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Initial Control flow graph problem:
  Start location: f1
      0: f1 -> f3 : B'=A, [ A>=1 ], cost: 1
      1: f3 -> f3 : B'=-1+B, [ B>=1 ], cost: 1
      2: f3 -> f3 : A'=-1+A, B'=-1+A, [ 0>=B && A>=2 ], cost: 1

Eliminating 2 self-loops for location f3
  Self-Loop 1 has the metering function: B, resulting in the new transition 3.
  Found this metering function when nesting loops: -1+A,
  and nested parallel self-loops 4 (outer loop) and 3 (inner loop), obtaining the new transitions: 5, 6.
  Found this metering function when nesting loops: -1+A,
  and nested parallel self-loops 4 (outer loop) and 3 (inner loop), obtaining the new transitions: 7, 8.
  Found this metering function when nesting loops: -1+B,
  Removing the self-loops: 1 2 4.
Adding an epsilon transition (to model nonexecution of the loops): 9.

Removed all Self-loops using metering functions (where possible):
  Start location: f1
      0: f1 -> f3 : B'=A, [ A>=1 ], cost: 1
      3: f3 -> [2] : B'=0, [ B>=1 ], cost: B
      5: f3 -> [2] : A'=1, B'=1, [ B>=1 && 0>=0 && A>=2 ], cost: -3/2+(-1+A)*A-1/2*(-1+A)^2+3/2*A
      6: f3 -> [2] : A'=1, B'=1, [ 0>=B && A>=2 && -1+A>=1 && 0>=0 && -1+A>=2 ], cost: -2-1/2*(-2+A)^2+(-2+A)*(-1+A)+3/2*A
      7: f3 -> [2] : A'=1, B'=0, [ 0>=B && A>=2 && -1+A>=1 ], cost: -1/2+(-1+A)*A-1/2*(-1+A)^2+1/2*A
      8: f3 -> [2] : A'=1, B'=0, [ B>=1 && 0>=0 && A>=2 && -1+A>=1 ], cost: -1/2+B+(-1+A)*A-1/2*(-1+A)^2+1/2*A
      9: f3 -> [2] : [], cost: 0


Applied chaining over branches and pruning:
  Start location: f1
     10: f1 -> [2] : B'=0, [ A>=1 && A>=1 ], cost: 1+A
     11: f1 -> [2] : A'=1, B'=1, [ A>=1 && A>=1 && 0>=0 && A>=2 ], cost: -1/2+(-1+A)*A-1/2*(-1+A)^2+3/2*A
     12: f1 -> [2] : A'=1, B'=0, [ A>=1 && A>=1 && 0>=0 && A>=2 && -1+A>=1 ], cost: 1/2+(-1+A)*A-1/2*(-1+A)^2+3/2*A


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: f1
     10: f1 -> [2] : B'=0, [ A>=1 && A>=1 ], cost: 1+A
     11: f1 -> [2] : A'=1, B'=1, [ A>=1 && A>=1 && 0>=0 && A>=2 ], cost: -1/2+(-1+A)*A-1/2*(-1+A)^2+3/2*A
     12: f1 -> [2] : A'=1, B'=0, [ A>=1 && A>=1 && 0>=0 && A>=2 && -1+A>=1 ], cost: 1/2+(-1+A)*A-1/2*(-1+A)^2+3/2*A


Computing complexity for remaining 3 transitions.

  Found configuration with infinitely models for cost: 1+A
  and guard: A>=1 && A>=1:
  A: Pos

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

  Found configuration with infinitely models for cost: -1/2+(-1+A)*A-1/2*(-1+A)^2+3/2*A
  and guard: A>=1 && A>=1 && 0>=0 && A>=2:
  A: Pos

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


The final runtime is determined by this resulting transition:
  Final Guard: A>=1 && A>=1 && 0>=0 && A>=2
  Final Cost:  -1/2+(-1+A)*A-1/2*(-1+A)^2+3/2*A

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),?)