Still editing...

Ok, I think this is the first one of these that I’ve attempted to answer and I may be repeating another answer. I haven’t read them yet. Anyway... we can turn the problem into a binary such. We’re going to use the property that each pair of corridors has a single dead end room connecting them. If we number the doors from 1 to 2001 then we notice that at the start of the corridor the doors 1-2, 3-4, 5-6, and so on will share a room. Always with the odd numbered corridor to the left and the even numbered corridor to the right. At the end of the corridor the doors 1996-1997, 1998-1999, 2000-2001 will share a room always with the even numbered corridor to the left and the odd numbered corridor to the right. We’ll call this property the “parity” of the dead end room. Odd parity is with the odd number on the left. Even parity is with the even number on the left. This is because at some point along the main corridor our escape route takes up one of the numbers on its own. This means that the next corridor shifts the parity along by one. Further to this we know that the escape route must be an odd numbered door. Otherwise we would have to have a dead end room that only has a single door back out of it. So, we can perform a binary search by using the following logic... Starting with the range 1-2001 pick an odd number half way along the range. (1001 in this case). Now enter the corridor and walk to the end. If the door is the escape we’re done. If not note which way the doors are arranged in the dead end room. If you enter on the right of the room then the escape route is a higher numbered corridor than the one you’re in. (And vice versa). So adjust the range up (for odd) or down (for even) based on the parity of the dead end room. 1-1000 for even and 1002-2001 for odd. Now repeats the steps using the new range. On average this is a binary search and so has speed of log(n) which is about 11. !So the length of time for this search would be ...

OK, lets look at the problem first.

I think we can...

turn the problem into a binary search!

To do this we first have to notice a few things about the problem. Each dead end room has two doors exiting it. One to the corridor that you came from and one to the adjacent corridor.

From this property we can apply this logic...

If we number the doors from 1 to 2001 (from west to east) then we notice that at the start of the corridor the doors 1-2, 3-4, 5-6, and so on will share a room. Always with the odd numbered corridor to the west and the even numbered corridor to the east. At the end of the corridor the doors 1996-1997, 1998-1999, 2000-2001 will share a room always with the even numbered corridor to the west and the odd numbered corridor to the east. This is because at some point along the main corridor our escape route takes up one of the numbers on its own. This means that the next corridor shifts the parity along by one. We’ll call this property the “parity” of the dead end room. Odd parity is with the odd number on the west. Even parity is with the even number on the west.

Further to this we know that...

...the escape route must be an odd numbered door. Otherwise we would have to have a dead end room to the west of the escape that only has a single door back out of it. i.e. there must be an even number of corridors to the west and east of the escape.

So, we can perform our search by using the following "algorithm"...

Starting with the range 1-2001 pick an odd number half way along the range. (1001 in this case). Now enter the corridor and walk to the end. If the door is the escape we’re done. If not note which way the doors are arranged in the dead end room. If you enter on the east of the room then the escape route is a higher numbered corridor than the one you’re in. (And vice versa). So adjust the range up (for odd) or down (for even) based on the parity of the dead end room. 1-1000 for even and 1002-2001 for odd. Now repeats the steps using the new range. If needed you may keep a not of the current range on your piece of paper but two numbers is not hard to remember :D

Performance of this escape...

This is a binary search (which I first thought we could use) and so has speed of log(n) which is about 11 (where n = 2001). So the length of time for this search would be 11 hours. 1 hour (to walk to the north end of each chosen corridor and back) * 11 (the number of corridors you would have to search).

So we can definitely escape in less than a week. It just takes half a day of lots of walking.

I don’t know how to do spoilers. Can someone please wrap this while I edit it. Sorry

Ok, I think this is the first one of these that I’ve attempted to answer and I may be repeating another answer. I haven’t read them yet.

Anyway... we can turn the problem into a binary such.

We’re going to use the property that each pair of corridors has a single dead end room connecting them.

If we number the doors from 1 to 2001 then we notice that at the start of the corridor the doors 1-2, 3-4, 5-6, and so on will share a room. Always with the odd numbered corridor to the left and the even numbered corridor to the right.

At the end of the corridor the doors 1996-1997, 1998-1999, 2000-2001 will share a room always with the even numbered corridor to the left and the odd numbered corridor to the right.

We’ll call this property the “parity” of the dead end room. Odd parity is with the odd number on the left. Even parity is with the even number on the left.

This is because at some point along the main corridor our escape route takes up one of the numbers on its own. This means that the next corridor shifts the parity along by one.

Further to this we know that the escape route must be an odd numbered door. Otherwise we would have to have a dead end room that only has a single door back out of it.

So, we can perform a binary search by using the following logic...

Starting with the range 1-2001 pick an odd number half way along the range. (1001 in this case). Now enter the corridor and walk to the end.

If the door is the escape we’re done. If not note which way the doors are arranged in the dead end room. If you enter on the right of the room then the escape route is a higher numbered corridor than the one you’re in. (And vice versa).

So adjust the range up (for odd) or down (for even) based on the parity of the dead end room. 1-1000 for even and 1002-2001 for odd.

Now repeats the steps using the new range.

On average this is a binary search and so has speed of log(n) which is about 11. So the length of time for this search would be ...

Still editing...

Ok, I think this is the first one of these that I’ve attempted to answer and I may be repeating another answer. I haven’t read them yet. Anyway... we can turn the problem into a binary such. We’re going to use the property that each pair of corridors has a single dead end room connecting them. If we number the doors from 1 to 2001 then we notice that at the start of the corridor the doors 1-2, 3-4, 5-6, and so on will share a room. Always with the odd numbered corridor to the left and the even numbered corridor to the right. At the end of the corridor the doors 1996-1997, 1998-1999, 2000-2001 will share a room always with the even numbered corridor to the left and the odd numbered corridor to the right. We’ll call this property the “parity” of the dead end room. Odd parity is with the odd number on the left. Even parity is with the even number on the left. This is because at some point along the main corridor our escape route takes up one of the numbers on its own. This means that the next corridor shifts the parity along by one. Further to this we know that the escape route must be an odd numbered door. Otherwise we would have to have a dead end room that only has a single door back out of it. So, we can perform a binary search by using the following logic... Starting with the range 1-2001 pick an odd number half way along the range. (1001 in this case). Now enter the corridor and walk to the end. If the door is the escape we’re done. If not note which way the doors are arranged in the dead end room. If you enter on the right of the room then the escape route is a higher numbered corridor than the one you’re in. (And vice versa). So adjust the range up (for odd) or down (for even) based on the parity of the dead end room. 1-1000 for even and 1002-2001 for odd. Now repeats the steps using the new range. On average this is a binary search and so has speed of log(n) which is about 11. !So the length of time for this search would be ...