Trying to load file: main.koat

Initial Control flow graph problem:
  Start location: f0
      0: f0 -> f7 : A'=8, B'=0, [], cost: 1
      1: f7 -> f7 : B'=1+B, C'=free_10+free_5, D'=free_15, E'=free_3+free_8, F'=free_13, G'=free_6+free_1, H'=free_11, Q'=free_16+free_4, J'=free_9, K'=free_16+free_10+free_4+free_5, L'=-free_16+free_10-free_4+free_5, M'=free_6+free_3+free_1+free_8, N'=-free_6+free_3-free_1+free_8, O'=-3196, P'=free_14, Q_1'=free_2, R'=free_12+free_7, S'=free_12+free, T'=free_12, [ 7>=B ], cost: 1
      4: f7 -> f62 : B'=0, [ B>=8 ], cost: 1
      2: f62 -> f62 : B'=1+B, C'=free_27+free_22, D'=free_32, E'=free_25+free_20, F'=free_30, G'=free_18+free_23, H'=free_28, Q'=free_33+free_21, J'=free_26, K'=free_33+free_27+free_22+free_21, L'=-free_33+free_27+free_22-free_21, M'=free_18+free_25+free_23+free_20, N'=-free_18+free_25-free_23+free_20, O'=-3196, P'=free_31, Q_1'=free_19, R'=free_24+free_29, S'=free_17+free_29, T'=free_29, [ 7>=B ], cost: 1
      3: f62 -> f118 : [ B>=8 ], cost: 1


Simplified the transitions:
  Start location: f0
      0: f0 -> f7 : A'=8, B'=0, [], cost: 1
      1: f7 -> f7 : B'=1+B, C'=free_10+free_5, D'=free_15, E'=free_3+free_8, F'=free_13, G'=free_6+free_1, H'=free_11, Q'=free_16+free_4, J'=free_9, K'=free_16+free_10+free_4+free_5, L'=-free_16+free_10-free_4+free_5, M'=free_6+free_3+free_1+free_8, N'=-free_6+free_3-free_1+free_8, O'=-3196, P'=free_14, Q_1'=free_2, R'=free_12+free_7, S'=free_12+free, T'=free_12, [ 7>=B ], cost: 1
      4: f7 -> f62 : B'=0, [ B>=8 ], cost: 1
      2: f62 -> f62 : B'=1+B, C'=free_27+free_22, D'=free_32, E'=free_25+free_20, F'=free_30, G'=free_18+free_23, H'=free_28, Q'=free_33+free_21, J'=free_26, K'=free_33+free_27+free_22+free_21, L'=-free_33+free_27+free_22-free_21, M'=free_18+free_25+free_23+free_20, N'=-free_18+free_25-free_23+free_20, O'=-3196, P'=free_31, Q_1'=free_19, R'=free_24+free_29, S'=free_17+free_29, T'=free_29, [ 7>=B ], cost: 1

Eliminating 1 self-loops for location f7
  Self-Loop 1 has the metering function: 8-B, resulting in the new transition 5.
  Removing the self-loops: 1.
Eliminating 1 self-loops for location f62
  Self-Loop 2 has the metering function: 8-B, resulting in the new transition 6.
  Removing the self-loops: 2.

Removed all Self-loops using metering functions (where possible):
  Start location: f0
      0: f0 -> f7 : A'=8, B'=0, [], cost: 1
      5: f7 -> [4] : B'=8, C'=free_10+free_5, D'=free_15, E'=free_3+free_8, F'=free_13, G'=free_6+free_1, H'=free_11, Q'=free_16+free_4, J'=free_9, K'=free_16+free_10+free_4+free_5, L'=-free_16+free_10-free_4+free_5, M'=free_6+free_3+free_1+free_8, N'=-free_6+free_3-free_1+free_8, O'=-3196, P'=free_14, Q_1'=free_2, R'=free_12+free_7, S'=free_12+free, T'=free_12, [ 7>=B ], cost: 8-B
      6: f62 -> [5] : B'=8, C'=free_27+free_22, D'=free_32, E'=free_25+free_20, F'=free_30, G'=free_18+free_23, H'=free_28, Q'=free_33+free_21, J'=free_26, K'=free_33+free_27+free_22+free_21, L'=-free_33+free_27+free_22-free_21, M'=free_18+free_25+free_23+free_20, N'=-free_18+free_25-free_23+free_20, O'=-3196, P'=free_31, Q_1'=free_19, R'=free_24+free_29, S'=free_17+free_29, T'=free_29, [ 7>=B ], cost: 8-B
      4: [4] -> f62 : B'=0, [ B>=8 ], cost: 1


Applied simple chaining:
  Start location: f0
      0: f0 -> [5] : A'=8, B'=8, C'=free_27+free_22, D'=free_32, E'=free_25+free_20, F'=free_30, G'=free_18+free_23, H'=free_28, Q'=free_33+free_21, J'=free_26, K'=free_33+free_27+free_22+free_21, L'=-free_33+free_27+free_22-free_21, M'=free_18+free_25+free_23+free_20, N'=-free_18+free_25-free_23+free_20, O'=-3196, P'=free_31, Q_1'=free_19, R'=free_24+free_29, S'=free_17+free_29, T'=free_29, [ 7>=0 && 8>=8 && 7>=0 ], cost: 18


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: f0
      0: f0 -> [5] : A'=8, B'=8, C'=free_27+free_22, D'=free_32, E'=free_25+free_20, F'=free_30, G'=free_18+free_23, H'=free_28, Q'=free_33+free_21, J'=free_26, K'=free_33+free_27+free_22+free_21, L'=-free_33+free_27+free_22-free_21, M'=free_18+free_25+free_23+free_20, N'=-free_18+free_25-free_23+free_20, O'=-3196, P'=free_31, Q_1'=free_19, R'=free_24+free_29, S'=free_17+free_29, T'=free_29, [ 7>=0 && 8>=8 && 7>=0 ], cost: 18


Computing complexity for remaining 1 transitions.

Found new complexity const, because: const cost.


The final runtime is determined by this resulting transition:
  Final Guard: 7>=0 && 8>=8 && 7>=0
  Final Cost:  18

Obtained the following complexity w.r.t. the length of the input n:
  Complexity class: const
  Complexity value: 0

WORST_CASE(Omega(1),?)