Author 
Topic: Coins on a table (Read 8843 times) 

Stefan Kneifel
Newbie
Gender:
Posts: 25


Coins on a table
« on: Aug 30^{th}, 2003, 2:10pm » 
Quote Modify

There are n coins lying on a table. Mr. A and Mr. B are playing the following game: Mr. A begins by removing an arbitrary number of coins, except that he is not allowed to remove all coins in the first move. From then on, every player removes one or more coins, but not more than twice as many as the preceding player has taken with his last move. The player who removes the last coin wins. 1.) Who will win the game, depending on n? 2.) What is the optimal strategy? Greetings from Switzerland Stefan


IP Logged 



Icarus
wu::riddles Moderator Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863


Re: Coins on a table
« Reply #1 on: Aug 30^{th}, 2003, 2:49pm » 
Quote Modify

Welcome to the Forum! Unfortunately this puzzle, with a different story, is one of the HARD Riddles. It's a good puzzle, though.


IP Logged 
"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? "  Anonymous



visitor
Guest

Actually, it's not quite the same riddle. Here you're allowed to take twice as many chips as the other person. In the thread discussing the original riddle, BNC raised this variation, but no one answered him. It looks more complicated. Here's my guess at the winning strategy: The starting positions that must lose grow by the Fibonacci series: If the starting number is any other number, either take off to reach the nearest Fibonacci number, or, when you can't take off that many, think of that number as 0 and consider the fibonacci series that starts at that point. Lather, rinse, repeat as necessary.


IP Logged 



Stefan Kneifel
Newbie
Gender:
Posts: 25


Re: Coins on a table
« Reply #3 on: Aug 30^{th}, 2003, 10:24pm » 
Quote Modify

on Aug 30^{th}, 2003, 2:49pm, Icarus wrote: Hmm.... sorry... I made a search with "Fibonacci" through the posts, and as I couldn't find this riddle, I posted it Besides that: Congratulations to "visitor"


IP Logged 



Icarus
wu::riddles Moderator Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863


Re: Coins on a table
« Reply #4 on: Aug 31^{st}, 2003, 11:18am » 
Quote Modify

on Aug 30^{th}, 2003, 5:50pm, visitor wrote:Actually, it's not quite the same riddle. Here you're allowed to take twice as many chips as the other person. In the thread discussing the original riddle, BNC raised this variation, but no one answered him. 
 Sorry  I missed that. With the different limit, it is a different puzzle. As is evidenced by the different solution you provided. My apologies, Stefan! And well done, visitor. (By the way, with all the posts you've put in (assuming they are all the work of the same person), why haven't you registered?)


IP Logged 
"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? "  Anonymous



Stefan Kneifel
Newbie
Gender:
Posts: 25


Re: Coins on a table
« Reply #5 on: Aug 31^{st}, 2003, 12:33pm » 
Quote Modify

For the curious: The riddle is due to R.E.Gaskell and M.J.Whinihan and was originally published in "Fibonacci Quarterly", 1 (December 1963), pp. 912.


IP Logged 



Barukh
Guest

I would suggest the following generalization: at every move, the player takes no more than L times as many coins as the preceding player. So far, we know the answer for L = 1 the first player looses on powers of 2, and L = 2 the first player looses on Fibonacci numbers.


IP Logged 



DH
Newbie
Gender:
Posts: 19


Re: Coins on a table
« Reply #7 on: Sep 2^{nd}, 2003, 11:01am » 
Quote Modify

on Sep 2^{nd}, 2003, 5:34am, Barukh wrote:I would suggest the following generalization: at every move, the player takes no more than L times as many coins as the preceding player. 
 I think that the losing configurations are going to get more and more complex but it's not hard to compute them using dynamic programming. I think it can be done in O(n) time and O(n) space.


IP Logged 



visitor
Guest

For L>2 the losing numbers seem to form a sort of a Fibonacci series, except for which two numbers in the series are added. For L=3 the losing numbers are 1,2,3,4,6,8,11,15,21,29,40,55... For L=4 1,2,3,4,5,7,9,12,15,19,24,31,40,52... For L=5 1,2,3,4,5,6,8,10,12,15,19,24,30,38,48,60...


IP Logged 



