details

property	value
status	complete
benchmark	eric.koat
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n005.star.cs.uiowa.edu
space	T2

run statistics

property	value
solver	AProVE
configuration	complexity
runtime (wallclock)	2.7412109375 seconds
cpu usage	5.297152714
max memory	3.03996928E8

stage attributes

key	value
output-size	25130
starexec-result	WORST_CASE(Omega(n^1), O(n^2))

output

/export/starexec/sandbox2/solver/bin/starexec_run_complexity /export/starexec/sandbox2/benchmark/theBenchmark.koat /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- WORST_CASE(Omega(n^1), O(n^2)) 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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^2). (0) CpxIntTrs (1) Koat Proof [FINISHED, 211 ms] (2) BOUNDS(1, n^2) (3) Loat Proof [FINISHED, 1120 ms] (4) BOUNDS(n^1, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: f1(A, B, C) -> Com_1(f2(A, B, B + 1)) :|: A >= B && A >= 1 && B >= 1 f2(A, B, C) -> Com_1(f3(A, B, C)) :|: B >= C + 1 f2(A, B, C) -> Com_1(f3(A, B, C)) :|: C >= B + 1 f3(A, B, C) -> Com_1(f2(A, B, 0)) :|: A + 1 >= 0 && C >= 1 && C >= A + 1 f3(A, B, C) -> Com_1(f2(A, B, C + 1)) :|: A >= C && C + 1 >= 0 The start-symbols are:[f1_3] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 20*ar_0^2 + 32*ar_0*ar_1 + 280*ar_0 + 260*ar_1 + 12*ar_1^2 + 576) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f1(ar_0, ar_1, ar_2) -> Com_1(f2(ar_0, ar_1, ar_1 + 1)) [ ar_0 >= ar_1 /\ ar_0 >= 1 /\ ar_1 >= 1 ] (Comp: ?, Cost: 1) f2(ar_0, ar_1, ar_2) -> Com_1(f3(ar_0, ar_1, ar_2)) [ ar_1 >= ar_2 + 1 ] (Comp: ?, Cost: 1) f2(ar_0, ar_1, ar_2) -> Com_1(f3(ar_0, ar_1, ar_2)) [ ar_2 >= ar_1 + 1 ] (Comp: ?, Cost: 1) f3(ar_0, ar_1, ar_2) -> Com_1(f2(ar_0, ar_1, 0)) [ ar_0 + 1 >= 0 /\ ar_2 >= 1 /\ ar_2 >= ar_0 + 1 ] (Comp: ?, Cost: 1) f3(ar_0, ar_1, ar_2) -> Com_1(f2(ar_0, ar_1, ar_2 + 1)) [ ar_0 >= ar_2 /\ ar_2 + 1 >= 0 ] (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(f1(ar_0, ar_1, ar_2)) [ 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) f1(ar_0, ar_1, ar_2) -> Com_1(f2(ar_0, ar_1, ar_1 + 1)) [ ar_0 >= ar_1 /\ ar_0 >= 1 /\ ar_1 >= 1 ] (Comp: ?, Cost: 1) f2(ar_0, ar_1, ar_2) -> Com_1(f3(ar_0, ar_1, ar_2)) [ ar_1 >= ar_2 + 1 ] (Comp: ?, Cost: 1) f2(ar_0, ar_1, ar_2) -> Com_1(f3(ar_0, ar_1, ar_2)) [ ar_2 >= ar_1 + 1 ] (Comp: ?, Cost: 1) f3(ar_0, ar_1, ar_2) -> Com_1(f2(ar_0, ar_1, 0)) [ ar_0 + 1 >= 0 /\ ar_2 >= 1 /\ ar_2 >= ar_0 + 1 ] (Comp: ?, Cost: 1) f3(ar_0, ar_1, ar_2) -> Com_1(f2(ar_0, ar_1, ar_2 + 1)) [ ar_0 >= ar_2 /\ ar_2 + 1 >= 0 ] (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(f1(ar_0, ar_1, ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Applied AI with 'oct' on problem 2 to obtain the following invariants: For symbol f2: X_1 - X_3 + 1 >= 0 /\ X_1 - X_2 >= 0 /\ X_2 - 1 >= 0 /\ X_1 + X_2 - 2 >= 0 /\ X_1 - 1 >= 0 For symbol f3: X_1 - X_3 + 1 >= 0 /\ X_1 - X_2 >= 0 /\ X_2 - 1 >= 0 /\ X_1 + X_2 - 2 >= 0 /\ X_1 - 1 >= 0 This yielded the following problem: 3: T: (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(f1(ar_0, ar_1, ar_2)) [ 0 <= 0 ] popout

output may be truncated. 'popout' for the full output.

job log

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actions

all output return to Compl Integ Trans Syste 26843