Find minimum values of a linear system over $mathbb N$












1












$begingroup$


I have the following linear equations:



begin{align}
p &= 2(a-c)+b-d\
q &= 2(e-g)+f-h\
a+c &= f+h\
b+d &= e+g\
end{align}



where $p$ and $q$ are known, and $a$, $b$, $c$, $d$, $e$, $f$, $g$, $h$ are unknown. Also $p$, $q$, $a$, $b$, $c$, $d$, $e$, $f$, $g$, $h$ are all natural numbers (including zeros).



Obviously, this is a linear system with $4$ equations and $8$ unknowns that has infinitely many solutions.



Is there a way to find the minimum values of $a$, $b$, $c$, $d$, $e$, $f$, $g$, $h$ that solve the system? What I really want is the minimum value of the sum $a+b+c+d$ (which also happens to be equal with $e+f+g+h$).






linear-algebra systems-of-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is there a typo? Your $p$ and $q$ are the same.
    $endgroup$
    – Karn Watcharasupat
    Dec 8 '18 at 12:25










  • $begingroup$
    Yes, I'm sorry for that, now it's fixed.
    $endgroup$
    – George Savvidis
    Dec 8 '18 at 12:32










  • $begingroup$
    Of course, this can be formulated as an integer linear programming problem, but I assume that's not the answer you're looking for. I've written an answer with a couple of ideas that don't require specialized software, but if you have have access to ILP software, that would be the first thing to try.
    $endgroup$
    – saulspatz
    Dec 8 '18 at 13:59
















1












$begingroup$


I have the following linear equations:



begin{align}
p &= 2(a-c)+b-d\
q &= 2(e-g)+f-h\
a+c &= f+h\
b+d &= e+g\
end{align}



where $p$ and $q$ are known, and $a$, $b$, $c$, $d$, $e$, $f$, $g$, $h$ are unknown. Also $p$, $q$, $a$, $b$, $c$, $d$, $e$, $f$, $g$, $h$ are all natural numbers (including zeros).



Obviously, this is a linear system with $4$ equations and $8$ unknowns that has infinitely many solutions.



Is there a way to find the minimum values of $a$, $b$, $c$, $d$, $e$, $f$, $g$, $h$ that solve the system? What I really want is the minimum value of the sum $a+b+c+d$ (which also happens to be equal with $e+f+g+h$).






linear-algebra systems-of-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is there a typo? Your $p$ and $q$ are the same.
    $endgroup$
    – Karn Watcharasupat
    Dec 8 '18 at 12:25










  • $begingroup$
    Yes, I'm sorry for that, now it's fixed.
    $endgroup$
    – George Savvidis
    Dec 8 '18 at 12:32










  • $begingroup$
    Of course, this can be formulated as an integer linear programming problem, but I assume that's not the answer you're looking for. I've written an answer with a couple of ideas that don't require specialized software, but if you have have access to ILP software, that would be the first thing to try.
    $endgroup$
    – saulspatz
    Dec 8 '18 at 13:59














1












1








1


1



$begingroup$


I have the following linear equations:



begin{align}
p &= 2(a-c)+b-d\
q &= 2(e-g)+f-h\
a+c &= f+h\
b+d &= e+g\
end{align}



where $p$ and $q$ are known, and $a$, $b$, $c$, $d$, $e$, $f$, $g$, $h$ are unknown. Also $p$, $q$, $a$, $b$, $c$, $d$, $e$, $f$, $g$, $h$ are all natural numbers (including zeros).



Obviously, this is a linear system with $4$ equations and $8$ unknowns that has infinitely many solutions.



Is there a way to find the minimum values of $a$, $b$, $c$, $d$, $e$, $f$, $g$, $h$ that solve the system? What I really want is the minimum value of the sum $a+b+c+d$ (which also happens to be equal with $e+f+g+h$).






linear-algebra systems-of-equations







share|cite|improve this question











$endgroup$




I have the following linear equations:



begin{align}
p &= 2(a-c)+b-d\
q &= 2(e-g)+f-h\
a+c &= f+h\
b+d &= e+g\
end{align}



where $p$ and $q$ are known, and $a$, $b$, $c$, $d$, $e$, $f$, $g$, $h$ are unknown. Also $p$, $q$, $a$, $b$, $c$, $d$, $e$, $f$, $g$, $h$ are all natural numbers (including zeros).



Obviously, this is a linear system with $4$ equations and $8$ unknowns that has infinitely many solutions.



Is there a way to find the minimum values of $a$, $b$, $c$, $d$, $e$, $f$, $g$, $h$ that solve the system? What I really want is the minimum value of the sum $a+b+c+d$ (which also happens to be equal with $e+f+g+h$).






linear-algebra systems-of-equations




linear-algebra systems-of-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 15:02









Rodrigo de Azevedo

12.9k41857




12.9k41857










asked Dec 8 '18 at 12:20









George SavvidisGeorge Savvidis

62




62












  • $begingroup$
    Is there a typo? Your $p$ and $q$ are the same.
    $endgroup$
    – Karn Watcharasupat
    Dec 8 '18 at 12:25










  • $begingroup$
    Yes, I'm sorry for that, now it's fixed.
    $endgroup$
    – George Savvidis
    Dec 8 '18 at 12:32










  • $begingroup$
    Of course, this can be formulated as an integer linear programming problem, but I assume that's not the answer you're looking for. I've written an answer with a couple of ideas that don't require specialized software, but if you have have access to ILP software, that would be the first thing to try.
    $endgroup$
    – saulspatz
    Dec 8 '18 at 13:59


















  • $begingroup$
    Is there a typo? Your $p$ and $q$ are the same.
    $endgroup$
    – Karn Watcharasupat
    Dec 8 '18 at 12:25










  • $begingroup$
    Yes, I'm sorry for that, now it's fixed.
    $endgroup$
    – George Savvidis
    Dec 8 '18 at 12:32










  • $begingroup$
    Of course, this can be formulated as an integer linear programming problem, but I assume that's not the answer you're looking for. I've written an answer with a couple of ideas that don't require specialized software, but if you have have access to ILP software, that would be the first thing to try.
    $endgroup$
    – saulspatz
    Dec 8 '18 at 13:59
















$begingroup$
Is there a typo? Your $p$ and $q$ are the same.
$endgroup$
– Karn Watcharasupat
Dec 8 '18 at 12:25




$begingroup$
Is there a typo? Your $p$ and $q$ are the same.
$endgroup$
– Karn Watcharasupat
Dec 8 '18 at 12:25












$begingroup$
Yes, I'm sorry for that, now it's fixed.
$endgroup$
– George Savvidis
Dec 8 '18 at 12:32




$begingroup$
Yes, I'm sorry for that, now it's fixed.
$endgroup$
– George Savvidis
Dec 8 '18 at 12:32












$begingroup$
Of course, this can be formulated as an integer linear programming problem, but I assume that's not the answer you're looking for. I've written an answer with a couple of ideas that don't require specialized software, but if you have have access to ILP software, that would be the first thing to try.
$endgroup$
– saulspatz
Dec 8 '18 at 13:59




$begingroup$
Of course, this can be formulated as an integer linear programming problem, but I assume that's not the answer you're looking for. I've written an answer with a couple of ideas that don't require specialized software, but if you have have access to ILP software, that would be the first thing to try.
$endgroup$
– saulspatz
Dec 8 '18 at 13:59










2 Answers
2






active

oldest

votes


















0












$begingroup$

This isn't a full solution, just an idea to get started. If we have a solution $(a,b,c,d,e,f,g,h),$ let $$
begin{align}
m&=min(a,c,f,h)\
n&=min(b,d,e,g)\
a'&=a-m\
c'&=c-m\
f'&=f-m\
h'&=h-m\
b'&=b-n\
d'&=d-n\
e'&=e-n\
g'&=g-n\
end{align}$$
Then $(a',b',c',d',e',f',g',h')$ is a solution in naturals and $$a'+b'+c+d'le a+b+c+d.$$ So, we can assume that one of $a,c,f,h$ is $0$ and one of $b,d,e,g$ is $0$. This gives $16$ systems of equations, but they only have $6$ variables instead of $8.$



The only other thing I've noticed so far is that we have $$
begin{align}
p&equiv b-dequiv b+dpmod{2}\
q&equiv f-hequiv f+hequiv a+cpmod{2}\
end{align}$$
So that
$$a+b+c+dequiv p+qpmod{2}$$



A possible approach is to try test values of $a+b+c+d$ adding it as another equation. The above fact eliminates half the test values. If you get a solution, you have an upper bound on the minimum. If the system doesn't have a solution, I don't see that you have a lower bound though.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Perhaps, at first, it would be useful to consider the relaxation of this problem, i.e. all variables are non-negative real! numbers. Then it is possible to use the simplex method with some kind of caution because $p,q$ are not given explicitly. For example, If I did not make a mistake then the solution of the relaxation LP problem is something like
    $$
    a=f=frac{2p-q}3, e=b=frac{2q-p}3, c=d=g=h=0
    $$
    and the corresponding minimum is
    $$
    frac{p+q}3.
    $$
    This looks true if $2p-q$ and $2q-p$ are both non-negative (if they also are divided by 3 then they are solutions of the original integer programming problem). if $2p-q$ or $2q-p$ can be negative then the solution is different...



    Another idea is to try to rewrite it as a binary integer programming problem...






    share|cite|improve this answer











    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      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%2f3031046%2ffind-minimum-values-of-a-linear-system-over-mathbb-n%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









      0












      $begingroup$

      This isn't a full solution, just an idea to get started. If we have a solution $(a,b,c,d,e,f,g,h),$ let $$
      begin{align}
      m&=min(a,c,f,h)\
      n&=min(b,d,e,g)\
      a'&=a-m\
      c'&=c-m\
      f'&=f-m\
      h'&=h-m\
      b'&=b-n\
      d'&=d-n\
      e'&=e-n\
      g'&=g-n\
      end{align}$$
      Then $(a',b',c',d',e',f',g',h')$ is a solution in naturals and $$a'+b'+c+d'le a+b+c+d.$$ So, we can assume that one of $a,c,f,h$ is $0$ and one of $b,d,e,g$ is $0$. This gives $16$ systems of equations, but they only have $6$ variables instead of $8.$



      The only other thing I've noticed so far is that we have $$
      begin{align}
      p&equiv b-dequiv b+dpmod{2}\
      q&equiv f-hequiv f+hequiv a+cpmod{2}\
      end{align}$$
      So that
      $$a+b+c+dequiv p+qpmod{2}$$



      A possible approach is to try test values of $a+b+c+d$ adding it as another equation. The above fact eliminates half the test values. If you get a solution, you have an upper bound on the minimum. If the system doesn't have a solution, I don't see that you have a lower bound though.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        This isn't a full solution, just an idea to get started. If we have a solution $(a,b,c,d,e,f,g,h),$ let $$
        begin{align}
        m&=min(a,c,f,h)\
        n&=min(b,d,e,g)\
        a'&=a-m\
        c'&=c-m\
        f'&=f-m\
        h'&=h-m\
        b'&=b-n\
        d'&=d-n\
        e'&=e-n\
        g'&=g-n\
        end{align}$$
        Then $(a',b',c',d',e',f',g',h')$ is a solution in naturals and $$a'+b'+c+d'le a+b+c+d.$$ So, we can assume that one of $a,c,f,h$ is $0$ and one of $b,d,e,g$ is $0$. This gives $16$ systems of equations, but they only have $6$ variables instead of $8.$



        The only other thing I've noticed so far is that we have $$
        begin{align}
        p&equiv b-dequiv b+dpmod{2}\
        q&equiv f-hequiv f+hequiv a+cpmod{2}\
        end{align}$$
        So that
        $$a+b+c+dequiv p+qpmod{2}$$



        A possible approach is to try test values of $a+b+c+d$ adding it as another equation. The above fact eliminates half the test values. If you get a solution, you have an upper bound on the minimum. If the system doesn't have a solution, I don't see that you have a lower bound though.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          This isn't a full solution, just an idea to get started. If we have a solution $(a,b,c,d,e,f,g,h),$ let $$
          begin{align}
          m&=min(a,c,f,h)\
          n&=min(b,d,e,g)\
          a'&=a-m\
          c'&=c-m\
          f'&=f-m\
          h'&=h-m\
          b'&=b-n\
          d'&=d-n\
          e'&=e-n\
          g'&=g-n\
          end{align}$$
          Then $(a',b',c',d',e',f',g',h')$ is a solution in naturals and $$a'+b'+c+d'le a+b+c+d.$$ So, we can assume that one of $a,c,f,h$ is $0$ and one of $b,d,e,g$ is $0$. This gives $16$ systems of equations, but they only have $6$ variables instead of $8.$



          The only other thing I've noticed so far is that we have $$
          begin{align}
          p&equiv b-dequiv b+dpmod{2}\
          q&equiv f-hequiv f+hequiv a+cpmod{2}\
          end{align}$$
          So that
          $$a+b+c+dequiv p+qpmod{2}$$



          A possible approach is to try test values of $a+b+c+d$ adding it as another equation. The above fact eliminates half the test values. If you get a solution, you have an upper bound on the minimum. If the system doesn't have a solution, I don't see that you have a lower bound though.






          share|cite|improve this answer









          $endgroup$



          This isn't a full solution, just an idea to get started. If we have a solution $(a,b,c,d,e,f,g,h),$ let $$
          begin{align}
          m&=min(a,c,f,h)\
          n&=min(b,d,e,g)\
          a'&=a-m\
          c'&=c-m\
          f'&=f-m\
          h'&=h-m\
          b'&=b-n\
          d'&=d-n\
          e'&=e-n\
          g'&=g-n\
          end{align}$$
          Then $(a',b',c',d',e',f',g',h')$ is a solution in naturals and $$a'+b'+c+d'le a+b+c+d.$$ So, we can assume that one of $a,c,f,h$ is $0$ and one of $b,d,e,g$ is $0$. This gives $16$ systems of equations, but they only have $6$ variables instead of $8.$



          The only other thing I've noticed so far is that we have $$
          begin{align}
          p&equiv b-dequiv b+dpmod{2}\
          q&equiv f-hequiv f+hequiv a+cpmod{2}\
          end{align}$$
          So that
          $$a+b+c+dequiv p+qpmod{2}$$



          A possible approach is to try test values of $a+b+c+d$ adding it as another equation. The above fact eliminates half the test values. If you get a solution, you have an upper bound on the minimum. If the system doesn't have a solution, I don't see that you have a lower bound though.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 8 '18 at 13:57









          saulspatzsaulspatz

          14.5k21329




          14.5k21329























              0












              $begingroup$

              Perhaps, at first, it would be useful to consider the relaxation of this problem, i.e. all variables are non-negative real! numbers. Then it is possible to use the simplex method with some kind of caution because $p,q$ are not given explicitly. For example, If I did not make a mistake then the solution of the relaxation LP problem is something like
              $$
              a=f=frac{2p-q}3, e=b=frac{2q-p}3, c=d=g=h=0
              $$
              and the corresponding minimum is
              $$
              frac{p+q}3.
              $$
              This looks true if $2p-q$ and $2q-p$ are both non-negative (if they also are divided by 3 then they are solutions of the original integer programming problem). if $2p-q$ or $2q-p$ can be negative then the solution is different...



              Another idea is to try to rewrite it as a binary integer programming problem...






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                Perhaps, at first, it would be useful to consider the relaxation of this problem, i.e. all variables are non-negative real! numbers. Then it is possible to use the simplex method with some kind of caution because $p,q$ are not given explicitly. For example, If I did not make a mistake then the solution of the relaxation LP problem is something like
                $$
                a=f=frac{2p-q}3, e=b=frac{2q-p}3, c=d=g=h=0
                $$
                and the corresponding minimum is
                $$
                frac{p+q}3.
                $$
                This looks true if $2p-q$ and $2q-p$ are both non-negative (if they also are divided by 3 then they are solutions of the original integer programming problem). if $2p-q$ or $2q-p$ can be negative then the solution is different...



                Another idea is to try to rewrite it as a binary integer programming problem...






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Perhaps, at first, it would be useful to consider the relaxation of this problem, i.e. all variables are non-negative real! numbers. Then it is possible to use the simplex method with some kind of caution because $p,q$ are not given explicitly. For example, If I did not make a mistake then the solution of the relaxation LP problem is something like
                  $$
                  a=f=frac{2p-q}3, e=b=frac{2q-p}3, c=d=g=h=0
                  $$
                  and the corresponding minimum is
                  $$
                  frac{p+q}3.
                  $$
                  This looks true if $2p-q$ and $2q-p$ are both non-negative (if they also are divided by 3 then they are solutions of the original integer programming problem). if $2p-q$ or $2q-p$ can be negative then the solution is different...



                  Another idea is to try to rewrite it as a binary integer programming problem...






                  share|cite|improve this answer











                  $endgroup$



                  Perhaps, at first, it would be useful to consider the relaxation of this problem, i.e. all variables are non-negative real! numbers. Then it is possible to use the simplex method with some kind of caution because $p,q$ are not given explicitly. For example, If I did not make a mistake then the solution of the relaxation LP problem is something like
                  $$
                  a=f=frac{2p-q}3, e=b=frac{2q-p}3, c=d=g=h=0
                  $$
                  and the corresponding minimum is
                  $$
                  frac{p+q}3.
                  $$
                  This looks true if $2p-q$ and $2q-p$ are both non-negative (if they also are divided by 3 then they are solutions of the original integer programming problem). if $2p-q$ or $2q-p$ can be negative then the solution is different...



                  Another idea is to try to rewrite it as a binary integer programming problem...







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 8 '18 at 15:16

























                  answered Dec 8 '18 at 14:52









                  AAKAAK

                  365




                  365






























                      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%2f3031046%2ffind-minimum-values-of-a-linear-system-over-mathbb-n%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

                      Ellipse (mathématiques)

                      Image
                      Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
                      Read more

                      Quarter-circle Tiles

                      Image
                      up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
                      Read more

                      Mont Emei

                      Image
                      Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
                      Read more