relabel_bse: fix incorrect detection of common APs
Based on a report by Jean Kreber. * spot/tl/relabel.cc (cut_points): Really connect children of Boolean operators using undirected edges, not directed ones. * tests/core/ltlrel.test: Add test cases. * NEWS: Mention the bug.
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3 changed files with 35 additions and 10 deletions
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@ -1,5 +1,5 @@
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// -*- coding: utf-8 -*-
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// Copyright (C) 2012-2016, 2018-2019 Laboratoire de Recherche et
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// Copyright (C) 2012-2016, 2018-2020 Laboratoire de Recherche et
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// Développement de l'Epita (LRDE).
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//
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// This file is part of Spot, a model checking library.
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@ -194,15 +194,15 @@ namespace spot
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// ╱ ╲ ╱ ╲
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// a─────b c─────d
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//
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// (The root node is also a cut-point, but we only consider Boolean
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// cut-points for relabeling.)
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// (The root node is also a cut point, but we only consider Boolean
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// cut points for relabeling.)
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//
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// On the other hand, (a & b) U (b & !c) has only one Boolean
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// cut-point which corresponds to the NOT operator:
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//
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// (a&b)U(b&!c)
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// ╱ ╲
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// a&b b&c
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// a&b b&!c
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// ╱ ╲ ╱ ╲
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// a─────b────!c
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// │
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@ -222,7 +222,7 @@ namespace spot
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// replace that node by a fresh atomic proposition.
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//
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// In the example above (a&b)U(b&!c), the last recursion
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// stop a, b, and !c, producing (p0&p1)U(p1&p2).
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// stops on a, b, and !c, producing (p0&p1)U(p1&p2).
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namespace
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{
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typedef std::vector<formula> succ_vec;
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@ -282,20 +282,23 @@ namespace spot
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if (sz > 1 && f.is_boolean())
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{
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// For Boolean nodes, connect all children in a
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// loop. This way the node can only be a cut-point
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// loop. This way the node can only be a cut point
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// if it separates all children from the reset of
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// the graph (not only one).
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formula pred = f[0];
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for (i = 1; i < sz; ++i)
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{
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formula next = f[i];
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// Note that we only add an edge in one
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// direction, because we are building a cycle
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// between all children anyway.
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// Note that we only add an edge in both directions,
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// as the cut point algorithm really need undirected
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// graphs. (We used to do only one direction, and
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// that turned out to be a bug.)
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g[pred].emplace_back(next);
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g[next].emplace_back(pred);
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pred = next;
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}
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g[pred].emplace_back(f[0]);
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g[f[0]].emplace_back(pred);
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}
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s.pop();
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}
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