/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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 + Arg_0 + -1 * Arg_1) + nat(Arg_0 + -1 * Arg_0 * Arg_1 + -1 * Arg_2 * Arg_0 + Arg_0^2 + Arg_2 * Arg_1 + -1 * Arg_2) + nat(Arg_0 + -1 * Arg_2) + nat(Arg_0 + -1 * Arg_1) + nat(1 + Arg_0 + -1 * Arg_1)).

(0) CpxIntTrs
(1) Loat Proof [FINISHED, 609 ms]
(2) BOUNDS(n^1, INF)


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

The start-symbols are:[start_3]


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(1) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: start

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

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

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

      3: eval2 -> eval1 : A'=-1+A, [ C>=A ], cost: 1

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



### Simplification by acceleration and chaining ###



Eliminated locations (on tree-shaped paths):

   Start location: start

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

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

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

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



Accelerating simple loops of location 0.

   Accelerating the following rules:

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

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

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



   Accelerated rule 5 with metering function A-B, yielding the new rule 8.

   Accelerated rule 6 with metering function -C+A, yielding the new rule 9.

   Accelerated rule 7 with metering function A-B, yielding the new rule 10.

   Removing the simple loops: 5 6 7.



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

   Start location: start

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

      9: eval1 -> eval1 : C'=A, [ A>=1+B && A>=1+C ], cost: -2*C+2*A

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

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



Chained accelerated rules (with incoming rules):

   Start location: start

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

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

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

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



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

   Start location: start

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

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

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



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: start

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

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

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



Computing asymptotic complexity for rule 11

   Solved the limit problem by the following transformations:

      Created initial limit problem:

      -C+A (+/+!), A-B (+/+!), 1+2*A-2*B (+) [not solved]



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



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

      resulting limit problem:

      [solved]



   Solution:

      C / 0

      A / 1

      B / -n

   Resulting cost 3+2*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+2*n

   Rule cost:   1+2*A-2*B

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



WORST_CASE(Omega(n^1),?)


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(2)
BOUNDS(n^1, INF)