
Are you a match for the Logic Fiend?!!
The challenge of the 12 eggs The chamber of 9 doors The challenge of the 9 eggs The 5 protistas on the run The 4 protistas on the run Shakeel's Fiendish Illusion You lucky muskrats. I'm amazed! I hope you are ready. THIS IS AN OUTRAGE Friday the 13th Three naked logical fiends, and yourself. the selfish, logical protistas a teaser.  Shakeel's Fiendish Illusion Friday, July 22, 2005 Well, well, well, why am I not surprised this dumpster has attracted so many more nincompoops? I have been banished long enough, and I'm sure everyone has been awaiting my return. And what better time to return than now. And what better way to celebrate my return than with a logic puzzle. Let's see if you rodents have grown brains in my long absence. (I highly doubt it.) Long ago, there existed a beautiful place called the Kingdom of Liars and Logic Fiends. One logic fiend, Shakeel, wanted to perform a fiendish illusion to impress the king. To do this, he needed to create a perfectly square grid, with an even number of rows and columns. Each grid would be occupied by a liar or a logic fiend, such that each person can say "All my adjacent grids are occupied by people of the opposite persuasion." No, you dimwit, diagonal grids are not considered to be adjacent. (In other words, if you can't work out this simple math, there is a maximum of 4 adjacent grids.) Shakeel had a couple more requirements that needed to be fulfilled for his illusion to work: * there must be at least 100 participants * at least 37% of the participants must be logic fiends What is the mininum number of participants needed for Shakeel's fiendish illusion? Solved by Monkeymeister for 50 Protista Points. 8 Comments. Hi Hi... ur nutang seems pretty cool. If its okay, i think i'll ad u as a friend. Well...bye! » Jinaiah on 20050722 02:14:34 Hey Well...bye! supersiton* » dean on 20050724 02:55:49 I have an answer If every adjacent cell has a person of the opposite persuasion, then the cells just alternate. For the minimum number of participants, 100, that would mean that there are 50 logic fiends, which is 50%. And 50% >= 37%. QED. » flanger001 on 20050728 03:30:50 re: flanger001 Well, well, well, finally someone brave enough to step up to the plate. Unfortunately, you failed to read between the "lies." MuHAhahahahAhAHA. » theLogicFiend on 20050728 04:09:42 yeah flanger im sorry but then the liars wouldnt really be liar's they would be telling the truth. but you can use math mr. logic fiend and the answer is 2500! » monkeymeister on 20050729 10:33:00 Monkeymeister wins!! Though you offered no explanation, I commend you. You will now join Wiggum in the hall of fame. » theLogicFiend on 20050729 11:27:59 Explanation (How I did it) X = Logic Fiends O = Liars R = Rows X O X O X O X O X O O X O X O X O X O X X O X O X O X O X O O X O X O X O X O X X O X O X O X O X O X = 50/100 X = (R^2)/2 O X O X O X O X O X X O X O X O X O X O O X O X O X O X O X X O X O X O X O X O O X O X O X O X O X Replace every other logic fiend in every other diagonal row (subtract (r^2)/8). X O O O X O O O X O O X O X O X O X O O O O X O O O X O X O O X O X O X O O O X Without borders: X O O O X O X O X O X = 37.5/100 X = [(R^2)/2][(R^2)/8] O X O X O O O X O O X = 3/8 O O X O X O X O X O O X O O O X O O O X X O X O X O X O X O O O O X O O O X O O With borders, subtract R/4 logic fiends to make sure the border spots have a liar to keep with the conditions. (R^2)(3/8)(R/4) = Maximum Logic Fiends Plug in even numbers for R from 10 until 50, which returns a result of 925 (37% of 2500). » le_battement on 20050729 12:56:32 Explanation (Not screwed up looking) X = Logic Fiends O = Liars R = Rows X O X O X O X O X O O X O X O X O X O X X O X O X O X O X O O X O X O X O X O X X O X O X O X O X O X = 50/100 X = (R^2)/2 O X O X O X O X O X X O X O X O X O X O O X O X O X O X O X X O X O X O X O X O O X O X O X O X O X Replace every other logic fiend in every other diagonal row (subtract (r^2)/8). X O O O X O O O X O O X O X O X O X O O O O X O O O X O X O O X O X O X O O O X Without borders: X O O O X O X O X O X = 37.5/100 X = [(R^2)/2][(R^2)/8] O X O X O O O X O O X = 3/8 O O X O X O X O X O O X O O O X O O O X X O X O X O X O X O O O O X O O O X O O With borders, subtract R/4 logic fiends to make sure the border spots have a liar to keep with the conditions. (R^2)(3/8)(R/4) = Maximum Logic Fiends Plug in even numbers for R from 10 until 50, which returns a result of 925 (37% of 2500). » le_battement on 20050729 12:57:47
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