details

property	value
status	complete
benchmark	adding-exp-growth3.koat
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n104.star.cs.uiowa.edu
space	KoAT-2014

run statistics

property	value
solver	AProVE
configuration	complexity
runtime (wallclock)	3.22342419624 seconds
cpu usage	4.555534453
max memory	2.67833344E8

stage attributes

key	value
output-size	11629
starexec-result	WORST_CASE(NON_POLY, ?)

output

/export/starexec/sandbox/solver/bin/starexec_run_complexity /export/starexec/sandbox/benchmark/theBenchmark.koat /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- WORST_CASE(NON_POLY, ?) 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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(EXP, EXP). (0) CpxIntTrs (1) Koat Proof [FINISHED, 28 ms] (2) BOUNDS(1, EXP) (3) Loat Proof [FINISHED, 333 ms] (4) BOUNDS(EXP, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: f(A, B, C) -> Com_1(g(A, 1, 0)) :|: TRUE g(A, B, C) -> Com_1(g1(A - 1, B, B)) :|: A > 0 g1(A, B, C) -> Com_1(g(A, C + B, C)) :|: TRUE g(A, B, C) -> Com_1(h(A, B, C)) :|: A <= 0 h(A, B, C) -> Com_1(h(A, B - 1, C)) :|: B > 0 The start-symbols are:[f_3] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, pow(2, ar_2) + 2*ar_2 + 2) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f(ar_0, ar_1, ar_2) -> Com_1(g(1, 0, ar_2)) (Comp: ?, Cost: 1) g(ar_0, ar_1, ar_2) -> Com_1(g1(ar_0, ar_0, ar_2 - 1)) [ ar_2 >= 1 ] (Comp: ?, Cost: 1) g1(ar_0, ar_1, ar_2) -> Com_1(g(ar_1 + ar_0, ar_1, ar_2)) (Comp: ?, Cost: 1) g(ar_0, ar_1, ar_2) -> Com_1(h(ar_0, ar_1, ar_2)) [ 0 >= ar_2 ] (Comp: ?, Cost: 1) h(ar_0, ar_1, ar_2) -> Com_1(h(ar_0 - 1, ar_1, ar_2)) [ ar_0 >= 1 ] (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(f(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) f(ar_0, ar_1, ar_2) -> Com_1(g(1, 0, ar_2)) (Comp: ?, Cost: 1) g(ar_0, ar_1, ar_2) -> Com_1(g1(ar_0, ar_0, ar_2 - 1)) [ ar_2 >= 1 ] (Comp: ?, Cost: 1) g1(ar_0, ar_1, ar_2) -> Com_1(g(ar_1 + ar_0, ar_1, ar_2)) (Comp: ?, Cost: 1) g(ar_0, ar_1, ar_2) -> Com_1(h(ar_0, ar_1, ar_2)) [ 0 >= ar_2 ] (Comp: ?, Cost: 1) h(ar_0, ar_1, ar_2) -> Com_1(h(ar_0 - 1, ar_1, ar_2)) [ ar_0 >= 1 ] (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(f(ar_0, ar_1, ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f) = 1 Pol(g) = 1 Pol(g1) = 1 Pol(h) = 0 Pol(koat_start) = 1 orients all transitions weakly and the transition g(ar_0, ar_1, ar_2) -> Com_1(h(ar_0, ar_1, ar_2)) [ 0 >= ar_2 ] popout

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job log

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actions

all output return to Compl Integ Trans Syste 26843