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The PEMDAS Paradox
https://plus.maths.org/content/pemdasparadox
<div class="field fieldnamefieldabsimg fieldtypeimage fieldlabelhidden"><div class="fielditems"><div class="fielditem even"><img class="imgresponsive" src="https://plus.maths.org/content/sites/plus.maths.org/files/abstractpics/%5Buid%5D/%5Bsitedate%5D/icon_13.png" width="100" height="100" alt="" /></div></div></div><div class="field fieldnamefieldauthor fieldtypetext fieldlabelinlinec clearfix fieldlabelinline"><div class="fieldlabel">By </div><div class="fielditems"><div class="fielditem even">David Linkletter</div></div></div><div class="field fieldnamebody fieldtypetextwithsummary fieldlabelhidden"><div class="fielditems"><div class="fielditem even"><p>
It looks trivial but it keeps going viral. What answer do you get when
you calculate <img src="/MI/8d2f0430a7d568c69a8a212e81edbe41/images/img0001.png" alt="$6\div 2(1+2)$" style="verticalalign:4px;
width:85px;
height:18px" class="math gen" />? This question has reached every corner of social media, and has had millions of people respond with two common answers: <img src="/MI/8d2f0430a7d568c69a8a212e81edbe41/images/img0002.png" alt="$1$" style="verticalalign:0px;
width:6px;
height:12px" class="math gen" /> and <img src="/MI/8d2f0430a7d568c69a8a212e81edbe41/images/img0003.png" alt="$9$" style="verticalalign:0px;
width:8px;
height:12px" class="math gen" />.
</p>
<p>You might think one half of those people are right and the
other half need to check their arithmetic. But it never plays out like
that; respondents on both sides defend their answers with
confidence. There have been no formal mathematical publications about
the problem, but a growing number of mathematicians can explain what's going on:
<img src="/MI/719e07b9a03b7d9110a113bec3bef3d5/images/img0001.png" alt="$6\div 2(1+2)$" style="verticalalign:4px;
width:85px;
height:18px" class="math gen" /> is not a <em>welldefined</em> expression.
</p>
<div class="centreimage" ><img src="/content/sites/plus.maths.org/files/articles/2019/pemdas/equals91.png"/><p style="maxwidth:400px"></p></div>
<! Rubbish image made by Rachel for Plus >
<p>
<em><a href="https://en.wikipedia.org/wiki/Welldefined">Welldefined</a></em> is an important term in maths. It essentially means that a certain input
always yields the same output. All maths teachers agree that <img src="/MI/0f4a991d38967bcf1d6f5fc2e9629df3/images/img0001.png" alt="$6\div (2(1+2)) = 1$" style="verticalalign:4px;
width:128px;
height:18px" class="math gen" />, and that <img src="/MI/0f4a991d38967bcf1d6f5fc2e9629df3/images/img0002.png" alt="$(6\div 2)(1+2) = 9$" style="verticalalign:4px;
width:128px;
height:18px" class="math gen" />. The extra parentheses (brackets) remove the ambiguity and those expressions are welldefined. Most other viral maths problems, such as <img src="/MI/0f4a991d38967bcf1d6f5fc2e9629df3/images/img0003.png" alt="$93\div 1/3 + 1$" style="verticalalign:4px;
width:109px;
height:16px" class="math gen" /> (see <a
href="https://www.iflscience.com/editorsblog/canyousolvemathproblemwentviraljapan/">here</a>), are welldefined, with one correct answer and one (or
more) common erroneous answer(s). But calculating the value of the
expression <img src="/MI/7f11d7cb8d92960813f0bfd187df0fc1/images/img0001.png" alt="$6\div 2(1+2)$" style="verticalalign:4px;
width:85px;
height:18px" class="math gen" /> is a matter of convention. Neither answer, <img src="/MI/7f11d7cb8d92960813f0bfd187df0fc1/images/img0002.png" alt="$1$" style="verticalalign:0px;
width:6px;
height:12px" class="math gen" /> nor <img src="/MI/7f11d7cb8d92960813f0bfd187df0fc1/images/img0003.png" alt="$9$" style="verticalalign:0px;
width:8px;
height:12px" class="math gen" />, is wrong; it depends on what you learned from your maths teacher.
</p>
<p>
The
order in which to perform mathematical operations is given by the various
mnemonics PEMDAS, BODMAS, BIDMAS and BEDMAS:
<ul>
<li><strong>P</strong> (or <strong>B</strong>): first calculate the value of expressions inside any
parentheses (brackets);</li>
<li><strong>E</strong> (or <strong>O</strong> or <strong>I</strong>): next calculate any
exponents (orders/indices);</li>
<li><strong>MD</strong> (or <strong>DM</strong>): next carry out any multiplications and
divisions, working from left to right;</li>
<li><strong>AS</strong>: and finally carry out any additions and
subtractions, working from left to right.</li>
</ul>
</p>
<p>
Two slightly different interpretations of PEMDAS (or BODMAS, etc)
have been taught around the world, and the PEMDAS Paradox highlights their difference. Both
sides are substantially popular and there is currently no standard for the convention worldwide.
So you can stop that Twitter discussion and rest assured that each of you might be correctly
remembering what you were taught – it's just that you were taught differently.
</p>
<h3>The two sides</h3>
<p>
Mechanically, the people on the "9" side – such as in the most
popular <a href="https://www.youtube.com/watch?v=URcUvFIUIhQ">YouTube video</a> on
this question – tend to calculate <img src="/MI/445c4561b9cafb7a21e9c2a371fee963/images/img0001.png" alt="$6\div 2(1+2) = 6 \div 2 \times 3 = 3\times 3 = 9$" style="verticalalign:4px;
width:263px;
height:18px" class="math gen" />, or perhaps they write it as <img src="/MI/445c4561b9cafb7a21e9c2a371fee963/images/img0002.png" alt="$6\div 2(1+2) = 6\div 2(3) = 3(3) = 9$" style="verticalalign:4px;
width:248px;
height:18px" class="math gen" />. People on this side tend to say that <img src="/MI/445c4561b9cafb7a21e9c2a371fee963/images/img0003.png" alt="$a(b)$" style="verticalalign:4px;
width:27px;
height:18px" class="math gen" /> can be replaced with <img src="/MI/445c4561b9cafb7a21e9c2a371fee963/images/img0004.png" alt="$a\times b$" style="verticalalign:0px;
width:36px;
height:11px" class="math gen" /> at any time. It can be reduced down to that: the teaching that "<img src="/MI/445c4561b9cafb7a21e9c2a371fee963/images/img0003.png" alt="$a(b)$" style="verticalalign:4px;
width:27px;
height:18px" class="math gen" /> is always interchangeable with <img src="/MI/445c4561b9cafb7a21e9c2a371fee963/images/img0004.png" alt="$a\times b$" style="verticalalign:0px;
width:36px;
height:11px" class="math gen" />" determines the PEMDAS Paradox's answer to be <img src="/MI/445c4561b9cafb7a21e9c2a371fee963/images/img0005.png" alt="$9$" style="verticalalign:0px;
width:8px;
height:12px" class="math gen" />.
</p><p>
On the "1" side, some people calculate <img src="/MI/0928d82f22339bffcb74a29241bf31e9/images/img0001.png" alt="$6\div 2(1+2) = 6\div 2(3) = 6\div 6 = 1$" style="verticalalign:4px;
width:255px;
height:18px" class="math gen" />, while others point out the distributive property, <img src="/MI/0928d82f22339bffcb74a29241bf31e9/images/img0002.png" alt="$6\div 2(1+2) = 6\div (2+4) = 6\div 6 = 1$" style="verticalalign:4px;
width:275px;
height:18px" class="math gen" />. The driving principle on this side is that implied multiplication via juxtaposition takes priority. This has been taught in maths classrooms around the world and is also a stated convention in some programming contexts. So here, the teaching that "<img src="/MI/0928d82f22339bffcb74a29241bf31e9/images/img0003.png" alt="$a(b)$" style="verticalalign:4px;
width:27px;
height:18px" class="math gen" /> is always interchangeable with <img src="/MI/0928d82f22339bffcb74a29241bf31e9/images/img0004.png" alt="$(ab)$" style="verticalalign:4px;
width:26px;
height:18px" class="math gen" />" determines the PEMDAS Paradox answer to be <img src="/MI/0928d82f22339bffcb74a29241bf31e9/images/img0005.png" alt="$1$" style="verticalalign:0px;
width:6px;
height:12px" class="math gen" />.
</p><p>
Mathematically, it's inconsistent to simultaneously believe that <img src="/MI/e6f67b4858233e9d20f6498ba1ee0d92/images/img0001.png" alt="$a(b)$" style="verticalalign:4px;
width:27px;
height:18px" class="math gen" /> is interchangeable with <img src="/MI/e6f67b4858233e9d20f6498ba1ee0d92/images/img0002.png" alt="$a\times b$" style="verticalalign:0px;
width:36px;
height:11px" class="math gen" /> and also that <img src="/MI/e6f67b4858233e9d20f6498ba1ee0d92/images/img0001.png" alt="$a(b)$" style="verticalalign:4px;
width:27px;
height:18px" class="math gen" /> is interchangeable with <img src="/MI/e6f67b4858233e9d20f6498ba1ee0d92/images/img0003.png" alt="$(ab)$" style="verticalalign:4px;
width:26px;
height:18px" class="math gen" />. Because then it follows that <img src="/MI/e6f67b4858233e9d20f6498ba1ee0d92/images/img0004.png" alt="$1 = 9$" style="verticalalign:0px;
width:37px;
height:12px" class="math gen" /> via the arguments in the preceding paragraphs. Arriving at that contradiction is logical, simply illustrating that we can't have both answers. It also illuminates the fact that neither of those interpretations are inherent to PEMDAS. Both are subtle additional rules which decide what to do with syntax oddities such as <img src="/MI/e6f67b4858233e9d20f6498ba1ee0d92/images/img0005.png" alt="$6\div 2(1+2)$" style="verticalalign:4px;
width:85px;
height:18px" class="math gen" />, and so, accepting neither of them yields the formal mathematical conclusion that <img src="/MI/e6f67b4858233e9d20f6498ba1ee0d92/images/img0005.png" alt="$6\div 2(1+2)$" style="verticalalign:4px;
width:85px;
height:18px" class="math gen" /> is not welldefined. This is also why you can't "correct" each other in a satisfying way: your methods are logically incompatible.
</p><p>
So the disagreement distills down to this: Does it feel like <img src="/MI/0e862e9c98dac54910b3823da3feb7fd/images/img0001.png" alt="$a(b)$" style="verticalalign:4px;
width:27px;
height:18px" class="math gen" /> should always be interchangeable with <img src="/MI/0e862e9c98dac54910b3823da3feb7fd/images/img0002.png" alt="$a\times b$" style="verticalalign:0px;
width:36px;
height:11px" class="math gen" />? Or does it feel like <img src="/MI/0e862e9c98dac54910b3823da3feb7fd/images/img0001.png" alt="$a(b)$" style="verticalalign:4px;
width:27px;
height:18px" class="math gen" /> should always be interchangeable with <img src="/MI/0e862e9c98dac54910b3823da3feb7fd/images/img0003.png" alt="$(ab)$" style="verticalalign:4px;
width:26px;
height:18px" class="math gen" />? You can't say both.
</p>
<div class="centreimage" ><img src="/content/sites/plus.maths.org/files/articles/2019/pemdas/calculators.png"/><p style="maxwidth:540px">(Image from <a href="https://www.quora.com/Whatis11148">Quora</a>)</p></div>
<! Reproduced according to their terms and conditions https://www.quora.com/about/tos >
<p>
In practice, many mathematicians and scientists <a href="https://slate.com/technology/2013/03/facebookmathproblemwhypemdasdoesntalwaysgiveaclearanswer.html">respond</a> to the problem by saying
"unclear syntax, needs more parentheses", and explain why it's ambiguous, which is essentially
the correct answer. An infamous <a href="https://gineersnow.com/students/calculatorsdifferentanswersonemathproblem">picture</a> shows two different Casio calculators sidebyside
given the input <img src="/MI/c7ad99080d43f516530cf10057991fd4/images/img0001.png" alt="$6\div 2(1+2)$" style="verticalalign:4px;
width:85px;
height:18px" class="math gen" /> and showing the two different answers. Though "syntax error" would arguably be the best answer a calculator should give for this problem, it's unsurprising that they try to reconcile the ambiguity, and that's ok. But for us humans, upon noting both conventions are followed by large slices of the world, we must conclude that <img src="/MI/c7ad99080d43f516530cf10057991fd4/images/img0001.png" alt="$6\div 2(1+2)$" style="verticalalign:4px;
width:85px;
height:18px" class="math gen" /> is currently not welldefined.
</p>
<h3>Support for both sides</h3>
<p>
It's a fact that <a href="https://www.google.com/search?source=hp&ei=AHTXIXMKMLSsAfS0oQDA&q=6/2(1+2)&oq=6/2&gs_l=psyab.3.0.0l10.4745.6522..8355...0.0..0.191.506.0j3......0....1..gwswiz.....0..0i131.QpYA59i3UqM">Google</a>, <a href="https://www.wolframalpha.com/input/?i=6%2F2(1%2B2)">Wolfram</a>, and many pocket calculators give the answer of 9.
Calculators' answers here are of course determined by their <A href="https://en.wikipedia.org/wiki/Calculator_input_methods">input methods</a>. Calculators obviously aren't the best judges for the PEMDAS Paradox. They simply
reflect the current disagreement on the problem: calculator programmers are largely aware of
this exact problem and already know that it's not standardised worldwide, so if maths teachers all
unified on an answer, then those programmers would follow.</p>
<p>
Consider <a href="https://www.wolframalpha.com">Wolfram Alpha</a>, the website that provides an <em>answer engine</em> (like a search engine, but rather than provide links to webpages, it provides answers to queries, particularly maths queries). It <a href="https://www.wolframalpha.com/input/?i=6%C3%B72(1%2B2)">interprets</a> <img src="/MI/e678ecec5c89af7dd5c019ec2f4110e4/images/img0001.png" alt="$6\div 2(1+2)$" style="verticalalign:4px;
width:85px;
height:18px" class="math gen" /> as <img src="/MI/e678ecec5c89af7dd5c019ec2f4110e4/images/img0002.png" alt="$9$" style="verticalalign:0px;
width:8px;
height:12px" class="math gen" />, <a href="https://www.wolframalpha.com/input/?i=6%C3%B72x">interprets</a> <img src="/MI/569f1c074739b5ed30aab59dada45730/images/img0001.png" alt="$6\div 2x$" style="verticalalign:0px;
width:45px;
height:12px" class="math gen" /> as <img src="/MI/569f1c074739b5ed30aab59dada45730/images/img0002.png" alt="$3x$" style="verticalalign:0px;
width:17px;
height:12px" class="math gen" />,
and <a href="https://www.wolframalpha.com/input/?i=y%3D1%2F3x">interprets</a> <img src="/MI/4d3fd4db79ee46f983a2daf2fe187d80/images/img0001.png" alt="$y=1/3x$" style="verticalalign:4px;
width:64px;
height:16px" class="math gen" /> as the line through the origin with slope onethird. All three are consistent with each other in a programming sense, but the latter two feel odd to many observers. Typically if someone jots down <img src="/MI/4d3fd4db79ee46f983a2daf2fe187d80/images/img0002.png" alt="$1/3x$" style="verticalalign:4px;
width:32px;
height:16px" class="math gen" />, they mean <img src="/MI/4d3fd4db79ee46f983a2daf2fe187d80/images/img0003.png" alt="$\frac{1}{3x}$" style="verticalalign:5px;
width:15px;
height:20px" class="math gen" />, and if they meant to say <img src="/MI/4d3fd4db79ee46f983a2daf2fe187d80/images/img0004.png" alt="$\frac{1}{3}x$" style="verticalalign:5px;
width:17px;
height:20px" class="math gen" />, they would have written <img src="/MI/4d3fd4db79ee46f983a2daf2fe187d80/images/img0005.png" alt="$x/3$" style="verticalalign:4px;
width:25px;
height:16px" class="math gen" />.
</p>
<div class="centreimage" ><img src="/content/sites/plus.maths.org/files/articles/2019/pemdas/graphboth.png"><p style="maxwidth:500px"></p></div>
<! Rubbish image by Rachel for Plus >
<p>
In contrast, input <img src="/MI/af70010e9998eb935be7f85c18e71085/images/img0001.png" alt="$y=\sin 3x$" style="verticalalign:3px;
width:71px;
height:15px" class="math gen" /> into Wolfram Alpha and it <a href="https://www.wolframalpha.com/input/?i=y%3Dsin3x">yields</a> the sinusoid <img src="/MI/1aec7ff997358cbb507c361a3044e47d/images/img0001.png" alt="$y=\sin (3x)$" style="verticalalign:4px;
width:79px;
height:18px" class="math gen" />, rather than the line through the origin with slope <img src="/MI/1aec7ff997358cbb507c361a3044e47d/images/img0002.png" alt="$\sin 3$" style="verticalalign:0px;
width:31px;
height:12px" class="math gen" />. This example deviates from the previous examples regarding the rule "<img src="/MI/1aec7ff997358cbb507c361a3044e47d/images/img0003.png" alt="$3x$" style="verticalalign:0px;
width:17px;
height:12px" class="math gen" /> is interchangeable with <img src="/MI/1aec7ff997358cbb507c361a3044e47d/images/img0004.png" alt="$3\times x$" style="verticalalign:0px;
width:37px;
height:12px" class="math gen" />", in favor of better capturing the obvious intent of the input. Wolfram is just an algorithm feebly trying to figure out the meaning of its sensory inputs. Kinda like our brains. Anyway, the input of <img src="/MI/1aec7ff997358cbb507c361a3044e47d/images/img0005.png" alt="$6/x3$" style="verticalalign:4px;
width:33px;
height:16px" class="math gen" /> gets <a href="https://www.wolframalpha.com/input/?i=y%3D6%2Fx3">interpreted</a> as "six over <img src="/MI/b964a5b85030732e3f65df2cd643219a/images/img0001.png" alt="$x$" style="verticalalign:0px;
width:9px;
height:7px" class="math gen" /> cubed", so clearly Wolfram is not the authority on rectifying ugly syntax.
</p>
<p>
On the "1" side, a recent excellent <a href="https://www.youtube.com/watch?v=lLCDca6dYpA">video</a> by Jenni Gorham, a maths tutor with a degree in
Physics, explains several realworld examples supporting that interpretation. She points
out numerous occasions in which scientists write <img src="/MI/fe47c4588826cd853a56f9b6dbb98944/images/img0001.png" alt="$a/bc$" style="verticalalign:4px;
width:31px;
height:16px" class="math gen" /> to mean <img src="/MI/fe47c4588826cd853a56f9b6dbb98944/images/img0002.png" alt="$\frac{a}{bc}$" style="verticalalign:5px;
width:12px;
height:17px" class="math gen" /> . Indeed, you'll find abundant examples of this in chemistry,
physics and maths textbooks. Ms. Gorham and I have
corresponded about the PEMDAS Paradox and she endorses formally calling the problem not
welldefined, while also pointing out the need for a consensus convention for the sake of
calculator programming. She argues the consensus answer should be 1 since the precedence
of implied multiplication by juxtaposition has been the convention in most of the world in these
formal contexts.
</p>
<div class="rightimage" style="maxwidth:350px"><img src="/content/sites/plus.maths.org/files/articles/2019/pemdas/brackets1.png">
<p></p>
<img src="/content/sites/plus.maths.org/files/articles/2019/pemdas/brackets9.png"><p></p></div>
<! Rubbish image by Rachel for Plus >
<h3>The big picture</h3>
<p>
It should be pointed out that conventions don't need to be unified. If two of my students
argued over whether the least natural number is 0 or 1, I wouldn’t call either of them wrong, nor
would I take issue with the lack of worldwide consensus on the matter. Wolfram <a href="https://www.wolframalpha.com/input/?i=what+is+the+least+natural+number%3F">knows</a> the
convention is split between two answers, and life goes on. If everyone who cares simply learns
that the PEMDAS Paradox also has two popular answers (and thus itself is not a welldefined
maths question), then that should be satisfactory.
</p>
<p>
Hopefully, after reading this article, it's satisfying to understand how a problem that looks so
basic has uniquely lingered. In real life you should use more parentheses and avoid ambiguity. And hopefully it’s not too troubling that maths teachers worldwide
appear to be split on this convention, as that’s not very rare and not really problematic, except
maybe to calculator programmers.
</p>
<div class="centreimage" ><img src="/content/sites/plus.maths.org/files/articles/2019/pemdas/usebrackets.png"><p style="maxwidth:350px"></p></div>
<! Rubbish image by Rachel for Plus >
<p><em>For readers not fully satisfied with the depth of this article, perhaps my previous <a href="https://drive.google.com/file/d/14hWlVcvsj2fugDaJQUWXX6Mv5d0Wla4/view">much
longer paper</a> won't disappoint. It goes further into detail justifying the formalities of the logical
consistency of the two methods, as well as the problem's history and my experience with it.</em></p>
<hr />
<h3>About the author</h3>
<div class="rightimage" style="maxwidth: 250px;"><img src="/content/sites/plus.maths.org/files/articles/2019/pemdas/davidlinkletter.jpeg" alt="David Linkletter" width="250" height="153" />
<p>David Linkletter</p>
</div>
<p><a href="https://www.unlv.edu/people/davidlinkletter">David Linkletter</a> is a graduate student working on a PhD in Pure Mathematics at the University of Nevada, Las Vegas, in the USA. His research is in set theory  large cardinals. He also teaches undergraduate classes at UNLV; his favourite class to teach is Discrete Maths.</p></div></div></div>
Mon, 17 Jun 2019 13:29:54 +0000
Rachel
7194 at https://plus.maths.org/content
https://plus.maths.org/content/pemdasparadox#comments

Helping men with early prostate cancer
https://plus.maths.org/content/helpingmenearlyprostratecancer
<div class="field fieldnamefieldabsimg fieldtypeimage fieldlabelhidden"><div class="fielditems"><div class="fielditem even"><img class="imgresponsive" src="https://plus.maths.org/content/sites/plus.maths.org/files/abstractpics/%5Buid%5D/%5Bsitedate%5D/predict_icon.png" width="100" height="100" alt="" /></div></div></div><div class="field fieldnamebody fieldtypetextwithsummary fieldlabelhidden"><div class="fielditems"><div class="fielditem even"><p>One in 6 men in the UK will be diagnosed with prostate cancer at some point in their lives: 47,000 get this news every year. A new online tool developed with the help of statisticians helps these men decide on their treatment options.</p>
<div class="rightimage" style="maxwidth: 400px;"><img src="/content/sites/plus.maths.org/files/news/2019/winton/predictprostate.png" alt="Predict: Prostate cancer" width="400" height="189" />
<p><a href="https://prostate.predict.nhs.uk/">Predict: Prostate Cancer</a> is an online tool to help patients decide on their treatment options.</p>
</div>
<p>More than half of men discover their prostate cancer at an early stage, before it has spread, and for them the choice is between jumping in to a big therapy such as surgery or radiotherapy versus waiting and monitoring the cancer to see if it grows. It's a difficult choice because the therapies have a high chance of side effects (including incontinence and erectile dysfunction) whilst many prostate cancers are slow growing and unlikely to cause a person's death if left alone. The trouble, of course, is knowing whether your cancer is one of the slowgrowing ones.</p>
<p>A team at the University of Cambridge and Addenbrooke's Hospital developed an algorithm that uses a database of outcomes for patients in the past along with the data from trials of surgery or radiotherapy to help determine more accurately what an individual's prognosis is and what the benefits and harms of trying treatment might be for them.</p>
<p>Once they had produced the algorithm, though, the question was how to communicate the results both to healthcare professionals and the patients.</p>
<h3>The psychology of numbers</h3>
<p>Increasingly, patients are (quite rightly) being involved in decisionmaking about their treatment options. In many cases, there is no simple "right answer" for what treatment is best – it's an individual decision depending on personal circumstances, and how much the potential harms and potential benefits mean to a person. In order to take part in this shared decisionmaking, patients need to have the potential harms and benefits of different options put to them in a balanced and clear way.</p>
<p>This is what the <a href="https://wintoncentre.maths.cam.ac.uk/" target="blank">Winton Centre for Risk and Evidence Communication</a> in the <a href="https://www.dpmms.cam.ac.uk/">Department of Pure Maths and Mathematical Statistics</a> at Cambridge work on.</p>
<p>The psychology of how we understand numbers is sometimes quite surprising and means that the way that we communicate them can have a huge impact on how they are perceived.</p>
<p>For example, when asked which was the bigger risk — 1 in 10, 1 in 100 or 1 in 1000 — a quarter of people got the answer wrong. In fact "1 in <em>x</em>" statements like that are very difficult for people to compare – they are counterintuitive: the bigger number represents the smaller risk. Instead it's much clearer to keep the denominator the same (eg 10 in 100 versus 1 in 100).</p>
<p>We also tend to view a number differently whether it is presented as a frequency (<em>x</em> out of 100) or a percentage. The statement "20 out of 100" brings to mind the 20 times when an event might happen, and makes it sound much more likely than the more abstract statement that there's a "20% chance" of it happening.</p>
<p>Graphics can help people visualise and compare numbers, but again they each have a different effect on whether they make a number look large or small. To give people an overall balanced view, then, presenting numbers in a range of different ways – evening out the bias associated with any one method – can help.</p>
<h3>Putting users at the centre</h3>
<p>When designing a way to communicate statistics, the Winton Centre team adopt a strategy called <em>usercentred design</em>, long used in industry to help design products that really work for people, but not often associated with academic outputs.</p>
<p>In order to develop an online tool to present what could be very difficult and upsetting survival statistics to recentlydiagnosed patients, the team work closely with patients, healthcare professionals and the public, holding focusgroups, interviews and then working on a series of increasingly more realistic designs for the site as well as doing largescale online testing to see what people understand from different forms of visualisation.</p>
<div class="centreimage"><img src="/content/sites/plus.maths.org/files/news/2019/winton/prostatepredictoutput.png" alt="Example output" width="600" height="416" />
<p style="maxwidth: 600px;">An example of a visualisation on Predict: Prostate cancer.</p>
</div>
<p>The Winton Centre is working on a number of such sites, for different health conditions, and were delighted to take on the task of designing the site to present the prostate cancer statistics. They were particularly pleased to be able to work on a way of representing the potential harms of the different therapy options – something that they haven't yet been able to do for their other, similar, <a href="https://breast.predict.nhs.uk/" target="blank">site for breast cancer</a> because the harms of treatments are often very poorly recorded and reported.</p>
<p><a href="https://prostate.predict.nhs.uk/">Predict: Prostate Cancer</a> launched in March 2019 and is already being used thousands of times a month all over the world. It offers five different ways of showing the data: as text, a table, a survival curve, bar charts and icon arrays. The website has been enthusiastically received both by doctors and patients – one patient group describing it as "game changing".</p>
<p>The Winton Centre team are monitoring usage of the site, including how people use each of the visualisations within it, and continuing to carry out tests to see what people understand from each. They have also just made the site nonwifidependent (it will work on a device even when the wifi goes down) an issue that clinicians raised, as hospital wifi can be very patchy.</p>
<p>The Centre will be continuing to refine the site over the coming months. They are starting to prepare it for translation into other languages so that it can be used more easily to help patients across the globe. In addition they are working on a sister tool for men with metastatic prostate cancer who have similarly difficult decisions to make about their treatment choices.</p>
<p>Hopefully these web tools will allow all the data that has been painstakingly collected on cancer outcomes and clinical trials over many decades inform the decisions of patients all over the world. It will let them take part in the process of deciding what approach to their cancer is right for them — this is part of the revolution in our approach to healthcare that is now underway.</p>
</div></div></div>
Tue, 28 May 2019 14:35:47 +0000
Marianne
7198 at https://plus.maths.org/content
https://plus.maths.org/content/helpingmenearlyprostratecancer#comments

A cute problem goes big
https://plus.maths.org/content/cuteproblemgoesbig
<div class="field fieldnamefieldabsimg fieldtypeimage fieldlabelhidden"><div class="fielditems"><div class="fielditem even"><img class="imgresponsive" src="https://plus.maths.org/content/sites/plus.maths.org/files/abstractpics/%5Buid%5D/%5Bsitedate%5D/zeta_icon.jpg" width="100" height="100" alt="" /></div></div></div><div class="field fieldnamebody fieldtypetextwithsummary fieldlabelhidden"><div class="fielditems"><div class="fielditem even"><p>A team of mathematicians have revived an old approach to solving the famous <em>Riemann hypothesis</em>. Their new result grew from a "cute toy problem" and provides further evidence that Riemann's tricky conjecture is indeed true.</p>
<div class="rightimage" style="maxwidth: 350px;"><img src="/content/sites/plus.maths.org/files/articles/2018/Ramanujan/full.jpeg" alt="Ken Ono" width="350" height="233"/>
<p>Ken Ono.</p>
</div>
<p>"In a surprisingly short proof, we've shown that an old, abandoned approach to the Riemann hypothesis should not have been forgotten," says <a href="http://www.mathcs.emory.edu/~ono/">Ken Ono</a>, a number theorist at Emory University and coauthor of the <a href="http://dx.doi.org/10.1073/pnas.1902572116">paper</a> published this week in PNAS. </p>
<p>"By simply formulating a proper framework for an old approach we've proven some new theorems, including a large chunk of a criterion which implies the Riemann hypothesis. And our general framework also opens approaches to other basic unanswered questions."</p>
<h3>The Riemann hypothesis</h3>
<p>The hypothesis debuted in an 1859 paper by German mathematician <a href="http://wwwhistory.mcs.stand.ac.uk/~history/Biographies/Riemann.html">Bernhard Riemann</a>. He noticed that the distribution of prime numbers along the number line is closely related to the zeroes of a mathematical function, which came to be called the <em>Riemann zeta function</em>. In mathematical terms, the Riemann hypothesis is the assertion that all of the nontrivial zeroes of the zeta function have real part 1/2.</p>
<p>"His hypothesis is a mouthful, but Riemann's motivation was simple," Ono says. "He wanted to count prime numbers."</p>
<p>Prime numbers are those whole numbers that are only divisible by 1 and themselves. The first few are 2, 3, 5, 7 and 11. "It's well known that there are infinitely many prime numbers, but they become rare, even by the time you get to the 100s," Ono explains. "In fact, out of the first 100,000 numbers, only 9,592 are prime numbers, or roughly 9.5 percent. And they rapidly become rarer from there. The probability of picking a number at random and having it be prime is zero. It almost never happens."</p>
<p>You can find out more about the Riemann hypothesis in <a href="/content/musicprimes"><em>The music of the primes</em></a>.</p>
<h3>The revived approach</h3>
<div class="leftimage" style="width: 230px;"><img src="/issue28/features/sautoy/young_Riemann.jpg" alt="Riemann" width="230" height="324" /><p>Riemann: we still don't know if he was right</p>
</div>
<p>The revived approach goes back to work of the mathematicians <a href="https://wwwhistory.mcs.stand.ac.uk/Biographies/Jensen.html">Johan Jensen</a> and <a href="http://wwwhistory.mcs.stand.ac.uk/history/Biographies/Polya.html">George Pólya</a> who formulated a criterion for confirming the Riemann hypothesis. Using the Riemann zeta function it's possible to construct an infinite family of mathematical functions, called <em>Jensen polynomials</em>. These are functions of complex numbers. If you can show that the values at which these functions are 0 are all real numbers, then you have automatically proved the Riemann hypothesis — it follows as a result.</p>
<p>The problem with this was that there are infinitely many Jensen polynomials. During the past 90 years only a handful of them have been shown to be hyperbolic, causing mathematicians to abandon the approach as too slow and unwieldy.</p>
<p>For the PNAS paper, the authors devised a conceptual framework that combines the polynomials in groups (they classified them by degree). This method enabled them to prove the criterion for all but finitely many polynomials in each group.</p>
<p>"The method has a shocking sense of being universal, in that it applies to problems that are seemingly unrelated," says <a href="https://math.vanderbilt.edu/rolenl/index.html">Larry Rolens</a>, a coauthor on the paper. "And at the same time, its proofs are easy to follow and understand. Some of the most beautiful insights in maths are ones that took a long time to realise, but once you see them, they appear simple and clear."</p>
<h3>Playing around</h3>
<div class="rightimage" style="width: 272px;"><img src="/issue28/features/sautoy/Landscape.jpg" alt="An imaginary landscape" /><p>This "landscape" is produced by the Riemann zeta function. The points at sea level are the zeroes of the function, and they are conjectured to all line up along a line.</p>
</div>
<p>The idea for the paper was sparked two years ago by a "toy problem" that Ono presented as a "gift" to entertain coauthor <A href="https://people.mpimbonn.mpg.de/zagier/">Don Zagier</a> during the leadup to a maths conference celebrating his 65th birthday.
Zagier described it as "a cute problem about the asymptotic behaviour of certain polynomials involving <em>Euler's partition function</em> [find out more in <a href="/content/os/issue42/features/wilson/index">this article</a>] which is an old love of mine and of Ken's — and of about pretty much any classical number theorist."</p><p>
"I found the problem intractable and I didn't really expect Don to get anywhere with it," Ono recalls. "But he thought the challenge was super fun and soon he had crafted a solution."
Ono's hunch was that the solution could be turned into a more general theory. That's what the mathematicians ultimately achieved.</p>
<p>"It's been a fun project to work on, a really creative process," says <a href="https://math.byu.edu/~mjgriffin/">Michael Griffin</a>, the fourth coauthor on the PNAS paper. "Maths at a research level is often more art than calculation and that was certainly the case here. It required us to look at an almost 100yearold idea of Jensen and Pólya in a new way."</p>
<p>Despite their work, the results don't rule out the possibility that the Riemann Hypothesis is false and the authors believe that a complete proof of the famous conjecture is still far off.</p>
<p><em>This article has been adapted from an <a href="https://esciencecommons.blogspot.com/2019/05/mathematiciansreviveabandoned.html">Emory University press release</a>.</em></p>
</div></div></div>
Thu, 23 May 2019 15:07:11 +0000
Marianne
7199 at https://plus.maths.org/content
https://plus.maths.org/content/cuteproblemgoesbig#comments

Maths in a minute: Escape velocity
https://plus.maths.org/content/mathsminuteescapevelocity
<div class="field fieldnamefieldabsimg fieldtypeimage fieldlabelhidden"><div class="fielditems"><div class="fielditem even"><img class="imgresponsive" src="https://plus.maths.org/content/sites/plus.maths.org/files/abstractpics/%5Buid%5D/%5Bsitedate%5D/earth_icon_2.jpg" width="100" height="100" alt="" /></div></div></div><div class="field fieldnamebody fieldtypetextwithsummary fieldlabelhidden"><div class="fielditems"><div class="fielditem even"><div class="rightimage" style="maxwidth: 300px;"><img src="/content/sites/plus.maths.org/files/articles/2019/escapevelocity/earth.jpg" alt="Flags" width="300" height="301" />
<p>Dying to get away?</p>
</div><! Image in public domain >
<p>When you jump up into the air you'll come back down to the Earth with a bump. That's not because the laws of nature categorically forbid you leaving the Earth, but because your jump isn't powerful enough to escape the Earth's gravitational field. To do that, the speed of your jump would have to exceed, or be equal to, the Earth's <em>escape velocity</em> which we can calculate quite easily as follows.</p>
<p>When you jump into the air your kinetic energy is </p><p><img src="/MI/9b8faf1d2d52983df6a9e6f795b8a9bb/images/img0001.png" alt="$E_ k=\frac{1}{2}mv^2,$" style="verticalalign:5px;
width:87px;
height:20px" class="math gen" /> </p><p>where <img src="/MI/9b8faf1d2d52983df6a9e6f795b8a9bb/images/img0002.png" alt="$m$" style="verticalalign:0px;
width:14px;
height:7px" class="math gen" /> is your mass and <img src="/MI/9b8faf1d2d52983df6a9e6f795b8a9bb/images/img0003.png" alt="$v$" style="verticalalign:0px;
width:8px;
height:7px" class="math gen" /> is your velocity. The potential energy you experience due to the Earth’s gravitational pull is </p><p><img src="/MI/9b8faf1d2d52983df6a9e6f795b8a9bb/images/img0004.png" alt="$E_ p = \frac{GMm}{r},$" style="verticalalign:5px;
width:86px;
height:20px" class="math gen" /> </p><p>where <img src="/MI/9b8faf1d2d52983df6a9e6f795b8a9bb/images/img0002.png" alt="$m$" style="verticalalign:0px;
width:14px;
height:7px" class="math gen" /> is again your mass, <img src="/MI/9b8faf1d2d52983df6a9e6f795b8a9bb/images/img0005.png" alt="$M$" style="verticalalign:0px;
width:18px;
height:11px" class="math gen" /> is the mass of the Earth and <img src="/MI/9b8faf1d2d52983df6a9e6f795b8a9bb/images/img0006.png" alt="$r$" style="verticalalign:0px;
width:8px;
height:7px" class="math gen" /> is the Earth’s radius. For you to be able to escape from Earth, we need </p><table id="a0000000002" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/9b8faf1d2d52983df6a9e6f795b8a9bb/images/img0007.png" alt="\[ E_ k \geq E_ p, \]" style="width:67px;
height:15px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><p> so </p><table id="a0000000003" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/9b8faf1d2d52983df6a9e6f795b8a9bb/images/img0008.png" alt="\[ \frac{1}{2}mv^2 \geq \frac{GMm}{r}. \]" style="width:117px;
height:34px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><p>Solving for velocity <img src="/MI/9b8faf1d2d52983df6a9e6f795b8a9bb/images/img0003.png" alt="$v$" style="verticalalign:0px;
width:8px;
height:7px" class="math gen" /> gives </p><table id="a0000000004" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/9b8faf1d2d52983df6a9e6f795b8a9bb/images/img0009.png" alt="\[ v \geq \sqrt {\frac{2GM}{r}}. \]" style="width:95px;
height:41px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><p>The Earth’s escape velocity, <img src="/MI/9b8faf1d2d52983df6a9e6f795b8a9bb/images/img0010.png" alt="$v_{Earth}$" style="verticalalign:2px;
width:45px;
height:9px" class="math gen" />, is defined to be the smallest velocity that allows an object to escape, so </p><table id="a0000000005" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/9b8faf1d2d52983df6a9e6f795b8a9bb/images/img0011.png" alt="\[ v_{Earth}= \sqrt {\frac{2GM}{r}}. \]" style="width:132px;
height:41px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><p>Filling in the values </p><table id="a0000000006" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/9b8faf1d2d52983df6a9e6f795b8a9bb/images/img0012.png" alt="\[ G\approx 6.67 \times \frac{1}{10^{11}} \frac{m^3}{kg\; s}, \]" style="width:158px;
height:39px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><table id="a0000000007" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/9b8faf1d2d52983df6a9e6f795b8a9bb/images/img0013.png" alt="\[ M\approx 5.98 \times 10^{24} kg, \]" style="width:142px;
height:18px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><p> and </p><table id="a0000000008" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/9b8faf1d2d52983df6a9e6f795b8a9bb/images/img0014.png" alt="\[ r\approx 6.38 \times 10^6 m, \]" style="width:123px;
height:18px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><p> gives </p><table id="a0000000009" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/9b8faf1d2d52983df6a9e6f795b8a9bb/images/img0015.png" alt="\[ v_{Earth}\approx \sqrt {\frac{2\times 6.67 \times 5.98 \times 10^{24}}{ 6.38 \times 10^6 \times 10^{11} } \frac{m^3kg}{m\; kg\; s} .} \]" style="width:306px;
height:50px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><p>Putting this into your calculator gives </p><table id="a0000000010" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/9b8faf1d2d52983df6a9e6f795b8a9bb/images/img0016.png" alt="\[ v_{Earth}\approx 11182 \frac{m}{s}, \]" style="width:132px;
height:30px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><p> which translates to <img src="/MI/9b8faf1d2d52983df6a9e6f795b8a9bb/images/img0017.png" alt="$671km/h.$" style="verticalalign:4px;
width:71px;
height:17px" class="math gen" /> </p><p>You can use the same calculation to work out the escape velocity of any spherical body, as long as you know its mass and radius. </p>
<p>Note that our formula for escape velocity is independent on the mass <img src="/MI/9e359eb28ace39d7914eee793d66a5e3/images/img0001.png" alt="$m$" style="verticalalign:0px;
width:14px;
height:7px" class="math gen" /> of the object that is trying to escape, as <img src="/MI/9e359eb28ace39d7914eee793d66a5e3/images/img0001.png" alt="$m$" style="verticalalign:0px;
width:14px;
height:7px" class="math gen" /> cancels out. So in theory you would need to achieve the same velocity to escape Earth as, say, an elephant. We should point out, however, that our calculation ignores the effect of air resistance which would effect you and the elephant differently. What is more, if you reached such a high velocity within the Earth's atmosphere, you would burn up. To avoid this, you or the elephant should first get yourself into an orbit in which the Earth's atmosphere is weak or nonexistent, and then accelerate to the escape velocity needed to escape from that orbit. Find out more <a href="https://en.wikipedia.org/wiki/Escape_velocity#Practical_considerations">here</a>.</p>
</div></div></div>
Wed, 22 May 2019 11:50:20 +0000
Marianne
7195 at https://plus.maths.org/content
https://plus.maths.org/content/mathsminuteescapevelocity#comments

Maths in a minute: The d'Hondt method
https://plus.maths.org/content/mathsminutedhondtmethod
<div class="field fieldnamefieldabsimg fieldtypeimage fieldlabelhidden"><div class="fielditems"><div class="fielditem even"><img class="imgresponsive" src="https://plus.maths.org/content/sites/plus.maths.org/files/abstractpics/%5Buid%5D/%5Bsitedate%5D/flags_icon.jpg" width="100" height="100" alt="" /></div></div></div><div class="field fieldnamebody fieldtypetextwithsummary fieldlabelhidden"><div class="fielditems"><div class="fielditem even"><p>The European Parliament elections this week once again turn the spotlight on the maths of democracy. The elections use proportional representation: the idea here is that a party that got x% of the vote should get x% of the seats that are up for grabs.</p>
<div class="rightimage" style="maxwidth: 350px;"><img src="/content/sites/plus.maths.org/files/articles/2019/EU/flags.jpg" alt="Flags" width="350" height="233" />
<p>The UK votes in the European Parliament elections on May 23, 2019. Photo: Daina Le Lardic, © European Union 2018  Source : EP.</p>
</div>
<p> An obvious advantage of proportional representation is that people are better represented than in a winnertakesall system. But unfortunately, it's not as straightforward as it may seem at first because percentages don't always translate into whole numbers. For example, if there are 600,000 voters in an election for 100 seats and three parties who each got 200,000 votes, then each party should get exactly one third of the seats. Since a third of 100 is 33.33 and politicians can't be carved up that's impossible.</p>
<p>To deal with this problem, we need an extra layer of complexity; a method to translate percentages into seats. One common way, used in the European Parliament elections, is the <em>d'Hondt method</em>. The idea is that one seat should "cost" a certain number of votes. Each party should be able to buy as many seats (from the total number of seats) as its vote money allows, because if a party can't get all it can pay for or gets more than it can pay for, it's not fairly represented. Once each party has bought all the seats it can, no seats should be left over.</p>
<p>Setting the correct price per seat to achieve this seems tricky, but there's an iterative process that delivers the desired result. You start off with giving the party with the largest number of votes one seat. Then, for each party, you work out the number</p>
<table id="a0000000002" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/1264bafa05fd6594a19a3e71dab715e6/images/img0001.png" alt="\[ N = \frac{V}{(s+1)}, \]" style="width:95px;
height:37px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><p>where <img src="/MI/1264bafa05fd6594a19a3e71dab715e6/images/img0002.png" alt="$V$" style="verticalalign:0px;
width:12px;
height:11px" class="math gen" /> is the number of votes the party got in total and <img src="/MI/1264bafa05fd6594a19a3e71dab715e6/images/img0003.png" alt="$s$" style="verticalalign:0px;
width:7px;
height:7px" class="math gen" /> the number of seats it already has (at the beginning of the process <img src="/MI/1264bafa05fd6594a19a3e71dab715e6/images/img0003.png" alt="$s$" style="verticalalign:0px;
width:7px;
height:7px" class="math gen" /> is <img src="/MI/1264bafa05fd6594a19a3e71dab715e6/images/img0004.png" alt="$0$" style="verticalalign:0px;
width:8px;
height:12px" class="math gen" /> for all but the largest party for which it is <img src="/MI/1264bafa05fd6594a19a3e71dab715e6/images/img0005.png" alt="$1$" style="verticalalign:0px;
width:6px;
height:12px" class="math gen" />). The second seat is given to the party with the highest <img src="/MI/1264bafa05fd6594a19a3e71dab715e6/images/img0006.png" alt="$N.$" style="verticalalign:0px;
width:18px;
height:11px" class="math gen" /> You then again work out each party’s <img src="/MI/1264bafa05fd6594a19a3e71dab715e6/images/img0007.png" alt="$N$" style="verticalalign:0px;
width:15px;
height:11px" class="math gen" /> with <img src="/MI/1264bafa05fd6594a19a3e71dab715e6/images/img0003.png" alt="$s$" style="verticalalign:0px;
width:7px;
height:7px" class="math gen" /> increased as appropriate. The party with the highest <img src="/MI/1264bafa05fd6594a19a3e71dab715e6/images/img0007.png" alt="$N$" style="verticalalign:0px;
width:15px;
height:11px" class="math gen" /> gets the third seat. You then continue in this vein until all seats are gone. See below for an example. </p>
<p>Since proportional representation always involves rounding up or down the number of seats a party should get, no system designed to deliver it is perfect. The d'Hondt system, for example, tends to overrepresent the larger parties compared to smaller ones. Which system you choose depends on which of several possible problems you're happiest to live with.</p>
<h3>An example</h3>
<p>Suppose there are 3 parties (A, B and C) competing for 5 seats, and a total of 60 voters. Suppose the result of the election
is as follows:
</p>
<table class="datatable">
<tr><td></td><td>Votes</td><td>% of votes</td></tr>
<tr><td>Party A</td><td>20</td><td>33.33%</td></tr>
<tr><td>Party B</td><td>15</td><td>25%</td></tr>
<tr><td>Party C</td><td>25</td><td>41.66%</td></tr></table>
<p>If we went for exact proportionality then party A would have to get 1.66 seats, party B would have to get 1.25 seats, and party C would have to get 2.083 seats. These aren't whole numbers, so exact proportionality is impossible. Let's use the d'Hondt method instead.</p>
<p>Party C has the largest number of votes, so it gets one seat to start with. We now have:
<table class="datatable"><tr><td></td><td>Seats</td><td><em>s</em></td><td><em>N</em></td></tr>
<tr><td>Party A</td><td>0</td><td>0</td><td>20/1=20</td></tr>
<tr><td>Party B</td><td>0</td><td>0</td><td>15/1=15</td></tr>
<tr><td>Party C</td><td>1</td><td>1</td><td>25/2=12.5</td></tr></tr></table>
The largest value of <em>N</em> is that of party A, so it gets one seat. We now have:
<table class="datatable"><tr><td></td><td>Seats</td><td><em>s</em></td><td><em>N</em></td></tr>
<tr><td>Party A</td><td>1</td><td>1</td><td>20/2=10</td></tr>
<tr><td>Party B</td><td>0</td><td>0</td><td>15/1=15</td></tr>
<tr><td>Party C</td><td>1</td><td>1</td><td>25/2=12.5</td></tr></tr></table>
The largest value of <em>N</em> is that of party B, so it gets one seat. We now have:
<table class="datatable"><tr><td></td><td>Seats</td><td><em>s</em></td><td><em>N</em></td></tr>
<tr><td>Party A</td><td>1</td><td>1</td><td>20/2=10</td></tr>
<tr><td>Party B</td><td>1</td><td>1</td><td>15/2=7.5</td></tr>
<tr><td>Party C</td><td>1</td><td>1</td><td>25/2=12.5</td></tr></tr></table>
The largest value of <em>N</em> is that of party C, so it gets another seat. We now have:
<table class="datatable"><tr><td></td><td>Seats</td><td><em>s</em></td><td><em>N</em></td></tr>
<tr><td>Party A</td><td>1</td><td>1</td><td>20/2=10</td></tr>
<tr><td>Party B</td><td>1</td><td>1</td><td>15/2=7.5</td></tr>
<tr><td>Party C</td><td>2</td><td>2</td><td>25/3=8.33</td></tr></tr></table>
The largest value of <em>N</em> is that of party A, so it also gets another seat. All five seats are now gone and the final seat allocation is:
<table class="datatable">
<tr><td></td><td>Seats</td><td>% of seats</td><td>% of seats under exact proportionality</td></tr>
<tr><td>Party A</td><td>2</td><td>40%</td><td>33.33%</td></tr>
<tr><td>Party B</td><td>1</td><td>20%</td><td>25%</td></tr>
<tr><td>Party C</td><td>2</td><td>40%</td><td>41.66</td></tr></table>
<p>This shows that the party with the smallest share of the vote, party B, is underrepresented, while party A is overrepresented and the representation of party C is just about right.</p>
<p>This is equivalent to "selling" each seat for either 9 or 10 votes. You can check for yourself that selling a seat for any other number of votes would lead to not all five seats being filled, or more than 5 being filled.</p>
<hr/>
<h3>About this article</h3> <div class="rightimage" style="maxwidth: 150px;"><img src="/content/sites/plus.maths.org/files/articles/2019/dating/cover.jpg" alt="Book cover" width="150" height="210" /><p></p>
</div>
<p>This article is based on a chapter from the new book <em><a href="https://amzn.to/2G6v65L">Understanding numbers</a></em> by the <em>Plus</em> Editors <a href="/content/people/index.html#rachel">Rachel Thomas</a> and <a href="/content/people/index.html#marianne">Marianne Freiberger</a>. The book will be published on April 11, 2019!</p>
</div></div></div>
Tue, 21 May 2019 09:27:38 +0000
Marianne
7197 at https://plus.maths.org/content
https://plus.maths.org/content/mathsminutedhondtmethod#comments

Mysterious 6174
https://plus.maths.org/content/mysteriousnumber61742
<div class="field fieldnamefieldabsimg fieldtypeimage fieldlabelhidden"><div class="fielditems"><div class="fielditem even"><img class="imgresponsive" src="https://plus.maths.org/content/sites/plus.maths.org/files/abstractpics/%5Buid%5D/%5Bsitedate%5D/icon_35.jpg" width="100" height="100" alt="" /></div></div></div><div class="field fieldnamebody fieldtypetextwithsummary fieldlabelhidden"><div class="fielditems"><div class="fielditem even"><p>
I want to let you in on one of our favourite mathematical mysteries... To get started, choose a four digit number where the digits are not all the same (that is not 1111, 2222,...). Then rearrange the digits to get the largest and smallest numbers these digits can make. Finally, subtract the
smallest number from the largest to get a new number, and carry on repeating the operation for each new number.
</p>
<p>
We'll show you what we mean with the number 2005. The maximum number we can make with these digits is 5200, and the minimum is 0025 or 25 (if one or more of the digits is zero, embed these in the left hand side of the minimum number). The subtractions are:</p>
<p>5200  0025 = 5175<br />
7551  1557 = 5994<br />
9954  4599 = 5355<br />
5553  3555 = 1998<br />
9981  1899 = 8082<br />
8820  0288 = 8532<br />
8532  2358 = 6174<br />
7641  1467 = 6174<br /></p>
</center>
<p>When we reach 6174 the process repeats itself, returning 6174 every time.
</p>
<div class="centreimage" style="maxwidth:500px"><img src="https://plus.maths.org/issue38/features/nishiyama/6174.jpg" alt="6174"></div>
<p>
Now try with your four digit number... what do you get? I bet you'll get to 6174, every time, no matter what number you chose! Try a few more and see if you believe me!
</p>
<p>
Do you think we always reach the mysterious number 6174? If we do, why do you think that happens? If this mystery piques your interest, then you can find out why in <a href="/content/mysteriousnumber6174">this excellent article</a> by Yutaka Nishiyama. This question has been intriguing <em>Plus</em> readers for years, and Yutaka's article remains one of our most popular articles, generating reams of comments, emails and discussions. And spoiler alert  6174 isn't the only number with these special numbers  but you'll have to try the same process with three digit numbers, or read the article, to find out!</p></div></div></div>
Fri, 17 May 2019 15:16:43 +0000
Rachel
7196 at https://plus.maths.org/content
https://plus.maths.org/content/mysteriousnumber61742#comments

No maths for Europe
https://plus.maths.org/content/democraticdilemmas
<div class="field fieldnamefieldabsimg fieldtypeimage fieldlabelhidden"><div class="fielditems"><div class="fielditem even"><img class="imgresponsive" src="https://plus.maths.org/content/sites/plus.maths.org/files/abstractpics/%5Buid%5D/%5Bsitedate%5D/eu_icon.jpg" width="100" height="100" alt="" /></div></div></div><div class="field fieldnamefieldauthor fieldtypetext fieldlabelinlinec clearfix fieldlabelinline"><div class="fieldlabel">By </div><div class="fielditems"><div class="fielditem even">Marianne Freiberger</div></div></div><div class="field fieldnamebody fieldtypetextwithsummary fieldlabelhidden"><div class="fielditems"><div class="fielditem even"><p>Think about democracy and you soon find yourself thinking about maths. The European Parliament, which will see elections this month, is no exception. The people of (probably) 28 EU member states need to be represented by politicians who fill a fixed number of parliamentary seats (currently 751). But how many seats should each member state get? And how should the allocation change whenever a new country joins — or an old one leaves?</p>
<div class="rightimage" style="maxwidth: 350px;"><img src="/content/sites/plus.maths.org/files/articles/2019/EU/hemicycle.jpg" alt="The European Parliament" width="350" height="233" />
<p>A plenary session of the European Parliament. Photo: Mathieu Cugnot, © European Union 2017  source:EP.</p>
</div>
<p>If you treat countries as units then you can give each of them the same number of seats and be done with it. But since member states are made up of people, who should all be represented equally, it makes sense for the number of seats a country gets to be proportional to the size of its population: more people mean more seats. The problem is that population sizes vary hugely — compare Germany's nearly 83 million to Malta's 460,300 — so strict proportionality would mean that smaller countries hardly get a foot in the door.</p>
<p>To avoid this, larger EU member states have agreed to be generous: they are happy to be slightly underrepresented in Parliament so that smaller member states can be better represented. This show of solidarity is known as <em>degressive representation</em>. The <a href="https://en.wikipedia.org/wiki/Treaty_of_Lisbon">Lisbon Treaty</a> of 2009 enshrined it into law. It also stipulates that no country should have fewer than 6 or more than 96 seats, and that the European Parliament should have no more than 751 seats in total. </p>
<p>
Up until now, whenever a new country joined the EU, seats were allocated through political bartering. Some sort of principled method, an algorithm, would obviously be better: it would be fairer and more transparent, and could be repeated every time a country joins (or leaves). Back in 2010, long before the Brexit era, <em>rapporteur</em> of the European Constitutional Affairs Committee
<a href="http://www.andrewduff.eu/en/">Andrew Duff</a> approached mathematicians at the University of Cambridge for a mathematical solution (find out more <a href="/content/formulaeurope">here</a>). The Brexit referendum made the problem more pressing: when (or if) Brexit happens, the UK's 73 seats will suddenly become vacant and the remaining composition of Parliament off kilter. </p>
<p>In response to Duff's initial approach a committee of mathematicians from around Europe was formed, working to three nonmathematical constraints. "The first was that the system should be durable with respect to changes in size and shape of the EU," <a href="http://www.statslab.cam.ac.uk/~grg/">Geoffrey Grimmett</a>, who established and chaired the committee, <a href="/content/formulaeurope">told us</a> in 2011. "The second was that it should be transparent, we should be able to explain it to people. And the third was that the system should be impartial to politics, it shouldn't favour particular groups or nations."</p>
<h3>The Cambridge Compromise</h3>
<p>Grimmett's committee reached the unanimous conclusion that seats should be allocated to member states according to a method they called the <em><a href="http://www.europarl.europa.eu/RegData/etudes/note/join/2011/432760/IPOLAFCO_NT(2011)432760_EN.pdf">Cambridge Compromise</a></em>. The idea is straightforward. To begin with, every member state, no matter how big or small, is given 5 <em>base seats</em>. That's extremely degressive because population sizes don't even enter the picture. The remaining seats are then allocated proportionately to population figures: if a state has a population of <img src="/MI/32e816c5df987f09c32f6d8a7d640583/images/img0001.png" alt="$p$" style="verticalalign:3px;
width:10px;
height:10px" class="math gen" />, then it gets <img src="/MI/32e816c5df987f09c32f6d8a7d640583/images/img0002.png" alt="$p/d$" style="verticalalign:4px;
width:26px;
height:16px" class="math gen" /> seats, for some divisor <img src="/MI/32e816c5df987f09c32f6d8a7d640583/images/img0003.png" alt="$d$" style="verticalalign:0px;
width:9px;
height:11px" class="math gen" /> (which we'll explain in a moment). For example, if the divisor <img src="/MI/32e816c5df987f09c32f6d8a7d640583/images/img0003.png" alt="$d$" style="verticalalign:0px;
width:9px;
height:11px" class="math gen" /> is chosen to be 1 million, then a country with a population of 10 million gets <table id="a0000000002" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/32e816c5df987f09c32f6d8a7d640583/images/img0004.png" alt="\[ 5+\frac{10,000,000}{1,000,000}=5+10=15 \]" style="width:219px;
height:38px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table> seats.
</p>
<div class="leftimage" style="maxwidth: 350px;"><img src="/content/sites/plus.maths.org/files/articles/2019/EU/grimmett.jpg" alt="The European Parliament" width="350" height="343" />
<p>Geoffrey Grimmett, Professor of Mathematical Statistics at the University of Cambridge.</p>
</div>
<p>
<p>The problem here is that in general you can’t be sure that <img src="/MI/3d999b7a4383fe01f1253de2d5cde8a7/images/img0001.png" alt="$p/d$" style="verticalalign:4px;
width:26px;
height:16px" class="math gen" /> is a whole number. "You can’t give a country, say, 63.3 MEPs," explained Grimmett. "MEPs simply can’t be cut up that way. So you need to decide how to round fractions to whole numbers." The mathematicians settled on upward rounding: when <img src="/MI/3d999b7a4383fe01f1253de2d5cde8a7/images/img0001.png" alt="$p/d$" style="verticalalign:4px;
width:26px;
height:16px" class="math gen" /> isn’t a whole number, then go for the nearest whole number that’s larger than <img src="/MI/3d999b7a4383fe01f1253de2d5cde8a7/images/img0002.png" alt="$p/d.$" style="verticalalign:4px;
width:30px;
height:16px" class="math gen" /> It’s been shown that this way of rounding gives a slight advantage to smaller member states over larger ones, something that fits well with degressivity. </p>
</p>
<p>Since every country gets 5 base seats the Cambridge Compromise ensures that no country ends up with fewer than 6 seats (rounding <img src="/MI/82015f5ea26f89b575ee477aa18adfc8/images/img0001.png" alt="$p/d$" style="verticalalign:4px;
width:26px;
height:16px" class="math gen" /> upward will always give you at least 1). If a country is allocated too many seats, you overrule the calculation and cap the number of seats at 96. As for the divisor <img src="/MI/82015f5ea26f89b575ee477aa18adfc8/images/img0002.png" alt="$d$" style="verticalalign:0px;
width:9px;
height:11px" class="math gen" />, it is chosen so that the
number of seats adds up to whatever total number of seats you want there to be in parliament (see <a href="/content/formulaeurope">here</a> for an example).</p>
<h3>The Power Compromise</h3>
<p>The Cambridge Compromise seems easy enough, but it comes with a hitch. If you allocate seats using this method, making sure not to exceed the 751 seat total, you find that some member states lose seats. This is a political nono, which is why Grimmett, with his colleagues <a href="https://www.math.uniaugsburg.de/htdocs/emeriti/pukelsheim/welcome.html">Friedrich Pukelsheim</a> and <a href="https://www.kaifriederike.de/">KaiFrederike Oelbermann</a> from Augsburg, Germany, came up with another method called the <em><a href="https://arxiv.org/pdf/1108.1315.pdf">Power Compromise</a></em>. </p>
<p>
This method again starts by giving every member state five base seats. But rather than allocating the remaining seats by calculating <img src="/MI/06578b8025bd1a25d265ac7ce12dbb7d/images/img0001.png" alt="$p/d$" style="verticalalign:4px;
width:26px;
height:16px" class="math gen" /> as above, we calculate <table id="a0000000002" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/06578b8025bd1a25d265ac7ce12dbb7d/images/img0002.png" alt="\[ p^ q/d \]" style="width:33px;
height:16px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table> for some exponent <img src="/MI/06578b8025bd1a25d265ac7ce12dbb7d/images/img0003.png" alt="$q$" style="verticalalign:3px;
width:8px;
height:10px" class="math gen" />. The exponent is less than 1 and is chosen, along with the divisor <img src="/MI/06578b8025bd1a25d265ac7ce12dbb7d/images/img0004.png" alt="$d$" style="verticalalign:0px;
width:9px;
height:11px" class="math gen" />, so that no country gets more than 96 seats and that the total number of seats is whichever number you want it to be. As before, fractional numbers are rounded upwards.
</p><p>
<p>For example, if we choose <img src="/MI/ecfe5edf5920bd8f1802d2f5e8866a44/images/img0001.png" alt="$q=0.9$" style="verticalalign:3px;
width:51px;
height:15px" class="math gen" /> and <img src="/MI/ecfe5edf5920bd8f1802d2f5e8866a44/images/img0002.png" alt="$d=1,000,000$" style="verticalalign:3px;
width:102px;
height:15px" class="math gen" />, then for a country with a population of 10 million we get </p><table id="a0000000002" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/ecfe5edf5920bd8f1802d2f5e8866a44/images/img0003.png" alt="\[ p^ q/d=\frac{10,000,000^{0.9}}{1,000,000}\approx 1.99, \]" style="width:214px;
height:39px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><p> which is rounded upwards to 2. Taking into account the 5 base seats, the country therefore gets <img src="/MI/ecfe5edf5920bd8f1802d2f5e8866a44/images/img0004.png" alt="$5+2=7$" style="verticalalign:1px;
width:67px;
height:14px" class="math gen" /> seats. </p>
</p>
<div class="rightimage" style="maxwidth: 350px;"><img src="/content/sites/plus.maths.org/files/articles/2019/EU/eu_building.jpg" alt="The European Parliament" width="350" height="232" />
<p>The European Parliament in Strasbourg where plenary sessions take place. Photo: Michel CHRISTEN, © European Union 2019  source:EP.</p>
</div>
<p>If this calculation has caused you to glaze over then you have already spotted one disadvantage of the Power Compromise. Power functions are harder to explain to people than straight proportionality and are also harder to interpret. Simply dividing the population size by <img src="/MI/6d5d9f207bf637aea176f12cad19aa30/images/img0001.png" alt="$d=1,000,000$" style="verticalalign:3px;
width:102px;
height:15px" class="math gen" /> tells you how many MEPs you need to have one of them per 1 million citizens. But dividing <img src="/MI/6d5d9f207bf637aea176f12cad19aa30/images/img0002.png" alt="$10,000,000^{0.9}$" style="verticalalign:3px;
width:96px;
height:17px" class="math gen" /> by <img src="/MI/6d5d9f207bf637aea176f12cad19aa30/images/img0001.png" alt="$d=1,000,000$" style="verticalalign:3px;
width:102px;
height:15px" class="math gen" /> tells you how many MEPs you need to have 1 per...what, exactly? </p>
<p>It turns out that the Power Compromise wouldn't cause any country to lose seats compared to the current allocation, as long as the parliament contains at least 723 seats. <! There's another advantage which, back in the prehistoric era before the Brexit referendum, seemed appealing (and may yet come into its own). As the number of EU member states increases, the exponent recommended by the Power Compromise method tends to <img src="/MI/bc3451161e0cedd966458874900a9ac1/images/img0001.png" alt="$q=1$" style="verticalalign:3px;
width:37px;
height:15px" class="math gen" /> — in other words, the method comes closer and close to the more intuitive Cambridge Compromise. So if the EU keeps growing, the problems with the Power Compromise will vanish by virtue of the maths. ></p>
<p>In January 2017, with a view to the May 2019 elections, the European Constitutional Affairs Committee (AFCO) hosted a workshop in Brussels at which it solicited advice from invited mathematicians.
The Cambridge Compromise and the Power Compromise were two methods presented at the workshop, alongside two variants developed by others: a modified version of the Power Compromise, and a method based on socalled <em>adjusted quotas</em>, which avoids seat losses for a parliament bigger than the permitted 751 seats. (You can find out more about both <a href="http://www.statslab.cam.ac.uk/~grg/papers/PukelsheimGrimmettDegressive.pdf">here</a>). </p>
<h3>What's actually going to happen?</h3>
<p>Maths appears mysterious to some, but politics is more so. After the workshop in January 2017 AFCO entered a lengthy deliberation period. As with anything else, speculation about Brexit scuppered progress considerably, though eventually AFCO took a practical approach: assume, as seemed highly likely at the time, that the UK won't be around for the 2019 to 2024 legislative period and come up with a brand new composition of the European Parliament which excludes the UK. If the UK ends up remaining after all, carry on with the current composition, and if it quits within the 2019 to 2024 period, perform a spontaneous switch from old to new. </p>
<div class="leftimage" style="maxwidth: 350px;"><img src="/content/sites/plus.maths.org/files/articles/2019/EU/may.jpg" alt="Theresa May and JeanClaude Juncker" width="350" height="233" />
<p>Theresa May and JeanClaude Juncker giving a press conference after their meeting on March 12, 2019. Photo: Daina Le Lardic, © European Union 2017  source:EP.</p>
</div>
<p>In the summer of 2017 AFCO indicated what it wanted for the new composition: no member state should lose seats, degressivity should be respected, and the total number of seats should be between 700 and 710. This blew the power comprise out of the water: for nobody to lose seats it needs a parliament of at least 723 seats. But Grimmett and his colleagues came up with a variant of the Cambridge Compromise which would fit the bill. Other participants of the workshop also suggested alternatives (find out more <a href="http://www.statslab.cam.ac.uk/~grg/papers/PukelsheimGrimmettDegressive.pdf">here</a>).</p>
<p>In January 2018, after deliberations behind closed doors, AFCO finally announced the new composition — and surprised everyone by not taking recourse to any kind of method or formula. This is disappointing not only in a mathematical sense. "AFCO was obliged to agree on a process which is objective, fair, durable, and transparent. It has not met this obligation beyond achieving degressivity," Grimmett and Pukelsheim write in <a href="http://www.statslab.cam.ac.uk/~grg/papers/PukelsheimGrimmettDegressive.pdf">this paper</a>. "Parliament missed this opportunity to proceed from the dark ages to an era of enlightenment." You can see the new composition, along with the current one, <a href="/content/democraticdilemmas#tables">below</a>.</p>
<p>Andrew Duff, who made initial contact with mathematicians at Cambridge, is angry too. He blames the <a href="http://www.epc.eu/documents/uploads/pub_8271_governeuropebetter.pdf?doc_id=1949">"unfamiliarity of most politicians with mathematics"</a> and pressure from member states
that are currently overprivileged in terms of seats for AFCO's failure to adopt a mathematical formula. In June 2018 the European Commission <a href="http://www.europarl.europa.eu/factsheets/en/sheet/20/theeuropeanparliamentorganisationandoperation">adopted the proposed new composition</a>. Perhaps, with all the headaches caused by Brexit, the maths proved too much for parliamentarians.</p>
<hr/>
<h3>Further information</h3>
<p>You can find links to relevant articles, reports and videos on the webpages <a href="http://www.statslab.cam.ac.uk/~grg/apportionment.html">Geoffrey Grimmett</a> and <a href="http://www.statslab.cam.ac.uk/~grg/apportionment.html">Friedrich Pukelsheim</a>.</p>
<hr/>
<a name="tables"></a>
<h3>The old and the new compositions of the European Parliament</h3>
<p>The following table gives the current (April 2019) seat distribution of the European Parliament, which still includes the UK, and the new composition which will come in force once the UK leaves.</p>
<table class="datatable"><th colspan="2">Current (20142019) </th><th>New, excluding the UK</th>
<tr><td>Germany</td><td>96</td><td>96</td></tr>
<tr><td>France</td><td>74</td><td>79</td></tr>
<tr><td>Italy</td><td>73</td><td>76</td></tr>
<tr><td>United Kingdom</td><td>73</td><td></td></tr>
<tr><td>Spain</td><td>54</td><td>59</td></tr>
<tr><td>Poland</td><td>51</td><td>52</td></tr>
<tr><td>Romania</td><td>32</td><td>33</td></tr>
<tr><td>Netherlands</td><td>26</td><td>29</td></tr>
<tr><td>Belgium</td><td>21</td><td>21</td></tr>
<tr><td>Czech Republic</td><td>21</td><td>21</td></tr>
<tr><td>Greece</td><td>21</td><td>21</td></tr>
<tr><td>Hungary</td><td>21</td><td>21</td></tr>
<tr><td>Portugal</td><td>21</td><td>21</td></tr>
<tr><td>Sweden</td><td>20</td><td>21</td></tr>
<tr><td>Austria</td><td>18</td><td>19</td></tr>
<tr><td>Bulgaria</td><td>17</td><td>17</td></tr>
<tr><td>Denmark</td><td>13</td><td>14</td></tr>
<tr><td>Finland</td><td>13</td><td>14</td></tr>
<tr><td>Slovakia</td><td>13</td><td>14</td></tr>
<tr><td>Ireland</td><td>11</td><td>13</td></tr>
<tr><td>Croatia</td><td>11</td><td>12</td></tr>
<tr><td>Lithuania</td><td>11</td><td>11</td></tr>
<tr><td>Latvia</td><td>8</td><td>8</td></tr>
<tr><td>Slovenia</td><td>8</td><td>8</td></tr>
<tr><td>Estonia</td><td>6</td><td>7</td></tr>
<tr><td>Cyprus</td><td>6</td><td>6</td></tr>
<tr><td>Luxembourg</td><td>6</td><td>6</td></tr>
<tr><td>Malta</td><td>6</td><td>6</td></tr>
</table>
<hr/>
<h3>About the author</h3>
<p><a href="/content/people/index.html#marianne">Marianne Freiberger</a> is Editor of <em>Plus</em>.</p></div></div></div>
Thu, 02 May 2019 16:15:33 +0000
Marianne
7192 at https://plus.maths.org/content
https://plus.maths.org/content/democraticdilemmas#comments

Phantom jams
https://plus.maths.org/content/phantomjams
<div class="field fieldnamefieldabsimg fieldtypeimage fieldlabelhidden"><div class="fielditems"><div class="fielditem even"><img class="imgresponsive" src="https://plus.maths.org/content/sites/plus.maths.org/files/abstractpics/%5Buid%5D/%5Bsitedate%5D/traffic_miami_icon.jpg" width="100" height="100" alt="" /></div></div></div><div class="field fieldnamefieldauthor fieldtypetext fieldlabelinlinec clearfix fieldlabelinline"><div class="fieldlabel">By </div><div class="fielditems"><div class="fielditem even">Marianne Freiberger and Rachel Thomas</div></div></div><div class="field fieldnamebody fieldtypetextwithsummary fieldlabelhidden"><div class="fielditems"><div class="fielditem even"><p>Traffic jams are a pain, especially when they happen for no apparent reason. Even when a road is only moderately full and there are no obstacles up ahead, traffic can bunch up, slow down, and even come to a complete halt. Traffic researchers have long been interested in such <em>phantom jams </em> — and mathematical models can tell us why they happen.</p>
<div class="rightimage" style="maxwidth: 320px;"><img src="/content/sites/plus.maths.org/files/articles/2019/traffic/traffic_miami.jpg" alt="Traffic" width="320" height="240" />
<p>Rush hour on IS 95 in Florida. Photo: <a href="https://commons.wikimedia.org/wiki/File:Miami_traffic_jam,_I95_North_rush_hour.jpg">B137</a>, <a href="https://creativecommons.org/licenses/bysa/4.0/deed.en">CC BYSA 4,0</a>.</p>
</div>
<p>
You can build a model based on the simple assumption that drivers vary their speed according to the distance between themselves and the vehicle in front.
If the distance decreases because the car in front brakes, a driver will slow down.
If the distance increases, a driver will speed up until they reach the speed they desire (or the speed dictated to them by a speed limit). This is exactly what happens in reality.</p>
<p>To keep things simple let's also assume that we have a number of cars travelling in a single lane around a ring. This is less realistic but means we can assume that the number of cars on the road stays the same, which makes for easier maths.</p>
<p>It's relatively straightforward to capture this situation using mathematical equations that express the acceleration of each car in terms of the headway in front of them and their current speed (see <a href="/content/phantomjams#maths">below</a> for some details). Given an initial speed and distribution of cars, the solutions to the equations will tell you how fast the cars will be travelling at any given time. </p>
<p>This toy model is so simple, you'd expect the traffic to settle down into an even flow going round and round the ring. Carfollowing models like this do indeed predict just that: there is a <em>uniform flow equilibrium</em>, which is stable in the sense that little variations in speed won't knock the equilibrium off kilter. That's exactly the kind of situation we'd want in real life, but we all know that in reality things can play out very differently. </p>
<h3>Chain reaction</h3>
<p>One crucial thing the model is missing is the fact that drivers aren't machines: while driving along they might be fiddling with the radio, talking to the kids on the back seat, or arguing with the satnav. In short, drivers don't react immediately to a change in the distance to the car in front. A little delay in their reaction could make quite a difference to how hard they need to brake.</p>
<p>
We can add this fact into the model by including a reaction time delay parameter. When you do this, the dynamical system does indeed change its overall nature: apart from the stable uniform flow behaviour, the equations now also admit a stopandgo wave that can persistently travel down the road in the opposite direction of the traffic flow, even if the density of traffic isn't all that bad. The crucial ingredient here are the drivers' delayed reactions to even minor disruptions of traffic, such as a lorry ahead changing lanes. It's those delays that can trigger the stopandgo wave. Our simple model has therefore delivered an explanation for phantom jams that previously wasn't obvious.</p>
<p>Traffic jam waves spreading backwards through the traffic have been predicted by other mathematical traffic models too. They have even been observed in a reallife experiment which reconstructs the simple setup we described above: it had 22 cars travelling around a ring at a constant speed of roughly 30kmh and even though traffic started out flowing smoothly, jam waves were soon triggered by little variations in speed. </p>
<iframe width="560" height="315" src="https://www.youtube.com/embed/Suugnp5C1M" frameborder="0" allow="accelerometer; autoplay; encryptedmedia; gyroscope; pictureinpicture" allowfullscreen></iframe>
<p>The good thing about mathematical models of traffic is that they can not only explain baffling phenomena, but also help us decide how to avoid them. The simple model above suggests that keeping your reaction time to a minimum can do more to avoid jams than you might think — perhaps driverless cars will help in this context. But models can also be used to figure out how things such as variable speed limits may help guide traffic out of a jam and back into uniform flow. The roads of the future should, and hopefully will, be built on maths.</p>
<a name="maths"></a>
<h3>A traffic model</h3>
<p>Suppose there are <img src="/MI/d5221bb551beb5038d1eb0eede5f0e7a/images/img0001.png" alt="$n$" style="verticalalign:0px;
width:10px;
height:7px" class="math gen" /> cars and write <img src="/MI/d5221bb551beb5038d1eb0eede5f0e7a/images/img0002.png" alt="$v_ i(t)$" style="verticalalign:4px;
width:30px;
height:18px" class="math gen" /> for the velocity of car <img src="/MI/d5221bb551beb5038d1eb0eede5f0e7a/images/img0003.png" alt="$i$" style="verticalalign:0px;
width:5px;
height:11px" class="math gen" /> at time <img src="/MI/d5221bb551beb5038d1eb0eede5f0e7a/images/img0004.png" alt="$t$" style="verticalalign:0px;
width:6px;
height:10px" class="math gen" />. Write <img src="/MI/d5221bb551beb5038d1eb0eede5f0e7a/images/img0005.png" alt="$h_ i(t)$" style="verticalalign:4px;
width:31px;
height:18px" class="math gen" /> for the distance between car <img src="/MI/d5221bb551beb5038d1eb0eede5f0e7a/images/img0006.png" alt="$i+1$" style="verticalalign:1px;
width:33px;
height:13px" class="math gen" /> and car <img src="/MI/d5221bb551beb5038d1eb0eede5f0e7a/images/img0003.png" alt="$i$" style="verticalalign:0px;
width:5px;
height:11px" class="math gen" /> at time <img src="/MI/d5221bb551beb5038d1eb0eede5f0e7a/images/img0004.png" alt="$t$" style="verticalalign:0px;
width:6px;
height:10px" class="math gen" /> so <img src="/MI/d5221bb551beb5038d1eb0eede5f0e7a/images/img0007.png" alt="$h_ i$" style="verticalalign:2px;
width:13px;
height:13px" class="math gen" /> is the headway of car <img src="/MI/d5221bb551beb5038d1eb0eede5f0e7a/images/img0003.png" alt="$i$" style="verticalalign:0px;
width:5px;
height:11px" class="math gen" />. Then <img src="/MI/d5221bb551beb5038d1eb0eede5f0e7a/images/img0005.png" alt="$h_ i(t)$" style="verticalalign:4px;
width:31px;
height:18px" class="math gen" /> should satisfy </p><table id="a0000000002" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/d5221bb551beb5038d1eb0eede5f0e7a/images/img0008.png" alt="\[ dh_ i(t)/dt = v_{i+1}(t)v_ i(t), \]" style="width:193px;
height:18px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><p> capturing how the headway between two cars changes as a result of the cars’ changes in velocity. In an <em>optimal velocity model</em> the acceleration <img src="/MI/d5221bb551beb5038d1eb0eede5f0e7a/images/img0009.png" alt="$dv_ i(t)/dt$" style="verticalalign:4px;
width:63px;
height:18px" class="math gen" /> of car <img src="/MI/d5221bb551beb5038d1eb0eede5f0e7a/images/img0003.png" alt="$i$" style="verticalalign:0px;
width:5px;
height:11px" class="math gen" /> is described by the differential equation </p><table id="a0000000003" class="equation" width="100%" cellspacing="0" cellpadding="7">
<tr>
<td style="width:40%"> </td>
<td><img src="/MI/d5221bb551beb5038d1eb0eede5f0e7a/images/img0010.png" alt="\[ dv_ i(t)/dt =\alpha \left(V(h_ i(t))v_ i(t)\right), \]" style="width:233px;
height:18px" class="math gen" /></td>
<td style="width:40%"> </td>
<td class="eqnnum" style="width:20%"> </td>
</tr>
</table><p> where <img src="/MI/d5221bb551beb5038d1eb0eede5f0e7a/images/img0011.png" alt="$\alpha $" style="verticalalign:0px;
width:10px;
height:7px" class="math gen" /> is a parameter called the <em>sensitivity</em> and <img src="/MI/d5221bb551beb5038d1eb0eede5f0e7a/images/img0012.png" alt="$V(h_ i(t))$" style="verticalalign:4px;
width:57px;
height:18px" class="math gen" /> is a function that represents the <em>optimal velocity</em> for a given amount of headway. <img src="/MI/d5221bb551beb5038d1eb0eede5f0e7a/images/img0013.png" alt="$V(h_ i)$" style="verticalalign:4px;
width:39px;
height:18px" class="math gen" /> could be any type of function that satisfies certain realistic conditions. For example, as the headway <img src="/MI/d5221bb551beb5038d1eb0eede5f0e7a/images/img0007.png" alt="$h_ i$" style="verticalalign:2px;
width:13px;
height:13px" class="math gen" /> in front of a car increases, the optimal velocity <img src="/MI/d5221bb551beb5038d1eb0eede5f0e7a/images/img0013.png" alt="$V(h_ i)$" style="verticalalign:4px;
width:39px;
height:18px" class="math gen" /> should also increase until it reaches the legal speed limit, reflecting that drivers want to speed up when they can. </p><p>The equation above shows that if the velocity <img src="/MI/d5221bb551beb5038d1eb0eede5f0e7a/images/img0014.png" alt="$v_ i$" style="verticalalign:2px;
width:12px;
height:9px" class="math gen" /> of car <img src="/MI/d5221bb551beb5038d1eb0eede5f0e7a/images/img0003.png" alt="$i$" style="verticalalign:0px;
width:5px;
height:11px" class="math gen" /> is smaller than the optimal velocity <img src="/MI/d5221bb551beb5038d1eb0eede5f0e7a/images/img0013.png" alt="$V(h_ i)$" style="verticalalign:4px;
width:39px;
height:18px" class="math gen" /> then the driver will accelerate, if it is larger then the driver will brake, and if the two are equal the drive will do neither. </p><p>(In both of the equations above, if <img src="/MI/d5221bb551beb5038d1eb0eede5f0e7a/images/img0015.png" alt="$i=n$" style="verticalalign:0px;
width:37px;
height:11px" class="math gen" /> we take <img src="/MI/d5221bb551beb5038d1eb0eede5f0e7a/images/img0006.png" alt="$i+1$" style="verticalalign:1px;
width:33px;
height:13px" class="math gen" /> to be <img src="/MI/d5221bb551beb5038d1eb0eede5f0e7a/images/img0016.png" alt="$1$" style="verticalalign:0px;
width:6px;
height:12px" class="math gen" />, reflecting the fact that cars are travelling around a ring.) </p><p>As it stands, the second equation means that drivers react to a change in their headway instantaneously by breaking or speeding up. But we can also include a reaction time <img src="/MI/d5221bb551beb5038d1eb0eede5f0e7a/images/img0017.png" alt="$T$" style="verticalalign:0px;
width:12px;
height:11px" class="math gen" /> to get the equation </p><p><img src="/MI/d5221bb551beb5038d1eb0eede5f0e7a/images/img0018.png" alt="$dv_ i(t)/dt =\alpha \left(V(h_ i(tT))v_ i(t)\right).$" style="verticalalign:4px;
width:264px;
height:18px" class="math gen" /><img src="/MI/d5221bb551beb5038d1eb0eede5f0e7a/images/img0019.png" alt="$$" style="verticalalign:0px;
width:1px;
height:1px" class="math gen" /></p><p>You can find out more in <a href="http://wwwpersonal.umich.edu/~orosz/articles/PREpublished.pdf">this paper</a> by Gábor Orosz, R. Eddie Wilson, and Bernd Krauskopf or <a href="https://royalsocietypublishing.org/doi/pdf/10.1098/rspa.2006.1660">this one</a> by Gábor Orosz and Gábor Stépán.</p>
<hr/>
<h3>About this article</h3><div class="rightimage" style="maxwidth: 150px;"><img src="/content/sites/plus.maths.org/files/articles/2019/dating/cover.jpg" alt="Book cover" width="150" height="210" /><p></p>
</div>
<p>This article is based on a chapter from the new book <em><a href="https://amzn.to/2G6v65L">Understanding numbers</a></em> by the <em>Plus</em> Editors <a href="/content/people/index.html#rachel">Rachel Thomas</a> and <a href="/content/people/index.html#marianne">Marianne Freiberger</a>. </p>
</div></div></div>
Thu, 18 Apr 2019 15:53:10 +0000
Marianne
7190 at https://plus.maths.org/content
https://plus.maths.org/content/phantomjams#comments

The power of snooker
https://plus.maths.org/content/powersnooker
<div class="field fieldnamefieldabsimg fieldtypeimage fieldlabelhidden"><div class="fielditems"><div class="fielditem even"><img class="imgresponsive" src="https://plus.maths.org/content/sites/plus.maths.org/files/abstractpics/%5Buid%5D/%5Bsitedate%5D/icon_12.png" width="100" height="100" alt="" /></div></div></div><div class="field fieldnamefieldauthor fieldtypetext fieldlabelinlinec clearfix fieldlabelinline"><div class="fieldlabel">By </div><div class="fielditems"><div class="fielditem even">Wim Hordijk</div></div></div><div class="field fieldnamebody fieldtypetextwithsummary fieldlabelhidden"><div class="fielditems"><div class="fielditem even"><div class="rightimage" style="maxwidth:350px"><img src="/content/sites/plus.maths.org/files/articles/2019/snooker/snooker.png"><p>Snookered... (Photo by <a href="https://nl.wikipedia.org/wiki/Snooker_(spelsituatie)#/media/File:Snookered_on_two_reds.jpg">Florian Albrecht</a> – <a href="https://creativecommons.org/licenses/bysa/3.0">CC BYSA 3.0</a>)</p></div>
<p>A particular mathematical relationship known as a <em>power law</em> has been observed in many daytoday situations, from the frequencies in which words are used in natural languages to the connectivity distribution in Facebook friendship networks. As it turns out, though, such a power law can also be found in snooker statistics. And if the amazing <a href="https://en.wikipedia.org/wiki/Ronnie_O%27Sullivan">Ronnie O'Sullivan</a> continues to produce centuries at the same rate, the mathematical correspondence will be even better!
</p>
<br class="brclear"/>
<h3>Power laws</h3>
<p>In mathematics, we say a given value <em>x</em> is <em>raised to the power k</em> if it is multiplied by itself <em>k</em> times. Common examples where such mathematical power functions are used are in the calculation of the area of a square that has sides of length <em>x</em>, which is equal to <em>x<sup>2</sup></em>, or the volume of a cube with sides of length <em>x</em>, which is <em>x<sup>3</sup></em>.
</p>
<div class="leftimage" style="maxwidth: 350px">
<img src="/content/sites/plus.maths.org/files/articles/2019/snooker/volume.jpg" width="500" alt="Volume graph">
<p>The volume (<em>y</em>, vertical axis) of a cube against the length of its sides (<em>x</em>, horizontal axis). The open circles represent actual measurements on different cubes. The solid line represents the mathematical relationship <em>y=x<sup>3</sup></em>, which forms a perfect fit to the observed data.</p>
</div>
<p>A <em>power law</em> is a mathematical relationship between two variables, say <em>x</em> and <em>y</em>, such that the value of one is directly related to the value of the other raised to a certain power, for example, <em>y=x<sup>a</sup></em> for some fixed value of <em>a</em>. This value <em>a</em> is called the <em>exponent</em> of the power law. For example, if we were to measure the volume (<em>y</em>) of many cubes with different lengths of their sides (<em>x</em>), we would find a perfectly fitting power law with an exponent <em>a</em>=3.
</p>
<p>Power law relationships are found in many common systems and processes, both natural and manmade. For example, power laws are observed in the frequency distribution of the magnitudes of earthquakes, or the sizes of cities in a given country. They also show up in the distribution of connections in networks like Facebook friendships, movie star coappearances, or the national electricity grid. And they occur in the frequencies of words used in natural languages, or even in computer code. (You can read more about power laws in <a href="/content/statisticslanguage">language</a> and <a href="/content/mathsminutepowernetworks">networks</a> on <em>Plus</em>.)</p>
<p>In the example above, the data is plotted on a <em>linear</em> scale. In other words, each point along an axis represents a value that is larger than the previous point by a fixed amount. However, it is also possible to plot the same data on a nonlinear scale. An example of such a nonlinear scale is a <em>logarithmic scale</em>, which is based on orders of magnitude rather than a linear increase. This means that the value represented by each point along an axis is the value of the previous point <em>multiplied</em> by a fixed amount.</p>
<div class="rightimage" style="maxwidth:350px">
<img src="/content/sites/plus.maths.org/files/articles/2019/snooker/logvolume.jpg" width="500" alt="Volume on a loglog scale">
<p>The volume (<em>y</em>, vertical axis) of a cube against the length of its sides (<em>x</em>, horizontal axis) using a logarithmic scale on both axes. The data points (open circles) now fall along a straight line representing a power law with exponent <em>a</em>=3.</p>
</div>
<p>An example of this can be found in the way that the strength of an earthquake is indicated. A magnitude three (M3) earthquake releases a certain amount of energy, which translates into the amount of shaking we feel. However, a magnitude four (M4) earthquake releases an amount of energy that is ten times larger than an M3 earthquake. Similarly, an M5 earthquake is again ten times as strong as an M4 earthquake, and thus one hundred times stronger than an M3 earthquake.</p>
<p>The figure to the right shows the same data for the volume of a cube, but using such a logarithmic scale on both axes, resulting in a socalled <em>loglog plot</em>. Mathematically, a logarithmic function is the inverse of a power function. As a consequence, a power law shows up as a straight line in a loglog plot, as this figure illustrates.</p>
<br class="brclear"/>
<h3>Finding the power</h3>
<p>Unfortunately, most realworld data does not behave as nicely as the volume of a cube. In reality, there is always some noise in the data due to imprecise measurements, random fluctuations, or missing data. An example is given in the figure below, which shows a loglog plot of the frequencies of words used in English (as measured over a large collection of different texts) against their rank, where the most frequent word ("the") has rank one, the second most frequent word ("be") rank two, and so on.</p>
<p>As the figure shows, the data points do not fall along a straight line perfectly, but still reasonably well. The solid straight line represents a power law that best fits the given data, which can be calculated using a standard statistical technique known as a <em>regression analysis</em>. (You can read more about regression analysis <a href="/content/mathsminutelinearregression">here</a>.) This best fit results in an exponent of <em>a</em>=0.92. The exponent is negative in this case, as the word frequency decreases with increasing rank.</p>
<div class="centreimage" style="maxwidth:500px">
<img src="/content/sites/plus.maths.org/files/articles/2019/snooker/language.jpg" width="500" alt="loglog plot of word ranks in English">
<p>The frequency of words used in English (vertical axis) against their rank (horizontal axis) using a logarithmic scale on both axes. The data points (open circles) closely follow a straight line, although not exactly.</p>
</div>
<p>So, the frequency of a word as used in the English language is roughly proportional to its rank raised to the power <em>a</em>=0.92. It may be difficult to imagine what it means to multiply a number by itself 0.92 times, but mathematically this is perfectly well defined. In other words, the exponent <em>a</em> in a power law does not always need to be a positive whole number.</p>
<p>Finally, the regression analysis that calculates the exponent that gives the best fit also provides a measure of accuracy, that is, how closely the data falls along a straight line. In the case of the volumes of cubes, as we saw above, the accuracy (or <em>fit</em>) is obviously 100%, as the data points fall exactly on the line. However, for the word frequencies the accuracy is slightly less: 98%. Still pretty good.</p>
<p>As these examples have shown, to find out whether a given data set follows a power law, we can simply present the data in a loglog plot, and calculate how closely it falls along a straight line. So let's give this a try with some snooker statistics.</p>
<h3>Power laws in snooker statistics</h3>
<p>Consider the <a href="http://snookerinfo.webs.com/100centuries" target="_blank">ranking list</a> of all professional snooker players who have made at least 100 centuries throughout their career. A <em>century break</em> is a score of at least 100 points within one visit to the table (i.e., without missing a shot). There are currently 68 players who have scored at least a "century of centuries". The table below shows the ten highest ranked players according to this statistic, of course topped by the amazing Ronnie O'Sullivan who recently scored his 1000<sup>th</sup> century, and still going strong!</p>
<div class="rightimage" style="maxwidth:350px">
<img src="/content/sites/plus.maths.org/files/articles/2019/snooker/ronnie.jpg" alt="Ronnie O'Sullivan">
<p>Ronnie O'Sullivan in action at the snooker table. (Image: <a href="https://commons.wikimedia.org/wiki/File:Ronnie_O%E2%80%99Sullivan_at_Snooker_German_Masters_(DerHexer)_20150206_08.jpg" target="_blank">DerHexer, CCBYSA 4.0</a>).</p>
</div>
<table class="datatable">
<tr><th>Rank</th><th>Player</th><th>Centuries</th></tr>
<tr><td>1</td><td>Ronnie O'Sullivan</td><td align="right">1008</td></tr>
<tr><td>2</td><td>Stephen Hendry</td><td align="right">775</td></tr>
<tr><td>3</td><td>John Higgins</td><td align="right">750</td>
<tr><td>4</td><td>Neil Robertson</td><td align="right">636</td></tr>
<tr><td>5</td><td>Judd Trump</td><td align="right">602</td></tr>
<tr><td>6</td><td>Mark Selby</td><td align="right">577</td></tr>
<tr><td>7</td><td>Ding Junhui</td><td align="right">501</td></tr>
<tr><td>8</td><td>Marco Fu</td><td align="right">493</td></tr>
<tr><td>9</td><td>Shaun Murphy</td><td align="right">479</td></tr>
<tr><td>10</td><td>Mark Williams</td><td align="right">464</td></tr>
</table>
<p>Now, if we plot the number of career centuries of these 68 players against their rank in this list, but in a loglog plot, we get the result as shown in the next figure. The straight line represents a power law that gives the best fit to the given data, resulting in an exponent of <em>a</em>=0.63.</p>
<div class="centreimage" style="maxwidth:500px">
<img src="/content/sites/plus.maths.org/files/articles/2019/snooker/centuries.jpg" width="500" alt="Loglog plot of career centuries versus rank">
<p>The number of career centuries (vertical axis) against the rank (horizontal axis). The straight line is a fitted power law with exponent <em>a</em>=0.63.</p>
</div>
<p>Note that, as with the word frequencies data, the fit it is not perfect, especially not at the top of the ranking (i.e., the data points in the top left of the plot). This is often the case, especially with ranking data that is still in the making. However, according to the regression analysis, the current fit still has an accuracy of 95%. And if Ronnie O'Sullivan produces a few more centuries, the fit will be even better!</p>
<p>In a similar way, we can look at the number of <a href="https://en.wikipedia.org/wiki/List_of_snooker_players_by_number_of_ranking_titles" target="_blank">ranking titles</a> of each player (the number of tournaments they've won which count towards the snooker world rankings). There are currently 26 players who have obtained at least three ranking titles throughout their career. The table below shows the ten highest ranked players according to this statistic.</p>
<table class="datatable">
<tr><th>Rank</th><th>Player</th><th>Titles</th></tr>
<tr><td>1</td><td>Ronnie O'Sullivan</td><td align="right">36</td></tr>
<tr><td>2</td><td>Stephen Hendry</td><td align="right">36</td></tr>
<tr><td>3</td><td>John Higgins</td><td align="right">30</td>
<tr><td>4</td><td>Steve Davis</td><td align="right">28</td></tr>
<tr><td>5</td><td>Mark Williams</td><td align="right">22</td></tr>
<tr><td>6</td><td>Neil Robertson</td><td align="right">16</td></tr>
<tr><td>7</td><td>Mark Selby</td><td align="right">15</td></tr>
<tr><td>8</td><td>Ding Junhui</td><td align="right">13</td></tr>
<tr><td>9</td><td>Judd Trump</td><td align="right">10</td></tr>
<tr><td>10</td><td>Jimmy White</td><td align="right">10</td></tr>
</table>
<p>If we plot the number of ranking titles of these 26 players against their rank in the list, again in a loglog plot, we get the result as shown in the figure below. The straight line once more represents a power law that gives the best fit to the data, resulting in an exponent of <em>a</em>=0.96. The accuracy of the fit is slightly less in this case (92%), mostly due to the top two players (Ronnie O'Sullivan and Stephen Hendry) having exactly the same number of ranking titles. If we would rank them together as a joint number 1, the fit would actually be excellent.</p>
<div class="centreimage" style="maxwidth:500px">
<img src="/content/sites/plus.maths.org/files/articles/2019/snooker/rankingtitles.jpg" width="500" alt="Loglog plot of titles versus rank of the player">
<p>The number of titles (vertical axis) against the rank (horizontal axis). The straight line is a fitted power law.</p>
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<p>Finally, if we add up the number of ranking titles of all players from the same country, we get the list shown in the following table</p>.
<table class="datatable">
<tr><th>Rank</th><th>Country</th><th>Titles</th></tr>
<tr><td>1</td><td>England</td><td align="right">170</td></tr>
<tr><td>2</td><td>Scotland</td><td align="right">78</td></tr>
<tr><td>3</td><td>Wales</td><td align="right">37</td>
<tr><td>4</td><td>Australia</td><td align="right">16</td></tr>
<tr><td>5</td><td>China</td><td align="right">14</td></tr>
<tr><td>6</td><td>Northern Ireland</td><td align="right">8</td></tr>
<tr><td>7</td><td>Republic of Ireland</td><td align="right">7</td></tr>
<tr><td>8</td><td>Thailand</td><td align="right">4</td></tr>
<tr><td>9</td><td>Hong Kong</td><td align="right">3</td></tr>
<tr><td>10</td><td>Canada</td><td align="right">3</td></tr>
</table>
<p>Plotting the number of titles per country against the rank in the
list, as a loglog plot, gives the result as shown in the next
figure. The straight line shows the power law that gives the best fit,
with an exponent of <em>a</em>=2.12. The accuracy is much better again in this case: 96%.</p>
<div class="centreimage" style="maxwidth:500px">
<img src="/content/sites/plus.maths.org/files/articles/2019/snooker/countries.jpg" width="500" alt="Number of titlse by country against the rank">
<p>The number of titles per country (vertical axis) against the rank (horizontal axis). The straight line is a fitted power law.</p>
</div>
<p>Snooker statistics seem to follow mathematical power laws, just like natural languages, earthquake occurrences, and many types of natural and manmade networks. Why should this be so? Scientists are still arguing about the significance of the occurrence of power laws. On the one hand, there is no particular reason to expect such behavior in many of these situations. On the other hand, the phenomenon seems to be so common that perhaps it has little meaning after all. Either way, it is interesting to see that snooker statistics do indeed follow precise mathematical laws with a high degree of accuracy. I'm curious to see if this year's world championships snooker, with its usual display of power shots, will also produce some additional power laws.</p>
<hr/>
<h3>About the author</h3>
<div class="rightimage" style="width: 200px;"><img src="/content/sites/plus.maths.org/files/articles/2016/altruism/wim.jpg" alt="Wim Hordijk" width="200" height="200" /></div>
<p>Wim Hordijk is a computer scientist currently on a fellowship at the <a href="http://ias.uva.nl/">Institute for Advanced Study</a> of the University of Amsterdam, The Netherlands. He has worked on many research and computing projects all over the world, mostly focusing on questions related to evolution and the origin of life. More information about his research can be found on his <a href="http://www.worldwidewanderings.net">website</a>.</p></div></div></div>
Tue, 16 Apr 2019 14:43:06 +0000
Rachel
7193 at https://plus.maths.org/content
https://plus.maths.org/content/powersnooker#comments

Maths in a minute: The Sydney Opera House
https://plus.maths.org/content/mathsminutesydneyoperahouse
<div class="field fieldnamefieldabsimg fieldtypeimage fieldlabelhidden"><div class="fielditems"><div class="fielditem even"><img class="imgresponsive" src="https://plus.maths.org/content/sites/plus.maths.org/files/abstractpics/%5Buid%5D/%5Bsitedate%5D/sydney_icon.jpg" width="100" height="100" alt="" /></div></div></div><div class="field fieldnamefieldauthor fieldtypetext fieldlabelinlinec clearfix fieldlabelinline"><div class="fieldlabel">By </div><div class="fielditems"><div class="fielditem even">Rachel Thomas</div></div></div><div class="field fieldnamebody fieldtypetextwithsummary fieldlabelhidden"><div class="fielditems"><div class="fielditem even"><p>On 29 January 1957, when Jørn Utzon's saillike sketches were announced as the winning design for the Sydney Opera House, Utzon had a problem — he didn't know exactly how he would build them. The problem still wasn't solved two years later when construction began on 2 March 1959. </p>
<div class="rightimage" style="maxwidth: 320px;"><img src="/content/sites/plus.maths.org/files/articles/2019/sydney/sydney.jpg" alt="Sydney Opera House" width="320" height="211"/>
<p>The Sydney Opera House. Photo: <a href="https://commons.wikimedia.org/wiki/File:Sydney_Opera_House,_botanic_gardens_1.jpg">AdamJ.W.C.</a>, <a href="https://creativecommons.org/licenses/bysa/2.5/deed.en">CC BYSA 2.5</a>.</p>
</div>
<p>Inspired by the harbour location, the young Danish architect had envisioned a series of sweeping curved shells. But in order to build these shells the shapes had to be described mathematically to accurately calculate all the loads and stresses on the building. When asked by the engineers to specify the curves Utzon bent a ruler to trace the curves he wanted. Over the next four years they tried various mathematical forms — ellipses, parabolas — to try to capture Utzon's design.</p>
<p>Finally, in October 1961, Utzon found the "key to the shells": every sail was formed from a wedge cut from a single sphere (and its mirror image). This ingenious solution not only provided a mathematical description of the roof of his design, it also solved all the problems of constructing such a complex structure. Previously each shell appeared to be different, and it would have been almost impossible, both in terms of engineering and finance, to construct these huge bespoke parts. Instead, with shells having a common spherical geometry (based on a sphere with radius of 246 feet), they could instead be constructed by standard parts that could be mass produced and then assembled.</p>
<p>Describing the design mathematically made it possible to create one of the most iconic buildings in the world. The mathematics of the design also exemplified Utzon's artistic vision, providing "full harmony between all the shapes in this fantastic complex".
</p>
<div class="centreimage"><img src="/content/sites/plus.maths.org/files/articles/2019/sydney/shells.jpg" alt="Spherical shells" width="640" height="479"/>
<p style="maxwidth: 640px;">The key to the shells, Sydney Opera House. </p>
</div>
<! Image in public domain >
<p> It is now standard practise for large buildings to be modelled mathematically in threedimensions on a computer, so that the effect of any small changes in architectural design, or practical construction (such as moving the position of piping or electrics), immediately cascade through the model of the building. This highlights any unforeseen clashes, and allows for the most economic use of materials and people during the construction process. Mathematical models also help make sure that buildings are energy efficient, ideally suited for their purpose and improve people's lives — the people using the buildings as well as those admiring their forms from the outside.</p>
<p>To find out more about maths and architecture, see</p>
<ul><li><a href="/content/mathsminutestpaulsdome">Maths in a minute: St Paul's dome</a></li>
<li><a href="gcontent/howvelodromefounditsform">How the velodrome found its form</a></li>
<li><a href="/content/perfectbuildingsmathsmodernarchitecture">Perfect buildings: The maths of modern architecture</a></li>
<li><a href="/content/catenarygoeswembley">The catenary goes to Wembley</a> (video)</li>
<li><a href="/content/node/7127">Stadium maths</a> (podcast).</li></ul>
<hr/>
<h3>About this article</h3><div class="rightimage" style="maxwidth: 150px;"><img src="/content/sites/plus.maths.org/files/articles/2019/dating/cover.jpg" alt="Book cover" width="150" height="210" /><p></p>
</div>
<p>This article is based on a chapter from the new book <em><a href="https://amzn.to/2G6v65L">Understanding numbers</a></em> by the <em>Plus</em> Editors <a href="/content/people/index.html#rachel">Rachel Thomas</a> and <a href="/content/people/index.html#marianne">Marianne Freiberger</a>. </p>
</div></div></div>
Fri, 12 Apr 2019 12:39:05 +0000
Marianne
7189 at https://plus.maths.org/content
https://plus.maths.org/content/mathsminutesydneyoperahouse#comments