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Hull-Dobell theorem for linear congruential algorithm


Fractional iteration of the Newton-approximation-formula: how to resolve one unknown parameter?Quick sort algorithm average case complexity analysisIn-place inverse of DFT?Generalized Josephus problemFinding the limit of the recursive sequence $r_n+1 = sqrt2 + r_n$What is the link between the quotient and the Bézout coefficients in the Extended Euclidean Algorithm?Find a linear-time algorithm for finding if element occurs n/4 timesSources and sinks for parabolic PDE algorithmWhen the integers got upset.Prob. 24, Chap. 5 in Baby Rudin: For $alpha>1$, let $f(x) = (x+alpha/x)/2$, $g(x) = (alpha+x)/(1+x)$ have $sqrtalpha$ as their only fixed point













0












$begingroup$


Hull-Dobell theorem says for $x_n+1= ax_n+c , mod, m$ if c ≠ 0, correctly chosen parameters allow a period equal to m, for all seed values. This will occur if and only if



  1. c and m are relatively prime


  2. if q is prime and q | p then q | a-1


  3. if 4 | m then 4 | a-1


if m is prime and a < p then the sequence must have the full period but take
a=4, c= 5, p=11 and $x_0 = 0$. Then the sequence will
be 5,3,6,7,0,5,3,6,7,0,5. (the period is 5)
but we don't have the full period, why? what is missing here?.










share|cite|improve this question











$endgroup$
















    0












    $begingroup$


    Hull-Dobell theorem says for $x_n+1= ax_n+c , mod, m$ if c ≠ 0, correctly chosen parameters allow a period equal to m, for all seed values. This will occur if and only if



    1. c and m are relatively prime


    2. if q is prime and q | p then q | a-1


    3. if 4 | m then 4 | a-1


    if m is prime and a < p then the sequence must have the full period but take
    a=4, c= 5, p=11 and $x_0 = 0$. Then the sequence will
    be 5,3,6,7,0,5,3,6,7,0,5. (the period is 5)
    but we don't have the full period, why? what is missing here?.










    share|cite|improve this question











    $endgroup$














      0












      0








      0





      $begingroup$


      Hull-Dobell theorem says for $x_n+1= ax_n+c , mod, m$ if c ≠ 0, correctly chosen parameters allow a period equal to m, for all seed values. This will occur if and only if



      1. c and m are relatively prime


      2. if q is prime and q | p then q | a-1


      3. if 4 | m then 4 | a-1


      if m is prime and a < p then the sequence must have the full period but take
      a=4, c= 5, p=11 and $x_0 = 0$. Then the sequence will
      be 5,3,6,7,0,5,3,6,7,0,5. (the period is 5)
      but we don't have the full period, why? what is missing here?.










      share|cite|improve this question











      $endgroup$




      Hull-Dobell theorem says for $x_n+1= ax_n+c , mod, m$ if c ≠ 0, correctly chosen parameters allow a period equal to m, for all seed values. This will occur if and only if



      1. c and m are relatively prime


      2. if q is prime and q | p then q | a-1


      3. if 4 | m then 4 | a-1


      if m is prime and a < p then the sequence must have the full period but take
      a=4, c= 5, p=11 and $x_0 = 0$. Then the sequence will
      be 5,3,6,7,0,5,3,6,7,0,5. (the period is 5)
      but we don't have the full period, why? what is missing here?.







      recursive-algorithms






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 29 at 22:14









      Bernard

      124k741116




      124k741116










      asked Mar 29 at 21:18









      Richard KimRichard Kim

      11




      11




















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