Timeline for Unique way to partition into two parts of equal weight
Current License: CC BY-SA 3.0
7 events
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| Jul 5, 2017 at 14:17 | history | edited | Tony Huynh | CC BY-SA 3.0 |
added explanation from comments into the body
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| Apr 28, 2010 at 4:20 | comment | added | Tony Huynh | Thanks Ewan. Using the same technique it is easy to prove that if X is an exact subset of some sequence S, then there is another sequence S' such that X is the unique exact subset of S'. For example, for sequences S of length 12, we can first multiply S by 10. Then we can arbitrarily choose an index x in X and subtract 5 from the corresponding member of 10S. We then can add 1 to the members of 10S corresponding to X -x to obtain S'. | |
| Apr 28, 2010 at 3:22 | vote | accept | Ewan Delanoy | ||
| Apr 27, 2010 at 20:21 | comment | added | Tony Huynh | Let X={1,2,5,7,10,12} and Y={3,4,6,8,9,11}. Then for any subset of $X$ we have that the sum of the corresponding sequence values are 0 (mod 100). However, for any proper subset of Y, the sum of the corresponding sequence values are non-zero (mod 100). Thus, X and Y are the only exact subsets. | |
| Apr 27, 2010 at 19:47 | comment | added | Ewan Delanoy | It's not quite clear to me that {1,2,5,7,10,12} is the only exact subset up to complementation. How did you construct your sequence ? | |
| Apr 27, 2010 at 19:25 | history | edited | Tony Huynh | CC BY-SA 2.5 |
added 3 characters in body
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| Apr 27, 2010 at 19:06 | history | answered | Tony Huynh | CC BY-SA 2.5 |