feat(difficulty): implement Monte Carlo simulation for accurate difficulty calculation
This commit is contained in:
@@ -53,35 +53,78 @@ export function generateRandomGrid(size, density = 0.5) {
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}
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export function calculateDifficulty(density, size = 10) {
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// Shannon Entropy: H(x) = -x*log2(x) - (1-x)*log2(1-x)
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// Normalized to 0-1 range (since max entropy at 0.5 is 1)
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// Data derived from Monte Carlo Simulation (Logical Solver)
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// Format: { size: [solved_pct_at_0.1, ..., solved_pct_at_0.9] }
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// Densities: 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9
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const SIM_DATA = {
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5: [89, 74, 74, 81, 97, 98, 99, 100, 100],
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10: [57, 20, 16, 54, 92, 100, 100, 100, 100],
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15: [37, 10, 2, 12, 68, 100, 100, 100, 100],
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20: [23, 3, 1, 2, 37, 100, 100, 100, 100],
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25: [16, 0, 0, 1, 19, 99, 100, 100, 100],
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30: [8, 0, 0, 0, 5, 99, 100, 100, 100],
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35: [6, 0, 0, 0, 4, 91, 100, 100, 100],
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40: [3, 0, 0, 0, 2, 91, 100, 100, 100],
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45: [2, 0, 0, 0, 1, 82, 100, 100, 100],
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50: [2, 0, 0, 0, 1, 73, 100, 100, 100],
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60: [0, 0, 0, 0, 0, 35, 100, 100, 100],
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70: [0, 0, 0, 0, 0, 16, 100, 100, 100],
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80: [0, 0, 0, 0, 0, 1, 100, 100, 100]
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};
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// Helper to get interpolated value from array
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const getSimulatedSolvedPct = (s, d) => {
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// Find closest sizes
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const sizes = Object.keys(SIM_DATA).map(Number).sort((a, b) => a - b);
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let sLower = sizes[0];
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let sUpper = sizes[sizes.length - 1];
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for (let i = 0; i < sizes.length - 1; i++) {
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if (s >= sizes[i] && s <= sizes[i+1]) {
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sLower = sizes[i];
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sUpper = sizes[i+1];
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break;
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}
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}
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// Clamp density to 0.1 - 0.9
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const dClamped = Math.max(0.1, Math.min(0.9, d));
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// Index in array: 0.1 -> 0, 0.9 -> 8
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const dIndex = (dClamped - 0.1) * 10;
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const dLowerIdx = Math.floor(dIndex);
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const dUpperIdx = Math.ceil(dIndex);
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const dFraction = dIndex - dLowerIdx;
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// Bilinear Interpolation
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// 1. Interpolate Density for Lower Size
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const rowLower = SIM_DATA[sLower];
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const valLower = rowLower[dLowerIdx] * (1 - dFraction) + (rowLower[dUpperIdx] || rowLower[dLowerIdx]) * dFraction;
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// 2. Interpolate Density for Upper Size
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const rowUpper = SIM_DATA[sUpper];
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const valUpper = rowUpper[dLowerIdx] * (1 - dFraction) + (rowUpper[dUpperIdx] || rowUpper[dLowerIdx]) * dFraction;
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// 3. Interpolate Size
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if (sLower === sUpper) return valLower;
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const sFraction = (s - sLower) / (sUpper - sLower);
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return valLower * (1 - sFraction) + valUpper * sFraction;
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};
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const solvedPct = getSimulatedSolvedPct(size, density);
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// Avoid log(0)
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if (density <= 0 || density >= 1) return { level: 'easy', value: 0 };
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const entropy = -density * Math.log2(density) - (1 - density) * Math.log2(1 - density);
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// Difficulty score combines entropy (complexity) and size (scale)
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// We use sqrt(size) to dampen the effect of very large grids,
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// ensuring that density still plays a major role.
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// Normalized against max size (80)
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const sizeFactor = Math.sqrt(size / 80);
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const score = entropy * sizeFactor * 100;
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const value = Math.round(score);
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// Difficulty Score: Inverse of Solved Percent
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// 100% Solved -> 0 Difficulty
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// 0% Solved -> 100 Difficulty
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const value = Math.round(100 - solvedPct);
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// Thresholds
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let level = 'easy';
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if (value >= 80) level = 'extreme';
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else if (value >= 60) level = 'hardest';
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else if (value >= 40) level = 'harder';
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else if (value >= 20) level = 'medium'; // Using 'medium' key if available, or we need to add it?
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// Wait, useI18n only has: easy, harder, hardest, extreme.
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// Let's stick to those keys but adjust ranges.
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if (value >= 75) level = 'extreme';
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else if (value >= 50) level = 'hardest';
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else if (value >= 25) level = 'harder';
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else level = 'easy';
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if (value >= 90) level = 'extreme'; // < 10% Solved
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else if (value >= 60) level = 'hardest'; // < 40% Solved
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else if (value >= 30) level = 'harder'; // < 70% Solved
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else level = 'easy'; // > 70% Solved
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return { level, value };
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}
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278
src/utils/solver.js
Normal file
278
src/utils/solver.js
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@@ -0,0 +1,278 @@
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/**
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* Represents the state of a cell in the solver.
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* -1: Unknown
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* 0: Empty
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* 1: Filled
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*/
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/**
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* Solves a single line (row or column) based on hints and current knowledge.
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* Uses the "Left-Right Overlap" algorithm to find common filled cells.
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* Also identifies definitely empty cells (reachable by no block).
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*
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* @param {number[]} currentLine - Array of -1, 0, 1
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* @param {number[]} hints - Array of block lengths
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* @returns {number[]} - Updated line (or null if contradiction/impossible - though shouldn't happen for valid puzzles)
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*/
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function solveLine(currentLine, hints) {
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const length = currentLine.length;
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// If no hints, all must be empty
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if (hints.length === 0 || (hints.length === 1 && hints[0] === 0)) {
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return Array(length).fill(0);
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}
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// Helper to check if a block can be placed at start index
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const canPlace = (line, start, blockSize) => {
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if (start + blockSize > line.length) return false;
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// Check if any cell in block is 0 (Empty) -> Invalid
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for (let i = start; i < start + blockSize; i++) {
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if (line[i] === 0) return false;
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}
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// Check boundaries (must be separated by empty or edge)
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if (start > 0 && line[start - 1] === 1) return false;
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if (start + blockSize < line.length && line[start + blockSize] === 1) return false;
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return true;
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};
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// 1. Calculate Left-Most Positions
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const leftPositions = [];
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let currentIdx = 0;
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for (let hIndex = 0; hIndex < hints.length; hIndex++) {
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const block = hints[hIndex];
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// Find first valid position
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while (currentIdx <= length - block) {
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if (canPlace(currentLine, currentIdx, block)) {
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// Verify we can fit remaining blocks
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// Simple heuristic: do we have enough space?
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// A full recursive check is better but slower.
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// For "Logical Solver" we assume valid placement is possible if we respect current constraints.
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// However, strictly, we need to know if there is *any* valid arrangement starting here.
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// Let's use a recursive check with memoization for "can fit rest".
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if (canFitRest(currentLine, currentIdx + block + 1, hints, hIndex + 1)) {
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leftPositions.push(currentIdx);
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currentIdx += block + 1; // Move past this block + 1 space
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break;
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}
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}
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currentIdx++;
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}
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if (leftPositions.length <= hIndex) return null; // Impossible
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}
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// 2. Calculate Right-Most Positions (by reversing line and hints)
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// This is symmetrical to Left-Most.
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// Instead of implementing reverse logic, we can just reverse inputs, run left-most, and reverse back.
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// But we need to respect the "currentLine" constraints which might be asymmetric.
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// Actually, "Right-Most" is just "Left-Most" on the reversed grid.
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const reversedLine = [...currentLine].reverse();
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const reversedHints = [...hints].reverse();
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const rightPositionsReversed = [];
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currentIdx = 0;
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for (let hIndex = 0; hIndex < reversedHints.length; hIndex++) {
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const block = reversedHints[hIndex];
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while (currentIdx <= length - block) {
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if (canPlace(reversedLine, currentIdx, block)) {
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if (canFitRest(reversedLine, currentIdx + block + 1, reversedHints, hIndex + 1)) {
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rightPositionsReversed.push(currentIdx);
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currentIdx += block + 1;
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break;
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}
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}
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currentIdx++;
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}
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if (rightPositionsReversed.length <= hIndex) return null;
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}
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// Convert reversed positions to actual indices
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// index in reversed = length - 1 - (original_index + block_size - 1)
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// original_start = length - 1 - (reversed_start + block_size - 1) = length - reversed_start - block_size
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const rightPositions = rightPositionsReversed.map((rStart, i) => {
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const block = reversedHints[i];
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return length - rStart - block;
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}).reverse();
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// 3. Intersect
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const newLine = [...currentLine];
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// Fill intersection
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for (let i = 0; i < hints.length; i++) {
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const l = leftPositions[i];
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const r = rightPositions[i];
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const block = hints[i];
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// If overlap exists: [r, l + block - 1]
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// Example: Block 5. Left: 2, Right: 4.
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// Left: ..XXXXX...
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// Right: ....XXXXX.
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// Overlap: ..XXX...
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// Indices: max(l, r) to min(l+block, r+block) - 1 ?
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// Range is [r, l + block - 1] (inclusive)
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if (r < l + block) {
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for (let k = r; k < l + block; k++) {
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newLine[k] = 1;
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}
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}
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}
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// Determine Empty cells?
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// A cell is empty if it is not covered by ANY block in ANY valid configuration.
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// This is harder with just L/R limits.
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// However, we can use the "Simple Glue" logic:
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// If a cell is outside the range [LeftLimit[i], RightLimit[i] + block] for ALL i, it's empty.
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// Wait, indices are not strictly partitioned. Block 1 could be at 0 or 10.
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// But logic dictates order.
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// Range of block i is [LeftPositions[i], RightPositions[i] + hints[i]].
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// If a cell k is not in ANY of these ranges, it is 0.
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// Mask of possible filled cells
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const possibleFilled = Array(length).fill(false);
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for (let i = 0; i < hints.length; i++) {
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for (let k = leftPositions[i]; k < rightPositions[i] + hints[i]; k++) {
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possibleFilled[k] = true;
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}
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}
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for (let k = 0; k < length; k++) {
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if (!possibleFilled[k]) {
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newLine[k] = 0;
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}
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}
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return newLine;
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}
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// Memoized helper for checking if hints fit
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const memo = new Map();
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function canFitRest(line, startIndex, hints, hintIndex) {
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// Optimization: If hints are empty, we just need to check if remaining line has no '1's
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if (hintIndex >= hints.length) {
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for (let i = startIndex; i < line.length; i++) {
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if (line[i] === 1) return false;
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}
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return true;
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}
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// Key for memoization (primitive approach)
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// In a full solver, we'd pass a cache. For single line, maybe overkill, but safe.
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// let key = `${startIndex}-${hintIndex}`;
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// Skipping memo for now as line lengths are small (<80) and recursion depth is low.
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const remainingLen = line.length - startIndex;
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// Min space needed: sum of hints + (hints.length - 1) spaces
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// Calculate lazily or precalc?
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let minSpace = 0;
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for(let i=hintIndex; i<hints.length; i++) minSpace += hints[i] + (i < hints.length - 1 ? 1 : 0);
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if (remainingLen < minSpace) return false;
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const block = hints[hintIndex];
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// Try to find *any* valid placement for this block
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// We only need ONE valid path to return true.
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for (let i = startIndex; i <= line.length - minSpace; i++) { // Optimization on upper bound?
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// Check placement
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let valid = true;
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// Block
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for (let k = 0; k < block; k++) {
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if (line[i+k] === 0) { valid = false; break; }
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}
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if (!valid) continue;
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// Boundary before (checked by loop start usually, but strictly:
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if (i > 0 && line[i-1] === 1) valid = false; // Should have been handled by caller or skip
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// Wait, the caller (loop) iterates i.
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// If i > startIndex, we implied space at i-1.
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// If line[i-1] is 1, we can't place here if we skipped it.
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// Actually, if we skip a '1', that's invalid.
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// So we can't just skip '1's.
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// Correct logic:
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// We iterate i. If we pass a '1' at index < i, that 1 is orphaned -> Invalid path.
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// So we can only scan forward as long as we don't skip a '1'.
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let skippedOne = false;
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for (let x = startIndex; x < i; x++) {
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if (line[x] === 1) { skippedOne = true; break; }
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}
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if (skippedOne) break; // Cannot go further right, we left a 1 behind.
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// Boundary after
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if (i + block < line.length && line[i+block] === 1) valid = false;
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if (valid) {
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// Recurse
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if (canFitRest(line, i + block + 1, hints, hintIndex + 1)) return true;
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}
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}
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return false;
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}
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/**
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* Solves the puzzle using logical iteration.
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* @param {number[][]} rowHints
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* @param {number[][]} colHints
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* @returns {object} { solvedGrid: number[][], percentSolved: number }
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*/
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export function solvePuzzle(rowHints, colHints) {
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const rows = rowHints.length;
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const cols = colHints.length;
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// Initialize grid with -1
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let grid = Array(rows).fill(null).map(() => Array(cols).fill(-1));
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let changed = true;
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let iterations = 0;
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const MAX_ITER = 100; // Safety break
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while (changed && iterations < MAX_ITER) {
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changed = false;
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iterations++;
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// Rows
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for (let r = 0; r < rows; r++) {
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const newLine = solveLine(grid[r], rowHints[r]);
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if (newLine) {
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for (let c = 0; c < cols; c++) {
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if (grid[r][c] !== newLine[c]) {
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grid[r][c] = newLine[c];
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changed = true;
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}
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}
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}
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}
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// Cols
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for (let c = 0; c < cols; c++) {
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const currentCol = grid.map(row => row[c]);
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const newCol = solveLine(currentCol, colHints[c]);
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if (newCol) {
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for (let r = 0; r < rows; r++) {
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if (grid[r][c] !== newCol[r]) {
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grid[r][c] = newCol[r];
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changed = true;
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}
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}
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}
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}
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}
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// Calculate solved %
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let solvedCount = 0;
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for (let r = 0; r < rows; r++) {
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for (let c = 0; c < cols; c++) {
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if (grid[r][c] !== -1) solvedCount++;
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}
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}
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return {
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solvedGrid: grid,
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percentSolved: (solvedCount / (rows * cols)) * 100
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};
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}
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