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
      5: f3 -> f4 : arg1'=arg1P_6, arg2'=arg2P_6, arg3'=arg3P_6, [ arg1>=0 && arg1==arg1P_6 && arg2==arg2P_6 && arg3==arg3P_6 ], cost: 1
      6: f3 -> f4 : arg1'=arg1P_7, arg2'=arg2P_7, arg3'=arg3P_7, [ arg2>=0 && arg1==arg1P_7 && arg2==arg2P_7 && arg3==arg3P_7 ], cost: 1
      8: f3 -> f8 : arg1'=arg1P_9, arg2'=arg2P_9, arg3'=arg3P_9, [ arg1<0 && arg2<0 && arg1==arg1P_9 && arg2==arg2P_9 && arg3==arg3P_9 ], cost: 1
      2: f4 -> f5 : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [ arg1==arg1P_3 && arg2==arg2P_3 && arg1==arg3P_3 ], cost: 1
      3: f5 -> f6 : arg1'=arg1P_4, arg2'=arg2P_4, arg3'=arg3P_4, [ arg1P_4==-1+arg2 && arg2==arg2P_4 && arg3==arg3P_4 ], cost: 1
      4: f6 -> f7 : arg1'=arg1P_5, arg2'=arg2P_5, arg3'=arg3P_5, [ arg2P_5==-1+arg3 && arg1==arg1P_5 && arg3==arg3P_5 ], cost: 1
      7: f7 -> f3 : arg1'=arg1P_8, arg2'=arg2P_8, arg3'=arg3P_8, [ arg1==arg1P_8 && arg2==arg2P_8 && arg3==arg3P_8 ], cost: 1
      9: __init -> f1 : arg1'=arg1P_10, arg2'=arg2P_10, arg3'=arg3P_10, [], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
      9: __init -> f1 : arg1'=arg1P_10, arg2'=arg2P_10, arg3'=arg3P_10, [], 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
      5: f3 -> f4 : arg1'=arg1P_6, arg2'=arg2P_6, arg3'=arg3P_6, [ arg1>=0 && arg1==arg1P_6 && arg2==arg2P_6 && arg3==arg3P_6 ], cost: 1
      6: f3 -> f4 : arg1'=arg1P_7, arg2'=arg2P_7, arg3'=arg3P_7, [ arg2>=0 && arg1==arg1P_7 && arg2==arg2P_7 && arg3==arg3P_7 ], cost: 1
      2: f4 -> f5 : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [ arg1==arg1P_3 && arg2==arg2P_3 && arg1==arg3P_3 ], cost: 1
      3: f5 -> f6 : arg1'=arg1P_4, arg2'=arg2P_4, arg3'=arg3P_4, [ arg1P_4==-1+arg2 && arg2==arg2P_4 && arg3==arg3P_4 ], cost: 1
      4: f6 -> f7 : arg1'=arg1P_5, arg2'=arg2P_5, arg3'=arg3P_5, [ arg2P_5==-1+arg3 && arg1==arg1P_5 && arg3==arg3P_5 ], cost: 1
      7: f7 -> f3 : arg1'=arg1P_8, arg2'=arg2P_8, arg3'=arg3P_8, [ arg1==arg1P_8 && arg2==arg2P_8 && arg3==arg3P_8 ], cost: 1
      9: __init -> f1 : arg1'=arg1P_10, arg2'=arg2P_10, arg3'=arg3P_10, [], 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
      5: f3 -> f4 : [ arg1>=0 ], cost: 1
      6: f3 -> f4 : [ arg2>=0 ], cost: 1
      2: f4 -> f5 : arg3'=arg1, [], cost: 1
      3: f5 -> f6 : arg1'=-1+arg2, [], cost: 1
      4: f6 -> f7 : arg2'=-1+arg3, [], cost: 1
      7: f7 -> f3 : [], cost: 1
      9: __init -> f1 : arg1'=arg1P_10, arg2'=arg2P_10, arg3'=arg3P_10, [], cost: 1

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: __init
      5: f3 -> f4 : [ arg1>=0 ], cost: 1
      6: f3 -> f4 : [ arg2>=0 ], cost: 1
     14: f4 -> f3 : arg1'=-1+arg2, arg2'=-1+arg1, arg3'=arg1, [], cost: 4
     11: __init -> f3 : arg1'=arg1P_1, arg2'=arg2P_2, arg3'=arg3P_10, [], cost: 3

Eliminated locations (on tree-shaped paths):
   Start location: __init
     15: f3 -> f3 : arg1'=-1+arg2, arg2'=-1+arg1, arg3'=arg1, [ arg1>=0 ], cost: 5
     16: f3 -> f3 : arg1'=-1+arg2, arg2'=-1+arg1, arg3'=arg1, [ arg2>=0 ], cost: 5
     11: __init -> f3 : arg1'=arg1P_1, arg2'=arg2P_2, arg3'=arg3P_10, [], cost: 3

Accelerating simple loops of location 2.
   Accelerating the following rules:
     15: f3 -> f3 : arg1'=-1+arg2, arg2'=-1+arg1, arg3'=arg1, [ arg1>=0 ], cost: 5
     16: f3 -> f3 : arg1'=-1+arg2, arg2'=-1+arg1, arg3'=arg1, [ arg2>=0 ], cost: 5

   Accelerated rule 15 with backward acceleration, yielding the new rule 17.
   Accelerated rule 16 with backward acceleration, yielding the new rule 18.
[accelerate] Nesting with 2 inner and 2 outer candidates
   Removing the simple loops: 15 16.

Accelerated all simple loops using metering functions (where possible):
   Start location: __init
     17: f3 -> f3 : arg1'=-2*k+arg1, arg2'=arg2-2*k, arg3'=1+arg2-2*k, [ k>=1 && 2-2*k+arg1>=0 && 1+arg2-2*k>=0 ], cost: 10*k
     18: f3 -> f3 : arg1'=-2*k_1+arg1, arg2'=arg2-2*k_1, arg3'=1+arg2-2*k_1, [ k_1>=1 && 2+arg2-2*k_1>=0 && 1-2*k_1+arg1>=0 ], cost: 10*k_1
     11: __init -> f3 : arg1'=arg1P_1, arg2'=arg2P_2, arg3'=arg3P_10, [], cost: 3

Chained accelerated rules (with incoming rules):
   Start location: __init
     11: __init -> f3 : arg1'=arg1P_1, arg2'=arg2P_2, arg3'=arg3P_10, [], cost: 3
     19: __init -> f3 : arg1'=-2*k+arg1P_1, arg2'=-2*k+arg2P_2, arg3'=1-2*k+arg2P_2, [ k>=1 && 2-2*k+arg1P_1>=0 && 1-2*k+arg2P_2>=0 ], cost: 3+10*k
     20: __init -> f3 : arg1'=arg1P_1-2*k_1, arg2'=-2*k_1+arg2P_2, arg3'=1-2*k_1+arg2P_2, [ k_1>=1 && 2-2*k_1+arg2P_2>=0 && 1+arg1P_1-2*k_1>=0 ], cost: 3+10*k_1

Removed unreachable locations (and leaf rules with constant cost):
   Start location: __init
     19: __init -> f3 : arg1'=-2*k+arg1P_1, arg2'=-2*k+arg2P_2, arg3'=1-2*k+arg2P_2, [ k>=1 && 2-2*k+arg1P_1>=0 && 1-2*k+arg2P_2>=0 ], cost: 3+10*k
     20: __init -> f3 : arg1'=arg1P_1-2*k_1, arg2'=-2*k_1+arg2P_2, arg3'=1-2*k_1+arg2P_2, [ k_1>=1 && 2-2*k_1+arg2P_2>=0 && 1+arg1P_1-2*k_1>=0 ], cost: 3+10*k_1

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: __init
     19: __init -> f3 : arg1'=-2*k+arg1P_1, arg2'=-2*k+arg2P_2, arg3'=1-2*k+arg2P_2, [ k>=1 && 2-2*k+arg1P_1>=0 && 1-2*k+arg2P_2>=0 ], cost: 3+10*k
     20: __init -> f3 : arg1'=arg1P_1-2*k_1, arg2'=-2*k_1+arg2P_2, arg3'=1-2*k_1+arg2P_2, [ k_1>=1 && 2-2*k_1+arg2P_2>=0 && 1+arg1P_1-2*k_1>=0 ], cost: 3+10*k_1

Computing asymptotic complexity for rule 19
   Resulting cost 0 has complexity: Unknown

Computing asymptotic complexity for rule 20
   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),?)