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### Day 2

### Part 1

### Part 3

### Part 3

The constraint solving algorithm

Outline

- Recall of Lesson 1.
- The verification algorithm.
- A worked example.
- Remarks.

Refreshing the memory

Syntax: Prolog-based

- terms.
- a, b, [A,b]*pk(A), …
- messages are terms
- variables: begin with uppercase (or _)
- events
- send(t)
- recv(t)

Specifying a Protocol: Roles

- Roles are lists of events.
- Variables are used as parameters.
- Roles of NS:

initiator(A,B,Na,Nb) = [ send([A,Na]*pk(B)),

recv([Na,Nb]*pk(A)),

send(Nb*pk(B))].

responder(A,B,Na,Nb) = [ recv([A,Na]*pk(B)),

send([Na,Nb]*pk(A)),

recv(Nb*pk(B))]).

Scenarios

- Used to specify a session.
- E.g.
- {initiator(a,B,na,Nb), responder(A,b,Na,nb), {recv(nb)} }
- Scenarios allow to specify:
- how many agents are present in a session
- for each agent
- its name
- what he “knows”
- a (non)-secrecy role, that can be used for checking secrecy.

Originator Assumption

- When putting together a scenario, make sure that if an agent has a variable (say X) as parameter THEN in the definition of the agent X must occur first in a recv event.
- Recall:
- initiator(A,B,Na,Nb) = [send([A,Na]*pk(B)),recv([Na,Nb]*pk(A)),send(Nb*pk(B))].
- Then, compare:
- {initiator(a,B,na,Nb), responder(A,b,Na,nb), {recv(nb)} }
- {initiator(a,b,na,Nb), responder(A,b,Na,nb), {recv(nb)} }
- How can we analyze the first scenario?

Solution

- Add a recv event in the role definition.
- New role definition
- initiator(A,B,Na,Nb)=[recv(B), send([A,Na]*pk(B)), recv([Na,Nb]*pk(A)), send(Nb*pk(B))].
- The scenario
- {initiator(a,B,na,Nb), responder(A,b,Na,nb), {recv(nb)} }
- Not particularly elegant, but sound and effective.

Exercise

- Why is it important:
- If the O.A. is not satisfied: then
- the result of the analysis can be incorrect (more later)
- the search space could increase dramatically (crash risks).
- Apparently many
- initiator(A,B,S,Na) = [recv([A,B]),send([A,Na]*pk(S))].
- Decide if the following scenarios satisfy O.A.:
- { initiator(a,B,s,na) }
- { initiator(a,B,s,Na) }
- { initiator(A,B,S,na) }
- { initiator(a,B,B,na) }

Solution

- yes
- no
- no
- yes

Preliminaries: Unification

- a substitution is a mapping from variables to terms
- {Xa}, {X Y, Y a}
- is a unifier of t and s iff t = s.
- e.g., p(X,a), q(r(Z),W), ={X r(Z), W a}
- or 2 ={X r(b),W a,Z b}
- is more general than 2
- two terms unify if they have a unifier
- then there exists a most general unifier (mgu)

Constraints

- A constraint is a pair:

m:T

- m is a message term, T is a list of terms.
- is called simple if m is a variable.
- intuitive meaning: “m is generable from T”
- The Constraint Store (CS) is a set of constraints.

the Verification Algorithm

- S: scenario
- CS: constraint store (initially empty)
- K: intruder’s knowledge.
- A step of the verification algorithm:
- choose the first event e from a non-empty role of S
- case 1) e = send(t)
- K := K U {t}; proceed
- case 2) e = recv(t)
- CS := CS U {t:K }
- if CS can be solved to CS’ with solution ,
- S := S; K:= K; CS := CS’;
- proceed
- otherwise, stop

What is solvable?

- CS can be solved to CS’ with solution if we can apply reduction rules to CS until we obtain CS’, where
- CS’ is empty or
- CS’ contains only simple constraints.

Synthesis reduction rules

- :rewriting step yielding substitution
- is the empty substitution
- Local rules:
- Pair: [m1,m2]:T m1:T , m2:T
- hash: h(m):T m:T
- penc: m*pk(a) :T m:T, a:T
- senc: m+k :T m:T, k:T
- sig: m*sk(e) m:T
- Global rule
- unify: {m:T,C1,…,Cn} {C1,…,Cn}
- provided that =mgu(m,t), tT

Analysis reduction rules (2)

- Affect the other side of the constraint.
- Local rules
- split: m:{[t1,t2]} T m:{t1,t2} T
- pdec: m:{t*pk(e)} :T m:{t} T
- sdec: m:T{t+k} k:T{t-k}, m:T{t,k}
- forget about this one
- used only for constructed symmetric keys
- Global rule
- ksub: {m:{t*k}T, C1, …, Cn}

{m:({t*k}T), C1, …, Cn}

- where =mgu(k,pk(e)), kpk(e)

the Result

- CS0 1 CS1 2… n CSn
- each time a constraint in CS is selected and a rule is applied to it
- The rewriting stops when
- CSn is empty or made of simple constraints
- CS is solved
- the composition of the substitutions is the result of the simplification: := 1 2 … n
- a constraint is selected that cannot be simplified
- CS is unsolvable
- there is no result (failure)

Properties

- Is it Confluent?
- No.
- Different reduction sequences are possible:
- in total: 4 sources of nondeterminism
- choice of the event in the algorithm.
- choice of the constraint to be reduced.
- choice of the rule to be applied.
- in the the analysis rules and in the unify rule there is the additional choice of the term in T to which the rule is applied.
- Full backtracking to preserve completeness.
- Local analysis reduction rules preserve confluence, and this can be used for optimization.

Example

Example

- Consider the scenario for NS with OA:

{initiator(a,B,na,Nb),responder(A,b,Na,nb),

{recv(nb)}}

- A possible interleaving:

recv([A,B]), send([a,na]*pk(B))

recv([A,Na]*pk(b)), send([Na,nb]*pk(A))

recv( [na,Nb]*pk(a)), send([Nb]*pk(B)),

recv(nb) ...

- We omit the events after recv(nb)

Example (cont)

recv([A,B]), send([a,na]*pk(B))

recv([A,Na]*pk(b)), send([Na,nb]*pk(A))

recv( [na,Nb]*pk(a)), send([Nb]*pk(B)),

recv(nb) ...

- find out what happens to the CS
- T = {a,b,e} (intruder knowledge)
- CS = {}

The run (1)

recv([A,B]), send([a,na]*pk(B)), recv([A,Na]*pk(b)), send([Na,nb]*pk(A)), recv( [na,Nb]*pk(a)), send([Nb]*pk(B)),

recv(nb) ...

- Before the step
- T = {a,b,e}
- CS = {}
- After (T does not change)
- CS = {[A,B]:{a,b,e}}
- By applying pair
- CS’ = {A:{a,b,e}, B:{a,b,e}}

The run (2)

send([a,na]*pk(B)), recv([A,Na]*pk(b)), send([Na,nb]*pk(A)), recv( [na,Nb]*pk(a)), send([Nb]*pk(B)), recv(nb) ...

- Before the step
- T = T0 = {a,b,e}
- CS = {A:{a,b,e}, B:{a,b,e}}
- After
- T = T1 = {a,b,e,[a,na]*pk(B)}

The run (3)

recv([A,Na]*pk(b)), send([Na,nb]*pk(A)), recv( [na,Nb]*pk(a)), send([Nb]*pk(B)), recv(nb) ...

- Before the step
- T = T1 = {a,b,e,[a,na]*pk(B)} (T0 = {a,b,e})
- CS = {A:{a,b,e}, B:{a,b,e}}
- After
- CS = {A:T0, B:T0, [A,Na]*pk(b):T1}
- by penc + pair
- CS = {A:T0, B:T0, A:T1, Na:T1, b:T1}
- by nif
- CS = {A:T0, B:T0, A:T1, Na:T1}

The run (4)

send([Na,nb]*pk(A)), recv( [na,Nb]*pk(a)), send([Nb]*pk(B)), recv(nb) ...

- Before the step
- T = T1 = {a,b,e,[a,na]*pk(B)}
- T0 = {a,b,e}
- CS = {A:T0, B:T0, A:T1, Na:T1}
- After
- T = T2 = T1 U {[Na,nb]*pk(A)
- T1 = {a,b,e,[a,na]*pk(B)}
- T0 = {a,b,e}
- CS is unchanged

The run (5)

recv( [na,Nb]*pk(a)), send([Nb]*pk(B)), recv(nb) ...

- Before
- T = T2 = T1 U {[Na,nb]*pk(A)
- T1 = {a,b,e,[a,na]*pk(B)}
- T0 = {a,b,e}
- CS = {A:T0, B:T0, A:T1, Na:T1}
- After
- CS= {A:T0, B:T0, A:T1, Na:T1,[na,Nb]*pk(a):T2}
- unify! (Na-> na , Nb -> nb and A-> a)
- The unification has to be applied to the rest…

The run (5.1)

recv([na,b]*pk(a)), send([nb]*pk(B)), recv(nb) ...

- After the unification:
- T = T2 = T1 U {[na,nb]*pk(a)
- T1 = {a,b,e,[a,na]*pk(B)}
- T0 = {a,b,e}
- CS= {a:T0, B:T0, a:T1, na:T1}
- Unify
- CS= {B:{a,b,e}, na: {a,b,e,[a,na]*pk(B)}}

The run (5.2)

recv([na,b]*pk(a)), send([nb]*pk(e)), recv(nb) ...

- After the unification:
- CS= {B:{a,b,e}, na: {a,b,e,[a,na]*pk(B)}}
- ksub (unification B -> e) + split
- CS= {e:{a,b,e}, na: {a,b,e,a,na}}
- unify twice, with empty answer
- CS = {}
- T = {[na,nb]*pk(a), a,b,e,[a,na]*pk(e)}

The run (5.3)

send([nb]*pk(e)), recv(nb) ...

- Before
- CS = {}
- T = {[na,nb]*pk(a), a,b,e,[a,na]*pk(e)}
- After
- CS = {}
- T = {[na,nb]*pk(a), a,b,e,[a,na]*pk(e), [nb]*pk(e)}

The run (5.4)

recv(nb) ...

- Before
- CS = {}
- T = {[na,nb]*pk(a), a,b,e,[a,na]*pk(e), [nb]*pk(e)}
- After
- CS = {nb:{[na,nb]*pk(a), a,b,e,[a,na]*pk(e), [nb]*pk(e)}}
- pdec
- CS = {nb:{[na,nb]*pk(a), a,b,e,[a,na]*pk(e),nb}}
- unify (empty substitution)
- CS = {} !!!

The solution substitution

- We ended up with an empty CS
- => the system has a solution
- in the process, reduction rules gave us the `solution substitution’

= {A->a,Na->na,Nb->b, B->e}

Considerations

Again, the O.A.

- Consider two roles:
- roleA(X) = { send(X) }
- roleB(Nb) = { recv(Nb) }
- and scenario { roleA(X), roleB(nb) }
- intruder knowledge: {a,b,e}
- constraints generated:
- nb : {X}
- solvable, by rule (unify)
- this is not what we want!

Laziness

- We stop simplifying a constraint when the lhs is a variable.
- This enforces a call-by-need mechanism.
- As long as the lhs is a variable the constraint is trivially solvable.
- If subsequent unification step instantiate the lhs of a constraint, then I check further if it can be solved.
- It would be silly to guess.

Another Example for Laziness

- Consider two roles:

roleA(X,A) = { recv(X), recv(X*pk(A)) }

roleB(Na,A) = { send(Na*pk(A)) }

- and this scenario:{roleA(X, b), roleB(na,b)}
- initial intruder knowledge: {a,b,e}
- there’s only one possible order:

send(na*pk(b)), recv(X), recv(X*pk(b))

Another Example for Laziness

- send(na*pk(b)), recv(X), recv(X*pk(b))
- two constraints are generated:
- X : {a,b,e}
- X*pk(b) : {a,b,e,na*pk(b)}
- by rule (unify):
- na : {a,b,e}
- not solvable!
- we did not know this after the first step.

Bibliography Remark

- The system is strongly based on that of Millen and Shmatikov [MS01]
- Various differences:
- Constraints checked “on the fly”
- Consider run also with unfinished roles (very important in practice)
- Few other minor things.

Exercises: decide if solvable and provide a substitution

- 1)
- Kab : {a,b,e}
- [na, Kab]+kas : {a,b,e} U {[na,a,b]+kas}
- secret : {a,b,e} U {secret+Kab}
- 2)
- Kab : {a,b,e}
- Kab+kas : {a,b,e,[na,a,b]+kas }
- secret : {a,b,e,[na,a,b]+kas } U { secret+Kab }
- 3)
- Y : {e}
- X*pk(Y) : {e, h(msg)}
- h([X,Y]) : {e, h(msg)}

Solutions

- 1. solution: Yes. subst. = {Kab -> [a,b]}
- it is a type flow attack
- solution: No. (the first two already)
- solution: Yes. subst: { X->h(msg), Y->e } or { X->e, Y->e }

Exercise for home

- Add one or more rules to the system so that it handles secret keys (signatures).

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