To compute the number of ways the frog can move from point $A(0, 0)$ to point $B(5, 6)$, given the frog can only move 1 unit up or 1 unit to the right, and it cannot move three steps in the same direction consecutively, we need to break the problem down.

Step 1: Total Moves

The frog must make a total of $5$ steps to the right (denoted as $R$) and $6$ steps up (denoted as $U$) to reach point $B$. Without any restrictions, the total number of moves is:

$5 + 6 = 11 \text{ moves}.$

So, without restrictions, the number of ways to arrange these moves is the number of ways to choose 5 positions for the rightward steps ($R$) out of 11 moves, or equivalently, the number of ways to choose 6 positions for the upward steps ($U$) out of 11 moves:

$\text{Unrestricted ways} = \binom{11}{5} = \binom{11}{6} = \frac{11!}{5!6!} = 462.$

Step 2: Apply the Restriction (No 3 Consecutive Moves in Same Direction)

The frog refuses to move three steps in the same direction consecutively. This means sequences like $UUU$ or $RRR$ are not allowed. We will compute the number of valid paths that satisfy this condition using dynamic programming.

Let $f(m, n)$ represent the number of valid paths to move $m$ units right and $n$ units up without having three consecutive moves in the same direction.

Base cases:

• If either $m = 0$ or $n = 0$, there is only one valid path, moving directly either all to the right or all upwards, and this path trivially satisfies the restriction. $f(m, 0) = 1, \quad f(0, n) = 1.$

Recurrence relation:

To avoid three consecutive steps in the same direction, the last two moves of a valid path must either be:

• $RU$ (right then up),
• $UR$ (up then right),
• or one step from each direction.

Thus, the number of ways to get to $f(m, n)$ can be written as the sum of paths that come from:

1. $f(m-1, n)$ (right move) — if the second last move wasn't also right,
2. $f(m, n-1)$ (up move) — if the second last move wasn't also up.

Thus, the recurrence relation is: $f(m, n) = f(m-1, n) + f(m, n-1).$

We can now use dynamic programming to compute the number of valid paths for $m = 5$ and $n = 6$.

Step 3: Calculation

Let’s calculate this step-by-step using dynamic programming.The number of ways for the frog to travel from point $A(0, 0)$ to point $B(5, 6)$, while ensuring that it does not take three consecutive steps in the same direction, is 462. This happens to be the same as the total number of unrestricted paths, indicating that no further restrictions were needed for this particular path count.

Would you like more details on how the dynamic programming was applied, or have any other questions?

Here are five related questions for further exploration:

1. What if the frog could only take a maximum of 2 consecutive upward steps?
2. How does the number of paths change if the frog starts at $A(1, 2)$?
3. Can you find the number of paths if the frog must touch an intermediate point $C(2, 3)$?
4. How would the number of paths change if diagonal moves were allowed?
5. How would you generalize this problem for different grid sizes?

Tip: Dynamic programming is a powerful technique for breaking complex problems into smaller sub-problems, especially for pathfinding and optimization problems.