/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 44 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(D, 0, C)) :|: TRUE
f8(A, B, C) -> Com_1(f8(A, B + 1, C)) :|: 9 >= B
f19(A, B, C) -> Com_1(f19(A, B, C + 1)) :|: 9 >= C
f19(A, B, C) -> Com_1(f29(A, B, C)) :|: C >= 10
f8(A, B, C) -> Com_1(f19(A, B, 0)) :|: B >= 10

The start-symbols are:[f0_3]


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(1) Koat Proof (FINISHED)
YES(?, 25)



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    f0(ar_0, ar_1, ar_2) -> Com_1(f8(d, 0, ar_2))

		(Comp: ?, Cost: 1)    f8(ar_0, ar_1, ar_2) -> Com_1(f8(ar_0, ar_1 + 1, ar_2)) [ 9 >= ar_1 ]

		(Comp: ?, Cost: 1)    f19(ar_0, ar_1, ar_2) -> Com_1(f19(ar_0, ar_1, ar_2 + 1)) [ 9 >= ar_2 ]

		(Comp: ?, Cost: 1)    f19(ar_0, ar_1, ar_2) -> Com_1(f29(ar_0, ar_1, ar_2)) [ ar_2 >= 10 ]

		(Comp: ?, Cost: 1)    f8(ar_0, ar_1, ar_2) -> Com_1(f19(ar_0, ar_1, 0)) [ ar_1 >= 10 ]

		(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(d, 0, ar_2))

		(Comp: ?, Cost: 1)    f8(ar_0, ar_1, ar_2) -> Com_1(f8(ar_0, ar_1 + 1, ar_2)) [ 9 >= ar_1 ]

		(Comp: ?, Cost: 1)    f19(ar_0, ar_1, ar_2) -> Com_1(f19(ar_0, ar_1, ar_2 + 1)) [ 9 >= ar_2 ]

		(Comp: ?, Cost: 1)    f19(ar_0, ar_1, ar_2) -> Com_1(f29(ar_0, ar_1, ar_2)) [ ar_2 >= 10 ]

		(Comp: ?, Cost: 1)    f8(ar_0, ar_1, ar_2) -> Com_1(f19(ar_0, ar_1, 0)) [ ar_1 >= 10 ]

		(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(f8) = 2

	Pol(f19) = 1

	Pol(f29) = 0

	Pol(koat_start) = 2

orients all transitions weakly and the transitions

	f8(ar_0, ar_1, ar_2) -> Com_1(f19(ar_0, ar_1, 0)) [ ar_1 >= 10 ]

	f19(ar_0, ar_1, ar_2) -> Com_1(f29(ar_0, ar_1, ar_2)) [ ar_2 >= 10 ]

strictly and produces the following problem:

3:	T:

		(Comp: 1, Cost: 1)    f0(ar_0, ar_1, ar_2) -> Com_1(f8(d, 0, ar_2))

		(Comp: ?, Cost: 1)    f8(ar_0, ar_1, ar_2) -> Com_1(f8(ar_0, ar_1 + 1, ar_2)) [ 9 >= ar_1 ]

		(Comp: ?, Cost: 1)    f19(ar_0, ar_1, ar_2) -> Com_1(f19(ar_0, ar_1, ar_2 + 1)) [ 9 >= ar_2 ]

		(Comp: 2, Cost: 1)    f19(ar_0, ar_1, ar_2) -> Com_1(f29(ar_0, ar_1, ar_2)) [ ar_2 >= 10 ]

		(Comp: 2, Cost: 1)    f8(ar_0, ar_1, ar_2) -> Com_1(f19(ar_0, ar_1, 0)) [ ar_1 >= 10 ]

		(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 + 10

	Pol(f19) = -V_2

	Pol(f29) = -V_2

	Pol(koat_start) = 10

orients all transitions weakly and the transition

	f8(ar_0, ar_1, ar_2) -> Com_1(f8(ar_0, ar_1 + 1, ar_2)) [ 9 >= ar_1 ]

strictly and produces the following problem:

4:	T:

		(Comp: 1, Cost: 1)     f0(ar_0, ar_1, ar_2) -> Com_1(f8(d, 0, ar_2))

		(Comp: 10, Cost: 1)    f8(ar_0, ar_1, ar_2) -> Com_1(f8(ar_0, ar_1 + 1, ar_2)) [ 9 >= ar_1 ]

		(Comp: ?, Cost: 1)     f19(ar_0, ar_1, ar_2) -> Com_1(f19(ar_0, ar_1, ar_2 + 1)) [ 9 >= ar_2 ]

		(Comp: 2, Cost: 1)     f19(ar_0, ar_1, ar_2) -> Com_1(f29(ar_0, ar_1, ar_2)) [ ar_2 >= 10 ]

		(Comp: 2, Cost: 1)     f8(ar_0, ar_1, ar_2) -> Com_1(f19(ar_0, ar_1, 0)) [ ar_1 >= 10 ]

		(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) = 10

	Pol(f19) = -V_3 + 10

	Pol(f29) = -V_3

	Pol(koat_start) = 10

orients all transitions weakly and the transition

	f19(ar_0, ar_1, ar_2) -> Com_1(f19(ar_0, ar_1, ar_2 + 1)) [ 9 >= ar_2 ]

strictly and produces the following problem:

5:	T:

		(Comp: 1, Cost: 1)     f0(ar_0, ar_1, ar_2) -> Com_1(f8(d, 0, ar_2))

		(Comp: 10, Cost: 1)    f8(ar_0, ar_1, ar_2) -> Com_1(f8(ar_0, ar_1 + 1, ar_2)) [ 9 >= ar_1 ]

		(Comp: 10, Cost: 1)    f19(ar_0, ar_1, ar_2) -> Com_1(f19(ar_0, ar_1, ar_2 + 1)) [ 9 >= ar_2 ]

		(Comp: 2, Cost: 1)     f19(ar_0, ar_1, ar_2) -> Com_1(f29(ar_0, ar_1, ar_2)) [ ar_2 >= 10 ]

		(Comp: 2, Cost: 1)     f8(ar_0, ar_1, ar_2) -> Com_1(f19(ar_0, ar_1, 0)) [ ar_1 >= 10 ]

		(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 25



Time: 0.081 sec (SMT: 0.074 sec)


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(2)
BOUNDS(1, 1)