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Mohamed G. Gouda                                		CS 386S
Spring 2009                                     		Quiz 9&10
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Consider a Firewall Decision Diagram (FDD) that has two fields F1 and F2 and 
assume that the domain of values for each field is the interval [0, 9]. This  
FDD can be defined by the following non-conflicting rules:

	F1 in [1, 4]		^ F2 in [3, 9] --> discard
	F1 in [1, 4]		^ F2 in [0, 2] --> accept
	F1 in [0, 0][5, 9]	^ F2 in [0, 9] --> discard    

Use this FDD as an example to illustrate that some FDD markings, that minimize 
the FDD load, are better than the others since they lead, after applying the 
redundancy removal step and the rule simplification step, to a firewall with a
smaller number of rules.
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The FDD is as follows:

F1
	-[0,0][5,9]->	F2	-[0,9]->	dis
  	(e1)			(e3)

	-[1,4]->	F2	-[0,2]->	acc
	(e2)			(e4)

				-[3,9]->	dis
				(e5)

e3 is already labeled 'All'. The algorithm chooses e1 over e2 to label 'All', 
as it leads to a reduction of load.

However, we now have a choice of whether to choose e4 or e5 to label 'All'. 
Both will be markings minimizing the FDD load.

Choosing e4 to mark 'All',
F1 in [1,4] ^ F2 in [3,9] -> discard
F1 in [1,4] ^ F2 in [0,9] --> accept
F1 in [0,9] ^ F2 in [0,9] --> discard

no rule is redundant, so the firewall has three rules.

Choosing e5 to mark 'All',
F1 in [1,4] ^ F2 in [0,2] -> accept
F1 in [1,4] ^ F2 in [0,9] --> discard
F1 in [0,9] ^ F2 in [0,9] --> discard

The second rule is redundant. We eliminate it to get
F1 in [1,4] ^ F2 in [0,2] -> accept
F1 in [0,9] ^ F2 in [0,9] --> discard

hence the firewall has fewer rules (2 as opposed to 3) even among labelings 
that minimize load.