Control Car (Easy Version)

Turn the count into a probability

There are $4^{(n+1)(m+1)}$ orientations, all equally likely. So instead of counting valid orientations directly, compute

$bad(n,m)$ = probability that the car exits the grid.

Then

$answer = (1 - bad(n,m)) * 4^{(n+1)(m+1)}$.

This is the clean way in. Direct counting gets nasty fast; probability lets us average over unrevealed wall directions locally.

Which walls matter

For a current cell:

• Its bottom edge is blocked iff the bottom-left corner points right or the bottom-right corner points left.
• Its right edge is blocked iff the top-right corner points down or the bottom-right corner points up.

The car tries down first. Only if down is blocked does it care about the right edge.

Why coordinates alone are not enough

Suppose the car enters a cell from above. The top edge had to be open, so the top-right corner is known not to point left. But whether it points down still matters, because that same corner can block the right edge.

Suppose the car enters from the left. Similarly, whether the bottom-left corner points right matters, because that can block the bottom edge.

So each state needs one bit of memory. That is the whole trick; miss this and the DP just lies to your face.

Let $i$ be the number of rows remaining including the current cell, and $j$ be the number of columns remaining including the current cell. Define bad-exit probabilities:

• $A_0(i,j)$: entered from above, and the top-right corner points down.
• $A_1(i,j)$: entered from above, and the top-right corner does not point down.
• $L_0(i,j)$: entered from the left, and the bottom-left corner points right.
• $L_1(i,j)$: entered from the left, and the bottom-left corner does not point right.

If $i=0$ or $j=0$, the car is already outside, so all four values are $1$.

Deriving the transitions

First take $A_0(i,j)$. The top-right corner points down, so the right edge is already blocked. The car can avoid stopping only by moving down. The bottom edge is open with probability $3/4 * 3/4 = 9/16$. Given it is open, the bottom-right corner is not left; among its three allowed directions, one is down and two are not. Therefore

$A_0(i,j)=3/16*A_0(i-1,j)+3/8*A_1(i-1,j)$.

For $A_1(i,j)$, the down move contributes the same first two terms. If down is blocked but right is open, the car moves right. Enumerating the bottom-left and bottom-right corners gives probability $1/16$ that the new bottom-left points right, and $5/16$ that it does not. So

$A_1(i,j)=3/16*A_0(i-1,j)+3/8*A_1(i-1,j)+1/16*L_0(i,j-1)+5/16*L_1(i,j-1)$.

For $L_0(i,j)$, the bottom-left corner points right, so the bottom edge is already blocked. The car can only move right. The right edge is open if the top-right is not down and the bottom-right is not up. If the bottom-right is right, the new state is $L_0$; if it is down or left, the new state is $L_1$:

$L_0(i,j)=3/16*L_0(i,j-1)+3/8*L_1(i,j-1)$.

For $L_1(i,j)$, the bottom-left corner does not point right. The car moves down if the bottom-right corner is not left. If that bottom-right corner is down, the next state is $A_0$; if it is up or right, the next state is $A_1$. If the bottom-right is left, down is blocked, and then right is open with probability $3/4$, producing $L_1$:

$L_1(i,j)=1/4*A_0(i-1,j)+1/2*A_1(i-1,j)+3/16*L_1(i,j-1)$.

The missing probability mass in these formulas is not forgotten. It is exactly the case where the car stops inside the grid, which contributes $0$ to bad-exit probability.

Starting state

At the top-left cell, pretend the car entered from above. The top-right corner is completely random: it points down with probability $1/4$, and not down with probability $3/4$. Hence

$bad(n,m)=1/4*A_0(n,m)+3/4*A_1(n,m)$.

Finally multiply the good probability by the total number of orientations:

$answer=(1-bad(n,m))*4^{(n+1)(m+1)}$.

All fractions are computed modulo $10^9+7$ using modular inverses. The DP only depends on the previous row and the current row, so memory is $O(m)$. Time is $O(nm)$ for the largest requested rectangle, which is fine for $5000*5000$. Tiny DP tax, huge payoff.