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"
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
 Nonoblock.


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)


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)
    }
}