Google Kickstart Wandering Robot wrong answer - c++

Question

Hi I wrote a short program to try to answer Google Kickstart 2020 Round B problem Wandering Robot. I essentially tried to implement the analysis provided by Google, and my solution seems to work on all the sample cases I've tried, but it never passes Google's first test set.
Here's my code, and below it the problem description and a short explanation of my approach. What I really want to know is what I'm doing wrong, why I'm not passing the first test set, more than anything.
#include <iostream>
#include<math.h>
using namespace std;
#include <iomanip>
int W, H, L, U, R, D;
const int len = 100000;
double facts [len];
void factorial(){
facts[0] = log2(1);
for(int i=2; i<= len; i++){
facts[i-1] = log2(i) + facts[i-2];
}
}
bool impossible(){
if((L==1 && R==W) || (U==1 && D==H)){
return true;
}
return false;
}
double pass(int x, int y, int s, int t, int i){
double num = log2(x+y);
num = facts[x+y-1];
double prob = x+y;
double tot = 0;
double comb = 0;
//cout << x << " " << s << " " << y << " " << t << endl;
while(x!=s && y!=t){
double denom = facts[x-1] + facts[y-1];
comb = num - denom - prob;
// cout << num << endl;
tot += pow(2, comb);
x+=i;
y-=i;
}
//lower diagonal, hit the bottom
//x+1 = number of moves it takes
//to hit last square
if(y==H){
double p = pow(2,comb);
tot -= p;
x-=i;
for(int j=1; j<=x; j++){
p -= pow(0.5, prob);
p += pow(0.5, prob-j-1);
}
tot += p;
}
//upper diagonal, hit the right side
else if(x==W){
double p = pow(2,comb);
tot -= pow(2, comb);
y+=i;
for(int j=1; j<=y; j++){
p -= pow(0.5, prob);
p += pow(0.5, prob-j-1);
}
tot += p;
}
return tot;
}
int main() {
int t;
cin >> t; // read t. cin knows that t is an int, so it reads it as such.
for (int i = 1; i <= t; ++i) {
cin >> W >> H >> L >> U >> R >> D;
double tot = 0;
if(impossible()){
cout << "Case #" << i << ": 0.0" << endl;
}
//upper diagonal
else if(L==1 || D==H){
factorial();
tot = pass(R, U-2, W, -1, 1);
cout << "Case #" << i << ": " << fixed << setprecision(9) << tot << endl;
}
//lower diagonal
else if(R==W || U==1){
factorial();
tot = pass(L-2, D, -1, H, -1);
cout << "Case #" << i << ": " << fixed << setprecision(9) << tot << endl;
}
//symmetric
else if(D==R && L==U){
factorial();
tot = pass(R, U-2, W, -1, 1);
tot *=2;
cout << "Case #" << i << ": " << fixed << setprecision(9) << tot << endl;
}
//call count both
else{
factorial();
tot = pass(R, U-2, W, -1, 1);
tot += pass(L-2, D,-1, H, -1);
cout << "Case #" << i << ": " << fixed << setprecision(9) << tot << endl;
}
}
return 0;
}
The way that I tried to solve it was to find the probability of passing through the green squares and the reciprocal squares spanning from the upper right corner of the hole to the right edge of the arena.