details

property	value
status	complete
benchmark	simple_pre3.koat
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n014.star.cs.uiowa.edu
space	T2

run statistics

property	value
solver	AProVE
configuration	complexity
runtime (wallclock)	1.56057596207 seconds
cpu usage	3.370203808
max memory	2.21806592E8

stage attributes

key	value
output-size	2471
starexec-result	WORST_CASE(?, O(1))

output

/export/starexec/sandbox2/solver/bin/starexec_run_complexity /export/starexec/sandbox2/benchmark/theBenchmark.koat /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- 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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(1, 1).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, 0 ms] (2) BOUNDS(1, 1) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: f8(A, B, C) -> Com_1(f12(2, 1, C)) :|: A >= 2 && A <= 2 f8(A, B, C) -> Com_1(f12(A, 0, C)) :|: 1 >= A f8(A, B, C) -> Com_1(f12(A, 0, C)) :|: A >= 3 f0(A, B, C) -> Com_1(f12(1, 1, 1)) :|: TRUE The start-symbols are:[f0_3] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 1) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f8(ar_0, ar_1, ar_2) -> Com_1(f12(2, 1, ar_2)) [ ar_0 = 2 ] (Comp: ?, Cost: 1) f8(ar_0, ar_1, ar_2) -> Com_1(f12(ar_0, 0, ar_2)) [ 1 >= ar_0 ] (Comp: ?, Cost: 1) f8(ar_0, ar_1, ar_2) -> Com_1(f12(ar_0, 0, ar_2)) [ ar_0 >= 3 ] (Comp: ?, Cost: 1) f0(ar_0, ar_1, ar_2) -> Com_1(f12(1, 1, 1)) (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 Testing for reachability in the complexity graph removes the following transitions from problem 1: f8(ar_0, ar_1, ar_2) -> Com_1(f12(2, 1, ar_2)) [ ar_0 = 2 ] f8(ar_0, ar_1, ar_2) -> Com_1(f12(ar_0, 0, ar_2)) [ 1 >= ar_0 ] f8(ar_0, ar_1, ar_2) -> Com_1(f12(ar_0, 0, ar_2)) [ ar_0 >= 3 ] We thus obtain the following problem: 2: T: (Comp: ?, Cost: 1) f0(ar_0, ar_1, ar_2) -> Com_1(f12(1, 1, 1)) (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 2 produces the following problem: 3: T: (Comp: 1, Cost: 1) f0(ar_0, ar_1, ar_2) -> Com_1(f12(1, 1, 1)) (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 1 Time: 0.007 sec (SMT: 0.006 sec) popout

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job log

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actions

all output return to Compl Integ Trans Syste 26843