C, 824 803 bytes
#define Z return
#define Y char*b
#define N --n
i,j,n,w,h,A,B,C,D,E,F,G,H;char c[9999],*r,*d;x(b)Y;{if(b<c||*b<35)Z;++n;*b^=1;x(b-1);x(b+1);x(b-w);x(b+w);}m(b,p,y)Y,*p;{d=b;if(!y)for(y=-1,--p;1[++p]&31;)d+=w;for(i=0;*p&31?!(*p&16>>i)||b[i]&1:0;++i>4?p+=y,b+=w,i=0:0);Z!(*p&31)?x(d),n:0;}a(b)Y;{for(j=n=0;j<w*h;++j)if(m(c+j,b,1)||m(c+j,b,0))Z n;Z 0;}f(Y){bzero(c,9999);for(h=0,b=strcpy(c,b);r=b,b=strchr(b+1,10);h++,w=b-r);for(A=2,r=1+"@_`^C@|T@^R@XO@XX`|FB@|PP@|DD@PXN@XHX@XPX`PPXL@XHHX@XLDD@XPPX`PPPXH@PXHHH@PPPPP@";*r;r+=A+=r[-1]/96)while(a(r));A=B=C=D=E=F=G=H=0;while(a("PX")||a("XH")) (n-=3)?N?N?N?0:++H:++G:++F:++C;while(a("^")||a("PPPP"))(n-=4)?N?N?0:++H:++G:++E;while(a("P"))N?N?N?N?N?N?0:++H:++G:++F:++D:++B:++A;Z H||(G&&A)||(F&&B+B+A>1)||(E&&A>1)||D>1||C>1||((D||C)*3+B*2+A>5)*(A>1||B>2||A*B);}
Note: Includes a bug fix (the previous entry falsely identified a tromino and two dominoes as forming a cube).
In the TIO driver code, there are more test cases and there's now a pass/fail tracker; hexomino test cases were updated with the expected pass/fail value in the label.
Try it online!
...and before explaining this in detail, it's worth a high level overview.
Basic Overview
This algorithm applies a pattern matcher to classify each polyomino it finds with a given pattern as its subset. As polyominoes are matched they are "unmarked", excluding them from pattern matching again. The initial classification given by the matcher is simply a count of the number of tiles in the polyomino.
The matcher is applied first to cull out all polyominoes that cannot be folded onto a cube; the classification of these polyominos is discarded. The match succeeds if these polyominoes appear within higher level ones; therefore, we only care about the smallest subset of "unfoldable" for each class. Shown here along with padded encodings are all such polyominoes (excluding their vertical reflections). The encoding uses bits 4-0 of each character to represent squares on each row:
[^C```] [XHX``] [PPPXH] [XHHX`] [PXN``] [|PP``]
####. ##... #.... ##... #.... ###..
...## .#... #.... .#... ##... #....
..... ##... #.... .#... .###. #....
..... ..... ##... ##... ..... .....
..... ..... .#... ..... ..... .....
[|FB``] [XPX``] [PPXL`] [XLDD`] [XPPX`] [|DD``]
###.. ##... #.... ##... ##... ###..
..##. #.... #.... .##.. #.... ..#..
...#. ##... ##... ..#.. #.... ..#..
..... ..... .##.. ..#.. ##... .....
..... ..... ..... ..... ..... .....
[|T```] [^R```] [PXHHH] [XO```] [_````] [PPPPP]
###.. ####. #.... ##... ##### #....
#.#.. #..#. ##... .#### ..... #....
..... ..... .#... ..... ..... #....
..... ..... .#... ..... ..... #....
..... ..... .#... ..... ..... #....
[XX```]
##...
##...
.....
.....
.....
Once these polyominoes are culled, we categorize the remaining polyominoes into relevant categories. It's worth noting that in almost all cases, one can just find polyominoes that remain (those foldable onto cubes) and simply search for sums of six. There are two exceptions:
- A corner tromino and a line tromino cannot form a cube
- A line tetromino and a domino cannot form a cube
In order to be able to accommodate this restriction we form 8 categories, from A-H: A for monominoes (lone tiles), B for dominoes, C for corner trominoes, D for line trominoes, E for line tetrominoes, F for other tetrominoes, G for pentominoes, and H for hexominoes. Anything not falling into one of these categories is just ignored. Counting polyominoes that fall into each category suffices.
At the end, we just return truthiness or falsiness based on a giant equation and these tabulations.
Ungolfed with comments
i,j,n,w,h,A,B,C,D,E,F,G,H;char c[9999],*r,*d;
x(b)char*b;{ // recursively unmarks polyomino pointed to by b.
if(b<c||*b<35)return;
++n; *b^=1; // Tabulates tiles in n as it goes.
x(b-1);x(b+1);x(b-w);x(b+w); // left/up/down/right
}
m(b,p,y)char*b,*p;{ // pattern match area b with pattern p, direction y.
// y=1 scans down; y=0 scans up.
d=b; // d tracks a tile in the currently matched pattern for unmarking.
// Note that all patterns are oriented to where "top-left" is a tile.
if(!y) // ...when scanning up, move p to the end, set y to -1 to count backward,
// and advance d to guarantee it points to a tile (now "bottom-left")
for(y=-1,--p;1[++p]&31;)d+=w;
// Match the pattern
for(i=0;*p&31?!(*p&16>>i)||b[i]&1:0;++i>4?p+=y,b+=w,i=0:0);
return !(*p&31) // If it matches...
? x(d),n // ...unmark/count total polyomino tiles and return the count
: 0;
}
a(b)char*b;{ // Scan for an occurrence of the pattern b.
for(j=n=0;j<w*h;++j)
if(m(c+j,b,1)||m(c+j,b,0)) // (short circuit) try down then up
return n;
return 0;
}
// This is our function. The parameter is a string containing the entire area,
// delimited by new lines. The algorithm assumes that this is a rectangular area.
// '#' is used for tiles; ' ' spaces.
f(char*b) {
bzero(c,9999); // Init categories, c buffer
for(h=0,b=strcpy(c,b);r=b,b=strchr(b+1,10);h++,w=b-r); // Find width/height
// Unmark all polyominoes that contain unfoldable subsets. This was
// compacted since the last version as follows. A tracks
// the current pattern's length; r[-1], usually terminator for the
// previous pattern, encodes whether the length increases; and o/c
// the patterns were sorted by length.
for(A=2,r=1+"@_`^C@|T@^R@XO@XX`|FB@|PP@|DD@PXN@XHX@XPX`PPXL@XHHX@XLDD@XPPX`PPPXH@PXHHH@PPPPP@";*r;r+=A+=r[-1]/96)
while(a(r));
A=B=C=D=E=F=G=H=0;
// Match corner trominoes now to ensure they go into C.
while(a("PX")||a("XH"))
(n-=3)
? --n
? --n
? --n
? 0 // More than 6 tiles? Ignore it.
: ++H // 6 tiles? It's an H.
: ++G // 5 tiles? It's a G.
: ++F // 4 tiles? It's an F.
: ++C; // only 3 tiles? It's a C.
// Now match line tetrominoes to ensure they go into E.
while(a("^")||a("PPPP"))
(n-=4)
? --n
? --n
? 0 // More than 6 tiles? Ignore it.
: ++H // 6 tiles? It's an H.
: ++G // 5 tiles? It's a G.
: ++E; // only 4 tiles? It's an E.
// Find all remaining tetrominoes ("P" is a monomino pattern)
while(a("P"))
--n
? --n
? --n
? --n
? --n
? --n
? 0 // More than 6 tiles? Ignore it.
: ++H // 6 tiles? It's an H.
: ++G // 5 tiles? It's a G.
: ++F // 4 tiles? It's an F.
: ++D // 3 tiles? It's a D.
: ++B // 2 tiles? It's a B.
: ++A; // only 1 tile? It's an A.
// Figure out if we can form a cube:
return H // Yes if we have a foldable hexomino
||(G&&A) // Yes if we have a foldable pentomino
// and a monomino
||(F&&B+B+A>1) // Yes if we have a foldable non-line tetromino
// and 2 other tiles (dominoes count twice).
// Either one domino or two monominoes will do.
||(E&&A>1) // Yes if we have a foldable line tetromino (E)
// and two monominoes (A). Note we can't make a
// cube with a line tetromino and a domino (B).
||D>1 // Yes if we have two line trominoes
||C>1 // Yes if we have two corner trominoes
||((D||C)*3+B*2+A>5)
// Any combination of trominoes, dominoes, monominoes>6,
// where trominoes are used at most once
// (via logical or)...
* (A>1||B>2||A*B)
// ...times this includer/excluder fudge factor
// that culls out the one non-working case;
// see table:
//
// Trominos Dominos Monomos Cube A>1 B>2 A*B
// 1 0 3+ yes Y N 0
// 1 1 1+ yes Y N 1
// 1 2 0 no N N 0
// 0+ 3 0+ yes Y Y 1
;
}
Why is the JS code snippet just printing additional hardcoded testcases? Why not just put them in the post haha? – Magic Octopus Urn – 2016-11-08T15:19:10.490
1@carusocomputing That was just a measure to prevent the post from being to cluttered. – flawr – 2016-11-08T19:07:14.737
Will there always be six pixels on? – Post Rock Garf Hunter – 2016-12-31T05:50:25.427
No, there might more or less on. – flawr – 2016-12-31T17:31:37.160
Can I receive input as an array of pieces rather than a single image containing all the pieces? – Blue – 2017-01-07T23:27:10.320
I don't quite understand what you mean, can you provide an example? – flawr – 2017-01-07T23:35:06.197
For test case 4,
[([1,0,1,1], 2, 2), ([1,1,0,1], 2, 2)]instead of([1,0,1,1,1,1,0,1], 4, 2). Each tuple represents the data with a list containing 1 for each piece followed by the width and the height. – Blue – 2017-01-08T01:17:39.6701@Blue Ah no, analyzing the input for pieces is part of the challenge. This input would simplify that quite a bit, so I would not allow that. Thank you for asking! – flawr – 2017-01-08T10:01:13.967