armjan Posted April 10, 2005 Report Share Posted April 10, 2005 (edited) ok, we need our top phd ppl on this, rules of game... 10 pencils, u r playing against an algorithm named Larry u can remove 1,2 or 3 pencils at each move last one left with a pencil loses, that simple! play quick game with Larry curious to c how many of u find a hack and exploit it. hint 1: don't be too greedy hint 2: what goes around, comes around hint 3: Larry is as good as u allow him to be Edited April 10, 2005 by armjan Quote Link to comment Share on other sites More sharing options...
vava Posted April 10, 2005 Report Share Posted April 10, 2005 Why does Larry always forfeit? Quote Link to comment Share on other sites More sharing options...
armjan Posted April 10, 2005 Author Report Share Posted April 10, 2005 (edited) Why does Larry always forfeit? style_images/master/snapback.png u'r good. u'r really good. larry only quits if he knows that u will win. it will be clear after the first move who the winner is, and so there is no point to continue; thus he quits. and there is only one way for him to lose. this version of larry is not equiped with a machine learning alg, and for this problem, it won't matter. So he is really dumb b/c once u found a hack, u can continue to exploit it. Edited April 10, 2005 by armjan Quote Link to comment Share on other sites More sharing options...
Azat Posted April 10, 2005 Report Share Posted April 10, 2005 this really has nothing to do with AI... But yes machines for sure can think Quote Link to comment Share on other sites More sharing options...
DominO123 Posted April 11, 2005 Report Share Posted April 11, 2005 (edited) u'r good. u'r really good. larry only quits if he knows that u will win. it will be clear after the first move who the winner is, and so there is no point to continue; thus he quits. and there is only one way for him to lose. this version of larry is not equiped with a machine learning alg, and for this problem, it won't matter. So he is really dumb b/c once u found a hack, u can continue to exploit it. style_images/master/snapback.png True, now can you tell why? Mathematcally speaking? Edited April 11, 2005 by QueBeceR Quote Link to comment Share on other sites More sharing options...
armjan Posted April 11, 2005 Author Report Share Posted April 11, 2005 True, now can you tell why? Mathematcally speaking? style_images/master/snapback.png tell what? Quote Link to comment Share on other sites More sharing options...
DominO123 Posted April 11, 2005 Report Share Posted April 11, 2005 tell what? style_images/master/snapback.png Why there is one way? Quote Link to comment Share on other sites More sharing options...
armjan Posted April 11, 2005 Author Report Share Posted April 11, 2005 (edited) Why there is one way? style_images/master/snapback.png for the sake of completeness, Larry is a version of a math game called nim. this is a very old problem that has been studied extensively. hand waivy proof for Larry: can be played with any # of objts, so let's work our way up and u will notice a pattern. let n=# of objts trivial case: n=2,3, or 4 easy. take all but one and u win since larry left w/ 1. n=5 if i take 1, 4 left, larry takes 3 =>Loss if i take 2, 3 left, larry takes 2 =>Loss if i take 3, 2 left, larry takes 1 =>Loss so i am hosed either way. i'll come back to this. any player reduced to this sad state will lose. n=6 if i take 1, 5 left, which by above means larry losses if i take 2, 4 left, larry takes 3 =>Loss if i take 3, 3 left, larry takes 2 =>Loss n=7 if i take 1, 6 left, larry takes 1 =>5 left, Loss, c above. if i take 2, 5 left, larry losses. if i take 3, 4 left, larry takes 3 =>Loss n=8 if i take 1, 7 left, larry takes 2 =>Loss if i take 2, 6 left, larry takes 1=>Loss if i take 3, 5 left, larry losses n= 9 if i take 1, 8 left, larry takes 3=>Loss if i take 2, 7 left, larry takes 2=>Loss if i take 3, 6 left, larry takes 1=>Loss n=10 if i take 1, 9 left, larry losses if i take 2, 8 left, larry takes 3=>Loss if i take 3, 7 left, larry takes 2=>Loss only way to win is to take 1, leaving larry with 9 which shown above to be loss. here is what we can take to the bank... in a sequence of natural #'s, every fourth number ie(1,5,9,...is a losing state). Charles Bouton stated a rigorous soln to NIM in general. the spirit of the proof was something to the order of the following... -express the number of objects in each pile in binary form -add the coefficients of the powers of 2. -then remove as many objs from a pile to leave sum of the coefficients of each power of 2 an even number -when opponent draws, they r bound to upset such an arrangement -must not leave an even number of piles containing one coin each. Edited April 11, 2005 by armjan Quote Link to comment Share on other sites More sharing options...
armjan Posted April 11, 2005 Author Report Share Posted April 11, 2005 (edited) this really has nothing to do with AI... But yes machines for sure can think style_images/master/snapback.png i got this problem from... a cognitive sciences research dept in AI. Larry, version of NIM, relates to game theory and so naturally, yes, game theory has to do with AI. some instructors choose to focus on AI from the perspective of search/find heuristics, but it is certainly not limited to that, ie. natural language processing for instance. I would agree that machines do what u tell them or the side-effects resulting but not more. Edited April 11, 2005 by armjan Quote Link to comment Share on other sites More sharing options...
Nakharar Posted April 11, 2005 Report Share Posted April 11, 2005 I knew this was flawed. Quote Link to comment Share on other sites More sharing options...
Sip Posted April 13, 2005 Report Share Posted April 13, 2005 Pretty easy but cool problem this NIM ... yah the game is over as soon as a player gets to a number of the form '4n+1' where n is a non negative integer. When your opponent gets to a '4n+1' type number, you just keep bringing her down using 4s .. If she subtracts x, you subtract 4-x and continue until she is left with 1. Solution: -------------- 1. Starting value is a 4n+1 number - Starting player loses. 2. Starting value is a 4n+2 number - Starting player must do -1. 3. Starting value is a 4n+3 number - Starting player must do -2. 4. Starting value is a 4n+4 number - Starting player must do -3. In 2, 3, and 4, any other move and it's a guaranteed loss for the starting player. Also note that 4n+5 just repeats since we can substitute n with n+1 and do the above. Problem solved. Quote Link to comment Share on other sites More sharing options...
Harut Posted April 13, 2005 Report Share Posted April 13, 2005 we used to play this game back in armenian 20 yrs ago with matchsticks... Quote Link to comment Share on other sites More sharing options...
ExtraHye Posted April 13, 2005 Report Share Posted April 13, 2005 we used to play this game back in armenian 20 yrs ago with matchsticks... style_images/master/snapback.png You played this game when you were 2? Quote Link to comment Share on other sites More sharing options...
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