WORST_CASE(Omega(1),?) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: __init 0: f1_0_main_ConstantStackPush -> f66_0_main_GE : arg1'=arg1P_1, arg2'=arg2P_1, arg3'=arg3P_1, [ arg1P_1<=arg1 && arg2>-1 && arg1>0 && arg1P_1>0 && 0==arg2P_1 && arg2==arg3P_1 ], cost: 1 1: f66_0_main_GE -> f66_0_main_GE : arg1'=arg1P_2, arg2'=arg2P_2, arg3'=arg3P_2, [ arg3>arg2 && arg3>-1 && arg1P_2<=arg1 && arg1>0 && arg1P_2>0 && 1+arg2==arg2P_2 && arg3==arg3P_2 ], cost: 1 2: __init -> f1_0_main_ConstantStackPush : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 2: __init -> f1_0_main_ConstantStackPush : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [], cost: 1 Simplified all rules, resulting in: Start location: __init 0: f1_0_main_ConstantStackPush -> f66_0_main_GE : arg1'=arg1P_1, arg2'=0, arg3'=arg2, [ arg1P_1<=arg1 && arg2>-1 && arg1>0 && arg1P_1>0 ], cost: 1 1: f66_0_main_GE -> f66_0_main_GE : arg1'=arg1P_2, arg2'=1+arg2, [ arg3>arg2 && arg3>-1 && arg1P_2<=arg1 && arg1>0 && arg1P_2>0 ], cost: 1 2: __init -> f1_0_main_ConstantStackPush : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [], cost: 1 ### Simplification by acceleration and chaining ### Accelerating simple loops of location 1. Accelerating the following rules: 1: f66_0_main_GE -> f66_0_main_GE : arg1'=arg1P_2, arg2'=1+arg2, [ arg3>arg2 && arg3>-1 && arg1P_2<=arg1 && arg1>0 && arg1P_2>0 ], cost: 1 Accelerated rule 1 with backward acceleration, yielding the new rule 3. [accelerate] Nesting with 1 inner and 1 outer candidates Removing the simple loops: 1. Accelerated all simple loops using metering functions (where possible): Start location: __init 0: f1_0_main_ConstantStackPush -> f66_0_main_GE : arg1'=arg1P_1, arg2'=0, arg3'=arg2, [ arg1P_1<=arg1 && arg2>-1 && arg1>0 && arg1P_1>0 ], cost: 1 3: f66_0_main_GE -> f66_0_main_GE : arg1'=arg1P_2, arg2'=arg3, [ arg3>-1 && arg1P_2<=arg1 && arg1>0 && arg1P_2>0 && -arg2+arg3>=1 ], cost: -arg2+arg3 2: __init -> f1_0_main_ConstantStackPush : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [], cost: 1 Chained accelerated rules (with incoming rules): Start location: __init 0: f1_0_main_ConstantStackPush -> f66_0_main_GE : arg1'=arg1P_1, arg2'=0, arg3'=arg2, [ arg1P_1<=arg1 && arg2>-1 && arg1>0 && arg1P_1>0 ], cost: 1 4: f1_0_main_ConstantStackPush -> f66_0_main_GE : arg1'=arg1P_2, arg3'=arg2, [ arg1>0 && arg1P_2>0 && arg2>=1 && arg1P_2<=arg1 ], cost: 1+arg2 2: __init -> f1_0_main_ConstantStackPush : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [], cost: 1 Removed unreachable locations (and leaf rules with constant cost): Start location: __init 4: f1_0_main_ConstantStackPush -> f66_0_main_GE : arg1'=arg1P_2, arg3'=arg2, [ arg1>0 && arg1P_2>0 && arg2>=1 && arg1P_2<=arg1 ], cost: 1+arg2 2: __init -> f1_0_main_ConstantStackPush : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [], cost: 1 Eliminated locations (on linear paths): Start location: __init 5: __init -> f66_0_main_GE : arg1'=arg1P_2, arg2'=arg2P_3, arg3'=arg2P_3, [ arg1P_3>0 && arg1P_2>0 && arg2P_3>=1 && arg1P_2<=arg1P_3 ], cost: 2+arg2P_3 ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: __init 5: __init -> f66_0_main_GE : arg1'=arg1P_2, arg2'=arg2P_3, arg3'=arg2P_3, [ arg1P_3>0 && arg1P_2>0 && arg2P_3>=1 && arg1P_2<=arg1P_3 ], cost: 2+arg2P_3 Computing asymptotic complexity for rule 5 Simplified the guard: 5: __init -> f66_0_main_GE : arg1'=arg1P_2, arg2'=arg2P_3, arg3'=arg2P_3, [ arg1P_2>0 && arg2P_3>=1 && arg1P_2<=arg1P_3 ], cost: 2+arg2P_3 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),?)