WORST_CASE(Omega(1),?) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: __init 0: f1_0_main_Load -> f145_0_main_LE : arg1'=arg1P_1, arg2'=arg2P_1, [ arg2P_1>-1 && arg2>-1 && arg1P_1>-1 && arg1>0 ], cost: 1 1: f145_0_main_LE -> f145_0_main_LE : arg1'=arg1P_2, arg2'=arg2P_2, [ arg22 && 1+arg1==arg1P_2 && 2*arg2==arg2P_2 ], cost: 1 2: __init -> f1_0_main_Load : arg1'=arg1P_3, arg2'=arg2P_3, [], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 2: __init -> f1_0_main_Load : arg1'=arg1P_3, arg2'=arg2P_3, [], cost: 1 Simplified all rules, resulting in: Start location: __init 0: f1_0_main_Load -> f145_0_main_LE : arg1'=arg1P_1, arg2'=arg2P_1, [ arg2P_1>-1 && arg2>-1 && arg1P_1>-1 && arg1>0 ], cost: 1 1: f145_0_main_LE -> f145_0_main_LE : arg1'=1+arg1, arg2'=2*arg2, [ arg22 ], cost: 1 2: __init -> f1_0_main_Load : arg1'=arg1P_3, arg2'=arg2P_3, [], cost: 1 ### Simplification by acceleration and chaining ### Accelerating simple loops of location 1. Accelerating the following rules: 1: f145_0_main_LE -> f145_0_main_LE : arg1'=1+arg1, arg2'=2*arg2, [ arg22 ], 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_Load -> f145_0_main_LE : arg1'=arg1P_1, arg2'=arg2P_1, [ arg2P_1>-1 && arg2>-1 && arg1P_1>-1 && arg1>0 ], cost: 1 3: f145_0_main_LE -> f145_0_main_LE : arg1'=arg1+k, arg2'=arg2*2^k, [ arg2>2 && k>=0 && arg2*2^(-1+k)<-1+arg1+k ], cost: k 2: __init -> f1_0_main_Load : arg1'=arg1P_3, arg2'=arg2P_3, [], cost: 1 Chained accelerated rules (with incoming rules): Start location: __init 0: f1_0_main_Load -> f145_0_main_LE : arg1'=arg1P_1, arg2'=arg2P_1, [ arg2P_1>-1 && arg2>-1 && arg1P_1>-1 && arg1>0 ], cost: 1 4: f1_0_main_Load -> f145_0_main_LE : arg1'=arg1P_1+k, arg2'=2^k*arg2P_1, [ arg2>-1 && arg1P_1>-1 && arg1>0 && arg2P_1>2 && k>=0 && 2^(-1+k)*arg2P_1<-1+arg1P_1+k ], cost: 1+k 2: __init -> f1_0_main_Load : arg1'=arg1P_3, arg2'=arg2P_3, [], cost: 1 Removed unreachable locations (and leaf rules with constant cost): Start location: __init 4: f1_0_main_Load -> f145_0_main_LE : arg1'=arg1P_1+k, arg2'=2^k*arg2P_1, [ arg2>-1 && arg1P_1>-1 && arg1>0 && arg2P_1>2 && k>=0 && 2^(-1+k)*arg2P_1<-1+arg1P_1+k ], cost: 1+k 2: __init -> f1_0_main_Load : arg1'=arg1P_3, arg2'=arg2P_3, [], cost: 1 Eliminated locations (on linear paths): Start location: __init 5: __init -> f145_0_main_LE : arg1'=arg1P_1+k, arg2'=2^k*arg2P_1, [ arg2P_3>-1 && arg1P_1>-1 && arg1P_3>0 && arg2P_1>2 && k>=0 && 2^(-1+k)*arg2P_1<-1+arg1P_1+k ], cost: 2+k ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: __init 5: __init -> f145_0_main_LE : arg1'=arg1P_1+k, arg2'=2^k*arg2P_1, [ arg2P_3>-1 && arg1P_1>-1 && arg1P_3>0 && arg2P_1>2 && k>=0 && 2^(-1+k)*arg2P_1<-1+arg1P_1+k ], cost: 2+k Computing asymptotic complexity for rule 5 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),?)