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Unbeatable game !?!
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| Acid Junkie |
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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| HappyToday |
| Nope.........not working out there. :D :happy2: |
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| victor |
ING AAAAAAAA
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.... :p ... :d |
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| Mosaic |
| mwhahaha, took a few tries, but I beat him |
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| marcus82 |
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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| RenderedDream |
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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| Acid Junkie |
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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| fathomless1 |
acid junkie is a genius
*I bow*:D |
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| SgtFoo |
I used that algorithm stuff and won!!!! yay!!!! but lost many times prior.
GENIUS, PURE GENIUS! |
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