Trying to load file: main.koat

Initial Control flow graph problem:
  Start location: f0
      0: f0 -> f9 : A'=0, B'=free, C'=0, [], cost: 1
      1: f9 -> f9 : C'=1+C, [ 49>=C ], cost: 1
      4: f9 -> f17 : A'=0, [ C>=50 ], cost: 1
      2: f17 -> f17 : A'=1+A, [ 49>=A ], cost: 1
      3: f17 -> f24 : [ A>=50 ], cost: 1


Simplified the transitions:
  Start location: f0
      0: f0 -> f9 : A'=0, B'=free, C'=0, [], cost: 1
      1: f9 -> f9 : C'=1+C, [ 49>=C ], cost: 1
      4: f9 -> f17 : A'=0, [ C>=50 ], cost: 1
      2: f17 -> f17 : A'=1+A, [ 49>=A ], cost: 1

Eliminating 1 self-loops for location f9
  Self-Loop 1 has the metering function: 50-C, resulting in the new transition 5.
  Removing the self-loops: 1.
Eliminating 1 self-loops for location f17
  Self-Loop 2 has the metering function: 50-A, 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 -> f9 : A'=0, B'=free, C'=0, [], cost: 1
      5: f9 -> [4] : C'=50, [ 49>=C ], cost: 50-C
      6: f17 -> [5] : A'=50, [ 49>=A ], cost: 50-A
      4: [4] -> f17 : A'=0, [ C>=50 ], cost: 1


Applied simple chaining:
  Start location: f0
      0: f0 -> [5] : A'=50, B'=free, C'=50, [ 49>=0 && 50>=50 && 49>=0 ], cost: 102


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: f0
      0: f0 -> [5] : A'=50, B'=free, C'=50, [ 49>=0 && 50>=50 && 49>=0 ], cost: 102


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
  Final Cost:  102

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

WORST_CASE(Omega(1),?)