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, [ arg2==arg2P_1 && arg3==arg3P_1 ], cost: 1 1: f2 -> f3 : arg1'=arg1P_2, arg2'=arg2P_2, arg3'=arg3P_2, [ arg1==arg1P_2 && arg3==arg3P_2 ], cost: 1 2: f3 -> f4 : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [ arg1==arg1P_3 && arg2==arg2P_3 ], cost: 1 13: f4 -> f5 : arg1'=arg1P_14, arg2'=arg2P_14, arg3'=arg3P_14, [ arg1>0 && arg2>0 && arg3>0 && arg1==arg1P_14 && arg2==arg2P_14 && arg3==arg3P_14 ], cost: 1 15: f4 -> f15 : arg1'=arg1P_16, arg2'=arg2P_16, arg3'=arg3P_16, [ arg3<=0 && arg1==arg1P_16 && arg2==arg2P_16 && arg3==arg3P_16 ], cost: 1 16: f4 -> f15 : arg1'=arg1P_17, arg2'=arg2P_17, arg3'=arg3P_17, [ arg1<=0 && arg1==arg1P_17 && arg2==arg2P_17 && arg3==arg3P_17 ], cost: 1 17: f4 -> f15 : arg1'=arg1P_18, arg2'=arg2P_18, arg3'=arg3P_18, [ arg2<=0 && arg1==arg1P_18 && arg2==arg2P_18 && arg3==arg3P_18 ], cost: 1 3: f6 -> f9 : arg1'=arg1P_4, arg2'=arg2P_4, arg3'=arg3P_4, [ arg1==arg1P_4 && arg3==arg2P_4 && arg3==arg3P_4 ], cost: 1 4: f9 -> f10 : arg1'=arg1P_5, arg2'=arg2P_5, arg3'=arg3P_5, [ arg2==arg2P_5 && arg3==arg3P_5 ], cost: 1 5: f10 -> f11 : arg1'=arg1P_6, arg2'=arg2P_6, arg3'=arg3P_6, [ arg3P_6==-1+arg1 && arg1==arg1P_6 && arg2==arg2P_6 ], cost: 1 11: f11 -> f8 : arg1'=arg1P_12, arg2'=arg2P_12, arg3'=arg3P_12, [ arg1==arg1P_12 && arg2==arg2P_12 && arg3==arg3P_12 ], cost: 1 6: f7 -> f12 : arg1'=arg1P_7, arg2'=arg2P_7, arg3'=arg3P_7, [ arg3P_7==-1+arg3 && arg1==arg1P_7 && arg2==arg2P_7 ], cost: 1 7: f12 -> f13 : arg1'=arg1P_8, arg2'=arg2P_8, arg3'=arg3P_8, [ arg2==arg2P_8 && arg3==arg3P_8 ], cost: 1 8: f13 -> f14 : arg1'=arg1P_9, arg2'=arg2P_9, arg3'=arg3P_9, [ arg2P_9==-1+arg1 && arg1==arg1P_9 && arg3==arg3P_9 ], cost: 1 12: f14 -> f8 : arg1'=arg1P_13, arg2'=arg2P_13, arg3'=arg3P_13, [ arg1==arg1P_13 && arg2==arg2P_13 && arg3==arg3P_13 ], cost: 1 9: f5 -> f6 : arg1'=arg1P_10, arg2'=arg2P_10, arg3'=arg3P_10, [ arg2>arg1 && arg1==arg1P_10 && arg2==arg2P_10 && arg3==arg3P_10 ], cost: 1 10: f5 -> f7 : arg1'=arg1P_11, arg2'=arg2P_11, arg3'=arg3P_11, [ arg2<=arg1 && arg1==arg1P_11 && arg2==arg2P_11 && arg3==arg3P_11 ], cost: 1 14: f8 -> f4 : arg1'=arg1P_15, arg2'=arg2P_15, arg3'=arg3P_15, [ arg1==arg1P_15 && arg2==arg2P_15 && arg3==arg3P_15 ], cost: 1 18: __init -> f1 : arg1'=arg1P_19, arg2'=arg2P_19, arg3'=arg3P_19, [], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 18: __init -> f1 : arg1'=arg1P_19, arg2'=arg2P_19, arg3'=arg3P_19, [], cost: 1 Removed unreachable and leaf rules: Start location: __init 0: f1 -> f2 : arg1'=arg1P_1, arg2'=arg2P_1, arg3'=arg3P_1, [ arg2==arg2P_1 && arg3==arg3P_1 ], cost: 1 1: f2 -> f3 : arg1'=arg1P_2, arg2'=arg2P_2, arg3'=arg3P_2, [ arg1==arg1P_2 && arg3==arg3P_2 ], cost: 1 2: f3 -> f4 : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [ arg1==arg1P_3 && arg2==arg2P_3 ], cost: 1 13: f4 -> f5 : arg1'=arg1P_14, arg2'=arg2P_14, arg3'=arg3P_14, [ arg1>0 && arg2>0 && arg3>0 && arg1==arg1P_14 && arg2==arg2P_14 && arg3==arg3P_14 ], cost: 1 3: f6 -> f9 : arg1'=arg1P_4, arg2'=arg2P_4, arg3'=arg3P_4, [ arg1==arg1P_4 && arg3==arg2P_4 && arg3==arg3P_4 ], cost: 1 4: f9 -> f10 : arg1'=arg1P_5, arg2'=arg2P_5, arg3'=arg3P_5, [ arg2==arg2P_5 && arg3==arg3P_5 ], cost: 1 5: f10 -> f11 : arg1'=arg1P_6, arg2'=arg2P_6, arg3'=arg3P_6, [ arg3P_6==-1+arg1 && arg1==arg1P_6 && arg2==arg2P_6 ], cost: 1 11: f11 -> f8 : arg1'=arg1P_12, arg2'=arg2P_12, arg3'=arg3P_12, [ arg1==arg1P_12 && arg2==arg2P_12 && arg3==arg3P_12 ], cost: 1 6: f7 -> f12 : arg1'=arg1P_7, arg2'=arg2P_7, arg3'=arg3P_7, [ arg3P_7==-1+arg3 && arg1==arg1P_7 && arg2==arg2P_7 ], cost: 1 7: f12 -> f13 : arg1'=arg1P_8, arg2'=arg2P_8, arg3'=arg3P_8, [ arg2==arg2P_8 && arg3==arg3P_8 ], cost: 1 8: f13 -> f14 : arg1'=arg1P_9, arg2'=arg2P_9, arg3'=arg3P_9, [ arg2P_9==-1+arg1 && arg1==arg1P_9 && arg3==arg3P_9 ], cost: 1 12: f14 -> f8 : arg1'=arg1P_13, arg2'=arg2P_13, arg3'=arg3P_13, [ arg1==arg1P_13 && arg2==arg2P_13 && arg3==arg3P_13 ], cost: 1 9: f5 -> f6 : arg1'=arg1P_10, arg2'=arg2P_10, arg3'=arg3P_10, [ arg2>arg1 && arg1==arg1P_10 && arg2==arg2P_10 && arg3==arg3P_10 ], cost: 1 10: f5 -> f7 : arg1'=arg1P_11, arg2'=arg2P_11, arg3'=arg3P_11, [ arg2<=arg1 && arg1==arg1P_11 && arg2==arg2P_11 && arg3==arg3P_11 ], cost: 1 14: f8 -> f4 : arg1'=arg1P_15, arg2'=arg2P_15, arg3'=arg3P_15, [ arg1==arg1P_15 && arg2==arg2P_15 && arg3==arg3P_15 ], cost: 1 18: __init -> f1 : arg1'=arg1P_19, arg2'=arg2P_19, arg3'=arg3P_19, [], 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 2: f3 -> f4 : arg3'=arg3P_3, [], cost: 1 13: f4 -> f5 : [ arg1>0 && arg2>0 && arg3>0 ], cost: 1 3: f6 -> f9 : arg2'=arg3, [], cost: 1 4: f9 -> f10 : arg1'=arg1P_5, [], cost: 1 5: f10 -> f11 : arg3'=-1+arg1, [], cost: 1 11: f11 -> f8 : [], cost: 1 6: f7 -> f12 : arg3'=-1+arg3, [], cost: 1 7: f12 -> f13 : arg1'=arg1P_8, [], cost: 1 8: f13 -> f14 : arg2'=-1+arg1, [], cost: 1 12: f14 -> f8 : [], cost: 1 9: f5 -> f6 : [ arg2>arg1 ], cost: 1 10: f5 -> f7 : [ arg2<=arg1 ], cost: 1 14: f8 -> f4 : [], cost: 1 18: __init -> f1 : arg1'=arg1P_19, arg2'=arg2P_19, arg3'=arg3P_19, [], cost: 1 ### Simplification by acceleration and chaining ### Eliminated locations (on linear paths): Start location: __init 13: f4 -> f5 : [ arg1>0 && arg2>0 && arg3>0 ], cost: 1 28: f5 -> f8 : arg1'=arg1P_5, arg2'=arg3, arg3'=-1+arg1P_5, [ arg2>arg1 ], cost: 5 29: f5 -> f8 : arg1'=arg1P_8, arg2'=-1+arg1P_8, arg3'=-1+arg3, [ arg2<=arg1 ], cost: 5 14: f8 -> f4 : [], cost: 1 21: __init -> f4 : arg1'=arg1P_1, arg2'=arg2P_2, arg3'=arg3P_3, [], cost: 4 Eliminated locations (on tree-shaped paths): Start location: __init 30: f4 -> f8 : arg1'=arg1P_5, arg2'=arg3, arg3'=-1+arg1P_5, [ arg1>0 && arg2>0 && arg3>0 && arg2>arg1 ], cost: 6 31: f4 -> f8 : arg1'=arg1P_8, arg2'=-1+arg1P_8, arg3'=-1+arg3, [ arg1>0 && arg2>0 && arg3>0 && arg2<=arg1 ], cost: 6 14: f8 -> f4 : [], cost: 1 21: __init -> f4 : arg1'=arg1P_1, arg2'=arg2P_2, arg3'=arg3P_3, [], cost: 4 Eliminated locations (on tree-shaped paths): Start location: __init 32: f4 -> f4 : arg1'=arg1P_5, arg2'=arg3, arg3'=-1+arg1P_5, [ arg1>0 && arg2>0 && arg3>0 && arg2>arg1 ], cost: 7 33: f4 -> f4 : arg1'=arg1P_8, arg2'=-1+arg1P_8, arg3'=-1+arg3, [ arg1>0 && arg2>0 && arg3>0 && arg2<=arg1 ], cost: 7 21: __init -> f4 : arg1'=arg1P_1, arg2'=arg2P_2, arg3'=arg3P_3, [], cost: 4 Accelerating simple loops of location 3. Accelerating the following rules: 32: f4 -> f4 : arg1'=arg1P_5, arg2'=arg3, arg3'=-1+arg1P_5, [ arg1>0 && arg2>0 && arg3>0 && arg2>arg1 ], cost: 7 33: f4 -> f4 : arg1'=arg1P_8, arg2'=-1+arg1P_8, arg3'=-1+arg3, [ arg1>0 && arg2>0 && arg3>0 && arg2<=arg1 ], cost: 7 [test] deduced pseudo-invariant -3-6*arg1P_5+5*arg2+5*arg3-4*arg1<=0, also trying 3+6*arg1P_5-5*arg2-5*arg3+4*arg1<=-1 [test] deduced pseudo-invariant -arg1P_5+arg3<=0, also trying arg1P_5-arg3<=-1 Failed to prove monotonicity of the guard of rule 32. [test] deduced pseudo-invariant 1-arg1P_8+arg2<=0, also trying -1+arg1P_8-arg2<=-1 [test] deduced pseudo-invariant 1-arg1P_8+arg2<=0, also trying -1+arg1P_8-arg2<=-1 Accelerated rule 33 with backward acceleration, yielding the new rule 34. [accelerate] Nesting with 2 inner and 2 outer candidates Accelerated all simple loops using metering functions (where possible): Start location: __init 32: f4 -> f4 : arg1'=arg1P_5, arg2'=arg3, arg3'=-1+arg1P_5, [ arg1>0 && arg2>0 && arg3>0 && arg2>arg1 ], cost: 7 33: f4 -> f4 : arg1'=arg1P_8, arg2'=-1+arg1P_8, arg3'=-1+arg3, [ arg1>0 && arg2>0 && arg3>0 && arg2<=arg1 ], cost: 7 34: f4 -> f4 : arg1'=arg1P_8, arg2'=-1+arg1P_8, arg3'=0, [ arg1>0 && arg2>0 && arg2<=arg1 && 1-arg1P_8+arg2<=0 && arg3>=1 ], cost: 7*arg3 21: __init -> f4 : arg1'=arg1P_1, arg2'=arg2P_2, arg3'=arg3P_3, [], cost: 4 Chained accelerated rules (with incoming rules): Start location: __init 21: __init -> f4 : arg1'=arg1P_1, arg2'=arg2P_2, arg3'=arg3P_3, [], cost: 4 35: __init -> f4 : arg1'=arg1P_5, arg2'=arg3P_3, arg3'=-1+arg1P_5, [ arg3P_3>0 ], cost: 11 36: __init -> f4 : arg1'=arg1P_8, arg2'=-1+arg1P_8, arg3'=-1+arg3P_3, [ arg3P_3>0 ], cost: 11 37: __init -> f4 : arg1'=arg1P_8, arg2'=-1+arg1P_8, arg3'=0, [ arg3P_3>=1 && 1<=-1+arg1P_8 ], cost: 4+7*arg3P_3 Removed unreachable locations (and leaf rules with constant cost): Start location: __init 37: __init -> f4 : arg1'=arg1P_8, arg2'=-1+arg1P_8, arg3'=0, [ arg3P_3>=1 && 1<=-1+arg1P_8 ], cost: 4+7*arg3P_3 ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: __init 37: __init -> f4 : arg1'=arg1P_8, arg2'=-1+arg1P_8, arg3'=0, [ arg3P_3>=1 && 1<=-1+arg1P_8 ], cost: 4+7*arg3P_3 Computing asymptotic complexity for rule 37 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),?)