Name for this Algorithm












2












$begingroup$


I've managed to prove a bunch of properties about this algorithm that I came up with. I'm now curious to know it's name to see what other people have done.



Given a number in base b $$N_0 = b N_X + N_L, qquad 0 leq N_L leq b-1 $$
And a divisor also in base b $$D = b D_X + D_L, qquad 0 leq D_L leq b-1 $$



($N_L$ and $D_L$ are the last digits of N and D)



Then compute $$N_{i+1} = left| begin{matrix} N_X & N_L \ D_X & D_L end{matrix} right|$$



I've shown this algorithm to a few lecturers at my uni and they don't know a name for it but suspect it is some sort of variation on the Euclidean Algorithm.



It can be used for divisibility tests (especially good in binary, if $D_L = 1$, or if $D$ is prime )



E.g. Does $9734$ divide by $31$



$N_X = 973$, $N_L = 4$



$D_X=3$, $D_L = 1$



$$N_0 = 9734$$



$$N_1 = left| begin{matrix} 973 & 4 \ 3 & 1 end{matrix} right| = 961$$
$$N_2 = left| begin{matrix} 96 & 1 \ 3 & 1 end{matrix} right| = 93$$
$$N_3 = left| begin{matrix} 9 & 3 \ 3 & 1 end{matrix} right| = 0$$



If the algorithm hits $0$, then $D|N$ (Under certain conditions, counter example N=48, D=36)



Also, if $D_L = 1$ the quotient appears



$$N_1 = left| begin{matrix} 973 & color{red}{4} \ 3 & 1 end{matrix} right| = 961$$
$$N_2 = left| begin{matrix} 96 & color{red}{1} \ 3 & 1 end{matrix} right| = 93$$
$$N_3 = left| begin{matrix} 9 & color{red}{3} \ 3 & 1 end{matrix} right| = 0$$



$9734 = color{red}{314} times 31$



Any help is greatly appreciated! Thanks, Ben






algorithms divisibility euclidean-algorithm







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    I've managed to prove a bunch of properties about this algorithm that I came up with. I'm now curious to know it's name to see what other people have done.



    Given a number in base b $$N_0 = b N_X + N_L, qquad 0 leq N_L leq b-1 $$
    And a divisor also in base b $$D = b D_X + D_L, qquad 0 leq D_L leq b-1 $$



    ($N_L$ and $D_L$ are the last digits of N and D)



    Then compute $$N_{i+1} = left| begin{matrix} N_X & N_L \ D_X & D_L end{matrix} right|$$



    I've shown this algorithm to a few lecturers at my uni and they don't know a name for it but suspect it is some sort of variation on the Euclidean Algorithm.



    It can be used for divisibility tests (especially good in binary, if $D_L = 1$, or if $D$ is prime )



    E.g. Does $9734$ divide by $31$



    $N_X = 973$, $N_L = 4$



    $D_X=3$, $D_L = 1$



    $$N_0 = 9734$$



    $$N_1 = left| begin{matrix} 973 & 4 \ 3 & 1 end{matrix} right| = 961$$
    $$N_2 = left| begin{matrix} 96 & 1 \ 3 & 1 end{matrix} right| = 93$$
    $$N_3 = left| begin{matrix} 9 & 3 \ 3 & 1 end{matrix} right| = 0$$



    If the algorithm hits $0$, then $D|N$ (Under certain conditions, counter example N=48, D=36)



    Also, if $D_L = 1$ the quotient appears



    $$N_1 = left| begin{matrix} 973 & color{red}{4} \ 3 & 1 end{matrix} right| = 961$$
    $$N_2 = left| begin{matrix} 96 & color{red}{1} \ 3 & 1 end{matrix} right| = 93$$
    $$N_3 = left| begin{matrix} 9 & color{red}{3} \ 3 & 1 end{matrix} right| = 0$$



    $9734 = color{red}{314} times 31$



    Any help is greatly appreciated! Thanks, Ben






    algorithms divisibility euclidean-algorithm







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      0



      $begingroup$


      I've managed to prove a bunch of properties about this algorithm that I came up with. I'm now curious to know it's name to see what other people have done.



      Given a number in base b $$N_0 = b N_X + N_L, qquad 0 leq N_L leq b-1 $$
      And a divisor also in base b $$D = b D_X + D_L, qquad 0 leq D_L leq b-1 $$



      ($N_L$ and $D_L$ are the last digits of N and D)



      Then compute $$N_{i+1} = left| begin{matrix} N_X & N_L \ D_X & D_L end{matrix} right|$$



      I've shown this algorithm to a few lecturers at my uni and they don't know a name for it but suspect it is some sort of variation on the Euclidean Algorithm.



      It can be used for divisibility tests (especially good in binary, if $D_L = 1$, or if $D$ is prime )



      E.g. Does $9734$ divide by $31$



      $N_X = 973$, $N_L = 4$



      $D_X=3$, $D_L = 1$



      $$N_0 = 9734$$



      $$N_1 = left| begin{matrix} 973 & 4 \ 3 & 1 end{matrix} right| = 961$$
      $$N_2 = left| begin{matrix} 96 & 1 \ 3 & 1 end{matrix} right| = 93$$
      $$N_3 = left| begin{matrix} 9 & 3 \ 3 & 1 end{matrix} right| = 0$$



      If the algorithm hits $0$, then $D|N$ (Under certain conditions, counter example N=48, D=36)



      Also, if $D_L = 1$ the quotient appears



      $$N_1 = left| begin{matrix} 973 & color{red}{4} \ 3 & 1 end{matrix} right| = 961$$
      $$N_2 = left| begin{matrix} 96 & color{red}{1} \ 3 & 1 end{matrix} right| = 93$$
      $$N_3 = left| begin{matrix} 9 & color{red}{3} \ 3 & 1 end{matrix} right| = 0$$



      $9734 = color{red}{314} times 31$



      Any help is greatly appreciated! Thanks, Ben






      algorithms divisibility euclidean-algorithm







      share|cite|improve this question









      $endgroup$




      I've managed to prove a bunch of properties about this algorithm that I came up with. I'm now curious to know it's name to see what other people have done.



      Given a number in base b $$N_0 = b N_X + N_L, qquad 0 leq N_L leq b-1 $$
      And a divisor also in base b $$D = b D_X + D_L, qquad 0 leq D_L leq b-1 $$



      ($N_L$ and $D_L$ are the last digits of N and D)



      Then compute $$N_{i+1} = left| begin{matrix} N_X & N_L \ D_X & D_L end{matrix} right|$$



      I've shown this algorithm to a few lecturers at my uni and they don't know a name for it but suspect it is some sort of variation on the Euclidean Algorithm.



      It can be used for divisibility tests (especially good in binary, if $D_L = 1$, or if $D$ is prime )



      E.g. Does $9734$ divide by $31$



      $N_X = 973$, $N_L = 4$



      $D_X=3$, $D_L = 1$



      $$N_0 = 9734$$



      $$N_1 = left| begin{matrix} 973 & 4 \ 3 & 1 end{matrix} right| = 961$$
      $$N_2 = left| begin{matrix} 96 & 1 \ 3 & 1 end{matrix} right| = 93$$
      $$N_3 = left| begin{matrix} 9 & 3 \ 3 & 1 end{matrix} right| = 0$$



      If the algorithm hits $0$, then $D|N$ (Under certain conditions, counter example N=48, D=36)



      Also, if $D_L = 1$ the quotient appears



      $$N_1 = left| begin{matrix} 973 & color{red}{4} \ 3 & 1 end{matrix} right| = 961$$
      $$N_2 = left| begin{matrix} 96 & color{red}{1} \ 3 & 1 end{matrix} right| = 93$$
      $$N_3 = left| begin{matrix} 9 & color{red}{3} \ 3 & 1 end{matrix} right| = 0$$



      $9734 = color{red}{314} times 31$



      Any help is greatly appreciated! Thanks, Ben






      algorithms divisibility euclidean-algorithm




      algorithms divisibility euclidean-algorithm






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 6 at 16:20









      Ben CrossleyBen Crossley

      840318




      840318






















          1 Answer
          1






          active

          oldest

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          0












          $begingroup$

          It's the special case $m = 10c!+!d, n = 10a!+!b $ of the following



          $quad bbox[5px,border:1px solid red]{(a,n) = 1,Rightarrow, (m,n) = (am!-!cn,n)} $ by $,(m,n)=(am,n) = (am!-!cn,n)$



          So $ (a,b)=1,Rightarrow, (10c!+!d,10a!+!b) = (ad!-!bc,10a!+!b), $ exactly as you observed.



          Remark $ $ Various well-known divisibity test rules are essentially special cases of this, e.g. the rule for divisibility by $7$ follows from the rule for $21$ given below (we negated $,ad! -! bc = 2d!-!c,$ below)



          $qquadqquad qquad , (10c!+!d,10(2)!+!1) = (c!-!2d, 10(2)!+!1)$



          ${rm similar to your} (10c!+!d,10(3)!+!1) = (c!-!3d, 10(3)!+!1) $ rule for $,31$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Do you know if there is a name for it? Or how I might find other work on it? I've used it to do some cool things with Gaussian integers and I highly doubt I'm the first person to play with this idea!
            $endgroup$
            – Ben Crossley
            Jan 7 at 23:28










          • $begingroup$
            @Ben I'm not aware of any common name for this. Things like this are folklore, e.g. occurring in various guises in divisibility tests. In number rings analogous results are often special cases of Hermite normal form algorithms for modules and related results, e.g. see here and here. If you give more details about your Gaussian integer results then I may be able to be more specific.
            $endgroup$
            – Bill Dubuque
            Jan 8 at 0:45










          • $begingroup$
            @Ben By the way, another angle is to view $,m = f(10),$ as a polynomial of degree $k$ in the radix. Then $$bmod 10a!+!b!:, 10equiv -b/a,Rightarrow,(f(10),10a!+!b) = (a^kf(-b/a),10a!+!b)qquad$$ and the algorithm can be viwed as a recursive scheme for evaluating $,a^kf(-b/a),$ (a homogeneous polynomial in $,a,b),,$ analogous to Horner evaluation and related results. If this is of interest let me know and I can elaborate.
            $endgroup$
            – Bill Dubuque
            Jan 8 at 0:46













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          1 Answer
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          0












          $begingroup$

          It's the special case $m = 10c!+!d, n = 10a!+!b $ of the following



          $quad bbox[5px,border:1px solid red]{(a,n) = 1,Rightarrow, (m,n) = (am!-!cn,n)} $ by $,(m,n)=(am,n) = (am!-!cn,n)$



          So $ (a,b)=1,Rightarrow, (10c!+!d,10a!+!b) = (ad!-!bc,10a!+!b), $ exactly as you observed.



          Remark $ $ Various well-known divisibity test rules are essentially special cases of this, e.g. the rule for divisibility by $7$ follows from the rule for $21$ given below (we negated $,ad! -! bc = 2d!-!c,$ below)



          $qquadqquad qquad , (10c!+!d,10(2)!+!1) = (c!-!2d, 10(2)!+!1)$



          ${rm similar to your} (10c!+!d,10(3)!+!1) = (c!-!3d, 10(3)!+!1) $ rule for $,31$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Do you know if there is a name for it? Or how I might find other work on it? I've used it to do some cool things with Gaussian integers and I highly doubt I'm the first person to play with this idea!
            $endgroup$
            – Ben Crossley
            Jan 7 at 23:28










          • $begingroup$
            @Ben I'm not aware of any common name for this. Things like this are folklore, e.g. occurring in various guises in divisibility tests. In number rings analogous results are often special cases of Hermite normal form algorithms for modules and related results, e.g. see here and here. If you give more details about your Gaussian integer results then I may be able to be more specific.
            $endgroup$
            – Bill Dubuque
            Jan 8 at 0:45










          • $begingroup$
            @Ben By the way, another angle is to view $,m = f(10),$ as a polynomial of degree $k$ in the radix. Then $$bmod 10a!+!b!:, 10equiv -b/a,Rightarrow,(f(10),10a!+!b) = (a^kf(-b/a),10a!+!b)qquad$$ and the algorithm can be viwed as a recursive scheme for evaluating $,a^kf(-b/a),$ (a homogeneous polynomial in $,a,b),,$ analogous to Horner evaluation and related results. If this is of interest let me know and I can elaborate.
            $endgroup$
            – Bill Dubuque
            Jan 8 at 0:46


















          0












          $begingroup$

          It's the special case $m = 10c!+!d, n = 10a!+!b $ of the following



          $quad bbox[5px,border:1px solid red]{(a,n) = 1,Rightarrow, (m,n) = (am!-!cn,n)} $ by $,(m,n)=(am,n) = (am!-!cn,n)$



          So $ (a,b)=1,Rightarrow, (10c!+!d,10a!+!b) = (ad!-!bc,10a!+!b), $ exactly as you observed.



          Remark $ $ Various well-known divisibity test rules are essentially special cases of this, e.g. the rule for divisibility by $7$ follows from the rule for $21$ given below (we negated $,ad! -! bc = 2d!-!c,$ below)



          $qquadqquad qquad , (10c!+!d,10(2)!+!1) = (c!-!2d, 10(2)!+!1)$



          ${rm similar to your} (10c!+!d,10(3)!+!1) = (c!-!3d, 10(3)!+!1) $ rule for $,31$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Do you know if there is a name for it? Or how I might find other work on it? I've used it to do some cool things with Gaussian integers and I highly doubt I'm the first person to play with this idea!
            $endgroup$
            – Ben Crossley
            Jan 7 at 23:28










          • $begingroup$
            @Ben I'm not aware of any common name for this. Things like this are folklore, e.g. occurring in various guises in divisibility tests. In number rings analogous results are often special cases of Hermite normal form algorithms for modules and related results, e.g. see here and here. If you give more details about your Gaussian integer results then I may be able to be more specific.
            $endgroup$
            – Bill Dubuque
            Jan 8 at 0:45










          • $begingroup$
            @Ben By the way, another angle is to view $,m = f(10),$ as a polynomial of degree $k$ in the radix. Then $$bmod 10a!+!b!:, 10equiv -b/a,Rightarrow,(f(10),10a!+!b) = (a^kf(-b/a),10a!+!b)qquad$$ and the algorithm can be viwed as a recursive scheme for evaluating $,a^kf(-b/a),$ (a homogeneous polynomial in $,a,b),,$ analogous to Horner evaluation and related results. If this is of interest let me know and I can elaborate.
            $endgroup$
            – Bill Dubuque
            Jan 8 at 0:46
















          0












          0








          0





          $begingroup$

          It's the special case $m = 10c!+!d, n = 10a!+!b $ of the following



          $quad bbox[5px,border:1px solid red]{(a,n) = 1,Rightarrow, (m,n) = (am!-!cn,n)} $ by $,(m,n)=(am,n) = (am!-!cn,n)$



          So $ (a,b)=1,Rightarrow, (10c!+!d,10a!+!b) = (ad!-!bc,10a!+!b), $ exactly as you observed.



          Remark $ $ Various well-known divisibity test rules are essentially special cases of this, e.g. the rule for divisibility by $7$ follows from the rule for $21$ given below (we negated $,ad! -! bc = 2d!-!c,$ below)



          $qquadqquad qquad , (10c!+!d,10(2)!+!1) = (c!-!2d, 10(2)!+!1)$



          ${rm similar to your} (10c!+!d,10(3)!+!1) = (c!-!3d, 10(3)!+!1) $ rule for $,31$






          share|cite|improve this answer











          $endgroup$



          It's the special case $m = 10c!+!d, n = 10a!+!b $ of the following



          $quad bbox[5px,border:1px solid red]{(a,n) = 1,Rightarrow, (m,n) = (am!-!cn,n)} $ by $,(m,n)=(am,n) = (am!-!cn,n)$



          So $ (a,b)=1,Rightarrow, (10c!+!d,10a!+!b) = (ad!-!bc,10a!+!b), $ exactly as you observed.



          Remark $ $ Various well-known divisibity test rules are essentially special cases of this, e.g. the rule for divisibility by $7$ follows from the rule for $21$ given below (we negated $,ad! -! bc = 2d!-!c,$ below)



          $qquadqquad qquad , (10c!+!d,10(2)!+!1) = (c!-!2d, 10(2)!+!1)$



          ${rm similar to your} (10c!+!d,10(3)!+!1) = (c!-!3d, 10(3)!+!1) $ rule for $,31$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 6 at 21:54

























          answered Jan 6 at 20:43









          Bill DubuqueBill Dubuque

          209k29191633




          209k29191633












          • $begingroup$
            Do you know if there is a name for it? Or how I might find other work on it? I've used it to do some cool things with Gaussian integers and I highly doubt I'm the first person to play with this idea!
            $endgroup$
            – Ben Crossley
            Jan 7 at 23:28










          • $begingroup$
            @Ben I'm not aware of any common name for this. Things like this are folklore, e.g. occurring in various guises in divisibility tests. In number rings analogous results are often special cases of Hermite normal form algorithms for modules and related results, e.g. see here and here. If you give more details about your Gaussian integer results then I may be able to be more specific.
            $endgroup$
            – Bill Dubuque
            Jan 8 at 0:45










          • $begingroup$
            @Ben By the way, another angle is to view $,m = f(10),$ as a polynomial of degree $k$ in the radix. Then $$bmod 10a!+!b!:, 10equiv -b/a,Rightarrow,(f(10),10a!+!b) = (a^kf(-b/a),10a!+!b)qquad$$ and the algorithm can be viwed as a recursive scheme for evaluating $,a^kf(-b/a),$ (a homogeneous polynomial in $,a,b),,$ analogous to Horner evaluation and related results. If this is of interest let me know and I can elaborate.
            $endgroup$
            – Bill Dubuque
            Jan 8 at 0:46




















          • $begingroup$
            Do you know if there is a name for it? Or how I might find other work on it? I've used it to do some cool things with Gaussian integers and I highly doubt I'm the first person to play with this idea!
            $endgroup$
            – Ben Crossley
            Jan 7 at 23:28










          • $begingroup$
            @Ben I'm not aware of any common name for this. Things like this are folklore, e.g. occurring in various guises in divisibility tests. In number rings analogous results are often special cases of Hermite normal form algorithms for modules and related results, e.g. see here and here. If you give more details about your Gaussian integer results then I may be able to be more specific.
            $endgroup$
            – Bill Dubuque
            Jan 8 at 0:45










          • $begingroup$
            @Ben By the way, another angle is to view $,m = f(10),$ as a polynomial of degree $k$ in the radix. Then $$bmod 10a!+!b!:, 10equiv -b/a,Rightarrow,(f(10),10a!+!b) = (a^kf(-b/a),10a!+!b)qquad$$ and the algorithm can be viwed as a recursive scheme for evaluating $,a^kf(-b/a),$ (a homogeneous polynomial in $,a,b),,$ analogous to Horner evaluation and related results. If this is of interest let me know and I can elaborate.
            $endgroup$
            – Bill Dubuque
            Jan 8 at 0:46


















          $begingroup$
          Do you know if there is a name for it? Or how I might find other work on it? I've used it to do some cool things with Gaussian integers and I highly doubt I'm the first person to play with this idea!
          $endgroup$
          – Ben Crossley
          Jan 7 at 23:28




          $begingroup$
          Do you know if there is a name for it? Or how I might find other work on it? I've used it to do some cool things with Gaussian integers and I highly doubt I'm the first person to play with this idea!
          $endgroup$
          – Ben Crossley
          Jan 7 at 23:28












          $begingroup$
          @Ben I'm not aware of any common name for this. Things like this are folklore, e.g. occurring in various guises in divisibility tests. In number rings analogous results are often special cases of Hermite normal form algorithms for modules and related results, e.g. see here and here. If you give more details about your Gaussian integer results then I may be able to be more specific.
          $endgroup$
          – Bill Dubuque
          Jan 8 at 0:45




          $begingroup$
          @Ben I'm not aware of any common name for this. Things like this are folklore, e.g. occurring in various guises in divisibility tests. In number rings analogous results are often special cases of Hermite normal form algorithms for modules and related results, e.g. see here and here. If you give more details about your Gaussian integer results then I may be able to be more specific.
          $endgroup$
          – Bill Dubuque
          Jan 8 at 0:45












          $begingroup$
          @Ben By the way, another angle is to view $,m = f(10),$ as a polynomial of degree $k$ in the radix. Then $$bmod 10a!+!b!:, 10equiv -b/a,Rightarrow,(f(10),10a!+!b) = (a^kf(-b/a),10a!+!b)qquad$$ and the algorithm can be viwed as a recursive scheme for evaluating $,a^kf(-b/a),$ (a homogeneous polynomial in $,a,b),,$ analogous to Horner evaluation and related results. If this is of interest let me know and I can elaborate.
          $endgroup$
          – Bill Dubuque
          Jan 8 at 0:46






          $begingroup$
          @Ben By the way, another angle is to view $,m = f(10),$ as a polynomial of degree $k$ in the radix. Then $$bmod 10a!+!b!:, 10equiv -b/a,Rightarrow,(f(10),10a!+!b) = (a^kf(-b/a),10a!+!b)qquad$$ and the algorithm can be viwed as a recursive scheme for evaluating $,a^kf(-b/a),$ (a homogeneous polynomial in $,a,b),,$ analogous to Horner evaluation and related results. If this is of interest let me know and I can elaborate.
          $endgroup$
          – Bill Dubuque
          Jan 8 at 0:46




















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          William S. Burroughs

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          William S. Burroughs From Wikipedia, the free encyclopedia   (Redirected from William Burroughs) Jump to navigation Jump to search For other people named William Burroughs, see William Burroughs (disambiguation). William S. Burroughs Burroughs in 1983 Born William Seward Burroughs II ( 1914-02-05 ) February 5, 1914 St. Louis, Missouri, United States Died August 2, 1997 (1997-08-02) (aged 83) Lawrence, Kansas, United States Pen name William Lee Occupation Author Alma mater Harvard University Genre Beat literature, paranoid fiction Literary movement Beat Generation, postmodernism Notable works Naked Lunch (1959), Junkie (1953) Spouse Ilse von Klapper (1937–1946) Joan Vollmer (1946–1951) Children William S. Burroughs Jr. Relatives William Seward Burroughs I, grandfather Ivy Lee, maternal uncle Signature William Seward Burroughs II ( / ˈ b ʌr oʊ z / ; February...
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          Eda skans

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          1924

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