# Nonogram solver

Nonogram solver
You are encouraged to solve this task according to the task description, using any language you may know.

A nonogram is a puzzle that provides numeric clues used to fill in a grid of cells, establishing for each cell whether it is filled or not. The puzzle solution is typically a picture of some kind.

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.

Example
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 4 problems below, read from a “nonogram_problems.txt” file that has this content (the blank lines are separators):

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

Extra credit: generate nonograms with unique solutions, of desired height and width.

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).

## C++

### The Solver

// A class to solve Nonogram (Hadje) Puzzles
// Nigel Galloway - January 23rd., 2017
template<uint _N, uint _G> class Nonogram {
enum class ng_val : char {X='#',B='.',V='?'};
template<uint _NG> struct N {
N() {}
N(std::vector<int> ni,const int l) : X{},B{},Tx{},Tb{},ng(ni),En{},gNG(l){}
std::bitset<_NG> X, B, T, Tx, Tb;
std::vector<int> ng;
int En, gNG;
void fn (const int n,const int i,const int g,const int e,const int l){
if (fe(g,l,false) and fe(g+l,e,true)){
if ((n+1) < ng.size()) {if (fe(g+e+l,1,false)) fn(n+1,i-e-1,g+e+l+1,ng[n+1],0);}
else {
if (fe(g+e+l,gNG-(g+e+l),false)){Tb &= T.flip(); Tx &= T.flip(); ++En;}
}}
if (l<=gNG-g-i-1) fn(n,i,g,e,l+1);
}
void fi (const int n,const bool g) {X.set(n,g); B.set(n, not g);}
ng_val fg (const int n) const{return (X.test(n))? ng_val::X : (B.test(n))? ng_val::B : ng_val::V;}
inline bool fe (const int n,const int i, const bool g){
for (int e = n;e<n+i;++e) if ((g and fg(e)==ng_val::B) or (!g and fg(e)==ng_val::X)) return false; else T[e] = g;
return true;
}
int fl (){
if (En == 1) return 1;
Tx.set(); Tb.set(); En=0;
fn(0,std::accumulate(ng.cbegin(),ng.cend(),0)+ng.size()-1,0,ng[0],0);
return En;
}}; // end of N
std::vector<N<_G>> ng;
std::vector<N<_N>> gn;
int En, zN, zG;
void setCell(uint n, uint i, bool g){ng[n].fi(i,g); gn[i].fi(n,g);}
public:
Nonogram(const std::vector<std::vector<int>>& n,const std::vector<std::vector<int>>& i,const std::vector<std::string>& g = {}) : ng{}, gn{}, En{}, zN(n.size()), zG(i.size()) {
for (int n=0; n<zG; n++) gn.push_back(N<_N>(i[n],zN));
for (int i=0; i<zN; i++) {
ng.push_back(N<_G>(n[i],zG));
if (i < g.size()) for(int e=0; e<zG or e<g[i].size(); e++) if (g[i][e]=='#') setCell(i,e,true);
}}
bool solve(){
int i{}, g{};
for (int l = 0; l<zN; l++) {
if ((g = ng[l].fl()) == 0) return false; else i+=g;
for (int i = 0; i<zG; i++) if (ng[l].Tx[i] != ng[l].Tb[i]) setCell (l,i,ng[l].Tx[i]);
}
for (int l = 0; l<zG; l++) {
if ((g = gn[l].fl()) == 0) return false; else i+=g;
for (int i = 0; i<zN; i++) if (gn[l].Tx[i] != gn[l].Tb[i]) setCell (i,l,gn[l].Tx[i]);
}
if (i == En) return false; else En = i;
if (i == zN+zG) return true; else return solve();
}
const std::string toStr() const {
std::ostringstream n;
for (int i = 0; i<zN; i++){for (int g = 0; g<zG; g++){n << static_cast<char>(ng[i].fg(g));}n<<std::endl;}
return n.str();
}};

// For the purpose of this task I provide a little code to read from a file in the required format
// Note though that Nonograms may contain blank lines and values greater than 24
int main(){
std::ifstream n ("nono.txt");
if (!n) {
std::cerr << "Unable to open nono.txt.\n";
exit(EXIT_FAILURE);
}
std::string i;
getline(n,i);
std::istringstream g(i);
std::string e;
std::vector<std::vector<int>> N;
while (g >> e) {
std::vector<int> G;
for (char l : e) G.push_back((int)l-64);
N.push_back(G);
}
getline(n,i);
std::istringstream gy(i);
std::vector<std::vector<int>> G;
while (gy >> e) {
std::vector<int> N;
for (char l : e) N.push_back((int)l-64);
G.push_back(N);
}
Nonogram<32,32> myN(N,G);
if (!myN.solve()) std::cout << "I don't believe that this is a nonogram!" << std::endl;
std::cout << "\n" << myN.toStr() << std::endl;
}

Output:
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

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

### Bonus GCHQ Xmas Puzzle

[GCHQ Xmas Puzzle] is a Nonogram. They say "We pre-shaded a few cells to help people get started. Without this, the puzzle would have been slightly ambiguous, though the error correction used in QR codes means that the URL would have been recovered anyway. As a small Easter egg, the pre-shaded cells spell out “GCHQ” in Morse code."

int main(){
const std::vector<std::vector<int>> Ngchq={{ 7,3,1, 1,7},
{ 1,1,2,2, 1,1},
{ 1,3,1,3,1,1, 3,1},
{ 1,3,1,1,6,1, 3,1},
{ 1,3,1,5,2,1, 3,1},
{ 1,1,2, 1,1},
{ 7,1,1,1,1, 1,7},
{ 3,3},
{1,2,3,1,1,3,1, 1,2},
{ 1,1,3,2, 1,1},
{ 4,1,4,2, 1,2},
{ 1,1,1,1,1,4, 1,3},
{ 2,1,1,1, 2,5},
{ 3,2,2,6, 3,1},
{ 1,9,1,1, 2,1},
{ 2,1,2,2, 3,1},
{ 3,1,1,1,1, 5,1},
{ 1,2, 2,5},
{ 7,1,2,1,1, 1,3},
{ 1,1,2,1,2, 2,1},
{ 1,3,1,4, 5,1},
{ 1,3,1,3,10,2},
{ 1,3,1,1, 6,6},
{ 1,1,2,1, 1,2},
{ 7,2,1, 2,5}};
const std::vector<std::vector<int>> Ggchq={{ 7,2,1,1,7},
{ 1,1,2,2,1,1},
{1,3,1,3,1,3,1,3,1},
{ 1,3,1,1,5,1,3,1},
{ 1,3,1,1,4,1,3,1},
{ 1,1,1,2,1,1},
{ 7,1,1,1,1,1,7},
{ 1,1,3},
{ 2,1,2,1,8,2,1},
{ 2,2,1,2,1,1,1,2},
{ 1,7,3,2,1},
{ 1,2,3,1,1,1,1,1},
{ 4,1,1,2,6},
{ 3,3,1,1,1,3,1},
{ 1,2,5,2,2},
{2,2,1,1,1,1,1,2,1},
{ 1,3,3,2,1,8,1},
{ 6,2,1},
{ 7,1,4,1,1,3},
{ 1,1,1,1,4},
{ 1,3,1,3,7,1},
{1,3,1,1,1,2,1,1,4},
{ 1,3,1,4,3,3},
{ 1,1,2,2,2,6,1},
{ 7,1,3,2,1,1}};

std::vector<std::string> n = {"",
"",
"",
"...##.......##.......#",
"",
"",
"",
"",
"......##..#...##..#",
"",
"",
"",
"",
"",
"",
"",
"......#....#....#...#",
"",
"",
"",
"",
"...##....##....#....##"};
Nonogram<25,25> myN(Ngchq,Ggchq,n);
if (!myN.solve()) std::cout << "I don't believe that this is a nonogram!" << std::endl;
std::cout << "\n" << myN.toStr() << std::endl;
}

Output:
#######.###...#.#.#######
#.....#.##.##.....#.....#
#.###.#.....###.#.#.###.#
#.###.#.#..######.#.###.#
#.###.#..#####.##.#.###.#
#.....#..##.......#.....#
#######.#.#.#.#.#.#######
........###...###........
#.##.###..#.#.###.#..#.##
#.#......###.##....#...#.
.####.#.####.##.#....##..
.#.#...#...#.#.####.#.###
..##..#.#.#......##.#####
...###.##.##.######.###.#
#.#########.#.#..##....#.
.##.#..##...##.###.....#.
###.#.#.#..#....#####.#..
........#...##.##...#####
#######.#..##...#.#.#.###
#.....#.##..#..##...##.#.
#.###.#...####..#####..#.
#.###.#.###.##########.##
#.###.#.#..######.######.
#.....#..##......#.#.##..
#######.##...#.##...#####

## C#

using System;
using System.Collections.Generic;
using static System.Linq.Enumerable;

public static class NonogramSolver
{
public static void Main2() {
foreach (var (x, y) in new [] {
("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")
})
{
Solve(x, y);
Console.WriteLine();
}
}

static void Solve(string rowLetters, string columnLetters) {
var r = rowLetters.Split(" ").Select(row => row.Select(s => s - 'A' + 1).ToArray()).ToArray();
var c = columnLetters.Split(" ").Select(column => column.Select(s => s - 'A' + 1).ToArray()).ToArray();
Solve(r, c);
}

static void Solve(int[][] rowRuns, int[][] columnRuns) {
int len = columnRuns.Length;
var rows = rowRuns.Select(row => Generate(len, row)).ToList();
var columns = columnRuns.Select(column => Generate(rowRuns.Length, column)).ToList();
Reduce(rows, columns);
foreach (var list in rows) {
if (list.Count != 1) Console.WriteLine(Repeat('?', len).Spaced());
}
}

static List<BitSet> Generate(int length, params int[] runs) {
var list = new List<BitSet>();
BitSet initial = BitSet.Empty;
int[] sums = new int[runs.Length];
sums[0] = 0;
for (int i = 1; i < runs.Length; i++) sums[i] = sums[i - 1] + runs[i - 1] + 1;
for (int r = 0; r < runs.Length; r++) initial = initial.AddRange(sums[r], runs[r]);
Generate(list, BitSet.Empty.Add(length), runs, sums, initial, 0, 0);
return list;
}

static void Generate(List<BitSet> result, BitSet max, int[] runs, int[] sums, BitSet current, int index, int shift) {
if (index == runs.Length) {
return;
}
while (current.Value < max.Value) {
Generate(result, max, runs, sums, current, index + 1, shift);
current = current.ShiftLeftAt(sums[index] + shift);
shift++;
}
}

static void Reduce(List<List<BitSet>> rows, List<List<BitSet>> columns) {
for (int count = 1; count > 0; ) {
foreach (var (rowIndex, row) in rows.WithIndex()) {
var allOn = row.Aggregate((a, b) => a & b);
var allOff = row.Aggregate((a, b) => a | b);
foreach (var (columnIndex, column) in columns.WithIndex()) {
count = column.RemoveAll(c => allOn.Contains(columnIndex) && !c.Contains(rowIndex));
count += column.RemoveAll(c => !allOff.Contains(columnIndex) && c.Contains(rowIndex));
}
}
foreach (var (columnIndex, column) in columns.WithIndex()) {
var allOn = column.Aggregate((a, b) => a & b);
var allOff = column.Aggregate((a, b) => a | b);
foreach (var (rowIndex, row) in rows.WithIndex()) {
count += row.RemoveAll(r => allOn.Contains(rowIndex) && !r.Contains(columnIndex));
count += row.RemoveAll(r => !allOff.Contains(rowIndex) && r.Contains(columnIndex));
}
}
}
}

static IEnumerable<(int index, T element)> WithIndex<T>(this IEnumerable<T> source) {
int i = 0;
foreach (T element in source) {
yield return (i++, element);
}
}

static string Reverse(this string s) {
char[] array = s.ToCharArray();
Array.Reverse(array);
return new string(array);
}

static string Spaced(this IEnumerable<char> s) => string.Join(" ", s);

struct BitSet //Unused functionality elided.
{
public static BitSet Empty => default;
public int Value => bits;

private BitSet(int bits) => this.bits = bits;

public BitSet Add(int item) => new BitSet(bits | (1 << item));
public BitSet AddRange(int start, int count) => new BitSet(bits | (((1 << (start + count)) - 1) - ((1 << start) - 1)));
public bool Contains(int item) => (bits & (1 << item)) != 0;
public BitSet ShiftLeftAt(int index) => new BitSet((bits >> index << (index + 1)) | (bits & ((1 << index) - 1)));
public override string ToString() => Convert.ToString(bits, 2);

public static BitSet operator &(BitSet a, BitSet b) => new BitSet(a.bits & b.bits);
public static BitSet operator |(BitSet a, BitSet b) => new BitSet(a.bits | b.bits);
}

}
Output:
. # # # . . . .
# # . # . . . .
. # # # . . # #
. . # # . . # #
. . # # # # # #
# . # # # # # .
# # # # # # . .
. . . . # . . .
. . . # # . . .

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

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

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

## Common Lisp

(defpackage :ac3
(:use :cl)
(:export :var
:domain
:satisfies-p
:constraint-possible-p
:ac3)
(:documentation "Implements the AC3 algorithm. Extend VAR with the variable
types for your particular problem and implement SATISFIES-P and
with unary constraints already satisfied and then pass them to AC3 in a list."
))

(in-package :ac3)

(defclass var ()
((domain :initarg :domain :accessor domain))
(:documentation "The base variable type from which all other
variables should extend."
))

(defgeneric satisfies-p (a b va vb)
(:documentation "Determine if constrainted variables A and B are
satisfied by the instantiation of their respective values VA and VB."
))

(defgeneric constraint-possible-p (a b)
(:documentation "Determine if variables A and B can even be
checked for a binary constraint."
))

(defun arc-reduce (a b)
"Assuming A and B truly form a constraint, prune all values
from A that do not satisfy any value in B. Return T if the domain
of A changed by any amount, NIL otherwise."

(let (change)
(setf (domain a)
(loop for va in (domain a)
when (loop for vb in (domain b)
do (when (satisfies-p a b va vb)
(return t))
finally (setf change t) (return nil))
collect va))
change))

(defun binary-constraint-p (a b)
"Check if variables A and B could form a constraint, then return T
if any of their values form a contradiction, NIL otherwise."

(when (constraint-possible-p a b)
(block found
(loop for va in (domain a)
do (loop for vb in (domain b)
do (unless (satisfies-p a b va vb)
(return-from found t)))))))

(defun ac3 (vars)
"Run the Arc Consistency 3 algorithm on the given set of variables.
Assumes unary constraints have already been satisfied."

;; Form a worklist of the constraints of every variable to every other variable.
(let ((worklist (loop for x in vars
append (loop for y in vars
when (and (not (eq x y))
(binary-constraint-p x y))
collect (cons x y)))))
;; Prune the worklist of satisfied arcs until it is empty.
(loop while worklist
do (destructuring-bind (x . y) (pop worklist)
(when (arc-reduce x y)
(if (domain x)
;; If the current arc's domain was reduced, then append any arcs it
;; is still constrained with to the end of the worklist, as they
;; need to be rechecked.
(setf worklist (nconc worklist (loop for z in vars
when (and (not (eq x z))
(not (eq y z))
(binary-constraint-p x z))
collect (cons z x))))
(error "No values left in ~a" x))))
finally (return vars))))

(defpackage :nonogram
(:use :cl :ac3)
(:documentation "Utilize the AC3 package to solve nonograms."))

(in-package :nonogram)

(defclass line (var)
((depth :initarg :depth :accessor depth))
(:documentation "A LINE is a variable that represents either a
column or row of cells and all of the permutations of values those
cells can assume"
))

(defmethod print-object ((o line) s)
(with-slots (depth domain) o
(format s ":depth ~a :domain ~a" depth domain))))

(defclass row (line) ())

(defclass col (line) ())

(defmethod satisfies-p ((a line) (b line) va vb)
(eq (aref va (depth b))
(aref vb (depth a))))

(defmethod constraint-possible-p ((a line) (b line))
(not (eq (type-of a) (type-of b))))

(defun make-line-domain (runs length &optional (start 0) acc)
"Enumerate all valid permutations of a line's values."
(if runs
(loop for i from start
to (- length
(reduce #'+ (cdr runs))
(length (cdr runs))
(car runs))
append (make-line-domain (cdr runs) length (+ 1 i (car runs)) (cons i acc)))
(list (reverse acc))))

(defun make-line (type runs depth length)
"Create and initialize a ROW or COL instance."
(make-instance
type :depth depth :domain
(loop for value in (make-line-domain runs length)
collect (let ((arr (make-array length :initial-element nil)))
(loop for pos in value
for run in runs
do (loop for i from pos below (+ pos run)
do (setf (aref arr i) t)))
arr))))

(defun make-lines (type run-set length)
"Initialize a set of lines."
(loop for runs across run-set
for depth from 0
collect (make-line type runs depth length)))

(defun nonogram (problem)
"Given a nonogram problem description, solve it and print the result."
(let* ((nrows (length (aref problem 0)))
(ncols (length (aref problem 1)))
(vars (ac3 (append (make-lines 'row (aref problem 0) ncols)
(make-lines 'col (aref problem 1) nrows)))))
(loop for var in vars
while (eq 'row (type-of var))
do (terpri)
(loop for cell across (car (domain var))
do (format t "~a " (if cell #\# #\.))))))

(defparameter *test-set*
'("C BA CB BB F AE F A B"
"AB CA AE GA E C D C"))

;; Helper functions to read and parse problems from a file.

(defun parse-word (word)
(map 'list (lambda (c) (1+ (- (char-code c) (char-code #\A)))) word))

(defun parse-line (line)
(map 'vector #'parse-word (uiop:split-string (string-upcase line))))

(defun parse-nonogram (rows columns)
(vector (parse-line rows)
(parse-line columns)))

(when (> (length (string-trim '(#\space) line)) 0)
(print line)
(return line)))))

(defun solve-from-file (file)
(handler-case
(with-open-file (s file)
(loop
(terpri)
(end-of-file ())))
Output:
CL-USER> (time (nonogram::solve-from-file "c:/Users/cro/Dropbox/Projects/rosetta-code/nonogram_problems.txt"))

"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"
. . . . . . . . . . . . . . . . . . . . # # # # #
. . # # . . . . . . . . . . . . . . # # # . . # #
. # # . . . . . . . . . . . . . . # # # # # . . #
# # . . . . . . . . . . . . . # # # # # # # # . .
# # . . . . # # # # # . # # # # # # # # # # # . .
# . # . . # # . . . . # . . . . # # # # # # . . .
# . . # # . . . . . # . . . . . . . # # # . . . .
# # . . . . . . . . # . . . . . . . . . . . . . #
. # # . . . . . # # # # # # . . . . . . . . . # #
. . # # # # # # # # # # # # # # # . . . . # # # #
. . . . . # # # # # # # # # # . . # # # # # # # #
. . . . # # . # . # # # # . # # # . . # # # # # #
. . . . . . . . # # # # # # # # # # # # # # # # #
. . . . . . . . # # # # # # # # # # # # # # # # #
. . . . . . . # # # # # # # # # # # # # # # # # #
. . . . . . . # . . . # # # # # # # # # # # # # #
. . . . . . . # . # . # # # # # # # # # # # # # #
. . . . . . . . # # # # # . . . # # # # # # # # #
. . . . . . . . . . . . . . . . . # # # # # # # #
. . . . . . . . . . . . . . . . . . # # # # # # #
Evaluation took:
0.906 seconds of real time
0.906250 seconds of total run time (0.890625 user, 0.015625 system)
100.00% CPU
1 form interpreted
59 lambdas converted
2,979,778,058 processor cycles
58,974,976 bytes consed

## D

Translation of: Python
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 @safe {
static int[][] genSeg(in int[][] o, in int sp) pure nothrow @safe {
if (o.empty)
return [[2].replicate(sp)];

typeof(return) result;
foreach (immutable x; 1 .. sp - o.length + 2)
foreach (const 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 - s.sum).map!dropOne;
}

/// Fix inevitable value of cells, and propagate.
void deduce(in int[][] hr, in int[][] vr) {
static int[] allowable(in int[][] row) pure nothrow @safe {
//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*/ @safe /*@nogc*/ {
return zip(a, b).all!(xy => xy[0] & xy[1]);
}

immutable int w = vr.length,
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[uint] modRows, modCols;
modCols = true.repeat.enumerate!uint.take(w).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*/ @safe {
const c = canDo.map!(x => x[n]).array;
cols[n] = cols[n].remove!(x => !fits(x, c)); // Throws.
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*/ @safe {
const c = canDo[n];
rows[n] = rows[n].remove!(x => !fits(x, c)); // Throws.
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.
m.each!(x => writefln("%-(%c %)", x.map!(i => "x#.?"[i])));
writeln;
}

while (modCols.length > 0) {
modCols.byKey.each!fixCol;
modCols = null;
modRows.byKey.each!fixRow;
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) {
immutable 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() {
immutable fn = "nonogram_problems.txt";

"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;
}
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.

## F#

(*
I define a discriminated union to provide Nonogram Solver functionality.
Nigel Galloway May 28th., 2016
*)

type N =
|X |B |V
static member fn n i =
let fn n i = [for g = 0 to i-n do yield Array.init (n+g) (fun e -> if e >= g then X else B)]
let rec fi n i = [
match n with
| h::t -> match t with
| [] -> for g in fn h i do yield Array.append g (Array.init (i-g.Length) (fun _ -> B))
| _ -> for g in fn h ((i-List.sum t)+t.Length) do for a in fi t (i-g.Length-1) do yield Array.concat[g;[|B|];a]
| [] -> yield Array.init i (fun _ -> B)
]
fi n i
static member fi n i = Array.map2 (fun n g -> match (n,g) with |X,X->X |B,B->B |_->V) n i
static member fg (n: N[]) (i: N[][]) g = n |> Seq.mapi (fun e n -> i.[e].[g] = n || i.[e].[g] = V) |> Seq.forall (fun n -> n)
static member fe (n: N[][]) = n|> Array.forall (fun n -> Array.forall (fun n -> n <> V) n)
static member fl n = n |> Array.Parallel.map (fun n -> Seq.reduce (fun n g -> N.fi n g) n)
static member fa (nga: list<N []>[]) ngb = Array.Parallel.mapi (fun i n -> List.filter (fun n -> N.fg n ngb i) n) nga
static member fo n i g e =
let na = N.fa n e
let ia = N.fl na
let ga = N.fa g ia
(na, ia, ga, (N.fl ga))
static member toStr n = match n with |X->"X"|B->"."|V->"?"
static member presolve ((na: list<N []>[]), (ga: list<N []>[])) =
let nb = N.fl na
let x = N.fa ga nb
let rec fn n i g e l =
let na,ia,ga,ea = N.fo n i g e
let el = ((Array.map (fun n -> List.length n) na), (Array.map (fun n -> List.length n) ga))
if ((fst el) = (fst l)) && ((snd el) = (snd l)) then (n,i,g,e,(Array.forall (fun n -> n = 1) (fst l))) else fn na ia ga ea el
fn na nb x (N.fl x) ((Array.map (fun n -> List.length n) na), (Array.map (fun n -> List.length n) ga))

For the purposes of this task I provide a little code to read the input from a file

let fe (n : array<string>) i = n |> Array.collect (fun n -> [|N.fn [for g in n -> ((int)g-64)] i|])
let fl (n : array<string>) (i : array<string>) = (fe n i.Length), (fe i n.Length)
let rFile =
try
use file = File.OpenText @"nonogram.txt"
with | _ -> printfn "Error reading file" ; None

This may be used:

let n,i,g,e,l = N.presolve rFile.Value
if l then i |> Array.iter (fun n -> n |> Array.iter (fun n -> printf "%s" (N.toStr n));printfn "") else printfn "No unique solution"

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

.XXX....
XX.X....
.XXX..XX
..XX..XX
..XXXXXX
X.XXXXX.
XXXXXX..
....X...
...XX...

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

..........XXXXXX....
........XXX.X..XXX..
...X..XXX...X....XXX
..XXX.XXXXXXXXXXXXXX
...X..X............X
..X.X.XX..........XX
XXXXX..XX........XX.
XXXXX...X........X..
XXXXX..XXX.XXX.XXX..
XXXXXXXX.XXX.XXX.XXX

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

....XXX.X...........
....XX.XXXX.X.......
....X.XXX.XXX.......
..XX.XXXX...........
.XXX.XXX.X....XXX...
XXX..XX.XX...X.XXX..
XX..XX.XX....XX.XX..
....XX.X.X..XX.X.X..
....X.XX.X...XXXX...
....X.X.XX.....XX...
.....XX.XX..XXXXXXXX
....XX.XX...XX..XXXX
....X.XX.XX.X...X..X
XXX..XXX.XXXXX.....X
X.X.XXX.X....X....XX
XX..XXX.X....XXX.XXX
.X.XXX.XX.XXXXXXXX..
.XXXX.XXX.XXXXXXXX..
...X.XXXX.XX.XXXXX..
...X.XXXX.XX...XX...
....XXXX..XX...XXXXX
...XXXXX.XXX...XXXXX
...XXXX.X..........X
..XXXX.XX...........
..XXX.XXX...........

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

....................XXXXX
..XX..............XXX..XX
.XX..............XXXXX..X
XX.............XXXXXXXX..
XX....XXXXX.XXXXXXXXXXX..
X.X..XX....X....XXXXXX...
X..XX.....X.......XXX....
XX........X.............X
.XX.....XXXXXX.........XX
..XXXXXXXXXXXXXXX....XXXX
.....XXXXXXXXXX..XXXXXXXX
....XX.X.XXXX.XXX..XXXXXX
........XXXXXXXXXXXXXXXXX
........XXXXXXXXXXXXXXXXX
.......XXXXXXXXXXXXXXXXXX
.......X...XXXXXXXXXXXXXX
.......X.X.XXXXXXXXXXXXXX
........XXXXX...XXXXXXXXX
.................XXXXXXXX
..................XXXXXXX

## Go

Translation of: Java
package main

import (
"fmt"
"strings"
)

type BitSet []bool

func (bs BitSet) and(other BitSet) {
for i := range bs {
if bs[i] && other[i] {
bs[i] = true
} else {
bs[i] = false
}
}
}

func (bs BitSet) or(other BitSet) {
for i := range bs {
if bs[i] || other[i] {
bs[i] = true
} else {
bs[i] = false
}
}
}

func iff(cond bool, s1, s2 string) string {
if cond {
return s1
}
return s2
}

func newPuzzle(data [2]string) {
rowData := strings.Fields(data[0])
colData := strings.Fields(data[1])
rows := getCandidates(rowData, len(colData))
cols := getCandidates(colData, len(rowData))

for {
numChanged := reduceMutual(cols, rows)
if numChanged == -1 {
fmt.Println("No solution")
return
}
if numChanged == 0 {
break
}
}

for _, row := range rows {
for i := 0; i < len(cols); i++ {
fmt.Printf(iff(row[0][i], "# ", ". "))
}
fmt.Println()
}
fmt.Println()
}

// collect all possible solutions for the given clues
func getCandidates(data []string, le int) [][]BitSet {
var result [][]BitSet
for _, s := range data {
var lst []BitSet
a := []byte(s)
sumBytes := 0
for _, b := range a {
sumBytes += int(b - 'A' + 1)
}
prep := make([]string, len(a))
for i, b := range a {
prep[i] = strings.Repeat("1", int(b-'A'+1))
}
for _, r := range genSequence(prep, le-sumBytes+1) {
bits := []byte(r[1:])
bitset := make(BitSet, len(bits))
for i, b := range bits {
bitset[i] = b == '1'
}
lst = append(lst, bitset)
}
result = append(result, lst)
}
return result
}

func genSequence(ones []string, numZeros int) []string {
le := len(ones)
if le == 0 {
return []string{strings.Repeat("0", numZeros)}
}
var result []string
for x := 1; x < numZeros-le+2; x++ {
skipOne := ones[1:]
for _, tail := range genSequence(skipOne, numZeros-x) {
result = append(result, strings.Repeat("0", x)+ones[0]+tail)
}
}
return result
}

/* If all the candidates for a row have a value in common for a certain cell,
then it's the only possible outcome, and all the candidates from the
corresponding column need to have that value for that cell too. The ones
that don't, are removed. The same for all columns. It goes back and forth,
until no more candidates can be removed or a list is empty (failure).
*/

func reduceMutual(cols, rows [][]BitSet) int {
countRemoved1 := reduce(cols, rows)
if countRemoved1 == -1 {
return -1
}
countRemoved2 := reduce(rows, cols)
if countRemoved2 == -1 {
return -1
}
return countRemoved1 + countRemoved2
}

func reduce(a, b [][]BitSet) int {
countRemoved := 0
for i := 0; i < len(a); i++ {
commonOn := make(BitSet, len(b))
for j := 0; j < len(b); j++ {
commonOn[j] = true
}
commonOff := make(BitSet, len(b))

// determine which values all candidates of a[i] have in common
for _, candidate := range a[i] {
commonOn.and(candidate)
commonOff.or(candidate)
}

// remove from b[j] all candidates that don't share the forced values
for j := 0; j < len(b); j++ {
fi, fj := i, j
for k := len(b[j]) - 1; k >= 0; k-- {
cnd := b[j][k]
if (commonOn[fj] && !cnd[fi]) || (!commonOff[fj] && cnd[fi]) {
lb := len(b[j])
copy(b[j][k:], b[j][k+1:])
b[j][lb-1] = nil
b[j] = b[j][:lb-1]
countRemoved++
}
}
if len(b[j]) == 0 {
return -1
}
}
}
return countRemoved
}

func main() {
p1 := [2]string{"C BA CB BB F AE F A B", "AB CA AE GA E C D C"}

p2 := [2]string{
"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",
}

p3 := [2]string{
"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",
}

p4 := [2]string{
"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",
}

for _, puzzleData := range [][2]string{p1, p2, p3, p4} {
newPuzzle(puzzleData)
}
}
Output:
. # # # . . . .
# # . # . . . .
. # # # . . # #
. . # # . . # #
. . # # # # # #
# . # # # # # .
# # # # # # . .
. . . . # . . .
. . . # # . . .

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

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

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

## Java

Works with: Java version 8
import java.util.*;
import static java.util.Arrays.*;
import static java.util.stream.Collectors.toList;

public class NonogramSolver {

static String[] p1 = {"C BA CB BB F AE F A B", "AB CA AE GA E C D C"};

static String[] p2 = {"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"};

static String[] p3 = {"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"};

static String[] p4 = {"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"};

public static void main(String[] args) {
for (String[] puzzleData : new String[][]{p1, p2, p3, p4})
newPuzzle(puzzleData);
}

static void newPuzzle(String[] data) {
String[] rowData = data[0].split("\\s");
String[] colData = data[1].split("\\s");

List<List<BitSet>> cols, rows;
rows = getCandidates(rowData, colData.length);
cols = getCandidates(colData, rowData.length);

int numChanged;
do {
numChanged = reduceMutual(cols, rows);
if (numChanged == -1) {
System.out.println("No solution");
return;
}
} while (numChanged > 0);

for (List<BitSet> row : rows) {
for (int i = 0; i < cols.size(); i++)
System.out.print(row.get(0).get(i) ? "# " : ". ");
System.out.println();
}
System.out.println();
}

// collect all possible solutions for the given clues
static List<List<BitSet>> getCandidates(String[] data, int len) {
List<List<BitSet>> result = new ArrayList<>();

for (String s : data) {

int sumChars = s.chars().map(c -> c - 'A' + 1).sum();
List<String> prep = stream(s.split(""))
.map(x -> repeat(x.charAt(0) - 'A' + 1, "1")).collect(toList());

for (String r : genSequence(prep, len - sumChars + 1)) {
char[] bits = r.substring(1).toCharArray();
BitSet bitset = new BitSet(bits.length);
for (int i = 0; i < bits.length; i++)
bitset.set(i, bits[i] == '1');
}
}
return result;
}

// permutation generator, translated from Python via D
static List<String> genSequence(List<String> ones, int numZeros) {
if (ones.isEmpty())
return asList(repeat(numZeros, "0"));

List<String> result = new ArrayList<>();
for (int x = 1; x < numZeros - ones.size() + 2; x++) {
List<String> skipOne = ones.stream().skip(1).collect(toList());
for (String tail : genSequence(skipOne, numZeros - x))
result.add(repeat(x, "0") + ones.get(0) + tail);
}
return result;
}

static String repeat(int n, String s) {
StringBuilder sb = new StringBuilder();
for (int i = 0; i < n; i++)
sb.append(s);
return sb.toString();
}

/* If all the candidates for a row have a value in common for a certain cell,
then it's the only possible outcome, and all the candidates from the
corresponding column need to have that value for that cell too. The ones
that don't, are removed. The same for all columns. It goes back and forth,
until no more candidates can be removed or a list is empty (failure). */

static int reduceMutual(List<List<BitSet>> cols, List<List<BitSet>> rows) {
int countRemoved1 = reduce(cols, rows);
if (countRemoved1 == -1)
return -1;

int countRemoved2 = reduce(rows, cols);
if (countRemoved2 == -1)
return -1;

return countRemoved1 + countRemoved2;
}

static int reduce(List<List<BitSet>> a, List<List<BitSet>> b) {
int countRemoved = 0;

for (int i = 0; i < a.size(); i++) {

BitSet commonOn = new BitSet();
commonOn.set(0, b.size());
BitSet commonOff = new BitSet();

// determine which values all candidates of ai have in common
for (BitSet candidate : a.get(i)) {
commonOn.and(candidate);
commonOff.or(candidate);
}

// remove from bj all candidates that don't share the forced values
for (int j = 0; j < b.size(); j++) {
final int fi = i, fj = j;

if (b.get(j).removeIf(cnd -> (commonOn.get(fj) && !cnd.get(fi))
|| (!commonOff.get(fj) && cnd.get(fi))))
countRemoved++;

if (b.get(j).isEmpty())
return -1;
}
}
return countRemoved;
}
}
. # # # . . . .
# # . # . . . .
. # # # . . # #
. . # # . . # #
. . # # # # # #
# . # # # # # .
# # # # # # . .
. . . . # . . .
. . . # # . . .

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

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

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

## Julia

using Base.Iterators

struct NonogramPuzzle
nrows::Int
ncols::Int
xhints::Vector{Vector{Int}}
yhints::Vector{Vector{Int}}
solutions:: Vector{Any}
NonogramPuzzle(xh, yh) = new(length(xh), length(yh), xh, yh, Vector{NTuple{4,Array{Int64,1}}}())
end

ycols2xrows(ycols) = [[ycols[i][j] for i in eachindex(ycols)] for j in eachindex(ycols[1])]

function hintsfromcol(rowvec, col, nrows)
hints = Vector{Int}()
hintrun = 0
for row in rowvec
if row[col] != 0
hintrun += 1
if col == nrows
push!(hints, hintrun)
end
elseif hintrun > 0
push!(hints, hintrun)
hintrun = 0
end
end
hints
end

function nonoblocks(hints, len)
minsized(arr) = vcat(map(x -> vcat(fill(1, x), [0]), arr)...)[1:end-1]
minlen(arr) = sum(arr) + length(arr) - 1
if isempty(hints)
return fill(0, len)
elseif minlen(hints) == len
return minsized(hints)
end
possibilities = Vector{Vector{Int}}()
for leftspace in 0:(len - minlen(hints))
header = vcat(fill(0, leftspace), fill(1, hints[1]), [0])
else
end
end
possibilities
end

function exclude!(xchoices, ychoices)
andvec(a) = findall(x -> x == 1, foldl((x, y) -> [x[i] & y[i] for i in 1:length(x)], a))
orvec(a) = findall(x -> x == 0, foldl((x, y) -> [x[i] | y[i] for i in 1:length(x)], a))
filterbyval!(arr, val, pos) = if !isempty(arr) filter!(x -> x[pos] == val, arr); end
ensurevecvec(arr::Vector{Vector{Int}}) = arr
ensurevecvec(arr::Vector{Int}) = [arr]
function excl!(choices, otherchoices)
for i in 1:length(choices)
if length(choices[i]) > 0
all1 = andvec(choices[i])
all0 = orvec(choices[i])
foreach(n -> filterbyval!(otherchoices[n], 1, i), all1)
foreach(n -> filterbyval!(otherchoices[n], 0, i), all0)
end
end
end
xclude!(x, y) = (excl!(x, y); x = map(ensurevecvec, x); y = map(ensurevecvec, y); (x, y))
xlen, ylen = sum(map(length, xchoices)), sum(map(length, ychoices))
while true
ychoices, xchoices = xclude!(ychoices, xchoices)
if any(isempty, xchoices)
return
end
xchoices, ychoices = xclude!(xchoices, ychoices)
if any(isempty, ychoices)
return
end
newxlen, newylen = sum(map(length, xchoices)), sum(map(length, ychoices))
if newxlen == xlen && newylen == ylen
return
end
xlen, ylen = newxlen, newylen
end
end

function trygrids(nonogram)
xchoices = [nonoblocks(nonogram.xhints[i], nonogram.ncols) for i in 1:nonogram.nrows]
ychoices = [nonoblocks(nonogram.yhints[i], nonogram.nrows) for i in 1:nonogram.ncols]
exclude!(xchoices, ychoices)
if all(x -> length(x) == 1, xchoices)
println("Unique solution.")
push!(nonogram.solutions, [x[1] for x in xchoices])
elseif all(x -> length(x) == 1, ychoices)
println("Unique solution.")
ycols = [y[1] for y in ychoices]
push!(nonogram.solutions, ycols2xrows(ycols))
else
println("Brute force: \$(prod(map(length, xchoices))) possibilities.")
for stack in product(xchoices...)
arr::Vector{Vector{Int}} = [i isa Vector ? i : [i] for i in stack]
if all(x -> length(x) == nonogram.ncols, arr) &&
all(y -> hintsfromcol(arr, y, nonogram.nrows) == nonogram.yhints[y], 1:nonogram.ncols)
push!(nonogram.solutions, arr)
end
end
nsoln = length(nonogram.solutions)
println(nsoln == 0 ? "No" : nsoln, " solutions.")
end
end

# The first puzzle below requires brute force, and the second has no solutions.
const testnonograms = """
B B A A
B B A A

B A A
A A A

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
"""

function processtestpuzzles(txt)
solutiontxt(a) = (s = ""; for r in a for c in r; s *= (c == 0 ? "." : "#") end; s *= "\n" end; s)
txtline2ints(s) = [[UInt8(ch - 'A' + 1) for ch in r] for r in split(s, r"\s+")]
linepairs = uppercase.(string.(split(txt, "\n\n")))
pcount = 0
for xyhints in linepairs
xh, yh = map(x -> txtline2ints(strip(x)), split(xyhints, "\n"))
nonogram = NonogramPuzzle(xh, yh)
println("\nPuzzle \$(pcount += 1):")
trygrids(nonogram)
foreach(x -> println(solutiontxt(x), "\n"), nonogram.solutions)
end
end

processtestpuzzles(testnonograms)

Puzzle 1:
Brute force: 144 possibilities.
2 solutions.
.##.
##..
#...
...#

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

Puzzle 2:
Brute force: 8 possibilities.
No solutions.

Puzzle 3:
Unique solution.
.###....
##.#....
.###..##
..##..##
..######
#.#####.
######..
....#...
...##...

Puzzle 4:
Unique solution.
..........######....
........###.#..###..
...#..###...#....###
..###.##############
...#..#............#
..#.#.##..........##
#####..##........##.
#####...#........#..
#####..###.###.###..
########.###.###.###

Puzzle 5:
Unique solution.
....###.#...........
....##.####.#.......
....#.###.###.......
..##.####...........
.###.###.#....###...
###..##.##...#.###..
##..##.##....##.##..
....##.#.#..##.#.#..
....#.##.#...####...
....#.#.##.....##...
.....##.##..########
....##.##...##..####
....#.##.##.#...#..#
###..###.#####.....#
#.#.###.#....#....##
##..###.#....###.###
.#.###.##.########..
.####.###.########..
...#.####.##.#####..
...#.####.##...##...
....####..##...#####
...#####.###...#####
...####.#..........#
..####.##...........
..###.###...........

Puzzle 6:
Unique solution.
....................#####
..##..............###..##
.##..............#####..#
##.............########..
##....#####.###########..
#.#..##....#....######...
#..##.....#.......###....
##........#.............#
.##.....######.........##
..###############....####
.....##########..########
....##.#.####.###..######
........#################
........#################
.......##################
.......#...##############
.......#.#.##############
........#####...#########
.................########
..................#######

## Kotlin

Translation of: Java
// version 1.2.0

import java.util.BitSet

typealias BitSets = List<MutableList<BitSet>>

val rx = Regex("""\s""")

fun newPuzzle(data: List<String>) {
val rowData = data[0].split(rx)
val colData = data[1].split(rx)
val rows = getCandidates(rowData, colData.size)
val cols = getCandidates(colData, rowData.size)

do {
val numChanged = reduceMutual(cols, rows)
if (numChanged == -1) {
println("No solution")
return
}
}
while (numChanged > 0)

for (row in rows) {
for (i in 0 until cols.size) {
print(if (row[0][i]) "# " else ". ")
}
println()
}
println()
}

// collect all possible solutions for the given clues
fun getCandidates(data: List<String>, len: Int): BitSets {
val result = mutableListOf<MutableList<BitSet>>()
for (s in data) {
val lst = mutableListOf<BitSet>()
val a = s.toCharArray()
val sumChars = a.sumBy { it - 'A' + 1 }
val prep = a.map { "1".repeat(it - 'A' + 1) }

for (r in genSequence(prep, len - sumChars + 1)) {
val bits = r.substring(1).toCharArray()
val bitset = BitSet(bits.size)
for (i in 0 until bits.size) bitset[i] = bits[i] == '1'
}
}
return result
}

fun genSequence(ones: List<String>, numZeros: Int): List<String> {
if (ones.isEmpty()) return listOf("0".repeat(numZeros))
val result = mutableListOf<String>()
for (x in 1 until numZeros - ones.size + 2) {
val skipOne = ones.drop(1)
for (tail in genSequence(skipOne, numZeros - x)) {
}
}
return result
}

/* If all the candidates for a row have a value in common for a certain cell,
then it's the only possible outcome, and all the candidates from the
corresponding column need to have that value for that cell too. The ones
that don't, are removed. The same for all columns. It goes back and forth,
until no more candidates can be removed or a list is empty (failure).
*/

fun reduceMutual(cols: BitSets, rows: BitSets): Int {
val countRemoved1 = reduce(cols, rows)
if (countRemoved1 == -1) return -1
val countRemoved2 = reduce(rows, cols)
if (countRemoved2 == -1) return -1
return countRemoved1 + countRemoved2
}

fun reduce(a: BitSets, b: BitSets): Int {
var countRemoved = 0
for (i in 0 until a.size) {
val commonOn = BitSet()
commonOn[0] = b.size
val commonOff = BitSet()

// determine which values all candidates of a[i] have in common
for (candidate in a[i]) {
commonOn.and(candidate)
commonOff.or(candidate)
}

// remove from b[j] all candidates that don't share the forced values
for (j in 0 until b.size) {
val fi = i
val fj = j
if (b[j].removeIf { cnd ->
(commonOn[fj] && !cnd[fi]) ||
(!commonOff[fj] && cnd[fi]) }) countRemoved++
if (b[j].isEmpty()) return -1
}
}
return countRemoved
}

val p1 = listOf("C BA CB BB F AE F A B", "AB CA AE GA E C D C")

val p2 = listOf(
"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"
)

val p3 = listOf(
"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"
)

val p4 = listOf(
"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"
)

fun main(args: Array<String>) {
for (puzzleData in listOf(p1, p2, p3, p4)) {
newPuzzle(puzzleData)
}
}
Output:
. # # # . . . .
# # . # . . . .
. # # # . . # #
. . # # . . # #
. . # # # # # #
# . # # # # # .
# # # # # # . .
. . . . # . . .
. . . # # . . .

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

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

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

## Perl

use strict;
use warnings;

my \$file = 'nonogram_problems.txt';
open my \$fd, '<', \$file or die "\$! opening \$file";

while(my \$row = <\$fd> )
{
\$row =~ /\S/ or next;
my \$column = <\$fd>;
my @rpats = makepatterns(\$row);
my @cpats = makepatterns(\$column);
my @rows = ( '.' x @cpats ) x @rpats;
for( my \$prev = ''; \$prev ne "@rows"; )
{
\$prev = "@rows";
try(\@rows, \@rpats);
my @cols = map { join '', map { s/.//; \$& } @rows } 0..\$#cpats;
try(\@cols, \@cpats);
@rows = map { join '', map { s/.//; \$& } @cols } 0..\$#rpats;
}
print "\n", "@rows" =~ /\./ ? "Failed\n" : map { tr/01/.#/r, "\n" } @rows;
}

sub try
{
my (\$lines, \$patterns) = @_;
for my \$i ( 0 .. \$#\$lines )
{
while( \$lines->[\$i] =~ /\./g )
{
for my \$try ( 0, 1 )
{
\$lines->[\$i] =~ s/.\G/\$try/r =~ \$patterns->[\$i] or
\$lines->[\$i] =~ s// 1 - \$try /e;
}
}
}
}

sub makepatterns {
map { qr/^\$_\$/ # convert strings to regex
} map { '[0.]*' # prepend static pattern
. join('[0.]+', # join with static pattern
map { "[1.]{\$_}" # require to match exactly 'n' times
} map { -64+ord # convert letter value to repetition count 'n'
} split // # for each letter in group
)
. '[0.]*' # append static pattern
} split ' ', shift; # for each letter grouping
}
Output:
.###....
##.#....
.###..##
..##..##
..######
#.#####.
######..
....#...
...##...

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

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

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

## Phix

Deduction only, no exhaustive search.

sequence x, y, grid
integer unsolved

function count_grid()
integer res = length(x)*length(y)
for i=1 to length(x) do
for j=1 to length(y) do
res -= grid[i][j]!='?'
end for
end for
return res
end function

for i=ms to me do
if mask[i]!=neat[i] then return 0 end if
end if
end for
return 1
end function

if length(blocks)=0 then
for i=mi to length(neat) do
neat[i] = ' '
end for
if length(res)=0 then
res = neat
else
for i=1 to length(neat) do
if neat[i]!=res[i] then
res[i] = '?'
end if
end for
end if
end if
else
integer b = blocks[1]
blocks = blocks[2..\$]
integer l = (sum(blocks)+length(blocks)-1),
e = length(neat)-l-b
for i=mi to e do
for j=i to i+b-1 do
neat[j] = '#'
end for
if i+b<=length(neat) then
neat[i+b] = ' '
end if
end if
neat[i] = ' '
end for
end if
return res
end function

end function

global function vmask(sequence source, integer column)
string res = repeat(' ',length(source))
for i=1 to length(source) do
res[i] = source[i][column]
end for
return res
end function

function logic()
integer wasunsolved = unsolved
for i=1 to length(x) do
grid[i] = inner(grid[i],x[i])
end for
for j=1 to length(y) do
for i=1 to length(tmp) do
grid[i][j] = tmp[i]
end for
end for
unsolved = count_grid()
return wasunsolved!=unsolved
end function

sequence tests=split("""
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""",'\n')
--Alternatively:
--integer fn = open("nonogram_problems.txt","r")
--tests = get_text(fn,GT_LF_STRIPPED)
--close(fn)

function unpack(string s)
sequence res = split(s)
for i=1 to length(res) do
string ri = res[i]
sequence r = {}
for j=1 to length(ri) do
r &= ri[j]-'A'+1
end for
res[i] = r
end for
return res
end function

for i=1 to length(tests) by 3 do
x = unpack(tests[i])
y = unpack(tests[i+1])
grid = repeat(repeat('?',length(y)),length(x))
unsolved = length(x)*length(y)

while unsolved do
if not logic() then
?"partial"
exit
end if
end while

puts(1,join(grid,"\n")&"\n")
end for
Output:
###
## #
###  ##
##  ##
######
# #####
######
#
##
######
### #  ###
#  ###   #    ###
### ##############
#  #            #
# # ##          ##
#####  ##        ##
#####   #        #
#####  ### ### ###
######## ### ### ###
### #
## #### #
# ### ###
## ####
### ### #    ###
###  ## ##   # ###
##  ## ##    ## ##
## # #  ## # #
# ## #   ####
# # ##     ##
## ##  ########
## ##   ##  ####
# ## ## #   #  #
###  ### #####     #
# # ### #    #    ##
##  ### #    ### ###
# ### ## ########
#### ### ########
# #### ## #####
# #### ##   ##
####  ##   #####
##### ###   #####
#### #          #
#### ##
### ###
#####
##              ###  ##
##              #####  #
##             ########
##    ##### ###########
# #  ##    #    ######
#  ##     #       ###
##        #             #
##     ######         ##
###############    ####
##########  ########
## # #### ###  ######
#################
#################
##################
#   ##############
# # ##############
#####   #########
########
#######

## Prolog

Works with: SWI-Prolog version version 6.5.3

module(clpfd) is written by Markus Triska
Solution written by Lars Buitinck

Module solve-nonogram.pl

/*
* 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).

File nonogram.pl, used to read data in a file.

nonogram :-
open('C:/Users/Utilisateur/Documents/Prolog/Rosetta/nonogram/nonogram.txt',
repeat,
compute_values(Line_1, [], [], Lines),
compute_values(Line_2, [], [], Columns),
nonogram(Lines, Columns) , nl, nl,
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).

## 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.

### Python 2

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]:
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]:
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():
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()
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. ...)

### Python 3

Above code altered to work with Python 3:

from functools import reduce

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 range(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 zip(a, b)], row)

def fits(a, b):
return all(x & y for x, y in zip(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]:
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]:
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 = list(map(allowable, rows))

# Initially mark all columns for update.
mod_rows, mod_cols = set(), set(range(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 range(w) for i in range(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 range(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(s, 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])

## Racket

[See Example:Nonogram solver/Racket for editing of this section]

There are a number of functional parts to this...

• parsing the files
• getting the data into a usable state
• rendering (for debug and prettiness)
• the actual solution of the Nonograms themselves
• as well as the ubiquitous test cases

So we'll have it here on its own page

#lang racket
;;; --------------------------------------------------------------------------------------------------
;;; Representaiton is done with bits: the bit patterns for a block being:
;;; -------------------------------------------------------------------------
;;; #b00 (0) - block is not in representation (also a terminator on the left)
;;; #b01 (1) - block is white
;;; #b10 (2) - block is black
;;; #b11 (3) - block is undecided
;;; None of the problems has > 32 columns, so 64-bits should be fine
;;; If we go above 64-bits, then a. we have a difficult problem
;;; b. racket will use bignums rather
;;; than fixnums
;;;
;;; A "blocks" is an integer formed of two-bit block codes (above)
;;;
;;; A "representation" is a sequence (list) of black stretch lengths; which need to be separated by at
;;; least one white between, and optionally prefixed and suffied with whites
;;;
;;; A "candidate" is a sequence (list) of <white-length [black-length white-length]...>, specifying
;;; one instance of a "representation".
;;;
;;; A "puzzle" is a sequence (vector) of blocks
;;; -- if the puzzle is <= 32 blocks wide, this could well be an fxvector (but ignore that
;;; possibility for now)
;;;
;;; "Options" is a sequence (list) of blocks
;;;
;;; white is often abbreviated (in variables etc. to W), black to "B"
;;; --------------------------------------------------------------------------------------------------
(module+ test (require rackunit))
(define *problems-file-name* "nonogram_problems.txt")

;;; --------------------------------------------------------------------------------------------------
;;; Parsing Input
;;; --------------------------------------------------------------------------------------------------
(define (letter-rep->number-rep c) (+ 1 (- (char->integer (char-upcase c)) (char->integer #\A))))

;; takes the letters representation, returns a list of list of numbers. The list returned is an
;; "option" - a list of ([white-width black-width] ... white-width)
(define (letters-rep->list²-rep s)
(for/list ((b (regexp-split #rx" +" s)))
(map letter-rep->number-rep (string->list b))))

[(? eof-object?) #f]
[(? string? l)
(vector (letters-rep->list²-rep l)

(module+ test
(check-equal? (map letter-rep->number-rep '(#\A #\a #\B #\C)) '(1 1 2 3))
(check-equal? (letters-rep->list²-rep "C BA CB BB F AE F A B")
'([3] [2 1] [3 2] [2 2] [6] [1 5] [6] [1] [2]))
(check-equal? (letters-rep->list²-rep "AB CA AE GA E C D C")
'([1 2] [3 1] [1 5] [7 1] [5] [3] [4] [3]))
#(([3] [2 1] [3 2] [2 2] [6] [1 5] [6] [1] [2])
([1 2] [3 1] [1 5] [7 1] [5] [3] [4] [3]))))

;;; --------------------------------------------------------------------------------------------------
;;; Generate Candidates
;;; --------------------------------------------------------------------------------------------------
(define (rep->candidates n-cells blacks)
(define (inr cells-remain bs leftmost?)
(define bs-l (sequence-length bs))
(define min-space-needed (- (apply + bs-l bs) 1))
(cond
[(null? bs) (list (list cells-remain))]
[(> min-space-needed cells-remain) null]
[else
(define initial-whites-min-size (if leftmost? 0 1))
(define intial-whites-range
(in-range initial-whites-min-size (add1 (- cells-remain min-space-needed))))
(for*/list ((intial-whites intial-whites-range)
(tl (in-list (inr (- cells-remain intial-whites (car bs)) (cdr bs) #f))))
(list* intial-whites (car bs) tl))]))
(inr n-cells blacks #t))

(module+ test
(check-match
(rep->candidates 5 '(1)) (list-no-order '(0 1 4) '(1 1 3) '(2 1 2) '(3 1 1) '(4 1 0)))
(check-match
(rep->candidates 5 '(1 1))
(list-no-order '(0 1 1 1 2) '(0 1 2 1 1) '(0 1 3 1 0) '(1 1 1 1 1) '(1 1 2 1 0) '(2 1 1 1 0))))

(define (make-Ws l) (for/fold ((rv 0)) ((_ l)) (+ (* 4 rv) #b01)))
(define (make-Bs l) (* 2 (make-Ws l)))
(module+ test
(check-eq? (make-Ws 0) #b00)
(check-eq? (make-Bs 0) #b00)
(check-eq? (make-Ws 1) #b01)
(check-eq? (make-Bs 1) #b10)
(check-eq? (make-Ws 3) #b010101)
(check-eq? (make-Bs 3) #b101010))

(define (candidate->blocks cand)
(define (inr cand rv)
(match cand
[(list (and W (app make-Ws Ws)) (and B (app make-Bs Bs)) r ...)
(inr r (+ (arithmetic-shift rv (* 2 (+ B W))) (arithmetic-shift Ws (* 2 B)) Bs))]
[(list (and W (app make-Ws Ws))) (+ (arithmetic-shift rv (* 2 W)) Ws)]))
(inr cand 0))

(module+ test
(check-eq? (candidate->blocks '(0)) 0)
(check-eq? (candidate->blocks '(1)) #b01)
(check-eq? (candidate->blocks '(1 1 1)) #b011001)
(check-equal?
(map candidate->blocks
'((0 1 1 1 2) (0 1 2 1 1) (0 1 3 1 0) (1 1 1 1 1) (1 1 2 1 0) (2 1 1 1 0)))
'(#b1001100101 #b1001011001 #b1001010110 #b0110011001 #b0110010110 #b0101100110)))

;; Given a (non-empty) sequence of blocks return a list of blocks which must be black, must be
;; white or have to be dertermined another way (through matching along the other axis).
(define (find-definite-blocks blocks)
(for/fold ((known (sequence-ref blocks 0))) ((merge blocks))
(bitwise-ior known merge)))

(module+ test
(check-eq? (find-definite-blocks '(#b010101 #b010110 #b100110)) #b110111)
)

;; returns the list of blocks (from options) that can be overlaid over the solution
;; this means that the following must hold false for all bits (if it holds false, we can do a zero?
;; test, which is easiser than an all-significant-bits-set? test:
;; pattern cand
;; 0 0 F
;; 0 1 T
;; 1 0 F
;; 1 1 F
;; (cand !bitwise-impl pattern) = !(pattern | !cand) = (!pattern & cand)
(define (filter-against-partial-solution part-sltn options)
(define not-part-sltn (bitwise-not part-sltn))
(define (option-fits? cand) (zero? (bitwise-and cand not-part-sltn)))
(filter option-fits? options))

(module+ test
(check-equal?
(filter-against-partial-solution #b011110 '(#b101010 #b010110 #b011010))
'(#b010110 #b011010)))

;;; --------------------------------------------------------------------------------------------------
;;; Rendering -- it's pretty tough to see what's going on, when you have no pictures!
;;; --------------------------------------------------------------------------------------------------
(define ((render-puzzle knil kons-W kons-B kons-_ compose-lines) pzl)
(define (render-blocks bs)
(define (inr bs acc)
(match (bitwise-and bs #b11)
[#b00 acc]
[#b01 (inr (arithmetic-shift bs -2) (kons-W acc))]
[#b10 (inr (arithmetic-shift bs -2) (kons-B acc))]
[#b11 (inr (arithmetic-shift bs -2) (kons-_ acc))]))
(inr bs knil))
(compose-lines (map render-blocks pzl)))

(define string-render-puzzle
(render-puzzle ""
(curry string-append ".")
(curry string-append "#")
(curry string-append "?")
(curryr string-join "\n")))

(module+ test
(check-equal? (string-render-puzzle '(#b101010 #b010101 #b111111)) "###\n...\n???"))

;;; We need to work on x and y blocks uniformly, so this will convert from one to t'other
;;; Rotates a sequence of blocks
(define (rotate-blocks x-blocks)
(define x-width- (integer-length (sequence-ref x-blocks 0)))
(define x-width (if (odd? x-width-) (add1 x-width-) x-width-))
;(printf "~a ~a" x-width x-blocks)
(for/list ((block-idx (in-range x-width 0 -2)))
(for/fold ((y-block 0))
((x-block x-blocks))
(+ (arithmetic-shift y-block 2)
(bitwise-bit-field x-block (- block-idx 2) block-idx)))))

(module+ test
(check-equal? (rotate-blocks '(#b1110 #b0111)) '(#b1101 #b1011))
(check-equal? (rotate-blocks '(#b0110 #b0111)) '(#b0101 #b1011)))

;;; --------------------------------------------------------------------------------------------------
;;; SOLVER (finally!):
;;; --------------------------------------------------------------------------------------------------
;;; solve the nonogram... both "solvers" return as values:
;;; solution-changed? did the solution change -- if not we either have a solution or as good a
;;; solution as the program can provide
;;; new-solution the newly-changed solution
;;; new-x-blocks x-blocks that are now available as candidates
;;; new-y-blocks y-blocks that are now available as candidates
(define (solved? blocks) (for/and ((b blocks)) (= (sequence-length b) 1)))

(define (solve-nonogram x-rep.y-rep) ; pair of reps as read from e.g. read-nonogram-from-file
(match-define (vector x-rep y-rep) x-rep.y-rep)
(define width (sequence-length y-rep))
(define height (sequence-length x-rep))
(define x-candidates (map (curry rep->candidates width) x-rep))
(define y-candidates (map (curry rep->candidates height) y-rep))
(define x-options (for/list ((cnds x-candidates)) (map candidate->blocks cnds)))
(define y-options (for/list ((cnds y-candidates)) (map candidate->blocks cnds)))

(define-values (solution complete?) (sub-solve-nonogram x-options y-options))
(unless complete? (displayln "INCOMPLETE SOLUTION"))
solution)

(define (sub-solve-nonogram x-options y-options)
(define known-x (map find-definite-blocks x-options))
(define known-y (map find-definite-blocks y-options))
(cond
[(solved? x-options) (values known-x #t)]
[else
(define new-y-options (map filter-against-partial-solution (rotate-blocks known-x) y-options))
(define new-x-options (map filter-against-partial-solution (rotate-blocks known-y) x-options))
(displayln (string-render-puzzle (map find-definite-blocks new-x-options)) (current-error-port))
(newline (current-error-port))
(if (and (equal? x-options new-x-options) (equal? y-options new-y-options))
(values known-x #f) ; oh... we can't get any further
(sub-solve-nonogram new-x-options new-y-options))]))

;;;; TESTING
(module+ test
(define chicken #<<EOS
.###....
##.#....
.###..##
..##..##
..######
#.#####.
######..
....#...
...##...
EOS
)
(check-equal?
chicken))

;;; IMAGE RENDERING
(require pict racket/gui/base)

(define *cell-size* 10)
(define ((paint-cell fill-colour) dc dx dy)
(define C (- *cell-size* 1))
(define old-brush (send dc get-brush))
(define old-pen (send dc get-pen))
(define path (new dc-path%))
(send path rectangle 0 0 C C)
(send* dc
(set-brush (new brush% [color fill-colour]))
(set-pen (new pen% [width 1] [color "black"]))
(draw-path path dx dy)
(set-brush old-brush)
(set-pen old-pen)))

(define (draw-cell fill-colour)
(dc (paint-cell fill-colour)
*cell-size* *cell-size*))

(define ((row-append-cell colour) row-so-far)
(hc-append (draw-cell colour) row-so-far))

(define image-render-puzzle
(render-puzzle
(blank)
(row-append-cell "white")
(row-append-cell "black")
(row-append-cell "yellow")
(λ (rows) (apply vc-append rows))))

(module+ test
; test though visual inspection... it's a style thing, really

;;; MAIN
(module+ main
(unless (directory-exists? "images") (make-directory "images"))

(call-with-input-file *problems-file-name*
(lambda (prt)
(let loop ((idx 1) (pzl (read-nonogram-description prt)) (collage (blank)))
(cond
[pzl
(define img (image-render-puzzle (solve-nonogram pzl)))
(send (pict->bitmap img) save-file (build-path "images" (format "nonogram-~a.png" idx))
'png)
(displayln (image-render-puzzle (solve-nonogram pzl)))
[else
(send (pict->bitmap collage) save-file (build-path "images" (format "nonogram-all.png"))
'png)
(displayln collage)])))))
Output:

I have little luck with attaching images. You may need to run the program in Racket and, if you can, attach some pictures here!

Ah... foo! Here's an external link to one I did earlier... [[1]]. Life's too short to wrestle with Rosetta Code...

After changing the:

(displayln (image-render-puzzle (solve-nonogram pzl)))

to

(displayln (string-render-puzzle (solve-nonogram pzl)))

at the end of the main module part, we get the following on standard out (stderr has been redirected to /dev/null)

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

## Rexx

Nonogram Solver/Rexx:

/* */
Parse Arg fn
Parse Var fn ou'.'
maxpn = 10000               /* maximum posibilites to check through */
output = ou'.out.txt'
/* read row/col values into rowpp. and colpp. arrays */
cc = linein(fn)
rows = words(cc)
dd = linein(fn)
cols = words(dd)
char = '0ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijk'
cntr = 0
Do i = 1 To rows
rowpp.i = CV(cc,i)
cntr = cntr + sum
End
cntc = 0
Do i = 1 To cols
colpp.i = CV(dd,i)
cntc = cntc + sum
End
If (cntr <> cntc)|(cntr = 0) Then Do
Say 'error Sum of rows <> sum of cols'
Exit 999
End
Say cntr 'colored cells'
ar = copies('-',rows*cols)
/* values are -=unknown .=blank @=Color */
/* PREFILL  array */
'erase' output
/**********COL PREFILL ************/
Do col = 1 To cols
r = colpp.col
Parse Var r z r
Do While r <> ''
Parse Var r q r
z = z + q + 1
End
result = copies('-',rows)
If z = rows Then result = FILL_LINE(colpp.col)
Else If z = 0 Then result = copies('.',rows)
Do row = 1 To rows
ar = overlay(substr(result,row,1),ar,(row-1)*cols+col)
End
End
/**********ROW PREFILL ************/
Do row = 1 To rows
c = rowpp.row
Parse Var c t c
Do While c <> ''
Parse Var c q c
t = t + q + 1
End
result = substr(ar,(row-1)*cols+1,cols)
If t = cols Then result = left(FILL_LINE(rowpp.row),cols)
Else If t = 0 Then result = copies('.',cols)
ar = overlay(result,ar,(row-1)*cols+1)
End
/********** ok here we loop ************/
cnttry = 1
nexttry = 2
next.cnttry = ar
sol = 0
Do label nextpos While cnttry < nexttry
Say 'trying' cnttry 'of' nexttry-1
ar = next.cnttry
cnttry = cnttry + 1
Do Until sar = ar
sar = ar
Do row = 1 To rows
/**********process rows ************/
rowcol = substr(ar,(row-1)*cols+1,cols)
pp = rowpp.row
If PROCESSROW() Then Iterate nextpos
Else ar = overlay(left(rowcol,cols),ar,(row-1)*cols+1)
End
Do col = 1 To cols
rowcol = ''
Do row = 1 To rows
rowcol = rowcol || substr(ar,(row-1)*cols+ col,1)
End
pp = colpp.col
If PROCESSROW() Then Iterate nextpos
Do row = 1 To rows
ar = overlay(substr(rowcol,row,1),ar,(row-1)*cols + col)
End
End
If pos('-',ar) = 0 Then Do       /* hurray we have a solution */
/* at this point we need to verify solution */
If CHECKBOARD() Then Iterate nextpos   /* too bad didn't match */
sol = sol + 1
Call LINEOUT output,'This is solution no:' sol
Call DUMPBOARD
Iterate nextpos
End
If sar = ar Then Do
fnd = pos('-',ar)
next.nexttry = overlay('.',ar,fnd)
nexttry = nexttry + 1
ar = overlay('@',ar,fnd)
End
End
End nextpos
If sol = 0 Then sol = 'No'
Say sol 'solutions found'
Exit

CHECKBOARD:
Do row = 1 To rows
/**********process rows ************/
rowcol = substr(ar,(row-1)*cols+1,cols)
pp = rowpp.row
If CHECKROW() Then Return 1
End
Do col = 1 To cols
rowcol = ''
Do row = 1 To rows
rowcol = rowcol || substr(ar,(row-1)*cols+ col,1)
End
pp = colpp.col
If CHECKROW() Then Return 1
End
Return 0                                               /* we did it */

CHECKROW:
len_item = length(rowcol)
st = 1
If pp = 0 Then Return rowcol <> copies('.',len_item)
Else If pp = len_item Then Return rowcol <> copies('@',len_item)
Do While (pp <> '') & (st <= len_item)
Parse Var pp p1 pp
of = pos('@',rowcol'@',st)
If of > len_item Then Return 1
If substr(rowcol,of,p1) <> copies('@',p1) Then Return 1
st = of + p1
If substr(rowcol'.',st,1) <> '.' Then Return 1
End
Return 0

DUMPBOARD:
Parse Arg qr
p = '..'
q = '..'
Do i = 1 To cols
n = right(i,2)
p = p left(n,1)
q = q right(n,1)
End
Call LINEOUT output, p
Call LINEOUT output, q
Do i = 1 To rows
o = right(i,2)
p = substr(ar,(i-1)*cols+1,cols)
Do j = 1 To cols
Parse Var p z +1 p
o = o z
End
Call LINEOUT output, o
End
Return

FILL_LINE:
Parse Arg items
oo = ''
Do While items <> ''
Parse Var items a items
oo = oo||copies('@',a)'.'
End
Return oo

CV:
Parse Arg cnts, rwcl
str = word(cnts,rwcl)
ret = ''
sum = 0
Do k = 1 To length(str)
this = pos(substr(str,k,1),char)-1
ret = ret this
sum = sum + this
End
Return space(ret)

PROCESSROW:                           /* rowcol pp in, rowcol pp of ol */
prerow = rowcol
len_item = length(rowcol)
If pos('-',rowcol) = 0 Then Do
pp = ''
Return 0
End
of = 1
kcnt = 0
/* reduce the left side with already populated values */
Do While (of < len_item) & (pp <> '')
kcnt = kcnt + 1
If kcnt > len_item Then Return 1
If substr(rowcol,of,1) = '.' Then Do
k = verify(substr(rowcol,of)'%','.')
of = of + k - 1
Iterate
End
nl = word(pp,1)
len = verify(substr(rowcol,of)'%','[email protected]') - 1
If len < nl Then Do
rowcol = overlay(copies('.',len),rowcol,of)
of = of + len
Iterate
End
If (len = nl) & (pos('@',substr(rowcol,of,nl))>0) Then Do
rowcol = overlay(copies('@',nl),rowcol,of)
of = of + nl
pp = subword(pp,2)
Iterate
End
If substr(rowcol,of,1) = '@' Then Do
rowcol = overlay(copies('@',nl)'.',rowcol,of)
of = of + nl
pp = subword(pp,2)
Iterate
End
Leave
End
/* reduce the right side with already populated values */
ofm = len_item + 1 - of
ol = 1
kcnt = 0
Do While (ol < ofm) & (pp <> '')
kcnt = kcnt + 1
If kcnt > len_item Then Return 1
revrow = reverse(rowcol)
If substr(revrow,ol,1) = '.' Then Do
k = verify(substr(revrow,ol)'%','.')
ol = ol + k - 1
Iterate
End
nl = word(pp,words(pp))
len = verify(substr(revrow,ol)'%','[email protected]') - 1
If len < nl Then Do
rowcol = overlay(copies('.',len),rowcol,len_item-ol-len+2)
ol = ol + len
Iterate
End
If (len = nl) & (pos('@',substr(revrow,ol,nl))>0) Then Do
rowcol = overlay(copies('@',nl),rowcol,len_item-ol-nl+2)
ol = ol + nl
pp = subword(pp,1,words(pp)-1)
Iterate
End
If substr(revrow,ol,1) = '@' Then Do
rowcol = overlay('.'copies('@',nl),rowcol,len_item-ol-nl+1)
ol = ol + nl
pp = subword(pp,1,words(pp)-1)
Iterate
End
Leave
End
If pp = 0 Then pp = ''
If pp = '' Then rowcol = changestr('-',rowcol,'.')
If pp <> '' Then Do
lv = len_item-of-ol+2
pos. = ''
pn = 0
pi = substr(rowcol,of,lv)
If (copies('-',length(pi)) = pi) Then Do
len = CNT(pp)
If (len + mx) <= lv Then Do
Return 0
End
End
/* oh oh need to check for posibilities */
Call TRY '',pp
If pn > maxpn Then Do
over = over + 1
Return 0
End
fnd = 0
fu = pos.1
Do z = 2 To pn
Do j = 1 To lv
If substr(fu,j,1) <> substr(pos.z,j,1) Then fu = overlay('-',fu,j)
End
End
Do z = 1 To lv
If substr(fu,z,1) <> '-' Then rowcol = overlay(substr(fu,z,1),rowcol,of+z-1)
End
End
Return 0
TRY: Procedure Expose pn pos. maxpn lv pi
Parse Arg prev,pp
If pp = '' Then Do
rem = substr(pi,length(prev)+1)
If translate(rem,'..','.-') <> copies('.',length(rem)) Then Return
prev = left(prev||copies('.',lv),lv)
pn = pn + 1
If pn > maxpn Then Return
pos.pn = prev
Return
End
Parse Var pp p1 pp
If length(prev)+p1 > lv Then Return
Do i = 0 To lv - length(prev)-p1
If translate(substr(pi,length(prev)+1,i),'..','.-') = copies('.',i) Then
If translate(substr(pi,length(prev)+i+1,p1),'@@','@-') = copies('@',p1) Then
If substr(pi,length(prev)+i+p1+1,1) <> '@' Then
Call TRY prev||copies('.',i)||copies('@',p1)'.',pp
End
Return
CNT: Procedure Expose mx
Parse Arg len items
mx = len
Do While items <> ''
Parse Var items ii items
len = len + ii + 1
If ii > mx Then mx = ii
End
Return len

Puzzle
C BA CB BB F AE F A B
AB CA AE GA E C D C
This is solution no: 1
..
.. 1 2 3 4 5 6 7 8
1 . @ @ @ . . . .
2 @ @ . @ . . . .
3 . @ @ @ . . @ @
4 . . @ @ . . @ @
5 . . @ @ @ @ @ @
6 @ . @ @ @ @ @ .
7 @ @ @ @ @ @ . .
8 . . . . @ . . .
9 . . . @ @ . . .

Puzzle
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
This is solution no: 1
..                   1 1 1 1 1 1 1 1 1 1 2
.. 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0
1 . . . . . . . . . . @ @ @ @ @ @ . . . .
2 . . . . . . . . @ @ @ . @ . . @ @ @ . .
3 . . . @ . . @ @ @ . . . @ . . . . @ @ @
4 . . @ @ @ . @ @ @ @ @ @ @ @ @ @ @ @ @ @
5 . . . @ . . @ . . . . . . . . . . . . @
6 . . @ . @ . @ @ . . . . . . . . . . @ @
7 @ @ @ @ @ . . @ @ . . . . . . . . @ @ .
8 @ @ @ @ @ . . . @ . . . . . . . . @ . .
9 @ @ @ @ @ . . @ @ @ . @ @ @ . @ @ @ . .
10 @ @ @ @ @ @ @ @ . @ @ @ . @ @ @ . @ @ @

Puzzle
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

This is solution no: 1
..                   1 1 1 1 1 1 1 1 1 1 2
.. 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0
1 . . . . @ @ @ . @ . . . . . . . . . . .
2 . . . . @ @ . @ @ @ @ . @ . . . . . . .
3 . . . . @ . @ @ @ . @ @ @ . . . . . . .
4 . . @ @ . @ @ @ @ . . . . . . . . . . .
5 . @ @ @ . @ @ @ . @ . . . . @ @ @ . . .
6 @ @ @ . . @ @ . @ @ . . . @ . @ @ @ . .
7 @ @ . . @ @ . @ @ . . . . @ @ . @ @ . .
8 . . . . @ @ . @ . @ . . @ @ . @ . @ . .
9 . . . . @ . @ @ . @ . . . @ @ @ @ . . .
10 . . . . @ . @ . @ @ . . . . . @ @ . . .
11 . . . . . @ @ . @ @ . . @ @ @ @ @ @ @ @
12 . . . . @ @ . @ @ . . . @ @ . . @ @ @ @
13 . . . . @ . @ @ . @ @ . @ . . . @ . . @
14 @ @ @ . . @ @ @ . @ @ @ @ @ . . . . . @
15 @ . @ . @ @ @ . @ . . . . @ . . . . @ @
16 @ @ . . @ @ @ . @ . . . . @ @ @ . @ @ @
17 . @ . @ @ @ . @ @ . @ @ @ @ @ @ @ @ . .
18 . @ @ @ @ . @ @ @ . @ @ @ @ @ @ @ @ . .
19 . . . @ . @ @ @ @ . @ @ . @ @ @ @ @ . .
20 . . . @ . @ @ @ @ . @ @ . . . @ @ . . .
21 . . . . @ @ @ @ . . @ @ . . . @ @ @ @ @
22 . . . @ @ @ @ @ . @ @ @ . . . @ @ @ @ @
23 . . . @ @ @ @ . @ . . . . . . . . . . @
24 . . @ @ @ @ . @ @ . . . . . . . . . . .
25 . . @ @ @ . @ @ @ . . . . . . . . . . .

Puzzle
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
This is solution no: 1
..                   1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2
.. 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5
1 . . . . . . . . . . . . . . . . . . . . @ @ @ @ @
2 . . @ @ . . . . . . . . . . . . . . @ @ @ . . @ @
3 . @ @ . . . . . . . . . . . . . . @ @ @ @ @ . . @
4 @ @ . . . . . . . . . . . . . @ @ @ @ @ @ @ @ . .
5 @ @ . . . . @ @ @ @ @ . @ @ @ @ @ @ @ @ @ @ @ . .
6 @ . @ . . @ @ . . . . @ . . . . @ @ @ @ @ @ . . .
7 @ . . @ @ . . . . . @ . . . . . . . @ @ @ . . . .
8 @ @ . . . . . . . . @ . . . . . . . . . . . . . @
9 . @ @ . . . . . @ @ @ @ @ @ . . . . . . . . . @ @
10 . . @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ . . . . @ @ @ @
11 . . . . . @ @ @ @ @ @ @ @ @ @ . . @ @ @ @ @ @ @ @
12 . . . . @ @ . @ . @ @ @ @ . @ @ @ . . @ @ @ @ @ @
13 . . . . . . . . @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @
14 . . . . . . . . @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @
15 . . . . . . . @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @
16 . . . . . . . @ . . . @ @ @ @ @ @ @ @ @ @ @ @ @ @
17 . . . . . . . @ . @ . @ @ @ @ @ @ @ @ @ @ @ @ @ @
18 . . . . . . . . @ @ @ @ @ . . . @ @ @ @ @ @ @ @ @
19 . . . . . . . . . . . . . . . . . @ @ @ @ @ @ @ @
20 . . . . . . . . . . . . . . . . . . @ @ @ @ @ @ @

Puzzle
GCAAG AABBAA ACACAACA ACAAFACA ACAEBACA AABAA GAAAAAG CC ABCAACAAB AACBAA DADBAB AAAAADAC BAAABE CBBFCA AIAABA BABBCA CAAAAEA ABBE GABAAAC AABABBA ACADEA ACACJB ACAAFF AABAAB GBABE
This is solution no: 1
..                   1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2
.. 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5
1 @ @ @ @ @ @ @ . @ @ @ . . . @ . @ . @ @ @ @ @ @ @
2 @ . . . . . @ . @ @ . @ @ . . . . . @ . . . . . @
3 @ . @ @ @ . @ . . . . . @ @ @ . @ . @ . @ @ @ . @
4 @ . @ @ @ . @ . @ . . @ @ @ @ @ @ . @ . @ @ @ . @
5 @ . @ @ @ . @ . . @ @ @ @ @ . @ @ . @ . @ @ @ . @
6 @ . . . . . @ . . @ @ . . . . . . . @ . . . . . @
7 @ @ @ @ @ @ @ . @ . @ . @ . @ . @ . @ @ @ @ @ @ @
8 . . . . . . . . @ @ @ . . . @ @ @ . . . . . . . .
9 @ . @ @ . @ @ @ . . @ . @ . @ @ @ . @ . . @ . @ @
10 @ . @ . . . . . . @ @ @ . @ @ . . . . @ . . . @ .
11 . @ @ @ @ . @ . @ @ @ @ . @ @ . @ . . . . @ @ . .
12 . @ . @ . . . @ . . . @ . @ . @ @ @ @ . @ . @ @ @
13 . . @ @ . . @ . @ . @ . . . . . . @ @ . @ @ @ @ @
14 . . . @ @ @ . @ @ . @ @ . @ @ @ @ @ @ . @ @ @ . @
15 @ . @ @ @ @ @ @ @ @ @ . @ . @ . . @ @ . . . . @ .
16 . @ @ . @ . . @ @ . . @ @ . . @ @ @ . . . . . @ .
17 @ @ @ . @ . @ . @ . . . . @ . . @ @ @ @ @ . @ . .
18 . . . . . . . . @ . . @ @ . . @ @ . . . @ @ @ @ @
19 @ @ @ @ @ @ @ . @ . . . @ @ . . @ . @ . @ . @ @ @
20 @ . . . . . @ . @ @ . . @ . . @ @ . . . @ @ . @ .
21 @ . @ @ @ . @ . . . @ @ @ @ . . @ @ @ @ @ . . @ .
22 @ . @ @ @ . @ . @ @ @ . @ @ @ @ @ @ @ @ @ @ . @ @
23 @ . @ @ @ . @ . @ . . @ @ @ @ @ @ . @ @ @ @ @ @ .
24 @ . . . . . @ . . @ @ . . . . . . @ . @ . @ @ . .
25 @ @ @ @ @ @ @ . @ @ . . . @ . @ @ . . . @ @ @ @ @
This is solution no: 2
..                   1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2
.. 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5
1 @ @ @ @ @ @ @ . @ @ @ . . . @ . @ . @ @ @ @ @ @ @
2 @ . . . . . @ . @ @ . @ @ . . . . . @ . . . . . @
3 @ . @ @ @ . @ . . . . . @ @ @ . @ . @ . @ @ @ . @
4 @ . @ @ @ . @ . @ . . @ @ @ @ @ @ . @ . @ @ @ . @
5 @ . @ @ @ . @ . . @ @ @ @ @ . @ @ . @ . @ @ @ . @
6 @ . . . . . @ . . @ @ . . . . . . . @ . . . . . @
7 @ @ @ @ @ @ @ . @ . @ . @ . @ . @ . @ @ @ @ @ @ @
8 . . . . . . . . @ @ @ . . . @ @ @ . . . . . . . .
9 @ . @ @ . @ @ @ . . @ . @ . @ @ @ . . @ . @ . @ @
10 @ . @ . . . . . . @ @ @ . @ @ . . . @ . . . . @ .
11 . @ @ @ @ . @ . @ @ @ @ . @ @ . @ . . . . @ @ . .
12 . @ . @ . . . @ . . . @ . @ . @ @ @ @ . @ . @ @ @
13 . . @ @ . . @ . @ . @ . . . . . . @ @ . @ @ @ @ @
14 . . . @ @ @ . @ @ . @ @ . @ @ @ @ @ @ . @ @ @ . @
15 @ . @ @ @ @ @ @ @ @ @ . @ . @ . . @ @ . . . . @ .
16 . @ @ . @ . . @ @ . . @ @ . . @ @ @ . . . . . @ .
17 @ @ @ . @ . @ . @ . . . . @ . . @ @ @ @ @ . @ . .
18 . . . . . . . . @ . . @ @ . . @ @ . . . @ @ @ @ @
19 @ @ @ @ @ @ @ . @ . . . @ @ . . @ . @ . @ . @ @ @
20 @ . . . . . @ . @ @ . . @ . . @ @ . . . @ @ . @ .
21 @ . @ @ @ . @ . . . @ @ @ @ . . @ @ @ @ @ . . @ .
22 @ . @ @ @ . @ . @ @ @ . @ @ @ @ @ @ @ @ @ @ . @ @
23 @ . @ @ @ . @ . @ . . @ @ @ @ @ @ . @ @ @ @ @ @ .
24 @ . . . . . @ . . @ @ . . . . . . @ . @ . @ @ . .
25 @ @ @ @ @ @ @ . @ @ . . . @ . @ @ . . . @ @ @ @ @
This is solution no: 3
..                   1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2
.. 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5
1 @ @ @ @ @ @ @ . @ @ @ . . . @ . @ . @ @ @ @ @ @ @
2 @ . . . . . @ . @ @ . @ @ . . . . . @ . . . . . @
3 @ . @ @ @ . @ . . . . . @ @ @ . @ . @ . @ @ @ . @
4 @ . @ @ @ . @ . @ . . @ @ @ @ @ @ . @ . @ @ @ . @
5 @ . @ @ @ . @ . . @ @ @ @ @ . @ @ . @ . @ @ @ . @
6 @ . . . . . @ . . @ @ . . . . . . . @ . . . . . @
7 @ @ @ @ @ @ @ . @ . @ . @ . @ . @ . @ @ @ @ @ @ @
8 . . . . . . . . @ @ @ . . . @ @ @ . . . . . . . .
9 @ . @ @ . @ @ @ . . @ . @ . @ @ @ . @ . . @ . @ @
10 @ . @ . . . . . . @ @ @ . @ @ . . . . @ . . . @ .
11 . @ @ @ @ . @ . @ @ @ @ . @ @ . @ . . . . @ @ . .
12 . @ . @ . . . @ . . . @ . @ . @ @ @ @ . @ . @ @ @
13 . . @ @ . . @ . @ . @ . . . . . . @ @ . @ @ @ @ @
14 . . . @ @ @ . @ @ . @ @ . @ @ @ @ @ @ . @ @ @ . @
15 @ . @ @ @ @ @ @ @ @ @ . @ . @ . . @ @ . . . . @ .
16 . @ @ . @ . . @ @ . . . @ @ . @ @ @ . . . . . @ .
17 @ @ @ . @ . @ . @ . . @ . . . . @ @ @ @ @ . @ . .
18 . . . . . . . . @ . . . @ @ . @ @ . . . @ @ @ @ @
19 @ @ @ @ @ @ @ . @ . . @ @ . . . @ . @ . @ . @ @ @
20 @ . . . . . @ . @ @ . . @ . . @ @ . . . @ @ . @ .
21 @ . @ @ @ . @ . . . @ @ @ @ . . @ @ @ @ @ . . @ .
22 @ . @ @ @ . @ . @ @ @ . @ @ @ @ @ @ @ @ @ @ . @ @
23 @ . @ @ @ . @ . @ . . @ @ @ @ @ @ . @ @ @ @ @ @ .
24 @ . . . . . @ . . @ @ . . . . . . @ . @ . @ @ . .
25 @ @ @ @ @ @ @ . @ @ . . . @ . @ @ . . . @ @ @ @ @
This is solution no: 4
..                   1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2
.. 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5
1 @ @ @ @ @ @ @ . @ @ @ . . . @ . @ . @ @ @ @ @ @ @
2 @ . . . . . @ . @ @ . @ @ . . . . . @ . . . . . @
3 @ . @ @ @ . @ . . . . . @ @ @ . @ . @ . @ @ @ . @
4 @ . @ @ @ . @ . @ . . @ @ @ @ @ @ . @ . @ @ @ . @
5 @ . @ @ @ . @ . . @ @ @ @ @ . @ @ . @ . @ @ @ . @
6 @ . . . . . @ . . @ @ . . . . . . . @ . . . . . @
7 @ @ @ @ @ @ @ . @ . @ . @ . @ . @ . @ @ @ @ @ @ @
8 . . . . . . . . @ @ @ . . . @ @ @ . . . . . . . .
9 @ . @ @ . @ @ @ . . @ . @ . @ @ @ . . @ . @ . @ @
10 @ . @ . . . . . . @ @ @ . @ @ . . . @ . . . . @ .
11 . @ @ @ @ . @ . @ @ @ @ . @ @ . @ . . . . @ @ . .
12 . @ . @ . . . @ . . . @ . @ . @ @ @ @ . @ . @ @ @
13 . . @ @ . . @ . @ . @ . . . . . . @ @ . @ @ @ @ @
14 . . . @ @ @ . @ @ . @ @ . @ @ @ @ @ @ . @ @ @ . @
15 @ . @ @ @ @ @ @ @ @ @ . @ . @ . . @ @ . . . . @ .
16 . @ @ . @ . . @ @ . . . @ @ . @ @ @ . . . . . @ .
17 @ @ @ . @ . @ . @ . . @ . . . . @ @ @ @ @ . @ . .
18 . . . . . . . . @ . . . @ @ . @ @ . . . @ @ @ @ @
19 @ @ @ @ @ @ @ . @ . . @ @ . . . @ . @ . @ . @ @ @
20 @ . . . . . @ . @ @ . . @ . . @ @ . . . @ @ . @ .
21 @ . @ @ @ . @ . . . @ @ @ @ . . @ @ @ @ @ . . @ .
22 @ . @ @ @ . @ . @ @ @ . @ @ @ @ @ @ @ @ @ @ . @ @
23 @ . @ @ @ . @ . @ . . @ @ @ @ @ @ . @ @ @ @ @ @ .
24 @ . . . . . @ . . @ @ . . . . . . @ . @ . @ @ . .
25 @ @ @ @ @ @ @ . @ @ . . . @ . @ @ . . . @ @ @ @ @