Timeline for Positive integer combination of non-negative integer vectors
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 13, 2017 at 12:47 | vote | accept | kakia | ||
| Oct 13, 2017 at 12:47 | vote | accept | kakia | ||
| Oct 13, 2017 at 12:47 | |||||
| Oct 13, 2017 at 12:46 | vote | accept | kakia | ||
| Oct 13, 2017 at 12:47 | |||||
| Oct 9, 2017 at 10:10 | answer | added | Fedor Petrov | timeline score: 5 | |
| Oct 9, 2017 at 0:46 | history | edited | Michael Hardy | CC BY-SA 3.0 |
added 3 characters in body
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| Oct 9, 2017 at 0:26 | answer | added | Gerhard Paseman | timeline score: -1 | |
| Oct 8, 2017 at 23:47 | answer | added | Gjergji Zaimi | timeline score: 5 | |
| Oct 8, 2017 at 23:11 | comment | added | kakia | @GerhardPaseman, you now proved $2n$ vectors are enough, I tried hard and didn't succeed to reduce it to $n$, that's why I asked here. I'd be grateful if you could help. Easy induction does not solve the problem, I believe. | |
| Oct 8, 2017 at 22:59 | comment | added | Gerhard Paseman | If you always get to pick the vectors, then this seems an easy consequence of some results either in modules, matroids, or both. Start by picking (k,0,0...,0), subtract off multiples of that, and you have reduced the problem one dimension. If the first coordinate is less than k, try (a,k-a,0...) or an obvious modification. This is probably a result relating to Smith normal form. Gerhard "It Should Be Inductively Easy" Paseman, 2017.10.08. | |
| Oct 8, 2017 at 22:42 | comment | added | kakia | It is not because I can find $2$ vectors, namely $(3,0)$ and $(1,2)$, so that $(8,4) = 2 \cdot (3,0) + 2\cdot (1,2)$. | |
| Oct 8, 2017 at 22:39 | comment | added | Martin Rubey | could you explain why for k=3, n=2, $(8,4) = 2 (3,0) + (2,1) + (0,3)$ is not a counterexample? | |
| Oct 8, 2017 at 22:09 | comment | added | Fedor Petrov | @kakia if we solve this case, we may decrease the number of vectors used in such a combination until it becomes less than $n$ | |
| Oct 8, 2017 at 21:54 | history | edited | kakia | CC BY-SA 3.0 |
less confusing
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| Oct 8, 2017 at 21:52 | comment | added | kakia | @FedorPetrov: why? | |
| Oct 8, 2017 at 21:48 | comment | added | kakia | @GerhardPaseman: I claim that I can always find at most $n$ vectors satisfying the above described property, maybe the English in the problem statement is not perfect. | |
| Oct 8, 2017 at 21:34 | comment | added | Gerhard Paseman | I am not sure how to interpret this problem. It seems false though. For example (240,120) is a combination of more than two vectors in more than two ways, e.g (30-t,t) for various choices of t. Perhaps more examples are needed to clarify, or show your proof for n=2. Gerhard "Maybe It's Not About Vectors" Paseman, 2017.10.08. | |
| Oct 8, 2017 at 21:34 | comment | added | Fedor Petrov | Also, we may suppose that $a$ is a positive integer linear combination of exactly $n+1$ vectors with sum of coordinates equal to $k$. | |
| Oct 8, 2017 at 20:52 | history | asked | kakia | CC BY-SA 3.0 |