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# Exercises - Fast Exponentiation and Fermat's Little Theorem

1. Find a number $a$ where $0 \le a \lt 73$ and $a \equiv 9^{794}\pmod{73}$

2. Solve $x^{86} \equiv 6 \pmod{29}$

3. Solve $x^{39} \equiv 3 \pmod{13}$

4. The statement $7^{1734250} \equiv 1660565\pmod{1734251}$ is true. Can you conclude 1734251 is a composite number?

5. Verify using fast exponentiation that the congruence $$129^{64026} \equiv 15179\pmod{64027}$$ is true. Can you conclude 64027 is a composite number?

6. Verify using fast exponentiation that the congruence $$2^{52632} \equiv 1\pmod{52633}$$ is true. Can you conclude 52633 is a prime number?

◆ ◆ ◆

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# Force upright Greek letters with isomath

The isomath package makes it possible to typeset upright Greek letters via \mathrm and \mathbf. Is it it possible to force all Greek letters to be typeset upright unless specified otherwise, e.g. to make it so that $$\Phi \Psi$$ delivers the same output as $$\mathrm{\Phi \Psi}$$? An ideal solution would be to load isomath with some option and if it optionally could make difference of upper and lower case Greek letters.

\documentclass{article}

\usepackage{isomath}

\begin{document}

$$\Phi \Psi$$% These are typeset in italic

$$\mathrm{\Phi \Psi}$$% These are typeset upright

\end{document}

• Just for reference, if XeLaTeX is an option, the package unicode-math provides an easy solution to this problem through the "math-style" option, documented at section 5 of the manual. Oct 15, 2011 at 22:00
• add \usepackage{upgreek}, then you can write \upPsi for example Jun 2, 2014 at 5:01
• @AndyBarbour - I found your solution to be the least impactive for what I was trying to solve. (meaning the best solution without other side-effects) Aug 10, 2019 at 19:12

You don't need the isomath package:

\DeclareMathSymbol{\Gamma}{\mathalpha}{operators}{0}
\DeclareMathSymbol{\Delta}{\mathalpha}{operators}{1}
\DeclareMathSymbol{\Theta}{\mathalpha}{operators}{2}
\DeclareMathSymbol{\Lambda}{\mathalpha}{operators}{3}
\DeclareMathSymbol{\Xi}{\mathalpha}{operators}{4}
\DeclareMathSymbol{\Pi}{\mathalpha}{operators}{5}
\DeclareMathSymbol{\Sigma}{\mathalpha}{operators}{6}
\DeclareMathSymbol{\Upsilon}{\mathalpha}{operators}{7}
\DeclareMathSymbol{\Phi}{\mathalpha}{operators}{8}
\DeclareMathSymbol{\Psi}{\mathalpha}{operators}{9}
\DeclareMathSymbol{\Omega}{\mathalpha}{operators}{10}

To get an italic Gamma, use \mathnormal{\Gamma}. Try

$\Gamma\mathnormal{\Gamma}\mathbf{\Gamma}$

The command \DeclareMathSymbol is used for assigning a meaning to characters or commands in math mode. The second argument contains the "kind" of symbol; \mathalpha means a symbol that obeys to the alphabet selection commands. The third argument is the alphabet used by default; the fourth is the slot in the font. All the standard declarations are in fontmath.ltx loaded at format creation time.

Notice that these declarations won't work if the default font encoding of the document is T1 instead of OT1, as operators refers to the ordinary Roman document font. One should define a new math alphabet, in this case.

How to do this when the document encoding is T1? One has to define a new symbol font:

\DeclareSymbolFont{otone}{OT1}{cmr}{m}{n}
\SetSymbolFont{bold}{otone}{OT1}{cmr}{m}{n}
\DeclareMathSymbol{\Gamma}{\mathalpha}{otone}{0}
\DeclareMathSymbol{\Delta}{\mathalpha}{otone}{1}
\DeclareMathSymbol{\Theta}{\mathalpha}{otone}{2}
\DeclareMathSymbol{\Lambda}{\mathalpha}{otone}{3}
\DeclareMathSymbol{\Xi}{\mathalpha}{otone}{4}
\DeclareMathSymbol{\Pi}{\mathalpha}{otone}{5}
\DeclareMathSymbol{\Sigma}{\mathalpha}{otone}{6}
\DeclareMathSymbol{\Upsilon}{\mathalpha}{otone}{7}
\DeclareMathSymbol{\Phi}{\mathalpha}{otone}{8}
\DeclareMathSymbol{\Psi}{\mathalpha}{otone}{9}
\DeclareMathSymbol{\Omega}{\mathalpha}{otone}{10}

Of course cmr can be changed into the document's main font family name, provided it contains Greek uppercase letters in its OT1 version; but it's just a matter of giving the correct slot numbers for the chosen font.

• I tried your solution but \Gamma produces italic output.
– N.N.
Oct 5, 2011 at 12:45
• @N.N. Please, show a minimal example, as mine works. The suggested code must go after loading isomath. Oct 5, 2011 at 12:59
• pastebin.com/pxXkGuce produces i.imgur.com/f92xj.png. I use Ubuntu's TeX Live 2009 if that matters.
– N.N.
Oct 5, 2011 at 13:03
• @N.N. Sorry! I copied the wrong code! Oct 5, 2011 at 13:06
• Cheers, it works now! Do you mind including an explanation of what the lines of \DeclareMathSymbol does, i.e. their meaning, in your answer? Also, if there is an isomath solution I'd be interested to learn about it.
– N.N.
Oct 5, 2011 at 13:10

isomath was written to set the math style to an 'ISO math style', as the author of the package puts it. Essentially, this means that it will be possible to typeset Greek letters (and non-Greek ones, for that matter) as bold italics, to denote vectors and matrices according to the ISO standard. It is possible to have upright letters, too. isomath requires OML font encoding. The isomath documentation (2011-01-14) states:

Currently only the mathdesign package provides upright fonts in OML encoding.

So, this package also needs to be loaded in order to obtain the correct fonts. LaTeX may otherwise display entirely different characters.

An example based on code snippets from the package documentation could be:

\documentclass[12pt,a4paper]{article}
\usepackage[utopia]{mathdesign}
\usepackage[OMLmathrm,OMLmathbf]{isomath} % options define which alphabets will be loaded, i.e. if bold face font is not necessary, OMLmathbf can be ommitted.
\begin{document}
$\Gamma$ $\mathrm{\Gamma}$ $\mathbf{\Gamma}$ $\mathbfit{\Gamma}$

$\pi$ $\mathrm{\pi}$ $\mathbf{\pi}$ $\mathbfit{\pi}$
\end{document}

The output will be:

It seems that the definition of the new math alphabets by isomath can lead to an error. To circumvent this, load the package with the option reuseMathAlphabets.

Known incompatibilities: fourierpackage

• Thanks for you answer. How does this answer my question for a way to force all Greek letters to be typeset upright unless specified otherwise, e.g. to make it so that $$\Phi \Psi$$ delivers the same output as $$\mathrm{\Phi \Psi}$$?
– N.N.
Oct 17, 2011 at 8:14
• Note that you can specify any of the three upright families from mathdesign without loading the package, with, e.g. \usepackage[OMLmathrm, rmdefault=mdput]{isomath} for Math Design Utopia. Dec 26, 2020 at 20:51

An older font with upright Greek letters is AMS Euler. Loading eulervm or eulerpx will also set your Latin letters upright, but some packages have an option like eulergreek to set only the Greek letters to Euler. The stix and stix2 packages support Greek letters in \mathrm. There is also an upright Greek font in fourier. None of these are compatible with isomath or listed in its manual because they use an encoding other than OML.

The lucidamatx package supports math-style=upright. The mathdesign package supports greeklowercase=upright.

The kpfonts and mathdesign packages support the commands \alphaup\Omegaup. The upgreek package supports \upalpha\upOmega. Both newtxmath and newpxmath support \upGamma as well as \Gammaup, etc. The fourier package supports \otheralpha\otherOmega.

If using isomath, you can either load mathdesign and then \usepackage[OMLmathrm]{isomath}, or get the math design fonts with, for example, \usepackage[OMLmathrm, rmdefault=mdput]{isomath} for Math Design Utopia, or load an upright OML math font with \SetMathAlphabet. See table 3 in the isomath manual for the upright families that come in OML. (If you load mathdesign, I would recommend you load erewhon, garamondx or XCharter afterward to fix some bugs with its text fonts.)

I would recommend you use unicode-math in LuaTeX when you can, and legacy math fonts when you have to. All OpenType math fonts contain upright Greek letters that you can use with, for example, \symup{\pi}, \muppi or \uppi. You can also give unicode-math the package option mathrm=sym to make \mathrm{\pi} an alias for \symup{\pi}. If you do so, you should \setoperatorfont and make sure to write \textnormal{iff}, not \mathrm{iff}, for words in math mode. Finally, you can give unicode-math the option math-style=upright.

In fact, the documentation of isomath, Section 2.2.1 (for the 2012-09-04 version), provides several alternatives for setting upright greek letters in math mode. The one I liked the most, because it did not require changing the fonts I was using, was option (e):

e) Use the text character with the alphabeta package from the lgrx bundle:

\usepackage{amsmath}
\usepackage{alphabeta}

and in the body

$u = 2 \text{\pi} r$

Edit: As noted by Davislor, it is safer to use \textnormal instead of \text to prevent italics in italicized environments, such as Theorems.

$u = 2 \textnormal{\pi} r$

• You want \textnormal here. \text{\pi} will set \pi as italic if the surrounding text is italic, such as inside a theorem statement. Dec 26, 2020 at 20:40


text:
## Sunday, September 06, 2009 ... /////

### Schellnhuber: West has exceeded quotas

In his previous life, Hans Joachim Schellnhuber used to be a fairly good theoretical physicist. For example, he would solve the Schrödinger equation with an almost periodic potential in 1983. He has spent a year or so as a postdoc at KITP in Santa Barbara (1981-82).

But the times have changed. For a couple of years, he has been the director of the Potsdam Institute for Climate Impact Research and the main German government's climate protection adviser. What he has just said for Spiegel, in

Industrialized nations are facing CO2 insolvency (click),
is just breathtaking and it helps me to understand how crazy political movements such as the Nazis or communists could have so easily taken over a nation that is as sensible as Germany. A few rotten steps in the hierarchy is enough for a loon to get to the very top. He is proposing the creation of a CO2 budget for every person on the planet, regardless whether they live in Berlin or Beijing. Let us allow him to speak:
Humankind has to limit itself to emit only fixed amount of carbon into the atmosphere until 2050. [...] Because the industrialized nations have already exceeded their quotas if you take into account past emissions. [...] With the current output you see that Germany, the US and other industrialized nations have either already used up their permissible quota, or will do so within the next few years. [...]

The industrialized nations are facing CO2 insolvency. This means that they have to notch up their efforts to reduce climate change, otherwise they will use up the CO2 budget actually designated to poorer countries and future generations.
Question: So industrialized nations would have to pay massive sums of money?
Yes. Up to €100 billion (\$142 billion) annually. If the richest sixth of the world's population were to pay this amount, each person would have to pay €100 per year. The West would give back part of the wealth it has taken from the South in the past centuries and be indebted to countries that are now amongst the poorest in the world. It would, however, have to be ensured that the poorer nations use the money for the proposes it is intended -- namely to help them to develop a greener economy.
Of course, Schellnhuber is not the first hardcore nutcase of this kind who has been saying such things, pretending that he is oh so smart. Many of you may remember Richard Feynman's popular book, Surely You're Joking, Mr Feynman, where he also described a crazy "interdisciplinary" conference where a similar "thinker" has been proposing the same "reparations" paid to the poor countries, based on the same assumptions that Mr Schellnhuber has used.

In order for me to save some time, let me just copy Feynman's entertaining description of the crazy conference he attended in the 1950s. The amount and basic types of pompous fools haven't changed: they have just taken over many institutions that apparently include the German government:
There was a special dinner at some point, and the head of the theology place, a very nice, very Jewish man, gave a speech. It was a good speech, and he was a very good speaker, so while it sounds crazy now, when I'm telling about it, at that time his main idea sounded completely obvious and true. He talked about the big differences in the welfare of various countries, which cause jealousy, which leads to conflict, and now that we have atomic weapons, any war and we're doomed, so therefore the right way out is to strive for peace by making sure there are no great differences from place to place, and since we have so much in the United States, we should give up nearly everything to the other countries until we're all even. Everybody was listening to this, and we were all full of sacrificial feeling, and all thinking we ought to do this. But I came back to my senses on the way home.

The next day one of the guys in our group said, "I think that speech last night was so good that we should all endorse it, and it should be the summary of our conference."

I started to say that the idea of distributing everything evenly is based on a theory that there's only X amount of stuff in the world, that somehow we took it away from the poorer countries in the first place, and therefore we should give it back to them. But this theory doesn't take into account the real reason for the differences between countries -- that is, the development of new techniques for growing food, the development of machinery to grow food and to do other things, and the fact that all this machinery requires the concentration of capital. It isn't the stuff, but the power to make the stuff, that is important. But I realize now that these people were not in science; they didn't understand it. They didn't understand technology; they didn't understand their time.

The conference made me so nervous that a girl I knew in New York had to calm me down. "Look," she said, "you're shaking! You've gone absolutely nuts! Just take it easy, and don't take it so seriously. Back away a minute and look at what it is." So I thought about the conference, how crazy it was, and it wasn't so bad. But if someone were to ask me to participate in something like that again, I'd shy away from it like mad -- I mean zero! No! Absolutely not! And I still get invitations for this kind of thing today.

When it came time to evaluate the conference at the end, the others told how much they got out of it, how successful it was, and so on. When they asked me, I said, "This conference was worse than a Rorschach test: There's a meaningless inkblot, and the others ask you what you think you see, but when you tell them, they start arguing with you!"

Even worse, at the end of the conference they were going to have another meeting, but this time the public would come, and the guy in charge of our group has the nerve to say that since we've worked out so much, there won't be any time for public discussion, so we'll just tell the public all the things we've worked out. My eyes bugged out: I didn't think we had worked out a damn thing!

Finally, when we were discussing the question of whether we had developed a way of having a dialogue among people of different disciplines -- our second basic "problem" -- I said that I noticed something interesting. Each of us talked about what we thought the "ethics of equality" was, from our own point of view, without paying any attention to the other guy's point of view. For example, the historian proposed that the way to understand ethical problems is to look historically at how they evolved and how they developed; the international lawyer suggested that the way to do it is to see how in fact people actually act in different situations and make their arrangements; the Jesuit priest was always referring to "the fragmentation of knowledge"; and I, as a scientist, proposed that we should isolate the problem in a way analogous to Galileo's techniques for experiments; and so on. "So, in my opinion," I said, "we had no dialogue at all. Instead, we had nothing but chaos!"

Of course I was attacked, from all around. "Don't you think that order can come from chaos?"

"Uh, well, as a general principle, or..." I didn't understand what to do with a question like "Can order come from chaos?" Yes, no, what of it?

There were a lot of fools at that conference -- pompous fools -- and pompous fools drive me up the wall. Ordinary fools are all right; you can talk to them, and try to help them out. But pompous fools -- guys who are fools and are covering it all over and impressing people as to how wonderful they are with all this hocus pocus -- THAT, I CANNOT STAND! An ordinary fool isn't a faker; an honest fool is all right. But a dishonest fool is terrible! And that's what I got at the conference, a bunch of pompous fools, and I got very upset. I'm not going to get upset like that again, so I won't participate in interdisciplinary conferences any more.
Feynman's book continues with a story involving the young rabbis whose main concern was whether electricity was fire.

I wonder how Feynman would feel if he had to be talking to not just a few nuts of this kind but e.g. to 2,500 similar nuts who would be moreover described by the media as good scientists, if not the best ones in the world. ;-) Good for him that he managed to die in time.

Hat tip: Marc Morano

A British religious bonus

The German readers could feel that they're superior because their government science adviser is the greatest nutcase among the world's government science advisers. ;-)

In order for me to fight against such a new wave of German supremacy, let us look into the United Kingdom.

Lord Robert May, Baron May of Oxford, has been the chief adviser to the British government for some time. Now he is the head of the British Science Association. According to The Guardian, The Telegraph, and others, God is obliged to fight against climate change, and so are the religious leaders. A deity can serve as a "punisher". Lord May will kindly accept the role of the commander-in-chief who will instruct God and all churches in the world what they should demand from the people.

More precisely, faith groups "could lead policing of social behavior". And if God fails to make a soul green, the soul may always be burned at stake. Welcome to the Postmodern Dark Ages, at least until Lord May is admitted to a mental asylum! ;-)

#### snail feedback (4) :

(I think your fast-comment section is having problems.)
OK ... I can understand worrying about extreme measures as such, etc. But if CO2 is a warming stimulus (from spectroscopic data), there is still as risk factor, right? Most around here wouldn't take the multiplying factor to be around three, but maybe around one, etc. But if they aren't sure, then neither is anyone else AFAICT. So what is the proper "risk" posture then? How much is worth doing? People want critics to come up with alternative plans and not just critiques.

damn Lubos - I'm gone a couple weeks and your fan club doubles.
How do you keep them all fed?

Neil' you're missing the main point. Warm is good. Period.
How do I underline with this thing?
Alright I know.

We use to want to save the rainforest. Co2 augmentation saved the rainforest. We have more rainforest then we know what to do with. It's creeping into Mexico we have so much.
We use to want to plant trees. We don't have to plant trees anymore. Trees are springing up all over the Boreal forest of their own accord.
We use to want rain. Now we get plenty of rain, more then we need.

We all like to see the girls in bikinis.

Warm is good.
Good doesn't need a risk factor.

Before we run around with our heads cut off, we should verify our data, verify our models.

Let's do something silly and make sure the models predict reality before we make decisions based off of them.

There are a lot of really intelligent people out there that think Global Climate Change is a big problem. I am pretty sure they are all going to wake up some day and wonder what the hell they were doing.

Lots of hand waving going on. Every once in a while an actual bit of science shows up. Just about everything else appears to be someone hyping something beyond the boundaries of rationality.

Acidifying the oceans means going to from a basic solution to a slightly less basic solution. (pH from 8.1 to maybe 7.9) To say this is acidification requires someone to be purposely misleading.

The real cause of the problem.... SUPERMAN.

Everyone wants to save the world just like Superman does.


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# Talk:Wizengamot

11,812pages on
this wiki

## Elphias Doge in Wizengamot?

Is Elphias Doge a member of Wizengamot? I don't recall that being mentioned anywhere. Yaquez 18:42, 2 February 2008 (UTC)

I thought he was present at one of the trials. Either Barty Jr's, Harry's, or it was mentioned in AD's obit, but i remember looking it up, maybe my mind is playing tricks on me? Mafalda Hopkirk 02:10, 3 February 2008 (UTC)
No i remember now, you know what it is, he is Special Advisor to the Wizengamot. Mafalda Hopkirk 02:18, 3 February 2008 (UTC)
Not that I don't believe you, but where precisely is that mentioned?Yaquez 14:42, 5 February 2008 (UTC)
For the moment all i know is that it's on HP Lexicon but i'm going to skim through some wizengamot book scenes and see what i can find. Mafalda Hopkirk 17:06, 5 February 2008 (UTC)

## Dolores Umbridge?

I thought that Umbridge was only there becuase she is Senior Undersecretary to the Minister, not an official member. Freakatone 18:03, 28 February 2008 (UTC)

She seems to have a lot of heft for just an undersecty so i assumed she was a member, anyway wasn't she also wearing a wizengamot robe? Maybe i'm wrong? Mafalda Hopkirk 00:25, 29 February 2008 (UTC)

## Pronunciation

How is this word pronounced, and would it be useful to add a pronunciation at the head? 70.181.171.159 21:28, 9 August 2008 (UTC)

I think it is pronounced like in the films (Wee-zan-AGA-mot)... -- Seth Cooper Owl Post 19:06, 22 February 2009 (UTC)
The Pronunciation Guide on the Scholastic website has it as "wiz-en-gam-ot." Starstuff (Owl me!) 03:28, 23 February 2009 (UTC)

## Behind the scenes

I don't understand that behind the scenes bit. First, it says "ladies and gentlemen", when in the film Fudge says "witches and wizards" and, second, I can't understand where is the mistake... -- Seth Cooper Owl Post 19:06, 22 February 2009 (UTC)

Seconded. I don't get it either. --MartectX 22:13, 7 May 2009 (UTC)
I think he says Wizenagamot instead of Wizengamot -Smonocco 22:47, 7 May 2009 (UTC)
Hmmm. Quite possible, yes... Does anyone have an source on the pronounciation? -- Seth Cooper (Owl Post) 22:56, 7 May 2009 (UTC)

## Main image

Shouldn't it be changed, as I believe it is an image of the Council of Magical Law? -Smonocco 16:31, December 14, 2010 (UTC)
Also, does the Wizengamot appear in G.O.F. (including the film) at all? -Smonocco 15:16, February 27, 2011 (UTC)

Done. I don't think it does, no. -- 21:12, March 2, 2011 (UTC)
Yes, I think Wizengamot does appear in Goblet of Fire. In Dumbledore's pensieve, when Harry sees Barty Crouch Jr.'s trial and Bellatrix's, and Karkaroff's. That's the Wizengamot, isn't it? Because if it was the Council of Magical Law, Dumbledore wouldn't have been there, right...?Allsevenbooks (talk) 15:14, July 13, 2014 (UTC)
No, that's the Council of Magical Law. --Hunnie Bunn (talk) 15:28, July 13, 2014 (UTC)

## Parliament?

Where is it said that the Wizengamot is a legislative body, or parliament? We've only seen the Wizengamot act in a judicial manner.

And where did it say the Wizengamot passed and sanctioned the Educational Decrees? Unless I'm mistaken, didn't Umbridge have free reign to enact the Decress as High Inquisitor?Apwbd150 03:00, June 26, 2011 (UTC)

It doesn't say that anywhere; the section is plain wrong. The Wizengamot isn't mentioned in any other context than as a court. There is, in fact, no legislative branch ever explicitly mentioned, Wizengamot or no. It just says 'The Mininstry of Magic'. We do know that laws are written in the Minstry Departments that concern them (e.g. Arthur Weasley wrote the Muggle Protection Act), but that's about it.

The decrees in OotP, specifically, were directly issued by Fudge. That is why they are decrees -- orders issued by the head of state. The Daily Prophet mentions a 'passing' in the following context:

In a surprise move last night the Ministry of Magic passed new legis-
lation giving itself an unprecedented level of control at Hogwarts School of
Witchcraft and Wizardry.

and

This is not the first time in recent weeks Fudge has used new laws to
effect improvements at the Wizarding school. As recently as August 30th
Educational Decree Twenty-two was passed, to ensure that, in the event of
the current headmaster being unable to provide a candidate for a teaching
post, the Ministry should select an appropriate person.

and

It is this last function that the Ministry has now formalized with the
passing of Educational Decree Twenty-three, which creates the new posi-
tion of ‘Hogwarts High Inquisitor.’

The first quote is already the point here: it's the Ministry doing the passing. Assuming the Wizengamot is part of the Ministry, that could be it's purpose, but that is nothing but an assumption.

79.213.76.128 16:16, July 2, 2012 (UTC)

After long reserch,I finally found out what type of hat the members of the Wizengamot wear in the film adaption of Order of the Phoneix and I was wondering if I can state it in the main article? Since the rules of canon say information from the films can be considered canon unless clearly stated otherwise in the novels and there's no mention of them wering these headwear should I go ahead and put it in or should I put it in the "Behind the scenes" section?Professor Ambrius (talk) 03:02, December 3, 2012 (UTC)

Hmm. I'd wait to get at least one other opinion, but I'd say that since it doesn't specifically say in the books that they don't wear it, it could be considered canon. Even so, it still seems like something for the Behind the scenes section, unless you can think of an in-universe way to put it in the main article that doesn't feel clunky. ProfessorTofty (talk) 04:19, December 3, 2012 (UTC)
It looks vaguely ecclesiastical to me, and the name of the hat in the Harry Potter universe may differ from that of the real world, so while I think that it would be very interesting to have the name of the hat in the article, I agree with Professor Tofty that a Behind the scenes comment would perhaps be the best place for it (at least until the name of the hat was confirmed in-universe). --xensyriaT 20:36, December 7, 2012 (UTC)

## WOMBAT

The Wizards' Ordinary Magic and Basic Aptitude Test says that 18% of wizards would like to see one of the following changed:

1. Average age lowered from 87
2. Proof of pure-blood status
3. Maximum of three years in post
4. Goblin representation

Does this therefore mean that the average age (I think Cornelius Fudge, being British Ministry of Magic, might possibly be excluded) is 87, that all of them are pure-blood (which would fit for Umbridge, Bones and Fudge), and have three years in post?

There is no evidence against these things being true, and indeed the question (What change would 18% of wizards like to see to the requirements for membership of the Wizengamot? (source: Ministry of Magic poll)) seems to imply that they are all true. --Hunnie Bunn (talk) 22:22, November 19, 2013 (UTC)

Indeed, they are all true. In fact, some of that information is already presented in articles (the "Wizengamot" article says "In recent times, the average age of a Wizengamot member was 87 years"). What I think we can gleam from the options is the following:
1. "Average age lowered from 87" - average age of Wizengamot members is 87 years old. However, this does not allow us to estimate the ages of any particular Wizengamot member (if we presume a Gaussian distribution for of the ages, we'd conclude that most members are around 87, but some could be far younger than that and others could be older).
2. "Proof of pure-blood status" - it seems that all members of the Wizengamot are pure-bloods (and are requested proof of blood status).
3. "Maximum of three years in post" - it would seem that no one can be a member of the Wizengamot for more than three years.
4. "Goblin representation" - this one is more ambiguously-worded, but I take it to mean (and the all-wizard "full Wizengamot" in Order of the Phoenix seems to support it) that Goblins are not represented at all in the Wizengamot. It comes as no surprise, really.
The thing with 2 and 3 is that, while we know it applies to regular members of the Wizengamot, I'm not sure they still hold true for ex officio members (that would include Fudge, Umbridge and Bones; respectively, Minister, Senior Undersecretary, and Head of the DMLE) — for instance, Nobby Leach (known to be a Muggle-born) was Minister for Magic, and it would seem that he would've belonged to the Wizengamot in virtue of being Minister. Yet, he was not a pure-blood.
Other thing that bothers me is that Dumbledore was Chief Warlock of the Wizengamot for a period of over three years (he was Chief Warlock in 1991 and was sacked in 1995 - that's 4 years right there).
Thoughts? -- 23:20, November 19, 2013 (UTC)
Last one first: I think perhaps the Chief Warlock serves for longer than the others (maybe it's just my school board, but here the principal has to serve five years as principal before they can do anything, while teachers can move after three - odd how close the numbers are).
I think that the Minister for Magic is exempt from the blood status, age and length in position regulations, given that they run the whole entire Ministry. It's nothing more than speculation, I know, but I think this is likely. My other thought would be that perhaps the policy was more recent and came in after Leach's imposition, but this is less likely.
As for the first point: assuming Dumbledore did count along with the others, he was a hundred and ten at the time; the way I was taught averages (probably improperly) was that the average was exactly between the highest and lowest number no matter what. If the way I was taught is right (I doubt it) then the Wizengamot members were all between 110 and 57...
Anyways, that's just my thoughts.--Hunnie Bunn (talk) 23:32, November 19, 2013 (UTC)
I disagree with your interpretations of 2 and 3, Seth, partially because they directly contradict canon. You've already pointed out the problem with 3, and, as for 2, a more direct contradiction is also with Dumbledore: he was half-blood, and never made any attempt to hide this. No, I think that the opposite would be true: that proof of pure-blood status and a maximum of three years in post were the changes people wanted to happen, not the things they wanted changed -- 1337star (Drop me a line!) 01:21, November 20, 2013 (UTC)
Hunnie Bunn: I'm afraid what you said about averages is, on the whole, incorrect. The average is not always exactly in the middle (perhaps you are mistaking it for the concept of median?)
See this example: let's say that we take five random Wizengamot members and their ages are 67, 100, 104, 104 and 110. The average would be calculated like this:
$\bar{x} = \sum_{i=1}^n\frac{x_i}{n} = \frac{x_1 + \cdots + x_n}{n}$, where the numerator ($x_i$) is the sum of all quantities (in this case, the sum of all ages), and the denominator ($n$) is the total number of quantities (total number of ages). That is:
$\bar{x} = \frac{(67+100+104+104+110)}{5} = \frac{485}{5} = 97$
Note that, according to your definition, the average would be the number exactly between 67 and 110, which would be 88,5, which of course, does not correspond to the average 97.
Either way, we would not know if Dumbledore is the oldest member of the Wizengamot (and therefore his age the highest number).
1337star: Or that, yes, that makes far much more sense. It does not go against canon, and provides a by far more likely interpretation of the stubby, ambiguous alternatives. So that would mean that no proof of blood status is required for membership, and that there isn't a limit (at least that we know of) for time in the post. -- 02:51, November 20, 2013 (UTC)


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How to find the longest orf from a transcriptome
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5 months ago
Luis999 ▴ 20

Hello, good day, sorry for the simplicity but, I have a super basic question. I have been trying to identify the longest ORF of a transcriptome for a long time, but from the previous failed attempts it seems that it is not clear to me how to do it. You see, I have to make a small ORF (30 to 300 nucleotide ORF) prediction of a transcriptome, but for that, I first need to identify the longest ORF, so what I did was translate the 6 possible reading frames of the transcriptome, to do this used the function transeq from EMBOSS

transeq inputFasta outputFast -frame=6

The above is an example that creates a fasta file with the 6 different reading frames, I did that and also generated a fasta file for each reading frame. Then I made a variety of scripts that did not work to detect the longest ORF in the variety of generated files. If anyone could tell me about some program and function to help me find this longest ORF or give me clues on how to make a proper script, I would greatly appreciate it. Thanks for your time and have a nice day!

longest EMBOSS ORF • 348 views
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Entering edit mode


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`ISRN Communications and NetworkingVolume 2012 (2012), Article ID 951290, 9 pagesdoi:10.5402/2012/951290`
Research Article

## Gain Improvement of Dual Band Antenna Based on Complementary Rectangular Split-Ring Resonator

Millimeter Wave Laboratory, Electrical and Electronic Engineering Department, Public University of Navarra, Arrosadía Campus, 31006 Pamplona, Spain

Received 7 October 2011; Accepted 9 November 2011

Academic Editors: C. Luxey and J. K. Muppala

Copyright © 2012 Noelia Ortiz et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

A simple and successful dual band patch linear polarized rectangular antenna design is presented. The dual band antenna is designed etching a complementary rectangular split-ring resonator in the patch of a conventional rectangular patch antenna. Furthermore, a parametric study shows the influence of the location of the CSRR particle on the radiation characteristics of the dual band antenna. Going further, a miniaturization of the conventional rectangular patch antenna and an enhancement of the complementary split-ring resonator resonance gain versus the location of the CSRR on the patch are achieved. The dual band antenna design has been made feasible due to the quasistatic resonance property of the complementary split-ring resonators. The simulated results are compared with measured data and good agreement is reported.

#### 1. Introduction

The possibility of obtaining media with simultaneously negative permeability and permittivity was hypothesized by Veselago in the late 1960s [1]. In spite of the interesting properties presented by such media, it was not until 2000 that the first experimental evidence of a medium with simultaneously negative permeability and permittivity was demonstrated [2]. The original medium proposed in [2] consists of a bulky combination of metal wires and split-ring resonators (SRRs) [3].

The SRR electromagnetic properties have been already analyzed in [4, 5]. This analysis shows that the SRR behaves as an LC resonant tank that can be excited by an external time-varying magnetic field applied parallel to the particle axis, thus producing a quasi-static resonant effect [4]. Therefore, the SRR has subwavelength dimensions at its quasi-static resonance, allowing very compact device designs. Up to now, these self-resonant particles have been used in the design of microwave filters in planar technology [6, 7]. However, in this paper, we have taken advantage of the complementary split-ring resonator (CSRR) concept [8] to design a miniaturized dual band patch antenna with vertical polarization, also studying how to improve radiation efficiency for the resonance produced by the CSRR in this kind of antennas. The CSRR is inspired on Babinet principle [9], and, as occurs with the SRR, it also exhibits a quasi-static resonance, which enables the particle to be electrically small [8, 9].

Up to now, the use of metamaterial concepts in practical miniaturized antennas is a very challenging research topic [1016] and the achieved results based on self-resonance particles as SRRs or CSRRs [16] exhibit low radiation efficiencies driving to low-gain antennas comparing to the results of the parametric study presented in this paper. In this sense, the excitation of a CSRR etched in the patch of a conventional patch antenna allows us to design dual band patch antennas. Going further, the presence of the CSRR etched in some positions of the path also produces a miniaturization of both patch antenna resonances, leading to miniaturized dual band antennas. Replacing the CSRR within a slot of its same external dimensions, the slot does not exhibit a resonance at the same frequency of the CSRR, but at higher frequencies and more than one slot should be placed in the patch depending on their position in order to achieve a dual band response. The resonance frequency of a rectangular slot on a dielectric is approximately given by , where , , , and are the speed of light in vacuum, dielectric relative permittivity, and the external dimensions of the rectangular iris, respectively. In opposition to the resonance frequency of a rectangular slot, the resonant frequency of the CSRR is much lower for the same physical size. Hence, it can be designed to exhibit a resonance at lower frequencies comparing to different shapes of slots that can also be etched in a conventional patch. The design presented in this paper gives an alternative solution to the existing dual band antenna designs [17], as the ones carried out by loading a rectangular patch antenna with a pair of bent slots or embedded step slots close to the patch nonradiating edges, or the ones done by spur lines or shorted microstrip antenna with rectangular patch. Overall, as it can be shown in this paper, the properties of the CSRR allow us to design dual band miniaturized antennas based on the anisotropic properties of the CSRR as indicated by the measurements of a fabricated prototype and the parametric studies of the presented design in opposition to dual band antennas produced by the radiation of slots, whose resonances are dependent on their physical length. The prototype has been designed to exhibit a dual band behaviour in two frequency bands in the range from 4 GHz to 5 GHz for wireless applications. Nowadays, there is a growing trend to integrate different wireless communication systems in one single user terminal as long as to reduce the overall size. Since all these systems work at different frequency bands, dual and multiband antennas with frequency ratios around 1.2 between different bands are desirable. For this application, the type of antennas presented in this paper are a good alternative as introducing different CSRRs on the patch multiband antennas can be obtained very easily taking as a starting point the dual band antennas presented in this paper. The dual band antenna design presented has been validated, and a parametric study of the CSRR location and its influence on antenna radiating characteristics is presented and analyzed.

#### 2. CSRR Excitation in a Rectangular Patch

The excitation of CSRRs has been usually driven by an incident electric field normal to the particle plane. In order to understand the excitation of these particles, let us consider the CSRR presented in Figures 1(b) and 1(c). Comparing the excitation of an SRR with a CSRR, the CSRR particle should be rotated 90° from the position of the SRR particle, as it is shown in Figure 1(a). Following the theoretical discussion shown in [4, 5], the operation of the SRR near its first resonance frequency obeys the effect of resonant polarizabilities, which gives the resonant magnetic and electric dipolar moments , , and as a function of the exciting field components , , and . Complementarily, using the Babinet principle [9], the CSRR can be excited by the incident complementary fields and , which are related to and by and by means of another set of resonant polarizabilities, thus given an electric dipole and magnetic dipoles and . Then, as seen in Figures 1(b) and 1(c), depending on the CSRR position inside the patch, it will be excited by incident electric field normal to the particle plane and by incident magnetic field tangent to the particle plane . For the excitation of the CSRR by the magnetic field , the CSRR should be rotated 90° in the patch as it is shown in Figure 1(c) comparing to the orientation of the CSRR in Figure 1(b).

Figure 1: (a) SRR and CSRR topologies relevant dimensions. (b) Configuration of dual band CSRR-rectangular patch antenna. CSRR excitation by incident electric field normal to the particle plane, . (c) Configuration of dual band CSRR-rectangular patch antenna. CSRR excitation by magnetic field tangent to the particle plane, , and by electric field normal to the particle plane, .

In order to show graphically the excitation of the CSRR by the existing fields inside a rectangular patch antenna, Figures 2 and 3 show the electric fields and surface current distributions at the resonance frequency of the CSRR for the locations of this particle according to the layouts of Figures 2(a) and 3(a). In Figure 2, the CSRR particle has been located for its proper excitation by electric field component, placed in one position inside the area where field distributions are higher. Otherwise, in Figure 3, the CSRR particle has been located where magnetic field distributions are more concentrated, for the best suitable CSRR excitation by magnetic field component. Though, for this case, the CSRR is also excited by . The simulations of the structures presented in this paper have been performed with the commercial finite-integration time-domain CST Microwave Studio Code.

Figure 2: (a) Top view of the layout for field and current distributions analysis in reference with Figure 1(b). (b) Simulated field distribution in the CSRR at its resonant frequency. (c) Simulated current distributions in the CSRR at its resonant frequency.
Figure 3: (a) Top view of the layout for field and current distributions analysis in reference with Figure 1(c). (b) Simulated field distribution in the CSRR at its resonant frequency. (c) Simulated current distributions in the CSRR at its resonant frequency.

#### 3. Parametric Study

The parametric study carried out in this work shows how the position of the CSRR (without changing its dimensions) has influence on the radiation efficiency of both antenna resonances for the orientation of the particle as it is in Figure 1(b). On this way, for some locations of the CSRR in the patch, radiation efficiencies up to 50% are achieved for both resonances. These results together with the measurement results of the prototype comparing to the simulated results show the usefulness of this kind of dual band patch antennas. The simulated antenna in the parametric study has the same dimensions as the fabricated one, excluding the placement of the CSRR inside the patch, which varies from to in the -axis direction and from to in the -axis direction. In Figure 1(b), the references of both Pos and parameters are specified. The substrate employed in the simulated and fabricated prototype is the commercially available Arlon 250-LX-0193-43-11 ( and thickness ). The physical width and length of the rectangular patch antenna are and , respectively [18]. Then, its resonance has been set around 5 GHz. The width of the micro strip line is 1.34 mm, corresponding to a characteristic impedance of . This line exhibits an offset from the centre of the patch antenna in order to match its reflection coefficient at its working frequency. The offset is ( and (see Figure 1(b)).

The CSRR particle has been designed to exhibit its quasi-static resonance frequency below the resonance frequency of the patch, obtaining more compact devices highlighting the advantages of the resonance properties of anisotropic particles comparing to other slots already used for dual band antenna designs. The radiation produced by conventional slots does not have the same origin comparing to the radiation produced by a CSRR particle, and their electric length should be longer comparing to the electric length of the CSRR. The necessary physical dimensions of the CSRR to achieve a radiation frequency below the resonant frequency of the designed rectangular patch have been calculated using the design formulas for SRR reported in [4], resulting in this case in , , and . In all the results presented in this paper, the first resonance of the dual band antenna is the one produced by the excitation of the CSRR, while the second resonance is produced by the conventional patch itself.

In Figures 4(a) and 4(b) radiation efficiencies as results from the parametric study are shown for both resonances. Figure 4(a) shows how the radiation efficiency of the resonance produced by the excitation of the CSRR increases while the anisotropic particle is placed at the radiating edges of the rectangular patch (Pos parameter values of −9 mm and 9 mm). However, there are some positions for the CSRR inside the patch where the radiation due to the CSRR is cancelled. These positions correspond to values around Pos parameter of 0 mm and values around this value. This means that when the CSRR is etched in the centre of the rectangular patch its resonance is cancelled leading to a single resonance antenna.

Figure 4: (a) Simulated radiation efficiency for the resonance produced by the CSRR. (b) Simulated radiation efficiency for the resonance produced by the rectangular patch antenna.

Figure 4(b) shows how the radiation efficiency of the resonance produced by the patch decreases for some positions when the CSRR is placed at the radiating edges of the conventional patch antenna. However, there are other positions at these edges where both resonances exhibit radiation efficiencies up to 50%. The position values where radiation efficiency has been set to zero mean that the resonance produced by the CSRR particle or by the patch has been cancelled as explained before. The results of the parametric study for radiation efficiencies of both resonances show that the results are not symmetric, these differences are due to the asymmetric microstrip line excitation of the rectangular patch.

In Figures 5(a) and 5(b) gain values for parameter values (, and 13), which drive to the highest radiation efficiencies for both resonances, are depicted. The discontinuities with no values in the curves of Figures 5(a), 6(a), and 6(b) are because there is no resonance of the CSRR for those positions.

Figure 5: (a) Simulated gain values for the resonance produced by the CSRR. (b) Simulated gain values for the resonance produced by the rectangular patch antenna.
Figure 6: (a) Parametric study results. and resonant frequencies. (b) Parametric study results. ratio.

For the locations where the CSRR is etched around the centre of the rectangular patch, , , and , both resonances, the one produced by the CSRR and the one produced by the patch, are shifted to lower frequencies, resulting in a miniaturization of both frequency bands of dual band patch antenna comparing to a conventional rectangular patch antenna of the same dimensions. By contrast, as the CSRR moves away from the centre of the patch towards the nonradiating edges (in direction for all its Pos parameter values), the resonance produced by the rectangular patch shifts to higher frequencies comparing to the resonant frequency of the conventional rectangular patch itself. In Figures 6(a) and 6(b) the resonant frequencies for the first and second resonances versus Pos parameter and ratio are shown for , , and parameter values, where and are the resonance produced by the CSRR and the one produced by the rectangular patch, respectively. The miniaturization ratio of this type of antennas based on this design is around 1.2, but it also depends on the position of the CSRR on the patch, as the resonance frequencies are shifted. This behaviour is clearly shown in Figures 6(a) and 6(b).

From Figures 6(a) and 6(b) the miniaturization factor has been calculated for two different positions of the CSRR on the patch. In the first case, for particle location parameters of and the miniaturization factor is 1.16, corresponding to a radiation efficiency of 45.37% and gain of 3.59%. For the second case, the position parameters values are and . In this case the resonance produced by the CSRR has been cancelled and a single-band antenna is achieved. The miniaturization factor is 1.1, corresponding to a radiation efficiency of 84.36% and gain of 6.56 dB.

#### 4. Experimental Results

In order to demonstrate the usefulness of this dual band antenna design, a prototype has been fabricated. The fabricated prototype has been chosen from the parametric study in a case where the radiation efficiency is low comparing to the highest values obtained of this parameter. This case has been chosen to validate the usefulness of this design in a worst case condition. The prototype has been fabricated using a laser drilling machine. The design parameters of the CSRR particle are , , and (the same ones of those of the CSRR used in the parametric study). In the fabricated prototype the CSRR has been placed at A point (see Figure 1(b)) being the coordinates of this point (X = −9.3 mm, Y = 15.54 mm).

In Figure 7 simulated and measured reflection coefficient results of the fabricated prototype are shown. For matching measurements data has been collected by using an HP8510 network analyzer. As it can be seen in Figure 7, there is a frequency shift of 162 MHz to lower frequencies for the lower resonance. The upper resonance presents a frequency shift of 84 MHz, shifted to lower frequencies. Although there is a frequency shift between simulated and measured results, the matching values achieved are properly predicted by simulations. The discrepancies between simulated and measured results are due to the manufacturing process as the CSRR manufacturing tolerances are critical, changing slightly its frequency resonance. In simulations, materials have been simulated considering their corresponding finite conductance and substrate has been simulated considering its dielectric losses. In Figure 8 a picture of the fabricated prototype is shown.

Figure 7: Simulated and measured reflection coefficient results.
Figure 8: Fabricated prototype.

In Figures 9(a) and 9(b) measured results for normalized gain radiation patterns for 0° and 90° phi cut planes for both resonant frequencies are shown. Besides, simulated results just for 90° phi cut are shown. No more simulated cuts are introduced to maintain the figures legible. These frequencies are  GHz and  GHz, the resonant frequencies produced by the CSRR particle and conventional patch, respectively. The radiation pattern measured results show the feasibility of this type of dual band antenna design. Both resonances show cross-polar levels around −20 dB for theta 0°.

Figure 9: (a) Measured normalized copolar and cross-polar gain radiation patterns for 0° and 90° phi cuts. Simulated normalized copular and cross-polar gain radiation pattern for 90° phi cut at  GHz. (b) Measured normalized copolar and cross-polar gain radiation patterns for 0° and 90° phi cuts. Simulated normalized copular and cross-polar gain radiation pattern for 90° phi cut at  GHz.

Table 1 shows a comparison between simulated and measured results of the fabricated prototype from impedance matching and radiation point of views. This table shows that both resonances have similar characteristics. From matching point of view, the parameters shown in Table 1 are reflection coefficient values for both resonant frequencies. From radiation point of view, parameters shown in Table 1 are peak directivity, peak gain, and radiation efficiency. Directivity values have been calculated [18] from measured gain radiation patterns for both resonances in order to calculate efficiency and verify the good agreement between simulated and measured data. Within the fabricated prototype the parametric study made in this work has been validated due to the good agreement between simulations and measured results. Although gain and radiation efficiency obtained by the resonance produced by the CSRR is low comparing to the second resonance in the fabricated prototype, these values are in accordance with the simulated results. Furthermore, it is remarkable that gain values obtained for the positions of the CSRR in the patch studied in this work are higher compared to previous works [16].

Table 1

Up to this point, in order to compare the performances of the dual band antenna topology presented in this paper with those of the same conventional patch antenna without a CSRR etched in its centre, in Table 2 measured data of the conventional rectangular patch antenna is shown.

Table 2

Comparing Tables 1 and 2, the designed prototype presents similar performances to those of the conventional patch antenna from matching point of view (effective bandwidth and reflection coefficient values). On the other side, from radiation point of view, the designed and fabricated dual band antenna resonances show lower radiation efficiencies as it has been mentioned before.

Finally, to understand in an oversimplified way the radiation mechanism of the proposed antenna, one needs to consider the pair of electric dipoles that are described in [9] and the effect of the finite ground plane. In a recent work a refined equivalent circuit model for the CSRR which explains more accurately the physical interpretation of the influence of reactive parameters [19] is presented. This circuit model will be developed in further works in order to take into account the radiation resistance and the internal coupling to the patch and ground plane.

#### 5. Conclusions

In this work a dual band patch antenna based on a CSRR has been proposed, studied, and successfully tested, demonstrating the feasibility of this type of dual band antennas, adding a miniaturization of patch dimensions for some locations of the CSRR inside the patch. A good agreement between simulated and measured results is shown. The parametric study shows the influence of the location of the CSRR on the patch on the radiation characteristics. The design of the dual patch antenna is simple as the only design parameters comparing to a conventional path antenna are the ones of the CSRR particle design parameters. Also, multiband antennas can be designed in a similar way by simply adding different CSRRs on the patch.

#### Acknowledgments

This work has been supported by the Spanish Government and EU Feder by the Contracts Consolider “Engineering Metamaterials” CSD2008-00066 and TEC2008-06871-C02-01.

#### References

1. V. G. Veselago, “The electrodynamics of substances with simultaneously negative values of $\epsilon$ and μ,” Soviet Physics Uspekhi, vol. 10, pp. 509–514, 1968.
2. D. R. Smith, W. J. Padilla, D. C. Vier, S. C. Nemat-Nasser, and S. Schultz, “Composite medium with simultaneously negative permeability and permittivity,” Physical Review Letters, vol. 84, no. 18, pp. 4184–4187, 2000.
3. J. B. Pendry, A. J. Holden, D. J. Robbins, and W. J. Stewart, “Magnetism from conductors and enhanced nonlinear phenomena,” IEEE Transactions on Microwave Theory and Techniques, vol. 47, no. 11, pp. 2075–2084, 1999.
4. R. Marqués, F. Mesa, J. Martel, and F. Medina, “Comparative analysis of edge- and broadside-coupled split ring resonators for metamaterial design—theory and experiments,” IEEE Transactions on Antennas and Propagation, vol. 51, no. 10, pp. 2572–2581, 2003.
5. R. Marqués, F. Medina, and R. Rafii-El-Idrissi, “Role of bianisotropy in negative, permeability and left-handed metamaterials,” Physical Review B, vol. 65, no. 14, pp. 1444401–1444406, 2002.
6. F. Martín, F. Falcone, J. Bonache, T. Lopetegi, R. Marqués, and M. Sorolla, “Miniaturized coplanar waveguide stopband filters based on multiple tuned split rin resonators,” IEEE Microwave and Wireless Components Letters, vol. 13, no. 12, pp. 511–513, 2003.
7. F. Falcone, F. Martín, J. Bonache, R. Marqués, T. Lopetegi, and M. Sorolla, “Left handed coplanar waveguide band pass filters based on Bi-layer split ring resonators,” IEEE Microwave and Wireless Components Letters, vol. 14, no. 1, pp. 10–12, 2004.
8. F. Falcone, T. Lopetegi, J. D. Baena, R. Marqués, F. Martín, and M. Sorolla, “Effective negative-ε stopband microstrip lines based on complementary split ring resonators,” IEEE Microwave and Wireless Components Letters, vol. 14, no. 6, pp. 280–282, 2004.
9. F. Falcone, T. Lopetegi, M. A. G. Laso et al., “Babinet principle applied to metasurface and metamaterial design,” Physical Review Letters. In press.
10. R. W. Ziolkowski and A. D. Kipple, “Application of double negative materials to increase the power radiated by electrically small antennas,” IEEE Transactions on Antennas and Propagation, vol. 51, no. 10, pp. 2626–2640, 2003.
11. F. Qureshi, M. A. Antoniades, and G. V. Eleftheriades, “A compact and low-profile metamaterial ring antenna with vertical polarization,” IEEE Antennas and Wireless Propagation Letters, vol. 4, no. 1, pp. 333–336, 2005.
12. R. K. Baee, G. Dadashzadeh, and F. G. Kharakhili, “Using of CSRR and its equivalent circuit model in size reduction of microstrip antenna,” in Proceedings of the Asia-Pacific Microwave Conference (APMC '07), pp. 1–4, 2007.
13. Y. Lee, S. Tse, Y. Hao, and C. G. Parini, “A compact microstrip antenna with improved bandwidth using complementary split-ring resonator (CSRR) loading,” in Proceedings of the IEEE Antennas and Propagation Society International Symposium, pp. 5431–5434, June 2007.
14. L. Meng, L. Mingzhi, and J. C. Tie, “Novel miniaturized dual band antenna design using complementary metamaterial,” in Proceedings of the International Workshop on Metamaterials (META '08), pp. 374–376, November 2008.
15. J. Liu, S. Gong, Y. Xu, X. Zhang, C. Feng, and N. Qi, “Compact printed ultra-wideband monopole antenna with dual band-notched characteristics,” Electronics Letters, vol. 44, no. 12, pp. 710–711, 2008.
16. H. Zhang, Y. Q. Li, X. Chen, Y. Q. Fu, and N. C. Yuan, “Design of circular/dual-frequency linear polarization antennas based on the anisotropic complementary split ring resonator,” IEEE Transactions on Antennas and Propagation, vol. 57, no. 10, Article ID 5196779, pp. 3352–3355, 2009.
17. K.-L. Wong, Compact and Broadband Microstrip Antennas, John Wiley & Sons, New York, NY, USA, 2002.
18. J. D. Kraus and R.J. Marhefka, Antennas for all Applications, McGraw-Hill, New York, NY, USA, 3rd edition, 2002.
19. F. Aznar, M. Gil, J. Bonache, and F. Martín, “Revising the equivalent circuit models of resonant-type metamaterial transmission lines,” in Proceedings of the IEEE MTT-S International Microwave Symposium Digest (MTT '08), pp. 323–326, June 2008.


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# Steiner trees and soap bubbles

March 11, 2020. This is a self-contained introduction to minimal networks (aka Steiner trees) in the plane. The first section discusses local geometric features, the second the structure of networks, and in the final section, what minimal networks can tell us about soap bubbles and vice versa.

### Contents

1. Introduction
2. Triangles
3. Graphs
4. Soap bubbles

## 1. Introduction

Suppose we want to join up three towns ($A$, $B$ and $C$) by rail. If I can use this railway to travel from one town to any other, we say that the rail network is connected. Two simple examples of connected networks are shown below.

Building railways is expensive, since we not only need to design and build the rail itself, but acquire the land beneath it. In contrast, stations are cheap: we just slap together some sidings, a platform, and a bench or two, and voila. To minimise cost, we should make the total length of the rail network as short as possible, adding extra stations if they help us do this. The resulting layout is called the minimal or Steiner network connecting $A$, $B$ and $C$.

We can consider the same problem for $n$ stations, and find the rail networks of smallest total length which connect them. Although finding the exact minimal network is difficult in general, we can derive some beautiful results about their geometry and layout. We step through the reasoning in bite-sized chunks, starting with the problem of connecting three cities.

Exercise 1 (choosing sides). Suppose $A$, $B$ and $C$ are separated by distances $AB = c$, $AC = b$ and $BC = a$. A triangular network consists of two sides of the triangle. Which ones should we choose?

Exercise 2 (triangle or trident). In Figure 1, we have two networks connecting the same towns: a triangular network, and a trident-shaped network with a hub station $D$ in the middle. Which is shorter? See if you can do better than either.

Hint. Measure lengths on the screen with a ruler. Simple but it works!

## 2. Triangles

The three-town problem is already surprising. The simplest network consists of two sides of the triangle, but it is not minimal; to minimise length, a trident-shaped network, with a new hub station $D$ in the centre, is optimal. The best place to put the hub station is called the Fermat point, which you can find defined in various places online. But for the rest of the post, we will only worry about when we should build a hub, rather than where it goes exactly. This will save us some hard trigonometry! The one exception will be an important special case: the equilateral triangle.

### 2.1. Equilateral triangles

Suppose that the towns $A$, $B$ and $C$ sit on the corners of an equilateral triangle of side length $d$. The triangular network has total length $L_\Lambda = 2d$.

It seems plausible that the trident network does better, but instead of checking with a ruler, let’s do some basic trigonometry.

Exercise 3 (equilateral trident). Show that the length of the trident network, with a hub in the centre of the triangle, is

$L_{\text{Y}} = \sqrt{3}d.$

Since $\sqrt{3} \approx 1.7 < 2$, the trident is indeed shorter than the triangle.

Hint. Use the grey triangle in Figure 2.

You might guess that the best place to put the hub station $D$ is right right in the centre, and in fact, we can prove this from symmetry. First of all, draw an axis of symmetry of the triangle, represented by the red line in Figure 3.

We are going to wiggle the hub side to side around this axis, along the dark blue line in Figure 3. From left-right symmetry, the length must either be a minimum or a maximum on the axis itself. Since the total nework length gets very large when we take the hub outside the triangle, it must be a minimum. So, for a minimal network, we must place $D$ on the red line.

But there are axes of symmetry associated with $B$ and $C$ as well, and all three intersect at the centre of the triangle. Since $D$ should lie on each of these lines, it must lie at the centre! This completes our proof. The minimal network is depicted in Figure 4.

### 2.2. Deforming the triangle

We are now going to take our solution to the equilateral triangle and slowly deform it. In other words, we imagine slowly moving the corners of the triangle so it is no longer equilateral. What will happen to the optimal position of the hub $D$? Since everything is being smoothly modified, the position of the hub should change smoothly as well. In Figure 5, the smooth paths of the corners are depicted in purple, and the corresponding smooth change of hub in green.

Since the hub position changes smoothly, it should stay inside the triangle for small deformations of the corners. But for triangles which are very far from equilateral, the hub might intersect one of the corners, and our trident network becomes a simpler triangular network, formed from two sides of the triangle.

We picture this in Figure 6 above. In this diagram, $B$ remains fixed in position, but $A$ and $C$ slowly lower and open out the angle of the triangle, with the optimal hub $D$ moving down as they do so. At some critical angle $\theta_\text{crit}$, it will coincide with $B$. The question is: what is the angle?

It turns out this critical angle is $\theta_\text{crit} = 120^\circ$. Although we won’t provide a watertight proof, we can give a plausibility argument. Let’s return to the equilateral triangle. Instead of adding a hub in the middle, suppose that $D$ is in fact a fourth city fixed in place. Clearly, the solution in Figure 4 is still optimal, since if we could add more hubs to reduce the total length, we could add more hubs to improve the network for the equilateral triangle! If I now remove any of the corner cities $A$, $B$ or $C$, the optimal network simply removes the corresponding leg of the trident (Figure 7).

Exercise 4 (cutting corners). Suppose that in Figure 7, we can add a new hub $E$ which reduces the length of the network. Explain how adding $E$ could reduce the length of the network in Figure 7, and thereby improve our solution for the equilateral triangle.

Note. This is an example of a proof by contradiction. To show something is false, we assume it is true and use it to derive a contradiction with known facts. We then reason backwards to conclude that it cannot be true!

Exercise 5 (critical isosceles). The argument above really only establishes that $\theta_\text{crit} \leq 120^\circ$. It is possible, in principle, that the triangular network becomes shorter at some angle $\theta_\text{crit} < 120^\circ$. In this exercise, we will show for an isosceles triangle that this is not the case.

Figure 8 shows the triangular network (blue lines) $ABC$, forming an angle of $120^\circ$. We now raise the two nodes $A$ and $B$ at the end, so that the angle $ABC$ is less than $120^\circ$. You can prove that the green and purple lines are shorter than the red lines, so that an interior hub $D$ yields a shorter network.

(a) Show using the law of cosines (or otherwise) that

$c^2 = a^2 + b^2 + ab.$

(b) From part (a), show that

$a + 2b < 2c.$

(b) Conclude that for an isosceles triangle $ABC$, the critical angle is $\theta_\text{crit} = 120^\circ$.

### 2.3. Hubs and spokes

All this work with triangles pays off with a powerful conclusion for minimal networks connecting any number of cities:

In any minimal network, all hubs have three incoming legs separated by angles of $120^\circ$.

The argument is beautiful and simple.

Proof. Our first step is to show that it is impossible for a hub to have edges separated by less than $120^\circ$. Suppose we have cities $A_1, A_2, \ldots, A_n$ connected by a minimal rail network. Also suppose there is a hub station $H$ with incoming rail lines separated by less than $\theta_\text{crit} = 120^\circ$, as on the left in Figure 11. There may be other incoming lines (the horizontal lines in Figure 11), but these will play no role in our proof and can be ignored.

We can build two new stations on these outgoing legs, $h_1$ and $h_2$, which are (for simplicity) the same distance from $H$.

It’s easy to see that every angle in this triangle is less than $120^\circ$, since the angles at $h_1$ and $h_2$ are at most $90^\circ$, and we’ve assume the angle at $H$ is less than $120^\circ$. Thus, from our work in the previous section, we know that the minimal network connecting $h_1$, $h_2$ and $H$ is not the triangle we have drawn, but a trident with another hub $h_3$ in the middle! This strictly decreases the length of the network (shown right in Figure 9), and hence, our original network must not have been minimal. There’s our contradiction!

So, we know that any hub must have spokes separated by at least $120^\circ$. How do we know that there are three, separated by exactly $120^\circ$? This is very simple. Suppose two lines enter $H$, separated by more than $120^\circ$. Then there can only be two incoming edges, joining $H$ to some cities $A$ and $B$, since any additional lines would have to be closer than $120^\circ$ to one of these lines, given the result we just proved. So, we have the situation depicted on the left of Figure 10:

Hopefully you can see what goes wrong. If there is a “kink” in the line, with separating angle $\theta \neq 180^\circ$, then we can obtain a shorter network be deleting $H$ and directly connecting $A$ and $B$. Once again, we have a contradiction!

Now, strictly speaking, we can have hubs with only two incoming edges, separated by $180^\circ$. But such a hub is always unnecessary, since all it does is sit on a straight line. In real life, hubs do cost some money (even if it is dramatically less than spokes) so we will never build unnecessary ones! If we always delete these useless hubs, we have the general result advertised above, namely that any hub in a minimal network has three equally spaced spokes.

Exercise 6 (outer rim). Our proof applies to hubs only, but the same style of argument allows you to deduce properties of the cities or fixed nodes $A_1, A_2, \ldots, A_n$. Prove the following properties:

(a) No incoming edges can be separated by less than $120^\circ$.

(b) The number of incoming edges is between $1$ and $3$.

(c) A city has three edges only if it lies at the centre of an equilateral triangle of nodes.

The upshot is that the fixed nodes can have from one to three edges, with three only in special circumstances.

Exercise 7 (easy polygons). Consider cities on the corners of a regular $n$-gon. Show that for $n \geq 6$, the minimal network is just the perimeter with one edge removed.

## 3. Graphs

In this section, we will learn more about the connectivity and layout of minimal networks. The geometry we learned in the last section will be essential!

### 3.1. Trees

In a minimal rail network, how many ways can I get from $A$ to $B$? If the network is connected, then the answer is at least one. That’s the definition of connected! But if the network is minimal, the answer is exactly one. If there is more than one way to get from $A$ to $B$, the network has unnecessary edges and can be pruned to get something shorter. We might worry that pruning redundant paths between $A$ and $B$ could disconnect other cities, but this is never the case. Figure 11 shows why.

Suppose $A$ and $B$ are connected by two paths, labelled $1$ and $2$. The green blob to the left is all the nodes whose paths to $B$ go through $A$ first, and similarly, nodes on the right connect to $A$ through $B$. If two nodes are in the same blob, such as $C$ and $E$, then pruning path $2$ has no effect on whether they are connected. If two nodes are in different blobs, like $C$ and $D$, they can still reach each other using path $1$. So, pruning redundant paths between $A$ and $B$ does not affect whether other nodes connect.

Exercise 8 (path pruning). Generalise the argument above to account for nodes that lie between $A$ and $B$. More precisely, these are nodes which can connect to either $A$ or $B$ without passing through the other, and schematically lie on paths $1$ or $2$.

Once we have completely pruned the network, there is only a single path connecting any two nodes $A$ and $B$. Such a network is called a tree because of the way it branches. For this reason, minimal networks are more often called Steiner trees. An example is shown in Figure 12.

Our reasoning shows that minimal networks are trees. The number of nodes in the network is $N = n + h$, where $n$ is the number of fixed nodes and $h$ counts the hubs added to make the network as short as possible. In Figure 12, for instance, we have $N = 10$. The number of edges $E = 9$ is one less than the number of nodes. This is not a coincidence, and for any tree, it turns out that $E = N - 1$. To prove this, we first need to know that (unlike real life), in graph theory, every tree has leaves.

Exercise 9 (trees and leaves). A leaf is a node with a single edge, e.g. the nodes $J$, $D$, $G$, and $I$ in Figure 12. Part (a) will show what we need, namely that every tree has a leaf, and parts (b) and (c) probe this question a little more deeply. To begin, we will choose a node at random (red in Figure 13) and count the number of steps to each other node. Since there is a unique path between two nodes in a tree, this is well-defined.

(a) Consider the node or nodes furthest from our selected node (orange nodes in Figure 15). Argue that these must be leaves. Hint. If they are not, what are the distances of the neighbours?

(b) Take the furthest node and now find the node or nodes furthest from it. Argue that these are also leaves, and conclude that every tree has at least two leaves.

(c) Show, using an example, that a tree need not have more than two leaves.

The procedure for finding leaves is illustrated in Figure 15. We start with an arbitrarily chosen node $C$, and calculate distances to all other nodes. The most distance node is $I$, which is indeed a leaf. Measuring from $I$, the most distant node is $J$, which is also a leaf.

We now know that every tree has a leaf. If we remove it, and the single edge connecting it to the rest of the graph, we decrease the number of nodes and edges by $1$, $N \to N- 1$ and $E \to E - 1$. We keep doing this until we are down to a single node. This has no edges at all, and since we decrease by $N-1$ nodes to obtain a single node, decreasing $E$ by the same amount gives $0$. Hence, for trees in general,

$E = N - 1.$

### 3.2. Bounding hubs

If we combine our geometric results from earlier with our result about trees, we can constrain the layout of minimal networks further. Recall that each of the $h$ hubs (once we delete the useless ones) has exactly three edges. Each of the $n$ fixed nodes has at least one edge to ensure it is connected to the rest of the network. Thus, the total number of edges $E$ obeys the inequality

$E \geq \frac{1}{2}\left(n + 3h\right).$

In diagrams,

The factor of $2$ occurs because we are counting edges twice in the leftmost and rightmost expression: each end of an edge is associated with a vertex, and we are counting the contribution at each vertex once. From the previous section, we know that $E = N - 1$. This gives

$\frac{1}{2}\left(n + 3h\right) \leq N - 1 = n +h - 1.$

After some algebra, we find that

$h \leq n - 2.$

The maximum number of hubs $h = n - 2$ is achieved when each fixed point has precisely one edge. In this case, the leaves of the graph are exactly the fixed nodes.

### 3.3. Tinkertoys

Combining the geometry and graph results gives us a simple tool for building minimal networks. For $n$ fixed nodes, pick $h = n-2$ hubs, with spokes emerging at angles of $120^\circ$, and connect them together to form a tree. We call this configuration of hubs and spokes a tinkertoy. It is almost rigid, but we can extend the spokes and legs. For $h = n-2$, the minimal network is always a tinkertoy, but see Exercise 10 for an example of a minimal network which is not.

Let’s illustrate the tinkertoy approach for fixed nodes on a rectangle. There is only one tinkertoy, connecting the closest external nodes in pairs, as shown in Figure 14.

The situation for a general quadrilateral is a bit trickier, but can be quickly solved by a computer fiddling with the tinkertoy.

Exercise 10 (minimal network on a rectangle). A rectangle has width $w$ and height $h$, with $w \leq h$. Show that the minimal network has length

$L = w + \sqrt{3}h.$

Exercise 11 (harder polygons). When $h= n-2$, you can use a single tinkertoy, but if $h < n - 2$, you will need multiple tinkertoys. Here we give examples of both.

(a) Find the minimal network connecting a regular hexagon with a node in the centre. (To be clear, this central node is fixed.) You will need multiple tinkertoys!

(b) Draw the tinkertoy for a regular pentagon, and schematically indicate what the minimal network looks like.

For a small number of fixed nodes, tinkertoys are useful. But how useful are they for many nodes? Let’s assume that fiddling with tinkertoys is a quick operation, and we can tell after some fixed time (independent of $n$) whether a particular tinkertoy can connect the fixed nodes. Unlike the cases we’ve look at so far, for large $n$, there are many different tinkertoys to try out. For instance, we show the four different tinkertoys for $h = 6$ (or $n = 8$) in Figure 16.

In Exercise 12, you’ll prove that there are exponentially many tinkertoys with $h$ hubs as $h$ gets large. A brute force approach, which simply checks each tinkertoy, will take an exponential amount of time, and since exponential get large very quickly, this approach is not computationally feasible for large $h$. Unfortunately, there are no algorithmic shortcuts! You can eliminate some tinkertoys from consideration, but as $n$ gets large, there will always be fixed nodes which require you to try an exponential number. This makes the mathematical solution of the minimal network effectively impossible to compute.

Exercise 12 (counting tinkertoys). Counting the precise number of tinkertoys $T_h$ for any number of hubs $h$ is difficult. But we can cheat in two different ways: show that $T_h$ is greater than some exponentially growing sequence, $S_h$; or use “physicist’s induction”, where we calculate for small $h$ and guess the rest of the sequence.

(a) Consider the set of “snake” tinkertoys, where hubs are connected in a single line. Show that, for $h \geq 3$ hubs, there are $S_h = 2^{n-3}$ snake tinkertoys. Since $T_h \geq S_h$, this demonstrates that the total number of tinkertoys grows exponentially.

(b) Calculate the number of tinkertoys from $h = 0$ to $h = 6$. You should be able to find the general sequence $T_h$ by looking for these numbers in the Online Encyclopedia of Integer Sequences. At large $h$, this sequence is indeed exponential, with

$T_h \approx \frac{2^{2n-4}}{\sqrt{\pi}h^{5/2}}.$

## 4. Soap bubbles

Humans are not the only players in the minimisation game. Nature also likes to minimise, and has far more power at her disposal than us mere mortals! While we struggle with tinkertoys, Nature can solve our problems, to good approximation, as effortlessly as a ball falls to the ground. We’ll start by thinking about soap bubble networks in general, and finish by describing soap bubble computers to help us find minimal networks.

### 4.1. Bubble networks

Bubbles are formed when a film of liquid separates two volumes of air. Surface tension tries to pull the bubble surface taut in all directions, which results in the minimisation of area. A soap bubble is spherical, for instance, since this is the smallest surface containing a fixed volume of air. If we attach two plexiglass plates together and dip them in soapy water, a two-dimensional network of bubbles with vertical walls will form. A single bubble will now be a cylinder, rather than a sphere, as in Figure 18.

Typically, bubbles will form closed loops and hence are not minimal networks in our sense, since we explicitly ruled out loops earlier. But bubbles are locally minimal, in the sense that surface tension will tug on bubble walls, and reconfigure small bubbles, until doing so cannot reduce the area. And the result we found earlier, that hubs have three equally separated spokes, is a local condition, since it applies in the small region around a node! This means that in bubble networks, the same property is true: bubble walls always meet in sets of three, separated by $120^\circ$. This even applies to curved bubble walls (which arise from pressure differences), since when you zoom in on a junction these curved walls look straight.

We can use this to explain why two-dimensional foams form hexagons! We don’t need any physics, just the result about hubs, some network theory, and a few reasonable assumptions about what happens as a foam network gets bigger. The main fact about network theory is Euler’s formula, governing the relationship between nodes $N$, edges $E$, and faces $F$ in a graph:

$N - E + F = 2.$

This relation is illustrated for a cube in Figure 19. Note that, for a graph, we count the whole region outside the network as a face! Exercise 14 guides you through a proof of this formula.

Now, for a two-dimensional foam, there are no “external” nodes. Each node is a hub, with three edges, so

$2E = 3N,$

where we include the factor of $2$ as usual because of double counting. Putting this into Euler’s formula, we can eliminate $N$ and find a relation between the number of faces and number of edges:

$\frac{2}{3} E - E + F = 2 \quad \Longrightarrow \quad 3F - E = 6.$

As advertised, we can treat the external face a little differently. Let’s call the internal faces $F’$, so that $F = F’ + 1$. Rewriting in terms of $F’$, we get $3F’ - E = 3$.

What has all this got to do with hexagons? Let’s introduce a number $F_s$ which counts the number of internal faces with $s$ sides, and $E_b$ stand for the number of edges of the outer face. Since we need at least one side to form a face, the total number of internal faces is

$F' = F_1 + F_2 + F_3 + \cdots \,.$

But since each edge is associated with two faces, we can also express edges in terms of these numbers:

$2E = E_b + 1 \cdot F_1 + 2 \cdot F_2 + \cdots + s\cdot F_s + \cdots \,.$

Since the $F_s$ only count the internal faces, and each external edge will be adjacent to one internal face. If we plug these expressions into $3F’ - E = 3$ and multiply everything by two, we finally get

$6 + E_b = 6F' - 2E + E_b = (6-1) \cdot F_1 + (6-2) \cdot F_2 + \cdots + (6-s)\cdot F_s + \cdots \,.$

On the LHS, we have a term involving the number of external edges $E_b$. On the RHS, we have the difference from hexagonality, which we’ll call $D_\text{hex}$. For a face with $s$ sides, $6-s$ tells you how different it is from a hexagon! So, more simply, we have

$D_\text{hex} = 6 + E_b.$

There are a few cute things we can learn from this expression. First of all, the LHS is positive, so $D_\text{hex} \geq 0$. The number of “small” faces $F_1, \ldots, F_5$ places a strong constraint on the number of “large” faces $F_7, F_8, \ldots$, and so on:

$5 \cdot F_1 + 4 \cdot F_2 + \cdots + 1\cdot F_5 \geq + 1 \cdot F_7 + 2\cdot F_8 + \cdots \,.$

So if I count the number of small faces, I can immediately figure out the maximum number of large faces!

But our main goal was to show that foams are approximately hexagonal, so let’s proceed. Suppose the foam has size $\sim L$, where $L$ is some length scale like the radius, and $\sim$ means “scales as”. Then, assuming bubbles have a typical size independent of $L$, the number of external edges $E_b \sim L$. For instance, if bubbles have average edge length $\ell$, independent of $L$, and we have a circular foam of radius $L$, then $E_b = (\pi/\ell)L$. It follows that, for large $L$,

$D_\text{hex} = 6 + E_b \sim L.$

The “density” of difference from hexagonality $d_\text{hex}$ is just the total difference divided by the area of the foam. This should scale as $L^2$, so that

$d_\text{hex} \sim \frac{D_\text{hex}}{L^2} \sim \frac{1}{L}.$

As the foam becomes larger, differences from hexagonality become increasingly rare! A typical bubble in a two-dimensional foam has six sides. You can repeat the argument, but for a set of contiguous bubbles in a larger foam, in Exercise 13.

Exercise 13 (bubble blobs). A “bubble blob” is a set of contiguous bubbles in a (two-dimensional) foam. Let $E_o$ denote the number of edges extending outward from the boundary, and $E_i$ the number extending inward.

(a) Explain why the difference from hexagonality is now

$D_\text{hex} = 6 + E_i - E_o.$

(b) Repeat the scaling argument above, and conclude that in a large blob, departures from hexagonality become harder and harder to find.

Exercise 14 (Euler’s formula). Our goal here will be to prove Euler’s famous formula:

$N - E + F = \chi = 2,$

where $N$ is the number of nodes, $E$ the number of edges, and $F$ the number of faces, in a network where edges do not cross. We have also defined $\chi = N - E + F$, called the Euler characteristic, for notational convenience. Although we are discussing this in the context of graphs, it holds in general for three-dimensional polyhedra like cubes and dodecahedra. In fact, the two are equivalent, since we can always flatten a polyhedron to get its net, which is a network.

First, we will establish Euler’s formula for networks made out of triangles.

(a) Show that a lone triangle in the plane obeys Euler’s formula. Remember, we count the area outside the triangle as a face!

(b) Suppose I have a network which obeys Euler’s formula. Add a triangle (two edges and a node) to an external edge, and explain why the Euler characteristic doesn’t change, $\Delta \chi = 0$. Conclude that the new network obeys Euler’s formula.

(c) Explain why any network composed entirely of triangles obeys Euler’s formula.

Now, we can generalise to any network without crossings.

(d) Consider a face, i.e. loop, in such a network. Describe a procedure to add edges so that the face is split into triangles.

(e) Show that, after your procedure in part (d), $\Delta \chi = 0$.

(f) Combine your results to conclude that any network without crossings obeys $\chi = 2$.

### 4.2. Soap bubble computers

Soap bubbles find local minima quickly because of the laws of physics. Can we somehow “hack” soap bubbles, and turn them into analogue computers which find Steiner trees? Sort of. Computer scientist Scott Aaronson has persuasively argued that hard problems (such as finding Steiner trees) cannot be solved quickly, in the general case, by any physical mechanism. So as a matter of physical principle, we should not expect soap bubbles to quickly find arbitrary minimal networks.

But this doesn’t prevent soap bubbles from giving quick and correct solutions to some problems, nor does it forbid them from approximating the answer. Indeed, there are quick approximation algorithms for regular computers, so we shouldn’t be surprised if soap bubbles can do the same. So, can we quickly (and perhaps approximately) find minimal networks by dipping them in soapy water? The answer is yes!

The key is to give the bubble walls something to hold onto (Figure 23). In a foam, they can only hold onto themselves, and we get the non-crossing bubble networks described above. But if we drill some screws through the plexiglass, these will act like fixed nodes, and walls can form between these screws and hubs arising from the junction of three bubble walls. This analogue computer isn’t perfect, since the soap bubbles can converge on a locally minimal tinkertoy which connects the screws, but is not globally minimal. But it really is a lucky dip, and you can sometimes strike the true minimum, particularly for smaller problems where only one tinkertoy works. See this video for a demonstration, and the classic What is Mathematics? for discussions of how soap bubble computers can be applied to other mathematically difficult problems.

Soap bubbles do not magically solve problems thought to be impossible. But they do demonstrate strikingly that physical objects are computers, and the laws of physics do computational work. To expand on this statement, let’s consider the traditional division of computers into hardware and software. Hardware is the physical substrate the computer is built on. It tells us which operations are basic, and therefore computationally cheap. Software is a layer of instruction and abstraction above the hardware, telling the computer how to represent data and manipulate it to do useful things.

In the soap bubble computer, the hardware is clearly the bubbles themselves, and local area minimisation a cheap operation by virtue of the laws of physics. The software is nothing more complicated than drilling screws, dipping, and waiting for bubbles to form. In terms of computers built out of physically interesting hardware, this is just the tip of the iceberg! One of the big technological revolutions now in the progress is the advent of quantum computing, leveraging the weirdness of quantum mechanics. Perhaps more surprisingly, there is a sense in which black holes, DNA, and animal brains are all analogue computers, like the soap bubbles, with certain cheap operations. There are more computers in heaven and earth than are dreamt of in our philosophy!

Exercise 15 (trains and soap bubbles). The first mathematician to consider minimal networks on four points was the great Carl Friedrich Gauss (1777 – 1855). In an 1836 letter to his friend, the astronomer Schuhmacher, Gauss asked:

How can a railway network of minimal length which connects the four German cities Bremen, Harburg, Hannover, and Braunschweig be created?

The cities are drawn in Figure 25:

Your task is simple: build a soap bubble computer, and use it solve the original rail network problem! Figure 25 also gives the GPS coordinates, which you may find useful in placing your screws.

#### Acknowledgments

I’d like to thank Rafael Haenel, Pedro Lopes and Haris Amiri for feedback, and Rafael particularly for his encouragement. Some of this material was tested at the UBC Physics Circle, as part of a workshop run by Geering Up and Diversifying Talent in Quantum Computing, so I would also like the students for their participation and feedback!


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Home > PC > Chapter 9 > Lesson 9.3.5 > Problem9-160

9-160.

Solve for $x$ in the simplest manner (no calculator or Quadratic Formula).

1. $5x^{2} = 15x$

1. Do NOT divide both sides by $x$. This will eliminate one of your two answers.
2. Instead, move all terms to one side and set = to $0$
Factor. Solve.

1. $2\left(x − 6\right)^{2} + 5 = 23$

$2\left(x − 6\right)^{2} = 18$
$\left(x − 6\right)^{2} = 9$
$x − 6 = ± 3$
$x = ±3 + 6$
$x = 9$ $x = 3$


text:
# First Law of Thermodynamics

...an internal energy function, $U$, exists that is a state function.

That is, $dU$ is an exact differential.

• Both mechanical work and heat have units of energy,
• But neither one is path independent: It is usually the case that$$\oint \delta Q \neq 0,$$ and $$\oint \delta W \neq 0.$$

### The hypothesis of energy conservation

1. We suspect that there is a unique, total energy, $E$, for any system (that is to say, it's a state function),
2. The total energy $E$ includes: kinetic energy, potential energy, internal energy ($U$), chemical energy, plus other forms of energy,
3. When heat $\Delta Q$ is added to a system, or when the system performs work $\Delta W$, the total energy, $E$, of a system changes according to: $$dE = dE_{\rm pot} + dE_{\rm K.E.} +dU+dE_{\rm chem} +...= \delta Q - \delta W.$$
4. It's possible to do work and/or transfer heat to a system with no changes in the non-$U$ forms of energy: K.E., potential, chemical, etc. When this is the case, we say the internal energy changes according to: $$dU = \delta Q - \delta W.$$

Convention:

$\delta Q \gt 0$ means that heat energy flows in to the system,

$\delta W \gt 0$ is the work that a system does on its surroundings, so this is an energy flow out of the system.

We write the change in internal energy as an exact differential because it is this particular combination of inexact differentials is exact, that is $$\oint dU =\oint$\delta Q - \delta W$ = 0.$$

### "Random" motion

The "kinetic energy" of an object or container of atoms only depends on the speed of its center of mass, $v_{\text{CM}}$. And yet we know that the atoms in the object/container are also in motion...

We shall see that $T\propto \text{KE}_\text{ave}$.

A portion of what we mean by "internal energy" is the kinetic energy of these atoms.

Play around with this gas property simulator to see how either heat or doing work on the system can cause the atoms to speed up.

For a monatomic gas, it is a pretty good approximation that *all* the internal energy is in the form of kinetic energy of the atoms. And under these circumstances, $U\propto T$ alone. [See problem 3.7]

What other kinds of internal energy might there be besides the kinetic energy of the atoms?

### Equivalence of mechanical work and heat

Joule's experiment (right) measured the equivalence of mechanical work and heat. In what sense?

The system of water at right is insulated, so no heat can flow in ($\delta Q=0$). All the mechanical energy $\delta W_{\rm dis}$ is dissipated in the water. So the dissipative work must have gone towards changing the internal energy. $$dU = -\delta W_{\rm dis}.$$

On the other hand, the same change in internal energy could be brought about by isolating the system mechanically and only supplying heat. $$dU = \delta Q$$

So, with these restrictions, a certain quantity of dissipative work, or that same quantity of heat energy will always change the system in the same way.

### Problem 3.8

A gas is contained in a cylinder fitted with a frictionless piston, and is taken from the state $a$ to the state $b$ along the path $acb$. In all, 80 J of heat flowed into the system, and the system did 30 J of work.

a.) How much heat flows into the system along $adb$ if the work done by the gas is 10 J?

### Adiabatic work is path independent

Normally, the amount of work to get from one state to another depends on the path taken.

But now imagine some adiabatically isolated system such that no heat can enter the system: It can only interact mechanically with its surroundings. Since $\delta Q=0$, the first law says for this particular case $$dU = -\delta W_{\rm ad}.$$

The work done by the system along one of the paths in the diagram shown is... $$\int_{\mathcal P} \delta W_{\rm ad} = -\int_{\mathcal P} dU = -\Delta U.$$

Since $U$ is a function of the thermodynamic state, it is path independent, so the adiabatic work for any process is path independent.

This also gives us a means of measuring changes in internal energy, by measuring the adiabatic work done.

Unfortunately, an adiabatic path does not necessarily exist between any two randomly chosen points.

When should work done on a system should be considered work and when it should be considered heat? (E.g. electrical work)

• Depends on system definition.
• Work done ultimately boils down to being able to lift a weight.

Problem - Carter 3.7: Two kilomoles of a monatomic ideal gas are at a temperature of 300 K. The gas expands reversibly and isothermally to twice its original volume. Calculate (a) the work done by the gas; (b) the heat supplied. (For such an ideal gas, $U\propto T$ alone, so $\Delta U=0$ in an isothermal process).


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# Special Talk - Saharon Shelah

Simplicity and universality

Fixing a complete first order theory T, countable for transparency, we had known quite well for which cardinals T has a saturated model. This depends on T of course - mainly of
whether it is stable/super-stable. But the older, precursor notion of having
a universal notion lead us to more complicated answer, quite partial so far, e.g
the strict order property and even SOP_4 lead to having "few cardinals"
(a case of GCH almost holds near the cardinal). Note  that eg GCH gives a complete
uninteresting answer and so is the situation eg in the Easton model.
It seems that necessarily the answer involves sufficient
conditions for non-existence of a universal model
(in ZFC) and consistency for additional existence.
We conjecture that simplicity of T is crucial in answering this.
We shall speak on  recent advances: a new criterion covers the simplest non-
simple theory, so called $T_feq$. We may also speak complementary results
on having a universal model for simple theories.

## Date:

Sun, 23/06/2019 - 16:00 to 18:00

## Location:

Manchester Building, Room 110


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\begin{frame}
\frametitle{Antiderivatives / Integrals}

\begin{exampleblock}{}
A particle moves in a straight line and has
\begin{itemize}
\item acceleration $a(t) = 6t + 4$
\item initial velocity is $v(0) = -6$cm/s
\item initial displacement is $s(0) = 9$cm
\end{itemize}
Find the position function $s(t)$.
\pause\medskip

The velocity is an antiderivative of the acceleration:
\begin{talign}
v(t) = \mpause[1]{3t^2} \mpause{+ 4t}\mpause{ + C}
\end{talign}
\pause\pause\pause\pause
As $v(0) = -6$cm/s, it follows that $C = \pause -6$.
\pause\medskip

The position function is an antiderivative of the velocity:
\begin{talign}
s(t) = \mpause[1]{t^3} \mpause{+ 2t^2}\mpause{ - 6t}\mpause{+ D}
\end{talign}
\pause\pause\pause\pause\pause
As $s(0) = 9$cm/s, it follows that $D = \pause 9$.
\pause\medskip

Thus the position function is:
\begin{talign}
s(t) = t^3 + 2t^2 - 6t + 9 \quad \text{in cm}
\end{talign}
\end{exampleblock}
\end{frame}


text:
### J-class weighted shifts on the space of bounded sequences of complex numbers

Costakis G, Manoussos A (2008)
Integral Equations and Operator Theory 62(2): 149-158.

Zeitschriftenaufsatz | Veröffentlicht | Englisch

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Autor*in
Costakis, George; Manoussos, AntoniosUniBi
Abstract / Bemerkung
We provide a characterization of $J$-class and $J^{mix}$-class unilateralweighted shifts on $l^{\infty}(\mathbb{N})$ in terms of their weight sequences.In contrast to the previously mentioned result we show that a bilateralweighted shift on $l^{\infty}(\mathbb{Z})$ cannot be a $J$-class operator.
Stichworte
hypercyclic operators; unilateral and bilateral weighted shifts; J(mix)-class operators; J-class operators
Erscheinungsjahr
2008
Zeitschriftentitel
Integral Equations and Operator Theory
Band
62
Ausgabe
2
Seite(n)
149-158
ISSN
0378-620X
eISSN
1420-8989
Page URI
https://pub.uni-bielefeld.de/record/2562311

### Zitieren

Costakis G, Manoussos A. J-class weighted shifts on the space of bounded sequences of complex numbers. Integral Equations and Operator Theory. 2008;62(2):149-158.
Costakis, G., & Manoussos, A. (2008). J-class weighted shifts on the space of bounded sequences of complex numbers. Integral Equations and Operator Theory, 62(2), 149-158. doi:10.1007/s00020-008-1621-6
Costakis, G., and Manoussos, A. (2008). J-class weighted shifts on the space of bounded sequences of complex numbers. Integral Equations and Operator Theory 62, 149-158.
Costakis, G., & Manoussos, A., 2008. J-class weighted shifts on the space of bounded sequences of complex numbers. Integral Equations and Operator Theory, 62(2), p 149-158.
G. Costakis and A. Manoussos, “J-class weighted shifts on the space of bounded sequences of complex numbers”, Integral Equations and Operator Theory, vol. 62, 2008, pp. 149-158.
Costakis, G., Manoussos, A.: J-class weighted shifts on the space of bounded sequences of complex numbers. Integral Equations and Operator Theory. 62, 149-158 (2008).
Costakis, George, and Manoussos, Antonios. “J-class weighted shifts on the space of bounded sequences of complex numbers”. Integral Equations and Operator Theory 62.2 (2008): 149-158.

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arXiv: 0704.3349