Timeline for number of non-negative integer solutions for a set of equations [closed]
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Feb 6, 2013 at 16:59 | comment | added | Barry Cipra | I'm just guessing what the OP has in mind, but if you swap the subscripts for $x_1$ and $x_2$ in my previous comments, there's a fairly obvious generalization. | |
| Feb 6, 2013 at 16:46 | comment | added | Barry Cipra | To remove any mystery from my previous comment, the $A$, $B$ and $C$ there come from writing the three main equations as $x_2 + x_3 = 2(2-x_1)$, $x_2 + x_5 = 2(2-x_4)$, and $x_3+x_5 = 2(2-x_6)$ and solving these for $x_2$, $x_3$, and $x_5$. | |
| Feb 6, 2013 at 16:39 | comment | added | Barry Cipra | I agree with quid, it's unclear what generalization the OP has in mind. The specific case can be solved writing $x_2=A+B-C$, $x_3=A-B+C$, and $x_5=-A+B+C$ with $A=2-x_1$, $B=2-x_4$, and $C=2-x_6$. The non-negativity constraints amount to saying $A,B,C$ are sides of a (possibly degenerate) triangle, each of length no greater than 2, and from this the 15 solutions found by Thomas Kahle are easily obtained. Even if there's an interesting generalization, it might be better asked at math.stackexchange.com as quid advised (but with too many e's in the url). | |
| Feb 6, 2013 at 16:35 | comment | added | Gerhard Paseman | Which reduces the problem to 5^3 cases or fewer, and then exploiting symmetry. Gerhard "Using Intelligent Brute Force Indeed" Paseman, 2013.02.06 | |
| Feb 6, 2013 at 13:29 | history | closed |
user9072 Felipe Voloch George Lowther Emil Jeřábek Lee Mosher |
too localized | |
| Feb 6, 2013 at 13:28 | comment | added | Barry Cipra | Note that the first equation can be ignored. It is the sum of the other three, divided by 2. | |
| Feb 6, 2013 at 10:56 | comment | added | user9072 | It's about 10^5 that's really very smallish (for a computer). But, even to solve it by hand in an ad-hoc way seems feasible (by slightly intelligent brute force). But more immportantly what do you mean by "you want to solve for n"? Is 3n and 2n instead of 6 and 4 (and perhaps n for 2 or not), or n variables, in which form precisely, or still something different. Please formulate a precise question. | |
| Feb 6, 2013 at 10:46 | comment | added | aaaaaa | 7^6 is "smallish" !!!!! anyway i want to solve for n. So, want a general approach | |
| Feb 6, 2013 at 9:34 | answer | added | Thomas Kahle | timeline score: 3 | |
| Feb 6, 2013 at 9:24 | comment | added | user9072 | Could you please provide some motivation for this particular problem. I mean this one is a bit smallish, so I do not see what should be the problem about solving this particular one. Even the simplest of all ideas (use 1. to find 0<= xi <=6 and check the 7^6 possibilities by brute force) is perfectly feasible. So if this is not an instance of a general problem or has a clear research motivation it is offtopic. You could ask it on math.stackeexchange.com though. | |
| Feb 6, 2013 at 9:17 | comment | added | Thomas Kahle | This is counting lattice points in polytopes. You can google for this beautiful theory. Concrete instances such as the one you are giving can be solved with the software Latte: math.ucdavis.edu/~latte | |
| Feb 6, 2013 at 9:12 | history | asked | aaaaaa | CC BY-SA 3.0 |