Timeline for Unique way to partition into two parts of equal weight
Current License: CC BY-SA 2.5
14 events
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| Jun 10, 2023 at 9:55 | history | edited | Tony Huynh |
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| Oct 4, 2013 at 11:56 | history | edited | user9072 |
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| Apr 28, 2010 at 11:27 | comment | added | damiano | The weights are the coeffs of a linear function. Rescaling the weights, it suffices to treat the case of rational coeffs, translating the coordinates, it suffices to consider the kernel. A subset I is exact if and only if evaluating the linear function on the vector of 1 and -1 determined by I you get zero. Thus you are trying to find the zeros of a linear function on the vertices of a cube and you want the solution to consist exactly of a given pair of opposite vertices. If there is a linear function vanishing on the given pair, then clearly there is one vanishing only on that pair. | |
| Apr 28, 2010 at 11:13 | answer | added | damiano | timeline score: 1 | |
| Apr 28, 2010 at 3:23 | history | edited | Ewan Delanoy | CC BY-SA 2.5 |
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| Apr 28, 2010 at 3:22 | vote | accept | Ewan Delanoy | ||
| Apr 28, 2010 at 3:20 | comment | added | Ewan Delanoy | @ damiano : I believe your linear-function-on-$ {\mathbb R}^{12} $ analogy only provides nondecreasing sequences instead of increasing sequences. If you insist that the sequence be increasing, you get an affine function instead of a linear function. Also, note that $ \lbrace 1,2,3,4,5,6 \rbrace $ can never be an exact subset. | |
| Apr 28, 2010 at 3:18 | comment | added | Ewan Delanoy | @ Benoît : I encountered this problem trying to construct nontrivial examples in the context of partitions-with-weights problems that appear in recreational mathematics. The subset $\lbrace 1,2,5,7,10,12 \rbrace$ was chosen ``as random as possible". | |
| Apr 27, 2010 at 19:30 | history | edited | Ewan Delanoy | CC BY-SA 2.5 |
correction : "unique up to complementation"
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| Apr 27, 2010 at 19:06 | answer | added | Tony Huynh | timeline score: 10 | |
| Apr 27, 2010 at 18:23 | comment | added | damiano | Having said this, it seems that you have a linear function on $R^{12}$ with integer coefficients and you are trying to find its zeros on the set of vertices of a cube and you would like the set of zeros to consist exactly of a pair of opposite vertices. This does not seem hard to achieve. | |
| Apr 27, 2010 at 18:09 | comment | added | damiano | Maybe I am missing something, but isn't the complement of an exact subset also an exact subset? | |
| Apr 27, 2010 at 17:27 | comment | added | Benoît Kloeckner | A motivation would be welcome. | |
| Apr 27, 2010 at 16:46 | history | asked | Ewan Delanoy | CC BY-SA 2.5 |