# The fastest Sudoku solver

## Winner found

It seems as if we have a winner! Unless anyone plans on contesting the world's current fastest Sudoku solver, user 53x15 wins with the staggeringly fast solver Tdoku. For anyone still working on their solvers, I'll still benchmark new submissions when I have time.

## The challenge

The goal of a game of Sudoku is to fill the board with the numbers 1-9, one in each cell, in such a way that each row, column and box only contains each number once. A very important aspect of a Sudoku puzzle is that there should only be one valid solution.

The goal of this challenge is simple, you should solve a set of Sudoku puzzles as fast as possible. However, you won't just be solving any old Sudoku, you'll be solving the very hardest Sudoku puzzles in existence, the 17-clue Sudokus. Here's an example:

## Rules

### Language

You're free to use any language. If I don't have a compiler installed for your language, you should be able to provide a set of command line instructions needed to install an environment where your script can be run on Linux.

### Benchmark machine

The benchmark will be run on a Dell XPS 9560, 2.8GHz Intel Core i7-7700HQ (3.8GHz boost) 4 cores, 8 threads, 16GB RAM. GTX 1050 4GB. The machine runs Ubuntu 19.04. Here's the uname output, for anyone interested.

Linux 5.0.0-25-generic #26-Ubuntu SMP Thu Aug 1 12:04:58 UTC 2019 x86_64 x86_64 x86_64 GNU/Linux


### Input

The input will be given as a file. It can be found here. The file contains 49151 Sudoku puzzles. The first line of the file is the number of puzzles, and every line after that is 81 characters long and represents a puzzle. The unknown cells are 0, and the known cells are 1-9.

Your program should be able to take the filename as an argument, or have the file input from STDIN, to facilitate manual checking of your solution. Please include an instruction for how your program takes input.

### Timing / scoring

From discussions in the comments, and some reflection, the scoring criteria has been changed to be the time of your entire program. Your program should produce the output file with the correct hash even during official scoring. This doesn't interfere with any existing solution, and doesn't change the rankings as they stand now. Any thoughts on the scoring system are appreciated.

If two solutions have similar scores for individual runs, I will run multiple benchmarks, and the average time will be the final score. If the average scores differ by less than 2%, I will consider it a draw.

If your solution takes longer than an hour to run, it will not be officially scored. In those cases, you are responsible for reporting the machine on which it ran, and your score. For an optimized solver, this should not be an issue.

EDIT: It was brought to my attention that while difficult, the problem set at hand is not the most difficult there is. If time is available, I'll try to benchmark the solutions presented here against the harder puzzle set, and add the score to each submission. However, this will not be an official scoring, and is just for fun.

### Verification

Your solution will be verified by a MD5/SHA256 checksum. Your script should be able to generate a file containing all puzzles and their solutions. However, the file will also be manually inspected, so don't try to get a hash collision. Your output file should match:

MD5: 41704fd7d8fd0723a45ffbb2dbbfa488
SHA256: 0bc8dda364db7b99f389b42383e37b411d9fa022204d124cb3c8959eba252f05

The file will be on the format:

<num_puzzles>
<unsolved_puzzle#1>,<solved_puzzle#1>
<unsolved_puzzle#2>,<solved_puzzle#2>
...
<unsolved_puzzle#n>,<solved_puzzle#n>


with a single trailing newline.

### What's not allowed

You are in no way allowed to hard-code solutions. Your algorithm should be applicable on any set of Sudoku puzzles, both easy and harder Sudokus. However, it is entirely fine if your solution is slow for easier puzzles.

You are not allowed to have a non-deterministic program. You are allowed to use a random number generator, but the seed of the generator should be fixed. This rule is to ensure that measurements are more precise, and have less variance. (Thanks to Peter Taylor for the tip)

You are not allowed to use any external resources or web requests during the runtime of your program. Everything should be self-contained. This does not apply to installed libraries and packages, which are allowed.

### Other info

If you want another test set to check your solution, here are 10000 easier Sudokus. Here are their solutions.

MD5: 3cb465ef6077c4fcab5bd6ae3bc50d62
SHA256: 0bc8dda364db7b99f389b42383e37b411d9fa022204d124cb3c8959eba252f05

If you have any questions, feel free to ask, and I'll try to clarify any misunderstandings.

• I have an APL+WIN solver but unless you have a copy of the interpreter on your machine you will have to count me out. For info your hard example took 30ms and the first easy example 16ms. – Graham Aug 23 at 17:24
• @Graham it took 30ms for all 49151 sudokus, or 30ms on average? – maxb Aug 23 at 17:55
• Sadly 30ms is for the hard example only. Unless this is worth pursuing I have only run the APL solver against your hard example and the first of the easy examples. If we can extrapolate from the hard example the we are looking at around 1500 seconds for the full set – Graham Aug 23 at 18:28
• Should the entries also be code golfed? Or... Can they be golfed, for the fun it? ;-) – The Matt Aug 24 at 3:28
• @TheMatt I'd prefer non-golfed, just so I can verify that nothing fishy is going on – maxb Aug 24 at 6:26

# C++ - 0.201s official score

Using Tdoku (code; design; benchmarks) gives these results:

~/tdoku$lscpu | grep Model.name Model name: Intel(R) Core(TM) i7-4930K CPU @ 3.40GHz ~/tdoku$ # build:
~/tdoku$CC=clang-8 CXX=clang++-8 ./BUILD.sh ~/tdoku$ clang -o solve example/solve.c build/libtdoku.a

~/tdoku$# adjust input format: ~/tdoku$ sed -e "s/0/./g" all_17_clue_sudokus.txt > all_17_clue_sudokus.txt.in

~/tdoku$# solve: ~/tdoku$ time ./solve 1 < all_17_clue_sudokus.txt.in > out.txt
real    0m0.241s
user    0m0.229s
sys     0m0.012s

~/tdoku$# adjust output format and sha256sum: ~/tdoku$ grep -v "^:0:$" out.txt | sed -e "s/:1:/,/" | tr . 0 | sha256sum 0bc8dda364db7b99f389b42383e37b411d9fa022204d124cb3c8959eba252f05 -  Tdoku has been optimized for hard Sudoku instances. But note, contrary to the problem statement, that 17 clue puzzles are far from the hardest Sudoku. Actually they're among the easiest, with the majority requiring no backtracking at all. See some of the other benchmark datasets in the Tdoku project for puzzles that are actually hard. Also note that while Tdoku is the fastest solver I'm aware of for hard puzzles, it's not the fastest for 17 clue puzzles. For these I think the fastest is this rust project, a derivative of JCZSolve, which was optimized for 17 clue puzzles during development. Depending on the platform it might be 5-25% faster than Tdoku for these puzzles. • Wow, that was an interesting read about the implementation and theory behind it. Before I started this challenge, I wanted to find state of the art solvers and datasets. I guess I didn't look hard enough. From popular "scientific" articles, the 17 clue puzzles were all that's ever talked about, so it was my assumption that those were the hardest. I'll try to run all submissions against the data sets presented in your article, and I'll benchmark your submission later today. Fantastic job! – maxb Aug 30 at 4:36 • Thanks! You see from the article that finding state-of-the-art solving took me on a long journey. :-) I get why people focus on 17 clue puzzles: the dataset is well known, well defined, complete or nearly so, moderately large, and hard for naive solvers. While it's interesting to study harder puzzles, hardness is tricky to formalize. e.g., do we mean subjectively or empirically hard for humans based on the techniques required? do we mean slow on average for a given solver under random permutations? Do we mean minimum backdoor size under a formula with chosen pigeonhole inferences? etc. – 53x15 Aug 30 at 21:14 # Node.js, 8.231s 6.735s official score Takes the file name as argument. The input file may already contain the solutions in the format described in the challenge, in which case the program will compare them with its own solutions. The results are saved in 'sudoku.log'. ### Code 'use strict'; const fs = require('fs'); const BLOCK = []; const BLOCK_NDX = []; const N_BIT = []; const ZERO = []; const BIT = []; console.time('Processing time'); init(); let filename = process.argv[2], puzzle = fs.readFileSync(filename).toString().split('\n'), len = puzzle.shift(), output = len + '\n'; console.log("File '" + filename + "': " + len + " puzzles"); // solve all puzzles puzzle.forEach((p, i) => { let sol, res; [ p, sol ] = p.split(','); if(p.length == 81) { if(!(++i % 2000)) { console.log((i * 100 / len).toFixed(1) + '%'); } if(!(res = solve(p))) { throw "Failed on puzzle " + i; } if(sol && res != sol) { throw "Invalid solution for puzzle " + i; } output += p + ',' + res + '\n'; } }); // results console.timeEnd('Processing time'); fs.writeFileSync('sudoku.log', output); console.log("MD5 = " + require('crypto').createHash('md5').update(output).digest("hex")); // initialization of lookup tables function init() { let ptr, x, y; for(x = 0; x < 0x200; x++) { N_BIT[x] = [0, 1, 2, 3, 4, 5, 6, 7, 8].reduce((s, n) => s + (x >> n & 1), 0); ZERO[x] = ~x & -~x; } for(x = 0; x < 9; x++) { BIT[1 << x] = x; } for(ptr = y = 0; y < 9; y++) { for(x = 0; x < 9; x++, ptr++) { BLOCK[ptr] = (y / 3 | 0) * 3 + (x / 3 | 0); BLOCK_NDX[ptr] = (y % 3) * 3 + x % 3; } } } // solver function solve(p) { let ptr, x, y, v, count = 81, m = Array(81).fill(-1), row = Array(9).fill(0), col = Array(9).fill(0), blk = Array(9).fill(0); // helper function to check and play a move function play(stack, x, y, n) { let p = y * 9 + x; if(~m[p]) { if(m[p] == n) { return true; } undo(stack); return false; } let msk, b; msk = 1 << n; b = BLOCK[p]; if((col[x] | row[y] | blk[b]) & msk) { undo(stack); return false; } count--; col[x] ^= msk; row[y] ^= msk; blk[b] ^= msk; m[p] = n; stack.push(x << 8 | y << 4 | n); return true; } // helper function to undo all moves on the stack function undo(stack) { stack.forEach(v => { let x = v >> 8, y = v >> 4 & 15, p = y * 9 + x, b = BLOCK[p]; v = 1 << (v & 15); count++; col[x] ^= v; row[y] ^= v; blk[b] ^= v; m[p] = -1; }); } // convert the puzzle into our own format for(ptr = y = 0; y < 9; y++) { for(x = 0; x < 9; x++, ptr++) { if(~(v = p[ptr] - 1)) { col[x] |= 1 << v; row[y] |= 1 << v; blk[BLOCK[ptr]] |= 1 << v; count--; m[ptr] = v; } } } // main recursive search function let res = (function search() { // success? if(!count) { return true; } let ptr, x, y, v, n, max, best, k, i, stack = [], dCol = Array(81).fill(0), dRow = Array(81).fill(0), dBlk = Array(81).fill(0), b, v0; // scan the grid: // - keeping track of where each digit can go on a given column, row or block // - looking for a cell with the fewest number of legal moves for(max = ptr = y = 0; y < 9; y++) { for(x = 0; x < 9; x++, ptr++) { if(m[ptr] == -1) { v = col[x] | row[y] | blk[BLOCK[ptr]]; n = N_BIT[v]; // abort if there's no legal move on this cell if(n == 9) { return false; } // update dCol[], dRow[] and dBlk[] for(v0 = v ^ 0x1FF; v0;) { b = v0 & -v0; dCol[x * 9 + BIT[b]] |= 1 << y; dRow[y * 9 + BIT[b]] |= 1 << x; dBlk[BLOCK[ptr] * 9 + BIT[b]] |= 1 << BLOCK_NDX[ptr]; v0 ^= b; } // update the cell with the fewest number of moves if(n > max) { best = { x : x, y : y, ptr: ptr, msk: v }; max = n; } } } } // play all forced moves (unique candidates on a given column, row or block) // and make sure that it doesn't lead to any inconsistency for(k = 0; k < 9; k++) { for(n = 0; n < 9; n++) { if(N_BIT[dCol[k * 9 + n]] == 1) { i = BIT[dCol[k * 9 + n]]; if(!play(stack, k, i, n)) { return false; } } if(N_BIT[dRow[k * 9 + n]] == 1) { i = BIT[dRow[k * 9 + n]]; if(!play(stack, i, k, n)) { return false; } } if(N_BIT[dBlk[k * 9 + n]] == 1) { i = BIT[dBlk[k * 9 + n]]; if(!play(stack, (k % 3) * 3 + i % 3, (k / 3 | 0) * 3 + (i / 3 | 0), n)) { return false; } } } } // if we've played at least one forced move, do a recursive call right away if(stack.length) { if(search()) { return true; } undo(stack); return false; } // otherwise, try all moves on the cell with the fewest number of moves while((v = ZERO[best.msk]) < 0x200) { col[best.x] ^= v; row[best.y] ^= v; blk[BLOCK[best.ptr]] ^= v; m[best.ptr] = BIT[v]; count--; if(search()) { return true; } count++; m[best.ptr] = -1; col[best.x] ^= v; row[best.y] ^= v; blk[BLOCK[best.ptr]] ^= v; best.msk ^= v; } return false; })(); return res ? m.map(n => n + 1).join('') : false; } // debugging function dump(m) { let x, y, c = 81, s = ''; for(y = 0; y < 9; y++) { for(x = 0; x < 9; x++) { s += (~m[y * 9 + x] ? (c--, m[y * 9 + x] + 1) : '-') + (x % 3 < 2 || x == 8 ? ' ' : ' | '); } s += y % 3 < 2 || y == 8 ? '\n' : '\n------+-------+------\n'; } console.log(c); console.log(s); }  ### Example output Tested on an Intel Core i7 7500U @ 2.70 GHz. • I might have to clear up the scoring. If you do anything in parallel, your score is still the sum of all individual solve times. You should calculate that sum and present it as your score. That way it's more about getting the code as fast as possible. The code can always parallelize across the 49151 puzzles, making that part trivial. I might change the scoring to be the total time of the program, and disallow multithreading. Or, perhaps multithreading should be a part of the challenge? – maxb Aug 24 at 21:04 • @maxb I see. I did not understand that your concern was about multi-threading. – Arnauld Aug 24 at 21:22 • Why is your solution so much faster than the other ones? – Anush Aug 28 at 18:54 • @Anush What I've called 'forced moves' in the code is what makes it significantly faster and is better known as hidden singles. (We could also look for hidden twins, triples, quads, etc. but I'm not sure it's really worth it, at least in Node.) – Arnauld Aug 28 at 19:58 • "I started looking at naked singles" careful with the phrasing :) – ngn Aug 29 at 8:20 # Python 3 (with dlx) 4min 46.870s official score (single core i7-3610QM here) Obviously beatable with a compiled language like C, and making use of threading, but it's a start... sudoku is a module I've placed on github (copied at the footer of this post) which uses dlx under the hood. #!/usr/bin/python import argparse import gc import sys from timeit import timeit from sudoku import Solver def getSolvers(filePath): solvers = [] with open(filePath, 'r') as inFile: for line in inFile: content = line.rstrip() if len(content) == 81 and content.isdigit(): solvers.append(Solver(content)) return solvers def solve(solvers): for solver in solvers: yield next(solver.genSolutions()) if __name__ == '__main__': parser = argparse.ArgumentParser(description='Time or print solving of some sudoku.') parser.add_argument('filePath', help='Path to the file containing proper sudoku on their own lines as 81 digits in row-major order with 0s as blanks') parser.add_argument('-p', '--print', dest='printEm', action='store_true', default=False, help='print solutions in the same fashion as the input') parser.add_argument('-P', '--pretty', dest='prettyPrintEm', action='store_true', default=False, help='print inputs and solutions formatted for human consumption') args = parser.parse_args() if args.printEm or args.prettyPrintEm: solvers = getSolvers(args.filePath) print(len(solvers)) for solver, solution in zip(solvers, solve(solvers)): if args.prettyPrintEm: print(solver) print(solution) else: print('{},{}'.format(solver.representation(noneCharacter='0'), solution.representation())) else: setup = '''\ from __main__ import getSolvers, solve, args, gc gc.disable() solvers = getSolvers(args.filePath)''' print(timeit("for solution in solve(solvers): pass", setup=setup, number=1))  ### Usage • Install Python 3 • Save sudoku.py somewhere on your path (from the git hub link or copy it from below) • Save the above code as testSolver.py somewhere on your path • Install dlx: python -m pip install dlx • Run it (by the way it consumes memory like it's going out of fashion) usage: testSolver.py [-h] [-p] [-P] filePath Time or print solving of some sudoku. positional arguments: filePath Path to the file containing proper sudoku on their own lines as 81 digits in row-major order with 0s as blanks optional arguments: -h, --help show this help message and exit -p, --print print solutions in the same fashion as the input -P, --pretty print inputs and solutions formatted for human consumption Pipe output as required in the challenge spec to a file if need be: python testSolver.py -p input_file_path > output_file_path sudoku.py (yes there are extra features here other than solving) import dlx from itertools import permutations, takewhile from random import choice, shuffle ''' A 9 by 9 sudoku solver. ''' _N = 3 _NSQ = _N**2 _NQU = _N**4 _VALID_VALUE_INTS = list(range(1, _NSQ + 1)) _VALID_VALUE_STRS = [str(v) for v in _VALID_VALUE_INTS] _EMPTY_CELL_CHAR = '·' # The following are mutually related by their ordering, and define ordering throughout the rest of the code. Here be dragons. # _CANDIDATES = [(r, c, v) for r in range(_NSQ) for c in range(_NSQ) for v in range(1, _NSQ + 1)] _CONSTRAINT_INDEXES_FROM_CANDIDATE = lambda r, c, v: [ _NSQ * r + c, _NQU + _NSQ * r + v - 1, _NQU * 2 + _NSQ * c + v - 1, _NQU * 3 + _NSQ * (_N * (r // _N) + c // _N) + v - 1] _CONSTRAINT_FORMATTERS = [ "R{0}C{1}" , "R{0}#{1}" , "C{0}#{1}" , "B{0}#{1}"] _CONSTRAINT_NAMES = [(s.format(a, b + (e and 1)), dlx.DLX.PRIMARY) for e, s in enumerate(_CONSTRAINT_FORMATTERS) for a in range(_NSQ) for b in range(_NSQ)] _EMPTY_GRID_CONSTRAINT_INDEXES = [_CONSTRAINT_INDEXES_FROM_CANDIDATE(r, c, v) for (r, c, v) in _CANDIDATES] # # The above are mutually related by their ordering, and define ordering throughout the rest of the code. Here be dragons. class Solver: def __init__(self, representation=''): if not representation or len(representation) != _NQU: self._complete = False self._NClues = 0 self._repr = [None]*_NQU # blank grid, no clues - maybe to extend to a generator by overriding the DLX column selection to be stochastic. else: nClues = 0 repr = [] for value in representation: if not value: repr.append(None) elif isinstance(value, int) and 1 <= value <= _NSQ: nClues += 1 repr.append(value) elif value in _VALID_VALUE_STRS: nClues += 1 repr.append(int(value)) else: repr.append(None) self._complete = nClues == _NQU self._NClues = nClues self._repr = repr def genSolutions(self, genSudoku=True, genNone=False, dlxColumnSelctor=None): ''' if genSudoku=False, generates each solution as a list of cell values (left-right, top-bottom) ''' if self._complete: yield self else: self._initDlx() dlxColumnSelctor = dlxColumnSelctor or dlx.DLX.smallestColumnSelector if genSudoku: for solution in self._dlx.solve(dlxColumnSelctor): yield Solver([v for (r, c, v) in sorted([self._dlx.N[i] for i in solution])]) elif genNone: for solution in self._dlx.solve(dlxColumnSelctor): yield else: for solution in self._dlx.solve(dlxColumnSelctor): yield [v for (r, c, v) in sorted([self._dlx.N[i] for i in solution])] def uniqueness(self, returnSolutionIfProper=False): ''' Returns: 0 if unsolvable; 1 (or the unique solution if returnSolutionIfProper=True) if uniquely solvable; or 2 if multiple possible solutions exist - a 'proper' sudoku is uniquely solvable. ''' slns = list(takewhile(lambda t: t[0] < 2, ((i, sln) for i, sln in enumerate(self.genSolutions(genSudoku=returnSolutionIfProper, genNone=not returnSolutionIfProper))))) uniqueness = len(slns) if returnSolutionIfProper and uniqueness == 1: return slns[0][1] else: return uniqueness def representation(self, asString=True, noneCharacter='.'): if asString: return ''.join([v and str(_VALID_VALUE_STRS[v - 1]) or noneCharacter for v in self._repr]) return self._repr[:] def __repr__(self): return display(self._repr) def _initDlx(self): self._dlx = dlx.DLX(_CONSTRAINT_NAMES) rowIndexes = self._dlx.appendRows(_EMPTY_GRID_CONSTRAINT_INDEXES, _CANDIDATES) for r in range(_NSQ): for c in range(_NSQ): v = self._repr[_NSQ * r + c] if v is not None: self._dlx.useRow(rowIndexes[_NQU * r + _NSQ * c + v - 1]) _ROW_SEPARATOR_COMPACT = '+'.join(['-' * (2 * _N + 1) for b in range(_N)])[1:-1] + '\n' _ROW_SEPARATOR = ' ·-' + _ROW_SEPARATOR_COMPACT[:-1] + '-·\n' _TOP_AND_BOTTOM = _ROW_SEPARATOR.replace('+', '·') _ROW_LABELS = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'J'] _COL_LABELS = ['1', '2', '3', '4', '5', '6', '7', '8', '9'] _COLS_LABEL = ' ' + ' '.join([i % _N == 0 and ' ' + l or l for i, l in enumerate(_COL_LABELS)]) + '\n' def display(representation, conversion=None, labelled=True): result = '' raw = [conversion[n or 0] for n in representation] if conversion else representation if labelled: result += _COLS_LABEL + _TOP_AND_BOTTOM rSep = _ROW_SEPARATOR else: rSep = _ROW_SEPARATOR_COMPACT for r in range(_NSQ): if r > 0 and r % _N == 0: result += rSep for c in range(_NSQ): if c % _N == 0: if c == 0: if labelled: result += _ROW_LABELS[r] + '| ' else: result += '| ' result += str(raw[_NSQ * r + c] or _EMPTY_CELL_CHAR) + ' ' if labelled: result += '|' result += '\n' if labelled: result += _TOP_AND_BOTTOM else: result = result[:-1] return result def permute(representation): ''' returns a random representation from the given representation's equivalence class ''' rows = [list(representation[i:i+_NSQ]) for i in range(0, _NQU, _NSQ)] rows = permuteRowsAndBands(rows) rows = [[r[i] for r in rows] for i in range(_NSQ)] rows = permuteRowsAndBands(rows) pNumbers = [str(i) for i in range(1, _NSQ + 1)] shuffle(pNumbers) return ''.join(''.join([pNumbers[int(v) - 1] if v.isdigit() and v != '0' else v for v in r]) for r in rows) def permuteRowsAndBands(rows): bandP = choice([x for x in permutations(range(_N))]) rows = [rows[_N * b + r] for b in bandP for r in range(_N)] for band in range(0, _NSQ, _N): rowP = choice([x for x in permutations([band + i for i in range(_N)])]) rows = [rows[rowP[i % _N]] if i // _N == band else rows[i] for i in range(_NSQ)] return rows def getRandomSolvedStateRepresentation(): return permute('126459783453786129789123456897231564231564897564897231312645978645978312978312645') def getRandomSudoku(): r = getRandomSolvedStateRepresentation() s = Solver(r) indices = list(range(len(r))) shuffle(indices) for i in indices: ns = Solver(s._repr[:i] + [None] + s._repr[i+1:]) if ns.uniqueness() == 1: s = ns return s if __name__ == '__main__': print('Some example useage:') inputRepresentation = '..3......4......2..8.12...6.........2...6...7...8.7.31.1.64.9..6.5..8...9.83...4.' print('>>> s = Solver({})'.format(inputRepresentation)) s = Solver(inputRepresentation) print('>>> s') print(s) print('>>> print(s.representation())') print(s.representation()) print('>>> print(display(s.representation(), labelled=False))') print(display(s.representation(), labelled=False)) print('>>> for solution in s.genSolutions(): solution') for solution in s.genSolutions(): print(solution) inputRepresentation2 = inputRepresentation[:2] + '.' + inputRepresentation[3:] print('>>> s.uniqueness()') print(s.uniqueness()) print('>>> s2 = Solver({}) # removed a clue; this has six solutions rather than one'.format(inputRepresentation2)) s2 = Solver(inputRepresentation2) print('>>> s2.uniqueness()') print(s2.uniqueness()) print('>>> for solution in s2.genSolutions(): solution') for solution in s2.genSolutions(): print(solution) print('>>> s3 = getRandomSudoku()') s3 = getRandomSudoku() print('>>> s3') print(s3) print('>>> for solution in s3.genSolutions(): solution') for solution in s3.genSolutions(): print(solution)  • Impressive for a Python solution, I'll try to benchmark it later today. – maxb Aug 24 at 8:31 • Thanks; and wow, so much faster there maxb! – Jonathan Allan Aug 24 at 13:42 • +1 for using dancing links – Anush Aug 25 at 4:32 # Python 3 + Z3 - 10min 45.657s official score about 1000s on my laptop. import time start = time.time() import z3.z3 as z3 import itertools import datetime import sys solver = z3.Solver() ceils = [[None] * 9 for i in range(9)] for row in range(9): for col in range(9): name = 'c' + str(row * 9 + col) ceil = z3.BitVec(name, 9) solver.add(z3.Or( ceil == 0b000000001, ceil == 0b000000010, ceil == 0b000000100, ceil == 0b000001000, ceil == 0b000010000, ceil == 0b000100000, ceil == 0b001000000, ceil == 0b010000000, ceil == 0b100000000 )) solver.add(ceil != 0) ceils[row][col] = ceil for i in range(9): for j in range(9): for k in range(9): if j == k: continue solver.add(ceils[i][j] & ceils[i][k] == 0) solver.add(ceils[j][i] & ceils[k][i] == 0) row, col = i // 3 * 3, i % 3 * 3 solver.add(ceils[row + j // 3][col + j % 3] & ceils[row + k // 3][col + k % 3] == 0) row_col = list(itertools.product(range(9), range(9))) lookup = { 1 << i: str(i + 1) for i in range(9) } def solve(line): global solver, output, row_col, ceils, lookup solver.push() for value, (row, col) in zip(line, row_col): val = ord(value) - 48 if val == 0: continue solver.add(ceils[row][col] == 1 << (val - 1)) output = [] if solver.check() == z3.sat: model = solver.model() for row in range(9): for col in range(9): val = model[ceils[row][col]].as_long() output.append(lookup[val]) solver.pop() return ''.join(output) count = int(input()) print(count) for i in range(count): if i % 1000 == 0: sys.stderr.write(str(i) + '\n') line = input() print(line + "," + solve(line)) end = time.time() sys.stderr.write(str(end - start))  Install dependency pip install z3-solver Run python3 solve.py < in.txt > out.txt I'm not sure how to improve its performance, since it just solved magically... • Quite impressive for a general constraint solver. My first implementation was way slower than this. Running a benchmark right now, I'll update the post once it's done. – maxb Aug 26 at 5:53 • @maxb just done some general clean up, and i believe there is no need to update the benchmark... – tsh Aug 26 at 7:03 # C - 2.228s 1.690s official score based on @Arnauld's #include<fcntl.h> #define O const #define R return #define S static #define$(x,y...)if(x){y;}
#define  W(x,y...)while(x){y;}
#define fi(x,y...)for(I i=0,_n=(x);i<_n;i++){y;}
#define fj(x,y...)for(I j=0,_n=(x);j<_n;j++){y;}
#define fp81(x...)for(I p=0;p<81;p++){x;}
#define  fq3(x...)for(I q=0;q<3;q++){x;}
#define fij9(x...){fi(9,fj(9,x))}
#define m0(x)m0_((V*)(x),sizeof(x));
#define popc(x)__builtin_popcount(x)
#define ctz(x)__builtin_ctz(x)
#include<sys/syscall.h>
#define sc(f,x...)({L u;asm volatile("syscall":"=a"(u):"0"(SYS_##f)x:"cc","rcx","r11","memory");u;})
#define sc1(f,x)    sc(f,,"D"(x))
#define sc2(f,x,y)  sc(f,,"D"(x),"S"(y))
#define sc3(f,x,y,z)sc(f,,"D"(x),"S"(y),"d"(z))
#define wr(a...)sc3(write,a)
#define op(a...)sc3( open,a)
#define cl(a...)sc1(close,a)
#define fs(a...)sc2(fstat,a)
#define ex(a...)sc1( exit,a)
#define mm(x,y,z,t,u,v)({register L r10 asm("r10")=t,r8 asm("r8")=u,r9 asm("r9")=v;\
(V*)sc(mmap,,"D"(x),"S"(y),"d"(z),"r"(r10),"r"(r8),"r"(r9));})
typedef void V;typedef char C;typedef short H;typedef int I;typedef long long L;
S C BL[81],KL[81],IJK[81][3],m[81],t_[81-17],*t;S H rcb[3][9],cnt;
S V*mc(V*x,O V*y,L n){C*p=x;O C*q=y;fi(n,*p++=*q++)R x;}S V m0_(C*p,L n){fi(n,*p++=0);}
S I undo(C*t0){cnt+=t-t0;W(t>t0,C p=*--t;H v=1<<m[p];fq3(rcb[q][IJK[p][q]]^=v)m[p]=-1)R 0;}
S I play(C p,H v){$(m[p]>=0,R 1<<m[p]==v)I w=0;fq3(w|=rcb[q][IJK[p][q]])$(w&v,R 0)cnt--;
fq3(rcb[q][IJK[p][q]]^=v);m[p]=ctz(v);*t++=p;R 1;}
S I f(){$(!cnt,R 1)C*t0=t;H max=0,bp,bv,d[9][9][4];m0(d); fij9(I p=i*9+j;$(m[p]<0,
I v=0;fq3(v|=rcb[q][IJK[p][q]])I w=v^511;$(!w,R 0)H g[]={1<<j,1<<i,1<<BL[p]}; do{I z=ctz(w);w&=w-1;fq3(d[IJK[p][q]][z][q]|=g[q]);}while(w); I n=popc(v);$(max<n,max=n;bp=p;bv=v)))
fij9(I u=d[i][j][0];$(popc(u)==1,I l=ctz(u);$(!play(   i*9+l ,1<<j),R undo(t0)))
u=d[i][j][1];$(popc(u)==1,I l=ctz(u);$(!play(   l*9+i ,1<<j),R undo(t0)))
u=d[i][j][2];$(popc(u)==1,I l=ctz(u);$(!play(KL[i*9+l],1<<j),R undo(t0))))
$(t-t0,R f()||undo(t0)) W(1,I v=1<<ctz(~bv);$(v>511,R 0)fq3(rcb[q][IJK[bp][q]]^=v)m[bp]=ctz(v);cnt--;$(f(),R 1) cnt++;m[bp]=-1;fq3(rcb[q][IJK[bp][q]]^=v)bv^=v) R 0;} asm(".globl _start\n_start:pop %rdi\nmov %rsp,%rsi\njmp main"); V main(I ac,C**av){$(ac!=2,ex(2))
fij9(I p=i*9+j;BL[p]=i%3*3+j%3;KL[p]=(i/3*3+j/3)*9+BL[p];IJK[p][0]=i;IJK[p][1]=j;IJK[p][2]=i/3*3+j/3)
I d=op(av[1],0,0);struct stat h;fs(d,&h);C*s0=mm(0,h.st_size,1,0x8002,d,0),*s=s0;cl(d); //in
C*r0=mm(0,2*h.st_size,3,0x22,-1,0),*r=r0; //out
I n=0;W(*s!='\n',n*=10;n+=*s++-'0')s++;mc(r,s0,s-s0);r+=s-s0;
fi(n,m0(rcb);cnt=81;t=t_;$(s[81]&&s[81]!='\n',ex(3))mc(r,s,81);r+=81;*r++=','; fp81(I v=m[p]=*s++-'1';$(v>=0,v=1<<v;fq3(rcb[q][IJK[p][q]]|=v)cnt--))
s++;$(!f(),ex(4))fp81(r[p]=m[p]+'1')r+=81;*r++='\n') wr(1,r0,r-r0);ex(0);}  compile and run: gcc -O3 -march=native -nostdlib -ffreestanding time ./a.out all_17_clue_sudokus.txt | md5sum  • Congratulations, you (and Arnauld) are in the lead by a lot right now. – maxb Aug 26 at 12:32 • @maxb i tried using more efficient i/o (direct syscalls without libc) but the effect wasn't as great as i hoped. i also tidied up the rest of the code. this should take away ~0.2s. do you mind re-scoring? – ngn Aug 27 at 20:36 • Of course, I'll try to get it done sometime today – maxb Aug 28 at 5:06 • I was also thinking about trying a RAMdisk for all I/O, just to see if that makes a difference. I doubt it'll make a huge difference, since reads and writes are sequential, and my SSD has a large enough cache to fit everything. – maxb Aug 28 at 7:24 • @maxb there probably won't be any difference at all. the second time you run the program, the input file will already be in ram anyway - in linux's filesystem cache. – ngn Aug 28 at 7:44 # C - 12min 28.374s official score runs for about 30m 15m on my i5-7200U and produces the correct md5 hash #include<stdio.h> #include<stdlib.h> #include<string.h> #include<sys/time.h> #define B break #define O const #define P printf #define R return #define S static #define$(x,y...)  if(x){y;}
#define E(x...)    else{x;}
#define W(x,y...)  while(x){y;}
#define fi(x,y...) for(I i=0,_n=(x);i<_n;i++){y;}
#define fj(x,y...) for(I j=0,_n=(x);j<_n;j++){y;}
typedef void V;typedef char C;typedef short H;typedef int I;typedef long long L;
S C h[81][20]; //h[i][0],h[i][1],..,h[i][19] are the squares that clash with square i
S H a[81]      //a[i]: bitmask of possible choices; initially one of 1<<0, 1<<1 .. 1<<8, or 511 (i.e. nine bits set)
,b[81];     //b[i]: negated bitmask of impossible chioces; once we know square i has value v, b[i] becomes ~(1<<v)
S I f(){ //f:recursive solver
I p=-1; //keep track of the popcount (number of 1 bits) in a
W(1,I q=0;                                         //simple non-recursive deductions:
fi(81,fj(20,a[i]&=b[h[i][j]])                  // a[i] must not share bits with its clashing squares
$(!(a[i]&a[i]-1),$(!a[i],R 0)b[i]=~a[i]) // if a[i] has one bit left, update b[i].  if a[i]=0, we have a contradiction
q+=__builtin_popcount(a[i]))             // compute new popcount
$(p==q,B)p=q;) // if the popcount of a[] changed, try to do more deductions I k=-1,mc=10;fi(81,$(b[i]==-1,I c=__builtin_popcount(a[i]);$(c<mc,k=i;mc=c;$(c==2,B)))) //find square with fewest options left
$(k==-1,R 1) //if there isn't any such, we're done - success! otherwise k is that square fi(9,$(a[k]&1<<i,H a0[81],b0[81];                                        //try different values for square k
memcpy(a0,a,81*sizeof(*a));memcpy(b0,b,81*sizeof(*b));  // save a and b
a[k]=1<<i;b[k]=~a[k];$(f(),R 1) // set square k and make a recursive call memcpy(a,a0,81*sizeof(*a));memcpy(b,b0,81*sizeof(*b)))) // restore a and b R 0;} S L tm(){struct timeval t;gettimeofday(&t,0);R t.tv_sec*1000000+t.tv_usec;} //current time in microseconds I main(){L t=0;I n;scanf("%d",&n);P("%d\n",n); fi(81,L l=0;fj(81,$(i!=j&&(i%9==j%9||i/9==j/9||(i/27==j/27&&i%9/3==j%9/3)),h[i][l++]=j))) //precompute h
fi(n,S C s[82];scanf("%s",s);printf("%s,",s);                        //i/o and loop over puzzles
fj(81,a[j]=s[j]=='0'?511:1<<(s[j]-'1');b[j]=s[j]=='0'?-1:~a[j]) //represent '1' .. '9' as 1<<0 .. 1<<8, and 0 as 511
t-=tm();I r=f();t+=tm();                                        //measure time only for the solving function
$(!r,P("can't solve\n");exit(1)) //shouldn't happen fj(81,s[j]=a[j]&a[j]-1?'0':'1'+__builtin_ctz(a[j])) //1<<0 .. 1<<8 to '1' .. '9' P("%s\n",s)) //output fflush(stdout);dprintf(2,"time:%lld microseconds\n",t);R 0;} //print self-measured time to stderr so it doesn't affect stdout's md5  compile (preferably with clang v6) and run: clang -O3 -march=native a.c time ./a.out <all_17_clue_sudokus.txt | tee o.txt | nl md5sum o.txt  • Why so ugly? This isn't code-golf! – Jonathan Allan Aug 23 at 22:00 • @JonathanAllan that's how i usually code (unless i'm in a team who prefer to do otherwise). it's beautiful :) – ngn Aug 23 at 22:01 • Haha, "beautiful", and easy to come back to in 6 months :p – Jonathan Allan Aug 23 at 22:06 • yes, actually. i've been doing this for a couple of years and i find it more efficient. in the apl world it's known as incunabulum style. with bloated code you move your eyes mostly vertically (unnatural and unfit for our landscape monitors) and scroll a lot. with tight code you can see all of it at once, so it's easier to find your way around it and to judge its complexity at a glance. – ngn Aug 23 at 22:13 • Is it a backtracking solution? I see two memcpy in there and some recursion going on. I'll try to verify it today. – maxb Aug 24 at 6:34 # Java - 4.056s official score The main idea of this is to never allocate memory when it is not needed. The only exception are primitives, which should be optimized by the compiler anyway. Everything else is stored as masks and arrays of operations done in each step, which can be undone when the recursion step is completed. About half of all sudokus are solved completely without backtracking, but if I push that number higher the overall time seems to be slower. I'm planning om rewriting this in C++ and optimize even further, but this solver is becoming a behemoth. I wanted to implement as much caching as possible, which lead to some issues. For example, if there are two cells on the same row which can only have the number 6, then we have reached an impossible case, and should return to the backtracking. But since I calculated all options in one sweep, and then placed numbers in cells with only one possibility, I didn't double check that I had placed a number in the same row just before. This lead to impossible solutions. With everything being contained in the arrays defined at the top, the memory usage of the actual solver is about 216kB. The main part of the memory usage comes from the array containing all the puzzles, and the I/O handlers in Java. EDIT: I have a version which is translated to C++ now, but it isn't vastly faster. The official time is around 3.5 seconds, which isn't a huge improvement. I think the main issue with my implementation is that I keep my masks as arrays rather than bitmasks. I'll try to analyze Arnauld's solution to see what can be done to improve it. import java.util.HashMap; import java.util.ArrayList; import java.util.Arrays; import java.io.IOException; import java.nio.charset.StandardCharsets; import java.io.BufferedReader; import java.io.InputStreamReader; import java.io.BufferedInputStream; import java.io.FileInputStream; import java.io.File; import java.io.PrintWriter; public class Sudoku { final private int[] unsolvedBoard; final private int[] solvedBoard; final private int[][] neighbors; final private int[][] cells; private static int[] clues; final private int[][] mask; final private int[] formattedMask; final private int[][] placedMask; final private boolean[][][] lineMask; final private int[] lineCounters; final private int[][] sectionCounters; final private int[][] sectionMask; private int easySolved; private boolean isEasy; private int totEasy; private int placedNumbers; public long totTime = 0; private boolean solutionFound; public long lastPrint; private boolean shouldPrint; private boolean isImpossible = false; public Sudoku() { mask = new int[81][9]; formattedMask = new int[81]; placedMask = new int[64][64]; lineMask = new boolean[64][81][9]; sectionCounters = new int[9][27]; sectionMask = new int[9][27]; lineCounters = new int[64]; neighbors = new int[81][20]; unsolvedBoard = new int[81]; solvedBoard = new int[81]; cells = new int[][] {{0 ,1 ,2 ,9 ,10,11,18,19,20}, {3 ,4 ,5 ,12,13,14,21,22,23}, {6 ,7 ,8 ,15,16,17,24,25,26}, {27,28,29,36,37,38,45,46,47}, {30,31,32,39,40,41,48,49,50}, {33,34,35,42,43,44,51,52,53}, {54,55,56,63,64,65,72,73,74}, {57,58,59,66,67,68,75,76,77}, {60,61,62,69,70,71,78,79,80}}; } final public long solveSudoku(int[] board, int clue) { long t1 = 0,t2 = 0; t1 = System.nanoTime(); System.arraycopy(board, 0, unsolvedBoard, 0, 81); System.arraycopy(board, 0, solvedBoard, 0, 81); placedNumbers = 0; solutionFound = false; isEasy = true; isImpossible = false; for (int[] i : mask) { Arrays.fill(i, 0); } for (boolean[][] i : lineMask) { for (boolean[] j : i) { Arrays.fill(j, false); } } for (int i = 0; i < 81; i++) { if (solvedBoard[i] != -1) { put(i, solvedBoard[i]); placedNumbers++; } } solve(0, 0); t2 = System.nanoTime(); easySolved += isEasy ? 1 : 0; if (solutionFound && placedNumbers == 81) { totTime += t2-t1; if (shouldPrint || t2-t1 > 5*1_000_000_000L) { System.out.print(String.format( "Solution from %2d clues found in %7s", clue, printTime(t1, t2) )); shouldPrint = false; if (t2-t1 > 1*1000_000_000L) { System.out.println(); display2(board, solvedBoard); } } } else { System.out.println("No solution"); display2(unsolvedBoard, solvedBoard); return -1; } return t2 - t1; } final private void solve(int v, int vIndex) { lineCounters[vIndex] = 0; int easyIndex = placeEasy(vIndex); if (isImpossible) { resetEasy(vIndex, easyIndex); resetLineMask(vIndex); return; } if (placedNumbers == 81) { solutionFound = true; return; } // if (true) { // return; // } // either get the next empty cell // while (v < 81 && solvedBoard[v] >= 0) { // v++; // } // or get the cell with the fewest options generateFormattedMasks(); int minOptions = 9; for (int i = 0; i < 81; i++) { int options = formattedMask[i] & 0xffff; if (options > 0 && options < minOptions) { minOptions = options; v = i; } if (options == 0 && solvedBoard[i] == -1) { isImpossible = true; } } if (!isImpossible) { for (int c = 0; c < 9; c++) { if (isPossible(v, c)) { isEasy = false; put(v, c); placedNumbers++; solve(v + 1, vIndex + 1); if (solutionFound) { return; } unput(v, c); placedNumbers--; } } } resetEasy(vIndex, easyIndex); resetLineMask(vIndex); } final private void resetEasy(int vIndex, int easyIndex) { for (int i = 0; i < easyIndex; i++) { int tempv2 = placedMask[vIndex][i]; int c2 = solvedBoard[tempv2]; unput(tempv2, c2); placedNumbers--; } } final private void resetLineMask(int vIndex) { if (lineCounters[vIndex] > 0) { for (int i = 0; i < 81; i++) { for (int c = 0; c < 9; c++) { if (lineMask[vIndex][i][c]) { enable(i, c); lineMask[vIndex][i][c] = false; } } } } isImpossible = false; } final private int placeEasy(int vIndex) { int easyIndex = 0; int lastPlaced = 0, tempPlaced = 0, easyplaced = 0; int iter = 0; while (placedNumbers > lastPlaced+1) { lastPlaced = placedNumbers; tempPlaced = 0; while (placedNumbers > tempPlaced + 5) { tempPlaced = placedNumbers; easyIndex = placeNakedSingles(vIndex, easyIndex); if (isImpossible) { return easyIndex; } } tempPlaced = 0; while (placedNumbers < 55*1 && placedNumbers > tempPlaced + 2) { tempPlaced = placedNumbers; easyIndex = placeHiddenSingles(vIndex, easyIndex); if (isImpossible) { return easyIndex; } } tempPlaced = 0; while (placedNumbers < 65*1 && placedNumbers > tempPlaced + 1) { tempPlaced = placedNumbers; easyIndex = placeNakedSingles(vIndex, easyIndex); if (isImpossible) { return easyIndex; } } if (iter < 2 && placedNumbers < 55*1) { checkNakedTriples(vIndex); } if (placedNumbers < 45*1) { checkNakedDoubles(vIndex); identifyLines(vIndex); } iter++; } return easyIndex; } final private int placeNakedSingles(int vIndex, int easyIndex) { generateFormattedMasks(); for (int tempv = 0; tempv < 81; tempv++) { int possibilities = formattedMask[tempv]; if ((possibilities & 0xffff) == 1) { possibilities >>= 16; int c = 0; while ((possibilities & 1) == 0) { possibilities >>= 1; c++; } if (isPossible(tempv, c)) { put(tempv, c); placedMask[vIndex][easyIndex++] = tempv; placedNumbers++; } else { isImpossible = true; return easyIndex; } } else if (possibilities == 0 && solvedBoard[tempv] == -1) { isImpossible = true; return easyIndex; } } return easyIndex; } final private int placeHiddenSingles(int vIndex, int easyIndex) { for (int[] i : sectionCounters) { Arrays.fill(i, 0); } for (int c = 0; c < 9; c++) { for (int v = 0; v < 81; v++) { if (isPossible(v, c)) { int cell = 3 * (v / 27) + ((v / 3) % 3); sectionCounters[c][v / 9]++; sectionCounters[c][9 + (v % 9)]++; sectionCounters[c][18 + cell]++; sectionMask[c][v / 9] = v; sectionMask[c][9 + (v % 9)] = v; sectionMask[c][18 + cell] = v; } } int v; for (int i = 0; i < 9; i++) { if (sectionCounters[c][i] == 1) { v = sectionMask[c][i]; if (isPossible(v, c)) { put(v, c); placedMask[vIndex][easyIndex++] = v; placedNumbers++; int cell = 3 * (v / 27) + ((v / 3) % 3); sectionCounters[c][9 + (v%9)] = 9; sectionCounters[c][18 + cell] = 9; } else { isImpossible = true; return easyIndex; } } } for (int i = 9; i < 18; i++) { if (sectionCounters[c][i] == 1) { v = sectionMask[c][i]; if (isPossible(v, c)) { put(v, c); placedMask[vIndex][easyIndex++] = v; int cell = 3 * (v / 27) + ((v / 3) % 3); placedNumbers++; sectionCounters[c][18 + cell]++; } else { isImpossible = true; return easyIndex; } } } for (int i = 18; i < 27; i++) { if (sectionCounters[c][i] == 1) { v = sectionMask[c][i]; if (isPossible(v, c)) { put(v, c); placedMask[vIndex][easyIndex++] = v; placedNumbers++; } else { isImpossible = true; return easyIndex; } } } } return easyIndex; } final private int getFormattedMask(int v) { if (solvedBoard[v] >= 0) { return 0; } int x = 0; int y = 0; for (int c = 8; c >= 0; c--) { x <<= 1; x += mask[v][c] == 0 ? 1 : 0; y += mask[v][c] == 0 ? 1 : 0; } x <<= 16; return x + y; } final private int getCachedMask(int v) { return formattedMask[v]; } final private void generateFormattedMasks() { for (int i = 0; i < 81; i++) { formattedMask[i] = getFormattedMask(i); } } final private void generateFormattedMasks(int[] idxs) { for (int i : idxs) { formattedMask[i] = getFormattedMask(i); } } final private void checkNakedDoubles(int vIndex) { generateFormattedMasks(); for (int i = 0; i < 81; i++) { int bitmask = formattedMask[i]; if ((bitmask & 0xffff) == 2) { for (int j = i+1; j < (i/9+1)*9; j++) { int bitmask_j = formattedMask[j]; if (bitmask == bitmask_j) { bitmask >>= 16; int c0, c1, k = 0; while ((bitmask & 1) == 0) { k++; bitmask >>= 1; } c0 = k; bitmask >>= 1; k++; while ((bitmask & 1) == 0) { k++; bitmask >>= 1; } c1 = k; for (int cell = (i/9)*9; cell < (i/9+1)*9; cell++) { if (cell != i && cell != j) { if (!lineMask[vIndex][cell][c0]) { disable(cell, c0); lineMask[vIndex][cell][c0] = true; lineCounters[vIndex]++; } if (!lineMask[vIndex][cell][c1]) { disable(cell, c1); lineMask[vIndex][cell][c1] = true; lineCounters[vIndex]++; } } } } } } } for (int idx = 0; idx < 81; idx++) { int i = (idx%9)*9 + idx/9; int bitmask = formattedMask[i]; if ((bitmask & 0xffff) == 2) { for (int j = i+9; j < 81; j += 9) { int bitmask_j = formattedMask[j]; if (bitmask == bitmask_j) { bitmask >>= 16; int c0, c1, k = 0; while ((bitmask & 1) == 0) { k++; bitmask >>= 1; } c0 = k; bitmask >>= 1; k++; while ((bitmask & 1) == 0) { k++; bitmask >>= 1; } c1 = k; for (int cell = i % 9; cell < 81; cell += 9) { if (cell != i && cell != j) { if (!lineMask[vIndex][cell][c0]) { disable(cell, c0); lineMask[vIndex][cell][c0] = true; lineCounters[vIndex]++; } if (!lineMask[vIndex][cell][c1]) { disable(cell, c1); lineMask[vIndex][cell][c1] = true; lineCounters[vIndex]++; } } } } } } } for (int idx = 0; idx < 9; idx++) { for (int i = 0; i < 9; i++) { int bitmask = formattedMask[cells[idx][i]]; if ((bitmask & 0xffff) == 2) { for (int j = i+1; j < 9; j++) { int bitmask_j = formattedMask[cells[idx][j]]; if (bitmask == bitmask_j) { bitmask >>= 16; int c0, c1, k = 0; while ((bitmask & 1) == 0) { k++; bitmask >>= 1; } c0 = k; bitmask >>= 1; k++; while ((bitmask & 1) == 0) { k++; bitmask >>= 1; } c1 = k; for (int cellIdx = 0; cellIdx < 9; cellIdx++) { if (cellIdx != i && cellIdx != j) { int cell = cells[idx][cellIdx]; if (!lineMask[vIndex][cell][c0]) { disable(cell, c0); lineMask[vIndex][cell][c0] = true; lineCounters[vIndex]++; } if (!lineMask[vIndex][cell][c1]) { disable(cell, c1); lineMask[vIndex][cell][c1] = true; lineCounters[vIndex]++; } } } } } } } } } final private void checkNakedTriples(int vIndex) { generateFormattedMasks(); for (int i = 0; i < 81; i++) { int bitmask = formattedMask[i]; if ((bitmask & 0xffff) == 3) { for (int j = i+1; j < (i/9+1)*9; j++) { int bitmask_j = formattedMask[j]; if (bitmask_j > 0 && bitmask == (bitmask | bitmask_j)) { for (int k = j+1; k < (i/9+1)*9; k++) { int bitmask_k = formattedMask[k]; if (bitmask_k > 0 && bitmask == (bitmask | bitmask_k)) { int bitmask_shifted = bitmask >> 16; int c0, c1, c2, l = 0; while ((bitmask_shifted & 1) == 0) { l++; bitmask_shifted >>= 1; } c0 = l; bitmask_shifted >>= 1; l++; while ((bitmask_shifted & 1) == 0) { l++; bitmask_shifted >>= 1; } c1 = l; bitmask_shifted >>= 1; l++; while ((bitmask_shifted & 1) == 0) { l++; bitmask_shifted >>= 1; } c2 = l; for (int cell = (i/9)*9; cell < (i/9+1)*9; cell++) { if (cell != i && cell != j && cell != k) { if (!lineMask[vIndex][cell][c0]) { disable(cell, c0); lineMask[vIndex][cell][c0] = true; lineCounters[vIndex]++; } if (!lineMask[vIndex][cell][c1]) { disable(cell, c1); lineMask[vIndex][cell][c1] = true; lineCounters[vIndex]++; } if (!lineMask[vIndex][cell][c2]) { disable(cell, c2); lineMask[vIndex][cell][c2] = true; lineCounters[vIndex]++; } } } } } } } } } for (int idx = 0; idx < 81; idx++) { int i = (idx%9)*9 + idx/9; int bitmask = formattedMask[i]; if ((bitmask & 0xffff) == 3) { for (int j = i+9; j < 81; j += 9) { int bitmask_j = formattedMask[j]; if (bitmask_j > 0 && bitmask == (bitmask | bitmask_j)) { for (int k = j+9; k < 81; k += 9) { int bitmask_k = formattedMask[k]; if (bitmask_k > 0 && bitmask == (bitmask | bitmask_k)) { int bitmask_shifted = bitmask >> 16; int c0, c1, c2, l = 0; while ((bitmask_shifted & 1) == 0) { l++; bitmask_shifted >>= 1; } c0 = l; bitmask_shifted >>= 1; l++; while ((bitmask_shifted & 1) == 0) { l++; bitmask_shifted >>= 1; } c1 = l; bitmask_shifted >>= 1; l++; while ((bitmask_shifted & 1) == 0) { l++; bitmask_shifted >>= 1; } c2 = l; for (int cell = i%9; cell < 81; cell += 9) { if (cell != i && cell != j && cell != k) { if (!lineMask[vIndex][cell][c0]) { disable(cell, c0); lineMask[vIndex][cell][c0] = true; lineCounters[vIndex]++; } if (!lineMask[vIndex][cell][c1]) { disable(cell, c1); lineMask[vIndex][cell][c1] = true; lineCounters[vIndex]++; } if (!lineMask[vIndex][cell][c2]) { disable(cell, c2); lineMask[vIndex][cell][c2] = true; lineCounters[vIndex]++; } } } } } } } } } for (int idx = 0; idx < 9; idx++) { for (int i = 0; i < 9; i++) { int bitmask = formattedMask[cells[idx][i]]; if ((bitmask & 0xffff) == 3) { for (int j = i+1; j < 9; j++) { int bitmask_j = formattedMask[cells[idx][j]]; if (bitmask_j > 0 && bitmask == (bitmask | bitmask_j)) { for (int k = j+1; k < 9; k++) { int bitmask_k = formattedMask[cells[idx][k]]; if (bitmask_k > 0 && bitmask == (bitmask | bitmask_k)) { int bitmask_shifted = bitmask >> 16; int c0, c1, c2, l = 0; while ((bitmask_shifted & 1) == 0) { l++; bitmask_shifted >>= 1; } c0 = l; bitmask_shifted >>= 1; l++; while ((bitmask_shifted & 1) == 0) { l++; bitmask_shifted >>= 1; } c1 = l; bitmask_shifted >>= 1; l++; while ((bitmask_shifted & 1) == 0) { l++; bitmask_shifted >>= 1; } c2 = l; for (int cellIdx = 0; cellIdx < 9; cellIdx++) { if (cellIdx != i && cellIdx != j && cellIdx != k) { int cell = cells[idx][cellIdx]; if (!lineMask[vIndex][cell][c0]) { disable(cell, c0); lineMask[vIndex][cell][c0] = true; lineCounters[vIndex]++; } if (!lineMask[vIndex][cell][c1]) { disable(cell, c1); lineMask[vIndex][cell][c1] = true; lineCounters[vIndex]++; } if (!lineMask[vIndex][cell][c2]) { disable(cell, c2); lineMask[vIndex][cell][c2] = true; lineCounters[vIndex]++; } } } } } } } } } } } final private void identifyLines(int vIndex) { int disabledLines = 0; int[][] tempRowMask = new int[3][9]; int[][] tempColMask = new int[3][9]; for (int i = 0; i < 9; i++) { for (int c = 0; c < 9; c++) { for (int j = 0; j < 3; j++) { tempRowMask[j][c] = 0; tempColMask[j][c] = 0; } for (int j = 0; j < 9; j++) { if (mask[cells[i][j]][c] == 0) { tempRowMask[j/3][c]++; tempColMask[j%3][c]++; } } int rowCount = 0; int colCount = 0; int rowIdx = -1, colIdx = -1; for (int j = 0; j < 3; j++) { if (tempRowMask[j][c] > 0) { rowCount++; rowIdx = j; } if (tempColMask[j][c] > 0) { colCount++; colIdx = j; } } if (rowCount == 1) { for (int j = (i/3)*3; j < (i/3 + 1)*3; j++) { if (j != i) { for (int k = rowIdx*3; k < (rowIdx+1)*3; k++) { int cell = cells[j][k]; if (!lineMask[vIndex][cell][c]) { disable(cell, c); lineMask[vIndex][cell][c] = true; lineCounters[vIndex]++; } } } } } if (colCount == 1) { for (int j = i % 3; j < 9; j += 3) { if (j != i) { for (int k = colIdx; k < 9; k += 3) { int cell = cells[j][k]; if (!lineMask[vIndex][cell][c]) { disable(cell, c); lineMask[vIndex][cell][c] = true; lineCounters[vIndex]++; } } } } } } } } final private boolean isPossible(int v, int c) { return mask[v][c] == 0; } final private int checkMask(int[][] neighbors, int v, int c) { int tempValue = 0; for (int n : neighbors[v]) { if (mask[n][c] > 0) { tempValue++; } } return tempValue; } final private void put(int v, int c) { solvedBoard[v] = c; for (int i : neighbors[v]) { mask[i][c]++; } for (int i = 0; i < 9; i++) { mask[v][i]++; } } final private void disable(int v, int c) { mask[v][c]++; } final private void unput(int v, int c) { solvedBoard[v] = -1; for (int i : neighbors[v]) { mask[i][c]--; } for (int i = 0; i < 9; i++) { mask[v][i]--; } } final private void enable(int v, int c) { // enables++; mask[v][c]--; } public String getString(int[] board) { StringBuilder s = new StringBuilder(); for (int i : board) { s.append(i+1); } return s.toString(); } public long getTime() { return totTime; } public static String printTime(long t1, long t2) { String unit = " ns"; if (t2-t1 > 10000) { unit = " us"; t1 /= 1000; t2 /= 1000; } if (t2-t1 > 10000) { unit = " ms"; t1 /= 1000; t2 /= 1000; } if (t2-t1 > 10000) { unit = " seconds"; t1 /= 1000; t2 /= 1000; } return (t2-t1) + unit; } public void display(int[] board) { for (int i = 0; i < 9; i++) { if (i % 3 == 0) { System.out.println("+-----+-----+-----+"); } for (int j = 0; j < 9; j++) { if (j % 3 == 0) { System.out.print("|"); } else { System.out.print(" "); } if (board[i*9+j] != -1) { System.out.print(board[i*9+j]+1); } else { System.out.print(" "); } } System.out.println("|"); } System.out.println("+-----+-----+-----+"); } public void display2(int[] board, int[] solved) { for (int i = 0; i < 9; i++) { if (i % 3 == 0) { System.out.println("+-----+-----+-----+ +-----+-----+-----+"); } for (int j = 0; j < 9; j++) { if (j % 3 == 0) { System.out.print("|"); } else { System.out.print(" "); } if (board[i*9+j] != -1) { System.out.print(board[i*9+j]+1); } else { System.out.print(" "); } } System.out.print("| "); for (int j = 0; j < 9; j++) { if (j % 3 == 0) { System.out.print("|"); } else { System.out.print(" "); } if (solved[i*9+j] != -1) { System.out.print(solved[i*9+j]+1); } else { System.out.print(" "); } } System.out.println("|"); } System.out.println("+-----+-----+-----+ +-----+-----+-----+"); } private boolean contains(int[] a, int v) { for (int i : a) { if (i == v) { return true; } } return false; } public void connect() { for (int i = 0; i < 81; i++) { for (int j = 0; j < 20; j++) { neighbors[i][j] = -1; } } int[] n_count = new int[81]; HashMap<Integer,ArrayList<Integer>> map = new HashMap<Integer,ArrayList<Integer>>(); for (int[] c: cells) { ArrayList<Integer> temp = new ArrayList<Integer>(); for (int v : c) { temp.add(v); } for (int v : c) { map.put(v,temp); } } for (int i = 0; i < 81; i++) { for (int j = (i/9)*9; j < (i/9)*9 + 9; j++) { if (i != j) { neighbors[i][n_count[i]++] = j; } } for (int j = i%9; j < 81; j += 9) { if (i != j) { neighbors[i][n_count[i]++] = j; } } for (int j : map.get(i)) { if (i != j) { if (!contains(neighbors[i], j)) { neighbors[i][n_count[i]++] = j; } } } } } public static int[][] getInput(String filename) { int[][] boards; try (BufferedInputStream in = new BufferedInputStream( new FileInputStream(filename))) { BufferedReader r = new BufferedReader( new InputStreamReader(in, StandardCharsets.UTF_8)); int n = Integer.valueOf(r.readLine()); boards = new int[n][81]; clues = new int[n]; for (int i = 0; i < n; i++) { for (int j = 0; j < 81; j++) { int x = r.read(); boards[i][j] = x - 49; clues[i] += x > 48 ? 1 : 0; } r.read(); } r.close(); } catch (IOException ex) { throw new RuntimeException(ex); } return boards; } private int getTotEasy() { return totEasy; } public String getSolution() { StringBuilder s = new StringBuilder(256); for (int i : unsolvedBoard) { s.append(i+1); } s.append(","); for (int i : solvedBoard) { s.append(i+1); } return s.toString(); } public static void main (String[] args) { long t0 = System.nanoTime(); Sudoku gc = new Sudoku(); File f; PrintWriter p; try { f = new File("sudoku_output.txt"); p = new PrintWriter(f); } catch (Exception e) { return; } if (args.length != 1) { System.out.println("Usage: java Sudoku <input_file>"); return; } int[][] boards = gc.getInput(args[0]); long tinp = System.nanoTime(); gc.connect(); long t1 = System.nanoTime(); p.println(boards.length); long maxSolveTime = 0; int maxSolveIndex = 0; long[] solveTimes = new long[boards.length]; for (int i = 0; i < boards.length; i++) { long tempTime = System.nanoTime(); if (tempTime - gc.lastPrint > 200_000_000 || i == boards.length - 1) { gc.shouldPrint = true; gc.lastPrint = tempTime; System.out.print(String.format( "\r(%7d/%7d) ", i+1, boards.length)); } long elapsed = gc.solveSudoku(boards[i], gc.clues[i]); if (elapsed == -1) { System.out.println("Impossible: " + i); } if (elapsed > maxSolveTime) { maxSolveTime = elapsed; maxSolveIndex = i; } solveTimes[i] = elapsed; p.println(gc.getSolution()); // break; } p.close(); long t2 = System.nanoTime(); Arrays.sort(solveTimes); System.out.println(); System.out.println("Median solve time: " + gc.printTime(0, solveTimes[boards.length/2])); System.out.println("Longest solve time: " + gc.printTime(0, maxSolveTime) + " for board " + maxSolveIndex); gc.display(boards[maxSolveIndex]); System.out.println(); System.out.println("Total time (including prints): " + gc.printTime(t0,t2)); System.out.println("Sudoku solving time: " + gc.printTime(0,gc.getTime())); System.out.println("Average time per board: " + gc.printTime(0,gc.getTime()/boards.length)); System.out.println("Number of one-choice digits per board: " + String.format("%.2f", gc.getTotEasy()/(double)boards.length)); System.out.println("Easily solvable boards: " + gc.easySolved); System.out.println("\nInput time: " + gc.printTime(t0,tinp)); System.out.println("Connect time: " + gc.printTime(tinp,t1)); try { Thread.sleep(10000); } catch (InterruptedException e) { } } }  # C++ with Minisat(2.2.1-5) - 11.735s official score This is nowhere near as fast as a specialized algorithm, but it's a different approach, an interesting point of reference, and easy to understand.$ clang++ -o solve -lminisat solver_minisat.cc

#include <minisat/core/Solver.h>

namespace {

using Minisat::Lit;
using Minisat::mkLit;
using namespace std;

struct SolverMiniSat {
Minisat::Solver solver;

SolverMiniSat() {
InitializeVariables();
InitializeCellOnnes();
}

// normal cell literals, of which we have 9*9*9
static Lit Literal(int row, int column, int value) {
return mkLit(value + 9 * (column + 9 * row), true);
}

// horizontal triad literals, of which we have 9*3*9, starting after the cell literals
static Lit HTriadLiteral(int row, int column, int value) {
int base = 81 * 9;
return mkLit(base + value + 9 * (column + 3 * row));
}

// vertical triad literals, of which we have 3*9*9, starting after the h_triad literals
static Lit VTriadLiteral(int row, int column, int value) {
int base = (81 + 27) * 9;
return mkLit(base + value + 9 * (row + 3 * column));
}

void InitializeVariables() {
for (int i = 0; i < 15 * 9 * 9; i++) {
solver.newVar();
}
}

// create an exactly-one constraint over a set of literals
void CreateOnne(const Minisat::vec<Minisat::Lit> &literals) {
for (int i = 0; i < literals.size() - 1; i++) {
for (int j = i + 1; j < literals.size(); j++) {
}
}
}

for (int i = 0; i < 9; i++) {
for (int j = 0; j < 3; j++) {
for (int value = 0; value < 9; value++) {
int j0 = j * 3 + 0, j1 = j * 3 + 1, j2 = j * 3 + 2;

}
}
}
}

for (int i = 0; i < 9; i++) {
for (int value = 0; value < 9; value++) {
Minisat::vec<Minisat::Lit> row;
CreateOnne(row);

Minisat::vec<Minisat::Lit> column;
CreateOnne(column);

Minisat::vec<Minisat::Lit> hbox;
hbox.push(HTriadLiteral(3 * (i / 3) + 0, i % 3, value));
hbox.push(HTriadLiteral(3 * (i / 3) + 1, i % 3, value));
hbox.push(HTriadLiteral(3 * (i / 3) + 2, i % 3, value));
CreateOnne(hbox);

Minisat::vec<Minisat::Lit> vbox;
vbox.push(VTriadLiteral(i % 3, 3 * (i / 3) + 0, value));
vbox.push(VTriadLiteral(i % 3, 3 * (i / 3) + 1, value));
vbox.push(VTriadLiteral(i % 3, 3 * (i / 3) + 2, value));
CreateOnne(vbox);
}
}
}

void InitializeCellOnnes() {
for (int row = 0; row < 9; row++) {
for (int column = 0; column < 9; column++) {
Minisat::vec<Minisat::Lit> literals;
for (int value = 0; value < 9; value++) {
literals.push(Literal(row, column, value));
}
CreateOnne(literals);
}
}
}

bool SolveSudoku(const char *input, char *solution, size_t *num_guesses) {
Minisat::vec<Minisat::Lit> assumptions;
for (int row = 0; row < 9; row++) {
for (int column = 0; column < 9; column++) {
char digit = input[row * 9 + column];
if (digit != '.') {
assumptions.push(Literal(row, column, digit - '1'));
}
}
}
solver.decisions = 0;
bool satisfied = solver.solve(assumptions);
if (satisfied) {
for (int row = 0; row < 9; row++) {
for (int column = 0; column < 9; column++) {
for (int value = 0; value < 9; value++) {
if (solver.model[value + 9 * (column + 9 * row)] ==
Minisat::lbool((uint8_t) 1)) {
solution[row * 9 + column] = value + '1';
}
}
}
}
}
*num_guesses = solver.decisions - 1;
return satisfied;
}
};

} //end anonymous namespace

int main(int argc, const char **argv) {
char *puzzle = NULL;
char solution[81];
size_t size, guesses;

SolverMiniSat solver;

while (getline(&puzzle, &size, stdin) != -1) {
int count = solver.SolveSudoku(puzzle, solution, &guesses);
printf("%.81s:%d:%.81s\n", puzzle, count, solution);
}
}