Trying to load file: main.koat

Initial Control flow graph problem:
  Start location: f0
      0: f0 -> f1 : A'=0, [], cost: 1
      1: f1 -> f1 : B'=-1+B, C'=free, [ B>=1 && free>=1 ], cost: 1
      2: f1 -> f1 : B'=-2+B, C'=free_1, [ B>=1 && 0>=free_1 ], cost: 1
      3: f1 -> f4 : C'=free_2, [ 0>=B ], cost: 1
      4: f4 -> f4 : A'=1, C'=free_3, [ C>=1 ], cost: 1
      5: f4 -> f4 : A'=2, C'=free_4, [ 0>=C ], cost: 1

Eliminating 2 self-loops for location f1
  Self-Loop 1 has the metering function: B, resulting in the new transition 6.
  Self-Loop 2 has the metering function: meter, resulting in the new transition 7.
  Removing the self-loops: 1 2.
Eliminating 2 self-loops for location f4
  Removing the self-loops: 4 5.
Adding an epsilon transition (to model nonexecution of the loops): 10.

Removed all Self-loops using metering functions (where possible):
  Start location: f0
      0: f0 -> f1 : A'=0, [], cost: 1
      6: f1 -> [3] : B'=0, C'=free, [ B>=1 && free>=1 ], cost: B
      7: f1 -> [3] : B'=B-2*meter, C'=free_1, [ B>=1 && 0>=free_1 && 2*meter==B ], cost: meter
      8: f4 -> [4] : A'=1, C'=free_3, [ C>=1 ], cost: 1
      9: f4 -> [4] : A'=2, C'=free_4, [ 0>=C ], cost: 1
     10: f4 -> [4] : [], cost: 0
      3: [3] -> f4 : C'=free_2, [ 0>=B ], cost: 1


Applied chaining over branches and pruning:
  Start location: f0
     11: f0 -> [3] : A'=0, B'=0, C'=free, [ B>=1 && free>=1 ], cost: 1+B
     12: f0 -> [3] : A'=0, B'=B-2*meter, C'=free_1, [ B>=1 && 0>=free_1 && 2*meter==B ], cost: 1+meter


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: f0
     11: f0 -> [3] : A'=0, B'=0, C'=free, [ B>=1 && free>=1 ], cost: 1+B
     12: f0 -> [3] : A'=0, B'=B-2*meter, C'=free_1, [ B>=1 && 0>=free_1 && 2*meter==B ], cost: 1+meter


Computing complexity for remaining 2 transitions.

  Found configuration with infinitely models for cost: 1+B
  and guard: B>=1 && free>=1:
  free: Pos, B: Pos

Found new complexity n^1, because: Found infinity configuration.


The final runtime is determined by this resulting transition:
  Final Guard: B>=1 && free>=1
  Final Cost:  1+B

Obtained the following complexity w.r.t. the length of the input n:
  Complexity class: n^1
  Complexity value: 1

WORST_CASE(Omega(n^1),?)