How to prove pairwise independence of a family of hash functions?
$begingroup$
I want to prove pairwise independence of a family of hash functions, but I don't know where to start.
Given the family of hash functions:
H with h(x) = a * x + b (mod M).
( Say h: U -> V, then: M is a prime and M >= IUI )
So how do I show that the family is pairwise independent. The definition is: The probability that h(u1) = v1 AND h(u2) = v2 is 1/M^2.
I can imagine that the solution has something to do with the modulo ring but I don't know how to get there.
Can someone help me?
Thanks a lot in advance!
probability probability-theory independence hash-function universal-property
asked Jul 9 '16 at 8:51
IpsiderIpsider
83
$endgroup$
add a comment |
$begingroup$
I want to prove pairwise independence of a family of hash functions, but I don't know where to start.
Given the family of hash functions:
H with h(x) = a * x + b (mod M).
( Say h: U -> V, then: M is a prime and M >= IUI )
So how do I show that the family is pairwise independent. The definition is: The probability that h(u1) = v1 AND h(u2) = v2 is 1/M^2.
I can imagine that the solution has something to do with the modulo ring but I don't know how to get there.
Can someone help me?
Thanks a lot in advance!
probability probability-theory independence hash-function universal-property
asked Jul 9 '16 at 8:51
IpsiderIpsider
83
$endgroup$
add a comment |
$begingroup$
I want to prove pairwise independence of a family of hash functions, but I don't know where to start.
Given the family of hash functions:
H with h(x) = a * x + b (mod M).
( Say h: U -> V, then: M is a prime and M >= IUI )
So how do I show that the family is pairwise independent. The definition is: The probability that h(u1) = v1 AND h(u2) = v2 is 1/M^2.
I can imagine that the solution has something to do with the modulo ring but I don't know how to get there.
Can someone help me?
Thanks a lot in advance!
probability probability-theory independence hash-function universal-property
asked Jul 9 '16 at 8:51
IpsiderIpsider
83
$endgroup$
I want to prove pairwise independence of a family of hash functions, but I don't know where to start.
Given the family of hash functions:
H with h(x) = a * x + b (mod M).
( Say h: U -> V, then: M is a prime and M >= IUI )
So how do I show that the family is pairwise independent. The definition is: The probability that h(u1) = v1 AND h(u2) = v2 is 1/M^2.
I can imagine that the solution has something to do with the modulo ring but I don't know how to get there.
Can someone help me?
Thanks a lot in advance!
probability probability-theory independence hash-function universal-property
probability probability-theory independence hash-function universal-property
asked Jul 9 '16 at 8:51
IpsiderIpsider
83
asked Jul 9 '16 at 8:51
IpsiderIpsider
83
asked Jul 9 '16 at 8:51
IpsiderIpsider
83
asked Jul 9 '16 at 8:51
IpsiderIpsider
83
asked Jul 9 '16 at 8:51
IpsiderIpsider
83
83
add a comment |
add a comment |
1 Answer
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$begingroup$
Well, we have $h(x) = a x + b pmod{M}$.
Since M is prime, it follows from Lagrange's Theorem that $a x + b$ has at most one solution modulo $m$. Therefore, for a fixed $a$ and $b$, each $x$ will be mapped to a different number in the range $[0 ... M-1]$, and hence, the probability of a specific $x$ to equal one specific element in this range is $1/M$. The logic is the same for $y$. So, the probability that $x$ is equal to one specific value $i$ and $y$ is equal to another specific value $j$ ($i$ can equal $j$) in the range $[0 ... M-1] = frac{1}{M} cdot frac{1}{M} = frac{1}{M^2}$, which is what is required to be shown.
I hope this helps.
edited Dec 12 '18 at 5:44
Xander Henderson
14.3k103554
answered Dec 12 '18 at 5:25
mosemosmosemos
262
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Well, we have $h(x) = a x + b pmod{M}$.
Since M is prime, it follows from Lagrange's Theorem that $a x + b$ has at most one solution modulo $m$. Therefore, for a fixed $a$ and $b$, each $x$ will be mapped to a different number in the range $[0 ... M-1]$, and hence, the probability of a specific $x$ to equal one specific element in this range is $1/M$. The logic is the same for $y$. So, the probability that $x$ is equal to one specific value $i$ and $y$ is equal to another specific value $j$ ($i$ can equal $j$) in the range $[0 ... M-1] = frac{1}{M} cdot frac{1}{M} = frac{1}{M^2}$, which is what is required to be shown.
I hope this helps.
edited Dec 12 '18 at 5:44
Xander Henderson
14.3k103554
answered Dec 12 '18 at 5:25
mosemosmosemos
262
$endgroup$
add a comment |
$begingroup$
Well, we have $h(x) = a x + b pmod{M}$.
Since M is prime, it follows from Lagrange's Theorem that $a x + b$ has at most one solution modulo $m$. Therefore, for a fixed $a$ and $b$, each $x$ will be mapped to a different number in the range $[0 ... M-1]$, and hence, the probability of a specific $x$ to equal one specific element in this range is $1/M$. The logic is the same for $y$. So, the probability that $x$ is equal to one specific value $i$ and $y$ is equal to another specific value $j$ ($i$ can equal $j$) in the range $[0 ... M-1] = frac{1}{M} cdot frac{1}{M} = frac{1}{M^2}$, which is what is required to be shown.
I hope this helps.
edited Dec 12 '18 at 5:44
Xander Henderson
14.3k103554
answered Dec 12 '18 at 5:25
mosemosmosemos
262
$endgroup$
add a comment |
$begingroup$
Well, we have $h(x) = a x + b pmod{M}$.
Since M is prime, it follows from Lagrange's Theorem that $a x + b$ has at most one solution modulo $m$. Therefore, for a fixed $a$ and $b$, each $x$ will be mapped to a different number in the range $[0 ... M-1]$, and hence, the probability of a specific $x$ to equal one specific element in this range is $1/M$. The logic is the same for $y$. So, the probability that $x$ is equal to one specific value $i$ and $y$ is equal to another specific value $j$ ($i$ can equal $j$) in the range $[0 ... M-1] = frac{1}{M} cdot frac{1}{M} = frac{1}{M^2}$, which is what is required to be shown.
I hope this helps.
edited Dec 12 '18 at 5:44
Xander Henderson
14.3k103554
answered Dec 12 '18 at 5:25
mosemosmosemos
262
$endgroup$
Well, we have $h(x) = a x + b pmod{M}$.
Since M is prime, it follows from Lagrange's Theorem that $a x + b$ has at most one solution modulo $m$. Therefore, for a fixed $a$ and $b$, each $x$ will be mapped to a different number in the range $[0 ... M-1]$, and hence, the probability of a specific $x$ to equal one specific element in this range is $1/M$. The logic is the same for $y$. So, the probability that $x$ is equal to one specific value $i$ and $y$ is equal to another specific value $j$ ($i$ can equal $j$) in the range $[0 ... M-1] = frac{1}{M} cdot frac{1}{M} = frac{1}{M^2}$, which is what is required to be shown.
I hope this helps.
edited Dec 12 '18 at 5:44
Xander Henderson
14.3k103554
answered Dec 12 '18 at 5:25
mosemosmosemos
262
edited Dec 12 '18 at 5:44
Xander Henderson
14.3k103554
edited Dec 12 '18 at 5:44
Xander Henderson
14.3k103554
edited Dec 12 '18 at 5:44
Xander Henderson
14.3k103554
14.3k103554
answered Dec 12 '18 at 5:25
mosemosmosemos
262
answered Dec 12 '18 at 5:25
mosemosmosemos
262
answered Dec 12 '18 at 5:25
mosemosmosemos
262
262
add a comment |
add a comment |
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