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 7: f4 -> f5 : arg1'=arg1P_8, arg2'=arg2P_8, arg3'=arg3P_8, [ arg2==arg3 && arg1==arg1P_8 && arg2==arg2P_8 && arg3==arg3P_8 ], cost: 1 8: f4 -> f6 : arg1'=arg1P_9, arg2'=arg2P_9, arg3'=arg3P_9, [ arg2 f6 : arg1'=arg1P_10, arg2'=arg2P_10, arg3'=arg3P_10, [ arg2>arg3 && arg1==arg1P_10 && arg2==arg2P_10 && arg3==arg3P_10 ], cost: 1 3: f8 -> f9 : arg1'=arg1P_4, arg2'=arg2P_4, arg3'=arg3P_4, [ arg1P_4==-1+arg2-arg3+arg1 && arg2==arg2P_4 && arg3==arg3P_4 ], cost: 1 5: f9 -> f5 : arg1'=arg1P_6, arg2'=arg2P_6, arg3'=arg3P_6, [ arg1==arg1P_6 && arg2==arg2P_6 && arg3==arg3P_6 ], cost: 1 4: f5 -> f8 : arg1'=arg1P_5, arg2'=arg2P_5, arg3'=arg3P_5, [ arg1>=0 && arg1==arg1P_5 && arg2==arg2P_5 && arg3==arg3P_5 ], cost: 1 6: f5 -> f10 : arg1'=arg1P_7, arg2'=arg2P_7, arg3'=arg3P_7, [ arg1<0 && arg1==arg1P_7 && arg2==arg2P_7 && arg3==arg3P_7 ], cost: 1 10: f10 -> f7 : arg1'=arg1P_11, arg2'=arg2P_11, arg3'=arg3P_11, [ arg1==arg1P_11 && arg2==arg2P_11 && arg3==arg3P_11 ], cost: 1 11: f6 -> f7 : arg1'=arg1P_12, arg2'=arg2P_12, arg3'=arg3P_12, [ arg1==arg1P_12 && arg2==arg2P_12 && arg3==arg3P_12 ], cost: 1 12: __init -> f1 : arg1'=arg1P_13, arg2'=arg2P_13, arg3'=arg3P_13, [], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 12: __init -> f1 : arg1'=arg1P_13, arg2'=arg2P_13, arg3'=arg3P_13, [], 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 7: f4 -> f5 : arg1'=arg1P_8, arg2'=arg2P_8, arg3'=arg3P_8, [ arg2==arg3 && arg1==arg1P_8 && arg2==arg2P_8 && arg3==arg3P_8 ], cost: 1 3: f8 -> f9 : arg1'=arg1P_4, arg2'=arg2P_4, arg3'=arg3P_4, [ arg1P_4==-1+arg2-arg3+arg1 && arg2==arg2P_4 && arg3==arg3P_4 ], cost: 1 5: f9 -> f5 : arg1'=arg1P_6, arg2'=arg2P_6, arg3'=arg3P_6, [ arg1==arg1P_6 && arg2==arg2P_6 && arg3==arg3P_6 ], cost: 1 4: f5 -> f8 : arg1'=arg1P_5, arg2'=arg2P_5, arg3'=arg3P_5, [ arg1>=0 && arg1==arg1P_5 && arg2==arg2P_5 && arg3==arg3P_5 ], cost: 1 12: __init -> f1 : arg1'=arg1P_13, arg2'=arg2P_13, arg3'=arg3P_13, [], 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 7: f4 -> f5 : [ arg2==arg3 ], cost: 1 3: f8 -> f9 : arg1'=-1+arg2-arg3+arg1, [], cost: 1 5: f9 -> f5 : [], cost: 1 4: f5 -> f8 : [ arg1>=0 ], cost: 1 12: __init -> f1 : arg1'=arg1P_13, arg2'=arg2P_13, arg3'=arg3P_13, [], cost: 1 ### Simplification by acceleration and chaining ### Eliminated locations (on linear paths): Start location: __init 18: f5 -> f5 : arg1'=-1+arg2-arg3+arg1, [ arg1>=0 ], cost: 3 16: __init -> f5 : arg1'=arg1P_1, arg2'=arg2P_2, arg3'=arg3P_3, [ arg2P_2==arg3P_3 ], cost: 5 Accelerating simple loops of location 6. Accelerating the following rules: 18: f5 -> f5 : arg1'=-1+arg2-arg3+arg1, [ arg1>=0 ], cost: 3 [test] deduced invariant arg2-arg3<=0 Accelerated rule 18 with non-termination, yielding the new rule 19. Accelerated rule 18 with backward acceleration, yielding the new rule 20. [accelerate] Nesting with 1 inner and 1 outer candidates Accelerated all simple loops using metering functions (where possible): Start location: __init 18: f5 -> f5 : arg1'=-1+arg2-arg3+arg1, [ arg1>=0 ], cost: 3 19: f5 -> [11] : [ arg2==1 && arg3==0 && arg1==0 ], cost: NONTERM 20: f5 -> f5 : arg1'=-k-k*arg3+arg1+arg2*k, [ arg2-arg3<=0 && k>=0 && 1+arg2*(-1+k)-k-arg3*(-1+k)+arg1>=0 ], cost: 3*k 16: __init -> f5 : arg1'=arg1P_1, arg2'=arg2P_2, arg3'=arg3P_3, [ arg2P_2==arg3P_3 ], cost: 5 Chained accelerated rules (with incoming rules): Start location: __init 16: __init -> f5 : arg1'=arg1P_1, arg2'=arg2P_2, arg3'=arg3P_3, [ arg2P_2==arg3P_3 ], cost: 5 21: __init -> f5 : arg1'=-1+arg1P_1, arg2'=arg3P_3, arg3'=arg3P_3, [ arg1P_1>=0 ], cost: 8 22: __init -> f5 : arg1'=arg1P_1-k, arg2'=arg3P_3, arg3'=arg3P_3, [ k>=0 && 1+arg1P_1-k>=0 ], cost: 5+3*k Removed unreachable locations (and leaf rules with constant cost): Start location: __init 22: __init -> f5 : arg1'=arg1P_1-k, arg2'=arg3P_3, arg3'=arg3P_3, [ k>=0 && 1+arg1P_1-k>=0 ], cost: 5+3*k ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: __init 22: __init -> f5 : arg1'=arg1P_1-k, arg2'=arg3P_3, arg3'=arg3P_3, [ k>=0 && 1+arg1P_1-k>=0 ], cost: 5+3*k Computing asymptotic complexity for rule 22 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),?)