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RationalWiki will be going through an upgrade cycle this weekend. This will cause some down time, details are at the tech blog. As a reminder, we try and add information to the tech blog during any outage, you can always refer to it for details or as a form of communication if the site is down. from Tmtoulouse (Talk), group Site wide (urgent) at 15:23, 18 April 2014 NextPrevious

# Millennium Prize Problems

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 Part of the series on 2+2=4 v - t - e

The Millennium Prize Problems area set of seven problems in mathematics that were stated by the Clay Mathematics Institute in 2000. A correct solution to any of the problems results in a US$1,000,000 prize (popularly known in the mathematics world as a Millennium Prize) being awarded by the institute. The Clay Mathematics Institute, or CMI, is a private, non-profit foundation, based in Providence, Rhode Island. As of 2013, six of the problems still remain unsolved. The Poincaré conjecture is the only Millennium Prize Problem to be resolved, with a solution published by Grigori Perelman in 2003. Perelman was also awarded the Fields medal, but he declined both awards, saying "I don't want to be on display like an animal in a zoo."[1] The list of problems was also acknowledged as a nod to the list of 23 problems presented by the German mathematician David Hilbert in 1900. The seven problems were presented in Paris in 2000 in the order given below. One of Hilbert's original problems (the Riemann hypothesis) is on the new list, and another (The Birch and Swinnerton-Dyer conjecture) follows on directly from the solution of the tenth Hilbert problem. And as exciting as the thought of winning a million bucks might be, if this is the first time you've heard of the Millennium Prizes, then there's a fairly strong chance you don't have the 10-15 years of formal mathematics study you'd probably need to even get started on solving one of these. ## Contents ## P versus NP "P versus NP" is a (very 20th century) problem which emerged from computer science. The problem was introduced in 1971 by Stephen Cook in his paper "The complexity of theorem proving procedures"[2] and is considered by many to be the most important open problem in computer science.[3] Obviously in order to understand the nature of the question, it is necessary to know what P and NP actually refer to. Both describe computer algorithms (repeating processes) which eventually reach a conclusion of some form. First, consider a computer program that is capable of "verifying" an answer to a given problem. For example, assume the problem is: "From a list of one million randomly chosen numbers, find a list of 1000 numbers that add up to exactly one million". If you go then through the list and choose the required subset of 1000 numbers, the program will easily be able to verify whether or not your list adds up to one million (using simple arithmetic). There is a linear relationship between the number of members in the set (1000) and the number of steps required to solve the problem (1000 individual additions). A linear relationship is called a 'polynomial' relationship, so we will call these "answer verification" programs P. Now consider the task of choosing the required subset of 1000 numbers in the first place. Currently there is no known way of quickly choosing all possible subsets, other than sheer brute force (ie. testing every possible combination, of which there are 2n-1 possibilities). If you add another member to the subset, the number of solutions increases from 2n to 2n+1, which is exponential or "non-polynomial" growth. So we will call this "problem solving" programs NP. The problem thus asks, is there some way of coming up with a nifty way of writing "problem solving" programs so that we can always avoid this "non-polynomial" growth? In other words, can there be a program that solves the problem in the same polynomial time as the verification program? If the answer is "yes", then $\scriptstyle P = NP$. If the answer is "no", then $\scriptstyle P \neq NP$ This is important because if $\scriptstyle P = NP$ then somewhere in "mathematics land" there is a quick and efficient way of cracking all those data security systems used by intelligence agencies and by the banks in financial transfers. This is generally regarded as a "bad thing" (at least by the banks). On the plus side, most mathematicians are quite confident that $\scriptstyle P \neq NP$. The problem is that no-one knows how to prove it yet. ## The Hodge conjecture Some of the Millenium problems defy easy description, and this is one of them. The official statement of this problem is as follows: The Hodge conjecture asserts that for particularly nice types of spaces called projective algebraic varieties, the pieces called Hodge cycles are actually (rational linear) combinations of geometric pieces called algebraic cycles..[4] In math jargon, this is: $\operatorname{Hdg}^k(X) = H^{2k}(X, \mathbf{Q}) \cap H^{k,k}(X).$ So basically, unless you already know what a projective complex manifold is, and what cohomology means, there is no way you'll ever understand what this problem is even about. Regardless, do let us know when you've solved it. ## The Poincaré conjecture (solved) How topology turns a coffee cup into a doughnut (torus) and back. The Poincare Conjecture was an important conjecture in the field of topology, although its importance wasn't apparent at the time it was proposed. Topology is the mathematics of deformation; it considers surfaces to be independent of their dimensions. Hence topologists are jokingly called "people who cannot tell the difference between a coffee cup and a donut" (because both are a continuous object with a single hole in the structure - see image). The Poincaré conjecture involves a claim about objects in multi-dimensional space, and whether (under certain conditions) the object can be deformed to a sphere. Poincaré believed it to be true when he stated it in 1904. It wasn't regarded as a particularly important conjecture at the time, but it subsequently acquired great notoriety chiefly because on several occasions during the 20th century someone would claim they had solved the problem, only to later find a flaw. By the 1980s all attempts to solve the problem had still failed, but had demonstrated that the problem had important implications in other areas of topology that Poincaré had never dreamed of. Hence it went from being an obscure comment in a 1904 paper to a major area of mathematical research. The final solution was published online to a maths forum in 2003 by a socially withdrawn Russian named Grigori Perelman. After three years of intense scrutiny from leading universities such as MIT in the USA and Sun Yat-sen in China, his work was accepted as definitive in 2006. How Perelman developed his thinking is a mystery as he refuses to talk to anyone, and there is speculation he may have ceased doing mathematics altogether. He was officially awarded the Millennium Prize in 2010, but he refused to accept it. ## The Riemann hypothesis The Riemann hypothesis was the eighth problem on Hilbert's list back in 1900, and is still regarded as an important unsolved problem a century later. It is the only problem to appear on both lists. A proof or disproof of this would have far-reaching implications in number theory, especially for the distribution of prime numbers. It also implies strong bounds on the growth of many other arithmetic functions. The Riemann hypothesis concerns the properties of a certain object called the Riemann zeta function. The conjecture claims that all nontrivial zeros of the analytic continuation of the Riemann zeta function have a real part of 1/2. ## Yang–Mills existence and mass gap The Yang–Mills problem isn't so much a problem in mathematics as it is an open issue in mathematical physics. It is fair to say that it is really complicated. Really, REALLY complicated. Regardless, here goes an attempt at a basic explanation. We have quantum physics, which in the 1960s evolved into the existence of the Standard Model, and by the 1980s this had been refined to incorporate quantum field theory (QFT). One aspect of QFT is the existence of gauge particles which move back and forth between other particles, creating what we call "force". The most widely accepted theoretical framework for explaining QFT is called Yang-Mills Theory, and (among many other things) Yang-Mills predicts that the lightest gauge particles will still have mass. All experimental evidence seems to agree with this. (To help you understand the naming of the problem, the difference between the energy of the vacuum and the lightest particle is called the "mass gap".) While this is all well and good, we're still a long way from being able to pin down what is exactly happening. The problem with QFT is it all happens in four (or more) dimensions, and there are dozens of different entities involved, so frankly, everyone is just a bit baffled. For an analogy, we're at the point in time where we know if we fire a catapult, the ball will land "over there somewhere", but we don't yet know calculus and so we aren't able to accurately predict where the ball will end up. So solving this will be the quantum theory equivalent of developing calculus. The solution should also predict that the lightest gauge particles have mass. ## Navier–Stokes existence and smoothness Of the seven Millennium Prize problems, the Navier–Stokes have the most practical applications, particularly in resolving the issue of fluid turbulence. The Navier–Stokes equations were developed in the 19th century and describe the motion of fluids. Despite having been around for quite some time, they are still poorly understood. The problem is to make progress toward a mathematical theory that will give insight into these equations. The official definition of the problem is: Prove or give a counter-example of the following statement: In three space dimensions and time, given an initial velocity field, there exists a vector velocity and a scalar pressure field, which are both smooth and globally defined, that solve the Navier–Stokes equations. ## The Birch and Swinnerton-Dyer conjecture The Birch and Swinnerton-Dyer conjecture also has a long history. Hilbert's tenth problem dealt with a more general type of equation, and in that case it was proven that there is no way to decide whether a given equation even has any solutions. This problem deals with a specific type of equation, those defining elliptic curves over the rational numbers. The conjecture is that there is a procedural way to determine whether such equations have a finite or infinite number of rational solutions. ## Footnotes 1. Russian maths genius Perelman urged to take$1m prize bbc.co.uk, Wednesday, 24 March 2010.
2. Cook, Stephen (1971). "The complexity of theorem proving procedures". Proceedings of the Third Annual ACM Symposium on Theory of Computing. pp. 151–158.
3. Lance Fortnow, The status of the P versus NP problem, Communications of the ACM 52 (2009), no. 9, pp. 78–86. Template:Doi
4. Hodge Conjecture, Clay Mathematics Institute


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You are currently browsing the category archive for the ‘MATH6000’ category.

“Straight-Line-Graph-Through-The-Origin”

The words of Mr Michael Twomey, physics teacher, in Coláiste an Spioraid Naoimh, I can still hear them.

There were two main reasons to produce this straight-line-graph-through-the-origin:

• to measure some quantity (e.g. acceleration due to gravity, speed of sound, etc.)
• to demonstrate some law of nature (e.g. Newton’s Second Law, Ohm’s Law, etc.)

We were correct to draw this straight-line-graph-through-the origin for measurement, but not always, perhaps, in my opinion, for the demonstration of laws of nature.

The purpose of this piece is to explore this in detail.

## Direct Proportion

Two variables $P$ and $Q$ are in direct proportion when there is some (real number) constant $k$ such that $P=k\cdot Q$.

Quadratics are ubiquitous in mathematics. For the purposes of this piece a quadratic is a real-valued function $q:\mathbb{R}\rightarrow \mathbb{R}$ of the form

$q(x)=ax^2+bx+c$,

where $a,\,b,\,c\in \mathbb{R}$ such that $a\neq 0$. There is a little bit more to be said — particularly about the differences between a quadratic and a quadratic function but for those this piece is addressed to (third level: non-maths; all second level), the distinction is unimportant.

## Geometry

The basic object we study is the square function, $s:\mathbb{R}\rightarrow \mathbb{R}$, $x\mapsto x^2$:

All quadratics look similar to $x^2$. If $a>0$ then the quadratic has this $\bigcup$ geometry. Otherwise it looks like $y=-x^2$ and has $\bigcap$ geometry

The geometry dictates that quadratics can have either zero, one or two real roots. A root of a function is an input $x$ such that $f(x)=0$. As the graph of a function is of the form $y=f(x)$, roots are such that $y=f(x)=0\Rightarrow y=0$, that is where the graph cuts the $x$-axis. With the geometry of quadratics they can cut the $x$-axis no times, once (like $s(x)=x^2$), or twice.

There are a number of ways of explaining why you cannot divide by zero. Here are my two favourites.

## Any Set of Numbers Collapses to a Single Number

How old are you? Zero years old.

How tall are you? Zero metres old.

How many teeth do you have? Zero.

How many Superbowls has Tom Brady won? Zero

Yep, if you allow division by zero you only end up with one number to measure everything with.

## Assessment 5

Assessment 5 will be on Tuesday, 6 December 2016.

The following  group will sit the assessment at 17:15 (arrive around 17:05):

• COMP1B-X

The following group will sit the assessment at 18:15 (arrive around 18:05)

• COMP1B-Y

Keep an eye on your CIT Blackboard and email for the latest and definitive assessment information.

## Week 12

We looked at graphing and mathematical modelling.

## Week 13

Two tutorials:

• COMP1B-Y: Monday 10:00 in F1.3
• COMP1B-X: Monday 15:00 in B251
• COMP1B-X and 1B-Y: Tuesday 14:00 in B225

## Study

Please feel free to ask me questions about the exercises via email or even better on this webpage.

## Student Resources

Please see the Student Resources tab on the top of this page for information on the Academic Learning Centre, etc.

You will need graph paper for the remaining MATH6000 classes. I advise that you gang up and buy graph paper together in the shop downstairs.

## Assessment 4 Results

If you require a result of 40% or more on Assessment 5, you will receive an email regarding your results on or before Wednesday. Otherwise you must wait until your results are released, in January.

## Assessment 5

Assessment 5 will be on Tuesday, 6 December 2016.

The following  group will sit the assessment at 17:15 (arrive around 17:05):

• COMP1B-X

The following group will sit the assessment at 18:15 (arrive around 18:05)

• COMP1B-Y

Keep an eye on your CIT Blackboard and email for the latest and definitive assessment information.

## Week 11

We began our study of coordinate geometry.

## Week 12

We will look at graphing and mathematical modelling.

## Assessment 4

Assessment 4 will be on this Wednesday, 23 November 2016.

The following  group will sit the assessment at 17:15 (arrive around 17:05):

• COMP1B-X

The following group will sit the assessment at 18:15 (arrive around 18:05)

• COMP1B-Y

Keep an eye on your CIT Blackboard and email for the latest and definitive assessment information.

## Week 10

We looked at scientific notation, unit conversion and logarithms.

## Week 11

We will finish our study of logarithms and begin our study of coordinate geometry.

## Study

Please feel free to ask me questions about the exercises via email or even better on this webpage.

## Student Resources

Please see the Student Resources tab on the top of this page for information on the Academic Learning Centre, etc.

## Assessment 4

Assessment 4 will be on this Wednesday, 23 November 2016.

The following  group will sit the assessment at 17:15 (arrive around 17:05):

• COMP1B-X

The following group will sit the assessment at 18:15 (arrive around 18:05)

• COMP1B-Y

Keep an eye on your CIT Blackboard and email for the latest and definitive assessment information.

## Week 9

We began looking at indices.

## Week 10

We will look at scientific notation, unit conversion, functions and logarithms.

## Assessment 3

Assessment 3 will be on the Wednesday of Week 9, 9 November 2016.

The following  group will sit the assessment at 17:15 (arrive around 17:05):

• COMP1B-X

The following group will sit the assessment at 18:15 (arrive around 18:05)

• COMP1B-Y

Keep an eye on your CIT Blackboard and email for the latest and definitive assessment information.

## Week 8

We looked at quadratic and simultaneous equations.

## Week 9

We will look at functions and begin looking at indices and scientific notation.

This post follows on from this post where the logic for the below is discussed. I am not going to define here what easy means!

Here is the strategy/guiding principle:

Fundamental Principle of Solving ‘Easy’ Equations

Identify what is difficult or troublesome about the equation and get rid of it. As long as you do the same thing to both numbers (the “Lhs” and the “Rhs”), the equation will be replaced by a simpler equation with the same solution.

I am emailing a link of this to everyone on the class list every week. If you are not receiving these emails or want to have them sent to another email address feel free to email me at jpmccarthymaths@gmail.com and I will add you to the mailing list.

## Assessment 3

Assessment 3 will be on the Wednesday of Week 9, 9 November 2016.

The following  group will sit the assessment at 17:15 (arrive around 17:05):

• COMP1B-X

The following group will sit the assessment at 18:15 (arrive around 18:05)

• COMP1B-Y

Keep an eye on your CIT Blackboard and email for the latest and definitive assessment information.

## Week 7

We looked at equations.

## Week 8

We will look at quadratic and simultaneous equations.

format: plain


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# Why does $…$ do the wrong thing at the top of a page?

Upon reading What are the differences between , [, align, equation and displaymath?, I decided I really wanted to understand what was going on inside $...$ and so I devised some tests. What I discovered confounds me. Why does $...$ do the wrong thing at the top of a page? The definition of $ goes out of its way to suppress additional vertical space by invoking \nointerlineskip, but then adds vertical space implicitly by making an empty box 60% the width of the line: \makebox[.6\linewidth]{}. I can see how this might be desirable in some cases, but it's certifiably the wrong thing to do for a displayed formula at the top of a page, as it causes a whole extra line's worth of unwanted space at the top. This happens whether the result of a forced page break or the result of natural break after a paragraph. Here is a minimal working example: \documentclass{article} \begin{document} \newpage y=x^2 \newpage \[y=x^2$ \end{document}  Notice how the second equation is much lower on the page than the first. Does anyone know why this is? Is it a flaw in $...$? Is it by design? Note that it also does this inside of (at the top of) a \parbox. - This is fixed with the loading of amsmath, since it redefines $ and $. – Werner Feb 8 '12 at 19:34 Aha! Now all my test-cases of ... vs. $...$ line up perfectly. Cool. – Todd Lehman Feb 8 '12 at 19:57

The amsmath package redefines the display math equation definitions of $ and $ which corrects this behaviour. So, all you need to do is use

\usepackage{amsmath}% http://ctan.org/pkg/amsmath

-
That's a pretty good reason always to load amsmath as part of a boilerplate! – Todd Lehman Feb 8 '12 at 20:41

The simple answer is unfortunately that the whole set of core LaTeX "math" environments are flawed (if not to say broken). Compare, for example, the result of $a=b$ with \begin{eqnarray} a&=&b \end{eqnarray} -- the spacing is all wrong.

Your example is just another instance of this, however in that case for some reason a deliberate design decision by Leslie (or so I believe from the code).

In any case, if doing any serious math the use of amsmath is the way to go; the basic implementations in core LaTeX should rather be avoid.

Now in case anybody asks why isn't this fixed? The answer is document typesetting compatibility with all documents produced in the last 25+ years that have been written with just core LaTeX. The fix is there since 1994 and it is called

\usepackage{amsmath}

-

You have two good answers already; I'd like to explain why you don't really have to care. Under usual circumstances there should be no displayed formula at the top of a page. From the TeXbook, page 189:

Plain TeX sets \predisplaypenalty=10000, because fine printers traditionally shun displayed formulas at the very top of a page.

The effect of \predisplaypenalty=10000 is that no page breaks are allowed before a displayed formula. Thus your premise is wrong in

This happens whether the result of a forced page break or the result of natural break after a paragraph.

Yes, after a forced page break it can happen, but "naturally" it doesn't happen unless you change \predisplaypenalty. Moreover, forced page breaks should only be used in special circumstances, and never before a displayed formula :-) I also wouldn't recommend starting a \parbox with a displayed formula.

-


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[This article was first published on R – Statistical Modeling, Causal Inference, and Social Science, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)
Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

This week, more Stan!

• Charles Margossian is rock star of the week, finishing off the algebraic solver math library fixture and getting all plumbed through Stan and documented. Now you can solve nonlinear sets of equations and get derivatives with the implicit function theory all as part of defining your log density. There is a chapter in the revised manual

• The entire Stan Development Team, spearheaded by Ben Goodrich needing fixes for RStan, is about to roll out Stan 2.17 along with the interfaces. Look for that to begin trickling out next week. This will fix some further install and error message reporting issues as well as include the algebraic solver. We are also working on moving things toward Stan 3 behind the scenes. We won’t just keep incrementing 2.x forever!

• Ben Goodrich fixed the inlining declarations in C++ to allow multiple Stan models to be linked or built in a single translation unit. This will be part of the 2.17 release.

• Sean Talts is still working on build issues after Travis changed some config and compilers changed everywhere disrupting our continuous integration.

• Sean is working with Michael Betancourt on the Cook-Gelman-Rubin diagnostic and have gotten to the bottom of the quantization errors (usingly on 1000 posterior draws and splitting into too many bins).

• Imad Ali is looking ahead to spatio-temporal models as he wraps up the spatial models in RStanArm.

• Yuling Yao and Aki Vehtari finished the stacking paper (like model averaging), adding references, etc.

• Yuling has also made some updates to the loo package that are coming soon.

• Andrew Gelman wrote papers (with me and Michael) about R-hat and effective sample size and a paper on how priors can only be understood in the context of the likelihood.
• Jonah Gabry, Ben and Imad have been working on the paper for priors based on $R^2$

• Andrew, Breck Baldwin and I are trying to put together some educational proposals for NSF to teach intro stats, and we’re writing more grants now that it’s grant season. All kinds of interesting things at NSF and NIH with spatial modeling.

• Jonah Gabry continues the ggplot-ification of the new Gelman and Hill book.

• Ben Goodrich has been working on multivariate effective sample size calculations.

• Breck Baldwin has been working with Michael on courses before Stan (and elsewhere).

• Jonah Gabry has been patching all the packagedown references for our doc and generally cleaning things up with cross-references and organization.

• Mitzi Morris finished adding a data qualifier to function arguments to allow propagation of data-only-ness as required by our ODE functions and now the algebraic solver.

• Dootika Vats was visiting Michael; she’s been working on multivariate effective sample sizes, which considers how bad things can be with linear combinations of parameters; scaling is quadratic, so there are practical issues. She has previously worked on efficient and robust effective sample size calculations which look promising for inclusion in Stan.

• Bob Carpenter has been working on exposing properties of variables in a Stan program in the C++ model at runtime and statically.

The post Stan Weekly Roundup, 11 August 2017 appeared first on Statistical Modeling, Causal Inference, and Social Science.


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# 4. (20pts) Consider a mixture of 50.0 mL of 0.100 M HCl and 50.0 mL of...

###### Question:

4. (20pts) Consider a mixture of 50.0 mL of 0.100 M HCl and 50.0 mL of 0.100 M acetic acid. Acetic acid has a K. of 1.8 x 10-5. a. Calculate the pH of both solutions before mixing. b. Construct an ICE table representative of this mixture. c. Determine the approximate pH of the solution. d. Determine the percent ionization of the acetic acid in this mixture.

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text:
# Counterparty credit exposure won't spark the next Lehman

## Curbing of riskiest exposures and shedding of assets means banks in far better shape 10 years on

The tenth anniversary of Lehman Brothers’ demise has brought with it speculation on the potential source of the next crisis. Evidence from recent risk disclosures suggests a future panic is unlikely to be triggered by the failure of a bulge-bracket dealer as in 2008. Not only are the biggest banks far smaller and less interconnected than in their pre-crisis heydays, they are also carrying less risk.

Data from the European Banking Authority shows the European Union’s largest banks collectively shed $3.2 trillion in assets in the five years to end-2017, with the top four – Barclays, Deutsche Bank, BNP Paribas and HSBC – alone accounting for$1.74 trillion of the shrinkage.

US banks have also slimmed down substantially, while evidence published by the Federal Reserve in July suggests mid-tier firms are curbing their growth to duck enhanced regulations linked to asset size.

That firms have taken their axe to balance sheets bloated by complex, hard-to-value assets and webs of bilateral derivatives to reduce their risks is old news. There are two post-crisis trends, however, that merit fresh comment.

First of all, dealers on both sides of the Atlantic have focused their asset curbing on those exposures whose required capital cannot be calculated using internal risk models.

Fourteen EU and Swiss global systemically important banks (G-Sibs) have cut credit exposures assessed using regulator-set standardised approaches at twice the speed of those governed by internal ratings-based approaches since 2013.

Meanwhile, of the eight US G-Sibs, all but two have now fallen below the so-called ‘Collins floor’, meaning their risks as measured by their own internal models are lower than those as gauged by the standardised approach.

That could concern regulators: the rapid growth in the proportion of banks’ credit assets capitalised using internal ratings-based approaches is precisely the sort of development that encouraged the Basel Committee on Banking Supervision to floor capital requirements at a set fraction of those dictated by the standardised approach – so as to prevent banks shrinking their prudential buffers according to their own views of the risks they are running.

###### Data from the European Banking Authority shows the European Union’s largest banks collectively shed \$3.2 trillion in assets in the five years to end-2017

The second trend is the sustained length of the post-crisis credit cycle upswing. EBA data shows firms’ estimates of corporate default risk continue to be on a downward slope, almost a decade into the economic recovery.

Elsewhere, counterparty credit risk as reported by UK and US banks continues to be spread across many thousands of entities considered to be at very low risk of collapse rather than concentrating among a handful of large firms teetering on the brink. Several factors may be at play here: counterparties may well be in fine fettle – or it may be that banks have simply junked riskier borrowers, leaving them to non-bank lenders.

Either way, the evidence suggests a counterparty credit calamity such as that triggered by Lehman is unlikely to shake the financial system anytime soon.

Editing by Tom Osborn


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# In a group of 10 people, the median height is 70 inches, the average

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In a group of 10 people, the median height is 70 inches, the average  [#permalink]

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11 Dec 2019, 01:53
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In a group of 10 people, the median height is 70 inches, the average (arithmetic mean) height is 70.5 inches, and the range of the heights is 12 inches. If an additional person who is 74 inches tall joins the group, which of the three statistics must change?

A. Average only
B. Median only
C. Range only
D. Average and median
E. Average and range

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Re: In a group of 10 people, the median height is 70 inches, the average  [#permalink]

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11 Dec 2019, 02:20
Bunuel wrote:
In a group of 10 people, the median height is 70 inches, the average (arithmetic mean) height is 70.5 inches, and the range of the heights is 12 inches. If an additional person who is 74 inches tall joins the group, which of the three statistics must change?

A. Average only
B. Median only
C. Range only
D. Average and median
E. Average and range

Explanation:

Median for 10 people = 70
Means 2 person in the middle is having average = 70
So it can be same height also, i.e. 70, 70 or more values
as another person with Height 74 is added in a group of 10 people, so Median will be 6th person in the group when the height is in ascending.
So it can be or cannot be changed.

The average must be changed.

Range, Highest - lowest in Height, will not be affected.

IMO-A
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Re: In a group of 10 people, the median height is 70 inches, the average  [#permalink]

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12 Dec 2019, 12:31
This is my reasoning, please correct me if I am wrong!

In a set of a odd number of integers the median will always be an integers as the middle number will be an integer, hence the median can not stay the same if it started out as a rational (70,5). The average must change unless the added number happens to correspond to the current average, in this question that is NOT the case, as such the average needs to change. As no alternative includes all three alternatives we need not consider the range. The range only has to change if the extreme values changes, in this case we do not have any information that regarding the extreme values thus we can not with certainty say that the range has changes.
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Re: In a group of 10 people, the median height is 70 inches, the average  [#permalink]

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15 Dec 2019, 19:38
2
Bunuel wrote:
In a group of 10 people, the median height is 70 inches, the average (arithmetic mean) height is 70.5 inches, and the range of the heights is 12 inches. If an additional person who is 74 inches tall joins the group, which of the three statistics must change?

A. Average only
B. Median only
C. Range only
D. Average and median
E. Average and range

Since the additional person’s height (74 in.) is not the same as the average height (70.5 in.), the new average height must differ from the original average height.

The original median height is the average height of the 5th and 6th tallest people, and the new median height is the height of the 6th tallest person. However, if the 5th and 6th tallest people have the same height, the new median height will be the same as the original median height.

If the additional person is not taller than the tallest person or shorter than the shortest person in the group of 10 people (i.e., his or her height is somewhere in the middle of the pack), the range will not change.

Thus, the only statistic that must change is the average.

Answer: A
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Re: In a group of 10 people, the median height is 70 inches, the average  [#permalink]

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03 Jan 2020, 09:43
Since mean and median are different, so we know that this is NOT an evenly spaced set.

There are 10 people

_ _ _ _ _ _ _ _ _ _

Now, we know that the range is 12 so the smallest term is X and the largest term is X+12.

Median = $$\frac{a_5 + a_6}{2} = 70$$

=$$a_5 + a_6 = 140$$

We take $$a_5$$ and $$a_6$$ as close as possible so we get $$a_5 = a_6 = 70$$

Our set as of now looks like

_ _ _ _ 70 70 _ _ _ _

When 74 is added the median changes from $$\frac{a_5 + a_6}{2}$$ to $$a_6 = 70$$

The median remains the same, but since a new term is added , which is far away from 70.5 , the average will change. We need to check if the range changes or not because if (smallest integer < 74 < greatest integer), then the range remains same else the range changes as well.

We are given that the average = 70.5 so Sum = 705 $$(70.5 \times 10)$$

Now we need to take 2 integers such that there is a difference of 12 between the two. Let's take the two integers as 64 and 76.

Till here, our set looks like

64 _ _ _ 70 70 _ _ _ 76

Let's maximize the value of 76 so that our set looks like

64 _ _ _ 70 70 76 76 76 76

We need to fill the 3 gaps and the remaining sum is $$705 - 76 \times 4 - 140 - 64 = 197$$
We will distribute 197 in the remaining three gaps such that each number is less than 69.

64 65 65 67 70 70 76 76 76 76

Now, if we add 74 to this set, the updated set becomes

64 65 65 67 70 70 74 76 76 76 76

The median, range remains same but the average changes.

Now, is it possible to take a number outside the range.

What if our set is

61 _ _ _ 70 70 _ _ _ 73

In this case is it possible to reach the sum 705? Let's check, we maximize the values so that our set looks like

61 70 70 70 70 70 73 73 73 73

The sum of this set becomes 703 so we need 2 extra and the way to do that is changing our smallest integer to 62 and largest integer as 74.
Now, when we add another 74 to the list the range will not change.

Hence, only the average changes.

OA, A

Bunuel wrote:
In a group of 10 people, the median height is 70 inches, the average (arithmetic mean) height is 70.5 inches, and the range of the heights is 12 inches. If an additional person who is 74 inches tall joins the group, which of the three statistics must change?

A. Average only
B. Median only
C. Range only
D. Average and median
E. Average and range
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In a group of 10 people, the median height is 70 inches, the average  [#permalink]

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06 Jan 2020, 14:27
In a group of 10 people, the median height is 70 inches, the average (arithmetic mean) height is 70.5 inches, and the range of the heights is 12 inches. If an additional person who is 74 inches tall joins the group, which of the three statistics must change?

Statistic question

Let's go in order.
The median does not necessarily changes as element 6 and element 7 could be identical.
The average must change since the new element introduced is higher than the original average.
Lastly, the range could also not chance as the highest element possible is higher than 74.

POE:

A. Average only
B. Median only
C. Range only
D. Average and median
E. Average and range[/quote]
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Re: In a group of 10 people, the median height is 70 inches, the average  [#permalink]

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06 Jan 2020, 15:07
why the 5th and 6th terms cannot be 69 and 71 inches?
Re: In a group of 10 people, the median height is 70 inches, the average   [#permalink] 06 Jan 2020, 15:07
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# Math Help - Trig ratios

1. ## Trig ratios

Hi forum,

I've got a question - What is tan^-1(cot(x))? My calculator gives me a senseless answer.

Thanks.

2. ## Re: Trig ratios

Originally Posted by qwertyuiop
Hi forum,

I've got a question - What is tan^-1(cot(x))? My calculator gives me a senseless answer.

Thanks.
1. Go back to the basics. Use a right triangle (see attachment)

2. $\tan(A)=\frac ab~\implies~\tan^{-1}\left(\frac ab\right) = A$

3. $\angle(A)+\angle(B)=\frac \pi2$

4. $\cot(A)=\frac ba = \tan(B)$

5. ... and now you are able to determine $\tan^{-1}(\cot(A))$

3. ## Re: Trig ratios

Ok... I see... kinda...

x=tan^-1(cot(x))
x=tan^-2(tan(pi/2-x))
x=pi-x

Yes?

4. ## Re: Trig ratios

Originally Posted by qwertyuiop
... What is tan^-1(cot(x))?
Originally Posted by qwertyuiop
Ok... I see... kinda...

x=tan^-1(cot(x))
...
No.

That's completely different from your original question!

You only can state that

$\tan^{-1}\left(\cot(x) \right) = \frac \pi2 - x$

5. ## Re: Trig ratios

Sorry, that's what i meant

Thank you!


text:
# How to overline only part of the text in subscript format

I can format h_y using the command $\overline{h_y}$, but I want to type h_y,max where y,max is subscript but the horizontal line of overline only cover h_y.

Thank you

• Probably not ideal, but what about $\overline{h_y}{}_{,max}$? – leandriis Jun 5 '19 at 7:13
• Can you explain what the notation should mean? I would suspect something like $(\overline{h_y})_{\max}$ would be more appropriate, but the context matters. – Andrew Swann Jun 5 '19 at 7:56

Though I think this notation might confuse the reader, here is a solution which makes the max part have zero width by using \mathrlap. The following \hphantom puts (nearly) the same amount of space back in.

\documentclass{article}
\usepackage{mathtools}

\begin{document}
$\overline{h_{y,\mathrm{max}}} = 1$
\medskip

$\overline{h_{y\mathrlap{,\mathrm{max}}}}_{\hphantom{,\mathrm{max}}} = 1$
\end{document}

\documentclass{article}
\usepackage{mathtools}
\begin{document}

$\overline{h_{y\mathrlap{,\max}}}$ \qquad foo

\end{document}

• Why \qquad? That's arbitrary. – egreg Jun 5 '19 at 22:54

The simplest strategy is to add an overline with the desired width (but occupying no horizontal space) and then typeset the symbol to be partially overlined.

\documentclass{article}
\usepackage{mathtools}

\newcommand{\partialoverline}[1]{%
\mathrlap{\overline{\phantom{#1}}}%
}

\begin{document}

$\partialoverline{h_y}h_{y,\max}$

\end{document}

• Great short macro code. – Sebastiano Jun 5 '19 at 23:05

My proposal is to use a spartan macro called \oversize.

\documentclass[a4paper,12pt]{article}
\newcommand{\oversize}[1]{\mkern-1.5mu\overline{\mkern-1.5mu#1\mkern-28mu}\mkern30mu}
\begin{document}
$\oversize{h_{y,\mathrm{max}}}= 1$
\end{document}

Changhing the values of two \mkern of \overline{\mkern0mu#1\mkern-10mu} you can increase the lenght of \overline. The last \mkern10mu it is only to bring closer to the code equality and 1. See the example below.

\documentclass[a4paper,12pt]{article}
\newcommand{\oversize}[1]{\mkern-1.5mu\overline{\mkern0mu#1\mkern-10mu}\mkern10mu}
\begin{document}
$\oversize{h_{y,\mathrm{max}}}= 1$
\end{document}


text:
# Tarski Monster group with prime 5

Does the Tarski Monster group with prime 5 exist? I know that for 2 and 3, the group does not exist, but what about 5?

• The Burnside problem en.wikipedia.org/wiki/Burnside's_problem for exponent 5 still seems to be open. – Primoz Aug 2 '13 at 7:23
• It is an open problem. It is not even known whether finitely generated groups of exponent 5 are finite. – Derek Holt Aug 2 '13 at 7:26
• Then in that case, i will try it. – D. N. Aug 2 '13 at 7:37
• @Deibor: Good choice! Finding a Tarski monster for $p=5$ is the surest way to get a Fields medal. – Mark Sapir Aug 2 '13 at 7:56
• Of course, you won't get a Fields medal even with a Tarski monster, if you are at least 40 years old on January 1st of the next year that is congruent to 2 mod 4. – S. Carnahan Aug 3 '13 at 8:05


text:
## Friday, June 23, 2017 ... /////

### Decoherence doesn't make observers unnecessary

Philip Ball wrote a long aeon.co essay titled Quantum Common Sense which argues that quantum mechanics isn't really "weird" and everything that was mysterious was basically explained... by decoherence.

I sympathize with the text to some extent and the extent could have approached 100% some 25 years ago. In particular, decoherence is a legitimate insight and quantum mechanics isn't weird when you look at it calmly. But there are lots of claims that Ball makes that I heavily disagree with, too.

OK, what is decoherence? And equally importantly, what decoherence isn't? Well, starting with the positive things, decoherence is a process that

1. allows one to calculate at what "point" in the parameter space, classical physics (gradually) becomes a decent approximate theory for a given physical system
2. puts severe constraints on the possible "basis of states" that may arise as "the states" after a measurement
3. eliminates the physical visibility of the complex phases in the probability amplitudes, so that the probability amplitudes may effectively be replaced with their absolute values
How does it work?

### Macron is a just a pseudo-Napoleon

A union in Europe should primarily be a supermarket

For years, we have known France as a little flaccid organ squeezed in between two giant German muscles.

While this description will remain accurate for many years, France's new left-wing star decided to change at least the perception of this reality among its gullible frogeaters so during an EU summit in Brussels, he painted himself as the new Napoleon who will be a tougher warrior against the evils of the world such as the United States of America (which has "partly disappeared from the world"), climate change (no comment), Islamic terrorism (he wants to fight it by smuggling millions of new terrorists into Europe), and especially the Central and Eastern Europe. He wants to be tougher on these Untermenschen than even the Germans.

Macron: Some political representatives of Eastern Europe have displayed a cynical attitude to the European Union. But the European Union isn't a supermarket. It is a fatal union [or common destiny].
That's both serious and amusing. I assure you that if the EU were really bound to be a "fatal union", the support in our part of Europe for the plans to dissolve it would surpass 90%. The word "fate" or "destiny" indicates that the people no longer have a control over the detailed events – which are being imposed on them. We surely don't want that. Indeed, whether a left-wing babbler likes it or not, our political representatives mostly view the future of the European Union – if the bloc survives at all – as a supermarket. They do so because most of their constituencies see it in this rational, pragmatic way. This word "supermarket" nicely summarizes several dimensions:
1. A free trade zone – which is the aspect of the European integration that is widely and most universally viewed as a benefit.
2. Maybe, a unified labor market – but the free movement of the European citizens in between the member countries is much more questionable and was the main reason behind Brexit, among other things.
3. Cherry-picking of the products, rules, and parts of the life that the European nations want to share with others, get from others, or otherwise coordinate or unify.
The third point is more general and the "supermarket" buzzword may very well mean exactly the same thing as "cherry-picking" I discussed previously. Free European nations should choose what is good for them and only codify and develop those policies at the EU level that seem good for them – just like a buyer in the supermarket picks what is good for her. If the nations find out that there isn't enough consensus that it is a benefit to share XY, then XY shouldn't be shared.

Years ago, even the EU acknowledged this common sense constructive principle. It was known as the principle of subsidiarity in the Eurospeak. But this principle was too rational and democratic so it is being gradually eliminated from the EU ideology.

Such a pragmatic, supermarket-like approach is a necessary condition for the European Union – or any entity in the world – to get improved as time goes by. The good things are being picked, the bad things are not. And the mechanisms that determine what is good and what is bad must be working well, they must allow the nations to have their say, and they must ultimately boil down to some democratic decisions of the European nations. If Mr Macron has a problem with these common sense things, it is very, very bad because he misunderstands absolutely everything about the human psychology and the progress of the economies and human societies.

## Thursday, June 22, 2017 ... /////

### Dwarf galaxies: gravity really, really is not entropic

Verlinde has already joined the community of fraudulent pseudoscientists who keep on "working" on something they must know to be complete rubbish

In the text Researchers Check Space-Time to See if It’s Made of Quantum Bits, the Quanta Magazine describes a fresh paper by Kris Pardo (Princeton U.)

Testing Emergent Gravity with Isolated Dwarf Galaxies
which tested some 2016 dark matter "application" of Erik Verlinde's completely wrong "entropic gravity" meme. Verlinde has irrationally linked his "entropic gravity" meme with some phenomenological, parameter-free fit for the behavior of galaxies. What a surprise, when this formula is compared to dwarf galaxies which are, you know, a bit smaller, it doesn't seem to work.

The maximum circular velocities are observed to reach up to 280 km/s but the predicted ones are at most 165 km/s. So it doesn't work, the model is falsified. This moment of the death of the model is where the discussion of the model should end and this is indeed where my discussion of the model ends.

## Wednesday, June 21, 2017 ... /////

### "Weak gravity" sometimes enforces "cosmic censorship"

Clarifying the truth value status of WGC, CCC, loop quantum gravity, and their relationships

Natalie Wolchover wrote a Quanta Magazine story

Where Gravity Is Weak and Naked Singularities Are Verboten
which is promoting some fun May 2017 work by Jorge Santos (Prof) and Toby Crisford (student, both Cambridge UK). I guess that just like in many other cases, the reason why a popular article was scheduled was that the February 2017 preprint (which has about 1 citation so far) has recently appeared in a prestigious printed journal:
Violating the Weak Cosmic Censorship Conjecture in Four-Dimensional Anti–de Sitter Space (PRL; arXiv)
They considered GR coupled to electromagnetism – the Einstein-Maxwell system – in $AdS_4$ and found a counterexample to the 1969 Cosmic Censorship Conjecture by Roger Penrose. It's the electromagnetic force that allows them to achieve the outcome (which Penrose considered forbidden) with the naked singularity. And they argue that exactly when the 2006 Weak Gravity Conjecture by ArkaniHamed-Motl-Nicolis-Vafa is obeyed, the cosmic censorship will be defended and the pathological singularity will be hidden within a black hole.

## Tuesday, June 20, 2017 ... /////

### Top green MEP wants to move Syria to Eastern Europe, one village at a time

Drang nach Osten reloaded, Mohammed edition

The story that attracted the greatest attention of the Czech readers today was an incredible interview (in German) with Ms Franziska "Ska" Keller, a co-chairwoman of the Greens in the European Parliament. The journalist was familiar with her proposal – which we haven't previously heard about.

Even though Ms Keller later lied about the content of the interview and claimed that the media have misreported it, the relevant part of the interview speaks a clear language:

Keller: ...The main problem is the lack of solidarity in between the member states. Governments rely on a treaty with Libya that should be solving our problems. This is naive. We need a system of shared asylum policies that guarantees the same standards everywhere and distributes the solidarity.

Journalist: You are proposing to relocate whole groups of refugees into a country – for example, a whole village should be moved to Latvia. Do you really believe that the Eastern Europeans will cooperate?

Keller: The idea about the Syrian village is just one of the tools that we could employ. For example in the situation when refugees don't want to go to a country where no other refugees have moved so far. People want to go to places where their compatriots already live which makes the integration easier. The refusal of the Czech Republic and others to accept the refugees violates the laws of the EU. That's why the European Commission has started proceedings against these countries.

Journalist: How would you convince critics that this Union has a future?

Keller: This Union has a future because the member states can't solve the problems facing the world separately. Globalization and the climate change but also migration, those don't have any solutions at the national level.
I am among those millions of people who just say Wow. Decades ago, everyone used to agree that the greens were loons. But Western Europe has been so disrupted that this loon is very powerful in the would-be Parliament of a bloc claiming to dominate a whole civilized continent.

### Overbye on "ominous silence" in particle physics

Dennis Overbye is a top science writer and his new text in the New York Times,

Yearning for New Physics at CERN, in a Post-Higgs Way
is also pretty good. He quotes various particle physicists, including fans of SUSY and critics of SUSY, and those give him different ideas about the probability that new physics is going to be observed in coming years, but he decides to make a conclusion in one direction, anyway: the silence in particle physics is ominous, the subtitle says.

In particular, Overbye talks about the silence after the $750\GeV$ hint faded away. It's been just a year or so when this bump disappeared. Should one be using ominous words such as "ominous" when there has been no similar intense wave of activity for a year? I don't think so.

## Monday, June 19, 2017 ... /////

### A reverse Muslim van attack in London?

At midnight, an event that was seemingly just another Islam-powered terrorist attack has taken place in London's Finsbury Park. A van ran over people. It seems that we have been there. No casualties are reported at this point but the event is "novel", anyway: The injured people are... worshipers from the Finsbury Park Mosque.

This adds a new dimension to the messy co-existence of Muslims and infidels in Western Europe.

The mosque is a Salafist-and-Muslim-Brotherhood-supporting one and has co-produced numerous Al-Qaeda-affiliated terrorists in the past such as one 9/11 culprit, one who got famous in the U.S. in December 2001, and numerous holy warriors against the U.S. forces in Afghanistan. Despite this productive publication record, the mosque keeps on working just fine. So it would make sense to conjecture that the van attack was organized by a private individual Western anti-terror activist who thinks that he needs to do something that his government fails to do. (Update: The suspect was named as Darren Osborne, 47.)

If that's the case, the incident clearly turns him or her into a terrorist as well but conflicts usually end up more symmetric than the words might suggest.

## Saturday, June 17, 2017 ... /////

### Danish "LIGO discovery is noise" paper is hogwash

All of them (and maybe lots or even all of LIGO folks) are missing that the two LIGO detectors aren't predicted to see signals that are exactly proportional to each other

At a Forbes blog, Sabine Hossenfelder uncritically promotes a combative Danish paper in her article

Was It All Just Noise? Independent Analysis Casts Doubt On LIGO's Detections
It is all about a fresh paper
On the time lags of the LIGO signals
by Andrew Jackson and pals (Creswell, von Hausegger, Liu, and Naselsky). They say that the LIGO discoveries could be noise or based on completely fake data (as Bulgarian crank Pentcho Valev helpfully says under Hossenfelder's article). ATdotDE and Telescoper claim to be agnostic.

Has everyone lost his or her mind? Please give me a break.

## Friday, June 16, 2017 ... /////

### Loop quantum gravity was Aryan physics of aether reloaded

Philipp Lenard won a physics Nobel prize and was widely regarded as a top ethnic German physicist as recently as in the 1930s and 1940s.

This fact sounds utterly bizarre today (but yes, aside from Heisenberg, Born, and Jordan, plus the founder of "quantum theory" Max Planck, all the German-sounding founders of quantum mechanics tended to be Swiss or Austrian etc. – maybe this weakness of Germany was affected by the enhanced anti-Semitism and other ideologies in that country). He got his Nobel for cathode rays. They were streams of electrons – I think that you need this to be explained because the very phrase doesn't sound important today. Moreover, J.J. Thomson, Johann Hittorf, and Eugen Goldstein were arguably more vital and earlier discoverers of the cathode ray, in the same way in which Conrad Röntgen was the actual discoverer of the X-rays, even though Lenard wanted to take credit for those, too.

As the "Genius" series on National Geographic reminds us, Lenard was a top Nazi hater of Einstein – and a top warrior against modern physics which he called "Jewish physics". In 2015, Bruce Hillman wrote the book The Man Who Stalked Einstein: How Nazi Scientist Philipp Lenard Changed the Course of History which is extremely interesting because it reveals that the Šmoit-style criticism of modern physics which I considered to be a recent phenomenon isn't new, after all. It's just the crackpottery of the Aryan physics reloaded.

### Czech Senate boss' Stalinist proposal faces a surprising backlash

Last night, before I went to bed, I saw an article about a proclamation by the social democratic chairman of the Czech Senate – the upper chamber of the Parliament – Mr Milan Štěch (I know he has been a top politician for many years but I don't remember a single thing he has done or said in his life so far), a former communist, who said:

I think that there exists a large enough set of people, the pseudo-entrepreneurs, who are working as self-employed people or entrepreneurs (i.e. as OSVČ, please get used to this acronym) and they are not effectively exploited from the viewpoint of the national economy. We should gradually double or triple the minimum payments (healthcare insurance and contributions to the pension funds) that these people make to the government so that tens of thousands of them will prefer to be hired by someone and become employees.
Well, yes, I was somewhat scared – not that the doubling or tripling would be totally devastating. It would be unpleasant, anyway. During the tenure of the current government, I got so used to this kind of left-wing intimidation directed against OSVČ and the increasing bureaucratic and other harassment of the people who lead businesses or who work independently to make their living that I assumed that there will be silence and this proposal may be gradually supported by lots of people and maybe adopted after the elections.

The number of extra bureaucratic burdens, forms, and bans that the current government introduced in recent years was substantial. The "verification reports" for the payers of the value-added tax. The Croatian-style "fiscalization" – mandatory real-time reporting of all cash transactions done by businesses through Internet-connected devices that every businessman has to buy and maintain. Most recently, just days ago, the smokers' super-liberal Czechia turned into another hell of prohibition when the almost complete permission to smoke in pubs was changed to a nearly complete ban. And there were other events like that.

### Do svidánija, khalif

How will Daesh deal with its constitutional crisis?

The Russian defense ministry announced that it believes that its May 28th strike has also killed Al-Baghdadi, the chieftain of the Islamic State. Al-Baghdadi, an intrinsically average terrorist, was elected the caliph i.e. the successor of Prophet Mohammed. If you want a good enough approximation what Mohammed looked like, Al-Baghdadi's mugshot is probably the best answer you may get.

If the reports are true, and I find it more likely than not that they are true, it is our duty to express deep condolences to those fellow terrorists who weren't killed along with their caliph (330 have been, including 30 commissars in the ISIS' version of the European Commission). And my condolences also go to John Kerry who was already previously deeply frustrated that the expansion of his ISIS allies stopped. George Soros is also sad but he should fund a cry-in room for the likes of Kerry now.

## Thursday, June 15, 2017 ... /////

### Wise politics is all about cherry-picking

In recent months and at several inequivalent places, I have repeatedly collided with an incredible meme promoted by the globalists. They say that we – their political opponents – must be bad because we want to "cherry-pick the good things". Wow, what an accusation.

Angela Merkel and numerous parrots on her side have pledged to block cherry-picking during Brexit negotiations. And yesterday, a Russian globalist Bloomberg writer tried to criticize Poland, Hungary, and Czechia because we're cherry-pickers, too.

There are probably other examples, too. And it's conceivable that the two examples above aren't independent of one another – the Russian writer could have just gotten the meme from Angela Merkel or someone in between.

Imagine that you feel somewhat hungry and you find yourself in a cherry orchard. What will you do? You should better find something to eat. There is some soil beneath your feet. There are rocks in it. And some sand. Then you observe wooden branches of the tree. And nicely green leaves. What will you do?

Well, you will pick the damn cherries. Oops, they're so yummy. If you are at least somewhat rational, you will ignore the rocks, grass, and leaves; instead, you will cherry-pick. In fact, most of the TRF readers are so smart that they will pick the red cherries, more precisely red cherries without caterpillars. And to show how brutally picky you are, you will probably spit out the cherry stones, I mean the pits, too. Is there a relationship between spit and pits? ;-)

What is the cherry-picking that I have just described? It's also called the "rational behavior". People (and more primitive animals) as well as politicians representing themselves, their parties, or their nations are doing it, too. You just want to eat or have the things that are good for you – and ignore or avoid the things that are bad or irrelevant for you.

## Wednesday, June 14, 2017 ... /////

### LHC null results haven't changed the qualitative big picture in HEP

Two years ago, I wrote about relaxions, a new way to create awkward theories – that could be said "to be rather natural according to some criteria but not all criteria" – which are capable of "explaining" the existence of large numbers in physics.

One starts with a large, but only logarithmically large, number of fields and assigns somewhat exotic values of charges under a $U(1)$ gauge group – and yes, it has to be an Abelian group which may be considered a damning flaw of the whole paradigm. Consequently, one finds that there exists a scalar boson with a periodic range of values whose periodicity is "exponentially large" in the number of elementary fields we have used.

Backreaction discusses those papers and some of their recent followups under the new brand, Clockworks. It is an OK idea – which is probably irrelevant in physics but has some chance to be relevant – but it can in no way be classified as the "#1 idea" of a decade or something big like that.

Instead of discussing the somewhat modest and vague idea again, let me express my disbelief about a general statement made at Backreaction.

## Tuesday, June 13, 2017 ... /////

### EU migrant quota infringement proceedings against PL, HU, CZ not to be taken seriously

In late 2015, multiculturalist activists within the European Union launched a new program to accelerate the Islamization of Europe – migrant quotas.

Every country was ordered to embrace roughly 1 new Muslim per 300 citizens – everyone was convinced that it was just the first testing number that would grow. Czechia, Slovakia, Hungary, and Romania voted against but because of the Treaty of Lisbon, a simple majority was enough to make this extremely sensitive vote "official". Poland approved the quota system at that time but a new government came just weeks later and it's very clear that this new government would surely join the red bloc above.

Czechia, Hungary, and Poland have officially stopped all the efforts to incorporate Muslim migrants because of security concerns (recent terrorist attacks), insurmountable logistic problems (they just can't sit at one place), impossibility to find reliable information about the migrants (most of them lie about their age as well as nationality), and the migrants' own lack of will to stay in our countries. More seriously, our nations and its political representatives realize that the very idea is wrong because these efforts encourage a further growth of the illegal immigration. If this migration is caused by a conflict outside Europe, the problems must be solved at the place of the conflict.

## Monday, June 12, 2017 ... /////

### CZ, NL resist Paris Agreement: ODS heroically filibusters

Reuters has reminded us that Juncker will work hard tomorrow to start legal proceedings against the Visegrád Group countries – although Slovakia is missing in his list because "it made some pledge in recent 12 months" (Slovakia just banned the registration of Islam as religion, thankfully, it's not viewed as a problem yet LOL) – because we won't participate in the intentional Islamization of the European continent. The Czech government has officially stopped all procedures that were meant to embrace some Muslim migrants (or, ideally, refugees) and it's probably similar in Poland and Hungary.

I have often felt jealous when only Poland and Hungary were mentioned as the beacons of the European culture's fight against these suicidal efforts. So at least Juncker himself has managed to return my homeland to the club, after our ambiguous PM Sobotka almost left us behind. ;-) Slovakia and Hungary have already sued the EU over the unacceptable decision on the quotas. For some strange reason, the government of the Czech nation – whose opinion is far clearer than the opinion of Slovaks and Hungarians – hasn't joined this lawsuit.

Needless to say, this decision will have a deep emotional impact on the EU's approval rate in our countries – the EU and its unelected officials will be hated even more strongly than so far (the comment sections under the Czech articles sound rather tough) – and on the other hand, they will only want us to pay a symbolic fine, like \$10 million, which makes absolutely no difference for any government's budget. So I think that Juncker really sucks as a politician, he is shooting himself in the foot. He doesn't seem to understand that he's losing his influence over a big part of the EU – and what is reasonable to do in such a situation. What he does must either end with his marginal loss of influence; or with some greater EU-Visegrád rift which would probably lead to a more widespread decay of the whole EU.

Now, I want to mention another pet cause of the EU's far left: the Paris climate agreement.

### Deep-learning the landscape

Two hep-th papers are "conceptual" today. First, Eva Silverstein wrote her report from the December 2015 German meeting with philosophers "Why Trust a Theory",

The dangerous irrelevance of string theory
"Dangerous irrelevance" isn't a special example of "irrelevance" in the colloquial sense. Instead, it's the dependence of physics on the laws that are valid at "higher than probed" energy scales. She discusses some totally standard technical questions that are being settled by the ongoing research of string cosmology – and explains why this research is an unquestionable example of the scientific method even when it comes to the seemingly most abstract questions such as the "existence of the landscape".

Yang-Hui He (London, Oxford, Tianjin) wrote something fascinating,
Deep-Learning the Landscape
He uses Wolfram Mathematica (the word "He" is correct even if the author is female! But in this case, he is not) to turn a computer to a hard-working string theorist. I've been calling for such advances for years but he actually did it. Using the machine learning functions, He can train His PC to find new patterns (methods to classify) in the "landscape data" as well as predict some properties of a compactification.


text:
# Valid Usage of Chi Squared Test

I have 2 distributions. One is a population distribution for a type of image. Another is a sample of an undetermined distribution. I want to compare the distributions to see if they are the same (or reject that they are different).

The two distributions I have are 3 dimensional and discrete, but fit within a 28x28x3 array.

To be more specific, my data are greyscale images. 2 of the dimensions are actually spacial dimensions of the image, and 1 is the intensity of the image.

I want to take the first 2 dimensions as coordinates, and then perform a $\chi^2$ test for goodness of fit, taking the intensity of the sample distribution at a given pixel as the observed, and the intensity of the the population distribution at the same pixel as the expected. Then I can conclude on whether the data I have observed are consistent with the distribution I have.

I would use a Kolmogorov–Smirnov Test, but that requires a continuous distribution, and my distributions are not continuous, only my variables are.

So is this a proper usage of the $\chi^2$ test for Goodness of Fit, or do I need to use a different method to compare these distributions?

## 1 Answer

The Chi-Squared goodness of fit test is generally used for counts of categorical variables. Counts are generally more than 0, but here white is encoded as 0. This means that in the expected case, we are dividing by 0, which generates a statistic of infinity. This is because in this test, observing something for which we expect no counts is considered impossible. But in my own circumstance, I want whitespace to be considered a valid value.

Therefore a Chi-Squared goodness of fit test is innappropriate for this problem.


text:
# If $AB=BA$, show that $B$ is diagonalizable.

We were asked to prove the following:

Let $A$ be an $n \times n$ matrix with $n$ distinct real eigenvalues. If $AB=BA$, show that $B$ is diagonalizable.

It was suggested I show that an eigenvector of $A$ is also an eigenvector of $B$. I am both having trouble doing this and failing to see how I would complete the proof after. Any help would be appreciated.

Thanks

-
This is (a weaker version of) part (b) of the following question: math.stackexchange.com/questions/236212/…. It is also a consequence of the statement in this question: math.stackexchange.com/questions/65012/… And this one: math.stackexchange.com/questions/46544/… –  Jonas Meyer Jan 24 '13 at 3:30
Hey Jonas, I still don't understand how it is that if $v$ is an eigenvalue of $A$ corresponding to some eigenvalue $\lambda$, that $A(Bv)= \lambda Bv$ implies that $v$ is also an eigenvector of $B$. I was wondering if you could clear that up for me. Thanks. –  Ernest Singleton Jan 24 '13 at 4:44

I still don't understand how it is that if $v$ is an eigenvector of $A$ corresponding to some eigenvalue $\lambda$, that $A(Bv)=\lambda Bv$ implies that $v$ is also an eigenvector of $B$.
If we see that this is true, then we will be able to conclude that $B$ is diagonalized by a basis of eigenvectors for $A$.
It is true because the eigenspace of $A$ for the eigenvalue $\lambda$, $\{x:Ax=\lambda x\}$, is one dimensional by the hypothesis that $A$ has $n$ distinct eigenvalues. The equation $A(Bv)=\lambda Bv$ says that $Bv$ is in this space, which is spanned by $v$. Therefore there exists a scalar $c$ such that $Bv=cv$.


text:
### Description

Dearboy, a goods victualer, now comes to a big problem, and he needs your help. In his sale area there are N shopkeepers (marked from 1 to N) which stocks goods from him.Dearboy has M supply places (marked from 1 to M), each provides K different kinds of goods (marked from 1 to K). Once shopkeepers order goods, Dearboy should arrange which supply place provide how much amount of goods to shopkeepers to cut down the total cost of transport.

It's known that the cost to transport one unit goods for different kinds from different supply places to different shopkeepers may be different. Given each supply places' storage of K kinds of goods, N shopkeepers' order of K kinds of goods and the cost to transport goods for different kinds from different supply places to different shopkeepers, you should tell how to arrange the goods supply to minimize the total cost of transport.

### Input

The input consists of multiple test cases. The first line of each test case contains three integers N, M, K $(0 < N, M, K < 50)$, which are described above. The next N lines give the shopkeepers' orders, with each line containing K integers (there integers are belong to [0, 3]), which represents the amount of goods each shopkeeper needs. The next M lines give the supply places' storage, with each line containing K integers (there integers are also belong to [0, 3]), which represents the amount of goods stored in that supply place.

Then come K integer matrices (each with the size N * M), the integer (this integer is belong to (0, 100)) at the i-th row, j-th column in the k-th matrix represents the cost to transport one unit of k-th goods from the j-th supply place to the i-th shopkeeper.

The input is terminated with three "0"s. This test case should not be processed.

### Output

For each test case, if Dearboy can satisfy all the needs of all the shopkeepers, print in one line an integer, which is the minimum cost; otherwise just output "-1".

1 3 3
1 1 1
0 1 1
1 2 2
1 0 1
1 2 3
1 1 1
2 1 1

1 1 1
3
2
20

0 0 0

4
-1

### Source

POJ Monthly--2005.07.31, Wang Yijie

### Code

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#define maxn 60
#define maxe 10010
#define INF 0x3f3f3f3f
#define DEBUG
using namespace std;
int n,m,k,cnt,ans,s,t,flag;
int need[maxn][maxn],provide[maxn][maxn],cost[maxn][maxn];

struct edge
{
int to,cap,val,next;
edge(int to=0,int cap=0,int val=0,int next=0):to(to),cap(cap),val(val),next(next) {}
}e[maxe];

{
char ch;
do
ch=getchar();
while((ch<'0'||ch>'9')&&ch!='-');
if(ch=='-') sign=-1,ch=getchar();
while(ch>='0'&&ch<='9')
{
ch=getchar();
}
}

inline void ins(int u,int v,int cap,int val)
{
}

inline void insert(int u,int v,int cap,int val)
{ ins(u,v,cap,val);ins(v,u,0,-val); }

void build_graph(int x)
{
s=0,t=m+n+1,cnt=1;
for(int i=1;i<=n;++i) insert(m+i,t, need[i][x],0);
for(int i=1;i<=m;++i)
{
for(int j=1;j<=n;++j)
insert(i,j+m,INF,cost[j][i]);
insert(s,i,provide[i][x],0);
}
}

bool SPFA()
{
memset(dis,0x3f,sizeof(dis));
queue<int> q;
q.push(s);
dis[s]=0,prev[s]=-1;
while(!q.empty())
{
int u=q.front();q.pop();
while(i)
{
if(e[i].cap && dis[u]+e[i].val<dis[e[i].to])
{
dis[e[i].to]=dis[u]+e[i].val;
prev[e[i].to]=u;
path[e[i].to]=i;
q.push(e[i].to);
}
i=e[i].next;
}
}
if(dis[t]==INF) return 0;
return 1;
}

int argument()
{
int w=INF,cost=0;
for(int i=t;prev[i]!=-1;i=prev[i])
w=min(w,e[path[i]].cap);
for(int i=t;prev[i]!=-1;i=prev[i])
{
e[path[i]].cap-=w;
e[path[i]^1].cap+=w;
cost+=w*e[path[i]].val;
}
return cost;
}

int MCMF()
{
int ans=0;
while(SPFA()) ans+=argument();
return ans;
}

void work()
{
ans=0,flag=0;
for(int i=1;i<=k;++i) need[0][i]=provide[0][i]=0;
for(int i=1;i<=n;++i)
for(int j=1;j<=k;++j)
for(int i=1;i<=m;++i)
for(int j=1;j<=k;++j)
for(int i=1;i<=k;++i)
{
if(provide[0][i]<need[0][i]) ans=-1,flag=1;
for(int l=1;l<=n;++l)
for(int j=1;j<=m;++j)
if(!flag)
{
build_graph(i);
ans+=MCMF();
}
}
printf("%d\n",ans);
}

int main()
{
while(scanf("%d%d%d",&n,&m,&k)==3 && n)
{
work();
}
return 0;
}

Responses


text:
# slick-proof-of-trick-for-counting-domino-tilings

The trick for rewriting the number of domino tilings of a simply-connected finite lattice region as the absolute value of the determinant of a matrix (due I believe to Kasteleyn and Percus, but if Fisher deserves credit for this too please set me straight!) hinges on a combinatorial lemma relating the number of vertical dominos in a tiling to the parity of the permutation-matrix associated with the tiling.

What's the most direct way to see this?

My favorite approach is to set up a distributive lattice structure on the set of tilings, and use it to show that every tiling may be obtained from every other by rotating 2-by-2 blocks, and then use this last fact to prove the lemma, but this is a bit indirect. I've seen a way to do it with Pick's Theorem, but I suspect that there's a simpler way. Right now I'm thinking about an approach that involves superimposing matchings to get unions of cycles, and arguing that the retiling associated with such a cycle can always be achieved by a succession of retilings of 2-by-2 blocks, but maybe there's an even simpler way.

The trick applies more generally to all bipartite planar graphs, and Kasteleyn showed that you could handle arbitrary planar graphs by this method (using Pfaffians), but for right now I'll be satisfied with an argument that applies in the case of dominos on a square grid. (As you might guess, the motivation behind my question is pedagogical; I'll be teaching a class on this material tomorrow.)

-
Can I come to your class? –  Anthony Quas Jan 23 '12 at 16:40
Anyone in with an interest in these topics is welcome to attend my UC Berkeley class "Topics in Random Spatial Processes", which meets in 5 Evans Hall from 9:30 to 11:00 on Tuesdays and Thursdays, with the proviso that if the room fills up, registered students have precedent over auditors. –  James Propp Jan 23 '12 at 18:38
This is just me rambling, but isn't tilings related to non-inttersecting paths, and then, you can use Gessel-Vienot-Lindelöf theorem. –  Per Alexandersson Aug 8 '14 at 9:31

$\def\RR{\mathbb{R}}$This isn't the right answer for your class, because it works for a level of generality you don't need, but last summer I found what I think is the right way to prove this result for a general bipartite planar graph or, working harder, for any planar graph and I thought I'd record it here. I'm sure this isn't original, but I didn't find it in a quick literature search.

Let $G$ be a bipartite graph. We'll define an immersion of $G$ to $\RR^2$ to be a map $\phi: G \to \RR^2$ which is injective except that $\phi$ is allowed to make edges cross transversely in their interiors; $\phi$ is not permitted to create triple crossings. For example, we can send the vertices of $G$ to generic points of $\RR^2$ and join them by line segments.

Assign a weight $x_e$ to each edge of $G$. For a perfect matching $M$ of $G$, define $\mathrm{cross}(M, \phi)$ to be the number of places where edges of $M$ cross. Define the matching polynomial of $(G, \phi)$ to be $$\mu(G, \phi) = \sum_M (-1)^{\mathrm{cross}(M, \phi)} \prod_{e \in M} x_e$$ where the sum is over all perfect matchings of $G$.

We claim that there is a way to take the adjacency matrix of $G$ and replace the entry corresponding to $e$ by $\pm x_e$ to make a matrix $A$ with $\mu(G, \phi) = \det A$. In particular, if $G$ is a planar graph, we can get the generating function for matchings with no signs.

Proof First, suppose that all of the black vertices lie on the line $y=0$ and all the white vertices lie on the line $y=1$, in the same order as the rows and columns of the matrix, and joined by line segments. Then we can put $x_e$ in all positions of $A$, and $(-1)^{\mathrm{cross}(\phi, M)}$ is the sign of the permutation $M$.

Now, consider varying $\phi$ while preserving $G$ and we claim that the result stays true. If we vary $\phi$ generically, we can get from any immersion to any other while passing through singularities of the following sorts:

• A triple crossing of edges. The same signs work on both sides of the triple crossing.

• Two edges become tangent. On one side of the singularity, they cross $m$ times, on the other side they cross $m+2$ times. The same signs work on both sides of the tangency.

• An edge $e$ forms a cusp. On one side of the cusp, it passes through itself, on the other side it doesn't. Negate the sign of $x_e$.

• $\phi$ sends a vertex into the interior of an edge $e$. Negate the sign of $x_e$.

So in every case, as the topology of $\phi$ changes, we can change the signs in the matrix. $\square$

The analogous argument works with general graphs and Pfaffians, starting with the vertices all lying on a circle.

-
So, so, so cool. –  John Wiltshire-Gordon Mar 9 at 16:05
Nice; I'll definitely include this proof the next time I teach a course on perfect matchings! (Though I am a bit fuzzy on how one would prove the topological lemma that asserts that these four kinds of singularities are all that can happen.) –  James Propp Mar 10 at 12:25
If all edges are straight line segments then it isn't too bad (and only types 1 and 4 occur). If $G$ has $n$ vertices, then all maps from $G$ to $\mathbb{R}^2$ can be parametrized by $\mathbb{R}^{2n}$. The immersions form a dense open subset, separated by hyperplane corresponding to singularities of types 1 and 4. The points corresponding to worse singularities are finitely many codimension two polytopes, so we can avoid them. –  David Speyer Mar 10 at 13:19
If you allow curvy edges (which I'd definitely prefer) then I have to admit that I don't exactly know how you make these transversality arguments either. The analogous claim for knot diagrams is that the Reidemester moves are enough. –  David Speyer Mar 10 at 13:21
The proof of Reidemeister's Theorem is famously incomplete; Reidemeister is sketchy in the original paper, and most other sources don't bother going through it. It's much easier to prove it for PL edges, and then approximate smooth edges by PL edges. –  Dylan Thurston Mar 10 at 17:24


text:
## FANDOM

10,825 Pages

So.....why do I like googology?

Why am I even here?

What am I doing?

And why?

Well.....first of all,since my oldest memories,I always remember myself loving mathematics.

When I was five I already knew how to multiply millions.

Everybody around me was facinated by that and from what I can remember they used to ask what that and that makes.

At the time I didn't really understand why everyone was so interasted in me answering questions.

My parents had big hopes for me.Nowdays it really doesn't seem like that's the case :(

When I first went to school,I thought it was the absolute worst thing in the intire universe.I didn't at all listen to what my teachers told me and never did my homework in time.I am left-handed and have dysgraphia,so it's always been hard for me to write well.It used to take me 4 hours to do a homework,because I couldn't concentrate and wrote extremely slowly.I wasn't what you would call a well-behaved student at all!On top of all I hated all teachers and never paid any atention in class.I was bad at sports and still am today.I was bad at learning languiges and hated history and geography.However,Math was allways my favorite!I loved doing it,while everyone else absolutely hated it!I don't remember ever struggling at a math question and ever complaining that it's difficult.

On tests I was never scared and in class I was only bored when what we were studying was too simple(even when everybody else claimed it's difficult).

I never understood why everybody in class had such poor logic.Whenever there were questions that were oversimplified,no one would listen to what I was saying and they would say the facepalmiest thing you can imagine.

So,a little bit of time passed and most things seemed better!

I was bullied less because I had stayed it that class for 5 years and had now a lot of friends who would defend me.

And I may have still been bad at sports,languiges,history..... I was good at science,many people told me I was artistic but most of all I was as good as ever im Maths!

My mom got a job in the european comission in JRC.So we had to leave to go to Italy.

In the beginning I was very exited,but as time passed I realised that I miss my actual home.

So at vacantions we would go to visit my friends and family in Bulgaria.

While I was going to school in Italy,I realised something:the mathematical education was BAD

We were studying stuff like 5x+2=7 x=?

AND MY CLASSMATES WERE STRUGGLING AT IT!!!

That got me really pissed off because in Bulgaria we got a wonderfull education.

We studied things like this:

And what we were studying here was A JOKE!

But then I accepted it and for 3 years evrybody tought I'm a genius or something(exept for the fact that since I did not know what we were doing in Bulgaria and would probably fail if I tried a test).

After a while things got better:

What we were doing was not so terably easy and I started going to extra lessons in the summer to cope whith what we were doing in Bulgaria.The teacher loved me because I standed out of the croud for loving math so much(and knowing the first 14 digits of Pi).I passed with a 9.5 and went back to Bulgaria after 6 years.

How is that googology related?

Here's how!

After doing that I searched in internet for the black hole fenomena,watched a video by a guy called Sharkee and saw the "What's the largest number" video in the recomended section.After watching it I was facinated By the sheer size of the numbers shown there.

Never thought big numbers are what was gonna take me over.I watched it again and again,and again......and it got more interasting every single time!

Then I had to go to a nearby village with my grandpa.While there instead of playing outside I stayed at home untill late night under the lamp thinking of how to make the largest number possible.I made a bunch of extentions to googol and I even invented my own method of naming systems.These are technicaly the first numbers I have ever coined.I made about 20-30 numbers.At that time they looked UNBELIEVEBLY HUGE!!! to me.

Now when I look back,they seem very,very tiny.The largest of all was about $$f_{\omega+2}(10^{100})$$.

But considering I only used powers to create them,that's quite impressive!

I might one day put those numbers in my site if I ever find the paper.

Now I have much more sufficient ways of making big numbers,but back then I didn't know how to go beyond simple stuff.

So what kept you here?

When I first came I was very shy.I saw the blog post feature and made my first notation.

I called it "Boris's Sezration notation".It was very unprofecional and even naive,but I'm still proud of it!

Overall isn't making mistakes what makes us learn?

It was this thrill of making notations stronger and stronger that kept me into this.And I learned a lot while doing it!

Now I can say I'm an all out Googologist!

People make weird facial expressions when they see me writing googological stuff in my googology notebook.For most people spending your free time doing this is madness.But not for me!

I love to spend my free time analising BHAN or other notations made by me or by others.

I don't think googology will ever get old.And I sincierly hope I keep beating myself! All throughout my life I've never been happy with myself.I kepp pushing myself forward and I hope one day people remember me as an actual googologist,not some dude trolling the website.


text:
## Mercedes-Benz SL: Well Thought-out Right Down to the Smallest Detail

By |2011-11-23T09:44:16+10:00November 23rd, 2011|Gay Car News|0 Comments

Before the year is out, Mercedes-Benz will present a completely redeveloped SL Roadster made for the first time almost entirely from aluminium. The full aluminium bodyshell of this newly developed vehicle weighs around 110 kilograms less than a comparable bodyshell constructed with the technology used for its predecessor. It thus reduces the overall weight of the new SL by [...]


text:
# How to generate, validate, and invalidate a set/list of numbers in O(1) time and space?

Imagine my server is generating "tokens" of some sort for a client on a regular basis. When a client asks for a token, the server responds with a new value (and any other supplemental information it wants to, like a "witness"). Later, the client will submit the token (and optionally the witness) and the server needs to be able to quickly determine if it is a token value that it (the server) has previously issued. The token does not need to be "secure", so a trivially forgeable token is okay. Ideally, the server can:

• Generate the tokens in O(1) time and space
• Validate that it issued the tokens in O(1) time and space

The most obvious solution to this problem as stated would be to simply use an incrementing counter as the token. The server stores a single integer and increments it when asked for a new value. Determining if the token has been issued by this server means simply ensuring that the token submitted by the client is <= than the server's currently stored value.

The twist: The server might crash and get restored from a backup, meaning that its counter value might regress to a previous value. (We'll call this the regression value). If this happens, any tokens previously issued that are less than or equal to its regression value should be considered valid, but any tokens previously issued with a value higher than its regression value should be considered invalid since they were not generated by "this" instance of the server. In short, we need to detect a "fork" in our token issuing.

For example: Imagine I issued tokens T0 to T20. Then, I restore the server to a backup right after it had issued T10. I want the server to continue to validate/recognize T0-T10, but no longer recognize T10-T20. Furthermore, I want to make sure that it never re-issues tokens that are identical to the previously issued T11-T20.

The restored server doesn't know what the previously highest (MAX) issued token was. (If it did, it could "mark" any token values as invalid if they had a value between its current value and its max value.)

We also need to avoid the restored server issuing the same token its previous incarnation did. The integer counter scheme would not be capable of such a guarantee since it is not sure what the previous maximum token issued was. Thus, it wouldn't know what value to safely increment its counter to. One way to solve this would be to have a separate "restoration counter" to note the number of times the server has been restored. But presumably if we were capable of saving this separate state somewhere, we could just store the previously maximum issued counter as well.

Another way to solve this problem would be to use timestamps as the token. Assuming an accurate clock, the restored server would be guaranteed to never issue the same value that its predecessor did. However, if we use a timestamp, the server no longer knows which specific timestamp tokens it ever issued in the past, unless it keeps track of a full list of them, bringing the time and space complexity well above O(1). (It could keep a list of the times it was restored from backup and reference this list when validating tokens, but this adds complexity to time and space for validation).

However, if there were a way to compactly store a list of N timestamps and an easy way to test a given timestamp for set membership, the problem would be easy to solve.

Options I have considered:

Option 1: Bloom filter: The server could use a bloom filter and add each number issued its filter. Unfortunately, the set membership test would be probabilistic, but I could make the filter big enough to reduce my probability of a false positive quite low. However, it seems like the addition of information other than just the token value in the form of a witness changes allows us to offload additional information to the clients that later ask for verification, and would allow us to do better somehow.

Option 2: A cryptographic accumulator, like an RSA accumulator. If each of the numbers I was generating were prime, I believe I could use an RSA accumulator to store a single accumulator value of constant size S, where S is significantly less bytes than storing a list of N. Each time I add a new prime to my set, I add it to the accumulator, then I generate its witness as well and ship both the number and the witness to my client. Later, the client would submit the integer it is testing and the witness, and I would be able to quickly determine if the number being submitted is in fact a member of my set or not. Possible problems: I need to be able to hash to a prime number deterministically. (Not the end of the world, but adds complexity.) I think I have to update my witness values as I add new values to the accumulator which adds time to the verification step. Lastly: My understanding of accumulators is rudimentary, and I'm not sure how large the accumulator needs to be in relation to the set being accumulated.

Option 3: I'm overthinking this tremendously.

** Related problems: **

• This seems a lot like having a lot of hash values in a Merkle tree or hash chain (blockchain), and wanting to be able to determine if a particular hash value were ever seen in the chain, without having to store every value that had been seen in the chain. I'm hopeful that with the additional concept of generating a "witness" value of some sort to be stored along with the number, the server can make a membership determination with much less overhead than having to store all of the numbers. (Coda uses techniques to keep its last chain in its blockchain deterministic in size, and offloads the full corresponding merkle tree to clients. https://eprint.iacr.org/2020/352.pdf)

• This feels similar to a vector commitment accumulator, where the numbers being committed to are in a given order, but I think committing to a set is simpler than committing to a vector.

• I think that this is the same problem as labeling a tree that looks like the following:

      A -> B -> C -> D -> E
\
-> F -> G
\
-> H

Given any two labels (L1, L2), and only two labels, determine if L1 is an ancestor of L2. There are labeling schemes that allow for this determination in constant time in significantly less than N bits per label, but I am not sure that it perfectly maps to this.