}
}
}
// The paper which we base on our difficulty ratings did not do this, so we simply always break right now.
// Note: For "proper" AC3, we wouldn't simply just loop, but only add the constraints to check again if a change was made.
// The result is the same, we might do a few more comparisons, but it is easier to implement.
break;
const {sudoku: newGrid, solvable} = ac3(grid);
if (!solvable) {
stack = rest;
continue;
}
grid = newGrid;

const isFilled = grid.every((row) => {
return row.every((cells) => {
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