Compl Integ Trans Syste 26843 pair #381744165

details

property	value
status	complete
benchmark	sqrt.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	patrs

run statistics

property	value
solver	AProVE
configuration	complexity
runtime (wallclock)	2.21578383446 seconds
cpu usage	4.482539358
max memory	2.7660288E8

stage attributes

key	value
output-size	3093
starexec-result	WORST_CASE(?, O(n^1))

output

/export/starexec/sandbox/solver/bin/starexec_run_complexity /export/starexec/sandbox/benchmark/theBenchmark.koat /export/starexec/sandbox/output/output_files


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WORST_CASE(?, O(n^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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, max(2, 2 + 5 * Arg_3)).

(0) CpxIntTrs
(1) Koat2 Proof [FINISHED, 349 ms]
(2) BOUNDS(1, max(2, 2 + 5 * Arg_3))


----------------------------------------

(0)
Obligation:
Complexity Int TRS consisting of the following rules:
sqrt(A, B, C, D) -> Com_1(f(0, 1, 1, D)) :|: TRUE
f(A, B, C, D) -> Com_1(f(A + 1, B + 2, C + B + 2, D)) :|: D >= C && B >= 0
f(A, B, C, D) -> Com_1(end(A, B, C, D)) :|: C >= D + 1

The start-symbols are:[sqrt_4]


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(1) Koat2 Proof (FINISHED)
YES( ?, max([2, 2+5*Arg_3]) {O(n)})



Initial Complexity Problem:

  Start:  sqrt  

  Program_Vars:  Arg_0, Arg_1, Arg_2, Arg_3  

  Temp_Vars:    

  Locations:  end, f, sqrt  

  Transitions:

  f(Arg_0,Arg_1,Arg_2,Arg_3) -> end(Arg_0,Arg_1,Arg_2,Arg_3):|:1 <= Arg_2 && 2 <= Arg_1+Arg_2 && Arg_1 <= Arg_2 && 1 <= Arg_0+Arg_2 && 1+Arg_0 <= Arg_2 && 1 <= Arg_1 && 1 <= Arg_0+Arg_1 && 1+Arg_0 <= Arg_1 && 0 <= Arg_0 && Arg_3+1 <= Arg_2

  f(Arg_0,Arg_1,Arg_2,Arg_3) -> f(Arg_0+1,Arg_1+2,Arg_2+Arg_1+2,Arg_3):|:1 <= Arg_2 && 2 <= Arg_1+Arg_2 && Arg_1 <= Arg_2 && 1 <= Arg_0+Arg_2 && 1+Arg_0 <= Arg_2 && 1 <= Arg_1 && 1 <= Arg_0+Arg_1 && 1+Arg_0 <= Arg_1 && 0 <= Arg_0 && Arg_2 <= Arg_3 && 0 <= Arg_1

  sqrt(Arg_0,Arg_1,Arg_2,Arg_3) -> f(0,1,1,Arg_3):|:  



Timebounds: 

  Overall timebound: max([2, 2+5*Arg_3]) {O(n)}

  1: f->f: max([0, 5*Arg_3]) {O(n)}

  2: f->end: 1 {O(1)}

  0: sqrt->f: 1 {O(1)}



Costbounds:

  Overall costbound: max([2, 2+5*Arg_3]) {O(n)}

  1: f->f: max([0, 5*Arg_3]) {O(n)}

  2: f->end: 1 {O(1)}

  0: sqrt->f: 1 {O(1)}



Sizebounds:

`Lower:

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

  1: f->f, Arg_1: 3 {O(1)}

  1: f->f, Arg_2: 4 {O(1)}

  1: f->f, Arg_3: 1 {O(1)}

  2: f->end, Arg_0: 0 {O(1)}

  2: f->end, Arg_1: 1 {O(1)}

  2: f->end, Arg_2: 1 {O(1)}

  2: f->end, Arg_3: min([1, Arg_3]) {O(n)}

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