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 4: f3 -> f4 : arg1'=arg1P_5, arg2'=arg2P_5, [ arg1>0 && arg2<0 && arg1==arg1P_5 && arg2==arg2P_5 ], cost: 1 6: f3 -> f7 : arg1'=arg1P_7, arg2'=arg2P_7, [ arg1<=0 && arg1==arg1P_7 && arg2==arg2P_7 ], cost: 1 7: f3 -> f7 : arg1'=arg1P_8, arg2'=arg2P_8, [ arg2>=0 && arg1==arg1P_8 && arg2==arg2P_8 ], cost: 1 2: f4 -> f5 : arg1'=arg1P_3, arg2'=arg2P_3, [ arg1P_3==arg2+arg1 && arg2==arg2P_3 ], cost: 1 3: f5 -> f6 : arg1'=arg1P_4, arg2'=arg2P_4, [ arg2P_4==-1+arg2 && arg1==arg1P_4 ], cost: 1 5: f6 -> f3 : arg1'=arg1P_6, arg2'=arg2P_6, [ arg1==arg1P_6 && arg2==arg2P_6 ], cost: 1 8: __init -> f1 : arg1'=arg1P_9, arg2'=arg2P_9, [], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 8: __init -> f1 : arg1'=arg1P_9, arg2'=arg2P_9, [], 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 4: f3 -> f4 : arg1'=arg1P_5, arg2'=arg2P_5, [ arg1>0 && arg2<0 && arg1==arg1P_5 && arg2==arg2P_5 ], cost: 1 2: f4 -> f5 : arg1'=arg1P_3, arg2'=arg2P_3, [ arg1P_3==arg2+arg1 && arg2==arg2P_3 ], cost: 1 3: f5 -> f6 : arg1'=arg1P_4, arg2'=arg2P_4, [ arg2P_4==-1+arg2 && arg1==arg1P_4 ], cost: 1 5: f6 -> f3 : arg1'=arg1P_6, arg2'=arg2P_6, [ arg1==arg1P_6 && arg2==arg2P_6 ], cost: 1 8: __init -> f1 : arg1'=arg1P_9, arg2'=arg2P_9, [], 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 4: f3 -> f4 : [ arg1>0 && arg2<0 ], cost: 1 2: f4 -> f5 : arg1'=arg2+arg1, [], cost: 1 3: f5 -> f6 : arg2'=-1+arg2, [], cost: 1 5: f6 -> f3 : [], cost: 1 8: __init -> f1 : arg1'=arg1P_9, arg2'=arg2P_9, [], cost: 1 ### Simplification by acceleration and chaining ### Eliminated locations (on linear paths): Start location: __init 13: f3 -> f3 : arg1'=arg2+arg1, arg2'=-1+arg2, [ arg1>0 && arg2<0 ], cost: 4 10: __init -> f3 : arg1'=arg1P_1, arg2'=arg2P_2, [], cost: 3 Accelerating simple loops of location 2. Accelerating the following rules: 13: f3 -> f3 : arg1'=arg2+arg1, arg2'=-1+arg2, [ arg1>0 && arg2<0 ], cost: 4 Accelerated rule 13 with backward acceleration, yielding the new rule 14. [accelerate] Nesting with 1 inner and 1 outer candidates Removing the simple loops: 13. Accelerated all simple loops using metering functions (where possible): Start location: __init 14: f3 -> f3 : arg1'=arg2*k-1/2*k^2+1/2*k+arg1, arg2'=arg2-k, [ arg2<0 && k>=0 && -1/2-1/2*(-1+k)^2+1/2*k+(-1+k)*arg2+arg1>0 ], cost: 4*k 10: __init -> f3 : arg1'=arg1P_1, arg2'=arg2P_2, [], cost: 3 Chained accelerated rules (with incoming rules): Start location: __init 10: __init -> f3 : arg1'=arg1P_1, arg2'=arg2P_2, [], cost: 3 15: __init -> f3 : arg1'=arg1P_1-1/2*k^2+1/2*k+arg2P_2*k, arg2'=arg2P_2-k, [ arg2P_2<0 && k>=0 && -1/2+arg1P_1-1/2*(-1+k)^2+(-1+k)*arg2P_2+1/2*k>0 ], cost: 3+4*k Removed unreachable locations (and leaf rules with constant cost): Start location: __init 15: __init -> f3 : arg1'=arg1P_1-1/2*k^2+1/2*k+arg2P_2*k, arg2'=arg2P_2-k, [ arg2P_2<0 && k>=0 && -1/2+arg1P_1-1/2*(-1+k)^2+(-1+k)*arg2P_2+1/2*k>0 ], cost: 3+4*k ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: __init 15: __init -> f3 : arg1'=arg1P_1-1/2*k^2+1/2*k+arg2P_2*k, arg2'=arg2P_2-k, [ arg2P_2<0 && k>=0 && -1/2+arg1P_1-1/2*(-1+k)^2+(-1+k)*arg2P_2+1/2*k>0 ], cost: 3+4*k Computing asymptotic complexity for rule 15 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),?)