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10 points to the man above!!

Is this another riddle?

Mr_Grue wrote:Is this another riddle?

I guess Ice.

In which case the question is probably what do you get instead of water when the ambient temperature drops below zero C?

Very cold?

Just to keep things moving, here's one for ya:

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.


By the way, 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.

Any takers ??

Sorry but is the table rotated randomly? Could you end up at the same holes you stated with in your second turn?

Jean Eugene Roberts wrote:Sorry but is the table rotated randomly? Could you end up at the same holes you stated with in your second turn?

The table is rotated randomly, so it could end up back in the same position again.

You choose any 2 holes, then it's up to you what you do with them ......
You can either:
1. Leave them both as they are.
2. Flip them both so they are the opposite way to which they started.
3. Leave one and flip the other.

Can I check something, moonbeam? If you start with all four of the table's holes closed (or open), that won't give an automatic win, will it? That is sort of pre-supposed by the question, but I want to make sure, especially as it goes on to say that whenever all four holes are open or closed, an alarm will sound.

Also, if I have assumed correctly and you really can start with all four holes closed (say), if the blind man does nothing on his first go, does he still win, or must he change at least one of the holes over in order to win?

If the holes are all open (or shut) to start with, then yeah, the blind-man wins on his first go without doing anything.

But what if all the holes are not open or shut??

You have to work out a sequence of moves that will guarantee that all the holes will be open (or shut) no matter what the starting position.

What's the minimum number of moves required to guarantee victory, irrespective of the starting position - and more importantly, what are the moves ??

But if they rotate randomly, there is no way of doing it in a minimal number of moves. It could technically keep hitting the same first two holes forever.

I'm really dense at these kind of riddles.

4 holes.

first go:
Left hand does bottom left right hand does bottom right. flips them both so they are the one way.

spun

second go:
Left hand does bottom left right hand does top right. flips both so they are the same as the first (you've now hit 3 different holes)

spun

third go:
left hand to top left right hand to bottom right. flip them

Not sure if it's right but it's the start of something anyway

Jean Eugene Roberts wrote:But if they rotate randomly, there is no way of doing it in a minimal number of moves. It could technically keep hitting the same first two holes forever.

Yes and no. (We're done with Bingo about 'holes' I trust.)

It's hard to explain without a picture, but the holes are in the four corners of a square table. You pick the two on the left side (top left and bottom left). However the table rotates, if next time you chose top left and bottom right, or top right and bottom left, you are guaranteed to get one different hole from the last round.

The same is true if you start on diagonals, then switch to a 'side'.

However, I've worked it through and, within three moves you can know that you have three holes that are either open or shut, and the last one in the opposite setting. (The other options will see the game end; this is the worst case scenario.)

Even switching your pattern (or sticking with the same one), it is theoretically possible that you will continually miss the last one that is needed to end the game.

All of which makes me think I've gone off on the wrong track entirely.

You're on the right lines:
first touch two adjacent lids and close them ..... then touch two diagonally opposite lids and close them. If you're not done, you now know that there can only be one open lid left.
However, like you say, you may never get to close this lid due to the randomness of the game.
Remember though, that the game can be won by all the lids being closed OR open, so if you can somehow manage to ........... oops I'm saying too much now ..........

Moonbeam got it! You win.