Nonoblock

From Rosetta Code
Task
Nonoblock
You are encouraged to solve this task according to the task description, using any language you may know.

Nonoblock is a chip off the old Nonogram puzzle.


Given
  • The number of cells in a row.
  • The size of each, (space separated), connected block of cells to fit in the row, in left-to right order.


Task
  • show all possible positions.
  • show the number of positions of the blocks for the following cases within the row.
  • show all output on this page.
  • use a "neat" diagram of the block positions.


Enumerate the following configurations
  1.   5   cells   and   [2, 1]   blocks
  2.   5   cells   and   []   blocks   (no blocks)
  3.   10   cells   and   [8]   blocks
  4.   15   cells   and   [2, 3, 2, 3]   blocks
  5.   5   cells   and   [2, 3]   blocks   (should give some indication of this not being possible)


Example

Given a row of five cells and a block of two cells followed by a block of one cell - in that order, the example could be shown as:

   |_|_|_|_|_| # 5 cells and [2, 1] blocks

And would expand to the following 3 possible rows of block positions:

   |A|A|_|B|_|
   |A|A|_|_|B|
   |_|A|A|_|B|


Note how the sets of blocks are always separated by a space.

Note also that it is not necessary for each block to have a separate letter. Output approximating

This:

                       |#|#|_|#|_|
                       |#|#|_|_|#|
                       |_|#|#|_|#|

This would also work:

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


An algorithm
  • Find the minimum space to the right that is needed to legally hold all but the leftmost block of cells (with a space between blocks remember).
  • The leftmost cell can legitimately be placed in all positions from the LHS up to a RH position that allows enough room for the rest of the blocks.
  • for each position of the LH block recursively compute the position of the rest of the blocks in the remaining space to the right of the current placement of the LH block.

(This is the algorithm used in the Nonoblock#Python solution).


Reference



11l

Translation of: Python
F nonoblocks([Int] &blocks, Int cells) -> [[(Int, Int)]]
   [[(Int, Int)]] r
   I blocks.empty | blocks[0] == 0
      r [+]= [(0, 0)]
   E
      assert(sum(blocks) + blocks.len - 1 <= cells, ‘Those blocks will not fit in those cells’)
      V (blength, brest) = (blocks[0], blocks[1..])
      V minspace4rest = sum(brest.map(b -> 1 + b))

      L(bpos) 0 .. cells - minspace4rest - blength
         I brest.empty
            r [+]= [(bpos, blength)]
         E
            V offset = bpos + blength + 1
            L(subpos) nonoblocks(&brest, cells - offset)
               V rest = subpos.map((bp, bl) -> (@offset + bp, bl))
               V vec = [(bpos, blength)] [+] rest
               r [+]= vec
   R r

F pblock(vec, cells)
   ‘Prettyprints each run of blocks with a different letter A.. for each block of filled cells’
   V vector = [‘_’] * cells
   L(bp_bl) vec
      V ch = L.index + ‘A’.code
      V (bp, bl) = bp_bl
      L(i) bp .< bp + bl
         vector[i] = I vector[i] == ‘_’ {Char(code' ch)} E Char(‘?’)
   R ‘|’vector.join(‘|’)‘|’

L(blocks, cells) [
      ([2, 1], 5),
      ([Int](), 5),
      ([8], 10),
      ([2, 3, 2, 3], 15)
      ]
   print("\nConfiguration:\n    #. ## #. cells and #. blocks".format(pblock([(Int, Int)](), cells), cells, blocks))
   print(‘  Possibilities:’)
   V nb = nonoblocks(&blocks, cells)
   L(vector) nb
      print(‘    ’pblock(vector, cells))
   print(‘  A total of #. Possible configurations.’.format(nb.len))
Output:

Configuration:
    |_|_|_|_|_| # 5 cells and [2, 1] blocks
  Possibilities:
    |A|A|_|B|_|
    |A|A|_|_|B|
    |_|A|A|_|B|
  A total of 3 Possible configurations.

Configuration:
    |_|_|_|_|_| # 5 cells and [] blocks
  Possibilities:
    |_|_|_|_|_|
  A total of 1 Possible configurations.

Configuration:
    |_|_|_|_|_|_|_|_|_|_| # 10 cells and [8] blocks
  Possibilities:
    |A|A|A|A|A|A|A|A|_|_|
    |_|A|A|A|A|A|A|A|A|_|
    |_|_|A|A|A|A|A|A|A|A|
  A total of 3 Possible configurations.

Configuration:
    |_|_|_|_|_|_|_|_|_|_|_|_|_|_|_| # 15 cells and [2, 3, 2, 3] blocks
  Possibilities:
    |A|A|_|B|B|B|_|C|C|_|D|D|D|_|_|
    |A|A|_|B|B|B|_|C|C|_|_|D|D|D|_|
    |A|A|_|B|B|B|_|C|C|_|_|_|D|D|D|
    |A|A|_|B|B|B|_|_|C|C|_|D|D|D|_|
    |A|A|_|B|B|B|_|_|C|C|_|_|D|D|D|
    |A|A|_|B|B|B|_|_|_|C|C|_|D|D|D|
    |A|A|_|_|B|B|B|_|C|C|_|D|D|D|_|
    |A|A|_|_|B|B|B|_|C|C|_|_|D|D|D|
    |A|A|_|_|B|B|B|_|_|C|C|_|D|D|D|
    |A|A|_|_|_|B|B|B|_|C|C|_|D|D|D|
    |_|A|A|_|B|B|B|_|C|C|_|D|D|D|_|
    |_|A|A|_|B|B|B|_|C|C|_|_|D|D|D|
    |_|A|A|_|B|B|B|_|_|C|C|_|D|D|D|
    |_|A|A|_|_|B|B|B|_|C|C|_|D|D|D|
    |_|_|A|A|_|B|B|B|_|C|C|_|D|D|D|
  A total of 15 Possible configurations.

Action!

DEFINE MAX_BLOCKS="10"
DEFINE NOT_FOUND="255"

BYTE FUNC GetBlockAtPos(BYTE p BYTE ARRAY blocks,pos INT count)
  INT i
  FOR i=0 TO count-1
  DO
    IF p>=pos(i) AND p<pos(i)+blocks(i) THEN
      RETURN (i)
    FI
  OD
RETURN (NOT_FOUND)

PROC PrintResult(BYTE cells BYTE ARRAY blocks,pos INT count)
  BYTE i,b

  Print("[")
  FOR i=0 TO cells-1
  DO
    b=GetBlockAtPos(i,blocks,pos,count)
    IF b=NOT_FOUND THEN
      Put('.)
    ELSE
      Put(b+'A)
    FI
  OD
  PrintE("]")
RETURN

BYTE FUNC LeftMostPos(BYTE cells BYTE ARRAY blocks,pos INT count,startFrom)
  INT i

  FOR i=startFrom TO count-1
  DO
    pos(i)=pos(i-1)+blocks(i-1)+1
    IF pos(i)+blocks(i)>cells THEN
      RETURN (0)
    FI
  OD
RETURN (1)

BYTE FUNC MoveToRight(BYTE cells BYTE ARRAY blocks,pos INT count,startFrom)
  pos(startFrom)==+1
  IF pos(startFrom)+blocks(startFrom)>cells THEN
    RETURN (0)
  FI
RETURN (LeftMostPos(cells,blocks,pos,count,startFrom+1))

PROC Process(BYTE cells BYTE ARRAY blocks INT count)
  BYTE ARRAY pos(MAX_BLOCKS)
  BYTE success
  INT current

  IF count=0 THEN
    PrintResult(cells,blocks,pos,count)
    RETURN
  FI

  pos(0)=0
  success=LeftMostPos(cells,blocks,pos,count,1)
  IF success=0 THEN
    PrintE("No solutions")
    RETURN
  FI
  current=count-1
  WHILE success
  DO
    PrintResult(cells,blocks,pos,count)
    DO
      success=MoveToRight(cells,blocks,pos,count,current)
      IF success THEN
        current=count-1
      ELSE
        current==-1
        IF current<0 THEN
          EXIT
        FI
      FI
    UNTIL success
    OD
  OD
RETURN

PROC Test(BYTE cells BYTE ARRAY blocks INT count)
  BYTE CH=$02FC ;Internal hardware value for last key pressed
  INT i
  
  PrintB(cells) Print(" cells [")
  FOR i=0 TO count-1
  DO
    PrintB(blocks(i))
    IF i<count-1 THEN
      Put(32)
    FI
  OD
  PrintE("]")

  Process(cells,blocks,count)

  PutE()
  PrintE("Press any key to continue...")
  DO UNTIL CH#$FF OD
  CH=$FF
  PutE()
RETURN

PROC Main()
  BYTE ARRAY t1=[2 1],t2=[],t3=[8],t4=[2 3 2 3],t5=[2 3]

  Test(5,t1,2)
  Test(5,t2,0)
  Test(10,t3,1)
  Test(15,t4,4)
  Test(5,t5,2)
RETURN
Output:

Screenshot from Atari 8-bit computer

5 cells [2 1]
[AA.B.]
[AA..B]
[.AA.B]

Press any key to continue...

5 cells []
[.....]

Press any key to continue...

10 cells [8]
[AAAAAAAA..]
[.AAAAAAAA.]
[..AAAAAAAA]

Press any key to continue...

15 cells [2 3 2 3]
[AA.BBB.CC.DDD..]
[AA.BBB.CC..DDD.]
[AA.BBB.CC...DDD]
[AA.BBB..CC.DDD.]
[AA.BBB..CC..DDD]
[AA.BBB...CC.DDD]
[AA..BBB.CC.DDD.]
[AA..BBB.CC..DDD]
[AA..BBB..CC.DDD]
[AA...BBB.CC.DDD]
[.AA.BBB.CC.DDD.]
[.AA.BBB.CC..DDD]
[.AA.BBB..CC.DDD]
[.AA..BBB.CC.DDD]
[..AA.BBB.CC.DDD]

Press any key to continue...

5 cells [2 3]
No solutions

Press any key to continue...

AutoHotkey

;-------------------------------------------
NonoBlock(cells, blocks){
    result := [], line := ""
    for i, v in blocks
        B .= v ", "
    output := cells " cells and [" Trim(B, ", ") "] blocks`n"
    
    if ((Arr := NonoBlockCreate(cells, blocks)) = "Error")
        return output "No Solution`n"	
    for i, v in arr
        line.= v ";"
    result[line] := true
    result := NonoBlockRecurse(Arr, result)
    output .= NonoBlockShow(result)
    return output
}
;-------------------------------------------
; create cells+1 size array, stack blocks to left with one gap in between
; gaps are represented by negative number
; stack extra gaps to far left
; for example : 6 cells and [2, 1] blocks
; returns [-2, 2, -1, 1, 0, 0, 0]
NonoBlockCreate(cells, blocks){
    Arr := [], B := blocks.Count()
    if !B									; no blocks
        return [0-cells, 0]
    for i, v in blocks{
        total += v
        Arr.InsertAt(1, blocks[B-A_Index+1])
        Arr.InsertAt(1, -1)
    }
    if (cells < total + B-1)				; not possible
        return "Error"
    Arr[1] := total + B-1 - cells
    loop % cells - Arr.Count() + 1
        Arr.Push(0)
    return Arr
}
;-------------------------------------------
; shift negative numbers from left to right recursively.
; preserve at least one gap between blocks.
; [-2, 2, -1, 1, 0, 0, 0]
; [-1, 2, -2, 1, 0, 0, 0]
NonoBlockRecurse(Arr, result, pos:= 1){
    i := pos-1
    while (i < Arr.count())
    {
        if ((B:=Arr[++i])>=0) || (B=-1 && i>1) 
            continue
        if (i=Arr.count()-1)
            return result
        Arr[i] := ++B, Arr[i+2] := Arr[i+2] -1
        result := NonoBlockRecurse(Arr.Clone(), result, i)
        line := []
        for k, v in Arr
            line.=v ";"
        result[line] := true
    }
    return result
}
;-------------------------------------------
; represent positve numbers by a block of "#", negative nubmers by a block of "."
NonoBlockShow(result){
    for line in result{
        i := A_Index
        nLine := ""
        for j, val in StrSplit(line, ";")
            loop % Abs(val)
                nLine .= val > 0 ? "#" : "."
        output .= nLine "`n"
    }
    Sort, output, U
    return output
}
;-------------------------------------------
Examples:
Results .= NonoBlock(5, [2, 1])		"------------`n"
Results .= NonoBlock(5, [])		"------------`n"
Results .= NonoBlock(10, [8])		"------------`n"
Results .= NonoBlock(15, [2, 3, 2, 3])	"------------`n"
Results .= NonoBlock(5, [2, 3])		"------------`n"
MsgBox, 262144, , % Results
return
Output:
---------------------------
5 cells and [2, 1] blocks
##.#.
##..#
.##.#
------------
5 cells and [] blocks
.....
------------
10 cells and [8] blocks
########..
.########.
..########
------------
15 cells and [2, 3, 2, 3] blocks
##.###.##.###..
##.###.##..###.
##.###.##...###
##.###..##.###.
##.###..##..###
##.###...##.###
##..###.##.###.
##..###.##..###
##..###..##.###
##...###.##.###
.##.###.##.###.
.##.###.##..###
.##.###..##.###
.##..###.##.###
..##.###.##.###
------------
5 cells and [2, 3] blocks
No Solution
------------

C

#include <stdio.h>
#include <string.h>

void nb(int cells, int total_block_size, int* blocks, int block_count,
        char* output, int offset, int* count) {
    if (block_count == 0) {
        printf("%2d  %s\n", ++*count, output);
        return;
    }
    int block_size = blocks[0];
    int max_pos = cells - (total_block_size + block_count - 1);
    total_block_size -= block_size;
    cells -= block_size + 1;
    ++blocks;
    --block_count;
    for (int i = 0; i <= max_pos; ++i, --cells) {
        memset(output + offset, '.', max_pos + block_size);
        memset(output + offset + i, '#', block_size);
        nb(cells, total_block_size, blocks, block_count, output,
           offset + block_size + i + 1, count);
    }
}

void nonoblock(int cells, int* blocks, int block_count) {
    printf("%d cells and blocks [", cells);
    for (int i = 0; i < block_count; ++i)
        printf(i == 0 ? "%d" : ", %d", blocks[i]);
    printf("]:\n");
    int total_block_size = 0;
    for (int i = 0; i < block_count; ++i)
        total_block_size += blocks[i];
    if (cells < total_block_size + block_count - 1) {
        printf("no solution\n");
        return;
    }
    char output[cells + 1];
    memset(output, '.', cells);
    output[cells] = '\0';
    int count = 0;
    nb(cells, total_block_size, blocks, block_count, output, 0, &count);
}

int main() {
    int blocks1[] = {2, 1};
    nonoblock(5, blocks1, 2);
    printf("\n");
    
    nonoblock(5, NULL, 0);
    printf("\n");

    int blocks2[] = {8};
    nonoblock(10, blocks2, 1);
    printf("\n");
    
    int blocks3[] = {2, 3, 2, 3};
    nonoblock(15, blocks3, 4);
    printf("\n");
    
    int blocks4[] = {2, 3};
    nonoblock(5, blocks4, 2);

    return 0;
}
Output:
5 cells and blocks [2, 1]:
 1  ##.#.
 2  ##..#
 3  .##.#

5 cells and blocks []:
 1  .....

10 cells and blocks [8]:
 1  ########..
 2  .########.
 3  ..########

15 cells and blocks [2, 3, 2, 3]:
 1  ##.###.##.###..
 2  ##.###.##..###.
 3  ##.###.##...###
 4  ##.###..##.###.
 5  ##.###..##..###
 6  ##.###...##.###
 7  ##..###.##.###.
 8  ##..###.##..###
 9  ##..###..##.###
10  ##...###.##.###
11  .##.###.##.###.
12  .##.###.##..###
13  .##.###..##.###
14  .##..###.##.###
15  ..##.###.##.###

5 cells and blocks [2, 3]:
no solution

C#

This solution uses a StringBuilder. Spaces are moved from right to left and the problem is then solved recursively.

using System;
using System.Linq;
using System.Text;

public static class Nonoblock
{
    public static void Main() {
        Positions(5, 2,1);
        Positions(5);
        Positions(10, 8);
        Positions(15, 2,3,2,3);
        Positions(5, 2,3);
    }

    public static void Positions(int cells, params int[] blocks) {
        if (cells < 0 || blocks == null || blocks.Any(b => b < 1)) throw new ArgumentOutOfRangeException();
        Console.WriteLine($"{cells} cells with [{string.Join(", ", blocks)}]");
        if (blocks.Sum() + blocks.Length - 1 > cells) {
            Console.WriteLine("No solution");
            return;
        }
        var spaces = new int[blocks.Length + 1];
        int total = -1;
        for (int i = 0; i < blocks.Length; i++) {
            total += blocks[i] + 1;
            spaces[i+1] = total;
        }
        spaces[spaces.Length - 1] = cells - 1;
        var sb = new StringBuilder(string.Join(".", blocks.Select(b => new string('#', b))).PadRight(cells, '.'));
        Iterate(sb, spaces, spaces.Length - 1, 0);
        Console.WriteLine();
    }

    private static void Iterate(StringBuilder output, int[] spaces, int index, int offset) {
        Console.WriteLine(output.ToString());
        if (index <= 0) return;
        int count = 0;
        while (output[spaces[index] - offset] != '#') {
            count++;
            output.Remove(spaces[index], 1);
            output.Insert(spaces[index-1], '.');
            spaces[index-1]++;
            Iterate(output, spaces, index - 1, 1);
        }
        if (offset == 0) return;
        spaces[index-1] -= count;
        output.Remove(spaces[index-1], count);
        output.Insert(spaces[index] - count, ".", count);
    }

}
Output:
5 cells with [2, 1]
##.#.
##..#
.##.#

5 cells with []
.....

10 cells with [8]
########..
.########.
..########

15 cells with [2, 3, 2, 3]
##.###.##.###..
##.###.##..###.
##.###..##.###.
##..###.##.###.
.##.###.##.###.
##.###.##...###
##.###..##..###
##..###.##..###
.##.###.##..###
##.###...##.###
##..###..##.###
.##.###..##.###
##...###.##.###
.##..###.##.###
..##.###.##.###

5 cells with [2, 3]
No solution

C++

#include <iomanip>
#include <iostream>
#include <algorithm>
#include <numeric>
#include <string>
#include <vector>

typedef std::pair<int, std::vector<int> > puzzle;

class nonoblock {
public:
    void solve( std::vector<puzzle>& p ) {
        for( std::vector<puzzle>::iterator i = p.begin(); i != p.end(); i++ ) {
            counter = 0;
            std::cout << " Puzzle: " << ( *i ).first << " cells and blocks [ ";
            for( std::vector<int>::iterator it = ( *i ).second.begin(); it != ( *i ).second.end(); it++ )
                std::cout << *it << " ";
            std::cout << "] ";
            int s = std::accumulate( ( *i ).second.begin(), ( *i ).second.end(), 0 ) + ( ( *i ).second.size() > 0 ? ( *i ).second.size() - 1 : 0 );
            if( ( *i ).first - s < 0 ) {
                std::cout << "has no solution!\n\n\n";
                continue;
            }
            std::cout << "\n Possible configurations:\n\n";
            std::string b( ( *i ).first, '-' );
            solve( *i, b, 0 );
            std::cout << "\n\n";
        } 
    }

private:
    void solve( puzzle p, std::string n, int start ) {
        if( p.second.size() < 1 ) {
            output( n );
            return;
        }
        std::string temp_string;
        int offset,
            this_block_size = p.second[0];

        int space_need_for_others = std::accumulate( p.second.begin() + 1, p.second.end(), 0 );
        space_need_for_others += p.second.size() - 1;

        int space_for_curr_block = p.first - space_need_for_others - std::accumulate( p.second.begin(), p.second.begin(), 0 );

        std::vector<int> v1( p.second.size() - 1 );
        std::copy( p.second.begin() + 1, p.second.end(), v1.begin() );
        puzzle p1 = std::make_pair( space_need_for_others, v1 );

        for( int a = 0; a < space_for_curr_block; a++ ) {
            temp_string = n;

            if( start + this_block_size > n.length() ) return;

            for( offset = start; offset < start + this_block_size; offset++ )
                temp_string.at( offset ) = 'o';

            if( p1.first ) solve( p1, temp_string, offset + 1 );
            else output( temp_string );

            start++;
        }
    }
    void output( std::string s ) {
        char b = 65 - ( s.at( 0 ) == '-' ? 1 : 0 );
        bool f = false;
        std::cout << std::setw( 3 ) << ++counter << "\t|";
        for( std::string::iterator i = s.begin(); i != s.end(); i++ ) {
            b += ( *i ) == 'o' && f ? 1 : 0;
            std::cout << ( ( *i ) == 'o' ? b : '_' ) << "|";
            f = ( *i ) == '-' ? true : false;
        }
        std::cout << "\n";
    }

    unsigned counter;
};

int main( int argc, char* argv[] )
{
    std::vector<puzzle> problems;
    std::vector<int> blocks; 
    blocks.push_back( 2 ); blocks.push_back( 1 );
    problems.push_back( std::make_pair( 5, blocks ) );
    blocks.clear();
    problems.push_back( std::make_pair( 5, blocks ) );
    blocks.push_back( 8 );
    problems.push_back( std::make_pair( 10, blocks ) );
    blocks.clear();
    blocks.push_back( 2 ); blocks.push_back( 3 );
    problems.push_back( std::make_pair( 5, blocks ) );
    blocks.push_back( 2 ); blocks.push_back( 3 );
    problems.push_back( std::make_pair( 15, blocks ) );

    nonoblock nn;
    nn.solve( problems );

    return 0;
}
Output:
 Puzzle: 5 cells and blocks [ 2 1 ]
 Possible configurations:

  1     |A|A|_|B|_|
  2     |A|A|_|_|B|
  3     |_|A|A|_|B|


 Puzzle: 5 cells and blocks [ ]
 Possible configurations:

  1     |_|_|_|_|_|


 Puzzle: 10 cells and blocks [ 8 ]
 Possible configurations:

  1     |A|A|A|A|A|A|A|A|_|_|
  2     |_|A|A|A|A|A|A|A|A|_|
  3     |_|_|A|A|A|A|A|A|A|A|


 Puzzle: 5 cells and blocks [ 2 3 ] has no solution!


 Puzzle: 15 cells and blocks [ 2 3 2 3 ]
 Possible configurations:

  1     |A|A|_|B|B|B|_|C|C|_|D|D|D|_|_|
  2     |A|A|_|B|B|B|_|C|C|_|_|D|D|D|_|
  3     |A|A|_|B|B|B|_|C|C|_|_|_|D|D|D|
  4     |A|A|_|B|B|B|_|_|C|C|_|D|D|D|_|
  5     |A|A|_|B|B|B|_|_|C|C|_|_|D|D|D|
  6     |A|A|_|B|B|B|_|_|_|C|C|_|D|D|D|
  7     |A|A|_|_|B|B|B|_|C|C|_|D|D|D|_|
  8     |A|A|_|_|B|B|B|_|C|C|_|_|D|D|D|
  9     |A|A|_|_|B|B|B|_|_|C|C|_|D|D|D|
 10     |A|A|_|_|_|B|B|B|_|C|C|_|D|D|D|
 11     |_|A|A|_|B|B|B|_|C|C|_|D|D|D|_|
 12     |_|A|A|_|B|B|B|_|C|C|_|_|D|D|D|
 13     |_|A|A|_|B|B|B|_|_|C|C|_|D|D|D|
 14     |_|A|A|_|_|B|B|B|_|C|C|_|D|D|D|
 15     |_|_|A|A|_|B|B|B|_|C|C|_|D|D|D|

D

Translation of: python
import std.stdio, std.array, std.algorithm, std.exception, std.conv,
       std.concurrency, std.range;

struct Solution { uint pos, len; }

Generator!(Solution[]) nonoBlocks(in uint[] blocks, in uint cells) {
    return new typeof(return)({
        if (blocks.empty || blocks[0] == 0) {
            yield([Solution(0, 0)]);
        } else {
            enforce(blocks.sum + blocks.length - 1 <= cells,
                    "Those blocks cannot fit in those cells.");
            immutable firstBl = blocks[0];
            const restBl = blocks.dropOne;

            // The other blocks need space.
            immutable minS = restBl.map!(b => b + 1).sum;

            // Slide the start position from left to max RH
            // index allowing for other blocks.
            foreach (immutable bPos; 0 .. cells - minS - firstBl + 1) {
                if (restBl.empty) {
                    // No other blocks to the right so just yield
                    // this one.
                    yield([Solution(bPos, firstBl)]);
                } else {
                    // More blocks to the right so create a sub-problem
                    // of placing the restBl blocks in the cells one
                    // space to the right of the RHS of this block.
                    immutable offset = bPos + firstBl + 1;
                    immutable newCells = cells - offset;

                    // Recursive call to nonoBlocks yields multiple
                    // sub-positions.
                    foreach (const subPos; nonoBlocks(restBl, newCells)) {
                        // Remove the offset from sub block positions.
                        auto rest = subPos.map!(sol => Solution(offset + sol.pos, sol.len));

                        // Yield this block plus sub blocks positions.
                        yield(Solution(bPos, firstBl) ~ rest.array);
                    }
                }
            }
        }
    });
}


/// Pretty prints each run of blocks with a
/// different letter for each block of filled cells.
string show(in Solution[] vec, in uint nCells) pure {
    auto result = ['_'].replicate(nCells);
    foreach (immutable i, immutable sol; vec)
        foreach (immutable j; sol.pos .. sol.pos + sol.len)
            result[j] = (result[j] == '_') ? to!char('A' + i) : '?';
    return '[' ~ result ~ ']';
}

void main() {
    static struct Problem { uint[] blocks; uint nCells; }

    immutable Problem[] problems = [{[2, 1], 5},
                                    {[], 5},
                                    {[8], 10},
                                    {[2, 3, 2, 3], 15},
                                    {[4, 3], 10},
                                    {[2, 1], 5},
                                    {[3, 1], 10},
                                    {[2, 3], 5}];

    foreach (immutable prob; problems) {
        writefln("Configuration (%d cells and %s blocks):",
                 prob.nCells, prob.blocks);
        show([], prob.nCells).writeln;
        "Possibilities:".writeln;
        auto nConfigs = 0;
        foreach (const sol; nonoBlocks(prob.tupleof)) {
            show(sol, prob.nCells).writeln;
            nConfigs++;
        }
        writefln("A total of %d possible configurations.", nConfigs);
        writeln;
    }
}
Output:
Configuration (5 cells and [2, 1] blocks):
[_____]
Possibilities:
[AA_B_]
[AA__B]
[_AA_B]
A total of 3 possible configurations.

Configuration (5 cells and [] blocks):
[_____]
Possibilities:
[_____]
A total of 1 possible configurations.

Configuration (10 cells and [8] blocks):
[__________]
Possibilities:
[AAAAAAAA__]
[_AAAAAAAA_]
[__AAAAAAAA]
A total of 3 possible configurations.

Configuration (15 cells and [2, 3, 2, 3] blocks):
[_______________]
Possibilities:
[AA_BBB_CC_DDD__]
[AA_BBB_CC__DDD_]
[AA_BBB_CC___DDD]
[AA_BBB__CC_DDD_]
[AA_BBB__CC__DDD]
[AA_BBB___CC_DDD]
[AA__BBB_CC_DDD_]
[AA__BBB_CC__DDD]
[AA__BBB__CC_DDD]
[AA___BBB_CC_DDD]
[_AA_BBB_CC_DDD_]
[_AA_BBB_CC__DDD]
[_AA_BBB__CC_DDD]
[_AA__BBB_CC_DDD]
[__AA_BBB_CC_DDD]
A total of 15 possible configurations.

Configuration (10 cells and [4, 3] blocks):
[__________]
Possibilities:
[AAAA_BBB__]
[AAAA__BBB_]
[AAAA___BBB]
[_AAAA_BBB_]
[_AAAA__BBB]
[__AAAA_BBB]
A total of 6 possible configurations.

Configuration (5 cells and [2, 1] blocks):
[_____]
Possibilities:
[AA_B_]
[AA__B]
[_AA_B]
A total of 3 possible configurations.

Configuration (10 cells and [3, 1] blocks):
[__________]
Possibilities:
[AAA_B_____]
[AAA__B____]
[AAA___B___]
[AAA____B__]
[AAA_____B_]
[AAA______B]
[_AAA_B____]
[_AAA__B___]
[_AAA___B__]
[_AAA____B_]
[_AAA_____B]
[__AAA_B___]
[__AAA__B__]
[__AAA___B_]
[__AAA____B]
[___AAA_B__]
[___AAA__B_]
[___AAA___B]
[____AAA_B_]
[____AAA__B]
[_____AAA_B]
A total of 21 possible configurations.

Configuration (5 cells and [2, 3] blocks):
[_____]
Possibilities:
object.Exception @nonoblock.d(17): Those blocks cannot fit in those cells.
----------------
0x0040AC17 in pure @safe void std.exception.bailOut(immutable(char)[], uint, const(char[]))
...

EchoLisp

;; size is the remaining # of cells
;; blocks is the list of remaining blocks size
;; cells is a stack where we push 0 = space or block size.
(define (nonoblock size blocks into: cells)
(cond
	((and (empty? blocks) (= 0 size)) (print-cells (stack->list cells)))
	
	((<= size 0) #f) ;; no hope - cut search
	((> (apply + blocks) size)  #f) ;; no hope - cut search
	
	(else
		(push cells 0) ;; space
		(nonoblock (1- size) blocks  cells)
		(pop cells)

	(when (!empty? blocks) 
		(when (stack-empty? cells) ;; first one (no space is allowed)
		(push cells (first blocks))
		(nonoblock  (- size (first blocks)) (rest blocks) cells)
		(pop cells))

		(push cells 0) ;; add space before
		(push cells (first blocks))
		(nonoblock  (- size (first blocks) 1) (rest blocks) cells)
		(pop cells)
		(pop cells)))))
	
(string-delimiter "")
(define block-symbs #( ?  📦 💣 💊  🍒 🌽 📘 📙 💰 🍯 ))

(define (print-cells cells)
	(writeln (string-append "|"
		(for/string ((cell cells)) 
			(if (zero? cell) "_"  
				(for/string ((i cell)) [block-symbs cell]))) "|")))
	
(define (task  nonotest)
	(for ((test nonotest)) 
		(define size (first test))
		(define blocks (second test))
		(printf "\n size:%d blocks:%d" size blocks)
		(if
			(> (+ (apply + blocks)(1- (length blocks))) size) 
				(writeln "❌ no solution for" size blocks)
			    (nonoblock size blocks (stack 'cells)))))
Output:
(define nonotest '((5 (2 1)) (5 ()) (10 (8)) (15 (2 3 2 3)) (5 (2 3))))
(task nonotest)

    size:5 blocks:(2 1)
    |💣💣__📦|    
    |💣💣_📦_|    
    |_💣💣_📦|    

    size:5 blocks:()
    |_____|    

    size:10 blocks:(8)
    |__💰💰💰💰💰💰💰💰|    
    |💰💰💰💰💰💰💰💰__|    
    |_💰💰💰💰💰💰💰💰_|    

    size:15 blocks:(2 3 2 3)
    |__💣💣_💊💊💊_💣💣_💊💊💊|    
    |💣💣___💊💊💊_💣💣_💊💊💊|    
    |💣💣__💊💊💊__💣💣_💊💊💊|    
    |💣💣__💊💊💊_💣💣__💊💊💊|    
    |💣💣__💊💊💊_💣💣_💊💊💊_|    
    |💣💣_💊💊💊___💣💣_💊💊💊|    
    |💣💣_💊💊💊__💣💣__💊💊💊|    
    |💣💣_💊💊💊__💣💣_💊💊💊_|    
    |💣💣_💊💊💊_💣💣___💊💊💊|    
    |💣💣_💊💊💊_💣💣__💊💊💊_|    
    |💣💣_💊💊💊_💣💣_💊💊💊__|    
    |_💣💣__💊💊💊_💣💣_💊💊💊|    
    |_💣💣_💊💊💊__💣💣_💊💊💊|    
    |_💣💣_💊💊💊_💣💣__💊💊💊|    
    |_💣💣_💊💊💊_💣💣_💊💊💊_|    

    size:5 blocks:(2 3)
    ❌ no solution for     5     (2 3)  

Elixir

Translation of: Ruby
defmodule Nonoblock do
  def solve(cell, blocks) do
    width = Enum.sum(blocks) + length(blocks) - 1
    if cell < width do
      raise "Those blocks will not fit in those cells"
    else
      nblocks(cell, blocks, "")
    end
  end
  
  defp nblocks(cell, _, position) when cell<=0, do:
    display(String.slice(position, 0..cell-1))
  defp nblocks(cell, blocks, position) when length(blocks)==0 or hd(blocks)==0, do:
    display(position <> String.duplicate(".", cell))
  defp nblocks(cell, blocks, position) do
    rest = cell - Enum.sum(blocks) - length(blocks) + 2
    [bl | brest] = blocks
    Enum.reduce(0..rest-1, 0, fn i,acc ->
      acc + nblocks(cell-i-bl-1, brest, position <> String.duplicate(".", i) <> String.duplicate("#",bl) <> ".")
    end)
  end
  
  defp display(str) do
    IO.puts nonocell(str)
    1                           # number of positions
  end
  
  def nonocell(str) do                  # "##.###..##" -> "|A|A|_|B|B|B|_|_|C|C|"
    slist = String.to_char_list(str) |> Enum.chunk_by(&(&1==?.)) |> Enum.map(&List.to_string(&1))
    chrs = Enum.map(?A..?Z, &List.to_string([&1]))
    result = nonocell_replace(slist, chrs, "")
             |> String.replace(".", "_")
             |> String.split("") |> Enum.join("|")
    "|" <> result
  end
  
  defp nonocell_replace([], _, result), do: result
  defp nonocell_replace([h|t], chrs, result) do
    if String.first(h) == "#" do
      [c | rest] = chrs
      nonocell_replace(t, rest, result <> String.replace(h, "#", c))
    else
      nonocell_replace(t, chrs, result <> h)
    end
  end
end

conf = [{ 5, [2, 1]},
        { 5, []},
        {10, [8]},
        {15, [2, 3, 2, 3]},
        { 5, [2, 3]}       ]
Enum.each(conf, fn {cell, blocks} ->
  try do
    IO.puts "Configuration:"
    IO.puts "#{Nonoblock.nonocell(String.duplicate(".",cell))} # #{cell} cells and #{inspect blocks} blocks"
    IO.puts "Possibilities:"
    count = Nonoblock.solve(cell, blocks)
    IO.puts "A total of #{count} Possible configurations.\n"
  rescue
    e in RuntimeError -> IO.inspect e
  end
end)
Output:
Configuration:
|_|_|_|_|_| # 5 cells and [2, 1] blocks
Possibilities:
|A|A|_|B|_|
|A|A|_|_|B|
|_|A|A|_|B|
A total of 3 Possible configurations.

Configuration:
|_|_|_|_|_| # 5 cells and [] blocks
Possibilities:
|_|_|_|_|_|
A total of 1 Possible configurations.

Configuration:
|_|_|_|_|_|_|_|_|_|_| # 10 cells and '\b' blocks
Possibilities:
|A|A|A|A|A|A|A|A|_|_|
|_|A|A|A|A|A|A|A|A|_|
|_|_|A|A|A|A|A|A|A|A|
A total of 3 Possible configurations.

Configuration:
|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_| # 15 cells and [2, 3, 2, 3] blocks
Possibilities:
|A|A|_|B|B|B|_|C|C|_|D|D|D|_|_|
|A|A|_|B|B|B|_|C|C|_|_|D|D|D|_|
|A|A|_|B|B|B|_|C|C|_|_|_|D|D|D|
|A|A|_|B|B|B|_|_|C|C|_|D|D|D|_|
|A|A|_|B|B|B|_|_|C|C|_|_|D|D|D|
|A|A|_|B|B|B|_|_|_|C|C|_|D|D|D|
|A|A|_|_|B|B|B|_|C|C|_|D|D|D|_|
|A|A|_|_|B|B|B|_|C|C|_|_|D|D|D|
|A|A|_|_|B|B|B|_|_|C|C|_|D|D|D|
|A|A|_|_|_|B|B|B|_|C|C|_|D|D|D|
|_|A|A|_|B|B|B|_|C|C|_|D|D|D|_|
|_|A|A|_|B|B|B|_|C|C|_|_|D|D|D|
|_|A|A|_|B|B|B|_|_|C|C|_|D|D|D|
|_|A|A|_|_|B|B|B|_|C|C|_|D|D|D|
|_|_|A|A|_|B|B|B|_|C|C|_|D|D|D|
A total of 15 Possible configurations.

Configuration:
|_|_|_|_|_| # 5 cells and [2, 3] blocks
Possibilities:
%RuntimeError{message: "Those blocks will not fit in those cells"}

Go

Translation of: Kotlin
package main

import (
    "fmt"
    "strings"
)

func printBlock(data string, le int) {
    a := []byte(data)
    sumBytes := 0
    for _, b := range a {
        sumBytes += int(b - 48)
    }
    fmt.Printf("\nblocks %c, cells %d\n", a, le)
    if le-sumBytes <= 0 {
        fmt.Println("No solution")
        return
    }
    prep := make([]string, len(a))
    for i, b := range a {
        prep[i] = strings.Repeat("1", int(b-48))
    }
    for _, r := range genSequence(prep, le-sumBytes+1) {
        fmt.Println(r[1:])
    }
}

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

func main() {
    printBlock("21", 5)
    printBlock("", 5)
    printBlock("8", 10)
    printBlock("2323", 15)
    printBlock("23", 5)
}
Output:
blocks [2 1], cells 5
11010
11001
01101

blocks [], cells 5
00000

blocks [8], cells 10
1111111100
0111111110
0011111111

blocks [2 3 2 3], cells 15
110111011011100
110111011001110
110111011000111
110111001101110
110111001100111
110111000110111
110011101101110
110011101100111
110011100110111
110001110110111
011011101101110
011011101100111
011011100110111
011001110110111
001101110110111

blocks [2 3], cells 5
No solution

J

Implementation:

nonoblock=:4 :0
  s=. 1+(1+x)-+/1+y
  pad=.1+(#~ s >+/"1)((1+#y)#s) #: i.s^1+#y
  ~.pad (_1}.1 }. ,. #&, 0 ,. 1 + i.@#@])"1]y,0
)

neat=: [: (#~ # $ 0 1"_)@": {&(' ',65}.a.)&.>

Task example:

   neat 5 nonoblock 2 1
AA B 
AA  B
 AA B
   neat 5 nonoblock ''
     
   neat 10 nonoblock 8
AAAAAAAA  
 AAAAAAAA 
  AAAAAAAA
   neat 15 nonoblock 2 3 2 3
AA BBB CC DDD  
AA BBB CC  DDD 
AA BBB CC   DDD
AA BBB  CC DDD 
AA BBB  CC  DDD
AA BBB   CC DDD
AA  BBB CC DDD 
AA  BBB CC  DDD
AA  BBB  CC DDD
AA   BBB CC DDD
 AA BBB CC DDD 
 AA BBB CC  DDD
 AA BBB  CC DDD
 AA  BBB CC DDD
  AA BBB CC DDD
   neat 5 nonoblock 2 3

Java

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

public class Nonoblock {

    public static void main(String[] args) {
        printBlock("21", 5);
        printBlock("", 5);
        printBlock("8", 10);
        printBlock("2323", 15);
        printBlock("23", 5);
    }

    static void printBlock(String data, int len) {
        int sumChars = data.chars().map(c -> Character.digit(c, 10)).sum();
        String[] a = data.split("");

        System.out.printf("%nblocks %s, cells %s%n", Arrays.toString(a), len);
        if (len - sumChars <= 0) {
            System.out.println("No solution");
            return;
        }

        List<String> prep = stream(a).filter(x -> !"".equals(x))
                .map(x -> repeat(Character.digit(x.charAt(0), 10), "1"))
                .collect(toList());

        for (String r : genSequence(prep, len - sumChars + 1))
            System.out.println(r.substring(1));
    }

    // permutation generator, translated from Python via D
    static List<String> genSequence(List<String> ones, int numZeros) {
        if (ones.isEmpty())
            return Arrays.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();
    }
}
blocks [2, 1], cells 5
11010
11001
01101

blocks [], cells 5
00000

blocks [8], cells 10
1111111100
0111111110
0011111111

blocks [2, 3, 2, 3], cells 15
110111011011100
110111011001110
110111011000111
110111001101110
110111001100111
110111000110111
110011101101110
110011101100111
110011100110111
110001110110111
011011101101110
011011101100111
011011100110111
011001110110111
001101110110111

blocks [2, 3], cells 5
No solution


JavaScript

const compose = (...fn) => (...x) => fn.reduce((a, b) => c => a(b(c)))(...x);
const inv = b => !b;
const arrJoin = str => arr => arr.join(str);
const mkArr = (l, f) => Array(l).fill(f);
const sumArr = arr => arr.reduce((a, b) => a + b, 0);
const sumsTo = val => arr => sumArr(arr) === val;
const zipper = arr => (p, c, i) => arr[i] ? [...p, c, arr[i]] : [...p, c];
const zip = (a, b) => a.reduce(zipper(b), []);
const zipArr = arr => a => zip(a, arr);
const hasInner = v => arr => arr.slice(1, -1).indexOf(v) >= 0;
const choose = (even, odd) => n => n % 2 === 0 ? even : odd;
const toBin = f => arr => arr.reduce(
    (p, c, i) => [...p, ...mkArr(c, f(i))], []);


const looper = (arr, max, acc = [[...arr]], idx = 0) => {
  if (idx !== arr.length) {
    const b = looper([...arr], max, acc, idx + 1)[0];
    if (b[idx] !== max) {
      b[idx] = b[idx] + 1;
      acc.push(looper([...b], max, acc, idx)[0]);
    }
  }
  return [arr, acc];
};

const gapPerms = (grpSize, numGaps, minVal = 0) => {
  const maxVal = numGaps - grpSize * minVal + minVal;
  return maxVal <= 0
      ? (grpSize === 2 ? [[0]] : [])
      : looper(mkArr(grpSize, minVal), maxVal)[1];
}

const test = (cells, ...blocks) => {
  const grpSize = blocks.length + 1;
  const numGaps = cells - sumArr(blocks);

  // Filter functions
  const sumsToTrg = sumsTo(numGaps);
  const noInnerZero = compose(inv, hasInner(0));

  // Output formatting
  const combine = zipArr([...blocks]);
  const choices = toBin(choose(0, 1));
  const output = compose(console.log, arrJoin(''), choices, combine);

  console.log(`\n${cells} cells. Blocks: ${blocks}`);
  gapPerms(grpSize, numGaps)
      .filter(noInnerZero)
      .filter(sumsToTrg)
      .map(output);
};

test(5, 2, 1);
test(5);
test(5, 5);
test(5, 1, 1, 1);
test(10, 8);
test(15, 2, 3, 2, 3);
test(10, 4, 3);
test(5, 2, 3);
Output:
5 cells. Blocks: 2,1
11010
11001
01101

5 cells. Blocks: 
00000

5 cells. Blocks: 5
11111

5 cells. Blocks: 1,1,1
10101

10 cells. Blocks: 8
1111111100
0111111110
0011111111

15 cells. Blocks: 2,3,2,3
110111011011100
110111011001110
110111011000111
110111001101110
110111001100111
110111000110111
110011101101110
110011101100111
110011100110111
110001110110111
011011101101110
011011101100111
011011100110111
011001110110111
001101110110111

10 cells. Blocks: 4,3
1111011100
1111001110
1111000111
0111101110
0111100111
0011110111

5 cells. Blocks: 2,3


Julia

minsized(arr) = join(map(x->"#"^x, arr), ".")
minlen(arr) = sum(arr) + length(arr) - 1
 
function sequences(blockseq, numblanks)
    if isempty(blockseq)
        return ["." ^ numblanks]
    elseif minlen(blockseq) == numblanks
        return minsized(blockseq)
    else    
        result = Vector{String}()
        allbuthead = blockseq[2:end]
        for leftspace in 0:(numblanks - minlen(blockseq))
            header = "." ^ leftspace * "#" ^ blockseq[1] * "."
            rightspace = numblanks - length(header)
            if isempty(allbuthead)
                push!(result, rightspace <= 0 ? header[1:numblanks] : header * "." ^ rightspace)            
            elseif minlen(allbuthead) == rightspace
                push!(result, header * minsized(allbuthead))
            else
                map(x -> push!(result, header * x), sequences(allbuthead, rightspace))
            end
        end
    end
    result
end

function nonoblocks(bvec, len)
    println("With blocks $bvec and $len cells:")
    len < minlen(bvec) ? println("No solution") : for seq in sequences(bvec, len) println(seq) end
end

nonoblocks([2, 1], 5)
nonoblocks(Vector{Int}([]), 5)
nonoblocks([8], 10)
nonoblocks([2, 3, 2, 3], 15)
nonoblocks([2, 3], 5)
Output:

With blocks [2, 1] and 5 cells:
##.#.
##..#
.##.#
With blocks Int64[] and 5 cells:
.....
With blocks [8] and 10 cells:
########..
.########.
..########
With blocks [2, 3, 2, 3] and 15 cells:
##.###.##.###..
##.###.##..###.
##.###.##...###
##.###..##.###.
##.###..##..###
##.###...##.###
##..###.##.###.
##..###.##..###
##..###..##.###
##...###.##.###
.##.###.##.###.
.##.###.##..###
.##.###..##.###
.##..###.##.###
..##.###.##.###
With blocks [2, 3] and 5 cells:
No solution

Kotlin

Translation of: Java
// version 1.2.0

fun printBlock(data: String, len: Int) {
    val a = data.toCharArray()
    val sumChars = a.map { it.toInt() - 48 }.sum()
    println("\nblocks ${a.asList()}, cells $len")
    if (len - sumChars <= 0) {
        println("No solution")
        return
    }
    val prep = a.map { "1".repeat(it.toInt() - 48) }
    for (r in genSequence(prep, len - sumChars + 1)) println(r.substring(1))
}

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)) {
            result.add("0".repeat(x) + ones[0] + tail)
        }
    }
    return result
}

fun main(args: Array<String>) {
    printBlock("21", 5)
    printBlock("", 5)
    printBlock("8", 10)
    printBlock("2323", 15)
    printBlock("23", 5)
}
Output:
blocks [2, 1], cells 5
11010
11001
01101

blocks [], cells 5
00000

blocks [8], cells 10
1111111100
0111111110
0011111111

blocks [2, 3, 2, 3], cells 15
110111011011100
110111011001110
110111011000111
110111001101110
110111001100111
110111000110111
110011101101110
110011101100111
110011100110111
110001110110111
011011101101110
011011101100111
011011100110111
011001110110111
001101110110111

blocks [2, 3], cells 5
No solution

Lua

local examples = {
	{5, {2, 1}},
	{5, {}},
	{10, {8}},
	{15, {2, 3, 2, 3}},
	{5, {2, 3}},
}

function deep (blocks, iBlock, freedom, str)
	if iBlock == #blocks then -- last
		for takenFreedom = 0, freedom do
			print (str..string.rep("0", takenFreedom) .. string.rep("1", blocks[iBlock]) .. string.rep("0", freedom - takenFreedom))
			total = total + 1
		end
	else
		for takenFreedom = 0, freedom do
			local str2 = str..string.rep("0", takenFreedom) .. string.rep("1", blocks[iBlock]) .. "0"
			deep (blocks, iBlock+1, freedom-takenFreedom, str2)
		end
	end
end

function main (cells, blocks) -- number, list
	local str = "	"
	print (cells .. ' cells and {' .. table.concat(blocks, ', ') .. '} blocks')
	local freedom = cells - #blocks + 1 -- freedom
	for iBlock = 1, #blocks do
		freedom = freedom - blocks[iBlock]
	end
	if #blocks == 0 then
		print ('no blocks')
		print (str..string.rep("0", cells))
		total = 1
	elseif freedom < 0 then
		print ('no solutions')
	else
		print ('Possibilities:')
		deep (blocks, 1, freedom, str)
	end
end

for i, example in ipairs (examples) do
	print ("\n--")
	total = 0
	main (example[1], example[2])
	print ('A total of ' .. total .. ' possible configurations.')
end
Output:
--
5 cells and {2, 1} blocks
Possibilities:
	11010
	11001
	01101
A total of 3 possible configurations.

--
5 cells and {} blocks
no blocks
	00000
A total of 1 possible configurations.

--
10 cells and {8} blocks
Possibilities:
	1111111100
	0111111110
	0011111111
A total of 3 possible configurations.

--
15 cells and {2, 3, 2, 3} blocks
Possibilities:
	110111011011100
	110111011001110
	110111011000111
	110111001101110
	110111001100111
	110111000110111
	110011101101110
	110011101100111
	110011100110111
	110001110110111
	011011101101110
	011011101100111
	011011100110111
	011001110110111
	001101110110111
A total of 15 possible configurations.

--
5 cells and {2, 3} blocks
no solutions
A total of 0 possible configurations.

M2000 Interpreter

Recursive

Module NonoBlock {
      Form 80,40
      Flush
      Print "Nonoblock" 
      Data 5, (2, 1)
      Data 5, (,)
      Data 10, (8,) 
      Data 15, (2,3,2,3)
      Data 5, (2,3)
      Def BLen(a$)=(Len(a$)-1)/2
      Function UseLetter(arr) {
            Dim Base 0, Res$(Len(arr))
            Link Res$() to Res()
            Def Ord$(a$)=ChrCode$(Chrcode(a$)+1)
            L$="A"
            i=each(arr)
            While i {
                  Res$(i^)=String$("|"+L$, Array(i))+"|"
                  L$=Ord$(L$)
            }
            =Res()
      }
      Count=0
      For i=1 to 5
      Read Cells, Blocks
      Blocks=UseLetter(Blocks)
      Print str$(i,"")+".", "Cells=";Cells, "", iF(len(Blocks)=0->("Empty",), Blocks)
      PrintRow( "|", Cells, Blocks, &Count)
      CheckCount()
      Next I
      Sub CheckCount()
            If count=0 Then Print " Impossible"
            count=0
      End Sub
      Sub PrintRow(Lpart$, Cells, Blocks, &Comp)
            If len(Blocks)=0 Then Comp++ :Print Format$("{0::-3} {1}", Comp, lpart$+String$("_|", Cells)):  Exit Sub
            If Cells<=0 Then Exit Sub
            Local TotalBlocksLength=0, Sep_Spaces=-1
            Local Block=Each(Blocks), block$
            While Block {
                  Block$=Array$(Block)    
                  TotalBlocksLength+=Blen(Block$)
                  Sep_Spaces++
            }
            Local MaxLengthNeed=TotalBlocksLength+Sep_Spaces
            If MaxLengthNeed>Cells Then Exit Sub
            block$=Array$(Car(Blocks))
            local temp=Blen(block$)
            block$=Mid$(Block$, 2)
            If Len(Blocks)>1 Then block$+="_|" :temp++
            PrintRow(Lpart$+block$, Cells-temp, Cdr(Blocks), &Comp)
            PrintRow(lpart$+String$("_|", 1), Cells-1,Blocks, &Comp)
      End Sub
}
NonoBlock
Output:
Nonoblock
1.    Cells=5 |A|A| |B| 
  1 |A|A|_|B|_|
  2 |A|A|_|_|B|
  3 |_|A|A|_|B|
2.    Cells=5 Empty
  1 |_|_|_|_|_|
3.    Cells=10 |A|A|A|A|A|A|A|A| 
  1 |A|A|A|A|A|A|A|A|_|_|
  2 |_|A|A|A|A|A|A|A|A|_|
  3 |_|_|A|A|A|A|A|A|A|A|
4.    Cells=15 |A|A| |B|B|B| |C|C| |D|D|D| 
  1 |A|A|_|B|B|B|_|C|C|_|D|D|D|_|_|
  2 |A|A|_|B|B|B|_|C|C|_|_|D|D|D|_|
  3 |A|A|_|B|B|B|_|C|C|_|_|_|D|D|D|
  4 |A|A|_|B|B|B|_|_|C|C|_|D|D|D|_|
  5 |A|A|_|B|B|B|_|_|C|C|_|_|D|D|D|
  6 |A|A|_|B|B|B|_|_|_|C|C|_|D|D|D|
  7 |A|A|_|_|B|B|B|_|C|C|_|D|D|D|_|
  8 |A|A|_|_|B|B|B|_|C|C|_|_|D|D|D|
  9 |A|A|_|_|B|B|B|_|_|C|C|_|D|D|D|
 10 |A|A|_|_|_|B|B|B|_|C|C|_|D|D|D|
 11 |_|A|A|_|B|B|B|_|C|C|_|D|D|D|_|
 12 |_|A|A|_|B|B|B|_|C|C|_|_|D|D|D|
 13 |_|A|A|_|B|B|B|_|_|C|C|_|D|D|D|
 14 |_|A|A|_|_|B|B|B|_|C|C|_|D|D|D|
 15 |_|_|A|A|_|B|B|B|_|C|C|_|D|D|D|
5.    Cells=5 |A|A| |B|B|B| 
 Impossible

Non Recursive

Module Nonoblock (n, m) {
      Print "Cells:",n," Blocks:",m
      Dim n(1 to n), m(1 to m), sp(1 to m*2), sk(1 to m*2), part(1 to m)
      queue=0      
      If m>0 Then {
            Print "Block Size:",
            For i=1 to m {
                  Read m(i)
                  Print m(i),
            }
            Print
            part(m)=m(m)      
            If m>1 Then {
                  For i=m-1 to 1 {
                        part(i)=m(i)+part(i+1)+1
                  }
            }
      }
      If part(1)>n Then {
          Print "Impossible"  
      } Else {
            p1=0
            l=0
            Counter=0
            While p1<=n-part(1) {
                  k=0
                  p=p1+1
                  For i=1 to n {
                        n(i)=0
                  }
                  flag=True
                  Repeat {
                        While k<m {
                              k++
                              l=0
                              While l<m(k) and p<=n {
                                    l++
                                    n(p)=1
                                    p++
                              }
                              If p<n Then {
                                    n(p)=0
                                    p++
                                   If k<m Then {
                                          If p+part(k+1)<n+1 Then {
                                                queue++
                                                sp(queue)=p
                                                sk(queue)=k
                                          }
                                    }
                              }
                        }
                        flag=True
                        If l=m(k)  Then {
                              counter++
                              Print Str$(counter,"0000  ");
                              For i=1 to n {
                                    Print n(i);" ";
                              }
                              Print
                              If queue>0 Then  {
                                    p=sp(queue)
                                    k=sk(queue)
                                    queue--
                                    For i=p to n {
                                          n(i)=0
                                    }
                                    p++
                                    If k<m Then {
                                          If p+part(k+1)<n+1 Then {
                                                queue++
                                                sp(queue)=p
                                               ' sk(queue)=k
                                       }
                                    }
                                    flag=False
                              } 
                        }
                  } Until flag
                  p1++
                  If k=0 Then Exit
            }
      }
}

Nonoblock 5,2,2,1
Nonoblock 5,0
Nonoblock 10,1,8
Nonoblock 15,4,2,3,2,3
Nonoblock 5,2,3,2

Mathematica/Wolfram Language

ClearAll[SpacesDistributeOverN, Possibilities]
SpacesDistributeOverN[s_, p_] := 
 Flatten[
  Permutations /@ (Join[#, ConstantArray[0, p - Length[#]]] & /@ 
     IntegerPartitions[s, p]), 1]
Possibilities[hint_, len_] := 
 Module[{p = hint, l = len, b = Length[hint], Spaces, out},
  Spaces = # + (Prepend[Append[ConstantArray[1, b - 1], 0], 
        0]) & /@ (SpacesDistributeOverN[l - Total@p - (b - 1), b + 1]);
  out = Flatten /@ ( 
     Riffle[#, Map[Table[1, {#}] &, p, {1}]] & /@ 
      Map[Table[0, {#}] &, Spaces, {2}]);
  StringJoin @@@ (out /. {0 -> ".", 1 -> "#"})
  ]
Possibilities[{}, len_] := Module[{},
  {StringJoin[ConstantArray[".", len]]}
  ]
Possibilities[{2, 1}, 5]
Possibilities[{}, 5]
Possibilities[{8}, 10]
Possibilities[{2, 3, 2, 3}, 15]
Possibilities[{2, 3}, 5]
Output:
{".##.#", "##..#", "##.#."}
{"....."}
{"..########", "########..", ".########."}
{"..##.###.##.###", "##...###.##.###", "##.###...##.###", "##.###.##...###", "##.###.##.###..", ".##..###.##.###", ".##.###..##.###", ".##.###.##..###", ".##.###.##.###.", "##..###..##.###", "##..###.##..###", "##..###.##.###.", "##.###..##..###", "##.###..##.###.", "##.###.##..###."}
{}

Nim

Translation of: Go
import math, sequtils, strformat, strutils


proc genSequence(ones: seq[string]; numZeroes: Natural): seq[string] =
  if ones.len == 0: return @[repeat('0', numZeroes)]
  for x in 1..(numZeroes - ones.len + 1):
    let skipOne = ones[1..^1]
    for tail in genSequence(skipOne, numZeroes - x):
      result.add repeat('0', x) & ones[0] & tail


proc printBlock(data: string; length: Positive) =

  let a = mapIt(data, ord(it) - ord('0'))
  let sumBytes = sum(a)

  echo &"\nblocks {($a)[1..^1]} cells {length}"
  if length - sumBytes <= 0:
    echo "No solution"
    return

  var prep: seq[string]
  for b in a: prep.add repeat('1', b)

  for r in genSequence(prep, length - sumBytes + 1):
    echo r[1..^1]


when isMainModule:
  printBlock("21", 5)
  printBlock("", 5)
  printBlock("8", 10)
  printBlock("2323", 15)
  printBlock("23", 5)
Output:
blocks [2, 1] cells 5
11010
11001
01101

blocks [] cells 5
00000

blocks [8] cells 10
1111111100
0111111110
0011111111

blocks [2, 3, 2, 3] cells 15
110111011011100
110111011001110
110111011000111
110111001101110
110111001100111
110111000110111
110011101101110
110011101100111
110011100110111
110001110110111
011011101101110
011011101100111
011011100110111
011001110110111
001101110110111

blocks [2, 3] cells 5
No solution

Pascal

A console application in Free Pascal, created with the Lazarus IDE.

With 15 cells and [2,3,2,3] blocks, it's a question of how to distribute 5 gap characters among 5 gaps (including the 2 gaps at the ends). To allow for the requirement that the 3 inner gaps must be strictly positive, we can reduce the size of each inner gap by 1, provided we remember to restore the deleted gap character when printing the result. Then 2 gap characters need to be distributed among 5 non-negative gaps. In general, for integers n > 0 and s, the task amounts to finding all arrays of n non-negative integers that sum to s. An iterative method is shown below.

program Nonoblock;
uses SysUtils;

// Working through solutions to the problem:
// Fill an array z[] with non-negative integers
//  whose sum is the passed-in integer s.
function GetFirstSolution( var z : array of integer;
                               s : integer) : boolean;
var
  j : integer;
begin
  result := (s >= 0) and (High(z) >= 0);  // failed if s < 0 or array is empty
  if result then begin // else initialize to solution 0, ..., 0, s
    j := High(z);  z[j] := s;
    while (j > 0) do begin
      dec(j);      z[j] := 0;
    end;
  end;
end;

// Next solution: return true for success, false if no more solutions.
// Solutions are generated in lexicographic order.
function GetNextSolution( var z : array of integer) : boolean;
var
  h, j : integer;
begin
  h := High(z);
  j := h; // find highest index j such that z[j] > 0.
  while (j > 0) and (z[j] = 0) do dec(j);
  result := (j > 0);   // if index is 0, or there is no such index, failed
  if result then begin // else update caller's array to give next solution
    inc(z[j - 1]);
    z[h] := z[j] - 1;
    if (j < h) then z[j] := 0;
  end;
end;

// Procedure to print solutions to nonoblock task on RosettaCode
procedure PrintSolutions( nrCells : integer;
                          blockSizes : array of integer);
const // cosmetic
  MARGIN = 4;
  GAP_CHAR = '.';
  BLOCK_CHAR = '#';
var
  sb : SysUtils.TStringBuilder;
  nrBlocks, blockSum, gapSum : integer;
  gapSizes : array of integer;
  i, nrSolutions : integer;
begin
  nrBlocks := Length( blockSizes);

  // Print a title, showing the number of cells and the block sizes
  sb := SysUtils.TStringBuilder.Create();
  sb.AppendFormat('%d cells; blocks [', [nrCells]);
  for i := 0 to nrBlocks - 1 do begin
    if (i > 0) then sb.Append(',');
    sb.Append( blockSizes[i]);
  end;
  sb.Append(']');
  WriteLn( sb.ToString());

  blockSum := 0; // total of block sizes
  for i := 0 to nrBlocks - 1 do inc( blockSum, blockSizes[i]);

  gapSum := nrCells - blockSum;
  // Except in the trivial case of no blocks,
  // we reduce the size of each inner gap by 1.
  if nrBlocks > 0 then dec( gapSum, nrBlocks - 1);

  // Work through all solutions and print them nicely.
  nrSolutions := 0;
  SetLength( gapSizes, nrBlocks + 1); // include the gap at each end
  if GetFirstSolution( gapSizes, gapSum) then begin
    repeat
      inc( nrSolutions);
      sb.Clear();
      sb.Append( ' ', MARGIN);
      for i := 0 to nrBlocks - 1 do begin
        sb.Append( GAP_CHAR, gapSizes[i]);
        // We reduced the inner gaps by 1; now we restore the deleted char.
        if (i > 0) then sb.Append( GAP_CHAR);
        sb.Append( BLOCK_CHAR, blockSizes[i]);
      end;
      sb.Append( GAP_CHAR, gapSizes[nrBlocks]);
      WriteLn( sb.ToString());
    until not GetNextSolution( gapSizes);
  end;
  sb.Free();
  WriteLn( SysUtils.Format( 'Number of solutions = %d', [nrSolutions]));
  WriteLn('');
end;

// Main program
begin
  PrintSolutions( 5, [2,1]);
  PrintSolutions( 5, []);
  PrintSolutions( 10, [8]);
  PrintSolutions( 15, [2,3,2,3]);
  PrintSolutions( 5, [2,3]);
end.
Output:
5 cells; blocks [2,1]
    ##.#.
    ##..#
    .##.#
Number of solutions = 3

5 cells; blocks []
    .....
Number of solutions = 1

10 cells; blocks [8]
    ########..
    .########.
    ..########
Number of solutions = 3

15 cells; blocks [2,3,2,3]
    ##.###.##.###..
    ##.###.##..###.
    ##.###.##...###
    ##.###..##.###.
    ##.###..##..###
    ##.###...##.###
    ##..###.##.###.
    ##..###.##..###
    ##..###..##.###
    ##...###.##.###
    .##.###.##.###.
    .##.###.##..###
    .##.###..##.###
    .##..###.##.###
    ..##.###.##.###
Number of solutions = 15

5 cells; blocks [2,3]
Number of solutions = 0

Perl

use strict;
use warnings;

while( <DATA> )
  {
  print "\n$_", tr/\n/=/cr;
  my ($cells, @blocks) = split;
  my $letter = 'A';
  $_ = join '.', map { $letter++ x $_ } @blocks;
  $cells < length and print("no solution\n"), next;
  $_ .= '.' x ($cells - length) . "\n";
  1 while print, s/^(\.*)\b(.*?)\b(\w+)\.\B/$2$1.$3/;
  }

__DATA__
5 2 1
5
10 8
15 2 3 2 3
5 2 3
Output:
5 2 1
=====
AA.B.
AA..B
.AA.B

5
=
.....

10 8
====
AAAAAAAA..
.AAAAAAAA.
..AAAAAAAA

15 2 3 2 3
==========
AA.BBB.CC.DDD..
AA.BBB.CC..DDD.
AA.BBB..CC.DDD.
AA..BBB.CC.DDD.
.AA.BBB.CC.DDD.
AA.BBB.CC...DDD
AA.BBB..CC..DDD
AA..BBB.CC..DDD
.AA.BBB.CC..DDD
AA.BBB...CC.DDD
AA..BBB..CC.DDD
.AA.BBB..CC.DDD
AA...BBB.CC.DDD
.AA..BBB.CC.DDD
..AA.BBB.CC.DDD

5 2 3
=====
no solution

Phix

with javascript_semantics
function nobr(sequence res, string neat, integer ni, integer ch, sequence blocks)
    if length(blocks)=0 then
        res = append(res,neat)
    else
        integer b = blocks[1]
        blocks = blocks[2..$]
        integer l = (sum(blocks)+length(blocks)-1)*2,
                e = length(neat)-l-b*2
        for i=ni to e by 2 do
            for j=i to i+b*2-2 by 2 do
                neat[j] = ch
            end for
            res = nobr(res,neat,i+b*2+2,ch+1,blocks)
            neat[i] = ' '
        end for
    end if
    return res
end function
 
function nonoblock(integer len, sequence blocks)
    string neat = "|"&join(repeat(' ',len),'|')&"|"
    return nobr({},neat,2,'A',blocks)
end function
 
sequence tests = {{5,{2,1}},
                  {5,{}},
                  {10,{8}},
                  {15,{2, 3, 2, 3}},
                  {10,{4, 3}},
                  {5,{2,1}},
                  {10,{3, 1}},
                  {5,{2, 3}}}
integer len
sequence blocks, res
for i=1 to length(tests) do
    {len,blocks} = tests[i]
    string ti = sprintf("%d cells with blocks %s",{len,sprint(blocks)})
    printf(1,"%s\n%s\n",{ti,repeat('=',length(ti))})
    res = nonoblock(len,blocks)
    if length(res)=0 then
        printf(1,"No solutions.\n")
    else
        for ri=1 to length(res) do
            printf(1,"%3d:  %s\n",{ri,res[ri]})
        end for
    end if
    printf(1,"\n")
end for
Output:
5 cells with blocks {2,1}
=========================
  1:  |A|A| |B| |
  2:  |A|A| | |B|
  3:  | |A|A| |B|

5 cells with blocks {}
======================
  1:  | | | | | |

10 cells with blocks {8}
========================
  1:  |A|A|A|A|A|A|A|A| | |
  2:  | |A|A|A|A|A|A|A|A| |
  3:  | | |A|A|A|A|A|A|A|A|

15 cells with blocks {2,3,2,3}
==============================
  1:  |A|A| |B|B|B| |C|C| |D|D|D| | |
  2:  |A|A| |B|B|B| |C|C| | |D|D|D| |
  3:  |A|A| |B|B|B| |C|C| | | |D|D|D|
  4:  |A|A| |B|B|B| | |C|C| |D|D|D| |
  5:  |A|A| |B|B|B| | |C|C| | |D|D|D|
  6:  |A|A| |B|B|B| | | |C|C| |D|D|D|
  7:  |A|A| | |B|B|B| |C|C| |D|D|D| |
  8:  |A|A| | |B|B|B| |C|C| | |D|D|D|
  9:  |A|A| | |B|B|B| | |C|C| |D|D|D|
 10:  |A|A| | | |B|B|B| |C|C| |D|D|D|
 11:  | |A|A| |B|B|B| |C|C| |D|D|D| |
 12:  | |A|A| |B|B|B| |C|C| | |D|D|D|
 13:  | |A|A| |B|B|B| | |C|C| |D|D|D|
 14:  | |A|A| | |B|B|B| |C|C| |D|D|D|
 15:  | | |A|A| |B|B|B| |C|C| |D|D|D|

10 cells with blocks {4,3}
==========================
  1:  |A|A|A|A| |B|B|B| | |
  2:  |A|A|A|A| | |B|B|B| |
  3:  |A|A|A|A| | | |B|B|B|
  4:  | |A|A|A|A| |B|B|B| |
  5:  | |A|A|A|A| | |B|B|B|
  6:  | | |A|A|A|A| |B|B|B|

5 cells with blocks {2,1}
=========================
  1:  |A|A| |B| |
  2:  |A|A| | |B|
  3:  | |A|A| |B|

10 cells with blocks {3,1}
==========================
  1:  |A|A|A| |B| | | | | |
  2:  |A|A|A| | |B| | | | |
  3:  |A|A|A| | | |B| | | |
  4:  |A|A|A| | | | |B| | |
  5:  |A|A|A| | | | | |B| |
  6:  |A|A|A| | | | | | |B|
  7:  | |A|A|A| |B| | | | |
  8:  | |A|A|A| | |B| | | |
  9:  | |A|A|A| | | |B| | |
 10:  | |A|A|A| | | | |B| |
 11:  | |A|A|A| | | | | |B|
 12:  | | |A|A|A| |B| | | |
 13:  | | |A|A|A| | |B| | |
 14:  | | |A|A|A| | | |B| |
 15:  | | |A|A|A| | | | |B|
 16:  | | | |A|A|A| |B| | |
 17:  | | | |A|A|A| | |B| |
 18:  | | | |A|A|A| | | |B|
 19:  | | | | |A|A|A| |B| |
 20:  | | | | |A|A|A| | |B|
 21:  | | | | | |A|A|A| |B|

5 cells with blocks {2,3}
=========================
No solutions.

Python

def nonoblocks(blocks, cells):
    if not blocks or blocks[0] == 0:
        yield [(0, 0)]
    else:
        assert sum(blocks) + len(blocks)-1 <= cells, \
            'Those blocks will not fit in those cells'
        blength, brest = blocks[0], blocks[1:]      # Deal with the first block of length
        minspace4rest = sum(1+b for b in brest)     # The other blocks need space
        # Slide the start position from left to max RH index allowing for other blocks.
        for bpos in range(0, cells - minspace4rest - blength + 1):
            if not brest:
                # No other blocks to the right so just yield this one.
                yield [(bpos, blength)]
            else:
                # More blocks to the right so create a *sub-problem* of placing
                # the brest blocks in the cells one space to the right of the RHS of 
                # this block.
                offset = bpos + blength +1
                nonoargs = (brest, cells - offset)  # Pre-compute arguments to nonoargs
                # recursive call to nonoblocks yields multiple sub-positions
                for subpos in nonoblocks(*nonoargs):
                    # Remove the offset from sub block positions
                    rest = [(offset + bp, bl) for bp, bl in subpos]
                    # Yield this block plus sub blocks positions
                    vec = [(bpos, blength)] + rest
                    yield vec

def pblock(vec, cells):
    'Prettyprints each run of blocks with a different letter A.. for each block of filled cells'
    vector = ['_'] * cells
    for ch, (bp, bl) in enumerate(vec, ord('A')):
        for i in range(bp, bp + bl):
            vector[i] = chr(ch) if vector[i] == '_' else'?'
    return '|' + '|'.join(vector) + '|'


if __name__ == '__main__':
    for blocks, cells in (
            ([2, 1], 5),
            ([], 5),
            ([8], 10),
            ([2, 3, 2, 3], 15),
           # ([4, 3], 10),
           # ([2, 1], 5),
           # ([3, 1], 10),
            ([2, 3], 5),
            ):
        print('\nConfiguration:\n    %s # %i cells and %r blocks' % (pblock([], cells), cells, blocks))        
        print('  Possibilities:')
        for i, vector in enumerate(nonoblocks(blocks, cells)):
            print('   ', pblock(vector, cells))
        print('  A total of %i Possible configurations.' % (i+1))
Output:
Configuration:
    |_|_|_|_|_| # 5 cells and [2, 1] blocks
  Possibilities:
    |A|A|_|B|_|
    |A|A|_|_|B|
    |_|A|A|_|B|
  A total of 3 Possible configurations.

Configuration:
    |_|_|_|_|_| # 5 cells and [] blocks
  Possibilities:
    |_|_|_|_|_|
  A total of 1 Possible configurations.

Configuration:
    |_|_|_|_|_|_|_|_|_|_| # 10 cells and [8] blocks
  Possibilities:
    |A|A|A|A|A|A|A|A|_|_|
    |_|A|A|A|A|A|A|A|A|_|
    |_|_|A|A|A|A|A|A|A|A|
  A total of 3 Possible configurations.

Configuration:
    |_|_|_|_|_|_|_|_|_|_|_|_|_|_|_| # 15 cells and [2, 3, 2, 3] blocks
  Possibilities:
    |A|A|_|B|B|B|_|C|C|_|D|D|D|_|_|
    |A|A|_|B|B|B|_|C|C|_|_|D|D|D|_|
    |A|A|_|B|B|B|_|C|C|_|_|_|D|D|D|
    |A|A|_|B|B|B|_|_|C|C|_|D|D|D|_|
    |A|A|_|B|B|B|_|_|C|C|_|_|D|D|D|
    |A|A|_|B|B|B|_|_|_|C|C|_|D|D|D|
    |A|A|_|_|B|B|B|_|C|C|_|D|D|D|_|
    |A|A|_|_|B|B|B|_|C|C|_|_|D|D|D|
    |A|A|_|_|B|B|B|_|_|C|C|_|D|D|D|
    |A|A|_|_|_|B|B|B|_|C|C|_|D|D|D|
    |_|A|A|_|B|B|B|_|C|C|_|D|D|D|_|
    |_|A|A|_|B|B|B|_|C|C|_|_|D|D|D|
    |_|A|A|_|B|B|B|_|_|C|C|_|D|D|D|
    |_|A|A|_|_|B|B|B|_|C|C|_|D|D|D|
    |_|_|A|A|_|B|B|B|_|C|C|_|D|D|D|
  A total of 15 Possible configurations.

Configuration:
    |_|_|_|_|_| # 5 cells and [2, 3] blocks
  Possibilities:
Traceback (most recent call last):
  File "C:/Users/Paddy/Google Drive/Code/nonoblocks.py", line 104, in <module>
    for i, vector in enumerate(nonoblocks(blocks, cells)):
  File "C:/Users/Paddy/Google Drive/Code/nonoblocks.py", line 60, in nonoblocks
    'Those blocks will not fit in those cells'
AssertionError: Those blocks will not fit in those cells

Racket

This implementation does not "error" on the impossible case.

Knowing that there are no solutions (empty result list) is good enough.

Also, the blocks are not identified. I suppose they could be easily enough, but in the nonogram task, these patterns are converted to bit-fields shortly after the nonoblock generation, and bits have no names (sad, but true).

#lang racket
(require racket/trace)

(define add1-to-car (match-lambda [(cons (app add1 p1) t) (cons p1 t)]))

;; inputs:
;;   cells  -- available cells
;;   blocks -- list of block widths
;; output:
;;   gap-block+gaps
;;   where gap-block+gaps is:
;;   (list gap)                            -- a single gap
;;   (list gap block-width gap-block+gaps) -- padding to left, a block, right hand side
(define (nonoblock cells blocks)
  (match* ((- cells (apply + (length blocks) -1 blocks)) #| padding available on both sides |# blocks)
    [(_ (list)) (list (list cells))] ; generates an empty list of padding
    
    [((? negative?) _) null] ; impossible to satisfy
    
    [((and avp
           ;; use add1 with in-range because we actually want from 0 to available-padding
           ;; without add1, in-range iterates from 0 to (available-padding - 1)
           (app add1 avp+1))
      (list block))
     (for/list ((l-pad (in-range 0 avp+1)))
       (define r-pad (- avp l-pad)) ; what remains goes to right
       (list l-pad block r-pad))]
    
    [((app add1 avp+1) (list block more-blocks ...))
     (for*/list ((l-pad (in-range 0 avp+1))
                 (cells-- (in-value (- cells block l-pad 1)))
                 (r-blocks (in-value (nonoblock cells-- more-blocks)))
                 (r-block (in-list r-blocks)))
       (list* l-pad block (add1-to-car r-block)))])) ; put a single space pad on left of r-block

(define (neat rslt)
  (define dots (curryr make-string #\.))
  (define Xes (curryr make-string #\X))
  (define inr
    (match-lambda
      [(list 0 (app Xes b) t ...)
       (string-append b (inr t))]
      [(list (app dots p) (app Xes b) t ...)
       (string-append p b (inr t))]
      [(list (app dots p)) p]))
  (define (neat-row r)
    (string-append "|" (inr r) "|"))
  (string-join (map neat-row rslt) "\n"))

(define (tst c b)
  (define rslt (nonoblock c b))
  (define rslt-l (length rslt))
  (printf "~a cells, ~a blocks => ~a~%~a~%" c b
          (match rslt-l
            [0 "impossible"]
            [1 "1 solution"]
            [(app (curry format "~a solutions") r) r])
          (neat rslt)))

(module+ test
  (tst  5 '[2 1])
  (tst  5 '[])
  (tst 10 '[8])
  (tst 15 '[2 3 2 3])
  (tst  5 '[2 3]))
Output:
5 cells, (2 1) blocks => 3 solutions
|XX.X.|
|XX..X|
|.XX.X|
5 cells, () blocks => 1 solution
|.....|
10 cells, (8) blocks => 3 solutions
|XXXXXXXX..|
|.XXXXXXXX.|
|..XXXXXXXX|
15 cells, (2 3 2 3) blocks => 15 solutions
|XX.XXX.XX.XXX..|
|XX.XXX.XX..XXX.|
|XX.XXX.XX...XXX|
|XX.XXX..XX.XXX.|
|XX.XXX..XX..XXX|
|XX.XXX...XX.XXX|
|XX..XXX.XX.XXX.|
|XX..XXX.XX..XXX|
|XX..XXX..XX.XXX|
|XX...XXX.XX.XXX|
|.XX.XXX.XX.XXX.|
|.XX.XXX.XX..XXX|
|.XX.XXX..XX.XXX|
|.XX..XXX.XX.XXX|
|..XX.XXX.XX.XXX|
5 cells, (2 3) blocks => impossible

Raku

(formerly Perl 6)

Translation of: Perl
for (5, [2,1]), (5, []), (10, [8]), (5, [2,3]), (15, [2,3,2,3]) -> ($cells, @blocks) {
    say $cells, ' cells with blocks: ', @blocks ?? join ', ', @blocks !! '∅';
    my $letter = 'A';
    my $row = join '.', map { $letter++ x $_ }, @blocks;
    say "no solution\n" and next if $cells < $row.chars;
    say $row ~= '.' x $cells - $row.chars;
    say $row while $row ~~ s/^^ (\.*) <|w> (.*?) <|w> (\w+) \.<!|w> /$1$0.$2/;
    say '';
}
Output:
5 cells with blocks: 2, 1
AA.B.
AA..B
.AA.B

5 cells with blocks: ∅
.....

10 cells with blocks: 8
AAAAAAAA..
.AAAAAAAA.
..AAAAAAAA

5 cells with blocks: 2, 3
no solution

15 cells with blocks: 2, 3, 2, 3
AA.BBB.CC.DDD..
AA.BBB.CC..DDD.
AA.BBB..CC.DDD.
AA..BBB.CC.DDD.
.AA.BBB.CC.DDD.
AA.BBB.CC...DDD
AA.BBB..CC..DDD
AA..BBB.CC..DDD
.AA.BBB.CC..DDD
AA.BBB...CC.DDD
AA..BBB..CC.DDD
.AA.BBB..CC.DDD
AA...BBB.CC.DDD
.AA..BBB.CC.DDD
..AA.BBB.CC.DDD

REXX

/*REXX program enumerates all possible configurations (or an error) for nonogram puzzles*/
             $.=;    $.1=  5   2 1
                     $.2=  5
                     $.3= 10   8
                     $.4= 15   2 3 2 3
                     $.5=  5   2 3
      do  i=1  while $.i\==''
      parse var  $.i   N  blocks                 /*obtain  N  and  blocks   from array. */
      N= strip(N);     blocks= space(blocks)     /*assign stripped   N   and   blocks.  */
      call nono                                  /*incoke NONO subroutine for heavy work*/
      end   /*i*/
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
nono: say copies('=', 70)                                 /*display seperator for title.*/
      say 'For '   N   " cells  and blocks of: "   blocks /*display the title for output*/
      z=                                                  /*assign starter value for Z. */
          do w=1  for words(blocks)                       /*process each of the blocks. */
          z= z copies('#', word(blocks,w) )               /*build a string for 1st value*/
          end   /*w*/                                     /*Z  now has a leading blank. */
      #= 1                                                /*number of positions (so far)*/
      z= translate( strip(z), ., ' ');   L= length(z)     /*change blanks to periods.   */
      if L>N  then do;   say '***error***  invalid blocks for number of cells.';   return
                   end
      @.0=;           @.1= z;         !.=0       /*assign default and the first position*/
      z= pad(z)                                  /*fill─out (pad) the value with periods*/

         do prepend=1  while words(blocks)\==0   /*process all the positions (leading .)*/
         new= . || @.prepend                     /*create positions with leading dots.  */
         if length(new)>N  then leave            /*Length is too long?  Then stop adding*/
         call add                                /*add position that has a leading dot. */
         end   /*prepend*/                       /* [↑]  prepend positions with dots.   */

         do   k=1  for N                         /*process each of the positions so far.*/
           do c=1  for N                         /*   "      "   "  "  position blocks. */
           if @.c==''  then iterate              /*if string is null,  skip the string. */
           p= loc(@.c, k)                        /*find location of block in position.  */
           if p==0 | p>=N  then iterate          /*Location zero or out─of─range?  Skip.*/
           new= strip( insert(., @.c, p),'T',.)  /*insert a dot and strip trailing dots.*/
           if strip(new,'T',.)=@.c  then iterate /*Is it the same value?  Then skip it. */
           if length(new)<=N  then call add      /*Is length OK?   Then add position.   */
           end   /*k*/
         end     /*c*/
      say
      say '─position─'  center("value", max(7, length(z) ), '─')  /*show hdr for output.*/

               do m=1  for #
               say center(m, 10)   pad(@.m)      /*display the index count and position.*/
               end   /*m*/
      return
/*──────────────────────────────────────────────────────────────────────────────────────*/
loc:  _=0; do arg(2); _=pos('#.',pad(arg(1)),_+1); if _==0  then return 0; end; return _+1
add:  if !.new==1  then return;  #= # + 1;     @.#= new;    !.new=1;    return
pad:  return  left( arg(1), N, .)
output   when using the default inputs:
======================================================================
For  5  cells  and blocks of:  2 1

─position─ ─value─
    1      ##.#.
    2      .##.#
    3      ##..#
======================================================================
For  5  cells  and blocks of:

─position─ ─value─
    1      .....
======================================================================
For  10  cells  and blocks of:  8

─position─ ──value───
    1      ########..
    2      .########.
    3      ..########
======================================================================
For  15  cells  and blocks of:  2 3 2 3

─position─ ─────value─────
    1      ##.###.##.###..
    2      .##.###.##.###.
    3      ..##.###.##.###
    4      ##..###.##.###.
    5      .##..###.##.###
    6      ##...###.##.###
    7      ##.###..##.###.
    8      .##.###..##.###
    9      ##..###..##.###
    10     ##.###...##.###
    11     ##.###.##..###.
    12     .##.###.##..###
    13     ##..###.##..###
    14     ##.###..##..###
    15     ##.###.##...###
======================================================================
For  5  cells  and blocks of:  2 3
***error***  invalid blocks for number of cells.

Ruby

Simple version:

def nonoblocks(cell, blocks)
  raise 'Those blocks will not fit in those cells' if cell < blocks.inject(0,:+) + blocks.size - 1
  nblock(cell, blocks, '', [])
end

def nblock(cell, blocks, position, result)
  if cell <= 0
    result << position[0..cell-1]
  elsif blocks.empty? or blocks[0].zero?
    result << position + '.' * cell
  else
    rest = cell - blocks.inject(:+) - blocks.size + 2
    bl, *brest = blocks
    rest.times.inject(result) do |res, i|
      nblock(cell-i-bl-1, brest, position + '.'*i + '#'*bl + '.', res)
    end
  end
end

conf = [[ 5, [2, 1]],
        [ 5, []],
        [10, [8]],
        [15, [2, 3, 2, 3]],
        [ 5, [2, 3]],      ]
conf.each do |cell, blocks|
  begin
    puts "#{cell} cells and #{blocks} blocks"
    result = nonoblocks(cell, blocks)
    puts result, result.size, ""
  rescue => e
    p e
  end
end
Output:
5 cells and [2, 1] blocks
##.#.
##..#
.##.#
3

5 cells and [] blocks
.....
1

10 cells and [8] blocks
########..
.########.
..########
3

15 cells and [2, 3, 2, 3] blocks
##.###.##.###..
##.###.##..###.
##.###.##...###
##.###..##.###.
##.###..##..###
##.###...##.###
##..###.##.###.
##..###.##..###
##..###..##.###
##...###.##.###
.##.###.##.###.
.##.###.##..###
.##.###..##.###
.##..###.##.###
..##.###.##.###
15

5 cells and [2, 3] blocks
#<RuntimeError: Those blocks will not fit in those cells>

Class version

The output form consulted the one of the python.

class NonoBlock
  def initialize(cell, blocks)
    raise 'Those blocks will not fit in those cells' if cell < blocks.inject(0,:+) + blocks.size - 1
    @result = []
    nonoblocks(cell, blocks, '')
  end
  
  def result(correct=true)
    correct ? @result.map(&:nonocell) : @result
  end
  
  private
  def nonoblocks(cell, blocks, position)
    if cell <= 0
      @result << position[0..cell-1]
    elsif blocks.empty? or blocks[0].zero?
      @result << position + '.' * cell
    else
      rest = cell - blocks.inject(0,:+) - blocks.size + 2
      bl, *brest = blocks
      rest.times do |i|
        nonoblocks(cell-i-bl-1, brest, position + '.'*i + '#'*bl + '.')
      end
    end
  end
end

class String
  def nonocell                  # "##.###..##" -> "|A|A|_|B|B|B|_|_|C|C|"
    chr = ('A'..'Z').each
    s = tr('.','_').gsub(/#+/){|sharp| chr.next * sharp.size}
    "|#{s.chars.join('|')}|"
  end
end

if __FILE__ == $0
  conf = [[ 5, [2, 1]],
          [ 5, []],
          [10, [8]],
          [15, [2, 3, 2, 3]],
          [ 5, [2, 3]]       ]
  conf.each do |cell, blocks|
    begin
      puts "Configuration:",
           "#{('.'*cell).nonocell} # #{cell} cells and #{blocks} blocks",
           "Possibilities:"
      result = NonoBlock.new(cell, blocks).result
      puts result,
           "A total of #{result.size} Possible configurations.", ""
    rescue => e
      p e
    end
  end
end
Output:
Configuration:
|_|_|_|_|_| # 5 cells and [2, 1] blocks
Possibilities:
|A|A|_|B|_|
|A|A|_|_|B|
|_|A|A|_|B|
A total of 3 Possible configurations.

Configuration:
|_|_|_|_|_| # 5 cells and [] blocks
Possibilities:
|_|_|_|_|_|
A total of 1 Possible configurations.

Configuration:
|_|_|_|_|_|_|_|_|_|_| # 10 cells and [8] blocks
Possibilities:
|A|A|A|A|A|A|A|A|_|_|
|_|A|A|A|A|A|A|A|A|_|
|_|_|A|A|A|A|A|A|A|A|
A total of 3 Possible configurations.

Configuration:
|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_| # 15 cells and [2, 3, 2, 3] blocks
Possibilities:
|A|A|_|B|B|B|_|C|C|_|D|D|D|_|_|
|A|A|_|B|B|B|_|C|C|_|_|D|D|D|_|
|A|A|_|B|B|B|_|C|C|_|_|_|D|D|D|
|A|A|_|B|B|B|_|_|C|C|_|D|D|D|_|
|A|A|_|B|B|B|_|_|C|C|_|_|D|D|D|
|A|A|_|B|B|B|_|_|_|C|C|_|D|D|D|
|A|A|_|_|B|B|B|_|C|C|_|D|D|D|_|
|A|A|_|_|B|B|B|_|C|C|_|_|D|D|D|
|A|A|_|_|B|B|B|_|_|C|C|_|D|D|D|
|A|A|_|_|_|B|B|B|_|C|C|_|D|D|D|
|_|A|A|_|B|B|B|_|C|C|_|D|D|D|_|
|_|A|A|_|B|B|B|_|C|C|_|_|D|D|D|
|_|A|A|_|B|B|B|_|_|C|C|_|D|D|D|
|_|A|A|_|_|B|B|B|_|C|C|_|D|D|D|
|_|_|A|A|_|B|B|B|_|C|C|_|D|D|D|
A total of 15 Possible configurations.

Configuration:
|_|_|_|_|_| # 5 cells and [2, 3] blocks
Possibilities:
#<RuntimeError: Those blocks will not fit in those cells>

Rust

Works with: Rust version 1.29.2
struct Nonoblock {
  width: usize,
  config: Vec<usize>,
  spaces: Vec<usize>,
}

impl Nonoblock {
  pub fn new(width: usize, config: Vec<usize>) -> Nonoblock {
    Nonoblock {
      width: width,
      config: config,
      spaces: Vec::new(),
    }
  }

  pub fn solve(&mut self) -> Vec<Vec<i32>> {
    let mut output: Vec<Vec<i32>> = Vec::new();
    self.spaces = (0..self.config.len()).fold(Vec::new(), |mut s, i| {
      s.push(match i {
        0 => 0,
        _ => 1,
      });
      s
    });
    if self.spaces.iter().sum::<usize>() + self.config.iter().sum::<usize>() <= self.width {
      'finished: loop {
        match self.spaces.iter().enumerate().fold((0, vec![0; self.width]), |mut a, (i, s)| {
            (0..self.config[i]).for_each(|j| a.1[a.0 + j + *s] = 1 + i as i32);
            return (a.0 + self.config[i] + *s, a.1);
          }) {
          (_, out) => output.push(out),
        }
        let mut i: usize = 1;
        'calc: loop {
          let len = self.spaces.len();
          if i > len {
            break 'finished;
          } else {
            self.spaces[len - i] += 1
          }
          if self.spaces.iter().sum::<usize>() + self.config.iter().sum::<usize>() > self.width {
            self.spaces[len - i] = 1;
            i += 1;
          } else {
            break 'calc;
          }
        }
      }
    }
    output
  }
}

fn main() {
  let mut blocks = [
    Nonoblock::new(5, vec![2, 1]),
    Nonoblock::new(5, vec![]),
    Nonoblock::new(10, vec![8]),
    Nonoblock::new(15, vec![2, 3, 2, 3]),
    Nonoblock::new(5, vec![2, 3]),
  ];

  for block in blocks.iter_mut() {
    println!("{} cells and {:?} blocks", block.width, block.config);
    println!("{}",(0..block.width).fold(String::from("="), |a, _| a + "=="));
    let solutions = block.solve();
    if solutions.len() > 0 {
      for solution in solutions.iter() {
        println!("{}", solution.iter().fold(String::from("|"), |s, f| s + &match f {
          i if *i > 0 => (('A' as u8 + ((*i - 1) as u8) % 26) as char).to_string(),
          _ => String::from("_"),
        }+ "|"));
      }
    } else {
      println!("No solutions. ");
    }
    println!();
  }
}
Output:
5 cells and [2, 1] blocks
===========
|A|A|_|B|_|
|A|A|_|_|B|
|_|A|A|_|B|

5 cells and [] blocks
===========
|_|_|_|_|_|

10 cells and [8] blocks
=====================
|A|A|A|A|A|A|A|A|_|_|
|_|A|A|A|A|A|A|A|A|_|
|_|_|A|A|A|A|A|A|A|A|

15 cells and [2, 3, 2, 3] blocks
===============================
|A|A|_|B|B|B|_|C|C|_|D|D|D|_|_|
|A|A|_|B|B|B|_|C|C|_|_|D|D|D|_|
|A|A|_|B|B|B|_|C|C|_|_|_|D|D|D|
|A|A|_|B|B|B|_|_|C|C|_|D|D|D|_|
|A|A|_|B|B|B|_|_|C|C|_|_|D|D|D|
|A|A|_|B|B|B|_|_|_|C|C|_|D|D|D|
|A|A|_|_|B|B|B|_|C|C|_|D|D|D|_|
|A|A|_|_|B|B|B|_|C|C|_|_|D|D|D|
|A|A|_|_|B|B|B|_|_|C|C|_|D|D|D|
|A|A|_|_|_|B|B|B|_|C|C|_|D|D|D|
|_|A|A|_|B|B|B|_|C|C|_|D|D|D|_|
|_|A|A|_|B|B|B|_|C|C|_|_|D|D|D|
|_|A|A|_|B|B|B|_|_|C|C|_|D|D|D|
|_|A|A|_|_|B|B|B|_|C|C|_|D|D|D|
|_|_|A|A|_|B|B|B|_|C|C|_|D|D|D|

5 cells and [2, 3] blocks
===========
No solutions.

Swift

import Foundation

func nonoblock(cells: Int, blocks: [Int]) {
    print("\(cells) cells and blocks \(blocks):")
    let totalBlockSize = blocks.reduce(0, +)
    if cells < totalBlockSize + blocks.count - 1 {
        print("no solution")
        return
    }

    func solve(cells: Int, index: Int, totalBlockSize: Int, offset: Int) {
        if index == blocks.count {
            count += 1
            print("\(String(format: "%2d", count))  \(String(output))")
            return
        }
        let blockSize = blocks[index]
        let maxPos = cells - (totalBlockSize + blocks.count - index - 1)
        let t = totalBlockSize - blockSize
        var c = cells - (blockSize + 1)
        for pos in 0...maxPos {
            fill(value: ".", offset: offset, count: maxPos + blockSize)
            fill(value: "#", offset: offset + pos, count: blockSize)
            solve(cells: c, index: index + 1, totalBlockSize: t,
                  offset: offset + blockSize + pos + 1)
            c -= 1
        }
    }

    func fill(value: Character, offset: Int, count: Int) {
        output.replaceSubrange(offset..<offset+count,
                               with: repeatElement(value, count: count))
    }
    
    var output: [Character] = Array(repeating: ".", count: cells)
    var count = 0
    solve(cells: cells, index: 0, totalBlockSize: totalBlockSize, offset: 0)
}

nonoblock(cells: 5, blocks: [2, 1])
print()

nonoblock(cells: 5, blocks: [])
print()

nonoblock(cells: 10, blocks: [8])
print()

nonoblock(cells: 15, blocks: [2, 3, 2, 3])
print()

nonoblock(cells: 5, blocks: [2, 3])
Output:
5 cells and blocks [2, 1]:
 1  ##.#.
 2  ##..#
 3  .##.#

5 cells and blocks []:
 1  .....

10 cells and blocks [8]:
 1  ########..
 2  .########.
 3  ..########

15 cells and blocks [2, 3, 2, 3]:
 1  ##.###.##.###..
 2  ##.###.##..###.
 3  ##.###.##...###
 4  ##.###..##.###.
 5  ##.###..##..###
 6  ##.###...##.###
 7  ##..###.##.###.
 8  ##..###.##..###
 9  ##..###..##.###
10  ##...###.##.###
11  .##.###.##.###.
12  .##.###.##..###
13  .##.###..##.###
14  .##..###.##.###
15  ..##.###.##.###

5 cells and blocks [2, 3]:
no solution

Tcl

Works with: Tcl version 8.6
Library: Tcllib (Package: generator)
Translation of: Python
package require Tcl 8.6
package require generator

generator define nonoblocks {blocks cells} {
    set sum [tcl::mathop::+ {*}$blocks]
    if {$sum == 0 || [lindex $blocks 0] == 0} {
	generator yield {{0 0}}
	return
    } elseif {$sum + [llength $blocks] - 1 > $cells} {
	error "those blocks will not fit in those cells"
    }

    set brest [lassign $blocks blen]
    for {set bpos 0} {$bpos <= $cells - $sum - [llength $brest]} {incr bpos} {
	if {![llength $brest]} {
	    generator yield [list [list $bpos $blen]]
	    return
	}
	set offset [expr {$bpos + $blen + 1}]
	generator foreach subpos [nonoblocks $brest [expr {$cells - $offset}]] {
	    generator yield [linsert [lmap b $subpos {
		lset b 0 [expr {[lindex $b 0] + $offset}]
	    }] 0 [list $bpos $blen]]
	}
    }
}

if {[info script] eq $::argv0} {
    proc pblock {cells {vec {}}} {
	set vector [lrepeat $cells "_"]
	set ch 64
	foreach b $vec {
	    incr ch
	    lassign $b bp bl
	    for {set i $bp} {$i < $bp + $bl} {incr i} {
		lset vector $i [format %c $ch]
	    }
	}
	return |[join $vector "|"]|
    }
    proc flist {items} {
	return [format "\[%s\]" [join $items ", "]]
    }
    foreach {blocks cells} {
	{2 1} 5
	{} 5
	{8} 10
	{2 3 2 3} 15
	{2 3} 5
    } {
	puts "\nConfiguration:"
	puts [format "%s # %d cells and %s blocks" \
		  [pblock $cells] $cells [flist $blocks]]
	puts "  Possibilities:"
	set i 0
	try {
	    generator foreach vector [nonoblocks $blocks $cells] {
		puts "    [pblock $cells $vector]"
		incr i
	    }
	    puts "  A total of $i possible configurations"
	} on error msg {
	    puts "    --> ERROR: $msg"
	}
    }
}

package provide nonoblock 1
Output:

Configuration:
|_|_|_|_|_| # 5 cells and [2, 1] blocks
  Possibilities:
    |A|A|_|B|_|
    |A|A|_|_|B|
    |_|A|A|_|B|
  A total of 3 possible configurations

Configuration:
|_|_|_|_|_| # 5 cells and [] blocks
  Possibilities:
    |_|_|_|_|_|
  A total of 1 possible configurations

Configuration:
|_|_|_|_|_|_|_|_|_|_| # 10 cells and [8] blocks
  Possibilities:
    |A|A|A|A|A|A|A|A|_|_|
    |_|A|A|A|A|A|A|A|A|_|
    |_|_|A|A|A|A|A|A|A|A|
  A total of 3 possible configurations

Configuration:
|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_| # 15 cells and [2, 3, 2, 3] blocks
  Possibilities:
    |A|A|_|B|B|B|_|C|C|_|D|D|D|_|_|
    |A|A|_|B|B|B|_|C|C|_|_|D|D|D|_|
    |A|A|_|B|B|B|_|C|C|_|_|_|D|D|D|
    |A|A|_|B|B|B|_|_|C|C|_|D|D|D|_|
    |A|A|_|B|B|B|_|_|C|C|_|_|D|D|D|
    |A|A|_|B|B|B|_|_|_|C|C|_|D|D|D|
    |A|A|_|_|B|B|B|_|C|C|_|D|D|D|_|
    |A|A|_|_|B|B|B|_|C|C|_|_|D|D|D|
    |A|A|_|_|B|B|B|_|_|C|C|_|D|D|D|
    |A|A|_|_|_|B|B|B|_|C|C|_|D|D|D|
    |_|A|A|_|B|B|B|_|C|C|_|D|D|D|_|
    |_|A|A|_|B|B|B|_|C|C|_|_|D|D|D|
    |_|A|A|_|B|B|B|_|_|C|C|_|D|D|D|
    |_|A|A|_|_|B|B|B|_|C|C|_|D|D|D|
    |_|_|A|A|_|B|B|B|_|C|C|_|D|D|D|
  A total of 15 possible configurations

Configuration:
|_|_|_|_|_| # 5 cells and [2, 3] blocks
  Possibilities:
    --> ERROR: those blocks will not fit in those cells

Wren

Translation of: Kotlin
Library: Wren-math
import "./math" for Nums

var genSequence // recursive
genSequence = Fn.new { |ones, numZeros|
    if (ones.isEmpty) return ["0" * numZeros]
    var result = []
    for (x in 1...numZeros - ones.count + 2) {
        var skipOne = ones[1..-1]
        for (tail in genSequence.call(skipOne, numZeros - x)) {
            result.add("0" * x + ones[0] + tail)
        }
    }
    return result
}

var printBlock = Fn.new { |data, len|
    var a = data.toList
    var sumChars = Nums.sum(a.map { |c| c.bytes[0] - 48 }.toList)
    System.print("\nblocks %(a), cells %(len)")
    if (len - sumChars <= 0) {
        System.print("No solution")
        return
    }
    var prep = a.map { |c| "1" * (c.bytes[0] - 48) }.toList
    for (r in genSequence.call(prep, len - sumChars + 1)) {
        System.print(r[1..-1])
    }
}

printBlock.call("21", 5)
printBlock.call("", 5)
printBlock.call("8", 10)
printBlock.call("2323", 15)
printBlock.call("23", 5)
Output:
blocks [2, 1], cells 5
11010
11001
01101

blocks [], cells 5
00000

blocks [8], cells 10
1111111100
0111111110
0011111111

blocks [2, 3, 2, 3], cells 15
110111011011100
110111011001110
110111011000111
110111001101110
110111001100111
110111000110111
110011101101110
110011101100111
110011100110111
110001110110111
011011101101110
011011101100111
011011100110111
011001110110111
001101110110111

blocks [2, 3], cells 5
No solution

zkl

Translation of: Python
fcn nonoblocks(blocks,cells){
   if(not blocks or blocks[0]==0) vm.yield( T(T(0,0)) );
   else{
      if(not ( blocks.sum(0) + blocks.len() -1<=cells ))
	 throw(Exception.AssertionError("Those blocks will not fit in those cells"));
      blength,brest:=blocks[0], blocks[1,*];      # Deal with the first block of length
      minspace4rest:=brest.reduce('+(1),0);       # The other blocks need space
      # Slide the start position from left to max RH index allowing for other blocks.
      foreach bpos in (cells - minspace4rest - blength +1){
         if(not brest) # No other blocks to the right so just yield this one.
	    vm.yield(T(T(bpos,blength)));
	 else{
	    # More blocks to the right so create a *sub-problem* of placing
	    # the brest blocks in the cells one space to the right of the RHS of 
	    # this block.
	    offset:=bpos + blength +1;
	    # recursive call to nonoblocks yields multiple sub-positions
	    foreach subpos in (Utils.Generator(nonoblocks,brest,cells - offset)){
	       # Remove the offset from sub block positions
	       rest:=subpos.pump(List,'wrap([(bp,bl)]){ T(offset + bp, bl) });
	       # Yield this block plus sub blocks positions
	       vm.yield(T( T(bpos,blength) ).extend(rest) );
	    }
	 }
      }
   }
}
# Pretty print each run of blocks with a different letter for each block of filled cells
fcn pblock(vec,cells){
   vector,ch:=cells.pump(List(),"_".copy), ["A".."Z"];
   vec.apply2('wrap([(a,b)]){ a.walker(b).pump(Void,vector.set.fp1(ch.next())) });
   String("|",vector.concat("|"),"|");
}
foreach blocks,cells in (T( T(T(2,1),5), T(T,5), T(T(8),10), T(T(2,3,2,3),15),
			    T(T(2,3),5) )){
   println("\nConfiguration:\n    %s # %d cells and %s blocks"
         .fmt(pblock(T,cells),cells,blocks));
   println("  Possibilities:");
   Utils.Generator(nonoblocks,blocks,cells).reduce('wrap(n,vector){
      println("    ",pblock(vector,cells));
      n+1
   },0)
   : println("  A total of %d possible configurations.".fmt(_));
}
Output:
Configuration:
    |_|_|_|_|_| # 5 cells and L(2,1) blocks
  Possibilities:
    |A|A|_|B|_|
    |A|A|_|_|B|
    |_|A|A|_|B|
  A total of 3 possible configurations.

Configuration:
    |_|_|_|_|_| # 5 cells and L() blocks
  Possibilities:
    |_|_|_|_|_|
  A total of 1 possible configurations.

Configuration:
    |_|_|_|_|_|_|_|_|_|_| # 10 cells and L(8) blocks
  Possibilities:
    |A|A|A|A|A|A|A|A|_|_|
    |_|A|A|A|A|A|A|A|A|_|
    |_|_|A|A|A|A|A|A|A|A|
  A total of 3 possible configurations.

Configuration:
    |_|_|_|_|_|_|_|_|_|_|_|_|_|_|_| # 15 cells and L(2,3,2,3) blocks
  Possibilities:
    |A|A|_|B|B|B|_|C|C|_|D|D|D|_|_|
    |A|A|_|B|B|B|_|C|C|_|_|D|D|D|_|
    |A|A|_|B|B|B|_|C|C|_|_|_|D|D|D|
    |A|A|_|B|B|B|_|_|C|C|_|D|D|D|_|
    |A|A|_|B|B|B|_|_|C|C|_|_|D|D|D|
    |A|A|_|B|B|B|_|_|_|C|C|_|D|D|D|
    |A|A|_|_|B|B|B|_|C|C|_|D|D|D|_|
    |A|A|_|_|B|B|B|_|C|C|_|_|D|D|D|
    |A|A|_|_|B|B|B|_|_|C|C|_|D|D|D|
    |A|A|_|_|_|B|B|B|_|C|C|_|D|D|D|
    |_|A|A|_|B|B|B|_|C|C|_|D|D|D|_|
    |_|A|A|_|B|B|B|_|C|C|_|_|D|D|D|
    |_|A|A|_|B|B|B|_|_|C|C|_|D|D|D|
    |_|A|A|_|_|B|B|B|_|C|C|_|D|D|D|
    |_|_|A|A|_|B|B|B|_|C|C|_|D|D|D|
  A total of 15 possible configurations.

Configuration:
    |_|_|_|_|_| # 5 cells and L(2,3) blocks
  Possibilities:
VM#2 caught this unhandled exception:
   AssertionError : Those blocks will not fit in those cells
   <stack traces deleted>