/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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 + Arg_1, 1 + Arg_1 + nat(1 + Arg_0), 0) + max(2, 3 + Arg_0)).

(0) CpxIntTrs
(1) Koat2 Proof [FINISHED, 198 ms]
(2) BOUNDS(1, max(1 + Arg_1, 1 + Arg_1 + nat(1 + Arg_0), 0) + max(2, 3 + Arg_0))
(3) Loat Proof [FINISHED, 326 ms]
(4) BOUNDS(n^1, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
l0(A, B) -> Com_1(l1(A, B)) :|: TRUE
l1(A, B) -> Com_1(l1(A - 1, B + 1)) :|: A >= 1
l1(A, B) -> Com_1(l2(A, B)) :|: 0 >= A
l2(A, B) -> Com_1(l2(A, B - 1)) :|: B >= 1

The start-symbols are:[l0_2]


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



Initial Complexity Problem:

  Start:  l0  

  Program_Vars:  Arg_0, Arg_1  

  Temp_Vars:    

  Locations:  l0, l1, l2  

  Transitions:

  0: l0->l1

  1: l1->l1

  2: l1->l2

  3: l2->l2  



Timebounds: 

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

  0: l0->l1: 1 {O(1)}

  1: l1->l1: max([0, 1+Arg_0]) {O(n)}

  2: l1->l2: 1 {O(1)}

  3: l2->l2: max([0, 1+max([Arg_1, Arg_1+max([0, 1+Arg_0])])]) {O(n)}



Costbounds:

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

  0: l0->l1: 1 {O(1)}

  1: l1->l1: max([0, 1+Arg_0]) {O(n)}

  2: l1->l2: 1 {O(1)}

  3: l2->l2: max([0, 1+max([Arg_1, Arg_1+max([0, 1+Arg_0])])]) {O(n)}



Sizebounds:

`Lower:

  0: l0->l1, Arg_0: Arg_0 {O(n)}

  0: l0->l1, Arg_1: Arg_1 {O(n)}

  1: l1->l1, Arg_0: 0 {O(1)}

  1: l1->l1, Arg_1: Arg_1 {O(n)}

  2: l1->l2, Arg_0: min([0, Arg_0]) {O(n)}

  2: l1->l2, Arg_1: Arg_1 {O(n)}

  3: l2->l2, Arg_0: min([0, Arg_0]) {O(n)}

  3: l2->l2, Arg_1: 0 {O(1)}

`Upper:

  0: l0->l1, Arg_0: Arg_0 {O(n)}

  0: l0->l1, Arg_1: Arg_1 {O(n)}

  1: l1->l1, Arg_0: Arg_0 {O(n)}

  1: l1->l1, Arg_1: Arg_1+max([0, 1+Arg_0]) {O(n)}

  2: l1->l2, Arg_0: 0 {O(1)}

  2: l1->l2, Arg_1: max([Arg_1, Arg_1+max([0, 1+Arg_0])]) {O(n)}

  3: l2->l2, Arg_0: 0 {O(1)}

  3: l2->l2, Arg_1: max([Arg_1, Arg_1+max([0, 1+Arg_0])]) {O(n)}


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(2)
BOUNDS(1, max(1 + Arg_1, 1 + Arg_1 + nat(1 + Arg_0), 0) + max(2, 3 + Arg_0))

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(3) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: l0

      0: l0 -> l1 : [], cost: 1

      1: l1 -> l1 : A'=-1+A, B'=1+B, [ A>=1 ], cost: 1

      2: l1 -> l2 : [ 0>=A ], cost: 1

      3: l2 -> l2 : B'=-1+B, [ B>=1 ], cost: 1



Checking for constant complexity:

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

      0: l0 -> l1 : [], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 1.

   Accelerating the following rules:

      1: l1 -> l1 : A'=-1+A, B'=1+B, [ A>=1 ], cost: 1



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

   Removing the simple loops: 1.



Accelerating simple loops of location 2.

   Accelerating the following rules:

      3: l2 -> l2 : B'=-1+B, [ B>=1 ], cost: 1



   Accelerated rule 3 with metering function B, yielding the new rule 5.

   Removing the simple loops: 3.



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

   Start location: l0

      0: l0 -> l1 : [], cost: 1

      2: l1 -> l2 : [ 0>=A ], cost: 1

      4: l1 -> l1 : A'=0, B'=A+B, [ A>=1 ], cost: A

      5: l2 -> l2 : B'=0, [ B>=1 ], cost: B



Chained accelerated rules (with incoming rules):

   Start location: l0

      0: l0 -> l1 : [], cost: 1

      6: l0 -> l1 : A'=0, B'=A+B, [ A>=1 ], cost: 1+A

      2: l1 -> l2 : [ 0>=A ], cost: 1

      7: l1 -> l2 : B'=0, [ 0>=A && B>=1 ], cost: 1+B



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

   Start location: l0

      0: l0 -> l1 : [], cost: 1

      6: l0 -> l1 : A'=0, B'=A+B, [ A>=1 ], cost: 1+A

      7: l1 -> l2 : B'=0, [ 0>=A && B>=1 ], cost: 1+B



Eliminated locations (on tree-shaped paths):

   Start location: l0

      8: l0 -> l2 : B'=0, [ 0>=A && B>=1 ], cost: 2+B

      9: l0 -> l2 : A'=0, B'=0, [ A>=1 && A+B>=1 ], cost: 2+2*A+B



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: l0

      8: l0 -> l2 : B'=0, [ 0>=A && B>=1 ], cost: 2+B

      9: l0 -> l2 : A'=0, B'=0, [ A>=1 && A+B>=1 ], cost: 2+2*A+B



Computing asymptotic complexity for rule 8

   Solved the limit problem by the following transformations:

      Created initial limit problem:

      1-A (+/+!), B (+/+!), 2+B (+) [not solved]



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



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

      resulting limit problem:

      [solved]



   Solution:

      A / 0

      B / n

   Resulting cost 2+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: 2+n

   Rule cost:   2+B

   Rule guard:  [ 0>=A && B>=1 ]



WORST_CASE(Omega(n^1),?)


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(4)
BOUNDS(n^1, INF)