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
edited Jan 5 at 16:37
amWhy
1
asked Jan 5 at 12:05
TheSalamanderTheSalamander
1839
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$begingroup$
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
edited Jan 5 at 16:37
amWhy
1
asked Jan 5 at 12:05
TheSalamanderTheSalamander
1839
$endgroup$
add a comment |
$begingroup$
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
edited Jan 5 at 16:37
amWhy
1
asked Jan 5 at 12:05
TheSalamanderTheSalamander
1839
$endgroup$
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
edited Jan 5 at 16:37
amWhy
1
asked Jan 5 at 12:05
TheSalamanderTheSalamander
1839
edited Jan 5 at 16:37
amWhy
1
asked Jan 5 at 12:05
TheSalamanderTheSalamander
1839
edited Jan 5 at 16:37
amWhy
1
edited Jan 5 at 16:37
amWhy
1
edited Jan 5 at 16:37
amWhy
1
1
asked Jan 5 at 12:05
TheSalamanderTheSalamander
1839
asked Jan 5 at 12:05
TheSalamanderTheSalamander
1839
asked Jan 5 at 12:05
TheSalamanderTheSalamander
1839
1839
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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 $0,ldots, 2^{n+1}-1$.
answered Jan 5 at 12:48
Hagen von EitzenHagen von Eitzen
282k23272506
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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 $0,ldots, 2^{n+1}-1$.
answered Jan 5 at 12:48
Hagen von EitzenHagen von Eitzen
282k23272506
$endgroup$
add a comment |
$begingroup$
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 $0,ldots, 2^{n+1}-1$.
answered Jan 5 at 12:48
Hagen von EitzenHagen von Eitzen
282k23272506
$endgroup$
add a comment |
$begingroup$
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 $0,ldots, 2^{n+1}-1$.
answered Jan 5 at 12:48
Hagen von EitzenHagen von Eitzen
282k23272506
$endgroup$
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 $0,ldots, 2^{n+1}-1$.
answered Jan 5 at 12:48
Hagen von EitzenHagen von Eitzen
282k23272506
answered Jan 5 at 12:48
Hagen von EitzenHagen von Eitzen
282k23272506
answered Jan 5 at 12:48
Hagen von EitzenHagen von Eitzen
282k23272506
answered Jan 5 at 12:48
Hagen von EitzenHagen von Eitzen
282k23272506
282k23272506
add a comment |
add a comment |
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