details

property	value
status	complete
benchmark	queue_1000.koat
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n010.star.cs.uiowa.edu
space	T2

run statistics

property	value
solver	AProVE
configuration	complexity
runtime (wallclock)	1.74385690689 seconds
cpu usage	3.893301706
max memory	3.076096E8

stage attributes

key	value
output-size	12992
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(1, 1).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, 119 ms] (2) BOUNDS(1, 1) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: f0(A, B, C) -> Com_1(f8(0, B, C)) :|: TRUE f8(A, B, C) -> Com_1(f14(A, A, C)) :|: 999 >= A && 999 >= D f8(A, B, C) -> Com_1(f14(A, A, C)) :|: 999 >= A f23(A, B, C) -> Com_1(f28(A, B, D)) :|: 999 >= A && 0 >= E + 1 f23(A, B, C) -> Com_1(f28(A, B, D)) :|: 999 >= A f23(A, B, C) -> Com_1(f23(A + 1, B, C)) :|: 999 >= A f28(A, B, C) -> Com_1(f23(A + 1, B, C)) :|: TRUE f28(A, B, C) -> Com_1(f23(A + 1, B, C)) :|: 998 >= D f23(A, B, C) -> Com_1(f38(A, B, C)) :|: A >= 1000 f8(A, B, C) -> Com_1(f8(A + 1, A, C)) :|: 999 >= A f14(A, B, C) -> Com_1(f8(A + 1, B, C)) :|: TRUE f14(A, B, C) -> Com_1(f8(A + 1, B, C)) :|: 998 >= D f8(A, B, C) -> Com_1(f23(0, B, C)) :|: A >= 1000 The start-symbols are:[f0_3] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 14005) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f0(ar_0, ar_1, ar_2) -> Com_1(f8(0, ar_1, ar_2)) (Comp: ?, Cost: 1) f8(ar_0, ar_1, ar_2) -> Com_1(f14(ar_0, ar_0, ar_2)) [ 999 >= ar_0 /\ 999 >= d ] (Comp: ?, Cost: 1) f8(ar_0, ar_1, ar_2) -> Com_1(f14(ar_0, ar_0, ar_2)) [ 999 >= ar_0 ] (Comp: ?, Cost: 1) f23(ar_0, ar_1, ar_2) -> Com_1(f28(ar_0, ar_1, d)) [ 999 >= ar_0 /\ 0 >= e + 1 ] (Comp: ?, Cost: 1) f23(ar_0, ar_1, ar_2) -> Com_1(f28(ar_0, ar_1, d)) [ 999 >= ar_0 ] (Comp: ?, Cost: 1) f23(ar_0, ar_1, ar_2) -> Com_1(f23(ar_0 + 1, ar_1, ar_2)) [ 999 >= ar_0 ] (Comp: ?, Cost: 1) f28(ar_0, ar_1, ar_2) -> Com_1(f23(ar_0 + 1, ar_1, ar_2)) (Comp: ?, Cost: 1) f28(ar_0, ar_1, ar_2) -> Com_1(f23(ar_0 + 1, ar_1, ar_2)) [ 998 >= d ] (Comp: ?, Cost: 1) f23(ar_0, ar_1, ar_2) -> Com_1(f38(ar_0, ar_1, ar_2)) [ ar_0 >= 1000 ] (Comp: ?, Cost: 1) f8(ar_0, ar_1, ar_2) -> Com_1(f8(ar_0 + 1, ar_0, ar_2)) [ 999 >= ar_0 ] (Comp: ?, Cost: 1) f14(ar_0, ar_1, ar_2) -> Com_1(f8(ar_0 + 1, ar_1, ar_2)) (Comp: ?, Cost: 1) f14(ar_0, ar_1, ar_2) -> Com_1(f8(ar_0 + 1, ar_1, ar_2)) [ 998 >= d ] (Comp: ?, Cost: 1) f8(ar_0, ar_1, ar_2) -> Com_1(f23(0, ar_1, ar_2)) [ ar_0 >= 1000 ] (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 Slicing away variables that do not contribute to conditions from problem 1 leaves variables [ar_0]. We thus obtain the following problem: 2: T: (Comp: 1, Cost: 0) koat_start(ar_0) -> Com_1(f0(ar_0)) [ 0 <= 0 ] (Comp: ?, Cost: 1) f8(ar_0) -> Com_1(f23(0)) [ ar_0 >= 1000 ] (Comp: ?, Cost: 1) f14(ar_0) -> Com_1(f8(ar_0 + 1)) [ 998 >= d ] (Comp: ?, Cost: 1) f14(ar_0) -> Com_1(f8(ar_0 + 1)) (Comp: ?, Cost: 1) f8(ar_0) -> Com_1(f8(ar_0 + 1)) [ 999 >= ar_0 ] popout

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job log

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actions

all output return to Compl Integ Trans Syste 26843