/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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(2 + 8 * Arg_0 + -8 * Arg_1, 10, -6 + -8 * Arg_1 + 8 * Arg_2)).

(0) CpxIntTrs
(1) Koat2 Proof [FINISHED, 125 ms]
(2) BOUNDS(1, max(2 + 8 * Arg_0 + -8 * Arg_1, 10, -6 + -8 * Arg_1 + 8 * Arg_2))
(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:
l0(A, B, C) -> Com_1(l1(A, B, C)) :|: TRUE
l1(A, B, C) -> Com_1(l1(A + 1, C, C)) :|: A + 1 <= B

The start-symbols are:[l0_3]


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(1) Koat2 Proof (FINISHED)
YES( ?, 2+8*max([1, max([-1+Arg_2-Arg_1, Arg_0-Arg_1])]) {O(n)})



Initial Complexity Problem:

  Start:  l0  

  Program_Vars:  Arg_0, Arg_1, Arg_2  

  Temp_Vars:    

  Locations:  l0, l1  

  Transitions:

  0: l0->l1

  1: l1->l1  



Timebounds: 

  Overall timebound: 2+8*max([1, max([-1+Arg_2-Arg_1, Arg_0-Arg_1])]) {O(n)}

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

  1: l1->l1: 1+8*max([1, max([-1+Arg_2-Arg_1, Arg_0-Arg_1])]) {O(n)}



Costbounds:

  Overall costbound: 2+8*max([1, max([-1+Arg_2-Arg_1, Arg_0-Arg_1])]) {O(n)}

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

  1: l1->l1: 1+8*max([1, max([-1+Arg_2-Arg_1, Arg_0-Arg_1])]) {O(n)}



Sizebounds:

`Lower:

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

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

  0: l0->l1, Arg_2: Arg_2 {O(n)}

  1: l1->l1, Arg_0: Arg_2 {O(n)}

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

  1: l1->l1, Arg_2: Arg_2 {O(n)}

`Upper:

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

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

  0: l0->l1, Arg_2: Arg_2 {O(n)}

  1: l1->l1, Arg_0: Arg_2 {O(n)}

  1: l1->l1, Arg_1: 1+Arg_1+8*max([1, max([-1+Arg_2-Arg_1, Arg_0-Arg_1])]) {O(n)}

  1: l1->l1, Arg_2: Arg_2 {O(n)}


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(2)
BOUNDS(1, max(2 + 8 * Arg_0 + -8 * Arg_1, 10, -6 + -8 * Arg_1 + 8 * Arg_2))

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



   Accelerated rule 1 with metering function -1+C-B (after strengthening guard), yielding the new rule 2.

   Removing the simple loops:.



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

   Start location: l0

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

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

      2: l1 -> l1 : A'=C, B'=-1+C, [ A>=1+B && C>=2+B ], cost: -1+C-B



Chained accelerated rules (with incoming rules):

   Start location: l0

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

      3: l0 -> l1 : A'=C, B'=1+B, [ A>=1+B ], cost: 2

      4: l0 -> l1 : A'=C, B'=-1+C, [ A>=1+B && C>=2+B ], cost: C-B



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

   Start location: l0

      4: l0 -> l1 : A'=C, B'=-1+C, [ A>=1+B && C>=2+B ], cost: C-B



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: l0

      4: l0 -> l1 : A'=C, B'=-1+C, [ A>=1+B && C>=2+B ], cost: C-B



Computing asymptotic complexity for rule 4

   Solved the limit problem by the following transformations:

      Created initial limit problem:

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



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



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

      resulting limit problem:

      [solved]



   Solution:

      C / n

      A / 0

      B / -1

   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:   C-B

   Rule guard:  [ A>=1+B && C>=2+B ]



WORST_CASE(Omega(n^1),?)


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(4)
BOUNDS(n^1, INF)