So that makes for 7270 squares which need to be packed and as 727 is prime I think that means the only rectangle worth trying is 10x727. And if we leave out the ones with holes we have 683 which is also prime so there I could try 10x683. Odd...
Wouldn't filling in the holes be considered cheating?
Interesting... though this particular decomino can simply be folded up as is into a dicube. Why this need to "cut" it up?
I suppose I was tackling this problem from a more topological point of view. Imagine we had a paper cut out of each net and we consider two squares to be connected if and only if they are connected by an edge (conrners don't count!). Then all 11 cube unfoldings are topologically equivalent to a filled-disc. I used this as a starting point to bring the analogy up to a dicube. That means every dicube net would require a starting square with 9 other squares attached by one side to a growing net. That means there are 9 connections between the 10 squares on the net; 9 places in the paper where the edge has not been cut away, but rather creased.
(In fact this is how I performed the search: I considered all the ways to choose 11 of the 20 edges of a dicube to cut away, and examined the resulting net. After collecting all the nets that were in one piece and did not overlap themselves, I created rotations and reflections and made an ordered list and removed duplicates. I technically never searched for nets that were formed by choosing less than 11 edges to cut, but these are a subset of the list I did find, since every net that has more than 9 remaining edges can have some cut away to form another legal net that is identical for your purposes)
However, whenever a 2x2 box of squares appears, the analogy creates 4 shapes which essentially make the letter C rotated 4 times. Clearly there is a place where two squares could have been fused together completing an O-shape. However, this changes the topology! The net is now topologically equivalent to the shadow of a donut. From your point of view, you do not care about how many "extra" cuts there are in the shape, so this argument doesn't really apply - I'm just pointing out that there are more ways of looking at this than just as a decomino
Makes me wonder... are there decominos which need to be cut before they can be folded into a dicube on your list?
Any decomino that could be cut without forming two disjoint shapes has at least 10 edges that have not been cut. Therefore it has at least one closed path. If this closed path is a 2x2 box, then no square of that 2x2 box can be one of the ends of the dicube. They all therefore must belong to two long faces, which does not require cutting. If this closed path is a 3x3 box with the 1x1 center missing (which I don't believe exists... but I haven't confirmed), then without loss of generality, the top left square will be cut from the top middle square. Then:
If the top left square is an end of the dicube, it coincides with the top right square.
If the bottom left square is an end of the dicube, then the top left and right square coincide.
If the bottom right square is an end of the dicube, then the left square and the top right square coincide.
If the top right square is an end of the dicube, it coincides with the top left square.
If any edge is an end of the dicube, then the opposite edge in the 3x3 box is the same end... that doesn't work...
Finally, if none of these squares are an end, then the bottom 1x3 must wrap around the long faces of the dicube, which means either the middle left AND middle right square are both the same end, or the top left AND top right are both the same end, so those each have a sort of double contradiction. In any case, no such closed path can form a dicube.
If this closed path is a 4x3 box with the 1x2 center missing, then all 10 squares must be used in making this shape, so this net is unique and since this net does not give a dicube (proof of that: check the list
), such a closed path cannot form a dicube. FINALLY, if this closed path is any other shape, it uses more than 10 squares and the dicube net may only have 10 squares. QED
I believe all the foldings of a cube can be turned into a cube with just folding.
Correct. Topological reasons force all cube nets to have exactly 5 uncut edges which is the minimum required to keep the net in one piece.
Here with the dicube it appears there may be two types of unfoldings.
Yes. I suppose this is exactly my point. You can consider at least these two types of unfoldings. One where you just look at the shape of the net and one where you consider additional cuts made into the net. For a cube, the results are the same. For the dicube, the second gives many more nets than the first. Understanding this was also pretty crucial to the way I implemented my search as well.
I did not consider the same net folding into a dicube two different ways. If such a net exists, it would only appear once in my list though.
I'm not sure such a thing exists either to be honest. The ends of the dicube are clearly different then the other 8 squares so I could assume there may be a net where a particular square in one folding is an end square and in another folding of the same net a "side" square. Maybe there is some reason which keeps this from occuring. I just can't prove they don't and I haven't spend much time looking for an example.
I thought the same thing here too, but BOTH of our intuitions are WRONG! Take a look at the following net and see if you can tell me why
(approximating the best I can here, the O and the _ are different widths... -_- ... it's supposed to be two plus-shapes fused together)
How is your cubing related problem coming along? I haven't heard much about it in a while but I assume its still in the works. Being vague but I think you'll know what I'm talking about.
Pretty fantastic actually
At least as good as I could hope for at this point. (and actually you're not the only one who is curious
) I'm not done, and I can't guarantee that my idea will work or will be complete by the time my thesis is due in May, but I give myself probably a 30% success probability at this point, and considering what it is, I'll definitely take that for now. Must keep at it....
Jeffery Mewtamer wrote:
I have to ask, do the ones that can be divided into two of the pentominoes that can be folded into open cubes, and the ones without this property form interesting subsets?
Are there any other interesting properties that can be used to divide these into subsets?
Good questions! I have no idea
Maybe you should look into and tell us what you find
I don't have time to look into it more, but I can get you started: I have attached the files DicubeNetsPents and DicubeNetsOther. The -Pents file contains only those dicube nets that cut up exactly 3 of the 4 edges "fusing" the two cubes of a dicube together, thereby dividing the dicube into two open cubes at the remaining edge. Remembering that I am including additional cuts, this DOES include two open-cube nets fused along an edge more than one cube long, for example two L-shapes fused along the long sides. The -Others file includes all the nets not included in the -Pents file. Is this equal to the set of all nets that can be formed by cutting up less than 3 of the 4 edges fusing the two cubes of a dicube? NO! That is because there are some nets that can be folded into a dicube in more than one way!
One such way is given above. I do not know how many of these exist, but I have found at least a subset of these. The -Doubles file contains every net that could have come from cutting up 3 of the 4 fusing edges OR from cutting up less than 3 of the 4 fusing edges. Two such methods give clearly different folds, but identical nets in some cases...
In any case, I can now say that yes, there are at least 42 dicube nets that can be folded into a dicube in more than one way: the 42 given in DicubeNetsDoubles.
(it really is the answer to everything....