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Start explaining Ullmann

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Théophile Bastian 2017-08-21 23:35:00 +02:00
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@ -420,7 +420,13 @@ VF2 algorithm~\cite{cordella2004sub}, published in 2004. This algorithm is
mostly Ullmann's algorithm, transcribed in a recursive writing, with the
addition of five heuristics. \qtodo{Why not use it then?}
\todo{Describe Ullmann}
Ullmann is a widely used and fast algorithm for this problem. It makes an
extensive use of adjacency matrix description of the graph, and the initial
article takes advantage of the representation of those matrices as bitsets to
make extensive use of bitwise operations.
The to-be-built permutation matrix is a $\card{needle} \times \card{haystack}$
matrix,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Performance}
@ -432,6 +438,8 @@ addition of five heuristics. \qtodo{Why not use it then?}
There were a few observed cases where the algorithm tends to be slower on
certain configurations.
\todo{More corner cases}
\paragraph{Split/merge trees.} A common pattern that tends to slow down the
algorithm is split/merge trees. Those patterns occur when one wants to merge
$n$ one bit wires into a single $n$ bits wire, or the other way around.
@ -444,9 +452,16 @@ tree of depth \eg{} 8, a node just below the root will need a signature of
order 7 to have a different signature than another one at the same depth. With
a signature of order up to 6, only other gates from the tree will be included
in the signature when going down in the tree; the exact same gates will be
included above the tree's root. Thus, nothing will differentiate one gate from another while
the boundary of the tree is not reached (assuming the gates below the tree's
leaves are not all the same; if so, more levels will be needed).
included above the tree's root. Thus, nothing will differentiate one gate from
another while the boundary of the tree is not reached (assuming the gates below
the tree's leaves are not all the same; if so, more levels will be needed).
As the notion of ``left child'' and ``right child'' cannot be used (since it
would rely on the order or description of the graph), there seems to be no good
way to discriminate those two nodes. Furthermore, the nodes are not totally
interchangeable: indeed, when checking for an equality between two such trees,
it does not matter which node is the left one; but once this is fixed, the
nodes on the layer below cannot be freely exchanged.
\todo{Figure describing the problem}