批处理之家 - Powered by Discuz! Board

2,6,1,9,3,5,7,4,8
8,4,9,6,1,7,5,3,2
7,5,3,8,2,4,6,9,1
9,8,5,3,4,2,1,6,7
6,7,2,5,9,1,3,8,4
3,1,4,7,8,6,9,2,5
1,9,8,2,5,3,4,7,6
4,2,6,1,7,9,8,5,3
5,3,7,4,6,8,2,1,9
Game level: 1, time used: 0.000s
8,3,6,9,4,7,5,2,1
2,5,1,8,3,6,4,9,7
4,7,9,1,2,5,8,3,6
9,6,8,2,5,1,3,7,4
1,2,7,4,8,3,6,5,9
5,4,3,6,7,9,2,1,8
6,1,4,3,9,2,7,8,5
3,9,5,7,6,8,1,4,2
7,8,2,5,1,4,9,6,3
Game level: 4, time used: 537.594s

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=info
解数独初版
523066680 2017-09
=cut
use IO::Handle;
use Time::HiRes qw/time/;
STDOUT->autoflush(1);
my @games =
qw/
001000008800010532750024090005302060670500004010780900098053406426179803537460219
503092000700000008006007310020600000065000730007043500000706102070000800400009000
700306000000000050060000018000081000000000000900000200000200900050400000080000007
030000001200806000000005000000000000000000650043070000600002080090000000000010003
010000200300800000000504000800090000000070120500000000000000000020130000400000005
/;
our @unsolve;
our $block = []; #区块引用
my $mat = [[]];
my $level = 0;
my $time_a;
#创建区块引用，以便于操作
make_block_refs( $block, $mat );
while ( $level <= $#games )
{
#字符串转矩阵
str_to_mat( $games[$level] , $mat );
$time_a = time();
#设置 @unsolve 数组，存储空缺元素的坐标
set_unsolve_array($mat);
solve($mat, 0);
print_mat($mat);
printf "Game level: %d, time used: %.3fs\n\n", $level+1, time() - $time_a;
$level++;
}
sub solve
{
my ($mat, $prev, $lv) = @_;
my @possible;
my ($row, $col);
my $current;
for my $i ( $prev .. $#unsolve )
{
($row, $col) = @{$unsolve[$i]};
if ( $mat->[$row][$col] == 0 )
{
$current = $i;
@possible = get_possible_num( $mat, $row, $col );
last;
}
}
if ( not defined $current ) { return 1 }
else
{
return 0 if ( $#possible < 0 )
}
my $res = 0;
for my $p ( @possible )
{
$mat->[ $row ][ $col ] = $p;
$res = solve($mat, $current+1, $lv+1);
last if ($res == 1);
}
#使对应单元恢复为0 否则影响递归判断
$mat->[$row][$col] = 0 if ($res == 0) ;
return $res;
}
sub get_possible_num
{
our ($block, %hash);
my ($mat, $row, $col) = @_;
my %possible = (1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9);
#排除元素
for my $n ( 0 .. 8 )
{
delete $possible{ $mat->[$n][$col] };
delete $possible{ $mat->[$row][$n] };
delete $possible{ ${$block->[$row/3][$col/3][$n]} };
}
return sort keys %possible;
}
sub get_possible_num_2
{
our ($block, %hash);
my ($mat, $row, $col) = @_;
my @possible = (0,1,2,3,4,5,6,7,8,9);
#区块坐标
my $blockrow = int($row/3);
my $blockcol = int($col/3);
#排除元素
for my $n ( 0 .. 9 )
{
$possible[ $mat->[$n][$col] ] = 0;
$possible[ $mat->[$row][$n] ] = 0;
$possible[ ${$block->[$blockrow][$blockcol][$n]} ] = 0;
}
return grep { $_ } @possible;
}
sub set_unsolve_array
{
our ( @unsolve );
my ($mat) = @_;
@unsolve = ();
for my $row ( 0..8 )
{
for my $col ( 0..8 )
{
if ( $mat->[$row][$col] == 0 )
{
push @unsolve, [ $row, $col, [get_possible_num( $mat, $row, $col )] ];
}
}
}
#根据可选数字的数量由少到多排序
#@unsolve = sort { $#{$a->[2]} <=> $#{$b->[2]} } @unsolve;
}
sub make_block_refs
{
my ($block, $mat) = @_;
#将数独的九个宫对应的引用分组保存
for my $r ( 0..2 )
{
for my $c ( 0..2 )
{
for my $rr ( $r*3 .. $r*3+2 )
{
for my $cc ( $c*3 .. $c*3+2 )
{
push @{ $block->[$r][$c] }, \$mat->[$rr][$cc];
}
}
}
}
}
sub str_to_mat
{
my ( $str, $mat ) = @_;
my $idx = 0;
for my $row ( 0 .. 8 )
{
for my $col ( 0 .. 8 )
{
$mat->[$row][$col] = substr( $str, $idx++, 1 );
}
}
}
sub print_mat
{
my ($mat) = @_;
grep { print join(",", @{$mat->[$_]} ),"\n" } ( 0..8 );
}

复制代码

=info
方案 - 遍历数字分布情况
523066680 2017-09
=cut
use File::Slurp;
use IO::Handle;
STDOUT->autoflush(1);
# my $gamedata = eval read_file("../sudoku_nd1.txt");
# grep { push @games, $gamedata->{$_} } sort keys %$gamedata;
@games =
qw/
001000008800010532750024090005302060670500004010780900098053406426179803537460219
503092000700000008006007310020600000065000730007043500000706102070000800400009000
700306000000000050060000018000081000000000000900000200000200900050400000080000007
030000001200806000000005000000000000000000650043070000600002080090000000000010003
010000200300800000000504000800090000000070120500000000000000000020130000400000005
/;
our @order;
our $answer_mat = [[]];
my $mat;
my $res;
my $allcase;
my $answer_str;
#坐标对应区块位置的表
our @blk = map { int($_ / 3) } ( 0..8 );
while ( $index <= $#games )
{
$mat = [[]];
$time_a = time();
str_to_mat( $games[$index] , $mat );
$allcase = [ map { [] } (0..9) ];
grep { possible_mat($mat, $_, $allcase->[$_], 0) } (1..9);
@order = sort { $#{$allcase->[$a]} <=> $#{$allcase->[$b]} } ( 0 .. 9 );
$res = recursive( $mat, 1 ); #起点为1，下标 0占位
if ($res == 1)
{
print_mat_inline( $answer_mat );
}
else { die "false\n" }
printf "Game index: %d, time used: %.3fs\n\n", $index, time() - $time_a;
$index++;
}
sub recursive
{
our (@order, $answer_mat);
my ( $mat, $lv ) = @_;
my @case;
if ( $lv > 9 )
{
$answer_mat = $mat;
return 1;
}
$target = $order[$lv];
possible_mat($mat, $target, \@case, 0);
my $t_mat = [[]];
my $res = 0;
for my $s ( @case )
{
str_to_mat( $s, $t_mat );
$res = recursive( $t_mat, $lv+1 );
last if ($res == 1);
}
return $res;
}
sub possible_mat
{
my ( $mat, $target, $aref, $lv ) = @_;
# level means row
my $str;
if ($lv == 9)
{
$count++;
push @$aref, mat_to_str($mat);
return 1;
}
my @cols = get_possible_column( $mat, $lv, $target );
my $res = 0;
my $ever;
for my $c ( @cols )
{
$ever = $mat->[$lv][$c];
$mat->[$lv][$c] = $target;
$res = possible_mat( $mat, $target, $aref, $lv+1 );
$mat->[$lv][$c] = $ever;
}
return $res;
}
sub get_possible_column
{
our @blk;
my ( $mat, $row, $target ) = @_;
my @cols = ( 0..8 );
for my $c ( 0..8 )
{
#如果当前行已经存在这个数字，则直接返回这个数字的位置。
if ( $mat->[$row][$c] == $target )
{
return ( $c );
}
elsif ( $mat->[$row][$c] != 0 )
{
$cols[$c] = -1;
}
for my $r ( 0..8 )
{
if ( $mat->[$r][$c] == $target )
{
$cols[$c] = -1;
if ( $blk[$r] == $blk[$row] )
{
$cols[ $blk[$c] * 3 + 0] = -1;
$cols[ $blk[$c] * 3 + 1] = -1;
$cols[ $blk[$c] * 3 + 2] = -1;
}
}
}
}
return grep { $_ != -1 } @cols;
}
sub str_to_mat
{
my ( $str, $mat ) = @_;
my $idx = 0;
for my $row ( 0 .. 8 )
{
for my $col ( 0 .. 8 )
{
$mat->[$row][$col] = substr( $str, $idx++, 1 );
}
}
}
sub mat_to_str
{
my ( $mat ) = @_;
return join("", map { join("", @{$mat->[$_]} ) } (0..8));
}
sub print_mat
{
my ($mat) = @_;
grep { print join(",", @{$mat->[$_]} ),"\n" } ( 0..8 );
}
sub print_mat_inline
{
my ($mat) = @_;
grep { print join("", @{$mat->[$_]} ),"" } ( 0..8 );
print "\n";
}

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class Soduku
def initialize(str)
@matrix = 9.times.map{[0]*9}
str.reverse! <-只加了一句，LEVEL4立马难度变低了
nums = str.split('').map(&:to_i)
...
end

复制代码

sort_seq = []
while
sort_seq.push @solve_seq[min_seq]
x = @solve_seq[min_seq][0]
y = @solve_seq[min_seq][1]
z = ((x / 3) * 3) + (y / 3)
@solve_seq.delete_at(min_seq)
break if @solve_seq.size == 0
min = 10
min_seq = -1
@solve_seq.each_with_index do |seq, i|
zz = ((seq[0] / 3) * 3) + (seq[1] / 3)
if seq[0] == x || seq[1] == y || zz == z
seq[2] -= 1
if seq[2] < min
min = seq[2]
min_seq = i
end
end
end
end
p sort_seq
@solve_seq = sort_seq

复制代码

<script>
var cntEmptyCell = 9 * 9;
var schPath = [];
var P = [[], [], [], [], [], [], [], [], []];
// i 行 j 列
for (let i = 0; i < 9; i++)
for (let j = 0; j < 9; j++) {
P[i][j] = {fill: 0, maybe: 0x1FF, choose: 9, block: (Math.floor(i / 3) * 3 + Math.floor(j / 3))};
}
var problems = [
'001000008800010532750024090005302060670500004010780900098053406426179803537460219',
'503092000700000008006007310020600000065000730007043500000706102070000800400009000',
'700306000000000050060000018000081000000000000900000200000200900050400000080000007',
'030000001200806000000005000000000000000000650043070000600002080090000000000010003',
'010000200300800000000504000800090000000070120500000000000000000020130000400000005',
'510000700076089000004000805400100007000030000090706010000460201047000300005000084'
];
var problem;
problem = problems[3];
const LOOPCNT_CTRL = 50000;
var LOOPCNT_OUTPUT_BEGIN = LOOPCNT_CTRL - 50;
var a_p = Array.from(problem);
var minchoose = 10, i_minchoose = 0, j_minchoose = 0;
// 题面初始化
for (let i = 0; i < 9; i++)
for (let j = 0; j < 9; j++)
if (a_p[i * 9 + j] - 0 != 0) {
P[i][j].maybe = P[i][j].fill = 1 << (a_p[i * 9 + j] - 1);
P[i][j].choose = 1;
cntEmptyCell -= 1;
for (let u = 0; u < 9; u++)
if ((u != i) && (P[u][j].fill == 0)) {
if ((P[u][j].maybe & P[i][j].fill) != 0) { // 按位与必须括起来, 它比不等号优先级低
P[u][j].choose -= 1;
if (P[u][j].choose < minchoose) {
minchoose = P[u][j].choose;
i_minchoose = u;
j_minchoose = j;
}
P[u][j].maybe &= ~(P[i][j].fill);
}
}
for (let v = 0; v < 9; v++) {
if ((v != j) && (P[i][v].fill == 0)) {
if ((P[i][v].maybe & P[i][j].fill) != 0) { // 按位与必须括起来, 它比不等号优先级低
P[i][v].choose -= 1;
if (P[i][v].choose < minchoose) {
minchoose = P[i][v].choose;
i_minchoose = i;
j_minchoose = v;
}
P[i][v].maybe &= ~(P[i][j].fill);
}
}
}
g = P[i][j].block;
// 3X3 块的左上角坐标
r = Math.floor(g / 3) * 3;
c = g % 3 * 3;
for (let u = 0; u < 3; u++)
for (let v = 0; v < 3; v++) {
if (!(r + u == i && c + v == j) && (P[r + u][c + v].fill == 0)) {
if ((P[r + u][c + v].maybe & P[i][j].fill) != 0) { // 按位与必须括起来, 它比不等号优先级低
P[r + u][c + v].choose -= 1;
if (P[r + u][c + v].choose < minchoose) {
minchoose = P[r + u][c + v].choose;
i_minchoose = r + u;
j_minchoose = c + v;
}
P[r + u][c + v].maybe &= ~(P[i][j].fill);
}
}
}
// console.log('F=' + a_p[i * 9 + j] + ', i=' + i + ', j=' + j + ', g=' + P[i][j].block + '\n');
}
console.log('题面初始化后\n');
console.log('cntEmptyCell=' + cntEmptyCell + '\n');
print_fill_intuitive();
// 初始化尝试填数
let try_num;
try_num = 1 << 8;
while ((try_num != 0) && ((P[i_minchoose][j_minchoose].maybe & try_num) == 0)) {
try_num >>= 1;
}
if (try_num == 0) {
throw 'error problem\n';
}
let maybe;
let fail_flag = false;
let success = false;
let loopcnt = 0;
// 节点状态:
// NEW_POS: 填数未失败,新位置填第一个试数
// NEXT_TRY: 失败换填下一个试数
// UNDO_POS: 所有试数失败撤消填数
let state_code = 'NEW_POS';
do {
loopcnt++;
if (fail_flag) {
// 撤消所有 SET_MAYBE
while (schPath[0].type == 'SET_MAYBE') {
P[schPath[0].row][schPath[0].col].maybe = schPath[0].oldMaybe;
P[schPath[0].row][schPath[0].col].choose += 1;
schPath.shift();
}
// 撤消 FILL_POS
// 若 try_num 不是最后一个可试数,
// try_num 取下一个可试数, fail_flag 设为 false
if (schPath[0].type == 'FILL_POS') {
i = schPath[0].row;
j = schPath[0].col;
P[i][j].fill = schPath[0].oldFill;
P[i][j].choose = schPath[0].oldChoose;
P[i][j].maybe = schPath[0].oldMaybe;
if (schPath[0].oldFill == 0)
cntEmptyCell += 1;
// 下一个可试的数
try_num = schPath[0].fillValue >> 1;
schPath.shift();
// 试探是否还有下一个可试填数
while ((try_num != 0) && ((P[i][j].maybe & try_num) == 0)) {
try_num >>= 1;
}
if (try_num != 0) {
fail_flag = false;
state_code = 'NEXT_TRY';
} else {
fail_flag = true;
state_code = 'UNDO_POS';
}
}
} else {
if (state_code == 'NEW_POS') {
if (cntEmptyCell == 0) {
// 成功, 输出结果并退出
console.log('\n\n' + 'SUCCESS! @ ' + 'loopcnt = ' + loopcnt + '\n\n' + 'problem and answer:\n\n' + PA_STR());
print_fill_intuitive();
// print_state();
success = true;
break;
}
minchoose = 10; // 准备搜索下一轮的最少选择位置
i = i_minchoose;
j = j_minchoose;
// 初始化 try_num
try_num = 1 << 8;
while ((try_num != 0) && ((P[i][j].maybe & try_num) == 0)) {
try_num >>= 1;
}
} else if (state_code == 'NEXT_TRY') {
}
if (try_num == 0) {
throw 'BUG: try_num == 0';
} else {
// TRYFILL
schPath.unshift({type: 'FILL_POS', oldFill: P[i][j].fill, row: i, col: j, fillValue: try_num, fillValue_fact: get_fill_intuitive(try_num), oldChoose: P[i][j].choose, oldMaybe: P[i][j].maybe});
if (P[i][j].fill == 0)
cntEmptyCell -= 1;
P[i][j].choose = 1;
P[i][j].maybe = try_num;
P[i][j].fill = try_num;
for (let u = 0; !fail_flag && (u < 9); u++)
if ((u != i) && (P[u][j].fill == 0)) {
if ((P[u][j].maybe & P[i][j].fill) != 0) { // 按位与必须括起来, 它比不等号优先级低
// 搜索栈压栈
schPath.unshift({type: 'SET_MAYBE', row: u, col: j, oldMaybe: P[u][j].maybe});
P[u][j].choose -= 1;
if (P[u][j].choose < minchoose) {
minchoose = P[u][j].choose;
i_minchoose = u;
j_minchoose = j;
}
maybe = P[u][j].maybe &= ~(P[i][j].fill);
if (maybe == 0) {
fail_flag = true;
break;
}
}
}
for (let v = 0; !fail_flag && (v < 9); v++) {
if ((v != j) && (P[i][v].fill == 0)) {
if ((P[i][v].maybe & P[i][j].fill) != 0) { // 按位与必须括起来, 它比不等号优先级低
// 搜索栈压栈
schPath.unshift({type: 'SET_MAYBE', row: i, col: v, oldMaybe: P[i][v].maybe});
P[i][v].choose -= 1;
if (P[i][v].choose < minchoose) {
minchoose = P[i][v].choose;
i_minchoose = i;
j_minchoose = v;
}
maybe = P[i][v].maybe &= ~(P[i][j].fill);
if (maybe == 0) {
fail_flag = true;
break;
}
}
}
}
if (!fail_flag) {
g = P[i][j].block;
// 3X3 块的左上角坐标
r = Math.floor(g / 3) * 3;
c = g % 3 * 3;
for (let u = 0; !fail_flag && (u < 3); u++)
for (let v = 0; !fail_flag && (v < 3); v++) {
if (!(r + u == i && c + v == j) && (P[r + u][c + v].fill == 0)) {
if ((P[r + u][c + v].maybe & P[i][j].fill) != 0) { // 按位与必须括起来, 它比不等号优先级低
// 搜索栈压栈
schPath.unshift({type: 'SET_MAYBE', row: r + u, col: c + v, oldMaybe: P[r + u][c + v].maybe});
P[r + u][c + v].choose -= 1;
if (P[r + u][c + v].choose < minchoose) {
minchoose = P[r + u][c + v].choose;
i_minchoose = r + u;
j_minchoose = c + v;
}
maybe = P[r + u][c + v].maybe &= ~(P[i][j].fill);
if (maybe == 0) {
fail_flag = true;
break;
}
}
}
}
}
for (let u = 0; u < 9; u++)
for (let v = 0; v < 9; v++) {
if (!(u == i && v == j) && P[u][v].fill == 0) {
if (P[u][v].choose < minchoose) {
minchoose = P[u][v].choose;
i_minchoose = u;
j_minchoose = v;
}
}
}
if (COUNT_EmptyCell() != cntEmptyCell) {
throw 'BUG: COUNT_EmptyCell = ' + COUNT_EmptyCell();
}
}
if (fail_flag) {
state_code = 'NEXT_TRY';
} else {
state_code = 'NEW_POS';
}
}
} while ((true || (loopcnt < LOOPCNT_CTRL)) && (schPath.length > 0 || (!fail_flag && state_code == 'NEXT_TRY' && try_num != 0)));
function print_choose_num() {
console.log('\n\n\n' + 'print_choose_num:' + '\n\n');
for (let i = 0; i < 9; i++) {
let line = i + ': ';
for (let j = 0; j < 9; j++) {
line += ((P[i][j].fill != 0) ? '+' : P[i][j].choose) + ' ';
}
console.log(line);
}
}
function print_maybe_bin() {
console.log('\n\n\n' + 'print_maybe_bin:' + '\n\n');
for (let i = 0; i < 9; i++) {
let line = i + ': ';
for (let j = 0; j < 9; j++) {
line += ('000000000' + P[i][j].maybe.toString(2)).slice(-9).replace(/0/g, '_') + ' ';
}
console.log(line);
}
}
function print_block_num() {
for (let i = 0; i < 9; i++) {
let line = i + ': ';
for (let j = 0; j < 9; j++) {
line += ', ' + P[i][j].block;
}
console.log(line);
}
}
function print_fill_bin() {
console.log('\n\n\n' + 'print_fill_bin:' + '\n\n');
for (let i = 0; i < 9; i++) {
let line = i + ': ';
for (let j = 0; j < 9; j++) {
line += ('000000000' + P[i][j].fill.toString(2)).slice(-9).replace(/0/g, '_') + ' ';
}
console.log(line);
}
}
function print_fill_intuitive() {
console.log('\n\n\n' + 'print_fill_intuitive:' + '\n\n');
for (let i = 0; i < 9; i++) {
let line = i + ': ';
for (let j = 0; j < 9; j++) {
line += get_fill_intuitive(P[i][j].fill) + ' ';
}
console.log(line);
}
}
function COUNT_EmptyCell() {
let sum = 0;
for (let i = 0; i < 9; i++)
for (let j = 0; j < 9; j++)
if (P[i][j].fill == 0)
sum++;
return sum;
}
function get_fill_intuitive(y) {
if (y == 0) {
return '_'
} else
return (Math.round(Math.log(y) / Math.log(2)) + 1);
}
function print_state() {
console.log('\n\n\n' + 'BEGIN print_state:' + '\n\n');
print_fill_intuitive();
print_fill_bin();
print_choose_num();
print_maybe_bin();
console.log('\n\n\n' + 'END print_state.' + '\n\n');
console.log('\n\n\n' + 'loopcnt = ' + loopcnt);
}
function print_choose_state() {
console.log('\n\n\n' + 'BEGIN print_choose_state:' + '\n\n');
print_choose_num();
print_maybe_bin();
console.log('\n\n\n' + 'END print_choose_state.' + '\n\n');
console.log('\n\n\n' + 'loopcnt = ' + loopcnt);
}
function PA_STR() {
let t = problem;
for (let i = 0; i < 9; i++)
for (let j = 0; j < 9; j++)
t += get_fill_intuitive(P[i][j].fill);
return t;
}
/*
搜索算法
数组存储
P 二维数组 9 行 X 9 列,
元素为对象引用方式: P[行号,列号]
{
fill: 填数 -- 0 为未填,否则为已填的数值,
maybe: 可选数的完全可能 -- 以2进制位表达,右至左数第1位表示可选1,第2位表示可选2,...第9位表示可选9, 当此属性值为0x1FF表示1~9全部可选,
当有任意一个位置的 maybe 为 0 时, 说明最后试填的数失败, 必须撤回
choose: 可选数的数目 -- 也就是 maybe 的所有二进制位上 '1' 的总数,
block: 区块号 -- 值范围[0..8]
};
cntEmptyCell -- 未填数单元格计数
解成功判断: if (cntEmptyCell == 0) 成功, 输出结果并退出
最大关联位置 -- 以当前线索简单判断最少可能选择的位置
节点状态:
NEW_POS: 填数未失败,新位置填第一个试数
NEXT_TRY: 失败换填下一个试数
UNDO_POS: 所有试数失败撤消填数
初始化题面并找出最大关联位置
节点状态 = NEW_POS
试填失败标志: fail_flag = false
:loop
if (fail_flag == true) {
撤消搜索栈顶的所有 SET_MAYBE 操作并弹栈搜索栈
撤消 FILL_POS 操作弹栈搜索栈
如果当前处理位置还有可试填的其他数值可能(同时设置好下一个试填数值),
fail_flag = false;
节点状态 = 'NEXT_TRY';
否则
fail_flag = true;
节点状态 = 'UNDO_POS';
} else {
if (节点状态 == 'NEW_POS') {
检测解出是否成功(成功则输出并退出程序)
操作位置设为最大关联位置
找出操作位置上第一个可选数值(0x100 --> 0x1 的次序)
}
执行 FILL_POS 操作并压栈搜索栈, 将同行同列同区块的 maybe 值作相应处理并压栈搜索栈, 找出所有未填数单元格中的最大关联位置
处理 maybe 值时如果发现无解 (任意一个位置的 maybe == 0), 则设置失败标志 fail_flag = true
}
当搜索栈未空或 (fail_flag 为假且节点状态 == 'NEXT_TRY' 且存在可试填数)
goto :loop
搜索栈结构:
结点: {type: FILL_POS 或 SET_MAYBE, row:行坐标, col:列坐标, oldMaybe: Maybe原值, fillValue}
type 为 FILL_POS 时, 有 fillValue 值
fillValue 可能的值: 从 2 ** 8 (256) 0X100 开始依次 >> 1 位, 直到 1, 不包括 0
*/
</script>

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sudoku_nd3.txt
//'7.56...19........89....8.....81.2.7...13..4.237...5.9.8.3..1.25...5....41........',
//'.....4...7.......2.....61.....2.......4...3..8..79.6....78....9.36........1......',
//'...3.1...49......638.......1.5..7.......9...8.........6...4...........5...3...71.',
sudoku_nd4.txt
//'......3......2.1..1.84.9.....7....2..3.6...7...6.7...8....5846.2........9..1.....',
//'.....53..6.....5.8...1...2.45.8...1.3.....4.5..2.........2.18....7.3..6..2...4...',
//'....9..3...5..2.4.7.94....13.1....98.4.9........8..6..2...13..7..36...151....9...',

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#!/usr/bin/env python3
# http://www.cs.mcgill.ca/~aassaf9/python/algorithm_x.html
# Author: Ali Assaf <ali.assaf.mail@gmail.com>
# License: GNU General Public License <http://www.gnu.org/licenses/>
from itertools import product
def solve_sudoku(size, grid):
R, C = size
N = R * C
X = ([("rc", rc) for rc in product(range(N), range(N))] +
[("rn", rn) for rn in product(range(N), range(1, N + 1))] +
[("cn", cn) for cn in product(range(N), range(1, N + 1))] +
[("bn", bn) for bn in product(range(N), range(1, N + 1))])
Y = dict()
for r, c, n in product(range(N), range(N), range(1, N + 1)):
b = (r // R) * R + (c // C) # Box number
Y[(r, c, n)] = [
("rc", (r, c)),
("rn", (r, n)),
("cn", (c, n)),
("bn", (b, n))]
X, Y = exact_cover(X, Y)
for i, row in enumerate(grid):
for j, n in enumerate(row):
if n:
select(X, Y, (i, j, n))
for solution in solve(X, Y, []):
for (r, c, n) in solution:
grid[r][c] = n
yield grid
def exact_cover(X, Y):
X = {j: set() for j in X}
for i, row in Y.items():
for j in row:
X[j].add(i)
return X, Y
def solve(X, Y, solution):
if not X:
yield list(solution)
else:
c = min(X, key=lambda c: len(X[c]))
for r in list(X[c]):
solution.append(r)
cols = select(X, Y, r)
for s in solve(X, Y, solution):
yield s
deselect(X, Y, r, cols)
solution.pop()
def select(X, Y, r):
cols = []
for j in Y[r]:
for i in X[j]:
for k in Y[i]:
if k != j:
X[k].remove(i)
cols.append(X.pop(j))
return cols
def deselect(X, Y, r, cols):
for j in reversed(Y[r]):
X[j] = cols.pop()
for i in X[j]:
for k in Y[i]:
if k != j:
X[k].add(i)
grid = [
[0,0,0,4,0,2,0,0,0],
[0,0,0,0,0,5,0,0,3],
[1,0,0,0,7,0,0,9,6],
[6,0,0,0,3,0,0,0,0],
[0,0,0,0,0,0,2,0,0],
[0,0,4,0,0,0,7,0,0],
[0,0,0,0,9,0,0,0,0],
[0,0,0,0,0,0,0,0,1],
[0,0,5,7,0,4,0,0,0]]
for solution in solve_sudoku((3, 3), grid):
print(*solution, sep='\n')

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