Is it always possible to fit these pieces in a square?












9














Consider all possible pairs of squares that can fit in a row of length $n$ where every square has a width of 1. If I have a large square of width $n$, can all such pairs of squares fit in the large square simultaneously?
It's hard to explain without an example so here is the case when $n=4$:



enter image description here



To the left are all the pairs of squares and to the right is a way for them to fit in the $n$ by $n$ square. Note that the pairs are not allowed to be moved horizontally but can be moved vertically.



It becomes a little harder but still possible when $n=5$:



enter image description here



What I'm interested in is the general case. Is it possible to fit all pairs in the square for any $n$?










share|cite|improve this question
























  • What does "all possible pieces (disconnected or connected) of width 2 and height of 1 that could fit in the square" mean? Can you show us how to generate the pieces for a given square?
    – John Douma
    Nov 18 '18 at 21:39












  • @JohnDouma Well a piece here simply consists of choosing two squares in a single row of length n. So the total number of pieces will be (n choose 2).
    – Pazzaz
    Nov 18 '18 at 21:46






  • 1




    I think you have to specify the problem more accurately. A triangle can have width $2$ and height $1$, for example, and that seems to be different from what you mean (and that is without thinking about figures which are not convex). I think you mean figures constructed of two $1times 1$ squares in a particular kind of configuration (as in your comment). And these sub-figures need not be connected. I do think your intended question is interesting, and worth asking, but the way you have asked it is misleading.
    – Mark Bennet
    Nov 18 '18 at 21:56










  • @MarkBennet I agree, I did leave out some important parts. I've edited the question some, it should be more clear now.
    – Pazzaz
    Nov 18 '18 at 22:14










  • Well, if possible, there will always be $n^2-2binom n2=n$ leftover spaces.
    – YiFan
    Nov 18 '18 at 23:44
















9














Consider all possible pairs of squares that can fit in a row of length $n$ where every square has a width of 1. If I have a large square of width $n$, can all such pairs of squares fit in the large square simultaneously?
It's hard to explain without an example so here is the case when $n=4$:



enter image description here



To the left are all the pairs of squares and to the right is a way for them to fit in the $n$ by $n$ square. Note that the pairs are not allowed to be moved horizontally but can be moved vertically.



It becomes a little harder but still possible when $n=5$:



enter image description here



What I'm interested in is the general case. Is it possible to fit all pairs in the square for any $n$?










share|cite|improve this question
























  • What does "all possible pieces (disconnected or connected) of width 2 and height of 1 that could fit in the square" mean? Can you show us how to generate the pieces for a given square?
    – John Douma
    Nov 18 '18 at 21:39












  • @JohnDouma Well a piece here simply consists of choosing two squares in a single row of length n. So the total number of pieces will be (n choose 2).
    – Pazzaz
    Nov 18 '18 at 21:46






  • 1




    I think you have to specify the problem more accurately. A triangle can have width $2$ and height $1$, for example, and that seems to be different from what you mean (and that is without thinking about figures which are not convex). I think you mean figures constructed of two $1times 1$ squares in a particular kind of configuration (as in your comment). And these sub-figures need not be connected. I do think your intended question is interesting, and worth asking, but the way you have asked it is misleading.
    – Mark Bennet
    Nov 18 '18 at 21:56










  • @MarkBennet I agree, I did leave out some important parts. I've edited the question some, it should be more clear now.
    – Pazzaz
    Nov 18 '18 at 22:14










  • Well, if possible, there will always be $n^2-2binom n2=n$ leftover spaces.
    – YiFan
    Nov 18 '18 at 23:44














9












9








9


3





Consider all possible pairs of squares that can fit in a row of length $n$ where every square has a width of 1. If I have a large square of width $n$, can all such pairs of squares fit in the large square simultaneously?
It's hard to explain without an example so here is the case when $n=4$:



enter image description here



To the left are all the pairs of squares and to the right is a way for them to fit in the $n$ by $n$ square. Note that the pairs are not allowed to be moved horizontally but can be moved vertically.



It becomes a little harder but still possible when $n=5$:



enter image description here



What I'm interested in is the general case. Is it possible to fit all pairs in the square for any $n$?










share|cite|improve this question















Consider all possible pairs of squares that can fit in a row of length $n$ where every square has a width of 1. If I have a large square of width $n$, can all such pairs of squares fit in the large square simultaneously?
It's hard to explain without an example so here is the case when $n=4$:



enter image description here



To the left are all the pairs of squares and to the right is a way for them to fit in the $n$ by $n$ square. Note that the pairs are not allowed to be moved horizontally but can be moved vertically.



It becomes a little harder but still possible when $n=5$:



enter image description here



What I'm interested in is the general case. Is it possible to fit all pairs in the square for any $n$?







recreational-mathematics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 18 '18 at 22:51









Jens

3,7382928




3,7382928










asked Nov 18 '18 at 21:21









Pazzaz

485




485












  • What does "all possible pieces (disconnected or connected) of width 2 and height of 1 that could fit in the square" mean? Can you show us how to generate the pieces for a given square?
    – John Douma
    Nov 18 '18 at 21:39












  • @JohnDouma Well a piece here simply consists of choosing two squares in a single row of length n. So the total number of pieces will be (n choose 2).
    – Pazzaz
    Nov 18 '18 at 21:46






  • 1




    I think you have to specify the problem more accurately. A triangle can have width $2$ and height $1$, for example, and that seems to be different from what you mean (and that is without thinking about figures which are not convex). I think you mean figures constructed of two $1times 1$ squares in a particular kind of configuration (as in your comment). And these sub-figures need not be connected. I do think your intended question is interesting, and worth asking, but the way you have asked it is misleading.
    – Mark Bennet
    Nov 18 '18 at 21:56










  • @MarkBennet I agree, I did leave out some important parts. I've edited the question some, it should be more clear now.
    – Pazzaz
    Nov 18 '18 at 22:14










  • Well, if possible, there will always be $n^2-2binom n2=n$ leftover spaces.
    – YiFan
    Nov 18 '18 at 23:44


















  • What does "all possible pieces (disconnected or connected) of width 2 and height of 1 that could fit in the square" mean? Can you show us how to generate the pieces for a given square?
    – John Douma
    Nov 18 '18 at 21:39












  • @JohnDouma Well a piece here simply consists of choosing two squares in a single row of length n. So the total number of pieces will be (n choose 2).
    – Pazzaz
    Nov 18 '18 at 21:46






  • 1




    I think you have to specify the problem more accurately. A triangle can have width $2$ and height $1$, for example, and that seems to be different from what you mean (and that is without thinking about figures which are not convex). I think you mean figures constructed of two $1times 1$ squares in a particular kind of configuration (as in your comment). And these sub-figures need not be connected. I do think your intended question is interesting, and worth asking, but the way you have asked it is misleading.
    – Mark Bennet
    Nov 18 '18 at 21:56










  • @MarkBennet I agree, I did leave out some important parts. I've edited the question some, it should be more clear now.
    – Pazzaz
    Nov 18 '18 at 22:14










  • Well, if possible, there will always be $n^2-2binom n2=n$ leftover spaces.
    – YiFan
    Nov 18 '18 at 23:44
















What does "all possible pieces (disconnected or connected) of width 2 and height of 1 that could fit in the square" mean? Can you show us how to generate the pieces for a given square?
– John Douma
Nov 18 '18 at 21:39






What does "all possible pieces (disconnected or connected) of width 2 and height of 1 that could fit in the square" mean? Can you show us how to generate the pieces for a given square?
– John Douma
Nov 18 '18 at 21:39














@JohnDouma Well a piece here simply consists of choosing two squares in a single row of length n. So the total number of pieces will be (n choose 2).
– Pazzaz
Nov 18 '18 at 21:46




@JohnDouma Well a piece here simply consists of choosing two squares in a single row of length n. So the total number of pieces will be (n choose 2).
– Pazzaz
Nov 18 '18 at 21:46




1




1




I think you have to specify the problem more accurately. A triangle can have width $2$ and height $1$, for example, and that seems to be different from what you mean (and that is without thinking about figures which are not convex). I think you mean figures constructed of two $1times 1$ squares in a particular kind of configuration (as in your comment). And these sub-figures need not be connected. I do think your intended question is interesting, and worth asking, but the way you have asked it is misleading.
– Mark Bennet
Nov 18 '18 at 21:56




I think you have to specify the problem more accurately. A triangle can have width $2$ and height $1$, for example, and that seems to be different from what you mean (and that is without thinking about figures which are not convex). I think you mean figures constructed of two $1times 1$ squares in a particular kind of configuration (as in your comment). And these sub-figures need not be connected. I do think your intended question is interesting, and worth asking, but the way you have asked it is misleading.
– Mark Bennet
Nov 18 '18 at 21:56












@MarkBennet I agree, I did leave out some important parts. I've edited the question some, it should be more clear now.
– Pazzaz
Nov 18 '18 at 22:14




@MarkBennet I agree, I did leave out some important parts. I've edited the question some, it should be more clear now.
– Pazzaz
Nov 18 '18 at 22:14












Well, if possible, there will always be $n^2-2binom n2=n$ leftover spaces.
– YiFan
Nov 18 '18 at 23:44




Well, if possible, there will always be $n^2-2binom n2=n$ leftover spaces.
– YiFan
Nov 18 '18 at 23:44










2 Answers
2






active

oldest

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6














It is always possible, we can place the $binom{n}{2}$ pairs in a $n times n$ square when $n$ is odd and in a $(n-1) times n$ rectangle when $n$ is even.



This problem is equivalent to the edge coloring problem for complete graph $K_n$. Look at wiki for the geometric intuition underlying following construction.





Let $[n]$ be a short hand for ${ 0, ldots, n-1 }$.



Index the set of possible pairs by $(i,j) in [n]^2$ with $i < j$.

Label rows and columns of the large square using numbers from $[n]$.



When $n$ is odd, place the pair $(i,j)$ at row $k$ of the large square where $i + j equiv k pmod n$.



If two pairs $(i_1,j_1)$, $(i_2,j_2)$ on same row intersect, then
one of the following happens
$$i_1 = i_2 lor i_1 = j_2lor j_1 = i_2 lor j_1 = j_2$$
Since $i_1 + j_1 equiv i_2 + j_2 pmod n$, we find
$$(i_1,j_1) = (i_2,j_2) pmod n lor (i_1,j_1) = (j_2,i_2) pmod n$$



Since $i_1,i_2,j_1,j_2 in [n]$ and $i_1 < j_1$, $i_2 < j_2$, we can rule out the second case. From this, we can deduce distinct pairs on some row are disjoint. This generate
a desired packing of the $binom{n}{2}$ pairs into a $n times n$ square.



When $n$ is even, $n - 1$ is odd.



Place those pair $(i,j) in [n-1]^2$ into row $k$ where $i + j equiv k pmod {n-1}$.
Notice




  • For each row $i in [n-1]$, the slot at column $2i pmod {n - 1}$ and $n-1$ is unused.

  • For any column $j in [n-1]$, one and only slot at row $i in [n-1]$ is unused.


For those pair $(i,j) in [n]^2 setminus [n-1]^2$ with $i < j$, we have $j = n$.
We can place the pair on row $k$ where $2k = i pmod {n-1}$. This will fill all the unused slots in the first $n-1$ rows and generate a desired packing of the $binom{n}{2}$ pairs into a $(n-1) times n$ rectangle.






share|cite|improve this answer































    1














    For 7×7, we have the following. X is an empty space, A through U are the 21 pairs of squares. For example, D D in the second row indicates that the {1,3} pair is used in that row, while the X in the seventh position means that spot is left empty.



    A A B B C X C



    D E D E F F X



    G H X I H I G



    J K K J X L L



    X M N O O M N



    P X Q R Q P R



    S T U X S U T
    .






    share|cite|improve this answer



















    • 1




      It looks like for odd n, the empty spots are all in different columns. In other words, by rearranging the rows you can create a blank diagonal.
      – Pazzaz
      Nov 19 '18 at 10:03











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    2 Answers
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    6














    It is always possible, we can place the $binom{n}{2}$ pairs in a $n times n$ square when $n$ is odd and in a $(n-1) times n$ rectangle when $n$ is even.



    This problem is equivalent to the edge coloring problem for complete graph $K_n$. Look at wiki for the geometric intuition underlying following construction.





    Let $[n]$ be a short hand for ${ 0, ldots, n-1 }$.



    Index the set of possible pairs by $(i,j) in [n]^2$ with $i < j$.

    Label rows and columns of the large square using numbers from $[n]$.



    When $n$ is odd, place the pair $(i,j)$ at row $k$ of the large square where $i + j equiv k pmod n$.



    If two pairs $(i_1,j_1)$, $(i_2,j_2)$ on same row intersect, then
    one of the following happens
    $$i_1 = i_2 lor i_1 = j_2lor j_1 = i_2 lor j_1 = j_2$$
    Since $i_1 + j_1 equiv i_2 + j_2 pmod n$, we find
    $$(i_1,j_1) = (i_2,j_2) pmod n lor (i_1,j_1) = (j_2,i_2) pmod n$$



    Since $i_1,i_2,j_1,j_2 in [n]$ and $i_1 < j_1$, $i_2 < j_2$, we can rule out the second case. From this, we can deduce distinct pairs on some row are disjoint. This generate
    a desired packing of the $binom{n}{2}$ pairs into a $n times n$ square.



    When $n$ is even, $n - 1$ is odd.



    Place those pair $(i,j) in [n-1]^2$ into row $k$ where $i + j equiv k pmod {n-1}$.
    Notice




    • For each row $i in [n-1]$, the slot at column $2i pmod {n - 1}$ and $n-1$ is unused.

    • For any column $j in [n-1]$, one and only slot at row $i in [n-1]$ is unused.


    For those pair $(i,j) in [n]^2 setminus [n-1]^2$ with $i < j$, we have $j = n$.
    We can place the pair on row $k$ where $2k = i pmod {n-1}$. This will fill all the unused slots in the first $n-1$ rows and generate a desired packing of the $binom{n}{2}$ pairs into a $(n-1) times n$ rectangle.






    share|cite|improve this answer




























      6














      It is always possible, we can place the $binom{n}{2}$ pairs in a $n times n$ square when $n$ is odd and in a $(n-1) times n$ rectangle when $n$ is even.



      This problem is equivalent to the edge coloring problem for complete graph $K_n$. Look at wiki for the geometric intuition underlying following construction.





      Let $[n]$ be a short hand for ${ 0, ldots, n-1 }$.



      Index the set of possible pairs by $(i,j) in [n]^2$ with $i < j$.

      Label rows and columns of the large square using numbers from $[n]$.



      When $n$ is odd, place the pair $(i,j)$ at row $k$ of the large square where $i + j equiv k pmod n$.



      If two pairs $(i_1,j_1)$, $(i_2,j_2)$ on same row intersect, then
      one of the following happens
      $$i_1 = i_2 lor i_1 = j_2lor j_1 = i_2 lor j_1 = j_2$$
      Since $i_1 + j_1 equiv i_2 + j_2 pmod n$, we find
      $$(i_1,j_1) = (i_2,j_2) pmod n lor (i_1,j_1) = (j_2,i_2) pmod n$$



      Since $i_1,i_2,j_1,j_2 in [n]$ and $i_1 < j_1$, $i_2 < j_2$, we can rule out the second case. From this, we can deduce distinct pairs on some row are disjoint. This generate
      a desired packing of the $binom{n}{2}$ pairs into a $n times n$ square.



      When $n$ is even, $n - 1$ is odd.



      Place those pair $(i,j) in [n-1]^2$ into row $k$ where $i + j equiv k pmod {n-1}$.
      Notice




      • For each row $i in [n-1]$, the slot at column $2i pmod {n - 1}$ and $n-1$ is unused.

      • For any column $j in [n-1]$, one and only slot at row $i in [n-1]$ is unused.


      For those pair $(i,j) in [n]^2 setminus [n-1]^2$ with $i < j$, we have $j = n$.
      We can place the pair on row $k$ where $2k = i pmod {n-1}$. This will fill all the unused slots in the first $n-1$ rows and generate a desired packing of the $binom{n}{2}$ pairs into a $(n-1) times n$ rectangle.






      share|cite|improve this answer


























        6












        6








        6






        It is always possible, we can place the $binom{n}{2}$ pairs in a $n times n$ square when $n$ is odd and in a $(n-1) times n$ rectangle when $n$ is even.



        This problem is equivalent to the edge coloring problem for complete graph $K_n$. Look at wiki for the geometric intuition underlying following construction.





        Let $[n]$ be a short hand for ${ 0, ldots, n-1 }$.



        Index the set of possible pairs by $(i,j) in [n]^2$ with $i < j$.

        Label rows and columns of the large square using numbers from $[n]$.



        When $n$ is odd, place the pair $(i,j)$ at row $k$ of the large square where $i + j equiv k pmod n$.



        If two pairs $(i_1,j_1)$, $(i_2,j_2)$ on same row intersect, then
        one of the following happens
        $$i_1 = i_2 lor i_1 = j_2lor j_1 = i_2 lor j_1 = j_2$$
        Since $i_1 + j_1 equiv i_2 + j_2 pmod n$, we find
        $$(i_1,j_1) = (i_2,j_2) pmod n lor (i_1,j_1) = (j_2,i_2) pmod n$$



        Since $i_1,i_2,j_1,j_2 in [n]$ and $i_1 < j_1$, $i_2 < j_2$, we can rule out the second case. From this, we can deduce distinct pairs on some row are disjoint. This generate
        a desired packing of the $binom{n}{2}$ pairs into a $n times n$ square.



        When $n$ is even, $n - 1$ is odd.



        Place those pair $(i,j) in [n-1]^2$ into row $k$ where $i + j equiv k pmod {n-1}$.
        Notice




        • For each row $i in [n-1]$, the slot at column $2i pmod {n - 1}$ and $n-1$ is unused.

        • For any column $j in [n-1]$, one and only slot at row $i in [n-1]$ is unused.


        For those pair $(i,j) in [n]^2 setminus [n-1]^2$ with $i < j$, we have $j = n$.
        We can place the pair on row $k$ where $2k = i pmod {n-1}$. This will fill all the unused slots in the first $n-1$ rows and generate a desired packing of the $binom{n}{2}$ pairs into a $(n-1) times n$ rectangle.






        share|cite|improve this answer














        It is always possible, we can place the $binom{n}{2}$ pairs in a $n times n$ square when $n$ is odd and in a $(n-1) times n$ rectangle when $n$ is even.



        This problem is equivalent to the edge coloring problem for complete graph $K_n$. Look at wiki for the geometric intuition underlying following construction.





        Let $[n]$ be a short hand for ${ 0, ldots, n-1 }$.



        Index the set of possible pairs by $(i,j) in [n]^2$ with $i < j$.

        Label rows and columns of the large square using numbers from $[n]$.



        When $n$ is odd, place the pair $(i,j)$ at row $k$ of the large square where $i + j equiv k pmod n$.



        If two pairs $(i_1,j_1)$, $(i_2,j_2)$ on same row intersect, then
        one of the following happens
        $$i_1 = i_2 lor i_1 = j_2lor j_1 = i_2 lor j_1 = j_2$$
        Since $i_1 + j_1 equiv i_2 + j_2 pmod n$, we find
        $$(i_1,j_1) = (i_2,j_2) pmod n lor (i_1,j_1) = (j_2,i_2) pmod n$$



        Since $i_1,i_2,j_1,j_2 in [n]$ and $i_1 < j_1$, $i_2 < j_2$, we can rule out the second case. From this, we can deduce distinct pairs on some row are disjoint. This generate
        a desired packing of the $binom{n}{2}$ pairs into a $n times n$ square.



        When $n$ is even, $n - 1$ is odd.



        Place those pair $(i,j) in [n-1]^2$ into row $k$ where $i + j equiv k pmod {n-1}$.
        Notice




        • For each row $i in [n-1]$, the slot at column $2i pmod {n - 1}$ and $n-1$ is unused.

        • For any column $j in [n-1]$, one and only slot at row $i in [n-1]$ is unused.


        For those pair $(i,j) in [n]^2 setminus [n-1]^2$ with $i < j$, we have $j = n$.
        We can place the pair on row $k$ where $2k = i pmod {n-1}$. This will fill all the unused slots in the first $n-1$ rows and generate a desired packing of the $binom{n}{2}$ pairs into a $(n-1) times n$ rectangle.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 19 '18 at 16:38

























        answered Nov 19 '18 at 16:23









        achille hui

        95.4k5130256




        95.4k5130256























            1














            For 7×7, we have the following. X is an empty space, A through U are the 21 pairs of squares. For example, D D in the second row indicates that the {1,3} pair is used in that row, while the X in the seventh position means that spot is left empty.



            A A B B C X C



            D E D E F F X



            G H X I H I G



            J K K J X L L



            X M N O O M N



            P X Q R Q P R



            S T U X S U T
            .






            share|cite|improve this answer



















            • 1




              It looks like for odd n, the empty spots are all in different columns. In other words, by rearranging the rows you can create a blank diagonal.
              – Pazzaz
              Nov 19 '18 at 10:03
















            1














            For 7×7, we have the following. X is an empty space, A through U are the 21 pairs of squares. For example, D D in the second row indicates that the {1,3} pair is used in that row, while the X in the seventh position means that spot is left empty.



            A A B B C X C



            D E D E F F X



            G H X I H I G



            J K K J X L L



            X M N O O M N



            P X Q R Q P R



            S T U X S U T
            .






            share|cite|improve this answer



















            • 1




              It looks like for odd n, the empty spots are all in different columns. In other words, by rearranging the rows you can create a blank diagonal.
              – Pazzaz
              Nov 19 '18 at 10:03














            1












            1








            1






            For 7×7, we have the following. X is an empty space, A through U are the 21 pairs of squares. For example, D D in the second row indicates that the {1,3} pair is used in that row, while the X in the seventh position means that spot is left empty.



            A A B B C X C



            D E D E F F X



            G H X I H I G



            J K K J X L L



            X M N O O M N



            P X Q R Q P R



            S T U X S U T
            .






            share|cite|improve this answer














            For 7×7, we have the following. X is an empty space, A through U are the 21 pairs of squares. For example, D D in the second row indicates that the {1,3} pair is used in that row, while the X in the seventh position means that spot is left empty.



            A A B B C X C



            D E D E F F X



            G H X I H I G



            J K K J X L L



            X M N O O M N



            P X Q R Q P R



            S T U X S U T
            .







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 19 '18 at 1:15

























            answered Nov 19 '18 at 1:02









            Oscar Lanzi

            12.1k12036




            12.1k12036








            • 1




              It looks like for odd n, the empty spots are all in different columns. In other words, by rearranging the rows you can create a blank diagonal.
              – Pazzaz
              Nov 19 '18 at 10:03














            • 1




              It looks like for odd n, the empty spots are all in different columns. In other words, by rearranging the rows you can create a blank diagonal.
              – Pazzaz
              Nov 19 '18 at 10:03








            1




            1




            It looks like for odd n, the empty spots are all in different columns. In other words, by rearranging the rows you can create a blank diagonal.
            – Pazzaz
            Nov 19 '18 at 10:03




            It looks like for odd n, the empty spots are all in different columns. In other words, by rearranging the rows you can create a blank diagonal.
            – Pazzaz
            Nov 19 '18 at 10:03


















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