/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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(3, 2 + Arg_0) + max(1, 1 + Arg_0)).

(0) CpxIntTrs
(1) Koat2 Proof [FINISHED, 147 ms]
(2) BOUNDS(1, max(3, 2 + Arg_0) + max(1, 1 + Arg_0))
(3) Loat Proof [FINISHED, 242 ms]
(4) BOUNDS(n^1, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
evalndecrstart(A) -> Com_1(evalndecrentryin(A)) :|: TRUE
evalndecrentryin(A) -> Com_1(evalndecrbb1in(A - 1)) :|: TRUE
evalndecrbb1in(A) -> Com_1(evalndecrbbin(A)) :|: A >= 2
evalndecrbb1in(A) -> Com_1(evalndecrreturnin(A)) :|: 1 >= A
evalndecrbbin(A) -> Com_1(evalndecrbb1in(A - 1)) :|: TRUE
evalndecrreturnin(A) -> Com_1(evalndecrstop(A)) :|: TRUE

The start-symbols are:[evalndecrstart_1]


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



Initial Complexity Problem:

  Start:  evalndecrstart  

  Program_Vars:  Arg_0  

  Temp_Vars:    

  Locations:  evalndecrbb1in, evalndecrbbin, evalndecrentryin, evalndecrreturnin, evalndecrstart, evalndecrstop  

  Transitions:

  evalndecrbb1in(Arg_0) -> evalndecrbbin(Arg_0):|:2 <= Arg_0

  evalndecrbb1in(Arg_0) -> evalndecrreturnin(Arg_0):|:Arg_0 <= 1

  evalndecrbbin(Arg_0) -> evalndecrbb1in(Arg_0-1):|:2 <= Arg_0

  evalndecrentryin(Arg_0) -> evalndecrbb1in(Arg_0-1):|:

  evalndecrreturnin(Arg_0) -> evalndecrstop(Arg_0):|:Arg_0 <= 1

  evalndecrstart(Arg_0) -> evalndecrentryin(Arg_0):|:  



Timebounds: 

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

  2: evalndecrbb1in->evalndecrbbin: max([0, Arg_0]) {O(n)}

  3: evalndecrbb1in->evalndecrreturnin: 1 {O(1)}

  4: evalndecrbbin->evalndecrbb1in: max([0, -1+Arg_0]) {O(n)}

  1: evalndecrentryin->evalndecrbb1in: 1 {O(1)}

  5: evalndecrreturnin->evalndecrstop: 1 {O(1)}

  0: evalndecrstart->evalndecrentryin: 1 {O(1)}



Costbounds:

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

  2: evalndecrbb1in->evalndecrbbin: max([0, Arg_0]) {O(n)}

  3: evalndecrbb1in->evalndecrreturnin: 1 {O(1)}

  4: evalndecrbbin->evalndecrbb1in: max([0, -1+Arg_0]) {O(n)}

  1: evalndecrentryin->evalndecrbb1in: 1 {O(1)}

  5: evalndecrreturnin->evalndecrstop: 1 {O(1)}

  0: evalndecrstart->evalndecrentryin: 1 {O(1)}



Sizebounds:

`Lower:

  2: evalndecrbb1in->evalndecrbbin, Arg_0: 2 {O(1)}

  3: evalndecrbb1in->evalndecrreturnin, Arg_0: min([1, -(1-Arg_0)]) {O(n)}

  4: evalndecrbbin->evalndecrbb1in, Arg_0: 1 {O(1)}

  1: evalndecrentryin->evalndecrbb1in, Arg_0: -1+Arg_0 {O(n)}

  5: evalndecrreturnin->evalndecrstop, Arg_0: min([1, -(1-Arg_0)]) {O(n)}

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

`Upper:

  2: evalndecrbb1in->evalndecrbbin, Arg_0: -1+Arg_0 {O(n)}

  3: evalndecrbb1in->evalndecrreturnin, Arg_0: 1 {O(1)}

  4: evalndecrbbin->evalndecrbb1in, Arg_0: -1+Arg_0 {O(n)}

  1: evalndecrentryin->evalndecrbb1in, Arg_0: -1+Arg_0 {O(n)}

  5: evalndecrreturnin->evalndecrstop, Arg_0: 1 {O(1)}

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


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(2)
BOUNDS(1, max(3, 2 + Arg_0) + max(1, 1 + Arg_0))

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(3) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: evalndecrstart

      0: evalndecrstart -> evalndecrentryin : [], cost: 1

      1: evalndecrentryin -> evalndecrbb1in : A'=-1+A, [], cost: 1

      2: evalndecrbb1in -> evalndecrbbin : [ A>=2 ], cost: 1

      3: evalndecrbb1in -> evalndecrreturnin : [ 1>=A ], cost: 1

      4: evalndecrbbin -> evalndecrbb1in : A'=-1+A, [], cost: 1

      5: evalndecrreturnin -> evalndecrstop : [], cost: 1



Removed unreachable and leaf rules:

   Start location: evalndecrstart

      0: evalndecrstart -> evalndecrentryin : [], cost: 1

      1: evalndecrentryin -> evalndecrbb1in : A'=-1+A, [], cost: 1

      2: evalndecrbb1in -> evalndecrbbin : [ A>=2 ], cost: 1

      4: evalndecrbbin -> evalndecrbb1in : A'=-1+A, [], cost: 1



### Simplification by acceleration and chaining ###



Eliminated locations (on linear paths):

   Start location: evalndecrstart

      6: evalndecrstart -> evalndecrbb1in : A'=-1+A, [], cost: 2

      7: evalndecrbb1in -> evalndecrbb1in : A'=-1+A, [ A>=2 ], cost: 2



Accelerating simple loops of location 2.

   Accelerating the following rules:

      7: evalndecrbb1in -> evalndecrbb1in : A'=-1+A, [ A>=2 ], cost: 2



   Accelerated rule 7 with metering function -1+A, yielding the new rule 8.

   Removing the simple loops: 7.



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

   Start location: evalndecrstart

      6: evalndecrstart -> evalndecrbb1in : A'=-1+A, [], cost: 2

      8: evalndecrbb1in -> evalndecrbb1in : A'=1, [ A>=2 ], cost: -2+2*A



Chained accelerated rules (with incoming rules):

   Start location: evalndecrstart

      6: evalndecrstart -> evalndecrbb1in : A'=-1+A, [], cost: 2

      9: evalndecrstart -> evalndecrbb1in : A'=1, [ -1+A>=2 ], cost: -2+2*A



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

   Start location: evalndecrstart

      9: evalndecrstart -> evalndecrbb1in : A'=1, [ -1+A>=2 ], cost: -2+2*A



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: evalndecrstart

      9: evalndecrstart -> evalndecrbb1in : A'=1, [ -1+A>=2 ], cost: -2+2*A



Computing asymptotic complexity for rule 9

   Solved the limit problem by the following transformations:

      Created initial limit problem:

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



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



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

      resulting limit problem:

      [solved]



   Solution:

      A / n

   Resulting cost -2+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+2*n

   Rule cost:   -2+2*A

   Rule guard:  [ -1+A>=2 ]



WORST_CASE(Omega(n^1),?)


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(4)
BOUNDS(n^1, INF)