Playing caroms
Written on 7:21 am by Vja Students
A blindfolded man is asked to sit in the front of a carom 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 carom 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.
Answere::
The blindfolded man requires 5 turns.
1. Open two adjacent holes.
2. Open two diagonal holes. Now atleast 3 holes are open. If 4th hole is also open, then you are done. If not, the 4th hole is close.
3. Check two diagonal holes.
If one is close, open it and all the holes are open.
If both are close, open any one hole. Now, two holes are open and two are close. The diagonal holes are in the opposite status i.e. in both the diagonals, one hole is open and one is close.
4. Check any two adjacent holes.
If both are open, close both of them. Now, all holes are close.
If both are close, open both of them. Now, all holes are open.
If one is open and one is close, invert them i.e. close the open hole and open the close hole. Now, the diagonal holes are in the same status i.e. two holes in one diagonal are open and in other are close.
5. Check any two diagonal holes.
If both are open, close both of them. Now, all holes are close.
If both are close, open both of them. Now, all holes are open.
