details

property	value
status	complete
benchmark	24.koat
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n079.star.cs.uiowa.edu
space	Beerendonk

run statistics

property	value
solver	AProVE
configuration	complexity
runtime (wallclock)	2.8144390583 seconds
cpu usage	5.031267489
max memory	2.58920448E8

stage attributes

key	value
output-size	13497
starexec-result	WORST_CASE(Omega(n^1), O(n^1))

output

/export/starexec/sandbox/solver/bin/starexec_run_complexity /export/starexec/sandbox/benchmark/theBenchmark.koat /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- WORST_CASE(Omega(n^1), O(n^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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1). (0) CpxIntTrs (1) Koat Proof [FINISHED, 22 ms] (2) BOUNDS(1, n^1) (3) Loat Proof [FINISHED, 611 ms] (4) BOUNDS(n^1, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: eval1(A, B) -> Com_1(eval2(A, B)) :|: A >= 1 && B >= 1 && A >= B + 1 eval1(A, B) -> Com_1(eval3(A, B)) :|: A >= 1 && B >= 1 && B >= A eval2(A, B) -> Com_1(eval2(A - 1, B)) :|: A >= 1 eval2(A, B) -> Com_1(eval1(A, B)) :|: 0 >= A eval3(A, B) -> Com_1(eval3(A, B - 1)) :|: B >= 1 eval3(A, B) -> Com_1(eval1(A, B)) :|: 0 >= B start(A, B) -> Com_1(eval1(A, B)) :|: TRUE The start-symbols are:[start_2] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, ar_0 + ar_1 + 7) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) eval1(ar_0, ar_1) -> Com_1(eval2(ar_0, ar_1)) [ ar_0 >= 1 /\ ar_1 >= 1 /\ ar_0 >= ar_1 + 1 ] (Comp: ?, Cost: 1) eval1(ar_0, ar_1) -> Com_1(eval3(ar_0, ar_1)) [ ar_0 >= 1 /\ ar_1 >= 1 /\ ar_1 >= ar_0 ] (Comp: ?, Cost: 1) eval2(ar_0, ar_1) -> Com_1(eval2(ar_0 - 1, ar_1)) [ ar_0 >= 1 ] (Comp: ?, Cost: 1) eval2(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1)) [ 0 >= ar_0 ] (Comp: ?, Cost: 1) eval3(ar_0, ar_1) -> Com_1(eval3(ar_0, ar_1 - 1)) [ ar_1 >= 1 ] (Comp: ?, Cost: 1) eval3(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1)) [ 0 >= ar_1 ] (Comp: ?, Cost: 1) start(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1)) (Comp: 1, Cost: 0) koat_start(ar_0, ar_1) -> Com_1(start(ar_0, ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 1 produces the following problem: 2: T: (Comp: 1, Cost: 1) eval1(ar_0, ar_1) -> Com_1(eval2(ar_0, ar_1)) [ ar_0 >= 1 /\ ar_1 >= 1 /\ ar_0 >= ar_1 + 1 ] (Comp: 1, Cost: 1) eval1(ar_0, ar_1) -> Com_1(eval3(ar_0, ar_1)) [ ar_0 >= 1 /\ ar_1 >= 1 /\ ar_1 >= ar_0 ] (Comp: ?, Cost: 1) eval2(ar_0, ar_1) -> Com_1(eval2(ar_0 - 1, ar_1)) [ ar_0 >= 1 ] (Comp: ?, Cost: 1) eval2(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1)) [ 0 >= ar_0 ] (Comp: ?, Cost: 1) eval3(ar_0, ar_1) -> Com_1(eval3(ar_0, ar_1 - 1)) [ ar_1 >= 1 ] (Comp: ?, Cost: 1) eval3(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1)) [ 0 >= ar_1 ] (Comp: 1, Cost: 1) start(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1)) (Comp: 1, Cost: 0) koat_start(ar_0, ar_1) -> Com_1(start(ar_0, ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(eval3) = 1 Pol(eval1) = 0 popout

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job log

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actions

all output return to Compl Integ Trans Syste 26843