/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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_1) + nat(Arg_0)).

(0) CpxIntTrs
(1) Koat2 Proof [FINISHED, 117 ms]
(2) BOUNDS(1, max(1, 1 + Arg_1) + nat(Arg_0))
(3) Loat Proof [FINISHED, 634 ms]
(4) BOUNDS(n^1, INF)


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

The start-symbols are:[start_2]


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



Initial Complexity Problem:

  Start:  start  

  Program_Vars:  Arg_0, Arg_1  

  Temp_Vars:    

  Locations:  eval, start  

  Transitions:

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

  eval(Arg_0,Arg_1) -> eval(Arg_0,Arg_1-1):|:1 <= Arg_0+Arg_1 && Arg_0 <= 0 && 1 <= Arg_1

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



Timebounds: 

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

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

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

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



Costbounds:

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

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

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

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



Sizebounds:

`Lower:

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

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

  1: eval->eval, Arg_0: min([0, Arg_0]) {O(n)}

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

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

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

`Upper:

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

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

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

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

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

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


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

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(3) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: start

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

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

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

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



Removed rules with unsatisfiable guard:

   Start location: start

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

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

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



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 0.

   Accelerating the following rules:

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

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



   Accelerated rule 0 with backward acceleration, yielding the new rule 4.

   Accelerated rule 0 with backward acceleration, yielding the new rule 5.

   Accelerated rule 1 with metering function A+B, yielding the new rule 6.

   Removing the simple loops: 0 1.



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

   Start location: start

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

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

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

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



Chained accelerated rules (with incoming rules):

   Start location: start

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

      7: start -> eval : A'=-B, [ A+B>=1 && A>=1 && 1-B>=1 ], cost: 1+A+B

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

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



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

   Start location: start

      7: start -> eval : A'=-B, [ A+B>=1 && A>=1 && 1-B>=1 ], cost: 1+A+B

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

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



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: start

      7: start -> eval : A'=-B, [ A+B>=1 && A>=1 && 1-B>=1 ], cost: 1+A+B

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

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



Computing asymptotic complexity for rule 7

   Solved the limit problem by the following transformations:

      Created initial limit problem:

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



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



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

      resulting limit problem:

      [solved]



   Solution:

      A / 2*n

      B / -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+A+B

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



WORST_CASE(Omega(n^1),?)


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(4)
BOUNDS(n^1, INF)