Mixing
Connecting points in an annulus with horizontal and vertical segments
Clash Royale CLAN TAG#URR8PPP
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A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 1.34
I guess the answer is to do with Eg 1.13:
For $D[0,1]$ or $D[0,3]$, maximum is 2.
For $D cdot [0,1]$ or $D cdot [0,3]$, maximum is 3, achieved when the points are on a line passing through 0. Edit: Upon reflection, I think maximum is 5
For $D[0,3] setminus D[0,2]$, maximum is I guess 5, achieved when, but not only when, the points are on a line passing through 0 and close to $D[0,3]$?
If right, then what's the justification please?
If wrong, why and how else can I approach this please?
Asked here but there are no posted answers: Find the maximum number of horizontal and vertical segments in $G$ needed to connect two points of $G$.
Related:
Prove that $A_r,s$ $=[zin mathbb C : r<|z-z_0|<s ]$ is path connected.
An annulus in $mathbb R^2$ is path connected
general-topology complex-analysis complex-numbers connectedness path-connected
asked Jul 31 at 2:33
BCLC
6,99821973
 |Â
show 2 more comments
up vote
1
down vote
favorite
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 1.34
I guess the answer is to do with Eg 1.13:
For $D[0,1]$ or $D[0,3]$, maximum is 2.
For $D cdot [0,1]$ or $D cdot [0,3]$, maximum is 3, achieved when the points are on a line passing through 0. Edit: Upon reflection, I think maximum is 5
For $D[0,3] setminus D[0,2]$, maximum is I guess 5, achieved when, but not only when, the points are on a line passing through 0 and close to $D[0,3]$?
If right, then what's the justification please?
If wrong, why and how else can I approach this please?
Asked here but there are no posted answers: Find the maximum number of horizontal and vertical segments in $G$ needed to connect two points of $G$.
Related:
Prove that $A_r,s$ $=[zin mathbb C : r<|z-z_0|<s ]$ is path connected.
An annulus in $mathbb R^2$ is path connected
general-topology complex-analysis complex-numbers connectedness path-connected
asked Jul 31 at 2:33
BCLC
6,99821973
The MAXIMUM number is unlimited.
– William Elliot
Jul 31 at 2:56
@WilliamElliot Okay why $infty$ please? I'm trying to see a pattern here: if you delete a point then max is 3 and then if you delete a concentric circle with radius 0.00001 then max is perhaps 5 and then if the deleted concentric circle has a certain large enough radius than the max is $infty$. Is the threshold perhaps that the deleted concentric circle has a radius greater than the difference of the two radii?
– BCLC
Jul 31 at 3:14
1
@WilliamElliot Why would it be unlimited? There certainly is a closed zig-zag made of finitely many h/v segments contained within that annulus, then for any two points one only needs to "connect" them to that.
– dxiv
Jul 31 at 4:04
1
@BCLC That number is certainly bounded. The zig-zag itself is independent of the choice of the two points, and each of the two points can reach the zig-zag with (at most) one segment.
– dxiv
Jul 31 at 4:34
1
@BCLC Too long for a comment, posted as an answer. Note the last "but" however.
– dxiv
Jul 31 at 5:17
 |Â
show 2 more comments
up vote
1
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favorite
up vote
1
down vote
favorite
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 1.34
I guess the answer is to do with Eg 1.13:
For $D[0,1]$ or $D[0,3]$, maximum is 2.
For $D cdot [0,1]$ or $D cdot [0,3]$, maximum is 3, achieved when the points are on a line passing through 0. Edit: Upon reflection, I think maximum is 5
For $D[0,3] setminus D[0,2]$, maximum is I guess 5, achieved when, but not only when, the points are on a line passing through 0 and close to $D[0,3]$?
If right, then what's the justification please?
If wrong, why and how else can I approach this please?
Asked here but there are no posted answers: Find the maximum number of horizontal and vertical segments in $G$ needed to connect two points of $G$.
Related:
Prove that $A_r,s$ $=[zin mathbb C : r<|z-z_0|<s ]$ is path connected.
An annulus in $mathbb R^2$ is path connected
general-topology complex-analysis complex-numbers connectedness path-connected
asked Jul 31 at 2:33
BCLC
6,99821973
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 1.34
I guess the answer is to do with Eg 1.13:
For $D[0,1]$ or $D[0,3]$, maximum is 2.
For $D cdot [0,1]$ or $D cdot [0,3]$, maximum is 3, achieved when the points are on a line passing through 0. Edit: Upon reflection, I think maximum is 5
For $D[0,3] setminus D[0,2]$, maximum is I guess 5, achieved when, but not only when, the points are on a line passing through 0 and close to $D[0,3]$?
If right, then what's the justification please?
If wrong, why and how else can I approach this please?
Asked here but there are no posted answers: Find the maximum number of horizontal and vertical segments in $G$ needed to connect two points of $G$.
Related:
Prove that $A_r,s$ $=[zin mathbb C : r<|z-z_0|<s ]$ is path connected.
An annulus in $mathbb R^2$ is path connected
general-topology complex-analysis complex-numbers connectedness path-connected
asked Jul 31 at 2:33
BCLC
6,99821973
edited Jul 31 at 14:31
asked Jul 31 at 2:33
BCLC
6,99821973
asked Jul 31 at 2:33
BCLC
6,99821973
asked Jul 31 at 2:33
BCLC
6,99821973
6,99821973
The MAXIMUM number is unlimited.
– William Elliot
Jul 31 at 2:56
@WilliamElliot Okay why $infty$ please? I'm trying to see a pattern here: if you delete a point then max is 3 and then if you delete a concentric circle with radius 0.00001 then max is perhaps 5 and then if the deleted concentric circle has a certain large enough radius than the max is $infty$. Is the threshold perhaps that the deleted concentric circle has a radius greater than the difference of the two radii?
– BCLC
Jul 31 at 3:14
1
@WilliamElliot Why would it be unlimited? There certainly is a closed zig-zag made of finitely many h/v segments contained within that annulus, then for any two points one only needs to "connect" them to that.
– dxiv
Jul 31 at 4:04
1
@BCLC That number is certainly bounded. The zig-zag itself is independent of the choice of the two points, and each of the two points can reach the zig-zag with (at most) one segment.
– dxiv
Jul 31 at 4:34
1
@BCLC Too long for a comment, posted as an answer. Note the last "but" however.
– dxiv
Jul 31 at 5:17
 |Â
show 2 more comments
The MAXIMUM number is unlimited.
– William Elliot
Jul 31 at 2:56
@WilliamElliot Okay why $infty$ please? I'm trying to see a pattern here: if you delete a point then max is 3 and then if you delete a concentric circle with radius 0.00001 then max is perhaps 5 and then if the deleted concentric circle has a certain large enough radius than the max is $infty$. Is the threshold perhaps that the deleted concentric circle has a radius greater than the difference of the two radii?
– BCLC
Jul 31 at 3:14
1
@WilliamElliot Why would it be unlimited? There certainly is a closed zig-zag made of finitely many h/v segments contained within that annulus, then for any two points one only needs to "connect" them to that.
– dxiv
Jul 31 at 4:04
1
@BCLC That number is certainly bounded. The zig-zag itself is independent of the choice of the two points, and each of the two points can reach the zig-zag with (at most) one segment.
– dxiv
Jul 31 at 4:34
1
@BCLC Too long for a comment, posted as an answer. Note the last "but" however.
– dxiv
Jul 31 at 5:17
The MAXIMUM number is unlimited.
– William Elliot
Jul 31 at 2:56
The MAXIMUM number is unlimited.
– William Elliot
Jul 31 at 2:56
@WilliamElliot Okay why $infty$ please? I'm trying to see a pattern here: if you delete a point then max is 3 and then if you delete a concentric circle with radius 0.00001 then max is perhaps 5 and then if the deleted concentric circle has a certain large enough radius than the max is $infty$. Is the threshold perhaps that the deleted concentric circle has a radius greater than the difference of the two radii?
– BCLC
Jul 31 at 3:14
@WilliamElliot Okay why $infty$ please? I'm trying to see a pattern here: if you delete a point then max is 3 and then if you delete a concentric circle with radius 0.00001 then max is perhaps 5 and then if the deleted concentric circle has a certain large enough radius than the max is $infty$. Is the threshold perhaps that the deleted concentric circle has a radius greater than the difference of the two radii?
– BCLC
Jul 31 at 3:14
1
1
@WilliamElliot Why would it be unlimited? There certainly is a closed zig-zag made of finitely many h/v segments contained within that annulus, then for any two points one only needs to "connect" them to that.
– dxiv
Jul 31 at 4:04
@WilliamElliot Why would it be unlimited? There certainly is a closed zig-zag made of finitely many h/v segments contained within that annulus, then for any two points one only needs to "connect" them to that.
– dxiv
Jul 31 at 4:04
1
1
@BCLC That number is certainly bounded. The zig-zag itself is independent of the choice of the two points, and each of the two points can reach the zig-zag with (at most) one segment.
– dxiv
Jul 31 at 4:34
@BCLC That number is certainly bounded. The zig-zag itself is independent of the choice of the two points, and each of the two points can reach the zig-zag with (at most) one segment.
– dxiv
Jul 31 at 4:34
1
1
@BCLC Too long for a comment, posted as an answer. Note the last "but" however.
– dxiv
Jul 31 at 5:17
@BCLC Too long for a comment, posted as an answer. Note the last "but" however.
– dxiv
Jul 31 at 5:17
 |Â
show 2 more comments
2 Answers
2
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2
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Exercise 1.34 is clearly meant to be a little brain teaser. It has nothing to do with complex analysis or topology.
Choose an $hin>bigl]2,3oversqrt2bigr[>$, e.g., $h:=2.06$, and draw the four lines $x=pm h$, $y=pm h$. Their four points of intersection are lying in the interior of the annulus $Omega$. Every point in $Omega$ can be joined within $Omega$ to one of these lines by a horizontal or vertical segment. It follows that any two points in $Omega$ can be joined with a chain of $leq5$ horizontal or vertical segments $sigma_isubsetOmega$.
answered Jul 31 at 9:11
Christian Blatter
163k7107305
Thanks Christian Blatter! Is this some standard topology thing, and may you please provide relevant reading material please?
– BCLC
Jul 31 at 14:28
1
(+1)clearly meant to be a little brain teaserIndeed. If the circles has radii $3,4$ (or $13,14$ for that matter), the problem would have suddenly become a lot more complicated.
– dxiv
Aug 1 at 0:01
Christian Blatter, @dxiv Ok then how does one come up with this answer please? Edit: Oh I saw dxiv's edit. Do you agree Christian Blatter?
– BCLC
Aug 1 at 5:09
add a comment |Â
up vote
1
down vote
(Following up on a previous comment, posting as an answer for the sake of the pic.)
It is possible to draw a $12$-side polygon with only h/v sides contained within the given annulus, and each of the two points can reach it with (at most) one segment. Therefore an upper bound to the answer is $,8,$, but this alone doesn't prove that it's the lowest upper bound i.e. the maximum.
[ EDIT ] Â The minimum is in fact $,5,$, as shown in Christian Blatter's answer.
A related, more general question, would be what is the required number of segments to connect any two points in an annulus by arbitrary line segments, without restricting them to be horizontal or vertical, only. A similar argument could show that the magic number is $,lfloor n/2 rfloor +3,$ where $,n,$ is the minimum integer such that a regular $n$-gon can be strictly entirely inside the annulus.
Since the ratio between of the radii of the inscribed vs. circumscribed circle to a regular $n$-gon is $,r / R = cos pi/n,$, it follows that for $,n=4,$ an annulus with $,1/2 le r/R lt 1 / sqrt2,$ would require $,lfloor 4/2rfloor + 3=5,$ segments (which is the answer here because $2/3 in left[1/2, 1/sqrt2right),$ indeed), for $,n=5,$ an annulus with $,1 / sqrt2 le r/R lt (1+sqrt5)/4,$ would also require $,5,$ segments (but those would no longer be just horizontal and vertical), for $,n=6,$ an annulus with $,(1+sqrt5)/4 le r/R lt sqrt3 / 2,$ would require $,6,$ segments etc.
answered Jul 31 at 5:16
dxiv
53.8k64796
1
Thanks dxiv! I won't accept answer in case someone has an idea for supremum
– BCLC
Jul 31 at 5:47
1
@BCLC Thanks. Fixed and clarified the last line now. And, yes, it's not an answer to what the maximum is, but the context of the comment I was following up on was just claiming that it's bounded for sure.
– dxiv
Jul 31 at 5:58
There is a minimum. There is no maximum because the zigs and zags can be made smaller and smaller, requiring more and more of them.
– William Elliot
Jul 31 at 7:34
dxiv, cross-examine @WilliamElliot ?
– BCLC
Jul 31 at 8:06
dxiv, @WilliamElliot, do you concur with Christian Blatter?
– BCLC
Jul 31 at 14:28
 |Â
show 3 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Exercise 1.34 is clearly meant to be a little brain teaser. It has nothing to do with complex analysis or topology.
Choose an $hin>bigl]2,3oversqrt2bigr[>$, e.g., $h:=2.06$, and draw the four lines $x=pm h$, $y=pm h$. Their four points of intersection are lying in the interior of the annulus $Omega$. Every point in $Omega$ can be joined within $Omega$ to one of these lines by a horizontal or vertical segment. It follows that any two points in $Omega$ can be joined with a chain of $leq5$ horizontal or vertical segments $sigma_isubsetOmega$.
answered Jul 31 at 9:11
Christian Blatter
163k7107305
Thanks Christian Blatter! Is this some standard topology thing, and may you please provide relevant reading material please?
– BCLC
Jul 31 at 14:28
1
(+1)clearly meant to be a little brain teaserIndeed. If the circles has radii $3,4$ (or $13,14$ for that matter), the problem would have suddenly become a lot more complicated.
– dxiv
Aug 1 at 0:01
Christian Blatter, @dxiv Ok then how does one come up with this answer please? Edit: Oh I saw dxiv's edit. Do you agree Christian Blatter?
– BCLC
Aug 1 at 5:09
add a comment |Â
up vote
2
down vote
Exercise 1.34 is clearly meant to be a little brain teaser. It has nothing to do with complex analysis or topology.
Choose an $hin>bigl]2,3oversqrt2bigr[>$, e.g., $h:=2.06$, and draw the four lines $x=pm h$, $y=pm h$. Their four points of intersection are lying in the interior of the annulus $Omega$. Every point in $Omega$ can be joined within $Omega$ to one of these lines by a horizontal or vertical segment. It follows that any two points in $Omega$ can be joined with a chain of $leq5$ horizontal or vertical segments $sigma_isubsetOmega$.
answered Jul 31 at 9:11
Christian Blatter
163k7107305
Thanks Christian Blatter! Is this some standard topology thing, and may you please provide relevant reading material please?
– BCLC
Jul 31 at 14:28
1
(+1)clearly meant to be a little brain teaserIndeed. If the circles has radii $3,4$ (or $13,14$ for that matter), the problem would have suddenly become a lot more complicated.
– dxiv
Aug 1 at 0:01
Christian Blatter, @dxiv Ok then how does one come up with this answer please? Edit: Oh I saw dxiv's edit. Do you agree Christian Blatter?
– BCLC
Aug 1 at 5:09
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Exercise 1.34 is clearly meant to be a little brain teaser. It has nothing to do with complex analysis or topology.
Choose an $hin>bigl]2,3oversqrt2bigr[>$, e.g., $h:=2.06$, and draw the four lines $x=pm h$, $y=pm h$. Their four points of intersection are lying in the interior of the annulus $Omega$. Every point in $Omega$ can be joined within $Omega$ to one of these lines by a horizontal or vertical segment. It follows that any two points in $Omega$ can be joined with a chain of $leq5$ horizontal or vertical segments $sigma_isubsetOmega$.
answered Jul 31 at 9:11
Christian Blatter
163k7107305
Exercise 1.34 is clearly meant to be a little brain teaser. It has nothing to do with complex analysis or topology.
Choose an $hin>bigl]2,3oversqrt2bigr[>$, e.g., $h:=2.06$, and draw the four lines $x=pm h$, $y=pm h$. Their four points of intersection are lying in the interior of the annulus $Omega$. Every point in $Omega$ can be joined within $Omega$ to one of these lines by a horizontal or vertical segment. It follows that any two points in $Omega$ can be joined with a chain of $leq5$ horizontal or vertical segments $sigma_isubsetOmega$.
answered Jul 31 at 9:11
Christian Blatter
163k7107305
edited Jul 31 at 14:58
answered Jul 31 at 9:11
Christian Blatter
163k7107305
answered Jul 31 at 9:11
Christian Blatter
163k7107305
answered Jul 31 at 9:11
Christian Blatter
163k7107305
163k7107305
Thanks Christian Blatter! Is this some standard topology thing, and may you please provide relevant reading material please?
– BCLC
Jul 31 at 14:28
1
(+1)clearly meant to be a little brain teaserIndeed. If the circles has radii $3,4$ (or $13,14$ for that matter), the problem would have suddenly become a lot more complicated.
– dxiv
Aug 1 at 0:01
Christian Blatter, @dxiv Ok then how does one come up with this answer please? Edit: Oh I saw dxiv's edit. Do you agree Christian Blatter?
– BCLC
Aug 1 at 5:09
add a comment |Â
Thanks Christian Blatter! Is this some standard topology thing, and may you please provide relevant reading material please?
– BCLC
Jul 31 at 14:28
1
(+1)clearly meant to be a little brain teaserIndeed. If the circles has radii $3,4$ (or $13,14$ for that matter), the problem would have suddenly become a lot more complicated.
– dxiv
Aug 1 at 0:01
Christian Blatter, @dxiv Ok then how does one come up with this answer please? Edit: Oh I saw dxiv's edit. Do you agree Christian Blatter?
– BCLC
Aug 1 at 5:09
Thanks Christian Blatter! Is this some standard topology thing, and may you please provide relevant reading material please?
– BCLC
Jul 31 at 14:28
Thanks Christian Blatter! Is this some standard topology thing, and may you please provide relevant reading material please?
– BCLC
Jul 31 at 14:28
1
1
(+1)
clearly meant to be a little brain teaser Indeed. If the circles has radii $3,4$ (or $13,14$ for that matter), the problem would have suddenly become a lot more complicated.– dxiv
Aug 1 at 0:01
(+1)
clearly meant to be a little brain teaser Indeed. If the circles has radii $3,4$ (or $13,14$ for that matter), the problem would have suddenly become a lot more complicated.– dxiv
Aug 1 at 0:01
Christian Blatter, @dxiv Ok then how does one come up with this answer please? Edit: Oh I saw dxiv's edit. Do you agree Christian Blatter?
– BCLC
Aug 1 at 5:09
Christian Blatter, @dxiv Ok then how does one come up with this answer please? Edit: Oh I saw dxiv's edit. Do you agree Christian Blatter?
– BCLC
Aug 1 at 5:09
add a comment |Â
up vote
1
down vote
(Following up on a previous comment, posting as an answer for the sake of the pic.)
It is possible to draw a $12$-side polygon with only h/v sides contained within the given annulus, and each of the two points can reach it with (at most) one segment. Therefore an upper bound to the answer is $,8,$, but this alone doesn't prove that it's the lowest upper bound i.e. the maximum.
[ EDIT ] Â The minimum is in fact $,5,$, as shown in Christian Blatter's answer.
A related, more general question, would be what is the required number of segments to connect any two points in an annulus by arbitrary line segments, without restricting them to be horizontal or vertical, only. A similar argument could show that the magic number is $,lfloor n/2 rfloor +3,$ where $,n,$ is the minimum integer such that a regular $n$-gon can be strictly entirely inside the annulus.
Since the ratio between of the radii of the inscribed vs. circumscribed circle to a regular $n$-gon is $,r / R = cos pi/n,$, it follows that for $,n=4,$ an annulus with $,1/2 le r/R lt 1 / sqrt2,$ would require $,lfloor 4/2rfloor + 3=5,$ segments (which is the answer here because $2/3 in left[1/2, 1/sqrt2right),$ indeed), for $,n=5,$ an annulus with $,1 / sqrt2 le r/R lt (1+sqrt5)/4,$ would also require $,5,$ segments (but those would no longer be just horizontal and vertical), for $,n=6,$ an annulus with $,(1+sqrt5)/4 le r/R lt sqrt3 / 2,$ would require $,6,$ segments etc.
answered Jul 31 at 5:16
dxiv
53.8k64796
1
Thanks dxiv! I won't accept answer in case someone has an idea for supremum
– BCLC
Jul 31 at 5:47
1
@BCLC Thanks. Fixed and clarified the last line now. And, yes, it's not an answer to what the maximum is, but the context of the comment I was following up on was just claiming that it's bounded for sure.
– dxiv
Jul 31 at 5:58
There is a minimum. There is no maximum because the zigs and zags can be made smaller and smaller, requiring more and more of them.
– William Elliot
Jul 31 at 7:34
dxiv, cross-examine @WilliamElliot ?
– BCLC
Jul 31 at 8:06
dxiv, @WilliamElliot, do you concur with Christian Blatter?
– BCLC
Jul 31 at 14:28
 |Â
show 3 more comments
up vote
1
down vote
(Following up on a previous comment, posting as an answer for the sake of the pic.)
It is possible to draw a $12$-side polygon with only h/v sides contained within the given annulus, and each of the two points can reach it with (at most) one segment. Therefore an upper bound to the answer is $,8,$, but this alone doesn't prove that it's the lowest upper bound i.e. the maximum.
[ EDIT ] Â The minimum is in fact $,5,$, as shown in Christian Blatter's answer.
A related, more general question, would be what is the required number of segments to connect any two points in an annulus by arbitrary line segments, without restricting them to be horizontal or vertical, only. A similar argument could show that the magic number is $,lfloor n/2 rfloor +3,$ where $,n,$ is the minimum integer such that a regular $n$-gon can be strictly entirely inside the annulus.
Since the ratio between of the radii of the inscribed vs. circumscribed circle to a regular $n$-gon is $,r / R = cos pi/n,$, it follows that for $,n=4,$ an annulus with $,1/2 le r/R lt 1 / sqrt2,$ would require $,lfloor 4/2rfloor + 3=5,$ segments (which is the answer here because $2/3 in left[1/2, 1/sqrt2right),$ indeed), for $,n=5,$ an annulus with $,1 / sqrt2 le r/R lt (1+sqrt5)/4,$ would also require $,5,$ segments (but those would no longer be just horizontal and vertical), for $,n=6,$ an annulus with $,(1+sqrt5)/4 le r/R lt sqrt3 / 2,$ would require $,6,$ segments etc.
answered Jul 31 at 5:16
dxiv
53.8k64796
1
Thanks dxiv! I won't accept answer in case someone has an idea for supremum
– BCLC
Jul 31 at 5:47
1
@BCLC Thanks. Fixed and clarified the last line now. And, yes, it's not an answer to what the maximum is, but the context of the comment I was following up on was just claiming that it's bounded for sure.
– dxiv
Jul 31 at 5:58
There is a minimum. There is no maximum because the zigs and zags can be made smaller and smaller, requiring more and more of them.
– William Elliot
Jul 31 at 7:34
dxiv, cross-examine @WilliamElliot ?
– BCLC
Jul 31 at 8:06
dxiv, @WilliamElliot, do you concur with Christian Blatter?
– BCLC
Jul 31 at 14:28
 |Â
show 3 more comments
up vote
1
down vote
up vote
1
down vote
(Following up on a previous comment, posting as an answer for the sake of the pic.)
It is possible to draw a $12$-side polygon with only h/v sides contained within the given annulus, and each of the two points can reach it with (at most) one segment. Therefore an upper bound to the answer is $,8,$, but this alone doesn't prove that it's the lowest upper bound i.e. the maximum.
[ EDIT ] Â The minimum is in fact $,5,$, as shown in Christian Blatter's answer.
A related, more general question, would be what is the required number of segments to connect any two points in an annulus by arbitrary line segments, without restricting them to be horizontal or vertical, only. A similar argument could show that the magic number is $,lfloor n/2 rfloor +3,$ where $,n,$ is the minimum integer such that a regular $n$-gon can be strictly entirely inside the annulus.
Since the ratio between of the radii of the inscribed vs. circumscribed circle to a regular $n$-gon is $,r / R = cos pi/n,$, it follows that for $,n=4,$ an annulus with $,1/2 le r/R lt 1 / sqrt2,$ would require $,lfloor 4/2rfloor + 3=5,$ segments (which is the answer here because $2/3 in left[1/2, 1/sqrt2right),$ indeed), for $,n=5,$ an annulus with $,1 / sqrt2 le r/R lt (1+sqrt5)/4,$ would also require $,5,$ segments (but those would no longer be just horizontal and vertical), for $,n=6,$ an annulus with $,(1+sqrt5)/4 le r/R lt sqrt3 / 2,$ would require $,6,$ segments etc.
answered Jul 31 at 5:16
dxiv
53.8k64796
(Following up on a previous comment, posting as an answer for the sake of the pic.)
It is possible to draw a $12$-side polygon with only h/v sides contained within the given annulus, and each of the two points can reach it with (at most) one segment. Therefore an upper bound to the answer is $,8,$, but this alone doesn't prove that it's the lowest upper bound i.e. the maximum.
[ EDIT ] Â The minimum is in fact $,5,$, as shown in Christian Blatter's answer.
A related, more general question, would be what is the required number of segments to connect any two points in an annulus by arbitrary line segments, without restricting them to be horizontal or vertical, only. A similar argument could show that the magic number is $,lfloor n/2 rfloor +3,$ where $,n,$ is the minimum integer such that a regular $n$-gon can be strictly entirely inside the annulus.
Since the ratio between of the radii of the inscribed vs. circumscribed circle to a regular $n$-gon is $,r / R = cos pi/n,$, it follows that for $,n=4,$ an annulus with $,1/2 le r/R lt 1 / sqrt2,$ would require $,lfloor 4/2rfloor + 3=5,$ segments (which is the answer here because $2/3 in left[1/2, 1/sqrt2right),$ indeed), for $,n=5,$ an annulus with $,1 / sqrt2 le r/R lt (1+sqrt5)/4,$ would also require $,5,$ segments (but those would no longer be just horizontal and vertical), for $,n=6,$ an annulus with $,(1+sqrt5)/4 le r/R lt sqrt3 / 2,$ would require $,6,$ segments etc.
answered Jul 31 at 5:16
dxiv
53.8k64796
edited Jul 31 at 23:49
answered Jul 31 at 5:16
dxiv
53.8k64796
answered Jul 31 at 5:16
dxiv
53.8k64796
answered Jul 31 at 5:16
dxiv
53.8k64796
53.8k64796
1
Thanks dxiv! I won't accept answer in case someone has an idea for supremum
– BCLC
Jul 31 at 5:47
1
@BCLC Thanks. Fixed and clarified the last line now. And, yes, it's not an answer to what the maximum is, but the context of the comment I was following up on was just claiming that it's bounded for sure.
– dxiv
Jul 31 at 5:58
There is a minimum. There is no maximum because the zigs and zags can be made smaller and smaller, requiring more and more of them.
– William Elliot
Jul 31 at 7:34
dxiv, cross-examine @WilliamElliot ?
– BCLC
Jul 31 at 8:06
dxiv, @WilliamElliot, do you concur with Christian Blatter?
– BCLC
Jul 31 at 14:28
 |Â
show 3 more comments
1
Thanks dxiv! I won't accept answer in case someone has an idea for supremum
– BCLC
Jul 31 at 5:47
1
@BCLC Thanks. Fixed and clarified the last line now. And, yes, it's not an answer to what the maximum is, but the context of the comment I was following up on was just claiming that it's bounded for sure.
– dxiv
Jul 31 at 5:58
There is a minimum. There is no maximum because the zigs and zags can be made smaller and smaller, requiring more and more of them.
– William Elliot
Jul 31 at 7:34
dxiv, cross-examine @WilliamElliot ?
– BCLC
Jul 31 at 8:06
dxiv, @WilliamElliot, do you concur with Christian Blatter?
– BCLC
Jul 31 at 14:28
1
1
Thanks dxiv! I won't accept answer in case someone has an idea for supremum
– BCLC
Jul 31 at 5:47
Thanks dxiv! I won't accept answer in case someone has an idea for supremum
– BCLC
Jul 31 at 5:47
1
1
@BCLC Thanks. Fixed and clarified the last line now. And, yes, it's not an answer to what the maximum is, but the context of the comment I was following up on was just claiming that it's bounded for sure.
– dxiv
Jul 31 at 5:58
@BCLC Thanks. Fixed and clarified the last line now. And, yes, it's not an answer to what the maximum is, but the context of the comment I was following up on was just claiming that it's bounded for sure.
– dxiv
Jul 31 at 5:58
There is a minimum. There is no maximum because the zigs and zags can be made smaller and smaller, requiring more and more of them.
– William Elliot
Jul 31 at 7:34
There is a minimum. There is no maximum because the zigs and zags can be made smaller and smaller, requiring more and more of them.
– William Elliot
Jul 31 at 7:34
dxiv, cross-examine @WilliamElliot ?
– BCLC
Jul 31 at 8:06
dxiv, cross-examine @WilliamElliot ?
– BCLC
Jul 31 at 8:06
dxiv, @WilliamElliot, do you concur with Christian Blatter?
– BCLC
Jul 31 at 14:28
dxiv, @WilliamElliot, do you concur with Christian Blatter?
– BCLC
Jul 31 at 14:28
 |Â
show 3 more comments
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The MAXIMUM number is unlimited.
– William Elliot
Jul 31 at 2:56
@WilliamElliot Okay why $infty$ please? I'm trying to see a pattern here: if you delete a point then max is 3 and then if you delete a concentric circle with radius 0.00001 then max is perhaps 5 and then if the deleted concentric circle has a certain large enough radius than the max is $infty$. Is the threshold perhaps that the deleted concentric circle has a radius greater than the difference of the two radii?
– BCLC
Jul 31 at 3:14
1
@WilliamElliot Why would it be unlimited? There certainly is a closed zig-zag made of finitely many h/v segments contained within that annulus, then for any two points one only needs to "connect" them to that.
– dxiv
Jul 31 at 4:04
1
@BCLC That number is certainly bounded. The zig-zag itself is independent of the choice of the two points, and each of the two points can reach the zig-zag with (at most) one segment.
– dxiv
Jul 31 at 4:34
1
@BCLC Too long for a comment, posted as an answer. Note the last "but" however.
– dxiv
Jul 31 at 5:17