/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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(1, 5 + 4 * Arg_1)).

(0) CpxIntTrs
(1) Koat2 Proof [FINISHED, 420 ms]
(2) BOUNDS(1, max(1, 5 + 4 * Arg_1))
(3) Loat Proof [FINISHED, 727 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, A)) :|: 0 >= A && B >= 1 && B <= 1
eval(A, B) -> Com_1(eval(A, A)) :|: B >= 1 && 1 + B >= 0 && B >= A + 1
eval(A, B) -> Com_1(eval(A, 0)) :|: A >= 1 && B >= 1 && B <= 1
eval(A, B) -> Com_1(eval(A, B - 1)) :|: B >= 1 && 1 + B >= 0 && A >= 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+2*2*max([0, 1+Arg_1]) {O(n)})



Initial Complexity Problem:

  Start:  start  

  Program_Vars:  Arg_0, Arg_1  

  Temp_Vars:    

  Locations:  eval, start  

  Transitions:

  0: eval->eval

  1: eval->eval

  2: eval->eval

  3: eval->eval

  4: start->eval  



Timebounds: 

  Overall timebound: 1+2*2*max([0, 1+Arg_1]) {O(n)}

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

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

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

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

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



Costbounds:

  Overall costbound: 1+2*2*max([0, 1+Arg_1]) {O(n)}

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

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

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

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

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



Sizebounds:

`Lower:

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

  0: eval->eval, Arg_1: Arg_0 {O(n)}

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

  1: eval->eval, Arg_1: Arg_0 {O(n)}

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

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

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

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

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

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

`Upper:

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

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

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

  1: eval->eval, Arg_1: Arg_0 {O(n)}

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

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

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

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

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

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


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(2)
BOUNDS(1, max(1, 5 + 4 * Arg_1))

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(3) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: start

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

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

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

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

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



Checking for constant complexity:

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

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



Simplified all rules, resulting in:

   Start location: start

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

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

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

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

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



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 0.

   Accelerating the following rules:

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

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

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

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



   Accelerated rule 0 with NONTERM (after strengthening guard), yielding the new rule 5.

   Found no metering function for rule 1.

   Accelerated rule 2 with metering function B, yielding the new rule 6.

   Accelerated rule 3 with metering function B, yielding the new rule 7.

   Nested simple loops 1 (outer loop) and 6 (inner loop) with metering function 1-A, resulting in the new rules: 8.

   Removing the simple loops: 1 2 3.



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

   Start location: start

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

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

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

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

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

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



Chained accelerated rules (with incoming rules):

   Start location: start

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

      9: start -> eval : B'=A, [ 0>=A && B==1 ], cost: 2

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

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



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

   Start location: start

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

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



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: start

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

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



Computing asymptotic complexity for rule 10

   Could not solve the limit problem.

   Resulting cost 0 has complexity: Unknown



Computing asymptotic complexity for rule 11

   Solved the limit problem by the following transformations:

      Created initial limit problem:

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



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



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

      resulting limit problem:

      [solved]



   Solution:

      A / 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+B

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



WORST_CASE(Omega(n^1),?)


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(4)
BOUNDS(n^1, INF)