Inconsistency resolving of safety and utility in access control

Policy inconsistencies may arise between safety and utility policies due to their opposite objectives. In this work we provide a formal examination of policy inconsistencies resolution for the coexistence of static separation-of-duty (SSoD) policies and strict availability (SA) policies. Firstly, we reduce the complexity of reasoning about policy inconsistencies by static pruning technique and minimal inconsistency cover set. Secondly, we present a systematic methodology for measuring safety loss and utility loss, and evaluate the safety-utility tradeoff for each choice. Thirdly, we present two prioritized-based resolutions to deal with policy inconsistencies based on safety-utility tradeoff. Finally, experiments show the effectiveness and efficiency of our approach.

The safety and utility policies are very important in an access control system for ensuring security and availability when performing a certain task. Safety policies are used to describe safety requirements which ensure that users who should not have access do not get access. Such focus on safety requirements probably stems from the fact that safety policies have been mostly viewed as a tool for restricting access. An example of the safety policy is a static separation-of-duty (SSoD) policy, which precludes any group of users from possessing too many permissions [1]. An equally important aspect of access control is the utility policies that enables access [2, 3]. In our previous work [4], we have introduced the notion of availability policies which is an example of an utility policy. In this paper, we introduce the notion of strict availability (SA) policies, which is also an example of utility policy that requires that the cooperation among at most a certain number of users is necessary to perform a task. Due to their opposite objectives, safety policies and utility policies can conflict with each other. For example, let p₁ and p₂ be two permissions, and u₁ and u₂ two users. Assume that an SSoD policy requires that neither u₁ nor u₂ possess all permissions in {p₁, p₂}. An SA policy requires both u₁ and u₂ possess all permissions in {p₁, p₂}. Clearly, the two policies cannot be satisfied simultaneously.

This paper examines this kind of conflict: policy inconsistencies that result from the incompatibility between safety policies and utility policies, especially for the coexistence of SSoD policies and SA policies. Policy inconsistencies differ from the traditional policy conflicts [5] in that the composition of safety and utility policies is never supposed to be inconsistent. That means policy inconsistencies are checked at compile-time to prevent the construction of any safety or utility policy that may conflict with each other. A policy inconsistency results in a policy compilation error. Hence, the resolution for policy inconsistencies is a policy design problem, whereas policy conflicts are resolved at runtime. In practice, the policy administrator may define many safety and utility policies and these policies may be inconsistent. However, it is not easy to detect and resolve these policy inconsistencies. Thus, it is very important to help the policy administrator to detect and resolve the policy inconsistencies at compile-time. The above discussion motivates the problem considered in this paper.

In our previous work [4], we have addressed the problem of consistency checking for the coexistence of safety and utility policies [4]. In this paper, we aim for providing a formal examination of policy inconsistency resolution for safety and utility policies, which can help the policy administrators to specify reasonable access control policies when both safety and utility policies coexist. Our contributions are as follows:

• We formally define the policy inconsistency for the coexistence of safety policies and utility policies.

• We describe a static pruning technique that aims to reduce the number of policies that need to be taken into account.

• We compute the minimal inconsistency cover set that is responsible for the policy inconsistencies; thus we only need to examine the minimum number of policies.

• We present a systematic methodology for measuring safety loss and utility loss, and evaluate the safety-utility tradeoff for each candidate resolution.

• We present two prioritized-based resolutions to deal with policy inconsistencies for safety and utility policies based on safety-utility tradeoff.

The remainder of this paper is organized as follows. Section 2 formally defines the policy inconsistency problem for the coexistence of safety policies and utility policies. Section 3 presents prioritized-based resolutions for policy inconsistencies. The evaluation and illustration of our approaches are given in Section 4. Section 5 discusses related work, and Section 6 concludes and discusses the future work.

We assume that there are two countably infinite sets in an access control state: U (the set of all possible users), and P (the set of all possible permissions). An access control state ε is a binary relation UP ⊆ U × P, which determines the set of permissions a user possesses. Note that by assuming that an access control state ε is given by a binary relation UP ⊆ U × P, we are not assuming permissions are directly assigned to users; rather, we assume only that one can calculate the relation UP from the access control state.

Safety policies are used to describe safety requirements which ensure that users who should not have access do not get access. A safety policy is specified by giving a predicate on sets of executions. If conditions on (users, permissions) are satisfied, then a set U of users are prohibited from covering a set P of permissions. One example of a safety policy is an static separation-of-duty (SSoD) policy. SSoD policy is considered as a fundamental principle of information security that has been widely used in business, industry, and government applications [6]. An SSoD policy typically constrains the assignment of permissions to users, which precludes any group of users from possessing too many permissions. We first reproduce the definitions of SSoD policies from [4].

Definition 1. An SSoD policy ensures that at least k users from a user set are required to perform a task that requires all these permissions. It is formally defined as

• P and U denote the set of permissions and the set of users, respectively.

• UP ⊆ U × P, is a user-permission assignment relation.

• auth_p _ε (u) = {p|(p ∈ P ) ⋀ ((u, p) ∈ UP)}.

• ∀(P, U, k) ∈ SSoD, ∀U' ⊆ U : |U' | < k ⇒ ∪_u∈U' auth_p _ε (u) ⊉ P.

where P = {p₁, ..., p _m }, U = {u₁, ..., u _n }, each p _i in P is a permission, u _j in U is a user, and m, n, and k are integers, such that 2 ≤ k ≤ min(m, n), where min returns the smaller value of the two. We write an SSoD policy as ssod <P, U, k>. An access control state ε satisfies an SSoD policy e = ssod <P, U, k>, which is denoted by sat _e (ε). And sat _E (ε) represents ε satisfies a set E of SSoD policies.

A utility policy is also specified by giving a predicate on sets of executions. If conditions on (users, permissions) are satisfied, then a set U of users are obligated to possess all the permissions in P. We now introduce the notion of strict availability (SA) policies, which is an example of utility policies that states properties about enabling access in access control. An SA policy requires that the cooperation among at most a certain number of users is necessary to perform a task.

Definition 2. An strict availability (SA) policy ensures that all size-t subsets of U are required to complete a task that requires all these permissions in P. It is formally defined as

• P and U denote the set of permissions and the set of users, respectively.

• UP ⊆ U × P, is a user-permission assignment relation.

• auth_p _ε (u) = {p|(p ∈ P) ⋀ ((u, p) ∈ UP)}.

• ∀(P, U, t) ∈ SA, ∀U' ⊆ U : |U' | = t ⇒ ∪_u∈U' auth_p _ε (u) ⊇ P.

Where P = {p₁, ..., p _m }, U = {u₁, ..., u _n }, each p _i in P is a permission, u _j in U is a user, and m, n, and t are integers, such that 1 ≤ t ≤ min(m, n), where min returns the smaller value of the two, and the variable t in size-t is used to represent the cardinality of a set. We write an SA policy as sa <P, U, t>. An access control state ε satisfies an SA policy f = sa <P, U, t>, which is denoted by sat _f (ε). And sat _F (ε) represents ε satisfies a set F of SA policies.

Definition 3. UCP (the Utility Checking Problem) is defined as follows: Given an access control state ε and a set F of SA policies, determining whether sat _F (ε) is true.

Theorem 1. UCP is in P.

PROOF. Given an access control state ε and a set F of SA policies, if for each SA policy f = sa(P, U, t) in F that sat _f (ε) is true, then sat _F (ε) is true. In the following, we prove that sat _f (ε) is true if and only if each permission p ∈ P is assigned to no less than (|U| + 1 - t) users in the user set U, where |U| represents the cardinality of set U.

For the "only if" part, sat _f (ε) being true means that the users in each size-t subsets of U together possess all the permissions in P. Suppose, for the sake of contradiction, that sat _f (ε) is true, and there exists a permission p ∈ P that is only assigned to (|U| - t) users in U. Then we can find a user set U' where |U| = t, and each users in U' do not possess p. Thus sat _f (ε) is false, and this contradicts the assumption; therefore, each permission p ∈ P must be assigned to no less than (|U| + 1 - t) users in U.

For the "if" part, if each permission p ∈ P is assigned to no less than (|U|+ 1 - t) users in U, then the users in each size-t user set U' will together possess p. Thus all the permissions in p will be covered by each size-t user set. In other word, the users in each size-t user set together are authorized for all permissions in P. Therefore, sat _f (ε) is true.

Together with the above discussions, we now give a linear-time algorithm for determining whether sat _F (ε) is true: For each SA policy sa <P, U, t> in F, and for each permission p ∈ P. One first computes the set of all users the permission p is a member of, and compares this number with (|U|+1-t). This algorithm has a time complexity of O(N _U N _P M), where N _U is the number of users in U, N _P the number of permission in P, and M is the number of SA policies.   □

An availability policy ap<P, U, t> ensures that there exists a size-t subset of U that the users in this subset are required to possess all these permissions in P[4]. We now show that sa<P, U, t> is at least as restrictive as ap<P, U, t>.

Definition 4. Let P₁and P₂be two policies. We say that P₁is at least as restrictive as P₂(denoted by P₁ ≽ P₂) if\forall \epsilon \left(sa{t}_{P_1}\left(\epsilon \right)\Rightarrow sa{t}_{P_2}\left(\epsilon \right)\right). When P₁ ≽ p₂but not P₂ ≽ p₁, we say that P₁is more restrictive than P₂(denoted by P₁ ≻ P₂). And when (P₁ ≽ p₂) ⋀ (P₂ ≽ p₁), we say P₁and P₂are equivalent (denoted by P₁ ≜ P₂).

By definition, the ≽ relation among all policies is a partial order. The ≻ relation among all policies is a quasi order.

Theorem 2. Given an SA policy f = sa<P, U, t>, and an availability policy g = ap<P, U, t>, f ≻ g if and only if |U| > t.

PROOF. For the "only if", We show that if f ≻ g then |U| > t. Suppose, for the sake of contradiction that |U| ≤ t. By Definition 2, t ≤ |U|, then |U| = t. For any access control state ε, if sat _g (ε) is true, then (∃U' ⊆ U) ⋀ (|U' | = t)[∪_u∈U'auth_p _ε (u) ⊇ P], and U' = U as |U| = t. Then ∃U' ⊆ U ⋀ |U' | = t(∪_u∈U'auth_p _ε (u) ⊇ P) has the same meaning as (∀U' ⊆ U) ⋀ (|U' | = t) (∪_u∈U'auth_p _ε (u) ⊇ P). That means P₁ ≜ P₂, which contradicts the assumption. Therefore, if f ≻ g then |U| > t.

For the "if" part, if |U| > t then f ≻ g. By Definition 2, for each access control state ε that satisfies f if and only if the users in all size-t subsets of U together possess all the permissions in P, Let U' is a subset of U, that the users in U' together possess all of the permissions in P, and |U' | = t, then ε satisfies ap<P, U, t>. Therefore, ∀ε(sat _f (ε) ⇒ sat _g (ε)), and f ≽ g. We construct a new state ε' that satisfies g but does not satisfy f as follows: assign all permissions in P to only one user u ∈ U, but do not assign any permissions in P to any other users in U. Then we can find a user set \left(U^{\prime }\subset U\right)\wedge \left(\left|U^{\prime }\right|=t\right)\left[{\bigcup }_{u\in U^{\prime }}aut{h}_{-}{p^{\prime }}_{\epsilon }\left(u\right)\supseteq P\right], and sat _g (ε') is true. However, for any user set U'' that (U'' ⊂ U) ⋀ (|U'' | = t) ⋀ (u ∉ U''), as {\bigcup }_{u\in U^{″}}aut{h}_{-}{p}_{\epsilon }^{\prime }\left(u\right)⊉P, sat _f (ε') is false. Therefore, if |U| > t, then f ≻ g.   □

Intuitively, SA policies are a natural complement to SSoD policies in access control. Neither SA nor SSoD by itself is sufficient to capture both safety and utility requirements. Without the utility requirement, an access control state can satisfy any SSoD policy if the state does not contain any user set that covers all the permissions needed to accomplish the sensitive task. Similarly, without the safety requirement, any SA policy can be satisfied by giving all permissions to all users, which allows each single user be able to accomplish any task. In many cases, it is desirable for an access control system to have both SSoD and SA policies. However, these policies may conflict with each other due to their opposite objectives. Therefore, a formal description of policy inconsistency is necessary to detect and resolve it.

Definition 5. CCP (the Consistency Checking Problem) is defined as follows: Given a set E of SSoD policies and a set F of SA policies, determining that whether there exists an access control state ε that sat _E (ε) ⋀ sat _F (ε) is true.

Corollary 1. CCP is coNP-complete.

PROOF. That CCP is coNP-complete follows directly from the fact that the problem of determining whether sat _E (ε) is true is coNP-complete (Theorem 1 in [4]), and the problem of determining whether sat _F (ε) is true is in P (Theorem 1).   □

Consider the following example of SSoD and SA policies. It is not easy to check whether the policies in the set Q is consistent.

Example 1. Consider a set Q of SSoD and SA policies as follows.

We now show that the above SSoD and SA policies are inconsistent. Given any access control state ε, if sa{t}_{f_2}\left(\epsilon \right) is true, that means p₂ and p₃ must be authorized to both u₂ and u₃. If sa{t}_{f_1}\left(\epsilon \right) is true, then p₁ must be authorized to either u₂ or u₃. If u₂ possesses p₁, u₂ will possess all of the permissions in {p₁, p₂, p₃}, which violates both e₁ and e₂. If u₃ possesses p₁, u₃ will possess all of the permissions in {p₁, p₂, p₃}, which violates both e₁. Therefore, there does not exist an access control state ε that satisfies all of the four policies in Q.

In general, there may be many policy inconsistencies in a large access control policy set. Thus the following issues should be considered: (1) A large number of policy inconsistencies are possible, but many of them may be the result of a small number of policies that apply to aggregates. The key is to figure out the minimum number of policies that are responsible for the policy inconsistencies. (2) Once all the inconsistencies are known, we must determine the appropriate resolutions with little effort to resolve them, and estimate their impact on the policies. Like traditional policy conflict resolution, the theoretical resolution of policy inconsistencies is basically the same: remove some policies in the policy set. The primary difficulty is to determine which policies should be removed, and the resolution addresses the inconsistency most effectively.

In this section, we provide a formal examination of policy inconsistencies resolution for the coexistence of SSoD and SA policies.

3.1. Reducing complexity

Once all the inconsistencies are known, we must find a way to resolve them. However, determining which policy to remove is difficult because there may be many policy inconsistencies. In order to simplify the resolution task, we consider as few policies as possible. Thus we reduce the complexity of reasoning about policy inconsistencies by the techniques of static pruning and minimal inconsistency cover set.

3.1.1. Static pruning

SSoD and SA policies can conflict with each other due to their opposite objectives. In general, not all SSoD or SA policies should be taken into account as they do not cause inconsistencies. The following theorem asserts that the special cases of SSoD(or SA) policies do not affect its compatibility with SA(or SSoD) policies. This enables us to remove them from our consideration. This greatly simplifies the problem.

Theorem 3. Let Q = {e₁, ..., e _m , f₁, ..., f _n }, where e _i = ssod <P _i , U _i , k _i > (1 ≤ i ≤ m),f_j=ap⟨P_j^{\prime },U_j^{\prime },t_j⟩\left(1\le j\le n\right). If ∃e _i ∈ Q[(|P _i - R| > 0) ⋁ (|U _i ∩ T| = 0)], whereR={\bigcup }_{j=1}^n{P}_j^{\prime },T={\bigcup }_{j=1}^n{U}_j^{\prime }, then let Q' = Q' - {e _i }; If, whereS={\bigcup }_{i=1}^m{U}_i,W={\bigcup }_{i=1}^m{P}_i, then let Q' = Q' - {f _j }. Q is consistent if and only if Q' is consistent.

PROOF. For the "only if" part, it is clear that if Q is consistent then Q' is consistent as Q' ⊆ Q.

For the "if" part, we show that if Q' is consistent then Q is consistent. Q' is consistent implies that there exists an access control state ε satisfies all policies in Q'. We now construct a new state ε' that satisfies both Q' and Q as follows: for each e _i ∈ Q/Q', where |P _i - R| > 0. Add all users in U _i to ε, but do not assign any permissions in P _i ∩ R. In this way, ε' satisfies e _i as no less than k _i users in U _i together having all permissions in P _i , and note that adding new users will not lead to inconsistency of policies in Q'. If |U _i ∩ T| = 0, not assigning any permission in P _i to any user in U _i will not lead to inconsistency of policies in Q', but the new state satisfies e _i . For each f _j ∈ Q/Q', where , add all users in U_j^{\prime } to ε, and assign all permissions in P_j^{\prime } to each user in U_j^{\prime }\cap S. Then there is at least one user u\in {U^{\prime }}_j/S in each size-t _j user set in U_j^{\prime }, as u has all the permissions in P_j^{\prime }, thus each size-t _j user set in U_j^{\prime } together having all the permissions in P_j^{\prime }. In this way, ε' satisfies f _j , and note that adding new users, and assigning permissions to these new users will not lead to violation of policies in Q'. If |P_j^{\prime }\cap W|=0, assigning any permissions in P_j^{\prime } to each user in U_j^{\prime } will not lead to inconsistency of policies in Q', and thus the new state ε' satisfies f _j . Therefore, Q is consistent if and only if Q' is consistent.   □

3.1.2. Minimal inconsistency cover set

There may exist many policy inconsistencies in a policy set which contains a large number of SSoD and SA policies. But many of these inconsistencies may result from only a small number of these policies, and they may be disjoint with each other. We find the minimal inconsistency cover set is the minimal number of policies that represent a policy inconsistency. Therefore, the key question is how to organize the policy inconsistencies, so as to examine the minimum number of policies that are responsible for all the inconsistencies.

Definition 6. We define a minimal inconsistency cover (MIC) set responsible for a policy inconsistency that includes the smallest number of policies.

Note that for a policy inconsistency, there might be several policy sets that are responsible for this inconsistency. By definition, we say that a set S is an MIC set, if there does not exist another set S' responsible for this inconsistency and S' ⊂ S. We have the following property for MIC.

Theorem 4. Given any two MIC sets A and B, let P _A denotes the union of permissions in all policies in A, and U _A denotes the union of users in all policies in A. P _B and U _B have the similar meanings. Then (P _A ∩ P _B = ∅) ⋁ (U _A ∩ U _B = ∅).

PROOF. We assume that (P _A ∩ P _B = ∅) ⋁ (U _A ∩ U _B = ∅) is false, then (P _A ∩ P _B ≠ ∅) ⋀ (U _A ∩ U _B ≠ ∅). There are four cases should be considered:

1. (1)

   Permissions and users for {e ₁ , ..., e _m } ⊆ A(m ≥ 1) and \left\{e_1^{\prime },...,e_n^{\prime }\right\}\subseteq B\left(n\ge 1\right) are shared;

2. (2)

   Permissions and users for {e ₁ , ..., e _m } ⊆ A(m ≥ 1) and {f ₁ , ..., f _n } ⊆ B(n ≥ 1) are shared;

3. (3)

   Permissions and users for {f ₁ , ..., f _m } ⊆ A(m ≥ 1) and \left\{f_1^{\prime },\dots ,f_n^{\prime }\right\}\subseteq B\left(n\ge 1\right) are shared;

4. (4)

   Permissions and users for {e ₁ , ..., e _m , f ₁ , ..., f _n } ⊆ A(m ≥ 1, n ≥ 1) and \left\{e_1^{\prime },\dots ,e_l^{\prime },f_1^{\prime },\dots ,f_k^{\prime }\right\}\subseteq B\left(l\ge 1,k\ge 1\right) are shared.

For case (1), there exists at least one permission p\in P_{\left\{e_1,\dots ,e_m\right\}}, but p does not belong to any other policies in A. By Theorem 3, {e₁, ..., e _m } does not affect the inconsistency of other permissions in A, and thus {e₁, ..., e _m } can be removed from A. This would contradict the assertion that A is an MIC set. Moreover, there exists at least one permission p\in P_{\left\{e_1^{\prime },\dots ,e_n^{\prime }\right\}}, but p does not belong to any other policies in B. Thus \left\{e_1^{\prime },...,e_n^{\prime }\right\} also can be removed from B. For case (2) and case (3), the proof is essentially the same as the case (1). It should be noted that there exists at least one user u belongs to the policies in {f₁, ..., f _n }, but u does not belong to any other policies in B. Thus {f₁, ..., f _n } should be removed from B by Theorem 3. For case (4), no policies can be removed from \left\{e_1,\dots ,e_m,f_1,\dots ,f_n\right\}\cup \left\{e_1^{\prime },\dots ,e_l^{\prime },f_1^{\prime },\dots ,f_k^{\prime }\right\}, which means these policies may conflict with each other due to their opposite objectives. Therefore, these policies should be included by only one MIC set. This would contradict the assertion that A and B are two MIC sets. Together with the above discussions, given any two MIC sets, that (P _A ∩ P _B = ∅) ⋁ (U _A ∩ U _B = ∅). □

We now give an algorithm to generate the MIC sets for an access control policy set. Algorithm 1 includes an underlying presumption that all SSoD and SA policies which do not cause policy inconsistencies have been removed from our consideration by using "static pruning" technique. Given a policy set Q, the algorithm first divides Q into several subsets by the step 1 to 20. By the step 21 to 27, the algorithm combines the different sets which share the permissions and users. This algorithm has a worst-case time complexity of O(mnMN), where m is the number of SSoD policies, n is the number of SA policies, M is the number of users, N is the number of permissions. The fact that CCP is intractable (coNP-complete) means that there exist difficult problem instances that take exponential time in the worst case, while efficient algorithms for CCP exist when the number of policies is not too large. MIC helps to reduce the complexity of reasoning about policy inconsistencies.

Example 2. Continuing from Example 1, we add four policies {e₃, e₄, f₃, f₄} to Q, Consider the combination of following SSoD and SA policies.

By Theorem 3, no policy can be removed from our consideration by static pruning. But the permissions in {p₄, p₅, p₆} and the users in {u₄, u₅, u₆} only exist in {e₃, e₄, f₃, f₄}, and the policies in {e₃, e₄, f₃, f₄} do not affect the consistency of {e₁, e₂, f₁, f₂}. By Algorithm 1, Q' can be divided into two policy set Q_1^{\prime }=\left\{e_1,e_2,f_1,f_2\right\}, and Q_2^{\prime }=\left\{e_3,e_4,f_3,f_4\right\}, such that each set is an MIC set. As shown in Example 1, the policies in Q_1^{\prime } are inconsistent. It is easy to find that the policies in Q_2^{\prime } are inconsistent, too. Continuing from Example 2, assume that there exist another two policies e₅ = ssod <p₁, p₂, p₄, p₅, p₆}, {u₁, u₂, u₃, u₄, u₅, u₆}, 3> and f₅ = sa <{p₁, p₂, p₃, p₄, p₅, p₆}, {u₁, u₂, u₄, u₆}, 3>, then the whole policies in {e₁, e₂, e₃, e₄, e₅, f₁, f₂, f₃, f₄, f₅} is only one MIC set.

3.2. Measuring the safety-utility tradeoff

Given an MIC set for a policy inconsistency. Often, there may exist many choices for resolving this inconsistency. An interesting question for them is "which choice is optimal?". Our methodology helps policy administrators answer this question.

Algorithm 1. ComputeMIC (Q)

Input: Q = {e₁, ..., e _m , f₁, ..., f _n }

Output: the MIC sets of Q : S₁, ..., S _x

1: initialize S₁ = ∅, i = 1, j = 1, k = 1;

2: while (i < m||j < n) do

3:    if\left(\left(P_{e_i}\cap P_{S_k}\ne \varnothing \right)\wedge \left(U_{e_i}\cap U_{S_k}\ne \varnothing \right)\right)then

4:       S _k = S _k ∪ e _i ;

5:       i + +;

6:    else

7:       k + +;

8:       continue;

9:    end if

10:       k = 1;

11:    if\left(\left(P_{f_j}\cap P_{S_k}\ne \varnothing \right)\wedge \left(U_{f_j}\cap U_{S_k}\ne \varnothing \right)\right)then

12:       S _k = S _k ∪ f _j ;

13:       j + +;

14:    else

15:       k++;

16:       continue;

17:    end if

18:    k = 1;

19: end while;

20: MIC(Q) ← S₁, ..., S _x ;

21: for S _k ∈ MIC(Q) do

22:    if\exists S_t\in MIC\left(Q\right)\left[\left(P_{S_t}\cap P_{S_k}\ne \varnothing \right)\wedge \left(U_{S_t}\cap U_{S_k}\ne \varnothing \right)\right]then

23:       MIC(Q) = MIC(Q) - S _t - S _k ;

24:       S _k = S _k ∪ S _t ;

25:       MIC(Q) ← S _k ;

26:    end if

27: end for

28: return MIC(Q).

Example 3. Let us consider the same policies as the one from Example 1. After removing some policies from Q, the rest of policies will be consistent with each other. For example, resolving the policy inconsistency has the following choices.

• Removing only one policy:{e₁}, {f₁}, or {f₂}.

• Removing two policies:{e₁, e₂}, {e₁, f₁}, {e₁, f₂}, {e₂, f₁}, {e₂, f₂}, or {f₁, f₂}.

• Removing three policies:{e₁, e₂, f₁}, {e₁, e₂, f₂}, {e₁, f₁, f₂}, or {e₂, f₁, f₂}.

Currently we lack a method for measuring the safety-utility tradeoff in policy inconsistency resolving. Removing SSoD policies result in safety loss for the whole safety requirement in Q. Similarly, Removing SA policies result in utility loss for the whole utility requirement in Q. Hence before making the choice, one must ensure that the safety loss and utility loss are limited to an acceptable level. To use our method, one must choose a measure for safety loss (S _loss ) and utility loss (U _loss ).

Definition 7. Let e₁and e₂be two SSoD policies, we say thatS_{loss}^{e_1}\ge S_{loss}^{e_2}if and only if e₁ ≽ e₂. AndS_{loss}^{e_1}>S_{loss}^{e_2}if and only if e₁ ≻ e₂.

Where S_{loss}^{e_1} denotes the safety loss caused by removing e₁. As is intuitive, choosing to remove the policy with higher restrictive will cause more safety (or utility) loss.

Theorem 5. For any SSoD policies e₁= ssod <P₁, U₁, k₁> and e₂= ssod <P₂, U₂, k₂>, e₁ ≻ e₂if and only if (U₁ ⊇ U₂) ⋀ (k₁≥ k₂ + | P₁ - P₂|).

PROOF. For the "if" part, given (U₁ ⊇ U₂) ⋀ (k₁≥ k₂ + | P₁ - P₂|), we show that \forall \epsilon \left(\neg sa{t}_{e_2}\left(\epsilon \right)\Rightarrow \neg sa{t}_{e_1}\left(\epsilon \right)\right). There are two cases for (U₁ ⊇ U₂) ⋀ (k₁≥ k₂ + | P₁ - P₂|): (1) P₁ ⊆ P₂, (2) P₁ ⊃ P₂. \neg sa{t}_{e_2}\left(\epsilon \right) being true means that there exist k₂-1 users in U₂ together having all the permissions in P₂. For case (1), there also exists k₂-1 users in U₁ together having all the permissions in P₁ as (P₁ ⊆ P₂) ⋀ (U₁ ⊇ U₂), and (k₁ ≥ k₂ + |P₁ - P₂|) ⇒ (k₁- 1) ≥ (k₂ - 1). Therefore, there exists k₁-1 users in U₁ together having all the permissions in P₁, in other words, \neg sa{t}_{e_1}\left(\epsilon \right) is true. For case (2), there also exist k₂-1 users in U₁ together having all the permissions in P₁ ∪ {P₂ - P₁} as (U₁ ⊇ U₂). At most |P₁ - P₂| users together having all the permissions in {P₂ - P₁}, and (k₁ ≥ k₂ + | P₁ - P₂|) ⇒ (k₂ - 1) ≤ (k₁ - 1) - |P₁ - P₂|. Thus there exists k₁-1 users in U₁ together having all the permissions in P₁, sa{t}_{e_1}\left(\epsilon \right) is also false. Therefore, \forall \epsilon \left(\neg sa{t}_{e_2}\left(\epsilon \right)\Rightarrow \neg sa{t}_{e_1}\left(\epsilon \right)\right) is true.

For the "only if" part, given e₁ ≽ e₂, we show that (U₁ ⊇ U₂) ⋀ (k₁ ≥ k₂ + |P₁ - P₂|) is true. Suppose, for the sake of contradiction, that ¬((U₁ ⊇ U₂) ⋀ (k₁ ≥ k₂ + |P₁ - P₂|)) is true. In other words, both U₁ ⊇ U₂ and k₁ ≥ k₂ + |P₁ - P₂| are false. Let e₁ and e₂ are two SSoD policies, where e₁ = ssod <P₁, U₁, k₁>, e₂ = ssod <P₂, U₂, k₂>. If U₁ ⊇ U₂ is false, then ∃u ∈ U₂/U₁. Assuming that sa{t}_{e_1}\left(\epsilon \right) is true, assign all the permissions in P₂ to u, and then sa{t}_{e_2}\left(\epsilon \right) is false as k₂> 1. Therefore, U₁ ⊇ U₂ is true. If k₁ ≥ k₂ + |P₁ - P₂| is false, then k₁< k₂ + |P₁ - P₂|. If P₁ ⊆ P₂, then k₁< k₂ ⇒ k₁ ≤ k₂ - 1. sa{t}_{e_1}\left(\epsilon \right) being true means that at least k₁ users in U₁ together having all the permissions in P₁. We assume that there exist k₁ users in U₁ together having all the permissions in P₁ in ε; then there exist k₂-1 users in U₂ together having all the permissions in P₂ as to ε (let U₁ = U₂, and these k₁ users also have all the permissions in {P₂ - P₁}), then sa{t}_{e_2}\left(\epsilon \right) is false. If P₁ ⊃ P₂, let k₁< k₂ + |P₁ - P₂|; given an access control state ε that sa{t}_{e_1}\left(\epsilon \right) is true, for each permission in {P₂ - P₁}, assign it to |P₁ - P₂| different users, and these users are not assigned any other permissions in P₁, and then k₁-|P₁ - P₂| users together having all the permissions in P₁. Therefore, there exist less than k₂ users in U₂ together having all the permissions in P₂ (let U₁ = U₂), and therefore, sa{t}_{e_2}\left(\epsilon \right) is false. This contradicts the assumption that e₁ ≽ e₂. Therefore, if e₁ ≽ e₂, then (U₁ ⊇ U₂) ⋀ (k₁ ≥ k₂ + |P₁ - P₂|).   □

Definition 8. Let f₁and f₂be two SA policies, we say thatU_{loss}^{f_1}\ge U_{loss}^{f_2}if and only if f₁ ≽ f₂. AndU_{loss}^{f_1}>U_{loss}^{f_2}if and only if f₁ ≻ f₂.

Theorem 6. For any SA policies f₁= sa <P₁, U₁, t₁> and f₂= sa <P₂, U₂, t₂>, f₁ ≽ f₂if and only if (P₁ ⊇ P₂) ⋀ (U₁ ⊇ U₂) ⋀ (t₁ ≤ t₂).

PROOF. For the "if" part, given (P₁ ⊇ P₂) ⋀ (U₁ ⊇ U₂) ⋀ (t₁ ≤ t₂), we show that \forall \epsilon \left(sa{t}_{f_1}\left(\epsilon \right)\Rightarrow sa{t}_{f_2}\left(\epsilon \right)\right) is true. sa{t}_{f_1}\left(\epsilon \right) being true means that any size-t₁ user set U_1^{\prime } from U₁ together having all the permissions in P₁. Since (P₁ ⊇ P₂) ⋀ (U₁ ⊇ U₂) ⋀ (t₁ ≤ t₂), for each U_1^{\prime }\subseteq U_2\subseteq U_1,{\bigcup }_{u\in U_1^{\prime }}aut{h}_{-}{p}_{\epsilon }\left(u\right)\supseteq P_1\supseteq P_2, and \left|U_1^{\prime }\right|=t_1\le t_2. Therefore, sa{t}_{f_2}\left(\epsilon \right) is also true.

For the "only if" part, given f₁ ≽ f₂, we show that (P₁ ⊇ P₂) ⋀ (U₁ ⊇ U₂) ⋀ (t₁ ≤ t₂) is true. Suppose, for the sake of contradiction, that ¬( P₁ ⊇ P₂) ⋀ (U₁ ⊇ U₂) ⋀ (t₁ ≤ t₂) is true, thus (P₁ ⊂ P₂) ⋁ (U₁ ⊂ U₂) ⋁ (t₁> t₂) is true, then ∃P ∈ P₂/P₁. Assuming that there exists an access control state ε, and sa{t}_{f_1}\left(\epsilon \right) is true. Let P be not assigned to any user in U₂, that does not affect sa{t}_{f_1}\left(\epsilon \right). But sa{t}_{f_2}\left(\epsilon \right) is false, as no size-t₂ user set from U₂ can together cover P₂. Thus the assumption is false, and P₁ ⊇ P₂ is true.

If U₁ ⊂ U₂ is true, then ∃u ∈ U₂/ U₁. We now can construct a state ε that makes sa{t}_{f_2}\left(\epsilon \right) true, but sa{t}_{f_1}\left(\epsilon \right) false. By Theorem 1, sat _f (ε) being true means that each size-t user sets from U cover the permission set P. The above discussion shown that P₁ ⊇ P₂ is true, and let t₁ = t₂. As |U₂| + 1 - t₂> |U₁| + 1 - t₁, sa{t}_{f_1}\left(\epsilon \right) is true, which contradicts the assumption, and thus U₁ ⊇ U₂ is true.

If t₁> t₂ is true, let f_1^{\prime }=sa⟨P_2,U_2,t_1⟩. As shown above, f_1\succcurlyeq f_1^{\prime }, such as for any state ε that \neg sa{t}_{f_1^{\prime }}\left(\epsilon \right)\Rightarrow \neg sa{t}_{f_1}\left(\epsilon \right). Thus we only need to construct a state ε that sa{t}_{f_2}\left(\epsilon \right)is true, but sa{t}_{f_1^{\prime }}\left(\epsilon \right) is false as follows. Find a size-t₁ user set U' ⊂ U₂, and partition P₂ into t₁ disjoint sets v_1,\dots ,v_{t_1}, such that the permissions in each set be assigned to each user in U', respectively. Without any one user in U' can not cover P₂. Since t₁> t₂, we can find a size-t₂ user set U'' ⊂ U' that the users in U'' do not together have all the permissions in P₂. In other words, sa{t}_{f_1^{\prime }}\left(\epsilon \right) is false, and sa{t}_{f_1}\left(\epsilon \right) is also false. This contradicts the assumption, and thus t₁ ≤ t₂ is true. Consequently, if f₁ ≽ f₂, then (P₁ ⊇ P₂) ⋀ (U₁ ⊇ U₂) ⋀ (t₁ ≤ t₂).   □

After computing the rank of S _loss for each SSoD policy and U _loss for each SA policy. A fundamental problem in inconsistency resolving is how to make the right tradeoff between safety and utility. However, it is inappropriate to directly compare safety with utility. The most important reason is that removing SSoD policies will increase the safety loss for the whole policies, but will not increase the utility gain. Similarly, removing SA policies will increase the utility loss for the whole policies, but will not increase the safety gain. For example, if we choose to remove {e₁, e₂} in Example 5, then S _loss = 100%, U _loss = 0%. And if we choose to remove {f₁, f₂}, then S _loss = 0%, U _loss = 100%.

If safety and utility cannot be directly compared, how should one consider them in a policy set for inconsistency resolution? For this, given a number of policy sets that are candidates for removing, for each of which we measure its safety loss S _loss and its utility loss U _loss . We can obtain a set of (S _loss , U _loss ) pairs, one for each set. An ideal (but unachievable) choice will have the smallest S _loss and U _loss . For this, we need to be able to compare two different (S _loss , U _loss ) pairs.

Definition 9. Given two pairs (S _loss , U _loss )₁, and (S _loss , U _loss )₂, we define (S _loss , U _loss )₁ ≤ (S _loss , U _loss )₂if and only if\left(S_{loss}^1\le S_{loss}^2\right)\wedge \left(U_{loss}^1\le U_{loss}^2\right). And (S _loss , U _loss )₁ < (S _loss , U _loss )₂if and only if.

Definition 10. Let A and B be two policy sets; removing A will caused (S _loss , U _loss ) _A , and removing B will caused (S _loss , U _loss ) _B . We say that the choice of removing A is at least as optimal as removing B (denoted by (S _loss , U _loss ) _A ⊵ (S _loss , U _loss ) _B ) if (S _loss , U _loss ) _A ≤ (S _loss , U _loss ) _B . And the the choice of removing A is better than removing B (denoted by (S _loss , U _loss ) _A ⊳ (S _loss , U _loss ) _B ) if (S _loss , U _loss ) _A < (S _loss , U _loss ) _B .

Example 4. Let us consider the following policy sets from Example 3 that can be removed to resolve the policy inconsistency. S₁ = {e₁}, S₂ = {f₁}, S₃ = {e₁, e₂}, S₄ = {f₁, f₂}, S₅ = {e₁, e₂, f₁}.

Obviously, , and . Thus S₁ and S₂ are two ideal choices to resolve the policy inconsistency.

3.3. Prioritized-based resolution

The notion of priority is very important in the study of knowledge based systems, since inconsistencies have a better chance to be resolved. The following subsections present two prioritized-based approaches to deal with policy inconsistencies. We first present the possibilistic logic approach, which selects one consistent subbase. And we then give the lexicographical inference approach, which selects several maximally consistent subbases [7]. We assume that knowledge bases Ψ are prioritized. Prioritized knowledge bases have the form Ψ = Ψ ^E ∪ Ψ ^F , where {\Psi }^E=S_1^E\mathsf{\text{U}}\cdots \mathsf{\text{U}}{S}_m^E, {\Psi }^F=S_1^F\mathsf{\text{U}}\cdots \mathsf{\text{U}}{S}_n^F, E and F denote all the SSoD and SA policies in the system, respectively. Formulas in S_i^E(or S_i^F) have the same level of priority and have higher priority than the ones in S_j^E(or S_j^F) where j > i. S_1^E (or S_1^F) contains the one which have the highest priority in Ψ, and S_m^E(or S_n^F) contains the one which have the lowest priority in Ψ.

3.3.1. Possibilistic logic approach

Possibilistic logic approach selects one suitable consistent prioritized sub-base of Ψ, whereas the other policies in complement set for the subbase of Ψ

Algorithm 2. GeneratePoss(Ψ)

Input: knowledge bases Ψ = Ψ ^E ∪ Ψ ^F

Output: Poss(Ψ)

1: initializePoss\left(\Psi \right)=S_1^E\cup S_1^F, i = 1, j = 1;

2: while (i ≤ m&&j ≤ n) do

3:    if Poss(Ψ) is inconsistent then

4:       Poss\left(\Psi \right)=Poss\left(\Psi \right)-S_i^E-S_j^F;

5:       ifPoss\left(\Psi \right)\cup S_i^E is consistent then

6:          Poss\left(\Psi \right)=Poss\left(\Psi \right)\cup S_i^E;

7:          i++;

8:       else

9:          fore\in S_i^Edo

10:             if Poss(Ψ) ∪ p is consistent then

11:                Poss(Ψ) = Poss(Ψ) ∪ p;

12:             end if

13:          end for

14:       end if

15:       ifPoss\left(\Psi \right)\cup S_i^E is consistent then

16:          Poss\left(\Psi \right)=Poss\left(\Psi \right)\cup S_j^F;

17:          j + +;

18:       else

19:          forf\in S_j^Fdo

20:             if Poss(Ψ) ∪ f is consistent then

21:                Poss(Ψ) = Poss(Ψ) ∪ f;

22:             end if

23:          end for

24:       end if

25:    else

26:       i++;

27:       j ++;

28:       Poss\left(\Psi \right)=Poss\left(\Psi \right)\cup S_i^E\cup S_j^F;

29:    end if

30: end while;

31: return Poss(Ψ).

should be removed. We should extract a subbase φ(Ψ) from Ψ, which is made of the first x-important and consistent strata(levels): φ(Ψ) = S₁ ∪ ⋯ ∪ S _x , such that S₁ ∪ ⋯ ∪ S _x is consistent, but S₁ ∪ ⋯ ∪ S_x+1 is inconsistent.

Definition 11. We define Poss(Ψ) as the set of the preferred consistent possibilistic subbase of Ψ : Poss(Ψ) = {A: A ⊆ Ψ is consistent and ∄B ⊆ Ψ is consistent where B ⊃ A}.

We now give an algorithm to compute the Poss(Ψ) for Ψ (shown in Algorithm 2). This algorithm iteratively adds the SSoD and SA policies with higher priority. Removal of the policies not in Poss(Ψ) is essential to satisfy the consistency for the other policies in Ψ. This algorithm has a best-case time complexity of O(mn), and a worst-case time complexity of O(mnM 2 ^N ), wherem is the number of SSoD policies, n is the number of SA policies, M is the number of users, and N is the number of permissions.

Example 5. Consider the combination of following SSoD and SA policies.

By Theorems 5 and 6, we can find that e₁ ≻ e₂, f₁ ≻ f₂. Thus Ψ = Ψ ^E ∪ Ψ ^F , where {\Psi }^E=S_1^E\cup S_2^E, {\Psi }^F=S_1^F\cup S_2^F, S_1^E=\left\{e_1\right\}, S_2^E=\left\{e_2\right\}, S_1^F=\left\{f_1\right\}, S_2^F=\left\{f_2,f_3\right\}. By Algorithm 2, Poss\left(\Psi \right)=S_1^E\cup S_1^F\cup S_2^E\cup \left\{f_2\right\}=\left\{e_1,e_2,f_1,f_2\right\}. Therefore, the removal of f₃ is an optimal choice to resolve the policy inconsistency.

3.3.2. Lexicographical inference approach

The possibilistic way of dealing with inconsistency is not entirely satisfactory since it only considers the first x-important consistent formulas having the highest priority. However, the less certain formulas may be not responsible for inconsistencies that should also be taken into account. The idea of lexicographical inference approach is to select not only one consistent subbase but several maximally consistent subbases. Obviously, the lexicographical inference is more expensive than the possibilistic logic.

Definition 12. A consistent subbase A ⊆ Ψ is said to be lexicographically preferred to a consistent subbase B ⊆ Ψ, denoted by A ⊳ _lex B, if there exists a level i(1 ≤ i ≤ m) and j(1 ≤ j ≤ n) such that:

Definition 13. We define Lex(Ψ) as the set of all preferred consistent lexicographical subbases of Ψ : Lex(Ψ) = {A: A ⊆ Ψ is consistent and ∄B ⊆ Ψ is consistent, B ⊳ _lex A}.

We now give an algorithm to generate Lex(Ψ) that covers all preferred consistent possibilistic subbases of Ψ. The algorithm is similar to Algorithm 2, but we add following improvements as follows. Given the knowledge bases Ψ = Ψ ^E ∪ Ψ ^F : if Poss\left(\Psi \right)\cup S_i^E or Poss\left(\Psi \right)\cup S_j^F is inconsistent, the algorithm does not stop (While in Algorithm 2, any policies in S_k^E, S_l^F will not be considered, where k > i, l > j), by repeatedly adding policies in S_k^E and S_l^F to Poss(Ψ). In the enumeration approach, the algorithm tries all possibilities. Eventually, the algorithm outputs all preferred consistent possibilistic subbases of Ψ, such as Lex(Ψ). In Example 4. There exists two lexicographically consistent subbases that A = {e₁, e₂, f₁, f₂}, B = {e₁, f₁, f₂, f₃}, then Lex(Ψ) = {A, B}.

Given the results shown in Section 3, we define the following approach to policy inconsistencies resolution.

1. 1.

   Removing SSoD and SA policies from our consideration which do not cause inconsistencies by static pruning.

2. 2.

   Generating MIC sets.

3. 3.

   Consistency checking for each MIC set.

4. 4.

   Extracting priorities based on safety-utility tradeoff.

5. 5.

   Employing possibilistic logic (or lexicographical inference)approach

4.1. Running example

We now give a running example to show the validity of our approach for policy inconsistency resolving.

Example 6. Considering the task of ordering and paying for goods given by Clark and Wilson[6], there are four steps: (1) ordering the goods and recording the details of the order; (2) recording the arrival of the invoice and verifying that the details on the invoice match the details on the order; (3) verifying that the goods have been received and the features of the goods match the details on the invoice; (4) authorizing payment to the supplier against the invoice. We add another two steps: (5) checking the status of the task, and (6) commenting on the task. We have a permission corresponding to each step in the task. The permission set is {order, goods, invoice, payment, check, comment}. Assuming that there are eight users {alice, bob, carl, doris, eric, fox, harris, george} who prepare to accomplish this task. The policy administrator may define many policies that require safety and utility properties in this example and these policies may be inconsistent. Thus it is very important to help the policy administrators to specify reasonable access control policies without inconsistencies. Assuming that the policy administrator defines the following policies.

We now implement the proposed approach to resolve the policy inconsistency problem in Q. Firstly, by Theorem 3, we find that e₄, e₅ and f₅ can be removed from our consideration. Let Q' = {e₁, e₂, e₃, f₁, f₂, f₃, f₄}, thus we only need to consider the policies in Q'. Secondly, by Algorithm 1, we can get two MIC sets: {e₁, e₂, f₁, f₂, f₃} and {e₃, f₄}. Let Q _A = {e₁, e₂, f₁, f₂, f₃}, Q _B = {e₃, f₄}. Thirdly, we check whether the policies in each MIC set are consistent, and find that the policies in Q _A are inconsistent, but the policies in Q _B are consistent. Thus we only need to resolve the policy inconsistency in Q _A (Section 4.2 will give a more detailed description of consistency checking approach). Fourthly, we measure the safety loss for each SSoD policy and the utility loss for each SA policy. Via Theorem 5, we find that e₁ ≻ e₂, f₁ ≻ f₂. Thus we can have the form for prioritized knowledge bases Ψ = Ψ ^E ∪ Ψ ^F (where {\Psi }^E=S_1^E\cup S_2^E, {\Psi }^F=S_1^F\cup S_2^F, S_1^E=\left\{e_1\right\}, S_2^E=\left\{e_2\right\}, S_1^F=\left\{f_1\right\}, S_2^F=\left\{f_2,f_3\right\}.). We give the method for computing the S _loss and U _loss for each SSoD and SA policy, respectively as follows:

• S_{loss}^e=\frac{rank\left(e\right)}{{\Sigma }_{\left\{e^{\prime }\in {\Psi }^E\right\}}rank\left(e^{\prime }\right)}

• U_{loss}^f=\frac{rank\left(f\right)}{{\Sigma }_{\left\{f^{\prime }\in {\Psi }^F\right\}}rank\left(f^{\prime }\right)}

Let rank(e₁) = 2, rank(e₂) = 1, rank(f₁) = 2, and rank(f₂) = rank(f₃) = 1. Thus S_{loss}^{e_1}=\frac{2}{2\times 1+1\times 1}\times 100\%\approx 66.7\%, S_{loss}^{e_2}\approx 33.3\%, U_{loss}^{f_1}=50\%, U_{loss}^{f_2}=U_{loss}^{f_3}=25\%. Lastly, we employ Algorithm 2 to generate possibilistic logic subbase Poss(Ψ) = {e₁, e₂, f₁, f₂}, and compute its safety-utility pair (S _loss , U _loss )_Poss(Ψ) = (0, 25%). We also generate Lex(Ψ) and find that there exist two lexicographically consistent subbases that Lex(Ψ) = {Q₁, Q₂}, where Q₁ = {e₁, e₂, f₁, f₂}, and Q₂ = {e₂, f₁, f₂, f₃}. {\left(S_{loss},U_{loss}\right)}_{Q_1}=\left(0,25\%\right), {\left(S_{loss},U_{loss}\right)}_{Q_2}=\left(66.7\%,0\%\right).

The results above can help the policy administrator to resolve the policy inconsistency by removing some policies, and can specify reasonable access control policies. For example, if the safety requirement is more critical than the utility requirement in this running example, the policy administrator can choose to remove f₃, as it causes no safety loss, but 25% utility loss. Otherwise, he can choose to remove e₁ where it causes about 66.7% safety loss, but no utility loss.

4.2. Performance evaluation

In order to understand the effectiveness of our approach, we have implemented two algorithms, and performed several experiments using the running example as shown in Section 4.1. One is called improved algorithm based on our approach as discussed in above sections (employ the possibilistic logic approach), whereas the other is called straightforward algorithm discussed based on consistency checking problem [4]. The implementation of these two algorithms was written in Java. Experiments were carried out on a machine with an Intel(R) Core(TM)2 Duo CPU T5750 running at 2.0 GHz, and with DDR2 2 GB 667 Mhz RAM, running Microsoft Windows XP Professional.

Straightforward algorithm

Each time a new SSoD (or SA) policy is generated by a policy administrator, the algorithm determines whether this policy is consistent with already existing policies. If the answer to the consistency checking problem is "yes", then the new SSoD (or SA) policy is allowed to be added to the access control system. Otherwise, it will be disallowed. Finally, the generated policies are consistent. We also add the following improvements that greatly reduce the running time.

1. (1)

   Removing SSoD and SA policies from our consideration which do not cause policy inconsistencies using "static pruning" technique.

2. (2)

   Reducing the number of access control states that need to be considered. Given an access control state ε, for each SA policy f = sa<P, U, t>, ε satisfies f if and only if for each size-t set of users from U such that these users together possessing all permissions in P. One only needs to compute the set of permissions of each size-t subsets of U, and check whether it is a superset of P. There exist C_{\left|U\right|}^t size-t user sets for U. If the return for the algorithm is "no", then we know that the state ε does not satisfy f, and thus need not to be considered. By Lemma 1, for the sake of "least privilege" principle, in order to ensure sat _f (ε) being true, we let each permission p ∈ P be assigned to only (|U| + 1 - t) users in U. This can greatly reduce the number of access control states that should be taken into consideration.

(3)Reduction to SAT: Given an SSoD policy e = ssod<P, U, k> and an access control state ε, we have shown that determining whether sat _e (ε) is true is coNP-complete problem [8]. Thus we can use the algorithms for SAT to solve this problem. The SAT solver we use is SAT4J [9]. The translation works are as follows. Given an SSoD policy e = ssod<P, U, k> and an access control state ε, for each u _i ∈ U, we have a propositional variable v _i . This variable is true if u _i is a member of size-(k-1) user set U' ⊆ U to cover all the permissions in P. Then we have the following two kinds of constraints. For each p ∈ P, let u_{i_1},u_{i_2},\dots ,u_{i_x} be the users who are authorized for p. We add the first constraint v_{i_1}+v_{i_2}+\cdots +v_{i_x}\ge 1, which ensures that all the permissions in P are covered by U'. There are |P| such constraints. Then we add the second constraint v₁ + v₂ + ⋯ + v _n ≤ k - 1(n = |U|), which ensures that |U' | ≤ k - 1. There is only one such constraint. If the return for the algorithm is "true", then we know that sat _e (ε) is false; otherwise, sat _e (ε) is true.