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Irredundant Cover After performing Expand, we have a prime cover without single cube containment now. We want to find a proper subset which is also a cover.

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Presentation on theme: "Irredundant Cover After performing Expand, we have a prime cover without single cube containment now. We want to find a proper subset which is also a cover."— Presentation transcript:

1 Irredundant Cover After performing Expand, we have a prime cover without single cube containment now. We want to find a proper subset which is also a cover. (irredundant)

2 Irredundant Cover Proposition: A set of cubes C covers a cube p if and only if C p is a tautology. proof: Cp = p =>(Cp) p = p p => C p p p = 1 => C p = 1 <= => C p = 1 => C P P = P => C P = P

3 Example Ex: p = C = C p = Check tautology(C p ) If (C p ) is a tautology, C covers p.

4 Divide & Conquer by Cofactor Op.

5 Property of Unate Function Proposition: A unate cover is a tautology if and only if it contains a row of 2’s. pf: (<=) trivial (=>) Assume the cover represent a monotone increasing function, the function contains 1’s and 2’s. The minterm (0,0,....0) must be covered. Unless the cover contains (2,2,....2), the minterm (0,0,...0) will not be covered.

6 Tautology Check 1. speed-up by unate variables Let F(x 1,x 2,...x n ), x 1 : positive unate –F(x 1,x 2,...x n ) = x 1 A(x 2,...x n ) + B(x 2,...x n ) A : terms with x 1 B : terms without x 1 F x1 = A + B F x1 ’ = B –If (F x1’ = B) is tautology, F x1 = A + B is a tautology. –If (F x1’ = B) is not a tautology, F is not a tautology. If x is positive unate, test if F is a tautology. F x’ is a tautology

7 Tautology Check 2. Other techniques – a row of 2’s, answer ‘Yes’ – a column of all 1’s or all 0’s, answer ‘No’ – compute an upper bound on the no. of minterms of on-set Ex: number of minterms n = 16 > 8 answer ‘No’ – n7, test by truth table

8 A Possible Irredundant Cover Algorithm?

9 Example of Irredundant Cover a b c d e F =

10 Example F covers C 3 = ? => Check tautology (F – C 3 ) C 3 a b c d e on-set count= on-set count= < 32, not tautology C 3 cannot be removed from the cover F covers C 7 = ? => Check tautology (F – C 7 ) C 7 a b c d e C 7 can be removed from the cover b b’ tautology

11 Slides shown below are for your reference.

12 Irredundant Cover Primes in F (1) relatively essential cube c of F (E) F - {c} is not a cover (2) redundant cubes –Totally redundant (R t ) a cube covered by relatively essential cubes and Don’t care –Partially redundant (R p )

13 Example F ={c 1,c 2,c 3 } E = {c 1,c 3 } R t = {c 2 } R p =  F ={c 1,c 2,c 3, c 4 } E = {c 1,c 4 } R t =  R p = {c 2,c 3 } Ex: c1c1 c2c2 c3c3 c1c1 c2c2 c4c4 c3c3

14 Irredundant Cover (1) Check whether cubes are relatively essential 1. tautology (F-c) C 2. If not tautology (F-c) c, c is relatively essential. (2) Check partially, totally redundant 1. tautology (ED) C (D : don’t care) 2. If yes, c istotally redundant If no, c ispartially redundant

15 Irredundant Cover (3) Minimal Irredundant (R p ) For each rR p, define the minimal set S such that (R p - S)ED is NOT a cover ex: R p = {c 4,c 3,c 2 } S 4 ={{c 4,c 3 }} S 3 = {{c 3,c 4 },{c 3,c 2 }} S 2 = {{c 2,c 3 }} Find a minimum subset of partially irredundant set such that combining with relatively essential cubes, it forms a cover. c1c1 c2c2 c4c4 c3c3 c5c5

16 Irredundant Cover Define a matrix B. Each rows corresponds to a S i B ij = 1 if R pj in S i 0 otherwise Ex: R p = {c 4,c 3,c 2 } S 4 = {{c 4,c 3 }} S 3 = {{c 3,c 4 },{c 3,c 2 }} S 2 = {{c 2,c 3 }} c 2 c 3 c 4 S S S S minimum column cover of B minimal irredundant cover

17 Irredundant Cover Two things: 1. find minimal set S 2. find minimal column covering

18 Find Minimal Set S For each r in R p, find all minimal sets S (R p - S)ED does not cover r Tautology((R p - S)ED) r is not a tautology. Let A = (R p ) r B = (ED) r AB1 determine all minimal subset S A such that (A - S)B1

19 Find Minimal Set S D =  A = (R p ) r taut(A) If A is a unate cover with (A 1), then there must be at least one row of 2’s in the cover A take S to be the set of all such cubes, then A - S1 Let E else binate select x j split A xj, A xj’ S xj taut(A xj ) S xj’ taut(A xj’ ) merge: S = S x S x’ (A-S)1(A x -S x )1 OR (A x’ - S x’ )1 ) (

20 Irredundant Cover Let A =(R p ) r B=(E D) r AB1 find subset SA such that (A - S)B1

21 Example ex: A =(R p ) r B =   x 4 x 5 x 6                              x4x4 x 4’ x5x5 x 5’ x5x5 {5,7} {5,6} {4,7} {4,6} {{5,7},{5,6}} {{4,7},{4,6}}


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