/export/starexec/sandbox2/solver/bin/starexec_run_complexity /export/starexec/sandbox2/benchmark/theBenchmark.koat /export/starexec/sandbox2/output/output_files


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WORST_CASE(Omega(n^1), O(n^1))
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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(3, 5 + Arg_0) + max(1, 3 + Arg_0)).

(0) CpxIntTrs
(1) Koat2 Proof [FINISHED, 137 ms]
(2) BOUNDS(1, max(3, 5 + Arg_0) + max(1, 3 + Arg_0))
(3) Loat Proof [FINISHED, 205 ms]
(4) BOUNDS(n^1, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
evaleasy2start(A) -> Com_1(evaleasy2entryin(A)) :|: TRUE
evaleasy2entryin(A) -> Com_1(evaleasy2bb1in(A)) :|: TRUE
evaleasy2bb1in(A) -> Com_1(evaleasy2bbin(A)) :|: A >= 1
evaleasy2bb1in(A) -> Com_1(evaleasy2returnin(A)) :|: 0 >= A
evaleasy2bbin(A) -> Com_1(evaleasy2bb1in(A - 1)) :|: TRUE
evaleasy2returnin(A) -> Com_1(evaleasy2stop(A)) :|: TRUE

The start-symbols are:[evaleasy2start_1]


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



Initial Complexity Problem:

  Start:  evaleasy2start  

  Program_Vars:  Arg_0  

  Temp_Vars:    

  Locations:  evaleasy2bb1in, evaleasy2bbin, evaleasy2entryin, evaleasy2returnin, evaleasy2start, evaleasy2stop  

  Transitions:

  2: evaleasy2bb1in->evaleasy2bbin

  3: evaleasy2bb1in->evaleasy2returnin

  4: evaleasy2bbin->evaleasy2bb1in

  1: evaleasy2entryin->evaleasy2bb1in

  5: evaleasy2returnin->evaleasy2stop

  0: evaleasy2start->evaleasy2entryin  



Timebounds: 

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

  2: evaleasy2bb1in->evaleasy2bbin: max([0, 2+Arg_0]) {O(n)}

  3: evaleasy2bb1in->evaleasy2returnin: 1 {O(1)}

  4: evaleasy2bbin->evaleasy2bb1in: max([0, 2+Arg_0]) {O(n)}

  1: evaleasy2entryin->evaleasy2bb1in: 1 {O(1)}

  5: evaleasy2returnin->evaleasy2stop: 1 {O(1)}

  0: evaleasy2start->evaleasy2entryin: 1 {O(1)}



Costbounds:

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

  2: evaleasy2bb1in->evaleasy2bbin: max([0, 2+Arg_0]) {O(n)}

  3: evaleasy2bb1in->evaleasy2returnin: 1 {O(1)}

  4: evaleasy2bbin->evaleasy2bb1in: max([0, 2+Arg_0]) {O(n)}

  1: evaleasy2entryin->evaleasy2bb1in: 1 {O(1)}

  5: evaleasy2returnin->evaleasy2stop: 1 {O(1)}

  0: evaleasy2start->evaleasy2entryin: 1 {O(1)}



Sizebounds:

`Lower:

  2: evaleasy2bb1in->evaleasy2bbin, Arg_0: 1 {O(1)}

  3: evaleasy2bb1in->evaleasy2returnin, Arg_0: min([0, Arg_0]) {O(n)}

  4: evaleasy2bbin->evaleasy2bb1in, Arg_0: 0 {O(1)}

  1: evaleasy2entryin->evaleasy2bb1in, Arg_0: Arg_0 {O(n)}

  5: evaleasy2returnin->evaleasy2stop, Arg_0: min([0, Arg_0]) {O(n)}

  0: evaleasy2start->evaleasy2entryin, Arg_0: Arg_0 {O(n)}

`Upper:

  2: evaleasy2bb1in->evaleasy2bbin, Arg_0: Arg_0 {O(n)}

  3: evaleasy2bb1in->evaleasy2returnin, Arg_0: 0 {O(1)}

  4: evaleasy2bbin->evaleasy2bb1in, Arg_0: Arg_0 {O(n)}

  1: evaleasy2entryin->evaleasy2bb1in, Arg_0: Arg_0 {O(n)}

  5: evaleasy2returnin->evaleasy2stop, Arg_0: 0 {O(1)}

  0: evaleasy2start->evaleasy2entryin, Arg_0: Arg_0 {O(n)}


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(2)
BOUNDS(1, max(3, 5 + Arg_0) + max(1, 3 + Arg_0))

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(3) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: evaleasy2start

      0: evaleasy2start -> evaleasy2entryin : [], cost: 1

      1: evaleasy2entryin -> evaleasy2bb1in : [], cost: 1

      2: evaleasy2bb1in -> evaleasy2bbin : [ A>=1 ], cost: 1

      3: evaleasy2bb1in -> evaleasy2returnin : [ 0>=A ], cost: 1

      4: evaleasy2bbin -> evaleasy2bb1in : A'=-1+A, [], cost: 1

      5: evaleasy2returnin -> evaleasy2stop : [], cost: 1



Checking for constant complexity:

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

      0: evaleasy2start -> evaleasy2entryin : [], cost: 1



Removed unreachable and leaf rules:

   Start location: evaleasy2start

      0: evaleasy2start -> evaleasy2entryin : [], cost: 1

      1: evaleasy2entryin -> evaleasy2bb1in : [], cost: 1

      2: evaleasy2bb1in -> evaleasy2bbin : [ A>=1 ], cost: 1

      4: evaleasy2bbin -> evaleasy2bb1in : A'=-1+A, [], cost: 1



### Simplification by acceleration and chaining ###



Eliminated locations (on linear paths):

   Start location: evaleasy2start

      6: evaleasy2start -> evaleasy2bb1in : [], cost: 2

      7: evaleasy2bb1in -> evaleasy2bb1in : A'=-1+A, [ A>=1 ], cost: 2



Accelerating simple loops of location 2.

   Accelerating the following rules:

      7: evaleasy2bb1in -> evaleasy2bb1in : A'=-1+A, [ A>=1 ], cost: 2



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

   Removing the simple loops: 7.



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

   Start location: evaleasy2start

      6: evaleasy2start -> evaleasy2bb1in : [], cost: 2

      8: evaleasy2bb1in -> evaleasy2bb1in : A'=0, [ A>=1 ], cost: 2*A



Chained accelerated rules (with incoming rules):

   Start location: evaleasy2start

      6: evaleasy2start -> evaleasy2bb1in : [], cost: 2

      9: evaleasy2start -> evaleasy2bb1in : A'=0, [ A>=1 ], cost: 2+2*A



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

   Start location: evaleasy2start

      9: evaleasy2start -> evaleasy2bb1in : A'=0, [ A>=1 ], cost: 2+2*A



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: evaleasy2start

      9: evaleasy2start -> evaleasy2bb1in : A'=0, [ A>=1 ], cost: 2+2*A



Computing asymptotic complexity for rule 9

   Solved the limit problem by the following transformations:

      Created initial limit problem:

      A (+/+!), 2+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:  [ A>=1 ]



WORST_CASE(Omega(n^1),?)


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(4)
BOUNDS(n^1, INF)