/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 2 + Arg_0 + -1 * Arg_1) + nat(2 + 2 * Arg_2 + -2 * Arg_3) + nat(1 + Arg_0 + -1 * Arg_1)).

(0) CpxIntTrs
(1) Koat2 Proof [FINISHED, 853 ms]
(2) BOUNDS(1, max(1, 2 + Arg_0 + -1 * Arg_1) + nat(2 + 2 * Arg_2 + -2 * Arg_3) + nat(1 + Arg_0 + -1 * Arg_1))
(3) Loat Proof [FINISHED, 922 ms]
(4) BOUNDS(n^1, INF)


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

The start-symbols are:[start_5]


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(1) Koat2 Proof (FINISHED)
YES( ?, 1+max([0, 1+Arg_0-Arg_1])+max([0, 1+Arg_2-Arg_3])+max([0, 1+Arg_2-Arg_3])+max([0, 1+Arg_0-Arg_1]) {O(n)})



Initial Complexity Problem:

  Start:  start  

  Program_Vars:  Arg_0, Arg_1, Arg_2, Arg_3, Arg_4  

  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+max([0, 1+Arg_0-Arg_1])+max([0, 1+Arg_2-Arg_3])+max([0, 1+Arg_2-Arg_3])+max([0, 1+Arg_0-Arg_1]) {O(n)}

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

  1: eval->eval: max([0, 1+Arg_2-Arg_3]) {O(n)}

  2: eval->eval: max([0, 1+Arg_2-Arg_3]) {O(n)}

  3: eval->eval: max([0, 1+Arg_0-Arg_1]) {O(n)}

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



Costbounds:

  Overall costbound: 1+max([0, 1+Arg_0-Arg_1])+max([0, 1+Arg_2-Arg_3])+max([0, 1+Arg_2-Arg_3])+max([0, 1+Arg_0-Arg_1]) {O(n)}

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

  1: eval->eval: max([0, 1+Arg_2-Arg_3]) {O(n)}

  2: eval->eval: max([0, 1+Arg_2-Arg_3]) {O(n)}

  3: eval->eval: max([0, 1+Arg_0-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_1 {O(n)}

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

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

  0: eval->eval, Arg_4: Arg_4 {O(n)}

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

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

  1: eval->eval, Arg_2: Arg_2 {O(n)}

  1: eval->eval, Arg_3: Arg_3 {O(n)}

  1: eval->eval, Arg_4: Arg_4 {O(n)}

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

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

  2: eval->eval, Arg_2: Arg_2 {O(n)}

  2: eval->eval, Arg_3: Arg_3 {O(n)}

  2: eval->eval, Arg_4: Arg_4 {O(n)}

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

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

  3: eval->eval, Arg_2: Arg_2 {O(n)}

  3: eval->eval, Arg_3: Arg_3 {O(n)}

  3: eval->eval, Arg_4: Arg_4 {O(n)}

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

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

  4: start->eval, Arg_2: Arg_2 {O(n)}

  4: start->eval, Arg_3: Arg_3 {O(n)}

  4: start->eval, Arg_4: Arg_4 {O(n)}

`Upper:

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

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

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

  0: eval->eval, Arg_3: Arg_3+max([0, 1+Arg_2-Arg_3]) {O(n)}

  0: eval->eval, Arg_4: Arg_4+max([0, 1+Arg_2-Arg_3])+max([0, 1+Arg_0-Arg_1]) {O(n)}

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

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

  1: eval->eval, Arg_2: Arg_2 {O(n)}

  1: eval->eval, Arg_3: Arg_3+max([0, 1+Arg_2-Arg_3]) {O(n)}

  1: eval->eval, Arg_4: Arg_4+max([0, 1+Arg_2-Arg_3])+max([0, 1+Arg_0-Arg_1]) {O(n)}

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

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

  2: eval->eval, Arg_2: Arg_2 {O(n)}

  2: eval->eval, Arg_3: max([Arg_3, Arg_3+max([0, 1+Arg_2-Arg_3])])+max([0, 1+Arg_2-Arg_3]) {O(n)}

  2: eval->eval, Arg_4: max([Arg_4, Arg_4+max([0, 1+Arg_2-Arg_3])+max([0, 1+Arg_0-Arg_1])])+max([0, 1+Arg_2-Arg_3]) {O(n)}

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

  3: eval->eval, Arg_1: max([Arg_1, Arg_1+max([0, 1+Arg_0-Arg_1])])+max([0, 1+Arg_0-Arg_1]) {O(n)}

  3: eval->eval, Arg_2: Arg_2 {O(n)}

  3: eval->eval, Arg_3: max([Arg_3, Arg_3+max([0, 1+Arg_2-Arg_3])]) {O(n)}

  3: eval->eval, Arg_4: max([Arg_4, Arg_4+max([0, 1+Arg_2-Arg_3])+max([0, 1+Arg_0-Arg_1])])+max([0, 1+Arg_0-Arg_1]) {O(n)}

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

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

  4: start->eval, Arg_2: Arg_2 {O(n)}

  4: start->eval, Arg_3: Arg_3 {O(n)}

  4: start->eval, Arg_4: Arg_4 {O(n)}


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(2)
BOUNDS(1, max(1, 2 + Arg_0 + -1 * Arg_1) + nat(2 + 2 * Arg_2 + -2 * Arg_3) + nat(1 + Arg_0 + -1 * 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'=1+B, E'=1+E, [ A>=1+B && C>=1+D ], cost: 1

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

      2: eval -> eval : D'=1+D, E'=1+E, [ B>=A && C>=1+D ], cost: 1

      3: eval -> eval : B'=1+B, E'=1+E, [ A>=1+B && D>=C ], 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



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 0.

   Accelerating the following rules:

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

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

      2: eval -> eval : D'=1+D, E'=1+E, [ B>=A && C>=1+D ], cost: 1

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



   Accelerated rule 0 with metering function A-B, yielding the new rule 5.

   Accelerated rule 1 with metering function C-D, yielding the new rule 6.

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

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

   Removing the simple loops: 0 1 2 3.



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

   Start location: start

      5: eval -> eval : B'=A, E'=A+E-B, [ A>=1+B && C>=1+D ], cost: A-B

      6: eval -> eval : D'=C, E'=C-D+E, [ A>=1+B && C>=1+D ], cost: C-D

      7: eval -> eval : D'=C, E'=C-D+E, [ B>=A && C>=1+D ], cost: C-D

      8: eval -> eval : B'=A, E'=A+E-B, [ A>=1+B && D>=C ], cost: A-B

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



Chained accelerated rules (with incoming rules):

   Start location: start

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

      9: start -> eval : B'=A, E'=A+E-B, [ A>=1+B && C>=1+D ], cost: 1+A-B

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

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

     12: start -> eval : B'=A, E'=A+E-B, [ A>=1+B && D>=C ], cost: 1+A-B



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

   Start location: start

      9: start -> eval : B'=A, E'=A+E-B, [ A>=1+B && C>=1+D ], cost: 1+A-B

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

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

     12: start -> eval : B'=A, E'=A+E-B, [ A>=1+B && D>=C ], cost: 1+A-B



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: start

      9: start -> eval : B'=A, E'=A+E-B, [ A>=1+B && C>=1+D ], cost: 1+A-B

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

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

     12: start -> eval : B'=A, E'=A+E-B, [ A>=1+B && D>=C ], cost: 1+A-B



Computing asymptotic complexity for rule 9

   Solved the limit problem by the following transformations:

      Created initial limit problem:

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



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



      applying transformation rule (C) using substitution {C==1,D==0,A==0,B==-n}

      resulting limit problem:

      [solved]



   Solution:

      C / 1

      D / 0

      A / 0

      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+A-B

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



WORST_CASE(Omega(n^1),?)


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(4)
BOUNDS(n^1, INF)