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hardest game ever


nozaki

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Guest imported_El Mamerro

Well, since you Photoshopped it, the least you could have done to make it somewhat believable was change the time between the two pictures, and maybe trim the sloppy black square you threw in there covering the guy's thumb. I dunno, just an idea.

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convert the number of pearls in a row into binary. add each binary number together, using regular 10-power rules. if the number is even, you are safe. if the number is odd, he has the upper hand. make sure to take out appropriate amounts so that when you add the binary amounts together you have an even number. ... there. not so hard, i did a final paper in mathematics on this damn game.

 

edit: once you start getting rows with one pearl only in them the rules change around ... so if you have 1,2,1 you wanna make it 1,1,1 which is odd added together ...

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Originally posted by El Mamerro@May 28 2005, 07:19 PM

Well, since you Photoshopped it, the least you could have done to make it somewhat believable was change the time between the two pictures, and maybe trim the sloppy black square you threw in there covering the guy's thumb. I dunno, just an idea.

 

Hilarious.

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Originally posted by KillPretty@May 29 2005, 03:06 AM

convert the number of pearls in a row into binary. add each binary number together, using regular 10-power rules. if the number is even, you are safe. if the number is odd, he has the upper hand. make sure to take out appropriate amounts so that when you add the binary amounts together you have an even number. ... there. not so hard, i did a final paper in mathematics on this damn game.

 

edit: once you start getting rows with one pearl only in them the rules change around ... so if you have 1,2,1 you wanna make it 1,1,1 which is odd added together ...

 

 

I'm fascinated by this. Could you by chance write out an example of how this works?

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Originally posted by Dirty_habiT+May 29 2005, 08:48 AM--><div class='quotetop'>QUOTE (Dirty_habiT - May 29 2005, 08:48 AM)</div><div class='quotemain'><!--QuoteBegin-KillPretty@May 29 2005, 03:06 AM

convert the number of pearls in a row into binary. add each binary number together, using regular 10-power rules. if the number is even, you are safe. if the number is odd, he has the upper hand. make sure to take out appropriate amounts so that when you add the binary amounts together you have an even number. ... there. not so hard, i did a final paper in mathematics on this damn game.

 

edit: once you start getting rows with one pearl only in them the rules change around ... so if you have 1,2,1 you wanna make it 1,1,1 which is odd added together ...

 

 

I'm fascinated by this. Could you by chance write out an example of how this works?

[/b]

 

lets say you have three rows with 3,5,7 pearls respectively. 3 in binary is: 011 (we're gonna use three place holders, its easier ...). 5 in binary is 101. 7 in binary is 111. 011 + 101 + 111 = 223. since 223 is odd, your on the spot. what you have to do is take away the right amount of pearls so that when you add the numbers up, it will be even. so, you take away one from row 3 leaving 6 which is 110 in binary. 011 + 101 + 110 = 222. now the computer is on the spot. make sure every turn you make you end up with an even sum. towards the end you should be able to figure out how to win without converting to binary and all that ish.

 

btw just use google if you cant figure out binary convertions (sp?).

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Originally posted by GucciCondom@May 30 2005, 05:09 AM

Mine isn't photoshopped. Notice the sad face instead of the winning grin.

 

To beat it, you just open two windows and start one window with him going first and then follow his moves back and forth through the two games. One will win, and one will lose.

^^^^^^^^^^^^^^ This works.

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Guest Ginger Bread Man

this is like paying a game of chess against a computer...the computer is using mathematical equations on a higher level than we are used to...fnd the equation and it is over

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