WORST_CASE(Omega(1),?) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: __init 0: f1 -> f2 : arg1'=arg1P_1, arg2'=arg2P_1, [ arg2==arg2P_1 ], cost: 1 1: f2 -> f3 : arg1'=arg1P_2, arg2'=arg2P_2, [ arg1==arg1P_2 ], cost: 1 7: f3 -> f4 : arg1'=arg1P_8, arg2'=arg2P_8, [ arg1>0 && arg1==arg1P_8 && arg2==arg2P_8 ], cost: 1 9: f3 -> f9 : arg1'=arg1P_10, arg2'=arg2P_10, [ arg1<=0 && arg1==arg1P_10 && arg2==arg2P_10 ], cost: 1 2: f5 -> f6 : arg1'=arg1P_3, arg2'=arg2P_3, [ arg2P_3==-1+arg2 && arg1==arg1P_3 ], cost: 1 4: f6 -> f4 : arg1'=arg1P_5, arg2'=arg2P_5, [ arg1==arg1P_5 && arg2==arg2P_5 ], cost: 1 3: f4 -> f5 : arg1'=arg1P_4, arg2'=arg2P_4, [ arg2>0 && arg1==arg1P_4 && arg2==arg2P_4 ], cost: 1 5: f4 -> f7 : arg1'=arg1P_6, arg2'=arg2P_6, [ arg2<=0 && arg1==arg1P_6 && arg2==arg2P_6 ], cost: 1 6: f7 -> f8 : arg1'=arg1P_7, arg2'=arg2P_7, [ arg1P_7==-1+arg1 && arg2==arg2P_7 ], cost: 1 8: f8 -> f3 : arg1'=arg1P_9, arg2'=arg2P_9, [ arg1==arg1P_9 && arg2==arg2P_9 ], cost: 1 10: __init -> f1 : arg1'=arg1P_11, arg2'=arg2P_11, [], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 10: __init -> f1 : arg1'=arg1P_11, arg2'=arg2P_11, [], cost: 1 Removed unreachable and leaf rules: Start location: __init 0: f1 -> f2 : arg1'=arg1P_1, arg2'=arg2P_1, [ arg2==arg2P_1 ], cost: 1 1: f2 -> f3 : arg1'=arg1P_2, arg2'=arg2P_2, [ arg1==arg1P_2 ], cost: 1 7: f3 -> f4 : arg1'=arg1P_8, arg2'=arg2P_8, [ arg1>0 && arg1==arg1P_8 && arg2==arg2P_8 ], cost: 1 2: f5 -> f6 : arg1'=arg1P_3, arg2'=arg2P_3, [ arg2P_3==-1+arg2 && arg1==arg1P_3 ], cost: 1 4: f6 -> f4 : arg1'=arg1P_5, arg2'=arg2P_5, [ arg1==arg1P_5 && arg2==arg2P_5 ], cost: 1 3: f4 -> f5 : arg1'=arg1P_4, arg2'=arg2P_4, [ arg2>0 && arg1==arg1P_4 && arg2==arg2P_4 ], cost: 1 5: f4 -> f7 : arg1'=arg1P_6, arg2'=arg2P_6, [ arg2<=0 && arg1==arg1P_6 && arg2==arg2P_6 ], cost: 1 6: f7 -> f8 : arg1'=arg1P_7, arg2'=arg2P_7, [ arg1P_7==-1+arg1 && arg2==arg2P_7 ], cost: 1 8: f8 -> f3 : arg1'=arg1P_9, arg2'=arg2P_9, [ arg1==arg1P_9 && arg2==arg2P_9 ], cost: 1 10: __init -> f1 : arg1'=arg1P_11, arg2'=arg2P_11, [], cost: 1 Simplified all rules, resulting in: Start location: __init 0: f1 -> f2 : arg1'=arg1P_1, [], cost: 1 1: f2 -> f3 : arg2'=arg2P_2, [], cost: 1 7: f3 -> f4 : [ arg1>0 ], cost: 1 2: f5 -> f6 : arg2'=-1+arg2, [], cost: 1 4: f6 -> f4 : [], cost: 1 3: f4 -> f5 : [ arg2>0 ], cost: 1 5: f4 -> f7 : [ arg2<=0 ], cost: 1 6: f7 -> f8 : arg1'=-1+arg1, [], cost: 1 8: f8 -> f3 : [], cost: 1 10: __init -> f1 : arg1'=arg1P_11, arg2'=arg2P_11, [], cost: 1 ### Simplification by acceleration and chaining ### Eliminated locations (on linear paths): Start location: __init 7: f3 -> f4 : [ arg1>0 ], cost: 1 15: f4 -> f4 : arg2'=-1+arg2, [ arg2>0 ], cost: 3 16: f4 -> f3 : arg1'=-1+arg1, [ arg2<=0 ], cost: 3 12: __init -> f3 : arg1'=arg1P_1, arg2'=arg2P_2, [], cost: 3 Accelerating simple loops of location 5. Accelerating the following rules: 15: f4 -> f4 : arg2'=-1+arg2, [ arg2>0 ], cost: 3 Accelerated rule 15 with backward acceleration, yielding the new rule 17. [accelerate] Nesting with 1 inner and 1 outer candidates Removing the simple loops: 15. Accelerated all simple loops using metering functions (where possible): Start location: __init 7: f3 -> f4 : [ arg1>0 ], cost: 1 16: f4 -> f3 : arg1'=-1+arg1, [ arg2<=0 ], cost: 3 17: f4 -> f4 : arg2'=0, [ arg2>=0 ], cost: 3*arg2 12: __init -> f3 : arg1'=arg1P_1, arg2'=arg2P_2, [], cost: 3 Chained accelerated rules (with incoming rules): Start location: __init 7: f3 -> f4 : [ arg1>0 ], cost: 1 18: f3 -> f4 : arg2'=0, [ arg1>0 && arg2>=0 ], cost: 1+3*arg2 16: f4 -> f3 : arg1'=-1+arg1, [ arg2<=0 ], cost: 3 12: __init -> f3 : arg1'=arg1P_1, arg2'=arg2P_2, [], cost: 3 Eliminated locations (on tree-shaped paths): Start location: __init 19: f3 -> f3 : arg1'=-1+arg1, [ arg1>0 && arg2<=0 ], cost: 4 20: f3 -> f3 : arg1'=-1+arg1, arg2'=0, [ arg1>0 && arg2>=0 ], cost: 4+3*arg2 12: __init -> f3 : arg1'=arg1P_1, arg2'=arg2P_2, [], cost: 3 Accelerating simple loops of location 2. Accelerating the following rules: 19: f3 -> f3 : arg1'=-1+arg1, [ arg1>0 && arg2<=0 ], cost: 4 20: f3 -> f3 : arg1'=-1+arg1, arg2'=0, [ arg1>0 && arg2>=0 ], cost: 4+3*arg2 Accelerated rule 19 with backward acceleration, yielding the new rule 21. Accelerated rule 20 with backward acceleration, yielding the new rule 22. [accelerate] Nesting with 2 inner and 2 outer candidates Removing the simple loops: 19 20. Accelerated all simple loops using metering functions (where possible): Start location: __init 21: f3 -> f3 : arg1'=0, [ arg2<=0 && arg1>=0 ], cost: 4*arg1 22: f3 -> f3 : arg1'=0, arg2'=0, [ arg2>=0 && arg1>=1 ], cost: 4*arg1 12: __init -> f3 : arg1'=arg1P_1, arg2'=arg2P_2, [], cost: 3 Chained accelerated rules (with incoming rules): Start location: __init 12: __init -> f3 : arg1'=arg1P_1, arg2'=arg2P_2, [], cost: 3 23: __init -> f3 : arg1'=0, arg2'=arg2P_2, [ arg2P_2<=0 && arg1P_1>=0 ], cost: 3+4*arg1P_1 24: __init -> f3 : arg1'=0, arg2'=0, [ arg1P_1>=1 ], cost: 3+4*arg1P_1 Removed unreachable locations (and leaf rules with constant cost): Start location: __init 23: __init -> f3 : arg1'=0, arg2'=arg2P_2, [ arg2P_2<=0 && arg1P_1>=0 ], cost: 3+4*arg1P_1 24: __init -> f3 : arg1'=0, arg2'=0, [ arg1P_1>=1 ], cost: 3+4*arg1P_1 ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: __init 23: __init -> f3 : arg1'=0, arg2'=arg2P_2, [ arg2P_2<=0 && arg1P_1>=0 ], cost: 3+4*arg1P_1 24: __init -> f3 : arg1'=0, arg2'=0, [ arg1P_1>=1 ], cost: 3+4*arg1P_1 Computing asymptotic complexity for rule 24 Resulting cost 0 has complexity: Unknown Computing asymptotic complexity for rule 23 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),?)