Packing rectangles: Does rotation ever help?Box stacking problemPlacing Axis-parallel rectangles on 2-D planeUpper bound for tetrahedron packing?A circle packing conjectureDoes list of distances define points uniquely?rational rotation vector and closed curvesSpace packing fraction of tetrahedron and octahedronTranslative packing constant strictly larger than lattice packing constantTiling a square with rectanglesCompose/decompose rotation matrix from/to plane of rotation and angleSome inequalities on chain of circle packing
Packing rectangles: Does rotation ever help?
Box stacking problemPlacing Axis-parallel rectangles on 2-D planeUpper bound for tetrahedron packing?A circle packing conjectureDoes list of distances define points uniquely?rational rotation vector and closed curvesSpace packing fraction of tetrahedron and octahedronTranslative packing constant strictly larger than lattice packing constantTiling a square with rectanglesCompose/decompose rotation matrix from/to plane of rotation and angleSome inequalities on chain of circle packing
$begingroup$
Dominic van der Zypen posed an interesting Box stacking problem.
This is a spin-off question.
Let a collection of rectangles $r_1,ldots,r_n$ be given by their side lengths in $mathbbR$.
Let $R$ be a rectangle of minimum area enclosing the rectangles arranged
in the plane without overlap (i.e., with disjoint interiors).
Q. Is there an example where not all the rectangles have sides aligned with
the sides of $R$?
In other words, where at least one rectangle's sides are not parallel to the
sides of $R$? Is it ever advantageous to "tilt" one or more rectangles to
achieve a minimal area?
mg.metric-geometry plane-geometry
asked Apr 27 at 18:44
Joseph O'RourkeJoseph O'Rourke
86.8k16241717
$endgroup$
add a comment |
$begingroup$
Dominic van der Zypen posed an interesting Box stacking problem.
This is a spin-off question.
Let a collection of rectangles $r_1,ldots,r_n$ be given by their side lengths in $mathbbR$.
Let $R$ be a rectangle of minimum area enclosing the rectangles arranged
in the plane without overlap (i.e., with disjoint interiors).
Q. Is there an example where not all the rectangles have sides aligned with
the sides of $R$?
In other words, where at least one rectangle's sides are not parallel to the
sides of $R$? Is it ever advantageous to "tilt" one or more rectangles to
achieve a minimal area?
mg.metric-geometry plane-geometry
asked Apr 27 at 18:44
Joseph O'RourkeJoseph O'Rourke
86.8k16241717
$endgroup$
17
$begingroup$
I think this should work. Consider 2 rectangles with r1 having dimensions 1x11 and r2 having area 10x100. Then the packings that have aligned sides have enclosing areas at least 1100, but if you "lean" r1 against the end of r2 then you can get an enclosing rectangle with area around 1025.
$endgroup$
– Yosemite Stan
Apr 27 at 19:06
$begingroup$
There was mention of some research showing how one could pack (1+ delta)n^2 unit squares in a square of side (1 + epsilon)n. Consider searching "Packing squares in a square. There is also "shipping a pool cue" in the diagonal of a packing box. Gerhard "Can't Name Any Names Yet" Paseman, 2019.04.27.
$endgroup$
– Gerhard Paseman
Apr 27 at 19:09
add a comment |
$begingroup$
Dominic van der Zypen posed an interesting Box stacking problem.
This is a spin-off question.
Let a collection of rectangles $r_1,ldots,r_n$ be given by their side lengths in $mathbbR$.
Let $R$ be a rectangle of minimum area enclosing the rectangles arranged
in the plane without overlap (i.e., with disjoint interiors).
Q. Is there an example where not all the rectangles have sides aligned with
the sides of $R$?
In other words, where at least one rectangle's sides are not parallel to the
sides of $R$? Is it ever advantageous to "tilt" one or more rectangles to
achieve a minimal area?
mg.metric-geometry plane-geometry
asked Apr 27 at 18:44
Joseph O'RourkeJoseph O'Rourke
86.8k16241717
$endgroup$
Dominic van der Zypen posed an interesting Box stacking problem.
This is a spin-off question.
Let a collection of rectangles $r_1,ldots,r_n$ be given by their side lengths in $mathbbR$.
Let $R$ be a rectangle of minimum area enclosing the rectangles arranged
in the plane without overlap (i.e., with disjoint interiors).
Q. Is there an example where not all the rectangles have sides aligned with
the sides of $R$?
In other words, where at least one rectangle's sides are not parallel to the
sides of $R$? Is it ever advantageous to "tilt" one or more rectangles to
achieve a minimal area?
mg.metric-geometry plane-geometry
mg.metric-geometry plane-geometry
asked Apr 27 at 18:44
Joseph O'RourkeJoseph O'Rourke
86.8k16241717
asked Apr 27 at 18:44
Joseph O'RourkeJoseph O'Rourke
86.8k16241717
asked Apr 27 at 18:44
Joseph O'RourkeJoseph O'Rourke
86.8k16241717
asked Apr 27 at 18:44
Joseph O'RourkeJoseph O'Rourke
86.8k16241717
asked Apr 27 at 18:44
Joseph O'RourkeJoseph O'Rourke
86.8k16241717
86.8k16241717
17
$begingroup$
I think this should work. Consider 2 rectangles with r1 having dimensions 1x11 and r2 having area 10x100. Then the packings that have aligned sides have enclosing areas at least 1100, but if you "lean" r1 against the end of r2 then you can get an enclosing rectangle with area around 1025.
$endgroup$
– Yosemite Stan
Apr 27 at 19:06
$begingroup$
There was mention of some research showing how one could pack (1+ delta)n^2 unit squares in a square of side (1 + epsilon)n. Consider searching "Packing squares in a square. There is also "shipping a pool cue" in the diagonal of a packing box. Gerhard "Can't Name Any Names Yet" Paseman, 2019.04.27.
$endgroup$
– Gerhard Paseman
Apr 27 at 19:09
add a comment |
17
$begingroup$
I think this should work. Consider 2 rectangles with r1 having dimensions 1x11 and r2 having area 10x100. Then the packings that have aligned sides have enclosing areas at least 1100, but if you "lean" r1 against the end of r2 then you can get an enclosing rectangle with area around 1025.
$endgroup$
– Yosemite Stan
Apr 27 at 19:06
$begingroup$
There was mention of some research showing how one could pack (1+ delta)n^2 unit squares in a square of side (1 + epsilon)n. Consider searching "Packing squares in a square. There is also "shipping a pool cue" in the diagonal of a packing box. Gerhard "Can't Name Any Names Yet" Paseman, 2019.04.27.
$endgroup$
– Gerhard Paseman
Apr 27 at 19:09
17
17
$begingroup$
I think this should work. Consider 2 rectangles with r1 having dimensions 1x11 and r2 having area 10x100. Then the packings that have aligned sides have enclosing areas at least 1100, but if you "lean" r1 against the end of r2 then you can get an enclosing rectangle with area around 1025.
$endgroup$
– Yosemite Stan
Apr 27 at 19:06
$begingroup$
I think this should work. Consider 2 rectangles with r1 having dimensions 1x11 and r2 having area 10x100. Then the packings that have aligned sides have enclosing areas at least 1100, but if you "lean" r1 against the end of r2 then you can get an enclosing rectangle with area around 1025.
$endgroup$
– Yosemite Stan
Apr 27 at 19:06
$begingroup$
There was mention of some research showing how one could pack (1+ delta)n^2 unit squares in a square of side (1 + epsilon)n. Consider searching "Packing squares in a square. There is also "shipping a pool cue" in the diagonal of a packing box. Gerhard "Can't Name Any Names Yet" Paseman, 2019.04.27.
$endgroup$
– Gerhard Paseman
Apr 27 at 19:09
$begingroup$
There was mention of some research showing how one could pack (1+ delta)n^2 unit squares in a square of side (1 + epsilon)n. Consider searching "Packing squares in a square. There is also "shipping a pool cue" in the diagonal of a packing box. Gerhard "Can't Name Any Names Yet" Paseman, 2019.04.27.
$endgroup$
– Gerhard Paseman
Apr 27 at 19:09
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
@YosemiteStan's example.
Detail: Tilt angle $=sin ^-1left(5
sqrtfrac261right) approx 65^circ$.
$endgroup$
2
$begingroup$
Actually, in the first packing the smaller rectangle should lie on the top of the larger one, giving the intended area of 11*100=1100.
$endgroup$
– Victor Protsak
Apr 27 at 19:51
$begingroup$
@VictorProtsak: Thanks; corrected.
$endgroup$
– Joseph O'Rourke
Apr 27 at 19:55
$begingroup$
The first packing is $100cdot (10+1) = 1100$, but I think the second packing is $left( frac1061(11+6sqrt22)+100 right)cdot 10approx 1064.2$, so the example is valid, but $1025$ from the original comment is not so precise. To me, it is hard to estimate visually by staring at the figure, which packing is better, but I can see that the "leaning" thin box is more upright than $45^circ$.
$endgroup$
– Jeppe Stig Nielsen
Apr 29 at 10:07
$begingroup$
@JeppeStigNielsen: "more upright": Yes, about $65^circ$. See added detail.
$endgroup$
– Joseph O'Rourke
Apr 29 at 11:52
1
$begingroup$
@JeppeStigNielsen: You can increase the length from $100$ to an arbitrarily large length, in which case it is clear that the bottom picture covers less area.
$endgroup$
– Tony Huynh
Apr 29 at 12:43
add a comment |
$begingroup$
The classic answer to this is a paper of Erdos and Graham 'On packing squares with equal squares'. Given a square of side $n+varepsilon$, where $0<varepsilon<1$, we can obviously fit in $n^2$ unit squares, and it's fairly trivial to check that if the squares are axis aligned this is best possible (count the squares intersecting vertical lines on the integers). Obviously, this means about $2varepsilon n$ area is going to waste.
But Erdos and Graham show one can cover asymptotically all but $n^7/11$ area, using skew angles - this is maybe more surprising than Yosemite Stan's example (which also works perfectly well).
answered Apr 27 at 20:51
user36212user36212
1,169810
$endgroup$
4
$begingroup$
However the question allows any rectangle and one can get pack a bunch of unit cubes in a rectangle with no wasted space.
$endgroup$
– Aaron Meyerowitz
Apr 28 at 7:42
2
$begingroup$
Square packing was my first thought as well, some nice pictures are available at www2.stetson.edu/~efriedma/squinsqu. However on second thought it's apparent that square packing is not relevant to this question. The minimal area rectangle for any collection of equal sized squares is trivially realized simply by stacking them all in a single line.
$endgroup$
– Brady Gilg
Apr 30 at 6:06
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
@YosemiteStan's example.
Detail: Tilt angle $=sin ^-1left(5
sqrtfrac261right) approx 65^circ$.
$endgroup$
2
$begingroup$
Actually, in the first packing the smaller rectangle should lie on the top of the larger one, giving the intended area of 11*100=1100.
$endgroup$
– Victor Protsak
Apr 27 at 19:51
$begingroup$
@VictorProtsak: Thanks; corrected.
$endgroup$
– Joseph O'Rourke
Apr 27 at 19:55
$begingroup$
The first packing is $100cdot (10+1) = 1100$, but I think the second packing is $left( frac1061(11+6sqrt22)+100 right)cdot 10approx 1064.2$, so the example is valid, but $1025$ from the original comment is not so precise. To me, it is hard to estimate visually by staring at the figure, which packing is better, but I can see that the "leaning" thin box is more upright than $45^circ$.
$endgroup$
– Jeppe Stig Nielsen
Apr 29 at 10:07
$begingroup$
@JeppeStigNielsen: "more upright": Yes, about $65^circ$. See added detail.
$endgroup$
– Joseph O'Rourke
Apr 29 at 11:52
1
$begingroup$
@JeppeStigNielsen: You can increase the length from $100$ to an arbitrarily large length, in which case it is clear that the bottom picture covers less area.
$endgroup$
– Tony Huynh
Apr 29 at 12:43
add a comment |
$begingroup$
@YosemiteStan's example.
Detail: Tilt angle $=sin ^-1left(5
sqrtfrac261right) approx 65^circ$.
$endgroup$
2
$begingroup$
Actually, in the first packing the smaller rectangle should lie on the top of the larger one, giving the intended area of 11*100=1100.
$endgroup$
– Victor Protsak
Apr 27 at 19:51
$begingroup$
@VictorProtsak: Thanks; corrected.
$endgroup$
– Joseph O'Rourke
Apr 27 at 19:55
$begingroup$
The first packing is $100cdot (10+1) = 1100$, but I think the second packing is $left( frac1061(11+6sqrt22)+100 right)cdot 10approx 1064.2$, so the example is valid, but $1025$ from the original comment is not so precise. To me, it is hard to estimate visually by staring at the figure, which packing is better, but I can see that the "leaning" thin box is more upright than $45^circ$.
$endgroup$
– Jeppe Stig Nielsen
Apr 29 at 10:07
$begingroup$
@JeppeStigNielsen: "more upright": Yes, about $65^circ$. See added detail.
$endgroup$
– Joseph O'Rourke
Apr 29 at 11:52
1
$begingroup$
@JeppeStigNielsen: You can increase the length from $100$ to an arbitrarily large length, in which case it is clear that the bottom picture covers less area.
$endgroup$
– Tony Huynh
Apr 29 at 12:43
add a comment |
$begingroup$
@YosemiteStan's example.
Detail: Tilt angle $=sin ^-1left(5
sqrtfrac261right) approx 65^circ$.
$endgroup$
@YosemiteStan's example.
Detail: Tilt angle $=sin ^-1left(5
sqrtfrac261right) approx 65^circ$.
edited Apr 29 at 11:51
community wiki
3 revs
Joseph O'Rourke
2
$begingroup$
Actually, in the first packing the smaller rectangle should lie on the top of the larger one, giving the intended area of 11*100=1100.
$endgroup$
– Victor Protsak
Apr 27 at 19:51
$begingroup$
@VictorProtsak: Thanks; corrected.
$endgroup$
– Joseph O'Rourke
Apr 27 at 19:55
$begingroup$
The first packing is $100cdot (10+1) = 1100$, but I think the second packing is $left( frac1061(11+6sqrt22)+100 right)cdot 10approx 1064.2$, so the example is valid, but $1025$ from the original comment is not so precise. To me, it is hard to estimate visually by staring at the figure, which packing is better, but I can see that the "leaning" thin box is more upright than $45^circ$.
$endgroup$
– Jeppe Stig Nielsen
Apr 29 at 10:07
$begingroup$
@JeppeStigNielsen: "more upright": Yes, about $65^circ$. See added detail.
$endgroup$
– Joseph O'Rourke
Apr 29 at 11:52
1
$begingroup$
@JeppeStigNielsen: You can increase the length from $100$ to an arbitrarily large length, in which case it is clear that the bottom picture covers less area.
$endgroup$
– Tony Huynh
Apr 29 at 12:43
add a comment |
2
$begingroup$
Actually, in the first packing the smaller rectangle should lie on the top of the larger one, giving the intended area of 11*100=1100.
$endgroup$
– Victor Protsak
Apr 27 at 19:51
$begingroup$
@VictorProtsak: Thanks; corrected.
$endgroup$
– Joseph O'Rourke
Apr 27 at 19:55
$begingroup$
The first packing is $100cdot (10+1) = 1100$, but I think the second packing is $left( frac1061(11+6sqrt22)+100 right)cdot 10approx 1064.2$, so the example is valid, but $1025$ from the original comment is not so precise. To me, it is hard to estimate visually by staring at the figure, which packing is better, but I can see that the "leaning" thin box is more upright than $45^circ$.
$endgroup$
– Jeppe Stig Nielsen
Apr 29 at 10:07
$begingroup$
@JeppeStigNielsen: "more upright": Yes, about $65^circ$. See added detail.
$endgroup$
– Joseph O'Rourke
Apr 29 at 11:52
1
$begingroup$
@JeppeStigNielsen: You can increase the length from $100$ to an arbitrarily large length, in which case it is clear that the bottom picture covers less area.
$endgroup$
– Tony Huynh
Apr 29 at 12:43
2
2
$begingroup$
Actually, in the first packing the smaller rectangle should lie on the top of the larger one, giving the intended area of 11*100=1100.
$endgroup$
– Victor Protsak
Apr 27 at 19:51
$begingroup$
Actually, in the first packing the smaller rectangle should lie on the top of the larger one, giving the intended area of 11*100=1100.
$endgroup$
– Victor Protsak
Apr 27 at 19:51
$begingroup$
@VictorProtsak: Thanks; corrected.
$endgroup$
– Joseph O'Rourke
Apr 27 at 19:55
$begingroup$
@VictorProtsak: Thanks; corrected.
$endgroup$
– Joseph O'Rourke
Apr 27 at 19:55
$begingroup$
The first packing is $100cdot (10+1) = 1100$, but I think the second packing is $left( frac1061(11+6sqrt22)+100 right)cdot 10approx 1064.2$, so the example is valid, but $1025$ from the original comment is not so precise. To me, it is hard to estimate visually by staring at the figure, which packing is better, but I can see that the "leaning" thin box is more upright than $45^circ$.
$endgroup$
– Jeppe Stig Nielsen
Apr 29 at 10:07
$begingroup$
The first packing is $100cdot (10+1) = 1100$, but I think the second packing is $left( frac1061(11+6sqrt22)+100 right)cdot 10approx 1064.2$, so the example is valid, but $1025$ from the original comment is not so precise. To me, it is hard to estimate visually by staring at the figure, which packing is better, but I can see that the "leaning" thin box is more upright than $45^circ$.
$endgroup$
– Jeppe Stig Nielsen
Apr 29 at 10:07
$begingroup$
@JeppeStigNielsen: "more upright": Yes, about $65^circ$. See added detail.
$endgroup$
– Joseph O'Rourke
Apr 29 at 11:52
$begingroup$
@JeppeStigNielsen: "more upright": Yes, about $65^circ$. See added detail.
$endgroup$
– Joseph O'Rourke
Apr 29 at 11:52
1
1
$begingroup$
@JeppeStigNielsen: You can increase the length from $100$ to an arbitrarily large length, in which case it is clear that the bottom picture covers less area.
$endgroup$
– Tony Huynh
Apr 29 at 12:43
$begingroup$
@JeppeStigNielsen: You can increase the length from $100$ to an arbitrarily large length, in which case it is clear that the bottom picture covers less area.
$endgroup$
– Tony Huynh
Apr 29 at 12:43
add a comment |
$begingroup$
The classic answer to this is a paper of Erdos and Graham 'On packing squares with equal squares'. Given a square of side $n+varepsilon$, where $0<varepsilon<1$, we can obviously fit in $n^2$ unit squares, and it's fairly trivial to check that if the squares are axis aligned this is best possible (count the squares intersecting vertical lines on the integers). Obviously, this means about $2varepsilon n$ area is going to waste.
But Erdos and Graham show one can cover asymptotically all but $n^7/11$ area, using skew angles - this is maybe more surprising than Yosemite Stan's example (which also works perfectly well).
answered Apr 27 at 20:51
user36212user36212
1,169810
$endgroup$
4
$begingroup$
However the question allows any rectangle and one can get pack a bunch of unit cubes in a rectangle with no wasted space.
$endgroup$
– Aaron Meyerowitz
Apr 28 at 7:42
2
$begingroup$
Square packing was my first thought as well, some nice pictures are available at www2.stetson.edu/~efriedma/squinsqu. However on second thought it's apparent that square packing is not relevant to this question. The minimal area rectangle for any collection of equal sized squares is trivially realized simply by stacking them all in a single line.
$endgroup$
– Brady Gilg
Apr 30 at 6:06
add a comment |
$begingroup$
The classic answer to this is a paper of Erdos and Graham 'On packing squares with equal squares'. Given a square of side $n+varepsilon$, where $0<varepsilon<1$, we can obviously fit in $n^2$ unit squares, and it's fairly trivial to check that if the squares are axis aligned this is best possible (count the squares intersecting vertical lines on the integers). Obviously, this means about $2varepsilon n$ area is going to waste.
But Erdos and Graham show one can cover asymptotically all but $n^7/11$ area, using skew angles - this is maybe more surprising than Yosemite Stan's example (which also works perfectly well).
answered Apr 27 at 20:51
user36212user36212
1,169810
$endgroup$
4
$begingroup$
However the question allows any rectangle and one can get pack a bunch of unit cubes in a rectangle with no wasted space.
$endgroup$
– Aaron Meyerowitz
Apr 28 at 7:42
2
$begingroup$
Square packing was my first thought as well, some nice pictures are available at www2.stetson.edu/~efriedma/squinsqu. However on second thought it's apparent that square packing is not relevant to this question. The minimal area rectangle for any collection of equal sized squares is trivially realized simply by stacking them all in a single line.
$endgroup$
– Brady Gilg
Apr 30 at 6:06
add a comment |
$begingroup$
The classic answer to this is a paper of Erdos and Graham 'On packing squares with equal squares'. Given a square of side $n+varepsilon$, where $0<varepsilon<1$, we can obviously fit in $n^2$ unit squares, and it's fairly trivial to check that if the squares are axis aligned this is best possible (count the squares intersecting vertical lines on the integers). Obviously, this means about $2varepsilon n$ area is going to waste.
But Erdos and Graham show one can cover asymptotically all but $n^7/11$ area, using skew angles - this is maybe more surprising than Yosemite Stan's example (which also works perfectly well).
answered Apr 27 at 20:51
user36212user36212
1,169810
$endgroup$
The classic answer to this is a paper of Erdos and Graham 'On packing squares with equal squares'. Given a square of side $n+varepsilon$, where $0<varepsilon<1$, we can obviously fit in $n^2$ unit squares, and it's fairly trivial to check that if the squares are axis aligned this is best possible (count the squares intersecting vertical lines on the integers). Obviously, this means about $2varepsilon n$ area is going to waste.
But Erdos and Graham show one can cover asymptotically all but $n^7/11$ area, using skew angles - this is maybe more surprising than Yosemite Stan's example (which also works perfectly well).
answered Apr 27 at 20:51
user36212user36212
1,169810
answered Apr 27 at 20:51
user36212user36212
1,169810
answered Apr 27 at 20:51
user36212user36212
1,169810
answered Apr 27 at 20:51
user36212user36212
1,169810
1,169810
4
$begingroup$
However the question allows any rectangle and one can get pack a bunch of unit cubes in a rectangle with no wasted space.
$endgroup$
– Aaron Meyerowitz
Apr 28 at 7:42
2
$begingroup$
Square packing was my first thought as well, some nice pictures are available at www2.stetson.edu/~efriedma/squinsqu. However on second thought it's apparent that square packing is not relevant to this question. The minimal area rectangle for any collection of equal sized squares is trivially realized simply by stacking them all in a single line.
$endgroup$
– Brady Gilg
Apr 30 at 6:06
add a comment |
4
$begingroup$
However the question allows any rectangle and one can get pack a bunch of unit cubes in a rectangle with no wasted space.
$endgroup$
– Aaron Meyerowitz
Apr 28 at 7:42
2
$begingroup$
Square packing was my first thought as well, some nice pictures are available at www2.stetson.edu/~efriedma/squinsqu. However on second thought it's apparent that square packing is not relevant to this question. The minimal area rectangle for any collection of equal sized squares is trivially realized simply by stacking them all in a single line.
$endgroup$
– Brady Gilg
Apr 30 at 6:06
4
4
$begingroup$
However the question allows any rectangle and one can get pack a bunch of unit cubes in a rectangle with no wasted space.
$endgroup$
– Aaron Meyerowitz
Apr 28 at 7:42
$begingroup$
However the question allows any rectangle and one can get pack a bunch of unit cubes in a rectangle with no wasted space.
$endgroup$
– Aaron Meyerowitz
Apr 28 at 7:42
2
2
$begingroup$
Square packing was my first thought as well, some nice pictures are available at www2.stetson.edu/~efriedma/squinsqu. However on second thought it's apparent that square packing is not relevant to this question. The minimal area rectangle for any collection of equal sized squares is trivially realized simply by stacking them all in a single line.
$endgroup$
– Brady Gilg
Apr 30 at 6:06
$begingroup$
Square packing was my first thought as well, some nice pictures are available at www2.stetson.edu/~efriedma/squinsqu. However on second thought it's apparent that square packing is not relevant to this question. The minimal area rectangle for any collection of equal sized squares is trivially realized simply by stacking them all in a single line.
$endgroup$
– Brady Gilg
Apr 30 at 6:06
add a comment |
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17
$begingroup$
I think this should work. Consider 2 rectangles with r1 having dimensions 1x11 and r2 having area 10x100. Then the packings that have aligned sides have enclosing areas at least 1100, but if you "lean" r1 against the end of r2 then you can get an enclosing rectangle with area around 1025.
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– Yosemite Stan
Apr 27 at 19:06
$begingroup$
There was mention of some research showing how one could pack (1+ delta)n^2 unit squares in a square of side (1 + epsilon)n. Consider searching "Packing squares in a square. There is also "shipping a pool cue" in the diagonal of a packing box. Gerhard "Can't Name Any Names Yet" Paseman, 2019.04.27.
$endgroup$
– Gerhard Paseman
Apr 27 at 19:09