Trying to load file: main.koat

Initial Control flow graph problem:
  Start location: start0
      0: start -> stop : D'=F, [ 0>=1+A && B==C && D==E && F==A ], cost: 1
      1: start -> lbl121 : B'=1, D'=-1+F, [ A>=0 && 1>=A && B==C && D==E && F==A ], cost: 1
      2: start -> lbl101 : B'=2, D'=F, [ A>=2 && B==C && D==E && F==A ], cost: 1
      3: lbl121 -> stop : [ A>=0 && B>=0 && B>=1 && 1+D==0 && F==A ], cost: 1
      4: lbl121 -> lbl121 : B'=1, D'=-1+D, [ D>=0 && 1>=D && A>=1+D && B>=1+D && B>=1 && 1+D>=0 && F==A ], cost: 1
      5: lbl121 -> lbl101 : B'=2, [ D>=2 && A>=1+D && B>=1+D && B>=1 && 1+D>=0 && F==A ], cost: 1
      7: lbl101 -> lbl121 : D'=-1+D, [ B>=D && B>=2 && 2*D>=2+B && A>=D && F==A ], cost: 1
      6: lbl101 -> lbl101 : B'=2*B, [ D>=1+B && B>=2 && 2*D>=2+B && A>=D && F==A ], cost: 1
      8: start0 -> start : B'=C, D'=E, F'=A, [], cost: 1


Simplified the transitions:
  Start location: start0
      1: start -> lbl121 : B'=1, D'=-1+F, [ A>=0 && 1>=A && B==C && D==E && F==A ], cost: 1
      2: start -> lbl101 : B'=2, D'=F, [ A>=2 && B==C && D==E && F==A ], cost: 1
      4: lbl121 -> lbl121 : B'=1, D'=-1+D, [ D>=0 && 1>=D && A>=1+D && B>=1+D && B>=1 && F==A ], cost: 1
      5: lbl121 -> lbl101 : B'=2, [ D>=2 && A>=1+D && B>=1+D && B>=1 && F==A ], cost: 1
      7: lbl101 -> lbl121 : D'=-1+D, [ B>=D && B>=2 && 2*D>=2+B && A>=D && F==A ], cost: 1
      6: lbl101 -> lbl101 : B'=2*B, [ D>=1+B && B>=2 && 2*D>=2+B && A>=D && F==A ], cost: 1
      8: start0 -> start : B'=C, D'=E, F'=A, [], cost: 1

Eliminating 1 self-loops for location lbl121
  Self-Loop 4 has the metering function: 1+D, resulting in the new transition 9.
  Removing the self-loops: 4.
Eliminating 1 self-loops for location lbl101
  Removing the self-loops: 6.
Adding an epsilon transition (to model nonexecution of the loops): 11.

Removed all Self-loops using metering functions (where possible):
  Start location: start0
      1: start -> lbl121 : B'=1, D'=-1+F, [ A>=0 && 1>=A && B==C && D==E && F==A ], cost: 1
      2: start -> lbl101 : B'=2, D'=F, [ A>=2 && B==C && D==E && F==A ], cost: 1
      9: lbl121 -> [5] : B'=1, D'=-1, [ D>=0 && 1>=D && A>=1+D && B>=1+D && B>=1 && F==A ], cost: 1+D
     10: lbl101 -> [6] : B'=2*B, [ D>=1+B && B>=2 && 2*D>=2+B && A>=D && F==A ], cost: 1
     11: lbl101 -> [6] : [], cost: 0
      8: start0 -> start : B'=C, D'=E, F'=A, [], cost: 1
      5: [5] -> lbl101 : B'=2, [ D>=2 && A>=1+D && B>=1+D && B>=1 && F==A ], cost: 1
      7: [6] -> lbl121 : D'=-1+D, [ B>=D && B>=2 && 2*D>=2+B && A>=D && F==A ], cost: 1


Applied chaining over branches and pruning:
  Start location: start0
     14: lbl121 -> [7] : B'=1, D'=-1, [ D>=0 && 1>=D && A>=1+D && B>=1+D && B>=1 && F==A ], cost: 1+D
     15: lbl101 -> lbl121 : B'=2*B, D'=-1+D, [ D>=1+B && B>=2 && 2*D>=2+B && A>=D && F==A && 2*B>=D && 2*B>=2 && 2*D>=2+2*B && A>=D && F==A ], cost: 2
     16: lbl101 -> lbl121 : D'=-1+D, [ B>=D && B>=2 && 2*D>=2+B && A>=D && F==A ], cost: 1
     12: start0 -> lbl121 : B'=1, D'=-1+A, F'=A, [ A>=0 && 1>=A && C==C && E==E && A==A ], cost: 2
     13: start0 -> lbl101 : B'=2, D'=A, F'=A, [ A>=2 && C==C && E==E && A==A ], cost: 2


Applied chaining over branches and pruning:
  Start location: start0
     14: lbl121 -> [7] : B'=1, D'=-1, [ D>=0 && 1>=D && A>=1+D && B>=1+D && B>=1 && F==A ], cost: 1+D
     12: start0 -> lbl121 : B'=1, D'=-1+A, F'=A, [ A>=0 && 1>=A && C==C && E==E && A==A ], cost: 2
     17: start0 -> lbl121 : B'=4, D'=-1+A, F'=A, [ A>=2 && C==C && E==E && A==A && A>=3 && 2>=2 && 2*A>=4 && A>=A && A==A && 4>=A && 4>=2 && 2*A>=6 && A>=A && A==A ], cost: 4
     18: start0 -> lbl121 : B'=2, D'=-1+A, F'=A, [ A>=2 && C==C && E==E && A==A && 2>=A && 2>=2 && 2*A>=4 && A>=A && A==A ], cost: 3


Applied chaining over branches and pruning:
  Start location: start0
     19: start0 -> [7] : B'=1, D'=-1, F'=A, [ A>=0 && 1>=A && C==C && E==E && A==A && -1+A>=0 && 1>=-1+A && A>=A && 1>=A && 1>=1 && A==A ], cost: 2+A
     20: start0 -> [7] : B'=1, D'=-1, F'=A, [ A>=2 && C==C && E==E && A==A && 2>=A && 2>=2 && 2*A>=4 && A>=A && A==A && -1+A>=0 && 1>=-1+A && A>=A && 2>=A && 2>=1 && A==A ], cost: 3+A


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: start0
     19: start0 -> [7] : B'=1, D'=-1, F'=A, [ A>=0 && 1>=A && C==C && E==E && A==A && -1+A>=0 && 1>=-1+A && A>=A && 1>=A && 1>=1 && A==A ], cost: 2+A
     20: start0 -> [7] : B'=1, D'=-1, F'=A, [ A>=2 && C==C && E==E && A==A && 2>=A && 2>=2 && 2*A>=4 && A>=A && A==A && -1+A>=0 && 1>=-1+A && A>=A && 2>=A && 2>=1 && A==A ], cost: 3+A


Computing complexity for remaining 2 transitions.

  Found configuration with infinitely models for cost: 2+A
  and guard: A>=0 && 1>=A && C==C && E==E && A==A && -1+A>=0 && 1>=-1+A && A>=A && 1>=A && 1>=1 && A==A:
  E: Both, C: Both, A: Const

Found new complexity const, because: Found infinity configuration.

  Found configuration with infinitely models for cost: 3+A
  and guard: A>=2 && C==C && E==E && A==A && 2>=A && 2>=2 && 2*A>=4 && A>=A && A==A && -1+A>=0 && 1>=-1+A && A>=A && 2>=A && 2>=1 && A==A:
  E: Both, C: Both, A: Const


The final runtime is determined by this resulting transition:
  Final Guard: A>=0 && 1>=A && C==C && E==E && A==A && -1+A>=0 && 1>=-1+A && A>=A && 1>=A && 1>=1 && A==A
  Final Cost:  3

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

WORST_CASE(Omega(1),?)