/export/starexec/sandbox/solver/bin/starexec_run_complexity /export/starexec/sandbox/benchmark/theBenchmark.koat /export/starexec/sandbox/output/output_files


--------------------------------------------------------------------------------


WORST_CASE(Omega(n^1), ?)
proof of /export/starexec/sandbox/benchmark/theBenchmark.koat
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, INF).

(0) CpxIntTrs
(1) Loat Proof [FINISHED, 434 ms]
(2) BOUNDS(n^1, INF)


----------------------------------------

(0)
Obligation:
Complexity Int TRS consisting of the following rules:
eval(A, B) -> Com_1(eval(A - 1, C)) :|: A >= 1 && B >= 1
eval(A, B) -> Com_1(eval(A, B - 1)) :|: A >= 1 && B >= 1
start(A, B) -> Com_1(eval(A, B)) :|: TRUE

The start-symbols are:[start_2]


----------------------------------------

(1) Loat Proof (FINISHED)


### Pre-processing the ITS problem ###



Initial linear ITS problem

   Start location: start

      0: eval -> eval : A'=-1+A, B'=free, [ A>=1 && B>=1 ], cost: 1

      1: eval -> eval : B'=-1+B, [ A>=1 && B>=1 ], cost: 1

      2: start -> eval : [], cost: 1



Checking for constant complexity:

   The following rule is satisfiable with cost >= 1, yielding constant complexity:

      2: start -> eval : [], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 0.

   Accelerating the following rules:

      0: eval -> eval : A'=-1+A, B'=free, [ A>=1 && B>=1 ], cost: 1

      1: eval -> eval : B'=-1+B, [ A>=1 && B>=1 ], cost: 1



   Accelerated rule 0 with metering function A (after strengthening guard), yielding the new rule 3.

   Accelerated rule 1 with metering function B, yielding the new rule 4.

   Removing the simple loops: 1.



Accelerated all simple loops using metering functions (where possible):

   Start location: start

      0: eval -> eval : A'=-1+A, B'=free, [ A>=1 && B>=1 ], cost: 1

      3: eval -> eval : A'=0, B'=free, [ A>=1 && B>=1 && free>=1 ], cost: A

      4: eval -> eval : B'=0, [ A>=1 && B>=1 ], cost: B

      2: start -> eval : [], cost: 1



Chained accelerated rules (with incoming rules):

   Start location: start

      2: start -> eval : [], cost: 1

      5: start -> eval : A'=-1+A, B'=free, [ A>=1 && B>=1 ], cost: 2

      6: start -> eval : A'=0, B'=free, [ A>=1 && B>=1 && free>=1 ], cost: 1+A

      7: start -> eval : B'=0, [ A>=1 && B>=1 ], cost: 1+B



Removed unreachable locations (and leaf rules with constant cost):

   Start location: start

      6: start -> eval : A'=0, B'=free, [ A>=1 && B>=1 && free>=1 ], cost: 1+A

      7: start -> eval : B'=0, [ A>=1 && B>=1 ], cost: 1+B



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: start

      6: start -> eval : A'=0, B'=free, [ A>=1 && B>=1 && free>=1 ], cost: 1+A

      7: start -> eval : B'=0, [ A>=1 && B>=1 ], cost: 1+B



Computing asymptotic complexity for rule 6

   Solved the limit problem by the following transformations:

      Created initial limit problem:

      free (+/+!), A (+/+!), B (+/+!), 1+A (+) [not solved]



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



      applying transformation rule (C) using substitution {free==n,A==n,B==n}

      resulting limit problem:

      [solved]



   Solution:

      free / n

      A / n

      B / n

   Resulting cost 1+n has complexity: Poly(n^1)



Found new complexity Poly(n^1).



Obtained the following overall complexity (w.r.t. the length of the input n):

   Complexity:  Poly(n^1)

   Cpx degree:  1

   Solved cost: 1+n

   Rule cost:   1+A

   Rule guard:  [ A>=1 && B>=1 && free>=1 ]



WORST_CASE(Omega(n^1),?)


----------------------------------------

(2)
BOUNDS(n^1, INF)