Trying to load file: main.koat

Initial Control flow graph problem:
  Start location: f999
      3: f1 -> f1 : A'=1+A, B'=-1+B, [ B>=1 ], cost: 1
      0: f1 -> f2 : A'=-1+A, [ A>=1 ], cost: 1
      4: f2 -> f1 : A'=1+A, B'=-1+B, [ B>=1 ], cost: 1
      1: f2 -> f2 : A'=-1+A, B'=1+B, [ A>=1 ], cost: 1
      2: f999 -> f1 : A'=1, B'=-1+B, [ B>=1 && A==0 ], cost: 1

Eliminating 1 self-loops for location f1
  Self-Loop 3 has the metering function: B, resulting in the new transition 5.
  Removing the self-loops: 3.
Eliminating 1 self-loops for location f2
  Self-Loop 1 has the metering function: A, resulting in the new transition 6.
  Removing the self-loops: 1.

Removed all Self-loops using metering functions (where possible):
  Start location: f999
      5: f1 -> [3] : A'=B+A, B'=0, [ B>=1 ], cost: B
      6: f2 -> [4] : A'=0, B'=B+A, [ A>=1 ], cost: A
      2: f999 -> f1 : A'=1, B'=-1+B, [ B>=1 && A==0 ], cost: 1
      0: [3] -> f2 : A'=-1+A, [ A>=1 ], cost: 1
      4: [4] -> f1 : A'=1+A, B'=-1+B, [ B>=1 ], cost: 1


Applied simple chaining:
  Start location: f999
      5: f1 -> f1 : A'=1, B'=-2+B+A, [ B>=1 && B+A>=1 && -1+B+A>=1 && -1+B+A>=1 ], cost: 1+2*B+A
      2: f999 -> f1 : A'=1, B'=-1+B, [ B>=1 && A==0 ], cost: 1

Eliminating 1 self-loops for location f1
  Removing the self-loops: 5.
Adding an epsilon transition (to model nonexecution of the loops): 8.

Removed all Self-loops using metering functions (where possible):
  Start location: f999
      7: f1 -> [5] : A'=1, B'=-2+B+A, [ B>=1 && -1+B+A>=1 ], cost: 1+2*B+A
      8: f1 -> [5] : [], cost: 0
      2: f999 -> f1 : A'=1, B'=-1+B, [ B>=1 && A==0 ], cost: 1


Applied chaining over branches and pruning:
  Start location: f999
      9: f999 -> [5] : A'=1, B'=-2+B, [ B>=1 && A==0 && -1+B>=1 && -1+B>=1 ], cost: 1+2*B


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: f999
      9: f999 -> [5] : A'=1, B'=-2+B, [ B>=1 && A==0 && -1+B>=1 && -1+B>=1 ], cost: 1+2*B


Computing complexity for remaining 1 transitions.

  Found configuration with infinitely models for cost: 1+2*B
  and guard: B>=1 && A==0 && -1+B>=1 && -1+B>=1:
  B: Pos, A: Both

Found new complexity n^1, because: Found infinity configuration.


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

Obtained the following complexity w.r.t. the length of the input n:
  Complexity class: n^1
  Complexity value: 1

WORST_CASE(Omega(n^1),?)