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| | <big>'''[[IAG 1H]] [[POW]] <nowiki>#</nowiki> 6: Linear Nim'''</big> | | <big>'''[[IAG 1H]] [[POW]] <nowiki>#</nowiki> 6: Linear Nim'''</big> |
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| − | {{Word|Image:Linear_Nim.doc}} | + | |
| − | {{PDF|Image:Linear Nim.pdf}}
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| − | ==Instructions==
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| − | *[[:Image:Linear Nim - Instructions 1.jpg|Page 1]]
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| − | *[[:Image:Linear Nim - Instructions 2.jpg|Page 2]]
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| − | ==Problem Statement==
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| − | Not necessary to do.
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| − | ==Process (Playing)==
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| − | I started playing Linear Nim with Michael Lewis. We started out just playing to try and find strategies. Our tallies are recorded below:
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| − | On my first try, I played going first and picking 3. Bad idea, I thought at first, I would always loose, but now I am thinking differently. Of course if the other person picks 3, I am left with 4. I later found out that was a bad idea. Picking 3 first is a bad idea because if the other player picks 3, your stuck with 4.
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| − | On my 2<sup>nd</sup> try, I started first at 2 and won. Later on, I found out that I should start with 2 going first. I figured out that I should stick my opponent with 4, and later I found out that it is equal to ''(maximum per turn)<nowiki>+</nowiki>1''. This is sort of a interim goal where if you meet this goal you will win. I think you are able to think easier when you are dealing with less numbers. I won this game by playing the strategy mentioned in the solution.
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| − | On the 3<sup>rd</sup> try, I let my opponent go first. He picked 3 which was the mistake I made on my 1<sup>st</sup> turn. I ended up winning.
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| − | On the fourth try, I tried to pick 1 first. I lost by being stuck with 4. I could have won by watching more closely. If my opponent picks 1 on his first turn there are 8 lines left. If I pick 1, my opponent picks 1, I can pick 1, 2, or 3. Either way my opponent picks the opposite (you 1-he 3) or ''<nowiki>[</nowiki>(maximum per turn)-(his pick)<nowiki>+</nowiki>1<nowiki>]</nowiki> ''and it<nowiki>’</nowiki>s my choice with 4 left. He wins. If he picks 2, I can pick 3, and he picks with 4 and looses. If he picks 3, then I can pick 2, and leave him with 4. Therefore, picking 1<sup> </sup>first lets your opponent control your fate.
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| − | The fifth time around I developed the winning strategy. See below.
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| − | I did not do anymore playing where I don<nowiki>’</nowiki>t go first, because I found my answer.
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| − | ===Work===
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| − | [[Image:Linear Nim - Work.jpg]]
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| − | ==Solution (of original)==
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| − | I found out in my 5<sup>th</sup> term the winning strategy for the original game. If I went first, and picked 2, I could win. If my opponent picked 3, thin I would pick 1 on my next term. There would be 4 left and it would be his turn. If it<nowiki>’</nowiki>s your turn and you have 4 ''(maximum per turn)<nowiki>+</nowiki>1)'' left, you lose. Now when I picked 2 first, and my opponent picks 2, I would take 2 also to leave him with 4. If he takes 1, I take 3 and He is again left with 4. This is my strategy.
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| − | If the other person goes first, hope they won<nowiki>’</nowiki>t pick 2, if so, your done. If he picks 1 first, pick 2, or 3 and you might lose. If he picks 3, pick 3 and force him with 4.
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| − | ==Extension (Generalizations)==
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| − | I am also seeing a pattern that leaving him with 4 <nowiki>[</nowiki>''(maximum per turn)<nowiki>+</nowiki>1''<nowiki>]</nowiki> requires I can also leave him with 8, which happens to be ''<nowiki>[</nowiki>2x(maximum per turn<nowiki>+</nowiki>1)<nowiki>]</nowiki>'' When I start at 10, I need to take 2 to get him to ''<nowiki>[</nowiki>2x(maximum per turn<nowiki>+</nowiki>1)<nowiki>]</nowiki>.'' I did not know that at the time, so we tried playing with 15 marks and keeping 3 at a time. I found that if you got the game down to 10 and your turn, it is just like playing the original game. I now also know that if I got him down to picking from 8 ''<nowiki>[</nowiki>2x(maximum per turn<nowiki>+</nowiki>1)<nowiki>]</nowiki>'' and my turn, I could also win. I suppose this will also work with all variations of X: ''<nowiki>[</nowiki>'''X'''<nowiki>*</nowiki>(maximum per turn<nowiki>+</nowiki>1)<nowiki>]</nowiki> '' in all variations of liner nim. I can also try and get him to 12 ''<nowiki>[</nowiki>3x(maximum per turn<nowiki>+</nowiki>1)<nowiki>]</nowiki>'', by going first and taking 3. I can now know that if I can go first, I will always win, unless ''<nowiki>[</nowiki>X<nowiki>*</nowiki>(maximum per turn<nowiki>+</nowiki>1)<nowiki>]</nowiki>'' is the starting number, where I would want him to go first.. This is because I can always take ''(initial number)-<nowiki>[</nowiki>X<nowiki>*</nowiki>(maximum per turn<nowiki>+</nowiki>1)<nowiki>]</nowiki>'', and catch him in a loop picking the opposite of what he picks, which equals, ''<nowiki>[</nowiki>(maximum per turn)-(his pick)<nowiki>+</nowiki>1<nowiki>]</nowiki>.''
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| − | I found out that trapping people at 4 was the same as ''(maximum per turn)<nowiki>+</nowiki>1'' when I to increased the maximum per turn to 4 and played with the initial number of 10. I ended up winning, however it would have been better to have known the strategy above.
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| − | ===Faux Program===
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| − | '''If initial number equals <nowiki>[</nowiki>X<nowiki>*</nowiki>(maximum per turn<nowiki>+</nowiki>1)<nowiki>]</nowiki>;'''
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| − | '''Then'''
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| − | Label A
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| − | Have Partner Go
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| − | Take (maximum per turn)-(his pick)<nowiki>+</nowiki>1
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| − | Goto A
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| − | '''Else'''
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| − | Go First
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| − | Take (initial number)-<nowiki>[</nowiki>X<nowiki>*</nowiki>(maximum per turn<nowiki>+</nowiki>1)
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| − | Have Partner Go
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| − | Take (maximum per turn)-(his pick)<nowiki>+</nowiki>1
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| − | Goto A
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| − | During Problem: X is equal to the number closest to <nowiki><</nowiki> (initial number)
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| − | ==Evaluation==
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| − | Not necessary to do.
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| | [[Category:IAG 1H]] | | [[Category:IAG 1H]] |
| | [[Category:POW]] | | [[Category:POW]] |
IMP Takedown
I was asked to take down POW solutions. Remember it's best to think about how to solve the problems on your own.
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