/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^2), O(n^2))
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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^2, nat(2 + Arg_1) + max(2, 4 + Arg_1) + nat(6 + 9 * Arg_1 + 3 * Arg_1^2) + nat(3 + 3 * Arg_1)).

(0) CpxIntTrs
(1) Koat2 Proof [FINISHED, 972 ms]
(2) BOUNDS(1, nat(2 + Arg_1) + max(2, 4 + Arg_1) + nat(6 + 9 * Arg_1 + 3 * Arg_1^2) + nat(3 + 3 * Arg_1))
(3) Loat Proof [FINISHED, 1339 ms]
(4) BOUNDS(n^2, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f0(A, B, C, D) -> Com_1(f4(0, B, C, D)) :|: TRUE
f4(A, B, C, D) -> Com_1(f7(A, B, A + 1, D)) :|: B >= A + 1
f7(A, B, C, D) -> Com_1(f7(A, B, C + 1, 0)) :|: B >= C + 1
f7(A, B, C, D) -> Com_1(f7(A, B - 1, C, E)) :|: B >= C + 1 && 0 >= E + 1
f7(A, B, C, D) -> Com_1(f7(A, B - 1, C, E)) :|: B >= C + 1 && E >= 1
f7(A, B, C, D) -> Com_1(f4(A + 1, B, C, D)) :|: C >= B
f4(A, B, C, D) -> Com_1(f19(A, B, C, D)) :|: A >= B

The start-symbols are:[f0_4]


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(1) Koat2 Proof (FINISHED)
YES( ?, 2*(max([0, (2+Arg_1)*(1+Arg_1)])+max([0, 1+Arg_1]))+max([0, 2+Arg_1])+max([2, 4+Arg_1])+max([0, (2+Arg_1)*(1+Arg_1)])+max([0, 1+Arg_1]) {O(n^2)})



Initial Complexity Problem:

  Start:  f0  

  Program_Vars:  Arg_0, Arg_1, Arg_2, Arg_3  

  Temp_Vars:  E  

  Locations:  f0, f19, f4, f7  

  Transitions:

  0: f0->f4

  6: f4->f19

  1: f4->f7

  5: f7->f4

  2: f7->f7

  3: f7->f7

  4: f7->f7  



Timebounds: 

  Overall timebound: 2*(max([0, (2+Arg_1)*(1+Arg_1)])+max([0, 1+Arg_1]))+max([0, 2+Arg_1])+max([2, 4+Arg_1])+max([0, (2+Arg_1)*(1+Arg_1)])+max([0, 1+Arg_1]) {O(n^2)}

  0: f0->f4: 1 {O(1)}

  1: f4->f7: max([0, 2+Arg_1]) {O(n)}

  6: f4->f19: 1 {O(1)}

  2: f7->f7: max([0, (2+Arg_1)*(1+Arg_1)])+max([0, 1+Arg_1]) {O(n^2)}

  3: f7->f7: max([0, (2+Arg_1)*(1+Arg_1)])+max([0, 1+Arg_1]) {O(n^2)}

  4: f7->f7: max([0, (2+Arg_1)*(1+Arg_1)])+max([0, 1+Arg_1]) {O(n^2)}

  5: f7->f4: max([0, 2+Arg_1]) {O(n)}



Costbounds:

  Overall costbound: 2*(max([0, (2+Arg_1)*(1+Arg_1)])+max([0, 1+Arg_1]))+max([0, 2+Arg_1])+max([2, 4+Arg_1])+max([0, (2+Arg_1)*(1+Arg_1)])+max([0, 1+Arg_1]) {O(n^2)}

  0: f0->f4: 1 {O(1)}

  1: f4->f7: max([0, 2+Arg_1]) {O(n)}

  6: f4->f19: 1 {O(1)}

  2: f7->f7: max([0, (2+Arg_1)*(1+Arg_1)])+max([0, 1+Arg_1]) {O(n^2)}

  3: f7->f7: max([0, (2+Arg_1)*(1+Arg_1)])+max([0, 1+Arg_1]) {O(n^2)}

  4: f7->f7: max([0, (2+Arg_1)*(1+Arg_1)])+max([0, 1+Arg_1]) {O(n^2)}

  5: f7->f4: max([0, 2+Arg_1]) {O(n)}



Sizebounds:

`Lower:

  0: f0->f4, Arg_0: 0 {O(1)}

  0: f0->f4, Arg_1: Arg_1 {O(n)}

  0: f0->f4, Arg_2: Arg_2 {O(n)}

  0: f0->f4, Arg_3: Arg_3 {O(n)}

  1: f4->f7, Arg_0: 0 {O(1)}

  1: f4->f7, Arg_1: Arg_1+-2*(max([0, (2+Arg_1)*(1+Arg_1)])+max([0, 1+Arg_1])) {O(n^2)}

  1: f4->f7, Arg_2: 1 {O(1)}

  6: f4->f19, Arg_0: 0 {O(1)}

  6: f4->f19, Arg_1: min([Arg_1, -(-(Arg_1)+2*(max([0, (2+Arg_1)*(1+Arg_1)])+max([0, 1+Arg_1])))]) {O(n^2)}

  6: f4->f19, Arg_2: min([1, Arg_2]) {O(n)}

  2: f7->f7, Arg_0: 0 {O(1)}

  2: f7->f7, Arg_1: Arg_1+-2*(max([0, (2+Arg_1)*(1+Arg_1)])+max([0, 1+Arg_1])) {O(n^2)}

  2: f7->f7, Arg_2: 1 {O(1)}

  2: f7->f7, Arg_3: 0 {O(1)}

  3: f7->f7, Arg_0: 0 {O(1)}

  3: f7->f7, Arg_1: Arg_1+-2*(max([0, (2+Arg_1)*(1+Arg_1)])+max([0, 1+Arg_1])) {O(n^2)}

  3: f7->f7, Arg_2: 1 {O(1)}

  4: f7->f7, Arg_0: 0 {O(1)}

  4: f7->f7, Arg_1: Arg_1+-2*(max([0, (2+Arg_1)*(1+Arg_1)])+max([0, 1+Arg_1])) {O(n^2)}

  4: f7->f7, Arg_2: 1 {O(1)}

  4: f7->f7, Arg_3: 1 {O(1)}

  5: f7->f4, Arg_0: 0 {O(1)}

  5: f7->f4, Arg_1: Arg_1+-2*(max([0, (2+Arg_1)*(1+Arg_1)])+max([0, 1+Arg_1])) {O(n^2)}

  5: f7->f4, Arg_2: 1 {O(1)}

`Upper:

  0: f0->f4, Arg_0: 0 {O(1)}

  0: f0->f4, Arg_1: Arg_1 {O(n)}

  0: f0->f4, Arg_2: Arg_2 {O(n)}

  0: f0->f4, Arg_3: Arg_3 {O(n)}

  1: f4->f7, Arg_0: max([0, 2+Arg_1]) {O(n)}

  1: f4->f7, Arg_1: Arg_1 {O(n)}

  1: f4->f7, Arg_2: max([1, 3+Arg_1]) {O(n)}

  6: f4->f19, Arg_0: max([0, 2+Arg_1]) {O(n)}

  6: f4->f19, Arg_1: Arg_1 {O(n)}

  6: f4->f19, Arg_2: max([1, max([Arg_2, max([max([0, (2+Arg_1)*(1+Arg_1)])+max([0, 1+Arg_1])+max([1, 3+Arg_1]), 3+Arg_1])])]) {O(n^2)}

  2: f7->f7, Arg_0: max([0, 2+Arg_1]) {O(n)}

  2: f7->f7, Arg_1: Arg_1 {O(n)}

  2: f7->f7, Arg_2: max([0, (2+Arg_1)*(1+Arg_1)])+max([0, 1+Arg_1])+max([1, 3+Arg_1]) {O(n^2)}

  2: f7->f7, Arg_3: 0 {O(1)}

  3: f7->f7, Arg_0: max([0, 2+Arg_1]) {O(n)}

  3: f7->f7, Arg_1: Arg_1 {O(n)}

  3: f7->f7, Arg_2: max([0, (2+Arg_1)*(1+Arg_1)])+max([0, 1+Arg_1])+max([1, 3+Arg_1]) {O(n^2)}

  3: f7->f7, Arg_3: -1 {O(1)}

  4: f7->f7, Arg_0: max([0, 2+Arg_1]) {O(n)}

  4: f7->f7, Arg_1: Arg_1 {O(n)}

  4: f7->f7, Arg_2: max([0, (2+Arg_1)*(1+Arg_1)])+max([0, 1+Arg_1])+max([1, 3+Arg_1]) {O(n^2)}

  5: f7->f4, Arg_0: max([0, 2+Arg_1]) {O(n)}

  5: f7->f4, Arg_1: Arg_1 {O(n)}

  5: f7->f4, Arg_2: max([1, max([max([0, (2+Arg_1)*(1+Arg_1)])+max([0, 1+Arg_1])+max([1, 3+Arg_1]), 3+Arg_1])]) {O(n^2)}


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(2)
BOUNDS(1, nat(2 + Arg_1) + max(2, 4 + Arg_1) + nat(6 + 9 * Arg_1 + 3 * Arg_1^2) + nat(3 + 3 * Arg_1))

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(3) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: f0

      0: f0 -> f4 : A'=0, [], cost: 1

      1: f4 -> f7 : C'=1+A, [ B>=1+A ], cost: 1

      6: f4 -> f19 : [ A>=B ], cost: 1

      2: f7 -> f7 : C'=1+C, D'=0, [ B>=1+C ], cost: 1

      3: f7 -> f7 : B'=-1+B, D'=free, [ B>=1+C && 0>=1+free ], cost: 1

      4: f7 -> f7 : B'=-1+B, D'=free_1, [ B>=1+C && free_1>=1 ], cost: 1

      5: f7 -> f4 : A'=1+A, [ C>=B ], cost: 1



Checking for constant complexity:

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

      0: f0 -> f4 : A'=0, [], cost: 1



Removed unreachable and leaf rules:

   Start location: f0

      0: f0 -> f4 : A'=0, [], cost: 1

      1: f4 -> f7 : C'=1+A, [ B>=1+A ], cost: 1

      2: f7 -> f7 : C'=1+C, D'=0, [ B>=1+C ], cost: 1

      3: f7 -> f7 : B'=-1+B, D'=free, [ B>=1+C && 0>=1+free ], cost: 1

      4: f7 -> f7 : B'=-1+B, D'=free_1, [ B>=1+C && free_1>=1 ], cost: 1

      5: f7 -> f4 : A'=1+A, [ C>=B ], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 2.

   Accelerating the following rules:

      2: f7 -> f7 : C'=1+C, D'=0, [ B>=1+C ], cost: 1

      3: f7 -> f7 : B'=-1+B, D'=free, [ B>=1+C && 0>=1+free ], cost: 1

      4: f7 -> f7 : B'=-1+B, D'=free_1, [ B>=1+C && free_1>=1 ], cost: 1



   Accelerated rule 2 with metering function -C+B, yielding the new rule 7.

   Accelerated rule 3 with metering function -C+B, yielding the new rule 8.

   Accelerated rule 4 with metering function -C+B, yielding the new rule 9.

   Removing the simple loops: 2 3 4.



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

   Start location: f0

      0: f0 -> f4 : A'=0, [], cost: 1

      1: f4 -> f7 : C'=1+A, [ B>=1+A ], cost: 1

      5: f7 -> f4 : A'=1+A, [ C>=B ], cost: 1

      7: f7 -> f7 : C'=B, D'=0, [ B>=1+C ], cost: -C+B

      8: f7 -> f7 : B'=C, D'=free, [ B>=1+C && 0>=1+free ], cost: -C+B

      9: f7 -> f7 : B'=C, D'=free_1, [ B>=1+C && free_1>=1 ], cost: -C+B



Chained accelerated rules (with incoming rules):

   Start location: f0

      0: f0 -> f4 : A'=0, [], cost: 1

      1: f4 -> f7 : C'=1+A, [ B>=1+A ], cost: 1

     10: f4 -> f7 : C'=B, D'=0, [ B>=2+A ], cost: -A+B

     11: f4 -> f7 : B'=1+A, C'=1+A, D'=free, [ B>=2+A && 0>=1+free ], cost: -A+B

     12: f4 -> f7 : B'=1+A, C'=1+A, D'=free_1, [ B>=2+A && free_1>=1 ], cost: -A+B

      5: f7 -> f4 : A'=1+A, [ C>=B ], cost: 1



Eliminated locations (on tree-shaped paths):

   Start location: f0

      0: f0 -> f4 : A'=0, [], cost: 1

     13: f4 -> f4 : A'=1+A, C'=1+A, [ B>=1+A && 1+A>=B ], cost: 2

     14: f4 -> f4 : A'=1+A, C'=B, D'=0, [ B>=2+A ], cost: 1-A+B

     15: f4 -> f4 : A'=1+A, B'=1+A, C'=1+A, D'=free, [ B>=2+A && 0>=1+free ], cost: 1-A+B

     16: f4 -> f4 : A'=1+A, B'=1+A, C'=1+A, D'=free_1, [ B>=2+A && free_1>=1 ], cost: 1-A+B



Accelerating simple loops of location 1.

   Simplified some of the simple loops (and removed duplicate rules).

   Accelerating the following rules:

     13: f4 -> f4 : A'=1+A, C'=1+A, [ 1+A-B==0 ], cost: 2

     14: f4 -> f4 : A'=1+A, C'=B, D'=0, [ B>=2+A ], cost: 1-A+B

     15: f4 -> f4 : A'=1+A, B'=1+A, C'=1+A, D'=free, [ B>=2+A && 0>=1+free ], cost: 1-A+B

     16: f4 -> f4 : A'=1+A, B'=1+A, C'=1+A, D'=free_1, [ B>=2+A && free_1>=1 ], cost: 1-A+B



   Accelerated rule 13 with metering function -A+B, yielding the new rule 17.

   Accelerated rule 14 with metering function -1-A+B, yielding the new rule 18.

   Found no metering function for rule 15.

   Found no metering function for rule 16.

   Removing the simple loops: 13 14.



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

   Start location: f0

      0: f0 -> f4 : A'=0, [], cost: 1

     15: f4 -> f4 : A'=1+A, B'=1+A, C'=1+A, D'=free, [ B>=2+A && 0>=1+free ], cost: 1-A+B

     16: f4 -> f4 : A'=1+A, B'=1+A, C'=1+A, D'=free_1, [ B>=2+A && free_1>=1 ], cost: 1-A+B

     17: f4 -> f4 : A'=B, C'=B, [ 1+A-B==0 ], cost: -2*A+2*B

     18: f4 -> f4 : A'=-1+B, C'=B, D'=0, [ B>=2+A ], cost: -3/2-1/2*(1+A-B)^2-3/2*A+A*(1+A-B)-(1+A-B)*B+3/2*B



Chained accelerated rules (with incoming rules):

   Start location: f0

      0: f0 -> f4 : A'=0, [], cost: 1

     19: f0 -> f4 : A'=1, B'=1, C'=1, D'=free, [ B>=2 && 0>=1+free ], cost: 2+B

     20: f0 -> f4 : A'=1, B'=1, C'=1, D'=free_1, [ B>=2 && free_1>=1 ], cost: 2+B

     21: f0 -> f4 : A'=B, C'=B, [ 1-B==0 ], cost: 1+2*B

     22: f0 -> f4 : A'=-1+B, C'=B, D'=0, [ B>=2 ], cost: -1/2+(-1+B)*B+3/2*B-1/2*(-1+B)^2



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

   Start location: f0

     19: f0 -> f4 : A'=1, B'=1, C'=1, D'=free, [ B>=2 && 0>=1+free ], cost: 2+B

     20: f0 -> f4 : A'=1, B'=1, C'=1, D'=free_1, [ B>=2 && free_1>=1 ], cost: 2+B

     21: f0 -> f4 : A'=B, C'=B, [ 1-B==0 ], cost: 1+2*B

     22: f0 -> f4 : A'=-1+B, C'=B, D'=0, [ B>=2 ], cost: -1/2+(-1+B)*B+3/2*B-1/2*(-1+B)^2



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f0

     19: f0 -> f4 : A'=1, B'=1, C'=1, D'=free, [ B>=2 && 0>=1+free ], cost: 2+B

     20: f0 -> f4 : A'=1, B'=1, C'=1, D'=free_1, [ B>=2 && free_1>=1 ], cost: 2+B

     21: f0 -> f4 : A'=B, C'=B, [ 1-B==0 ], cost: 1+2*B

     22: f0 -> f4 : A'=-1+B, C'=B, D'=0, [ B>=2 ], cost: -1/2+(-1+B)*B+3/2*B-1/2*(-1+B)^2



Computing asymptotic complexity for rule 19

   Solved the limit problem by the following transformations:

      Created initial limit problem:

      -1+B (+/+!), 2+B (+), -free (+/+!) [not solved]



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



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

      resulting limit problem:

      [solved]



   Solution:

      free / -n

      B / n

   Resulting cost 2+n has complexity: Poly(n^1)



Found new complexity Poly(n^1).



Computing asymptotic complexity for rule 22

   Solved the limit problem by the following transformations:

      Created initial limit problem:

      -1+3/2*B+1/2*B^2 (+), -1+B (+/+!) [not solved]



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



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

      resulting limit problem:

      [solved]



   Solution:

      B / n

   Resulting cost -1+3/2*n+1/2*n^2 has complexity: Poly(n^2)



Found new complexity Poly(n^2).



Obtained the following overall complexity (w.r.t. the length of the input n):

   Complexity:  Poly(n^2)

   Cpx degree:  2

   Solved cost: -1+3/2*n+1/2*n^2

   Rule cost:   -1/2+(-1+B)*B+3/2*B-1/2*(-1+B)^2

   Rule guard:  [ B>=2 ]



WORST_CASE(Omega(n^2),?)


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(4)
BOUNDS(n^2, INF)