Nonoblock: Difference between revisions
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Added Swift solution |
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=========== |
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No solutions. |
No solutions. |
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</pre> |
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=={{header|Swift}}== |
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<lang swift>import Foundation |
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class Nonoblock { |
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let cells: Int |
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let blocks: [Int] |
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var positions: [Int] |
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var count: Int |
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private init(cells: Int, blocks: [Int]) { |
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self.cells = cells |
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self.blocks = blocks |
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positions = Array(repeating: 0, count: blocks.count) |
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count = 0 |
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} |
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class func printSolutions(cells: Int, blocks: [Int]) { |
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print("\(cells) cells and blocks \(blocks):") |
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var totalBlockSize = 0 |
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for blockSize in blocks { |
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totalBlockSize += blockSize |
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} |
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if cells < totalBlockSize + blocks.count - 1 { |
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print("no solution") |
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return |
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} |
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let nb = Nonoblock(cells: cells, blocks: blocks) |
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nb.solve(cells: cells, index: 0, totalBlockSize: totalBlockSize) |
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} |
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private func solve(cells: Int, index: Int, totalBlockSize: Int) { |
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if index == blocks.count { |
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count += 1 |
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printSolution() |
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return |
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} |
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let blockSize = blocks[index] |
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let maxPos = cells - (totalBlockSize + blocks.count - index - 1) |
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let t = totalBlockSize - blockSize |
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var c = cells - (blockSize + 1) |
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for pos in 0...maxPos { |
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positions[index] = pos |
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solve(cells: c, index: index + 1, totalBlockSize: t) |
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c -= 1 |
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} |
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} |
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private func printSolution() { |
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var str = String(format: "%2d ", count) |
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var c = 0 |
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for i in 0..<blocks.count { |
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var n = positions[i] |
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if i > 0 { |
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n += 1 |
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} |
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if n > 0 { |
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str.append(String(repeating: ".", count: n)) |
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c += n |
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} |
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str.append(String(repeating: "#", count: blocks[i])) |
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c += blocks[i] |
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} |
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if c < cells { |
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str.append(String(repeating: ".", count: cells - c)) |
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} |
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print(str) |
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} |
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} |
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Nonoblock.printSolutions(cells: 5, blocks: [2, 1]) |
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print() |
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Nonoblock.printSolutions(cells: 5, blocks: []) |
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print() |
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Nonoblock.printSolutions(cells: 10, blocks: [8]) |
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print() |
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Nonoblock.printSolutions(cells: 15, blocks: [2, 3, 2, 3]) |
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print() |
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Nonoblock.printSolutions(cells: 5, blocks: [2, 3])</lang> |
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{{out}} |
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<pre> |
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5 cells and blocks [2, 1]: |
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1 ##.#. |
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2 ##..# |
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3 .##.# |
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5 cells and blocks []: |
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1 ..... |
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10 cells and blocks [8]: |
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1 ########.. |
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2 .########. |
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3 ..######## |
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15 cells and blocks [2, 3, 2, 3]: |
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1 ##.###.##.###.. |
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2 ##.###.##..###. |
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3 ##.###.##...### |
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4 ##.###..##.###. |
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5 ##.###..##..### |
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6 ##.###...##.### |
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7 ##..###.##.###. |
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8 ##..###.##..### |
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9 ##..###..##.### |
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10 ##...###.##.### |
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11 .##.###.##.###. |
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12 .##.###.##..### |
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13 .##.###..##.### |
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14 .##..###.##.### |
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15 ..##.###.##.### |
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5 cells and blocks [2, 3]: |
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no solution |
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</pre> |
</pre> |
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Revision as of 14:28, 2 January 2021
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
- 5 cells and [2, 1] blocks
- 5 cells and [] blocks (no blocks)
- 10 cells and [8] blocks
- 15 cells and [2, 3, 2, 3] blocks
- 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
- The blog post Nonogram puzzle solver (part 1) Inspired this task and donated its Nonoblock#Python solution.
C
<lang 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;
}</lang>
- 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. <lang csharp>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); }
}</lang>
- 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++
<lang cpp>
- 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;
} </lang>
- 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
<lang d>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; }
}</lang>
- 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
<lang scheme>
- 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))))) </lang>
- 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
<lang elixir>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)</lang>
- 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
<lang go>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)
}</lang>
- 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:
<lang J>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.)&.></lang>
Task example:
<lang J> neat 5 nonoblock 2 1 │A│A│ │B│ │ │A│A│ │ │B│ │ │A│A│ │B│
neat 5 nonoblock
│ │ │ │ │ │
neat 10 nonoblock 8
│A│A│A│A│A│A│A│A│ │ │ │ │A│A│A│A│A│A│A│A│ │ │ │ │A│A│A│A│A│A│A│A│
neat 15 nonoblock 2 3 2 3
│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│
neat 5 nonoblock 2 3
</lang>
Java
<lang java>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(); }
}</lang>
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
<lang 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); </lang>
- 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
<lang 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)
</lang>
- 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
<lang scala>// 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 $len") if (len - sumChars <= 0) { println("No solution") return } val prep = a.map { "1".repeat(it.toInt() - 48) } for (r in genSequence(prep, len - sumChars + 1)) println(r.substring(1))
}
fun genSequence(ones: List<String>, numZeros: Int): List<String> {
if (ones.isEmpty()) return listOf("0".repeat(numZeros)) val result = mutableListOf<String>() for (x in 1 until numZeros - ones.size + 2) { val skipOne = ones.drop(1) for (tail in genSequence(skipOne, numZeros - x)) { result.add("0".repeat(x) + ones[0] + tail) } } return result
}
fun main(args: Array<String>) {
printBlock("21", 5) printBlock("", 5) printBlock("8", 10) printBlock("2323", 15) printBlock("23", 5)
}</lang>
- Output:
blocks [2, 1], cells 5 11010 11001 01101 blocks [], cells 5 00000 blocks [8], cells 10 1111111100 0111111110 0011111111 blocks [2, 3, 2, 3], cells 15 110111011011100 110111011001110 110111011000111 110111001101110 110111001100111 110111000110111 110011101101110 110011101100111 110011100110111 110001110110111 011011101101110 011011101100111 011011100110111 011001110110111 001101110110111 blocks [2, 3], cells 5 No solution
M2000 Interpreter
Recursive
<lang M2000 Interpreter> Module NonoBlock {
Form 80,40 Flush Print "Nonoblock" Data 5, (2, 1) Data 5, (,) Data 10, (8,) Data 15, (2,3,2,3) Data 5, (2,3) Def BLen(a$)=(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 </lang>
- 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
<lang M2000 Interpreter> 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$(counter,"0000 "); For i=1 to n { Print n(i);" "; } Print If queue>0 Then { p=sp(queue) k=sk(queue) queue-- For i=p to n { n(i)=0 } p++ If k<m Then { If p+part(k+1)<n+1 Then { queue++ sp(queue)=p ' sk(queue)=k } } flag=False } } } Until flag p1++ If k=0 Then Exit } }
}
Nonoblock 5,2,2,1 Nonoblock 5,0 Nonoblock 10,1,8 Nonoblock 15,4,2,3,2,3 Nonoblock 5,2,3,2 </lang>
Perl
<lang perl>use strict; use warnings;
while( )
{ print "\n$_", 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.$3/; }
__DATA__ 5 2 1 5 10 8 15 2 3 2 3 5 2 3</lang>
- Output:
5 2 1 ===== AA.B. AA..B .AA.B 5 = ..... 10 8 ==== AAAAAAAA.. .AAAAAAAA. ..AAAAAAAA 15 2 3 2 3 ========== AA.BBB.CC.DDD.. AA.BBB.CC..DDD. AA.BBB..CC.DDD. AA..BBB.CC.DDD. .AA.BBB.CC.DDD. AA.BBB.CC...DDD AA.BBB..CC..DDD AA..BBB.CC..DDD .AA.BBB.CC..DDD AA.BBB...CC.DDD AA..BBB..CC.DDD .AA.BBB..CC.DDD AA...BBB.CC.DDD .AA..BBB.CC.DDD ..AA.BBB.CC.DDD 5 2 3 ===== no solution
Phix
<lang Phix>function nobr(sequence res, string neat, integer ni, integer ch, sequence blocks)
if length(blocks)=0 then res = append(res,neat) else integer b = blocks[1] blocks = blocks[2..$] 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</lang>
- 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
<lang 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))</lang>
- 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>#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]))</lang>
- 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)
<lang perl6>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 ;
}</lang>
- 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
<lang 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, .)</lang>
- 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: <lang ruby>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</lang>
- 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. <lang ruby>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</lang>
- 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
<lang rust>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!(); }
}</lang>
- 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
<lang swift>import Foundation
class Nonoblock {
let cells: Int let blocks: [Int] var positions: [Int] var count: Int
private init(cells: Int, blocks: [Int]) { self.cells = cells self.blocks = blocks positions = Array(repeating: 0, count: blocks.count) count = 0 }
class func printSolutions(cells: Int, blocks: [Int]) { print("\(cells) cells and blocks \(blocks):") var totalBlockSize = 0 for blockSize in blocks { totalBlockSize += blockSize } if cells < totalBlockSize + blocks.count - 1 { print("no solution") return } let nb = Nonoblock(cells: cells, blocks: blocks) nb.solve(cells: cells, index: 0, totalBlockSize: totalBlockSize) }
private func solve(cells: Int, index: Int, totalBlockSize: Int) { if index == blocks.count { count += 1 printSolution() 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 { positions[index] = pos solve(cells: c, index: index + 1, totalBlockSize: t) c -= 1 } }
private func printSolution() { var str = String(format: "%2d ", count) var c = 0 for i in 0..<blocks.count { var n = positions[i] if i > 0 { n += 1 } if n > 0 { str.append(String(repeating: ".", count: n)) c += n } str.append(String(repeating: "#", count: blocks[i])) c += blocks[i] } if c < cells { str.append(String(repeating: ".", count: cells - c)) } print(str) }
}
Nonoblock.printSolutions(cells: 5, blocks: [2, 1]) print()
Nonoblock.printSolutions(cells: 5, blocks: []) print()
Nonoblock.printSolutions(cells: 10, blocks: [8]) print()
Nonoblock.printSolutions(cells: 15, blocks: [2, 3, 2, 3]) print()
Nonoblock.printSolutions(cells: 5, blocks: [2, 3])</lang>
- 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
<lang tcl>package require Tcl 8.6 package require generator
generator define nonoblocks {blocks cells} {
set sum [tcl::mathop::+ {*}$blocks] if {$sum == 0 || [lindex $blocks 0] == 0} {
generator yield Template:0 0 return
} elseif {$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</lang>
- 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
<lang ecmascript>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)</lang>
- 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
<lang zkl>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("|"),"|");
}</lang> <lang zkl>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(_));
}</lang>
- 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>