Trying to load file: main.koat

Initial Control flow graph problem:
  Start location: f0
      0: f0 -> f17 : A'=0, B'=free_1, C'=free, D'=0, [], cost: 1
      1: f17 -> f17 : D'=1+D, [ 49>=D ], cost: 1
     16: f17 -> f27 : E'=0, [ D>=50 ], cost: 1
      2: f27 -> f27 : E'=1+E, [ 49>=E ], cost: 1
     15: f27 -> f37 : F'=0, [ E>=50 ], cost: 1
      3: f37 -> f37 : F'=1+F, [ 49>=F ], cost: 1
     14: f37 -> f45 : A'=0, [ F>=50 ], cost: 1
      4: f45 -> f45 : A'=1+A, [ 49>=A ], cost: 1
     13: f45 -> f55 : G'=0, [ A>=50 ], cost: 1
      5: f55 -> f55 : G'=1+G, [ 49>=G ], cost: 1
     12: f55 -> f65 : H'=0, [ G>=50 ], cost: 1
      6: f65 -> f65 : H'=1+H, [ 49>=H ], cost: 1
     11: f65 -> f75 : Q'=0, [ H>=50 ], cost: 1
      7: f75 -> f75 : Q'=1+Q, [ 49>=Q ], cost: 1
     10: f75 -> f83 : A'=0, [ Q>=50 ], cost: 1
      8: f83 -> f83 : A'=1+A, [ 49>=A ], cost: 1
      9: f83 -> f93 : [ A>=50 ], cost: 1


Simplified the transitions:
  Start location: f0
      0: f0 -> f17 : A'=0, B'=free_1, C'=free, D'=0, [], cost: 1
      1: f17 -> f17 : D'=1+D, [ 49>=D ], cost: 1
     16: f17 -> f27 : E'=0, [ D>=50 ], cost: 1
      2: f27 -> f27 : E'=1+E, [ 49>=E ], cost: 1
     15: f27 -> f37 : F'=0, [ E>=50 ], cost: 1
      3: f37 -> f37 : F'=1+F, [ 49>=F ], cost: 1
     14: f37 -> f45 : A'=0, [ F>=50 ], cost: 1
      4: f45 -> f45 : A'=1+A, [ 49>=A ], cost: 1
     13: f45 -> f55 : G'=0, [ A>=50 ], cost: 1
      5: f55 -> f55 : G'=1+G, [ 49>=G ], cost: 1
     12: f55 -> f65 : H'=0, [ G>=50 ], cost: 1
      6: f65 -> f65 : H'=1+H, [ 49>=H ], cost: 1
     11: f65 -> f75 : Q'=0, [ H>=50 ], cost: 1
      7: f75 -> f75 : Q'=1+Q, [ 49>=Q ], cost: 1
     10: f75 -> f83 : A'=0, [ Q>=50 ], cost: 1
      8: f83 -> f83 : A'=1+A, [ 49>=A ], cost: 1

Eliminating 1 self-loops for location f17
  Self-Loop 1 has the metering function: 50-D, resulting in the new transition 17.
  Removing the self-loops: 1.
Eliminating 1 self-loops for location f27
  Self-Loop 2 has the metering function: 50-E, resulting in the new transition 18.
  Removing the self-loops: 2.
Eliminating 1 self-loops for location f37
  Self-Loop 3 has the metering function: 50-F, resulting in the new transition 19.
  Removing the self-loops: 3.
Eliminating 1 self-loops for location f45
  Self-Loop 4 has the metering function: 50-A, resulting in the new transition 20.
  Removing the self-loops: 4.
Eliminating 1 self-loops for location f55
  Self-Loop 5 has the metering function: 50-G, resulting in the new transition 21.
  Removing the self-loops: 5.
Eliminating 1 self-loops for location f65
  Self-Loop 6 has the metering function: 50-H, resulting in the new transition 22.
  Removing the self-loops: 6.
Eliminating 1 self-loops for location f75
  Self-Loop 7 has the metering function: 50-Q, resulting in the new transition 23.
  Removing the self-loops: 7.
Eliminating 1 self-loops for location f83
  Self-Loop 8 has the metering function: 50-A, resulting in the new transition 24.
  Removing the self-loops: 8.

Removed all Self-loops using metering functions (where possible):
  Start location: f0
      0: f0 -> f17 : A'=0, B'=free_1, C'=free, D'=0, [], cost: 1
     17: f17 -> [10] : D'=50, [ 49>=D ], cost: 50-D
     18: f27 -> [11] : E'=50, [ 49>=E ], cost: 50-E
     19: f37 -> [12] : F'=50, [ 49>=F ], cost: 50-F
     20: f45 -> [13] : A'=50, [ 49>=A ], cost: 50-A
     21: f55 -> [14] : G'=50, [ 49>=G ], cost: 50-G
     22: f65 -> [15] : H'=50, [ 49>=H ], cost: 50-H
     23: f75 -> [16] : Q'=50, [ 49>=Q ], cost: 50-Q
     24: f83 -> [17] : A'=50, [ 49>=A ], cost: 50-A
     16: [10] -> f27 : E'=0, [ D>=50 ], cost: 1
     15: [11] -> f37 : F'=0, [ E>=50 ], cost: 1
     14: [12] -> f45 : A'=0, [ F>=50 ], cost: 1
     13: [13] -> f55 : G'=0, [ A>=50 ], cost: 1
     12: [14] -> f65 : H'=0, [ G>=50 ], cost: 1
     11: [15] -> f75 : Q'=0, [ H>=50 ], cost: 1
     10: [16] -> f83 : A'=0, [ Q>=50 ], cost: 1


Applied simple chaining:
  Start location: f0
      0: f0 -> [17] : A'=50, B'=free_1, C'=free, D'=50, E'=50, F'=50, G'=50, H'=50, Q'=50, [ 49>=0 && 50>=50 && 49>=0 && 50>=50 && 49>=0 && 50>=50 && 49>=0 && 50>=50 && 49>=0 && 50>=50 && 49>=0 && 50>=50 && 49>=0 && 50>=50 && 49>=0 ], cost: 408


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: f0
      0: f0 -> [17] : A'=50, B'=free_1, C'=free, D'=50, E'=50, F'=50, G'=50, H'=50, Q'=50, [ 49>=0 && 50>=50 && 49>=0 && 50>=50 && 49>=0 && 50>=50 && 49>=0 && 50>=50 && 49>=0 && 50>=50 && 49>=0 && 50>=50 && 49>=0 && 50>=50 && 49>=0 ], cost: 408


Computing complexity for remaining 1 transitions.

Found new complexity const, because: const cost.


The final runtime is determined by this resulting transition:
  Final Guard: 49>=0 && 50>=50 && 49>=0 && 50>=50 && 49>=0 && 50>=50 && 49>=0 && 50>=50 && 49>=0 && 50>=50 && 49>=0 && 50>=50 && 49>=0 && 50>=50 && 49>=0
  Final Cost:  408

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

WORST_CASE(Omega(1),?)