Add computational hiding for 5d
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two different graphs to the same output which corresponds to solving the
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CGI problem.
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Computational hiding: Two commitments are computationally
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indistinguishable from each other as long as the adversary is PPT.
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Otherwise, the adversary could distinguish between the two distribution
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ensembles.
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\item If $G_{ch}=\phi_{ch}(G)$ and $G'=\psi(G)$, it follows that
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$G=\phi_{ch}^{-1}(G_{ch})$ and therefore $G'=\psi(\phi_{ch}^{-1}(G_{ch}))$
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so the verifier will always accept.
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