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from IPython.display import display
import spot
spot.setup(show_default='.bans')
This notebook demonstrates how to use the decompose_scc() function to split an automaton in up to three automata capturing different behaviors. This is based on the paper Strength-based decomposition of the property Büchi automaton for faster model checking (TACAS'13).
This page uses the Python bindings, but the same decompositions can be performed from the shell using autfilt and its --decompose-scc option.
Let's define the following strengths of accepting SCCs:
The strengths strong, stricly inherently weak, and inherently terminal define a partition of all accepting SCCs. The following Büchi automaton has 4 SCCs, and its 3 accepting SCCs show an example of each strength.
Note: the reason we use the word inherently is that the weak and terminal properties are usually defined syntactically: an accepting SCC would be weak if all its transitions belong to the same acceptance sets. This syntactic criterion is a sufficient condition for an accepting SCC to not have any rejecting cycle, but it is not necessary. Hence a weak SCC is inherently weak; but while an inherently weak SCC is not necessarily weak, it can be modified to be weak without alterning the langage.
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aut = spot.translate('(Ga -> Gb) W c')
aut
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The decompose_strength() function takes an automaton, and a string specifying which strength to preserve.
The letters used for this specification are as follows:
t: (inherently) terminalw: (strictly inherently) weaks: strongFor instance if we want to preserve only the strictly inherently weak part of this automaton, we should get only the SCC with the self-loop on $b$, and the SCC above it so that we can reach it. However the SCC above is not stricly weak, so it should not accept any word in the new automaton.
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spot.decompose_scc(aut, 'w')
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Similarly, we can extract all the behaviors captured by the inherently terminal part of the automaton:
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spot.decompose_scc(aut, 't')
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Here is the strong part:
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strong = spot.decompose_scc(aut, 's'); strong
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The union of these three automata recognize the same language as the original automaton.
The application proposed in the aforementioned TACAS'13 paper is for parallelizing model checking. Instead of testing the emptiness of the product between aut and a system, one could test the emptiness 3 products in parallel: each with a sub-automaton of different strength. Model checking using weak and terminal automata can be done with much more simpler emptiness check procedures than needed for the general case. So in effect, we have isolated the "hard" (i.e. strong) part of the original automaton in a smaller automaton, and we only need to use a full-fledged emptiness check for this case.
An additional bonus is that it is possible that the simplification algorithms will do a better job at simplifying the sub-automata than at simplifying the original aut. For instance here the strong automaton can be further simplified:
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strong.postprocess('small')
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for opt in ('sw', 'st', 'wt'):
a = spot.decompose_scc(aut, opt)
a.set_name("option: " + opt)
display(a)
There is nothing that prevents the above decomposition to work with other types of acceptance.
The following Rabin automaton was generated with
ltldo -f '(Ga -> Gb) W c' 'ltl2dstar --ltl2nba=spin:ltl2tgba@-Ds' -H | autfilt -H --merge-transitions
(The autfilt -H --merge-transitions pass is just here to reduce the size of the file and make the automaton more readable.)
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aut = spot.automaton("""
HOA: v1
States: 9
Start: 2
AP: 3 "a" "b" "c"
acc-name: Rabin 2
Acceptance: 4 (Fin(0) & Inf(1)) | (Fin(2) & Inf(3))
properties: trans-labels explicit-labels state-acc complete
properties: deterministic
--BODY--
State: 0 {2}
[0&!2] 0
[0&2] 1
[!0&!2] 5
[!0&2] 6
State: 1 {2}
[0] 1
[!0] 6
State: 2 {2}
[0&!1&!2] 3
[0&1&!2] 4
[!0&!2] 5
[2] 6
State: 3 {1 2}
[0&!2] 0
[0&2] 1
[!0&!2] 5
[!0&2] 6
State: 4 {1 2}
[0&!1&!2] 0
[0&!1&2] 1
[!0&!2] 5
[!0&2] 6
[0&1&!2] 7
[0&1&2] 8
State: 5 {1 2}
[0&!1&!2] 0
[!0&!2] 5
[2] 6
[0&1&!2] 7
State: 6 {1 2}
[t] 6
State: 7 {3}
[0&!1&!2] 0
[0&!1&2] 1
[!0&!2] 5
[!0&2] 6
[0&1&!2] 7
[0&1&2] 8
State: 8 {3}
[0&!1] 1
[!0] 6
[0&1] 8
--END--
""")
aut
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Let's decompose it into three strengths:
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for (name, opt) in (('terminal', 't'), ('strictly weak', 'w'), ('strong', 's')):
a = spot.decompose_scc(aut, opt)
a.set_name(name)
display(a)
Note how the two weak automata (i.e., stricly weak and terminal) are now using a Büchi acceptance condition (because that is sufficient for weak automata) while the strong automaton inherited the original acceptance condition.
When extracting multiple strengths and one of the strength is strong, we preserve the original acceptance. For instance extracting strong and inherently terminal gives the following automaton, where only stricly inherently weak SCCs have become rejecting.
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spot.decompose_scc(aut, "st")
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The weak automata seem to be good candidates for further simplification. Let's add a call to postprocess() to out decomposition loop, trying to preserve the determinism and state-based acceptance of the original automaton.
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for (name, opt) in (('inherently terminal', 't'), ('strictly inherently weak', 'w'), ('strong', 's')):
a = spot.decompose_scc(aut, opt).postprocess('deterministic', 'SBAcc')
a.set_name(name)
display(a)
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aut = spot.automaton("""
HOA: v1
States: 8
Start: 7
AP: 3 "a" "b" "c"
acc-name: Streett 2
Acceptance: 4 (Fin(0) | Inf(1)) & (Fin(2) | Inf(3))
properties: trans-labels explicit-labels state-acc complete
properties: deterministic
--BODY--
State: 0 {2}
[0&1] 0
[0&!1] 3
[!0] 4
State: 1 {2}
[0&1&2] 0
[0&1&!2] 1
[0&!1&!2] 2
[0&!1&2] 3
[!0&2] 4
[!0&!2] 7
State: 2 {2}
[0&1&!2] 1
[0&!1&!2] 2
[0&2] 3
[!0&2] 4
[!0&!2] 7
State: 3 {0 3}
[0] 3
[!0] 4
State: 4 {1 3}
[t] 4
State: 5 {3}
[0&!1] 3
[!0] 4
[0&1] 5
State: 6 {3}
[0&!1&!2] 2
[0&!1&2] 3
[!0&2] 4
[0&1&2] 5
[0&1&!2] 6
[!0&!2] 7
State: 7 {3}
[0&!1&!2] 2
[2] 4
[0&1&!2] 6
[!0&!2] 7
--END--
""")
aut
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for (name, opt) in (('inherently terminal', 't'), ('strictly inherently weak', 'w'), ('strong', 's')):
a = spot.decompose_strength(aut, opt)
a.set_name(name)
display(a)
The subtlety of Streett acceptance is that if a path that does not visit any accepting set infinitely often is accepting. So when disabling SCCs, we must be careful to label them with a combination of rejecting acceptance sets.
This is easy to understand using an example. In the following extraction of the strong and inherently terminal parts, the rejecting SCCs (that were either rejecting or strictly inherently weak originally) have been labeled by the same acceptance sets, to ensure that they are rejected.
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spot.decompose_scc(aut, 'st')
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When the acceptance condition is always satisfiable, all non-trivial SCCs are accepting, and inherently weak.
This include acceptances like Acceptance: 0 t, but also trickier ones like Acceptance: 1 Inf(0) | Fin(0) that you can make as complex as you fancy.
Acceptance: 0 tThis occur frequently whant translating LTL formulas that are safety properties:
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aut = spot.translate('(Gb|c) R a', 'any'); aut
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# There is no strong part for this automaton
assert spot.decompose_scc(aut, 's') is None
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for opt in ('w', 't'):
display(spot.decompose_scc(aut, opt))
If we try to extract multiple strengths and include the (empty) strong part, this request will simply be ignored:
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spot.decompose_scc(aut, 'st')
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Note that the above is exactly the output of decompose_strength(aut, 't'). The 's' flag was actively ignored. If 's' had not been ignored an the automaton processed as if its strong part had to be preserved, the original acceptance conditions would have been used, and this would have prevented the disabling of the initial SCC.
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aut = spot.automaton("""
HOA: v1
States: 4
Start: 0
AP: 1 "a"
Acceptance: 1 Inf(0) | Fin(0)
--BODY--
State: 0
[0] 0
[!0] 1
State: 1
[0] 1
[!0] 2 {0}
State: 2
[0] 1
[!0] 3
State: 3
[t] 3 {0}
--END--
""")
aut
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By our definitions, SCC $\{0\}$ and $\{1,2\}$ are inherently weak, and SCC $\{3\}$ is terminal.
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for (name, opt) in (('terminal', 't'), ('strictly weak', 'w'), ('strong', 's'), ('all strengths', 'swt')):
a = spot.decompose_scc(aut, opt)
if a:
a.set_name(name)
display(a)
else:
print("no output for " + name)
decompose_scc() by SCC numberDecompose SCC can also be called by SCC numbers.
The example below show the different SCC numbers and the state they contains, before extracting the sub-automaton containing SCC 1 and 2 (i.e., anything leading to states 1 and 4 of the original automaton). This example also shows that when an scc_info is available for to automaton to decompose, it can be passed to decompose_scc() in lieu of the automaton: doing so is faster because decompose_scc() does not need to rebuild this object.
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aut = spot.translate('(Ga -> Gb) W c')
si = spot.scc_info(aut)
for scc in range(si.scc_count()):
print("SCC #{} containts states {}".format(scc, list(si.states_of(scc))))
display(aut)
spot.decompose_scc(si, '1,2')
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If an SCC number N is prefixed by a, it signifies that we want to extract the Nth accepting SCC. In the above example SCC 2 is rejecting so SCC a2 denotes SCC 3.
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spot.decompose_scc(si, 'a2')
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