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Image redundant wire identification. Speaker: Yung-Chih Chen. Image equivalence. Two steps Split fanouts Remove image redundant wires Image redundant wire identification. Split fanouts. a1. a. a1. a2. a2. ( a1,a2 ) = {(1,0), (0,1)} does not result in illegal images.
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Image redundant wire identification Speaker: Yung-Chih Chen
Image equivalence • Two steps • Split fanouts • Remove image redundant wires • Image redundant wire identification
Split fanouts a1 a a1 a2 a2 (a1,a2)={(1,0), (0,1)} does not result in illegal images
Split fanouts (cont’d) • How to check if (a1, a2)=(value1, value2) results in illegal images Circuit 1 S 1
Theorem • Image G is illegal, if and only if image G leads to a conflict on a • <= • =>
Proof • => • G is obtained by setting a1!=a2 and MAs M for some nodes • G is an illegal image => doesn’t exist MAs M’ when a1=a2 such that G can be obtained • Therefore, a1=a2 =>~ G, G=> a1!=a2
Remove image redundant wires a O1 O1 b b O2 O2 c c
How to identify image redundant wires • Suppose image set G are obtained by setting a=cv • If G can be obtained by setting a=ncv as well, a is an image redundant wires
Theorem • a is an image redundant wire, if and only if image G in the new circuit (a=ncv) leads to a=cv • <= • =>
Probability for redundant wire identification and RAR 21845 a g2 13107 257 b 3855 4095 c g1 O1 255 d 511 511 g4 g3 O2
Probability for redundant wire identification and RAR 21845 a g2 13107 4369 b 3855 4095 c g1 O1 255 d 511 4607 g4 g3 O2
Probability for redundant wire identification and RAR 21845 a g2 257 13107 b 3855 4095 c g1 O1 255 d 511 511 g4 g3 O2 511 & 4095 = 511