spiraling out of control.... -凯发k8网页登录

created by matt fig

{"relationships":[{"relationshiptype":"http://schemas.mathworks.com/matlab/code/2013/relationships/document","targetmode":"","relationshipid":"rid1","target":"/matlab/document.xml"},{"relationshiptype":"http://schemas.mathworks.com/matlab/code/2013/relationships/output","targetmode":"","relationshipid":"rid2","target":"/matlab/output.xml"}],"parts":[{"parturi":"/matlab/document.xml","relationship":[],"contenttype":"application/vnd.mathworks.matlab.code.document xml","content":"<?xml version=\"1.0\" encoding=\"utf-8\"?>\n<w:document xmlns:w=\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\"><w:body><w:p><w:ppr><w:pstyle w:val=\"text\"/></w:ppr><w:r><w:t>imagine a square with corners at the origin and at (1,1). an interesting spiral can be created by tracing the path of a moving point that starts at the origin and intersects the leftmost opposite edge at a point (f,1), where f is the fraction of the length of the intersected edge. the point then continues on in this way, always heading for the leftmost opposite side of its path and intersecting the segment at a point which is fraction f of its length. for example, if f = 1/2, the first few points are:</w:t></w:r></w:p><w:p><w:ppr><w:pstyle w:val=\"code\"/></w:ppr><w:r><w:t><![cdata[p = [0 0;0.5 1;1 0.5;0.5 0;0.25 0.5;.75 .75];\nplot(p(:,1),p(:,2),'*-') \naxis equal]]></w:t></w:r></w:p><w:p><w:ppr><w:pstyle w:val=\"text\"/></w:ppr><w:r><w:t>if f = 1/3, then the first few positions of the particle are:</w:t></w:r></w:p><w:p><w:ppr><w:pstyle w:val=\"code\"/></w:ppr><w:r><w:t><![cdata[p = [0 0;1/3 1;1 2/3;2/3 0;1/9 1/3;5/9 8/9];]]></w:t></w:r></w:p><w:p><w:ppr><w:pstyle w:val=\"text\"/></w:ppr><w:r><w:t>write a function that takes two inputs and returns two outputs. the first input argument, n, is the nth location of the particle. the second input argument is the fraction f, where f is on (0,1). the first returned argument is a two element row vector that has the coordinates of the nth location of the particle as [x,y]. the second returned argument is the limiting value of the particles location as n-->infinity. for example,</w:t></w:r></w:p><w:p><w:ppr><w:pstyle w:val=\"code\"/></w:ppr><w:r><w:t><![cdata[[p,l] = my_spiral(3,1/2)]]></w:t></w:r></w:p><w:p><w:ppr><w:pstyle w:val=\"text\"/></w:ppr><w:r><w:t>returns</w:t></w:r></w:p><w:p><w:ppr><w:pstyle w:val=\"code\"/></w:ppr><w:r><w:t><![cdata[p = [1 0.5]]]></w:t></w:r></w:p><w:p><w:ppr><w:pstyle w:val=\"text\"/></w:ppr><w:r><w:t>and</w:t></w:r></w:p><w:p><w:ppr><w:pstyle w:val=\"code\"/></w:ppr><w:r><w:t><![cdata[l = [?x? ?y?]]]></w:t></w:r></w:p><w:p><w:ppr><w:pstyle w:val=\"text\"/></w:ppr><w:r><w:t>where both values of l are doubles accurate to 100*eps.</w:t></w:r></w:p></w:body></w:document>"},{"parturi":"/matlab/output.xml","contenttype":"text/xml","content":"<?xml version=\"1.0\" encoding=\"utf-8\" standalone=\"no\" ?><embeddedoutputs><metadata><evaluationstate>manual</evaluationstate><layoutstate>code</layoutstate><outputstatus>ready</outputstatus></metadata><outputarray type=\"array\"/><regionarray type=\"array\"/></embeddedoutputs>"}]}

imagine a square with corners at the origin and at (1,1). an interesting spiral can be created by tracing the path of a moving point that starts at the origin and intersects the leftmost opposite edge at a point (f,1), where f is the fraction of the length of the intersected edge. the point then continues on in this way, always heading for the leftmost opposite side of its path and intersecting the segment at a point which is fraction f of its length. for example, if f = 1/2, the first few points are:<pre class="language-matlab">p = [0 0;0.5 1;1 0.5;0.5 0;0.25 0.5;.75 .75];
plot(p(:,1),p(:,2),'*-') 
axis equal
</pre>if f = 1/3, then the first few positions of the particle are:<pre class="language-matlab">p = [0 0;1/3 1;1 2/3;2/3 0;1/9 1/3;5/9 8/9];
</pre>write a function that takes two inputs and returns two outputs. the first input argument, n, is the nth location of the particle. the second input argument is the fraction f, where f is on (0,1). the first returned argument is a two element row vector that has the coordinates of the nth location of the particle as [x,y]. the second returned argument is the limiting value of the particles location as n-->infinity. for example,<pre class="language-matlab">[p,l] = my_spiral(3,1/2)
</pre>returns<pre class="language-matlab">p = [1 0.5] 
</pre>and<pre class="language-matlab">l = [?x? ?y?] 
</pre>where both values of l are doubles accurate to 100*eps.

imagine a square with corners at the origin and at (1,1).  an interesting spiral can be created by tracing the path of a moving point that starts at the origin and intersects the leftmost opposite edge at a point (f,1), where f is the fraction of the length of the intersected edge.  the point then continues on in this way, always heading for the leftmost opposite side of its path and intersecting the segment at a point which is fraction f of its length.  for example, if f = 1/2, the first few points are:
  p = [0 0;0.5 1;1 0.5;0.5 0;0.25 0.5;.75 .75];
  plot(p(:,1),p(:,2),'*-') 
  axis equal
if f = 1/3, then the first few positions of the particle are:
  p = [0 0;1/3 1;1 2/3;2/3 0;1/9 1/3;5/9 8/9];
write a function that takes two inputs and returns two outputs. the first input argument, n, is the nth location of the particle. the second input argument is the fraction f, where f is on (0,1). the first returned argument is a two element row vector that has the coordinates of the nth location of the particle as [x,y]. the second returned argument is the limiting value of the particles location as n-->infinity. for example,
  [p,l] = my_spiral(3,1/2)
returns
  p = [1 0.5] 
and
  l = [?x? ?y?] 
where both values of l are doubles accurate to 100*eps.

solution stats

last submitted on jun 02, 2023

last 200 solutions

problem comments

2 comments

2 comments

on 29 feb 2012

i can solve the simpler test cases, but how do you handle the eps stuff without too much error or crashing the program?

on 13 dec 2021

good question

solution comments

problem recent solvers

problem tags

community treasure hunt

find the treasures in matlab central and discover how the community can help you!

start hunting!

spiraling out of control.... -凯发k8网页登录

solution stats

last 200 solutions

problem comments

solution comments

problem recent solvers

suggested problems

more from this author

problem tags

community treasure hunt

spiraling out of control.... -凯发k8网页登录

solution stats

last 200 solutions

problem comments

solution comments

problem recent solvers

suggested problems

more from this author

problem tags

community treasure hunt

players

wechat