Trying to load file: main.koat Initial Control flow graph problem: Start location: f0 0: f0 -> f10001 : [], cost: 1 11: f0 -> f10001 : B'=1, [], cost: 1 15: f0 -> f10001 : A'=B, D'=1, [], cost: 1 16: f0 -> f10001 : A'=B, B'=1, D'=1, [], cost: 1 10: f0 -> f2 : B'=2, [], cost: 1 13: f0 -> f12 : A'=B, B'=2, [], cost: 1 12: f0 -> f110 : A'=1, B'=1, [], cost: 1 2: f2 -> f10001 : [], cost: 1 1: f2 -> f2 : [], cost: 1 3: f2 -> f1200 : A'=B, [], cost: 1 22: f1200 -> f10001 : D'=1, [], cost: 1 9: f1200 -> f1200 : [], cost: 1 4: f2200 -> f10000 : C'=0, [ C==0 ], cost: 1 14: f12 -> f10001 : D'=1, [], cost: 1 5: f12 -> f12 : [], cost: 1 17: f100 -> f10001 : D'=1, [], cost: 1 6: f100 -> f110 : B'=1, [], cost: 1 18: f110 -> f10001 : D'=1, [], cost: 1 7: f110 -> f120 : B'=2, [], cost: 1 19: f120 -> f10001 : D'=1, [], cost: 1 8: f120 -> f120 : [], cost: 1 21: f1000 -> f10001 : D'=1, [], cost: 1 23: f1000 -> f10001 : B'=1, D'=1, [], cost: 1 20: f1000 -> f1200 : B'=2, [], cost: 1 Simplified the transitions: Start location: f0 10: f0 -> f2 : B'=2, [], cost: 1 13: f0 -> f12 : A'=B, B'=2, [], cost: 1 12: f0 -> f110 : A'=1, B'=1, [], cost: 1 1: f2 -> f2 : [], cost: 1 3: f2 -> f1200 : A'=B, [], cost: 1 9: f1200 -> f1200 : [], cost: 1 5: f12 -> f12 : [], cost: 1 7: f110 -> f120 : B'=2, [], cost: 1 8: f120 -> f120 : [], cost: 1 Eliminating 1 self-loops for location f2 Self-Loop 1 has unbounded runtime, resulting in the new transition 24. Removing the self-loops: 1. Eliminating 1 self-loops for location f1200 Self-Loop 9 has unbounded runtime, resulting in the new transition 25. Removing the self-loops: 9. Eliminating 1 self-loops for location f12 Self-Loop 5 has unbounded runtime, resulting in the new transition 26. Removing the self-loops: 5. Eliminating 1 self-loops for location f120 Self-Loop 8 has unbounded runtime, resulting in the new transition 27. Removing the self-loops: 8. Removed all Self-loops using metering functions (where possible): Start location: f0 10: f0 -> f2 : B'=2, [], cost: 1 13: f0 -> f12 : A'=B, B'=2, [], cost: 1 12: f0 -> f110 : A'=1, B'=1, [], cost: 1 24: f2 -> [11] : [], cost: INF 25: f1200 -> [12] : [], cost: INF 26: f12 -> [13] : [], cost: INF 7: f110 -> f120 : B'=2, [], cost: 1 27: f120 -> [14] : [], cost: INF 3: [11] -> f1200 : A'=B, [], cost: 1 Applied simple chaining: Start location: f0 10: f0 -> [12] : A'=2, B'=2, [], cost: INF 13: f0 -> [13] : A'=B, B'=2, [], cost: INF 12: f0 -> [14] : A'=1, B'=2, [], cost: INF Final control flow graph problem, now checking costs for infinitely many models: Start location: f0 10: f0 -> [12] : A'=2, B'=2, [], cost: INF 13: f0 -> [13] : A'=B, B'=2, [], cost: INF 12: f0 -> [14] : A'=1, B'=2, [], cost: INF Computing complexity for remaining 3 transitions. Found new complexity INF, because: INF sat. The final runtime is determined by this resulting transition: Final Guard: Final Cost: INF Obtained the following complexity w.r.t. the length of the input n: Complexity class: INF Complexity value: INF WORST_CASE(INF,?)