/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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))


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(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)}

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

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

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

  0: sqrt->f, Arg_3: Arg_3 {O(n)}

`Upper:

  1: f->f, Arg_0: max([0, 5*Arg_3]) {O(n)}

  1: f->f, Arg_1: max([1, 1+10*Arg_3]) {O(n)}

  1: f->f, Arg_2: max([1, 2^(5*Arg_3)])*max([1, 1+5*Arg_3]) {O(EXP)}

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

  2: f->end, Arg_0: max([0, 5*Arg_3]) {O(n)}

  2: f->end, Arg_1: max([1, max([1, 1+10*Arg_3])]) {O(n)}

  2: f->end, Arg_2: max([1, max([1, 2^(5*Arg_3)])*max([1, 1+5*Arg_3])]) {O(EXP)}

  2: f->end, Arg_3: Arg_3 {O(n)}

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

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

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

  0: sqrt->f, Arg_3: Arg_3 {O(n)}


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(2)
BOUNDS(1, max(2, 2 + 5 * Arg_3))