## An $n \times n$ grid of squares contains $n^2$ ants, one ant per square.

All ants decide to move simultaneously to an adjacent square (usually $4$ possibilities, except for ants on the edge of the grid or at the corners).

We define $f(n)$ to be the number of ways this can happen without any ants ending on the same square and without any two ants crossing the same edge between two squares.

You are given that $f(4) = 88$.

Find $f(10)$.

### This is a difficult type of mathematics problem typically referred to as a “derangement” problem in combinatorics. A derangement is a permutation of elements of a set, such that no element appears in its original position (here the ants on squares).

However, finding the exact value of f(10) (the number of ways to scramble 10×10 ants without any ant ending on their original position and without any two crossing the same edge) is not straightforward or simple at all. This is because it has high computational complexity and might require significant computational resources to solve directly.

One possible method would be to systematically go through all permutations and eliminate the ones where at least one ant ends at its original position or two ants cross the same edge. A brute force algorithm that does this would have to analyze 100 factorial (100!) permutations, which is an astronomically high amount and isn’t feasible with current technology.

In conclusion, this problem might not be solvable by hand or with a simple answer because it is computationally intensive.

The exact value of `f(10)` would likely only be found using advanced counting algorithms, possibly involving the usage of supercomputers. If you are studying mathematics at an advanced level, it could be possible to create an approximation or to find a pattern from `f(1)` to `f(9)`, but this would also likely require high-level mathematical techniques and a lot of computational power.

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