coloring 3x3 chessboard with two colors… The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)how many unique patterns exist for a $Ntimes N$ gridColorings of a $3times3$ chessboardCheckerboard coloring problemChessboard CombinatoricsProbability of $ 7$ white chessboard squares without neighboursCombinatorics arrangement on chessboardmaximum nonattacking black and white queens on infinite chessboardSimple chessboard exerciseThree knights on a 3x3 chess boardProblem in restricted 3 edge coloring$4times 4$ Chessboard ColoringsA computer screen shows a 98 × 98 chessboard, colored in the usual way.

Finding the path in a graph from A to B then back to A with a minimum of shared edges

How to politely respond to generic emails requesting a PhD/job in my lab? Without wasting too much time

What force causes entropy to increase?

He got a vote 80% that of Emmanuel Macron’s

Is there a writing software that you can sort scenes like slides in PowerPoint?

Wall plug outlet change

Is it ethical to upload a automatically generated paper to a non peer-reviewed site as part of a larger research?

Is above average number of years spent on PhD considered a red flag in future academia or industry positions?

What's the point in a preamp?

What information about me do stores get via my credit card?

Mortgage adviser recommends a longer term than necessary combined with overpayments

Why can't devices on different VLANs, but on the same subnet, communicate?

Do working physicists consider Newtonian mechanics to be "falsified"?

Segmentation fault output is suppressed when piping stdin into a function. Why?

How can I define good in a religion that claims no moral authority?

Arduino Pro Micro - switch off LEDs

How did the audience guess the pentatonic scale in Bobby McFerrin's presentation?

What can I do if neighbor is blocking my solar panels intentionally?

Simulation of a banking system with an Account class in C++

How did passengers keep warm on sail ships?

Derivation tree not rendering

Do warforged have souls?

Did the new image of black hole confirm the general theory of relativity?

Make it rain characters



coloring 3x3 chessboard with two colors…



The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)how many unique patterns exist for a $Ntimes N$ gridColorings of a $3times3$ chessboardCheckerboard coloring problemChessboard CombinatoricsProbability of $ 7$ white chessboard squares without neighboursCombinatorics arrangement on chessboardmaximum nonattacking black and white queens on infinite chessboardSimple chessboard exerciseThree knights on a 3x3 chess boardProblem in restricted 3 edge coloring$4times 4$ Chessboard ColoringsA computer screen shows a 98 × 98 chessboard, colored in the usual way.










0












$begingroup$


Consider a 3x3 chessboard with 9 elements. The elements are colored with black and white paints.



The task is to find the number of different chessboards of this type exist.



I don't know how to start with this problem....










share|cite|improve this question









$endgroup$







  • 1




    $begingroup$
    Does the orientation of the board matter? For instance, is the coloring that has a black square in the upper left corner and all the others squares white to be counted as different from the coloring that has a black square in the lower left corner and all the other squares white? Or are they to be counted as just one coloring, because the board can be rotated to turn one into the other?
    $endgroup$
    – Brian M. Scott
    Dec 13 '13 at 23:12










  • $begingroup$
    Draw a shape! And try to color it. Check if anything is ambiguous for you in the problem statement.
    $endgroup$
    – hhsaffar
    Dec 13 '13 at 23:12










  • $begingroup$
    This MSE link shows a solution that takes symmetries into account.
    $endgroup$
    – Marko Riedel
    Dec 13 '13 at 23:15










  • $begingroup$
    HINT: If the orientation of the board does matter, you’re really just counting the number of ways of choosing a subset of the $9$ elements to be colored black. How many subsets are there?
    $endgroup$
    – Brian M. Scott
    Dec 13 '13 at 23:20















0












$begingroup$


Consider a 3x3 chessboard with 9 elements. The elements are colored with black and white paints.



The task is to find the number of different chessboards of this type exist.



I don't know how to start with this problem....










share|cite|improve this question









$endgroup$







  • 1




    $begingroup$
    Does the orientation of the board matter? For instance, is the coloring that has a black square in the upper left corner and all the others squares white to be counted as different from the coloring that has a black square in the lower left corner and all the other squares white? Or are they to be counted as just one coloring, because the board can be rotated to turn one into the other?
    $endgroup$
    – Brian M. Scott
    Dec 13 '13 at 23:12










  • $begingroup$
    Draw a shape! And try to color it. Check if anything is ambiguous for you in the problem statement.
    $endgroup$
    – hhsaffar
    Dec 13 '13 at 23:12










  • $begingroup$
    This MSE link shows a solution that takes symmetries into account.
    $endgroup$
    – Marko Riedel
    Dec 13 '13 at 23:15










  • $begingroup$
    HINT: If the orientation of the board does matter, you’re really just counting the number of ways of choosing a subset of the $9$ elements to be colored black. How many subsets are there?
    $endgroup$
    – Brian M. Scott
    Dec 13 '13 at 23:20













0












0








0





$begingroup$


Consider a 3x3 chessboard with 9 elements. The elements are colored with black and white paints.



The task is to find the number of different chessboards of this type exist.



I don't know how to start with this problem....










share|cite|improve this question









$endgroup$




Consider a 3x3 chessboard with 9 elements. The elements are colored with black and white paints.



The task is to find the number of different chessboards of this type exist.



I don't know how to start with this problem....







combinatorics






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 13 '13 at 23:09









user108696user108696

112




112







  • 1




    $begingroup$
    Does the orientation of the board matter? For instance, is the coloring that has a black square in the upper left corner and all the others squares white to be counted as different from the coloring that has a black square in the lower left corner and all the other squares white? Or are they to be counted as just one coloring, because the board can be rotated to turn one into the other?
    $endgroup$
    – Brian M. Scott
    Dec 13 '13 at 23:12










  • $begingroup$
    Draw a shape! And try to color it. Check if anything is ambiguous for you in the problem statement.
    $endgroup$
    – hhsaffar
    Dec 13 '13 at 23:12










  • $begingroup$
    This MSE link shows a solution that takes symmetries into account.
    $endgroup$
    – Marko Riedel
    Dec 13 '13 at 23:15










  • $begingroup$
    HINT: If the orientation of the board does matter, you’re really just counting the number of ways of choosing a subset of the $9$ elements to be colored black. How many subsets are there?
    $endgroup$
    – Brian M. Scott
    Dec 13 '13 at 23:20












  • 1




    $begingroup$
    Does the orientation of the board matter? For instance, is the coloring that has a black square in the upper left corner and all the others squares white to be counted as different from the coloring that has a black square in the lower left corner and all the other squares white? Or are they to be counted as just one coloring, because the board can be rotated to turn one into the other?
    $endgroup$
    – Brian M. Scott
    Dec 13 '13 at 23:12










  • $begingroup$
    Draw a shape! And try to color it. Check if anything is ambiguous for you in the problem statement.
    $endgroup$
    – hhsaffar
    Dec 13 '13 at 23:12










  • $begingroup$
    This MSE link shows a solution that takes symmetries into account.
    $endgroup$
    – Marko Riedel
    Dec 13 '13 at 23:15










  • $begingroup$
    HINT: If the orientation of the board does matter, you’re really just counting the number of ways of choosing a subset of the $9$ elements to be colored black. How many subsets are there?
    $endgroup$
    – Brian M. Scott
    Dec 13 '13 at 23:20







1




1




$begingroup$
Does the orientation of the board matter? For instance, is the coloring that has a black square in the upper left corner and all the others squares white to be counted as different from the coloring that has a black square in the lower left corner and all the other squares white? Or are they to be counted as just one coloring, because the board can be rotated to turn one into the other?
$endgroup$
– Brian M. Scott
Dec 13 '13 at 23:12




$begingroup$
Does the orientation of the board matter? For instance, is the coloring that has a black square in the upper left corner and all the others squares white to be counted as different from the coloring that has a black square in the lower left corner and all the other squares white? Or are they to be counted as just one coloring, because the board can be rotated to turn one into the other?
$endgroup$
– Brian M. Scott
Dec 13 '13 at 23:12












$begingroup$
Draw a shape! And try to color it. Check if anything is ambiguous for you in the problem statement.
$endgroup$
– hhsaffar
Dec 13 '13 at 23:12




$begingroup$
Draw a shape! And try to color it. Check if anything is ambiguous for you in the problem statement.
$endgroup$
– hhsaffar
Dec 13 '13 at 23:12












$begingroup$
This MSE link shows a solution that takes symmetries into account.
$endgroup$
– Marko Riedel
Dec 13 '13 at 23:15




$begingroup$
This MSE link shows a solution that takes symmetries into account.
$endgroup$
– Marko Riedel
Dec 13 '13 at 23:15












$begingroup$
HINT: If the orientation of the board does matter, you’re really just counting the number of ways of choosing a subset of the $9$ elements to be colored black. How many subsets are there?
$endgroup$
– Brian M. Scott
Dec 13 '13 at 23:20




$begingroup$
HINT: If the orientation of the board does matter, you’re really just counting the number of ways of choosing a subset of the $9$ elements to be colored black. How many subsets are there?
$endgroup$
– Brian M. Scott
Dec 13 '13 at 23:20










2 Answers
2






active

oldest

votes


















3












$begingroup$

Suppose that rotating them does not matter, then we have two colors for each square and
so a total of $2^9$ possibillities.
For the other case, (possibillities that can be made by rotation of another possibillity are considered the same) we can use group theory. I will asume here that the bottom of the board is just black. Because we have $4$ possible roations and no mirroring (because the bottom of the board is black) we get as symmetry group $C_4$, the cyclic group of order $4$. No we will check the orbis of the squares after rotations. This will give us the following:enter image description here



So for the total possibillities we get (by The Counting Theorem):
$$
frac1sum_gin G |X^g| = \
frac1C_4 (1cdot 2^9 + 4 cdot 2^3 + 2 cdot 2^5+ 4 cdot 2^3)= \
frac14(1cdot 2^9 + 4 cdot 2^3 + 2 cdot 2^5+ 4 cdot 2^3)= 160.
$$
Where $X$ is the set of colourings of the chessboard and $X^g$ is the subset of $X$ consisting of those points which are left fixed by the element $g in G$.
So this will give us a total of $160$ chessboards,
If the back of the chessboard is also in account, for example see-through, please add this to the question. This will give the symmetric group $D_4$ and some extra possobillities, for if you want to work it out.






share|cite|improve this answer











$endgroup$




















    1












    $begingroup$

    If you consider rotations and reflections to be the same.



    Number of $ntimes n$ binary matrices under action of dihedral group of the square $D_4$.



    http://oeis.org/A054247



    $$a(n)=begincasesdfrac18left(2^n^2+2cdot2^n^2/4+3cdot2^n^2/2+2cdot2^(n^2+n)/2right) & n= 0pmod2 \
    dfrac18left(2^n^2+2cdot2^(n^2+3)/4+2^(n^2+1)/2+4cdot2^(n^2+n)/2right)& n= 1pmod2endcases$$

    beginalign*
    a(3)&=,dfrac18left(2^3^2+2cdot2^(3^2+3)/4+2^(3^2+1)/2+4cdot2^(3^2+3)/2right)\
    &=,dfrac18left(2^9+2cdot2^3+2^5+4cdot2^6right)=dfrac2^9+2^4+2^5+2^88=,quadlargecolorred102
    endalign*






    share|cite|improve this answer









    $endgroup$













      Your Answer








      StackExchange.ready(function()
      var channelOptions =
      tags: "".split(" "),
      id: "69"
      ;
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function()
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled)
      StackExchange.using("snippets", function()
      createEditor();
      );

      else
      createEditor();

      );

      function createEditor()
      StackExchange.prepareEditor(
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader:
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      ,
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      );



      );













      draft saved

      draft discarded


















      StackExchange.ready(
      function ()
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f606031%2fcoloring-3x3-chessboard-with-two-colors%23new-answer', 'question_page');

      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Suppose that rotating them does not matter, then we have two colors for each square and
      so a total of $2^9$ possibillities.
      For the other case, (possibillities that can be made by rotation of another possibillity are considered the same) we can use group theory. I will asume here that the bottom of the board is just black. Because we have $4$ possible roations and no mirroring (because the bottom of the board is black) we get as symmetry group $C_4$, the cyclic group of order $4$. No we will check the orbis of the squares after rotations. This will give us the following:enter image description here



      So for the total possibillities we get (by The Counting Theorem):
      $$
      frac1sum_gin G |X^g| = \
      frac1C_4 (1cdot 2^9 + 4 cdot 2^3 + 2 cdot 2^5+ 4 cdot 2^3)= \
      frac14(1cdot 2^9 + 4 cdot 2^3 + 2 cdot 2^5+ 4 cdot 2^3)= 160.
      $$
      Where $X$ is the set of colourings of the chessboard and $X^g$ is the subset of $X$ consisting of those points which are left fixed by the element $g in G$.
      So this will give us a total of $160$ chessboards,
      If the back of the chessboard is also in account, for example see-through, please add this to the question. This will give the symmetric group $D_4$ and some extra possobillities, for if you want to work it out.






      share|cite|improve this answer











      $endgroup$

















        3












        $begingroup$

        Suppose that rotating them does not matter, then we have two colors for each square and
        so a total of $2^9$ possibillities.
        For the other case, (possibillities that can be made by rotation of another possibillity are considered the same) we can use group theory. I will asume here that the bottom of the board is just black. Because we have $4$ possible roations and no mirroring (because the bottom of the board is black) we get as symmetry group $C_4$, the cyclic group of order $4$. No we will check the orbis of the squares after rotations. This will give us the following:enter image description here



        So for the total possibillities we get (by The Counting Theorem):
        $$
        frac1sum_gin G |X^g| = \
        frac1C_4 (1cdot 2^9 + 4 cdot 2^3 + 2 cdot 2^5+ 4 cdot 2^3)= \
        frac14(1cdot 2^9 + 4 cdot 2^3 + 2 cdot 2^5+ 4 cdot 2^3)= 160.
        $$
        Where $X$ is the set of colourings of the chessboard and $X^g$ is the subset of $X$ consisting of those points which are left fixed by the element $g in G$.
        So this will give us a total of $160$ chessboards,
        If the back of the chessboard is also in account, for example see-through, please add this to the question. This will give the symmetric group $D_4$ and some extra possobillities, for if you want to work it out.






        share|cite|improve this answer











        $endgroup$















          3












          3








          3





          $begingroup$

          Suppose that rotating them does not matter, then we have two colors for each square and
          so a total of $2^9$ possibillities.
          For the other case, (possibillities that can be made by rotation of another possibillity are considered the same) we can use group theory. I will asume here that the bottom of the board is just black. Because we have $4$ possible roations and no mirroring (because the bottom of the board is black) we get as symmetry group $C_4$, the cyclic group of order $4$. No we will check the orbis of the squares after rotations. This will give us the following:enter image description here



          So for the total possibillities we get (by The Counting Theorem):
          $$
          frac1sum_gin G |X^g| = \
          frac1C_4 (1cdot 2^9 + 4 cdot 2^3 + 2 cdot 2^5+ 4 cdot 2^3)= \
          frac14(1cdot 2^9 + 4 cdot 2^3 + 2 cdot 2^5+ 4 cdot 2^3)= 160.
          $$
          Where $X$ is the set of colourings of the chessboard and $X^g$ is the subset of $X$ consisting of those points which are left fixed by the element $g in G$.
          So this will give us a total of $160$ chessboards,
          If the back of the chessboard is also in account, for example see-through, please add this to the question. This will give the symmetric group $D_4$ and some extra possobillities, for if you want to work it out.






          share|cite|improve this answer











          $endgroup$



          Suppose that rotating them does not matter, then we have two colors for each square and
          so a total of $2^9$ possibillities.
          For the other case, (possibillities that can be made by rotation of another possibillity are considered the same) we can use group theory. I will asume here that the bottom of the board is just black. Because we have $4$ possible roations and no mirroring (because the bottom of the board is black) we get as symmetry group $C_4$, the cyclic group of order $4$. No we will check the orbis of the squares after rotations. This will give us the following:enter image description here



          So for the total possibillities we get (by The Counting Theorem):
          $$
          frac1sum_gin G |X^g| = \
          frac1C_4 (1cdot 2^9 + 4 cdot 2^3 + 2 cdot 2^5+ 4 cdot 2^3)= \
          frac14(1cdot 2^9 + 4 cdot 2^3 + 2 cdot 2^5+ 4 cdot 2^3)= 160.
          $$
          Where $X$ is the set of colourings of the chessboard and $X^g$ is the subset of $X$ consisting of those points which are left fixed by the element $g in G$.
          So this will give us a total of $160$ chessboards,
          If the back of the chessboard is also in account, for example see-through, please add this to the question. This will give the symmetric group $D_4$ and some extra possobillities, for if you want to work it out.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 14 '13 at 0:54

























          answered Dec 14 '13 at 0:12









          user112167user112167

          1,2621715




          1,2621715





















              1












              $begingroup$

              If you consider rotations and reflections to be the same.



              Number of $ntimes n$ binary matrices under action of dihedral group of the square $D_4$.



              http://oeis.org/A054247



              $$a(n)=begincasesdfrac18left(2^n^2+2cdot2^n^2/4+3cdot2^n^2/2+2cdot2^(n^2+n)/2right) & n= 0pmod2 \
              dfrac18left(2^n^2+2cdot2^(n^2+3)/4+2^(n^2+1)/2+4cdot2^(n^2+n)/2right)& n= 1pmod2endcases$$

              beginalign*
              a(3)&=,dfrac18left(2^3^2+2cdot2^(3^2+3)/4+2^(3^2+1)/2+4cdot2^(3^2+3)/2right)\
              &=,dfrac18left(2^9+2cdot2^3+2^5+4cdot2^6right)=dfrac2^9+2^4+2^5+2^88=,quadlargecolorred102
              endalign*






              share|cite|improve this answer









              $endgroup$

















                1












                $begingroup$

                If you consider rotations and reflections to be the same.



                Number of $ntimes n$ binary matrices under action of dihedral group of the square $D_4$.



                http://oeis.org/A054247



                $$a(n)=begincasesdfrac18left(2^n^2+2cdot2^n^2/4+3cdot2^n^2/2+2cdot2^(n^2+n)/2right) & n= 0pmod2 \
                dfrac18left(2^n^2+2cdot2^(n^2+3)/4+2^(n^2+1)/2+4cdot2^(n^2+n)/2right)& n= 1pmod2endcases$$

                beginalign*
                a(3)&=,dfrac18left(2^3^2+2cdot2^(3^2+3)/4+2^(3^2+1)/2+4cdot2^(3^2+3)/2right)\
                &=,dfrac18left(2^9+2cdot2^3+2^5+4cdot2^6right)=dfrac2^9+2^4+2^5+2^88=,quadlargecolorred102
                endalign*






                share|cite|improve this answer









                $endgroup$















                  1












                  1








                  1





                  $begingroup$

                  If you consider rotations and reflections to be the same.



                  Number of $ntimes n$ binary matrices under action of dihedral group of the square $D_4$.



                  http://oeis.org/A054247



                  $$a(n)=begincasesdfrac18left(2^n^2+2cdot2^n^2/4+3cdot2^n^2/2+2cdot2^(n^2+n)/2right) & n= 0pmod2 \
                  dfrac18left(2^n^2+2cdot2^(n^2+3)/4+2^(n^2+1)/2+4cdot2^(n^2+n)/2right)& n= 1pmod2endcases$$

                  beginalign*
                  a(3)&=,dfrac18left(2^3^2+2cdot2^(3^2+3)/4+2^(3^2+1)/2+4cdot2^(3^2+3)/2right)\
                  &=,dfrac18left(2^9+2cdot2^3+2^5+4cdot2^6right)=dfrac2^9+2^4+2^5+2^88=,quadlargecolorred102
                  endalign*






                  share|cite|improve this answer









                  $endgroup$



                  If you consider rotations and reflections to be the same.



                  Number of $ntimes n$ binary matrices under action of dihedral group of the square $D_4$.



                  http://oeis.org/A054247



                  $$a(n)=begincasesdfrac18left(2^n^2+2cdot2^n^2/4+3cdot2^n^2/2+2cdot2^(n^2+n)/2right) & n= 0pmod2 \
                  dfrac18left(2^n^2+2cdot2^(n^2+3)/4+2^(n^2+1)/2+4cdot2^(n^2+n)/2right)& n= 1pmod2endcases$$

                  beginalign*
                  a(3)&=,dfrac18left(2^3^2+2cdot2^(3^2+3)/4+2^(3^2+1)/2+4cdot2^(3^2+3)/2right)\
                  &=,dfrac18left(2^9+2cdot2^3+2^5+4cdot2^6right)=dfrac2^9+2^4+2^5+2^88=,quadlargecolorred102
                  endalign*







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 25 at 5:34









                  D.MatthewD.Matthew

                  414




                  414



























                      draft saved

                      draft discarded
















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid


                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.

                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function ()
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f606031%2fcoloring-3x3-chessboard-with-two-colors%23new-answer', 'question_page');

                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      Popular posts from this blog

                      How should I support this large drywall patch? Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?How do I cover large gaps in drywall?How do I keep drywall around a patch from crumbling?Can I glue a second layer of drywall?How to patch long strip on drywall?Large drywall patch: how to avoid bulging seams?Drywall Mesh Patch vs. Bulge? To remove or not to remove?How to fix this drywall job?Prep drywall before backsplashWhat's the best way to fix this horrible drywall patch job?Drywall patching using 3M Patch Plus Primer

                      random experiment with two different functions on unit interval Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Random variable and probability space notionsRandom Walk with EdgesFinding functions where the increase over a random interval is Poisson distributedNumber of days until dayCan an observed event in fact be of zero probability?Unit random processmodels of coins and uniform distributionHow to get the number of successes given $n$ trials , probability $P$ and a random variable $X$Absorbing Markov chain in a computer. Is “almost every” turned into always convergence in computer executions?Stopped random walk is not uniformly integrable

                      Lowndes Grove History Architecture References Navigation menu32°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661132°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661178002500"National Register Information System"Historic houses of South Carolina"Lowndes Grove""+32° 48' 6.00", −79° 57' 58.00""Lowndes Grove, Charleston County (260 St. Margaret St., Charleston)""Lowndes Grove"The Charleston ExpositionIt Happened in South Carolina"Lowndes Grove (House), Saint Margaret Street & Sixth Avenue, Charleston, Charleston County, SC(Photographs)"Plantations of the Carolina Low Countrye