Trying to load file: main.koat Initial Control flow graph problem: Start location: f0 0: f47 -> f47 : [], cost: 1 1: f49 -> f52 : [], cost: 1 2: f11 -> f47 : C'=0, [ A>=B ], cost: 1 6: f11 -> f47 : C'=0, F'=free_1, G'=free_2, H'=free_3, Q'=free, J'=C, K'=free_1, L'=free_1, [ free_1>=1 && B>=1+A ], cost: 1 9: f11 -> f11 : A'=1+A, D'=1, F'=free_15, G'=free_16, H'=free_17, Q'=free_14, J'=C, K'=free_15, L'=free_15, M'=free_15, N'=O, P'=1, Q_1'=1, R'=0, [ 0>=free_15 && B>=1+A ], cost: 1 7: f11 -> f35 : D'=free_6, F'=free_5, G'=free_7, H'=free_8, Q'=free_4, J'=C, K'=free_5, L'=free_5, M'=free_5, N'=O, O'=0, P'=free_6, Q_1'=free_6, R'=0, [ B>=1+A && 0>=free_5 && free_6>=2 ], cost: 1 8: f11 -> f35 : D'=free_11, F'=free_10, G'=free_12, H'=free_13, Q'=free_9, J'=C, K'=free_10, L'=free_10, M'=free_10, N'=O, O'=0, P'=free_11, Q_1'=free_11, R'=0, [ B>=1+A && 0>=free_10 && 0>=free_11 ], cost: 1 3: f35 -> f47 : C'=0, [ D>=3 ], cost: 1 4: f35 -> f47 : C'=0, [ 1>=D ], cost: 1 5: f35 -> f47 : C'=0, D'=2, E'=F, [ D==2 ], cost: 1 10: f0 -> f11 : S'=0, T'=0, [], cost: 1 Simplified the transitions: Start location: f0 0: f47 -> f47 : [], cost: 1 2: f11 -> f47 : C'=0, [ A>=B ], cost: 1 6: f11 -> f47 : C'=0, F'=free_1, G'=free_2, H'=free_3, Q'=free, J'=C, K'=free_1, L'=free_1, [ free_1>=1 && B>=1+A ], cost: 1 9: f11 -> f11 : A'=1+A, D'=1, F'=free_15, G'=free_16, H'=free_17, Q'=free_14, J'=C, K'=free_15, L'=free_15, M'=free_15, N'=O, P'=1, Q_1'=1, R'=0, [ 0>=free_15 && B>=1+A ], cost: 1 7: f11 -> f35 : D'=free_6, F'=free_5, G'=free_7, H'=free_8, Q'=free_4, J'=C, K'=free_5, L'=free_5, M'=free_5, N'=O, O'=0, P'=free_6, Q_1'=free_6, R'=0, [ B>=1+A && 0>=free_5 && free_6>=2 ], cost: 1 8: f11 -> f35 : D'=free_11, F'=free_10, G'=free_12, H'=free_13, Q'=free_9, J'=C, K'=free_10, L'=free_10, M'=free_10, N'=O, O'=0, P'=free_11, Q_1'=free_11, R'=0, [ B>=1+A && 0>=free_10 && 0>=free_11 ], cost: 1 3: f35 -> f47 : C'=0, [ D>=3 ], cost: 1 4: f35 -> f47 : C'=0, [ 1>=D ], cost: 1 5: f35 -> f47 : C'=0, D'=2, E'=F, [ D==2 ], cost: 1 10: f0 -> f11 : S'=0, T'=0, [], cost: 1 Eliminating 1 self-loops for location f47 Self-Loop 0 has unbounded runtime, resulting in the new transition 11. Removing the self-loops: 0. Eliminating 1 self-loops for location f11 Self-Loop 9 has the metering function: B-A, resulting in the new transition 12. Removing the self-loops: 9. Removed all Self-loops using metering functions (where possible): Start location: f0 11: f47 -> [6] : [], cost: INF 12: f11 -> [7] : A'=B, D'=1, F'=free_15, G'=free_16, H'=free_17, Q'=free_14, J'=C, K'=free_15, L'=free_15, M'=free_15, N'=O, P'=1, Q_1'=1, R'=0, [ 0>=free_15 && B>=1+A ], cost: B-A 3: f35 -> f47 : C'=0, [ D>=3 ], cost: 1 4: f35 -> f47 : C'=0, [ 1>=D ], cost: 1 5: f35 -> f47 : C'=0, D'=2, E'=F, [ D==2 ], cost: 1 10: f0 -> f11 : S'=0, T'=0, [], cost: 1 2: [7] -> f47 : C'=0, [ A>=B ], cost: 1 6: [7] -> f47 : C'=0, F'=free_1, G'=free_2, H'=free_3, Q'=free, J'=C, K'=free_1, L'=free_1, [ free_1>=1 && B>=1+A ], cost: 1 7: [7] -> f35 : D'=free_6, F'=free_5, G'=free_7, H'=free_8, Q'=free_4, J'=C, K'=free_5, L'=free_5, M'=free_5, N'=O, O'=0, P'=free_6, Q_1'=free_6, R'=0, [ B>=1+A && 0>=free_5 && free_6>=2 ], cost: 1 8: [7] -> f35 : D'=free_11, F'=free_10, G'=free_12, H'=free_13, Q'=free_9, J'=C, K'=free_10, L'=free_10, M'=free_10, N'=O, O'=0, P'=free_11, Q_1'=free_11, R'=0, [ B>=1+A && 0>=free_10 && 0>=free_11 ], cost: 1 Applied simple chaining: Start location: f0 11: f47 -> [6] : [], cost: INF 3: f35 -> f47 : C'=0, [ D>=3 ], cost: 1 4: f35 -> f47 : C'=0, [ 1>=D ], cost: 1 5: f35 -> f47 : C'=0, D'=2, E'=F, [ D==2 ], cost: 1 10: f0 -> [7] : A'=B, D'=1, F'=free_15, G'=free_16, H'=free_17, Q'=free_14, J'=C, K'=free_15, L'=free_15, M'=free_15, N'=O, P'=1, Q_1'=1, R'=0, S'=0, T'=0, [ 0>=free_15 && B>=1+A ], cost: 1+B-A 2: [7] -> f47 : C'=0, [ A>=B ], cost: 1 6: [7] -> f47 : C'=0, F'=free_1, G'=free_2, H'=free_3, Q'=free, J'=C, K'=free_1, L'=free_1, [ free_1>=1 && B>=1+A ], cost: 1 7: [7] -> f35 : D'=free_6, F'=free_5, G'=free_7, H'=free_8, Q'=free_4, J'=C, K'=free_5, L'=free_5, M'=free_5, N'=O, O'=0, P'=free_6, Q_1'=free_6, R'=0, [ B>=1+A && 0>=free_5 && free_6>=2 ], cost: 1 8: [7] -> f35 : D'=free_11, F'=free_10, G'=free_12, H'=free_13, Q'=free_9, J'=C, K'=free_10, L'=free_10, M'=free_10, N'=O, O'=0, P'=free_11, Q_1'=free_11, R'=0, [ B>=1+A && 0>=free_10 && 0>=free_11 ], cost: 1 Applied chaining over branches and pruning: Start location: f0 11: f47 -> [6] : [], cost: INF 13: f0 -> f47 : A'=B, C'=0, D'=1, F'=free_15, G'=free_16, H'=free_17, Q'=free_14, J'=C, K'=free_15, L'=free_15, M'=free_15, N'=O, P'=1, Q_1'=1, R'=0, S'=0, T'=0, [ 0>=free_15 && B>=1+A && B>=B ], cost: 2+B-A 14: f0 -> [8] : A'=B, D'=1, F'=free_15, G'=free_16, H'=free_17, Q'=free_14, J'=C, K'=free_15, L'=free_15, M'=free_15, N'=O, P'=1, Q_1'=1, R'=0, S'=0, T'=0, [ 0>=free_15 && B>=1+A ], cost: 1+B-A 15: f0 -> [9] : A'=B, D'=1, F'=free_15, G'=free_16, H'=free_17, Q'=free_14, J'=C, K'=free_15, L'=free_15, M'=free_15, N'=O, P'=1, Q_1'=1, R'=0, S'=0, T'=0, [ 0>=free_15 && B>=1+A ], cost: 1+B-A 16: f0 -> [10] : A'=B, D'=1, F'=free_15, G'=free_16, H'=free_17, Q'=free_14, J'=C, K'=free_15, L'=free_15, M'=free_15, N'=O, P'=1, Q_1'=1, R'=0, S'=0, T'=0, [ 0>=free_15 && B>=1+A ], cost: 1+B-A Applied simple chaining: Start location: f0 13: f0 -> [6] : A'=B, C'=0, D'=1, F'=free_15, G'=free_16, H'=free_17, Q'=free_14, J'=C, K'=free_15, L'=free_15, M'=free_15, N'=O, P'=1, Q_1'=1, R'=0, S'=0, T'=0, [ 0>=free_15 && B>=1+A && B>=B ], cost: INF 14: f0 -> [8] : A'=B, D'=1, F'=free_15, G'=free_16, H'=free_17, Q'=free_14, J'=C, K'=free_15, L'=free_15, M'=free_15, N'=O, P'=1, Q_1'=1, R'=0, S'=0, T'=0, [ 0>=free_15 && B>=1+A ], cost: 1+B-A 15: f0 -> [9] : A'=B, D'=1, F'=free_15, G'=free_16, H'=free_17, Q'=free_14, J'=C, K'=free_15, L'=free_15, M'=free_15, N'=O, P'=1, Q_1'=1, R'=0, S'=0, T'=0, [ 0>=free_15 && B>=1+A ], cost: 1+B-A 16: f0 -> [10] : A'=B, D'=1, F'=free_15, G'=free_16, H'=free_17, Q'=free_14, J'=C, K'=free_15, L'=free_15, M'=free_15, N'=O, P'=1, Q_1'=1, R'=0, S'=0, T'=0, [ 0>=free_15 && B>=1+A ], cost: 1+B-A Final control flow graph problem, now checking costs for infinitely many models: Start location: f0 13: f0 -> [6] : A'=B, C'=0, D'=1, F'=free_15, G'=free_16, H'=free_17, Q'=free_14, J'=C, K'=free_15, L'=free_15, M'=free_15, N'=O, P'=1, Q_1'=1, R'=0, S'=0, T'=0, [ 0>=free_15 && B>=1+A && B>=B ], cost: INF 14: f0 -> [8] : A'=B, D'=1, F'=free_15, G'=free_16, H'=free_17, Q'=free_14, J'=C, K'=free_15, L'=free_15, M'=free_15, N'=O, P'=1, Q_1'=1, R'=0, S'=0, T'=0, [ 0>=free_15 && B>=1+A ], cost: 1+B-A 15: f0 -> [9] : A'=B, D'=1, F'=free_15, G'=free_16, H'=free_17, Q'=free_14, J'=C, K'=free_15, L'=free_15, M'=free_15, N'=O, P'=1, Q_1'=1, R'=0, S'=0, T'=0, [ 0>=free_15 && B>=1+A ], cost: 1+B-A 16: f0 -> [10] : A'=B, D'=1, F'=free_15, G'=free_16, H'=free_17, Q'=free_14, J'=C, K'=free_15, L'=free_15, M'=free_15, N'=O, P'=1, Q_1'=1, R'=0, S'=0, T'=0, [ 0>=free_15 && B>=1+A ], cost: 1+B-A Computing complexity for remaining 4 transitions. Found new complexity INF, because: INF sat. The final runtime is determined by this resulting transition: Final Guard: 0>=free_15 && B>=1+A && B>=B 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,?)