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 8: f2 -> f3 : arg1'=arg1P_9, arg2'=arg2P_9, [ arg1>=0 && arg1==arg1P_9 && arg2==arg2P_9 ], cost: 1 10: f2 -> f10 : arg1'=arg1P_11, arg2'=arg2P_11, [ arg1<0 && arg1==arg1P_11 && arg2==arg2P_11 ], cost: 1 1: f3 -> f4 : arg1'=arg1P_2, arg2'=arg2P_2, [ arg1P_2==1+arg1 && arg2==arg2P_2 ], cost: 1 2: f4 -> f5 : arg1'=arg1P_3, arg2'=arg2P_3, [ arg1==arg1P_3 && 1==arg2P_3 ], cost: 1 4: f5 -> f6 : arg1'=arg1P_5, arg2'=arg2P_5, [ arg1>=arg2 && arg1==arg1P_5 && arg2==arg2P_5 ], cost: 1 6: f5 -> f8 : arg1'=arg1P_7, arg2'=arg2P_7, [ arg1 f7 : arg1'=arg1P_4, arg2'=arg2P_4, [ arg2P_4==1+arg2 && arg1==arg1P_4 ], cost: 1 5: f7 -> f5 : arg1'=arg1P_6, arg2'=arg2P_6, [ arg1==arg1P_6 && arg2==arg2P_6 ], cost: 1 7: f8 -> f9 : arg1'=arg1P_8, arg2'=arg2P_8, [ arg1P_8==-2+arg1 && arg2==arg2P_8 ], cost: 1 9: f9 -> f2 : arg1'=arg1P_10, arg2'=arg2P_10, [ arg1==arg1P_10 && arg2==arg2P_10 ], cost: 1 11: __init -> f1 : arg1'=arg1P_12, arg2'=arg2P_12, [], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 11: __init -> f1 : arg1'=arg1P_12, arg2'=arg2P_12, [], cost: 1 Removed unreachable and leaf rules: Start location: __init 0: f1 -> f2 : arg1'=arg1P_1, arg2'=arg2P_1, [ arg2==arg2P_1 ], cost: 1 8: f2 -> f3 : arg1'=arg1P_9, arg2'=arg2P_9, [ arg1>=0 && arg1==arg1P_9 && arg2==arg2P_9 ], cost: 1 1: f3 -> f4 : arg1'=arg1P_2, arg2'=arg2P_2, [ arg1P_2==1+arg1 && arg2==arg2P_2 ], cost: 1 2: f4 -> f5 : arg1'=arg1P_3, arg2'=arg2P_3, [ arg1==arg1P_3 && 1==arg2P_3 ], cost: 1 4: f5 -> f6 : arg1'=arg1P_5, arg2'=arg2P_5, [ arg1>=arg2 && arg1==arg1P_5 && arg2==arg2P_5 ], cost: 1 6: f5 -> f8 : arg1'=arg1P_7, arg2'=arg2P_7, [ arg1 f7 : arg1'=arg1P_4, arg2'=arg2P_4, [ arg2P_4==1+arg2 && arg1==arg1P_4 ], cost: 1 5: f7 -> f5 : arg1'=arg1P_6, arg2'=arg2P_6, [ arg1==arg1P_6 && arg2==arg2P_6 ], cost: 1 7: f8 -> f9 : arg1'=arg1P_8, arg2'=arg2P_8, [ arg1P_8==-2+arg1 && arg2==arg2P_8 ], cost: 1 9: f9 -> f2 : arg1'=arg1P_10, arg2'=arg2P_10, [ arg1==arg1P_10 && arg2==arg2P_10 ], cost: 1 11: __init -> f1 : arg1'=arg1P_12, arg2'=arg2P_12, [], cost: 1 Simplified all rules, resulting in: Start location: __init 0: f1 -> f2 : arg1'=arg1P_1, [], cost: 1 8: f2 -> f3 : [ arg1>=0 ], cost: 1 1: f3 -> f4 : arg1'=1+arg1, [], cost: 1 2: f4 -> f5 : arg2'=1, [], cost: 1 4: f5 -> f6 : [ arg1>=arg2 ], cost: 1 6: f5 -> f8 : [ arg1 f7 : arg2'=1+arg2, [], cost: 1 5: f7 -> f5 : [], cost: 1 7: f8 -> f9 : arg1'=-2+arg1, [], cost: 1 9: f9 -> f2 : [], cost: 1 11: __init -> f1 : arg1'=arg1P_12, arg2'=arg2P_12, [], cost: 1 ### Simplification by acceleration and chaining ### Eliminated locations (on linear paths): Start location: __init 14: f2 -> f5 : arg1'=1+arg1, arg2'=1, [ arg1>=0 ], cost: 3 17: f5 -> f5 : arg2'=1+arg2, [ arg1>=arg2 ], cost: 3 18: f5 -> f2 : arg1'=-2+arg1, [ arg1 f2 : arg1'=arg1P_1, arg2'=arg2P_12, [], cost: 2 Accelerating simple loops of location 4. Accelerating the following rules: 17: f5 -> f5 : arg2'=1+arg2, [ arg1>=arg2 ], cost: 3 Accelerated rule 17 with backward acceleration, yielding the new rule 19. [accelerate] Nesting with 1 inner and 1 outer candidates Removing the simple loops: 17. Accelerated all simple loops using metering functions (where possible): Start location: __init 14: f2 -> f5 : arg1'=1+arg1, arg2'=1, [ arg1>=0 ], cost: 3 18: f5 -> f2 : arg1'=-2+arg1, [ arg1 f5 : arg2'=1+arg1, [ 1-arg2+arg1>=0 ], cost: 3-3*arg2+3*arg1 12: __init -> f2 : arg1'=arg1P_1, arg2'=arg2P_12, [], cost: 2 Chained accelerated rules (with incoming rules): Start location: __init 14: f2 -> f5 : arg1'=1+arg1, arg2'=1, [ arg1>=0 ], cost: 3 20: f2 -> f5 : arg1'=1+arg1, arg2'=2+arg1, [ arg1>=0 ], cost: 6+3*arg1 18: f5 -> f2 : arg1'=-2+arg1, [ arg1 f2 : arg1'=arg1P_1, arg2'=arg2P_12, [], cost: 2 Eliminated locations (on tree-shaped paths): Start location: __init 21: f2 -> f2 : arg1'=-1+arg1, arg2'=2+arg1, [ arg1>=0 ], cost: 9+3*arg1 12: __init -> f2 : arg1'=arg1P_1, arg2'=arg2P_12, [], cost: 2 Accelerating simple loops of location 1. Accelerating the following rules: 21: f2 -> f2 : arg1'=-1+arg1, arg2'=2+arg1, [ arg1>=0 ], cost: 9+3*arg1 Accelerated rule 21 with backward acceleration, yielding the new rule 22. [accelerate] Nesting with 1 inner and 1 outer candidates Removing the simple loops: 21. Accelerated all simple loops using metering functions (where possible): Start location: __init 22: f2 -> f2 : arg1'=-1, arg2'=2, [ 1+arg1>=1 ], cost: 21/2+21/2*arg1+3*(1+arg1)*arg1-3/2*(1+arg1)^2 12: __init -> f2 : arg1'=arg1P_1, arg2'=arg2P_12, [], cost: 2 Chained accelerated rules (with incoming rules): Start location: __init 12: __init -> f2 : arg1'=arg1P_1, arg2'=arg2P_12, [], cost: 2 23: __init -> f2 : arg1'=-1, arg2'=2, [ 1+arg1P_1>=1 ], cost: 25/2+21/2*arg1P_1+3*arg1P_1*(1+arg1P_1)-3/2*(1+arg1P_1)^2 Removed unreachable locations (and leaf rules with constant cost): Start location: __init 23: __init -> f2 : arg1'=-1, arg2'=2, [ 1+arg1P_1>=1 ], cost: 25/2+21/2*arg1P_1+3*arg1P_1*(1+arg1P_1)-3/2*(1+arg1P_1)^2 ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: __init 23: __init -> f2 : arg1'=-1, arg2'=2, [ 1+arg1P_1>=1 ], cost: 25/2+21/2*arg1P_1+3*arg1P_1*(1+arg1P_1)-3/2*(1+arg1P_1)^2 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),?)