Secret sharing over reals - constructing a (k,n) threshold scheme

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After following a discussion that Shamir's Secret Sharing scheme cannot be used to share a real number as secret, I came across the paper "Secret Sharing Over Infinite Domains" - B. Chor and E. Kushilevitz
The above paper describes a method for sharing a real number as a secret, I quote from Section 4 (note: this paper may be accessed freely)




We first introduce a (k,k) secret-sharing scheme which distributes a
secret a taken from the interval [0,1). We use the Lebesgue
measure on [0,1) Choose independently, with a uniform distribution,
k-1 real numbers, $s_1$,.., $s_k-1$ in the interval [0,1). 2)
Choose $s_k$ $in$ [0,1) which satisfies $s_1$ +...+ $s_k-1$
+$s_k$ = a (mod 1). The proof that this is indeed a secret-sharing scheme is similar to the proof of its analogue in the finite case.

For introducing a (k ,n) secret-sharing scheme for every k $leq$ n,
we observe that the same technique described in [BL] works here as
well.




I can see how this (k,k) threshold scheme works. However, I am having some issues with the (k,n) threshold scheme - I've tried to look at Generalized Secret Sharing and Monotone Functions which is referred to above as BL (note - this paper can also be accessed freely.) I don't see how this paper helps me construct a (k,n) threshold scheme.

Any help would be appreciated!










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    up vote
    1
    down vote

    favorite












    After following a discussion that Shamir's Secret Sharing scheme cannot be used to share a real number as secret, I came across the paper "Secret Sharing Over Infinite Domains" - B. Chor and E. Kushilevitz
    The above paper describes a method for sharing a real number as a secret, I quote from Section 4 (note: this paper may be accessed freely)




    We first introduce a (k,k) secret-sharing scheme which distributes a
    secret a taken from the interval [0,1). We use the Lebesgue
    measure on [0,1) Choose independently, with a uniform distribution,
    k-1 real numbers, $s_1$,.., $s_k-1$ in the interval [0,1). 2)
    Choose $s_k$ $in$ [0,1) which satisfies $s_1$ +...+ $s_k-1$
    +$s_k$ = a (mod 1). The proof that this is indeed a secret-sharing scheme is similar to the proof of its analogue in the finite case.

    For introducing a (k ,n) secret-sharing scheme for every k $leq$ n,
    we observe that the same technique described in [BL] works here as
    well.




    I can see how this (k,k) threshold scheme works. However, I am having some issues with the (k,n) threshold scheme - I've tried to look at Generalized Secret Sharing and Monotone Functions which is referred to above as BL (note - this paper can also be accessed freely.) I don't see how this paper helps me construct a (k,n) threshold scheme.

    Any help would be appreciated!










    share|improve this question







    New contributor




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





















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      After following a discussion that Shamir's Secret Sharing scheme cannot be used to share a real number as secret, I came across the paper "Secret Sharing Over Infinite Domains" - B. Chor and E. Kushilevitz
      The above paper describes a method for sharing a real number as a secret, I quote from Section 4 (note: this paper may be accessed freely)




      We first introduce a (k,k) secret-sharing scheme which distributes a
      secret a taken from the interval [0,1). We use the Lebesgue
      measure on [0,1) Choose independently, with a uniform distribution,
      k-1 real numbers, $s_1$,.., $s_k-1$ in the interval [0,1). 2)
      Choose $s_k$ $in$ [0,1) which satisfies $s_1$ +...+ $s_k-1$
      +$s_k$ = a (mod 1). The proof that this is indeed a secret-sharing scheme is similar to the proof of its analogue in the finite case.

      For introducing a (k ,n) secret-sharing scheme for every k $leq$ n,
      we observe that the same technique described in [BL] works here as
      well.




      I can see how this (k,k) threshold scheme works. However, I am having some issues with the (k,n) threshold scheme - I've tried to look at Generalized Secret Sharing and Monotone Functions which is referred to above as BL (note - this paper can also be accessed freely.) I don't see how this paper helps me construct a (k,n) threshold scheme.

      Any help would be appreciated!










      share|improve this question







      New contributor




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











      After following a discussion that Shamir's Secret Sharing scheme cannot be used to share a real number as secret, I came across the paper "Secret Sharing Over Infinite Domains" - B. Chor and E. Kushilevitz
      The above paper describes a method for sharing a real number as a secret, I quote from Section 4 (note: this paper may be accessed freely)




      We first introduce a (k,k) secret-sharing scheme which distributes a
      secret a taken from the interval [0,1). We use the Lebesgue
      measure on [0,1) Choose independently, with a uniform distribution,
      k-1 real numbers, $s_1$,.., $s_k-1$ in the interval [0,1). 2)
      Choose $s_k$ $in$ [0,1) which satisfies $s_1$ +...+ $s_k-1$
      +$s_k$ = a (mod 1). The proof that this is indeed a secret-sharing scheme is similar to the proof of its analogue in the finite case.

      For introducing a (k ,n) secret-sharing scheme for every k $leq$ n,
      we observe that the same technique described in [BL] works here as
      well.




      I can see how this (k,k) threshold scheme works. However, I am having some issues with the (k,n) threshold scheme - I've tried to look at Generalized Secret Sharing and Monotone Functions which is referred to above as BL (note - this paper can also be accessed freely.) I don't see how this paper helps me construct a (k,n) threshold scheme.

      Any help would be appreciated!







      secret-sharing






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          1 Answer
          1






          active

          oldest

          votes

















          up vote
          4
          down vote



          accepted










          This $(k,n)$ scheme works, but isn't very interesting.



          Effectively, it is:



          • For each set of $k$ participants out of $n$, construct a $(k,k)$ threshold scheme, and distribute those shares to the participants in the set.

          For example, in a $(2, 3)$ scheme, if $z$ is the secret, we'd generate $binom32 = 3$ indepedent $(2,2)$ threshold schemes $(r_1, z-r_1 bmod 1), (r_2, z-r_2 bmod 1), (r_3, z-r_3 bmod 1)$, and distribute to the three share holders the shares:



          $$(r_1, z-r_2 bmod 1)$$
          $$(r_2, z-r_3 bmod 1)$$
          $$(r_3, z-r_1 bmod 1)$$



          It works, as:



          • For any set of $k$ share holders, they can reconstruct the secret, as there is a $(k,k)$ threshold scheme with those share holders with all the shares. In the above example, the first two share holders jointly know both $z-r_2 bmod 1$ and $r_2$, allowing them to reconstruct $z$.


          • For any set of $k-1$ share holders, they learn nothing of the secret; for any of the $(k,k)$ threshold schemes, there will always be at least 1 missing share, and so they can learn nothing.


          It's quite straight-forward; the biggest issue is that this requires distributing $binomn-1k-1$ independent values to each share holder; this is rather large if you're trying to implement a $(500,1000)$ access structure.






          share|improve this answer




















          • thank you so much for your in-depth answer; really appreciate the clarity of your example along with the generalization of the solution. Really makes me wish I was also in a position to be able to answer questions on Stack Exchange - hopefully, I will get there one day!
            – Rahul Mathur
            2 hours ago










          • (500,1000) is "rather large" he says... 1.35144120472718284758e+299 to be approximate. I wish we had an understatement badge. Either that or I'm gonna need a poncho for all the sarcasm dripping off that statement.
            – Ken Goss
            20 mins ago











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          4
          down vote



          accepted










          This $(k,n)$ scheme works, but isn't very interesting.



          Effectively, it is:



          • For each set of $k$ participants out of $n$, construct a $(k,k)$ threshold scheme, and distribute those shares to the participants in the set.

          For example, in a $(2, 3)$ scheme, if $z$ is the secret, we'd generate $binom32 = 3$ indepedent $(2,2)$ threshold schemes $(r_1, z-r_1 bmod 1), (r_2, z-r_2 bmod 1), (r_3, z-r_3 bmod 1)$, and distribute to the three share holders the shares:



          $$(r_1, z-r_2 bmod 1)$$
          $$(r_2, z-r_3 bmod 1)$$
          $$(r_3, z-r_1 bmod 1)$$



          It works, as:



          • For any set of $k$ share holders, they can reconstruct the secret, as there is a $(k,k)$ threshold scheme with those share holders with all the shares. In the above example, the first two share holders jointly know both $z-r_2 bmod 1$ and $r_2$, allowing them to reconstruct $z$.


          • For any set of $k-1$ share holders, they learn nothing of the secret; for any of the $(k,k)$ threshold schemes, there will always be at least 1 missing share, and so they can learn nothing.


          It's quite straight-forward; the biggest issue is that this requires distributing $binomn-1k-1$ independent values to each share holder; this is rather large if you're trying to implement a $(500,1000)$ access structure.






          share|improve this answer




















          • thank you so much for your in-depth answer; really appreciate the clarity of your example along with the generalization of the solution. Really makes me wish I was also in a position to be able to answer questions on Stack Exchange - hopefully, I will get there one day!
            – Rahul Mathur
            2 hours ago










          • (500,1000) is "rather large" he says... 1.35144120472718284758e+299 to be approximate. I wish we had an understatement badge. Either that or I'm gonna need a poncho for all the sarcasm dripping off that statement.
            – Ken Goss
            20 mins ago















          up vote
          4
          down vote



          accepted










          This $(k,n)$ scheme works, but isn't very interesting.



          Effectively, it is:



          • For each set of $k$ participants out of $n$, construct a $(k,k)$ threshold scheme, and distribute those shares to the participants in the set.

          For example, in a $(2, 3)$ scheme, if $z$ is the secret, we'd generate $binom32 = 3$ indepedent $(2,2)$ threshold schemes $(r_1, z-r_1 bmod 1), (r_2, z-r_2 bmod 1), (r_3, z-r_3 bmod 1)$, and distribute to the three share holders the shares:



          $$(r_1, z-r_2 bmod 1)$$
          $$(r_2, z-r_3 bmod 1)$$
          $$(r_3, z-r_1 bmod 1)$$



          It works, as:



          • For any set of $k$ share holders, they can reconstruct the secret, as there is a $(k,k)$ threshold scheme with those share holders with all the shares. In the above example, the first two share holders jointly know both $z-r_2 bmod 1$ and $r_2$, allowing them to reconstruct $z$.


          • For any set of $k-1$ share holders, they learn nothing of the secret; for any of the $(k,k)$ threshold schemes, there will always be at least 1 missing share, and so they can learn nothing.


          It's quite straight-forward; the biggest issue is that this requires distributing $binomn-1k-1$ independent values to each share holder; this is rather large if you're trying to implement a $(500,1000)$ access structure.






          share|improve this answer




















          • thank you so much for your in-depth answer; really appreciate the clarity of your example along with the generalization of the solution. Really makes me wish I was also in a position to be able to answer questions on Stack Exchange - hopefully, I will get there one day!
            – Rahul Mathur
            2 hours ago










          • (500,1000) is "rather large" he says... 1.35144120472718284758e+299 to be approximate. I wish we had an understatement badge. Either that or I'm gonna need a poncho for all the sarcasm dripping off that statement.
            – Ken Goss
            20 mins ago













          up vote
          4
          down vote



          accepted







          up vote
          4
          down vote



          accepted






          This $(k,n)$ scheme works, but isn't very interesting.



          Effectively, it is:



          • For each set of $k$ participants out of $n$, construct a $(k,k)$ threshold scheme, and distribute those shares to the participants in the set.

          For example, in a $(2, 3)$ scheme, if $z$ is the secret, we'd generate $binom32 = 3$ indepedent $(2,2)$ threshold schemes $(r_1, z-r_1 bmod 1), (r_2, z-r_2 bmod 1), (r_3, z-r_3 bmod 1)$, and distribute to the three share holders the shares:



          $$(r_1, z-r_2 bmod 1)$$
          $$(r_2, z-r_3 bmod 1)$$
          $$(r_3, z-r_1 bmod 1)$$



          It works, as:



          • For any set of $k$ share holders, they can reconstruct the secret, as there is a $(k,k)$ threshold scheme with those share holders with all the shares. In the above example, the first two share holders jointly know both $z-r_2 bmod 1$ and $r_2$, allowing them to reconstruct $z$.


          • For any set of $k-1$ share holders, they learn nothing of the secret; for any of the $(k,k)$ threshold schemes, there will always be at least 1 missing share, and so they can learn nothing.


          It's quite straight-forward; the biggest issue is that this requires distributing $binomn-1k-1$ independent values to each share holder; this is rather large if you're trying to implement a $(500,1000)$ access structure.






          share|improve this answer












          This $(k,n)$ scheme works, but isn't very interesting.



          Effectively, it is:



          • For each set of $k$ participants out of $n$, construct a $(k,k)$ threshold scheme, and distribute those shares to the participants in the set.

          For example, in a $(2, 3)$ scheme, if $z$ is the secret, we'd generate $binom32 = 3$ indepedent $(2,2)$ threshold schemes $(r_1, z-r_1 bmod 1), (r_2, z-r_2 bmod 1), (r_3, z-r_3 bmod 1)$, and distribute to the three share holders the shares:



          $$(r_1, z-r_2 bmod 1)$$
          $$(r_2, z-r_3 bmod 1)$$
          $$(r_3, z-r_1 bmod 1)$$



          It works, as:



          • For any set of $k$ share holders, they can reconstruct the secret, as there is a $(k,k)$ threshold scheme with those share holders with all the shares. In the above example, the first two share holders jointly know both $z-r_2 bmod 1$ and $r_2$, allowing them to reconstruct $z$.


          • For any set of $k-1$ share holders, they learn nothing of the secret; for any of the $(k,k)$ threshold schemes, there will always be at least 1 missing share, and so they can learn nothing.


          It's quite straight-forward; the biggest issue is that this requires distributing $binomn-1k-1$ independent values to each share holder; this is rather large if you're trying to implement a $(500,1000)$ access structure.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 2 hours ago









          poncho

          87.1k2130220




          87.1k2130220











          • thank you so much for your in-depth answer; really appreciate the clarity of your example along with the generalization of the solution. Really makes me wish I was also in a position to be able to answer questions on Stack Exchange - hopefully, I will get there one day!
            – Rahul Mathur
            2 hours ago










          • (500,1000) is "rather large" he says... 1.35144120472718284758e+299 to be approximate. I wish we had an understatement badge. Either that or I'm gonna need a poncho for all the sarcasm dripping off that statement.
            – Ken Goss
            20 mins ago

















          • thank you so much for your in-depth answer; really appreciate the clarity of your example along with the generalization of the solution. Really makes me wish I was also in a position to be able to answer questions on Stack Exchange - hopefully, I will get there one day!
            – Rahul Mathur
            2 hours ago










          • (500,1000) is "rather large" he says... 1.35144120472718284758e+299 to be approximate. I wish we had an understatement badge. Either that or I'm gonna need a poncho for all the sarcasm dripping off that statement.
            – Ken Goss
            20 mins ago
















          thank you so much for your in-depth answer; really appreciate the clarity of your example along with the generalization of the solution. Really makes me wish I was also in a position to be able to answer questions on Stack Exchange - hopefully, I will get there one day!
          – Rahul Mathur
          2 hours ago




          thank you so much for your in-depth answer; really appreciate the clarity of your example along with the generalization of the solution. Really makes me wish I was also in a position to be able to answer questions on Stack Exchange - hopefully, I will get there one day!
          – Rahul Mathur
          2 hours ago












          (500,1000) is "rather large" he says... 1.35144120472718284758e+299 to be approximate. I wish we had an understatement badge. Either that or I'm gonna need a poncho for all the sarcasm dripping off that statement.
          – Ken Goss
          20 mins ago





          (500,1000) is "rather large" he says... 1.35144120472718284758e+299 to be approximate. I wish we had an understatement badge. Either that or I'm gonna need a poncho for all the sarcasm dripping off that statement.
          – Ken Goss
          20 mins ago











          Rahul Mathur is a new contributor. Be nice, and check out our Code of Conduct.









           

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