/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^2, max(1, 3 + 2 * Arg_0) + nat(3 * Arg_0) + nat(Arg_0 * max(3 * Arg_0, -3) + max(3 * Arg_0, -3))).

(0) CpxIntTrs
(1) Koat2 Proof [FINISHED, 228 ms]
(2) BOUNDS(1, max(1, 3 + 2 * Arg_0) + nat(3 * Arg_0) + nat(Arg_0 * max(3 * Arg_0, -3) + max(3 * Arg_0, -3)))
(3) Loat Proof [FINISHED, 416 ms]
(4) BOUNDS(n^2, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
eval1(A, B) -> Com_1(eval2(A, 0)) :|: A >= 0
eval2(A, B) -> Com_1(eval2(A, B + 1)) :|: A >= 1 + B
eval2(A, B) -> Com_1(eval1(A - 1, B)) :|: B >= A
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])+max([0, 3*Arg_0])+max([0, (1+Arg_0)*max([-3, 3*Arg_0])]) {O(n^2)})



Initial Complexity Problem:

  Start:  start  

  Program_Vars:  Arg_0, Arg_1  

  Temp_Vars:    

  Locations:  eval1, eval2, start  

  Transitions:

  eval1(Arg_0,Arg_1) -> eval2(Arg_0,0):|:0 <= Arg_0

  eval2(Arg_0,Arg_1) -> eval1(Arg_0-1,Arg_1):|:Arg_1 <= Arg_0 && 0 <= Arg_1 && 0 <= Arg_0+Arg_1 && 0 <= Arg_0 && Arg_0 <= Arg_1

  eval2(Arg_0,Arg_1) -> eval2(Arg_0,Arg_1+1):|:Arg_1 <= Arg_0 && 0 <= Arg_1 && 0 <= Arg_0+Arg_1 && 0 <= Arg_0 && 1+Arg_1 <= Arg_0

  start(Arg_0,Arg_1) -> eval1(Arg_0,Arg_1):|:  



Timebounds: 

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

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

  1: eval2->eval2: max([0, 3*Arg_0])+max([0, (1+Arg_0)*max([-3, 3*Arg_0])]) {O(n^2)}

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

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



Costbounds:

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

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

  1: eval2->eval2: max([0, 3*Arg_0])+max([0, (1+Arg_0)*max([-3, 3*Arg_0])]) {O(n^2)}

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

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



Sizebounds:

`Lower:

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

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

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

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

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

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

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

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

`Upper:

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

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

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

  1: eval2->eval2, Arg_1: max([0, 3*Arg_0])+max([0, (1+Arg_0)*max([-3, 3*Arg_0])]) {O(n^2)}

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

  2: eval2->eval1, Arg_1: max([0, max([0, 3*Arg_0])+max([0, (1+Arg_0)*max([-3, 3*Arg_0])])]) {O(n^2)}

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

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


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

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(3) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: start

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

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

      2: eval2 -> eval1 : A'=-1+A, [ B>=A ], cost: 1

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



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 1.

   Accelerating the following rules:

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



   Accelerated rule 1 with metering function A-B, yielding the new rule 4.

   Removing the simple loops: 1.



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

   Start location: start

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

      2: eval2 -> eval1 : A'=-1+A, [ B>=A ], cost: 1

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

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



Chained accelerated rules (with incoming rules):

   Start location: start

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

      5: eval1 -> eval2 : B'=A, [ A>=1 ], cost: 1+A

      2: eval2 -> eval1 : A'=-1+A, [ B>=A ], cost: 1

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



Eliminated locations (on tree-shaped paths):

   Start location: start

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

      7: eval1 -> eval1 : A'=-1+A, B'=A, [ A>=1 ], cost: 2+A

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



Accelerating simple loops of location 0.

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

   Accelerating the following rules:

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

      7: eval1 -> eval1 : A'=-1+A, B'=A, [ A>=1 ], cost: 2+A



   Accelerated rule 6 with metering function 1+A, yielding the new rule 8.

   Accelerated rule 7 with metering function A, yielding the new rule 9.

   Removing the simple loops: 6 7.



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

   Start location: start

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

      9: eval1 -> eval1 : A'=0, B'=1, [ A>=1 ], cost: 5/2*A+1/2*A^2

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



Chained accelerated rules (with incoming rules):

   Start location: start

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

     10: start -> eval1 : A'=-1, B'=0, [ -A==0 ], cost: 3+2*A

     11: start -> eval1 : A'=0, B'=1, [ A>=1 ], cost: 1+5/2*A+1/2*A^2



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

   Start location: start

     10: start -> eval1 : A'=-1, B'=0, [ -A==0 ], cost: 3+2*A

     11: start -> eval1 : A'=0, B'=1, [ A>=1 ], cost: 1+5/2*A+1/2*A^2



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: start

     10: start -> eval1 : A'=-1, B'=0, [ -A==0 ], cost: 3+2*A

     11: start -> eval1 : A'=0, B'=1, [ A>=1 ], cost: 1+5/2*A+1/2*A^2



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+5/2*A+1/2*A^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 1+1/2*n^2+5/2*n 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+1/2*n^2+5/2*n

   Rule cost:   1+5/2*A+1/2*A^2

   Rule guard:  [ A>=1 ]



WORST_CASE(Omega(n^2),?)


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(4)
BOUNDS(n^2, INF)