Timeline for Unique way to partition into two parts of equal weight
Current License: CC BY-SA 2.5
4 events
| when toggle format | what | by | license | comment | |
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| Apr 28, 2010 at 17:05 | comment | added | Tony Huynh | A sequence $(x_1, \dots, x_n)$ is less than or equal to another sequence $(y_1, \dots, y_m)$ if and only if there is an increasing map $f:[n] \to [m]$ such that $x_i \leq y_{f(i)}$ for all $i \in [n]$. So, it is clear that if $(X) \leq (Y)$, then neither $X$ nor $Y$ can be exact. I believe the converse is true as well. | |
| Apr 28, 2010 at 15:50 | comment | added | damiano | I am pretty sure that there is a way of handling the general case, but I just have not thought about it. I do not know what is the "usual ordering of finite sequences"... | |
| Apr 28, 2010 at 14:47 | comment | added | Tony Huynh | Can't we handle the general case where exact sequences don't necessarily have half the indices? That is, for some subset X of [n], let (X) be the sequence of elements of X in increasing order. Letting Y by [n]-X, we have that X and Y are both exact for some sequence of length n if and only if (X) and (Y) are incomparable under the usual ordering of finite sequences (not necessarily of the same length) of integers. | |
| Apr 28, 2010 at 11:13 | history | answered | damiano | CC BY-SA 2.5 |