55
Explanation:
38
37
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:
33
Strategy:
How this works:
33
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. (...
33
The Algoritm is :
Then
Here is the complete solution from Achim Flammenkamp Ph.D.
There are totally 57 solutions
32
Because:
According to Wolfram-Alpha, there are
One possible solution is:
A list (and images!) of all
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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 ...
31
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 ...
30
I believe this works as a short proof.
29
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
27
62 is the optimal result, I wrote an (admittedly somewhat hackneyed (and in retrospect not very good)) brute force program in Java which can be found here. It returns that 2 is the smallest number of uncovered squares possible using four queens.
In short, it places four queens on a board, scans the board for each position, and outputs the case with the ...
27
I hope I didn't make any mistakes:
Edit : Replace queens by king
26
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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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.
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 ...
23
Here is mine with passive kings, some minutes late :
22
Answer:
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While (as other answers proved it) it is impossible to solve this just by using knights, the problem can still be solved.
Matthew says "As I want to play white, my first move isn't really a secret. So I tell you what my first move would be, and you open with the move what you would have moved as an answer. After that I make the move I said, and then we can ...
21
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 ...
21
You need at least 16 Moves.
Let's make the task visually more simple. The initial board is:
a4 b4 c4
a3 b3 c3
a2 b2 c2
a1 b1 c1
We cut it into 12 cells and connect only those, which are separated exactly by one move of a knight. Easy to check that the result is the following:
c4 - a3 - c2 - a1
| | | |
b2 b1 b4 b3
...
20
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 ...
19
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.
19
Let's just go straight to the $m\times n$ board.
If $n$ is even, start in the bottom left corner. Go up all the way, right 1, down almost all the way (just 1 block from the bottom), right 1, up all the way, and so on. As we are going up at column 1 and we are alternating between up and down, at column $n$ we will be going down. This time go all the way down ...
18
The answer to the first question is:
Two rocks are sufficient to prevent the snake from winning.
Proof: Use the standard black-and-white checkerboard coloring, and put the rocks on two white squares (while obeying the rules on placing the rocks).
Then there are 43 empty white squares and 45 empty black squares.
Since the snake alternately moves between ...
18
It is not possible. The area of a $10 \times 10$ checkerboard is $100$, so it takes $25$ T pieces to have the same area. The checkerboard has the same number of red and black squares, but each piece covers three of one color and one of the other. $25$ pieces cannot cover $50$ squares of each color, the most even they can get is $51-49$
18
This is a pretty common puzzle.
Warm up Answer:
Advanced Answer Explanation:
17
Answer: 48639.
Prove: Let's number moves like this:
A) one square to the right
B) one square up
C) diagonally one square up and to the right
We can consider set of moves, that leads to the other corner. For example, CCCCCACB.
It is obvious that for any set like this the numbers of A-moves and B-moves must be always the same: $N_A=N_B$. Also $N_A+N_B+...
17
I give a construction of 92 moves and prove it optimal.
As rand al'thor notes, 84 moves forward are inevitable. So, we'll only count "excess moves" that do not move a pawn forward.
Each white pawn is in a column opposed against a black pawn that blocks its way. The two cannot pass each other, so each column requires an excess move to unblock it by moving ...
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