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


--------------------------------------------------------------------------------


WORST_CASE(?, O(1))
proof of /export/starexec/sandbox/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 28 ms]
(2) BOUNDS(1, 1)


----------------------------------------

(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f2(A, B, C, D) -> Com_1(f2(1 + A, 1 + B, -(1) + C, D)) :|: TRUE
f1(A, B, C, D) -> Com_1(f2(A, B, C, D)) :|: A >= C && B >= 1 + A && C >= 1 + B
f1(A, B, C, D) -> Com_1(f300(A, B, C, E)) :|: C >= 1 + B && B >= 1 + A && C >= 1 + A
f1(A, B, C, D) -> Com_1(f300(A, B, C, E)) :|: B >= 1 + A && B >= C
f1(A, B, C, D) -> Com_1(f300(A, B, C, E)) :|: A >= B

The start-symbols are:[f1_4]


----------------------------------------

(1) Koat Proof (FINISHED)
YES(?, 3)



Initial complexity problem:

1:	T:

		(Comp: ?, Cost: 1)    f2(ar_0, ar_1, ar_2, ar_3) -> Com_1(f2(ar_0 + 1, ar_1 + 1, ar_2 - 1, ar_3))

		(Comp: ?, Cost: 1)    f1(ar_0, ar_1, ar_2, ar_3) -> Com_1(f2(ar_0, ar_1, ar_2, ar_3)) [ ar_0 >= ar_2 /\ ar_1 >= ar_0 + 1 /\ ar_2 >= ar_1 + 1 ]

		(Comp: ?, Cost: 1)    f1(ar_0, ar_1, ar_2, ar_3) -> Com_1(f300(ar_0, ar_1, ar_2, e)) [ ar_2 >= ar_1 + 1 /\ ar_1 >= ar_0 + 1 /\ ar_2 >= ar_0 + 1 ]

		(Comp: ?, Cost: 1)    f1(ar_0, ar_1, ar_2, ar_3) -> Com_1(f300(ar_0, ar_1, ar_2, e)) [ ar_1 >= ar_0 + 1 /\ ar_1 >= ar_2 ]

		(Comp: ?, Cost: 1)    f1(ar_0, ar_1, ar_2, ar_3) -> Com_1(f300(ar_0, ar_1, ar_2, e)) [ ar_0 >= ar_1 ]

		(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_2, ar_3) -> Com_1(f1(ar_0, ar_1, ar_2, ar_3)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Testing for reachability in the complexity graph removes the following transitions from problem 1:

	f2(ar_0, ar_1, ar_2, ar_3) -> Com_1(f2(ar_0 + 1, ar_1 + 1, ar_2 - 1, ar_3))

	f1(ar_0, ar_1, ar_2, ar_3) -> Com_1(f2(ar_0, ar_1, ar_2, ar_3)) [ ar_0 >= ar_2 /\ ar_1 >= ar_0 + 1 /\ ar_2 >= ar_1 + 1 ]

We thus obtain the following problem:

2:	T:

		(Comp: ?, Cost: 1)    f1(ar_0, ar_1, ar_2, ar_3) -> Com_1(f300(ar_0, ar_1, ar_2, e)) [ ar_0 >= ar_1 ]

		(Comp: ?, Cost: 1)    f1(ar_0, ar_1, ar_2, ar_3) -> Com_1(f300(ar_0, ar_1, ar_2, e)) [ ar_1 >= ar_0 + 1 /\ ar_1 >= ar_2 ]

		(Comp: ?, Cost: 1)    f1(ar_0, ar_1, ar_2, ar_3) -> Com_1(f300(ar_0, ar_1, ar_2, e)) [ ar_2 >= ar_1 + 1 /\ ar_1 >= ar_0 + 1 /\ ar_2 >= ar_0 + 1 ]

		(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_2, ar_3) -> Com_1(f1(ar_0, ar_1, ar_2, ar_3)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Repeatedly propagating knowledge in problem 2 produces the following problem:

3:	T:

		(Comp: 1, Cost: 1)    f1(ar_0, ar_1, ar_2, ar_3) -> Com_1(f300(ar_0, ar_1, ar_2, e)) [ ar_0 >= ar_1 ]

		(Comp: 1, Cost: 1)    f1(ar_0, ar_1, ar_2, ar_3) -> Com_1(f300(ar_0, ar_1, ar_2, e)) [ ar_1 >= ar_0 + 1 /\ ar_1 >= ar_2 ]

		(Comp: 1, Cost: 1)    f1(ar_0, ar_1, ar_2, ar_3) -> Com_1(f300(ar_0, ar_1, ar_2, e)) [ ar_2 >= ar_1 + 1 /\ ar_1 >= ar_0 + 1 /\ ar_2 >= ar_0 + 1 ]

		(Comp: 1, Cost: 0)    koat_start(ar_0, ar_1, ar_2, ar_3) -> Com_1(f1(ar_0, ar_1, ar_2, ar_3)) [ 0 <= 0 ]

	start location:	koat_start

	leaf cost:	0



Complexity upper bound 3



Time: 0.023 sec (SMT: 0.021 sec)


----------------------------------------

(2)
BOUNDS(1, 1)