Trying to load file: main.koat

Initial Control flow graph problem:
  Start location: start
      0: eval -> eval : B'=A, [ 0>=A && B==1 ], cost: 1
      1: eval -> eval : B'=A, [ B>=1 && 1+B>=0 && B>=1+A ], cost: 1
      2: eval -> eval : B'=0, [ A>=1 && B==1 ], cost: 1
      3: eval -> eval : B'=-1+B, [ B>=1 && 1+B>=0 && A>=B ], cost: 1
      4: start -> eval : [], cost: 1


Simplified the transitions:
  Start location: start
      0: eval -> eval : B'=A, [ 0>=A && B==1 ], cost: 1
      1: eval -> eval : B'=A, [ B>=1 && B>=1+A ], cost: 1
      2: eval -> eval : B'=0, [ A>=1 && B==1 ], cost: 1
      3: eval -> eval : B'=-1+B, [ B>=1 && A>=B ], cost: 1
      4: start -> eval : [], cost: 1

Eliminating 4 self-loops for location eval
  Self-Loop 2 has the metering function: B, resulting in the new transition 7.
  Self-Loop 3 has the metering function: B, resulting in the new transition 8.
  Found this metering function when nesting loops: meter,
  Found this metering function when nesting loops: 1-A,
  and nested parallel self-loops 6 (outer loop) and 7 (inner loop), obtaining the new transitions: 9.
  Removing the self-loops: 0 1 2 3 6.
Adding an epsilon transition (to model nonexecution of the loops): 10.

Removed all Self-loops using metering functions (where possible):
  Start location: start
      5: eval -> [2] : B'=A, [ 0>=A && B==1 ], cost: 1
      7: eval -> [2] : B'=0, [ A>=1 && B==1 ], cost: B
      8: eval -> [2] : B'=0, [ B>=1 && A>=B ], cost: B
      9: eval -> [2] : B'=0, [ B>=1 && B>=1+A && A>=1 && A==1 ], cost: 1-(-1+A)*A-A
     10: eval -> [2] : [], cost: 0
      4: start -> eval : [], cost: 1


Applied chaining over branches and pruning:
  Start location: start
     12: start -> [2] : B'=0, [ A>=1 && B==1 ], cost: 1+B
     13: start -> [2] : B'=0, [ B>=1 && A>=B ], cost: 1+B
     14: start -> [2] : B'=0, [ B>=1 && B>=1+A && A>=1 && A==1 ], cost: 2-(-1+A)*A-A


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: start
     12: start -> [2] : B'=0, [ A>=1 && B==1 ], cost: 1+B
     13: start -> [2] : B'=0, [ B>=1 && A>=B ], cost: 1+B
     14: start -> [2] : B'=0, [ B>=1 && B>=1+A && A>=1 && A==1 ], cost: 2-(-1+A)*A-A


Computing complexity for remaining 3 transitions.

Found new complexity const, because: const cost.

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

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


The final runtime is determined by this resulting transition:
  Final Guard: B>=1 && A>=B
  Final Cost:  1+B

Obtained the following complexity w.r.t. the length of the input n:
  Complexity class: n^1
  Complexity value: 1

WORST_CASE(Omega(n^1),?)