Trying to load file: main.koat

Initial Control flow graph problem:
  Start location: f0
      0: f0 -> f5 : [], cost: 1
      4: f0 -> f5 : D'=1, E'=1, [ 15>=free_4 ], cost: 1
      1: f0 -> f12 : A'=free, [ 15>=free_1 ], cost: 1
      6: f12 -> f5 : A'=0, D'=0, E'=0, H'=free_5, Q'=free_5, J'=0, K'=0, [ A==0 ], cost: 1
      2: f12 -> f20 : B'=free_2, C'=free_2, [ 0>=1+A ], cost: 1
      3: f12 -> f20 : B'=free_3, C'=free_3, [ A>=1 ], cost: 1
      5: f20 -> f12 : A'=-1+A, F'=0, G'=0, [], cost: 1


Simplified the transitions:
  Start location: f0
      1: f0 -> f12 : A'=free, [], cost: 1
      2: f12 -> f20 : B'=free_2, C'=free_2, [ 0>=1+A ], cost: 1
      3: f12 -> f20 : B'=free_3, C'=free_3, [ A>=1 ], cost: 1
      5: f20 -> f12 : A'=-1+A, F'=0, G'=0, [], cost: 1


Applied chaining over branches and pruning:
  Start location: f0
      1: f0 -> f12 : A'=free, [], cost: 1
      7: f12 -> f12 : A'=-1+A, B'=free_2, C'=free_2, F'=0, G'=0, [ 0>=1+A ], cost: 2
      8: f12 -> f12 : A'=-1+A, B'=free_3, C'=free_3, F'=0, G'=0, [ A>=1 ], cost: 2

Eliminating 2 self-loops for location f12
  Self-Loop 7 has unbounded runtime, resulting in the new transition 9.
  Self-Loop 8 has the metering function: A, resulting in the new transition 10.
  Removing the self-loops: 7 8.

Removed all Self-loops using metering functions (where possible):
  Start location: f0
      1: f0 -> f12 : A'=free, [], cost: 1
      9: f12 -> [4] : [ 0>=1+A ], cost: INF
     10: f12 -> [4] : A'=0, B'=free_3, C'=free_3, F'=0, G'=0, [ A>=1 ], cost: 2*A


Applied chaining over branches and pruning:
  Start location: f0
     11: f0 -> [4] : A'=free, [ 0>=1+free ], cost: INF
     12: f0 -> [4] : A'=0, B'=free_3, C'=free_3, F'=0, G'=0, [ free>=1 ], cost: 1+2*free


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: f0
     11: f0 -> [4] : A'=free, [ 0>=1+free ], cost: INF
     12: f0 -> [4] : A'=0, B'=free_3, C'=free_3, F'=0, G'=0, [ free>=1 ], cost: 1+2*free


Computing complexity for remaining 2 transitions.

Found new complexity INF, because: INF sat.


The final runtime is determined by this resulting transition:
  Final Guard: 0>=1+free
  Final Cost:  INF

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

WORST_CASE(INF,?)