Trying to load file: main.koat

Initial Control flow graph problem:
  Start location: l0
      0: l0 -> l1 : A'=0, [], cost: 1
      1: l1 -> l1 : A'=1+A, B'=-1+B, [ B>=1 ], cost: 1
      2: l1 -> l2 : C'=A, [ 0>=B ], cost: 1
      3: l2 -> l3 : D'=C, [ C>=1 ], cost: 1
      5: l3 -> l2 : C'=-1+C, [ 0>=D && C>=1 ], cost: 1
      4: l3 -> l3 : D'=-1+D, [ D>=1 && C>=1 ], cost: 1

Eliminating 1 self-loops for location l1
  Self-Loop 1 has the metering function: B, resulting in the new transition 6.
  Removing the self-loops: 1.
Eliminating 1 self-loops for location l3
  Self-Loop 4 has the metering function: D, resulting in the new transition 7.
  Removing the self-loops: 4.

Removed all Self-loops using metering functions (where possible):
  Start location: l0
      0: l0 -> l1 : A'=0, [], cost: 1
      6: l1 -> [4] : A'=B+A, B'=0, [ B>=1 ], cost: B
      3: l2 -> l3 : D'=C, [ C>=1 ], cost: 1
      7: l3 -> [5] : D'=0, [ D>=1 && C>=1 ], cost: D
      2: [4] -> l2 : C'=A, [ 0>=B ], cost: 1
      5: [5] -> l2 : C'=-1+C, [ 0>=D && C>=1 ], cost: 1


Applied simple chaining:
  Start location: l0
      0: l0 -> l2 : A'=B, B'=0, C'=B, [ B>=1 && 0>=0 ], cost: 2+B
      3: l2 -> l2 : C'=-1+C, D'=0, [ C>=1 && C>=1 && C>=1 && 0>=0 && C>=1 ], cost: 2+C

Eliminating 1 self-loops for location l2
  Self-Loop 3 has the metering function: C, resulting in the new transition 8.
  Removing the self-loops: 3.

Removed all Self-loops using metering functions (where possible):
  Start location: l0
      0: l0 -> l2 : A'=B, B'=0, C'=B, [ B>=1 && 0>=0 ], cost: 2+B
      8: l2 -> [6] : C'=0, D'=0, [ C>=1 ], cost: 5/2*C+1/2*C^2


Applied simple chaining:
  Start location: l0
      0: l0 -> [6] : A'=B, B'=0, C'=0, D'=0, [ B>=1 && 0>=0 && B>=1 ], cost: 2+7/2*B+1/2*B^2


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: l0
      0: l0 -> [6] : A'=B, B'=0, C'=0, D'=0, [ B>=1 && 0>=0 && B>=1 ], cost: 2+7/2*B+1/2*B^2


Computing complexity for remaining 1 transitions.

  Found configuration with infinitely models for cost: 2+7/2*B+1/2*B^2
  and guard: B>=1 && 0>=0 && B>=1:
  B: Pos

Found new complexity n^2, because: Found infinity configuration.


The final runtime is determined by this resulting transition:
  Final Guard: B>=1 && 0>=0 && B>=1
  Final Cost:  2+7/2*B+1/2*B^2

Obtained the following complexity w.r.t. the length of the input n:
  Complexity class: n^2
  Complexity value: 2

WORST_CASE(Omega(n^2),?)