Trying to load file: main.koat

Initial Control flow graph problem:
  Start location: f2
      0: f300 -> f300 : A'=-99+A, B'=0, [ 0>=1+A && 1+B==0 ], cost: 1
      1: f300 -> f300 : A'=1+A, B'=1+B, [ 0>=1+A && B>=0 ], cost: 1
      2: f300 -> f300 : A'=1+A, B'=1+B, [ 0>=1+A && 0>=2+B ], cost: 1
      3: f300 -> f1 : C'=free, [ A>=0 ], cost: 1
      4: f2 -> f300 : [], cost: 1


Simplified the transitions:
  Start location: f2
      0: f300 -> f300 : A'=-99+A, B'=0, [ 0>=1+A && 1+B==0 ], cost: 1
      1: f300 -> f300 : A'=1+A, B'=1+B, [ 0>=1+A && B>=0 ], cost: 1
      2: f300 -> f300 : A'=1+A, B'=1+B, [ 0>=1+A && 0>=2+B ], cost: 1
      4: f2 -> f300 : [], cost: 1

Eliminating 3 self-loops for location f300
  Self-Loop 0 has the metering function: -B, resulting in the new transition 5.
  Self-Loop 1 has the metering function: -A, resulting in the new transition 6.
  Self-Loop 7 has the metering function: -A, resulting in the new transition 10.
  Self-Loop 8 has the metering function: -1-B, resulting in the new transition 11.
  Found this metering function when nesting loops: meter,
  Found this metering function when nesting loops: meter_1,
  Found this metering function when nesting loops: meter_2,
  Removing the self-loops: 0 1 2 7 8.
Adding an epsilon transition (to model nonexecution of the loops): 12.

Removed all Self-loops using metering functions (where possible):
  Start location: f2
      5: f300 -> [3] : A'=99*B+A, B'=0, [ 0>=1+A && 1+B==0 ], cost: -B
      6: f300 -> [3] : A'=0, B'=B-A, [ 0>=1+A && B>=0 ], cost: -A
      9: f300 -> [3] : A'=1+A, B'=1+B, [ 0>=1+A && 0>=2+B ], cost: 1
     10: f300 -> [3] : A'=0, B'=B-A, [ 0>=1+A && 0>=2+B && A>B ], cost: -A
     11: f300 -> [3] : A'=-1-B+A, B'=-1, [ 0>=1+A && 0>=2+B && B>A ], cost: -1-B
     12: f300 -> [3] : [], cost: 0
      4: f2 -> f300 : [], cost: 1


Applied chaining over branches and pruning:
  Start location: f2
     13: f2 -> [3] : A'=99*B+A, B'=0, [ 0>=1+A && 1+B==0 ], cost: 1-B
     14: f2 -> [3] : A'=0, B'=B-A, [ 0>=1+A && B>=0 ], cost: 1-A
     16: f2 -> [3] : A'=0, B'=B-A, [ 0>=1+A && 0>=2+B && A>B ], cost: 1-A
     17: f2 -> [3] : A'=-1-B+A, B'=-1, [ 0>=1+A && 0>=2+B && B>A ], cost: -B


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: f2
     13: f2 -> [3] : A'=99*B+A, B'=0, [ 0>=1+A && 1+B==0 ], cost: 1-B
     14: f2 -> [3] : A'=0, B'=B-A, [ 0>=1+A && B>=0 ], cost: 1-A
     16: f2 -> [3] : A'=0, B'=B-A, [ 0>=1+A && 0>=2+B && A>B ], cost: 1-A
     17: f2 -> [3] : A'=-1-B+A, B'=-1, [ 0>=1+A && 0>=2+B && B>A ], cost: -B


Computing complexity for remaining 4 transitions.

Found new complexity const, because: const cost.

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

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


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

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),?)