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Title: Mosquitoes and new repellent Post by Altamira_64 on Jun 15th, 2012, 12:15am A team of biologists has been working on a new repellent to face the increasing number of mosquitoes in an area of Africa. They have been working with the combination of various substances in different percentages and each sample was placed in a test tube and then tested on a mosquito. They then waited for one hour to see if the mosquito would die. All test tubes with the various repellent samples were kept aside, just in case they were needed again. The first 499 attempts and absolutely no effect on the mosquito but the 500th killed it after about one hour. They started jumping of joy and dancing for their success. Unfortunately, in the meantime, a careless assistant mixed up the test tubes, so in effect they did not know which one had the correct fluid. Moreover, there were only 10 mosquitoes left and also the deadline for the end of the experiment was in 1,5 hour. Given that they would need about half an hour to prepare a new experiment, how can they determine which test-tube contains the correct sample, within the remaining one hour? |
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Title: Re: Mosquitoes and new repellent Post by towr on Jun 15th, 2012, 12:25am Sound like the criminal cupbearer riddle (http://www.ocf.berkeley.edu/~wwu/riddles/hard.shtml#criminalCupbearers) (forum thread (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_hard;action=display;num=1027806173;start=0)). |
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Title: Re: Mosquitoes and new repellent Post by Grimbal on Jun 15th, 2012, 4:22pm If there are only 10 mosquitoes left, just kill them and you won't need the repellent any more. ::) |
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Title: Re: Mosquitoes and new repellent Post by Altamira_64 on Jun 16th, 2012, 12:22am So let's start by saying that we divide the tubes in groups of 2 4 8 16 32 64 128 and the remaining 246. We start by giving each mosquito one drop from every tube in each group and wait for one hour to see the results. Then what? |
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Title: Re: Mosquitoes and new repellent Post by towr on Jun 16th, 2012, 2:13am Assign each vial a binary number, and each mosquito a bit in the binary number. Give each mosquito a drop from each vial where its bit is one. The live and dead mosquitoes after one hour spell the binary number of the correct vial. |
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Title: Re: Mosquitoes and new repellent Post by Altamira_64 on Jun 16th, 2012, 3:40am Wow! So, to start with, here is the list of the 500 vials in binary format: 000000001 000000010 000000011 000000100 000000101 000000110 000000111 000001000 000001001 000001010 000001011 000001100 000001101 000001110 000001111 000010000 000010001 000010010 000010011 000010100 000010101 000010110 000010111 000011000 000011001 000011010 000011011 000011100 000011101 000011110 000011111 000100000 000100001 000100010 000100011 000100100 000100101 000100110 000100111 000101000 000101001 000101010 000101011 000101100 000101101 000101110 000101111 000110000 000110001 000110010 000110011 000110100 000110101 000110110 000110111 000111000 000111001 000111010 000111011 000111100 000111101 000111110 000111111 001000000 001000001 001000010 001000011 001000100 001000101 001000110 001000111 001001000 001001001 001001010 001001011 001001100 001001101 001001110 001001111 001010000 001010001 001010010 001010011 001010100 001010101 001010110 001010111 001011000 001011001 001011010 001011011 001011100 001011101 001011110 001011111 001100000 001100001 001100010 001100011 001100100 001100101 001100110 001100111 001101000 001101001 001101010 001101011 001101100 001101101 001101110 001101111 001110000 001110001 001110010 001110011 001110100 001110101 001110110 001110111 001111000 001111001 001111010 001111011 001111100 001111101 001111110 001111111 010000000 010000001 010000010 010000011 010000100 010000101 010000110 010000111 010001000 010001001 010001010 010001011 010001100 010001101 010001110 010001111 010010000 010010001 010010010 010010011 010010100 010010101 010010110 010010111 010011000 010011001 010011010 010011011 010011100 010011101 010011110 010011111 010100000 010100001 010100010 010100011 010100100 010100101 010100110 010100111 010101000 010101001 010101010 010101011 010101100 010101101 010101110 010101111 010110000 010110001 010110010 010110011 010110100 010110101 010110110 010110111 010111000 010111001 010111010 010111011 010111100 010111101 010111110 010111111 011000000 011000001 011000010 011000011 011000100 011000101 011000110 011000111 011001000 011001001 011001010 011001011 011001100 011001101 011001110 011001111 011010000 011010001 011010010 011010011 011010100 011010101 011010110 011010111 011011000 011011001 011011010 011011011 011011100 011011101 011011110 011011111 011100000 011100001 011100010 011100011 011100100 011100101 011100110 011100111 011101000 011101001 011101010 011101011 011101100 011101101 011101110 011101111 011110000 011110001 011110010 011110011 011110100 011110101 011110110 011110111 011111000 011111001 011111010 011111011 011111100 011111101 011111110 011111111 100000000 100000001 100000010 100000011 100000100 100000101 100000110 100000111 100001000 100001001 100001010 100001011 100001100 100001101 100001110 100001111 100010000 100010001 100010010 100010011 100010100 100010101 100010110 100010111 100011000 100011001 100011010 100011011 100011100 100011101 100011110 100011111 100100000 100100001 100100010 100100011 100100100 100100101 100100110 100100111 100101000 100101001 100101010 100101011 100101100 100101101 100101110 100101111 100110000 100110001 100110010 100110011 100110100 100110101 100110110 100110111 100111000 100111001 100111010 100111011 100111100 100111101 100111110 100111111 101000000 101000001 101000010 101000011 101000100 101000101 101000110 101000111 101001000 101001001 101001010 101001011 101001100 101001101 101001110 101001111 101010000 101010001 101010010 101010011 101010100 101010101 101010110 101010111 101011000 101011001 101011010 101011011 101011100 101011101 101011110 101011111 101100000 101100001 101100010 101100011 101100100 101100101 101100110 101100111 101101000 101101001 101101010 101101011 101101100 101101101 101101110 101101111 101110000 101110001 101110010 101110011 101110100 101110101 101110110 101110111 101111000 101111001 101111010 101111011 101111100 101111101 101111110 101111111 110000000 110000001 110000010 110000011 110000100 110000101 110000110 110000111 110001000 110001001 110001010 110001011 110001100 110001101 110001110 110001111 110010000 110010001 110010010 110010011 110010100 110010101 110010110 110010111 110011000 110011001 110011010 110011011 110011100 110011101 110011110 110011111 110100000 110100001 110100010 110100011 110100100 110100101 110100110 110100111 110101000 110101001 110101010 110101011 110101100 110101101 110101110 110101111 110110000 110110001 110110010 110110011 110110100 110110101 110110110 110110111 110111000 110111001 110111010 110111011 110111100 110111101 110111110 110111111 111000000 111000001 111000010 111000011 111000100 111000101 111000110 111000111 111001000 111001001 111001010 111001011 111001100 111001101 111001110 111001111 111010000 111010001 111010010 111010011 111010100 111010101 111010110 111010111 111011000 111011001 111011010 111011011 111011100 111011101 111011110 111011111 111100000 111100001 111100010 111100011 111100100 111100101 111100110 111100111 111101000 111101001 111101010 111101011 111101100 111101101 111101110 111101111 111110000 111110001 111110010 111110011 111110100 So you are saying that I assign the first bit to mosquito number 1, the second to number 2 etc, so in effect I give mosquito number 1 one drop from vials 1, 3, 5, 7 etc, mosquito number 2 one drop from 2, 3, 6, 7, 10, 11 etc and so on. Then, if, for example, the correct repellent is in vial 27, the mosquitoes 1, 2, 4 and 5 will die. Is this it? So we also have one extra mosquito as bonus gift! (because we only need 9 for this method). |
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Title: Re: Mosquitoes and new repellent Post by towr on Jun 16th, 2012, 4:11am Yup, that's it. You could use the final mosquito to check for errors, using it as a parity bit :P (With a few more you could do an error-correcting code.) |
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Title: Re: Mosquitoes and new repellent Post by Altamira_64 on Jun 17th, 2012, 1:52am Thanks for the assistance :) |
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