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Open and shut lids .........
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A blindfolded man is asked to sit in front of a carrom board. The holes of the board are open and 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.
Note that you can only open/close the holes that you touch, i.e. once you touch two holes, you can only open or close these holes (or leave them be) � you cannot alter the other holes until you touch them in a later round.
After he has done it, the carrom board is rotated and again brought to some position. The man is again not aware of the position of the holes, which are open or closed.
You need to devise a strategy that will ensure that, irrespective of the starting position of the open/closed holes, the holes will either be all open or all closed after "n" moves. What is the minimum value for "n", that will guarantee, whatever the starting position of the board, after "n" moves, they will be either be all closed, or all open?
Note that whenever all the holes are either open or closed, there will be an alarm so that the blindfolded man will know that he has won.
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.
Note that you can only open/close the holes that you touch, i.e. once you touch two holes, you can only open or close these holes (or leave them be) � you cannot alter the other holes until you touch them in a later round.
After he has done it, the carrom board is rotated and again brought to some position. The man is again not aware of the position of the holes, which are open or closed.
You need to devise a strategy that will ensure that, irrespective of the starting position of the open/closed holes, the holes will either be all open or all closed after "n" moves. What is the minimum value for "n", that will guarantee, whatever the starting position of the board, after "n" moves, they will be either be all closed, or all open?
Note that whenever all the holes are either open or closed, there will be an alarm so that the blindfolded man will know that he has won.
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However, cannot help you Gizmonster as I've never played the game. I've only heard of it because my mother in law used to sell the boards via ebay!
However, cannot help you Gizmonster as I've never played the game. I've only heard of it because my mother in law used to sell the boards via ebay!
You don't need to know what a Carrom board is, or how to play the game.
Basically, you have a rotating table with 4 holes - each hole having a lid that opens and closes. You are blindfolded and the table is set up in a random way, such that any of the holes can be open or closed. You do not know the initial state of the table, i.e. which holes are open and which holes are closed.
You are then sat down (blindfolded) in front of the table and you can touch any 2 of the 4 holes. You can then leave 1 or both of the holes that you have just touched, or you can open or close them - it's up to you. The table is then rotated to a new random position and again you touch 2 holes and can either leave them as they are, or you can open or close them.
Your job is to try and ensure that all of the holes will end up open or shut, in as few a moves as possible, no matter what the starting condition of the table.
Basically, you have a rotating table with 4 holes - each hole having a lid that opens and closes. You are blindfolded and the table is set up in a random way, such that any of the holes can be open or closed. You do not know the initial state of the table, i.e. which holes are open and which holes are closed.
You are then sat down (blindfolded) in front of the table and you can touch any 2 of the 4 holes. You can then leave 1 or both of the holes that you have just touched, or you can open or close them - it's up to you. The table is then rotated to a new random position and again you touch 2 holes and can either leave them as they are, or you can open or close them.
Your job is to try and ensure that all of the holes will end up open or shut, in as few a moves as possible, no matter what the starting condition of the table.
If anyone's struggling grasping the concept of the puzzle - just ignore the fact that it's a Carrom board ...... imagine a rotating circular table with 4 holes, all with lids ...........
What's the least number of turns that will be required, to ensure that the holes will either all be closed, or all be open - irrespective of the initial starting position ??
What's the least number of turns that will be required, to ensure that the holes will either all be closed, or all be open - irrespective of the initial starting position ??
I'm still confused.
Let's call the 4 holes A, B, C and D. After each turn am I presented with two adjacent holes (not of my choice) viz A&B, B&C, C&D or D&A? Or can I select opposite holesA&C or B&D? Can I select one (random) hole and then having ascertained its state (open or closed) then select another hole?
My problem is that if I just get adjacent holes and these are random, then I may get A&B consistently and be unable ever to affect the states of C or D.
Let's call the 4 holes A, B, C and D. After each turn am I presented with two adjacent holes (not of my choice) viz A&B, B&C, C&D or D&A? Or can I select opposite holesA&C or B&D? Can I select one (random) hole and then having ascertained its state (open or closed) then select another hole?
My problem is that if I just get adjacent holes and these are random, then I may get A&B consistently and be unable ever to affect the states of C or D.
The minimum value for 'n' has to be 2, under the condition given below.
If the four holes are labelled A, B, C and D and you are presented with A and B first (with A on the left), set them both either open. If the table is then turned 90 degrees clockwise, the blindfolded man then only has to set the two holes on his right to open to complete the challenge. Hence a minimum value on 2 for 'n'.
Obviously of the board turners can turn the board in any direction and turn it any number of degrees the value for 'n' goes up, and I'm not up to that kind of math !
If the four holes are labelled A, B, C and D and you are presented with A and B first (with A on the left), set them both either open. If the table is then turned 90 degrees clockwise, the blindfolded man then only has to set the two holes on his right to open to complete the challenge. Hence a minimum value on 2 for 'n'.
Obviously of the board turners can turn the board in any direction and turn it any number of degrees the value for 'n' goes up, and I'm not up to that kind of math !
@ dundurn - you can pick any 2 holes you wish - you are not restricted to the 2 holes in front of you.
You must also select 2 holes at a time - you cannot pick one hole, then determine its state and then pick another .........
@ Huderon - the minimum value for "n" would be 2 in your case, but like you say - you're not aware of the initial state of the board and you're also not aware of where the board will end up after each spin ........
@ sara3 - was that just a random guess ?? (it's wrong by the way) ..........
You must also select 2 holes at a time - you cannot pick one hole, then determine its state and then pick another .........
@ Huderon - the minimum value for "n" would be 2 in your case, but like you say - you're not aware of the initial state of the board and you're also not aware of where the board will end up after each spin ........
@ sara3 - was that just a random guess ?? (it's wrong by the way) ..........
I think 6. The ABCD terms in what follows may be constant between moves but don't have to be - it's unimportant;
o is open, s is shut, x is unknown
1 Pick two adjacent lids - open both AoBoCxDx
2 Pick two opposite lids - if one is shut open it
assuming no bell position has to be AoBoCoDs
3 pick two adjacent lids - if one shut open it - success - however if both open shut one (doesn't matter which)position now either 1) AoBoCsDs or 2) AoBsCoDs
4 Pick two opposite lids - if both open, shut, if both shut, open, success, position was 2)
- if different leave alone, posistion 1)
5 Pick two adjacent lids - if both open, shut, if both shut, open, success
- if different swap, i.e. shut the open one and open the shut one position now AoBsCoDs as in 2) above
6 Pick two opposite lids - if both open, shut, if both shut, open, success
o is open, s is shut, x is unknown
1 Pick two adjacent lids - open both AoBoCxDx
2 Pick two opposite lids - if one is shut open it
assuming no bell position has to be AoBoCoDs
3 pick two adjacent lids - if one shut open it - success - however if both open shut one (doesn't matter which)position now either 1) AoBoCsDs or 2) AoBsCoDs
4 Pick two opposite lids - if both open, shut, if both shut, open, success, position was 2)
- if different leave alone, posistion 1)
5 Pick two adjacent lids - if both open, shut, if both shut, open, success
- if different swap, i.e. shut the open one and open the shut one position now AoBsCoDs as in 2) above
6 Pick two opposite lids - if both open, shut, if both shut, open, success
Counter-intuitive but will this work?
1 Pick two adjacent lids - open both AoBoCxDx
2 Pick two diagonally opposite lids - if one is shut open it
assuming no bell position has to be AoBoCoDs
3 Pick two diagonally opposite lids - if one shut open it - success - however if both open shut one (doesn't matter which)position now AoBoCsDs
4 Pick two adjacent lids and swap, i.e. shut the open one and open the shut one position now AoBsCoDs
5 Pick two opposite lids - if both open, shut, if both shut, open, success
1 Pick two adjacent lids - open both AoBoCxDx
2 Pick two diagonally opposite lids - if one is shut open it
assuming no bell position has to be AoBoCoDs
3 Pick two diagonally opposite lids - if one shut open it - success - however if both open shut one (doesn't matter which)position now AoBoCsDs
4 Pick two adjacent lids and swap, i.e. shut the open one and open the shut one position now AoBsCoDs
5 Pick two opposite lids - if both open, shut, if both shut, open, success
We got a winner ............
Your solution is the same as mine except I start off by closing my lids and you open yours - makes no difference.
My solution (same as yours, just worded differently):
1. Pick a diagonal pair and close them.
2. Pick an adjacent pair and close them. (If you're not done, there's just one open hole).
3. Pick a diagonal pair. If one of them is open, close it and you're done. If both of them are closed, open one and the two open holes are now adjacent.
4. Pick an adjacent pair and flip them. If they were the same you're done; if not, then the open holes are now diagonally across from each other.
5. Pick a diagonal pair and flip them.
Your solution is the same as mine except I start off by closing my lids and you open yours - makes no difference.
My solution (same as yours, just worded differently):
1. Pick a diagonal pair and close them.
2. Pick an adjacent pair and close them. (If you're not done, there's just one open hole).
3. Pick a diagonal pair. If one of them is open, close it and you're done. If both of them are closed, open one and the two open holes are now adjacent.
4. Pick an adjacent pair and flip them. If they were the same you're done; if not, then the open holes are now diagonally across from each other.
5. Pick a diagonal pair and flip them.
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