Timeline for System of boolean equations, Satisfiability
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 27, 2014 at 16:39 | vote | accept | Valera | ||
| Oct 27, 2014 at 16:27 | answer | added | Emil Jeřábek | timeline score: 3 | |
| Oct 27, 2014 at 16:05 | answer | added | user90909 | timeline score: 2 | |
| Oct 27, 2014 at 15:50 | comment | added | Valera | This is not a weighted MaxSat. But I would be happy to be wrong, if you can show me how to reduce my problem to the equivalent MaxSat instance. | |
| Oct 27, 2014 at 15:48 | history | edited | Valera | CC BY-SA 3.0 |
added 3 characters in body
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| Oct 27, 2014 at 15:44 | comment | added | joro | If you want to satisfy as many as possible, I believe this is weighted maxsat. | |
| Oct 27, 2014 at 15:21 | comment | added | Sam Hopkins | You should make it clearer that you are not trying to solve this system (which may be inconsistent), but rather satisfy as many (in some sense) of the equations as possible. | |
| Oct 27, 2014 at 15:19 | comment | added | Valera | Please see the updated version | |
| Oct 27, 2014 at 15:19 | history | edited | Valera | CC BY-SA 3.0 |
added 199 characters in body
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| Oct 27, 2014 at 15:15 | comment | added | joro | If $b_i=0$ this implies all involved $x_i$ are zero, so set them to zero and eliminate them from the the other equations. If equations remains, pick any solution to each equation by setting one $x$ to $1$. | |
| Oct 27, 2014 at 15:13 | comment | added | joro | @RamirodelaVega I am not sure this is solution, since you might get contradiction $x_i \ne x_i$. | |
| Oct 27, 2014 at 14:33 | comment | added | Valera | $ik\in\{1,\dots,M\}$, where $M$ is number of variables. | |
| Oct 27, 2014 at 14:22 | history | asked | Valera | CC BY-SA 3.0 |