How to prove pairwise independence of a family of hash functions?












1












$begingroup$


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







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    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







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      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







      share|cite|improve this question









      $endgroup$




      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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jul 9 '16 at 8:51









      IpsiderIpsider

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      83






















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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.






          share|cite|improve this 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.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              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.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                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.






                share|cite|improve this answer











                $endgroup$



                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.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 12 '18 at 5:44









                Xander Henderson

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                14.3k103554










                answered Dec 12 '18 at 5:25









                mosemosmosemos

                262




                262






























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