Difference between revisions of "Talk:Eight options with six sides"
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== This seems too easy == | == This seems too easy == | ||
− | Determining the <i>least</i> number of rolls he <i>will</i> have to make is so trivial I wouldn't even call it a puzzle. Determining the <i>greatest</i> number of rolls he <i>may</i> have to make is a slightly less trivial, but still very simple. One can optimise either of these values | + | Determining the <i>least</i> number of rolls he <i>will</i> have to make is so trivial I wouldn't even call it a puzzle. Determining the <i>greatest</i> number of rolls he <i>may</i> have to make is a slightly less trivial, but still very simple. One can optimise either of these values. Perhaps one for [[:Category:Easy puzzles]]? |
As far as I can tell, I'm not missing anything. [[User:Zerrakhi|Zerrakhi]] 03:43, 22 October 2010 (EDT) | As far as I can tell, I'm not missing anything. [[User:Zerrakhi|Zerrakhi]] 03:43, 22 October 2010 (EDT) | ||
: Hmmm. I'm not seeing this as easy. Perhaps I am missing something. That or I worded the question in an unclear way. Could you tell me what you think the answer is? That would help me understand what is going on here. [[User:Oscarlevin|Oscarlevin]] 15:03, 18 November 2010 (EST) | : Hmmm. I'm not seeing this as easy. Perhaps I am missing something. That or I worded the question in an unclear way. Could you tell me what you think the answer is? That would help me understand what is going on here. [[User:Oscarlevin|Oscarlevin]] 15:03, 18 November 2010 (EST) | ||
+ | |||
+ | :: '''To optimise the LEAST number of rolls he WILL have to make:''' | ||
+ | |||
+ | :: Let N be the least integer such that 6^N >= D where D is the number of dishes. | ||
+ | :: Of the 6^N possible permutations of rolled numbers, allocate a certain number to each dish and leave the rest blank. | ||
+ | :: The number of permutations allocated to each dish is the greatest integer A such that A <= (6^N)/D. | ||
+ | :: Roll N dice. If the permutation that's rolled is allocated a dish, select that dish. Otherwise roll again. | ||
+ | :: With this method, an infinite number of rolls MAY be needed, but the least number of rolls that WILL be needed is simply N. | ||
+ | :: If D = 8 then N = 2 and A = 4. | ||
+ | |||
+ | :: '''To optimise the MOST number of rolls he MAY have to make:''' | ||
+ | |||
+ | :: Let N be the least integer such that 6^N is a multiple of D. | ||
+ | :: Allocate permutations arbitrarily to dishes, as above except that now every permutation is allocated a dish (there are no blank spaces). | ||
+ | :: With this method, exactly N rolls are needed, no more no less. | ||
+ | :: If D = 8 then N = 3 and A = 27 (because 6 ^ 3 = 8 * 27). | ||
+ | :: The least easy bit is characterising the values of D such that N exists, but that's not part of the main puzzle. | ||
+ | |||
+ | :: [[User:Zerrakhi|Zerrakhi]] 21:40, 23 December 2010 (EST) |
Revision as of 20:40, 23 December 2010
This seems too easy
Determining the least number of rolls he will have to make is so trivial I wouldn't even call it a puzzle. Determining the greatest number of rolls he may have to make is a slightly less trivial, but still very simple. One can optimise either of these values. Perhaps one for Category:Easy puzzles?
As far as I can tell, I'm not missing anything. Zerrakhi 03:43, 22 October 2010 (EDT)
- Hmmm. I'm not seeing this as easy. Perhaps I am missing something. That or I worded the question in an unclear way. Could you tell me what you think the answer is? That would help me understand what is going on here. Oscarlevin 15:03, 18 November 2010 (EST)
- To optimise the LEAST number of rolls he WILL have to make:
- Let N be the least integer such that 6^N >= D where D is the number of dishes.
- Of the 6^N possible permutations of rolled numbers, allocate a certain number to each dish and leave the rest blank.
- The number of permutations allocated to each dish is the greatest integer A such that A <= (6^N)/D.
- Roll N dice. If the permutation that's rolled is allocated a dish, select that dish. Otherwise roll again.
- With this method, an infinite number of rolls MAY be needed, but the least number of rolls that WILL be needed is simply N.
- If D = 8 then N = 2 and A = 4.
- To optimise the MOST number of rolls he MAY have to make:
- Let N be the least integer such that 6^N is a multiple of D.
- Allocate permutations arbitrarily to dishes, as above except that now every permutation is allocated a dish (there are no blank spaces).
- With this method, exactly N rolls are needed, no more no less.
- If D = 8 then N = 3 and A = 27 (because 6 ^ 3 = 8 * 27).
- The least easy bit is characterising the values of D such that N exists, but that's not part of the main puzzle.
- Zerrakhi 21:40, 23 December 2010 (EST)