RobPratt
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As in my answer to My Mother's Dish Collection, I used a nonlinear optimization solver, with variables $x_i$, $y_i$ to represent the coordinates of the trees. The constraints are: \begin{align} 0 \le ...

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The maximum is 1 2 2 3 3 . 4 4 5 5 1 1 2 3 6 6 . 4 5 7 8 8 . 9 6 10 10 . 7 7 8 11 12 9 9 10 13 14 15 15 11 11 12 12 . 13 13 14 14 15 16 16 17 17 18 19 20 20 21 ...

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I'll get things started with

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I used integer linear programming as follows. Let $P$ be the set of pieces, with number $n_p$ of pieces available: $n_\text{king}=1, n_\text{bishop}=n_\text{knight}=n_\text{rook}=2, n_\text{queen}=9$....

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If the number of people per table is $p$, then to cover each pair we must have $$\frac{600}{p} \binom{p}{2} \ge \binom{30}{2},$$ which implies that $p \ge \lceil 49/20 \rceil = 3$. The following set ...

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You are looking for a partition of $40$ with the minimum number of parts that is a common refinement of all $154$ partitions of $40$ into at most $3$ parts. You can satisfy all $154$ scenarios with ...

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Here are results for small $n$: \begin{matrix} n &amp; \text{minimum expected number of rolls} \\ \hline 1 &amp; 1 \\ 2 &amp; 6 \\ 3 &amp; 63/8 = 7.875 \\ 4 &amp; 1388/143 \approx 9.706 \\ 5 &amp; ...

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Here's the unique solution, obtained via integer linear programming:

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Still trying various integer linear programming formulations. Along the way, I found that if you ignore the sudoku constraints, you can fit 9 knight paths. Not an answer, but I wanted to share the ...

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You asked for 6 queens, but the maximum is at least

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Via integer linear programming, the maximum for knights is The maximum for queens is at least Other maxima are

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For kings, $N=1$ yields a maximum of

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The maximum number of coins is with a minimum total value of achieved by The integer linear programming solution approach I used might be of interest. Let nonnegative integer decision variable $... View answer 2 answers votes 611 views Accepted answer 4 votes The upper bound for$K_3(13,3)$here yields This guarantees$10$correct out of the first$13$questions, so we can surely do better by considering all$15$questions. A lower bound is the sphere ... View answer 1 answers votes 79 views Accepted answer 4 votes Another solution, with smallest possible maximum entry subject to minimizing the sum: If you ignore the sum, the smallest possible maximum entry is smaller by$1$: View answer 1 answers votes 226 views 4 votes I used integer linear programming (and, sorry, a computer): The solution is unique up to rotation and reflection. View answer 1 answers votes 112 views Accepted answer 4 votes The minimum is View answer 4 answers votes 215 views Accepted answer 4 votes Here's a symmetric solution with View answer 1 answers votes 171 views Accepted answer 4 votes There are 52 edges, so that is a lower bound. There are six odd-degree nodes. If you choose the two middle ones to be the endpoints of the overall path, the other four can be paired up with distance ... View answer 4 answers votes 2k views 4 votes You can solve the problem via integer linear programming as follows. Let$N=\{1,\dots,12\}$, and let$r_i$and$c_j$be the required row and column sums, respectively. For$(i,j)\in N \times N$, let ... View answer 3 answers votes 670 views 4 votes You can solve the problem via integer linear programming as follows. Let$n$be the number of coins, and let$k$be the number of extra coins the knight can use. For$b \in \{1,\dots,n\}$, let ... View answer 4 answers votes 506 views 4 votes I confirm @AlexeyBurdin's count of 470 solutions, which I obtained via integer linear programming as follows. Let$S=\{2, 4, 5, 6, 7, 8, 10, 11, 12, 13\}$be the set of desired sums. Let binary ... View answer 1 answers votes 132 views 4 votes Al Zimmermann's Programming Contests match this description. The name includes &quot;programming&quot; but does not require it: You can enter whether you use a computer, manual calculations, or tea ... View answer 3 answers votes 320 views Accepted answer 4 votes Indeed it can be solved as a traveling salesman problem on 13 nodes, but brute force is not required. First define an undirected graph with 145 nodes, one per box, with an edge between each pair of ... View answer 1 answers votes 166 views Accepted answer 4 votes View answer 2 answers vote 326 views Accepted answer 3 votes Via mixed integer linear programming, I found a solution that uses$76$meetings and has a total waste of$4230 + 1900 + 45 = 6175\$: 1 : 1516 3432 2 : 303 363 2339 3120 3400 3 : 4134 4 : 476 836 1567 ...

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