Trying to load file: main.koat

Initial Control flow graph problem:
  Start location: f3
      0: f1 -> f1 : B'=1+B, C'=D, D'=free, E'=D, F'=free_1, G'=B, [ A>=1+B && B>=0 ], cost: 1
      1: f1 -> f4 : A'=free_8, B'=free_2, C'=free_4, D'=free_3, E'=free_10, H'=free_5, Q'=free_7, J'=free_6, K'=free_9, L'=C, [ B>=A && B>=0 && free_2>=free_5 && free_5>=2 ], cost: 1
      3: f3 -> f1 : A'=free_26, B'=2, C'=free_23, D'=free_22, E'=free_23, H'=free_26, Q'=free_23, M'=free_24, N'=free_25, [ free_26>=2 ], cost: 1
      2: f3 -> f4 : A'=free_13, B'=free_17, C'=free_11, D'=free_16, E'=free_12, H'=free_18, Q'=free_20, J'=free_19, K'=free_15, L'=0, M'=free_14, [ 0>=free_18 && 0>=free_21 ], cost: 1
      4: f3 -> f4 : A'=free_33, B'=free_27, C'=free_29, D'=free_28, E'=free_35, H'=1, Q'=free_32, J'=free_31, K'=free_34, L'=D, M'=free_30, [], cost: 1


Simplified the transitions:
  Start location: f3
      0: f1 -> f1 : B'=1+B, C'=D, D'=free, E'=D, F'=free_1, G'=B, [ A>=1+B && B>=0 ], cost: 1
      3: f3 -> f1 : A'=free_26, B'=2, C'=free_23, D'=free_22, E'=free_23, H'=free_26, Q'=free_23, M'=free_24, N'=free_25, [ free_26>=2 ], cost: 1

Eliminating 1 self-loops for location f1
  Self-Loop 0 has the metering function: -B+A, resulting in the new transition 5.
  Removing the self-loops: 0.

Removed all Self-loops using metering functions (where possible):
  Start location: f3
      5: f1 -> [3] : B'=A, C'=free, D'=free, E'=free, F'=free_1, G'=-1+A, [ A>=1+B && B>=0 ], cost: -B+A
      3: f3 -> f1 : A'=free_26, B'=2, C'=free_23, D'=free_22, E'=free_23, H'=free_26, Q'=free_23, M'=free_24, N'=free_25, [ free_26>=2 ], cost: 1


Applied simple chaining:
  Start location: f3
      3: f3 -> [3] : A'=free_26, B'=free_26, C'=free, D'=free, E'=free, F'=free_1, G'=-1+free_26, H'=free_26, Q'=free_23, M'=free_24, N'=free_25, [ free_26>=2 && free_26>=3 && 2>=0 ], cost: -1+free_26


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: f3
      3: f3 -> [3] : A'=free_26, B'=free_26, C'=free, D'=free, E'=free, F'=free_1, G'=-1+free_26, H'=free_26, Q'=free_23, M'=free_24, N'=free_25, [ free_26>=2 && free_26>=3 && 2>=0 ], cost: -1+free_26


Computing complexity for remaining 1 transitions.

  Found configuration with infinitely models for cost: -1+free_26
  and guard: free_26>=2 && free_26>=3 && 2>=0:
  free_26: Pos

Found new complexity INF, because: Found infinity configuration.


The final runtime is determined by this resulting transition:
  Final Guard: free_26>=2 && free_26>=3 && 2>=0
  Final Cost:  -1+free_26

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

WORST_CASE(INF,?)