/export/starexec/sandbox2/solver/bin/starexec_run_complexity /export/starexec/sandbox2/benchmark/theBenchmark.koat /export/starexec/sandbox2/output/output_files


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WORST_CASE(?, O(1))
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 28 ms]
(2) BOUNDS(1, 1)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f0(A, B, C) -> Com_1(f8(0, 10, 0)) :|: TRUE
f8(A, B, C) -> Com_1(f8(A + 2, B, C + 1)) :|: B >= C + 1
f8(A, B, C) -> Com_1(f6(A, B, C)) :|: 2 * B >= A + 1 && C >= B
f8(A, B, C) -> Com_1(f6(A, B, C)) :|: A >= 2 * B && C >= B

The start-symbols are:[f0_3]


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(1) Koat Proof (FINISHED)
YES(?, 13)



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    f0(Ar_0, Ar_1, Ar_2) -> Com_1(f8(0, 10, 0))

		(Comp: ?, Cost: 1)    f8(Ar_0, Ar_1, Ar_2) -> Com_1(f8(Ar_0 + 2, Ar_1, Ar_2 + 1)) [ Ar_1 >= Ar_2 + 1 ]

		(Comp: ?, Cost: 1)    f8(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Ar_0, Ar_1, Ar_2)) [ 2*Ar_1 >= Ar_0 + 1 /\ Ar_2 >= Ar_1 ]

		(Comp: ?, Cost: 1)    f8(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= 2*Ar_1 /\ Ar_2 >= Ar_1 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Repeatedly propagating knowledge in problem 1 produces the following problem:

2:	T:

		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_2) -> Com_1(f8(0, 10, 0))

		(Comp: ?, Cost: 1)    f8(Ar_0, Ar_1, Ar_2) -> Com_1(f8(Ar_0 + 2, Ar_1, Ar_2 + 1)) [ Ar_1 >= Ar_2 + 1 ]

		(Comp: ?, Cost: 1)    f8(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Ar_0, Ar_1, Ar_2)) [ 2*Ar_1 >= Ar_0 + 1 /\ Ar_2 >= Ar_1 ]

		(Comp: ?, Cost: 1)    f8(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= 2*Ar_1 /\ Ar_2 >= Ar_1 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(f0) = 1

	Pol(f8) = 1

	Pol(f6) = 0

	Pol(koat_start) = 1

orients all transitions weakly and the transitions

	f8(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Ar_0, Ar_1, Ar_2)) [ 2*Ar_1 >= Ar_0 + 1 /\ Ar_2 >= Ar_1 ]

	f8(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= 2*Ar_1 /\ Ar_2 >= Ar_1 ]

strictly and produces the following problem:

3:	T:

		(Comp: 1, Cost: 1)    f0(Ar_0, Ar_1, Ar_2) -> Com_1(f8(0, 10, 0))

		(Comp: ?, Cost: 1)    f8(Ar_0, Ar_1, Ar_2) -> Com_1(f8(Ar_0 + 2, Ar_1, Ar_2 + 1)) [ Ar_1 >= Ar_2 + 1 ]

		(Comp: 1, Cost: 1)    f8(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Ar_0, Ar_1, Ar_2)) [ 2*Ar_1 >= Ar_0 + 1 /\ Ar_2 >= Ar_1 ]

		(Comp: 1, Cost: 1)    f8(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= 2*Ar_1 /\ Ar_2 >= Ar_1 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(f0) = 10

	Pol(f8) = V_2 - V_3

	Pol(f6) = V_2 - V_3

	Pol(koat_start) = 10

orients all transitions weakly and the transition

	f8(Ar_0, Ar_1, Ar_2) -> Com_1(f8(Ar_0 + 2, Ar_1, Ar_2 + 1)) [ Ar_1 >= Ar_2 + 1 ]

strictly and produces the following problem:

4:	T:

		(Comp: 1, Cost: 1)     f0(Ar_0, Ar_1, Ar_2) -> Com_1(f8(0, 10, 0))

		(Comp: 10, Cost: 1)    f8(Ar_0, Ar_1, Ar_2) -> Com_1(f8(Ar_0 + 2, Ar_1, Ar_2 + 1)) [ Ar_1 >= Ar_2 + 1 ]

		(Comp: 1, Cost: 1)     f8(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Ar_0, Ar_1, Ar_2)) [ 2*Ar_1 >= Ar_0 + 1 /\ Ar_2 >= Ar_1 ]

		(Comp: 1, Cost: 1)     f8(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= 2*Ar_1 /\ Ar_2 >= Ar_1 ]

		(Comp: 1, Cost: 0)     koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Complexity upper bound 13



Time: 0.048 sec (SMT: 0.040 sec)


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(2)
BOUNDS(1, 1)