There are 10 caves arranged along the circumference of a circle There is a fox in one of the caves

But in a case - if hunters are staying in caves at night then they will catch fox in 3 days.
11 years agoHelpfull: Yes(8) No(3)

To understand the question imagine a circle with 10 boxes around its circumference and number them from 1 to 10.
Lets take the worst case and find out the minimum number of days.

The fox is staying at cave number 10 and hunters are staying at cave no (4 and 5) or at cave no (5 and 6)

DAY Fox Location Hunter Location
1 10 (4,5)
2 9 (3,6)
3 10 (2,7)
4 9 (1,8)
5--> The hunters will catch the fox either he switch to cave 10 or stay at cave 9(ignoring the case that he can switch to cave 1 because hunter is already there and coming from cave 1).
11 years agoHelpfull: Yes(6) No(0)
Infifinite
11 years agoHelpfull: Yes(5) No(2)
Minimum number of days=1 day (because hunters inspect of his choice then it is also possible 1st day only )
Maximum number of days=infinite
11 years agoHelpfull: Yes(5) No(5)
9...
consider both the hunters are in cave 1 and the fox is in cave 10.Now consider it as a relative speed questn now both the hunters cover 2 caves with in one day while the fox will cover only one cave so in general hunters have to travel 9 caves and both the fox and the hunters are running with the relative spee of 1(2-1).
so time=9/1=9
11 years agoHelpfull: Yes(1) No(6)
the best strategy to catch the fox in minimum no of times surely.so i think if we assume the worst case when hunter are just on oppsite side suppose no 1 hunter at 5 no2 hunter at6 and fox at 1 then if h1 move left each night and h2 move right each night then surely at 4th day or before that they can catch the fox. but by this strategy they can catch in 4 dats sure.
11 years agoHelpfull: Yes(1) No(3)
if the hunters reside in the caves at night then 5(taking that the hunters start at the same point and go in opposite direction) , otherwise INFINITE
11 years agoHelpfull: Yes(0) No(2)
6 days
suppose hunter search from one at number 4 and other at 9
now they can serch one from anticlockwise other is clock wise in 2 days if they not found then they start again in one is anti and one is clock wise thwen it will take 4 days more bt gaurateed searched
11 years agoHelpfull: Yes(0) No(2)
follow that ques ans..
Puzzle: Consider five holes in a line. One of them is occupied by a fox. Each night, the fox moves to a neighboring hole, either to the left or to the right. Each morning, you get to inspect a hole of your choice. What strategy would ensure that the fox is eventually caught?

Solution: Let holes be numbered 1 thru 5. Inspecting the holes in any of the following sequences suffices:
2, 3, 4, 2, 3, 4
2, 3, 4, 4, 3, 2
4, 3, 2, 2, 3, 4
4, 3, 2, 4, 3, 2
Explanation for sequence 2, 3, 4, 2, 3, 4: Let F denote the set of holes where the fox might be hiding. On any morning, the fox is either in an even numbered hole or an odd numbered hole. So on the first morning, either F = {1, 3, 5} or F = {2, 4}. If F = {2, 4}, then the following sequence of inspections suffices to catch the fox: 2, 3, 4. However, if the fox was not caught, then F must have equalled {1, 3, 5} on the first morning, so F must equal {2, 4} on the fourth morning. Therefore, repeating the sequence 2, 3, 4 from the fourth day onwards would suffice to catch the fox.
Another explanation to convince us that the sequence 2, 3, 4, 2, 3, 4 suffices to catch the fox: Let F denote the set of holes where the fox might be hiding. Initially, F = {1, 2, 3, 4, 5}. If the fox is not in hole 2, it must move so that F = {2, 3, 4, 5}. If the fox is not in hole 3, it must move so that to F = {1, 3, 4, 5}. If the fox is not in hole 4, it must move so that F = {2, 4}. If the fox is not in hole 2, it must move so that F = {3, 5}. If the fox is not in hole 3, it must move so that F = {4}.
7 years agoHelpfull: Yes(0) No(0)

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