XOR of 6 numbers in pair of 3












-1












$begingroup$


I have this question where u are given 6 numbers with indices 1,2,3,4,5,6 and u are given the xor of these numbers in pair of 3.



$$x_1oplus x_2⊕x_3=a\
x_2⊕x_3⊕x_4=b\
x_3⊕x_4⊕x_5=c\
x_4⊕x_5⊕x_6=d\
x_5⊕x_6⊕x_1=e\
x_6⊕x_1⊕x_2=f$$



and we need to find all the six numbers. What i tried was to take xor of these triplets to get xor of any other pair (which worked for me for 4 and 5 numbers) but this method is not working for 6 numbers.



my attempt for n=4.



$$x_1⊕x_2⊕x_3=a\
x_2⊕x_3⊕x_4=b\
x_3⊕x_4⊕x_1=c\
x_4⊕x_1oplus x_2=d$$



$a⊕b⊕c$ gives me the 3rd number. Similarly we can get the rest of the numbers, but i cant figure out a solution for 6 numbers.






elementary-number-theory logic







share|cite|improve this question











$endgroup$

















    -1












    $begingroup$


    I have this question where u are given 6 numbers with indices 1,2,3,4,5,6 and u are given the xor of these numbers in pair of 3.



    $$x_1oplus x_2⊕x_3=a\
    x_2⊕x_3⊕x_4=b\
    x_3⊕x_4⊕x_5=c\
    x_4⊕x_5⊕x_6=d\
    x_5⊕x_6⊕x_1=e\
    x_6⊕x_1⊕x_2=f$$



    and we need to find all the six numbers. What i tried was to take xor of these triplets to get xor of any other pair (which worked for me for 4 and 5 numbers) but this method is not working for 6 numbers.



    my attempt for n=4.



    $$x_1⊕x_2⊕x_3=a\
    x_2⊕x_3⊕x_4=b\
    x_3⊕x_4⊕x_1=c\
    x_4⊕x_1oplus x_2=d$$



    $a⊕b⊕c$ gives me the 3rd number. Similarly we can get the rest of the numbers, but i cant figure out a solution for 6 numbers.






    elementary-number-theory logic







    share|cite|improve this question











    $endgroup$















      -1












      -1








      -1





      $begingroup$


      I have this question where u are given 6 numbers with indices 1,2,3,4,5,6 and u are given the xor of these numbers in pair of 3.



      $$x_1oplus x_2⊕x_3=a\
      x_2⊕x_3⊕x_4=b\
      x_3⊕x_4⊕x_5=c\
      x_4⊕x_5⊕x_6=d\
      x_5⊕x_6⊕x_1=e\
      x_6⊕x_1⊕x_2=f$$



      and we need to find all the six numbers. What i tried was to take xor of these triplets to get xor of any other pair (which worked for me for 4 and 5 numbers) but this method is not working for 6 numbers.



      my attempt for n=4.



      $$x_1⊕x_2⊕x_3=a\
      x_2⊕x_3⊕x_4=b\
      x_3⊕x_4⊕x_1=c\
      x_4⊕x_1oplus x_2=d$$



      $a⊕b⊕c$ gives me the 3rd number. Similarly we can get the rest of the numbers, but i cant figure out a solution for 6 numbers.






      elementary-number-theory logic







      share|cite|improve this question











      $endgroup$




      I have this question where u are given 6 numbers with indices 1,2,3,4,5,6 and u are given the xor of these numbers in pair of 3.



      $$x_1oplus x_2⊕x_3=a\
      x_2⊕x_3⊕x_4=b\
      x_3⊕x_4⊕x_5=c\
      x_4⊕x_5⊕x_6=d\
      x_5⊕x_6⊕x_1=e\
      x_6⊕x_1⊕x_2=f$$



      and we need to find all the six numbers. What i tried was to take xor of these triplets to get xor of any other pair (which worked for me for 4 and 5 numbers) but this method is not working for 6 numbers.



      my attempt for n=4.



      $$x_1⊕x_2⊕x_3=a\
      x_2⊕x_3⊕x_4=b\
      x_3⊕x_4⊕x_1=c\
      x_4⊕x_1oplus x_2=d$$



      $a⊕b⊕c$ gives me the 3rd number. Similarly we can get the rest of the numbers, but i cant figure out a solution for 6 numbers.






      elementary-number-theory logic




      elementary-number-theory logic






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 12 '18 at 8:59









      Ivan Neretin

      8,93021535




      8,93021535










      asked Dec 12 '18 at 8:30









      bigibigi

      1




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          3 Answers
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          $begingroup$

          Guide:



          We have



          $$x_1 oplus x_2 oplus x_3 oplus x_4 oplus x_5 oplus x_6= aoplus d= b oplus e = c oplus f$$



          Hence, to be feasible, we require $$aoplus d= b oplus e = c oplus f.$$



          Suppose the system above holds, to solve the system, we first choose our desired $x_5$ and $x_6$ and solve for $x_4, x_3, x_2, x_1$ in that order.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            You can't do that.



            Say, our xors of triplets are $(a,b,c,d,e,f)=(0,0,0,0,0,0)$. Can we guess what the numbers were?



            Why, sure: $(x_1,x_2,x_3,x_4,x_5,x_6)=(0,0,0,0,0,0)$.



            But wait, there is another solution: $(x_1,x_2,x_3,x_4,x_5,x_6)=(1,2,3,1,2,3)$. (Don't take my word for it! Check it yourself, make sure it works.)



            And there are many more: $(x_1,x_2,x_3,x_4,x_5,x_6)=(3,4,7,3,4,7)$... and so on.



            How can that be, might you ask? After all, we were able to solve the system for $n=4$ and $5$, weren't we?



            Well, it just turns out there is something peculiar about the number $6$ that makes it behave differently from $4$ and $5$. What is this trait I leave for you to discover.






            share|cite|improve this answer









            $endgroup$





















              -1












              $begingroup$

              for $6$ numbers you can try:
              begin{align*}
              x_1⊕x_2⊕x_3&=a\
              x_1⊕x_2⊕x_4&=b\
              x_3⊕x_4⊕x_5&=c\
              x_3⊕x_4⊕x_6&=d\
              x_5⊕x_6⊕x_1&=e\
              x_6⊕x_5⊕x_2&=f
              end{align*}
              there is actually a pattern, let me give another example:
              for $9$ numbers you can try:
              begin{align*}
              x_1⊕x_2⊕x_3&=a\
              x_1⊕x_2⊕x_4&=b\
              x_3⊕x_4⊕x_5&=c\
              x_3⊕x_4⊕x_6&=d\
              x_5⊕x_6⊕x_7&=e\
              x_6⊕x_7⊕x_8&=f\
              x_7⊕x_8⊕x_9&=g\
              x_8⊕x_9⊕x_1&=h\
              x_9⊕x_5⊕x_2&=i
              end{align*}






              share|cite|improve this answer











              $endgroup$














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                3 Answers
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                3 Answers
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                0












                $begingroup$

                Guide:



                We have



                $$x_1 oplus x_2 oplus x_3 oplus x_4 oplus x_5 oplus x_6= aoplus d= b oplus e = c oplus f$$



                Hence, to be feasible, we require $$aoplus d= b oplus e = c oplus f.$$



                Suppose the system above holds, to solve the system, we first choose our desired $x_5$ and $x_6$ and solve for $x_4, x_3, x_2, x_1$ in that order.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  Guide:



                  We have



                  $$x_1 oplus x_2 oplus x_3 oplus x_4 oplus x_5 oplus x_6= aoplus d= b oplus e = c oplus f$$



                  Hence, to be feasible, we require $$aoplus d= b oplus e = c oplus f.$$



                  Suppose the system above holds, to solve the system, we first choose our desired $x_5$ and $x_6$ and solve for $x_4, x_3, x_2, x_1$ in that order.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    Guide:



                    We have



                    $$x_1 oplus x_2 oplus x_3 oplus x_4 oplus x_5 oplus x_6= aoplus d= b oplus e = c oplus f$$



                    Hence, to be feasible, we require $$aoplus d= b oplus e = c oplus f.$$



                    Suppose the system above holds, to solve the system, we first choose our desired $x_5$ and $x_6$ and solve for $x_4, x_3, x_2, x_1$ in that order.






                    share|cite|improve this answer









                    $endgroup$



                    Guide:



                    We have



                    $$x_1 oplus x_2 oplus x_3 oplus x_4 oplus x_5 oplus x_6= aoplus d= b oplus e = c oplus f$$



                    Hence, to be feasible, we require $$aoplus d= b oplus e = c oplus f.$$



                    Suppose the system above holds, to solve the system, we first choose our desired $x_5$ and $x_6$ and solve for $x_4, x_3, x_2, x_1$ in that order.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 12 '18 at 8:46









                    Siong Thye GohSiong Thye Goh

                    103k1468120




                    103k1468120























                        0












                        $begingroup$

                        You can't do that.



                        Say, our xors of triplets are $(a,b,c,d,e,f)=(0,0,0,0,0,0)$. Can we guess what the numbers were?



                        Why, sure: $(x_1,x_2,x_3,x_4,x_5,x_6)=(0,0,0,0,0,0)$.



                        But wait, there is another solution: $(x_1,x_2,x_3,x_4,x_5,x_6)=(1,2,3,1,2,3)$. (Don't take my word for it! Check it yourself, make sure it works.)



                        And there are many more: $(x_1,x_2,x_3,x_4,x_5,x_6)=(3,4,7,3,4,7)$... and so on.



                        How can that be, might you ask? After all, we were able to solve the system for $n=4$ and $5$, weren't we?



                        Well, it just turns out there is something peculiar about the number $6$ that makes it behave differently from $4$ and $5$. What is this trait I leave for you to discover.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          You can't do that.



                          Say, our xors of triplets are $(a,b,c,d,e,f)=(0,0,0,0,0,0)$. Can we guess what the numbers were?



                          Why, sure: $(x_1,x_2,x_3,x_4,x_5,x_6)=(0,0,0,0,0,0)$.



                          But wait, there is another solution: $(x_1,x_2,x_3,x_4,x_5,x_6)=(1,2,3,1,2,3)$. (Don't take my word for it! Check it yourself, make sure it works.)



                          And there are many more: $(x_1,x_2,x_3,x_4,x_5,x_6)=(3,4,7,3,4,7)$... and so on.



                          How can that be, might you ask? After all, we were able to solve the system for $n=4$ and $5$, weren't we?



                          Well, it just turns out there is something peculiar about the number $6$ that makes it behave differently from $4$ and $5$. What is this trait I leave for you to discover.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            You can't do that.



                            Say, our xors of triplets are $(a,b,c,d,e,f)=(0,0,0,0,0,0)$. Can we guess what the numbers were?



                            Why, sure: $(x_1,x_2,x_3,x_4,x_5,x_6)=(0,0,0,0,0,0)$.



                            But wait, there is another solution: $(x_1,x_2,x_3,x_4,x_5,x_6)=(1,2,3,1,2,3)$. (Don't take my word for it! Check it yourself, make sure it works.)



                            And there are many more: $(x_1,x_2,x_3,x_4,x_5,x_6)=(3,4,7,3,4,7)$... and so on.



                            How can that be, might you ask? After all, we were able to solve the system for $n=4$ and $5$, weren't we?



                            Well, it just turns out there is something peculiar about the number $6$ that makes it behave differently from $4$ and $5$. What is this trait I leave for you to discover.






                            share|cite|improve this answer









                            $endgroup$



                            You can't do that.



                            Say, our xors of triplets are $(a,b,c,d,e,f)=(0,0,0,0,0,0)$. Can we guess what the numbers were?



                            Why, sure: $(x_1,x_2,x_3,x_4,x_5,x_6)=(0,0,0,0,0,0)$.



                            But wait, there is another solution: $(x_1,x_2,x_3,x_4,x_5,x_6)=(1,2,3,1,2,3)$. (Don't take my word for it! Check it yourself, make sure it works.)



                            And there are many more: $(x_1,x_2,x_3,x_4,x_5,x_6)=(3,4,7,3,4,7)$... and so on.



                            How can that be, might you ask? After all, we were able to solve the system for $n=4$ and $5$, weren't we?



                            Well, it just turns out there is something peculiar about the number $6$ that makes it behave differently from $4$ and $5$. What is this trait I leave for you to discover.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 12 '18 at 8:46









                            Ivan NeretinIvan Neretin

                            8,93021535




                            8,93021535























                                -1












                                $begingroup$

                                for $6$ numbers you can try:
                                begin{align*}
                                x_1⊕x_2⊕x_3&=a\
                                x_1⊕x_2⊕x_4&=b\
                                x_3⊕x_4⊕x_5&=c\
                                x_3⊕x_4⊕x_6&=d\
                                x_5⊕x_6⊕x_1&=e\
                                x_6⊕x_5⊕x_2&=f
                                end{align*}
                                there is actually a pattern, let me give another example:
                                for $9$ numbers you can try:
                                begin{align*}
                                x_1⊕x_2⊕x_3&=a\
                                x_1⊕x_2⊕x_4&=b\
                                x_3⊕x_4⊕x_5&=c\
                                x_3⊕x_4⊕x_6&=d\
                                x_5⊕x_6⊕x_7&=e\
                                x_6⊕x_7⊕x_8&=f\
                                x_7⊕x_8⊕x_9&=g\
                                x_8⊕x_9⊕x_1&=h\
                                x_9⊕x_5⊕x_2&=i
                                end{align*}






                                share|cite|improve this answer











                                $endgroup$


















                                  -1












                                  $begingroup$

                                  for $6$ numbers you can try:
                                  begin{align*}
                                  x_1⊕x_2⊕x_3&=a\
                                  x_1⊕x_2⊕x_4&=b\
                                  x_3⊕x_4⊕x_5&=c\
                                  x_3⊕x_4⊕x_6&=d\
                                  x_5⊕x_6⊕x_1&=e\
                                  x_6⊕x_5⊕x_2&=f
                                  end{align*}
                                  there is actually a pattern, let me give another example:
                                  for $9$ numbers you can try:
                                  begin{align*}
                                  x_1⊕x_2⊕x_3&=a\
                                  x_1⊕x_2⊕x_4&=b\
                                  x_3⊕x_4⊕x_5&=c\
                                  x_3⊕x_4⊕x_6&=d\
                                  x_5⊕x_6⊕x_7&=e\
                                  x_6⊕x_7⊕x_8&=f\
                                  x_7⊕x_8⊕x_9&=g\
                                  x_8⊕x_9⊕x_1&=h\
                                  x_9⊕x_5⊕x_2&=i
                                  end{align*}






                                  share|cite|improve this answer











                                  $endgroup$
















                                    -1












                                    -1








                                    -1





                                    $begingroup$

                                    for $6$ numbers you can try:
                                    begin{align*}
                                    x_1⊕x_2⊕x_3&=a\
                                    x_1⊕x_2⊕x_4&=b\
                                    x_3⊕x_4⊕x_5&=c\
                                    x_3⊕x_4⊕x_6&=d\
                                    x_5⊕x_6⊕x_1&=e\
                                    x_6⊕x_5⊕x_2&=f
                                    end{align*}
                                    there is actually a pattern, let me give another example:
                                    for $9$ numbers you can try:
                                    begin{align*}
                                    x_1⊕x_2⊕x_3&=a\
                                    x_1⊕x_2⊕x_4&=b\
                                    x_3⊕x_4⊕x_5&=c\
                                    x_3⊕x_4⊕x_6&=d\
                                    x_5⊕x_6⊕x_7&=e\
                                    x_6⊕x_7⊕x_8&=f\
                                    x_7⊕x_8⊕x_9&=g\
                                    x_8⊕x_9⊕x_1&=h\
                                    x_9⊕x_5⊕x_2&=i
                                    end{align*}






                                    share|cite|improve this answer











                                    $endgroup$



                                    for $6$ numbers you can try:
                                    begin{align*}
                                    x_1⊕x_2⊕x_3&=a\
                                    x_1⊕x_2⊕x_4&=b\
                                    x_3⊕x_4⊕x_5&=c\
                                    x_3⊕x_4⊕x_6&=d\
                                    x_5⊕x_6⊕x_1&=e\
                                    x_6⊕x_5⊕x_2&=f
                                    end{align*}
                                    there is actually a pattern, let me give another example:
                                    for $9$ numbers you can try:
                                    begin{align*}
                                    x_1⊕x_2⊕x_3&=a\
                                    x_1⊕x_2⊕x_4&=b\
                                    x_3⊕x_4⊕x_5&=c\
                                    x_3⊕x_4⊕x_6&=d\
                                    x_5⊕x_6⊕x_7&=e\
                                    x_6⊕x_7⊕x_8&=f\
                                    x_7⊕x_8⊕x_9&=g\
                                    x_8⊕x_9⊕x_1&=h\
                                    x_9⊕x_5⊕x_2&=i
                                    end{align*}







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    edited Dec 14 '18 at 14:09









                                    Tianlalu

                                    3,27521238




                                    3,27521238










                                    answered Dec 14 '18 at 13:38









                                    glennmarkglennmark

                                    1




                                    1






























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                                        Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

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                                        up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
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                                        Is anime1.com a legal site for watching anime?

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                                        up vote 4 down vote favorite 1 Other anime streaming sites are blocking my IP because I'm Iranian while anime1.com isn't. I wanted to know if it's legal to watch/stream anime from this site. Thanks. anime-production share | improve this question edited Nov 27 at 22:38 W. Are 1,599 2 23 asked Nov 27 at 20:36 mhnajafi 23 3 add a comment ...
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