Trying to load file: main.koat

Initial Control flow graph problem:
  Start location: f0
      0: f0 -> f4 : A'=0, [], cost: 1
      1: f4 -> f4 : A'=1+A, [ 1>=A ], cost: 1
      5: f4 -> f10 : B'=0, [ A>=2 ], cost: 1
      2: f10 -> f10 : B'=1+B, [ 1>=B ], cost: 1
      3: f10 -> f18 : [ B>=2 && 0>=1+free ], cost: 1
      4: f10 -> f18 : [ B>=2 ], cost: 1


Simplified the transitions:
  Start location: f0
      0: f0 -> f4 : A'=0, [], cost: 1
      1: f4 -> f4 : A'=1+A, [ 1>=A ], cost: 1
      5: f4 -> f10 : B'=0, [ A>=2 ], cost: 1
      2: f10 -> f10 : B'=1+B, [ 1>=B ], cost: 1

Eliminating 1 self-loops for location f4
  Self-Loop 1 has the metering function: 2-A, resulting in the new transition 6.
  Removing the self-loops: 1.
Eliminating 1 self-loops for location f10
  Self-Loop 2 has the metering function: 2-B, resulting in the new transition 7.
  Removing the self-loops: 2.

Removed all Self-loops using metering functions (where possible):
  Start location: f0
      0: f0 -> f4 : A'=0, [], cost: 1
      6: f4 -> [4] : A'=2, [ 1>=A ], cost: 2-A
      7: f10 -> [5] : B'=2, [ 1>=B ], cost: 2-B
      5: [4] -> f10 : B'=0, [ A>=2 ], cost: 1


Applied simple chaining:
  Start location: f0
      0: f0 -> [5] : A'=2, B'=2, [ 1>=0 && 2>=2 && 1>=0 ], cost: 6


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: f0
      0: f0 -> [5] : A'=2, B'=2, [ 1>=0 && 2>=2 && 1>=0 ], cost: 6


Computing complexity for remaining 1 transitions.

Found new complexity const, because: const cost.


The final runtime is determined by this resulting transition:
  Final Guard: 1>=0 && 2>=2 && 1>=0
  Final Cost:  6

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

WORST_CASE(Omega(1),?)