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
Depth-first graph search: <lang c>#include <stdio.h>
- include <stdlib.h>
- include <string.h>
- include <ctype.h>
int *board, **fixed, top = 0, w, h, start_x, start_y;
- define cell(x, y) board[(y) * w + (x)]
void make_board(int x, int y, char *s) { int *b;
w = x, h = y;
fixed = calloc(sizeof(int *), w * h + 1); b = board = calloc(sizeof(int), w * h);
for (y = 0; y < w * h; y++, b++) { while (isspace(*s)) s++;
switch (*s) { case '.': *b = -1; case '_': break; default: fixed[*b = strtol(s, 0, 10)] = b; if (*b == 1) start_x = y % w, start_y = y / w; if (*b > top) top = *b; }
while (!isspace(*s)) s++; } }
void show_board(char *s) { int i, j, c;
printf("\n%s:\n", s); for (i = 0; i < h; i++) { for (j = 0; j < w; j++) { c = cell(j, i); printf(!c ? " __" : c == -1 ? " " : " %2d", c); } putchar('\n'); } }
int fill(int x, int y, int n) { int i, j, *b;
if (n > top) return 1;
if (x < 0 || x >= w || y < 0 || y >= h || (*(b = &(cell(x, y))) && *b != n) || (fixed[n] && fixed[n] != b)) return 0; *b = n; for (i = -1; i <= 1; i++) for (j = -1; j <= 1; j++) if (fill(x + j, y + i, n + 1)) 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 __");
show_board("Before"); fill(start_x, start_y, 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
<lang d>import std.stdio, std.string, std.conv, std.array, std.algorithm,
std.exception;
struct Hidato {
enum : int { empty = -1, unknown_board = -2 } enum int[2] unknown = [-3, -3]; alias typeof(unknown)[] KnownT;
int board[][]; immutable KnownT known; immutable int board_max, path_start_c, path_start_r;
this(in int nr, in int nc, in string input) /*pure*/ in { assert(nr > 0 && nc > 0); assert(input.split().length == nr * nc); } body { board = new typeof(board)(nr, nc); auto known_mutable = new KnownT(nr * nc); known_mutable[] = unknown; foreach (r, row; std.range.chunks(input.split(), nc).array()) { foreach (c, item; row) { switch (item) { case ".": // empty board[r][c] = empty; break; case "__": // unknown board[r][c] = unknown_board; break; default: // known immutable n = to!int(item); enforce(n > 0, "Path numbers must be > 0"); enforce(n <= nr * nc, "Too high path number"); board[r][c] = n; known_mutable[n] = [r, c]; if (n == 1) { path_start_c = c; path_start_r = r; } board_max = max(board_max, n); } } }
known = assumeUnique(known_mutable); // Not verified }
bool solve() pure nothrow { /*static*/ bool fill(in int r, in int c, in int n) pure nothrow { if (n > board_max) return true;
if (c < 0 || c >= board[0].length || r < 0 || r >= board.length) return false;
if ((board[r][c] != unknown_board && board[r][c] != n) || (known[n] != unknown && 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] = unknown_board; return false; }
return fill(path_start_r, path_start_c, 1); }
string toString() const { immutable form = "%" ~ text(text(board_max).length + 1) ~ "s";
string[] rows; foreach (row; board) { string row_str; foreach (c; row) { switch (c) { case unknown_board: row_str ~= format(form, "__"); break; case empty: row_str ~= format(form, " "); break; default: row_str ~= format(form, c); } } rows ~= row_str; } return rows.join("\n"); }
}
void main() {
auto hi = Hidato(8, 8, "__ 33 35 __ __ . . . __ __ 24 22 __ . . . __ __ __ 21 __ __ . . __ 26 __ 13 40 11 . . 27 __ __ __ 9 __ 1 . . . __ __ 18 __ __ . . . . . __ 7 __ __ . . . . . . 5 __");
writeln("Problem:\n", hi); hi.solve(); writeln("Solution:\n", hi);
}</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
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>
Produces:
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
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>class Hidato:
EMPTY = -1 UNKNOWN = None UNKNOWN_BOARD = -2
def __init__(self, nr, nc, input): assert nr > 0 and nc > 0 self.board = [[Hidato.UNKNOWN_BOARD] * nc for _ in xrange(nr)] self.known = [Hidato.UNKNOWN for _ in xrange(nr * nc + 1)] self.board_max = -1
items = input.split() assert len(items) == nr * nc chunks = (items[i : i+nc] for i in xrange(0, len(items), nc)) for r, row in enumerate(chunks): for c, item in enumerate(row): if item == ".": # empty self.board[r][c] = Hidato.EMPTY continue if item == "__": # unknown continue else: # known n = int(item) assert n > 0, "Path numbers must be > 0" assert n <= nr * nc, "Too high path number" self.board[r][c] = n self.known[n] = (r, c) if n == 1: self.path_start_c = c self.path_start_r = r self.board_max = max(self.board_max, n)
def solve(self): def fill(r, c, n): if n > self.board_max: return True
if (c < 0 or c >= len(self.board[0]) or r < 0 or r >= len(self.board)): return False
if ((self.board[r][c] != Hidato.UNKNOWN_BOARD and self.board[r][c] != n) or (self.known[n] != Hidato.UNKNOWN and self.known[n] != (r, c))): return False
self.board[r][c] = n for i in xrange(-1, 2): for j in xrange(-1, 2): if fill(r + i, c + j, n + 1): return True
self.board[r][c] = Hidato.UNKNOWN_BOARD return False
return fill(self.path_start_r, self.path_start_c, 1)
def __repr__(self): form = "%" + str(len(str(self.board_max)) + 1) + "s" rows = [] for row in self.board: row_str = "" for c in row: if c == Hidato.UNKNOWN_BOARD: row_str += form % "__"; continue if c == Hidato.EMPTY: row_str += form % " "; continue else: row_str += form % c rows.append(row_str) return "\n".join(rows)
def main():
hi = Hidato(8, 8, """__ 33 35 __ __ . . . __ __ 24 22 __ . . . __ __ __ 21 __ __ . . __ 26 __ 13 40 11 . . 27 __ __ __ 9 __ 1 . . . __ __ 18 __ __ . . . . . __ 7 __ __ . . . . . . 5 __""")
print "Problem:\n", hi hi.solve() print "Solution:\n", hi
main()</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
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