/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 26 ms]
(2) BOUNDS(1, 1)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f0(A, B, C, D) -> Com_1(f4(A, E, C, D)) :|: A >= 10
f0(A, B, C, D) -> Com_1(f0(1 + A, B, A, D)) :|: 9 >= A
f1(A, B, C, D) -> Com_1(f0(1, B, C, D)) :|: 9 >= E && A >= 0 && A <= 0
f2(A, B, C, D) -> Com_1(f0(2, B, C, 2)) :|: 9 >= A
f3(A, B, C, D) -> Com_1(f0(0, B, C, D)) :|: TRUE

The start-symbols are:[f3_4]


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(1) Koat Proof (FINISHED)
YES(?, 12)



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f4(Ar_0, Fresh_0, Ar_2, Ar_3)) [ Ar_0 >= 10 ]

		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0 + 1, Ar_1, Ar_0, Ar_3)) [ 9 >= Ar_0 ]

		(Comp: ?, Cost: 1)    f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(1, Ar_1, Ar_2, Ar_3)) [ 9 >= E /\ Ar_0 = 0 ]

		(Comp: ?, Cost: 1)    f2(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(2, Ar_1, Ar_2, 2)) [ 9 >= Ar_0 ]

		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(0, Ar_1, Ar_2, Ar_3))

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f3(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



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

	f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(1, Ar_1, Ar_2, Ar_3)) [ 9 >= E /\ Ar_0 = 0 ]

	f2(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(2, Ar_1, Ar_2, 2)) [ 9 >= Ar_0 ]

We thus obtain the following problem:

2:	T:

		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f4(Ar_0, Fresh_0, Ar_2, Ar_3)) [ Ar_0 >= 10 ]

		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0 + 1, Ar_1, Ar_0, Ar_3)) [ 9 >= Ar_0 ]

		(Comp: ?, Cost: 1)    f3(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(0, Ar_1, Ar_2, Ar_3))

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f3(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Repeatedly propagating knowledge in problem 2 produces the following problem:

3:	T:

		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f4(Ar_0, Fresh_0, Ar_2, Ar_3)) [ Ar_0 >= 10 ]

		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0 + 1, Ar_1, Ar_0, Ar_3)) [ 9 >= Ar_0 ]

		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(0, Ar_1, Ar_2, Ar_3))

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f3(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(f0) = 1

	Pol(f4) = 0

	Pol(f3) = 1

	Pol(koat_start) = 1

orients all transitions weakly and the transition

	f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f4(Ar_0, Fresh_0, Ar_2, Ar_3)) [ Ar_0 >= 10 ]

strictly and produces the following problem:

4:	T:

		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f4(Ar_0, Fresh_0, Ar_2, Ar_3)) [ Ar_0 >= 10 ]

		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0 + 1, Ar_1, Ar_0, Ar_3)) [ 9 >= Ar_0 ]

		(Comp: 1, Cost: 1)    f3(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(0, Ar_1, Ar_2, Ar_3))

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f3(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(f0) = -V_1 + 10

	Pol(f4) = -V_1

	Pol(f3) = 10

	Pol(koat_start) = 10

orients all transitions weakly and the transition

	f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0 + 1, Ar_1, Ar_0, Ar_3)) [ 9 >= Ar_0 ]

strictly and produces the following problem:

5:	T:

		(Comp: 1, Cost: 1)     f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f4(Ar_0, Fresh_0, Ar_2, Ar_3)) [ Ar_0 >= 10 ]

		(Comp: 10, Cost: 1)    f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0 + 1, Ar_1, Ar_0, Ar_3)) [ 9 >= Ar_0 ]

		(Comp: 1, Cost: 1)     f3(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(0, Ar_1, Ar_2, Ar_3))

		(Comp: 1, Cost: 0)     koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f3(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Complexity upper bound 12



Time: 0.044 sec (SMT: 0.037 sec)


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(2)
BOUNDS(1, 1)