# 13 Caves and a Thief Puzzle

There are 13 caves arranged in a circle. There is a thief hiding in one of the caves. Each day the the thief can move to any one of of the caves that is adjacent to the cave in which he was staying the previous day. And each day, you are allowed to enter any two caves of your choice.

What is the minimum number of days to guarantee in which you can catch the thief?

**Note:**

1. Thief may or may not move to adjacent cave.

2. You can check any two caves, not necessarily be adjacent.

Answer is 12 days.

**How?**

Let’s assume the thief is in cave 1 and going clockwise and you start searching from cave 13 and 12 on your first day.

Cave 13 and 11 on second day,

Cave 13 and 10 on third day and so on till Cave 13 and 1 on 12th day.

So basically the idea is to check Cave no.13 everyday so that if thief tries to go anti clockwise you immediately catch it and if he goes clockwise you will catch him in maximum 12 days (including the case where he remains in Cave 1).

How does this “guarantee” that “you can catch the thief”? What if the thief moves into a cave you’ve already checked? (Suppose, on Nth day, you checked Cave 13 and Cave 7, when the thief was in Cave 6. Next day, you would check C13 and C6, whereas the thief moves in to C7, so that you won’t see him.) Or, what if Mr. Thief decided to move into a cave you just checked, when you were in the other cave you planned for the day? You can’t be checking inside TWO caves at the SAME time, right?

There is definitely no “guaranty” to catch the thieve.

Even when you know he is in cave 2 there is a 1/3 change he will not be catched the next move.

If you check 1 and 3 he might have stayed in cave 2.

If you check 1 and 2 he might have moved to cave 3.

If you check 2 and 3 he might have moved to cave 1.

there doesn’t exist any strategy if the thief is allowed to stay in the cave. (the ‘may’ condition of Note 1), if it is sure that he will definitely move in the adjacent cave then there exists a strategy.

In the worst case, thief can be caught in maximum of 24 searches. Strategy is as follows:

Fix one of the cave you are going to search every day, no matter what, (say it is cave no. 13). Now the problem reduces to searching of a linear block of caves and you can search one cave at a time. So any particular day, thief can either be in an even numbered cave E = {2,4,6,8,10,12} or an odd numbered cave O = {1,3,5,7,9,11}. Now, use this strategy to search 2-3-4-5-6-7-8-9-10-11-12-12-11-10-9-8-7-6-5-4-3-2 . This strategy guarantees to find the thief.

First day of checking (best case)

worst case?

How 1 day?

1 day is the best of the best case…. but that doesn’t guarantee that you will most definitely catch him.

So, hence the answer is 12 days.

7 days..

He can go in clock and anti clockwise together.. lets say he check… 7 and 8 one day.. then 6 and 9 ..then 5 and 10 ..then 4 and 11 then 3 and 12 …. 2 and 13 .. either he would encounter in any one of these cases.. worst sceneario is when thief is found in the last attempt of checking in cave 1 (left still to check) .. and hence 7 days are sufficient to determine.

If any loophole is there…i welcome the suggestion

Check out this strategy of mine!

Strategy:

Start from 1 and 2 in first day,

go to 3 and 4 on the second day,

5 and 6 on the following day and so on..

Worst case scenario:

Assume thief is in 13th cave. In the worst case, thief will always move behind you. Good news is you can move 2 steps, he can move just one step. At most you need 12 days to find the thief. 🙂

### Your Answer

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