Trying to load file: main.koat

Initial Control flow graph problem:
  Start location: f0
      0: f18 -> f18 : B'=1+B, [ A>=1+B ], cost: 1
      5: f18 -> f24 : B'=0, [ B>=A ], cost: 1
      1: f24 -> f24 : B'=1+B, [ A>=1+B ], cost: 1
      4: f24 -> f31 : B'=0, [ B>=A ], cost: 1
      2: f31 -> f31 : B'=1+B, [ A>=1+B ], cost: 1
      3: f31 -> f39 : [ B>=A ], cost: 1
      6: f0 -> f18 : A'=10, B'=0, C'=10, D'=free_1, E'=10, F'=free, [], cost: 1


Simplified the transitions:
  Start location: f0
      0: f18 -> f18 : B'=1+B, [ A>=1+B ], cost: 1
      5: f18 -> f24 : B'=0, [ B>=A ], cost: 1
      1: f24 -> f24 : B'=1+B, [ A>=1+B ], cost: 1
      4: f24 -> f31 : B'=0, [ B>=A ], cost: 1
      2: f31 -> f31 : B'=1+B, [ A>=1+B ], cost: 1
      6: f0 -> f18 : A'=10, B'=0, C'=10, D'=free_1, E'=10, F'=free, [], cost: 1

Eliminating 1 self-loops for location f18
  Self-Loop 0 has the metering function: -B+A, resulting in the new transition 7.
  Removing the self-loops: 0.
Eliminating 1 self-loops for location f24
  Self-Loop 1 has the metering function: -B+A, resulting in the new transition 8.
  Removing the self-loops: 1.
Eliminating 1 self-loops for location f31
  Self-Loop 2 has the metering function: -B+A, resulting in the new transition 9.
  Removing the self-loops: 2.

Removed all Self-loops using metering functions (where possible):
  Start location: f0
      7: f18 -> [5] : B'=A, [ A>=1+B ], cost: -B+A
      8: f24 -> [6] : B'=A, [ A>=1+B ], cost: -B+A
      9: f31 -> [7] : B'=A, [ A>=1+B ], cost: -B+A
      6: f0 -> f18 : A'=10, B'=0, C'=10, D'=free_1, E'=10, F'=free, [], cost: 1
      5: [5] -> f24 : B'=0, [ B>=A ], cost: 1
      4: [6] -> f31 : B'=0, [ B>=A ], cost: 1


Applied simple chaining:
  Start location: f0
      6: f0 -> [7] : A'=10, B'=10, C'=10, D'=free_1, E'=10, F'=free, [ 10>=1 && 10>=10 && 10>=1 && 10>=10 && 10>=1 ], cost: 33


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: f0
      6: f0 -> [7] : A'=10, B'=10, C'=10, D'=free_1, E'=10, F'=free, [ 10>=1 && 10>=10 && 10>=1 && 10>=10 && 10>=1 ], cost: 33


Computing complexity for remaining 1 transitions.

Found new complexity const, because: const cost.


The final runtime is determined by this resulting transition:
  Final Guard: 10>=1 && 10>=10 && 10>=1 && 10>=10 && 10>=1
  Final Cost:  33

Obtained the following complexity w.r.t. the length of the input n:
  Complexity class: const
  Complexity value: 0

WORST_CASE(Omega(1),?)