Trying to load file: main.koat

Initial Control flow graph problem:
  Start location: f300
      0: f2 -> f2 : A'=-1+A, B'=free, [ free>=1 && A>=2 ], cost: 1
      1: f2 -> f2 : A'=-1+A, B'=free_1, [ 0>=1+free_1 && A>=2 ], cost: 1
      4: f2 -> f2 : B'=0, [ A>=1 ], cost: 1
      2: f2 -> f1 : A'=-1+A, B'=free_3, C'=free_2, [ free_3>=1 && 1>=A ], cost: 1
      3: f2 -> f1 : A'=-1+A, B'=free_5, C'=free_4, [ 0>=1+free_5 && 1>=A ], cost: 1
      5: f2 -> f1 : B'=0, C'=free_6, [ 0>=A ], cost: 1
      6: f300 -> f2 : [], cost: 1


Simplified the transitions:
  Start location: f300
      0: f2 -> f2 : A'=-1+A, B'=free, [ free>=1 && A>=2 ], cost: 1
      1: f2 -> f2 : A'=-1+A, B'=free_1, [ 0>=1+free_1 && A>=2 ], cost: 1
      4: f2 -> f2 : B'=0, [ A>=1 ], cost: 1
      6: f300 -> f2 : [], cost: 1

Eliminating 3 self-loops for location f2
  Self-Loop 0 has the metering function: -1+A, resulting in the new transition 7.
  Self-Loop 1 has the metering function: -1+A, resulting in the new transition 8.
  Self-Loop 4 has unbounded runtime, resulting in the new transition 9.
  Removing the self-loops: 0 1 4.

Removed all Self-loops using metering functions (where possible):
  Start location: f300
      7: f2 -> [3] : A'=1, B'=free, [ free>=1 && A>=2 ], cost: -1+A
      8: f2 -> [3] : A'=1, B'=free_1, [ 0>=1+free_1 && A>=2 ], cost: -1+A
      9: f2 -> [3] : [ A>=1 ], cost: INF
      6: f300 -> f2 : [], cost: 1


Applied chaining over branches and pruning:
  Start location: f300
     10: f300 -> [3] : A'=1, B'=free, [ free>=1 && A>=2 ], cost: A
     11: f300 -> [3] : A'=1, B'=free_1, [ 0>=1+free_1 && A>=2 ], cost: A
     12: f300 -> [3] : [ A>=1 ], cost: INF


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: f300
     10: f300 -> [3] : A'=1, B'=free, [ free>=1 && A>=2 ], cost: A
     11: f300 -> [3] : A'=1, B'=free_1, [ 0>=1+free_1 && A>=2 ], cost: A
     12: f300 -> [3] : [ A>=1 ], cost: INF


Computing complexity for remaining 3 transitions.

  Found configuration with infinitely models for cost: A
  and guard: free>=1 && A>=2:
  free: Pos, A: Pos

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

Found new complexity INF, because: INF sat.


The final runtime is determined by this resulting transition:
  Final Guard: A>=1
  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,?)