Timeline for Number of integer solutions of a linear equation under constraints
Current License: CC BY-SA 4.0
32 events
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| Oct 9, 2019 at 22:19 | comment | added | Gerry Myerson | A question with no research angle, and an answer with 20 edits. | |
| S Oct 9, 2019 at 21:50 | history | suggested | Charles Valente | CC BY-SA 4.0 |
The last formula, for the example given, lacks the alternation of the signal of the terms. I just had to change two "+" to "-", but the edition must be at least 6 characters, so I added a term to the summation.
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| Oct 9, 2019 at 18:06 | review | Suggested edits | |||
| S Oct 9, 2019 at 21:50 | |||||
| Sep 4, 2018 at 20:45 | vote | accept | Satya Prakash | ||
| Sep 4, 2018 at 17:00 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Sep 4, 2018 at 15:56 | comment | added | Konstantinos Kanakoglou | yes you are right: $x_i>m_i \Leftrightarrow x_i\geq m_i+1$ and then apply prop. 1, to get the $(k-1)$ in the $N(q_i)$ formula ... etc | |
| Sep 4, 2018 at 9:37 | comment | added | Satya Prakash | In $N(q_1)$, you have $(k-1)$. I think this because the strict inequality in $x_1>m_1$. Please confirm. | |
| Sep 2, 2018 at 23:54 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Sep 2, 2018 at 22:06 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Sep 2, 2018 at 21:21 | comment | added | Konstantinos Kanakoglou | Apply the formula given above to get: $N(q_2)=\binom{13+(3-1)-1-6-4-3}{3-1}=\binom{1}{2}=0$ | |
| Sep 2, 2018 at 20:29 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Sep 2, 2018 at 20:14 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Sep 2, 2018 at 20:09 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Sep 2, 2018 at 13:53 | comment | added | Satya Prakash | $(m_1,m_2,m_3) = (6,3,3)$, $(n_1,n_2,n_3) = (8,4,4)$ and $N = 13$. Then how to compute $N(q_2)$ ? | |
| Sep 2, 2018 at 13:35 | vote | accept | Satya Prakash | ||
| Sep 2, 2018 at 13:50 | |||||
| Aug 31, 2018 at 18:00 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Aug 31, 2018 at 14:58 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Aug 31, 2018 at 5:09 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Aug 31, 2018 at 2:46 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Aug 31, 2018 at 2:27 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Aug 31, 2018 at 1:59 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Aug 31, 2018 at 1:52 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Aug 29, 2018 at 1:07 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Aug 27, 2018 at 22:47 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Aug 27, 2018 at 14:29 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Aug 27, 2018 at 13:34 | comment | added | Konstantinos Kanakoglou | But if you need an upper bound, you can for example compute (applying the formula of the example) the number of solutions of the problem satisfying the constraints $1\leq x_i\leq m$, where $m=\max \{ m_i \}$. This will obviously be an upper bound but i am not sure if it will be sharp enough for your purposes. | |
| Aug 27, 2018 at 13:14 | comment | added | Konstantinos Kanakoglou | i am not sure what do you need the upper bound for: following the method decribed in the answer you can get the exact number of solutions and not just an upper bound. | |
| Aug 27, 2018 at 12:57 | comment | added | Satya Prakash | Thank you for your comment. Actually, I have to compare the cardinality of the solutions set of this problem with that of some other set. Is it possible to get a sharp upper bound to the cardinality of this set? | |
| Aug 27, 2018 at 12:28 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Aug 27, 2018 at 11:54 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Aug 27, 2018 at 4:47 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Aug 27, 2018 at 4:24 | history | answered | Konstantinos Kanakoglou | CC BY-SA 4.0 |