/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) CpxIntTrs
(1) Koat Proof [FINISHED, 19 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(f9(0, D, 0)) :|: TRUE
f9(A, B, C) -> Com_1(f9(A, B, C + 1)) :|: 49 >= C
f17(A, B, C) -> Com_1(f17(A + 1, B, C)) :|: 49 >= A
f17(A, B, C) -> Com_1(f24(A, B, C)) :|: A >= 50
f9(A, B, C) -> Com_1(f17(0, B, C)) :|: C >= 50

The start-symbols are:[f0_3]


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(1) Koat Proof (FINISHED)
YES(?, 105)



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    f0(ar_0, ar_1, ar_2) -> Com_1(f9(0, d, 0))

		(Comp: ?, Cost: 1)    f9(ar_0, ar_1, ar_2) -> Com_1(f9(ar_0, ar_1, ar_2 + 1)) [ 49 >= ar_2 ]

		(Comp: ?, Cost: 1)    f17(ar_0, ar_1, ar_2) -> Com_1(f17(ar_0 + 1, ar_1, ar_2)) [ 49 >= ar_0 ]

		(Comp: ?, Cost: 1)    f17(ar_0, ar_1, ar_2) -> Com_1(f24(ar_0, ar_1, ar_2)) [ ar_0 >= 50 ]

		(Comp: ?, Cost: 1)    f9(ar_0, ar_1, ar_2) -> Com_1(f17(0, ar_1, ar_2)) [ ar_2 >= 50 ]

		(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(f9(0, d, 0))

		(Comp: ?, Cost: 1)    f9(ar_0, ar_1, ar_2) -> Com_1(f9(ar_0, ar_1, ar_2 + 1)) [ 49 >= ar_2 ]

		(Comp: ?, Cost: 1)    f17(ar_0, ar_1, ar_2) -> Com_1(f17(ar_0 + 1, ar_1, ar_2)) [ 49 >= ar_0 ]

		(Comp: ?, Cost: 1)    f17(ar_0, ar_1, ar_2) -> Com_1(f24(ar_0, ar_1, ar_2)) [ ar_0 >= 50 ]

		(Comp: ?, Cost: 1)    f9(ar_0, ar_1, ar_2) -> Com_1(f17(0, ar_1, ar_2)) [ ar_2 >= 50 ]

		(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) = 2

	Pol(f9) = 2

	Pol(f17) = 1

	Pol(f24) = 0

	Pol(koat_start) = 2

orients all transitions weakly and the transitions

	f9(ar_0, ar_1, ar_2) -> Com_1(f17(0, ar_1, ar_2)) [ ar_2 >= 50 ]

	f17(ar_0, ar_1, ar_2) -> Com_1(f24(ar_0, ar_1, ar_2)) [ ar_0 >= 50 ]

strictly and produces the following problem:

3:	T:

		(Comp: 1, Cost: 1)    f0(ar_0, ar_1, ar_2) -> Com_1(f9(0, d, 0))

		(Comp: ?, Cost: 1)    f9(ar_0, ar_1, ar_2) -> Com_1(f9(ar_0, ar_1, ar_2 + 1)) [ 49 >= ar_2 ]

		(Comp: ?, Cost: 1)    f17(ar_0, ar_1, ar_2) -> Com_1(f17(ar_0 + 1, ar_1, ar_2)) [ 49 >= ar_0 ]

		(Comp: 2, Cost: 1)    f17(ar_0, ar_1, ar_2) -> Com_1(f24(ar_0, ar_1, ar_2)) [ ar_0 >= 50 ]

		(Comp: 2, Cost: 1)    f9(ar_0, ar_1, ar_2) -> Com_1(f17(0, ar_1, ar_2)) [ ar_2 >= 50 ]

		(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) = 50

	Pol(f9) = -V_3 + 50

	Pol(f17) = -V_3

	Pol(f24) = -V_3

	Pol(koat_start) = 50

orients all transitions weakly and the transition

	f9(ar_0, ar_1, ar_2) -> Com_1(f9(ar_0, ar_1, ar_2 + 1)) [ 49 >= ar_2 ]

strictly and produces the following problem:

4:	T:

		(Comp: 1, Cost: 1)     f0(ar_0, ar_1, ar_2) -> Com_1(f9(0, d, 0))

		(Comp: 50, Cost: 1)    f9(ar_0, ar_1, ar_2) -> Com_1(f9(ar_0, ar_1, ar_2 + 1)) [ 49 >= ar_2 ]

		(Comp: ?, Cost: 1)     f17(ar_0, ar_1, ar_2) -> Com_1(f17(ar_0 + 1, ar_1, ar_2)) [ 49 >= ar_0 ]

		(Comp: 2, Cost: 1)     f17(ar_0, ar_1, ar_2) -> Com_1(f24(ar_0, ar_1, ar_2)) [ ar_0 >= 50 ]

		(Comp: 2, Cost: 1)     f9(ar_0, ar_1, ar_2) -> Com_1(f17(0, ar_1, ar_2)) [ ar_2 >= 50 ]

		(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) = 50

	Pol(f9) = 50

	Pol(f17) = -V_1 + 50

	Pol(f24) = -V_1

	Pol(koat_start) = 50

orients all transitions weakly and the transition

	f17(ar_0, ar_1, ar_2) -> Com_1(f17(ar_0 + 1, ar_1, ar_2)) [ 49 >= ar_0 ]

strictly and produces the following problem:

5:	T:

		(Comp: 1, Cost: 1)     f0(ar_0, ar_1, ar_2) -> Com_1(f9(0, d, 0))

		(Comp: 50, Cost: 1)    f9(ar_0, ar_1, ar_2) -> Com_1(f9(ar_0, ar_1, ar_2 + 1)) [ 49 >= ar_2 ]

		(Comp: 50, Cost: 1)    f17(ar_0, ar_1, ar_2) -> Com_1(f17(ar_0 + 1, ar_1, ar_2)) [ 49 >= ar_0 ]

		(Comp: 2, Cost: 1)     f17(ar_0, ar_1, ar_2) -> Com_1(f24(ar_0, ar_1, ar_2)) [ ar_0 >= 50 ]

		(Comp: 2, Cost: 1)     f9(ar_0, ar_1, ar_2) -> Com_1(f17(0, ar_1, ar_2)) [ ar_2 >= 50 ]

		(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 105



Time: 0.095 sec (SMT: 0.089 sec)


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(2)
BOUNDS(1, 1)