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, arg3'=arg3P_1, [ arg1==arg1P_1 && arg3==arg3P_1 ], cost: 1 1: f2 -> f3 : arg1'=arg1P_2, arg2'=arg2P_2, arg3'=arg3P_2, [ arg1==arg1P_2 && arg2==arg2P_2 ], cost: 1 2: f3 -> f4 : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [ 0==arg1P_3 && arg2==arg2P_3 && arg3==arg3P_3 ], cost: 1 14: f4 -> f5 : arg1'=arg1P_15, arg2'=arg2P_15, arg3'=arg3P_15, [ arg2>0 && arg3>0 && arg1==arg1P_15 && arg2==arg2P_15 && arg3==arg3P_15 ], cost: 1 15: f4 -> f6 : arg1'=arg1P_16, arg2'=arg2P_16, arg3'=arg3P_16, [ arg2<=0 && arg1==arg1P_16 && arg2==arg2P_16 && arg3==arg3P_16 ], cost: 1 16: f4 -> f6 : arg1'=arg1P_17, arg2'=arg2P_17, arg3'=arg3P_17, [ arg3<=0 && arg1==arg1P_17 && arg2==arg2P_17 && arg3==arg3P_17 ], cost: 1 3: f9 -> f12 : arg1'=arg1P_4, arg2'=arg2P_4, arg3'=arg3P_4, [ arg1==arg1P_4 && arg3==arg2P_4 && arg3==arg3P_4 ], cost: 1 7: f12 -> f11 : arg1'=arg1P_8, arg2'=arg2P_8, arg3'=arg3P_8, [ arg1==arg1P_8 && arg2==arg2P_8 && arg3==arg3P_8 ], cost: 1 4: f10 -> f13 : arg1'=arg1P_5, arg2'=arg2P_5, arg3'=arg3P_5, [ arg2P_5==-1+arg2 && arg1==arg1P_5 && arg3==arg3P_5 ], cost: 1 8: f13 -> f11 : arg1'=arg1P_9, arg2'=arg2P_9, arg3'=arg3P_9, [ arg1==arg1P_9 && arg2==arg2P_9 && arg3==arg3P_9 ], cost: 1 5: f8 -> f9 : arg1'=arg1P_6, arg2'=arg2P_6, arg3'=arg3P_6, [ arg2>arg3 && arg1==arg1P_6 && arg2==arg2P_6 && arg3==arg3P_6 ], cost: 1 6: f8 -> f10 : arg1'=arg1P_7, arg2'=arg2P_7, arg3'=arg3P_7, [ arg2<=arg3 && arg1==arg1P_7 && arg2==arg2P_7 && arg3==arg3P_7 ], cost: 1 9: f11 -> f14 : arg1'=arg1P_10, arg2'=arg2P_10, arg3'=arg3P_10, [ arg1P_10==1+arg1 && arg2==arg2P_10 && arg3==arg3P_10 ], cost: 1 12: f14 -> f5 : arg1'=arg1P_13, arg2'=arg2P_13, arg3'=arg3P_13, [ arg1==arg1P_13 && arg2==arg2P_13 && arg3==arg3P_13 ], cost: 1 10: f5 -> f8 : arg1'=arg1P_11, arg2'=arg2P_11, arg3'=arg3P_11, [ arg2<0 && arg1==arg1P_11 && arg2==arg2P_11 && arg3==arg3P_11 ], cost: 1 11: f5 -> f8 : arg1'=arg1P_12, arg2'=arg2P_12, arg3'=arg3P_12, [ arg2>0 && arg1==arg1P_12 && arg2==arg2P_12 && arg3==arg3P_12 ], cost: 1 13: f5 -> f15 : arg1'=arg1P_14, arg2'=arg2P_14, arg3'=arg3P_14, [ arg2==0 && arg1==arg1P_14 && arg2==arg2P_14 && arg3==arg3P_14 ], cost: 1 17: f15 -> f7 : arg1'=arg1P_18, arg2'=arg2P_18, arg3'=arg3P_18, [ arg1==arg1P_18 && arg2==arg2P_18 && arg3==arg3P_18 ], cost: 1 18: f6 -> f7 : arg1'=arg1P_19, arg2'=arg2P_19, arg3'=arg3P_19, [ arg1==arg1P_19 && arg2==arg2P_19 && arg3==arg3P_19 ], cost: 1 19: __init -> f1 : arg1'=arg1P_20, arg2'=arg2P_20, arg3'=arg3P_20, [], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 19: __init -> f1 : arg1'=arg1P_20, arg2'=arg2P_20, arg3'=arg3P_20, [], cost: 1 Removed unreachable and leaf rules: Start location: __init 0: f1 -> f2 : arg1'=arg1P_1, arg2'=arg2P_1, arg3'=arg3P_1, [ arg1==arg1P_1 && arg3==arg3P_1 ], cost: 1 1: f2 -> f3 : arg1'=arg1P_2, arg2'=arg2P_2, arg3'=arg3P_2, [ arg1==arg1P_2 && arg2==arg2P_2 ], cost: 1 2: f3 -> f4 : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [ 0==arg1P_3 && arg2==arg2P_3 && arg3==arg3P_3 ], cost: 1 14: f4 -> f5 : arg1'=arg1P_15, arg2'=arg2P_15, arg3'=arg3P_15, [ arg2>0 && arg3>0 && arg1==arg1P_15 && arg2==arg2P_15 && arg3==arg3P_15 ], cost: 1 3: f9 -> f12 : arg1'=arg1P_4, arg2'=arg2P_4, arg3'=arg3P_4, [ arg1==arg1P_4 && arg3==arg2P_4 && arg3==arg3P_4 ], cost: 1 7: f12 -> f11 : arg1'=arg1P_8, arg2'=arg2P_8, arg3'=arg3P_8, [ arg1==arg1P_8 && arg2==arg2P_8 && arg3==arg3P_8 ], cost: 1 4: f10 -> f13 : arg1'=arg1P_5, arg2'=arg2P_5, arg3'=arg3P_5, [ arg2P_5==-1+arg2 && arg1==arg1P_5 && arg3==arg3P_5 ], cost: 1 8: f13 -> f11 : arg1'=arg1P_9, arg2'=arg2P_9, arg3'=arg3P_9, [ arg1==arg1P_9 && arg2==arg2P_9 && arg3==arg3P_9 ], cost: 1 5: f8 -> f9 : arg1'=arg1P_6, arg2'=arg2P_6, arg3'=arg3P_6, [ arg2>arg3 && arg1==arg1P_6 && arg2==arg2P_6 && arg3==arg3P_6 ], cost: 1 6: f8 -> f10 : arg1'=arg1P_7, arg2'=arg2P_7, arg3'=arg3P_7, [ arg2<=arg3 && arg1==arg1P_7 && arg2==arg2P_7 && arg3==arg3P_7 ], cost: 1 9: f11 -> f14 : arg1'=arg1P_10, arg2'=arg2P_10, arg3'=arg3P_10, [ arg1P_10==1+arg1 && arg2==arg2P_10 && arg3==arg3P_10 ], cost: 1 12: f14 -> f5 : arg1'=arg1P_13, arg2'=arg2P_13, arg3'=arg3P_13, [ arg1==arg1P_13 && arg2==arg2P_13 && arg3==arg3P_13 ], cost: 1 10: f5 -> f8 : arg1'=arg1P_11, arg2'=arg2P_11, arg3'=arg3P_11, [ arg2<0 && arg1==arg1P_11 && arg2==arg2P_11 && arg3==arg3P_11 ], cost: 1 11: f5 -> f8 : arg1'=arg1P_12, arg2'=arg2P_12, arg3'=arg3P_12, [ arg2>0 && arg1==arg1P_12 && arg2==arg2P_12 && arg3==arg3P_12 ], cost: 1 19: __init -> f1 : arg1'=arg1P_20, arg2'=arg2P_20, arg3'=arg3P_20, [], cost: 1 Simplified all rules, resulting in: Start location: __init 0: f1 -> f2 : arg2'=arg2P_1, [], cost: 1 1: f2 -> f3 : arg3'=arg3P_2, [], cost: 1 2: f3 -> f4 : arg1'=0, [], cost: 1 14: f4 -> f5 : [ arg2>0 && arg3>0 ], cost: 1 3: f9 -> f12 : arg2'=arg3, [], cost: 1 7: f12 -> f11 : [], cost: 1 4: f10 -> f13 : arg2'=-1+arg2, [], cost: 1 8: f13 -> f11 : [], cost: 1 5: f8 -> f9 : [ arg2>arg3 ], cost: 1 6: f8 -> f10 : [ arg2<=arg3 ], cost: 1 9: f11 -> f14 : arg1'=1+arg1, [], cost: 1 12: f14 -> f5 : [], cost: 1 10: f5 -> f8 : [ arg2<0 ], cost: 1 11: f5 -> f8 : [ arg2>0 ], cost: 1 19: __init -> f1 : arg1'=arg1P_20, arg2'=arg2P_20, arg3'=arg3P_20, [], cost: 1 ### Simplification by acceleration and chaining ### Eliminated locations (on linear paths): Start location: __init 26: f8 -> f11 : arg2'=arg3, [ arg2>arg3 ], cost: 3 27: f8 -> f11 : arg2'=-1+arg2, [ arg2<=arg3 ], cost: 3 28: f11 -> f5 : arg1'=1+arg1, [], cost: 2 10: f5 -> f8 : [ arg2<0 ], cost: 1 11: f5 -> f8 : [ arg2>0 ], cost: 1 23: __init -> f5 : arg1'=0, arg2'=arg2P_1, arg3'=arg3P_2, [ arg2P_1>0 && arg3P_2>0 ], cost: 5 Eliminated locations (on tree-shaped paths): Start location: __init 28: f11 -> f5 : arg1'=1+arg1, [], cost: 2 29: f5 -> f11 : arg2'=arg3, [ arg2<0 && arg2>arg3 ], cost: 4 30: f5 -> f11 : arg2'=-1+arg2, [ arg2<0 && arg2<=arg3 ], cost: 4 31: f5 -> f11 : arg2'=arg3, [ arg2>0 && arg2>arg3 ], cost: 4 32: f5 -> f11 : arg2'=-1+arg2, [ arg2>0 && arg2<=arg3 ], cost: 4 23: __init -> f5 : arg1'=0, arg2'=arg2P_1, arg3'=arg3P_2, [ arg2P_1>0 && arg3P_2>0 ], cost: 5 Eliminated locations (on tree-shaped paths): Start location: __init 33: f5 -> f5 : arg1'=1+arg1, arg2'=arg3, [ arg2<0 && arg2>arg3 ], cost: 6 34: f5 -> f5 : arg1'=1+arg1, arg2'=-1+arg2, [ arg2<0 && arg2<=arg3 ], cost: 6 35: f5 -> f5 : arg1'=1+arg1, arg2'=arg3, [ arg2>0 && arg2>arg3 ], cost: 6 36: f5 -> f5 : arg1'=1+arg1, arg2'=-1+arg2, [ arg2>0 && arg2<=arg3 ], cost: 6 23: __init -> f5 : arg1'=0, arg2'=arg2P_1, arg3'=arg3P_2, [ arg2P_1>0 && arg3P_2>0 ], cost: 5 Accelerating simple loops of location 11. Accelerating the following rules: 33: f5 -> f5 : arg1'=1+arg1, arg2'=arg3, [ arg2<0 && arg2>arg3 ], cost: 6 34: f5 -> f5 : arg1'=1+arg1, arg2'=-1+arg2, [ arg2<0 && arg2<=arg3 ], cost: 6 35: f5 -> f5 : arg1'=1+arg1, arg2'=arg3, [ arg2>0 && arg2>arg3 ], cost: 6 36: f5 -> f5 : arg1'=1+arg1, arg2'=-1+arg2, [ arg2>0 && arg2<=arg3 ], cost: 6 Failed to prove monotonicity of the guard of rule 33. Accelerated rule 34 with non-termination, yielding the new rule 37. Failed to prove monotonicity of the guard of rule 35. Accelerated rule 36 with backward acceleration, yielding the new rule 38. [accelerate] Nesting with 3 inner and 3 outer candidates Removing the simple loops: 34 36. Accelerated all simple loops using metering functions (where possible): Start location: __init 33: f5 -> f5 : arg1'=1+arg1, arg2'=arg3, [ arg2<0 && arg2>arg3 ], cost: 6 35: f5 -> f5 : arg1'=1+arg1, arg2'=arg3, [ arg2>0 && arg2>arg3 ], cost: 6 37: f5 -> [16] : [ arg2<0 && arg2<=arg3 ], cost: NONTERM 38: f5 -> f5 : arg1'=arg2+arg1, arg2'=0, [ arg2<=arg3 && arg2>=0 ], cost: 6*arg2 23: __init -> f5 : arg1'=0, arg2'=arg2P_1, arg3'=arg3P_2, [ arg2P_1>0 && arg3P_2>0 ], cost: 5 Chained accelerated rules (with incoming rules): Start location: __init 23: __init -> f5 : arg1'=0, arg2'=arg2P_1, arg3'=arg3P_2, [ arg2P_1>0 && arg3P_2>0 ], cost: 5 39: __init -> f5 : arg1'=1, arg2'=arg3P_2, arg3'=arg3P_2, [ arg3P_2>0 ], cost: 11 40: __init -> f5 : arg1'=arg2P_1, arg2'=0, arg3'=arg3P_2, [ arg2P_1>0 && arg3P_2>0 && arg2P_1<=arg3P_2 ], cost: 5+6*arg2P_1 Removed unreachable locations (and leaf rules with constant cost): Start location: __init 40: __init -> f5 : arg1'=arg2P_1, arg2'=0, arg3'=arg3P_2, [ arg2P_1>0 && arg3P_2>0 && arg2P_1<=arg3P_2 ], cost: 5+6*arg2P_1 ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: __init 40: __init -> f5 : arg1'=arg2P_1, arg2'=0, arg3'=arg3P_2, [ arg2P_1>0 && arg3P_2>0 && arg2P_1<=arg3P_2 ], cost: 5+6*arg2P_1 Computing asymptotic complexity for rule 40 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),?)