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Exercise 1: Task 7

Verfasst: 9. Feb 2009 15:35
von Xelord
In the solution is the size of the private key with \(2^{20} \cdot 2 \cdot 160^2\) denoted. Is this correct ?
In my documents i have the formula \(2^{H} \cdot n^2\), so what does the two mean ?

Re: Exercise 1: Task 7

Verfasst: 10. Feb 2009 15:56
von Patr0rc
The \(2n^2\) is the right size for one OTS secret key I think. (So \(2^H (2n^2)\) is correct at all - like the solution says.)
Getting to the example (discussed one lecture before your mentioned notes) may solve your problem:
Having the
1st bit being "0" you have (0,1,1), being "1" you have (1,0,0),
2nd bit being "0" you have (1,1,1), being "1" you have (1,0,0), and
3rd bit being "0" you have (1,0,1), being "1" you have (0,1,1).
So for each bit you have to store 2 possible "vectors" of size n which for n bits gets you \(2n^2\).
Does this help you getting it?

Re: Exercise 1: Task 7

Verfasst: 10. Feb 2009 17:54
von Xelord
Yeah, i get it. Thanks