Mathemagical Box--A Mind Experiment

[InfiniteNow]

The question of how many different prints is not yet resolved.:doh:

 

Any progress since last post?

 

 

 

It has six sides. With each turn you can pivot on axis to one of, at most, four sides.

 

Are there other rules? Are you asking for a scenario where each side can only be touched to the paper once?

 

 

Originally posted by Turtle:

...Every "side" is different & the carving began at the opening & I worked out in a spiral. I ignored the edges as boundries & wrapped the pattern right over them.

 

You also have it right to name the "sides" & now since you see we have 6 different such "sides", I have these names for them.

 

• Face 1 or F1 is the large "side" with the opening you see in the photo
  
• Side 2 or S2 is the long thin "side" you see in the photo
  
• End 1 or E1 is the top smallest "side" you see in the photo
  
• Face 2 or F2 is the other large "side" which is in back & not seen in the photo
  
• Side 1 or S1 is the long thin "side" you can't see in the photo
  
• End 2 or E2 is small "side" the box is sitting on(which of course you can't see in the photo)
  

 

The name of the print in the picture is thus F1S2E1F2S1E2.

 

Last for this post, the carving technique is called 'chip carving' & more specifically I used what is called a '6 cut chip'. I proceded by laying out 1 'chip' & then carving it, laying out another chip next to it & carving it, & so on round the box. 6 years of off & on carving to completely cover the box with carving.

:D

[Turtle]

Any progress since last post?

 

 

 

It has six sides. With each turn you can pivot on axis to one of, at most, four sides.

 

Are there other rules? Are you asking for a scenario where each side can only be touched to the paper once?

 

 

Originally posted by Turtle:

[ATTACH]1558[/ATTACH]

 

no new progress. i seem to have misplaced my decision tree solution attempt, but i'll find it & get the scan posted.

 

yes, each side may touch just once, if it touches at all. as you say, on the first turn/roll, one has four choices, however on the second turn (since you can't go back to start) you have only 3 possible next moves. here's another photo with more of the prints. :D :doh:

[InfiniteNow]

yes, each side may touch just once, if it touches at all.

Okay... it's that "at all" part on which I need to cogitate. :doh:

[Turtle]

Okay... it's that "at all" part on which I need to cogitate. :cup::cup:

 

indeed. it is an emergent property. took me by surprise i must say. it turns out that certain turn orders isolate a side so that it can't be reached without going back. here's my tree. one advantage of minimalism is that if i have something, i don't have far to look for it. :hihi: :clock:

[InfiniteNow]

Well, for one, you must add an additional "possibility" at each and every branch/stem since there is the possibility of stopping. So, instead of just F1 and S1, you can have F1, S1, and STOP... that sort of thing.

 

I assume also you are interested in the permutations of just one starting position? Or, must we, once we arrive at an answer, multiply that by six to represent the different possible starting positions?

[Turtle]

Well, for one, you must add an additional "possibility" at each and every branch/stem since there is the possibility of stopping. So, instead of just F1 and S1, you can have F1, S1, and STOP... that sort of thing.

 

I assume also you are interested in the permutations of just one starting position? Or, must we, once we arrive at an answer, multiply that by six to represent the different possible starting positions?

 

intermediate stopping is not an option in my circumstance. if a turn is possible, it is taken. the path is ended only when no more turns are possible.

 

my tree is for starting from 1 'side' only, but all other 5 sides have the same tree structure so i didn't draw them. so yes, multiply by 6 >> 6 *24 = 144

 

however, because turning any print up-side down (or exchanging start for finish if you will) gives another print from the count, i divide 144 by 2 to arrive at my answer of 72 possible different paths, 12 of which have path-length 5. :cup: :clock:

[InfiniteNow]

I was going to suggest 82 possibilities for each starting position (I'd show my work, but don't have scanner here), hence a total of 82 * 6 = 492.

 

Then, you threw in another caveat about it not being a simple N becomes n-1 possibilities, but N becomes n-(1-(unreachable side(s))) possibilities...

 

 

Hmmmfff... I have a plane ride next week. If someone doesn't post the answer by then, I'll spend some time on the trip drawing on some napkins.

 

I've looked for the formula, and can tell you how many possible combinations of a rubix cube there are, and how many ways you can organize the letters in a word, but this one currently escapes my ability to reframe the issue into proper terms.

 

Btw... math and irish whiskey not so complimentary. :cup::clock:

[Turtle]

I was going to suggest 82 possibilities for each starting position (I'd show my work, but don't have scanner here), hence a total of 82 * 6 = 492.

 

Then, you threw in another caveat about it not being a simple N becomes n-1 possibilities, but N becomes n-(1-(unreachable side(s))) possibilities...

 

Hmmmfff... I have a plane ride next week. If someone doesn't post the answer by then, I'll spend some time on the trip drawing on some napkins.

 

I've looked for the formula, and can tell you how many possible combinations of a rubix cube there are, and how many ways you can organize the letters in a word, but this one currently escapes my ability to reframe the issue into proper terms.

 

Btw... math and irish whiskey not so complimentary. :cup: :hihi:

 

acknowledged. :clock: what you need now is a pipe.:cup:

 

even though my question is couched in terms that permutation/combination formulae often provide answers for, i.e. "how many different ways____?", they don't work here as far as i have found. this is why i chose to use a decision tree. in the strictest sense, i didn't 'throw in' the caveat; it threw me off. the tree itself does not solve the problem either, but is only a map after the fact of the actual experiments of rolling the box.

 

the problem is as much geometric as it is counting, and trying to follow it in your head gedanken style is beyond the ken of most people (myself included). make a box, label it, and roll, roll, roll in za hay...

 

anyone else have an opinion on the matter? :cup: :cup:

[InfiniteNow]

42

[Turtle]

42

 

that's only the answer to life. :rolleyes: if you have some work to support it i'll be glad to have a look, but i am still going with my 72 for the answer to how many different prints i can make with the box. :shade: ;)

[CraigD]

If you ink up all 6 sides of the box & on a large sheet of paper place the box on any side you choose. Now without moving the box completely from the papaer, roll it to an adjacent side. Now roll it again to a different adjacent side & so on until you have a continuous print of the box. I have to wait for my friend to take a picture of one of the prints.

___Now the question is, how many different such prints can you (I) make? I have a method to solve this in mind, but I never actually did. I don't know the answer. So, if you can tell me a method & answer I suspect is right, I'll try to check it.

If you start each pattern in the center of a large sheet of paper, count patterns that are just shifted and rotated images of one another as unique, and use each face no more or less than 1 time, 960 patterns are possible. If you count patterns in which some faces are not used, 3156. Excluding patters that are different due to the 4 possible orientations of the starting face, these numbers are divided by 4 to become 240 and 789

 

The edge lengths of the polyhedron (the box) doesn’t matter, but the number of sides it has (6) does.

 

Things get more interesting if you allow a face to be used more than 1 time. They also get more complicated to calculate to count, unless you count patterns with the same shape and faces, but different multiple inkings of the same face – “rollback” patterns. In that case, allowing a face to be used up to 2 times produced 943488 patterns that use each face exactly twice, and 3785112 that don’t.

[Turtle]

If you start each pattern in the center of a large sheet of paper, count patterns that are just shifted and rotated images of one another as unique, and use each face no more or less than 1 time, 960 patterns are possible. If you count patterns in which some faces are not used, 3156. Excluding patters that are different due to the 4 possible orientations of the starting face, these numbers are divided by 4 to become 240 and 789

 

The edge lengths of the polyhedron (the box) doesn’t matter, but the number of sides it has (6) does.

 

Things get more interesting if you allow a face to be used more than 1 time. They also get more complicated to calculate to count, unless you count patterns with the same shape and faces, but different multiple inkings of the same face – “rollback” patterns. In that case, allowing a face to be used up to 2 times produced 943488 patterns that use each face exactly twice, and 3785112 that don’t.

 

OK. The orientation of the starting face doesn't matter, because it does not change how many faces the box can roll to from start (always 4).

 

I agree the edge length doesn't matter to the geometry; however it does help in visualizing the 3 different sets of opposite sides.

 

Your answers seem too high - 240 and 789 - based on the 20 prints I already made. How did you make the count? by experiment? some calculation? Can you see an error to correct in the tree I posted?

 

I'll put the photos of the prints in the Gallery & come back to edit them in. Thanks Craig. :Glasses:

[img=http://hypography.com/gallery/displayimage.php?imageid=1800]http://hypography.com/gallery/displayimage.php?imageid=1800[/img]

[CraigD]

Your answers seem too high - 240 and 789 - based on the 20 prints I already made. How did you make the count? by experiment? some calculation? Can you see an error to correct in the tree I posted?

I just wrote a small program to roll a cube every possible way that had each face down no more than one time.

 

Starting on face “1” and counting only when each face prints 1 time, this generates the following:

  1. 1,2,3,4,5,6
 2. 1,2,3,6,5,4
 3. 1,2,4,3,5,6
 4. 1,2,4,3,6,5
 5. 1,2,4,5,3,6
 6. 1,2,4,5,6,3
 7. 1,2,6,3,4,5
 8. 1,2,6,3,5,4
 9. 1,2,6,5,3,4
10. 1,2,6,5,4,3
11. 1,4,2,3,5,6
12. 1,4,2,3,6,5
13. 1,4,2,6,3,5
14. 1,4,2,6,5,3
15. 1,4,3,2,6,5
16. 1,4,3,5,6,2
17. 1,4,5,3,2,6
18. 1,4,5,3,6,2
19. 1,4,5,6,2,3
20. 1,4,5,6,3,2
21. 1,5,3,4,2,6
22. 1,5,3,6,2,4
23. 1,5,4,2,3,6
24. 1,5,4,2,6,3
25. 1,5,4,3,2,6
26. 1,5,4,3,6,2
27. 1,5,6,2,3,4
28. 1,5,6,2,4,3
29. 1,5,6,3,2,4
30. 1,5,6,3,4,2
31. 1,6,2,3,4,5
32. 1,6,2,3,5,4
33. 1,6,2,4,3,5
34. 1,6,2,4,5,3
35. 1,6,3,2,4,5
36. 1,6,3,5,4,2
37. 1,6,5,3,2,4
38. 1,6,5,3,4,2
39. 1,6,5,4,2,3
40. 1,6,5,4,3,2

Repeating this for each face results in 240 patterns.

 

I only notices this thread when infi’s “math and wiskey” post bumped it, and haven’t read all its posts yet, so haven’t compared my list to yours, Turtle.

 

PS: This is the MUMPS code I used. Most of it (XTMBME(1)) isn’t code at all, but a diagram of adjacent cube faces

k C s (C,A)=0 f  s A=$o(XTMBME(1,+A))_",0" q:'A  x XTMBME(2) ;XTMBME: hypography.com/forums/social-sciences/1606-mathemagical-box-mind-experiment.html stuff
;XTMBME(1,1,2)
;XTMBME(1,1,4)
;XTMBME(1,1,5)
;XTMBME(1,1,6)
;XTMBME(1,2,1)
;XTMBME(1,2,3)
;XTMBME(1,2,4)
;XTMBME(1,2,6)
;XTMBME(1,3,2)
;XTMBME(1,3,4)
;XTMBME(1,3,5)
;XTMBME(1,3,6)
;XTMBME(1,4,1)
;XTMBME(1,4,2)
;XTMBME(1,4,3)
;XTMBME(1,4,5)
;XTMBME(1,5,1)
;XTMBME(1,5,3)
;XTMBME(1,5,4)
;XTMBME(1,5,6)
;XTMBME(1,6,1)
;XTMBME(1,6,2)
;XTMBME(1,6,3)
;XTMBME(1,6,5)
f  w A,! s L=$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(C(A)) C=C+1,C(A)="" w C,". " 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

[Turtle]

I just wrote a small program to roll a cube every possible way that had each face down no more than one time.

 

Starting on face “1” and counting only when each face prints 1 time, this generates the following:

  1. 1,2,3,4,5,6
 2. 1,2,3,6,5,4
 3. 1,2,4,3,5,6
 4. 1,2,4,3,6,5
 5. 1,2,4,5,3,6
 6. 1,2,4,5,6,3
 7. 1,2,6,3,4,5
 8. 1,2,6,3,5,4
 9. 1,2,6,5,3,4
10. 1,2,6,5,4,3
11. 1,4,2,3,5,6
12. 1,4,2,3,6,5
13. 1,4,2,6,3,5
14. 1,4,2,6,5,3
15. 1,4,3,2,6,5
16. 1,4,3,5,6,2
17. 1,4,5,3,2,6
18. 1,4,5,3,6,2
19. 1,4,5,6,2,3
20. 1,4,5,6,3,2
21. 1,5,3,4,2,6
22. 1,5,3,6,2,4
23. 1,5,4,2,3,6
24. 1,5,4,2,6,3
25. 1,5,4,3,2,6
26. 1,5,4,3,6,2
27. 1,5,6,2,3,4
28. 1,5,6,2,4,3
29. 1,5,6,3,2,4
30. 1,5,6,3,4,2
31. 1,6,2,3,4,5
32. 1,6,2,3,5,4
33. 1,6,2,4,3,5
34. 1,6,2,4,5,3
35. 1,6,3,2,4,5
36. 1,6,3,5,4,2
37. 1,6,5,3,2,4
38. 1,6,5,3,4,2
39. 1,6,5,4,2,3
40. 1,6,5,4,3,2

Repeating this for each face results in 240 patterns.

 

I only notices this thread when infi’s “math and wiskey” post bumped it, and haven’t read all its posts yet, so haven’t compared my list to yours, Turtle.

 

OK Hopefully when you read the thread you will see why some in your list don't belong (I haven't checked which) because in actually rolling the box without lifting to make the prints, some faces aren't available with a roll even though they haven't been printed. Does that make sense?I think you are the right track however as when I divide by your 240 by my 72 I get 3.333333. Interesting mathematically. :)

 

Did you look at my tree in post #38?

 

http://hypography.com/forums/attachment.php?attachmentid=1560&d=1182391241[/img]

I think we can transpose your 1 through 6 to my face/edge/side terminology and compare our lists. Understand that my proposed answer is based on really making the prints and observing how possibilities for rolling the box without lifting to print. I think there are 60 different patterns of length 6 (all 6 faces can be printed) and 12 different patterns of length 5 (a face can't be reached).

 

If you need me too, I can make some short videos to illustrate the problem? :doh: :hihi:

[InfiniteNow]

In sum, you have a rectangular cuboid (hence six sides).

 

You roll it, pivoting from one of those sides.

 

You cannot touch the same side down more than once. After it's touched, it's out of play.

 

When rolled, not each of the remaining five (or less, if it's no longer the first roll) sides are available to pivot toward.

 

You can either explore those remaining sides which are available to you (not yet stamped and available from the side you're on) or stop entirely.

 

How many different possibilities are there?

[Turtle]

I think we can transpose your 1 through 6 to my face/edge/side terminology and compare our lists. Understand that my proposed answer is based on really making the prints and observing how possibilities for rolling the box without lifting to print. I think there are 60 different patterns of length 6 (all 6 faces can be printed) and 12 different patterns of length 5 (a face can't be reached).

 

If you need me too, I can make some short videos to illustrate the problem? :) :hihi:

 

First, it helps if you actually get a box, rather than try to gedanken this. A cube is fine if that's what you have.

 

Let Craig's 1 = my F1

___Craig's 2 = my S1

___Craig's 3 = my E1

___Craig's 4 = my F2

___Craig's 5 = my S2

___Craig's 6 = my E2

 

Now start with #1 face down and roll to #2 along their common edge, then roll to #3 along the common edge of #2 & #3, then roll to #4, then #5, then finally #6. This is Craig's first path in his list, and the 4th in my list below the tree.

 

Now try to follow the same procedure with Craig's second path in his list, #1, #2, #3, #6, #5, #4. See that when you arrive on face #3 that #6 does not share a common edge as it is straight up and opposite the down #3? It does not belong in the list. Gettin it yet?

 

Will take me a bit to make a vid. make a box and try it. Think about the edges that are common to any given faces. ;) :doh:

[CraigD]

Did you look at my tree in post #38?

…

I think we can transpose your 1 through 6 to my face/edge/side terminology and compare our lists.

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.

 

As I read you tree, our results agree. You appear to find 40 strings of 6 faces for each starting face, as did I.

 

You also find 8 strings of 5 faces for each starting face. I didn’t list my 5-face strings in my first post. I find 48 – however, 40 of them are simple the 40 6-face strings, before the last roll of the cube. Subtracting those, I find the following 8:

  1. 1,2,3,5,4
 2. 1,2,3,5,6
 3. 1,4,3,6,2
 4. 1,4,3,6,5
 5. 1,5,3,2,4
 6. 1,5,3,2,6
 7. 1,6,3,4,2
 8. 1,6,3,4,5

which agrees with your tree.

 

:hihi: I can’t figure out where you’re getting the “72” figure from, though. :doh: