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ISBN-10: 1593398298

ISBN-13: 9781593398293

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For instance, the above fact explains why a maximal block of a Boolean equation system can be solved by complementing the block, and then using exactly the same solution method as for minimal blocks. To see this, consider the following example as an application of Lemma 36. Example 37 Consider the Boolean equation system E below, with only maximal equations: (νx1 = x2 ∧ x3 )(νx2 = x3 ∨ x4 )(νx3 = x2 ∨ x4 )(νx4 = 0). In order to solve system E , we first take its complement E given below: (µx1 = x2 ∨ x3 )(µx2 = x3 ∧ x4 )(µx3 = x2 ∧ x4 )(µx4 = 1).

Divide the system E into blocks by calculating the maximal strongly connected components of GE . 3. Topologically sort GE into blocks B1 , B2 , . . , Bm ; here blocks are numbered so that if a block Bi depends on a block Bj then i < j . 4. Beginning with Bm , process each block Bi in turn by performing the following steps: (a) Generate a subsystem E of E containing all equations of E whose left-hand sides are from Bi . These equations are modified by replacing each occurrence of all variables xj outside the block Bi by a constant 0 or 1 (according to the already known solution to xj ), and then propagating the constants using the rules to simplify the equations of E .

If it is odd then player ∀ wins the play π. 2 BACKGROUND 21 A strategy for player x is a function σx : Vx → V that tells player x which choice to make depending on the current construction of a play. A strategy σx for player x is called a winning strategy, if x wins every play by using σx , no matter how the other player plays. Given a parity game G and a strategy σ∃ for player ∃ we write G|σ for the parity game that is induced by σ∃ on G. Formally, G|σ = (V, E ∩ (V∀ × V ∪ {(v, σ∃ (v)) | v ∈ V∃ }), v0 , Ω).

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Colors by illustrations, page layouts and processing by Repro India Ltd.


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