details

property	value
status	complete
benchmark	random1d.koat
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n101.star.cs.uiowa.edu
space	SAS10

run statistics

property	value
solver	AProVE
configuration	complexity
runtime (wallclock)	2.1645181179 seconds
cpu usage	4.843937253
max memory	3.1162368E8

stage attributes

key	value
output-size	23379
starexec-result	WORST_CASE(Omega(n^1), O(n^1))

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^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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1). (0) CpxIntTrs (1) Koat Proof [FINISHED, 203 ms] (2) BOUNDS(1, n^1) (3) Loat Proof [FINISHED, 522 ms] (4) BOUNDS(n^1, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: start(A, B, C, D, E, F) -> Com_1(stop(A, B, C, D, E, F)) :|: 0 >= A && B >= C && B <= C && D >= A && D <= A && E >= F && E <= F start(A, B, C, D, E, F) -> Com_1(lbl101(A, 2, C, D, 1, F)) :|: A >= 1 && B >= C && B <= C && D >= A && D <= A && E >= F && E <= F start(A, B, C, D, E, F) -> Com_1(lbl101(A, 2, C, D, -(1), F)) :|: A >= 1 && B >= C && B <= C && D >= A && D <= A && E >= F && E <= F lbl101(A, B, C, D, E, F) -> Com_1(stop(A, B, C, D, E, F)) :|: E + A >= 0 && A >= 1 && A >= E && B >= A + 1 && B <= A + 1 && D >= A && D <= A lbl101(A, B, C, D, E, F) -> Com_1(lbl101(A, 1 + B, C, D, 1 + E, F)) :|: A >= B && E + B >= 1 && A + 1 >= B && B >= 2 && B >= E + 1 && D >= A && D <= A lbl101(A, B, C, D, E, F) -> Com_1(lbl101(A, 1 + B, C, D, E - 1, F)) :|: A >= B && E + B >= 1 && A + 1 >= B && B >= 2 && B >= E + 1 && D >= A && D <= A start0(A, B, C, D, E, F) -> Com_1(start(A, C, C, A, F, F)) :|: TRUE The start-symbols are:[start0_6] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 4*ar_0 + 17) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) start(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(stop(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5)) [ 0 >= ar_0 /\ ar_1 = ar_2 /\ ar_3 = ar_0 /\ ar_4 = ar_5 ] (Comp: ?, Cost: 1) start(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(lbl101(ar_0, 2, ar_2, ar_3, 1, ar_5)) [ ar_0 >= 1 /\ ar_1 = ar_2 /\ ar_3 = ar_0 /\ ar_4 = ar_5 ] (Comp: ?, Cost: 1) start(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(lbl101(ar_0, 2, ar_2, ar_3, -1, ar_5)) [ ar_0 >= 1 /\ ar_1 = ar_2 /\ ar_3 = ar_0 /\ ar_4 = ar_5 ] (Comp: ?, Cost: 1) lbl101(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(stop(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5)) [ ar_4 + ar_0 >= 0 /\ ar_0 >= 1 /\ ar_0 >= ar_4 /\ ar_1 = ar_0 + 1 /\ ar_3 = ar_0 ] (Comp: ?, Cost: 1) lbl101(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(lbl101(ar_0, ar_1 + 1, ar_2, ar_3, ar_4 + 1, ar_5)) [ ar_0 >= ar_1 /\ ar_4 + ar_1 >= 1 /\ ar_0 + 1 >= ar_1 /\ ar_1 >= 2 /\ ar_1 >= ar_4 + 1 /\ ar_3 = ar_0 ] (Comp: ?, Cost: 1) lbl101(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(lbl101(ar_0, ar_1 + 1, ar_2, ar_3, ar_4 - 1, ar_5)) [ ar_0 >= ar_1 /\ ar_4 + ar_1 >= 1 /\ ar_0 + 1 >= ar_1 /\ ar_1 >= 2 /\ ar_1 >= ar_4 + 1 /\ ar_3 = ar_0 ] (Comp: ?, Cost: 1) start0(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(start(ar_0, ar_2, ar_2, ar_0, ar_5, ar_5)) (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(start0(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5)) [ 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) start(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(stop(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5)) [ 0 >= ar_0 /\ ar_1 = ar_2 /\ ar_3 = ar_0 /\ ar_4 = ar_5 ] (Comp: 1, Cost: 1) start(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(lbl101(ar_0, 2, ar_2, ar_3, 1, ar_5)) [ ar_0 >= 1 /\ ar_1 = ar_2 /\ ar_3 = ar_0 /\ ar_4 = ar_5 ] (Comp: 1, Cost: 1) start(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(lbl101(ar_0, 2, ar_2, ar_3, -1, ar_5)) [ ar_0 >= 1 /\ ar_1 = ar_2 /\ ar_3 = ar_0 /\ ar_4 = ar_5 ] (Comp: ?, Cost: 1) lbl101(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(stop(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5)) [ ar_4 + ar_0 >= 0 /\ ar_0 >= 1 /\ ar_0 >= ar_4 /\ ar_1 = ar_0 + 1 /\ ar_3 = ar_0 ] (Comp: ?, Cost: 1) lbl101(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(lbl101(ar_0, ar_1 + 1, ar_2, ar_3, ar_4 + 1, ar_5)) [ ar_0 >= ar_1 /\ ar_4 + ar_1 >= 1 /\ ar_0 + 1 >= ar_1 /\ ar_1 >= 2 /\ ar_1 >= ar_4 + 1 /\ ar_3 = ar_0 ] (Comp: ?, Cost: 1) lbl101(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(lbl101(ar_0, ar_1 + 1, ar_2, ar_3, ar_4 - 1, ar_5)) [ ar_0 >= ar_1 /\ ar_4 + ar_1 >= 1 /\ ar_0 + 1 >= ar_1 /\ ar_1 >= 2 /\ ar_1 >= ar_4 + 1 /\ ar_3 = ar_0 ] (Comp: 1, Cost: 1) start0(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(start(ar_0, ar_2, ar_2, ar_0, ar_5, ar_5)) (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5) -> Com_1(start0(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(start) = 1 Pol(stop) = 0 popout

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job log

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actions

all output return to Compl Integ Trans Syste 26843