Mathemagical Box--A Mind Experiment

[Turtle]

Originally Posted by CraigD

My apologies for not explaining my arbitary labeling scheme. Unfolded, my cube looks like this:

  6
1 2 3
  4
  5

I believe my numbers map to your terms in several possible ways, including: 1=E1, 2=F1, 3=E2, 4=S1, 5=F2, 6=S2.

aha!! i recognized from past experience that such an unfolding of polyhedra is called a 'planar net' so i went a garglin for those beasts. Seems that just for the cube(oid) there are 11 possible such planar nets. who knew?

anyway, if a method is good enough for Craig then it's usually better enough for me. I haven't changed my guess or reasoning yet, but I haven't fully examined the planar nets i have attached below for the icosahedron and the dodecahedron; i reserve the right to change my mind.

Here's the links where i got 'em:
Dodecahedron -- from Wolfram MathWorld
Icosahedron -- from Wolfram MathWorld

good readin on planar nets. >> Net -- from Wolfram MathWorld

[Turtle]

Seems I might have well as pluralized 'box' in the title. No matter, I think Mathemagical well enough includes the Platonic solids.

All that just to preface that I noticed today that the tetrahedron is the only of the 5 that has no opposite parallel faces. Maybe Bucky told me and I just forgot.

PS I think technically I ought to qualify that the octahedron has parallel sides, but they aren't opposite.

Now back to our irregularly scheduled challenge. >>

Originally Posted by Craig

I repeat the challenge, inviting all numeric/geometric puzzle enthusiast to place your wagers on the regular hexahedron (cube), dodecahedron, or icosahedron for which can produce the most patterns per Turtle’s polyhedron rolling procedure.

We already know the octahedron (8 faces), at 240/120, is beaten by the cube at 288/144, and can likely guess the poor tetrahedron (4 faces) isn’t in the running.

I’ve generated the numbers for the 5 Platonic solids, but won’t post anything until enough time has passed for everyone to put their intuitive, logical, or lazy-I-can-program-so-let-a-computer-do-my-deep-thinking skills to the test.

[CraigD]

Originally Posted by CraigD

I repeat the challenge, inviting all numeric/geometric puzzle enthusiast to place your wagers on the regular hexahedron (cube), dodecahedron, or icosahedron for which can produce the most patterns per Turtle’s polyhedron rolling procedure.

We already know the octahedron (8 faces), at 240/120, is beaten by the cube at 288/144, and can likely guess the poor tetrahedron (4 faces) isn’t in the running.

I’ve generated the numbers for the 5 Platonic solids, but won’t post anything until enough time has passed for everyone to put their intuitive, logical, or lazy-I-can-program-so-let-a-computer-do-my-deep-thinking skills to the test.

A couple of days have passed, since the challenge was issued. Here’s the promised data for the 5 platonic solids:

       Term    Total
#      count*  count*  Total  End    # faces / term count
faces  #faces  #faces  Count  Count  4 5  6  7  8   9    10   11   12   13  14  15  16  17  18  19  20
4      24      64      16     6      6    
6      288     798     133    48       8  40    
8      240     896     112    30          6  6  18
12     214560  595152  49596  17880       10 60 290 1120 3400 6680 6320
20     41600   167180  8359   2080              4   4    8    28   40   72  164 240 308 424 420 260 108

”Term count” means count of patterns where no more roles can be made. “Total count” includes all patterns. Since the polyhedra are all regular, the number of unique patterns can be found by multiplying these counts by the number of faces. “# faces / term count” gives the number of counts for patterns of various lengths – notice that the 4-sided tetrahedron can’t fail to use all its faces, while for polyhedra with more tha 8 sides, most of the patterns can’t get to all the faces, but some do.

So, Turtle’s reasoned guess

Originally Posted by Turtle

Just wingin' it here as I roll my icosahedral and dodecaheral poly-dice in my hand. I am going to suggest that the dodecahedron has more paths than either the cuboid or the icosahedron. My only reason is that even though the icosahedron has more faces, they are all triangular and so have at most 2 possible turns along the way once the initial turn is made. The dodecahedron on the other hand, with its faces all pentagons, has as many as 4 possible edges to roll over after the initial move. Fewer places to go, but more ways to get to them I guess.

is right on. The 12-faced dodecahedron has many makes many more patterns than the 20-faced icosahedron, and, at 17880/5=3576 patterns for each starting face and first move, more than a human being is likely to be willing or able to count.

PS: Here’s the MUMPS code used to get the data (entering the net diagram of each polyhedron and formatting the data into the table above was done manually):

f  r "[1]Load poyhedron net [2]Count 1 [3]Count all [4]List [5]Totals:",R,! q:R=""  x $g(XTMBME(0,R)) ;XTMBME: solve hypography.com/forums/social-sciences/1606-mathemagical-box-mind-experiment.html
k XTMBME(1) f  r A x:'A "s Q=$na(XTMBME(1)) f  s Q=$Q(@Q) q:$qs(Q,1)-1  w "" ;"",Q,!" q:'A  s B=0,D="" x "f  s B=$o(XTMBME(1,A,B)) q:'B  w D,""-"",B,! s D=A" w D f  r "-",B,! q:'B  w A s (XTMBME(1,A,B),XTMBME(1,B,A))="" ;XTMBME(0,1): load polyhedron net diagram
k ^XD("XTMBME","C") s (C,A)=0 s A="1,0" x XTMBME(2) ;XTMBME(0,2): single count
k ^XD("XTMBME","C") s (C,A)=0 f  s A=$o(XTMBME(1,+A))_",0" q:'A  x XTMBME(2) ;XTMBME(0,3): count all
k WF x XTMBME(0,4,1) ;XTMBME(0,4): list
k CL,CL0 s CL=0,A="" f  s A=$o(^XD("XTMBME","C",A)) q:'A  s L=$l(A,","),B=$p(A,",",L),I="",CL0(L)=$g(CL0(L))+1 f  s I=$o(XTMBME(1,B,I)) q:","_A_","'[(","_I_",")&I  i 'I s CL=CL+1,CL(L)=$g(CL(L))+1 w:$g(WF,1) CL,". ",A,! q  ;XTMBME(0,4,1)
s WF=0 x XTMBME(0,4,1) w "#Strings:",C," #Terminal strings:",CL x XTMBME(0,5,1),XTMBME(0,5,2) w ! ;XTMBME(0,5): list
w !,"Length.#strings:" s L="" f  s L=$o(CL0(L)) q:'L  w L,".",CL0(L)," " ;XTMBME(0,5,1)
w !,"Length.#terminal strings:" s L="" f  s L=$o(CL(L)) q:'L  w L,".",CL(L)," " ;XTMBME(0,5,2)
f  s L=$l(A,",") q:L<2  s B=$o(XTMBME(1,$p(A,",",L-1),$p(A,",",L))),I=0 f  s I=$o(XTMBME(2,I)) q:'I  x XTMBME(2,I) i  q  ;XTMBME(2)
i 'B s A=$P(A,",",1,L-1) s:'$d(^XD("XTMBME","C",A)) C=C+1,^XD("XTMBME","C",A)="" q  ;XTMBME(2,1): no more sides
i (","_A_",")[(","_B_",") s $p(A,",",L)=B q  ;XTMBME(2,2): side already inked
s $p(A,",",L)=B_",0"  ;XTMBME(2,3): next side
 ;XTMBME(1,1,2)
 ;XTMBME(1,1,4)
 ;XTMBME(1,1,5)
 ;XTMBME(1,2,1)
 ;XTMBME(1,2,3)
 ;XTMBME(1,2,6)
 ;XTMBME(1,3,2)
 ;XTMBME(1,3,4)
 ;XTMBME(1,3,7)
 ;XTMBME(1,4,1)
 ;XTMBME(1,4,3)
 ;XTMBME(1,4,8)
 ;XTMBME(1,5,1)
 ;XTMBME(1,5,6)
 ;XTMBME(1,5,8)
 ;XTMBME(1,6,2)
 ;XTMBME(1,6,5)
 ;XTMBME(1,6,7)
 ;XTMBME(1,7,3)
 ;XTMBME(1,7,6)
 ;XTMBME(1,7,8)
 ;XTMBME(1,8,4)
 ;XTMBME(1,8,5)
 ;XTMBME(1,8,7)

The code has the net diagram (XTMBME(1)) for a cube – anything else must be manually entered.

[CraigD]

The previous post lists the number of prints that can be rolled out with the 5 regular polyhedra. However, irregular polyhedra can be more interesting, and are as easy for the program used above (XTMBME) to calculate as the regular ones.

For example, a regular 9-sided octahedron has 240 patterns (or 120, if you don’t count a pattern and its reverse twice), fewer than the 4-sided cube’s 288. However, the irregular octahedron pictured in the attached thumbnail has 6836 patterns.

A new challenge, then, is to find the non-regular polyhedron of 20 or fewer sides that can produce the most prints. I don’t know the answer, but can run XTMBME for any give polyhedron, so can quickly tell that one has more prints than another.

Note that just changing the edge lengths of a polyhedron doesn’t change the number of prints it produces – it’s necessary to change the network diagram describing the edges shared by the faces.