A Circular Diversion
March 27, 2013 9:25 AM Subscribe
The Circle Drawing Experiment. You've seen competitive circle drawing (previously). Now try your own hand (mouse?) at drawing a freehand circle. Bonus: cats.
Wulfhere: "Bonus: Who can get the lowest score?"
A short straight line will get you 0.00%. For some reason, I can't get the cat to change, though.
posted by Apropos of Something at 9:31 AM on March 27, 2013
A short straight line will get you 0.00%. For some reason, I can't get the cat to change, though.
posted by Apropos of Something at 9:31 AM on March 27, 2013
97.84, with a mouse! My years spent earning my Ph.D. have paid off!
posted by Elementary Penguin at 9:31 AM on March 27, 2013
posted by Elementary Penguin at 9:31 AM on March 27, 2013
A kinda straight line resulted in 1.94%. That will be my Wolfram metric of the day.
posted by Foci for Analysis at 9:32 AM on March 27, 2013
posted by Foci for Analysis at 9:32 AM on March 27, 2013
Score = 96.41%
First try.
On a Magic Trackpad I got just this week.
posted by Thorzdad at 9:32 AM on March 27, 2013
First try.
On a Magic Trackpad I got just this week.
posted by Thorzdad at 9:32 AM on March 27, 2013
On the other hand, I got 11.95% by drawing a starburst.
On Preview: Drawing a two dimensional figure seems like cheating :(
posted by Elementary Penguin at 9:33 AM on March 27, 2013
On Preview: Drawing a two dimensional figure seems like cheating :(
posted by Elementary Penguin at 9:33 AM on March 27, 2013
I got NaN%. Which competition do I win?
posted by edd at 9:33 AM on March 27, 2013 [4 favorites]
posted by edd at 9:33 AM on March 27, 2013 [4 favorites]
it's much easier to get in the high 90s than i thought it would be. i saw a couple different cats. also, it seems that my scores are better when i draw much larger circles.
posted by rude.boy at 9:35 AM on March 27, 2013 [1 favorite]
posted by rude.boy at 9:35 AM on March 27, 2013 [1 favorite]
I can get 93% but might do better if I wasn't having to dodge all the crap on my desk while trying to draw. I wonder if larger or smaller circles are easier to draw accurately?
posted by TedW at 9:49 AM on March 27, 2013
posted by TedW at 9:49 AM on March 27, 2013
97.18% on my first try with a trackpad on a MacBook Pro. It didn't look so good, but I guess it was statistically sound. :)
posted by pmbuko at 9:49 AM on March 27, 2013
posted by pmbuko at 9:49 AM on March 27, 2013
The worse I drew, the closer to 100% I got. So I drew a really bad potato-shaped thing in about 2 seconds and somehow achieved over 98%. A decent-looking circle got me 88%
It's not working too well.
posted by pipeski at 9:50 AM on March 27, 2013
It's not working too well.
posted by pipeski at 9:50 AM on March 27, 2013
It seemed like it wasn't so much the worse I drew, but the faster I drew, that got me the highest scores. I stopped at 98.81.
posted by louche mustachio at 9:53 AM on March 27, 2013
posted by louche mustachio at 9:53 AM on March 27, 2013
99.35% with a high precision mouse and still getting better. Draw a large circle. Draw quickly with wrist motion. Let go of the mouse button before completing the circle and let it finish for you.
posted by Xoc at 9:54 AM on March 27, 2013
posted by Xoc at 9:54 AM on March 27, 2013
98.72 as well, for a circle I would personally rate as "mediocre". The secret seems to be to draw large, and let go a bit early -- even a small amount of overlap at the end = much lower score. Just let the circle close itself with a flat (but short) segment.
posted by Fnarf at 9:57 AM on March 27, 2013
posted by Fnarf at 9:57 AM on March 27, 2013
99.08 with a mouse after six tries.
(90% on first quick try without much effort)
posted by mikepop at 9:59 AM on March 27, 2013
(90% on first quick try without much effort)
posted by mikepop at 9:59 AM on March 27, 2013
I tried this several times and a circle with a big obvious flat lumpy edge scored 97% and what looked to me like a near perfect circle scored in the mid 80s.
I don't think this thing's mathematics of circularity based on perimiter and area bears much resemblance to my intuitive judgements of circularity.
posted by edheil at 9:59 AM on March 27, 2013
I don't think this thing's mathematics of circularity based on perimiter and area bears much resemblance to my intuitive judgements of circularity.
posted by edheil at 9:59 AM on March 27, 2013
I drew a friggin' donkey and it scored at 47% , any tiny straight line scores at 20% or so...
This is sort of a crap thing...
posted by HuronBob at 10:00 AM on March 27, 2013 [2 favorites]
This is sort of a crap thing...
posted by HuronBob at 10:00 AM on March 27, 2013 [2 favorites]
99.37 after following the advice to let the circle close itself.
posted by mikepop at 10:01 AM on March 27, 2013
posted by mikepop at 10:01 AM on March 27, 2013
99.43 but I've seen all the cats now, so losing motivation.
posted by mikepop at 10:04 AM on March 27, 2013 [1 favorite]
posted by mikepop at 10:04 AM on March 27, 2013 [1 favorite]
Yeah, uh, I drew a square and got 85%. So I don't feel so good about my 95% when I tried to draw a circle.
posted by Justinian at 10:05 AM on March 27, 2013 [1 favorite]
posted by Justinian at 10:05 AM on March 27, 2013 [1 favorite]
Bonus: cats
Was anyone else hoping for a cat drawing bonus round?
posted by Ned G at 10:07 AM on March 27, 2013
Was anyone else hoping for a cat drawing bonus round?
posted by Ned G at 10:07 AM on March 27, 2013
Error based on a least squares fit would be a better measure of "circle-ness". As it is, any mostly-convex blob scores pretty high.
posted by qxntpqbbbqxl at 10:11 AM on March 27, 2013
posted by qxntpqbbbqxl at 10:11 AM on March 27, 2013
How I would test for it:
-- make an axis-aligned bounding rectangle that just encompasses the points.
-- compare width vs. height of rectangle for a general level of "trueness"
-- find the center of the bounding rectangle.
-- average the width and height to come up with the circle radius
-- at some arbitrary sampling of headings, measure the difference between where a circle at that radius should fall, and where your line actually is. (If your line doubled back, you don't even have a convex shape.)
-- alternatively, score for accuracy on an ellipse, and use the difference between radii to adjust your score in some way
-- come up with some kind of bullshit scoring mechanism to summarize.
posted by Foosnark at 10:12 AM on March 27, 2013
-- make an axis-aligned bounding rectangle that just encompasses the points.
-- compare width vs. height of rectangle for a general level of "trueness"
-- find the center of the bounding rectangle.
-- average the width and height to come up with the circle radius
-- at some arbitrary sampling of headings, measure the difference between where a circle at that radius should fall, and where your line actually is. (If your line doubled back, you don't even have a convex shape.)
-- alternatively, score for accuracy on an ellipse, and use the difference between radii to adjust your score in some way
-- come up with some kind of bullshit scoring mechanism to summarize.
posted by Foosnark at 10:12 AM on March 27, 2013
"Your score...
Perimeter = 0.0 units
Area = 0.0 units squared
A perfect circle with this perimeter would be 0.0 units squared
Score = NaN%
BTW, you a score as close to 100% as possible. The happiness of the cat reflects your score."
Apparently I need to hold down the mouse.
posted by Lemurrhea at 10:21 AM on March 27, 2013
Perimeter = 0.0 units
Area = 0.0 units squared
A perfect circle with this perimeter would be 0.0 units squared
Score = NaN%
BTW, you a score as close to 100% as possible. The happiness of the cat reflects your score."
Apparently I need to hold down the mouse.
posted by Lemurrhea at 10:21 AM on March 27, 2013
When I went slow and really tried to make it perfect, I got 86%. When I made one as quickly as possible, I got 94%. Hmph.
posted by ThePinkSuperhero at 10:26 AM on March 27, 2013
posted by ThePinkSuperhero at 10:26 AM on March 27, 2013
To minimize your score, draw a straight line. This is also harder than it looks - I can't get below about 3%, despite the presence of a built-in grid to tell me what to do.
posted by madcaptenor at 10:27 AM on March 27, 2013
posted by madcaptenor at 10:27 AM on March 27, 2013
I got 2.58% by pressing CTRL - about ten times before drawing a straight line across the screen, but I can't do it consistently.
posted by Elementary Penguin at 10:39 AM on March 27, 2013
posted by Elementary Penguin at 10:39 AM on March 27, 2013
Draw quickly with wrist motion.
Years of art classes taught me: don't move your wrist. Draw with your arm. Arms are better at circles than wrists are.
posted by shakespeherian at 10:40 AM on March 27, 2013
Years of art classes taught me: don't move your wrist. Draw with your arm. Arms are better at circles than wrists are.
posted by shakespeherian at 10:40 AM on March 27, 2013
Apparently I need to hold down the mouse.
Make sure to hold it down with your arm, not your wrist. I just learned that.
posted by Panjandrum at 11:10 AM on March 27, 2013
Make sure to hold it down with your arm, not your wrist. I just learned that.
posted by Panjandrum at 11:10 AM on March 27, 2013
I drew a rudimentary penis.
posted by GallonOfAlan at 11:49 AM on March 27, 2013 [1 favorite]
posted by GallonOfAlan at 11:49 AM on March 27, 2013 [1 favorite]
We all did, GallonofAlan, we all did.
posted by Justinian at 12:20 PM on March 27, 2013 [1 favorite]
posted by Justinian at 12:20 PM on March 27, 2013 [1 favorite]
100% and you get rewarded with this guy
I changed the line s = (s * 100) - 100; in the JavaScript so that s = 0; and I always get 100%. I'll go hang my head in shame now.
posted by inflatablekiwi at 1:46 PM on March 27, 2013
I changed the line s = (s * 100) - 100; in the JavaScript so that s = 0; and I always get 100%. I'll go hang my head in shame now.
posted by inflatablekiwi at 1:46 PM on March 27, 2013
I drew an angry cat and got 70.49%
posted by changeling at 2:47 PM on March 27, 2013
posted by changeling at 2:47 PM on March 27, 2013
Yeah, the calculation for "perfect" appears to take the length of your line, assume a circle with that perimeter, then compare that circle's area to the area of what you drew.
The text says that, but the score seems to come from comparing the perimeters. (For example, I just got "Perimeter = 865.2 units. Area = 57613.5 units squared. A perfect circle with this perimeter would be 59571.7 units squared. Score = 98.34%." But the ratio of areas is 57613.5/59571.7≈0.967, which doesn't match the score, while the perimeter of a circle with area 57613.5 would be sqrt(4*pi*57613.5)≈850.9, and 850.9/865.2≈0.983, which does match the score.) (Or maybe he's taking the square root of the ratio of the areas, which would come out the same.)
If that is indeed what the code is doing, then there's no intrinsic preference for larger figures: if you scale your figure up by a factor of 2, its perimeter goes up by a factor of 2, but so does the perimeter of the circle you'd compare it to, so the ratio (and hence the score) is unchanged.
The theoretical justification for this method of measuring circularity, by the way, is the isoperimetric inequality.
posted by stebulus at 3:19 PM on March 27, 2013
The text says that, but the score seems to come from comparing the perimeters. (For example, I just got "Perimeter = 865.2 units. Area = 57613.5 units squared. A perfect circle with this perimeter would be 59571.7 units squared. Score = 98.34%." But the ratio of areas is 57613.5/59571.7≈0.967, which doesn't match the score, while the perimeter of a circle with area 57613.5 would be sqrt(4*pi*57613.5)≈850.9, and 850.9/865.2≈0.983, which does match the score.) (Or maybe he's taking the square root of the ratio of the areas, which would come out the same.)
If that is indeed what the code is doing, then there's no intrinsic preference for larger figures: if you scale your figure up by a factor of 2, its perimeter goes up by a factor of 2, but so does the perimeter of the circle you'd compare it to, so the ratio (and hence the score) is unchanged.
The theoretical justification for this method of measuring circularity, by the way, is the isoperimetric inequality.
posted by stebulus at 3:19 PM on March 27, 2013
Yeah, uh, I drew a square and got 85%.
The theoretical result for a square would be sqrt(pi/4)≈0.886, so, yeah.
posted by stebulus at 3:24 PM on March 27, 2013
The theoretical result for a square would be sqrt(pi/4)≈0.886, so, yeah.
posted by stebulus at 3:24 PM on March 27, 2013
I used a transparent glass on the screen as a guide and got a 94%. The highest I could get freehand was about 90% by making the circle really small.
posted by codacorolla at 3:50 PM on March 27, 2013
posted by codacorolla at 3:50 PM on March 27, 2013
I think the reason it favours drawing quickly is if you draw slowly your hand has more time to wobble, increasing the length of the perimeter.
Like odinsdream and Foosnark talked about, I think something based around average distance from the centre would evaluate it more in line with what you'd intuitively expect.
posted by RobotHero at 5:00 PM on March 27, 2013
Like odinsdream and Foosnark talked about, I think something based around average distance from the centre would evaluate it more in line with what you'd intuitively expect.
posted by RobotHero at 5:00 PM on March 27, 2013
Like odinsdream and Foosnark talked about, I think something based around average distance from the centre would evaluate it more in line with what you'd intuitively expect.
Arbitrary shapes don't have centres. The nice thing about a metric based on perimeter and area is that you don't have to try to figure out what "centre" means for shapes which don't really have centres, or which have more than one reasonable candidate for that title. (odinsdream said "midpoint", but I'm not sure what that means for things that aren't line segments. Foosnark suggested using the centre of the bounding axis-aligned rectangle, but that would give different answers for rotated copies of the same shape, which seems pretty unnatural.)
If the problem is just that the numbers seem unintuitively high then transforming the scale to magnify the difference between circles and near-circles might work better than cooking up ad hoc measures of circularity. For example, you could report x/(1-x), where x is the score that the site gives; then the shapes which the site reports as between 95% and 100% would be spread out between 19 and ∞. A line segment would score 0, a square would score roughly 7.8, and a perfect circle would score ∞.
And, you know, just because a metric doesn't accord with our intuition doesn't mean the metric is wrong. Sure, it would be interesting to try to figure out how people assess circularity, to try to capture our intuition in a precise quantitative form, but that's not the only interesting topic here. The geometry of the isoperimetric inequality speaks for itself, and maybe we should listen to what it has to say, rather than being disappointed that it's not just repeating our preconceptions.
posted by stebulus at 5:52 PM on March 27, 2013
Arbitrary shapes don't have centres. The nice thing about a metric based on perimeter and area is that you don't have to try to figure out what "centre" means for shapes which don't really have centres, or which have more than one reasonable candidate for that title. (odinsdream said "midpoint", but I'm not sure what that means for things that aren't line segments. Foosnark suggested using the centre of the bounding axis-aligned rectangle, but that would give different answers for rotated copies of the same shape, which seems pretty unnatural.)
If the problem is just that the numbers seem unintuitively high then transforming the scale to magnify the difference between circles and near-circles might work better than cooking up ad hoc measures of circularity. For example, you could report x/(1-x), where x is the score that the site gives; then the shapes which the site reports as between 95% and 100% would be spread out between 19 and ∞. A line segment would score 0, a square would score roughly 7.8, and a perfect circle would score ∞.
And, you know, just because a metric doesn't accord with our intuition doesn't mean the metric is wrong. Sure, it would be interesting to try to figure out how people assess circularity, to try to capture our intuition in a precise quantitative form, but that's not the only interesting topic here. The geometry of the isoperimetric inequality speaks for itself, and maybe we should listen to what it has to say, rather than being disappointed that it's not just repeating our preconceptions.
posted by stebulus at 5:52 PM on March 27, 2013
Simple it should be the centre of the circle that I wanted to draw. :)
posted by RobotHero at 6:30 PM on March 27, 2013 [1 favorite]
posted by RobotHero at 6:30 PM on March 27, 2013 [1 favorite]
For a rather crude Pacman on a trackpad: 68.14%
Non-Pacman circles on a trackpad: 60s ~ 90s
Using a tablet pen: 99.11%
I'm not sure if I should try this with a trackpoint. Never really got used to using one.
posted by aroweofshale at 6:45 PM on March 27, 2013
Non-Pacman circles on a trackpad: 60s ~ 90s
Using a tablet pen: 99.11%
I'm not sure if I should try this with a trackpoint. Never really got used to using one.
posted by aroweofshale at 6:45 PM on March 27, 2013
stebulus: "Arbitrary shapes don't have centres."
They have centers of gravity, which might work for this.
posted by Joakim Ziegler at 10:04 PM on March 27, 2013
They have centers of gravity, which might work for this.
posted by Joakim Ziegler at 10:04 PM on March 27, 2013
Yeah, centre of gravity is a good candidate.
odinsdream, your parallel lines idea is a very good one, which also nicely avoids having to pick a centre. The distance between the bounding parallel lines is called the "width" in that direction; if it's the same in all directions, we say the figure "has constant width". Surprisingly, there are figures which are not circles but do have constant width, such as the Reuleaux triangle.
(Incidentally, if you take the average of the widths in all directions, you get the "mean width", which for convex two-dimensional shapes is basically the same as the perimeter. In three dimensions and up, though, this is a different quantity, and there's an analogue of the isoperimetric inequality — of all figures with the same volume, the ball has the smallest mean width — which could also be used to quantify how close a shape is to a ball.)
(Notions based on width don't distinguish between a figure and its convex hull, which might or might not be a desirable property.)
posted by stebulus at 6:50 AM on March 28, 2013
odinsdream, your parallel lines idea is a very good one, which also nicely avoids having to pick a centre. The distance between the bounding parallel lines is called the "width" in that direction; if it's the same in all directions, we say the figure "has constant width". Surprisingly, there are figures which are not circles but do have constant width, such as the Reuleaux triangle.
(Incidentally, if you take the average of the widths in all directions, you get the "mean width", which for convex two-dimensional shapes is basically the same as the perimeter. In three dimensions and up, though, this is a different quantity, and there's an analogue of the isoperimetric inequality — of all figures with the same volume, the ball has the smallest mean width — which could also be used to quantify how close a shape is to a ball.)
(Notions based on width don't distinguish between a figure and its convex hull, which might or might not be a desirable property.)
posted by stebulus at 6:50 AM on March 28, 2013
The rolling idea won't work.
On further reflection, it can be adjusted a bit to work: the circle is the only curve of constant width which is also centrally symmetric (i.e., symmetric under reflection in a point). So you could combine a metric of constant width (say, the variance of the width) with a metric of central symmetry to get a metric of circularity. It's a bit ad hoc, though.
posted by stebulus at 8:45 AM on March 28, 2013
On further reflection, it can be adjusted a bit to work: the circle is the only curve of constant width which is also centrally symmetric (i.e., symmetric under reflection in a point). So you could combine a metric of constant width (say, the variance of the width) with a metric of central symmetry to get a metric of circularity. It's a bit ad hoc, though.
posted by stebulus at 8:45 AM on March 28, 2013
Interesting. Considering all these is making me think about what constitutes "circleness."
My own non-mathematical way of visualizing it was:
1) Calculate the area of the shape, same as they are currently doing
2) Create a circle of the same area
3) Position it over the shape
4) Calculate the ratio of area where the two shapes overlap to the area they don't overlap.
But of course you now have the problem of finding the optimized position for comparison in step 3.
The rolling idea is an interesting idea. It would ignore concavities, so for example a Pac Man or a pizza with a slice missing would score the same as if the missing piece were just a flat segment, and a 5-pointed star would score the same as pentagon. Whether that's a problem would depend on how "circleness" needs to be evaluated.
posted by RobotHero at 12:40 PM on March 28, 2013
My own non-mathematical way of visualizing it was:
1) Calculate the area of the shape, same as they are currently doing
2) Create a circle of the same area
3) Position it over the shape
4) Calculate the ratio of area where the two shapes overlap to the area they don't overlap.
But of course you now have the problem of finding the optimized position for comparison in step 3.
The rolling idea is an interesting idea. It would ignore concavities, so for example a Pac Man or a pizza with a slice missing would score the same as if the missing piece were just a flat segment, and a 5-pointed star would score the same as pentagon. Whether that's a problem would depend on how "circleness" needs to be evaluated.
posted by RobotHero at 12:40 PM on March 28, 2013
But of course you now have the problem of finding the optimized position for comparison in step 3.
If you don't mind replacing the given figure with its convex hull (i.e., ignoring concavities, as you said), then I think that wouldn't be too bad, actually. If you fix a convex shape and slide a circle around over top of it, then the volume of their overlap is a ½-concave function of the position of the circle (a fact first proved in 1950 by Fáry and Rédei, as far as I know). That's good because it means that a "hill-climbing" algorithm will work — making small shifts that increase the volume of the overlap will lead you to the global maximum position, with no possibility of getting trapped in a local-but-not-global maximum.
(This is not at all true if your set is nonconvex.)
posted by stebulus at 2:22 PM on March 28, 2013
If you don't mind replacing the given figure with its convex hull (i.e., ignoring concavities, as you said), then I think that wouldn't be too bad, actually. If you fix a convex shape and slide a circle around over top of it, then the volume of their overlap is a ½-concave function of the position of the circle (a fact first proved in 1950 by Fáry and Rédei, as far as I know). That's good because it means that a "hill-climbing" algorithm will work — making small shifts that increase the volume of the overlap will lead you to the global maximum position, with no possibility of getting trapped in a local-but-not-global maximum.
(This is not at all true if your set is nonconvex.)
posted by stebulus at 2:22 PM on March 28, 2013
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Bonus: Who can get the lowest score?
posted by Wulfhere at 9:27 AM on March 27, 2013