Time-Storage Trade-Offs for Cryptographically-Enforced Access - - PowerPoint PPT Presentation

Time-Storage Trade-Offs for Cryptographically-Enforced Access Control

Jason Crampton

Information Security Group Royal Holloway, University of London

ESORICS 2011

“Traditional” Access Control

SECRET

?

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“Traditional” Access Control

SECRET

?

SECRET

Allow

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“Traditional” Access Control

SECRET

?

SECRET

Deny

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“Traditional” Access Control

SECRET

?

SECRET Policy Enforcement Policy

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Cryptographically-Enforced Access Control

Allow

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Cryptographically-Enforced Access Control

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Graph-Based Authorization Policies

top secret secret classified unclassified

SECRET

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Graph-Based Authorization Policies

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A Generic Single-Key Enforcement Mechanism

◮ For every y that is reachable from x,

encrypt κ(y) using κ(x)

❞ a

Encκ(b)(κ(a))

❞ b

Encκ(c)(κ(b))

❞ c

Encκ(d)(κ(c))

❞ d

Time-Storage Trade-Offs/Crampton/ESORICS 2011 5/20

A Generic Single-Key Enforcement Mechanism

◮ For every y that is reachable from x,

encrypt κ(y) using κ(x)

◮ Alternatively, for every edge (x, y), encrypt

κ(y) using κ(x)

❞ a

Encκ(b)(κ(a))

❞ b

Encκ(c)(κ(b))

❞ c

Encκ(d)(κ(c))

❞ d

Time-Storage Trade-Offs/Crampton/ESORICS 2011 5/20

A Generic Single-Key Enforcement Mechanism

◮ For every y that is reachable from x,

encrypt κ(y) using κ(x)

◮ Alternatively, for every edge (x, y), encrypt

κ(y) using κ(x)

◮ Clearly, there are trade-offs between the

number of keys that need to be encrypted and the number of key derivation

• perations performed by a user

❞ a

Encκ(b)(κ(a))

❞ b

Encκ(c)(κ(b))

❞ c

Encκ(d)(κ(c))

❞ d

Time-Storage Trade-Offs/Crampton/ESORICS 2011 5/20

Space-Time Trade-Offs

◮ Let (x, y) ∈ Eenf if and only if κ(y) is encrypted using κ(x) ◮ Given an authorization graph Gauth = (V , Eauth), we say

Eenf ⊆ V × V is policy-enforcing if and only if E ∗

auth = E ∗ enf

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Space-Time Trade-Offs

◮ Let (x, y) ∈ Eenf if and only if κ(y) is encrypted using κ(x) ◮ Given an authorization graph Gauth = (V , Eauth), we say

Eenf ⊆ V × V is policy-enforcing if and only if E ∗

auth = E ∗ enf

|Eauth| = 12; diameter = 3

Time-Storage Trade-Offs/Crampton/ESORICS 2011 6/20

Space-Time Trade-Offs

◮ Let (x, y) ∈ Eenf if and only if κ(y) is encrypted using κ(x) ◮ Given an authorization graph Gauth = (V , Eauth), we say

Eenf ⊆ V × V is policy-enforcing if and only if E ∗

auth = E ∗ enf

|Eauth| = 12; diameter = 3 |Eenf | = 25; diameter = 1

Time-Storage Trade-Offs/Crampton/ESORICS 2011 6/20

Two Interesting Graph-Based Policies

Dn

◮ The diamond represents the natural authorization policy for scalable

multi-media formats such as MPEG-4 and JPEG2000 (Crampton et al.)

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Two Interesting Graph-Based Policies

Dn Tn

◮ The diamond represents the natural authorization policy for scalable

multi-media formats such as MPEG-4 and JPEG2000 (Crampton et al.)

◮ The triangle represents a common policy for temporal access control

(Atallah et al., De Santis et al.)

Time-Storage Trade-Offs/Crampton/ESORICS 2011 7/20

Two Interesting Graph-Based Policies

Dn Tn

◮ The diamond represents the natural authorization policy for scalable

multi-media formats such as MPEG-4 and JPEG2000 (Crampton et al.)

◮ The triangle represents a common policy for temporal access control

(Atallah et al., De Santis et al.)

Problem: What trade-offs are possible for Dn and Tn?

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The Master Theorem

Theorem

Given an enforcing set of edges for Da and Db, we can construct an enforcing set of edges for Dab such that e♦(ab) = a2e♦(b) + b2e♦(a) and d♦(ab) = d♦(a) + d♦(b)

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The Master Theorem

Theorem

Given an enforcing set of edges for Da and Db, we can construct an enforcing set of edges for Dab such that e♦(ab) = a2e♦(b) + b2e♦(a) and d♦(ab) = d♦(a) + d♦(b)

Corollary

Given an enforcing set of edges for Da, we can construct an enforcing set of edges for D2a such that e♦(2a) = 4a2 + 4e♦(a) and d♦(2a) = 2 + d♦(a)

Time-Storage Trade-Offs/Crampton/ESORICS 2011 8/20

The Master Theorem

Theorem

Given an enforcing set of edges for Da and Db, we can construct an enforcing set of edges for Dab such that e♦(ab) = a2e♦(b) + b2e♦(a) and d♦(ab) = d♦(a) + d♦(b)

Corollary

Given an enforcing set of edges for Da, we can construct an enforcing set of edges for D2a such that e♦(2a) = 4a2 + 4e♦(a) and d♦(2a) = 2 + d♦(a)

Corollary

Given an enforcing set of edges for Da, we can construct an enforcing set of edges for Da2 such that e♦(a2) = 2a2e♦(a) and d♦(a2) = 2d♦(a)

Time-Storage Trade-Offs/Crampton/ESORICS 2011 8/20

Sketch Proof: Number of Edges

Db

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Sketch Proof: Number of Edges

Db

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Sketch Proof: Number of Edges

Db

e♦(ab) = b2e♦(a) +

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Sketch Proof: Number of Edges

Db

e♦(ab) = b2e♦(a) + a2e♦(b)

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Sketch Proof: Key Derivation

Db

d♦(ab) =

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Sketch Proof: Key Derivation

Db

d♦(ab) = d♦(a) +

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Sketch Proof: Key Derivation

Db

d♦(ab) = d♦(a) + d♦(b)

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Decomposition of Dn into Chains

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Decomposition of Dn into Chains

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Decomposition of D4 into 2-Hop Chains

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Decomposition of D4 into 2-Hop Chains

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Constructions for D4

e♦ = 24 e♦ = 32 e♦ = 36 e♦ = 40 d♦ = 6 d♦ = 4 d♦ = 3 d♦ = 2

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Constructions for Dn

◮ Using constructions for D4 as the base case, we can obtain

constructions for D16, D256,. . . ,D22k e♦(n) d♦(n)

3 4n2 log n

6 log log n n2 log n 4 log log n

9 8n2 log n

3 log log n

5 4n2 log n

2 log log n

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Constructions for Tn

Theorem

Given a set of enforcing edges for Ta, Tb, Da and Db such that d△(a) < d♦(a) and d△(b) < d♦(b), there exists a set of enforcing edges for Tab such that e△(ab) < b2e△(a) + 1 2a(a + 1)e♦(b) and d△(ab) d♦(a) + d♦(b) − 2

Theorem

Given a set of enforcing edges for Ta and Tb there exists a set of enforcing edges for Tab such that e△(ab) = a2e△(b) + b2e△(a) and d△(ab) = max {d△(a) + d△(b), d△(a) + 2d△(b) − 1}

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Proof Intuition (1)

Db

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Proof Intuition (1)

Db

e△(ab) = b2e△(a)

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Proof Intuition (1)

Db

e△(ab) = b2e△(a) + 1 2a(a + 1)e♦(b)

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Proof Intuition (2)

Db

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Proof Intuition (2)

Db

e△(ab) = b2e△(a) + ae△(b) + 1 2a(a − 1)e♦(b)

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Proof Intuition (2)

Db

e△(ab) = b2e△(a) + ae△(b) + a(a − 1)e△(b)

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Proof Intuition (2)

Db

e△(ab) = b2e△(a) + a2e△(b)

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Related Work

e♦(m) d♦(m) 2n2 log n 4 n2 log n 4 log log n n2 log n 2 log n 4n(n − 1) 4(√n − 1) (a) Our results for Dm e△(n) d△(n) dSFB O

• n2 log n log log n
• 3

ABF O

• n2 log n
• 4

ABF O

• n2

O (log∗ n) dSFB O

• n2 log n
• O (log∗ n)

dSFB O

• n2

O (log n log∗ n) (b) Existing results for Tm

Time-Storage Trade-Offs/Crampton/ESORICS 2011 18/20

Potential Future Work

◮ Investigate the interplay between schemes for triangles and diamonds

further

◮ Investigate whether existing schemes for chains can be leveraged to

improve the trade-offs

◮ Investigate whether additional trade-offs are possible if users are given

multiple keys

◮ What about pyramids?

Time-Storage Trade-Offs/Crampton/ESORICS 2011 19/20

References

M.J. Atallah, M. Blanton, and K.B. Frikken. Incorporating temporal capabilities in existing key management schemes. In Proceedings of ESORICS 2007.

• J. Crampton.

Practical constructions for the efficient cryptographic enforcement of interval-based access control policies. ACM Transactions on Information and System Security, 2011.

• J. Crampton, R. Daud, and K.M. Martin.

Constructing key assignment schemes from chain partitions. Proceedings of Data and Applications Security XXIV, 2010.

• A. De Santis, A.L. Ferrara, and B. Masucci.

New constructions for provably-secure time-bound hierarchical key assignment schemes. Theoretical Computer Science, 2008.

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