/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(n^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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, n^1).

(0) CpxIntTrs
(1) Koat Proof [FINISHED, 37 ms]
(2) BOUNDS(1, n^1)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f4(A, B, C, D, E, F, G) -> Com_1(f4(A + B, B, C, D, E, F, G)) :|: A >= 0
f4(A, B, C, D, E, F, G) -> Com_1(f6(A, B, 0, 0, 0, 0, 0)) :|: 0 >= A + 1
f5(A, B, C, D, E, F, G) -> Com_1(f4(A, B, C, D, E, F, G)) :|: 0 >= B + 1
f5(A, B, C, D, E, F, G) -> Com_1(f6(A, B, 0, 0, 0, 0, 0)) :|: B >= 0

The start-symbols are:[f5_7]


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(1) Koat Proof (FINISHED)
YES(?, Ar_0 + Ar_1 + 3)



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f4(Ar_0 + Ar_1, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ Ar_0 >= 0 ]

		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f6(Ar_0, Ar_1, 0, 0, 0, 0, 0)) [ 0 >= Ar_0 + 1 ]

		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ 0 >= Ar_1 + 1 ]

		(Comp: ?, Cost: 1)    f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f6(Ar_0, Ar_1, 0, 0, 0, 0, 0)) [ Ar_1 >= 0 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Repeatedly propagating knowledge in problem 1 produces the following problem:

2:	T:

		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f4(Ar_0 + Ar_1, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ Ar_0 >= 0 ]

		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f6(Ar_0, Ar_1, 0, 0, 0, 0, 0)) [ 0 >= Ar_0 + 1 ]

		(Comp: 1, Cost: 1)    f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ 0 >= Ar_1 + 1 ]

		(Comp: 1, Cost: 1)    f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f6(Ar_0, Ar_1, 0, 0, 0, 0, 0)) [ Ar_1 >= 0 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(f4) = 1

	Pol(f6) = 0

	Pol(f5) = 1

	Pol(koat_start) = 1

orients all transitions weakly and the transition

	f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f6(Ar_0, Ar_1, 0, 0, 0, 0, 0)) [ 0 >= Ar_0 + 1 ]

strictly and produces the following problem:

3:	T:

		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f4(Ar_0 + Ar_1, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ Ar_0 >= 0 ]

		(Comp: 1, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f6(Ar_0, Ar_1, 0, 0, 0, 0, 0)) [ 0 >= Ar_0 + 1 ]

		(Comp: 1, Cost: 1)    f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ 0 >= Ar_1 + 1 ]

		(Comp: 1, Cost: 1)    f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f6(Ar_0, Ar_1, 0, 0, 0, 0, 0)) [ Ar_1 >= 0 ]

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Applied AI with 'oct' on problem 3 to obtain the following invariants:

  For symbol f4: -X_2 - 1 >= 0





This yielded the following problem:

4:	T:

		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ 0 <= 0 ]

		(Comp: 1, Cost: 1)    f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f6(Ar_0, Ar_1, 0, 0, 0, 0, 0)) [ Ar_1 >= 0 ]

		(Comp: 1, Cost: 1)    f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ 0 >= Ar_1 + 1 ]

		(Comp: 1, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f6(Ar_0, Ar_1, 0, 0, 0, 0, 0)) [ -Ar_1 - 1 >= 0 /\ 0 >= Ar_0 + 1 ]

		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f4(Ar_0 + Ar_1, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ -Ar_1 - 1 >= 0 /\ Ar_0 >= 0 ]

	start location:	koat_start

	leaf cost:	0



A polynomial rank function with

	Pol(koat_start) = V_1 - V_2

	Pol(f5) = V_1 - V_2

	Pol(f6) = V_1 - V_2

	Pol(f4) = V_1 - V_2

orients all transitions weakly and the transition

	f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f4(Ar_0 + Ar_1, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ -Ar_1 - 1 >= 0 /\ Ar_0 >= 0 ]

strictly and produces the following problem:

5:	T:

		(Comp: 1, Cost: 0)              koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ 0 <= 0 ]

		(Comp: 1, Cost: 1)              f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f6(Ar_0, Ar_1, 0, 0, 0, 0, 0)) [ Ar_1 >= 0 ]

		(Comp: 1, Cost: 1)              f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ 0 >= Ar_1 + 1 ]

		(Comp: 1, Cost: 1)              f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f6(Ar_0, Ar_1, 0, 0, 0, 0, 0)) [ -Ar_1 - 1 >= 0 /\ 0 >= Ar_0 + 1 ]

		(Comp: Ar_0 + Ar_1, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f4(Ar_0 + Ar_1, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ -Ar_1 - 1 >= 0 /\ Ar_0 >= 0 ]

	start location:	koat_start

	leaf cost:	0



Complexity upper bound Ar_0 + Ar_1 + 3



Time: 0.120 sec (SMT: 0.100 sec)


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(2)
BOUNDS(1, n^1)