Deriv Compl Full Rewri 33144 pair #381922413

details

property	value
status	complete
benchmark	3-matchbox.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n078.star.cs.uiowa.edu
space	Secret_06_SRS

run statistics

property	value
solver	tct 2018-07-13
configuration	tct_dc
runtime (wallclock)	294.033385038 seconds
cpu usage	764.737379707
max memory	4.5520896E8

stage attributes

key	value
output-size	11754
starexec-result	WORST_CASE(?,O(n^5))

output

/export/starexec/sandbox/solver/bin/starexec_run_tct_dc /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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WORST_CASE(?,O(n^5))
* Step 1: DecomposeCP WORST_CASE(?,O(n^5))
    + Considered Problem:
        - Strict TRS:
            a(b(c(x1))) -> c(b(a(x1)))
            b(a(a(a(x1)))) -> a(a(a(b(x1))))
            c(b(a(a(x1)))) -> a(a(b(c(x1))))
            c(c(b(b(x1)))) -> b(b(c(c(x1))))
        - Signature:
            {a/1,b/1,c/1} / {}
        - Obligation:
             derivational complexity wrt. signature {a,b,c}
    + Applied Processor:
        DecomposeCP {onSelectionCP_ = any strict-rules, withBoundCP_ = RelativeMul, withCP_ = NaturalMI {miDimension = 3, miDegree = 3, miKind = Algebraic, uargs = NoUArgs, urules = NoURules, selector = Nothing}}
    + Details:
        We first use the processor NaturalMI {miDimension = 3, miDegree = 3, miKind = Algebraic, uargs = NoUArgs, urules = NoURules, selector = Nothing} to orient following rules strictly:
          
          b(a(a(a(x1)))) -> a(a(a(b(x1))))
        The Processor induces the complexity certificate TIME (?,O(n^3))
        SPACE(?,?)
        
        Observe that Problem (R) is non-size-increasing. Once the complexity of (R) has been assessed, it suffices
        to consider only rules whose complexity has not been estimated in (R) resulting in the following Problem
        (S). Overall the certificate is obtained by multiplication.
        Problem (S)
          - Strict TRS:
              a(b(c(x1))) -> c(b(a(x1)))
              c(b(a(a(x1)))) -> a(a(b(c(x1))))
              c(c(b(b(x1)))) -> b(b(c(c(x1))))
          - Signature:
              {a/1,b/1,c/1} / {}
          - Obligation:
               derivational complexity wrt. signature {a,b,c}
** Step 1.a:1: NaturalMI WORST_CASE(?,O(n^3))
    + Considered Problem:
        - Strict TRS:
            a(b(c(x1))) -> c(b(a(x1)))
            b(a(a(a(x1)))) -> a(a(a(b(x1))))
            c(b(a(a(x1)))) -> a(a(b(c(x1))))
            c(c(b(b(x1)))) -> b(b(c(c(x1))))
        - Signature:
            {a/1,b/1,c/1} / {}
        - Obligation:
             derivational complexity wrt. signature {a,b,c}
    + Applied Processor:
        NaturalMI {miDimension = 3, miDegree = 3, miKind = Algebraic, uargs = NoUArgs, urules = NoURules, selector = Just first alternative for decompose on any strict-rules}
    + Details:
        We apply a matrix interpretation of kind triangular matrix interpretation:
        
        Following symbols are considered usable:
          all
        TcT has computed the following interpretation:
          p(a) = [1 0 1]      [0]
                 [0 1 0] x1 + [1]
                 [0 0 1]      [0]
          p(b) = [1 1 0]      [0]
                 [0 1 0] x1 + [0]
                 [0 0 1]      [0]
          p(c) = [1 0 0]      [0]
                 [0 1 0] x1 + [0]
                 [0 0 1]      [1]
        
        Following rules are strictly oriented:
        b(a(a(a(x1)))) = [1 1 3]      [3]
                         [0 1 0] x1 + [3]
                         [0 0 1]      [0]
                       > [1 1 3]      [0]
                         [0 1 0] x1 + [3]
                         [0 0 1]      [0]
                       = a(a(a(b(x1))))  
        
        
        Following rules are (at-least) weakly oriented:
           a(b(c(x1))) =  [1 1 1]      [1]
                          [0 1 0] x1 + [1]
                          [0 0 1]      [1]
                       >= [1 1 1]      [1]
                          [0 1 0] x1 + [1]
                          [0 0 1]      [1]
                       =  c(b(a(x1)))     
        
        c(b(a(a(x1)))) =  [1 1 2]      [2]
                          [0 1 0] x1 + [2]
                          [0 0 1]      [1]
                       >= [1 1 2]      [2]
                          [0 1 0] x1 + [2]
                          [0 0 1]      [1]
                       =  a(a(b(c(x1))))  
        
        c(c(b(b(x1)))) =  [1 2 0]      [0]
                          [0 1 0] x1 + [0]
                          [0 0 1]      [2]
                       >= [1 2 0]      [0]
                          [0 1 0] x1 + [0]

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