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559 \noindent for $z$. Again that is easy, but it does not allow |
559 \noindent for $z$. Again that is easy, but it does not allow |
560 us to know $r_i$, because then we would again need to solve |
560 us to know $r_i$, because then we would again need to solve |
561 a modular logarithm problem. Let us call an $h_i$ which was |
561 a modular logarithm problem. Let us call an $h_i$ which was |
562 solved the easy way as \emph{bogus}. Alice has to produce |
562 solved the easy way as \emph{bogus}. Alice has to produce |
563 bogus $h_i$ for all bits that are going to be $1$ in advance! |
563 bogus $h_i$ for all bits that are going to be $1$ in advance! |
564 This means she has to guess all the bits correctly. (Yes?) |
564 This means she has to guess all the bits correctly. (Yes? |
|
565 I let you think about this.) |
565 |
566 |
566 Let us see what happens if she guesses wrongly: Suppose the |
567 Let us see what happens if she guesses wrongly: Suppose the |
567 bit $b_i = 1$ where she thought she will get a 0. Then she has |
568 bit $b_i = 1$ where she thought she will get a 0. Then she has |
568 already sent an $h_i$ and $h_j$ and now must find an $s_i$ |
569 already sent an $h_i$ and $h_j$ and now must find an $s_i$ |
569 such that |
570 such that |