Fast and efficient phase conflict detection and correction in standard ...

. Hence, starting from Ò ½, the node phases will flip an odd number oftimes in . Therefore, the node Ò ½ will be assigned two different phases,which is impossible. Hence our assumption that , whose correspondinglayout Ä is phase-assignable, has a conflict cycle is wrong.()Assume does not contain any conflict cycles. WLOG, assume is connected (if is not connected, the proof can be done for each of connectedcomponents of ). Assume further that Ä is not phase-assignable.Then, there exists at least two shifters × ½ and × ¾ that do not satisfy theconditions for phase assignment. Let Ò ½ and Ò ¾ be the edge shifter nodescorresponding to × ½ and × ¾, respectively. Let the edge connecting them bedenoted as . Let be a cycle that contains and let Ò ¼ be an intermediatenode in (any node other than Ò ½ and Ò ¾).There are two possible cases:1. Shifters × ½ and × ¾ are on opposite sides of a critical feature and havesame color: Since the phases are only changed across feature edges,if Ò ½ has the same phase as Ò ¼, then there must be an even numberof feature edges from Ò ¼ to Ò ½. By assumption, Ò ¾ has the samephase as Ò ½, and hence as Ò ¼, Thus, there must be an even numberof feature edges from Ò ¼ to Ò ¾. Then, must contain an odd numberof feature edges and hence is a conflict cycle.2. Overlapping shifters × ½ and × ¾ have different colors: If Ò ½ and Ò ¼have the same phase, then there must be an even number of featureedges on the path from Ò ½ to Ò ¼. By assumption, Ò ¾ has an oppositephase from Ò ½ and hence an opposite phase from Ò ¼. Thus, the pathfrom Ò ¼ to Ò ¾ must have an odd number of feature edges. Hence must contain an odd number of feature edges and hence is aconflict cycle.This contradicts our initial assumption that has no conflict cycles.Hence, our assumption that Ä is not phase-assignable is wrong. ÙØDefinition 4 Two faces are neighboring faces if they share at least onecommon edge.Definition 5 Two neighboring faces can form a merged face by deletingat least one common edge.Lemma 1 The parity of the number of feature edges of the merged face oftwo neighboring faces is equal to the parity of the sum of the numbers offeature edges of two faces.Proof: Let the two faces have Ñ ½ and Ñ ¾ feature edges and they share Ñ ¿feature edges. Then the merged face has Ñ ½ · Ñ ¾ ¾Ñ ¿ feature edges,which has the same parity as Ñ ½ · Ñ ¾.ÙØLemma 2 A planar embedded graph has no conflict cycles if and onlyif all faces are legal.Proof: For a planar embedded graph, any cycle is the result of mergingÒ faces. If all faces are legal, we know that the number of feature edgesin the merged face is even from Lemma 1. Therefore, by definition, thegraph has no conflict cycles. It is obvious that if the original graph has noconflict cycles, then every face is legal (by definition).ÙØFigure 8. Deleting all common edges (in this case,only one) results in a merged face.Theorem 2 Removing an odd number of edges from every conflict faceand an even number of edges from every legal face will generate a graphwith no conflict cycles.Proof: Assume that an odd number of edges are removed from all conflictfaces and an even number of edges are removed from every legal face of156. As shown in Figure 8, the deletion of one or more common edgesresults in the creation of a merged face. Let ¼ be the graph obtained afterthe edge deletion. Any faceËin ¼must be the result of merging a set Ë offaces in . LetË Ë ½ ˾,whereË ½ is the set of all conflict faces in Ëand Ë ¾ is the set of all legal faces in Ë. The sum of the numbers of deletededges of all the faces in Ë must be even, since any deleted edge belongs totwo faces of Ë and is counted twice. The sum of the numbers of deletededges of all the faces in Ë ¾ is even since an even number of edges areremoved from any legal face. Hence, the sum of the numbers of deletededges of all the faces in Ë ½ should also be even. Since an odd number ofedges are removed from any conflict face, the number of faces in Ë ½ mustbe even. Thus, the sum of the numbers of feature edges of all the facesin Ë ½ is even. It is obvious that the sum of the numbers of feature edgesof all the faces in Ë ¾ is even since every legal face has an even number offeature edges. Therefore, the sum of the numbers of feature edges of allthe faces in Ë is even. By Lemma 1, the number of feature edges of themerged face is also even. So the merged face is legal. This is true of anyface in ¼ . Thus, ¼ has no conflict cycles (Lemma 2). ÙØTheorem 3 The T-join problem for a graph ´Î Û Ìµ,where Ì denotes the set of all conflict nodes in Î , can be reducedto a minimum-weighted perfect matching on the gadget graph ´Î ¼ ¼ Û ¼ µ, constructed using Algorithm Construct Gadget.Proof: () Mapping perfect matching solution of the gadget graph to avalid solution of the T-join problem on : For any node Ú ¾ whosegadget is Ú,1. If Ø Ú is matched within Ú, ¾ Ë;2. If Ø Ú is not matched within Ú,then ¾ Ë;3. If Ú is matched within Ú,then ¾ Ë;4. If Ú is not matched within Ú,then ¾ Ë;The set Ë thus constructed is a valid solution to the T-join problem. Assume ghost nodes are matched within Ú, ghost nodes are not matchedwith Ú, true nodes are matched within Ú and true nodes are notmatched within Ú. In any perfect matching solution, the number of nodesmatched within Ú, ´ · µ, is even. The parity of ´ · µ is the sameas the parity of ´ · · · µ, whose parity is the same as the parity of´·µ. Here, ´·µ is the number of edges ¾ Ë and ´·µ is the numberof ghost nodes in Ú. For conflict nodes, we require that the parity of thenumber of ghost nodes in Ú, ´·µ, to be different from the parity of thenode degree. Therefore, the parity of the number of edges ¾ Ë, ´ · µ,is also different from the parity of the node degree. Hence, the number ofedges ¾ Ë is odd for a conflict node. Similarly, it can be shown that forlegal node, the number of edges ¾ Ë is even. Therefore, the solution Ë isa valid solution of the T-join problem.() Mapping a solution Ë of the T-join problem to a solution of theperfect matching problem of can be done as follows: For any node Ú inthe dual graph , divide the edges connecting node Ú into four sets:¯ Ë ½ Ú ¾ Ú and ¾ ˯ Ë ¾ Ú ¾ Ú and ¾ ˯ Ë ¿ Ø Ú ¾ Ú and ¾ ˯ Ë Ø Ú ¾ Ú and ¾ ËLet the cardinality of Ë ½, Ë ¾, Ë ¿ and Ë be , , and , respectively.The ghost nodes and the true nodes are matched within Ú and theremaining nodes are matched outside Ú. Since Ë is a valid solution ofthe T-join problem, the parity of ´ · µ is the same as the parity of thenumber of ghost nodes ´·µ (by construction). Thus, ´´·µ·´·µµ ´ · µ·¾ is even. Hence, ´ · µ is even, which is the number of nodesto be matched within Ú. This is always possible since Ú is a completegraph.ÙØ