Coin Puzzle

A

alex_bell

Well I've got loads of these puzzles from a mailing list so here's one I enjoyed.

You and I are playing a game. We've got a perfectly circular table, and an
unlimited supply of 10p pieces. We each take it in turns to place a coin on
the table, taking care that no coin overlaps another (touching is fine), and no
coin can stick out over the edge of the table. The first person to break
either of these rules loses. What strategy must you adopt in order to be
guaranteed to beat me every time?

The answer is below but you will need to highlight the text to be able to read it.

The main part is this:

A circle is perfectly symmetrical, and in order to win, you must place a coin
in the diametrically opposite position on the table. So, if I put one in the
north-east corner, you put one south-west - the same distance from the edge,
etc.

The second part is:

The only way to ensure this strategy works is to go first - you have to be able
to place a coin in the center, in order to make it that you get the one spot
that is non-symmetrical.
 
M

munagada

Well I've got loads of these puzzles from a mailing list so here's one I enjoyed.



The answer is below but you will need to highlight the text to be able to read it.

The main part is this:

A circle is perfectly symmetrical, and in order to win, you must place a coin
in the diametrically opposite position on the table. So, if I put one in the
north-east corner, you put one south-west - the same distance from the edge,
etc.

The second part is:

The only way to ensure this strategy works is to go first - you have to be able
to place a coin in the center, in order to make it that you get the one spot
that is non-symmetrical.
Thanks for the puzzle. Let us say the radius of the table is 10inch, and the coin radius is 0.4 inch then your can place 625 coins, if the coin radius is 0.5 inch, you can place 400 coins, if 0.6 radius -- 277 coins; 0.7 radius -- 204 coins; these numbers are odd and even. The strategy should be choosing the "turn" first or next, based on whether total number of coins can be placed is EVEN or ODD.
 
G

Geoff Withnell

Thanks for the puzzle. Let us say the radius of the table is 10inch, and the coin radius is 0.4 inch then your can place 625 coins, if the coin radius is 0.5 inch, you can place 400 coins, if 0.6 radius -- 277 coins; 0.7 radius -- 204 coins; these numbers are odd and even. The strategy should be choosing the "turn" first or next, based on whether total number of coins can be placed is EVEN or ODD.

Not necessarily. The coins may not be placed in the most compact possible configuration. The reason the strategy works is this. Assume I am the frist player and have placed my initial coin in the exact center. All the possible coin positions have a symetrically opposite position on the other side of the circle. This is true regardless of whether the position is part of a packed distribution or not, and regardless of whether the maximum number of coins is odd or even. The players are very unlikely to place the coins perfectly for the packed configuration in any case. So by always placing my coin in the other position of the pair (from where my opponent placed his), I maintain the status of all possible positions being paired. Eventually there will be only one pair left. My opponent will place his coin in one of the positions, and i will place my coin in the other, and I win.

Geoff Withnell
 

Tim Folkerts

Trusted Information Resource
Thanks for the puzzle. Let us say the radius of the table is 10inch, and the coin radius is 0.4 inch then your can place 625 coins, if the coin radius is 0.5 inch, you can place 400 coins, if 0.6 radius -- 277 coins; 0.7 radius -- 204 coins; these numbers are odd and even. The strategy should be choosing the "turn" first or next, based on whether total number of coins can be placed is EVEN or ODD.


If these were square coins on a square table, then this would be true. But since there is always some open space (about 10% under optimal conditions) between circles. And much of teh area around the edge will not be covered (since part of the coin would have to be off the edge of the table). This means you would end up with far fewer coins than (radius of table)^2 / (radius of coin)^2.


And as Geoff pointed out, there is no requirement to put the coins in optimal locations.


Tim F
 
Top Bottom