You start with this board. You're the White. Freedom Index of initial arrangment of your chess pieces is equal to 20 (each pawn has two moves, each knight has two moves)

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Your goal is to have as much freedom of movement for your chess pieces as possible, i.e. do your best to get high Freedom Index. By this I mean that you must arrange your chess pieces in such way, that when you will count number of moves each of your chess pieces has and then sum said numbers, then resulting sum (also known as Freedom Index) must be as high as you can get.You can create such arrangment only by consequently moving your chess pieces in accordance with chess rules (i.e. each chess pieces can move only as it can move in chess), with exception that castling is forbidden and consequently doesn't count when Freedom Index is calculated. You can make as many moves as you want.Your answer must contain picture of your final arrangment of chess pieces. Pawn promotion is allowed. Use of computer for finding an arrangmenet is allowed, assuming that it follows the rules of the puzzle (in particular, no castling and bishops must be walking on different colors). Also, you should give Freedom Index of your arrangmenet in your post.

I will select answer with the highest Freedom Index, compared to other answers.

P.S. The upper bound for Freedom Index is 321. The lower bound for Freedom Index is 0 (i.e.you stalemated yourself)

  • $\begingroup$ @bobble Yes, upper bound. Suppose we promoted each pawn to queen. Let's also suppose that each chess piece will get maximum possible freedom of movement that it can get in chess at all, not accounting how we can achieve it while having other pieces on the board (i.e. as if it was placed in the center of empty board). 9*27+1*8+2*14+2*13+2*8=321. It can't get higher than that, but this is likely an overestimate. $\endgroup$ Commented Oct 3, 2020 at 16:53
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    $\begingroup$ "..as long as computer keeps history of its moves" seems like a bizarre limitation. Since we are promoting all the pawns anyway, and there's no limit on how many moves we can make, any position with a plausible piece composition (16 pieces including one king, at least one bishop on a dark square, etc) is reachable. $\endgroup$
    – Bass
    Commented Oct 3, 2020 at 18:05
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    $\begingroup$ @Bass Okay, I dropped this requirement $\endgroup$ Commented Oct 4, 2020 at 2:41

1 Answer 1


222: \begin{matrix}R_{11} & . & . & . & . & . & . & R_{10} \\. & K_7 & . & . & . & Q_{18} & . & . \\. & . & . & Q_{24} & . & . & . & . \\. & Q_{18} & . & . & . & . & Q_{20} & . \\. & . & . & . & Q_{24} & . & . & . \\. & . & Q_{21} & . & . & . & . & Q_{16} \\Q_{15} & . & . & . & . & Q_{18} & . & . \\. & . & B_5 & B_7 & N_4 & N_4 & . & . \\\end{matrix}

I used integer linear programming as follows. Let $P$ be the set of pieces, with number $n_p$ of pieces available: $n_\text{king}=1, n_\text{bishop}=n_\text{knight}=n_\text{rook}=2, n_\text{queen}=9$. Let $C=\{1,\dots,8\}^2$ be the set of cells. For each piece $p\in P$ and cell $(i,j)\in C$, let $N_{p,i,j} \subseteq C$ be the set of neighboring cells with respect to possible moves of $p$. For $p\in P$, $(i,j)\in C$, and $(i_2,j_2)\in N_{p,i,j}$, let $B_{p,i,j,i_2,j_2} \subseteq C$ be the set of cells strictly between $(i,j)$ and $(i_2,j_2)$. Let binary decision variable $x_{p,i,j}$ indicate whether piece $p$ occupies cell $(i,j)$. Let binary decision variable $m_{p,i,j,i_2,j_2}$ indicate whether piece $p$ occupies cell $(i,j)$ and can move to cell $(i_2,j_2)$. The problem is to maximize $$\sum_{p\in P} \sum_{(i,j)\in C} \sum_{(i_2,j_2)\in N_{p,i,j}} m_{p,i,j,i_2,j_2}$$ subject to \begin{align} \sum_{p\in P} x_{p,i,j} &\le 1 &&\text{for $(i,j)\in C$} \tag1\\ \sum_{(i,j)\in C} x_{p,i,j} &\le n_p &&\text{for $p\in P$} \tag2\\ m_{p,i,j,i_2,j_2} &\le x_{p,i,j} &&\text{for $p\in P, (i,j)\in C, (i_2,j_2)\in N_{p,i,j}$} \tag3\\ m_{p,i,j,i_2,j_2} &\le 1-\sum_{p_2} x_{p_2,i_2,j_2} &&\text{for $p\in P, (i,j)\in C, (i_2,j_2)\in N_{p,i,j}$} \tag4\\ m_{p,i,j,i_2,j_2} &\le 1-\sum_{p_2} x_{p_2,i_3,j_3} &&\text{for $p\in P, (i,j)\in C, (i_2,j_2)\in N_{p,i,j}, (i_3,j_3)\in B_{p,i,j,i_2,j_2}$} \tag5 \\ \sum_{\substack{(i,j) \in C:\\ \mod(i+j,2) = r}} x_{\text{bishop},i,j} &\le 1 &&\text{for $r \in \{0,1\}$} \tag6 \end{align} Constraint $(1)$ places at most one piece per cell. Constraint $(2)$ places at most $n_p$ copies of piece $p$. Constraint $(3)$ enforces $m_{p,i,j,i_2,j_2} = 1 \implies x_{p,i,j} = 1$. Constraint $(4)$ enforces $m_{p,i,j,i_2,j_2} = 1 \implies x_{p_2,i_2,j_2} = 0$. Constraint $(5)$ enforces $m_{p,i,j,i_2,j_2} = 1 \implies x_{p_2,i_3,j_3} = 0$. Constraint $(6)$ enforces at most one bishop per color.

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    $\begingroup$ @bobble It should be trivial to construct one. The position has the correct number of pieces, the bishops are on opposite coloured squares, and we have unlimited moves to spare. $\endgroup$
    – Bass
    Commented Oct 3, 2020 at 20:08
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    $\begingroup$ I'm with Rob and @Bass on this one, list of moves is too trivial ro enforce. $\endgroup$ Commented Oct 3, 2020 at 21:55
  • $\begingroup$ Took me some trying to one-up this to realize just how good it is. $\endgroup$ Commented Oct 4, 2020 at 17:14
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    $\begingroup$ Unless I made an error, it is optimal. $\endgroup$
    – RobPratt
    Commented Oct 11, 2020 at 18:35
  • $\begingroup$ @RobPratt How do you know that it's optimal? Please, add information about how you came up with the solution in your answer. $\endgroup$ Commented Nov 2, 2020 at 10:58

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