Sharing a Burger Between Three

There are several ways to divide a burger evenly between three people.

Probably the best is to cut it radially (like a pizza). It can be tricky working out exactly where to make the cuts, but if you can pull it off, then all the pieces will be roughly identical (topping distribution notwithstanding).

But for the sake of arguing, lets say you want to divide the burger by making two parallel cuts: Where, then, do you make the cuts so that all three people get the same amount of burger?

NB/ This gets quite maths-heavy, so if you're not interested in that sort of thing, feel free to skip right to the end for the solution.



Geometry

For simplicity, we're going to consider the burger as a circle, and make the cuts so that each chunk has the same area. The two cuts are going to be the same distance from, and parallel to, the central axis of the burger, so we only need to consider the position of one of the cuts.

Here's the set-up

We work out the area of the cut-off as the area of the circular segment, minus the area of the triangle.


- Aside: Radians

Radians are basically an alternative way of measuring angles. For maths and physics they're generally more useful than degrees.

They're relatively easy - there are 2pi radians in a full circle, so 2pi radians = 360 degrees

1 radian = 180/pi = 57.3 degrees
1 degree = pi/180 = 0.017 radians, etc.

*    *    *

Back to the circle; with angle x in radians, the area of the circular segment is

The area of a triangle is half base times height..


- Aside: Area of the Triangle

We start by splitting the triangle down the middle, so that we have two identical right angle triangles

The height, l = r*cos(x/2)

The base, b = 2*(r*sin(x/2))

So the area of the triangle is (l*b)/2 = r^2 sin(x/2)cos(x/2)

Finally, use the identity sin(2x) = 2sin(x)cos(x)
to get

*    *    *

So the area of the cut-off is

and it needs to equal a third the area of the circle = 1/3 pi r^2.

So first, we need to find x satisfying

or

Once we have a value for x, we find where to make the cut from

Intermission

The thing about this equation is it doesn't have an exact, analytical solution - to find the solution you have to use numerical methods. Well, I say you have to use numerical methods; these days you can just type the equation into WolframAlpha, and you'll get a solution like *snaps fingers*

Which is nice. I even have the WolframAlpha app on my phone. But when I thought up this question I was on holiday in Sherwood forest, where there was literally no mobile singal.

So that was out of the question. And since I'm not in the habit of carrying a scientific calculator around with me, I was stuck with the basic calculator on my phone. It looks like this:

No trig functions, no square roots, no pi button. It doesn't even do brackets, or have a memory function. Luckily, I am in the habit of carrying around a notepad and pen.

Anyway, there are two ways of working this out with only a basic calculator. The first is 'easier', but only if you know some stuff, and the numbers happen to be nice (in this case, they kind of are). The second is harder, in that it requires more number crunching, but it'll work with any numbers, and can be more precise.

Again, feel free to skip to the solution if you're not interested in the gritty details.



Method One

First of all, here's a graph of the two sides of the equation

hand-drawn with Skitch

We want to find the point at which the two graphs cross. We can see that that happens somewhere between 2pi/3 and pi (120 and 180 degrees). So, lets make a guess that it's exactly halfway between these two values: 5pi/6 (150 degrees).

For the right hand side of the equation: 5pi/6 - 2pi/3 = 0.52

NB/ I'm using pi=3.1416 (rounded to 4 decimal places). If you prefer, you could use the approximation 22/7. The result should be roughly the same.

For the left hand side of the equation, we need to work out sin(5pi/6)

At A-level, we were expected to memorise sin() and cos() of angles 0, 30, 45, 60, 90, and 180 (degrees). We were also expected to know the formulas for sin() and cos() of sums of angles. For sin(), it works like

Why is this important? Well 150 degrees = 180 - 30 (5pi/6 rads = pi - pi/6)

So

And since 0.5 is pretty close to 0.52 - less than 5% error - we can accept the convenience of that answer and say it's close enough.

So our approximate value of x is 5pi/6 = 2.618

[Incidentally, the identity for sin(2x) is just a special case of the above, with a=b=x; i.e. sin(2x) = sin(x+x) = 2sin(x)cos(x)]


Now we just need to work out l/r = cos(x/2) = cos(5pi/12)

For this one, 5pi/12 rads = 75 degrees = 45 + 30, so we can use

So

And we just have to evaluate that. But we don't have a square root button. Now, I just happen to know that sqrt(3) ~ 1.73 and sqrt(2) ~ 1.41.

But I'm just weird like that. Let's say you don't. How do you work it out?


- Aside: Square Roots

There are several ways of working out square roots with just basic operators. For two easy examples:

The first is 'Trial and Improvement' - pick a number, square it, does that give the right answer? If not, pick another number based on whether the last guess was too big or too small.

For example: sqrt(3)

1.5 -> 2.25 -> too small
1.7 -> 2.89 -> too small
1.8 -> 3.24 -> too big
1.75 -> 3.0625 -> too big
1.73 -> 2.9929 -> too small
1.74 -> 3.0276 -> too big
1.735 -> 3.010225 -> too big
1.7325 -> 3.00155625 -> too big
1.732 -> 2.999824 -> too small
etc.

The second method is the "Babylonian Method". It's more systematic, and can converge to the correct answer quicker than guessing. But it can be irritating if your calculator doesn't have a memory function.

It uses the recurrence relation

Basically, you make a guess xn. Divide the number you want to square root (S) by xn. If xn is lower than the actual square root, then S/xn will be greater than it. That means the actual root will be between xn and S/xn, so we make the next guess xn+1 the average of these two values. Repeat until x is sufficiently accurate.

For example: sqrt(2)

x0 = 1.5 -> 2/1.5 = 1.33
x1 = (1.5 + 1.33)/2 = 1.4166.. -> 2/1.4167 = 1.41176..
x2 = (1.4167 + 1.41176..)/2 = 1.41421.. -> 2/1.41421 = 1.41421..

*    *    *

Whatever way you do it, you repeat the process until you get the degree of accuracy you're happy with.

You should get the answer around l/r = 0.259



Method Two

We go back to the equation sin(x) = x - 2pi/3

We still have to find the solution numerically, we still don't have a calculator with a sin() function, and this time the numbers don't work out nicely.

So, the question is, how do we calculate sin(x)?


- Aside: Taylor Expansion

The Taylor Expansion of a function is a way of fitting a polynomial (sums of powers) to a more complicated function. It works like this

Basically, it gives a way of converting a function we can't calculate into an infinite sum of powers of x, which we can calculate.

It's usually expanded around the origin (x0=0), since the equations work out neater. But you can do it around any point, x0=a. This is useful if the value you are trying to calculate is far from x=0. The closer x is to x0=a, the quicker the sum converges.

Even though the expansion is an infinite sum, it's usually sufficient to just take the first few terms, since each additional term makes a smaller and smaller contribution to the sum.

So the trick is working out how many terms you need to include to get some desired level of accuracy.

*    *    *

In this case, I'm going to use the Taylor Expansion of sin(x) around x0=pi, since the approximate value (2.6) is nearer to pi than 0.

Here's what the expansion looks like

So, how many terms do we need to include?

Here's what the graph looks like for different numbers of terms

plotted with WolframAlpha

For a value around 2.6, it can be shown that including the first two terms is correct to ~3 decimal places; the first three terms is correct to ~5 decimal places; the first four terms to ~7 decimal places, etc.

So I would probably go to the third term (for 5dp), but only take the result to 3dp.

NB/ We shouldn't get too hung up on getting an extremely accurate value for x, since we're already getting rounding errors from the factors of pi in the expansion. Also, since we're calculating x to 5dp, we should use pi=3.14159

That means we want to solve

which can't be solved exactly.

So, for finding the correct value (without WolframAlpha), we can use any root-finding method. For what it's worth, I used Trial and Improvement; the other methods are easier with a computer.

But, note that the function is decreasing

So if the guess gives a value greater than zero, you need to increase the value of x (and vice versa).

If you run through all that (I won't go into detail), it gives a value around x=2.605


Alternatively, you could expand around x0=5pi/6 (if you know/can work-out sin and cos of 150 deg without a calculator).

In this case you'd only need up to the term in x^2 (correct to ~4dp). Using this expansion would mean solving a quadratic equation, which is easy. But using this expansion can introduce more rounding errors from the factors of sqrt(3). It's a matter of preference, I guess. The answer should be about the same.


Finally, we need to calculate cos(x/2)

Again, we use the Taylor Expansion to calculate cos(). In this case, we're doing the expansion around x0=0; the expansion is

In this case, you just keep adding terms until the result remains approximately constant to some desired degree of accuracy (3pd).

This gives a value around l/r = 0.265



So What is the Real Answer?

Once I got to somewhere where I could get at WolframAlpha, I checked the real numbers; here are the results:

The approximation of x from Method One (2.618) is an over estimate by ~0.5%, which is relatively acceptable. The approximation from Method Two (2.605) is correct to 3 decimal places, which is definitely acceptable.

And for the value of l/r

From Method One (0.259), the approximation is an under estimate by 2%, and correct to 2 decimal places, so is probably acceptable. The approximation from Method Two (0.265) is, again, correct to 3 decimal places. So that is also acceptable.

So, if the numbers happen to be convenient and you know some trigonometry, you're probably as well using Method One. If not, or if you just want more accuracy, then go for Method Two.



Applying the Results

The results are actually quite nice, in terms of practical application (dividing up a burger). The ratio of the radius (0.265) being close to one quarter, you find the cuts like this

That is, find the central axis, then find the (imaginary) line halfway between the centre and the edge - make the cut halfway between the centre and this imaginary line (maybe cut an extra hair's breadth towards the edge). Repeat on the other side.

Easy.


So, now you know. Obviously, all this applies to dividing any circular thing evenly between three people. You could probably even adapt the methods for sharing between even more people.

And in theory, You could do all this with just pen and paper (no calculator). Though you probably wouldn't want to. I know I wouldn't..


Oatzy.


[Wow, I really managed to stretch that one out.]

Some Twitter Infographics

I did some stuff like this before. And I figured, while I was updating my network graphs, why not update some of the other graphics?

And it helps that I worked out how to easily extract data from Twitter (see previous blog). The code is here. Again, rate limits apply.


Who Do I Follow?

This is one of the ones I did before - collect together the bios of the people I follow, then make a word cloud (using Wordle)

Basically, I follow a bunch of geeks and writers. Who like 'things'. So really, same as a year and a half ago.

I would point out though that 6 of the people I follow don't have bios, and about 7 just have lyrics.

Data here.


Who Tweets the Most?

These rates are worked out as (total tweets posted)/(total days online). Obviously, the actually post rate will vary over different time scales..

Bubble chart (made with ManyEyes) - bubbles sized by tweet rate (the numbers on some of the bubbles).

The graph below gives a better idea of relative rates, and 'rankings' (click to embiggen)

The blue line is actual values.

The orange is a logarithmic trend-line. It's a pretty good fit (R2=0.95); and, loosely speaking, it means ~70% of the tweets in my timeline come from ~30% of the people I follow. [cf: Pareto Principle]

You get similar log-shaped graphs when you split up the genders.

Full data here.


Chattiest Gender?

You can read all the explanation, caveats, etc. in the previous posts (here and here). I'm just going to go straight into the data.

I follow 27 men and 21 women (excluding celebrities, etc.). The stats are as follow:

Men:
Average = 6.21 tweets/day
Standard Deviation = 6.47

Women:
Average = 13.79 tweets/day
Standard Deviation = 14.27

For clarity, here's a  boxplot (made in R)

Basically, the women tweet more on average, and their rates are more spread out than for the men. In fact, roughly three quarters of the men tweet less than half of the women. Also, there's one outlier in the female group.

This is similar to what we found last time; although the women's average and spread aren't quite as high (average: 13.79 vs 19.21), and the men's average has increased slightly (6.21 vs 5.29).

If you take the ratio of the averages, the women tweet 2.15 times as much as the men. But maybe I just follow particularly chatty women..

Here's treemap (ManyEyes), which should give you a better idea of the gender balance (boxes sized by tweet rate)

Specifically, the graphic above is 62.5% purple (female).

Data here.


Where in the World Are My Followers?

The site I used last time doesn't seem to exist anymore. So I'm using MapMyFollowers instead. As the name suggests, these are my followers, rather than just the people I follow. Nonetheless..

Mostly in the UK and the US. As you'd probably expect.

I will point out though, some of the locations are a little suspect. Some people haven't made their location available so aren't included, and others seem to be in countries they couldn't possibly be in. But it's the best we can do.

Here's a zoom in on the UK

What Do I Tweet?

Made with Wordle, with data from TweetStats.

Words are sized by how often I tweet them; and by extension, @usernames are sized by how often I tweet those people.

In fact, here are the people I 'mention' the most (TweetStats)

Couldn't get a good source on who @replies me. That was one of the things Twoolr used to do..


When Do I Tweet?

Twoolr used to be awesome for Twitter statistics. But sadly, when they left beta, they started charging. And their free service went to shit. Luckily, I found TweetStats. Weirdly, it doesn't need you to log-in or anything, but somehow it can pull data on (nearly) all your tweets - beyond the 3,200 limit. Strange.

Here's some more graphs

Basically, I tweet most on a Friday and Saturday, and at around 1-2pm.

And I've never tweeted at 5am. But that's probably because I'm always asleep at 5am

Except that one time I got really drunk. (SleepBot)


How Much Do I Tweet?

This is another one I used to go to Twoolr for. And, to be fair, I still could. But that only goes as far back as April '10, and its graphics aren't as clear. Here's TweetStats again

Like I said before, I didn't tweet much in my first year. In fact, I only posted 36 tweets in all of 2009.

Now, the one problem with TweetStats is that 5 month gap in 2010. Why is this significant? Well, I was definitely tweeting during that time. In fact, by my estimates, over those 5 months I posted 5,724 tweets (~37tweets/day). So those 5 months account for 43% of all my tweets.

See, the thing is, in 2010, I was out of university, single, and unemployed. I posted a total 8,823 tweets - 24tweets/day. Since I've been back at university, that number's dropped to 11tweets/day.

That lull in Summer 2011 was when I was spending all my time on Tumblr and watching classic Doctor Who. Incidentally, I haven't posted on Tumblr since the start of September '11. It's terribly addictive, you see. I wouldn't recommend it; unless you're addicted to Doctor Who and Sherlock, and have lots of time on your hands..


So yeah.


Oatzy.


[Self-indulgent statistics, and pretty illustrations.]

Friend Network Evolution

Back in February 2009, I created my Twitter account, upon the insistence of my then-girlfriend. I didn't get it. Back then, Facebook was where it was at, I didn't really get Twitter's appeal. I was pretty much just following the handful of people I knew in real life, and Stephen Fry.

So I didn't use it much. I'd pop up every now and then, post a couple tweets and give up on it again. At one point, I even developed an irrational dislike of it - whenever I saw a site had a "follow us on Twitter" button, it irked me for some reason.

But at some point, towards the end of 2009/early 2010, I gave it yet another try. I don't know why. And even when I started using it, I was resistant; still half-heartedly hating it. But what was different this time, is I started chatting with people, and I was introduced to new people.

People who don't get Twitter think it's just that thing where you can tell people when you're eating a sandwich. It's not. It's the people that make Twitter. (Tweens and arseholes notwithstanding.)

But I'm going off on a tangent.


By August 2010, I was well into Twitter - I was posting around 30 tweets per day, and I had around 30 friends*. And back then, I decided I wanted to see what my friends network looked like. So I broke out Python and the Twitter API, I pulled data, and I made the graph. Here's an updated version of it.

[click to embiggen]

Fairly small, and tidy, and relatively uncomplicated. The bulk on the right is the people I knew in real life (from school, etc.) with a few strands of new acquaintances. Note how tightly packed and interconnected they are. To the left is mostly people I met through Twitter - and in particular, through PkmnTrainerJ.

(In case you hadn't figured it out, the node and label sizes are proportional to number of connections.)

By December 2010, I decided to have a look again.

Again, this is an update of the version I originally posted; and in this case, I've tried to arrange it so that key people stay in approximately the same place.

So you still vaguely have that left-right divide, but now there's much more mixing in the middle. I'd made some new friends, but more interesting is the people who were already in the graph who formed new connections with others in my graph.

I'd also like to draw your attention to shinelikestars_ (formerly shinelikestars6) - take a look at the previous graph, can you spot him? From 2 shared connections to 8 in the space of two months. I don't think there's any sort of point I'm trying to make here. I'm just pointing it out 'cause it's interesting.


And for the next year and a half I didn't do any data collecting. It became too labourious - Twitter changed its API, so that my old code didn't work, and I had to do everything by hand.

So, the latest graph was March 2011 (technical details below). As you can imagine, a lot can happen in a year and a half.

First of all, the new people add, and the old people removed. But more importantly, look how much tighter, and how much more 'segmented' the graph is.

There are now three major groups, loosely centred on the three most connected of my friends.

On the far right are, again, the people I knew in real life. In particular, note how little that group has changed since the first graph.

In the middle, we have 'Shiney's People' - people I was introduced to by shinelikestars_. And on the left are the people I was introduced to by PkmnTrainerJ.

The smaller groups circled in red are cliques - smaller subgroups that, at least from my point of view, form their own little groupings, where (almost) everyone is interconnected. The bottom left 'clique', for example, is my parents and big sister.

And I suspect, if you were to extend the graph beyond my network, you would find that those cliques are just parts of larger interconnected groups.

In case you were wondering, PkmnTrainerJ and SallyBembridge are most connected, both with degree 14. shinelikestars_ is next most, with degree 10.

Notable disappearing nodes - Benjidoom, who deleted his account, then created a new, private one (benjirino); and AimlessAmy, who is a long story.

I should also point out that the people I follow who aren't friends with anyone else in my network do not appear in the pictured graphs. Not that they aren't as cool, they just don't join onto the graph.

* I use friend here to mean people who I follow and who follow me back. Though I would probably consider all the people in my current network (including those not pictured) friends to some degree.


Technical stuff

You can read details on how I collected the data before, in the previous blogs. But, as I say, those methods don't work anymore.

For this run, I read up on the API, and found some bits that don't need authentication to grab and manipulate.

First, you can grab a list of a user's friends with this URL

https://api.twitter.com/1/friends/ids.xml?screen_name=<username>

This will give you a list of the friends ID numbers, so you also need to use this to grab usernames

https://api.twitter.com/1/users/lookup.xml?user_id=<idnumber>

There is also a URL to check if a user follows another user

https://api.twitter.com/1/friendships/exists.xml?user_id_a=<idnumber1>&user_id_b=<idnumber2>

Which works through the browser, but I couldn't get to work in my code. So in place, I used the site DoesFollow.com; partly because it uses the URL scheme  

DoesFollow.com/user1/user2.

Which is very convenient. Though I do worry all the requests might be putting strain on that site's server.

So, putting all that together with a bit of Python, you get something like this.


A few important points:

1) It will take a while to run. I have ~50 friends, and it took well over an hour to pull all the data. In terms of computational complexity, it's O(n^2), but each of those operations takes a significant amount of time.

2) Twitter has an API limit of 150 requests per hour. The number of API requests the code will make is ~ the number of friends being looked up. I think. Which means, if you have more than 100 or so friends, this code probably won't work. Sorry. There might be a way around it, but I don't know how.

3) Obviously, this doesn't work on protected accounts. So for those people you will have to grab data by hand. Though it's not too bad for a small enough number of people.

If you do want to use the code, I've made it so you just have to change the username at the top, and run it. You will need to install Python though.

For creating the graphs, I previously use ManyEyes. But I moved to using Gephi, because it allows for more customising. The output from the code is a text file with a list of name pairs, which you can import directly into Gephi. It will build the graph for you, and then you're free to play as you like.


Aaand... Yeah, I think that's about it.


Oatzy.


[shinelikestars6 lost his red circle, on account of he isn't my nemesis anymore.]

So What Was the Best Day To Go Shopping?

Alright, let's be done with this.

Just a quick reminder - what I did was collect Foursquare check-in data for various shopping centres around the UK, in the hope that the data might show something interesting.

Previous blog posts on this data collecting - Best Day to Go Shopping, Panic Saturday, Christmas Eve.

Anyway, I've been collecting data for over 3 months now. And that seems like quite enough.

Here's a graph of (normalised) averaged check-ins on each day of the week for 4 periods:

DecAv (blue) is 21st Nov 2011 to 18th Dec 2011
ChrAv (grey) is 19th Dec 2011 to 1st Jan 2012
JanAv (orange) is 5th Jan 2012 to 2nd Feb 2012
FebAv (green) is 6th Feb 2012 to 4th Mar 2012

Aside from the two weeks either side of Christmas (grey) - when people, apparently, did their shopping more midweek - the pattern is basically the same.

For further clarity, here's the average of those three averages (excluding Christmas)

And here is the order of days, from least to most busy ('relative busyness' in brackets):

1) Wednesday (1.00)
2) Monday (1.01)
3) Tuesday (1.03)
4) Thursday (1.11)
5) Sunday (1.15)
6) Friday (1.24)
7) Saturday (1.78)

Note that the differences between Monday, Tuesday, and Wednesday are not statistically significant - they're essentially the same, and are likely to be as busy as each other/not noticeably different.

So, to answer the title question - Monday, Tuesday, and Wednesday are the best days to go shopping. At least, in as much as they're the days shopping centres are likely to be least busy. And, as you'd expect, Saturday is, by far, the worst/most busy.

And the last thing to point out is that these are the averages over 20 shopping centres for a ~3 month period - numbers for specific locations, and at different times (eg holidays) are likely to deviate from the averages.

And, basically, that's that.

If you're interested, you can see the raw check-in data here.


Oatzy.


[That was definitely worth the effort.]

Simple Harmonic Sleep

So I started using SleepBot Tracker to track my sleep. This is what it's looking like so far

Basically, I'm averaging about 8 hours, except on those two days near the start where I had to get up early for exams.

Now, being a physics student, the graph reminded me of that for damped simple harmonic motion (starting from 20/01, ignoring the first 3 points). And being a crazy person, I decided to try and model my sleep as such.

So what's simple harmonic motion?

Simple harmonic motion is a type of periodic motion with a restoring force directly proportional to the the system's displacement from equilibrium.

For example, a pendulum is a simple harmonic oscillator - it has periodic motion, its equilibrium is the lowest point of the swing, and the restoring force is gravity.

Now, one could argue that sleep is SHM-like - we have some typical sleep length (equilibrium), and if we get too little sleep, then we'll tend to sleep more in response, and vice versa (restoring force). But it's a dubious analogy at best.

A damped harmonic oscillator is one with damping, which tends to reduce the amplitude of oscillations. So, like air resistance in the case of the pendulum, which eventually causes it to stop swinging.

I'm not sure what the sleep-based analogy for damping would be.


There's a standard equation for defining a (weakly) damped harmonic oscillator. It looks like this:

Where:- A0 is the initial displacement, the e bit is the decaying term, gamma is the damping coefficient (which determines how quickly the oscillations decay), cos() is the oscillating term, omega is the (damped) frequency of oscillation, t is time (in days), phi is the phase shift, and C is the equilibrium amplitude.

So working out the variables from the data, the model equation for my sleep looks something like this

And the graph looks like this

And to prove I'm not entirely crazy, here are the real values and the model values plotted together

Not a bad fit, right? [error 0.19]

In fact, you might notice the real data is still oscillating a little. But the equation outlined above tends to a constant amplitude of 8.3 (no more oscillations).

To account for this, we could add a baseline oscillation term

And fitting the data again, here's what the graph looks like

Arguably, a slightly better fit. [error 0.16]

So yeah. Basically, I'm procrastinating..


Oatzy.


[Just gotta be careful not to hit resonant sleepquency]

Christmas Eve

The run up to Christmas is over, and hopefully everyone has their Christmas shopping done. So now seems like as good a time as any to look at what the crowds are doing:

Now that is interesting. I think it is, anyway.

So first of all, the weekday spikes. It seems like a significant number of people left their Christmas shopping to the last week. Now fair enough if they've only just finished work/university/whatever.

Notice also the general week-on-week trend for the weekdays.

But more importantly than all that, look at today - Christmas Eve.


Anti-Crowds

Everyone 'knows' that going shopping on Christmas Eve is suicidal. It's a well known cliché. But that's the thing - everyone thinks it's going to be insanely busy, so a lot of people avoid shopping centres.

But then, this means the shopping centres end up practically deserted (by Christmas standards). And it's not necessarily the result of what I mentioned last time - I was out there, and I saw the lack of crowds.

It's an interesting, and previously studied phenomenon.

I won't go into too much detail, cause it's past midnight on Christmas Eve. But if you're interested in this effect, and in this sort of thing in general, I would suggest reading up on Complexity Theory. In particular, I'd recommend this book, which covers this exact subject in one of the early chapters.


Of course, if we were looking at supermarkets, that would be a different story..


Oatzy.


[Merry Christmas!]