Trying to load file: main.koat

Initial Control flow graph problem:
  Start location: f0
      0: f0 -> f4 : A'=0, [], cost: 1
      1: f4 -> f7 : C'=1+A, [ B>=1+A ], cost: 1
      6: f4 -> f19 : [ A>=B ], cost: 1
      5: f7 -> f4 : A'=1+A, [ C>=B ], cost: 1
      2: f7 -> f7 : C'=1+C, D'=0, [ B>=1+C ], cost: 1
      3: f7 -> f7 : B'=-1+B, D'=free, [ B>=1+C && 0>=1+free ], cost: 1
      4: f7 -> f7 : B'=-1+B, D'=free_1, [ B>=1+C && free_1>=1 ], cost: 1


Simplified the transitions:
  Start location: f0
      0: f0 -> f4 : A'=0, [], cost: 1
      1: f4 -> f7 : C'=1+A, [ B>=1+A ], cost: 1
      5: f7 -> f4 : A'=1+A, [ C>=B ], cost: 1
      2: f7 -> f7 : C'=1+C, D'=0, [ B>=1+C ], cost: 1
      3: f7 -> f7 : B'=-1+B, D'=free, [ B>=1+C && 0>=1+free ], cost: 1
      4: f7 -> f7 : B'=-1+B, D'=free_1, [ B>=1+C && free_1>=1 ], cost: 1

Eliminating 3 self-loops for location f7
  Self-Loop 2 has the metering function: B-C, resulting in the new transition 7.
  Self-Loop 3 has the metering function: B-C, resulting in the new transition 8.
  Self-Loop 4 has the metering function: B-C, resulting in the new transition 9.
  Removing the self-loops: 2 3 4.

Removed all Self-loops using metering functions (where possible):
  Start location: f0
      0: f0 -> f4 : A'=0, [], cost: 1
      1: f4 -> f7 : C'=1+A, [ B>=1+A ], cost: 1
      7: f7 -> [4] : C'=B, D'=0, [ B>=1+C ], cost: B-C
      8: f7 -> [4] : B'=C, D'=free, [ B>=1+C && 0>=1+free ], cost: B-C
      9: f7 -> [4] : B'=C, D'=free_1, [ B>=1+C && free_1>=1 ], cost: B-C
      5: [4] -> f4 : A'=1+A, [ C>=B ], cost: 1


Applied chaining over branches and pruning:
  Start location: f0
      0: f0 -> f4 : A'=0, [], cost: 1
     10: f4 -> [4] : C'=B, D'=0, [ B>=1+A && B>=2+A ], cost: B-A
     11: f4 -> [4] : B'=1+A, C'=1+A, D'=free, [ B>=1+A && B>=2+A && 0>=1+free ], cost: B-A
     12: f4 -> [4] : B'=1+A, C'=1+A, D'=free_1, [ B>=1+A && B>=2+A && free_1>=1 ], cost: B-A
      5: [4] -> f4 : A'=1+A, [ C>=B ], cost: 1


Applied chaining over branches and pruning:
  Start location: f0
      0: f0 -> f4 : A'=0, [], cost: 1
     13: f4 -> f4 : A'=1+A, C'=B, D'=0, [ B>=1+A && B>=2+A && B>=B ], cost: 1+B-A
     14: f4 -> f4 : A'=1+A, B'=1+A, C'=1+A, D'=free, [ B>=1+A && B>=2+A && 0>=1+free && 1+A>=1+A ], cost: 1+B-A
     15: f4 -> f4 : A'=1+A, B'=1+A, C'=1+A, D'=free_1, [ B>=1+A && B>=2+A && free_1>=1 && 1+A>=1+A ], cost: 1+B-A

Eliminating 3 self-loops for location f4
  Self-Loop 13 has the metering function: -1+B-A, resulting in the new transition 16.
  Removing the self-loops: 13 14 15.
Adding an epsilon transition (to model nonexecution of the loops): 19.

Removed all Self-loops using metering functions (where possible):
  Start location: f0
      0: f0 -> f4 : A'=0, [], cost: 1
     16: f4 -> [5] : A'=-1+B, C'=B, D'=0, [ B>=2+A ], cost: -3/2+3/2*B+B*(-1+B-A)-1/2*(-1+B-A)^2-3/2*A-(-1+B-A)*A
     17: f4 -> [5] : A'=1+A, B'=1+A, C'=1+A, D'=free, [ B>=2+A && 0>=1+free ], cost: 1+B-A
     18: f4 -> [5] : A'=1+A, B'=1+A, C'=1+A, D'=free_1, [ B>=2+A && free_1>=1 ], cost: 1+B-A
     19: f4 -> [5] : [], cost: 0


Applied chaining over branches and pruning:
  Start location: f0
     20: f0 -> [5] : A'=-1+B, C'=B, D'=0, [ B>=2 ], cost: -1/2-1/2*(-1+B)^2+3/2*B+B*(-1+B)
     21: f0 -> [5] : A'=1, B'=1, C'=1, D'=free, [ B>=2 && 0>=1+free ], cost: 2+B
     22: f0 -> [5] : A'=1, B'=1, C'=1, D'=free_1, [ B>=2 && free_1>=1 ], cost: 2+B


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: f0
     20: f0 -> [5] : A'=-1+B, C'=B, D'=0, [ B>=2 ], cost: -1/2-1/2*(-1+B)^2+3/2*B+B*(-1+B)
     21: f0 -> [5] : A'=1, B'=1, C'=1, D'=free, [ B>=2 && 0>=1+free ], cost: 2+B
     22: f0 -> [5] : A'=1, B'=1, C'=1, D'=free_1, [ B>=2 && free_1>=1 ], cost: 2+B


Computing complexity for remaining 3 transitions.

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

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


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

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),?)