Trying to load file: main.koat

Initial Control flow graph problem:
  Start location: l0
      0: l0 -> l1 : A'=0, [], cost: 1
      1: l1 -> l2 : C'=0, D'=0, [ B>0 ], cost: 1
      4: l1 -> l3 : [ B<=0 ], cost: 1
      3: l2 -> l1 : A'=D+A, B'=-1+B, [ C>=B ], cost: 1
      2: l2 -> l2 : C'=1+C, D'=C+D, [ C<B ], cost: 1
      5: l3 -> l3 : A'=-1+A, [ A>0 ], cost: 1

Eliminating 1 self-loops for location l2
  Self-Loop 2 has the metering function: B-C, resulting in the new transition 6.
  Removing the self-loops: 2.
Eliminating 1 self-loops for location l3
  Self-Loop 5 has the metering function: A, resulting in the new transition 7.
  Removing the self-loops: 5.

Removed all Self-loops using metering functions (where possible):
  Start location: l0
      0: l0 -> l1 : A'=0, [], cost: 1
      1: l1 -> l2 : C'=0, D'=0, [ B>0 ], cost: 1
      4: l1 -> l3 : [ B<=0 ], cost: 1
      6: l2 -> [4] : C'=B, D'=-1+1/2*B+(B-C)*C-1/2*C+1/2*(B-C)^2+D, [ C<B ], cost: B-C
      7: l3 -> [5] : A'=0, [ A>0 ], cost: A
      3: [4] -> l1 : A'=D+A, B'=-1+B, [ C>=B ], cost: 1


Applied simple chaining:
  Start location: l0
      0: l0 -> l1 : A'=0, [], cost: 1
      1: l1 -> l1 : A'=-1+1/2*B+1/2*B^2+A, B'=-1+B, C'=B, D'=-1+1/2*B+1/2*B^2, [ B>0 && 0<B && B>=B ], cost: 2+B
      4: l1 -> [5] : A'=0, [ B<=0 && A>0 ], cost: 1+A

Eliminating 1 self-loops for location l1
  Self-Loop 1 has the metering function: B, resulting in the new transition 8.
  Removing the self-loops: 1.

Removed all Self-loops using metering functions (where possible):
  Start location: l0
      0: l0 -> l1 : A'=0, [], cost: 1
      8: l1 -> [6] : A'=1-7/6*B+1/6*B^3+A, B'=0, C'=1, D'=0, [ B>0 ], cost: 5/2*B+1/2*B^2
      4: [6] -> [5] : A'=0, [ B<=0 && A>0 ], cost: 1+A


Applied simple chaining:
  Start location: l0
      0: l0 -> [5] : A'=0, B'=0, C'=1, D'=0, [ B>0 && 0<=0 && 1-7/6*B+1/6*B^3>0 ], cost: 3+4/3*B+1/2*B^2+1/6*B^3


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: l0
      0: l0 -> [5] : A'=0, B'=0, C'=1, D'=0, [ B>0 && 0<=0 && 1-7/6*B+1/6*B^3>0 ], cost: 3+4/3*B+1/2*B^2+1/6*B^3


Computing complexity for remaining 1 transitions.

  Found configuration with infinitely models for cost: 3+4/3*B+1/2*B^2+1/6*B^3
  and guard: B>0 && 0<=0 && 1-7/6*B+1/6*B^3>0:
  B: Pos

Found new complexity n^3, because: Found infinity configuration.


The final runtime is determined by this resulting transition:
  Final Guard: B>0 && 0<=0 && 1-7/6*B+1/6*B^3>0
  Final Cost:  3+4/3*B+1/2*B^2+1/6*B^3

Obtained the following complexity w.r.t. the length of the input n:
  Complexity class: n^3
  Complexity value: 3

WORST_CASE(Omega(n^3),?)