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
- 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.
11l
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
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[])) ...
EasyLang
func$ rep s$ n .
for i to n
r$ &= s$
.
return r$
.
func$[] genseq ones$[] nzeros .
if len ones$[] = 0
return [ rep "0" nzeros ]
.
for x = 1 to nzeros - len ones$[] + 1
skip1$[] = [ ]
for i = 2 to len ones$[]
skip1$[] &= ones$[i]
.
h$[] = genseq skip1$[] (nzeros - x)
for tail$ in h$[]
r$[] &= rep "0" x & ones$[1] & tail$
.
.
return r$[]
.
proc block data$ le . .
a[] = number strchars data$
for b in a[]
nbytes += b
.
print "blocks " & a[] & " cells " & le
if le - nbytes <= 0
print "No solution"
return
.
for b in a[]
prep$[] &= rep "1" b
.
for r$ in genseq prep$[] (le - nbytes + 1)
print substr r$ 2 99999
.
print ""
.
block "21" 5
block "" 5
block "8" 10
block "2323" 15
block "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
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
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"}
FreeBASIC
Function minsized(arr() As Integer) As String
Dim result As String = ""
For i As Integer = Lbound(arr) To Ubound(arr)
result &= String(arr(i), "#")
If i < Ubound(arr) Then result &= "."
Next
Return result
End Function
Function minlen(arr() As Integer) As Integer
Dim sum As Integer = 0
For i As Integer = Lbound(arr) To Ubound(arr)
sum += arr(i)
Next
Return sum + Ubound(arr)
End Function
Sub sequences(blockseq() As Integer, numblanks As Integer, result() As String)
Dim As Integer i, leftspace, rightspace, allbuthead()
Dim As String header
If Ubound(blockseq) = -1 Then
Redim result(0)
result(0) = String(numblanks, ".")
Elseif minlen(blockseq()) = numblanks Then
Redim result(0)
result(0) = minsized(blockseq())
Else
Redim allbuthead(Ubound(blockseq) - 1)
For i = 1 To Ubound(blockseq)
allbuthead(i - 1) = blockseq(i)
Next
For leftspace = 0 To (numblanks - minlen(blockseq()))
header = String(leftspace, ".") & String(blockseq(0), "#") & "."
rightspace = numblanks - Len(header)
If Ubound(allbuthead) = -1 Then
If rightspace <= 0 Then
Redim Preserve result(Ubound(result) + 1)
result(Ubound(result)) = Left(header, numblanks)
Else
Redim Preserve result(Ubound(result) + 1)
result(Ubound(result)) = header & String(rightspace, ".")
End If
Elseif minlen(allbuthead()) = rightspace Then
Redim Preserve result(Ubound(result) + 1)
result(Ubound(result)) = header & minsized(allbuthead())
Else
Dim As String subresult()
sequences(allbuthead(), rightspace, subresult())
For i= Lbound(subresult) To Ubound(subresult)
Redim Preserve result(Ubound(result) + 1)
result(Ubound(result)) = header & subresult(i)
Next
End If
Next
End If
End Sub
Sub nonoblocks(bvec() As Integer, longi As Integer)
Dim As Integer i
Dim As String seq()
Print !"\nWith blocks [ ";
For i = Lbound(bvec) To Ubound(bvec)
Print Chr(8); bvec(i); " ";
Next i
Print Chr(8); " ] and "; longi; " cells:"
If longi < minlen(bvec()) Then
Print "No solution"
Else
sequences(bvec(), longi, seq())
For i = Lbound(seq) To Ubound(seq)
Print seq(i)
Next
End If
End Sub
Dim bvec1(1) As Integer = {2, 1}
Dim bvec2() As Integer
Dim bvec3(0) As Integer = {8}
Dim bvec4(3) As Integer = {2, 3, 2, 3}
Dim bvec5(1) As Integer = {2, 3}
nonoblocks(bvec1(), 5)
nonoblocks(bvec2(), 5)
nonoblocks(bvec3(), 10)
nonoblocks(bvec4(), 15)
nonoblocks(bvec5(), 5)
Sleep
- Output:
With blocks [ 2 1 ] and 5 cells: ##.#. ##..# .##.# With blocks [ ] 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
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)
}
- 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
│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
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();
}
}
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
jq
Adapted from Wren
Works with jq, the C implementation of jq
Works with gojq, the Go implementation of jq
Works with jaq, the Rust implementation of jq
def sum(stream): reduce stream as $x (0; . + $x);
def genSequence($ones; $numZeros):
if $ones|length == 0 then "." * $numZeros
else range(1; $numZeros - ($ones|length) + 2) as $x
| genSequence($ones[1:]; $numZeros - $x) as $tail
| "." * $x + $ones[0] + $tail
end;
def printBlock($data; $len):
sum($data | explode[] | . - 48) as $sumChars
| "\nblocks \($data), cells \($len)",
(if $len - $sumChars <= 0
then "No solution"
else ( $data | explode | map( "1" * (. - 48) ) ) as $prep
| genSequence($prep; $len - $sumChars + 1)[1:]
end) ;
printBlock( "21"; 5),
printBlock( ""; 5),
printBlock( "8"; 10),
printBlock("2323"; 15),
printBlock( "23"; 5)- Output:
blocks 21, cells 5 11.1. 11..1 .11.1 blocks , cells 5 ..... blocks 8, cells 10 11111111.. .11111111. ..11111111 blocks 2323, cells 15 11.111.11.111.. 11.111.11..111. 11.111.11...111 11.111..11.111. 11.111..11..111 11.111...11.111 11..111.11.111. 11..111.11..111 11..111..11.111 11...111.11.111 .11.111.11.111. .11.111.11..111 .11.111..11.111 .11..111.11.111 ..11.111.11.111 blocks 23, cells 5 No solution
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
// 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$(counter,"0000 ");
For i=1 to n {
Print n(i);" ";
}
Print
If queue>0 Then {
p=sp(queue)
k=sk(queue)
queue--
For i=p to n {
n(i)=0
}
p++
If k<m Then {
If p+part(k+1)<n+1 Then {
queue++
sp(queue)=p
' sk(queue)=k
}
}
flag=False
}
}
} Until flag
p1++
If k=0 Then Exit
}
}
}
Nonoblock 5,2,2,1
Nonoblock 5,0
Nonoblock 10,1,8
Nonoblock 15,4,2,3,2,3
Nonoblock 5,2,3,2Mathematica /Wolfram Language
ClearAll[SpacesDistributeOverN, Possibilities]
SpacesDistributeOverN[s_, p_] :=
Flatten[
Permutations /@ (Join[#, ConstantArray[0, p - Length[#]]] & /@
IntegerPartitions[s, p]), 1]
Possibilities[hint_, len_] :=
Module[{p = hint, l = len, b = Length[hint], Spaces, out},
Spaces = # + (Prepend[Append[ConstantArray[1, b - 1], 0],
0]) & /@ (SpacesDistributeOverN[l - Total@p - (b - 1), b + 1]);
out = Flatten /@ (
Riffle[#, Map[Table[1, {#}] &, p, {1}]] & /@
Map[Table[0, {#}] &, Spaces, {2}]);
StringJoin @@@ (out /. {0 -> ".", 1 -> "#"})
]
Possibilities[{}, len_] := Module[{},
{StringJoin[ConstantArray[".", len]]}
]
Possibilities[{2, 1}, 5]
Possibilities[{}, 5]
Possibilities[{8}, 10]
Possibilities[{2, 3, 2, 3}, 15]
Possibilities[{2, 3}, 5]
- Output:
{".##.#", "##..#", "##.#."}
{"....."}
{"..########", "########..", ".########."}
{"..##.###.##.###", "##...###.##.###", "##.###...##.###", "##.###.##...###", "##.###.##.###..", ".##..###.##.###", ".##.###..##.###", ".##.###.##..###", ".##.###.##.###.", "##..###..##.###", "##..###.##..###", "##..###.##.###.", "##.###..##..###", "##.###..##.###.", "##.###.##..###."}
{}
Nim
import math, sequtils, strformat, strutils
proc genSequence(ones: seq[string]; numZeroes: Natural): seq[string] =
if ones.len == 0: return @[repeat('0', numZeroes)]
for x in 1..(numZeroes - ones.len + 1):
let skipOne = ones[1..^1]
for tail in genSequence(skipOne, numZeroes - x):
result.add repeat('0', x) & ones[0] & tail
proc printBlock(data: string; length: Positive) =
let a = mapIt(data, ord(it) - ord('0'))
let sumBytes = sum(a)
echo &"\nblocks {($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)
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
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
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
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
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>