WORST_CASE(Omega(1),?) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: __init 0: f1_0_main_Load -> f145_0_main_EQ : arg1'=arg1P_1, arg2'=arg2P_1, [ arg2P_1>-1 && arg2>-1 && arg1P_1>-1 && arg1>0 ], cost: 1 1: f145_0_main_EQ -> f145_0_main_EQ : arg1'=arg1P_2, arg2'=arg2P_2, [ arg2>arg1 && 1+arg1==arg1P_2 && arg2==arg2P_2 ], cost: 1 2: f145_0_main_EQ -> f145_0_main_EQ : arg1'=arg1P_3, arg2'=arg2P_3, [ arg2 f1_0_main_Load : arg1'=arg1P_4, arg2'=arg2P_4, [], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 3: __init -> f1_0_main_Load : arg1'=arg1P_4, arg2'=arg2P_4, [], cost: 1 Simplified all rules, resulting in: Start location: __init 0: f1_0_main_Load -> f145_0_main_EQ : arg1'=arg1P_1, arg2'=arg2P_1, [ arg2P_1>-1 && arg2>-1 && arg1P_1>-1 && arg1>0 ], cost: 1 1: f145_0_main_EQ -> f145_0_main_EQ : arg1'=1+arg1, [ arg2>arg1 ], cost: 1 2: f145_0_main_EQ -> f145_0_main_EQ : arg2'=1+arg2, [ arg2 f1_0_main_Load : arg1'=arg1P_4, arg2'=arg2P_4, [], cost: 1 ### Simplification by acceleration and chaining ### Accelerating simple loops of location 1. Accelerating the following rules: 1: f145_0_main_EQ -> f145_0_main_EQ : arg1'=1+arg1, [ arg2>arg1 ], cost: 1 2: f145_0_main_EQ -> f145_0_main_EQ : arg2'=1+arg2, [ arg2 f145_0_main_EQ : arg1'=arg1P_1, arg2'=arg2P_1, [ arg2P_1>-1 && arg2>-1 && arg1P_1>-1 && arg1>0 ], cost: 1 4: f145_0_main_EQ -> f145_0_main_EQ : arg1'=arg2, [ arg2-arg1>=0 ], cost: arg2-arg1 5: f145_0_main_EQ -> f145_0_main_EQ : arg2'=arg1, [ -arg2+arg1>=0 ], cost: -arg2+arg1 3: __init -> f1_0_main_Load : arg1'=arg1P_4, arg2'=arg2P_4, [], cost: 1 Chained accelerated rules (with incoming rules): Start location: __init 0: f1_0_main_Load -> f145_0_main_EQ : arg1'=arg1P_1, arg2'=arg2P_1, [ arg2P_1>-1 && arg2>-1 && arg1P_1>-1 && arg1>0 ], cost: 1 6: f1_0_main_Load -> f145_0_main_EQ : arg1'=arg2P_1, arg2'=arg2P_1, [ arg2P_1>-1 && arg2>-1 && arg1P_1>-1 && arg1>0 && -arg1P_1+arg2P_1>=0 ], cost: 1-arg1P_1+arg2P_1 7: f1_0_main_Load -> f145_0_main_EQ : arg1'=arg1P_1, arg2'=arg1P_1, [ arg2P_1>-1 && arg2>-1 && arg1P_1>-1 && arg1>0 && arg1P_1-arg2P_1>=0 ], cost: 1+arg1P_1-arg2P_1 3: __init -> f1_0_main_Load : arg1'=arg1P_4, arg2'=arg2P_4, [], cost: 1 Removed unreachable locations (and leaf rules with constant cost): Start location: __init 6: f1_0_main_Load -> f145_0_main_EQ : arg1'=arg2P_1, arg2'=arg2P_1, [ arg2P_1>-1 && arg2>-1 && arg1P_1>-1 && arg1>0 && -arg1P_1+arg2P_1>=0 ], cost: 1-arg1P_1+arg2P_1 7: f1_0_main_Load -> f145_0_main_EQ : arg1'=arg1P_1, arg2'=arg1P_1, [ arg2P_1>-1 && arg2>-1 && arg1P_1>-1 && arg1>0 && arg1P_1-arg2P_1>=0 ], cost: 1+arg1P_1-arg2P_1 3: __init -> f1_0_main_Load : arg1'=arg1P_4, arg2'=arg2P_4, [], cost: 1 Eliminated locations (on tree-shaped paths): Start location: __init 8: __init -> f145_0_main_EQ : arg1'=arg2P_1, arg2'=arg2P_1, [ arg2P_1>-1 && arg2P_4>-1 && arg1P_1>-1 && arg1P_4>0 && -arg1P_1+arg2P_1>=0 ], cost: 2-arg1P_1+arg2P_1 9: __init -> f145_0_main_EQ : arg1'=arg1P_1, arg2'=arg1P_1, [ arg2P_1>-1 && arg2P_4>-1 && arg1P_1>-1 && arg1P_4>0 && arg1P_1-arg2P_1>=0 ], cost: 2+arg1P_1-arg2P_1 ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: __init 8: __init -> f145_0_main_EQ : arg1'=arg2P_1, arg2'=arg2P_1, [ arg2P_1>-1 && arg2P_4>-1 && arg1P_1>-1 && arg1P_4>0 && -arg1P_1+arg2P_1>=0 ], cost: 2-arg1P_1+arg2P_1 9: __init -> f145_0_main_EQ : arg1'=arg1P_1, arg2'=arg1P_1, [ arg2P_1>-1 && arg2P_4>-1 && arg1P_1>-1 && arg1P_4>0 && arg1P_1-arg2P_1>=0 ], cost: 2+arg1P_1-arg2P_1 Computing asymptotic complexity for rule 8 Simplified the guard: 8: __init -> f145_0_main_EQ : arg1'=arg2P_1, arg2'=arg2P_1, [ arg2P_4>-1 && arg1P_1>-1 && arg1P_4>0 && -arg1P_1+arg2P_1>=0 ], cost: 2-arg1P_1+arg2P_1 Resulting cost 0 has complexity: Unknown Computing asymptotic complexity for rule 9 Resulting cost 0 has complexity: Unknown Obtained the following overall complexity (w.r.t. the length of the input n): Complexity: Constant Cpx degree: 0 Solved cost: 1 Rule cost: 1 Rule guard: [] WORST_CASE(Omega(1),?)