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How can a node establish pairwise shared key with other nodes using its own polynomial share together with other's public values?

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0

$begingroup$

A server has a symmetric bivariate polynomial $ F(x, y) = sum_i,j=0^t-1a_i,jx^iy^j$ $in GF(p)[X, Y] $ of degree $t-1$. For simpliciy, $ F(x, y) = a_0,0+a_1,0 x+a_0,1y+ a_1,1xy$ mod $p$, where $a_0, 1 = a_1,0$ and $p$ is a large prime number.

• From the origional polynomial $ F(x, y)$, the server generates $n$ univariate polynomial shares as $f_x_i(y)=F(x_i,y)$ , where 1 ≤ $i$ ≤ $n$, and $x_i$ is private (noboday knows it except the server) and distinct number for each node $i$.




• Each node $i$ has a distinct public number $r_i$, where 1 ≤ $i$ ≤ $n$ and every node else knows this number as it's public

Now, Let's say $node$ $i$ has got its own share $f_x_i(y)$ and $node$ $j$ has got $f_x_j(y)$ , where $x_i neq x_j$, as said earlier.

The question is:

Utilizing the symmetric property of the origional bivariate polynomail and Using their own polynomial shares together with other's public numbers, how can $node$ $i$ and $node$ $j$ ,$i neq j$, establish a pairwise shared key/value such that: $ k = f_x_i(r_j)$ =$f_x_j(r_i)$ holds???

$Hint$: node $i$ evaluates its own polynomial share using node $j$'s public number and node $j$ evaluates its own polynomial share using node $i$'s public number. Is there any trick to make them get the same value?? and How??

I appreciate your help

modular-arithmetic finite-fields cryptography

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edited Mar 14 at 9:07

Jyrki Lahtonen

110k13171386

asked Mar 14 at 6:31

A. AZEMiA. AZEMi

11

New contributor

A. AZEMi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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0

$begingroup$

A server has a symmetric bivariate polynomial $ F(x, y) = sum_i,j=0^t-1a_i,jx^iy^j$ $in GF(p)[X, Y] $ of degree $t-1$. For simpliciy, $ F(x, y) = a_0,0+a_1,0 x+a_0,1y+ a_1,1xy$ mod $p$, where $a_0, 1 = a_1,0$ and $p$ is a large prime number.

• From the origional polynomial $ F(x, y)$, the server generates $n$ univariate polynomial shares as $f_x_i(y)=F(x_i,y)$ , where 1 ≤ $i$ ≤ $n$, and $x_i$ is private (noboday knows it except the server) and distinct number for each node $i$.




• Each node $i$ has a distinct public number $r_i$, where 1 ≤ $i$ ≤ $n$ and every node else knows this number as it's public

Now, Let's say $node$ $i$ has got its own share $f_x_i(y)$ and $node$ $j$ has got $f_x_j(y)$ , where $x_i neq x_j$, as said earlier.

The question is:

Utilizing the symmetric property of the origional bivariate polynomail and Using their own polynomial shares together with other's public numbers, how can $node$ $i$ and $node$ $j$ ,$i neq j$, establish a pairwise shared key/value such that: $ k = f_x_i(r_j)$ =$f_x_j(r_i)$ holds???

$Hint$: node $i$ evaluates its own polynomial share using node $j$'s public number and node $j$ evaluates its own polynomial share using node $i$'s public number. Is there any trick to make them get the same value?? and How??

I appreciate your help

modular-arithmetic finite-fields cryptography

share|cite|improve this question

edited Mar 14 at 9:07

Jyrki Lahtonen

110k13171386

asked Mar 14 at 6:31

A. AZEMiA. AZEMi

11

New contributor

A. AZEMi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

A server has a symmetric bivariate polynomial $ F(x, y) = sum_i,j=0^t-1a_i,jx^iy^j$ $in GF(p)[X, Y] $ of degree $t-1$. For simpliciy, $ F(x, y) = a_0,0+a_1,0 x+a_0,1y+ a_1,1xy$ mod $p$, where $a_0, 1 = a_1,0$ and $p$ is a large prime number.

• From the origional polynomial $ F(x, y)$, the server generates $n$ univariate polynomial shares as $f_x_i(y)=F(x_i,y)$ , where 1 ≤ $i$ ≤ $n$, and $x_i$ is private (noboday knows it except the server) and distinct number for each node $i$.




• Each node $i$ has a distinct public number $r_i$, where 1 ≤ $i$ ≤ $n$ and every node else knows this number as it's public

Now, Let's say $node$ $i$ has got its own share $f_x_i(y)$ and $node$ $j$ has got $f_x_j(y)$ , where $x_i neq x_j$, as said earlier.

The question is:

Utilizing the symmetric property of the origional bivariate polynomail and Using their own polynomial shares together with other's public numbers, how can $node$ $i$ and $node$ $j$ ,$i neq j$, establish a pairwise shared key/value such that: $ k = f_x_i(r_j)$ =$f_x_j(r_i)$ holds???

$Hint$: node $i$ evaluates its own polynomial share using node $j$'s public number and node $j$ evaluates its own polynomial share using node $i$'s public number. Is there any trick to make them get the same value?? and How??

I appreciate your help

modular-arithmetic finite-fields cryptography

share|cite|improve this question

edited Mar 14 at 9:07

Jyrki Lahtonen

110k13171386

asked Mar 14 at 6:31

A. AZEMiA. AZEMi

11

New contributor

A. AZEMi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Mar 14 at 9:07

Jyrki Lahtonen

110k13171386

asked Mar 14 at 6:31

A. AZEMiA. AZEMi

11

New contributor

A. AZEMi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Mar 14 at 9:07

Jyrki Lahtonen

110k13171386

asked Mar 14 at 6:31

A. AZEMiA. AZEMi

11

New contributor

A. AZEMi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Mar 14 at 9:07

Jyrki Lahtonen

110k13171386

edited Mar 14 at 9:07

Jyrki Lahtonen

110k13171386

asked Mar 14 at 6:31

A. AZEMiA. AZEMi

11

New contributor

A. AZEMi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 14 at 6:31

A. AZEMiA. AZEMi

11