Optimal set of rectangle sizes to pack arbitrary rectangle?












4












$begingroup$


I'm looking to build a set of wooden storage boxes of various standard sizes for storing small objects.



I would like to choose a set of "optimal" box sizes (outside dimensions) for filling arbitrary rectangular spaces.



I define "optimal" here as:




  • minimize wasted space after packing;


  • minimize different box sizes; and


  • minimize the number of boxes required to fill the space, assuming the definition of some largest box dimension.



In 1 dimension this is straightforward: a set of "boxes" of sizes { 16, 8, 4, 2, 1 } can fill any 1-dimensional space with less than 1 unit of waste with my definition of "optimal" as above.



For 2-dimensional spaces, I don't know how to solve this. It seems like there should be some geometric answer to this, but I'm not versed enough in packing problems to know where to start looking.



I was wondering if anybody on here could help solve this, or point the way to an answer?



Thanks!






geometry packing-problem







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$endgroup$








  • 2




    $begingroup$
    Hi & welcome. I'm not sure if there is an answer for $2$ (or higher) dimensional spaces which will always simultaneously minimize all $3$ of your definitions for "optimal". If this is the case, can you rank your optimal requirements from most to least important?
    $endgroup$
    – John Omielan
    Jan 13 at 1:41










  • $begingroup$
    Excellent question. I'd say, in order, they'd be: (1 - most important) minimize wasted space, (2) minimize distinct box sizes, (3) minimize number of boxes to fill space. Thanks!
    $endgroup$
    – Shawn Vincent
    Jan 14 at 2:50


















4












$begingroup$


I'm looking to build a set of wooden storage boxes of various standard sizes for storing small objects.



I would like to choose a set of "optimal" box sizes (outside dimensions) for filling arbitrary rectangular spaces.



I define "optimal" here as:




  • minimize wasted space after packing;


  • minimize different box sizes; and


  • minimize the number of boxes required to fill the space, assuming the definition of some largest box dimension.



In 1 dimension this is straightforward: a set of "boxes" of sizes { 16, 8, 4, 2, 1 } can fill any 1-dimensional space with less than 1 unit of waste with my definition of "optimal" as above.



For 2-dimensional spaces, I don't know how to solve this. It seems like there should be some geometric answer to this, but I'm not versed enough in packing problems to know where to start looking.



I was wondering if anybody on here could help solve this, or point the way to an answer?



Thanks!






geometry packing-problem







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Hi & welcome. I'm not sure if there is an answer for $2$ (or higher) dimensional spaces which will always simultaneously minimize all $3$ of your definitions for "optimal". If this is the case, can you rank your optimal requirements from most to least important?
    $endgroup$
    – John Omielan
    Jan 13 at 1:41










  • $begingroup$
    Excellent question. I'd say, in order, they'd be: (1 - most important) minimize wasted space, (2) minimize distinct box sizes, (3) minimize number of boxes to fill space. Thanks!
    $endgroup$
    – Shawn Vincent
    Jan 14 at 2:50
















4












4








4





$begingroup$


I'm looking to build a set of wooden storage boxes of various standard sizes for storing small objects.



I would like to choose a set of "optimal" box sizes (outside dimensions) for filling arbitrary rectangular spaces.



I define "optimal" here as:




  • minimize wasted space after packing;


  • minimize different box sizes; and


  • minimize the number of boxes required to fill the space, assuming the definition of some largest box dimension.



In 1 dimension this is straightforward: a set of "boxes" of sizes { 16, 8, 4, 2, 1 } can fill any 1-dimensional space with less than 1 unit of waste with my definition of "optimal" as above.



For 2-dimensional spaces, I don't know how to solve this. It seems like there should be some geometric answer to this, but I'm not versed enough in packing problems to know where to start looking.



I was wondering if anybody on here could help solve this, or point the way to an answer?



Thanks!






geometry packing-problem







share|cite|improve this question









$endgroup$




I'm looking to build a set of wooden storage boxes of various standard sizes for storing small objects.



I would like to choose a set of "optimal" box sizes (outside dimensions) for filling arbitrary rectangular spaces.



I define "optimal" here as:




  • minimize wasted space after packing;


  • minimize different box sizes; and


  • minimize the number of boxes required to fill the space, assuming the definition of some largest box dimension.



In 1 dimension this is straightforward: a set of "boxes" of sizes { 16, 8, 4, 2, 1 } can fill any 1-dimensional space with less than 1 unit of waste with my definition of "optimal" as above.



For 2-dimensional spaces, I don't know how to solve this. It seems like there should be some geometric answer to this, but I'm not versed enough in packing problems to know where to start looking.



I was wondering if anybody on here could help solve this, or point the way to an answer?



Thanks!






geometry packing-problem




geometry packing-problem






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 13 at 1:17









Shawn VincentShawn Vincent

1233




1233








  • 2




    $begingroup$
    Hi & welcome. I'm not sure if there is an answer for $2$ (or higher) dimensional spaces which will always simultaneously minimize all $3$ of your definitions for "optimal". If this is the case, can you rank your optimal requirements from most to least important?
    $endgroup$
    – John Omielan
    Jan 13 at 1:41










  • $begingroup$
    Excellent question. I'd say, in order, they'd be: (1 - most important) minimize wasted space, (2) minimize distinct box sizes, (3) minimize number of boxes to fill space. Thanks!
    $endgroup$
    – Shawn Vincent
    Jan 14 at 2:50
















  • 2




    $begingroup$
    Hi & welcome. I'm not sure if there is an answer for $2$ (or higher) dimensional spaces which will always simultaneously minimize all $3$ of your definitions for "optimal". If this is the case, can you rank your optimal requirements from most to least important?
    $endgroup$
    – John Omielan
    Jan 13 at 1:41










  • $begingroup$
    Excellent question. I'd say, in order, they'd be: (1 - most important) minimize wasted space, (2) minimize distinct box sizes, (3) minimize number of boxes to fill space. Thanks!
    $endgroup$
    – Shawn Vincent
    Jan 14 at 2:50










2




2




$begingroup$
Hi & welcome. I'm not sure if there is an answer for $2$ (or higher) dimensional spaces which will always simultaneously minimize all $3$ of your definitions for "optimal". If this is the case, can you rank your optimal requirements from most to least important?
$endgroup$
– John Omielan
Jan 13 at 1:41




$begingroup$
Hi & welcome. I'm not sure if there is an answer for $2$ (or higher) dimensional spaces which will always simultaneously minimize all $3$ of your definitions for "optimal". If this is the case, can you rank your optimal requirements from most to least important?
$endgroup$
– John Omielan
Jan 13 at 1:41












$begingroup$
Excellent question. I'd say, in order, they'd be: (1 - most important) minimize wasted space, (2) minimize distinct box sizes, (3) minimize number of boxes to fill space. Thanks!
$endgroup$
– Shawn Vincent
Jan 14 at 2:50






$begingroup$
Excellent question. I'd say, in order, they'd be: (1 - most important) minimize wasted space, (2) minimize distinct box sizes, (3) minimize number of boxes to fill space. Thanks!
$endgroup$
– Shawn Vincent
Jan 14 at 2:50












2 Answers
2






active

oldest

votes


















2












$begingroup$

I'm not sure if this is optimal, but it works.



My collection optimises "no wasted space", and it also minimises the number of boxes required in the sense that "the most efficient tiling of any space" is obtained by using the biggest boxes first. However, you may feel that there are too many boxes.



For $2$-dimensional space, I use $(x,y)$ to denote rectangles with side lengths $x$ and $y$.





  1. ${(1,1), (1,2), (1,2), (2,2)}$ will fill all spaces (with integer sides) up to $3times 3$ with no gaps (with the greedy algorithm)


  2. ${(1,1),(1,2),(1,2),(1,4),(1,4),(2,2),(2,4),(2,4),(4,4)}$ will fill all spaces with integer sides up to $7times 7$ with no gaps (with the greedy algorithm)


More generally, the collections of boxes are defined by:



one of each box of size $(x,x)$,
two of each box of size $(x,y)$ with $xneq y$, where $x,y in {1,2,4,8,...}$.



I conjecture that the collection of boxes generated in this way will always tile larger 2d spaces with no gaps using the greedy algorithm. (I have verified this for the two cases above, but not for e.g. $15times 15$, $31times 31$,...)



By "using the greedy algorithm", I mean that if you attempt to tile some space, you just have to start with the biggest box, then use the second biggest box, etc.



I suspect that there maybe a more optimal solution which makes use of golden-ratio-eqsue maths.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Although you don't need all the properties of this particular partition
    based on triangular numbers, you
    might consider using it, or an analogous partition:



             


             

    Image from Wikipedia.





    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      I'm not sure if this is optimal, but it works.



      My collection optimises "no wasted space", and it also minimises the number of boxes required in the sense that "the most efficient tiling of any space" is obtained by using the biggest boxes first. However, you may feel that there are too many boxes.



      For $2$-dimensional space, I use $(x,y)$ to denote rectangles with side lengths $x$ and $y$.





      1. ${(1,1), (1,2), (1,2), (2,2)}$ will fill all spaces (with integer sides) up to $3times 3$ with no gaps (with the greedy algorithm)


      2. ${(1,1),(1,2),(1,2),(1,4),(1,4),(2,2),(2,4),(2,4),(4,4)}$ will fill all spaces with integer sides up to $7times 7$ with no gaps (with the greedy algorithm)


      More generally, the collections of boxes are defined by:



      one of each box of size $(x,x)$,
      two of each box of size $(x,y)$ with $xneq y$, where $x,y in {1,2,4,8,...}$.



      I conjecture that the collection of boxes generated in this way will always tile larger 2d spaces with no gaps using the greedy algorithm. (I have verified this for the two cases above, but not for e.g. $15times 15$, $31times 31$,...)



      By "using the greedy algorithm", I mean that if you attempt to tile some space, you just have to start with the biggest box, then use the second biggest box, etc.



      I suspect that there maybe a more optimal solution which makes use of golden-ratio-eqsue maths.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        I'm not sure if this is optimal, but it works.



        My collection optimises "no wasted space", and it also minimises the number of boxes required in the sense that "the most efficient tiling of any space" is obtained by using the biggest boxes first. However, you may feel that there are too many boxes.



        For $2$-dimensional space, I use $(x,y)$ to denote rectangles with side lengths $x$ and $y$.





        1. ${(1,1), (1,2), (1,2), (2,2)}$ will fill all spaces (with integer sides) up to $3times 3$ with no gaps (with the greedy algorithm)


        2. ${(1,1),(1,2),(1,2),(1,4),(1,4),(2,2),(2,4),(2,4),(4,4)}$ will fill all spaces with integer sides up to $7times 7$ with no gaps (with the greedy algorithm)


        More generally, the collections of boxes are defined by:



        one of each box of size $(x,x)$,
        two of each box of size $(x,y)$ with $xneq y$, where $x,y in {1,2,4,8,...}$.



        I conjecture that the collection of boxes generated in this way will always tile larger 2d spaces with no gaps using the greedy algorithm. (I have verified this for the two cases above, but not for e.g. $15times 15$, $31times 31$,...)



        By "using the greedy algorithm", I mean that if you attempt to tile some space, you just have to start with the biggest box, then use the second biggest box, etc.



        I suspect that there maybe a more optimal solution which makes use of golden-ratio-eqsue maths.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          I'm not sure if this is optimal, but it works.



          My collection optimises "no wasted space", and it also minimises the number of boxes required in the sense that "the most efficient tiling of any space" is obtained by using the biggest boxes first. However, you may feel that there are too many boxes.



          For $2$-dimensional space, I use $(x,y)$ to denote rectangles with side lengths $x$ and $y$.





          1. ${(1,1), (1,2), (1,2), (2,2)}$ will fill all spaces (with integer sides) up to $3times 3$ with no gaps (with the greedy algorithm)


          2. ${(1,1),(1,2),(1,2),(1,4),(1,4),(2,2),(2,4),(2,4),(4,4)}$ will fill all spaces with integer sides up to $7times 7$ with no gaps (with the greedy algorithm)


          More generally, the collections of boxes are defined by:



          one of each box of size $(x,x)$,
          two of each box of size $(x,y)$ with $xneq y$, where $x,y in {1,2,4,8,...}$.



          I conjecture that the collection of boxes generated in this way will always tile larger 2d spaces with no gaps using the greedy algorithm. (I have verified this for the two cases above, but not for e.g. $15times 15$, $31times 31$,...)



          By "using the greedy algorithm", I mean that if you attempt to tile some space, you just have to start with the biggest box, then use the second biggest box, etc.



          I suspect that there maybe a more optimal solution which makes use of golden-ratio-eqsue maths.






          share|cite|improve this answer









          $endgroup$



          I'm not sure if this is optimal, but it works.



          My collection optimises "no wasted space", and it also minimises the number of boxes required in the sense that "the most efficient tiling of any space" is obtained by using the biggest boxes first. However, you may feel that there are too many boxes.



          For $2$-dimensional space, I use $(x,y)$ to denote rectangles with side lengths $x$ and $y$.





          1. ${(1,1), (1,2), (1,2), (2,2)}$ will fill all spaces (with integer sides) up to $3times 3$ with no gaps (with the greedy algorithm)


          2. ${(1,1),(1,2),(1,2),(1,4),(1,4),(2,2),(2,4),(2,4),(4,4)}$ will fill all spaces with integer sides up to $7times 7$ with no gaps (with the greedy algorithm)


          More generally, the collections of boxes are defined by:



          one of each box of size $(x,x)$,
          two of each box of size $(x,y)$ with $xneq y$, where $x,y in {1,2,4,8,...}$.



          I conjecture that the collection of boxes generated in this way will always tile larger 2d spaces with no gaps using the greedy algorithm. (I have verified this for the two cases above, but not for e.g. $15times 15$, $31times 31$,...)



          By "using the greedy algorithm", I mean that if you attempt to tile some space, you just have to start with the biggest box, then use the second biggest box, etc.



          I suspect that there maybe a more optimal solution which makes use of golden-ratio-eqsue maths.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 14 at 2:04









          HarambeHarambe

          6,06821843




          6,06821843























              2












              $begingroup$

              Although you don't need all the properties of this particular partition
              based on triangular numbers, you
              might consider using it, or an analogous partition:



                       


                       

              Image from Wikipedia.





              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Although you don't need all the properties of this particular partition
                based on triangular numbers, you
                might consider using it, or an analogous partition:



                         


                         

                Image from Wikipedia.





                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Although you don't need all the properties of this particular partition
                  based on triangular numbers, you
                  might consider using it, or an analogous partition:



                           


                           

                  Image from Wikipedia.





                  share|cite|improve this answer









                  $endgroup$



                  Although you don't need all the properties of this particular partition
                  based on triangular numbers, you
                  might consider using it, or an analogous partition:



                           


                           

                  Image from Wikipedia.






                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 14 at 1:32









                  Joseph O'RourkeJoseph O'Rourke

                  18.2k350113




                  18.2k350113






























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