WORST_CASE(INF,?) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: __init 0: f1_0_main_Load -> f127_0_eq_LE : arg1'=arg1P_1, arg2'=arg2P_1, [ arg1P_1>-1 && arg2>1 && arg2P_1>-1 && arg1>0 ], cost: 1 1: f127_0_eq_LE -> f127_0_eq_LE : arg1'=arg1P_2, arg2'=arg2P_2, [ arg2>0 && arg1>0 && -1+arg1 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 -> f127_0_eq_LE : arg1'=arg1P_1, arg2'=arg2P_1, [ arg1P_1>-1 && arg2>1 && arg2P_1>-1 && arg1>0 ], cost: 1 1: f127_0_eq_LE -> f127_0_eq_LE : arg1'=-1+arg1, arg2'=-1+arg2, [ arg2>0 && arg1>0 ], 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: f127_0_eq_LE -> f127_0_eq_LE : arg1'=-1+arg1, arg2'=-1+arg2, [ arg2>0 && arg1>0 ], cost: 1 Accelerated rule 1 with metering function arg2 (after adding arg1>=arg2), yielding the new rule 3. Accelerated rule 1 with metering function arg1 (after adding arg1<=arg2), yielding the new rule 4. Removing the simple loops: 1. Accelerated all simple loops using metering functions (where possible): Start location: __init 0: f1_0_main_Load -> f127_0_eq_LE : arg1'=arg1P_1, arg2'=arg2P_1, [ arg1P_1>-1 && arg2>1 && arg2P_1>-1 && arg1>0 ], cost: 1 3: f127_0_eq_LE -> f127_0_eq_LE : arg1'=arg1-arg2, arg2'=0, [ arg2>0 && arg1>0 && arg1>=arg2 ], cost: arg2 4: f127_0_eq_LE -> f127_0_eq_LE : arg1'=0, arg2'=-arg1+arg2, [ arg2>0 && arg1>0 && arg1<=arg2 ], cost: arg1 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 -> f127_0_eq_LE : arg1'=arg1P_1, arg2'=arg2P_1, [ arg1P_1>-1 && arg2>1 && arg2P_1>-1 && arg1>0 ], cost: 1 5: f1_0_main_Load -> f127_0_eq_LE : arg1'=-arg2P_1+arg1P_1, arg2'=0, [ arg2>1 && arg1>0 && arg2P_1>0 && arg1P_1>0 && arg1P_1>=arg2P_1 ], cost: 1+arg2P_1 6: f1_0_main_Load -> f127_0_eq_LE : arg1'=0, arg2'=arg2P_1-arg1P_1, [ arg2>1 && arg1>0 && arg2P_1>0 && arg1P_1>0 && arg1P_1<=arg2P_1 ], cost: 1+arg1P_1 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 5: f1_0_main_Load -> f127_0_eq_LE : arg1'=-arg2P_1+arg1P_1, arg2'=0, [ arg2>1 && arg1>0 && arg2P_1>0 && arg1P_1>0 && arg1P_1>=arg2P_1 ], cost: 1+arg2P_1 6: f1_0_main_Load -> f127_0_eq_LE : arg1'=0, arg2'=arg2P_1-arg1P_1, [ arg2>1 && arg1>0 && arg2P_1>0 && arg1P_1>0 && arg1P_1<=arg2P_1 ], cost: 1+arg1P_1 2: __init -> f1_0_main_Load : arg1'=arg1P_3, arg2'=arg2P_3, [], cost: 1 Eliminated locations (on tree-shaped paths): Start location: __init 7: __init -> f127_0_eq_LE : arg1'=-arg2P_1+arg1P_1, arg2'=0, [ arg2P_3>1 && arg1P_3>0 && arg2P_1>0 && arg1P_1>0 && arg1P_1>=arg2P_1 ], cost: 2+arg2P_1 8: __init -> f127_0_eq_LE : arg1'=0, arg2'=arg2P_1-arg1P_1, [ arg2P_3>1 && arg1P_3>0 && arg2P_1>0 && arg1P_1>0 && arg1P_1<=arg2P_1 ], cost: 2+arg1P_1 ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: __init 7: __init -> f127_0_eq_LE : arg1'=-arg2P_1+arg1P_1, arg2'=0, [ arg2P_3>1 && arg1P_3>0 && arg2P_1>0 && arg1P_1>0 && arg1P_1>=arg2P_1 ], cost: 2+arg2P_1 8: __init -> f127_0_eq_LE : arg1'=0, arg2'=arg2P_1-arg1P_1, [ arg2P_3>1 && arg1P_3>0 && arg2P_1>0 && arg1P_1>0 && arg1P_1<=arg2P_1 ], cost: 2+arg1P_1 Computing asymptotic complexity for rule 7 Solved the limit problem by the following transformations: Created initial limit problem: arg2P_1 (+/+!), -1+arg2P_3 (+/+!), 2+arg2P_1 (+), 1-arg2P_1+arg1P_1 (+/+!), arg1P_1 (+/+!), arg1P_3 (+/+!) [not solved] removing all constraints (solved by SMT) resulting limit problem: [solved] applying transformation rule (C) using substitution {arg2P_1==n,arg2P_3==n,arg1P_1==1+n,arg1P_3==1} resulting limit problem: [solved] Solution: arg2P_1 / n arg2P_3 / n arg1P_1 / 1+n arg1P_3 / 1 Resulting cost 2+n has complexity: Unbounded Found new complexity Unbounded. Obtained the following overall complexity (w.r.t. the length of the input n): Complexity: Unbounded Cpx degree: Unbounded Solved cost: 2+n Rule cost: 2+arg2P_1 Rule guard: [ arg2P_3>1 && arg1P_3>0 && arg2P_1>0 && arg1P_1>0 && arg1P_1>=arg2P_1 ] WORST_CASE(INF,?)