details

property	value
status	complete
benchmark	consts1.koat
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n034.star.cs.uiowa.edu
space	T2

run statistics

property	value
solver	AProVE
configuration	complexity
runtime (wallclock)	1.73174500465 seconds
cpu usage	3.58346048
max memory	2.39706112E8

stage attributes

key	value
output-size	2649
starexec-result	WORST_CASE(?, O(1))

output

/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(1, 1).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, 9 ms] (2) BOUNDS(1, 1) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: f0(A) -> Com_1(f1(300)) :|: TRUE f1(A) -> Com_1(f1(A - 1)) :|: A >= 102 f1(A) -> Com_1(f1(A - 1)) :|: 100 >= A The start-symbols are:[f0_1] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 200) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f0(ar_0) -> Com_1(f1(300)) (Comp: ?, Cost: 1) f1(ar_0) -> Com_1(f1(ar_0 - 1)) [ ar_0 >= 102 ] (Comp: ?, Cost: 1) f1(ar_0) -> Com_1(f1(ar_0 - 1)) [ 100 >= ar_0 ] (Comp: 1, Cost: 0) koat_start(ar_0) -> Com_1(f0(ar_0)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Testing for reachability in the complexity graph removes the following transition from problem 1: f1(ar_0) -> Com_1(f1(ar_0 - 1)) [ 100 >= ar_0 ] We thus obtain the following problem: 2: T: (Comp: ?, Cost: 1) f1(ar_0) -> Com_1(f1(ar_0 - 1)) [ ar_0 >= 102 ] (Comp: ?, Cost: 1) f0(ar_0) -> Com_1(f1(300)) (Comp: 1, Cost: 0) koat_start(ar_0) -> Com_1(f0(ar_0)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 2 produces the following problem: 3: T: (Comp: ?, Cost: 1) f1(ar_0) -> Com_1(f1(ar_0 - 1)) [ ar_0 >= 102 ] (Comp: 1, Cost: 1) f0(ar_0) -> Com_1(f1(300)) (Comp: 1, Cost: 0) koat_start(ar_0) -> Com_1(f0(ar_0)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f1) = V_1 - 101 Pol(f0) = 199 Pol(koat_start) = 199 orients all transitions weakly and the transition popout

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actions

all output return to Compl Integ Trans Syste 26843