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


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WORST_CASE(Omega(n^1), O(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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, max(1, 2 + Arg_0)).

(0) CpxIntTrs
(1) Koat2 Proof [FINISHED, 124 ms]
(2) BOUNDS(1, max(1, 2 + Arg_0))
(3) Loat Proof [FINISHED, 227 ms]
(4) BOUNDS(n^1, INF)


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

The start-symbols are:[start_2]


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(1) Koat2 Proof (FINISHED)
YES( ?, max([1, 2+Arg_0]) {O(n)})



Initial Complexity Problem:

  Start:  start  

  Program_Vars:  Arg_0, Arg_1  

  Temp_Vars:    

  Locations:  eval, start  

  Transitions:

  0: eval->eval

  1: start->eval  



Timebounds: 

  Overall timebound: max([1, 2+Arg_0]) {O(n)}

  0: eval->eval: max([0, 1+Arg_0]) {O(n)}

  1: start->eval: 1 {O(1)}



Costbounds:

  Overall costbound: max([1, 2+Arg_0]) {O(n)}

  0: eval->eval: max([0, 1+Arg_0]) {O(n)}

  1: start->eval: 1 {O(1)}



Sizebounds:

`Lower:

  0: eval->eval, Arg_0: 0 {O(1)}

  0: eval->eval, Arg_1: 0 {O(1)}

  1: start->eval, Arg_0: Arg_0 {O(n)}

  1: start->eval, Arg_1: Arg_1 {O(n)}

`Upper:

  0: eval->eval, Arg_0: Arg_0 {O(n)}

  0: eval->eval, Arg_1: Arg_1 {O(n)}

  1: start->eval, Arg_0: Arg_0 {O(n)}

  1: start->eval, Arg_1: Arg_1 {O(n)}


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(2)
BOUNDS(1, max(1, 2 + Arg_0))

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(3) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: start

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

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



Checking for constant complexity:

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

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



   Accelerated rule 0 with metering function B (after adding A>=B), yielding the new rule 2.

   Accelerated rule 0 with metering function A (after adding A<=B), yielding the new rule 3.

   Removing the simple loops: 0.



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

   Start location: start

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

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

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



Chained accelerated rules (with incoming rules):

   Start location: start

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

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

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



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

   Start location: start

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

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



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: start

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

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



Computing asymptotic complexity for rule 4

   Simplified the guard:

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

   Solved the limit problem by the following transformations:

      Created initial limit problem:

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



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



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

      resulting limit problem:

      [solved]



   Solution:

      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+B

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



WORST_CASE(Omega(n^1),?)


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(4)
BOUNDS(n^1, INF)