Inverse Functions One-Way OnlyDate: 06/01/99 at 12:07:03 From: Al Hewitt Subject: One-way function Hello, I am trying to figure out whether the following problem has a solution. To verify two functions are inverses of one another, you need to show that (f of g)(x) = x and that (g of f)(x) = x. I would like to know if there is a problem that exists that when composed, the two functions work one way, but not the other; i.e.: (f of g)(x) = x, but (g of f)(x) does not equal x. If such a problem does exist, could you please show me an example? Thank you, Al Date: 06/01/99 at 15:03:39 From: Doctor Rob Subject: Re: One-way function Yes, such examples exist. Let S be the set of all infinite sequences of real numbers S = {s:N->R}, where N are the natural numbers and R are the real numbers. Let f(s)(n) = s(n+1), and g(s)(n) = s(n-1) for n > 1, and g(s)(1) = 0. f is a left-shift (end off) by 1 place, g is a right-shift (zero fill) by 1 place. Then (f[g(s)])(n) = g(s)(n+1) = s(n) for all n, (g[f(s)])(n) = g(s)(n-1) = s(n) for all n > 1, = 0 for n = 1. So if you pick a sequence s with s(1) different from 0, the result of a right-shift followed by a left shift is not the same as the result of a left-shift followed by a right shift. In one case, you will get s back, but in the other, you get s with s(1) replaced by 0. Thus f*g is the identity on S, but g*f is not. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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