Attacker who has access to all secrets of n or less

Info icon This preview shows pages 54–56. Sign up to view the full content.

View Full Document Right Arrow Icon
attacker (who has access to all secrets of n or less nodes) cannot determine any illegitimate shared secret. However, with access to secrets of more than n nodes, the attacker can discover all pairwise secrets. For an ( n, p )-secure probabilistic KPS, an attacker who has access to all secrets of n randomly chosen nodes can compute only a fraction p of all illegitimate pairwise secrets. Alternately, an attacker with access to secrets of n nodes can discover any illegitimate pairwise secret with probability p . Note that as long as p is low (say 2 64 ), it is computationally infeasible for the attacker to even determine which illegitimate pairwise secret can be exposed using the secrets pooled from n nodes. For such probabilistic KPS 0 p ( n ) 1 is a monotonic (and increasing) function of n . Deterministic n -secure KPSs can actually be considered as special cases of ( n, p )-secure KPSs where p takes only binary values (0 or 1). Mathematically p ( x ) = 0 x n = 1 x > n Deterministic KPSs 0 x = 0 p ( x ) x < x 1 x → ∞ Probabilistic KPSs (4) Copyright © 2010. World Scientific Publishing Company. All rights reserved. May not be reproduced in any form without permission from the publisher, except fair uses permitted under U.S. or applicable copyright law. EBSCO Publishing : eBook Collection (EBSCOhost) - printed on 2/16/2016 3:46 AM via CGC-GROUP OF COLLEGES (GHARUAN) AN: 340572 ; Beyah, Raheem, Corbett, Cherita, McNair, Janise.; Security in Ad Hoc and Sensor Networks Account: ns224671
Image of page 54

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
42 M. Ramkumar Thus, while deterministic KPSs fail in a catastrophic manner, probabilistic KPSs fail gracefully. In this chapter we restrict ourselves to one representative scheme from each class - Blom’s deterministic KPS 10 and the ( n, p )-secure key subset and symmetric certificates (KSSC) 12 scheme. 3.2.2. Blom’s Deterministic Scheme The symmetric key generation scheme (SKGS) proposed by Blom 10 is based on maximum distance separation (MDS) codes over a finite-field - for ex- ample, the finite-field formed by the set of integers Z q = { 0 , 1 , . . ., q 1 } (where q is a prime). The KDC chooses (1) a public primitive element α Z q ; (2) ( n + 1) × ( n + 1) symmetric matrix D with ( n +1 2 ) independent values (secrets) chosen randomly from Z q . Corresponding to a node with ID A Z q the KDC computes g A = [ g A (0) , g A (1) , . . . , g A ( n )] T , g A ( i ) = α i × A . S A = Dg A = d A = [ s A 0 , s A 1 , . . . s A n ] T (5) The k = n + 1 secrets S A are then provided to A over a secure channel. Nodes A and B (with secrets d A = [ s A 0 · · · s A n ] T and d B = [ s B 0 · · · s B n ] T respectively) can compute K AB = ( d A ) T g B = ( d B ) T g A (as D is a sym- metric matrix). In other words, A and B compute K AB as K AB = n i =0 s A i α iB mod q by A n i =0 s B i α iA mod q by B (6) An n -secure SKGS is unconditionally secure as long as n or less nodes pool their secrets together. However, an attacker who has access to secrets assigned to more than n nodes can solve for all P = ( n +1 2 ) secrets chosen by the KDC, and thus compute any pairwise secret.
Image of page 55
Image of page 56
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern