Mathemagical Box--A Mind Experiment

[Turtle]

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:

 

I can't figure out where you think I have 40??? My complete list of paths under the tree is only 24 paths? then, 24 paths from each face, so 24*6=144...but!!, because each print is two if you make the ending face the starting face without changing the interior moves, so divide 144 by 2 to get 72.

 

Note. I expanded the tree for only 1 starting face.

 

the labeling is arbitrary as long as you stick with it from the git-go.

 

I have made a small vid & will start uploading to Youtube now. be patient, there is no set time for it to be available after an upload. ;) :)

[Turtle]

the gods are smiling on us...me anyway. :doh: vid loaded faster than i've ever seen. This is the print F1, S1, E1, F2, S2, E2. the numbers on the box reflect my transposition to Craig's numbers 1 through 6. :) :hihi:

 

YouTube - Printing Mathemagical Box http://www.youtube.com/watch?v=QPg2_58-xSM

[CraigD]

I can't figure out where you think I have 40???

I agree that there are 10 six-sided + 2 five-sided, can't make another roll = 12 total strings for each starting side and first side rolled to. There are 4 possible sides to roll to on the first roll, so there are 40 six-sided + 8 five-sided = 48 total strings for each starting side. There are 6 starting sides, so 240 six-sided + 48 five-sided = 288 total stings. Dividing by 2 to consider strings like “1,2,3,4,5,6” and “6,5,4,3,2,1” the same gives 120 six-sided +24 five-sided = 144 total strings.

My complete list of paths under the tree is only 24 paths? then, 24 paths from each face, so 24*6=144...but!!, because each print is two if you make the ending face the starting face without changing the interior moves, so divide 144 by 2 to get 72.

I think there's a problem with the “24”. Your post #38 tree (and my program) should agree that there are 48 paths from each face.

 

From the list under the diagram in the image, there are 12 paths beginning F1S1, so there should be 4 * 12 = 48 beginning F1, 48 * 6 = 288 beginning with any side. Dividing by 2 to consider strings like “1,2,3,4,5,6” and “6,5,4,3,2,1” the same, 288 / 2 = 144.

 

I can’t figure out where you’re getting the last division by 2 to give 144 / 2 = 72, and suspect it’s one division by 2 too many.

 

Here’s the complete 288-string list (ignoring entries where the last face has a greater number than the first will transform it into the 144-string list), generated by this nifty little 1-line program:

 s C=0,A="" f  s A=$o(C(A)) q:'A  s B=$p(A,",",$l(A,",")),I="" f  s I=$o(XTMBME(1,B,I)) q:A'[i  i 'I s C=C+1 w $j(C,2),". ",A,! q[/code]
(which requires the C array created by the code in [post=182768]post #47[/post]).[code] 1. 1,2,3,4,5,6
2. 1,2,3,5,4
3. 1,2,3,5,6
4. 1,2,3,6,5,4
5. 1,2,4,3,5,6
6. 1,2,4,3,6,5
7. 1,2,4,5,3,6
8. 1,2,4,5,6,3
9. 1,2,6,3,4,5
10. 1,2,6,3,5,4
11. 1,2,6,5,3,4
12. 1,2,6,5,4,3
13. 1,4,2,3,5,6
14. 1,4,2,3,6,5
15. 1,4,2,6,3,5
16. 1,4,2,6,5,3
17. 1,4,3,2,6,5
18. 1,4,3,5,6,2
19. 1,4,3,6,2
20. 1,4,3,6,5
21. 1,4,5,3,2,6
22. 1,4,5,3,6,2
23. 1,4,5,6,2,3
24. 1,4,5,6,3,2
25. 1,5,3,2,4
26. 1,5,3,2,6
27. 1,5,3,4,2,6
28. 1,5,3,6,2,4
29. 1,5,4,2,3,6
30. 1,5,4,2,6,3
31. 1,5,4,3,2,6
32. 1,5,4,3,6,2
33. 1,5,6,2,3,4
34. 1,5,6,2,4,3
35. 1,5,6,3,2,4
36. 1,5,6,3,4,2
37. 1,6,2,3,4,5
38. 1,6,2,3,5,4
39. 1,6,2,4,3,5
40. 1,6,2,4,5,3
41. 1,6,3,2,4,5
42. 1,6,3,4,2
43. 1,6,3,4,5
44. 1,6,3,5,4,2
45. 1,6,5,3,2,4
46. 1,6,5,3,4,2
47. 1,6,5,4,2,3
48. 1,6,5,4,3,2
49. 2,1,4,3,5,6
50. 2,1,4,3,6,5
51. 2,1,4,5,3,6
52. 2,1,4,5,6,3
53. 2,1,5,3,4
54. 2,1,5,3,6
55. 2,1,5,4,3,6
56. 2,1,5,6,3,4
57. 2,1,6,3,4,5
58. 2,1,6,3,5,4
59. 2,1,6,5,3,4
60. 2,1,6,5,4,3
61. 2,3,4,1,5,6
62. 2,3,4,1,6,5
63. 2,3,4,5,1,6
64. 2,3,4,5,6,1
65. 2,3,5,1,4
66. 2,3,5,1,6
67. 2,3,5,4,1,6
68. 2,3,5,6,1,4
69. 2,3,6,1,4,5
70. 2,3,6,1,5,4
71. 2,3,6,5,1,4
72. 2,3,6,5,4,1
73. 2,4,1,5,3,6
74. 2,4,1,5,6,3
75. 2,4,1,6,3,5
76. 2,4,1,6,5,3
77. 2,4,3,5,1,6
78. 2,4,3,5,6,1
79. 2,4,3,6,1,5
80. 2,4,3,6,5,1
81. 2,4,5,1,6,3
82. 2,4,5,3,6,1
83. 2,4,5,6,1
84. 2,4,5,6,3
85. 2,6,1,4,3,5
86. 2,6,1,4,5,3
87. 2,6,1,5,3,4
88. 2,6,1,5,4,3
89. 2,6,3,4,1,5
90. 2,6,3,4,5,1
91. 2,6,3,5,1,4
92. 2,6,3,5,4,1
93. 2,6,5,1,4,3
94. 2,6,5,3,4,1
95. 2,6,5,4,1
96. 2,6,5,4,3
97. 3,2,1,4,5,6
98. 3,2,1,5,4
99. 3,2,1,5,6
100. 3,2,1,6,5,4
101. 3,2,4,1,5,6
102. 3,2,4,1,6,5
103. 3,2,4,5,1,6
104. 3,2,4,5,6,1
105. 3,2,6,1,4,5
106. 3,2,6,1,5,4
107. 3,2,6,5,1,4
108. 3,2,6,5,4,1
109. 3,4,1,2,6,5
110. 3,4,1,5,6,2
111. 3,4,1,6,2
112. 3,4,1,6,5
113. 3,4,2,1,5,6
114. 3,4,2,1,6,5
115. 3,4,2,6,1,5
116. 3,4,2,6,5,1
117. 3,4,5,1,2,6
118. 3,4,5,1,6,2
119. 3,4,5,6,1,2
120. 3,4,5,6,2,1
121. 3,5,1,2,4
122. 3,5,1,2,6
123. 3,5,1,4,2,6
124. 3,5,1,6,2,4
125. 3,5,4,1,2,6
126. 3,5,4,1,6,2
127. 3,5,4,2,1,6
128. 3,5,4,2,6,1
129. 3,5,6,1,2,4
130. 3,5,6,1,4,2
131. 3,5,6,2,1,4
132. 3,5,6,2,4,1
133. 3,6,1,2,4,5
134. 3,6,1,4,2
135. 3,6,1,4,5
136. 3,6,1,5,4,2
137. 3,6,2,1,4,5
138. 3,6,2,1,5,4
139. 3,6,2,4,1,5
140. 3,6,2,4,5,1
141. 3,6,5,1,2,4
142. 3,6,5,1,4,2
143. 3,6,5,4,1,2
144. 3,6,5,4,2,1
145. 4,1,2,3,5,6
146. 4,1,2,3,6,5
147. 4,1,2,6,3,5
148. 4,1,2,6,5,3
149. 4,1,5,3,2,6
150. 4,1,5,3,6,2
151. 4,1,5,6,2,3
152. 4,1,5,6,3,2
153. 4,1,6,2,3,5
154. 4,1,6,3,2
155. 4,1,6,3,5
156. 4,1,6,5,3,2
157. 4,2,1,5,3,6
158. 4,2,1,5,6,3
159. 4,2,1,6,3,5
160. 4,2,1,6,5,3
161. 4,2,3,5,1,6
162. 4,2,3,5,6,1
163. 4,2,3,6,1,5
164. 4,2,3,6,5,1
165. 4,2,6,1,5,3
166. 4,2,6,3,5,1
167. 4,2,6,5,1
168. 4,2,6,5,3
169. 4,3,2,1,5,6
170. 4,3,2,1,6,5
171. 4,3,2,6,1,5
172. 4,3,2,6,5,1
173. 4,3,5,1,2,6
174. 4,3,5,1,6,2
175. 4,3,5,6,1,2
176. 4,3,5,6,2,1
177. 4,3,6,1,2
178. 4,3,6,1,5
179. 4,3,6,2,1,5
180. 4,3,6,5,1,2
181. 4,5,1,2,3,6
182. 4,5,1,2,6,3
183. 4,5,1,6,2,3
184. 4,5,1,6,3,2
185. 4,5,3,2,1,6
186. 4,5,3,2,6,1
187. 4,5,3,6,1,2
188. 4,5,3,6,2,1
189. 4,5,6,1,2,3
190. 4,5,6,2,1
191. 4,5,6,2,3
192. 4,5,6,3,2,1
193. 5,1,2,3,4
194. 5,1,2,3,6
195. 5,1,2,4,3,6
196. 5,1,2,6,3,4
197. 5,1,4,2,3,6
198. 5,1,4,2,6,3
199. 5,1,4,3,2,6
200. 5,1,4,3,6,2
201. 5,1,6,2,3,4
202. 5,1,6,2,4,3
203. 5,1,6,3,2,4
204. 5,1,6,3,4,2
205. 5,3,2,1,4
206. 5,3,2,1,6
207. 5,3,2,4,1,6
208. 5,3,2,6,1,4
209. 5,3,4,1,2,6
210. 5,3,4,1,6,2
211. 5,3,4,2,1,6
212. 5,3,4,2,6,1
213. 5,3,6,1,2,4
214. 5,3,6,1,4,2
215. 5,3,6,2,1,4
216. 5,3,6,2,4,1
217. 5,4,1,2,3,6
218. 5,4,1,2,6,3
219. 5,4,1,6,2,3
220. 5,4,1,6,3,2
221. 5,4,2,1,6,3
222. 5,4,2,3,6,1
223. 5,4,2,6,1
224. 5,4,2,6,3
225. 5,4,3,2,1,6
226. 5,4,3,2,6,1
227. 5,4,3,6,1,2
228. 5,4,3,6,2,1
229. 5,6,1,2,3,4
230. 5,6,1,2,4,3
231. 5,6,1,4,2,3
232. 5,6,1,4,3,2
233. 5,6,2,1,4,3
234. 5,6,2,3,4,1
235. 5,6,2,4,1
236. 5,6,2,4,3
237. 5,6,3,2,1,4
238. 5,6,3,2,4,1
239. 5,6,3,4,1,2
240. 5,6,3,4,2,1
241. 6,1,2,3,4,5
242. 6,1,2,3,5,4
243. 6,1,2,4,3,5
244. 6,1,2,4,5,3
245. 6,1,4,2,3,5
246. 6,1,4,3,2
247. 6,1,4,3,5
248. 6,1,4,5,3,2
249. 6,1,5,3,2,4
250. 6,1,5,3,4,2
251. 6,1,5,4,2,3
252. 6,1,5,4,3,2
253. 6,2,1,4,3,5
254. 6,2,1,4,5,3
255. 6,2,1,5,3,4
256. 6,2,1,5,4,3
257. 6,2,3,4,1,5
258. 6,2,3,4,5,1
259. 6,2,3,5,1,4
260. 6,2,3,5,4,1
261. 6,2,4,1,5,3
262. 6,2,4,3,5,1
263. 6,2,4,5,1
264. 6,2,4,5,3
265. 6,3,2,1,4,5
266. 6,3,2,1,5,4
267. 6,3,2,4,1,5
268. 6,3,2,4,5,1
269. 6,3,4,1,2
270. 6,3,4,1,5
271. 6,3,4,2,1,5
272. 6,3,4,5,1,2
273. 6,3,5,1,2,4
274. 6,3,5,1,4,2
275. 6,3,5,4,1,2
276. 6,3,5,4,2,1
277. 6,5,1,2,3,4
278. 6,5,1,2,4,3
279. 6,5,1,4,2,3
280. 6,5,1,4,3,2
281. 6,5,3,2,1,4
282. 6,5,3,2,4,1
283. 6,5,3,4,1,2
284. 6,5,3,4,2,1
285. 6,5,4,1,2,3
286. 6,5,4,2,1
287. 6,5,4,2,3
288. 6,5,4,3,2,1

I call that last program “nifty”, because not only is it short, but it makes no assumptions about the number of sides of the polyhedron used to roll out the print. It also proves exhaustively that there are no prints of fewer than 5 sides where no more rolls were possible.

 

By describing a different polyhedron in the XTMBME(1) array, it can be run for any polyhedron. Running it for an octahedron (8 triangular sides) produces surprising results – although you’d expect more possible prints, there are actually fewer: 144 of all 8 sides, 48 of only 7, and 48 of only 6, for a total of 240 (which could be divided by 2 if “1,2,3,4,8,7,6,5” and “5,6,7,8,4,3,2,1” etc. are considered one print). Here they are:

 1. 1,2,3,4,8,5,6,7
2. 1,2,3,4,8,7,6,5
3. 1,2,3,7,6,5,8,4
4. 1,2,3,7,8,4
5. 1,2,3,7,8,5,6
6. 1,2,6,5,8,4,3,7
7. 1,2,6,5,8,7,3,4
8. 1,2,6,7,3,4,8,5
9. 1,2,6,7,8,4,3
10. 1,2,6,7,8,5
11. 1,4,3,2,6,5,8,7
12. 1,4,3,2,6,7,8,5
13. 1,4,3,7,6,2
14. 1,4,3,7,6,5,8
15. 1,4,3,7,8,5,6,2
16. 1,4,8,5,6,2,3,7
17. 1,4,8,5,6,7,3,2
18. 1,4,8,7,3,2,6,5
19. 1,4,8,7,6,2,3
20. 1,4,8,7,6,5
21. 1,5,6,2,3,4,8,7
22. 1,5,6,2,3,7,8,4
23. 1,5,6,7,3,2
24. 1,5,6,7,3,4,8
25. 1,5,6,7,8,4,3,2
26. 1,5,8,4,3,2,6,7
27. 1,5,8,4,3,7,6,2
28. 1,5,8,7,3,2,6
29. 1,5,8,7,3,4
30. 1,5,8,7,6,2,3,4
31. 2,1,4,3,7,6,5,8
32. 2,1,4,3,7,8,5,6
33. 2,1,4,8,5,6,7,3
34. 2,1,4,8,7,3
35. 2,1,4,8,7,6,5
36. 2,1,5,6,7,3,4,8
37. 2,1,5,6,7,8,4,3
38. 2,1,5,8,4,3,7,6
39. 2,1,5,8,7,3,4
40. 2,1,5,8,7,6
41. 2,3,4,1,5,6,7,8
42. 2,3,4,1,5,8,7,6
43. 2,3,4,8,5,1
44. 2,3,4,8,5,6,7
45. 2,3,4,8,7,6,5,1
46. 2,3,7,6,5,1,4,8
47. 2,3,7,6,5,8,4,1
48. 2,3,7,8,4,1,5,6
49. 2,3,7,8,5,1,4
50. 2,3,7,8,5,6
51. 2,6,5,1,4,3,7,8
52. 2,6,5,1,4,8,7,3
53. 2,6,5,8,4,1
54. 2,6,5,8,4,3,7
55. 2,6,5,8,7,3,4,1
56. 2,6,7,3,4,1,5,8
57. 2,6,7,3,4,8,5,1
58. 2,6,7,8,4,1,5
59. 2,6,7,8,4,3
60. 2,6,7,8,5,1,4,3
61. 3,2,1,4,8,5,6,7
62. 3,2,1,4,8,7,6,5
63. 3,2,1,5,6,7,8,4
64. 3,2,1,5,8,4
65. 3,2,1,5,8,7,6
66. 3,2,6,5,1,4,8,7
67. 3,2,6,5,8,4,1
68. 3,2,6,5,8,7
69. 3,2,6,7,8,4,1,5
70. 3,2,6,7,8,5,1,4
71. 3,4,1,2,6,5,8,7
72. 3,4,1,2,6,7,8,5
73. 3,4,1,5,6,2
74. 3,4,1,5,6,7,8
75. 3,4,1,5,8,7,6,2
76. 3,4,8,5,1,2,6,7
77. 3,4,8,5,6,2,1
78. 3,4,8,5,6,7
79. 3,4,8,7,6,2,1,5
80. 3,4,8,7,6,5,1,2
81. 3,7,6,2,1,4,8,5
82. 3,7,6,2,1,5,8,4
83. 3,7,6,5,1,2
84. 3,7,6,5,1,4,8
85. 3,7,6,5,8,4,1,2
86. 3,7,8,4,1,2,6,5
87. 3,7,8,4,1,5,6,2
88. 3,7,8,5,1,2,6
89. 3,7,8,5,1,4
90. 3,7,8,5,6,2,1,4
91. 4,1,2,3,7,6,5,8
92. 4,1,2,3,7,8,5,6
93. 4,1,2,6,5,8,7,3
94. 4,1,2,6,7,3
95. 4,1,2,6,7,8,5
96. 4,1,5,6,2,3,7,8
97. 4,1,5,6,7,3,2
98. 4,1,5,6,7,8
99. 4,1,5,8,7,3,2,6
100. 4,1,5,8,7,6,2,3
101. 4,3,2,1,5,6,7,8
102. 4,3,2,1,5,8,7,6
103. 4,3,2,6,5,1
104. 4,3,2,6,5,8,7
105. 4,3,2,6,7,8,5,1
106. 4,3,7,6,2,1,5,8
107. 4,3,7,6,5,1,2
108. 4,3,7,6,5,8
109. 4,3,7,8,5,1,2,6
110. 4,3,7,8,5,6,2,1
111. 4,8,5,1,2,3,7,6
112. 4,8,5,1,2,6,7,3
113. 4,8,5,6,2,1
114. 4,8,5,6,2,3,7
115. 4,8,5,6,7,3,2,1
116. 4,8,7,3,2,1,5,6
117. 4,8,7,3,2,6,5,1
118. 4,8,7,6,2,1,5
119. 4,8,7,6,2,3
120. 4,8,7,6,5,1,2,3
121. 5,1,2,3,4,8,7,6
122. 5,1,2,3,7,6
123. 5,1,2,3,7,8,4
124. 5,1,2,6,7,3,4,8
125. 5,1,2,6,7,8,4,3
126. 5,1,4,3,2,6,7,8
127. 5,1,4,3,7,6,2
128. 5,1,4,3,7,8
129. 5,1,4,8,7,3,2,6
130. 5,1,4,8,7,6,2,3
131. 5,6,2,1,4,3,7,8
132. 5,6,2,1,4,8,7,3
133. 5,6,2,3,4,1
134. 5,6,2,3,4,8,7
135. 5,6,2,3,7,8,4,1
136. 5,6,7,3,2,1,4,8
137. 5,6,7,3,4,1,2
138. 5,6,7,3,4,8
139. 5,6,7,8,4,1,2,3
140. 5,6,7,8,4,3,2,1
141. 5,8,4,1,2,3,7,6
142. 5,8,4,1,2,6,7,3
143. 5,8,4,3,2,1
144. 5,8,4,3,2,6,7
145. 5,8,4,3,7,6,2,1
146. 5,8,7,3,2,1,4
147. 5,8,7,3,2,6
148. 5,8,7,3,4,1,2,6
149. 5,8,7,6,2,1,4,3
150. 5,8,7,6,2,3,4,1
151. 6,2,1,4,3,7,8,5
152. 6,2,1,4,8,5
153. 6,2,1,4,8,7,3
154. 6,2,1,5,8,4,3,7
155. 6,2,1,5,8,7,3,4
156. 6,2,3,4,1,5,8,7
157. 6,2,3,4,8,5,1
158. 6,2,3,4,8,7
159. 6,2,3,7,8,4,1,5
160. 6,2,3,7,8,5,1,4
161. 6,5,1,2,3,4,8,7
162. 6,5,1,2,3,7,8,4
163. 6,5,1,4,3,2
164. 6,5,1,4,3,7,8
165. 6,5,1,4,8,7,3,2
166. 6,5,8,4,1,2,3,7
167. 6,5,8,4,3,2,1
168. 6,5,8,4,3,7
169. 6,5,8,7,3,2,1,4
170. 6,5,8,7,3,4,1,2
171. 6,7,3,2,1,4,8,5
172. 6,7,3,2,1,5,8,4
173. 6,7,3,4,1,2
174. 6,7,3,4,1,5,8
175. 6,7,3,4,8,5,1,2
176. 6,7,8,4,1,2,3
177. 6,7,8,4,1,5
178. 6,7,8,4,3,2,1,5
179. 6,7,8,5,1,2,3,4
180. 6,7,8,5,1,4,3,2
181. 7,3,2,1,4,8,5,6
182. 7,3,2,1,5,6
183. 7,3,2,1,5,8,4
184. 7,3,2,6,5,1,4,8
185. 7,3,2,6,5,8,4,1
186. 7,3,4,1,2,6,5,8
187. 7,3,4,1,5,6,2
188. 7,3,4,1,5,8
189. 7,3,4,8,5,1,2,6
190. 7,3,4,8,5,6,2,1
191. 7,6,2,1,4,3
192. 7,6,2,1,4,8,5
193. 7,6,2,1,5,8,4,3
194. 7,6,2,3,4,1,5,8
195. 7,6,2,3,4,8,5,1
196. 7,6,5,1,2,3,4,8
197. 7,6,5,1,4,3,2
198. 7,6,5,1,4,8
199. 7,6,5,8,4,1,2,3
200. 7,6,5,8,4,3,2,1
201. 7,8,4,1,2,3
202. 7,8,4,1,2,6,5
203. 7,8,4,1,5,6,2,3
204. 7,8,4,3,2,1,5,6
205. 7,8,4,3,2,6,5,1
206. 7,8,5,1,2,3,4
207. 7,8,5,1,2,6
208. 7,8,5,1,4,3,2,6
209. 7,8,5,6,2,1,4,3
210. 7,8,5,6,2,3,4,1
211. 8,4,1,2,3,7,6,5
212. 8,4,1,2,6,5
213. 8,4,1,2,6,7,3
214. 8,4,1,5,6,2,3,7
215. 8,4,1,5,6,7,3,2
216. 8,4,3,2,1,5,6,7
217. 8,4,3,2,6,5,1
218. 8,4,3,2,6,7
219. 8,4,3,7,6,2,1,5
220. 8,4,3,7,6,5,1,2
221. 8,5,1,2,3,4
222. 8,5,1,2,3,7,6
223. 8,5,1,2,6,7,3,4
224. 8,5,1,4,3,2,6,7
225. 8,5,1,4,3,7,6,2
226. 8,5,6,2,1,4,3,7
227. 8,5,6,2,3,4,1
228. 8,5,6,2,3,7
229. 8,5,6,7,3,2,1,4
230. 8,5,6,7,3,4,1,2
231. 8,7,3,2,1,4
232. 8,7,3,2,1,5,6
233. 8,7,3,2,6,5,1,4
234. 8,7,3,4,1,2,6,5
235. 8,7,3,4,1,5,6,2
236. 8,7,6,2,1,4,3
237. 8,7,6,2,1,5
238. 8,7,6,2,3,4,1,5
239. 8,7,6,5,1,2,3,4
240. 8,7,6,5,1,4,3,2 

Anyone care to make an intuitive guess if a dodecahedron (12 sides) will produce more or less prints than an icosahedron (20 sides)?

[Turtle]

I agree that there are 10 six-sided + 2 five-sided, can't make another roll = 12 total strings for each starting side and first side rolled to. There are 4 possible sides to roll to on the first roll, so there are 40 six-sided + 8 five-sided = 48 total strings for each starting side. There are 6 starting sides, so 240 six-sided + 48 five-sided = 288 total stings. Dividing by 2 to consider strings like “1,2,3,4,5,6” and “6,5,4,3,2,1” the same gives 120 six-sided +24 five-sided = 144 total strings.I think there's a problem with the “24”. Your post #38 tree (and my program) should agree that there are 48 paths from each face.

 

From the list under the diagram in the image, there are 12 paths beginning F1S1, so there should be 4 * 12 = 48 beginning F1, 48 * 6 = 288 beginning with any side. Dividing by 2 to consider strings like “1,2,3,4,5,6” and “6,5,4,3,2,1” the same, 288 / 2 = 144.

 

I can’t figure out where you’re getting the last division by 2 to give 144 / 2 = 72, and suspect it’s one division by 2 too many.

 

Eureka!! Thank you Craig! The bolded is what I was missing in my own tree. I was not multiplying by 4. :cup: :omg: Now that you have me on track, let's see if I can do the same for you.

 

I divide by 2 at the end because every print is actually 2. In my video, after the ending on E2, I could restart on E2, then retrace the same path and end on F1. So from a print perspective, F1,S1,E1,F2,S2,E2 is the same as E2,S2,F2,E1,S1,F1. Make sense to you?

 

 

This then gives us 240/2 = 120, 6-length paths, and 48/2 = 24, 5-length paths. Are we now all on the same path? :doh: :cup:

[Pyrotex]

Well I'm still in limbo as to how many different prints I can make, but I saw another box problem here at Hypography with a bug crawling inside that Pyrotex solved nicely with showing the room unfolded. Any chance Pyrotex (et al) of confirming or falsifying my tentative answer of 72 different prints?...:lol:

The answer is... :eek::eek::eek:

 

48. :hihi: But not quite!

 

Assuming all 6 sides are UNIQUE, then you get 48 starting with side 1, 48 different prints starting with side 2, etc.

 

for a grand total of 48 x 6 = 288 !!

 

Now, of that 48 (starting at side 1) only 40 of those can involve all 6 sides in the print without lifting the box and flipping it over. See the attached diagram.

 

Therefore, if all 6 sides are UNIQUE and you insist on rolling for all 6 sides per print without flipping, then there are 40 x 6 = 240 unique prints.

 

Pyro the Perculator

[Turtle]

The answer is... :eek::eek::eek:

 

48. :hihi:

 

:eek: :eek:

This then gives us 240/2 = 120, 6-length paths, and 48/2 = 24, 5-length paths. Are we now all on the same path?

 

:eek: :eek: :doh: what a maroon that Turtle can be. :hyper:

 

Final answer...again...:hihi: 24*4=96 & then 96*6=576, and then 576/2=288 possible different paths. Of the 288, 4*4=16 & then 16*6=96, and 96/2=48 of the paths have length 5.

 

Let's see thens...only took me 10 years to sort this one out. :lol: Thanks everyone. :)

 

...Anyone care to make an intuitive guess if a dodecahedron (12 sides) will produce more or less prints than an icosahedron (20 sides)?

Not even... unless you'll spot me 20 years. :hyper: :lol:

[Pyrotex]

:hyper: :eek:...Let's see thens...only took me 10 years to sort this one out. :eek: Thanks everyone. :hyper:...

You're welcome. But don't be so hard on yourself. This was a VERY difficult problem in 3D geometry AND permutations!

 

Hell, it took me over two hours, and that includes 3 re-starts.

Whew!!! :hihi::eek_big::lol:

 

Pyro

[CraigD]

Anyone care to make an intuitive guess if a dodecahedron (12 sides) will produce more or less prints than an icosahedron (20 sides)? :phones:

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. :cup:

[Turtle]

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.

...

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. :cup:

 

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. :cup: Thanks Craig. :phones:

[Turtle]

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. :cup: who knew? :cup:

 

anyway, if a method is good enough for Craig then it's usually better enough for me. :boy_hug: 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

 

:phones:

[Turtle]

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

 

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. :eek2:

 

Now back to our irregularly scheduled challenge. >>

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]

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

Just wingin' it here as I roll my icosahedral and dodecaheral poly-dice in my hand. :cup: 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. :hihi: 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=$o(XTMBME(1,A,:)) q:'B  w D,""-"",B,! s D=A" w D f  r "-",B,! q:'B  w A s (XTMBME(1,A,B),XTMBME(1,B,A))="" ;XTMBME(0,1): load polyhedron net diagram
k ^XD("XTMBME","C") s (C,A)=0 s A="1,0" x XTMBME(2) ;XTMBME(0,2): single count
k ^XD("XTMBME","C") s (C,A)=0 f  s A=$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.