Does $O(log n)$ (space) equality testing of $n$-bit integers using $f(x) = x mod p$ fingerprinting work...












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I found the following lecture from CS271 at UC Berkeley interesting and was taking a look at some of the examples, namely 3.1 on the first page: Link



Here is a summary of the problem: Alice has a $n$-bit integer $a$ and Bob has a $n$-bit integer $b$. Can Alice send Bob $O(log n)$ bits such that Bob can verify whether or not $a = b$ with high probability?



Here is the algorithm: Alice picks prime number $p$ uniformly at random from ${2,...,cn text{ln} n}$ and computes $F(a) = a text{mod} p$. Alice sends $F(a)$ and $p$ to Bob. Bob computes $F(b) = b text{mod} p$ and checks whether or not $F(a) = F(b)$. The algorithm returns true if they are equal, and false if not (with a small probability of failure of $frac{1}{c} + o(1)$ if $F(a) neq F(b)$ due to hash collision.



My question is whether or not you can extend the algorithm to work for negative integers as well and still maintain that Pr[error] = $frac{1}{c} + o(1)$. Intuitively, if $a$ and $b$ are both negative it should work just fine but I am unsure of the case where one of $a$ or $b$ is negative and the other is positive.



Would the algorithm work if you extend the problem such that $a$ and $b$ have can be negative as well?






number-theory elementary-number-theory algorithms modular-arithmetic cryptography







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    1












    $begingroup$


    I found the following lecture from CS271 at UC Berkeley interesting and was taking a look at some of the examples, namely 3.1 on the first page: Link



    Here is a summary of the problem: Alice has a $n$-bit integer $a$ and Bob has a $n$-bit integer $b$. Can Alice send Bob $O(log n)$ bits such that Bob can verify whether or not $a = b$ with high probability?



    Here is the algorithm: Alice picks prime number $p$ uniformly at random from ${2,...,cn text{ln} n}$ and computes $F(a) = a text{mod} p$. Alice sends $F(a)$ and $p$ to Bob. Bob computes $F(b) = b text{mod} p$ and checks whether or not $F(a) = F(b)$. The algorithm returns true if they are equal, and false if not (with a small probability of failure of $frac{1}{c} + o(1)$ if $F(a) neq F(b)$ due to hash collision.



    My question is whether or not you can extend the algorithm to work for negative integers as well and still maintain that Pr[error] = $frac{1}{c} + o(1)$. Intuitively, if $a$ and $b$ are both negative it should work just fine but I am unsure of the case where one of $a$ or $b$ is negative and the other is positive.



    Would the algorithm work if you extend the problem such that $a$ and $b$ have can be negative as well?






    number-theory elementary-number-theory algorithms modular-arithmetic cryptography







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I found the following lecture from CS271 at UC Berkeley interesting and was taking a look at some of the examples, namely 3.1 on the first page: Link



      Here is a summary of the problem: Alice has a $n$-bit integer $a$ and Bob has a $n$-bit integer $b$. Can Alice send Bob $O(log n)$ bits such that Bob can verify whether or not $a = b$ with high probability?



      Here is the algorithm: Alice picks prime number $p$ uniformly at random from ${2,...,cn text{ln} n}$ and computes $F(a) = a text{mod} p$. Alice sends $F(a)$ and $p$ to Bob. Bob computes $F(b) = b text{mod} p$ and checks whether or not $F(a) = F(b)$. The algorithm returns true if they are equal, and false if not (with a small probability of failure of $frac{1}{c} + o(1)$ if $F(a) neq F(b)$ due to hash collision.



      My question is whether or not you can extend the algorithm to work for negative integers as well and still maintain that Pr[error] = $frac{1}{c} + o(1)$. Intuitively, if $a$ and $b$ are both negative it should work just fine but I am unsure of the case where one of $a$ or $b$ is negative and the other is positive.



      Would the algorithm work if you extend the problem such that $a$ and $b$ have can be negative as well?






      number-theory elementary-number-theory algorithms modular-arithmetic cryptography







      share|cite|improve this question











      $endgroup$




      I found the following lecture from CS271 at UC Berkeley interesting and was taking a look at some of the examples, namely 3.1 on the first page: Link



      Here is a summary of the problem: Alice has a $n$-bit integer $a$ and Bob has a $n$-bit integer $b$. Can Alice send Bob $O(log n)$ bits such that Bob can verify whether or not $a = b$ with high probability?



      Here is the algorithm: Alice picks prime number $p$ uniformly at random from ${2,...,cn text{ln} n}$ and computes $F(a) = a text{mod} p$. Alice sends $F(a)$ and $p$ to Bob. Bob computes $F(b) = b text{mod} p$ and checks whether or not $F(a) = F(b)$. The algorithm returns true if they are equal, and false if not (with a small probability of failure of $frac{1}{c} + o(1)$ if $F(a) neq F(b)$ due to hash collision.



      My question is whether or not you can extend the algorithm to work for negative integers as well and still maintain that Pr[error] = $frac{1}{c} + o(1)$. Intuitively, if $a$ and $b$ are both negative it should work just fine but I am unsure of the case where one of $a$ or $b$ is negative and the other is positive.



      Would the algorithm work if you extend the problem such that $a$ and $b$ have can be negative as well?






      number-theory elementary-number-theory algorithms modular-arithmetic cryptography




      number-theory elementary-number-theory algorithms modular-arithmetic cryptography






      share|cite|improve this question















      share|cite|improve this question













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      edited Jan 5 at 16:37









      amWhy

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      asked Jan 5 at 12:05









      TheSalamanderTheSalamander

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          $begingroup$

          The sign is essentially just an extra bit, i.e., performing the experiment for tha number range $-2^n,ldots, 2^n-1$ is the same as doing it for $0,ldots, 2^{n+1}-1$.






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            $begingroup$

            The sign is essentially just an extra bit, i.e., performing the experiment for tha number range $-2^n,ldots, 2^n-1$ is the same as doing it for $0,ldots, 2^{n+1}-1$.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              The sign is essentially just an extra bit, i.e., performing the experiment for tha number range $-2^n,ldots, 2^n-1$ is the same as doing it for $0,ldots, 2^{n+1}-1$.






              share|cite|improve this answer









              $endgroup$
















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                $begingroup$

                The sign is essentially just an extra bit, i.e., performing the experiment for tha number range $-2^n,ldots, 2^n-1$ is the same as doing it for $0,ldots, 2^{n+1}-1$.






                share|cite|improve this answer









                $endgroup$



                The sign is essentially just an extra bit, i.e., performing the experiment for tha number range $-2^n,ldots, 2^n-1$ is the same as doing it for $0,ldots, 2^{n+1}-1$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 5 at 12:48









                Hagen von EitzenHagen von Eitzen

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