/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, nat(2 + 2 * Arg_0 + -2 * Arg_1) + max(2, 3 + Arg_0 + -1 * Arg_1)).

(0) CpxIntTrs
(1) Koat2 Proof [FINISHED, 634 ms]
(2) BOUNDS(1, nat(2 + 2 * Arg_0 + -2 * Arg_1) + max(2, 3 + Arg_0 + -1 * Arg_1))
(3) Loat Proof [FINISHED, 678 ms]
(4) BOUNDS(n^1, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f2(A, B, C, D) -> Com_1(f300(A, B, C, D)) :|: TRUE
f300(A, B, C, D) -> Com_1(f300(1 + A, B, E, D)) :|: E >= 1 && B >= 1 + A
f300(A, B, C, D) -> Com_1(f300(1 + A, B, E, D)) :|: 0 >= E + 1 && B >= 1 + A
f300(A, B, C, D) -> Com_1(f300(A, -(1) + B, 0, D)) :|: B >= 1 + A
f300(A, B, C, D) -> Com_1(f1(A, B, C, E)) :|: A >= B

The start-symbols are:[f2_4]


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(1) Koat2 Proof (FINISHED)
YES( ?, max([0, 1+Arg_0-Arg_1])+max([2, 3+Arg_0-Arg_1])+max([0, 1+Arg_0-Arg_1]) {O(n)})



Initial Complexity Problem:

  Start:  f2  

  Program_Vars:  Arg_0, Arg_1, Arg_2, Arg_3  

  Temp_Vars:  E  

  Locations:  f1, f2, f300  

  Transitions:

  0: f2->f300

  4: f300->f1

  1: f300->f300

  2: f300->f300

  3: f300->f300  



Timebounds: 

  Overall timebound: max([0, 1+Arg_0-Arg_1])+max([2, 3+Arg_0-Arg_1])+max([0, 1+Arg_0-Arg_1]) {O(n)}

  0: f2->f300: 1 {O(1)}

  1: f300->f300: max([0, 1+Arg_0-Arg_1]) {O(n)}

  2: f300->f300: max([0, 1+Arg_0-Arg_1]) {O(n)}

  3: f300->f300: max([0, 1+Arg_0-Arg_1]) {O(n)}

  4: f300->f1: 1 {O(1)}



Costbounds:

  Overall costbound: max([0, 1+Arg_0-Arg_1])+max([2, 3+Arg_0-Arg_1])+max([0, 1+Arg_0-Arg_1]) {O(n)}

  0: f2->f300: 1 {O(1)}

  1: f300->f300: max([0, 1+Arg_0-Arg_1]) {O(n)}

  2: f300->f300: max([0, 1+Arg_0-Arg_1]) {O(n)}

  3: f300->f300: max([0, 1+Arg_0-Arg_1]) {O(n)}

  4: f300->f1: 1 {O(1)}



Sizebounds:

`Lower:

  0: f2->f300, Arg_0: Arg_0 {O(n)}

  0: f2->f300, Arg_1: Arg_1 {O(n)}

  0: f2->f300, Arg_2: Arg_2 {O(n)}

  0: f2->f300, Arg_3: Arg_3 {O(n)}

  1: f300->f300, Arg_0: Arg_0-max([0, 1+Arg_0-Arg_1]) {O(n)}

  1: f300->f300, Arg_1: Arg_1 {O(n)}

  1: f300->f300, Arg_2: 1 {O(1)}

  1: f300->f300, Arg_3: Arg_3 {O(n)}

  2: f300->f300, Arg_0: Arg_0-max([0, 1+Arg_0-Arg_1]) {O(n)}

  2: f300->f300, Arg_1: Arg_1 {O(n)}

  2: f300->f300, Arg_3: Arg_3 {O(n)}

  3: f300->f300, Arg_0: Arg_0-max([0, 1+Arg_0-Arg_1]) {O(n)}

  3: f300->f300, Arg_1: Arg_1 {O(n)}

  3: f300->f300, Arg_2: 0 {O(1)}

  3: f300->f300, Arg_3: Arg_3 {O(n)}

  4: f300->f1, Arg_0: min([Arg_0, -(-(Arg_0)+max([0, 1+Arg_0-Arg_1]))]) {O(n)}

  4: f300->f1, Arg_1: Arg_1 {O(n)}

`Upper:

  0: f2->f300, Arg_0: Arg_0 {O(n)}

  0: f2->f300, Arg_1: Arg_1 {O(n)}

  0: f2->f300, Arg_2: Arg_2 {O(n)}

  0: f2->f300, Arg_3: Arg_3 {O(n)}

  1: f300->f300, Arg_0: Arg_0 {O(n)}

  1: f300->f300, Arg_1: Arg_1+2*max([0, 1+Arg_0-Arg_1]) {O(n)}

  1: f300->f300, Arg_3: Arg_3 {O(n)}

  2: f300->f300, Arg_0: Arg_0 {O(n)}

  2: f300->f300, Arg_1: Arg_1+2*max([0, 1+Arg_0-Arg_1]) {O(n)}

  2: f300->f300, Arg_2: -1 {O(1)}

  2: f300->f300, Arg_3: Arg_3 {O(n)}

  3: f300->f300, Arg_0: Arg_0 {O(n)}

  3: f300->f300, Arg_1: Arg_1+2*max([0, 1+Arg_0-Arg_1]) {O(n)}

  3: f300->f300, Arg_2: 0 {O(1)}

  3: f300->f300, Arg_3: Arg_3 {O(n)}

  4: f300->f1, Arg_0: Arg_0 {O(n)}

  4: f300->f1, Arg_1: max([Arg_1, Arg_1+2*max([0, 1+Arg_0-Arg_1])]) {O(n)}


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(2)
BOUNDS(1, nat(2 + 2 * Arg_0 + -2 * Arg_1) + max(2, 3 + Arg_0 + -1 * Arg_1))

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(3) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: f2

      0: f2 -> f300 : [], cost: 1

      1: f300 -> f300 : B'=1+B, C'=free, [ free>=1 && A>=1+B ], cost: 1

      2: f300 -> f300 : B'=1+B, C'=free_1, [ 0>=1+free_1 && A>=1+B ], cost: 1

      3: f300 -> f300 : A'=-1+A, C'=0, [ A>=1+B ], cost: 1

      4: f300 -> f1 : D'=free_2, [ B>=A ], cost: 1



Checking for constant complexity:

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

      0: f2 -> f300 : [], cost: 1



Removed unreachable and leaf rules:

   Start location: f2

      0: f2 -> f300 : [], cost: 1

      1: f300 -> f300 : B'=1+B, C'=free, [ free>=1 && A>=1+B ], cost: 1

      2: f300 -> f300 : B'=1+B, C'=free_1, [ 0>=1+free_1 && A>=1+B ], cost: 1

      3: f300 -> f300 : A'=-1+A, C'=0, [ A>=1+B ], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 1.

   Accelerating the following rules:

      1: f300 -> f300 : B'=1+B, C'=free, [ free>=1 && A>=1+B ], cost: 1

      2: f300 -> f300 : B'=1+B, C'=free_1, [ 0>=1+free_1 && A>=1+B ], cost: 1

      3: f300 -> f300 : A'=-1+A, C'=0, [ A>=1+B ], cost: 1



   Accelerated rule 1 with metering function A-B, yielding the new rule 5.

   Accelerated rule 2 with metering function A-B, yielding the new rule 6.

   Accelerated rule 3 with metering function A-B, yielding the new rule 7.

   Removing the simple loops: 1 2 3.



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

   Start location: f2

      0: f2 -> f300 : [], cost: 1

      5: f300 -> f300 : B'=A, C'=free, [ free>=1 && A>=1+B ], cost: A-B

      6: f300 -> f300 : B'=A, C'=free_1, [ 0>=1+free_1 && A>=1+B ], cost: A-B

      7: f300 -> f300 : A'=B, C'=0, [ A>=1+B ], cost: A-B



Chained accelerated rules (with incoming rules):

   Start location: f2

      0: f2 -> f300 : [], cost: 1

      8: f2 -> f300 : B'=A, C'=free, [ free>=1 && A>=1+B ], cost: 1+A-B

      9: f2 -> f300 : B'=A, C'=free_1, [ 0>=1+free_1 && A>=1+B ], cost: 1+A-B

     10: f2 -> f300 : A'=B, C'=0, [ A>=1+B ], cost: 1+A-B



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

   Start location: f2

      8: f2 -> f300 : B'=A, C'=free, [ free>=1 && A>=1+B ], cost: 1+A-B

      9: f2 -> f300 : B'=A, C'=free_1, [ 0>=1+free_1 && A>=1+B ], cost: 1+A-B

     10: f2 -> f300 : A'=B, C'=0, [ A>=1+B ], cost: 1+A-B



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f2

      8: f2 -> f300 : B'=A, C'=free, [ free>=1 && A>=1+B ], cost: 1+A-B

      9: f2 -> f300 : B'=A, C'=free_1, [ 0>=1+free_1 && A>=1+B ], cost: 1+A-B

     10: f2 -> f300 : A'=B, C'=0, [ A>=1+B ], cost: 1+A-B



Computing asymptotic complexity for rule 8

   Solved the limit problem by the following transformations:

      Created initial limit problem:

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



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



      applying transformation rule (C) using substitution {free==1,A==0,B==-n}

      resulting limit problem:

      [solved]



   Solution:

      free / 1

      A / 0

      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-B

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



WORST_CASE(Omega(n^1),?)


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(4)
BOUNDS(n^1, INF)