tag:blogger.com,1999:blog-4115025577315673827.post8150741765604671711..comments2020-06-29T15:37:49.459+05:30Comments on CSE Blog - quant, math, computer science puzzles: Bottom DollarPratik Poddarhttp://www.blogger.com/profile/11577606981573330954noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-4115025577315673827.post-55510687366801105142010-01-23T23:42:32.927+05:302010-01-23T23:42:32.927+05:30@ashishwrites you also assumed that the two people...@ashishwrites you also assumed that the two people would form a group and play "together"<br />@ashishwrites ... @Nikhil.. I have already stated the assumption that: "both of the other players will play to their best personal advantage"<br /><br />So, Giridhar's solution is correct. :)<br /><br />You can see a more detailed solution <a href="http://plus.maths.org/issue36/puzzle/solution.html" rel="nofollow">here</a><br /><br />Thanx.Pratik Poddarhttps://www.blogger.com/profile/11577606981573330954noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-17011197084534999922010-01-23T18:01:18.300+05:302010-01-23T18:01:18.300+05:30Girdhar's solution & mine solution differ ...Girdhar's solution & mine solution differ because of our assumptions about game.<br /><br />He has taken it to be a strict combinatorial game (more precisely a general k player impartial subtraction nim )<br /><br />I on other hand have made assumptions about "behaviour" of slim (he wont play if he is drawing ) & that of shady ( that they essentially form a team ).<br /><br />& I think ,both are right under the domain of respective assumptions.NGhttps://www.blogger.com/profile/00137167173637522039noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-58583526006426690932010-01-22T08:44:22.493+05:302010-01-22T08:44:22.493+05:30here's another approach
i assert that shady w...here's another approach<br /><br />i assert that shady wil be able to offer slim 6 stones no matter what slim does in his first turn<br /><br />15 -> x -> y -> 6<br /><br />now for every value of x(slim's effort) between 10 and 14, there is a value of y (shady's effort) which can lead to 6 for slim's second turn in which case he loses<br /><br />so there's no way slim can winashishwriteshttps://www.blogger.com/profile/15883971335728638493noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-55050885417966164482010-01-21T09:51:36.550+05:302010-01-21T09:51:36.550+05:30Let us call our guy A.
He can win, i guess.
L: lo...Let us call our guy A.<br />He can win, i guess.<br /><br />L: looses<br />W: Wins<br />I: Indifferet. ( neither wins nor looses )<br />S = Stones left.<br /><br />If S <=5 , we know it is a winning position.<br />If S = 6, whatever we do next guy will win.<br />If S = 7, we pick 1 stone, just to remain indifferent. In this case A remains indifferent,B will loose, C will win.<br />If 8 <= S <= 12 ,we pick (S-7) stones. In all these cases, A wins, B remains indifferent, C looses.<br />if S = 13 , A looses, B wins.<br />if S = 14, A can remain indifferent by picking 1 stone, B looses, C wins.<br />if S = 15, A wins by picking 1 stone.Giridharhttps://www.blogger.com/profile/06086734346209959881noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-6819960914851483992010-01-20T18:59:14.254+05:302010-01-20T18:59:14.254+05:30I THINK our man ( lets call him slim ) should not ...I THINK our man ( lets call him slim ) should not play.<br /><br />Solution:<br />Slim clearly wont play if he has to pay his dollar. Also he wont play if he neither has to pay one nor gets one ( Afterall in time he wastes playing this , he could woo a pretty girl at counter to lend him a ride ) . So he plays only if he can win.<br /><br />Now our shady gentlemen know that if slim plays - he tries to win only (& is not interested in any draw for him ) So our gentlemen essentially form a team (lets call it shady ). So we have a match between slim & shady. Slim can take 1 to 5 coins at once , while shady can take 2 to 10 coins at once. ( they would also take 1 if only one coin was left ).<br /><br />So in maths , we have a finite , impartial nim with Slim having move set 1 to 5 & shady having move set 2 to 10 (& 1 if needed ). Every position of game is either winning or losing & by inductive way we can find which positions are winning for whom.<br /><br />As it happens 15 is not a winning position for Slim ( Whatever he chooses , he would get 6 in his next turn & then he would lose ).NGhttps://www.blogger.com/profile/00137167173637522039noreply@blogger.com