Nonogram solver

Nonogram solver is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Each row and column of a rectangular grid is annotated with the lengths of its distinct runs of occupied cells. Using only these lengths you should find one valid configuration of empty and occupied cells (or show a failure message):

Problem:                 Solution:

. . . . . . . .  3       . # # # . . . .  3
. . . . . . . .  2 1     # # . # . . . .  2 1
. . . . . . . .  3 2     . # # # . . # #  3 2
. . . . . . . .  2 2     . . # # . . # #  2 2
. . . . . . . .  6       . . # # # # # #  6
. . . . . . . .  1 5     # . # # # # # .  1 5
. . . . . . . .  6       # # # # # # . .  6
. . . . . . . .  1       . . . . # . . .  1
. . . . . . . .  2       . . . # # . . .  2
1 3 1 7 5 3 4 3          1 3 1 7 5 3 4 3
2 1 5 1                  2 1 5 1

The problem above could be represented by two lists of lists:

x = [[3], [2,1], [3,2], [2,2], [6], [1,5], [6], [1], [2]]
y = [[1,2], [3,1], [1,5], [7,1], [5], [3], [4], [3]]

A more compact representation of the same problem uses strings, where the letters represent the numbers, A=1, B=2, etc:

x = "C BA CB BB F AE F A B"
y = "AB CA AE GA E C D C"

For this task try to solve the problems read from a "nonogram_problems.txt" file, copied from this:

C BA CB BB F AE F A B
AB CA AE GA E C D C

F CAC ACAC CN AAA AABB EBB EAA ECCC HCCC
D D AE CD AE A DA BBB CC AAB BAA AAB DA AAB AAA BAB AAA CD BBA DA

CA BDA ACC BD CCAC CBBAC BBBBB BAABAA ABAD AABB BBH BBBD ABBAAA CCEA AACAAB BCACC ACBH DCH ADBE ADBB DBE ECE DAA DB CC
BC CAC CBAB BDD CDBDE BEBDF ADCDFA DCCFB DBCFC ABDBA BBF AAF BADB DBF AAAAD BDG CEF CBDB BBB FC

E BCB BEA BH BEK AABAF ABAC BAA BFB OD JH BADCF Q Q R AN AAN EI H G
E CB BAB AAA AAA AC BB ACC ACCA AGB AIA AJ AJ ACE AH BAF CAG DAG FAH FJ GJ ADK ABK BL CM

More info:
http://en.wikipedia.org/wiki/Nonogram

This task is the problem n.98 of the "99 Prolog Problems" by Werner Hett (also thanks to Paul Singleton for the idea and the examples):
https://sites.google.com/site/prologsite/prolog-problems

Some Haskell solutions:
http://www.haskell.org/haskellwiki/99_questions/Solutions/98
http://twanvl.nl/blog/haskell/Nonograms

PicoLisp solution:
http://picolisp.com/5000/!wiki?99p98

Bonus Problem: generate nonograms with unique solutions, of desired height and width.

D

Translation of: Python

<lang d>import std.stdio, std.range, std.file, std.algorithm, std.string;

/// Create all patterns of a row or col that match given runs. auto genRow(in int w, in int[] s) pure nothrow {

   static int[][] genSeg(in int[][] o, in int sp) pure nothrow {
       if (o.empty)
           return [[2].replicate(sp)];

       typeof(return) result;
       foreach (immutable x; 1 .. sp - o.length + 2)
           foreach (tail; genSeg(o[1 .. $], sp - x))
               result ~= [2].replicate(x) ~ o[0] ~ tail;
       return result;
   }

   const ones = s.map!(i => [1].replicate(i)).array;
   return genSeg(ones, w + 1 - sum(cast(int[])s)) //**
          .map!(x => x[1 .. $]);

}

/// Fix inevitable value of cells, and propagate. void deduce(in int[][] hr, in int[][] vr) {

   static int[] allowable(in int[][] row) pure /*nothrow*/ {
       //return row.dropOne.fold!q{ a[] |= b[] }(row[0].dup);
       return reduce!q{ a[] |= b[] }(row[0].dup, row.dropOne);
   }

   static bool fits(in int[] a, in int[] b) pure /*nothrow*/ {
       return zip(a, b).all!(xy => xy[0] & xy[1]);
   }

   immutable int w = vr.length;
   immutable int h = hr.length;
   auto rows = hr.map!(x => genRow(w, x).array).array;
   auto cols = vr.map!(x => genRow(h, x).array).array;
   auto canDo = rows.map!allowable.array;

   // Initially mark all columns for update.
   bool[int] modRows, modCols;
   modCols = w.iota.zip(true.repeat).assocArray;

   /// See if any value a given column is fixed; if so,
   /// mark its corresponding row for future fixup.
   void fixCol(in int n) /*nothrow*/ {
       const c = canDo.map!(x => x[n]).array;
       cols[n] = cols[n].remove!(x => !fits(x, c));
       foreach (immutable i, immutable x; allowable(cols[n]))
           if (x != canDo[i][n]) {
               modRows[i] = true;
               canDo[i][n] &= x;
           }
   }

   /// Ditto, for rows.
   void fixRow(in int n) /*nothrow*/ {
       const c = canDo[n];
       rows[n] = rows[n].remove!(x => !fits(x, c));
       foreach (immutable i, immutable x; allowable(rows[n]))
           if (x != canDo[n][i]) {
               modCols[i] = true;
               canDo[n][i] &= x;
           }
   }

   void showGram(in int[][] m) {
       // If there's 'x', something is wrong.
       // If there's '?', needs more work.
       foreach (const x; m)
           writefln("%-(%c %)", x.map!(i => "x#.?"[i]));
       writeln;
   }

   while (modCols.length > 0) {
       foreach (immutable i; modCols.byKey)
           fixCol(i);
       modCols = null;
       foreach (immutable i; modRows.byKey)
           fixRow(i);
       modRows = null;
   }

   if (cartesianProduct(h.iota, w.iota)
       .all!(ij => canDo[ij[0]][ij[1]] == 1 ||
                   canDo[ij[0]][ij[1]] == 2))
       "Solution would be unique".writeln;
   else
       "Solution may not be unique, doing exhaustive search:".writeln;

   // We actually do exhaustive search anyway. Unique
   // solution takes no time in this phase anyway.
   auto out_ = new const(int)[][](h);

   uint tryAll(in int n = 0) {
       if (n >= h) {
           foreach (immutable j; 0 .. w)
               if (!cols[j].canFind(out_.map!(x => x[j]).array))
                   return 0;
           showGram(out_);
           return 1;
       }
       typeof(return) sol = 0;
       foreach (const x; rows[n]) {
           out_[n] = x;
           sol += tryAll(n + 1);
       }
       return sol;
   }

   immutable n = tryAll;
   switch (n) {
       case 0:
           "No solution.".writeln;
           break;
       case 1:
           "Unique solution.".writeln;
           break;
       default:
           writeln(n, " solutions.");
           break;
   }
   writeln;

}

void solve(in string p, in bool showRuns=true) {

   const s = p.splitLines.map!(l => l.split.map!(w =>
               w.map!(c => int(c - 'A' + 1)).array).array).array;
               //w.map!(c => c - 'A' + 1))).to!(int[][][]);

   if (showRuns) {
       writeln("Horizontal runs: ", s[0]);
       writeln("Vertical runs: ", s[1]);
   }
   deduce(s[0], s[1]);

}

void main() {

   // Read problems from file.
   immutable fn = "nonogram_problems.txt";
   foreach (p; fn.readText.split("\n\n").filter!(p => !p.strip.empty))
       p.strip.solve;

   "Extra example not solvable by deduction alone:".writeln;
   "B B A A\nB B A A".solve;

   "Extra example where there is no solution:".writeln;
   "B A A\nA A A".solve;

}</lang>

Output:

Horizontal runs: [[3], [2, 1], [3, 2], [2, 2], [6], [1, 5], [6], [1], [2]]
Vertical runs: [[1, 2], [3, 1], [1, 5], [7, 1], [5], [3], [4], [3]]
Solution would be unique
. # # # . . . .
# # . # . . . .
. # # # . . # #
. . # # . . # #
. . # # # # # #
# . # # # # # .
# # # # # # . .
. . . . # . . .
. . . # # . . .

Unique solution.

Horizontal runs: [[6], [3, 1, 3], [1, 3, 1, 3], [3, 14], [1, 1, 1], [1, 1, 2, 2], [5, 2, 2], [5, 1, 1], [5, 3, 3, 3], [8, 3, 3, 3]]
Vertical runs: [[4], [4], [1, 5], [3, 4], [1, 5], [1], [4, 1], [2, 2, 2], [3, 3], [1, 1, 2], [2, 1, 1], [1, 1, 2], [4, 1], [1, 1, 2], [1, 1, 1], [2, 1, 2], [1, 1, 1], [3, 4], [2, 2, 1], [4, 1]]
Solution would be unique
. . . . . . . . . . # # # # # # . . . .
. . . . . . . . # # # . # . . # # # . .
. . . # . . # # # . . . # . . . . # # #
. . # # # . # # # # # # # # # # # # # #
. . . # . . # . . . . . . . . . . . . #
. . # . # . # # . . . . . . . . . . # #
# # # # # . . # # . . . . . . . . # # .
# # # # # . . . # . . . . . . . . # . .
# # # # # . . # # # . # # # . # # # . .
# # # # # # # # . # # # . # # # . # # #

Unique solution.

Horizontal runs: [[3, 1], [2, 4, 1], [1, 3, 3], [2, 4], [3, 3, 1, 3], [3, 2, 2, 1, 3], [2, 2, 2, 2, 2], [2, 1, 1, 2, 1, 1], [1, 2, 1, 4], [1, 1, 2, 2], [2, 2, 8], [2, 2, 2, 4], [1, 2, 2, 1, 1, 1], [3, 3, 5, 1], [1, 1, 3, 1, 1, 2], [2, 3, 1, 3, 3], [1, 3, 2, 8], [4, 3, 8], [1, 4, 2, 5], [1, 4, 2, 2], [4, 2, 5], [5, 3, 5], [4, 1, 1], [4, 2], [3, 3]]
Vertical runs: [[2, 3], [3, 1, 3], [3, 2, 1, 2], [2, 4, 4], [3, 4, 2, 4, 5], [2, 5, 2, 4, 6], [1, 4, 3, 4, 6, 1], [4, 3, 3, 6, 2], [4, 2, 3, 6, 3], [1, 2, 4, 2, 1], [2, 2, 6], [1, 1, 6], [2, 1, 4, 2], [4, 2, 6], [1, 1, 1, 1, 4], [2, 4, 7], [3, 5, 6], [3, 2, 4, 2], [2, 2, 2], [6, 3]]
Solution would be unique
. . . . # # # . # . . . . . . . . . . .
. . . . # # . # # # # . # . . . . . . .
. . . . # . # # # . # # # . . . . . . .
. . # # . # # # # . . . . . . . . . . .
. # # # . # # # . # . . . . # # # . . .
# # # . . # # . # # . . . # . # # # . .
# # . . # # . # # . . . . # # . # # . .
. . . . # # . # . # . . # # . # . # . .
. . . . # . # # . # . . . # # # # . . .
. . . . # . # . # # . . . . . # # . . .
. . . . . # # . # # . . # # # # # # # #
. . . . # # . # # . . . # # . . # # # #
. . . . # . # # . # # . # . . . # . . #
# # # . . # # # . # # # # # . . . . . #
# . # . # # # . # . . . . # . . . . # #
# # . . # # # . # . . . . # # # . # # #
. # . # # # . # # . # # # # # # # # . .
. # # # # . # # # . # # # # # # # # . .
. . . # . # # # # . # # . # # # # # . .
. . . # . # # # # . # # . . . # # . . .
. . . . # # # # . . # # . . . # # # # #
. . . # # # # # . # # # . . . # # # # #
. . . # # # # . # . . . . . . . . . . #
. . # # # # . # # . . . . . . . . . . .
. . # # # . # # # . . . . . . . . . . .

Unique solution.

Horizontal runs: [[5], [2, 3, 2], [2, 5, 1], [2, 8], [2, 5, 11], [1, 1, 2, 1, 6], [1, 2, 1, 3], [2, 1, 1], [2, 6, 2], [15, 4], [10, 8], [2, 1, 4, 3, 6], [17], [17], [18], [1, 14], [1, 1, 14], [5, 9], [8], [7]]
Vertical runs: [[5], [3, 2], [2, 1, 2], [1, 1, 1], [1, 1, 1], [1, 3], [2, 2], [1, 3, 3], [1, 3, 3, 1], [1, 7, 2], [1, 9, 1], [1, 10], [1, 10], [1, 3, 5], [1, 8], [2, 1, 6], [3, 1, 7], [4, 1, 7], [6, 1, 8], [6, 10], [7, 10], [1, 4, 11], [1, 2, 11], [2, 12], [3, 13]]
Solution would be unique
. . . . . . . . . . . . . . . . . . . . # # # # #
. . # # . . . . . . . . . . . . . . # # # . . # #
. # # . . . . . . . . . . . . . . # # # # # . . #
# # . . . . . . . . . . . . . # # # # # # # # . .
# # . . . . # # # # # . # # # # # # # # # # # . .
# . # . . # # . . . . # . . . . # # # # # # . . .
# . . # # . . . . . # . . . . . . . # # # . . . .
# # . . . . . . . . # . . . . . . . . . . . . . #
. # # . . . . . # # # # # # . . . . . . . . . # #
. . # # # # # # # # # # # # # # # . . . . # # # #
. . . . . # # # # # # # # # # . . # # # # # # # #
. . . . # # . # . # # # # . # # # . . # # # # # #
. . . . . . . . # # # # # # # # # # # # # # # # #
. . . . . . . . # # # # # # # # # # # # # # # # #
. . . . . . . # # # # # # # # # # # # # # # # # #
. . . . . . . # . . . # # # # # # # # # # # # # #
. . . . . . . # . # . # # # # # # # # # # # # # #
. . . . . . . . # # # # # . . . # # # # # # # # #
. . . . . . . . . . . . . . . . . # # # # # # # #
. . . . . . . . . . . . . . . . . . # # # # # # #

Unique solution.

Extra example not solvable by deduction alone:
Horizontal runs: [[2], [2], [1], [1]]
Vertical runs: [[2], [2], [1], [1]]
Solution may not be unique, doing exhaustive search:
# # . .
# # . .
. . # .
. . . #

# # . .
# # . .
. . . #
. . # .

. # # .
# # . .
# . . .
. . . #

3 solutions.

Extra example where there is no solution:
Horizontal runs: [[2], [1], [1]]
Vertical runs: [[1], [1], [1]]
Solution may not be unique, doing exhaustive search:
No solution.

The output is the same as the Python entry. The run-time with ldc2 compiler is about 0.29 seconds.

Prolog

Works with SWI-Prolog version 6.5.3
module(clpfd) is written by Markus Triska
Solution written by Lars Buitinck
Module solve-nonogram.pl <lang Prolog>/*

• Nonogram/paint-by-numbers solver in SWI-Prolog. Uses CLP(FD),
• in particular the automaton/3 (finite-state/RE) constraint.
• Copyright (c) 2011 Lars Buitinck.
• Do with this code as you like, but don't remove the copyright notice.
• /

- use_module(library(clpfd)).

nono(RowSpec, ColSpec, Grid) :- rows(RowSpec, Grid), transpose(Grid, GridT), rows(ColSpec, GridT).

rows([], []). rows([C|Cs], [R|Rs]) :- row(C, R), rows(Cs, Rs).

row(Ks, Row) :- sum(Ks, #=, Ones), sum(Row, #=, Ones), arcs(Ks, Arcs, start, Final), append(Row, [0], RowZ), automaton(RowZ, [source(start), sink(Final)], [arc(start,0,start) | Arcs]).

% Make list of transition arcs for finite-state constraint. arcs([], [], Final, Final). arcs([K|Ks], Arcs, CurState, Final) :- gensym(state, NextState), ( K == 0 -> Arcs = [arc(CurState,0,CurState), arc(CurState,0,NextState) | Rest], arcs(Ks, Rest, NextState, Final) ; Arcs = [arc(CurState,1,NextState) | Rest], K1 #= K-1, arcs([K1|Ks], Rest, NextState, Final)).

make_grid(Grid, X, Y, Vars) :- length(Grid,X), make_rows(Grid, Y, Vars).

make_rows([], _, []). make_rows([R|Rs], Len, Vars) :- length(R, Len), make_rows(Rs, Len, Vars0), append(R, Vars0, Vars).

print([]). print([R|Rs]) :- print_row(R), print(Rs).

print_row([]) :- nl. print_row([X|R]) :- ( X == 0 -> write(' ') ; write('x')), print_row(R).

nonogram(Rows, Cols) :- length(Rows, X), length(Cols, Y), make_grid(Grid, X, Y, Vars), nono(Rows, Cols, Grid), label(Vars), print(Grid).

</lang> File nonogram.pl, used to read data in a file. <lang Prolog>nonogram :- open('C:/Users/Utilisateur/Documents/Prolog/Rosetta/nonogram/nonogram.txt', read, In, []), repeat, read_line_to_codes(In, Line_1), read_line_to_codes(In, Line_2), compute_values(Line_1, [], [], Lines), compute_values(Line_2, [], [], Columns), nonogram(Lines, Columns) , nl, nl, read_line_to_codes(In, end_of_file), close(In).

compute_values([], Current, Tmp, R) :- reverse(Current, R_Current), reverse([R_Current | Tmp], R).

compute_values([32 | T], Current, Tmp, R) :- !, reverse(Current, R_Current), compute_values(T, [], [R_Current | Tmp], R).

compute_values([X | T], Current, Tmp, R) :- V is X - 64, compute_values(T, [V | Current], Tmp, R). </lang>

Python

First fill cells by deduction, then search through all combinations. It could take up a huge amount of storage, depending on the board size. <lang python>from itertools import izip

def gen_row(w, s):

   """Create all patterns of a row or col that match given runs."""
   def gen_seg(o, sp):
       if not o:
           return [[2] * sp]
       return [[2] * x + o[0] + tail
               for x in xrange(1, sp - len(o) + 2)
               for tail in gen_seg(o[1:], sp - x)]

   return [x[1:] for x in gen_seg([[1] * i for i in s], w + 1 - sum(s))]

def deduce(hr, vr):

   """Fix inevitable value of cells, and propagate."""
   def allowable(row):
       return reduce(lambda a, b: [x | y for x, y in izip(a, b)], row)

   def fits(a, b):
       return all(x & y for x, y in izip(a, b))

   def fix_col(n):
       """See if any value in a given column is fixed;
       if so, mark its corresponding row for future fixup."""
       c = [x[n] for x in can_do]
       cols[n] = [x for x in cols[n] if fits(x, c)]
       for i, x in enumerate(allowable(cols[n])):
           if x != can_do[i][n]:
               mod_rows.add(i)
               can_do[i][n] &= x

   def fix_row(n):
       """Ditto, for rows."""
       c = can_do[n]
       rows[n] = [x for x in rows[n] if fits(x, c)]
       for i, x in enumerate(allowable(rows[n])):
           if x != can_do[n][i]:
               mod_cols.add(i)
               can_do[n][i] &= x

   def show_gram(m):
       # If there's 'x', something is wrong.
       # If there's '?', needs more work.
       for x in m:
           print " ".join("x#.?"[i] for i in x)
       print

   w, h = len(vr), len(hr)
   rows = [gen_row(w, x) for x in hr]
   cols = [gen_row(h, x) for x in vr]
   can_do = map(allowable, rows)

   # Initially mark all columns for update.
   mod_rows, mod_cols = set(), set(xrange(w))

   while mod_cols:
       for i in mod_cols:
           fix_col(i)
       mod_cols = set()
       for i in mod_rows:
           fix_row(i)
       mod_rows = set()

   if all(can_do[i][j] in (1, 2) for j in xrange(w) for i in xrange(h)):
       print "Solution would be unique" # but could be incorrect!
   else:
       print "Solution may not be unique, doing exhaustive search:"

   # We actually do exhaustive search anyway. Unique solution takes
   # no time in this phase anyway, but just in case there's no
   # solution (could happen?).
   out = [0] * h

   def try_all(n = 0):
       if n >= h:
           for j in xrange(w):
               if [x[j] for x in out] not in cols[j]:
                   return 0
           show_gram(out)
           return 1
       sol = 0
       for x in rows[n]:
           out[n] = x
           sol += try_all(n + 1)
       return sol

   n = try_all()
   if not n:
       print "No solution."
   elif n == 1:
       print "Unique solution."
   else:
       print n, "solutions."
   print

def solve(p, show_runs=True):

   s = [[[ord(c) - ord('A') + 1 for c in w] for w in l.split()]
        for l in p.splitlines()]
   if show_runs:
       print "Horizontal runs:", s[0]
       print "Vertical runs:", s[1]
   deduce(s[0], s[1])

def main():

   # Read problems from file.
   fn = "nonogram_problems.txt"
   for p in (x for x in open(fn).read().split("\n\n") if x):
       solve(p)

   print "Extra example not solvable by deduction alone:"
   solve("B B A A\nB B A A")

   print "Extra example where there is no solution:"
   solve("B A A\nA A A")

main()</lang>

Output:

Horizontal runs: [[3], [2, 1], [3, 2], [2, 2], [6], [1, 5], [6], [1], [2]]
Vertical runs: [[1, 2], [3, 1], [1, 5], [7, 1], [5], [3], [4], [3]]
Solution would be unique
. # # # . . . .
# # . # . . . .
. # # # . . # #
. . # # . . # #
. . # # # # # #
# . # # # # # .
# # # # # # . .
. . . . # . . .
. . . # # . . .

Unique solution

(... etc. ...)

Racket

See: Example:Nonogram solver/Racket