/export/starexec/sandbox/solver/bin/starexec_run_complexity /export/starexec/sandbox/benchmark/theBenchmark.koat /export/starexec/sandbox/output/output_files


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WORST_CASE(?, O(1))
proof of /export/starexec/sandbox/benchmark/theBenchmark.koat
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(1, 1).

(0) CpxIntTrs
(1) Koat Proof [FINISHED, 9 ms]
(2) BOUNDS(1, 1)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f0(A) -> Com_1(f1(300)) :|: TRUE
f1(A) -> Com_1(f1(A - 1)) :|: A >= 102
f1(A) -> Com_1(f1(A - 1)) :|: 100 >= A

The start-symbols are:[f0_1]


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(1) Koat Proof (FINISHED)
YES(?, 200)



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    f0(Ar_0) -> Com_1(f1(300))

		(Comp: ?, Cost: 1)    f1(Ar_0) -> Com_1(f1(Ar_0 - 1)) [ Ar_0 >= 102 ]

		(Comp: ?, Cost: 1)    f1(Ar_0) -> Com_1(f1(Ar_0 - 1)) [ 100 >= Ar_0 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0) -> Com_1(f0(Ar_0)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Testing for reachability in the complexity graph removes the following transition from problem 1:

	f1(Ar_0) -> Com_1(f1(Ar_0 - 1)) [ 100 >= Ar_0 ]

We thus obtain the following problem:

2:	T:

		(Comp: ?, Cost: 1)    f1(Ar_0) -> Com_1(f1(Ar_0 - 1)) [ Ar_0 >= 102 ]

		(Comp: ?, Cost: 1)    f0(Ar_0) -> Com_1(f1(300))

		(Comp: 1, Cost: 0)    koat_start(Ar_0) -> Com_1(f0(Ar_0)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Repeatedly propagating knowledge in problem 2 produces the following problem:

3:	T:

		(Comp: ?, Cost: 1)    f1(Ar_0) -> Com_1(f1(Ar_0 - 1)) [ Ar_0 >= 102 ]

		(Comp: 1, Cost: 1)    f0(Ar_0) -> Com_1(f1(300))

		(Comp: 1, Cost: 0)    koat_start(Ar_0) -> Com_1(f0(Ar_0)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(f1) = V_1 - 101

	Pol(f0) = 199

	Pol(koat_start) = 199

orients all transitions weakly and the transition

	f1(Ar_0) -> Com_1(f1(Ar_0 - 1)) [ Ar_0 >= 102 ]

strictly and produces the following problem:

4:	T:

		(Comp: 199, Cost: 1)    f1(Ar_0) -> Com_1(f1(Ar_0 - 1)) [ Ar_0 >= 102 ]

		(Comp: 1, Cost: 1)      f0(Ar_0) -> Com_1(f1(300))

		(Comp: 1, Cost: 0)      koat_start(Ar_0) -> Com_1(f0(Ar_0)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Complexity upper bound 200



Time: 0.052 sec (SMT: 0.040 sec)


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(2)
BOUNDS(1, 1)