Trying to load file: main.koat

Initial Control flow graph problem:
  Start location: f1
      1: f0 -> f0 : C'=-1+C, D'=free_2, E'=0, F'=free_1, [ A>=1 && C>=3 ], cost: 1
      2: f0 -> f0 : A'=-1+A, C'=free_4, D'=free_5, E'=free_3, [ 0>=1+free_3 && A>=1 ], cost: 1
      3: f0 -> f0 : A'=-1+A, C'=free_7, D'=free_8, E'=free_6, [ free_6>=1 && A>=1 ], cost: 1
      0: f0 -> f2 : B'=free, [ 0>=A ], cost: 1
      4: f1 -> f0 : [], cost: 1


Simplified the transitions:
  Start location: f1
      1: f0 -> f0 : C'=-1+C, D'=free_2, E'=0, F'=free_1, [ A>=1 && C>=3 ], cost: 1
      2: f0 -> f0 : A'=-1+A, C'=free_4, D'=free_5, E'=free_3, [ 0>=1+free_3 && A>=1 ], cost: 1
      3: f0 -> f0 : A'=-1+A, C'=free_7, D'=free_8, E'=free_6, [ free_6>=1 && A>=1 ], cost: 1
      4: f1 -> f0 : [], cost: 1

Eliminating 3 self-loops for location f0
  Self-Loop 1 has the metering function: -2+C, resulting in the new transition 5.
  Self-Loop 2 has the metering function: A, resulting in the new transition 6.
  Self-Loop 3 has the metering function: A, resulting in the new transition 7.
  Found this metering function when nesting loops: -1+A,
  and nested parallel self-loops 2 (outer loop) and 5 (inner loop), obtaining the new transitions: 8, 9.
  Found this metering function when nesting loops: -1+A,
  and nested parallel self-loops 3 (outer loop) and 5 (inner loop), obtaining the new transitions: 10, 11.
  Removing the self-loops: 1 2 3.

Removed all Self-loops using metering functions (where possible):
  Start location: f1
      5: f0 -> [3] : C'=2, D'=free_2, E'=0, F'=free_1, [ A>=1 && C>=3 ], cost: -2+C
      6: f0 -> [3] : A'=0, C'=free_4, D'=free_5, E'=free_3, [ 0>=1+free_3 && A>=1 ], cost: A
      7: f0 -> [3] : A'=0, C'=free_7, D'=free_8, E'=free_6, [ free_6>=1 && A>=1 ], cost: A
      8: f0 -> [3] : A'=1, C'=2, D'=free_2, E'=0, F'=free_1, [ 0>=1+free_3 && A>=1 && -1+A>=1 && free_4>=3 ], cost: 1+free_4*(-1+A)-A
      9: f0 -> [3] : A'=1, C'=2, D'=free_2, E'=0, F'=free_1, [ A>=1 && C>=3 && 0>=1+free_3 && A>=1 && -1+A>=1 && free_4>=3 ], cost: -1+free_4*(-1+A)+C-A
     10: f0 -> [3] : A'=1, C'=2, D'=free_2, E'=0, F'=free_1, [ free_6>=1 && A>=1 && -1+A>=1 && free_7>=3 ], cost: 1+free_7*(-1+A)-A
     11: f0 -> [3] : A'=1, C'=2, D'=free_2, E'=0, F'=free_1, [ A>=1 && C>=3 && free_6>=1 && A>=1 && -1+A>=1 && free_7>=3 ], cost: -1+free_7*(-1+A)+C-A
      4: f1 -> f0 : [], cost: 1


Applied chaining over branches and pruning:
  Start location: f1
     12: f1 -> [3] : C'=2, D'=free_2, E'=0, F'=free_1, [ A>=1 && C>=3 ], cost: -1+C
     13: f1 -> [3] : A'=0, C'=free_4, D'=free_5, E'=free_3, [ 0>=1+free_3 && A>=1 ], cost: 1+A
     14: f1 -> [3] : A'=0, C'=free_7, D'=free_8, E'=free_6, [ free_6>=1 && A>=1 ], cost: 1+A
     16: f1 -> [3] : A'=1, C'=2, D'=free_2, E'=0, F'=free_1, [ A>=1 && C>=3 && 0>=1+free_3 && A>=1 && -1+A>=1 && free_4>=3 ], cost: free_4*(-1+A)+C-A
     18: f1 -> [3] : A'=1, C'=2, D'=free_2, E'=0, F'=free_1, [ A>=1 && C>=3 && free_6>=1 && A>=1 && -1+A>=1 && free_7>=3 ], cost: free_7*(-1+A)+C-A


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: f1
     12: f1 -> [3] : C'=2, D'=free_2, E'=0, F'=free_1, [ A>=1 && C>=3 ], cost: -1+C
     13: f1 -> [3] : A'=0, C'=free_4, D'=free_5, E'=free_3, [ 0>=1+free_3 && A>=1 ], cost: 1+A
     14: f1 -> [3] : A'=0, C'=free_7, D'=free_8, E'=free_6, [ free_6>=1 && A>=1 ], cost: 1+A
     16: f1 -> [3] : A'=1, C'=2, D'=free_2, E'=0, F'=free_1, [ A>=1 && C>=3 && 0>=1+free_3 && A>=1 && -1+A>=1 && free_4>=3 ], cost: free_4*(-1+A)+C-A
     18: f1 -> [3] : A'=1, C'=2, D'=free_2, E'=0, F'=free_1, [ A>=1 && C>=3 && free_6>=1 && A>=1 && -1+A>=1 && free_7>=3 ], cost: free_7*(-1+A)+C-A


Computing complexity for remaining 5 transitions.

  Found configuration with infinitely models for cost: -1+C
  and guard: A>=1 && C>=3:
  C: Pos, A: Pos

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

  Found configuration with infinitely models for cost: free_4*(-1+A)+C-A
  and guard: A>=1 && C>=3 && 0>=1+free_3 && A>=1 && -1+A>=1 && free_4>=3:
  free_4: Const, C: Pos, free_3: Neg, A: Const

  Found configuration with infinitely models for cost: free_7*(-1+A)+C-A
  and guard: A>=1 && C>=3 && free_6>=1 && A>=1 && -1+A>=1 && free_7>=3:
  free_7: Const, C: Pos, free_6: Pos, A: Const


The final runtime is determined by this resulting transition:
  Final Guard: A>=1 && C>=3
  Final Cost:  -1+C

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),?)