Nonoblock
Nonoblock is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
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.
The task is to
- show all possible positions
- and the number of positions
of the blocks for the following cases within the row. On this page. Using 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 1 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:
|#|#|_|#|_| |#|#|_|_|#| |_|#|#|_|#|
Or even 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.
D
<lang d>import std.stdio, std.array, std.algorithm, std.exception, std.conv;
struct Solution { uint pos, len; }
struct NonoBlocks {
const uint[] blocks; const uint cells;
int opApply(int delegate(ref const Solution[]) dg) {
int result;
if (blocks.empty || blocks[0] == 0) {
const vec = [Solution(0, 0)];
result = dg(vec);
if (result) return result;
} else {
enforce(blocks.sum + blocks.length - 1 <= cells,
"Those blocks cannot fit in those cells.");
immutable firstBl = blocks[0];
const restBl = blocks[1 .. $];
// 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.
const vec = [Solution(bPos, firstBl)];
result = dg(vec);
if (result) return result;
} 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.
const rest = subPos.map!(sol => Solution(offset + sol.pos,
sol.len)).array;
// Yield this block plus sub blocks positions.
const vec = [Solution(bPos, firstBl)] ~ rest;
result = dg(vec);
if (result) return result;
}
}
}
}
return result; }
}
/// 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[])) ...
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