Counting possible coordinate moves with recursion

I'm trying to write a method that does the following:

public class GridCounting {
    /** Returns the total number of possible routes (paths) from
     * (x,y) to (tx,ty).
     * There are only three valid kinds of moves:
     *  Increment x by one.
     *  Increment x by two.
     *  Increment y by one.
     *
     *  Hint: You'll need to two base cases.
     */
    public static int count(int x,int y, int tx, int ty) {
        if (x==tx && y==ty)
            return 1;
        if (x>tx || y>ty)
            return 0;
        if (x<tx)
            return 1 + count(x+1, y, tx, ty);
        if (x<tx) //if x is still less, then we can do another move
            return 1 + count(x+2, y, tx, ty);
        if (y<ty){
            return 1 + count(x, y+1, tx, ty);
        }
        return 0;
    }
}

The trouble that I'm having is that I'm always off by +1. It expects 4 but I give it 5. The confusing part is that if I feed the function cou开发者_如何学Pythonnt(10,15,10,15), then that still counts as 1 move. I dont know how to account for this.

Also y++ then x++ counts as 1 move, and x++ then y++ counts as another move. Edit: Fixed code:

public static int count(int x,int y, int tx, int ty) {
    if (x==tx && y==ty)
        return 1;
    if (x>tx || y>ty)
        return 0;
    if (x<tx)
        return count(x+1, y, tx, ty) + count(x+2, y, tx, ty) + count(x,y+1,tx,ty);
    if (y<ty) {
        return count(x, y+1, tx, ty); // what does this do?
    }
    return 0;
}

The trouble that I'm having is that I'm always off by +1. It expects 4 but I give it 5.
In position (x, y) we, generally speaking, have three choices (x+=1, x+=2, y+=1). Each of them produces separate paths and we need to find how many paths there're in total.
So, this part is wrong

        if (x<tx)
            return 1+ count(x+1, y, tx, ty);
        if (x<tx)
            return 1+ count(x+2, y, tx, ty);
        if (y<ty){
            return 1+ count(x, y+1, tx, ty);
        }
        return 0;

That was a hint, let me know if you need more.

Also y++ then x++ counts as 1 move, and x++ then y++ counts as another move.
Well, that sounds like two different paths to me.

edit
Ok, return count(x + 1, y, tx, ty) + count(x + 2, y, tx, ty) + count(x, y + 1, tx, ty); is all you need.
If there're 3 paths starting with first move x + 1, 2 paths starting with first move x + 2 and 2 paths starting with first move y + 1, then, obviously, there're 3 + 2 + 2 paths total.

You sure this line is right?

if (x==tx && y==ty)
  return 1;

It seems like it should return zero.

On a phone so I can't test it.

if (x==tx && y==ty) return 1; doesn't sound correct. If you're already at the destination, there are no more routes.

Maybe I not understanding the rules, but I think it's that first base case that's causing your problem. If you're already at the location then you're not making a move.