64
These should do it:
Just to show another example:
56
Explanation:
50
I think this arrangement works for the bonus question:
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I think this will do it
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Yes. The minimum number of pieces required is 5.
5 queens can be places such that they cover every space on the board, as in the following example:
There are 12 such arrangements, along with rotation and reflection of each of them.
Edit: The above proves that 5 queens is enough, but it doesn't prove that 4 queens isn't enough. According to this ...
38
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Strategy:
How this works:
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Proof:
Initially, there are an even number of knights on white squares (namely, there are two of them, at b1 and g8).
Every time a knight moves, the number of knights on white squares either increases by one (if a knight on a black square moves) or decreases by one (if a knight on a white square moves).
Either way, the parity changes each turn. Thus, if ...
36
There are more
ones.
Proof:
34
Because:
According to Wolfram-Alpha, there are
One possible solution is:
A list (and images!) of all
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Looks like:
Thanks to @Gamow's comment, this number's maximality can be proved
by self-contradiction of the assumption that it is not maximal.
Any more dominos would cover all 64 squares.
Assumption to be disproved: All squares can be covered with dominos.
A. As the top left corner must be covered, start with a horizontal domino
there. (...
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I'm not trying to solve the puzzle, I'm just interested in how many solutions there are, since the OP claims he doesn't know. I brute forced it with a program.
There are
First of all, there are 32432400 configurations, not taking rotations and reflections into account. Since the board has 16 squares, if we were to place the two kings anywhere, we'd be ...
34
The smallest sum that can be achieved is
Because
Reasoning
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The Algoritm is :
Then
Here is the complete solution from Achim Flammenkamp Ph.D.
There are totally 57 solutions
30
I believe this works as a short proof.
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Give these names to all the squares:
163
4 8
725
Each number can only be accessed by way of the numbers before and after it (where 8 wraps around to 1). That means they form a loop. Since they can never pass each other up on the loop, their relative ordering cannot change. Therefore it is impossible.
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There are
Proof:
Now
I bet the ratio of good to bad is
I wrote a little Python program (possibly buggy, so apply some skepticism, but I've tested the bit most likely to have bugs and it seems OK) to estimate the ratio by generating lots of random boards and counting good and bad ones. The result is
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First solution - 50 moves
Second solution - 59 moves
Current solution - 66 moves (beaten by Retudin - 70 moves and Rewan Demontay - 139 moves)
Moves:
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I found a solution that uses 16 moves.
After exhaustively checking that there is no solution in 14 moves, I conclude that 16 moves is optimal, because after any odd number of moves the number of white and black squares occupied by knights cannot be equal.
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I hope I didn't make any mistakes:
Edit : Replace queens by king
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We can work backwards to figure this out:
Conclusion:
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This indeed is an old puzzle. One possible source (but certainly not the first one) is:
Andy Liu: Two Applications of a Hamming Code
The College Mathematics Journal 40, (Jan 2009), pp. 2-5
The trick on the $2^k\times2^k$ board is to associate each of the $2^{2k}$ squares with a unique binary number with $2k$ bits. (Note that the number of squares ...
26
Suppose that we have a chessboard with the desired properties.
Find the greatest number in each row. Out of these numbers, let the smallest be $m_i$ in row $i$.
Find the smallest number in each row. Out of these numbers, let the largest be $n_j$ in row $j$.
Note the following:
Row $i$ contains only numbers that are at most $m_i$.
Row $j$ contains only ...
25
The rules of the question state that:
On your final grid, a letter (actually several is also mathematically possible) will be more frequent than all the other letters. Your aim is that this letter appears as much as possible.
If we interpret this to mean that all letters appearing with the greatest frequency contribute to the score then it is possible to ...
24
This type of chess puzzle is known as a domination problem, and as @Xynariz points out, only five queens are needed for the 8x8 board. It's also interesting to note that five queens are also sufficient for the 9x9, 10x10, and 11x11 boards, as shown by the following diagram taken from a Russian chess-puzzle book found here.
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I have a computer program for solving packing problems, and found a way to use it to solve this problem. One of the solutions it found is below:
Note that this is very close to the attempted solution in the question, as it differs only in the top left quadrant.
Modelling the problem in this way will not find all solutions (I'd need to allow other tile ...
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Here is mine with passive kings, some minutes late :
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No.
Every move the pawn makes it switches from a white to a black square or vice versa. Therefore it must either touch an equal number of white and black squares with an even number of moves, or one more square of either color with an odd number of moves. (A move including initially placing it on the board.)
Because there are two more white squares than ...
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Answer:
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