Trying to load file: main.koat

Initial Control flow graph problem:
  Start location: evalspeedpldi3start
      0: evalspeedpldi3start -> evalspeedpldi3entryin : [], cost: 1
      1: evalspeedpldi3entryin -> evalspeedpldi3returnin : [ 0>=A ], cost: 1
      2: evalspeedpldi3entryin -> evalspeedpldi3returnin : [ A>=B ], cost: 1
      3: evalspeedpldi3entryin -> evalspeedpldi3bb5in : C'=0, D'=0, [ A>=1 && B>=1+A ], cost: 1
     10: evalspeedpldi3returnin -> evalspeedpldi3stop : [], cost: 1
      5: evalspeedpldi3bb5in -> evalspeedpldi3returnin : [ D>=B ], cost: 1
      4: evalspeedpldi3bb5in -> evalspeedpldi3bb2in : [ B>=1+D ], cost: 1
      6: evalspeedpldi3bb2in -> evalspeedpldi3bb3in : [ A>=1+C ], cost: 1
      7: evalspeedpldi3bb2in -> evalspeedpldi3bb4in : [ C>=A ], cost: 1
      8: evalspeedpldi3bb3in -> evalspeedpldi3bb5in : C'=1+C, [], cost: 1
      9: evalspeedpldi3bb4in -> evalspeedpldi3bb5in : C'=0, D'=1+D, [], cost: 1


Simplified the transitions:
  Start location: evalspeedpldi3start
      0: evalspeedpldi3start -> evalspeedpldi3entryin : [], cost: 1
      3: evalspeedpldi3entryin -> evalspeedpldi3bb5in : C'=0, D'=0, [ A>=1 && B>=1+A ], cost: 1
      4: evalspeedpldi3bb5in -> evalspeedpldi3bb2in : [ B>=1+D ], cost: 1
      6: evalspeedpldi3bb2in -> evalspeedpldi3bb3in : [ A>=1+C ], cost: 1
      7: evalspeedpldi3bb2in -> evalspeedpldi3bb4in : [ C>=A ], cost: 1
      8: evalspeedpldi3bb3in -> evalspeedpldi3bb5in : C'=1+C, [], cost: 1
      9: evalspeedpldi3bb4in -> evalspeedpldi3bb5in : C'=0, D'=1+D, [], cost: 1


Applied simple chaining:
  Start location: evalspeedpldi3start
      0: evalspeedpldi3start -> evalspeedpldi3bb5in : C'=0, D'=0, [ A>=1 && B>=1+A ], cost: 2
      4: evalspeedpldi3bb5in -> evalspeedpldi3bb2in : [ B>=1+D ], cost: 1
      6: evalspeedpldi3bb2in -> evalspeedpldi3bb5in : C'=1+C, [ A>=1+C ], cost: 2
      7: evalspeedpldi3bb2in -> evalspeedpldi3bb5in : C'=0, D'=1+D, [ C>=A ], cost: 2


Applied chaining over branches and pruning:
  Start location: evalspeedpldi3start
      0: evalspeedpldi3start -> evalspeedpldi3bb5in : C'=0, D'=0, [ A>=1 && B>=1+A ], cost: 2
     11: evalspeedpldi3bb5in -> evalspeedpldi3bb5in : C'=1+C, [ B>=1+D && A>=1+C ], cost: 3
     12: evalspeedpldi3bb5in -> evalspeedpldi3bb5in : C'=0, D'=1+D, [ B>=1+D && C>=A ], cost: 3

Eliminating 2 self-loops for location evalspeedpldi3bb5in
  Self-Loop 11 has the metering function: -C+A, resulting in the new transition 13.
  Found this metering function when nesting loops: -1+B-D,
  and nested parallel self-loops 14 (outer loop) and 13 (inner loop), obtaining the new transitions: 15, 16.
  Found this metering function when nesting loops: -1-C+A,
  Removing the self-loops: 11 12 14.
Adding an epsilon transition (to model nonexecution of the loops): 17.

Removed all Self-loops using metering functions (where possible):
  Start location: evalspeedpldi3start
      0: evalspeedpldi3start -> evalspeedpldi3bb5in : C'=0, D'=0, [ A>=1 && B>=1+A ], cost: 2
     13: evalspeedpldi3bb5in -> [8] : C'=A, [ B>=1+D && A>=1+C ], cost: -3*C+3*A
     15: evalspeedpldi3bb5in -> [8] : C'=A, D'=-1+B, [ B>=1+D && C>=A && B>=2+D && A>=1 ], cost: -3+3*B+3*(-1+B-D)*A-3*D
     16: evalspeedpldi3bb5in -> [8] : C'=A, D'=-1+B, [ B>=1+D && A>=1+C && B>=1+D && A>=A && B>=2+D && A>=1 ], cost: -3+3*B+3*(-1+B-D)*A-3*C-3*D+3*A
     17: evalspeedpldi3bb5in -> [8] : [], cost: 0


Applied chaining over branches and pruning:
  Start location: evalspeedpldi3start
     18: evalspeedpldi3start -> [8] : C'=A, D'=0, [ A>=1 && B>=1+A && B>=1 && A>=1 ], cost: 2+3*A
     19: evalspeedpldi3start -> [8] : C'=A, D'=-1+B, [ A>=1 && B>=1+A && B>=1 && A>=1 && B>=1 && A>=A && B>=2 && A>=1 ], cost: -1+3*(-1+B)*A+3*B+3*A


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: evalspeedpldi3start
     18: evalspeedpldi3start -> [8] : C'=A, D'=0, [ A>=1 && B>=1+A && B>=1 && A>=1 ], cost: 2+3*A
     19: evalspeedpldi3start -> [8] : C'=A, D'=-1+B, [ A>=1 && B>=1+A && B>=1 && A>=1 && B>=1 && A>=A && B>=2 && A>=1 ], cost: -1+3*(-1+B)*A+3*B+3*A


Computing complexity for remaining 2 transitions.

  Found configuration with infinitely models for cost: 2+3*A
  and guard: A>=1 && B>=1+A && B>=1 && A>=1:
  B: Pos, A: Pos, where: B > A

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

  Found configuration with infinitely models for cost: -1+3*(-1+B)*A+3*B+3*A
  and guard: A>=1 && B>=1+A && B>=1 && A>=1 && B>=1 && A>=A && B>=2 && A>=1:
  B: Pos, A: Pos, where: B > A

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


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