Mathematicians calculate the true protagonist in ‘Game of Thrones’

Turns out that analyzing terrorist networks isnt the only practical use for crunching numbers on network science.

At Macalester College in St. Paul, Minnesota, associate professor of mathematics Andrew J. Beveridge and student Jie Shan have used math to solve one of lifes truly puzzling mysteries: Who is the real main character in the HBO series Game of Thrones?

Screengrab from

In their research, appropriately titled Network of Thrones,Beveridge and Shan found that the answer lies somewhere between Tyrion Lannister, the possibly dead Jon Snow, and Sansa Starkshocking.

Though the answer isnt completely revolutionary, the math behind it is impressive.

Basing their literary cast off the third Game of Thrones novel A Storm of Swords, the mathematicians calculated their findings by using network science, a branch of applied graph theory that brings together traditions from many disciplinesincluding sociology, economics, physics, computer science, and mathematics.

Through network science, Beveridge and Shan determined each characters clout with a series of factors including the number of connections they have, interactions between those connections, and the level of importance their connected characters have.

This is a fanciful application of network science, Beveridge told digital news outlet Quartz. But its the kind of accessible application that shows what mathematics is all about, which is finding and explaining patterns.

Whats next for the future of network science? Perhaps it could be determining the most powerful Kardashian family member, or even discovering the most influential parody account on Twitter.

Max Fleishman

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How Long Would It Take You To Fall Through Earth?

Just like when you were a kid and you were digging on the beach, imagine if there was a tunnel directly through Earth. If you jumped in, how long would it take to reach the other side?

Life Noggin has created this 3-minute animation showing all the mathematics and physics you need to answer this question that has probably bothered you since you were four.

To make it a bit easier, lets assume the same density all throughout, the world is perfectly spherical, and the hole has no air in it. Even then, theres that annoying little fundamental force known as gravity. When we start our fall, the mass of Earth pulls us towards its center, but what happens when we reach the center and go beyond it?

Check out the video for all the answers.

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I Still Miss You, And Thats Okay

When I lost you, I knew it was going to take me awhile until I completely forget you. But I didnt know it would take me this long. It took me six months worth of coffee in the morning and six months worth of tearful nights. Six months of wondering where it all went wrong; asking myself how could I have missed the target; losing my mind thinking was I ever good enough? Will I ever be?

Every day, when I pick up my guitar, all I can hear is the melody of your voice singing me to sleep at three oclock in the morning when my anxiety wont let me rest. Every day, when I pick up a mug of coffee, all I can picture is your lips resting on white ceramic with a tiny smirk as you savor the black coffee and tell me how you like coffee better than tea.

Every day, all I think about are the littlest things you used to do and Im reminded how theyre not just little things to me.

Every day I remember you, I think of you, I hear you, and I see you; and each day after every day, I still miss you. Even to this day even after ten months of that rainy night I told you I was done.So I guess Im not completely done after all. Not even three months ago when I met him.

Three months ago, thats after those six months of anguish, and crying myself to sleep, right? Youve always been the one better in Mathematics and Science. You probably calculated that we wouldnt last long and knew we didnt have the Chemistry.My apologies, I was the artsy one. All I can see then was the galaxy in your eyes and the pigment of the paints we used to color every moment we had each other.

Youll always be the better one in Math, the brighter one in Science. But when I met him, it didnt matter if you were the better one at being calculative and logical. Because I realized he was the better one between the two of you.

On the seventh month, I realized I still miss you.

Not because I still felt the same way. I missed you still because I craved the idea of telling you what happened to me each day. I missed you still because you were always there to listen to me. I missed you still because you calmed me down. I missed you still because you were my best friend.

The eight month was spent comparing you to him not him to you. Even during bad days when I see him, I picture you beside him and compare your outfit, your posture, your hand movements, your smile, your eyes, your nose, and your lips. You had manly in your outfit, he had comfort in his. You had handsome in your smile, he had charm in his. You had strength written all over your actions, he had safety in his. You had the galaxy in your eyes, he had the trees, beaches, birds in his. You had perfect written all over you from head to toe, but from head to toe he was perfect.

And on the ninth month, I realized I didnt need the universe, I only needed him to be my universe.

On the tenth month, I was aware that I still missed you and I knew that was okay. Because he taught me that you were a sweet memory. The same way he taught me how to love again.

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Coding on tape – computer science A-level 1970s style – BBC News

Image copyright The National Museum of Computing
Image caption Few schools had their own computers in the early 1970s – but by the end of the decade many began to invest in models such as the Elliott 903

The chances of walking into a UK school without any computer equipment in today’s world are practically zero.

But 45 years ago things were very different.

Staff at the AQA exam board have been trawling their archive and have found some early 1970s computer science papers, which highlight how much the subject has changed over the intervening period.

Back then, schools offering computer science A-level had to prove they had access to a computer, according to a syllabus from the time.

In 1970 computers were a rarity and pupils would have to visit machines in nearby universities or businesses, said AQA’s computing qualifications manager Steven Kenny, because the cost of a school owning one was “prohibitive”.

“They cost tens of thousands of pounds and were filing cabinet sized,” he explained.

Paper tape

Students would write programmes in longhand and if they were lucky, their school would have the facilities to punch the code on to paper tapes.

Otherwise the longhand notes would have to be sent off to be converted on to tape.

The students would then take the code with them when they visited the computer.

“They would spool the paper on to the computer and the code would appear on the screen if they had got it right. If it was wrong they would get an error message,” said Mr Kenny.

This meant they would have to take it back to school, work out was wrong and correct it, he explained.

“It was often the same even at universities in those days.”

Image copyright Picasa
Image caption Computer code was punched on to paper tape and then loaded on to the computer to be read
Image copyright AQA
Image caption Schools had to prove they had access to a computer
Image copyright AQA
Image caption Parts of the paper seem quaint by today’s standards

Kevin Murrell, co-founder of The National Museum of Computing, said even a comma or a full stop out of place could mean a programme would fail once it was put into the computer.

He described himself as “mortified” by the difficulty of the mathematics required by the 1971 A-level.

When he showed the paper to his staff, they “went white as a sheet”, he told the BBC.

The paper required candidates to break everything down into binary or “machine code”, which is “very hard work and not something that anyone would be asked to do today” he said.

“We write in languages that are akin to English which are converted into binary by the machine. For 99.9% of programmers, maths isn’t an issue at all because modern computers do so much more.”

In the 1970s machines were slow and students’ access was so limited they had to be very precise about the programmes they wrote.

Today’s programmers can afford to be less precise as “the sheer brute force of modern computers gets you through”.

Mr Murrell did O-level computer science in 1976 using a keyboard and printer connected by phone to an Open University computer.

“You would key in your programme and it would run remotely at the OU. I can’t tell you how exciting that was!”

Another generation

The price of computers gradually dropped with a Research Machines computer costing around 3,000 by the mid 1970s and a Commodore PET around 1,000 after that.

Another generation of coders learned their craft on machines like the Sinclair Spectrum and the BBC Micro which appeared in the early 1980s.

Mr Kenny remembers the arrival of a computer at his inner city comprehensive school in 1979.

“Pupils had very limited access – only a few were allowed to use it.”

All this had an effect on the exam questions.

“Because people didn’t have access to computers, a lot of it had to be focused on theory.

“There was a big focus on logical and computational thinking, using flow charts and algorithms to break instructions down into the smallest logical steps for a computer to follow.

“The difference now is that there is a lot more creativity involved: for example designing apps for use in daily life.”

However, one surprise was how many of the questions are still relevant, said Mr Kenny.

“About half the questions could probably be put into an A-level paper today with a bit more context.”

The big differences were in questions focused on the practical use of computers – some of which now seem quaint.

For example candidates were asked to “Write an essay on the use of computers in one of the following (a) airlines (b) banking (c) local government or (d) wholesale trade”.

Another question required them to prepare a flow chart to record 12,000 ticket sales between four destinations on the same railway line over a 30-day period.

“We wouldn’t phrase this the same way or use the same example but some of the same processes would be needed by programmers today,” said Mr Murrell.

Mr Kenny suggested today’s equivalent question would be about “the internet of things”, where, for example, a computerised fridge can detect when groceries are running low and order a delivery without the householder being involved.


By the early 1980s people began to buy home computers and Mr Kenny admitted his younger self used to wonder why people would use them.

Image copyright Picasa
Image caption A 1980s BBC Micro in use at The National Museum of Computing

“They would have been game-players or enthusiastic coders – much more geeky than now,” he suggests before adding that some people might have used them for “numbers and data”, particularly home finances.

He adds: “It’s easy for us to answer that question today because computers are everywhere. You have to remember that in 1971 not many people had even seen a computer, let alone used one.”

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Mathmetician solves puzzle of parking lots


A Mathematician has discovered a simple trick which could revolutionize car parking.

Professor David Percy, of Salford University, was frustrated by the poor geometry of car parks, which generally feature parking spaces arranged at a 90 degree angle to the access lane.

He performed some simple math and found out that one tweak could make it much easier for motorists to maneuver their motor into the space whilst simultaneously allowing more automobiles to fit into a car park.

All planners need to do is place the bays at a 45 degree angle, which cuts down the turning circle required and therefore needs a smaller access lane, freeing up more space for parking.

Professor Percys eureka moment came when his universitys car park was repainted, sticking to exactly the same layout.

This traditional conformity set me thinking, he wrote in Mathematics Today.

To understand his plan, you need to look at the space required to fit a car into a bay thats angled at 90 degrees.

Maneuvering a car into traditional bays requires a wide access road, whilst anyone the 45 degree bays will be able to swing in their car without needing as much space.

Instead of rectangular parking bays I figured if they are diagonal you might save space and fit more cars into a car park, he added.

If his recommendations are implemented in a car park which can currently fit 500 cars, another 119 should be able to fit into the same space.

For a 45 degree bay angle it was a 23 percent saving, he continued,

For 36 degrees it was 34 percent, but the difference is marginal and its easier to draw lines at 45 degrees.

His plans only work in larger car parks, because the useless space at the edge of bays negates the effects of the space-saving lanes.

More auto news from Sun Motors

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See If You Can Solve These 6 Fiendish Brain-Teasers Written By NSA Employees

Sure, between thespying on millions of Americans and thesitting on critical security vulnerabilities for years, the National Security Agency (NSA) has a pretty bad reputation.

But have you heard about itsbrain-teasers?

The US spy agencyemploys some of America’s best and brightest as spooks and code-breakers, so it’s no surprise that its employees might have an interest in riddles and puzzles.

Every month, the NSA publishes on its website a brain-teaser written by an employee that members of the public can try their hand at.

One month it’s a maths challenge created by an applied research mathematician; the next it’s a logic puzzle by a systems engineer. They’re all published in what the NSA calls its “Puzzle Periodical.”

“Intelligence. It’s the ability to think abstractly. Challenge the unknown. Solve the impossible. NSA employees work on some of the world’s most demanding and exhilarating high-tech engineering challenges. Applying complex algorithms and expressing difficult cryptographic problems in terms of mathematics is part of the work NSA employees do every day,” the NSA says on its website.

We’ve rounded up a sixof the most interesting brain-teasers below. So take a read, and see if you can out-smart the NSA’s most fiendishriddlers!

1. Here’s a relatively easy one to start off with, from July 2016:

Submitted by Sean A., NSA Applied Mathematician

content-1471883922-heres-a-relatively-eaOn a rainy summer day, brothers Dylan and Austin spend the day playing games and competing for prizes as their grandfather watches nearby. After winning two chess matches, three straight hands of poker and five rounds of ping-pong, Austin decides to challenge his brother, Dylan, to a final winner-take-all competition. Dylan clears the kitchen table and Austin grabs an old coffee can of quarters that their dad keeps on the counter.

The game seems simple as explained by Austin. The brothers take turns placing a quarter flatly on the top of the square kitchen table. Whoever is the first one to not find a space on his turn loses. The loser has to give his brother tonights dessert. Right before the game begins, Austin arrogantly asks Dylan, Do you want to go first or second?

Dylan turns to his grandfather for advice. The grandfather knows that Dylan is tired of losing every game to his brother. What does he whisper to Dylan?

Image credit:MoneyBlogNewz/Flickr (CC)

And here’s the solution:

Dylan should go first. By doing this, Dylan can guarantee a win by playing to a deliberate strategy. On his first turn, he can place a quarter right on the center of the table. Because the table is symmetric, whenever Austin places a quarter on the table, Dylan simply “mirrors” his brothers placement around the center quarter when it is his turn. For example, if Austin places a quarter near a corner of the table, Dylan can place one on the opposite corner. This strategy ensures that even when Austin finds an open space, so can Dylan. As a result, Dylan gains victory, since Austin will run out of free space first!

2. This one, from June 2016, requires a bit more math.

Submitted by Robert B., NSA Applied Mathematician

content-1471884144-this-one-from-june-20Following their latest trip, the 13 pirates of the ship, SIGINTIA, gather at their favorite tavern to discuss how to divvy up their plunder of gold coins. After much debate, Captain Code Breaker says, Argggg, it must be evenly distributed amongst all of us. Argggg. Hence, the captain begins to pass out the coins one by one as each pirate anxiously awaits her reward. However, when the captain gets close to the end of the pile, she realizes there are three extra coins.

After a brief silence, one of the pirates says, I deserve an extra coin because I loaded the ship while the rest of you slept. Another pirate states, Well, I should have an extra coin because I did all the cooking. Eventually, a brawl ensues over who should get the remaining three coins. The tavern keeper, annoyed by the chaos, kicks out a pirate who has broken a table and who is forced to return her coins. Then the tavern owner yells, Keep the peace or all of you must go!

The pirates return to their seats and the captain, left with only 12 total pirates, continues to distribute the coins – one for you, one for you. Now, as the pile is almost depleted, she realizes that there are five extra coins. Immediately, the pirates again argue over the five extra coins. The captain, fearing that they will be kicked out, grabs the angriest pirate and ushers her out of the tavern with no compensation. With only 11 pirates left, she resumes distribution. As the pile nears depletion, she sees that there wont be any extra coins. The captain breathes a sigh of relief. No arguments occur and everyone goes to bed in peace.

If there were less than 1,000 coins, how many did the pirates have to divvy up?

Image credit: Disney

The answer is 341.

There are actually infinite answers to the problem, but only one number if the answer is under 1,000. This puzzle is an example of modular arithmetic and the Chinese Remainder Theorem.

The smallest solution under 1,000 for this problem is 341 coins, and the answer is found by working backwards. To find it, we first note that with 11 pirates the coins divided evenly; hence, the number of coins is in the list:

11, 22, 33, 44, 55, 66, 77, 88, 99, 110, 121, 132, 143

What happens if we take these numbers and divide them among 12 pirates? How many coins would be left over? Well, we want 5 coins to be left over after dividing by 12. Hence, we reduce the list above to:

77, 209, 341, 473

These numbers divide by 11 evenly and have 5 left over when divided by 12. Now we take these remaining numbers and divide them by 13 until we find the number that gives 3 extra coins left over. Hence, 341 coins.

3. Here’s a logic puzzle from August 2015. You need to figure two things out.

Puzzle created by Roger B., Cryptanalytic Mathematician, NSA

content-1471884232-heres-a-logic-puzzle-Nadine is having a party, and she has invited three friends Aaron, Doug, and Maura. The three of them make the following statements on the days leading up to the party:

Two Days Before the Party:
Aaron: Doug is going to the party.
Doug: Maura is not going to the party.
Maura: Aaron will go to the party if, and only if, I do.

The Day Before the Party:
Aaron: Maura will go to the party if, and only if, I don’t go.
Doug: An even number out of the three of us are going to the party.
Maura: Aaron is going to the party.

The Day of the Party:
Aaron: It is not yet 2018.
Doug: Aaron will go to the party if, and only if, I do.
Maura: At least one of the three of us is not going to the party.

Nadine also knows that out of Aaron, Doug, and Maura:
One of them never lies.
A different one of them lies on days of the month that are divisible by 2, but is otherwise truthful.
The remaining one of them lies on days of the month that are divisible by 3, but is otherwise truthful.

(1) Can you figure out who is going to attend?
(2) Can you figure out on what day, month, and year the party will be held, assuming it takes place in the future?

Image credit:YouTube/blackboxberlin

Stuck? The answers are Doug and Aaron, and March 1, 2016. Here’s how you get there …

(1) Attendees solution:

The first step is to realize that the date rules imply that nobody may lie on two consecutive days.

If Maura’s third statement is false, then everyone is going to the party and Doug’s first two statements are lies, which is impossible. Therefore Maura’s third statement is true, and at least one person is not going to the party.

If Doug’s second statement is false, then (from the above) exactly one person is going to the party. This can’t happen without making Doug’s first or third statement also false, which would have Doug lying on two consecutive daysan impossibility. Therefore Doug’s second statement is true, and there are an even number of people going to the party.

If Doug is not going to the party, then Aaron’s first statement is false, so his second statement must be true. This would result in an odd number of people at the party, which we have shown is not the case. Therefore Doug is going to the party.

Since an even number of people are going to the party, only one of Aaron or Maura is going, making Maura’s first statement false. Therefore her second statement must be true, and Aaron is going to the party while Maura is not. The attendees are Doug and Aaron.

(2) Date solution:

Having established the attendees as Doug and Aaron, now we know that the only lies are Maura’s first statement and potentially Aaron’s third statement.

If the three days do not cross a month boundary, then either the second date or both the first and third dates would be divisible by 2, but nobody is available to lie with that pattern. Therefore either the second or third day is the first of a month.

If the day before the party was the first of a month, then the day of the party is the second of a month; Aaron would have to be the one who lies on dates divisible by 2. Then, for Maura but not Aaron to lie on the first day, it would have to be divisible by 3 but not 2. This is never true of the last day of a month. Therefore, the day before the party is not the first of a month, so the day of the party itself must be the first day of a month.

This makes the day before the party the last day of a month, and since nobody lies on that day it must not be divisible by 2.

This means two days before the party, the date is divisible by 2, so it must not also be divisible by 3 or there would be two liars on that day. The only way this can happen two days before the end of a month is when that day is February 28 of a leap year.

Since nobody lies on the first of a month, Aaron’s third statement is true and it is not yet 2018.

Finally, since the only leap year before 2018 is 2016, we conclude that the party is being held onMarch 1, 2016.

4. This one (April 2016) is simpler but not any easier.

Puzzle created by Andy F., Applied Research Mathematician, NSA

content-1471884342-this-one-april-2016-iMel has four weights. He weighs them two at a time in all possible pairs and finds that his pairs of weights total 6, 8, 10, 12, 14, and 16 pounds. How much do they each weigh individually?

Note: There is not one unique answer to this problem, but there is a finite number of solutions.

Image credit:Doug Pensinger/Getty Images

And the answer is …

There are exactly two possible answers: Mel’s weights can be 1, 5, 7, and 9 pounds, or they can be 2, 4, 6, and 10 pounds. No other combinations are possible.


Let the weights be a, b, c, and d, sorted such that a < b < c < d. We can chain inequalities to get a + b < a + c < a + d, b + c < b + d < c + d. Thus, a + b = 6, a + c = 8, b + d = 14, and c + d = 16. But we don’t know if a + d = 10 and b + c = 12 or the other way around. This is how we get two solutions. If a + d = 10, we get 1, 5, 7, and 9; if b + c = 10, we get 2, 4, 6, and 10.

More on the Problem

Where this problem really gets weird is that the number of solutions depends on the number of weights. For example, if Mel has three weights and knows the weight of all possible pairs, then there is only one possible solution for the individual weights. The same is true if he has five weights.

But now suppose that Mel has eight weights, and the sums of pairs are 8, 10, 12, 14, 16, 16, 18, 18, 20, 20, 22, 22, 24, 24, 24, 24, 26, 26, 28, 28, 30, 30, 32, 32, 34, 36, 38, and 40. Now what are the individual weights?

This time, there are three solutions:

  • 1, 7, 9, 11, 13, 15, 17, 23
  • 2, 6, 8, 10, 14, 16, 18, 22
  • 3, 5, 7, 11, 13, 17, 19, 21

5. October 2015 was another logic puzzler, with a school theme. Take a look:

Puzzle created by Ben E., Applied Research Mathematician, NSA

Kurt, a math professor, has to leave for a conference. At the airport, he realizes he forgot to find a substitute for the class he was teaching today! Before shutting his computer off for the flight, he sends an email: “Can one of you cover my class today? I’ll bake a pie for whomever can do it.” He sends the email to Julia, Michael, and Mary Ellen, his three closest friends in the math department, and boards the plane.

As Kurt is well-known for his delicious pies, Julia, Michael, and Mary Ellen are each eager to substitute for him. Julia, as department chair, knows which class Kurt had to teach, but she doesn’t know the time or building. Michael plays racquetball with Kurt so he knows what time Kurt teaches, but not the class or building. Mary Ellen helped Kurt secure a special projector for his class, so she knows what building Kurt’s class is in, but not the actual class or the time.

Julia, Michael, and Mary Ellen get together to figure out which class it is, and they all agree that the first person to figure out which class it is gets to teach it (and get Kurt’s pie). Unfortunately the college’s servers are down, so Julia brings a master list of all math classes taught that day. After crossing off each of their own classes, they are left with the following possibilities:

  • Calc 1 at 9 in North Hall
  • Calc 2 at noon in West Hall
  • Calc 1 at 3 in West Hall
  • Calc 1 at 10 in East Hall
  • Calc 2 at 10 in North Hall
  • Calc 1 at 10 in South Hall
  • Calc 1 at 10 in North Hall
  • Calc 2 at 11 in East Hall
  • Calc 3 at noon in West Hall
  • Calc 2 at noon in South Hall

After looking the list over, Julia says, “Does anyone know which class it is?” Michael and Mary Ellen immediately respond, “Well, you don’t.” Julia asks, “Do you?” Michael and Mary Ellen both shake their heads. Julia then smiles and says, “I do now. I hope he bakes me a chocolate peanut butter pie.”

Which class does Kurt need a substitute for?

The answer is Calc 2 at 10 in North Hall. But why?

1) Since Julia only knows the class name, the only way she could immediately know is if it was Calc 3. Since Michael and Mary Ellen both know that Julia doesn’t know, that means they know the class isn’t Calc 3. Since Michael only knows the time, that means the class can’t be at noon. Because Mary Ellen only knows the building, that means the building can’t be West Hall. That leaves only the following possibilities:

  • Calc 1 at 9 in North Hall
  • Calc 1 at 10 in East Hall
  • Calc 2 at 10 in North Hall
  • Calc 1 at 10 in South Hall
  • Calc 1 at 10 in North Hall
  • Calc 2 at 11 in East Hall

2) Since Michael doesn’t know which class it is, that means the time can’t be 9 or 11. Since Mary Ellen doesn’t know either, the class can’t be in South Hall. That leaves only three possibilities:

  • Calc 1 at 10 in East Hall
  • Calc 2 at 10 in North Hall
  • Calc 1 at 10 in North Hall

3) At this point, Julia now says she knows the answer. Since there are two Calc 1 classes, it must be that Julia knows the class is Calc 2. Thus, the class is Calc 2 at 10 in North Hall.

6. Last one: Can you figure out Charlie’s birthday? (July 2015)

Puzzle created by Stephen C., Applied Research Mathematician, NSA

After observing Albert and Bernard determine Cheryl’s birthday, Charlie decides he wants to play. He presents a list of 14 possible dates for his birthday to Albert, Bernard and Cheryl.

  • Apr 14, 1999
  • Feb 19, 2000
  • Mar 14, 2000
  • Mar 15, 2000
  • Apr 16, 2000
  • Apr 15, 2000
  • Feb 15, 2001
  • Mar 15, 2001
  • Apr 14, 2001
  • Apr 16, 2001
  • May 14, 2001
  • May 16, 2001
  • May 17, 2001
  • Feb 17, 2002

He then announces that he is going to tell Albert the month, Bernard the day, and Cheryl the year.

After he tells them, Albert says, “I don’t know Charlie’s birthday, but neither does Bernard.”

Bernard then says, “That is true, but Cheryl also does not know Charlie’s birthday.”

Cheryl says, “Yes and Albert still has not figured out Charlie’s birthday.”

Bernard then replies, “Well, now I know his birthday.”

At this point, Albert says, “Yes, we all know it now.”

What is Charlie’s birthday?

Image credit:Will Clayton/Flickr (CC)

The answer is Apr 16, 2000, and the rationale is …

When Albert claims that Bernard does not know Charlie’s birthday, he is saying that he knows that the correct day occurs more than once in the list. In other words, he is saying that Charlie’s birthday is not Feb 19, 2000, and the only way he could know that is that he knows that the month is not February.

So by making this claim Albert has reduced the list for everyone to:

  • Apr 14, 1999
  • Mar 14, 2000
  • Mar 15, 2000
  • Apr 16, 2000
  • Apr 15, 2000
  • Mar 15, 2001
  • Apr 14, 2001
  • Apr 16, 2001
  • May 14, 2001
  • May 16, 2001
  • May 17, 2001

When Bernard says that it is true that he does not know Charlie’s birthday, it tells everyone that even with the restricted list the correct day occurs more than once in the list. So everyone can thus eliminate May 17, 2001.

Furthermore, the claim that Cheryl also does not know Charlie’s birthday is a claim that Bernard knows that the year occurs more than once on the remaining list.

This rules out Apr 14, 1999, and since Bernard could only rule this out by knowing that the day was not 14, everyone can further reduce the list to:

  • Mar 15, 2000
  • Apr 16, 2000
  • Apr 15, 2000
  • Mar 15, 2001
  • Apr 16, 2001
  • May 16, 2001

When Cheryl says that Albert still has not figured out Charlie’s birthday, she is telling everyone that given the new list the month occurs more than once and thus rules out May 16, 2001. This tells everyone that Cheryl knows that the year is not 2001 and the list can be reduced to:

  • Mar 15, 2000
  • Apr 16, 2000
  • Apr 15, 2000

When Bernard then claims to know the date he is saying that the day occurs only once among the 3 remaining choices, thus telling everyone that Charlie’s birthday is Apr 16, 2000.

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‘Game Of Thrones’ Is All About This Character, According To Math

Who is the real star of “Game of Thrones?”

Set across the sprawling Seven Kingdoms and beyond, it can be tricky to pinpoint the books’ and subsequent hit TV show’s most important character.

But mathematicians from Macalester College in Saint Paul, Minnesota, think they’ve figured it out. And the answer is…

Yes. Tyrion Lannister — portrayed in the HBO show by Peter Dinklage — is reportedly the glue binding the entire fantasy epic together.

Macalester College’s associate professor of mathematics Andrew J. Beveridge and undergraduate student Jie Shan applied “network science” to the third installment (A Storm of Swords) of author George R.R. Martin’s epic “A Song Of Ice and Fire” novel series.

“We opted for this volume because the main narrative has matured, with the characters scattered geographically and enmeshed in their own social circles,” they wrote in their paper, published on the Mathematical Association of America’s website last week.

They analyzed the interconnection between all the characters, and linked them together every time they appeared within 15 words of one another. This diagram is the end result:

Tyrion appears to be the principal character. Jon Snow, portrayed by Kit Harington in the TV show, is close behind.

Perhaps surprisingly, Sansa Stark (played by Sophie Turner) is also in the running. “Other players are aware of her value as a Stark heir and they repeatedly use her as a pawn in their plays for power. If she can develop her cunning, then she can capitalize on her network importance to dramatic effect,” the researchers wrote.

Daenerys Targaryen, played by Emilia Clarke, was also deemed important — but not so much as the others, due to the way her character appears to be isolated.

“Game of Thrones” Season 6 premieres Sunday, April 24, on HBO.

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The Maths Behind Impossible” Never-Repeating Patterns

The Conversation

Remember the graph paper you used at school, the kind thats covered with tiny squares? Its the perfect illustration of what mathematicians call a periodic tiling of space, with shapes covering an entire area with no overlap or gap. If we moved the whole pattern by the length of a tile (translated it) or rotated it by 90 degrees, we will get the same pattern. Thats because in this case, the whole tiling has the same symmetry as a single tile. But imagine tiling a bathroom with pentagons instead of squares its impossible, because the pentagons wont fit together without leaving gaps or overlapping one another.

Patterns (made up of tiles) and crystals (made up of atoms or molecules) are typically periodic like a sheet of graph paper and have related symmetries. Among all possible arrangements, these regular arrangements are preferred in nature because they are associated with the least amount of energy required to assemble them. In fact weve only known that non-periodic tiling, which creates never-repeating patterns, can exist in crystals for a couple of decades. Now my colleagues and I have made a model that can help understand how this is expressed.

In the 1970s, physicist Roger Penrose discovered that it was possible to make a pattern from two different shapes with the angles and sides of a pentagon. This looks the same when rotated through 72-degree angles, meaning that if you turn it 360 degrees full circle, it looks the same from five different angles. We see that many small patches of patterns are repeated many times in this pattern. For example in the graphic below, the five-pointed orange star is repeated over and over again. But in each case these stars are surrounded by different shapes, which implies that the whole pattern never repeats in any direction. Therefore this graphic is an example of a pattern that has rotational symmetry but no translational symmetry.

image-20160811-11006-1xfss0s.gifPenrose tiling.PrzemekMajewski,CC BY-SA

Things get more complicated in three dimensions. In the 1980s, Dan Schechtman observed an aluminium-manganese alloy with a non-periodic pattern in all directions that still had rotational symmetry when rotated by the same 72-degree angle. Until then, crystals that had no translational symmetry but possessed rotational symmetry were in fact inconceivable and many scientists did not believe this result. In fact, this was one of those rare occasions when the definition of what is a crystal had to be altered because of a new discovery. In accordance, these crystals are now called quasicrystals.

Irrational number

The never-repeating pattern of a quasicrystal arises from the irrational number at the heart of its construction. In a regular pentagon, the ratio of the side length of the five-pointed star you can inscribe on the inside of a pentagon, to the side of the actual pentagon is the famous irrational number phi (not to be confused with pi), which is about 1.618. This number is also known as the golden ratio (and it also satisfies the relation phi = 1+1/phi). Consequently, when a quasicrystal is constructed with tiles that are derived from a pentagon like the ones Penrose used we observe rotational symmetry at 72-degree angles.


Quasicrystal lattice structure. Author provided

We see this five-fold symmetry both in the image of the quasicrystal as the ten radial lines around the central red dot (above), and also in the scale model of the central part of the quasicrystal made with Zometool (below). In the model, it helps to think of the white balls to be the locations where we would find the particles/atoms of the crystal structure and the red and yellow rods to indicate bonds between particles, that represent the shapes and symmetries of the structure.

In our recent publication, we identified the two traits that a system must have in order to form a 3D quasicrystal. The first is that patterns at two different sizes (length-scale) which are at an appropriate irrational ratio (like phi) both occur in the system. And second that these can influence each other strongly. In addition to the never-repeating quasicrystal patterns, this model can also form other observed regular crystal structures such as hexagons, body-centered cubes and so on. Such a model makes it possible to explore the competition between all these different patterns and to identify the conditions under which quasicrystals will be formed in nature.


The structure of a quasicrystal. Author provided

The mathematics behind how such never-repeating patterns are created is very useful in understanding how they are formed and even in designing them with specific properties. That is why we at the University of Leeds, along with colleagues at other institutions, are fascinated with research into such questions.

However, this research isnt just a conceptual mathematical idea (although the mathematics behind it is addictive) it has great promise for many practical applications, including making very efficient quasicrystal lasers. This is because, when periodic crystal patterns are used in a laser, a low-power laser beam is created by the symmetry of the repeating pattern. Having defects in the crystal pattern or alternatively using a never-repeating quasicrystal pattern at the output end of a laser, makes it possible to create an efficient laser beam with high peak output power. In other applications, some researchers are even considering the reflective finishes that quasicrystals might create if added to household paint.The Conversation

Priya Subramanian, Research Fellow Applied Mathematics, University of Leeds

This article was originally published on The Conversation. Read the original article.

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Can you solve it? Are you a puzzle Olympian?

Get those neurons out the blocks

Hello guzzlers,

Today, two puzzles inspired by the Olympics:

1) The 100m final

Early this morning Usain Bolt won gold in the 100m.

Lets assume that Bolt won the race by 10m from the second placed runner, Justin Gatlin. And now lets run the final again, but this time Bolt will start 10m behind the start line. If both Bolt and Gatlin run at the same constant speed in the second race as they did in the first, who wins?

2) The mystery games

(For the purposes of this puzzle, any resemblance to real persons, living or dead, and Olympic sports, actual or obsolete, is purely coincidental.)

Nafissatou, Jessica, and Brianne are taking part in an athletics competition that involves at least two events. In each event the winner gets G points, second placed gets S points and third place gets B points, where G, S and B are whole numbers, nonzero, and G> S>B. No event is tied. Nafissatou scores 22 points in total. Jessica and Brianne score 9 points each in total. Jessica wins the 100m hurdles. Who is second in the javelin? How many events are there?

Ill open comments at noon, and post the answers at 5pm BST. Please dont post the answer in the comments before 5pm because this spoils it for the many people who want to work out the answers themselves. Thanks!

I post a puzzle here on a Monday every two weeks. If you want to propose a puzzle for this column, please email me Id love to hear it.

Im the author of three popular maths books including Alexs Adventures in Numberland. My new book Football School: Where Football Explains the World is for children and out on Sept 1. You can check me out on Twitter, Facebook, Google+, my personal website or my Guardian maths blog.

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