Trying to load file: main.koat

Initial Control flow graph problem:
  Start location: evalwhile2start
      0: evalwhile2start -> evalwhile2entryin : [], cost: 1
      1: evalwhile2entryin -> evalwhile2bb4in : A'=B, [], cost: 1
      2: evalwhile2bb4in -> evalwhile2bb2in : C'=B, [ A>=1 ], cost: 1
      3: evalwhile2bb4in -> evalwhile2returnin : [ 0>=A ], cost: 1
      4: evalwhile2bb2in -> evalwhile2bb1in : [ C>=1 ], cost: 1
      5: evalwhile2bb2in -> evalwhile2bb3in : [ 0>=C ], cost: 1
      8: evalwhile2returnin -> evalwhile2stop : [], cost: 1
      6: evalwhile2bb1in -> evalwhile2bb2in : C'=-1+C, [], cost: 1
      7: evalwhile2bb3in -> evalwhile2bb4in : A'=-1+A, [], cost: 1


Simplified the transitions:
  Start location: evalwhile2start
      0: evalwhile2start -> evalwhile2entryin : [], cost: 1
      1: evalwhile2entryin -> evalwhile2bb4in : A'=B, [], cost: 1
      2: evalwhile2bb4in -> evalwhile2bb2in : C'=B, [ A>=1 ], cost: 1
      4: evalwhile2bb2in -> evalwhile2bb1in : [ C>=1 ], cost: 1
      5: evalwhile2bb2in -> evalwhile2bb3in : [ 0>=C ], cost: 1
      6: evalwhile2bb1in -> evalwhile2bb2in : C'=-1+C, [], cost: 1
      7: evalwhile2bb3in -> evalwhile2bb4in : A'=-1+A, [], cost: 1


Applied simple chaining:
  Start location: evalwhile2start
      0: evalwhile2start -> evalwhile2bb4in : A'=B, [], cost: 2
      2: evalwhile2bb4in -> evalwhile2bb2in : C'=B, [ A>=1 ], cost: 1
      5: evalwhile2bb2in -> evalwhile2bb4in : A'=-1+A, [ 0>=C ], cost: 2
      4: evalwhile2bb2in -> evalwhile2bb2in : C'=-1+C, [ C>=1 ], cost: 2

Eliminating 1 self-loops for location evalwhile2bb2in
  Self-Loop 4 has the metering function: C, resulting in the new transition 9.
  Removing the self-loops: 4.

Removed all Self-loops using metering functions (where possible):
  Start location: evalwhile2start
      0: evalwhile2start -> evalwhile2bb4in : A'=B, [], cost: 2
      2: evalwhile2bb4in -> evalwhile2bb2in : C'=B, [ A>=1 ], cost: 1
      9: evalwhile2bb2in -> [8] : C'=0, [ C>=1 ], cost: 2*C
      5: [8] -> evalwhile2bb4in : A'=-1+A, [ 0>=C ], cost: 2


Applied simple chaining:
  Start location: evalwhile2start
      0: evalwhile2start -> evalwhile2bb4in : A'=B, [], cost: 2
      2: evalwhile2bb4in -> evalwhile2bb4in : A'=-1+A, C'=0, [ A>=1 && B>=1 && 0>=0 ], cost: 3+2*B

Eliminating 1 self-loops for location evalwhile2bb4in
  Self-Loop 2 has the metering function: A, resulting in the new transition 10.
  Removing the self-loops: 2.

Removed all Self-loops using metering functions (where possible):
  Start location: evalwhile2start
      0: evalwhile2start -> evalwhile2bb4in : A'=B, [], cost: 2
     10: evalwhile2bb4in -> [9] : A'=0, C'=0, [ A>=1 && B>=1 ], cost: 2*B*A+3*A


Applied simple chaining:
  Start location: evalwhile2start
      0: evalwhile2start -> [9] : A'=0, C'=0, [ B>=1 && B>=1 ], cost: 2+3*B+2*B^2


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: evalwhile2start
      0: evalwhile2start -> [9] : A'=0, C'=0, [ B>=1 && B>=1 ], cost: 2+3*B+2*B^2


Computing complexity for remaining 1 transitions.

  Found configuration with infinitely models for cost: 2+3*B+2*B^2
  and guard: B>=1 && 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 && B>=1
  Final Cost:  2+3*B+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),?)