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Richard Green
Works at University of Colorado Boulder
Attended University College, Oxford
Lives in Longmont, Colorado
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Richard Green

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Average pace and the Universal Chord Theorem

If you run a three-mile race at an average pace of six minutes per mile, it will always be the case that you ran for one consecutive mile in exactly six minutes. But if you run a 3.1-mile race at an average pace of six minutes per mile, it does not necessarily follow that you ran a consecutive mile in exactly six minutes.

The key issue here is not the units of measurement, but whether or not the total length of the race is an integer multiple of the length of the subinterval of interest. Assuming for simplicity that this subinterval is one mile, then the following two theorems can be proved, as explained in the recent expository paper Average pace and horizontal chords by Keith Burns, Orit Davidovich and Diana Davis (

Theorem 1. If the length of the race is an integer number of miles, then there must be some mile-long interval of the race that was run at exactly the average pace of the whole race.

Theorem 2. If, on the other hand, the length of the race is not an integer number of miles, then it is possible that the race was run in such a way that no mile-long interval was run at exactly the overall average pace.

Of course, it is always possible that a mile-long interval of any race is run at exactly the overall average pace; for example, if the entire race were run at a constant speed. What the second result is asserting is that, if the race is not an integer number of miles long, it is possible to find a (smooth) position function for the race so that no mile-long interval is run at the average pace.

The diagram in the bottom right, which comes from the paper, shows an example of the situation covered by Theorem 2. The situation is motivated by two athletic word records: (a) Molly Huddle, who ran a 12km race in a time of 37:49, an average of 3:09 per kilometer; and (b) Mary Keitany, who ran a half marathon (21.1km) in a time of 65:50, an average of 3:07 per kilometer. It is tempting to conclude that Keitany must have run a consecutive 12km stretch of the race at an average pace of 3 minutes and 7 seconds per kilometer, thus beating Huddle's record. However, Theorem 2 shows that this is not necessarily true, because 21.1km is not an integer multiple of 12km. More precisely, the diagram shows a way this could happen: if there is a long, slow stretch in the middle of the race, then every consecutive stretch of 12km will contain the entire slow section of the race and will therefore be run at less than the average pace.

Theorem 1 can be proved with basic calculus, and it was known to the French physicist and mathematician André-Marie Ampère (of electric current fame) as early as 1806. The idea behind the proof is that if a race is n miles long and is subdivided into n consecutive miles, then at least one of these miles was run at greater than equal to the average pace, and at least one of them was run at less than or equal to the average pace. The result then follows by applying the Intermediate Value Theorem.

Theorem 2 was first proved by Paul Lévy in 1934. Lévy considered an equivalent form of the problem, in which the diagram shown is sheared vertically so that the beginning and end of the race both sit on the horizontal axis. A mile-long stretch of the race that is run at average pace then shows up as a horizontal chord of unit length, which partly explains why this result is sometimes known as the Universal Chord Theorem.

The paper also considers some generalizations of Lévy's result, including one proved by Heinz Hopf in 1937. Hopf proved that a set of positive real numbers is the horizontal chord set of a function if and only if the complement of the set is a topologically open set that is additively closed. What Hopf's result means for the athletics scenario is as follows. Given the position function of a race of length L (like the one in the diagram), consider the set S consisting of the lengths of all the distance subintervals of the race that were run at exactly the overall average pace; one of these numbers will be L itself. Hopf's theorem then shows that S is a topologically closed set, meaning that it has an open complement. The theorem also shows that the sum of two lengths that are not in S is also not in S. This forces all the numbers L/n to lie in S whenever n is a positive integer, which is the statement of Theorem 1.

Relevant links

Historical mathematicians mentioned in this post:é-Marie_Ampèreévy_(mathematician)

Intermediate value theorem:

Photo credit unknown. I found it via Google Images, whose best guess is 5k runners.

The Lonely Runner Conjecture is a problem that superficially sounds like the one in this post, but in fact it is much harder. Here are two posts by me about it:

#mathematics #sciencesunday #spnetwork arXiv:1507.00871
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+Boris Borcic, the lonely runner conjecture, which I mentioned in the post, is a slightly related result about cyclic motion.
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The mathematics of Boggle logic puzzles

This picture shows a logic puzzle based on the popular word game Boggle. The object of the game is to place the fourteen letters shown at the bottom into the grid in such a way that the grid spells out each of the ten words in the list on the right. The words must be constructed from the letters of sequentially adjacent squares, where adjacent refers to squares that are horizontal, vertical or diagonal neighbours, and where squares may not be reused.

It turns out that Boggle logic puzzles have mathematically interesting aspects; for example, they are related to the subgraph isomorphism problem, which is an example of an NP-complete problem. The recent paper 10 Questions about Boggle Logic Puzzles by Jonathan Needleman ( gives a survey of what is known and proposes a number (ten!) of related problems.

The paper starts with what seems like an easy challenge: construct a filling of a 3 by 3 Boggle grid that contains each of the words ACT, APE, ATE, COP, END and OLD. It is clear that the nine letters in the grid must be ACDELNOPT, but the puzzle is a lot harder than I thought it would be, and this is partly because there are only six words in the list.

If B is a Boggle grid that has been filled in with letters, Needleman defines the set W(B) to be the set of all words in the board B that are at least two letters long. Here, the term words refers to strings of letters that may or may not be valid English words. The paper also makes the simplifying assumption that each letter appearing in the grid should appear only once, like in the grid in the picture, although the paper also discusses how this second assumption may be removed.

Two boards B and B' are said to be equivalent if they produce the same list of words. It can be shown that if two boards are equivalent, it must be the case that they differ from each other only by mirror reflection or by rotation by a multiple of a right angle. [Precise statement for mathematicians: the group of automorphisms of the adjacency graph of the Boggle grid is dihedral of order 8.]

The formal definition of a Boggle logic puzzle is as a list of words P satisfying two conditions: (1) P is a subset of W(B) for some board B, and (2) if P is a subset of W(B') for some other board B', then B and B' are equivalent. For example, the assertion that “ACT, APE, ATE, COP, END, OLD” is a Boggle logic puzzle for n=3 is the claim that (1) there exists a 3x3 Boggle board that spells out all of these six words and (2) up to symmetry, there is no other Boggle board with this property. The puzzle in the picture has two letters already filled in. As well as making the puzzle easier, this also provides enough information to break the symmetry of the solution and give a unique answer.

Two of the main themes studied in the paper are minimal solutions and maximal non-solutions of Boggle logic puzzles. Theorem 3.1 says something about the minimal solutions of a 3x3 grid. What it proves is that a puzzle consisting only of three-letter words, on a board with no repeated letters, must contain at least six words. Recall the “ACT, APE, ATE, COP, END, OLD” puzzle from earlier, which contains six three-letter words and no repeated letters. The theorem then says that any list of only five three-letter words cannot possibly lead to a unique solution. One of the open questions in the paper concerns 3x3 grids with no repeated letters whose puzzles contain only two-letter words. It can be proved that one would need at least 11 two-letter words to achieve a unique solution, but it is not known if this bound is sharp, because the smallest known example of such a puzzle contains 12 two-letter words.

The maximal non-solutions arise from lists of words that lead to a puzzle that is as inefficiently long as possible. To see what this means, suppose that you have played a game of Boggle and produced a long list of words. Is it possible to reconstruct the board (up to symmetry) using only the words you have written down? How long does your list of words have to be before this is inevitable? Even on a 3x3 grid, the number turns out to be surprisingly large: one needs 137 distinct three-letter words (out of a possible total of 160) to guarantee that the puzzle can be reconstructed uniquely. For four-letter words, the number is 377 words (out of a possible total of 496).

Relevant links

The game Boggle was designed by Allan Turoff. It was originally manufactured by Parker Brothers, and is now manufactured by Hasbro. More information on the game is here:

Image source:

Details of the subgraph isomorphism problem:

Boggle logic puzzles are reminiscent of the game Sudoku. In 2012, Gary McGuire, Bastian Tugemann and Gilles Civario proved that the smallest possible number of clues on a standard Sudoku board that can lead to a unique solution is 17. +Richard Elwes has included this result in a recent blog post about his personal top 10 mathematical achievements of the last 5ish years. You can find this post at (I've abbreviated the URL because it is rather long.)

#mathematics #scienceeveryday #spnetwork arXiv:1506.04173
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I have some more time at the moment so I'm hoping to write a few more substantive mathematical posts. Here's the next one in my Mathematics collection:
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Christopher Wren and John Wallis

During my visit to Cambridge last month, I took this picture of Emmanuel College, where I stayed for one night. The building shown here is the chapel, which was designed by the English architect Sir Christopher Wren (1632–1723).

Wren designed many other buildings in England that are still in good condition. One of these is the Sheldonian Theatre in Oxford, which is remarkable for having a large flat roof. The roof of the theatre is supported Wren considered supporting the roof of the theatre by an ingenious network of wooden beams designed by the mathematician John Wallis (1616–1703), who was an alumnus of Emmanuel College, Cambridge. Each beam in the network is supported at each end, either by another beam or by the side of the building, and each beam in turn supports two others. 

Wallis was a calculating genius, and one night calculated the square root of a 53 digit number in his head. In order to design the network of beams to support the flat roof, Wallis had to solve a 25 by 25 system of linear equations by hand, some of which involved ratios such as 3088694/340167. The result was a structurally sound pattern of beams that stays intact without the use of glue or screws at the joints.

The Wikipedia page on Christopher Wren ( has many more pictures of buildings he designed. These include the library at Trinity College, Cambridge, which I visited with my host +Timothy Gowers, but which has a clear no-photography policy. The library has an original Winnie-the-Pooh manuscript on display, because the author A.A. Milne and his son, Christopher Robin Milne, both studied at Trinity College, where they were both mathematicians.

Other relevant links

The Sheldonian Theatre:

Wikipedia on John Wallis:

An article on Wallis's design of the Sheldonian Theatre roof, showing the supporting pattern of beams:

Emmanuel College, Cambridge:,_Cambridge

#mathematics #architecture  
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Very informative and very amusing!...Loved the comments especially those of Sean Walker...As I live in Germany I can only confirm his astute observations and not at all offensive!...Do so miss the British sense of humour!!..Have you two ever thought of becoming a comedy duo?lol...Seriously,I'm an Actor/Singer and wannabe Director (if the Jobcentre stop being so difficult)!
I think you would be a hit...Very Pythonesque...Did I get that right?...Hardly ever speak English anymore!...Just discovered you and will "follow" you (both) from now on!!
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For #floralfriday, here's a Camellia flower from my mother's garden in Southampton (UK).

More information on Camellias can be found here:
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Thanks, +Sean Walker! The problem is that I am intensely busy at the moment, and this isn't going to get any better until mid-August. As a result, there are hardly any posts to see.
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A Curious Property of 82000

The number 82000 in base 10 is equal to 10100000001010000 in base 2, 11011111001 in base 3, 110001100 in base 4, and 10111000 in base 5. It is the smallest integer bigger than 1 whose expressions in bases 2, 3, 4, and 5 all consist entirely of zeros and ones.

What is remarkable about this property is how much the situation changes if we alter the question slightly. The smallest number bigger than 1 whose base 2, 3, and 4 representations consist of zeros and ones is 4. If we ask the same question for bases up to 3, the answer is 3, and for bases up to 2, the answer is 2. The question does not make sense for base 1, which is what leads to the sequence in the picture: [undefined], 2, 3, 4, 82000.

The graphic comes from a blog post by Thomas Oléron Evans. Most of the post discusses the intriguing problem of finding the next term in this sequence, and whether the next term even exists. In other words, does there exist an integer greater than 1 whose representations in bases 2, 3, 4, 5, and 6 all consist entirely of zeros and ones? 

The number 82000 does not satisfy these conditions, because the representation of this number in base 6 is 1431344. This means that the next number in the sequence, if it exists, must be some number bigger than 82000 whose representations in bases 2, 3, 4, and 5 all consist entirely of zeros and ones. Unfortunately, even these weaker conditions are very difficult to satisfy. An exhaustive search has been carried out up to 3125 digits in base 5 and no solution exists in this range. 

The upshot of this is that, if the next term in the sequence exists, it must have more than 2184 digits in base 10. (The 2184 comes from multiplying 3125 by the base 10 logarithm of 5.) However, there is also no known proof that the next term in the sequence does not exist.

Relevant links

Thomas Oléron Evans's blog post has much more discussion of this problem, at

Details of the exhaustive search can be found in the notes to the sequence in the On-Line Encyclopedia of Integer Sequences.

There is a nice online number base converter tool at

#mathematics #sciencesunday  
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10000000....(82000) times zero onwords
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I took this picture of an apple blossom with my phone during my morning dog walk last Monday. It has been unusually wet here recently, as you might guess from the picture.


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i share only with myself certain posts. now that i know what is a collection, i may group them in a private collection.
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Why your friends, on average, have more friends than you do

The friendship paradox is the observation that your friends, on average, have more friends than you do. This phenomenon, which was first observed by the sociologist Scott L. Feld in 1991, is mathematically provable, even though it contradicts most people's intuition that they have more friends than their friends do.

Wikipedia gives a nice intuitive explanation for this phenomenon: People with more friends are more likely to be your friend in the first place; that is, they have a higher propensity to make friends in the first place. However, it is also possible to explain the phenomenon using graph theory and mathematical statistics. I give an outline of the mathematical proof at the end of this post for those who are interested, but the upshot is that if we look at everybody's numbers of friends, and these numbers have a mean of μ and a variance of σ^2, then the average number of friends that an average friend has is μ + (σ^2/μ), which will be greater than μ assuming that someone has at least one friend and that not everyone has the same number of friends.

The recent paper The Majority Illusion in Social Networks ( by Kristina Lerman, Xiaoran Yan, and Xin-Zeng Wu explores some phenomena that are related to the friendship paradox. The authors explain how, under certain conditions, the structure of a social network can make it appear to an individual that certain types of behaviour are far more common than they actually are.

The diagram here, which comes from the paper, illustrates this point. It shows two social networks, (a) and (b), each containing 14 individuals. In each case, three vertices are marked in red; let's suppose that these correspond to the heavy drinkers in the group. In social network (a), the heavy drinkers are three of the most popular people, and the configuration of the network means that each of the other eleven individuals observes that at least half of their friends are heavy drinkers. This will lead these eleven people to think that heavy drinking is common in their society, when in fact it is not: only 3/14 of the group are heavy drinkers. In social network (b), there are the same number of heavy drinkers, but they are not particularly popular, and nobody in the group will have heavy drinkers as most of their friends. 

Network (a) is experiencing what the authors call the majority illusion, whereas network (b) is not. The illusion will tend to occur when the behaviour in question is correlated with having many friends. The paper shows that the illusion is likely to be more prevalent in disassortative networks, which means networks in which people have less tendency to be friends with people like themselves. Observe that in network (b), many of the non-heavy drinkers are friends with each other, whereas in network (a), they are not. This suggests that network (a) is more disassortative, and thus more susceptible to the majority illusion.

The authors also study the phenomenon using three real-world data sets: (a) the coauthorship network of high energy physicists (HepTh), (b) the social network Digg, studying only the mutual-following links, and (c) the network representing the links between political blogs. It turns out that networks (a) and (b) are assortative, and (c) is disassortative. They also look at the the case of Erdős–Rényi-type networks, which can be thought of as random. 

As the paper points out, the friendship paradox has real life applications. For example, if one is monitoring a contagious outbreak, it is more efficient to monitor random friends of random people than it is to monitor random people. This is because the friends are more likely to be better connected, are more likely to get sick earlier, and are more likely to infect more people once sick. The reason for this has to do with the fact that these attributes are positively correlated with having many friends. 

If you're wondering why your coauthors are on average cited more often than you are, or why your sexual partners on average have had more sexual partners than you have, now you know. I found out about this paper via my Facebook friend +Paul Mitchener, who has more Facebook friends than I do.

Relevant links

Wikipedia's page on the Friendship Paradox contains much of the information in this post, including a sketch of the proof given below:

Wikipedia on assortativity:

Here's another post by me about the mathematics of social networks, in which I explain why it is impossible for everyone on Google+ to have more than 5000 followers. It provoked a surprisingly hostile reaction:

A post by me about Erdős and Rényi's construction of the random graph:

Appendix: Mathematical proof of the friendship paradox

Assume for simplicity that friendship is a symmetric relation: in other words, that whenever A is a friend of B, then B is also a friend of A. We can then model a friendship network with an undirected graph G, with a set of vertices V and a set of edges E. Each vertex v in V represents an individual, and each edge e in E connects a pair of individuals who are friends. For each vertex v in V, the number d(v) (the degree of v) is the number of edges connected to v; in other words, the number of friends v has. 

The average number of friends of everyone in the network is then given by summing d(v) over all vertices v of V, and then dividing by |V|, the total number of people. Using basic graph theory, this number, μ, can be shown to be equal to 2 times |E| divided by |V|, where |E| is the number of edges.

In order to find the number of friends that a typical friend has, one first chooses a random edge of E (which represents a pair of friends) and one of the two endpoints of E (representing one of the pair of friends); the degree of this latter vertex is the number of friends that a friend has.  Summing these degrees over all possible choices amounts to summing d(v)^2 over all possible vertices, and since the number of choices is 2 times |E|, it follows that the average number of friends a friend has is the sum of d(v)^2 divided by 2 times |E|. Using the formula for μ above, it follows that μ times the average number of friends that a friend has is equal to the average value of d(v)^2. However, the average value of d(v)^2 is also equal, by basic mathematical statistics, to the sum of the square of the mean of the d(v) plus the variance of the d(v). The result follows from this.

#mathematics #scienceeveryday #spnetwork arXiv:1506.03022
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For #floralfriday, here are some azalea flowers from my mother's garden in Southampton (UK). 

More information about azaleas can be found at Google Images suggests Iggy Azalea as a related search term, but I'm pretty sure they aren't related, at least not closely.
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The Gamma Function and Fractal Factorials!

This fractal image by Thomas Oléron Evans was created by using iterations of the Gamma function, which is a continuous version of the factorial function.

If n is a positive integer, the factorial of n, n!, is defined to be the product of all the integers from 1 up to n; for example, 4!=1x2x3x4=24. It is clear from the definition that (n+1)! is the product of n+1 and n!, but it is not immediately clear what the “right” way is to extend the factorial function to non-integer values.

If t is a complex number with a positive real part, the Gamma function Γ(t) is defined by integrating the function x^{t–1}e^{–x} from x=0 to infinity. It is a straightforward exercise using integration by parts and mathematical induction to prove that if n is a positive integer, then Γ(n) is equal to (n–1)!, the factorial of (n–1). Since Γ(1)=1, this gives a justification (there are many others) that the factorial of zero is 1.

Using a technique called analytic continuation, the Gamma function can then be extended to all complex numbers except negative integers and zero. The resulting function, Γ(t), is infinitely differentiable, except at the nonpositive integers, where it has simple poles; the latter are the same kind of singularity that the function f(x)=1/x has at x=0. A particularly nice property of the Gamma function is that it satisfies Γ(t+1)=tΓ(t), which extends the recursive property n!=n(n–1)! satisfied by factorials. It is therefore natural to define the factorial of a complex number z by z!=Γ(z+1).

At first, it may not seem very likely that iterating the complex factorial could produce interesting fractals. If n is an integer that is at least 3, then taking repeated factorials of n will produce a sequence that tends to infinity very quickly. However, if one starts with certain complex numbers, such as 1–i, repeated applications of the complex factorial behave very differently. It turns out that (1–i)! is approximately 0.653–0.343i, and taking factorials five times, we find that (1–i)!!!!! is approximately 0.991–0.003i. This suggests that iterated factorials of 1–i  may produce a sequence that converges to 1.

It turns out that if one takes repeated factorials of almost any complex number, we either obtain a sequence that converges to 1 (as in the case of 1–i) or a sequence that diverges to infinity (as in the case of 3). However, it is not possible to take factorials of negative integers, and there are some rare numbers, like z=2, that are solutions of z!=z and do not exhibit either type of behaviour.

By plotting the points that diverge to infinity in one colour, and the points that converge to 1 in a different colour, fractal patterns emerge. The image shown here uses an ad hoc method of colouring points to indicate the rate of convergence or divergence. The points that converge to 1 are coloured from red (fast convergence) to yellow (slow convergence), and the points that diverge to infinity are coloured from green (slow divergence) to blue (fast divergence)

Relevant links

Thomas Oléron Evans discusses these fractals in detail in a blog post ( which contains this image and many others. He (and I) would be interested in knowing if these fractals have been studied before.

The applications of the Gamma function in mathematics are extensive. Wikipedia has much more information about the function here:

This post appears in my Mathematics collection at

#mathematics #sciencesunday  

Various recent posts by me
Camellia flower:
Horse chestnut tree:
A Curious Property of 82000:
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For #treetuesday, here's a picture I took during my recent trip to Cambridge (UK). My host, +Timothy Gowers, identified this tree as a horse chestnut. In the autumn, the horse chestnut sheds its distinctive large nut-like seeds, which are called conkers.

Just behind the tree is King's College, Cambridge, where the pioneering computer scientist Alan Turing was an undergraduate in the early 1930s.

Relevant links
The horse chestnut (Aesculus hippocastanum):

King's College, Cambridge:'s_College,_Cambridge

Alan Turing:
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For #caturday, here's my sister's British Blue Shorthair cat, Aslan. You can find out more about the breed here:
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Wow - how many plusses?! I can only imagine how much preparation your clever maths posts take and this photo of a gorgeous tubby cat goes bananas. Clever name, he looks just like a lion +Richard Green
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Penrose Land Cover by Daniel P. Huffman

This land cover map of the continental United States was produced by Daniel P. Huffman using Penrose tiles.

A more traditional way to do this would be to use the technique of hexagonal binning, which achieves a similar result by using hexagonal cells, as in a honeycomb. It is possible to tile the entire plane using either identical hexagonal tiles or Penrose tiles. A key difference between the two is that the hexagons will produce a tiling with full translational symmetry, whereas the Penrose tiles will not.

Daniel Huffman recently remarked that “Penrose tilings are the new hexbins”. You can find this map, and some others of the same type, on Huffman's Twitter page:

Wikipedia seems not to have a good description of hexbins, but cartographer Zachary Forest Johnson wrote a nice blog post about them a few years ago, which you can find here:

Wikipedia has more on Penrose tilings here:

I have posted about Penrose tilings and related tilings several times, for example here:

#mathematics #cartography #sciencesunday

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professor of mathematics
  • University of Colorado Boulder
    Professor of Mathematics, present
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  • Oxford University
    Postdoctoral Research Assistant, 1995 - 1997
  • Lancaster University
    Lecturer in Pure Mathematics, 1997 - 2003
  • Colorado State University
    Visiting Assistant Professor of Mathematics, 2001 - 2001
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Longmont, Colorado
Winnipeg, Manitoba, Canada - Southampton, England - Middlesbrough, England - Nairobi, Kenya - Zomba, Malawi - Manzini, Swaziland - Oxford, England - Newbury, England - Leamington Spa, England - Coventry, England - Lancaster, England - Fort Collins, Colorado, USA
mathematician and father of twins
I'm a professor of mathematics at the University of Colorado Boulder, interested mainly in algebra and combinatorics. 

Some of the best posts in my Google+ stream are those with the #mathematics tag.  My aim in these is to explain both new and historically interesting mathematical ideas to a general audience, and to produce posts that can be enjoyed on various levels.  My posts with the most reshares are often posts of this type.

Some of my non-mathematical posts are commentary on other people's photography and works of art, or various photos I took.

I am unlikely to give +1s to the following material:

(1) extreme sports;

(2) pictures of spiders;

(3) pornography;

(4) posts that can't spell “its”.

My main aim on Google+ is to post about difficult topics and still get decent amounts of engagement. I try to keep the quality of my posts as high as I can, and I'm not interested in trying to get vast numbers of +1s. Sometimes I post pictures of cats and dogs, but they're usually photographs I took. I don't always post gifs, but when I do, they're often gifs I created, either from scratch or from someone else's video. I don't usually post poorly executed or out of focus pictures, however amusing they are. Whatever I post, I try to include interesting commentary on it, and I'm careful to try to assign credit. I never want someone to feel that I stole their work with my post.

In the 90s, I wrote a Multi-User Dungeon called "Island."  I was once described by Stephen Fry as "sick".

Some other topics I may post about on occasion include the Beatles, British comedy, chess, linguistics, music theory, science in general, Star Trek, and Tolkien.
Bragging rights
I'm the author of "Combinatorics of Minuscule Representations", a book published by Cambridge University Press. I have graduated six PhD students.
  • University College, Oxford
    BA (MA), mathematics, 1989 - 1992
  • University of Warwick
    MSc/PhD, mathematics, 1992 - 1995
  • St Bernadette's RC Primary School, Middlesbrough
  • Kestrel Manor School, Nairobi
  • Sir Harry Johnston Primary School, Zomba
Basic Information
February 10
Other names
R.M. Green