Nonogram solver
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.
You are encouraged to solve this task according to the task description, using any language you may know.
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"
- Task
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).
- Related tasks
- See also
- Arc Consistency Algorithm
- http://www.haskell.org/haskellwiki/99_questions/Solutions/98 (Haskell)
- http://twanvl.nl/blog/haskell/Nonograms (Haskell)
- http://picolisp.com/5000/!wiki?99p98 (PicoLisp)
C++
The Solver
<lang cpp> // 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; // The Board rowwise
std::vector<N<_N>> gn; // The Board colwise
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 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();
}};
</lang>
The Task
<lang cpp> // 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;
} </lang>
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." <lang cpp> 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;
} </lang>
- Output:
#######.###...#.#.####### #.....#.##.##.....#.....# #.###.#.....###.#.#.###.# #.###.#.#..######.#.###.# #.###.#..#####.##.#.###.# #.....#..##.......#.....# #######.#.#.#.#.#.####### ........###...###........ #.##.###..#.#.###.#..#.## #.#......###.##....#...#. .####.#.####.##.#....##.. .#.#...#...#.#.####.#.### ..##..#.#.#......##.##### ...###.##.##.######.###.# #.#########.#.#..##....#. .##.#..##...##.###.....#. ###.#.#.#..#....#####.#.. ........#...##.##...##### #######.#..##...#.#.#.### #.....#.##..#..##...##.#. #.###.#...####..#####..#. #.###.#.###.##########.## #.###.#.#..######.######. #.....#..##......#.#.##.. #######.##...#.##...#####
Common Lisp
<lang 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 CONSTRAINT-POSSIBLE-P for your variables. Initialize the DOMAIN of your variables 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)
(print-unreadable-object (o s :type t)
(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)))
(defun read-until-line (stream)
(loop (let ((line (read-line stream)))
(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)
(nonogram (parse-nonogram (read-until-line s)
(read-until-line s)))))
(end-of-file ())))</lang>
- 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
<lang d>import std.stdio, std.range, std.file, std.algorithm, std.string;
/// Create all patterns of a row or col that match given runs. auto genRow(in int w, in int[] s) pure nothrow @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() {
// Read problems from file.
immutable fn = "nonogram_problems.txt";
fn.readText.split("\n\n").filter!(p => !p.strip.empty).each!(p => p.strip.solve);
"Extra example not solvable by deduction alone:".writeln; "B B A A\nB B A A".solve;
"Extra example where there is no solution:".writeln; "B A A\nA A A".solve;
}</lang>
- Output:
Horizontal runs: [[3], [2, 1], [3, 2], [2, 2], [6], [1, 5], [6], [1], [2]] Vertical runs: [[1, 2], [3, 1], [1, 5], [7, 1], [5], [3], [4], [3]] Solution would be unique . # # # . . . . # # . # . . . . . # # # . . # # . . # # . . # # . . # # # # # # # . # # # # # . # # # # # # . . . . . . # . . . . . . # # . . . Unique solution. Horizontal runs: [[6], [3, 1, 3], [1, 3, 1, 3], [3, 14], [1, 1, 1], [1, 1, 2, 2], [5, 2, 2], [5, 1, 1], [5, 3, 3, 3], [8, 3, 3, 3]] Vertical runs: [[4], [4], [1, 5], [3, 4], [1, 5], [1], [4, 1], [2, 2, 2], [3, 3], [1, 1, 2], [2, 1, 1], [1, 1, 2], [4, 1], [1, 1, 2], [1, 1, 1], [2, 1, 2], [1, 1, 1], [3, 4], [2, 2, 1], [4, 1]] Solution would be unique . . . . . . . . . . # # # # # # . . . . . . . . . . . . # # # . # . . # # # . . . . . # . . # # # . . . # . . . . # # # . . # # # . # # # # # # # # # # # # # # . . . # . . # . . . . . . . . . . . . # . . # . # . # # . . . . . . . . . . # # # # # # # . . # # . . . . . . . . # # . # # # # # . . . # . . . . . . . . # . . # # # # # . . # # # . # # # . # # # . . # # # # # # # # . # # # . # # # . # # # Unique solution. Horizontal runs: [[3, 1], [2, 4, 1], [1, 3, 3], [2, 4], [3, 3, 1, 3], [3, 2, 2, 1, 3], [2, 2, 2, 2, 2], [2, 1, 1, 2, 1, 1], [1, 2, 1, 4], [1, 1, 2, 2], [2, 2, 8], [2, 2, 2, 4], [1, 2, 2, 1, 1, 1], [3, 3, 5, 1], [1, 1, 3, 1, 1, 2], [2, 3, 1, 3, 3], [1, 3, 2, 8], [4, 3, 8], [1, 4, 2, 5], [1, 4, 2, 2], [4, 2, 5], [5, 3, 5], [4, 1, 1], [4, 2], [3, 3]] Vertical runs: [[2, 3], [3, 1, 3], [3, 2, 1, 2], [2, 4, 4], [3, 4, 2, 4, 5], [2, 5, 2, 4, 6], [1, 4, 3, 4, 6, 1], [4, 3, 3, 6, 2], [4, 2, 3, 6, 3], [1, 2, 4, 2, 1], [2, 2, 6], [1, 1, 6], [2, 1, 4, 2], [4, 2, 6], [1, 1, 1, 1, 4], [2, 4, 7], [3, 5, 6], [3, 2, 4, 2], [2, 2, 2], [6, 3]] Solution would be unique . . . . # # # . # . . . . . . . . . . . . . . . # # . # # # # . # . . . . . . . . . . . # . # # # . # # # . . . . . . . . . # # . # # # # . . . . . . . . . . . . # # # . # # # . # . . . . # # # . . . # # # . . # # . # # . . . # . # # # . . # # . . # # . # # . . . . # # . # # . . . . . . # # . # . # . . # # . # . # . . . . . . # . # # . # . . . # # # # . . . . . . . # . # . # # . . . . . # # . . . . . . . . # # . # # . . # # # # # # # # . . . . # # . # # . . . # # . . # # # # . . . . # . # # . # # . # . . . # . . # # # # . . # # # . # # # # # . . . . . # # . # . # # # . # . . . . # . . . . # # # # . . # # # . # . . . . # # # . # # # . # . # # # . # # . # # # # # # # # . . . # # # # . # # # . # # # # # # # # . . . . . # . # # # # . # # . # # # # # . . . . . # . # # # # . # # . . . # # . . . . . . . # # # # . . # # . . . # # # # # . . . # # # # # . # # # . . . # # # # # . . . # # # # . # . . . . . . . . . . # . . # # # # . # # . . . . . . . . . . . . . # # # . # # # . . . . . . . . . . . Unique solution. Horizontal runs: [[5], [2, 3, 2], [2, 5, 1], [2, 8], [2, 5, 11], [1, 1, 2, 1, 6], [1, 2, 1, 3], [2, 1, 1], [2, 6, 2], [15, 4], [10, 8], [2, 1, 4, 3, 6], [17], [17], [18], [1, 14], [1, 1, 14], [5, 9], [8], [7]] Vertical runs: [[5], [3, 2], [2, 1, 2], [1, 1, 1], [1, 1, 1], [1, 3], [2, 2], [1, 3, 3], [1, 3, 3, 1], [1, 7, 2], [1, 9, 1], [1, 10], [1, 10], [1, 3, 5], [1, 8], [2, 1, 6], [3, 1, 7], [4, 1, 7], [6, 1, 8], [6, 10], [7, 10], [1, 4, 11], [1, 2, 11], [2, 12], [3, 13]] Solution would be unique . . . . . . . . . . . . . . . . . . . . # # # # # . . # # . . . . . . . . . . . . . . # # # . . # # . # # . . . . . . . . . . . . . . # # # # # . . # # # . . . . . . . . . . . . . # # # # # # # # . . # # . . . . # # # # # . # # # # # # # # # # # . . # . # . . # # . . . . # . . . . # # # # # # . . . # . . # # . . . . . # . . . . . . . # # # . . . . # # . . . . . . . . # . . . . . . . . . . . . . # . # # . . . . . # # # # # # . . . . . . . . . # # . . # # # # # # # # # # # # # # # . . . . # # # # . . . . . # # # # # # # # # # . . # # # # # # # # . . . . # # . # . # # # # . # # # . . # # # # # # . . . . . . . . # # # # # # # # # # # # # # # # # . . . . . . . . # # # # # # # # # # # # # # # # # . . . . . . . # # # # # # # # # # # # # # # # # # . . . . . . . # . . . # # # # # # # # # # # # # # . . . . . . . # . # . # # # # # # # # # # # # # # . . . . . . . . # # # # # . . . # # # # # # # # # . . . . . . . . . . . . . . . . . # # # # # # # # . . . . . . . . . . . . . . . . . . # # # # # # # Unique solution. Extra example not solvable by deduction alone: Horizontal runs: [[2], [2], [1], [1]] Vertical runs: [[2], [2], [1], [1]] Solution may not be unique, doing exhaustive search: # # . . # # . . . . # . . . . # # # . . # # . . . . . # . . # . . # # . # # . . # . . . . . . # 3 solutions. Extra example where there is no solution: Horizontal runs: [[2], [1], [1]] Vertical runs: [[1], [1], [1]] Solution may not be unique, doing exhaustive search: No solution.
The output is the same as the Python entry. The run-time with ldc2 compiler is about 0.29 seconds.
F#
<lang fsharp> (* 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))
</lang> For the purposes of this task I provide a little code to read the input from a file <lang fsharp> 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" Some(fl (file.ReadLine().Split ' ') (file.ReadLine().Split ' ')) with | _ -> printfn "Error reading file" ; None
</lang> This may be used: <lang fsharp> 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" </lang>
- 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
Java
<lang java>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) {
List<BitSet> lst = new LinkedList<>();
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');
lst.add(bitset);
}
result.add(lst);
}
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;
}
}</lang>
. # # # . . . . # # . # . . . . . # # # . . # # . . # # . . # # . . # # # # # # # . # # # # # . # # # # # # . . . . . . # . . . . . . # # . . . . . . . . . . . . . # # # # # # . . . . . . . . . . . . # # # . # . . # # # . . . . . # . . # # # . . . # . . . . # # # . . # # # . # # # # # # # # # # # # # # . . . # . . # . . . . . . . . . . . . # . . # . # . # # . . . . . . . . . . # # # # # # # . . # # . . . . . . . . # # . # # # # # . . . # . . . . . . . . # . . # # # # # . . # # # . # # # . # # # . . # # # # # # # # . # # # . # # # . # # # . . . . # # # . # . . . . . . . . . . . . . . . # # . # # # # . # . . . . . . . . . . . # . # # # . # # # . . . . . . . . . # # . # # # # . . . . . . . . . . . . # # # . # # # . # . . . . # # # . . . # # # . . # # . # # . . . # . # # # . . # # . . # # . # # . . . . # # . # # . . . . . . # # . # . # . . # # . # . # . . . . . . # . # # . # . . . # # # # . . . . . . . # . # . # # . . . . . # # . . . . . . . . # # . # # . . # # # # # # # # . . . . # # . # # . . . # # . . # # # # . . . . # . # # . # # . # . . . # . . # # # # . . # # # . # # # # # . . . . . # # . # . # # # . # . . . . # . . . . # # # # . . # # # . # . . . . # # # . # # # . # . # # # . # # . # # # # # # # # . . . # # # # . # # # . # # # # # # # # . . . . . # . # # # # . # # . # # # # # . . . . . # . # # # # . # # . . . # # . . . . . . . # # # # . . # # . . . # # # # # . . . # # # # # . # # # . . . # # # # # . . . # # # # . # . . . . . . . . . . # . . # # # # . # # . . . . . . . . . . . . . # # # . # # # . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . # # # # # . . # # . . . . . . . . . . . . . . # # # . . # # . # # . . . . . . . . . . . . . . # # # # # . . # # # . . . . . . . . . . . . . # # # # # # # # . . # # . . . . # # # # # . # # # # # # # # # # # . . # . # . . # # . . . . # . . . . # # # # # # . . . # . . # # . . . . . # . . . . . . . # # # . . . . # # . . . . . . . . # . . . . . . . . . . . . . # . # # . . . . . # # # # # # . . . . . . . . . # # . . # # # # # # # # # # # # # # # . . . . # # # # . . . . . # # # # # # # # # # . . # # # # # # # # . . . . # # . # . # # # # . # # # . . # # # # # # . . . . . . . . # # # # # # # # # # # # # # # # # . . . . . . . . # # # # # # # # # # # # # # # # # . . . . . . . # # # # # # # # # # # # # # # # # # . . . . . . . # . . . # # # # # # # # # # # # # # . . . . . . . # . # . # # # # # # # # # # # # # # . . . . . . . . # # # # # . . . # # # # # # # # # . . . . . . . . . . . . . . . . . # # # # # # # # . . . . . . . . . . . . . . . . . . # # # # # # #
Prolog
module(clpfd) is written by Markus Triska
Solution written by Lars Buitinck
Module solve-nonogram.pl <lang Prolog>/*
- Nonogram/paint-by-numbers solver in SWI-Prolog. Uses CLP(FD),
- in particular the automaton/3 (finite-state/RE) constraint.
- Copyright (c) 2011 Lars Buitinck.
- Do with this code as you like, but don't remove the copyright notice.
- /
- - use_module(library(clpfd)).
nono(RowSpec, ColSpec, Grid) :- rows(RowSpec, Grid), transpose(Grid, GridT), rows(ColSpec, GridT).
rows([], []). rows([C|Cs], [R|Rs]) :- row(C, R), rows(Cs, Rs).
row(Ks, Row) :- sum(Ks, #=, Ones), sum(Row, #=, Ones), arcs(Ks, Arcs, start, Final), append(Row, [0], RowZ), automaton(RowZ, [source(start), sink(Final)], [arc(start,0,start) | Arcs]).
% Make list of transition arcs for finite-state constraint. arcs([], [], Final, Final). arcs([K|Ks], Arcs, CurState, Final) :- gensym(state, NextState), ( K == 0 -> Arcs = [arc(CurState,0,CurState), arc(CurState,0,NextState) | Rest], arcs(Ks, Rest, NextState, Final) ; Arcs = [arc(CurState,1,NextState) | Rest], K1 #= K-1, arcs([K1|Ks], Rest, NextState, Final)).
make_grid(Grid, X, Y, Vars) :-
length(Grid,X),
make_rows(Grid, Y, Vars).
make_rows([], _, []). make_rows([R|Rs], Len, Vars) :- length(R, Len), make_rows(Rs, Len, Vars0), append(R, Vars0, Vars).
print([]). print([R|Rs]) :- print_row(R), print(Rs).
print_row([]) :- nl. print_row([X|R]) :- ( X == 0 -> write(' ') ; write('x')), print_row(R).
nonogram(Rows, Cols) :- length(Rows, X), length(Cols, Y), make_grid(Grid, X, Y, Vars), nono(Rows, Cols, Grid), label(Vars), print(Grid).</lang> File nonogram.pl, used to read data in a file. <lang Prolog>nonogram :- open('C:/Users/Utilisateur/Documents/Prolog/Rosetta/nonogram/nonogram.txt', read, In, []), repeat, read_line_to_codes(In, Line_1), read_line_to_codes(In, Line_2), compute_values(Line_1, [], [], Lines), compute_values(Line_2, [], [], Columns), nonogram(Lines, Columns) , nl, nl, read_line_to_codes(In, end_of_file), close(In).
compute_values([], Current, Tmp, R) :- reverse(Current, R_Current), reverse([R_Current | Tmp], R).
compute_values([32 | T], Current, Tmp, R) :- !, reverse(Current, R_Current), compute_values(T, [], [R_Current | Tmp], R).
compute_values([X | T], Current, Tmp, R) :- V is X - 64, compute_values(T, [V | Current], Tmp, R).</lang>
Python
First fill cells by deduction, then search through all combinations. It could take up a huge amount of storage, depending on the board size. <lang python>from itertools import izip
def gen_row(w, s):
"""Create all patterns of a row or col that match given runs."""
def gen_seg(o, sp):
if not o:
return [[2] * sp]
return [[2] * x + o[0] + tail
for x in xrange(1, sp - len(o) + 2)
for tail in gen_seg(o[1:], sp - x)]
return [x[1:] for x in gen_seg([[1] * i for i in s], w + 1 - sum(s))]
def deduce(hr, vr):
"""Fix inevitable value of cells, and propagate."""
def allowable(row):
return reduce(lambda a, b: [x | y for x, y in izip(a, b)], row)
def fits(a, b):
return all(x & y for x, y in izip(a, b))
def fix_col(n):
"""See if any value in a given column is fixed;
if so, mark its corresponding row for future fixup."""
c = [x[n] for x in can_do]
cols[n] = [x for x in cols[n] if fits(x, c)]
for i, x in enumerate(allowable(cols[n])):
if x != can_do[i][n]:
mod_rows.add(i)
can_do[i][n] &= x
def fix_row(n):
"""Ditto, for rows."""
c = can_do[n]
rows[n] = [x for x in rows[n] if fits(x, c)]
for i, x in enumerate(allowable(rows[n])):
if x != can_do[n][i]:
mod_cols.add(i)
can_do[n][i] &= x
def show_gram(m):
# If there's 'x', something is wrong.
# If there's '?', needs more work.
for x in m:
print " ".join("x#.?"[i] for i in x)
print
w, h = len(vr), len(hr) rows = [gen_row(w, x) for x in hr] cols = [gen_row(h, x) for x in vr] can_do = map(allowable, rows)
# Initially mark all columns for update. mod_rows, mod_cols = set(), set(xrange(w))
while mod_cols:
for i in mod_cols:
fix_col(i)
mod_cols = set()
for i in mod_rows:
fix_row(i)
mod_rows = set()
if all(can_do[i][j] in (1, 2) for j in xrange(w) for i in xrange(h)):
print "Solution would be unique" # but could be incorrect!
else:
print "Solution may not be unique, doing exhaustive search:"
# We actually do exhaustive search anyway. Unique solution takes # no time in this phase anyway, but just in case there's no # solution (could happen?). out = [0] * h
def try_all(n = 0):
if n >= h:
for j in xrange(w):
if [x[j] for x in out] not in cols[j]:
return 0
show_gram(out)
return 1
sol = 0
for x in rows[n]:
out[n] = x
sol += try_all(n + 1)
return sol
n = try_all()
if not n:
print "No solution."
elif n == 1:
print "Unique solution."
else:
print n, "solutions."
print
def solve(p, show_runs=True):
s = [[[ord(c) - ord('A') + 1 for c in w] for w in l.split()]
for l in p.splitlines()]
if show_runs:
print "Horizontal runs:", s[0]
print "Vertical runs:", s[1]
deduce(s[0], s[1])
def main():
# Read problems from file.
fn = "nonogram_problems.txt"
for p in (x for x in open(fn).read().split("\n\n") if x):
solve(p)
print "Extra example not solvable by deduction alone:"
solve("B B A A\nB B A A")
print "Extra example where there is no solution:"
solve("B A A\nA A A")
main()</lang>
- Output:
Horizontal runs: [[3], [2, 1], [3, 2], [2, 2], [6], [1, 5], [6], [1], [2]] Vertical runs: [[1, 2], [3, 1], [1, 5], [7, 1], [5], [3], [4], [3]] Solution would be unique . # # # . . . . # # . # . . . . . # # # . . # # . . # # . . # # . . # # # # # # # . # # # # # . # # # # # # . . . . . . # . . . . . . # # . . . Unique solution (... etc. ...)
Racket
[See Example:Nonogram solver/Racket for editing of this section]