Compl Integ Trans Syste 26843 pair #381744196

details

property	value
status	complete
benchmark	sect5-len.koat
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n048.star.cs.uiowa.edu
space	KoAT-2013

run statistics

property	value
solver	AProVE
configuration	complexity
runtime (wallclock)	1.74592399597 seconds
cpu usage	3.72660525
max memory	2.48287232E8

stage attributes

key	value
output-size	4380
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


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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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, max(2, 2 + Arg_1)).

(0) CpxIntTrs
(1) Koat2 Proof [FINISHED, 54 ms]
(2) BOUNDS(1, max(2, 2 + Arg_1))
(3) Loat Proof [FINISHED, 145 ms]
(4) BOUNDS(n^1, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
l0(A, B) -> Com_1(l1(0, B)) :|: TRUE
l1(A, B) -> Com_1(l1(A + 1, B - 1)) :|: B >= 1
l1(A, B) -> Com_1(l2(A, B)) :|: 0 >= B

The start-symbols are:[l0_2]


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(1) Koat2 Proof (FINISHED)
YES( ?, max([2, 2+Arg_1]) {O(n)})



Initial Complexity Problem:

  Start:  l0  

  Program_Vars:  Arg_0, Arg_1  

  Temp_Vars:    

  Locations:  l0, l1, l2  

  Transitions:

  l0(Arg_0,Arg_1) -> l1(0,Arg_1):|:

  l1(Arg_0,Arg_1) -> l1(Arg_0+1,Arg_1-1):|:0 <= Arg_0 && 1 <= Arg_1

  l1(Arg_0,Arg_1) -> l2(Arg_0,Arg_1):|:0 <= Arg_0 && Arg_1 <= 0  



Timebounds: 

  Overall timebound: max([2, 2+Arg_1]) {O(n)}

  0: l0->l1: 1 {O(1)}

  1: l1->l1: max([0, Arg_1]) {O(n)}

  2: l1->l2: 1 {O(1)}



Costbounds:

  Overall costbound: max([2, 2+Arg_1]) {O(n)}

  0: l0->l1: 1 {O(1)}

  1: l1->l1: max([0, Arg_1]) {O(n)}

  2: l1->l2: 1 {O(1)}



Sizebounds:

`Lower:

  0: l0->l1, Arg_0: 0 {O(1)}

  0: l0->l1, Arg_1: Arg_1 {O(n)}

  1: l1->l1, Arg_0: 1 {O(1)}

  1: l1->l1, Arg_1: 0 {O(1)}

  2: l1->l2, Arg_0: 0 {O(1)}

  2: l1->l2, Arg_1: min([0, Arg_1]) {O(n)}

`Upper:

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