i see. actually, it IS possible to win such a game. it's just a typical example of something called nim. whether ure gonna win or lose a nim game depends on what moves u make. if u always make the right moves, u will win no matter what moves your opponent makes. so in this case we have a 3-4-5-6 structure of pearls, which means if u want tt win, u will have to start FIRST. all u need to do is to keep making the correct moves and u will win no matter what. next, if u took any math/computer logic courses that have something to do with algorithms, u would learn that applying binomial calcualtions would make it much easier for u to win/understand a nim.

in this case,

3 = 11
4 = 100
5 = 101
6 = 110
--------
1=1
2=10 u will need these during the game



okay, i'm not going to go into deep explanations as to why the following works (go take an algorithms course), i'll just give u the essentials. u need to find the initial set-up by adding binary numbers togerther as though u were using decimal numeration. in this case: 110+101+100+11 = 322. your goal is to always make moves that would yield a remaining number all digits of which are EVEN except for the last move, using the above system for your calculations. for example, u can easily make 222=110+101+11 out of 322 with one move. so u just need to get rid of 100 (4) to accomplish this. so just remove 4 pearls from any row. when i played, i removed the entire row 4. the guy then removed only 1 ball from row 5, amking the remaining number equal 110+101+1=212. so i wanted it to be 202. 202 -> 101+100+1 -> 5+4+1. i removed 2 pearls from row 6. he removed 1 pearl from row 6, making it 101+11+1=113 and so on...


summary of moves:

me 322 -> 222

he makes 212

me 212 -> 202

he makes 113

me 113 -> 22

he makes 13

last move, therefore u have to leave 3 (odd) and not 2

13-> 3 = 1+1+1

HE LOST:

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Last edited by Noisician on May-18-2003 at 02:33

Nope.........not working out there.

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but hey !!! this is what the ****** does sometimes in the middle... he'll make it EVEN!!!!! instead of keeping it odd he makes it even...

not kidding... but when he kept it odd i raped him.... LOL thanks for the detailed algorithm!!!

i have a course in my third year on algorithms.... ... :d

mwhahaha, took a few tries, but I beat him

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//..

i suck

well, it doesn't really matter if u go first or not...

i've won more times letting him go first than i going first

it's all dependent on your last 3 rows of pearls and their composition

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that /&$%$%#$(=)=)=??())?$$%#$ son of a )(/%(%$&%//&#&%()()
if he laughs again i'm gonna kill him!!!!
i understood acidjunkies explanation but i don't know how to translate that into moves..

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okay, here'are 4 possible scenarios that can happen to u if u start by removing the entire row of 4 balls.

ooo
xxxx
ooooo
oooooo (4 balls out after the first move)

scenario 1

me 322 -> 222 (6+5+3)
the guy 222 -> 221 (6+5+2)
me 221 -> 220 (6+4+2)
the guy 220 -> 121 (4+3+2)
me 121 -> 22 (3+2+1)
the guy 22 -> 13 (3+1+1)
me (last move) 13 -> 3 (1+1+1)

scenario 2

me 322 -> 222 (6+5+3)
the guy 222 -> 212 (6+5+1)
me 212 -> 202 (5+4+1)
the guy 202 -> 113 (5+3+1)
me 113 -> 22 (3+2+1)
the guy 22 -> 13 (3+1+1)
me (last move) 13 -> 3 (1+1+1)

scenario 3

me 322 -> 222 (6+5+3)
the guy 222 -> 221 (6+4+3)
me 221 -> 220 (6+4+2)
the guy 220 -> 121 (4+3+2)
me 121 -> 22 (3+2+1)
the guy 22 -> 13 (3+1+1)
me (last move) 13 -> 3 (1+1+1)

scenario 4

me 322 -> 222 (6+5+3)
the guy 222 -> 212 (5+4+3)
me 212 -> 202 (5+4+1)
the guy 202 -> 113 (5+3+1)
me 113 -> 22 (3+2+1)
the guy 22 -> 13 (3+1+1)
me (last move) 13 -> 3 (1+1+1)

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Last edited by Noisician on May-18-2003 at 14:45

acid junkie is a genius
*I bow*

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I used that algorithm stuff and won!!!! yay!!!! but lost many times prior.

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