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 && arg2==arg2P_1 && 0==arg3P_1 ], cost: 1
      1: f2 -> f3 : arg1'=arg1P_2, arg2'=arg2P_2, arg3'=arg3P_2, [ 0==arg1P_2 && arg2==arg2P_2 && arg3==arg3P_2 ], cost: 1
      4: f3 -> f4 : arg1'=arg1P_5, arg2'=arg2P_5, arg3'=arg3P_5, [ arg1<100 && arg1==arg1P_5 && arg2==arg2P_5 && arg3==arg3P_5 ], cost: 1
      6: f3 -> f7 : arg1'=arg1P_7, arg2'=arg2P_7, arg3'=arg3P_7, [ arg1>=100 && arg1==arg1P_7 && arg2==arg2P_7 && arg3==arg3P_7 ], cost: 1
      2: f4 -> f5 : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [ arg3P_3==1+arg3 && arg1==arg1P_3 && arg2==arg2P_3 ], cost: 1
      3: f5 -> f6 : arg1'=arg1P_4, arg2'=arg2P_4, arg3'=arg3P_4, [ arg1P_4==1+arg1 && arg2==arg2P_4 && arg3==arg3P_4 ], cost: 1
      5: f6 -> f3 : arg1'=arg1P_6, arg2'=arg2P_6, arg3'=arg3P_6, [ arg1==arg1P_6 && arg2==arg2P_6 && arg3==arg3P_6 ], cost: 1
      7: f7 -> f8 : arg1'=arg1P_8, arg2'=arg2P_8, arg3'=arg3P_8, [ arg1==arg1P_8 && 5==arg2P_8 && arg3==arg3P_8 ], cost: 1
     10: f8 -> f9 : arg1'=arg1P_11, arg2'=arg2P_11, arg3'=arg3P_11, [ arg2<21 && arg1==arg1P_11 && arg2==arg2P_11 && arg3==arg3P_11 ], cost: 1
     12: f8 -> f12 : arg1'=arg1P_13, arg2'=arg2P_13, arg3'=arg3P_13, [ arg2>=21 && arg1==arg1P_13 && arg2==arg2P_13 && arg3==arg3P_13 ], cost: 1
      8: f9 -> f10 : arg1'=arg1P_9, arg2'=arg2P_9, arg3'=arg3P_9, [ arg3P_9==1+arg3 && arg1==arg1P_9 && arg2==arg2P_9 ], cost: 1
      9: f10 -> f11 : arg1'=arg1P_10, arg2'=arg2P_10, arg3'=arg3P_10, [ arg2P_10==3+arg2 && arg1==arg1P_10 && arg3==arg3P_10 ], 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
     13: __init -> f1 : arg1'=arg1P_14, arg2'=arg2P_14, arg3'=arg3P_14, [], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
     13: __init -> f1 : arg1'=arg1P_14, arg2'=arg2P_14, arg3'=arg3P_14, [], 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 && arg2==arg2P_1 && 0==arg3P_1 ], cost: 1
      1: f2 -> f3 : arg1'=arg1P_2, arg2'=arg2P_2, arg3'=arg3P_2, [ 0==arg1P_2 && arg2==arg2P_2 && arg3==arg3P_2 ], cost: 1
      4: f3 -> f4 : arg1'=arg1P_5, arg2'=arg2P_5, arg3'=arg3P_5, [ arg1<100 && arg1==arg1P_5 && arg2==arg2P_5 && arg3==arg3P_5 ], cost: 1
      6: f3 -> f7 : arg1'=arg1P_7, arg2'=arg2P_7, arg3'=arg3P_7, [ arg1>=100 && arg1==arg1P_7 && arg2==arg2P_7 && arg3==arg3P_7 ], cost: 1
      2: f4 -> f5 : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [ arg3P_3==1+arg3 && arg1==arg1P_3 && arg2==arg2P_3 ], cost: 1
      3: f5 -> f6 : arg1'=arg1P_4, arg2'=arg2P_4, arg3'=arg3P_4, [ arg1P_4==1+arg1 && arg2==arg2P_4 && arg3==arg3P_4 ], cost: 1
      5: f6 -> f3 : arg1'=arg1P_6, arg2'=arg2P_6, arg3'=arg3P_6, [ arg1==arg1P_6 && arg2==arg2P_6 && arg3==arg3P_6 ], cost: 1
      7: f7 -> f8 : arg1'=arg1P_8, arg2'=arg2P_8, arg3'=arg3P_8, [ arg1==arg1P_8 && 5==arg2P_8 && arg3==arg3P_8 ], cost: 1
     10: f8 -> f9 : arg1'=arg1P_11, arg2'=arg2P_11, arg3'=arg3P_11, [ arg2<21 && arg1==arg1P_11 && arg2==arg2P_11 && arg3==arg3P_11 ], cost: 1
      8: f9 -> f10 : arg1'=arg1P_9, arg2'=arg2P_9, arg3'=arg3P_9, [ arg3P_9==1+arg3 && arg1==arg1P_9 && arg2==arg2P_9 ], cost: 1
      9: f10 -> f11 : arg1'=arg1P_10, arg2'=arg2P_10, arg3'=arg3P_10, [ arg2P_10==3+arg2 && arg1==arg1P_10 && arg3==arg3P_10 ], 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
     13: __init -> f1 : arg1'=arg1P_14, arg2'=arg2P_14, arg3'=arg3P_14, [], cost: 1

Simplified all rules, resulting in:
   Start location: __init
      0: f1 -> f2 : arg3'=0, [], cost: 1
      1: f2 -> f3 : arg1'=0, [], cost: 1
      4: f3 -> f4 : [ arg1<100 ], cost: 1
      6: f3 -> f7 : [ arg1>=100 ], cost: 1
      2: f4 -> f5 : arg3'=1+arg3, [], cost: 1
      3: f5 -> f6 : arg1'=1+arg1, [], cost: 1
      5: f6 -> f3 : [], cost: 1
      7: f7 -> f8 : arg2'=5, [], cost: 1
     10: f8 -> f9 : [ arg2<21 ], cost: 1
      8: f9 -> f10 : arg3'=1+arg3, [], cost: 1
      9: f10 -> f11 : arg2'=3+arg2, [], cost: 1
     11: f11 -> f8 : [], cost: 1
     13: __init -> f1 : arg1'=arg1P_14, arg2'=arg2P_14, arg3'=arg3P_14, [], cost: 1

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: __init
     17: f3 -> f8 : arg2'=5, [ arg1>=100 ], cost: 2
     19: f3 -> f3 : arg1'=1+arg1, arg3'=1+arg3, [ arg1<100 ], cost: 4
     22: f8 -> f8 : arg2'=3+arg2, arg3'=1+arg3, [ arg2<21 ], cost: 4
     15: __init -> f3 : arg1'=0, arg2'=arg2P_14, arg3'=0, [], cost: 3

Accelerating simple loops of location 2.
   Accelerating the following rules:
     19: f3 -> f3 : arg1'=1+arg1, arg3'=1+arg3, [ arg1<100 ], cost: 4

   Accelerated rule 19 with backward acceleration, yielding the new rule 23.
[accelerate] Nesting with 1 inner and 1 outer candidates
   Removing the simple loops: 19.

Accelerating simple loops of location 7.
   Accelerating the following rules:
     22: f8 -> f8 : arg2'=3+arg2, arg3'=1+arg3, [ arg2<21 ], cost: 4

   Accelerated rule 22 with backward acceleration, yielding the new rule 24.
[accelerate] Nesting with 1 inner and 1 outer candidates
   Removing the simple loops: 22.

Accelerated all simple loops using metering functions (where possible):
   Start location: __init
     17: f3 -> f8 : arg2'=5, [ arg1>=100 ], cost: 2
     23: f3 -> f3 : arg1'=100, arg3'=100+arg3-arg1, [ 100-arg1>=0 ], cost: 400-4*arg1
     24: f8 -> f8 : arg2'=arg2+3*k_1, arg3'=k_1+arg3, [ k_1>=0 && -3+arg2+3*k_1<21 ], cost: 4*k_1
     15: __init -> f3 : arg1'=0, arg2'=arg2P_14, arg3'=0, [], cost: 3

Chained accelerated rules (with incoming rules):
   Start location: __init
     17: f3 -> f8 : arg2'=5, [ arg1>=100 ], cost: 2
     26: f3 -> f8 : arg2'=5+3*k_1, arg3'=k_1+arg3, [ arg1>=100 && k_1>=0 && 2+3*k_1<21 ], cost: 2+4*k_1
     15: __init -> f3 : arg1'=0, arg2'=arg2P_14, arg3'=0, [], cost: 3
     25: __init -> f3 : arg1'=100, arg2'=arg2P_14, arg3'=100, [], cost: 403

Removed unreachable locations (and leaf rules with constant cost):
   Start location: __init
     26: f3 -> f8 : arg2'=5+3*k_1, arg3'=k_1+arg3, [ arg1>=100 && k_1>=0 && 2+3*k_1<21 ], cost: 2+4*k_1
     15: __init -> f3 : arg1'=0, arg2'=arg2P_14, arg3'=0, [], cost: 3
     25: __init -> f3 : arg1'=100, arg2'=arg2P_14, arg3'=100, [], cost: 403

Eliminated locations (on tree-shaped paths):
   Start location: __init
     27: __init -> f8 : arg1'=100, arg2'=5+3*k_1, arg3'=100+k_1, [ k_1>=0 && 2+3*k_1<21 ], cost: 405+4*k_1

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: __init
     27: __init -> f8 : arg1'=100, arg2'=5+3*k_1, arg3'=100+k_1, [ k_1>=0 && 2+3*k_1<21 ], cost: 405+4*k_1

Computing asymptotic complexity for rule 27
   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),?)