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


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


WORST_CASE(Omega(n^1), 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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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(n^1, max(3, 3 + 2 * Arg_1)).

(0) CpxIntTrs
(1) Koat2 Proof [FINISHED, 138 ms]
(2) BOUNDS(1, max(3, 3 + 2 * Arg_1))
(3) Loat Proof [FINISHED, 215 ms]
(4) BOUNDS(n^1, INF)


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

(0)
Obligation:
Complexity Int TRS consisting of the following rules:
div(A, B) -> Com_1(end(A, B)) :|: 0 >= A
div(A, B) -> Com_1(end(A, B)) :|: A >= B
div(A, B) -> Com_1(div(A, B - A)) :|: B >= A + 1 && A >= 1
start(A, B) -> Com_1(div(A, B)) :|: TRUE

The start-symbols are:[start_2]


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

(1) Koat2 Proof (FINISHED)
YES( ?, max([3, 3+2*Arg_1]) {O(n)})



Initial Complexity Problem:

  Start:  start  

  Program_Vars:  Arg_0, Arg_1  

  Temp_Vars:    

  Locations:  div, end, start  

  Transitions:

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

  div(Arg_0,Arg_1) -> end(Arg_0,Arg_1):|:Arg_0 <= 0

  div(Arg_0,Arg_1) -> end(Arg_0,Arg_1):|:Arg_1 <= Arg_0

  start(Arg_0,Arg_1) -> div(Arg_0,Arg_1):|:  



Timebounds: 

  Overall timebound: max([3, 3+2*Arg_1]) {O(n)}

  0: div->end: 1 {O(1)}

  1: div->end: 1 {O(1)}

  2: div->div: max([0, 2*Arg_1]) {O(n)}

  3: start->div: 1 {O(1)}



Costbounds:

  Overall costbound: max([3, 3+2*Arg_1]) {O(n)}

  0: div->end: 1 {O(1)}

  1: div->end: 1 {O(1)}

  2: div->div: max([0, 2*Arg_1]) {O(n)}

  3: start->div: 1 {O(1)}



Sizebounds:

`Lower:

  0: div->end, Arg_0: Arg_0 {O(n)}

  0: div->end, Arg_1: Arg_1 {O(n)}

  1: div->end, Arg_0: min([1, Arg_0]) {O(n)}

  1: div->end, Arg_1: min([1, Arg_1]) {O(n)}

  2: div->div, Arg_0: 1 {O(1)}

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

  3: start->div, Arg_0: Arg_0 {O(n)}

  3: start->div, Arg_1: Arg_1 {O(n)}

`Upper:

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

  0: div->end, Arg_1: Arg_1 {O(n)}

  1: div->end, Arg_0: Arg_0 {O(n)}

  1: div->end, Arg_1: Arg_1 {O(n)}

  2: div->div, Arg_0: Arg_0 {O(n)}

  2: div->div, Arg_1: Arg_1 {O(n)}

  3: start->div, Arg_0: Arg_0 {O(n)}

  3: start->div, Arg_1: Arg_1 {O(n)}


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

(2)
BOUNDS(1, max(3, 3 + 2 * Arg_1))

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

(3) Loat Proof (FINISHED)


### Pre-processing the ITS problem ###



Initial linear ITS problem

   Start location: start

      0: div -> end : [ 0>=A ], cost: 1

      1: div -> end : [ A>=B ], cost: 1

      2: div -> div : B'=-A+B, [ B>=1+A && A>=1 ], cost: 1

      3: start -> div : [], cost: 1



Removed unreachable and leaf rules:

   Start location: start

      2: div -> div : B'=-A+B, [ B>=1+A && A>=1 ], cost: 1

      3: start -> div : [], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 0.

   Accelerating the following rules:

      2: div -> div : B'=-A+B, [ B>=1+A && A>=1 ], cost: 1



   Accelerated rule 2 with backward acceleration, yielding the new rule 4.

   Removing the simple loops: 2.



Accelerated all simple loops using metering functions (where possible):

   Start location: start

      4: div -> div : B'=-k*A+B, [ B>=1+A && A>=1 && k>0 && -(-1+k)*A+B>=1+A ], cost: k

      3: start -> div : [], cost: 1



Chained accelerated rules (with incoming rules):

   Start location: start

      3: start -> div : [], cost: 1

      5: start -> div : B'=-k*A+B, [ B>=1+A && A>=1 && k>0 && -(-1+k)*A+B>=1+A ], cost: 1+k



Removed unreachable locations (and leaf rules with constant cost):

   Start location: start

      5: start -> div : B'=-k*A+B, [ B>=1+A && A>=1 && k>0 && -(-1+k)*A+B>=1+A ], cost: 1+k



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: start

      5: start -> div : B'=-k*A+B, [ B>=1+A && A>=1 && k>0 && -(-1+k)*A+B>=1+A ], cost: 1+k



Computing asymptotic complexity for rule 5

   Simplified the guard:

      5: start -> div : B'=-k*A+B, [ A>=1 && k>0 && -(-1+k)*A+B>=1+A ], cost: 1+k

   Solved the limit problem by the following transformations:

      Created initial limit problem:

      -(-1+k)*A-A+B (+/+!), k (+/+!), 1+k (+), A (+/+!) [not solved]



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



      applying transformation rule (C) using substitution {k==n,A==1,B==2*n}

      resulting limit problem:

      [solved]



   Solved the limit problem by the following transformations:

      Created initial limit problem:

      -(-1+k)*A-A+B (+/+!), k (+/+!), 1+k (+), A (+/+!) [not solved]



      applying transformation rule (C) using substitution {A==1}

      resulting limit problem:

      1 (+/+!), -k+B (+/+!), k (+/+!), 1+k (+) [not solved]



      applying transformation rule (B), deleting 1 (+/+!)

      resulting limit problem:

      -k+B (+/+!), k (+/+!), 1+k (+) [not solved]



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



      applying transformation rule (C) using substitution {k==n,B==1+n}

      resulting limit problem:

      [solved]



   Solution:

      k / n

      A / 1

      B / 2*n

   Resulting cost 1+n has complexity: Poly(n^1)



Found new complexity Poly(n^1).



Obtained the following overall complexity (w.r.t. the length of the input n):

   Complexity:  Poly(n^1)

   Cpx degree:  1

   Solved cost: 1+n

   Rule cost:   1+k

   Rule guard:  [ A>=1 && k>0 && -(-1+k)*A+B>=1+A ]



WORST_CASE(Omega(n^1),?)


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

(4)
BOUNDS(n^1, INF)