Need a algorithm
- To: mathgroup at smc.vnet.net
- Subject: [mg33857] Need a algorithm
- From: "Juan" <erfa11 at hotmail.com>
- Date: Sat, 20 Apr 2002 02:49:47 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Hi, I ask for help to find an algorithm to resolve this question: There is 36 numbers T={1,2,3,…,36}, and we play 7 numbers, for example : w={3,9,11,19,22,27,33}, then we compare w with the price , for example : p={2,7,11,22,30,31,33}, and in this case we have 3 goods: {11,22,33}. One way to play is to buy for example 1000 w (7 random numbers from T, 1000 times), and to compare each with p, to see if you have been lucky. Another way if to choose 12 numbers from T , for example: q={8,11,13,14,16,22,23,28,31,32,34,35}, and to play for all combinations: Binomial[12,7]=792. If you have q, and you want to get at least k rights, assuming that the numbers in the prize p are in q, then you don’t need to buy all the 792 w. I have bought that for k=3 and I got only 12 w. W= {{8,11,13,14,16,22,23},{8,11,13,14,28,31,32},{8,11,16,22,28,31,34}, {8,11,23,31,32,34,35},{8,13,14,22,32,34,35},{8,13,16,23,28,32,34},{8,14,16,23,28,31,35},{11,13,14,16,31,34,35}, {11,13,22,23,28,34,35},{11,14,16,22,28,32,35},{13,16,22,23,31,32,35},{14,22,23,28,31,32,34}} (This is one of the solutions, and you get at least 3 rights). If you want at least k=5 rights then you have to select more combinations of the all 792. Knowing T=Range[36], q=(n numbers from T) and k<7. How do we get W? Thanks. Juan _________________________________________________________________ MSN Photos es la manera más sencilla de compartir e imprimir sus fotos: http://photos.msn.com/support/worldwide.aspx