Nonogram solver: Difference between revisions
Small formatting changes in the Python entry |
More small changes in the Python entry |
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rows = [gen_row(w, x) for x in hr] |
rows = [gen_row(w, x) for x in hr] |
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cols = [gen_row(h, x) for x in vr] |
cols = [gen_row(h, x) for x in vr] |
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can_do = |
can_do = map(allowable, rows) |
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# Initially mark all columns for update. |
# Initially mark all columns for update. |
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| Line 498: | Line 498: | ||
mod_rows = set() |
mod_rows = set() |
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# Return if solution is unique. |
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if all(can_do[i][j] in (1, 2) for j in xrange(w) for i in xrange(h)): |
if all(can_do[i][j] in (1, 2) for j in xrange(w) for i in xrange(h)): |
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print "Solution would be unique" |
print "Solution would be unique" # but could be incorrect! |
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else: |
else: |
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print "Solution may not be unique, doing exhaustive search:" |
print "Solution may not be unique, doing exhaustive search:" |
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| Line 550: | Line 549: | ||
solve("B B A A\nB B A A") |
solve("B B A A\nB B A A") |
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print "Extra example where there |
print "Extra example where there is no solution:" |
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solve("B A A\nA A A") |
solve("B A A\nA A A") |
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Revision as of 18:48, 6 April 2014
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
<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[][][] genPad(in int w, in int n) pure nothrow {
if (n) {
typeof(return) result;
foreach (immutable l; 1 .. w - n + 1)
foreach (r; genPad(w - l, n - 1))
result ~= [[2].replicate(l)] ~ r;
return result;
} else
return [[[2].replicate(w)]];
}
static int[] interlace(in int[][] z, in int[][] o) pure nothrow {
assert(z.length == o.length + 1);
//return iota(z.length + o.length) // #9148.
// .map!(i => (i % 2) ? o[i / 2] : z[i / 2])
// .array;
typeof(return) result;
foreach (immutable i; 0 .. z.length + o.length)
result ~= (i % 2) ? o[i / 2] : z[i / 2];
return result;
}
const ones = s.map!(i => [1].replicate(i)).array;
return genPad(w - sum(cast(int[])s) + 2, s.length) //**
.map!(x => interlace(x, ones)[1 .. $ - 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; bool[int] fixable; // A set. 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;
/// See if any value a given column is fixed; if so,
/// mark its corresponding row for future fixup.
void fixCol(in int n) /*nothrow*/ {
fixable.remove(-n - 1);
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])) {
immutable v = x & canDo[i][n];
if (v == canDo[i][n])
continue;
fixable[i + 1] = true;
canDo[i][n] = v;
}
}
/// Ditto, for rows.
void fixRow(in int n) /*nothrow*/ {
fixable.remove(n + 1);
const c = canDo[n];
rows[n] = rows[n].remove!(x => !fits(x, c));
foreach (immutable i, immutable x; allowable(rows[n])) {
immutable v = x & canDo[n][i];
if (v == canDo[n][i])
continue;
fixable[-i - 1] = true;
canDo[n][i] = v;
}
}
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("-".replicate(2 * vr.length - 1));
}
// Initially mark all columns for update.
foreach (immutable i; 0 .. w)
fixable[-i - 1] = true;
while (fixable.length) {
immutable i = fixable.byKey.front;
if (i < 0)
fixCol(-i - 1);
else
fixRow(i - 1);
}
showGram(canDo);
// Return if solution is unique.
if (cartesianProduct(h.iota, w.iota)
.all!(ij => canDo[ij[0]][ij[1]] == 1 ||
canDo[ij[0]][ij[1]] == 2))
return;
"Solution not unique, doing exhaustive search:".writeln; auto out_ = new const(int)[][](h);
void tryAll(in int n = 0) {
if (n >= h) {
foreach (immutable j; 0 .. w) {
if (!cols[j].canFind(out_.map!(x => x[j]).array))
return;
}
showGram(out_);
return;
}
foreach (const x; rows[n]) {
out_[n] = x;
tryAll(n + 1);
}
}
tryAll;
}
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]);
writeln('\n');
}
void main() {
// Read problems from file.
immutable fn = "nonogram_problems.txt";
foreach (p; fn.readText.split("\n\n").filter!(p => !p.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]] . # # # . . . . # # . # . . . . . # # # . . # # . . # # . . # # . . # # # # # # # . # # # # # . # # # # # # . . . . . . # . . . . . . # # . . . --------------- 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]] . . . . . . . . . . # # # # # # . . . . . . . . . . . . # # # . # . . # # # . . . . . # . . # # # . . . # . . . . # # # . . # # # . # # # # # # # # # # # # # # . . . # . . # . . . . . . . . . . . . # . . # . # . # # . . . . . . . . . . # # # # # # # . . # # . . . . . . . . # # . # # # # # . . . # . . . . . . . . # . . # # # # # . . # # # . # # # . # # # . . # # # # # # # # . # # # . # # # . # # # --------------------------------------- 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]] . . . . # # # . # . . . . . . . . . . . . . . . # # . # # # # . # . . . . . . . . . . . # . # # # . # # # . . . . . . . . . # # . # # # # . . . . . . . . . . . . # # # . # # # . # . . . . # # # . . . # # # . . # # . # # . . . # . # # # . . # # . . # # . # # . . . . # # . # # . . . . . . # # . # . # . . # # . # . # . . . . . . # . # # . # . . . # # # # . . . . . . . # . # . # # . . . . . # # . . . . . . . . # # . # # . . # # # # # # # # . . . . # # . # # . . . # # . . # # # # . . . . # . # # . # # . # . . . # . . # # # # . . # # # . # # # # # . . . . . # # . # . # # # . # . . . . # . . . . # # # # . . # # # . # . . . . # # # . # # # . # . # # # . # # . # # # # # # # # . . . # # # # . # # # . # # # # # # # # . . . . . # . # # # # . # # . # # # # # . . . . . # . # # # # . # # . . . # # . . . . . . . # # # # . . # # . . . # # # # # . . . # # # # # . # # # . . . # # # # # . . . # # # # . # . . . . . . . . . . # . . # # # # . # # . . . . . . . . . . . . . # # # . # # # . . . . . . . . . . . --------------------------------------- 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]] . . . . . . . . . . . . . . . . . . . . # # # # # . . # # . . . . . . . . . . . . . . # # # . . # # . # # . . . . . . . . . . . . . . # # # # # . . # # # . . . . . . . . . . . . . # # # # # # # # . . # # . . . . # # # # # . # # # # # # # # # # # . . # . # . . # # . . . . # . . . . # # # # # # . . . # . . # # . . . . . # . . . . . . . # # # . . . . # # . . . . . . . . # . . . . . . . . . . . . . # . # # . . . . . # # # # # # . . . . . . . . . # # . . # # # # # # # # # # # # # # # . . . . # # # # . . . . . # # # # # # # # # # . . # # # # # # # # . . . . # # . # . # # # # . # # # . . # # # # # # . . . . . . . . # # # # # # # # # # # # # # # # # . . . . . . . . # # # # # # # # # # # # # # # # # . . . . . . . # # # # # # # # # # # # # # # # # # . . . . . . . # . . . # # # # # # # # # # # # # # . . . . . . . # . # . # # # # # # # # # # # # # # . . . . . . . . # # # # # . . . # # # # # # # # # . . . . . . . . . . . . . . . . . # # # # # # # # . . . . . . . . . . . . . . . . . . # # # # # # # ------------------------------------------------- Extra example not solvable by deduction alone: Horizontal runs: [[2], [2], [1], [1]] Vertical runs: [[2], [2], [1], [1]] ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ------- Solution not unique, doing exhaustive search: # # . . # # . . . . # . . . . # ------- # # . . # # . . . . . # . . # . ------- . # # . # # . . # . . . . . . # ------- Extra example where there is no solution: Horizontal runs: [[2], [1], [1]] Vertical runs: [[1], [1], [1]] ? # ? ? . ? ? . ? ----- Solution not unique, doing exhaustive search:
The run-time with ldc2 compiler is about 0.50 seconds. The output is the same as the Python entry.
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. ...)