/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 9 ms]
(2) BOUNDS(1, 1)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f0(A, B, C, D, E, F, G, H, I, J) -> Com_1(f5(K, 0, 0, D, E, F, G, H, I, J)) :|: TRUE
f5(A, B, C, D, E, F, G, H, I, J) -> Com_1(f5(A, B + 1, C + 1, 1, E, F, G, H, I, J)) :|: 15 >= C
f5(A, B, C, D, E, F, G, H, I, J) -> Com_1(f5(A, B, C + 1, 0, E, F, G, H, I, J)) :|: 15 >= C
f5(A, B, C, D, E, F, G, H, I, J) -> Com_1(f27(A, B, C, D, B, B, K, L, L, L)) :|: C >= 16

The start-symbols are:[f0_10]


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(1) Koat Proof (FINISHED)
YES(?, 34)



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    f0(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6, ar_7, ar_8, ar_9) -> Com_1(f5(k, 0, 0, ar_3, ar_4, ar_5, ar_6, ar_7, ar_8, ar_9))

		(Comp: ?, Cost: 1)    f5(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6, ar_7, ar_8, ar_9) -> Com_1(f5(ar_0, ar_1 + 1, ar_2 + 1, 1, ar_4, ar_5, ar_6, ar_7, ar_8, ar_9)) [ 15 >= ar_2 ]

		(Comp: ?, Cost: 1)    f5(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6, ar_7, ar_8, ar_9) -> Com_1(f5(ar_0, ar_1, ar_2 + 1, 0, ar_4, ar_5, ar_6, ar_7, ar_8, ar_9)) [ 15 >= ar_2 ]

		(Comp: ?, Cost: 1)    f5(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6, ar_7, ar_8, ar_9) -> Com_1(f27(ar_0, ar_1, ar_2, ar_3, ar_1, ar_1, k, l, l, l)) [ ar_2 >= 16 ]

		(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6, ar_7, ar_8, ar_9) -> Com_1(f0(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6, ar_7, ar_8, ar_9)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Slicing away variables that do not contribute to conditions from problem 1 leaves variables [ar_2].

We thus obtain the following problem:

2:	T:

		(Comp: 1, Cost: 0)    koat_start(ar_2) -> Com_1(f0(ar_2)) [ 0 <= 0 ]

		(Comp: ?, Cost: 1)    f5(ar_2) -> Com_1(f27(ar_2)) [ ar_2 >= 16 ]

		(Comp: ?, Cost: 1)    f5(ar_2) -> Com_1(f5(ar_2 + 1)) [ 15 >= ar_2 ]

		(Comp: ?, Cost: 1)    f5(ar_2) -> Com_1(f5(ar_2 + 1)) [ 15 >= ar_2 ]

		(Comp: ?, Cost: 1)    f0(ar_2) -> Com_1(f5(0))

	start location:	koat_start

	leaf cost:	0



Repeatedly propagating knowledge in problem 2 produces the following problem:

3:	T:

		(Comp: 1, Cost: 0)    koat_start(ar_2) -> Com_1(f0(ar_2)) [ 0 <= 0 ]

		(Comp: ?, Cost: 1)    f5(ar_2) -> Com_1(f27(ar_2)) [ ar_2 >= 16 ]

		(Comp: ?, Cost: 1)    f5(ar_2) -> Com_1(f5(ar_2 + 1)) [ 15 >= ar_2 ]

		(Comp: ?, Cost: 1)    f5(ar_2) -> Com_1(f5(ar_2 + 1)) [ 15 >= ar_2 ]

		(Comp: 1, Cost: 1)    f0(ar_2) -> Com_1(f5(0))

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(koat_start) = 1

	Pol(f0) = 1

	Pol(f5) = 1

	Pol(f27) = 0

orients all transitions weakly and the transition

	f5(ar_2) -> Com_1(f27(ar_2)) [ ar_2 >= 16 ]

strictly and produces the following problem:

4:	T:

		(Comp: 1, Cost: 0)    koat_start(ar_2) -> Com_1(f0(ar_2)) [ 0 <= 0 ]

		(Comp: 1, Cost: 1)    f5(ar_2) -> Com_1(f27(ar_2)) [ ar_2 >= 16 ]

		(Comp: ?, Cost: 1)    f5(ar_2) -> Com_1(f5(ar_2 + 1)) [ 15 >= ar_2 ]

		(Comp: ?, Cost: 1)    f5(ar_2) -> Com_1(f5(ar_2 + 1)) [ 15 >= ar_2 ]

		(Comp: 1, Cost: 1)    f0(ar_2) -> Com_1(f5(0))

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(koat_start) = 16

	Pol(f0) = 16

	Pol(f5) = -V_1 + 16

	Pol(f27) = -V_1

orients all transitions weakly and the transition

	f5(ar_2) -> Com_1(f5(ar_2 + 1)) [ 15 >= ar_2 ]

strictly and produces the following problem:

5:	T:

		(Comp: 1, Cost: 0)     koat_start(ar_2) -> Com_1(f0(ar_2)) [ 0 <= 0 ]

		(Comp: 1, Cost: 1)     f5(ar_2) -> Com_1(f27(ar_2)) [ ar_2 >= 16 ]

		(Comp: 16, Cost: 1)    f5(ar_2) -> Com_1(f5(ar_2 + 1)) [ 15 >= ar_2 ]

		(Comp: 16, Cost: 1)    f5(ar_2) -> Com_1(f5(ar_2 + 1)) [ 15 >= ar_2 ]

		(Comp: 1, Cost: 1)     f0(ar_2) -> Com_1(f5(0))

	start location:	koat_start

	leaf cost:	0



Complexity upper bound 34



Time: 0.065 sec (SMT: 0.063 sec)


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(2)
BOUNDS(1, 1)