/export/starexec/sandbox/solver/bin/starexec_run_complexity /export/starexec/sandbox/benchmark/theBenchmark.koat /export/starexec/sandbox/output/output_files


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MAYBE
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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(1, INF).

(0) CpxIntTrs
(1) Loat Proof [FINISHED, 631 ms]
(2) BOUNDS(1, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
evalfstart(A, B) -> Com_1(evalfentryin(A, B)) :|: TRUE
evalfentryin(A, B) -> Com_1(evalfbb3in(B, A)) :|: A >= 1 && B >= A + 1
evalfbb3in(A, B) -> Com_1(evalfreturnin(A, B)) :|: 0 >= A
evalfbb3in(A, B) -> Com_1(evalfbb4in(A, B)) :|: A >= 1
evalfbb4in(A, B) -> Com_1(evalfbbin(A, B)) :|: 0 >= C + 1
evalfbb4in(A, B) -> Com_1(evalfbbin(A, B)) :|: C >= 1
evalfbb4in(A, B) -> Com_1(evalfreturnin(A, B)) :|: TRUE
evalfbbin(A, B) -> Com_1(evalfbb1in(A, B)) :|: B >= A + 1
evalfbbin(A, B) -> Com_1(evalfbb2in(A, B)) :|: A >= B
evalfbb1in(A, B) -> Com_1(evalfbb3in(A + 1, B)) :|: TRUE
evalfbb2in(A, B) -> Com_1(evalfbb3in(A - B, B)) :|: TRUE
evalfreturnin(A, B) -> Com_1(evalfstop(A, B)) :|: TRUE

The start-symbols are:[evalfstart_2]


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(1) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: evalfstart

      0: evalfstart -> evalfentryin : [], cost: 1

      1: evalfentryin -> evalfbb3in : A'=B, B'=A, [ A>=1 && B>=1+A ], cost: 1

      2: evalfbb3in -> evalfreturnin : [ 0>=A ], cost: 1

      3: evalfbb3in -> evalfbb4in : [ A>=1 ], cost: 1

      4: evalfbb4in -> evalfbbin : [ 0>=1+free ], cost: 1

      5: evalfbb4in -> evalfbbin : [ free_1>=1 ], cost: 1

      6: evalfbb4in -> evalfreturnin : [], cost: 1

      7: evalfbbin -> evalfbb1in : [ B>=1+A ], cost: 1

      8: evalfbbin -> evalfbb2in : [ A>=B ], cost: 1

      9: evalfbb1in -> evalfbb3in : A'=1+A, [], cost: 1

     10: evalfbb2in -> evalfbb3in : A'=A-B, [], cost: 1

     11: evalfreturnin -> evalfstop : [], cost: 1



Checking for constant complexity:

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

      0: evalfstart -> evalfentryin : [], cost: 1



Removed unreachable and leaf rules:

   Start location: evalfstart

      0: evalfstart -> evalfentryin : [], cost: 1

      1: evalfentryin -> evalfbb3in : A'=B, B'=A, [ A>=1 && B>=1+A ], cost: 1

      3: evalfbb3in -> evalfbb4in : [ A>=1 ], cost: 1

      4: evalfbb4in -> evalfbbin : [ 0>=1+free ], cost: 1

      5: evalfbb4in -> evalfbbin : [ free_1>=1 ], cost: 1

      7: evalfbbin -> evalfbb1in : [ B>=1+A ], cost: 1

      8: evalfbbin -> evalfbb2in : [ A>=B ], cost: 1

      9: evalfbb1in -> evalfbb3in : A'=1+A, [], cost: 1

     10: evalfbb2in -> evalfbb3in : A'=A-B, [], cost: 1



Simplified all rules, resulting in:

   Start location: evalfstart

      0: evalfstart -> evalfentryin : [], cost: 1

      1: evalfentryin -> evalfbb3in : A'=B, B'=A, [ A>=1 && B>=1+A ], cost: 1

      3: evalfbb3in -> evalfbb4in : [ A>=1 ], cost: 1

      5: evalfbb4in -> evalfbbin : [], cost: 1

      7: evalfbbin -> evalfbb1in : [ B>=1+A ], cost: 1

      8: evalfbbin -> evalfbb2in : [ A>=B ], cost: 1

      9: evalfbb1in -> evalfbb3in : A'=1+A, [], cost: 1

     10: evalfbb2in -> evalfbb3in : A'=A-B, [], cost: 1



### Simplification by acceleration and chaining ###



Eliminated locations (on linear paths):

   Start location: evalfstart

     12: evalfstart -> evalfbb3in : A'=B, B'=A, [ A>=1 && B>=1+A ], cost: 2

     13: evalfbb3in -> evalfbbin : [ A>=1 ], cost: 2

     14: evalfbbin -> evalfbb3in : A'=1+A, [ B>=1+A ], cost: 2

     15: evalfbbin -> evalfbb3in : A'=A-B, [ A>=B ], cost: 2



Eliminated locations (on tree-shaped paths):

   Start location: evalfstart

     12: evalfstart -> evalfbb3in : A'=B, B'=A, [ A>=1 && B>=1+A ], cost: 2

     16: evalfbb3in -> evalfbb3in : A'=1+A, [ A>=1 && B>=1+A ], cost: 4

     17: evalfbb3in -> evalfbb3in : A'=A-B, [ A>=1 && A>=B ], cost: 4



Accelerating simple loops of location 2.

   Accelerating the following rules:

     16: evalfbb3in -> evalfbb3in : A'=1+A, [ A>=1 && B>=1+A ], cost: 4

     17: evalfbb3in -> evalfbb3in : A'=A-B, [ A>=1 && A>=B ], cost: 4



   Accelerated rule 16 with metering function -A+B, yielding the new rule 18.

   Found no metering function for rule 17.

   Removing the simple loops: 16.



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

   Start location: evalfstart

     12: evalfstart -> evalfbb3in : A'=B, B'=A, [ A>=1 && B>=1+A ], cost: 2

     17: evalfbb3in -> evalfbb3in : A'=A-B, [ A>=1 && A>=B ], cost: 4

     18: evalfbb3in -> evalfbb3in : A'=B, [ A>=1 && B>=1+A ], cost: -4*A+4*B



Chained accelerated rules (with incoming rules):

   Start location: evalfstart

     12: evalfstart -> evalfbb3in : A'=B, B'=A, [ A>=1 && B>=1+A ], cost: 2

     19: evalfstart -> evalfbb3in : A'=-A+B, B'=A, [ A>=1 && B>=1+A && B>=1 ], cost: 6



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

   Start location: evalfstart

     <empty>



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: evalfstart

     <empty>



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

   Complexity:  Constant

   Cpx degree:  0

   Solved cost: 1

   Rule cost:   1

   Rule guard:  []



WORST_CASE(Omega(1),?)


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(2)
BOUNDS(1, INF)