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Initial Control flow graph problem:
  Start location: f0
      0: f4 -> f4 : B'=1+B, [ A>=1+B ], cost: 1
      1: f4 -> f4 : A'=1+A, B'=0, [ C>=2+A && B>=A ], cost: 1
      2: f0 -> f4 : A'=0, B'=0, [ C>=1 ], cost: 1

Eliminating 2 self-loops for location f4
  Self-Loop 0 has the metering function: -B+A, resulting in the new transition 3.
  Found this metering function when nesting loops: -1+C-A,
  and nested parallel self-loops 4 (outer loop) and 3 (inner loop), obtaining the new transitions: 5, 6.
  Removing the self-loops: 0 1 4.
Adding an epsilon transition (to model nonexecution of the loops): 7.

Removed all Self-loops using metering functions (where possible):
  Start location: f0
      3: f4 -> [2] : B'=A, [ A>=1+B ], cost: -B+A
      5: f4 -> [2] : A'=-1+C, B'=-1+C, [ C>=2+A && B>=A && 1+A>=1 ], cost: -3/2+1/2*(-1+C-A)^2+(-1+C-A)*A+3/2*C-3/2*A
      6: f4 -> [2] : A'=-1+C, B'=-1+C, [ A>=1+B && C>=2+A && A>=A && 1+A>=1 ], cost: -3/2-B+1/2*(-1+C-A)^2+(-1+C-A)*A+3/2*C-1/2*A
      7: f4 -> [2] : [], cost: 0
      2: f0 -> f4 : A'=0, B'=0, [ C>=1 ], cost: 1


Applied chaining over branches and pruning:
  Start location: f0
      8: f0 -> [2] : A'=-1+C, B'=-1+C, [ C>=1 && C>=2 && 0>=0 && 1>=1 ], cost: -1/2+1/2*(-1+C)^2+3/2*C


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: f0
      8: f0 -> [2] : A'=-1+C, B'=-1+C, [ C>=1 && C>=2 && 0>=0 && 1>=1 ], cost: -1/2+1/2*(-1+C)^2+3/2*C


Computing complexity for remaining 1 transitions.

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

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


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

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),?)