Arrange n tiles in a rectangle pattern












1














Is there a way to find how to arrange n tiles in order to form a rectangle that is closest to a square? As I result I am looking for the dimension of that rectangle if it exists.



For example, here are the results I expect from 1 to 9.




  1. a 1 by 1 square, fairly simple.

  2. a 1 by 2 rectangle.

  3. a 1 by 3 rectangle.

  4. a 2 by 2 square.

  5. this is impossible because a 1 by 5 rectangle is thinner than the previous shape.

  6. a 2 by 3 rectangle.

  7. this is impossible.

  8. a 2 by 4 rectangle.

  9. a 3 by 3 square.


I am only looking at n smaller than 100 but would rather not have to hardcode everything.






geometry square-numbers rectangles







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    1














    Is there a way to find how to arrange n tiles in order to form a rectangle that is closest to a square? As I result I am looking for the dimension of that rectangle if it exists.



    For example, here are the results I expect from 1 to 9.




    1. a 1 by 1 square, fairly simple.

    2. a 1 by 2 rectangle.

    3. a 1 by 3 rectangle.

    4. a 2 by 2 square.

    5. this is impossible because a 1 by 5 rectangle is thinner than the previous shape.

    6. a 2 by 3 rectangle.

    7. this is impossible.

    8. a 2 by 4 rectangle.

    9. a 3 by 3 square.


    I am only looking at n smaller than 100 but would rather not have to hardcode everything.






    geometry square-numbers rectangles







    share|cite|improve this question







    New contributor




    Alex Laquerre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.























      1












      1








      1







      Is there a way to find how to arrange n tiles in order to form a rectangle that is closest to a square? As I result I am looking for the dimension of that rectangle if it exists.



      For example, here are the results I expect from 1 to 9.




      1. a 1 by 1 square, fairly simple.

      2. a 1 by 2 rectangle.

      3. a 1 by 3 rectangle.

      4. a 2 by 2 square.

      5. this is impossible because a 1 by 5 rectangle is thinner than the previous shape.

      6. a 2 by 3 rectangle.

      7. this is impossible.

      8. a 2 by 4 rectangle.

      9. a 3 by 3 square.


      I am only looking at n smaller than 100 but would rather not have to hardcode everything.






      geometry square-numbers rectangles







      share|cite|improve this question







      New contributor




      Alex Laquerre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      Is there a way to find how to arrange n tiles in order to form a rectangle that is closest to a square? As I result I am looking for the dimension of that rectangle if it exists.



      For example, here are the results I expect from 1 to 9.




      1. a 1 by 1 square, fairly simple.

      2. a 1 by 2 rectangle.

      3. a 1 by 3 rectangle.

      4. a 2 by 2 square.

      5. this is impossible because a 1 by 5 rectangle is thinner than the previous shape.

      6. a 2 by 3 rectangle.

      7. this is impossible.

      8. a 2 by 4 rectangle.

      9. a 3 by 3 square.


      I am only looking at n smaller than 100 but would rather not have to hardcode everything.






      geometry square-numbers rectangles




      geometry square-numbers rectangles






      share|cite|improve this question







      New contributor




      Alex Laquerre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question







      New contributor




      Alex Laquerre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question






      New contributor




      Alex Laquerre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked 2 hours ago









      Alex Laquerre

      61




      61




      New contributor




      Alex Laquerre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      Alex Laquerre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Alex Laquerre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















          1 Answer
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          4














          Although your problem statement doesn't explicitly state it, your examples indicate that you want the value of the lower of the $2$ factors to never decrease. Thus, each increase of the lower factor among your examples occurs at a square, such as $1$ being $1 text{ by } 1$, $4$ being $2 text{ by } 2$ and $9$ being $3 text{ by } 3$.



          If this is a correct assumption on my part, and there are no other unstated conditions, then each next value which is not "impossible" would be the integers which are multiples of the current "low" factor until you encounter the square of the next higher integer. Thus, continuing from what you have written, the next valid values would be $12 = 3 times 4$ and $15 = 3 times 5$ for a lower value of $3$ before you hit the next square at $16 = 4 times 4$. From there, the next values would be $20 = 4 times 5$ and $24 = 4 times 6$ before you encounter $25 = 5 times 5$. You can then continue like this until you reach $100$ or most any other limit you care to set.



          To check for any specific value representing a valid rectangle, including what its shape would be, or if it's impossible, you need to determine what the largest perfect square is less or than or equal to your number. This can be done by using the integer part of the square root. For example, the square root of $143$ is $11.9ldots$ so the factor to check is $11$. As $143 = 11 times 13$, it is a valid rectangle. However, $142$ down to $133$ will not work, with $132 = 11 times 12$ being the next smaller value which would work.



          Stated more succinctly using mathematical terminology, for any positive integer $n$, determine $k = lfloor sqrt{n} rfloor$. If $k$ divides $n$, then the rectangle is $k times frac{n}{k}$, else it's impossible.






          share|cite|improve this answer























          • No, you're completely right. However, how could, given n of let's say 94, I know what's the shape without calculating every number by hand?
            – Alex Laquerre
            2 hours ago






          • 2




            Just determine the smallest perfect square less than or equal to the value. For $94$, that is $81 = 9 times 9$. Next, check if $9$ is a factor of $94$. It isn't in this case, so $94$ would be considered "impossible" by your algorithm. I hope this makes sense.
            – John Omielan
            2 hours ago








          • 1




            What I gather from John's answer is that, given $n$, you find $d$, the integer part of the square root of $n$, and then you check whether $d$ divides $n$. If it does, then $d$ by $n/d$ is what you want. If it doesn't, then that $n$ doesn't work.
            – Gerry Myerson
            2 hours ago






          • 1




            Stated as an algorithm, given $n$ compute $k=lfloor sqrt n rfloor$. If $k$ divides $n$ the rectangle is $k times frac nk$. If $k$ does not divide $n$ it is impossible. Good answer.
            – Ross Millikan
            1 hour ago








          • 1




            You are welcome to edit it into your answer, which will make it more accessible to readers.
            – Ross Millikan
            1 hour ago











          Your Answer





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          Although your problem statement doesn't explicitly state it, your examples indicate that you want the value of the lower of the $2$ factors to never decrease. Thus, each increase of the lower factor among your examples occurs at a square, such as $1$ being $1 text{ by } 1$, $4$ being $2 text{ by } 2$ and $9$ being $3 text{ by } 3$.



          If this is a correct assumption on my part, and there are no other unstated conditions, then each next value which is not "impossible" would be the integers which are multiples of the current "low" factor until you encounter the square of the next higher integer. Thus, continuing from what you have written, the next valid values would be $12 = 3 times 4$ and $15 = 3 times 5$ for a lower value of $3$ before you hit the next square at $16 = 4 times 4$. From there, the next values would be $20 = 4 times 5$ and $24 = 4 times 6$ before you encounter $25 = 5 times 5$. You can then continue like this until you reach $100$ or most any other limit you care to set.



          To check for any specific value representing a valid rectangle, including what its shape would be, or if it's impossible, you need to determine what the largest perfect square is less or than or equal to your number. This can be done by using the integer part of the square root. For example, the square root of $143$ is $11.9ldots$ so the factor to check is $11$. As $143 = 11 times 13$, it is a valid rectangle. However, $142$ down to $133$ will not work, with $132 = 11 times 12$ being the next smaller value which would work.



          Stated more succinctly using mathematical terminology, for any positive integer $n$, determine $k = lfloor sqrt{n} rfloor$. If $k$ divides $n$, then the rectangle is $k times frac{n}{k}$, else it's impossible.






          share|cite|improve this answer























          • No, you're completely right. However, how could, given n of let's say 94, I know what's the shape without calculating every number by hand?
            – Alex Laquerre
            2 hours ago






          • 2




            Just determine the smallest perfect square less than or equal to the value. For $94$, that is $81 = 9 times 9$. Next, check if $9$ is a factor of $94$. It isn't in this case, so $94$ would be considered "impossible" by your algorithm. I hope this makes sense.
            – John Omielan
            2 hours ago








          • 1




            What I gather from John's answer is that, given $n$, you find $d$, the integer part of the square root of $n$, and then you check whether $d$ divides $n$. If it does, then $d$ by $n/d$ is what you want. If it doesn't, then that $n$ doesn't work.
            – Gerry Myerson
            2 hours ago






          • 1




            Stated as an algorithm, given $n$ compute $k=lfloor sqrt n rfloor$. If $k$ divides $n$ the rectangle is $k times frac nk$. If $k$ does not divide $n$ it is impossible. Good answer.
            – Ross Millikan
            1 hour ago








          • 1




            You are welcome to edit it into your answer, which will make it more accessible to readers.
            – Ross Millikan
            1 hour ago
















          4














          Although your problem statement doesn't explicitly state it, your examples indicate that you want the value of the lower of the $2$ factors to never decrease. Thus, each increase of the lower factor among your examples occurs at a square, such as $1$ being $1 text{ by } 1$, $4$ being $2 text{ by } 2$ and $9$ being $3 text{ by } 3$.



          If this is a correct assumption on my part, and there are no other unstated conditions, then each next value which is not "impossible" would be the integers which are multiples of the current "low" factor until you encounter the square of the next higher integer. Thus, continuing from what you have written, the next valid values would be $12 = 3 times 4$ and $15 = 3 times 5$ for a lower value of $3$ before you hit the next square at $16 = 4 times 4$. From there, the next values would be $20 = 4 times 5$ and $24 = 4 times 6$ before you encounter $25 = 5 times 5$. You can then continue like this until you reach $100$ or most any other limit you care to set.



          To check for any specific value representing a valid rectangle, including what its shape would be, or if it's impossible, you need to determine what the largest perfect square is less or than or equal to your number. This can be done by using the integer part of the square root. For example, the square root of $143$ is $11.9ldots$ so the factor to check is $11$. As $143 = 11 times 13$, it is a valid rectangle. However, $142$ down to $133$ will not work, with $132 = 11 times 12$ being the next smaller value which would work.



          Stated more succinctly using mathematical terminology, for any positive integer $n$, determine $k = lfloor sqrt{n} rfloor$. If $k$ divides $n$, then the rectangle is $k times frac{n}{k}$, else it's impossible.






          share|cite|improve this answer























          • No, you're completely right. However, how could, given n of let's say 94, I know what's the shape without calculating every number by hand?
            – Alex Laquerre
            2 hours ago






          • 2




            Just determine the smallest perfect square less than or equal to the value. For $94$, that is $81 = 9 times 9$. Next, check if $9$ is a factor of $94$. It isn't in this case, so $94$ would be considered "impossible" by your algorithm. I hope this makes sense.
            – John Omielan
            2 hours ago








          • 1




            What I gather from John's answer is that, given $n$, you find $d$, the integer part of the square root of $n$, and then you check whether $d$ divides $n$. If it does, then $d$ by $n/d$ is what you want. If it doesn't, then that $n$ doesn't work.
            – Gerry Myerson
            2 hours ago






          • 1




            Stated as an algorithm, given $n$ compute $k=lfloor sqrt n rfloor$. If $k$ divides $n$ the rectangle is $k times frac nk$. If $k$ does not divide $n$ it is impossible. Good answer.
            – Ross Millikan
            1 hour ago








          • 1




            You are welcome to edit it into your answer, which will make it more accessible to readers.
            – Ross Millikan
            1 hour ago














          4












          4








          4






          Although your problem statement doesn't explicitly state it, your examples indicate that you want the value of the lower of the $2$ factors to never decrease. Thus, each increase of the lower factor among your examples occurs at a square, such as $1$ being $1 text{ by } 1$, $4$ being $2 text{ by } 2$ and $9$ being $3 text{ by } 3$.



          If this is a correct assumption on my part, and there are no other unstated conditions, then each next value which is not "impossible" would be the integers which are multiples of the current "low" factor until you encounter the square of the next higher integer. Thus, continuing from what you have written, the next valid values would be $12 = 3 times 4$ and $15 = 3 times 5$ for a lower value of $3$ before you hit the next square at $16 = 4 times 4$. From there, the next values would be $20 = 4 times 5$ and $24 = 4 times 6$ before you encounter $25 = 5 times 5$. You can then continue like this until you reach $100$ or most any other limit you care to set.



          To check for any specific value representing a valid rectangle, including what its shape would be, or if it's impossible, you need to determine what the largest perfect square is less or than or equal to your number. This can be done by using the integer part of the square root. For example, the square root of $143$ is $11.9ldots$ so the factor to check is $11$. As $143 = 11 times 13$, it is a valid rectangle. However, $142$ down to $133$ will not work, with $132 = 11 times 12$ being the next smaller value which would work.



          Stated more succinctly using mathematical terminology, for any positive integer $n$, determine $k = lfloor sqrt{n} rfloor$. If $k$ divides $n$, then the rectangle is $k times frac{n}{k}$, else it's impossible.






          share|cite|improve this answer














          Although your problem statement doesn't explicitly state it, your examples indicate that you want the value of the lower of the $2$ factors to never decrease. Thus, each increase of the lower factor among your examples occurs at a square, such as $1$ being $1 text{ by } 1$, $4$ being $2 text{ by } 2$ and $9$ being $3 text{ by } 3$.



          If this is a correct assumption on my part, and there are no other unstated conditions, then each next value which is not "impossible" would be the integers which are multiples of the current "low" factor until you encounter the square of the next higher integer. Thus, continuing from what you have written, the next valid values would be $12 = 3 times 4$ and $15 = 3 times 5$ for a lower value of $3$ before you hit the next square at $16 = 4 times 4$. From there, the next values would be $20 = 4 times 5$ and $24 = 4 times 6$ before you encounter $25 = 5 times 5$. You can then continue like this until you reach $100$ or most any other limit you care to set.



          To check for any specific value representing a valid rectangle, including what its shape would be, or if it's impossible, you need to determine what the largest perfect square is less or than or equal to your number. This can be done by using the integer part of the square root. For example, the square root of $143$ is $11.9ldots$ so the factor to check is $11$. As $143 = 11 times 13$, it is a valid rectangle. However, $142$ down to $133$ will not work, with $132 = 11 times 12$ being the next smaller value which would work.



          Stated more succinctly using mathematical terminology, for any positive integer $n$, determine $k = lfloor sqrt{n} rfloor$. If $k$ divides $n$, then the rectangle is $k times frac{n}{k}$, else it's impossible.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 30 mins ago

























          answered 2 hours ago









          John Omielan

          72418




          72418












          • No, you're completely right. However, how could, given n of let's say 94, I know what's the shape without calculating every number by hand?
            – Alex Laquerre
            2 hours ago






          • 2




            Just determine the smallest perfect square less than or equal to the value. For $94$, that is $81 = 9 times 9$. Next, check if $9$ is a factor of $94$. It isn't in this case, so $94$ would be considered "impossible" by your algorithm. I hope this makes sense.
            – John Omielan
            2 hours ago








          • 1




            What I gather from John's answer is that, given $n$, you find $d$, the integer part of the square root of $n$, and then you check whether $d$ divides $n$. If it does, then $d$ by $n/d$ is what you want. If it doesn't, then that $n$ doesn't work.
            – Gerry Myerson
            2 hours ago






          • 1




            Stated as an algorithm, given $n$ compute $k=lfloor sqrt n rfloor$. If $k$ divides $n$ the rectangle is $k times frac nk$. If $k$ does not divide $n$ it is impossible. Good answer.
            – Ross Millikan
            1 hour ago








          • 1




            You are welcome to edit it into your answer, which will make it more accessible to readers.
            – Ross Millikan
            1 hour ago


















          • No, you're completely right. However, how could, given n of let's say 94, I know what's the shape without calculating every number by hand?
            – Alex Laquerre
            2 hours ago






          • 2




            Just determine the smallest perfect square less than or equal to the value. For $94$, that is $81 = 9 times 9$. Next, check if $9$ is a factor of $94$. It isn't in this case, so $94$ would be considered "impossible" by your algorithm. I hope this makes sense.
            – John Omielan
            2 hours ago








          • 1




            What I gather from John's answer is that, given $n$, you find $d$, the integer part of the square root of $n$, and then you check whether $d$ divides $n$. If it does, then $d$ by $n/d$ is what you want. If it doesn't, then that $n$ doesn't work.
            – Gerry Myerson
            2 hours ago






          • 1




            Stated as an algorithm, given $n$ compute $k=lfloor sqrt n rfloor$. If $k$ divides $n$ the rectangle is $k times frac nk$. If $k$ does not divide $n$ it is impossible. Good answer.
            – Ross Millikan
            1 hour ago








          • 1




            You are welcome to edit it into your answer, which will make it more accessible to readers.
            – Ross Millikan
            1 hour ago
















          No, you're completely right. However, how could, given n of let's say 94, I know what's the shape without calculating every number by hand?
          – Alex Laquerre
          2 hours ago




          No, you're completely right. However, how could, given n of let's say 94, I know what's the shape without calculating every number by hand?
          – Alex Laquerre
          2 hours ago




          2




          2




          Just determine the smallest perfect square less than or equal to the value. For $94$, that is $81 = 9 times 9$. Next, check if $9$ is a factor of $94$. It isn't in this case, so $94$ would be considered "impossible" by your algorithm. I hope this makes sense.
          – John Omielan
          2 hours ago






          Just determine the smallest perfect square less than or equal to the value. For $94$, that is $81 = 9 times 9$. Next, check if $9$ is a factor of $94$. It isn't in this case, so $94$ would be considered "impossible" by your algorithm. I hope this makes sense.
          – John Omielan
          2 hours ago






          1




          1




          What I gather from John's answer is that, given $n$, you find $d$, the integer part of the square root of $n$, and then you check whether $d$ divides $n$. If it does, then $d$ by $n/d$ is what you want. If it doesn't, then that $n$ doesn't work.
          – Gerry Myerson
          2 hours ago




          What I gather from John's answer is that, given $n$, you find $d$, the integer part of the square root of $n$, and then you check whether $d$ divides $n$. If it does, then $d$ by $n/d$ is what you want. If it doesn't, then that $n$ doesn't work.
          – Gerry Myerson
          2 hours ago




          1




          1




          Stated as an algorithm, given $n$ compute $k=lfloor sqrt n rfloor$. If $k$ divides $n$ the rectangle is $k times frac nk$. If $k$ does not divide $n$ it is impossible. Good answer.
          – Ross Millikan
          1 hour ago






          Stated as an algorithm, given $n$ compute $k=lfloor sqrt n rfloor$. If $k$ divides $n$ the rectangle is $k times frac nk$. If $k$ does not divide $n$ it is impossible. Good answer.
          – Ross Millikan
          1 hour ago






          1




          1




          You are welcome to edit it into your answer, which will make it more accessible to readers.
          – Ross Millikan
          1 hour ago




          You are welcome to edit it into your answer, which will make it more accessible to readers.
          – Ross Millikan
          1 hour ago










          Alex Laquerre is a new contributor. Be nice, and check out our Code of Conduct.










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          Alex Laquerre is a new contributor. Be nice, and check out our Code of Conduct.












          Alex Laquerre is a new contributor. Be nice, and check out our Code of Conduct.
















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          Insert data from modal to MySQL (multiple modal on website)

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          0 I have problem with inserting data to MySQL from modal. My modal: <a href="#" class="badge badge-pill badge-success">6 komentarzy</a> <a href="#" class="badge badge-pill badge-danger">brak komentarzy</a> <a data-toggle="modal" href="#add_desk_comm_{$desk_['desk_id']}" data-target="#add_desk_comm_{$desk_['desk_id']}" class="ediiit">(dodaj)</a> <div class="modal fade" id="add_desk_comm_{$desk_['desk_id']}" tabindex="-1" role="dialog" aria-labelledby="edit_printer" aria-hidden="true"> <div class="modal-dialog" role="document"> ...
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