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On the Correctness Criteria of Fine-Grained Access Control in Relational Databases Qihua Wang, Ting Yu, Ninghui Li Jorge Lobo, Elisa Bertino Keith Irwin, Ji-Won Byun Outline Introduction Correctness Criteria A Fine-Grained Access


  1. On the Correctness Criteria of Fine-Grained Access Control in Relational Databases Qihua Wang, Ting Yu, Ninghui Li Jorge Lobo, Elisa Bertino Keith Irwin, Ji-Won Byun

  2. Outline � Introduction � Correctness Criteria � A Fine-Grained Access Control Solution � Implementation and Experiments � Conclusions

  3. Introduction � What is fine-grained access control? � Row-level or cell-level access control � In contrast to table-level � Why fine-grained access control? � Privacy: access respects individual preferences � How to implement? Applications � Application-level � Database-level Database � Hard to bypass � Consistency between various applications

  4. Introduction � Existing DB-Level approaches � VPD in Oracle � Label-based access control in DB2 � Limiting disclosure in Hippocratic databases � Fine-grained access control affects query results � No formal notion of correctness � Could lead to incorrect or misleading query results

  5. Example ID Name Age Phone C001 Linda 32 11111 C002 Mary 29 22222 C003 Nick NULL 34 33333 C004 Jack 21 44444 C005 Mary 30 55555 NULL � Q 1 = SELECT Name, Phone FROM T � Q 2 = SELECT Name, Phone FROM T WHERE Age ≥ 25 � Q = Q 1 – Q 2 � Select information of customers younger than 25

  6. Example � Q 1 = SELECT Name, Phone FROM T Name Phone Linda 11111 Mary 22222 Nick 33333 Jack 44444 Mary NULL

  7. Example ID Name Age Phone C001 Linda 32 11111 C002 Mary 29 22222 C003 Nick NULL 33333 C004 Jack 21 44444 C005 Mary 30 NULL � Q 2 = SELECT Name, Phone FROM T WHERE Age ≥ 25 Name Phone Linda 11111 Mary 22222 Mary NULL

  8. Example � Q = Q 1 – Q 2 Name Phone Linda 11111 Name Phone Name Phone _ Mary 22222 Linda 11111 = Nick 33333 Nick 33333 Mary 22222 Jack 44444 Jack 44444 Mary NULL Mary NULL

  9. Example ID Name Age Phone C001 Linda 32 11111 C002 Mary 29 22222 C003 Nick 34 33333 C004 Jack 21 44444 C005 Mary 30 55555 � Q 1 = SELECT Name, Phone FROM T � Q 2 = SELECT Name, Phone FROM T WHERE Age ≥ 25 � Q = Q 1 – Q 2 � Select information of customers younger than 25

  10. Example � Without fine-grained access control Name Phone Jack 44444 � With fine-grained access control Name Phone Nick 33333 Jack 44444

  11. Outline � Introduction � Correctness Criteria � A Solution � Implementation and Experiments � Conclusions

  12. Intuitive Explanation � Sound � Be consistent with when there is no access control � Secure � Do not leak information not allowed by policy � Maximum � Return as much correct information as allowed by policy

  13. Formal Definitions � D : Database � P : Disclosure policy � Determine what information may be disclosed � Defines an equivalence relation among database states � D ≡ P D ’ Name Age Phone Name Age Phone ≡ P Alice 25 111 Alice 33 111 Bob 30 888 Bob 30 666

  14. Formal Definitions � R : Relation � A cell may take the value unauthorized � A tuple is subsumed by another: t 1 t 2 � < x 1 … x n > < y 1 … y n > if and only if x i = y i or x i = unauthorized � E.g. <Alice, unauthorized> <Alice, 28> � A relation is subsumed by another: R1 R2 � Exists a mapping f : R 1 R 2 � For every tuple t in R 1 , t f(t)

  15. Formal Definitions � R : Relation � Q : Query � A : Query processing algorithm that takes disclosure policy into account � A(D,P,Q) : Answer to Q on D with policy P � S : Standard query processing algorithm � S(D,Q) : Answer to Q on D without access control

  16. Sound � May return less information due to access control � Should not return wrong information that is not in standard answer Name Phone Name Phone Nick NULL Jack 44444 Jack 44444

  17. Secure � Answer does not depend on information that is not disclosed by policy � Implies stronger security guarantee � Multi-user collusion resistance � Multi-query resistance

  18. Maximum Given any ( D, P, Q ), for any relation R such that We have � No other sound and secure answer that contains more information than the answer returned by A

  19. Correctness Criteria � Any query processing algorithm that provides fine-grained access control should be sound and secure, and strive to be maximum. � Many existing approaches are � Secure � Not sound � Not maximum � Too little information is returned in certain cases

  20. Outline � Introduction � Correctness Criteria � A Solution � Implementation and Experiments � Conclusions

  21. Solution � A sound query evaluation algorithm � Low evaluation Q – : tuples definitely correct � High evaluation Q – : tuples possibly correct – � Q 1 – Q 2 is evaluated as Q 1– – Q 2 � A variable-based labeling mechanism � Use variables instead of NULL to hide information � Secure � Preserves more information

  22. Variable-Based Labeling Mechanism � Existing approaches: replace every piece of unauthorized information with NULL � Too much information is lost � Unknown: NULL = 100?, NULL = NULL? Name Age NULL 25 Alice Q = SELECT Name FROM T WHERE Age = Age Result is an EMPTY relation!

  23. Variable-Based Labeling Mechanism � Information useful in query evaluation without leaking concrete value � A cell equals to itself � Cells in primary key take different values � Certain linkages through foreign key � Information of the same person stored in two tables so as to comply with normal forms � Our approach: replace unauthorized information with variables

  24. Two Types of Variables � Type-1 variable: v � Variable is equivalent to itself � True: v 1 = v 1 , v 2 = v 2 (in contrast to NULL ≠ NULL) � Unknown when compared with other variables or constants � Unknown: v 1 = v 2 ?, v 1 = 100? � Type-2 variable: < name , domain > � In the same domain, compare names � True: < a , 1> = < a , 1>, < a , 1> ≠ < b , 1> � Otherwise, unknown � Unknown: < a , 1> = < a , 2>?, < a , 1> ≠ < b , 2>? � Unknown: < a , 1> = v 1 ?, < a , 1> = 100?

  25. Example Based tables SSN Occupation 1111 Student SSN Name Age 1111 Waiter 1111 Alice 19 2222 Professor 2222 Bob 35 3333 Secretary 3333 Carol 19 3333 Dancer Our approach Traditional labeling approach SSN SSN Occupation Occupation SSN Name Age SSN Name Age NULL < a ,1> Student Student < a ,1> NULL Waiter Waiter < a ,1> Alice v 1 NULL Alice NULL < b ,1> NULL Professor Professor < b ,1> Bob 35 NULL Bob 35 NULL < c ,1> Secretary Secretary < c ,1> Carol v 2 NULL Carol NULL < c ,1> NULL Dancer Dancer

  26. Variable-Based Labeling Mechanism � Provides security � Variables hide concrete values � Makes it possible to return more information � Strive for maximum � Does not deal with sound

  27. A Sound Query Evaluation Algorithm � Low evaluation: Q - � Contains tuples that are definitely correct � High evaluation: Q - � Contains tuples that are possibly correct � Tuples < x 1 ,… x n > and < y 1 ,… y n > are compatible if it is possible make to them identical by setting the values of variables � Different type-2 variables in the same domain must have different values

  28. A Sound Query Evaluation Algorithm � Q = R : Q – = Q – = L(R) � Q = σ c Q 1 : Q – = σ c Q 1 – and Q – = σ c V IsUn(c) Q 1– � Q = π a1 … Q 1 : Q – = π a1 … Q 1 – and Q – = π a1 … Q 1 – � Q = Q 1 × Q 2 : Q – = Q 1 – × Q 2 – and Q – = Q 1 – × Q 2 – � Q = Q 1 U Q 2 : Q – = Q 1 – U Q 2 – and Q - = Q 1 – U Q 2 – � Q = Q 1 – Q 2 – � Q – contains all tuples t in Q 1 – such that no tuple in Q 2 is compatible with t – � Intuitively, Q – = Q 1 – – Q 2 � Q – contains all tuples that are in Q 1 – but not in Q 2 – � Intuitively, Q – = Q 1 – – Q 2 –

  29. A Sound and Secure Solution Given any query Q � Perform variable-based labeling 1. Compute and return Q – 2. Sound and secure � Returns at least as much information as existing � algorithms for fine-grained access control

  30. Example ID Name Age Phone C001 Linda 32 11111 C002 Mary 29 22222 v 1 C003 Nick 34 33333 C004 Jack 21 44444 C005 Mary 30 55555 v 3 � Q 1 = SELECT Name, Phone FROM T � Q 2 = SELECT Name, Phone FROM T WHERE Age ≥ 25 � Q 3 = SELECT Name, Phone FROM T WHERE Age < 30 � Q = Q 1 – (Q 2 – Q 3 ) � Select information of customers younger than 30

  31. Example � Given Q = Q 1 – (Q 2 – Q 3 ), compute Q – � Compute Q 1– � Compute (Q 2 – Q 3 ) – – and Q 3 – � Compute Q 2

  32. Example � Q 1 = SELECT Name, Phone FROM T � Q 1– : Name Phone Linda 11111 Mary 22222 Nick 33333 Jack 44444 v 3 Mary

  33. Example ID Name Age Phone C001 Linda 32 11111 C002 Mary 29 22222 C003 Nick v 1 33333 C004 Jack 21 44444 C005 Mary 30 v 3 � Q 2 = SELECT Name, Phone FROM T WHERE Age ≥ 25 – : � Q 2 Name Phone Linda 11111 Mary 22222 Nick 33333 Mary v 3

  34. Example ID Name Age Phone C001 Linda 32 11111 C002 Mary 29 22222 C003 Nick v 1 33333 C004 Jack 21 44444 C005 Mary 30 v 3 � Q 3 = SELECT Name, Phone FROM T WHERE Age < 30 � Q 3– : Name Phone Mary 22222 Jack 44444

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