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Unformatted text preview: March 17, 2004 Physics 681481; CS 483: Discussion of #4 I. (a) According to Eq. (1) in Assignment #4, the probability of finding 2 linearly inde pendent vectors (doing arithmetic modulo 2) among a random set of three 3vectors of 0s and 1s orthogonal to the vector 111, is q = ( 1 1 4 )( 1 1 8 ) = 21 32 . ( n = 3 , x = 0) (1) To check this note that the four vectors orthogonal to 111 are 000, 110, 101, and 011. Any two distinct nonzero vectors from this set are linearly independent, so the number of the 4 3 = 64 different trios without two linearly independent vectors, is just the number of trios containing at most one kind of nonzero vector. To enumerate these note that we can have three zero vectors (000) in just one way; we can have one nonzero vector and two zero vectors in 9 different ways (the nonzero vector can be either the first, second or third vector in the trio, and it can have one of the three forms (110), (101), or (011)); we can also have two identical nonzero vectors and one zero vector in 9 different ways (the zero...
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This note was uploaded on 09/28/2008 for the course PHYS 481 taught by Professor Anon during the Spring '05 term at Cornell.
 Spring '05
 ANON
 Physics

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