A Froggy Game of Logic

The key idea here is, that you never want to be in a position where two frogs of the same color are next to each other.

As the frogs have to alternate with each other and jump over each other, that makes $n*n=n^2$ jumps. The $n+n= 2n$ are the non-jump slides that the frogs move.

The generalization is that for $n$ frogs on either side it will take $n^2+2n$ or $(n+1)^2 -1$ moves.

The average number of spaces a frog advances is n+1. There are 2n frogs. So the total number of spaces advanced is 2n(n+1). The frogs advance one space for every step, so they would need 2n(n+1) moves, but every jump the frog advances 2 spaces, so we save ourselves a move. The number of jumps is n*n=n^2 (since there is one jump for each green-blue frog pair). So assuming that it's possible to switch the frogs, the number of moves must be 2n(n+1)-n^2=n^2+2n.

Now we just need to show that it is possible to switch the frogs. I've tried to show this below in the n=5 case (not showing every move). We can use the same approach for other values of n.

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Etcetera.

For each frog, one can consider two variables. The first variable, let's call this $a$ , is the distance a frog has to move to end up at it's final position. The second one, $b$ , is the number of frogs it has to jump over. Now, let $T$ be the sum of $a-b/2$ over all frogs. If a frog moves a single unit forward, a will decrease and b remain constant, so T will decrease by one. If a frog jumps over another frog, a will decrease by two, but b will decrease by one for both the moving frog, and the frog it jumped over, resulting also in T decreasing by one. By consequence, T decreases by one every step. As a starts at (n+1) for every frog and b at n, T starts at $2n(n+1) - n^2= n(n+2)$ . As it ends at 0, it can only finish in $n(n+2)$ moves, or 35 if n=5

Let n be the no. of Frogs on each side.

formula that applies to is n(n+2)

therefore,5(5+2)=3*7=35