Solve a Hidato puzzle
You are encouraged to solve this task according to the task description, using any language you may know.
The task is to write a program which solves Hidato puzzles.
The rules are:
- You are given a grid with some numbers placed in it. The other squares in the grid will be blank.
- The grid is not necessarily rectangular.
- The grid may have holes in it.
- The grid is always connected.
- The number “1” is always present, as is another number that is equal to the number of squares in the grid. Other numbers are present so as to force the solution to be unique.
- The aim is to place a natural number in each blank square so that in the sequence of numbered squares from “1” upwards, each square is in the wp:Moore neighborhood of the squares immediately before and after it in the sequence (except for the first and last squares, of course, which only have one-sided constraints).
- Thus, if the grid was overlaid on a chessboard, a king would be able to make legal moves along the path from first to last square in numerical order.
- A square may only contain one number.
- In a proper Hidato puzzle, the solution is unique.
For example the following problem
has the following solution, with path marked on it:
C
A fancier and less understandable version at Solve a Hidato puzzle/C extra.
Depth-first graph search: <lang c>#include <stdio.h>
- include <stdlib.h>
- include <ctype.h>
int *board, **fixed, top = 0, w, h;
- define cell(x, y) (board + (((y) + 1) * (w + 2) + (x) + 1))
void make_board(int x, int y, char *s) { int *b;
w = x, h = y; x = (2 + w) * (2 + h);
fixed = calloc(sizeof(int*), x); board = malloc(sizeof(int) * x);
while (x--) board[x] = -1;
for (y = 0; y < h; y++) for (x = 0; x < w; x++) { b = cell(x, y);
while (isspace(*s)) s++;
switch (*s) { case '_': *b = 0; case '.': break; default: fixed[*b = strtol(s, 0, 10)] = b; if (*b > top) top = *b; }
while (*s && !isspace(*s)) s++; } }
void show_board(char *s) { int i, j, c;
printf("\n%s:\n", s);
for (i = 0; i < h; i++, putchar('\n')) for (j = 0; j < w; j++) { c = *(cell(j, i)); printf(!c ? " __" : c == -1 ? " " : " %2d", c); } }
int fill(int *b, int n) { int i, j;
if (n > top) return 1;
if((*b && *b != n) || (fixed[n] && fixed[n] != b)) return 0;
for (*b = n++, i = -1; i <= 1; i++) for (j = -1; j <= 1; j ++) if (fill(b + i * (w + 2) + j, n)) return 1;
return *b = 0; }
int main() { make_board( 8,8, " __ 33 35 __ __ .. .. .." " __ __ 24 22 __ .. .. .." " __ __ __ 21 __ __ .. .." " __ 26 __ 13 40 11 .. .." " 27 __ __ __ 9 __ 1 .." " . . __ __ 18 __ __ .." " . .. . . __ 7 __ __" " . .. .. .. . . 5 __" /* 3, 3, " . 4 ." " _ 7 _" " 1 _ _" */ /* 50, 3, " 1 _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . 74" " . . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ ." " . . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ ." */ );
show_board("Before"); fill(fixed[1], 1); show_board("After"); /* "40 lbs in two weeks!" */
return 0; }</lang>
- Output:
Before:
__ 33 35 __ __
__ __ 24 22 __
__ __ __ 21 __ __
__ 26 __ 13 40 11
27 __ __ __ 9 __ 1
__ __ 18 __ __
__ 7 __ __
5 __
After:
32 33 35 36 37
31 34 24 22 38
30 25 23 21 12 39
29 26 20 13 40 11
27 28 14 19 9 10 1
15 16 18 8 2
17 7 6 3
5 4
D
More C-Style Version
This version retains some of the characteristics of the original C version. It uses global variables, it doesn't enforce immutability and purity. This style is faster to write for prototypes, short programs or less important code, but in larger programs you usually want more strictness to avoid some bugs and increase long-term maintainability.
<lang d>import std.stdio, std.array, std.conv, std.algorithm;
int[][] board; int[] given, start;
void setup(string s) {
auto lines = s.split("\n");
auto cols = lines[0].split().length;
auto rows = lines.length;
given.length = 0;
board = new int[][](rows + 2, cols + 2);
foreach (i; 0 .. rows + 2)
board[i][] = -1;
foreach (r, row; lines) {
foreach (c, cell; row.split()) {
switch (cell) {
case "__" : board[r + 1][c + 1] = 0; break;
case "." : break;
default :
int val = to!int(cell);
board[r + 1][c + 1] = val;
given ~= val;
if (val == 1) start = [r + 1, c + 1];
}
}
}
given.sort();
}
bool solve(int r, int c, int n, int next = 0) {
if (n > given[$-1]) return true;
if (board[r][c] && board[r][c] != n)
return false;
if (board[r][c] == 0 && given[next] == n)
return false;
int back;
if (board[r][c] == n) {
next++;
back = n;
}
board[r][c] = n;
foreach (i; -1 .. 2)
foreach (j; -1 .. 2)
if (solve(r + i, c + j, n + 1, next))
return true;
board[r][c] = back; return false;
}
void printBoard() {
foreach (r; 0 .. board.length) {
foreach (c; board[r])
writef(c == -1 ? " " : !c ? "__ " : "%2d ", c);
write('\n');
}
}
void main() {
auto hi = "__ 33 35 __ __ . . .
__ __ 24 22 __ . . .
__ __ __ 21 __ __ . .
__ 26 __ 13 40 11 . .
27 __ __ __ 9 __ 1 .
. . __ __ 18 __ __ .
. . . . __ 7 __ __
. . . . . . 5 __";
setup(hi); printBoard(); solve(start[0], start[1], 1); printBoard();
}</lang>
- Output:
__ 33 35 __ __
__ __ 24 22 __
__ __ __ 21 __ __
__ 26 __ 13 40 11
27 __ __ __ 9 __ 1
__ __ 18 __ __
__ 7 __ __
5 __
32 33 35 36 37
31 34 24 22 38
30 25 23 21 12 39
29 26 20 13 40 11
27 28 14 19 9 10 1
15 16 18 8 2
17 7 6 3
5 4
Stronger Version
This version uses a little stronger typing, performs tests a run-time with contracts, it doesn't use global variables, it enforces immutability and purity where possible, and produces a correct text output for both larger ad small boards. This style is more fit for larger programs, or when you want the code to be less bug-prone or a little more efficient. <lang d>import std.stdio, std.string, std.conv, std.array, std.algorithm,
std.exception;
struct Hidato {
alias int Cell;
enum : Cell { emptyCell = -2, unknownCell = -1 }
alias int[2][Cell] KnownT; immutable KnownT known; immutable Cell boardMax; Cell[][] board;
this(in string input) /*pure*/ nothrow in {
assert(!input.empty);
} out {
immutable nRows = board.length;
immutable nCols = board[0].length;
assert(boardMax > 0 && boardMax <= nRows * nCols);
assert(known.length >= 2 && known.length <= nRows * nCols);
assert(1 in known && boardMax in known);
foreach (row; board)
foreach (cell; row)
assert(cell == Hidato.emptyCell ||
cell == Hidato.unknownCell ||
(cell >= 1 && cell <= nRows * nCols));
foreach (n, rc; known) {
assert(n > 0 && n <= boardMax);
assert(rc[0] >= 0 && rc[0] < nRows);
assert(rc[1] >= 0 && rc[1] < nCols);
}
} body {
bool[Cell] pathSeen; // a set
KnownT knownMutable;
const lines = input.splitLines();
immutable nCols = lines[0].split().length;
foreach (int r, row; lines) {
assert(row.split().length == nCols,
text("Wrong cols n.: ", row.split().length));
auto boardRow = new typeof(board[0])(nCols);
foreach (int c, cell; row.split()) {
switch (cell) {
case ".":
boardRow[c] = Hidato.emptyCell;
break;
case "_":
boardRow[c] = Hidato.unknownCell;
break;
default: // known
immutable val = to!Cell(cell);
enforce(val > 0, "Path numbers must be > 0.");
enforce(val !in pathSeen,
text("Duplicated path number: ", val));
pathSeen[val] = true;
boardRow[c] = val;
knownMutable[val] = [r, c];
boardMax = max(boardMax, val);
}
}
board ~= boardRow;
}
known = assumeUnique(knownMutable); // Not verified }
bool solve() pure nothrow in {
assert(1 in known);
} body {
/*static*/ bool fill(in int r, in int c, in Cell n)
pure nothrow {
if (n > boardMax)
return true;
if (c < 0 || c >= board[0].length ||
r < 0 || r >= board.length)
return false;
if ((board[r][c] != Hidato.unknownCell &&
board[r][c] != n) ||
(n in known && known[n] != [r, c]))
return false;
board[r][c] = n;
foreach (i; -1 .. 2)
foreach (j; -1 .. 2)
if (fill(r + i, c + j, n + 1))
return true;
board[r][c] = Hidato.unknownCell;
return false;
}
return fill(known[1][0], known[1][1], 1); }
string toString() const {
immutable d = [Hidato.emptyCell: ".",
Hidato.unknownCell: "_"];
immutable form = "%" ~ text(text(boardMax).length + 1) ~ "s";
string result;
foreach (row; board) {
foreach (c; row)
result ~= xformat(form, d.get(c, text(c)));
result ~= "\n";
}
return result;
}
}
void solveHidato(in string problem) {
auto hi = Hidato(problem);
writeln("Problem:\n", hi);
hi.solve();
writeln("Solution:\n", hi);
writeln();
}
void main() {
solveHidato(" _ 33 35 _ _ . . .
_ _ 24 22 _ . . .
_ _ _ 21 _ _ . .
_ 26 _ 13 40 11 . .
27 _ _ _ 9 _ 1 .
. . _ _ 18 _ _ .
. . . . _ 7 _ _
. . . . . . 5 _");
solveHidato(". 4 .
_ 7 _
1 _ _");
solveHidato(
"1 _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . 74
. . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . . . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ ."
); }</lang>
- Output:
Problem: _ 33 35 _ _ . . . _ _ 24 22 _ . . . _ _ _ 21 _ _ . . _ 26 _ 13 40 11 . . 27 _ _ _ 9 _ 1 . . . _ _ 18 _ _ . . . . . _ 7 _ _ . . . . . . 5 _ Solution: 32 33 35 36 37 . . . 31 34 24 22 38 . . . 30 25 23 21 12 39 . . 29 26 20 13 40 11 . . 27 28 14 19 9 10 1 . . . 15 16 18 8 2 . . . . . 17 7 6 3 . . . . . . 5 4 Problem: . 4 . _ 7 _ 1 _ _ Solution: . 4 . 3 7 5 1 2 6 Problem: 1 _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . 74 . . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . . . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . Solution: 1 2 3 . . 8 9 . . 14 15 . . 20 21 . . 26 27 . . 32 33 . . 38 39 . . 44 45 . . 50 51 . . 56 57 . . 62 63 . . 68 69 . . 74 . . 4 . 7 . 10 . 13 . 16 . 19 . 22 . 25 . 28 . 31 . 34 . 37 . 40 . 43 . 46 . 49 . 52 . 55 . 58 . 61 . 64 . 67 . 70 . 73 . . . . 5 6 . . 11 12 . . 17 18 . . 23 24 . . 29 30 . . 35 36 . . 41 42 . . 47 48 . . 53 54 . . 59 60 . . 65 66 . . 71 72 .
Mathprog
<lang mathprog>/*Hidato.mathprog, part of KuKu by Nigel Galloway
Find a solution to a Hidato problem
Nigel_Galloway@operamail.com April 1st., 2011
- /
param ZBLS; param ROWS; param COLS; param D := 1; set ROWSR := 1..ROWS; set COLSR := 1..COLS; set ROWSV := (1-D)..(ROWS+D); set COLSV := (1-D)..(COLS+D); param Iz{ROWSR,COLSR}, integer, default 0; set ZBLSV := 1..(ZBLS+1); set ZBLSR := 1..ZBLS;
var BR{ROWSV,COLSV,ZBLSV}, binary;
void0{r in ROWSV, z in ZBLSR,c in (1-D)..0}: BR[r,c,z] = 0; void1{r in ROWSV, z in ZBLSR,c in (COLS+1)..(COLS+D)}: BR[r,c,z] = 0; void2{c in COLSV, z in ZBLSR,r in (1-D)..0}: BR[r,c,z] = 0; void3{c in COLSV, z in ZBLSR,r in (ROWS+1)..(ROWS+D)}: BR[r,c,z] = 0; void4{r in ROWSV,c in (1-D)..0}: BR[r,c,ZBLS+1] = 1; void5{r in ROWSV,c in (COLS+1)..(COLS+D)}: BR[r,c,ZBLS+1] = 1; void6{c in COLSV,r in (1-D)..0}: BR[r,c,ZBLS+1] = 1; void7{c in COLSV,r in (ROWS+1)..(ROWS+D)}: BR[r,c,ZBLS+1] = 1;
Izfree{r in ROWSR, c in COLSR, z in ZBLSR : Iz[r,c] = -1}: BR[r,c,z] = 0; Iz1{Izr in ROWSR, Izc in COLSR, r in ROWSR, c in COLSR, z in ZBLSR : Izr=r and Izc=c and Iz[Izr,Izc]=z}: BR[r,c,z] = 1;
rule1{z in ZBLSR}: sum{r in ROWSR, c in COLSR} BR[r,c,z] = 1; rule2{r in ROWSR, c in COLSR}: sum{z in ZBLSV} BR[r,c,z] = 1; rule3{r in ROWSR, c in COLSR, z in ZBLSR}: BR[0,0,z+1] + BR[r-1,c-1,z+1] + BR[r-1,c,z+1] + BR[r-1,c+1,z+1] + BR[r,c-1,z+1] + BR[r,c+1,z+1] + BR[r+1,c-1,z+1] + BR[r+1,c,z+1] + BR[r+1,c+1,z+1] - BR[r,c,z] >= 0;
solve;
for {r in ROWSR} {
for {c in COLSR} {
printf " %2d", sum{z in ZBLSR} BR[r,c,z]*z;
}
printf "\n";
} data;
param ROWS := 8; param COLS := 8; param ZBLS := 40; param Iz: 1 2 3 4 5 6 7 8 :=
1 . 33 35 . . -1 -1 -1 2 . . 24 22 . -1 -1 -1 3 . . . 21 . . -1 -1 4 . 26 . 13 40 11 -1 -1 5 27 . . . 9 . 1 -1 6 -1 -1 . . 18 . . -1 7 -1 -1 -1 -1 . 7 . . 8 -1 -1 -1 -1 -1 -1 5 . ; end;</lang>
- Output:
GLPSOL: GLPK LP/MIP Solver, v4.47 Parameter(s) specified in the command line: --math Hidato.mathprog ... INTEGER OPTIMAL SOLUTION FOUND Time used: 0.0 secs Memory used: 6.4 Mb (6712828 bytes) 32 33 35 36 37 0 0 0 31 34 24 22 38 0 0 0 30 25 23 21 12 39 0 0 29 26 20 13 40 11 0 0 27 28 14 19 9 10 1 0 0 0 15 16 18 8 2 0 0 0 0 0 17 7 6 3 0 0 0 0 0 0 5 4 Model has been successfully processed
Perl
<lang perl>use strict; use List::Util 'max';
our (@grid, @known, $n);
sub show_board { for my $r (@grid) { print map(!defined($_) ? ' ' : $_ ? sprintf("%3d", $_) : ' __' , @$r), "\n" } }
sub parse_board { @grid = map{[map(/^_/ ? 0 : /^\./ ? undef: $_, split ' ')]} split "\n", shift(); for my $y (0 .. $#grid) { for my $x (0 .. $#{$grid[$y]}) { $grid[$y][$x] > 0 and $known[$grid[$y][$x]] = "$y,$x"; } } $n = max(map { max @$_ } @grid); }
sub neighbors { my ($y, $x) = @_; my @out; for ( [-1, -1], [-1, 0], [-1, 1], [ 0, -1], [ 0, 1], [ 1, -1], [ 1, 0], [ 1, 1]) { my $y1 = $y + $_->[0]; my $x1 = $x + $_->[1]; next if $x1 < 0 || $y1 < 0; next unless defined $grid[$y1][$x1]; push @out, "$y1,$x1"; } @out }
sub try_fill { my ($v, $coord) = @_; return 1 if $v > $n;
my ($y, $x) = split ',', $coord; my $old = $grid[$y][$x];
return if $old && $old != $v; return if exists $known[$v] and $known[$v] ne $coord;
$grid[$y][$x] = $v; print "\033[0H"; show_board();
try_fill($v + 1, $_) && return 1 for neighbors($y, $x);
$grid[$y][$x] = $old; return }
parse_board
- ". 4 .
- _ 7 _
- 1 _ _";
- " 1 _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . 74
- . . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _ . _
- . . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _ . . _ _
- ";
"__ 33 35 __ __ .. .. .. . __ __ 24 22 __ .. .. .. . __ __ __ 21 __ __ .. .. . __ 26 __ 13 40 11 .. .. . 27 __ __ __ 9 __ 1 .. . . . __ __ 18 __ __ .. . . .. . . __ 7 __ __ . . .. .. .. . . 5 __ .";
print "\033[2J";
try_fill(1, $known[1]);</lang>
- Output:
32 33 35 36 37
31 34 24 22 38
30 25 23 21 12 39
29 26 20 13 40 11
27 28 14 19 9 10 1
15 16 18 8 2
17 7 6 3
5 4
Prolog
Works with SWI-Prolog and library(clpfd) written by Markus Triska.
Puzzle solved is from the Wilkipedia page : http://en.wikipedia.org/wiki/Hidato
<lang Prolog>:- use_module(library(clpfd)).
hidato :- init1(Li), % skip first blank line init2(1, 1, 10, Li), my_write(Li).
init1(Li) :-
Li = [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, A, 33, 35, B, C, 0, 0, 0, 0,
0, D, E, 24, 22, F, 0, 0, 0, 0,
0, G, H, I, 21, J, K, 0, 0, 0,
0, L, 26, M, 13, 40, 11, 0, 0, 0,
0, 27, N, O, P, 9, Q, 1, 0, 0,
0, 0, 0, R, S, 18, T, U, 0, 0,
0, 0, 0, 0, 0, V, 7, W, X, 0,
0, 0, 0, 0, 0, 0, 0, 5, Y, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
LV = [ A, 33, 35, B, C, D, E, 24, 22, F, G, H, I, 21, J, K, L, 26, M, 13, 40, 11, 27, N, O, P, 9, Q, 1, R, S, 18, T, U, V, 7, W, X, 5, Y],
LV ins 1..40,
all_distinct(LV).
% give the constraints % Stop before the last line init2(_N, Col, Max_Col, _L) :- Col is Max_Col - 1.
% skip zeros init2(N, Lig, Col, L) :- I is N + Lig * Col, element(I, L, 0), !, V is N+1, ( V > Col -> N1 = 1, Lig1 is Lig + 1; N1 = V, Lig1 = Lig), init2(N1, Lig1, Col, L).
% skip first column
init2(1, Lig, Col, L) :-
!,
init2(2, Lig, Col, L) .
% skip last column init2(Col, Lig, Col, L) :- !, Lig1 is Lig+1, init2(1, Lig1, Col, L).
% V5 V3 V6 % V1 V V2 % V7 V4 V8 % general case init2(N, Lig, Col, L) :- I is N + Lig * Col, element(I, L, V),
I1 is I - 1, I2 is I + 1, I3 is I - Col, I4 is I + Col, I5 is I3 - 1, I6 is I3 + 1, I7 is I4 - 1, I8 is I4 + 1,
maplist(compute_BI(L, V), [I1,I2,I3,I4,I5,I6,I7,I8], VI, BI),
sum(BI, #=, SBI),
( ((V #= 1 #\/ V #= 40) #/\ SBI #= 1) #\/ (V #\= 1 #/\ V #\= 40 #/\ SBI #= 2)) #<==> 1,
labeling([ffc, enum], [V | VI]),
N1 is N+1, init2(N1, Lig, Col, L).
compute_BI(L, V, I, VI, BI) :- element(I, L, VI), VI #= 0 #==> BI #= 0, ( VI #\= 0 #/\ (V - VI #= 1 #\/ VI - V #= 1)) #<==> BI.
% display the result my_write([0, A, B, C, D, E, F, G, H, 0 | T]) :- maplist(my_write_1, [A, B, C, D, E, F, G, H]), nl, my_write(T).
my_write([]).
my_write_1(0) :- write(' ').
my_write_1(X) :- writef('%3r', [X]).</lang>
- Output:
?- hidato.
32 33 35 36 37
31 34 24 22 38
30 25 23 21 12 39
29 26 20 13 40 11
27 28 14 19 9 10 1
15 16 18 8 2
17 7 6 3
5 4
true
Python
<lang python>board = [] given = [] start = None
def setup(s):
global board, given, start lines = s.splitlines() ncols = len(lines[0].split()) nrows = len(lines) board = [[-1] * (ncols + 2) for _ in xrange(nrows + 2)]
for r, row in enumerate(lines):
for c, cell in enumerate(row.split()):
if cell == "__" :
board[r + 1][c + 1] = 0
continue
elif cell == ".":
continue # -1
else:
val = int(cell)
board[r + 1][c + 1] = val
given.append(val)
if val == 1:
start = (r + 1, c + 1)
given.sort()
def solve(r, c, n, next=0):
if n > given[-1]:
return True
if board[r][c] and board[r][c] != n:
return False
if board[r][c] == 0 and given[next] == n:
return False
back = 0
if board[r][c] == n:
next += 1
back = n
board[r][c] = n
for i in xrange(-1, 2):
for j in xrange(-1, 2):
if solve(r + i, c + j, n + 1, next):
return True
board[r][c] = back
return False
def print_board():
d = {-1: " ", 0: "__"}
bmax = max(max(r) for r in board)
form = "%" + str(len(str(bmax)) + 1) + "s"
for r in board[1:-1]:
print "".join(form % d.get(c, str(c)) for c in r[1:-1])
hi = """\ __ 33 35 __ __ . . . __ __ 24 22 __ . . . __ __ __ 21 __ __ . . __ 26 __ 13 40 11 . . 27 __ __ __ 9 __ 1 .
. . __ __ 18 __ __ . . . . . __ 7 __ __ . . . . . . 5 __"""
setup(hi) print_board() solve(start[0], start[1], 1) print print_board()</lang>
- Output:
__ 33 35 __ __
__ __ 24 22 __
__ __ __ 21 __ __
__ 26 __ 13 40 11
27 __ __ __ 9 __ 1
__ __ 18 __ __
__ 7 __ __
5 __
32 33 35 36 37
31 34 24 22 38
30 25 23 21 12 39
29 26 20 13 40 11
27 28 14 19 9 10 1
15 16 18 8 2
17 7 6 3
5 4
Tcl
<lang tcl>proc init {initialConfiguration} {
global grid max filled
set max 1
set y 0
foreach row [split [string trim $initialConfiguration "\n"] "\n"] {
set x 0 set rowcontents {} foreach cell $row { if {![string is integer -strict $cell]} {set cell -1} lappend rowcontents $cell set max [expr {max($max, $cell)}] if {$cell > 0} { dict set filled $cell [list $y $x] } incr x } lappend grid $rowcontents incr y
}
}
proc findseps {} {
global max filled
set result {}
for {set i 1} {$i < $max-1} {incr i} {
if {[dict exists $filled $i]} { for {set j [expr {$i+1}]} {$j <= $max} {incr j} { if {[dict exists $filled $j]} { if {$j-$i > 1} { lappend result [list $i $j [expr {$j-$i}]] } break } } }
} return [lsort -integer -index 2 $result]
}
proc makepaths {sep} {
global grid filled
lassign $sep from to len
lassign [dict get $filled $from] y x
set result {}
foreach {dx dy} {-1 -1 -1 0 -1 1 0 -1 0 1 1 -1 1 0 1 1} {
discover [expr {$x+$dx}] [expr {$y+$dy}] [expr {$from+1}] $to \ [list [list $from $x $y]] $grid
} return $result
} proc discover {x y n limit path model} {
global filled
# Check for illegal
if {[lindex $model $y $x] != 0} return
upvar 1 result result
lassign [dict get $filled $limit] ly lx
# Special case
if {$n == $limit-1} {
if {abs($x-$lx)<=1 && abs($y-$ly)<=1 && !($lx==$x && $ly==$y)} { lappend result [lappend path [list $n $x $y] [list $limit $lx $ly]] } return
}
# Check for impossible
if {abs($x-$lx) > $limit-$n || abs($y-$ly) > $limit-$n} return
# Recursive search
lappend path [list $n $x $y]
lset model $y $x $n
incr n
foreach {dx dy} {-1 -1 -1 0 -1 1 0 -1 0 1 1 -1 1 0 1 1} {
discover [expr {$x+$dx}] [expr {$y+$dy}] $n $limit $path $model
}
}
proc applypath {path} {
global grid filled
puts "Found unique path for [lindex $path 0 0] -> [lindex $path end 0]"
foreach cell [lrange $path 1 end-1] {
lassign $cell n x y lset grid $y $x $n dict set filled $n [list $y $x]
}
}
proc printgrid {} {
global grid max
foreach row $grid {
foreach cell $row { puts -nonewline [format " %*s" [string length $max] [expr { $cell==-1 ? "." : $cell }]] } puts ""
}
}
proc solveHidato {initialConfiguration} {
init $initialConfiguration
set limit [llength [findseps]]
while {[llength [set seps [findseps]]] && [incr limit -1]>=0} {
foreach sep $seps { if {[llength [set paths [makepaths $sep]]] == 1} { applypath [lindex $paths 0] break } }
} puts "" printgrid
}</lang> Demonstrating (dots are “outside” the grid, and zeroes are the cells to be filled in): <lang tcl>solveHidato "
0 33 35 0 0 . . .
0 0 24 22 0 . . .
0 0 0 21 0 0 . .
0 26 0 13 40 11 . .
27 0 0 0 9 0 1 .
. . 0 0 18 0 0 .
. . . . 0 7 0 0
. . . . . . 5 0
"</lang>
- Output:
Found unique path for 5 -> 7 Found unique path for 7 -> 9 Found unique path for 9 -> 11 Found unique path for 11 -> 13 Found unique path for 33 -> 35 Found unique path for 18 -> 21 Found unique path for 1 -> 5 Found unique path for 35 -> 40 Found unique path for 22 -> 24 Found unique path for 24 -> 26 Found unique path for 27 -> 33 Found unique path for 13 -> 18 32 33 35 36 37 . . . 31 34 24 22 38 . . . 30 25 23 21 12 39 . . 29 26 20 13 40 11 . . 27 28 14 19 9 10 1 . . . 15 16 18 8 2 . . . . . 17 7 6 3 . . . . . . 5 4
More complex cases are solvable with an extended version of this code, though that has more onerous version requirements.