Trying to load file: main.koat Initial Control flow graph problem: Start location: f3 0: f1 -> f1 : B'=1+B, C'=D, D'=free, E'=D, F'=free_1, G'=B, [ A>=1+B && B>=0 ], cost: 1 1: f1 -> f4 : A'=free_5, B'=free_8, C'=free_4, D'=free_3, E'=free_10, H'=free_2, Q'=free_11, J'=free_7, K'=free_6, L'=free_9, M'=C, [ B>=A && B>=0 && free_8>=free_2 && free_2>=2 ], cost: 1 3: f3 -> f1 : A'=free_26, B'=2, C'=free_25, D'=free_28, E'=free_25, H'=free_26, Q'=free_29, J'=free_25, N'=free_24, O'=free_29, P'=free_27, [ free_26>=2 ], cost: 1 2: f3 -> f4 : A'=free_22, B'=free_18, C'=free_19, D'=free_17, E'=free_13, H'=free_15, Q'=free_14, J'=free_21, K'=free_20, L'=free_16, M'=O, N'=free_12, [ 0>=free_15 && 0>=free_23 ], cost: 1 4: f3 -> f4 : A'=free_33, B'=free_37, C'=free_32, D'=free_35, E'=free_31, H'=1, Q'=free_40, J'=free_39, K'=free_38, L'=free_34, M'=D, N'=free_30, O'=free_36, [], cost: 1 Simplified the transitions: Start location: f3 0: f1 -> f1 : B'=1+B, C'=D, D'=free, E'=D, F'=free_1, G'=B, [ A>=1+B && B>=0 ], cost: 1 3: f3 -> f1 : A'=free_26, B'=2, C'=free_25, D'=free_28, E'=free_25, H'=free_26, Q'=free_29, J'=free_25, N'=free_24, O'=free_29, P'=free_27, [ free_26>=2 ], cost: 1 Eliminating 1 self-loops for location f1 Self-Loop 0 has the metering function: -B+A, resulting in the new transition 5. Removing the self-loops: 0. Removed all Self-loops using metering functions (where possible): Start location: f3 5: f1 -> [3] : B'=A, C'=free, D'=free, E'=free, F'=free_1, G'=-1+A, [ A>=1+B && B>=0 ], cost: -B+A 3: f3 -> f1 : A'=free_26, B'=2, C'=free_25, D'=free_28, E'=free_25, H'=free_26, Q'=free_29, J'=free_25, N'=free_24, O'=free_29, P'=free_27, [ free_26>=2 ], cost: 1 Applied simple chaining: Start location: f3 3: f3 -> [3] : A'=free_26, B'=free_26, C'=free, D'=free, E'=free, F'=free_1, G'=-1+free_26, H'=free_26, Q'=free_29, J'=free_25, N'=free_24, O'=free_29, P'=free_27, [ free_26>=2 && free_26>=3 && 2>=0 ], cost: -1+free_26 Final control flow graph problem, now checking costs for infinitely many models: Start location: f3 3: f3 -> [3] : A'=free_26, B'=free_26, C'=free, D'=free, E'=free, F'=free_1, G'=-1+free_26, H'=free_26, Q'=free_29, J'=free_25, N'=free_24, O'=free_29, P'=free_27, [ free_26>=2 && free_26>=3 && 2>=0 ], cost: -1+free_26 Computing complexity for remaining 1 transitions. Found configuration with infinitely models for cost: -1+free_26 and guard: free_26>=2 && free_26>=3 && 2>=0: free_26: Pos Found new complexity INF, because: Found infinity configuration. The final runtime is determined by this resulting transition: Final Guard: free_26>=2 && free_26>=3 && 2>=0 Final Cost: -1+free_26 Obtained the following complexity w.r.t. the length of the input n: Complexity class: INF Complexity value: INF WORST_CASE(INF,?)