/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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 + 3 * Arg_0) + nat(1 + Arg_0) + nat(Arg_1)).

(0) CpxIntTrs
(1) Koat2 Proof [FINISHED, 126 ms]
(2) BOUNDS(1, max(1, 1 + 3 * Arg_0) + nat(1 + Arg_0) + nat(Arg_1))
(3) Loat Proof [FINISHED, 407 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
eval2(A, B) -> Com_1(eval2(A, B - 1)) :|: A >= 1 && B >= 1
eval2(A, B) -> Com_1(eval1(A - 1, B)) :|: A >= 1 && 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+max([0, 3*Arg_0])+max([0, 1+Arg_0])+max([0, Arg_1]) {O(n)})



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,Arg_1):|:1 <= Arg_0

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

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

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



Timebounds: 

  Overall timebound: 1+max([0, 3*Arg_0])+max([0, 1+Arg_0])+max([0, Arg_1]) {O(n)}

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

  1: eval2->eval2: max([0, Arg_1]) {O(n)}

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

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



Costbounds:

  Overall costbound: 1+max([0, 3*Arg_0])+max([0, 1+Arg_0])+max([0, Arg_1]) {O(n)}

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

  1: eval2->eval2: max([0, Arg_1]) {O(n)}

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

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



Sizebounds:

`Lower:

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

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

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

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

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

  2: eval2->eval1, Arg_1: min([0, Arg_1]) {O(n)}

  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: max([0, Arg_1]) {O(n)}

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

  1: eval2->eval2, Arg_1: max([0, Arg_1]) {O(n)}

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

  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)}


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

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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 ], cost: 1

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

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



   Accelerated rule 1 with metering function 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 : [ A>=1 ], cost: 1

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

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

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



Chained accelerated rules (with incoming rules):

   Start location: start

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

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

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

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



Eliminated locations (on tree-shaped paths):

   Start location: start

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

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

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



Accelerating simple loops of location 0.

   Accelerating the following rules:

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

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



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

   Found no metering function for rule 7.

   Removing the simple loops: 6.



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

   Start location: start

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

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

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



Chained accelerated rules (with incoming rules):

   Start location: start

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

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

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



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

   Start location: start

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

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



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: start

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

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



Computing asymptotic complexity for rule 9

   Solved the limit problem by the following transformations:

      Created initial limit problem:

      3+B (+), A (+/+!), 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 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+B

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



WORST_CASE(Omega(n^1),?)


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(4)
BOUNDS(n^1, INF)