/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, max(1, 1 + Arg_0 + -1 * Arg_1 + Arg_2)).

(0) CpxIntTrs
(1) Koat2 Proof [FINISHED, 26 ms]
(2) BOUNDS(1, max(1, 1 + Arg_0 + -1 * Arg_1 + Arg_2))
(3) Loat Proof [FINISHED, 234 ms]
(4) BOUNDS(n^1, INF)


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

The start-symbols are:[start_3]


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



Initial Complexity Problem:

  Start:  start  

  Program_Vars:  Arg_0, Arg_1, Arg_2  

  Temp_Vars:    

  Locations:  eval, start  

  Transitions:

  eval(Arg_0,Arg_1,Arg_2) -> eval(Arg_0,Arg_1+Arg_2,Arg_2):|:Arg_1 <= Arg_0 && 1 <= Arg_2

  start(Arg_0,Arg_1,Arg_2) -> eval(Arg_0,Arg_1,Arg_2):|:  



Timebounds: 

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

  0: eval->eval: max([0, Arg_0+Arg_2-Arg_1]) {O(n)}

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



Costbounds:

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

  0: eval->eval: max([0, Arg_0+Arg_2-Arg_1]) {O(n)}

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



Sizebounds:

`Lower:

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

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

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

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

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

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

`Upper:

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

  0: eval->eval, Arg_1: max([Arg_2, max([Arg_2, Arg_1])])+max([0, Arg_2*(Arg_0+Arg_2-Arg_1)]) {O(n^2)}

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

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

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

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


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

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(3) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: start

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

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



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 0.

   Accelerating the following rules:

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



   Accelerated rule 0 with backward acceleration, yielding the new rule 2.

   Removing the simple loops: 0.



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

   Start location: start

      2: eval -> eval : B'=C*k+B, [ A>=B && C>=1 && k>0 && A>=C*(-1+k)+B ], cost: k

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



Chained accelerated rules (with incoming rules):

   Start location: start

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

      3: start -> eval : B'=C*k+B, [ A>=B && C>=1 && k>0 && A>=C*(-1+k)+B ], cost: 1+k



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

   Start location: start

      3: start -> eval : B'=C*k+B, [ A>=B && C>=1 && k>0 && A>=C*(-1+k)+B ], cost: 1+k



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: start

      3: start -> eval : B'=C*k+B, [ A>=B && C>=1 && k>0 && A>=C*(-1+k)+B ], cost: 1+k



Computing asymptotic complexity for rule 3

   Simplified the guard:

      3: start -> eval : B'=C*k+B, [ C>=1 && k>0 && A>=C*(-1+k)+B ], cost: 1+k

   Solved the limit problem by the following transformations:

      Created initial limit problem:

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



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



      applying transformation rule (C) using substitution {C==1,A==0,k==n,B==-2*n}

      resulting limit problem:

      [solved]



   Solved the limit problem by the following transformations:

      Created initial limit problem:

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



      applying transformation rule (C) using substitution {C==1}

      resulting limit problem:

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



      applying transformation rule (C) using substitution {A==C*(-1+k)+B}

      resulting limit problem:

      1 (+/+!), 1+k (+), 2+C*(-1+k)-k (+/+!), k (+/+!) [not solved]



      applying transformation rule (B), deleting 1 (+/+!)

      resulting limit problem:

      1+k (+), 2+C*(-1+k)-k (+/+!), k (+/+!) [not solved]



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



      applying transformation rule (C) using substitution {C==2,k==n}

      resulting limit problem:

      [solved]



   Solved the limit problem by the following transformations:

      Created initial limit problem:

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



      applying transformation rule (C) using substitution {A==C*(-1+k)+B}

      resulting limit problem:

      1 (+/+!), C (+/+!), 1+k (+), k (+/+!) [not solved]



      applying transformation rule (B), deleting 1 (+/+!)

      resulting limit problem:

      C (+/+!), 1+k (+), k (+/+!) [not solved]



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



      applying transformation rule (C) using substitution {C==1,k==n}

      resulting limit problem:

      [solved]



   Solved the limit problem by the following transformations:

      Created initial limit problem:

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



      applying transformation rule (C) using substitution {C==1}

      resulting limit problem:

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



      applying transformation rule (B), deleting 1 (+/+!)

      resulting limit problem:

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



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



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

      resulting limit problem:

      [solved]



   Solution:

      C / 1

      A / 0

      k / n

      B / -2*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+k

   Rule guard:  [ C>=1 && k>0 && A>=C*(-1+k)+B ]



WORST_CASE(Omega(n^1),?)


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(4)
BOUNDS(n^1, INF)