A blindfolded man is asked to sit in the front of a carrom board.
The holes of the board are shut with lids in random order, i.e. any
number of all the four holes can be shut or open. Now the man is
supposed to touch any two holes at a time and can do the following.
* Open the closed hole.
* Close the open hole.
* Let the hole be as it is.
After he has done it, the carrom board is rotated and again brought
to some position. The man is again not aware of what are the holes
which are open or closed.
How many minimum number of turns does the blindfolded man require
to either open all the holes or close all the holes?
Note that whenever all the holes are either open or close, there
will be an alarm so that the blindfolded man will know that he has won.
6 comments:
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The Man will require TWO chances.
It will go this way,
First, Lets say the worst case is that all the 4 holes are opened. Then when he is offered the first time, then he will check one of the corners and close if they are open. Now he is sure that out of 4 holes at least 2 of them are open.
When he is offered again then he will check for one of the corner holes, if closed, he will check second corner holes and so on.
19225.....40054
Well, it asks for the minimum number of moves. i.e. best case scenario.
So, the best case is that all the holes are open or closed at the beginning, and the buzzer will sound.
Thus the minimum no. of moves is 0
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