/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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(1, 3 + 2 * Arg_0) + nat(2 + 2 * Arg_1) + nat(1 + 2 * Arg_1) + nat(1 + 2 * Arg_0)).

(0) CpxIntTrs
(1) Koat2 Proof [FINISHED, 516 ms]
(2) BOUNDS(1, max(1, 3 + 2 * Arg_0) + nat(2 + 2 * Arg_1) + nat(1 + 2 * Arg_1) + nat(1 + 2 * Arg_0))
(3) Loat Proof [FINISHED, 606 ms]
(4) BOUNDS(n^1, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
eval1(A, B) -> Com_1(eval2(A, B)) :|: A >= 1 && B >= 1 && A >= B + 1
eval1(A, B) -> Com_1(eval3(A, B)) :|: A >= 1 && B >= 1 && B >= A
eval2(A, B) -> Com_1(eval2(A - 1, B)) :|: A >= 1
eval2(A, B) -> Com_1(eval1(A, B)) :|: 0 >= A
eval3(A, B) -> Com_1(eval3(A, B - 1)) :|: B >= 1
eval3(A, B) -> Com_1(eval1(A, B)) :|: 0 >= B
start(A, B) -> Com_1(eval1(A, B)) :|: TRUE

The start-symbols are:[start_2]


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(1) Koat2 Proof (FINISHED)
YES( ?, 1+2*max([0, 1+Arg_0])+2*max([0, 1+Arg_1])+max([0, 1+2*Arg_1])+max([0, 1+2*Arg_0]) {O(n)})



Initial Complexity Problem:

  Start:  start  

  Program_Vars:  Arg_0, Arg_1  

  Temp_Vars:    

  Locations:  eval1, eval2, eval3, start  

  Transitions:

  0: eval1->eval2

  1: eval1->eval3

  3: eval2->eval1

  2: eval2->eval2

  5: eval3->eval1

  4: eval3->eval3

  6: start->eval1  



Timebounds: 

  Overall timebound: 1+2*max([0, 1+Arg_0])+2*max([0, 1+Arg_1])+max([0, 1+2*Arg_1])+max([0, 1+2*Arg_0]) {O(n)}

  0: eval1->eval2: max([0, 1+2*Arg_0]) {O(n)}

  1: eval1->eval3: max([0, 1+Arg_1]) {O(n)}

  2: eval2->eval2: max([0, 1+Arg_0]) {O(n)}

  3: eval2->eval1: max([0, 1+Arg_0]) {O(n)}

  4: eval3->eval3: max([0, 1+Arg_1]) {O(n)}

  5: eval3->eval1: max([0, 1+2*Arg_1]) {O(n)}

  6: start->eval1: 1 {O(1)}



Costbounds:

  Overall costbound: 1+2*max([0, 1+Arg_0])+2*max([0, 1+Arg_1])+max([0, 1+2*Arg_1])+max([0, 1+2*Arg_0]) {O(n)}

  0: eval1->eval2: max([0, 1+2*Arg_0]) {O(n)}

  1: eval1->eval3: max([0, 1+Arg_1]) {O(n)}

  2: eval2->eval2: max([0, 1+Arg_0]) {O(n)}

  3: eval2->eval1: max([0, 1+Arg_0]) {O(n)}

  4: eval3->eval3: max([0, 1+Arg_1]) {O(n)}

  5: eval3->eval1: max([0, 1+2*Arg_1]) {O(n)}

  6: start->eval1: 1 {O(1)}



Sizebounds:

`Lower:

  0: eval1->eval2, Arg_0: 2 {O(1)}

  0: eval1->eval2, Arg_1: 1 {O(1)}

  1: eval1->eval3, Arg_0: 1 {O(1)}

  1: eval1->eval3, Arg_1: 1 {O(1)}

  2: eval2->eval2, Arg_0: 0 {O(1)}

  2: eval2->eval2, Arg_1: 1 {O(1)}

  3: eval2->eval1, Arg_0: 0 {O(1)}

  3: eval2->eval1, Arg_1: 1 {O(1)}

  4: eval3->eval3, Arg_0: 1 {O(1)}

  4: eval3->eval3, Arg_1: 0 {O(1)}

  5: eval3->eval1, Arg_0: 1 {O(1)}

  5: eval3->eval1, Arg_1: 0 {O(1)}

  6: start->eval1, Arg_0: Arg_0 {O(n)}

  6: start->eval1, Arg_1: Arg_1 {O(n)}

`Upper:

  0: eval1->eval2, Arg_0: Arg_0 {O(n)}

  0: eval1->eval2, Arg_1: Arg_1 {O(n)}

  1: eval1->eval3, Arg_0: Arg_0 {O(n)}

  1: eval1->eval3, Arg_1: Arg_1 {O(n)}

  2: eval2->eval2, Arg_0: Arg_0 {O(n)}

  2: eval2->eval2, Arg_1: Arg_1 {O(n)}

  3: eval2->eval1, Arg_0: 0 {O(1)}

  3: eval2->eval1, Arg_1: Arg_1 {O(n)}

  4: eval3->eval3, Arg_0: Arg_0 {O(n)}

  4: eval3->eval3, Arg_1: Arg_1 {O(n)}

  5: eval3->eval1, Arg_0: Arg_0 {O(n)}

  5: eval3->eval1, Arg_1: 0 {O(1)}

  6: start->eval1, Arg_0: Arg_0 {O(n)}

  6: start->eval1, Arg_1: Arg_1 {O(n)}


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(2)
BOUNDS(1, max(1, 3 + 2 * Arg_0) + nat(2 + 2 * Arg_1) + nat(1 + 2 * Arg_1) + nat(1 + 2 * Arg_0))

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(3) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: start

      0: eval1 -> eval2 : [ A>=1 && B>=1 && A>=1+B ], cost: 1

      1: eval1 -> eval3 : [ A>=1 && B>=1 && B>=A ], cost: 1

      2: eval2 -> eval2 : A'=-1+A, [ A>=1 ], cost: 1

      3: eval2 -> eval1 : [ 0>=A ], cost: 1

      4: eval3 -> eval3 : B'=-1+B, [ B>=1 ], cost: 1

      5: eval3 -> eval1 : [ 0>=B ], cost: 1

      6: start -> eval1 : [], cost: 1



Checking for constant complexity:

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

      6: start -> eval1 : [], cost: 1



Simplified all rules, resulting in:

   Start location: start

      0: eval1 -> eval2 : [ B>=1 && A>=1+B ], cost: 1

      1: eval1 -> eval3 : [ A>=1 && B>=1 && B>=A ], cost: 1

      2: eval2 -> eval2 : A'=-1+A, [ A>=1 ], cost: 1

      3: eval2 -> eval1 : [ 0>=A ], cost: 1

      4: eval3 -> eval3 : B'=-1+B, [ B>=1 ], cost: 1

      5: eval3 -> eval1 : [ 0>=B ], cost: 1

      6: start -> eval1 : [], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 1.

   Accelerating the following rules:

      2: eval2 -> eval2 : A'=-1+A, [ A>=1 ], cost: 1



   Accelerated rule 2 with metering function A, yielding the new rule 7.

   Removing the simple loops: 2.



Accelerating simple loops of location 2.

   Accelerating the following rules:

      4: eval3 -> eval3 : B'=-1+B, [ B>=1 ], cost: 1



   Accelerated rule 4 with metering function B, yielding the new rule 8.

   Removing the simple loops: 4.



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

   Start location: start

      0: eval1 -> eval2 : [ B>=1 && A>=1+B ], cost: 1

      1: eval1 -> eval3 : [ A>=1 && B>=1 && B>=A ], cost: 1

      3: eval2 -> eval1 : [ 0>=A ], cost: 1

      7: eval2 -> eval2 : A'=0, [ A>=1 ], cost: A

      5: eval3 -> eval1 : [ 0>=B ], cost: 1

      8: eval3 -> eval3 : B'=0, [ B>=1 ], cost: B

      6: start -> eval1 : [], cost: 1



Chained accelerated rules (with incoming rules):

   Start location: start

      0: eval1 -> eval2 : [ B>=1 && A>=1+B ], cost: 1

      1: eval1 -> eval3 : [ A>=1 && B>=1 && B>=A ], cost: 1

      9: eval1 -> eval2 : A'=0, [ B>=1 && A>=1+B && A>=1 ], cost: 1+A

     10: eval1 -> eval3 : B'=0, [ A>=1 && B>=1 && B>=A ], cost: 1+B

      3: eval2 -> eval1 : [ 0>=A ], cost: 1

      5: eval3 -> eval1 : [ 0>=B ], cost: 1

      6: start -> eval1 : [], cost: 1



Eliminated locations (on tree-shaped paths):

   Start location: start

     11: eval1 -> eval1 : A'=0, [ B>=1 && A>=1+B && A>=1 ], cost: 2+A

     12: eval1 -> eval1 : B'=0, [ A>=1 && B>=1 && B>=A ], cost: 2+B

      6: start -> eval1 : [], cost: 1



Accelerating simple loops of location 0.

   Accelerating the following rules:

     11: eval1 -> eval1 : A'=0, [ B>=1 && A>=1+B && A>=1 ], cost: 2+A

     12: eval1 -> eval1 : B'=0, [ A>=1 && B>=1 && B>=A ], cost: 2+B



   Accelerated rule 11 with NONTERM (after strengthening guard), yielding the new rule 13.

   Accelerated rule 12 with NONTERM (after strengthening guard), yielding the new rule 14.

   Removing the simple loops:.



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

   Start location: start

     11: eval1 -> eval1 : A'=0, [ B>=1 && A>=1+B && A>=1 ], cost: 2+A

     12: eval1 -> eval1 : B'=0, [ A>=1 && B>=1 && B>=A ], cost: 2+B

     13: eval1 -> [6] : [ B>=1 && A>=1+B && A>=1 && 0>=1+B ], cost: NONTERM

     14: eval1 -> [6] : [ A>=1 && B>=1 && B>=A && 0>=A ], cost: NONTERM

      6: start -> eval1 : [], cost: 1



Chained accelerated rules (with incoming rules):

   Start location: start

      6: start -> eval1 : [], cost: 1

     15: start -> eval1 : A'=0, [ B>=1 && A>=1+B && A>=1 ], cost: 3+A

     16: start -> eval1 : B'=0, [ A>=1 && B>=1 && B>=A ], cost: 3+B



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

   Start location: start

     15: start -> eval1 : A'=0, [ B>=1 && A>=1+B && A>=1 ], cost: 3+A

     16: start -> eval1 : B'=0, [ A>=1 && B>=1 && B>=A ], cost: 3+B



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: start

     15: start -> eval1 : A'=0, [ B>=1 && A>=1+B && A>=1 ], cost: 3+A

     16: start -> eval1 : B'=0, [ A>=1 && B>=1 && B>=A ], cost: 3+B



Computing asymptotic complexity for rule 15

   Solved the limit problem by the following transformations:

      Created initial limit problem:

      A-B (+/+!), 3+A (+), B (+/+!) [not solved]



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



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

      resulting limit problem:

      [solved]



   Solution:

      A / n

      B / 1

   Resulting cost 3+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: 3+n

   Rule cost:   3+A

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



WORST_CASE(Omega(n^1),?)


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(4)
BOUNDS(n^1, INF)