Trying to load file: main.koat

Initial Control flow graph problem:
  Start location: f0
      0: f0 -> f8 : A'=free, B'=0, [], cost: 1
      1: f8 -> f8 : B'=1+B, [ 9>=B ], cost: 1
      4: f8 -> f19 : C'=0, [ B>=10 ], cost: 1
      2: f19 -> f19 : C'=1+C, [ 9>=C ], cost: 1
      3: f19 -> f29 : [ C>=10 ], cost: 1


Simplified the transitions:
  Start location: f0
      0: f0 -> f8 : A'=free, B'=0, [], cost: 1
      1: f8 -> f8 : B'=1+B, [ 9>=B ], cost: 1
      4: f8 -> f19 : C'=0, [ B>=10 ], cost: 1
      2: f19 -> f19 : C'=1+C, [ 9>=C ], cost: 1

Eliminating 1 self-loops for location f8
  Self-Loop 1 has the metering function: 10-B, resulting in the new transition 5.
  Removing the self-loops: 1.
Eliminating 1 self-loops for location f19
  Self-Loop 2 has the metering function: 10-C, 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 -> f8 : A'=free, B'=0, [], cost: 1
      5: f8 -> [4] : B'=10, [ 9>=B ], cost: 10-B
      6: f19 -> [5] : C'=10, [ 9>=C ], cost: 10-C
      4: [4] -> f19 : C'=0, [ B>=10 ], cost: 1


Applied simple chaining:
  Start location: f0
      0: f0 -> [5] : A'=free, B'=10, C'=10, [ 9>=0 && 10>=10 && 9>=0 ], cost: 22


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: f0
      0: f0 -> [5] : A'=free, B'=10, C'=10, [ 9>=0 && 10>=10 && 9>=0 ], cost: 22


Computing complexity for remaining 1 transitions.

Found new complexity const, because: const cost.


The final runtime is determined by this resulting transition:
  Final Guard: 9>=0 && 10>=10 && 9>=0
  Final Cost:  22

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

WORST_CASE(Omega(1),?)