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 3 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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          1 Answer
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          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
















          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 38 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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          Verónica Boquete

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          Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl ...
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          Ida-Boy-Ed-Garten

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          Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
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