text:
# "Guess the number" Problem on Turing machines

I am currently learning about the concept of Turing Machines and trying to relate it with my knowledge on the application of the Binary Search algorithm.

The problem I am working on is to write an algorithm that can find an unknown natural number between two given natural numbers by guessing.

A solution would look like the following code in JavaScript.

// fixed structure "generateAsk" is always provided to the problem
let guess = window.prompt("Guess: ");
return function(x) {
return ((x < guess) && "<") || ((x > guess) && ">") || "="
};
};

let low  = window.prompt("Low: ")
, high = window.prompt("High: ");

// uses Binary Search to search the unknown value
// and returns the unknown value when found or -1 otherwise
}

let result = find_unknown_number(low, high, ask)
console.log(result);

My question is not meant to ask for how to solve this problem, but to ask the practical use of any algorithm that can be introduced above.

To be clearer, as I know, mostly any program can be debugged. Isn't anyone able to just stop the program written above, during its execution, with a debugger and then just filter the body of the "ask" function parameter and extract the unknown value?

The way my question is related to the Turing Machines is that, for what I know, the Turing Machines receive input on a tape (so any character introduced on the tape may very well be "visible" from the outside during any execution phase of the algorithm).

Therefore, the questions that I present to you are:

How the implementation of a solution to this problem on a Turing Machine would not be pointless from the perspective in which one can stop the program, read the "ask" function body by reading the corresponding tokens on the Turing Machine and then return the "unknown" value?

Is there any concept of "privacy" of tokens in the world of Turing Machines or some structure that can be used to implement "privacy" in some specific tokens written on a Turing Machine?

• IMO, such considerations are alien to the Turing Machine theory. In fact you can define a machine the way you want, for instance by stating that the tape can only be seen by the reading head. Feb 17 at 16:48
• As I said, privacy is irrelevant to TM, so there is no reference. As if you were asking the color of a TM. Feb 17 at 17:42
• Ignore what Yves Daoust said, as it is completely bogus. As in JohnL.'s answer, such restrictions are trivially captured by the very common concept of oracle TMs. And you may also be interested in reading about Arthur-Merlin games in the future. Feb 18 at 12:18
• @coolest10, I am afraid that you are using this website the wrong way by adding new question in the same question/answers page, especially with so many existing answers. This website is designed for one question and answers to that question, one of which can become accepted and all of which can be upvoted/downvoted. This is not a discussion forum nor a chat room. When you have a further question, the best way is usually to open a new question page, in which you can use links to relate to previous questions of yours. Feb 19 at 8:23
• @coolest10: Are you familiar enough with the rigorous definition of TMs? If so, then move on to the definition of oracle TMs as mentioned by JohnL., and note that your guess-number problem can easily be represented by having ask be an oracle TM $A$, in which case find_unknown_number would be represented by a TM $F$ relative to ask. So the guess is not in the input tape of $F$, since it is inside the oracle $A$. There are other natural extensions of the TM model to support encrypted tapes, but oracle TMs are clearly the right way the view this guess-number problem. Feb 19 at 9:18

Yes, it is pointless and absurd to implement an algorithm to "guess the number" using the most common kind of Turing machine, whose head can read any cell on the tape, since, as you pointed out, there is no way to enforce the condition that the algorithm/the machine should not read "the number", which must be a part of the input.

Of course, that does not mean an algorithm that can "guess the number" (efficiently) without reading the number is not useful. For example, the binary search algorithm is used in many places in real life such as git-bisect or binary-search routines in the built-in libraries of many programming languages.

Rather, that just means that kind of Turing machine is not the right model of computation for that particular problem.

### What is a model of computation?

In computer science ... a model of computation is a model which describes how an output of a mathematical function is computed given an input. A model describes how units of computations, memories, and communications are organized.

### Which model of computation is for "guess the number"?

One suitable choice for "guess the number" is an oracle machine.

An oracle machine ... can be visualized as a Turing machine with a black box, called an oracle, which is able to solve certain problems in a single operation.

A version of the problem of "guess the number" can be stated as the following.

A number has been chosen between integer $$low$$ and integer $$high$$. There is an oracle which, given integer $$k$$, answers the question "is the chosen number smaller than $$k$$?". How can we design an oracle-machine with that oracle that will find the chosen number given input $$low$$ and $$high$$? The less number of times the oracle is used, the better the oracle-machine is.

Beside enabling a solution, the oracle above is used to, in your words, ensure 'the 'privacy' of tokens".

### There are many models of computation

There are many different kinds of computational problems. There are also many models of computation. A problem/proposition/argument makes sense only in some models of computation.

Among all models of computation, (that common kind of) Turing machine is the simplest that "can" compute anything that is "effectively computable", basically. That is why you are directed to learn it. However, that does not imply we should/may express and solve every algorithmic problem with it. For another example, it does not make much sense to implement quantum computation on a conventional Turing machine.

### Exercise. What is the most popular model of computation?

By "the most popular model of computation", I mean the one that is used, implicitly or explicitly, when big $$O$$-notations representing time-complexity or space-complexity are mentioned on almost all millions of web pages on programming. It is certainly not Turing machine.

• Please do not place exercises in answers. This is not the place to give exercise, now I am dealing with the question "wat is the mode popular model of computation?" and I did not have a course on computer science in a long time. Please do not introduce new unanswered exercises in answers. Feb 20 at 14:55
• @Cheiron, exercises are purely optional for you. My answer is complete without the exercise. Please ignore the exercise if you do not want to reinforce what is being taught here. Be content with yourself once you have understood the answer even if you missed the exercise. Feb 20 at 15:05
• If part of your answer can be ignored without loss of information, should that part of your answer not be removed? I just do not agree with the tone here, what if I conclude that the most popular model of computation must be the DFA (any other model requires something infinite after all!) I walk away with wrong information from your answer. Is that really what we want? Feb 20 at 15:13
• @Cheiron, yes and no. Let us chat at this chat room, an unanswered exercise is included in an answer. Feb 20 at 16:22
• @JohnL. re: most popular model of computation: is "register machine" the answer you're looking for? Feb 21 at 1:31

In casual words, a Turing machine can do anything computational, but nothing else; especially there's no IO. No keyboard, no mouse. It just takes a tape as input and runs a predefined program on it.

This means that you cannot use window.prompt or similar (anything that asks for user input) in the middle of your program, and therefore there is no equivalent of "Guess the number" program for TM. So, if you want to stick to the TM as the model of computation, you basically need a different problem that can be solved with binary search, without user input in the middle of the program.

Some examples of such problems include:

• An array of integers is given in strictly increasing order. Determine if a given number exists in the array.
• An array of integers is given, so that all odd numbers come before all even numbers. Find the first even number in the array.
• Given a polynomial and a range (upper and lower limits of the value of x), find a zero of the polynomial within the given range.

Here is how this kind of program could be useful. Suppose that you have a graph, and you are trying to find its chromatic number $$\chi$$, that is, the minimum number of colors needed to color the vertices of the graph so that each edge connects vertices of different colors.

Given a target number of colors $$C$$, you can attempt to color the graph using $$C$$ colors in various ways. This corresponds to an algorithm for answering the question is $$\chi \leq C$$?''. Using binary search, you can use such an algorithm to find $$\chi$$ exactly.

In this example, $$\chi$$ isn't written anywhere. Rather, we can check whether $$\chi \leq C$$ for any given $$C$$. This suffices for finding $$\chi$$ using only $$\log n$$ queries of this form, where $$n$$ is the number of vertices.

• As I can understand, in your case, the chromatic variable would be deductible from the graph you design, so it would not need to be given as input. The only difference I see between the 2 problems exposed (yours and mine) is that in my case, the "guess" variable could vary in the low-high interval, whereas in your case, the chromatic number "x" is fixed i.e. there can only be one minimum number of colors per graph. Please correct me if I am wrong and try to refer to the answer of @YvesDaoust above + tell me your opinion on the correctness of his statement. Thank you very much! Feb 17 at 16:59
• I think you have misunderstood the point of your exercise. The exercise describes how to compute a value $x$ given that you can answer questions of the form "$x \leq b?$", and this is useful since sometimes there is a natural algorithm for the latter but no direct algorithm for the former. The problem is framed using an artificial ploy, but this should be ignored rather than given the spotlight. Feb 17 at 17:04

Is there any concept of "privacy" of tokens in the world of Turing Machines or some structure that can be used to implement "privacy" in some specific tokens written on a Turing Machine?

There is a concept of obfuscation. A software code is obfuscated to make it difficult to analyze the behavior of the program, even if the source code of the program is given. This concept is mostly applied to practical programming languages or machine codes but can be applied to more computation models, including Turing machines.

A theoretical concept of obfuscation is black-box obfuscation. Informally speaking, with such obfuscation, no one with a "debugger" can inspect the program to get more information than just running the program. It is proven no such scheme exists.

The way I understand your question, you are wondering why there is a need to do some extra computation (binary search in your case) if the TM can simply read the input, and it will already know the number to be searched, thus performing binary search is not necessary. This might not always be the case.

Consider, the situation, such that the input is the hash value $$x'$$ of the number that the TM will guess, instead of the actual number $$x$$. Here we assume that the hashing algorithm used to produce $$x'$$, is built internally to the TM, that is the TM can perform a hash. Now, as you might imagine, simply reading $$x'$$ will not help the TM to know what $$x$$ is. We will force it to actually guess. Although at this point, binary search might not be possible since the TM will not be able to know if the number is less than or greater than the middle value of the range. But instead it must perform the guessing brute-force, trying all values from the range. That is, the TM will start with the lowest value, compute its hash, test if it is equal with the input. If it is, then the TM guessed the number, otherwise it moves to the next number in the range and repeat until it found the right value or exhaust the range.

As a side note, the idea here is very much similar to what you would expect from log-in systems. Even though they can validate the password you enter when logging-in, they actually do not know your actual password, since they only store the hash of the password you initially set.

As @jaxad0127 in the comment suggested, you can change the hash function to be some other function $$f()$$ that is difficult to invert but preserves order. That is, for any two numbers $$x_1$$ and $$x_2$$ such that $$x_1 \lt x_2$$, it is always guaranteed that $$f(x_1) \lt f(x_2)$$. This would now allow you to perform binary search. But still, reading the input $$x'=f(x)$$ will not give the TM any idea as to what the $$x$$ is and would be better-off performing binary search.

• Instead of a "traditional" hash, use a math function that is strictly increasing on the given domain and difficult to invert. Like a 5th degree polynomial. This preserves the needed relationship for a binary search. Feb 18 at 1:52
• Preserving a relationship makes a solvable problem easier.
– Rob
Feb 18 at 15:20

When you find X, you don't just find X but also any other information in the same location as X.

Imagine you have n pages, on each page is written a name and address. The pages are sorted alphabetically by name and numbered from 1 to n.

You wish to find the address for Drew Lee.

You could start at page 1 and just go page by page, but that will take a really long time if n is large.

Or you could do binary search, go to n/2, check if the name on page n/2 is greater than or less than Drew Lee, and then proceed...

So you are trying to find X where X is the number of the page that has Drew Lee on it. You can't just say you already know the name Drew Lee, because you actually need the page (to get the address). But the page isn't anywhere in the question to "peek" at, you can only find the page by searching.

This is a real problem that is really solved all the time, although usually by using a library/database/etc. that already has binary search (or even faster searches) implemented. (But a database isn't a Turing Machine, I hear you cry... Yes, but the search algorithm itself functionally is, you have some static input--the search term, the information to search--and it does it's thing and gives you some output.)


text:
# beamer font size

latex beamer change font size table. a DIN-A0 size … to the preamble after the beamer font size option described above will produce a presentation that looks very different from the default (with no other changes required! Starbeamer.ttf. Welcome to the beamer (latex) theme gallery. That was because I didn't know the themes and for each one I had to recompile the whole presentation and see how it looked. online LaTeX editor with autocompletion, highlighting and 400 math symbols. The t makes the body of each slide top-justified rather than vertically centered. The font size in the editor can be changed in the left hand menu. The picture below compares the three LaTeX standard font sizes. Enables math mode in code comments. Ad by Balpirick Studio. It is an extension of the LaTeX beamer and the a0poster classes in order to create LaTeX posters, e.g. The other font … I also read that footers can be added (if included in the file definition 4.2.3). Default Font Size. The font can also be changed for a specific element in the document. ): Figure 3: Two column slide with the Madrid Beamer theme. Having such a small font size makes it very easy to create slides full of text that are never … Beamer Tutorial About Beamer More Advantages of Beamer The layout, colors, and fonts used in a presentation can easily be changed globally, but you also have control over the most minute detail. From here, scroll down to find the Font Size option to change the font size in the editor. The font size was changed to Larger (\Large in the latex source). rulecolor. The default font size of latex corresponds to somewhere around 20pt in the same scale. This demo font is for PERSONAL USE ONLY! I am wondering the method to decrease the size of the font on the title of the block. An overview about the size and commands you will find here even for other documentclasses. LaTeX poster with beamerposter - beamer - a0poster. Introduction baposter a0poster beamerposter Background Basics beamerposter EXAMPLE CODE “documentclass[final,hyperref={pdfpagelabels=false}]{beamer} The standard type sizes in Latex are: tiny, scriptsize, footnotesize, small, normalsize, large, Large, LARGE, huge, Huge To change the size within the math environment, try the following  \\mbox{\\Huge … Export (png, jpg, gif, svg, pdf) and save & share with note system Using a different font size on a global level will affect all normal-sized text as well as the size … The default font size for Latex is 10pt. \documentclass { article } \usepackage [T1] { fontenc } \usepackage [utf8] { inputenc } \usepackage { geometry } \geometry { textwidth=7cm } \usepackage { tgbonum } \begin { document } This document is a sample document to test font families and font typefaces. 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Beamerv3.0Guide BeamerStructure BasicCode BasicCodeI Beamerclassloadingwiththemes \documentclass[slidestop,compress,mathserif]{beamer} %\usepackage[bars]{beamerthemetree} % Beamer theme v 2.2 How can I change the options to control the size (or font) of R output? In the example, {\huge huge font size} declares that the text inside the braces must be formatted in a huge font size. For … Viewer … You don’t need many packages, as lots of them (like color) are included by default with Beamer. 4.3 Beamer presentation. Math Fonts with Beamer. If you are not satisfied with the font themes that ship with Beamer, you can use any font family you want for your Beamer document. To create a Beamer presentation from R Markdown, you specify the beamer_presentation output format in the YAML metadata of your document. ﬁles, added grid mode option, renamed beamer speciﬁc variables, added font size normalization (scale=1.0 is now default for all DIN-A(n) sizes) 20/24. 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Can someone provide me with the information for changing the size of the font of the … I have a frame in beamer with an enumerated list where I increased the font size of the text. I searched for samples of those themes but … However the default size is too large, therefore the output is clipped and can't be seen entirely. It's also possible to put the table of contents at the beginning of each section and highlight the title of the current section. I'm preparing a beamer presentation using RMarkdown and want to include lm output in it. \begin{frame} \frametitle{There Is No Largest Prime Number} \framesubtitle{The proof uses \textit{reductio ad absurdum}.} The ﬁnal output is typically a.pdf ﬁle. Ask Question Asked 4 years, 6 months ago. To change the basic font size used all the way through your document, put either "11pt" or "12pt" in your \documentclass line. 1This option disables Beamer’s font substitution mechanism. Font families. Font size. 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The paper types with their dimensions are given below: letterpaper (11 x 8.5 in) legalpaper (14 x 8.5 in) a5paper (5.8 x 8.3 in) a4paper (8.3 x 11.7 in) executivepaper (10.5 x 7.25 in) b5paper (25 x 17.6 cm) LATEX TYPE SIZE. Viewed 6k times 11. Seems like I can use {.smaller} to change it for the whole slide. The 12pt makes the default font a bit bigger, but not actually twelve point. The font family you intend to use has to be available as a package. Active 2 years, 3 months ago. The default font size or type size that Tex produces is 10pt. Trending Fonts Home > Starbeamer Font Hello, you seem to have JavaScript turned off. If you still don't understand what I'm talking about, this is probably a problem you will never encounter because you aren't using high quality fonts. linenos Enables line numbers. Every time I wanted to pick a theme for a presentation I ended up spending a lot of time. Open an example in Overleaf. 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format: plain


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Initial Stages 2021

Jan 10 – 15, 2021
Weizmann Institute of Science
Asia/Jerusalem timezone
See you at IS2023 in Copenhagen in June 2023

Influence of fluctuating initial-state shape deformations in ultra-central collisions

Jan 13, 2021, 7:45 PM
1h 30m
Patio (vDLCC)

Patio

vDLCC

bullet talk (poster) The initial stages of heavy-ion collisions

Speaker

Peifeng Liu (Stony Brook University)

Description

It is a long standing puzzle that hydrodynamic calculations give larger
elliptic flow for ultra-central Pb+Pb events than experiments. One explanation
is that sampling from a single particle distribution in Glauber model generates
too much shape fluctuation, which can be expected to be modified by NN
correlations. We study the fluctuation of shape-deformation as currently
implemented in MC-Glauber-like models and its consequences on the magnitude of
eccentricities. We show that for both deformed and spherical species, there are
large fluctuations in the deformation that emerge event-by-event. For each
nucleonic configuration, we characterize the deformation with parameters
$\beta$ and $\gamma$, calculated using quadruple moments. The second order
eccentricity $\varepsilon_2$ shows a strong correlation with the E-by-E
deformation $\beta$'s for ultra-central collisions. We further show, via
acoustic scaling, with a moderate reduction of $\beta$ by rescaling, the
eccentricities we get then agree with the measured values of $v_n$, for all
experimentally available centrality bins. Therefore, the model provides
eccentricities that describe experimental data, both where geometry dominates
and where fluctuation dominates. We further show that this modification has important implication for the novel observable, the dependence of mean $p_T$ of $v_2$, for ultra-central collisions.

Primary author

Peifeng Liu (Stony Brook University)

Co-author

Roy Lacey (Stony Brook University)

Presentation materials

 peifeng liu is21.mp4 peifeng liu is21.pdf


text:
## This Crazy Dangerous Part Of I-35 Is Getting SIDEWALKS (Yes, Sidewalks)

Címkék: , , 2015 / 12 / 03

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Címkék: , , , , , , 2015 / 11 / 22

## It’s Official: I-35 In Austin Is The Most Congested Road In Texas

Címkék: , , 2015 / 11 / 10

## Now It’s Up To You: Which One Of These People Will Be Austin’s Next Mayor?

Címkék: , , , , , , , , , , , , , , , , , , , , , , , , , 2014 / 10 / 20

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## Watch Austin’s Old Timers Discuss ‘The Traffic Problem’ in the 1960s

Címkék: , , , , , , , , 2014 / 07 / 11


text:
# Applying to US Universities for Admission ¾«Ñ¡

ÒÑÓÐ 18213 ´ÎÔÄ¶Á 2007-10-17 19:40 |ÏµÍ³·ÖÀà:º£Íâ¹Û²ì

It is that time of the year again for admission application to US
universities. Offered below are some tips from the viewpoint of
someone who sits on admission committees.

By
Y.C. Ho

While one does best to help, one cannot customize answers to all
your questions. Otherwise, one will have no time to do anything
else.
2.Financial support for graduate study in the US, whether
scholarship or assistantship is primarily due to individual professors
who invest in you ~US$60,000/year (This is the cost of support a student for one year. He gets this from his research contract or from his department including overhead expenses). Consequently, He likes to have some return on the investment (ROI). There are two ways to get ROI: The first approach is to have a large project that requires a team.. The professor then employs master degree students to do small, not so advanced, but necessary work. He promotes the student to ph.d status if s/he shows promise. Such positions are available at large universities, e.g., UIUC (U. of Illinois at Urbana-Champaigne), MIT, Berkeley, etc. The second approach is to admit the students directly to ph.d program and invest$120k in him/her for the first two years
to study without getting any return. The student finally begins
to produce in the third and fourth year. There is a danger here
since the admission may be a mistake, or worse, student may leave
for better opportunities after getting his master degrees in two
years. In this case the professor have lost ~120K without getting
any return. While US is a free country there is the issue of
loyalty and obligation. I know more than one professor who swear
that he will never admit another Chinese student again.  Thus you
don¡¯t want to burn the bridge after you have crossed it for those
behind you. Unfortunately, such unethical ¡°get ahead at any cost
¡± attitude are not infrequent among Chinese students.  Remember
in a foreign country reflect on your Motherland.

3.  Tips on application: Unless previously primed, an application
usually gets five minutes of attention before decision to deny or
further consideration. A professor must read at least 50-100
applications. Thus you must make your application stand out .
Try to correspond directly with the professor. Be specific in
you request and desires. Show that you have invested time and
time, then the professor may reciprocate thinking you are
genuinely serious. Open ended question such as ¡°please tell me
everything I need to know to apply¡± are impossible for a busy
person to answer faithfully. A good application, or for that
matter a letter/message, should always reader-centered and NOT
self-centered.
4.General requirements for admission to good universities:
TOEFL score over 600. GRE above 95%. Graduate from a well known
school in China. Recommendation letters preferably should come
from writer known to or calibrated by the person reading your
application. US university are aware of the problem of false
recommendation letters and/or student-composed- professor-signed
letters.

http://blog.sciencenet.cn/blog-1565-9164.html

ÉÏÒ»Æª£ºProximity to Fame
ÏÂÒ»Æª£ºBeijing sights and thoughts- III

Êý¾Ý¼ÓÔØÖÐ...


text:
Dielectric in a Capacitor

1. Feb 23, 2016

mattbeatlefreak

1. The problem statement, all variables and given/known data
You want to maintain a potential difference of 5000 V between the plates of a parallel-plate capacitor.
What is the minimum value of the plate separation if the space between the plates is completely filled by a slab of mica?

2. Relevant equations
V=Ed (parallel plate capacitor)
Dielectric constant κ = 5 for mica
κ=V0/Vd
E=σ/ε0

3. The attempt at a solution
I know that the voltage of the capacitor needs to stay constant, so since the voltage decreases by a factor of five, the distance must increase by a factor of five. However, the dielectric also lowers the field. Whenever I try to approach the problem, I end up getting stuck because I do not know the charge density or any charge values.
Any help would be greatly appreciated!

2. Feb 23, 2016

Staff: Mentor

Besides the dielectric constant κ, dielectrics have another property that describes their ability to resist electric breakdown, which is what occurs when the applied electric field becomes strong enough to free the bound electrons in the dielectric material. Essentially, it's when its insulating property is overcome and current can flow through the dielectric. Usually the capacitor is destroyed when this happens.

So, look up the term "dielectric strength".

3. Feb 23, 2016

mattbeatlefreak

Thank you! As a follow up question, I am supposed to find the ratio of the maximum charges that could be put on a plate of the capacitor in part B to the capacitor in part A. Part A used mica, and part B used barium titanate as the dielectric. I know qfree = Vd*Cd. If we use the same capacitors as in the previous parts, then their voltage would be the same (5000 V). This leads me to believe that the only thing that would change the charge would be the capacitance. Since the κ for mica is 5 and the κ for barium titanate is 1200, I did 1200/5, which is 240. However, the program wants only one significant digit in the answer, and 200 was incorrect. What am I doing wrong here?

4. Feb 23, 2016

Staff: Mentor

Will the spacing of the plates be the same for both capacitors? (Did you take the available dielectric strengths into account as a contributing factor to the difference in capacity?)

5. Feb 23, 2016

mattbeatlefreak

So then, you can write the capacitance with a dielectric as C=κε0A/d. Do you assume that the plate area is the same for both? If so then I get that the charge q on the capacitor plates is q=κV/d. I then used the dielectric constants for each, used 5000 V in both, and used their respective distances found from the previous problems. Doing this I get a ratio of 120. Is this method correct?

6. Feb 23, 2016

Staff: Mentor

I would presume so. It seems that only the dielectric and plate separation are changed.
Strictly speaking it's not true, since the actual charge depends upon the area and the constant $\epsilon_o$. Replace the "equals"with "proportional to" and I'd agree. Since you're looking for a ratio then that's fine.
It looks correct. I can't confirm your result because you haven't supplied all the data that you've used for the materials involved.

7. Feb 23, 2016

mattbeatlefreak

It worked out, thanks again for all the help!


text:
# Thread: The use of division in simplifying.

1. ## The use of division in simplifying.

I'm having a hard time trying to understand the use of how to deal with division in simplifying.

For instance with multiplication if you had (-6n) X 3n you take the numbers and the letters seperately so -6 X 3 woudl be -18 and n X n would be n squared.

But with 3x / 12 3 divided by 12 would be 0.25 and x divided by 12 would be what? the answer given is x/4 and I can kind of see this if i think about it as a rational number and then cancel both parts by 3 to give x/4 but this is just guess work on my part as there's no mention of this in the chapter I'm working on.

Also in 8x / 4x with is the answer 2 and not 2x?

Thanks.

2. Confusing, isn't it?

You are to divide only aples by apples, oranges by oranges, etc.
Numbers by numbers, letters by letters.

3x / 12

Numbers by numbers:
3 / 12
= 1 / 4

Letters by letters:
x /....what? there is no x in the denomitor.
So you cannot divide x.
so leave it there in the numerator.

Hence,
3x / 12
= 1*x / 4

-------------------------------------------

8x / 4x

Numbers by numbers:
8 / 4 = 2

Letters by letters:
x / x = 1

So,
8x / 4x
= 2*1

3. I'm still not sure I understand. You seemed to treat the first as a fraction that needed simplifying but in the second you seemed to do something else. In the second example I was thinking of it again terms of a fraction meaning the 8 and 4 would reduce to a 2 and 1 and the x's would cancel each other out meaning you would be left with 2/1 which can be further simplified as 2.

As I said i'm taking a leap that these should be considered as fractions in order to simplify them. Is this right?

Assuming it is i'm also having a problem understanding at what point in the simplifing process the idea of dividing by a negative number in either the lower or upper part of the fraction alters the other part. For instance 9e squared /(-12e cubed) is it that as soon as I simplify both numbers that 9 becomes -3? and is it just convention that mean the answer is expressed with both the numerator and denominator as positive number with a single minus sign before the whole fraction or is there a deeper reason for this?

Another exampel of this is (-a squared) / (-a) squared. here you have -a x -a / -a x -a. If I follow the normal rules here they all cancel each other leaving me with 0. however the answer is -1?

Again thanks for the help.

4. $\frac {9 e^2}{-12 e^3}$

Apples divided by apples, oranges divided by oranges.

9 = 3x3, 12 = 3x4. There's a factor of 3 in both of them so you can cancel the 3 out. Putting something on the bottom of a fraction is like saying "divided by" that number. So it's like saying:

$9 \times e^2 \div (-12) \div e^3$

which is like

$3 \times 3 \times e \times e \div (-4) \div 3 \div e \div e \div e$

Dividing is "undoing" times, so times 3 is "undone" by "divided by 3".

(Going from 9/(-12) to -3, what you've done is subtracted which ain't right.)

So you got 3 on the top (one left out of the two) and on the bottom you got -4 and e, so you end up as:

$\frac {3}{-4e}$

5. $
\frac {3}{-4e}
$

is the answer I came to too. But the answer I have in the back of the book has a minus sign in the middle of the franction and then positve numbers on both the numerator and denominator implying the whole fraction is negative. Is this wrong?

Also i've just been reading up on this and a different method i've found is divide the coefficents and then and subtract the exponents. But when I apply this to the first problem I posted I end up with 2x and not 2 which is the answer I've been given.

[/COLOR][/FONT][/COLOR]

6. Originally Posted by Alyosha
But with 3x / 12 3 divided by 12 would be 0.25 and x divided by 12 would be what? the answer given is x/4 and I can kind of see this if i think about it as a rational number and then cancel both parts by 3 to give x/4 but this is just guess work on my part as there's no mention of this in the chapter I'm working on.

Also in 8x / 4x with is the answer 2 and not 2x?
The biggest problem I see is that you are looking for shortcut rules. don't do that. Do what is right, first.

The next biggest problem is that no one seems to be talking about the "Identity Element" or the "Reciprocal". You should know this...

$\frac{2}{2} = 1$

$\frac{x}{x} = 1$, if x is NOT zero.

$\frac{frog}{frog} = 1$

$\frac{3x}{12}\;=\;\frac{3*x}{3*4}\;=\;\frac{3}{3}* \frac{x}{4}\;=\;1*\frac{x}{4}\;=\;\frac{x}{4}$

There is just no confusion forging through in this way.

Similarly,

$\frac{8x}{4x}\;=\;\frac{2*4x}{4x}\;=\;2*\frac{4x}{ 4x}\;=\;2*1\;=\;2$

And again, I just don't see the confusion.

I blame your teachers for showing you how to muddle through rather than insisting that you understand what it is you are doing. You should have studied properties such as the Associative Property of Multiplication. These things may seem dry and arid when first encountered, but now is the time to drink deeply from their well of rigorous water. Let the camels muddle through the sand.

7. Originally Posted by Alyosha
$
\frac {3}{-4e}
$

is the answer I came to too. But the answer I have in the back of the book has a minus sign in the middle of the franction and then positve numbers on both the numerator and denominator implying the whole fraction is negative. Is this wrong?

[/color][/font][/color]

No it's not wrong. Sorry, I glossed over the minus.

$\frac {1}{-1} = \frac{-1}{1} = - \frac {1}{1} = -1$

It honestly doesn't matter where you put that minus sign, as long at it's over at the left hand side ...

8. but wouldn't
$
- \frac {1}{1} = -1
$

mean that both numbers were negative so that -1 divided by -1 would be +1?

I'm not really sure I understand what it is you're doing here

$
\frac{3x}{12}\;=\;\frac{3*x}{3*4}\;=\;\frac{3}{3}* \frac{x}{4}\;=\;1*\frac{x}{4}\;=\;\frac{x}{4}
$

are you creating the 4 * 3 with the common factor or finding a number that is equal to the coefficent so that you can create a seperate fraction will equal 1? meaning that if you had 6 * x you would have 6 * 2?

$
\frac{8x}{4x}\;=\;\frac{2*4x}{4x}\;=\;2*\frac{4x}{ 4x}\;=\;2*1\;=\;2
$

here I can you've split the 8x to a common factor sum of 4 * 2x but I can't see why you've then multiplied both side of the fraction by 2.

9. No, the minus sign applies to the whole thing.

$- \frac 1 1 = -\left({\frac 1 1}\right) = -1$

Minus divided by plus = minus.

Plus divided by minus = plus.

Minus divided by minus = plus.

Now $\frac {-1}{-1} = - \left({-\frac {1}{1}}\right) = \frac 1 1 = 1$

10. Originally Posted by Alyosha
I'm not really sure I understand what it is you're doing here

$\frac{3x}{12}\;=\;\frac{3*x}{3*4}\;=\;\frac{3}{3}* \frac{x}{4}\;=\;1*\frac{x}{4}\;=\;\frac{x}{4}$

$\frac{8x}{4x}\;=\;\frac{2*4x}{4x}\;=\;2*\frac{4x}{ 4x}\;=\;2*1\;=\;2$
I don't want to say that you are overthinking it, but you are thinking about it in a NONuseful way. It is NOT as complicated as you seem to be making it.

The expressions I have written represent a progression of the process from left to right. Take each pair of smaller expressions (one equal sign and the two expressions surrounding it) and see how it works.

$\frac{3x}{12}\;=\;\frac{3*x}{3*4}$

The ONLY thing that happened here is that 12 = 3*4. That's it. It is neither shocking nor difficult. I also rewrote the numerator with an explicit multiplication symbol, just to make it look more like the denominator. This is only cosmetic, not magic mathematics.

Now the next one.

$\frac{3*x}{3*4}\;=\;\frac{3}{3}*\frac{x}{4}$

Here we used properties of fractions. What is the rule for multiplying fractions. "Multiply numerators and multiply denominators". Look at this until you see that this rule has been applied in reverse. Given the right hand side, can you get the left hand side? This should also ring some bells concerning the cummutative and associative properties of multiplication.

Next one.

$\frac{3}{3}*\frac{x}{4}\;=\;1*\frac{x}{4}$

Here, we applied the principle of the multiplicative inverse. 3/3 = 1. That's all that occurred.

Finally

$1*\frac{x}{4}\;=\;\frac{x}{4}$

We applied the property of multiplicative identity. Multiply things by 1 and nothing happens?

The whole process is the application of the most fundamental principles. If you think it is new material, then you have not learned the truth. All the stuff you have been learning since 1st or 2nd grade should be brought to bear on such problems. I am not making fun of you, I am just emphasizing that this is NOT new material. You already know the pieces of the puzzle. Just put them together. None of it should be a surprise.

You do the other one. Explain how to get across each equal sign.

11. In the second example if I'm using the inverse of the idea of multiplying the numerator and denominator by the same number, mean dividing them by the same number, or common factor, which would be in this case 2. Wouldn't that leave 2 * 2x that equals 4x? as you've only seemed to divide the numerator and left the denominator alone.

12. A common factor does not have to be a number, it may also be a variable, in this case $x$.

$\frac {8x}{4x}$ yes, indeed it has a common factor of 2. And it's got another common factor of 2. There's two common factors of 2. This is because $8=2\times2\times2$ and $4=2 \times 2$. So you can look upon your $\frac {8x}{4x}$ as being $\frac {2 \times 2 \times 2 \times x}{2 \times 2 \times x}$

If it makes it easier, you can also write it as:
$\frac {2 \times 2 \times 2 \times x}{1 \times 2 \times 2 \times x}$

because multiplying something by 1 doesn't make any difference to it.

So by the rule of multiplying fractions, but backwards):

$\frac {2 \times 2 \times 2 \times x}{1 \times 2 \times 2 \times x} = \frac {2}{1} \times \frac {2}{2} \times \frac {2}{2} \times \frac {x}{x}$

As you can see, all but the first bit equal 1, leaving that last lonely little 2 all out on its own all unloved.

13. Originally Posted by Alyosha
In the second example if I'm using the inverse of the idea of multiplying the numerator and denominator by the same number, mean dividing them by the same number, or common factor, which would be in this case 2. Wouldn't that leave 2 * 2x that equals 4x? as you've only seemed to divide the numerator and left the denominator alone.
You are NOT multiplying numerator or denominator by anything.

You are NOT dividing by the same number anywhere.

You SHOULD BE applying the properties with which you should be familiar.

If you are NOT familiar with these properties, it is likely that you are in the wrong class and you should back up and get a better background.

Time for a Placement Exam!

1) What property is demonstrated here?

(a*b)*c = a*(b*c)

2) What property is demonstrated here?

a*0 = 0

3) What property is demonstrated here?

b*c = c*b

4) What property is demonstrated here?

a*1 = a

5) Find the Prime Factors of 12.

6) What property is demonstrated here? (For a <> 0)

a*(1/a) = 1

Six quick problems that should be second nature by now. Get 100% or I will be discouraged.

14. 1) The idea that you can multiplying things in any order

2) that anything times 0 = 0

3) same as 1

4) that anything *1 = 1

5) 2 X 2 x 3

6) assuming this (For a <> 0) means a is greater or lesser than 0, it's that any number multiplied by it's reciprocal is 1.

15. Originally Posted by Alyosha
1) The idea that you can multiplying things in any order

2) that anything times 0 = 0

3) same as 1

4) that anything *1 = 1

5) 2 X 2 x 3

6) assuming this (For a <> 0) means a is greater or lesser than 0, it's that any number multiplied by it's reciprocal is 1.
Not bad. I am officially not discouraged. They do have names. You should learn them.

1) Associateive Property of Multiplication
2) Zero Property of Multiplication
3) Commutative Property of Multiplication
4) Identity Property of Multiplication
5) You totally got this one. Good work.
6) Inverse Property of Multiplication

"<>" is code for "Not Equal To". It's a little clunky, but it is in rather general usage.

Okay, now that we have these properties firmly in our minds, we can drag through one of these simplifications.

We are presented with $\frac{8x}{4x}$.

The fundamental idea is to find ways to utilize the Inverse Property of multiplication. In this way, we can change very complicated expression in to much simpler expressions.

First, then, we use the Associative Property of Multiplication in the Numerator. Rather than group everything together, we'll regroup a chunk out.

$\frac{8x}{4x}\;=\;\frac{2*4x}{4x}$

Nothing else happened. Just that one property. I picked this particular application of a property because it left in the numerator something tantalizingly close to what is in the denominator. I may have played around with the prime factorization a bit until I found something that looked right.

Second, we're going to use the Associative Property of Multiplication again.

$\frac{8x}{4x}\;=\;\frac{2*4x}{4x}\;=\;2*\frac{4x}{ 4x}$

Before, you said I multiplied everything by 2. This simply is not the case. I have only changed the order of things a bit. Rather than doing multiplication in the numerator first, this expression may indicate doing something with the other two factors first.

Now, we're going to hope that x is NOT zero (and by extension that 4x is not zero) so that we can use the Inverse Property of Multiplication. This allows us to see that a/a = 1 as long as a is not zero. With this insight, we can deal with the fraction.

$\frac{8x}{4x}\;=\;\frac{2*4x}{4x}\;=\;2*\frac{4x}{ 4x}\;=\;2*1$

Notice, please, that the application of the fundamental inverse principle is ALL that occurred.

One last thing, the Identity Property of Multiplication

$\frac{8x}{4x}\;=\;\frac{2*4x}{4x}\;=\;2*\frac{4x}{ 4x}\;=\;2*1\;=\;2$

Again, just one step at a time. Look long and hard at each step. Don't try to complicate it. You need to get good at this so you can do it quickly.


text:
# Exact Definition of Riemann Integral

1. Nov 27, 2007

### Mr.Brown

Hi I recently stumbled upon this:
I know that the Riemann Integral is defined for every piecewise continouus curve.
But now suppose you´re asked the following:

you are given $$f(x,y)=\frac{xy^3}{(x^2+y^2)^2}$$ with additional Definition
f(0,0)=0. ( It´s a textbook problem :) )

Now surely f(x,y) is smooth everywhere except for the origin. So I tried to show that f(x,y) goes to 0 in the origin but it doesn´t goes to 1/4(easy to show by setting x=y because f(x,y)=1/4 for x=y).

So the book asks to proof that the function $$g(y)=\int_{0}^{1}f(x,y)dx$$ is well defined on the real line.
I would intuitively say no because of the discontinuity at the origin but otherwise the origin is a set of measure zero.
So how does that definition go ?

2. Nov 27, 2007

### SiddharthM

you say approaching the origin via y=x gives you 1/4, remember that limits in n-dimensions need to give the same answer no matter WHICH way you goto the origin. Now, i haven't worked with several variables in a while so i'm not going to attempt it although looking at the denomintator it looks like moving to polar coordinates would simplify matters.

the precise definition of reimann integration can be found on wikipedia, the darboux definition is the most practical for analytic purposes but in this case all you need to know is that the single discontinuity at the origin will NOT prevent the integral from existing. Regarding measure you are correct, a function f is integrable on a set E iff f is continuous almost everywhere on E.

hope this helps

3. Nov 27, 2007

### SiddharthM

oh yeah i forgot to say, try approaching the origin via the parabola y=x^2.

4. Nov 27, 2007

### LukeD

As you know, f(x,y) (with y held constant) is continuous except when x=y=0. This is a single point, so it is of measure zero. What else needs to be satisfied for a function to be Riemann integrable on a set [a,b]?

As for the exact definition of the Riemann Integral, I don't think it'll help you here, what you really want is the Riemann Lebesgue Theorem, but
A Riemann Sum
$$R(f,P,T) = \sum_1^n f(\hat x_i)\Delta x_i$$ where $$P=\{x_1, \dotsc, x_n, x_{n+1}\}$$ is a partition of the interval [a,b] (in this sense, a partition is an ordered set of points in the interval [a,b] with $$x_1 = a$$ and $$x_{n+1} = b$$) and $$T = \{\hat x_1, \dotsc, \hat x_n\}$$ is a set of points interspersed between the partition points of P (i.e. $$x_1 \leq \hat x_1 \leq x_2 \leq \dotsb \leq x_n \leq \hat x_n \leq x_{n+1}$$)

We defined $$\Delta x_i = x_{i+1} - x_i$$

The mesh of P is the largest $$\Delta x_i$$

The Riemann Integral is the limit of the Riemann Sums as the mesh of P goes to 0

In other words, although you learn in Calculus that the Riemann Sum is defined as dividing the interval up evenly and then picking a point at one of the end points of your divisions, and then the Riemann Integral is defined by the limit as the size of the divisions go to 0, the actual definition of a Riemann Sum allows you to divide the interval up however you want and place points anywhere in the divisions. You still take the integral to be the limit as the size of the divisions goes to 0.

Anyway, like I said, this is probably completely useless to you. Look up the Riemann Lebesgue Theorem.

Last edited: Nov 27, 2007
5. Nov 27, 2007

### LukeD

Not entirely true, you're missing something. The function $$f(x) = 1/x$$ is not integrable on any interval containing 0 even though it is only discontinuous at 0.

Edit: Mr. Brown: also keep in mind that because your function f is continuous everywhere except at (0,0), you only need to worry about g(0)

Last edited: Nov 27, 2007
6. Nov 27, 2007

Since it hasn't been mentioned yet, I just wanted to point out that h_y(x) = f(x,y) is actually continuous at 0 (and everywhere else). For nonzero y, the denominator is nonzero. For y = 0, h_y(x) = 0 everywhere. I think for the given problem, that's all he needed to show.

Correction.. $$g(y)=\int_{0}^{1}h_y(x)dx$$ is well defined because h_y is continuous for fixed y.

Last edited: Nov 27, 2007
7. Nov 27, 2007


text:
# Box topology defines a topological vector space?

Is the set of all real summable sequences, endowed with the box topology, a topological vector space?

Formally, I am interested in $X=\{(a_1,\ldots):a_n \in \mathbb{R} \ \forall n, \ \sum a_n < \infty\}$ endowed with the box topology.

I think that this is the case as vector addition and scalar multiplication both seem continuous to me.

But, there is at least one place which suggests that the box topology generally does not define a topological vector space:

Why is the box topology on a product of complete topological vector spaces complete?

The problem is that multiplication by $x \in X$ is not continuous from $\mathbb R$ to $X$ if $x$ has infinitely many nonzero terms. For example, take box neighbourhood $U = \prod_{n=1}^\infty (1/n^2-1/n^4, 1/n^2+1/n^4)$ of $x$ with $x_n = 1/n^2$. There is no $\delta > 0$ such that $(1-\delta, 1+\delta) x \subseteq U$, as this would require $\delta < 1/n^2$ for all $n$.
• Well, can you? ${}$ Jan 18, 2018 at 4:01
• Hi, I see it now. Thanks very much. Is there an interesting finest topology? More specifically, I am interested in a topology where I have a topological vector space and the set of positive sequences $(a_1,...)$ s.t $\frac{a_1}{a_2}>r_1$, $\frac{a_2}{a_3}>r_2$,... is open where the $r_i$ are some constants between 0 and 1.


text:
# Tag Info

5

One of the key properties we want out of PRNGs is that they produce results that follow a desired probability distribution. Nearly all of the time we want our PRNGs to produce data with uniform distribution—all of the possible output values should have equal probability. So let's assume that we are given a seq sequence whose values are samples drawn from a ...

2

Given any finite sequence $(a_0,a_1,\ldots,a_{N-1})$ we may be interested in generating exactly that sequence, or generating the minimal degree (periodic) sequence which has that sequence as its' starting bits. Trivially, the circulating shift register of length $L$, initially loaded by $(a_0,\ldots,a_{L-1})$, will generate this sequence, by implementing ...

2

If this can code can help you understanding the Linear Shift Register. (^ represents Exclusive OR) byte[] register = new byte[] { 0, 1, 0, 1 }; byte[] plain = new byte[] { 1, 0, 1, 0, 1, 0, 1, 0, 1, 1 }; ; byte[] cipher = new byte[plain.Length]; for (int i = 0; i < plain.Length; i++) { //plaintext ^ ...

2

Read the drawing assuming: each of the four (about) square boxes $a_j$ initially (at $t=0$) holds one of the four bits of $S_0$ (in reading order, unless otherwise stated); information follows arrows (almost) instantly, except when getting into a square box where that is delayed until $t$ has grown by $1$ (in time unit or clock period); notice this ...

3

The problem with answering this question as posed is that one would need to have a definition of "cryptographic quality" that applies to 1/f noise. You'd need to talk about something like how many data points an intelligent adversary would need to observe to be able to distinguish your pseudorandom 1/f noise generator from truly random 1/f noise. I suspect ...

2

If you want the encryption to be information-theoretically secure, then you need an information-theoretically secure RNG. And therein lies the problem—how do you establish that a given RNG is information-theoretically secure? Science may say that there are some types of events that are physically unpredictable, but that by itself is insufficient to get a ...

0

Short answer: @Stephan Touset's answer is of course correct: if this is just for your learning, then it doesn't matter, use whatever's convenient. Anything that's not this should be fine: *Mandatory XKCD The longer answer is that philosophers like to debate whether "truly random", as you put it, actually exists, and if it does, does it apply to ...

4

You're never going to release this, right? If so, it doesn't matter. Use a pseudorandom stream from /dev/urandom and "pretend" that it's truly random for the sake of learning about the concepts involved.

2

If, for example, the PRNG can only be seeded with $2^{32}$ bits, then it cannot possibly produce all the $52!$ distinct permutations. In addition, even if we could meet the 226 bits needed for a 52-card deck, the distribution may not be uniform. This is true, but a good PRNG (even non-cryptographic) will give you a random sample from a uniform distribution....

1

I have modified this Ruby script to not require public keys and to allow direct (r,s) input. Funnily enough, I stumbled on this thread (created two hours ago as of writing) on Google searching for the exact same problem. Note that I am using Secp192r1 (due to a CTF challenge), so feel free to modify that to your likings: require 'ecdsa' msghash1_hex = '...

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 the inverse ackerman functions JmsNxn Long Time Fellow Posts: 481 Threads: 86 Joined: Dec 2010 04/14/2011, 08:48 PM (This post was last modified: 04/14/2011, 08:54 PM by JmsNxn.) Every one here has heard of the ackermann function; we'll define it as: $\{x,\, y,\, u\, \epsilon\, R^{+}\}$ $A(x,\, y,\, u)\, =\, x\, \{u\}\, y$ I'm interested in itss inverses though. There would be three inverses therefore, the first one with relation to x: $R(x,\, y,\, u)\, =\, x\, \}u\{\, y$ which means R for roots, and }u{ is taken to mean: }0{ is subtraction, }1{ division, }2{ roots, }3{ super roots, etc etc and then are next inverse is taken with relation to y $L(x,\, y,\, u)\, =\, u;\log_x(y)$ where 0;log_x(y) = y-x, 1;log_x(y) = y/x, and 2;log_x(y) = $\log_x(y)$ and now, the final inverse, and definitely the most interesting inverse is the inverse taken with relation to u $I(x, y, u) = ?$ if $A(x, y, I(x, y, u)) = u)$ this question is the equivalent of asking $3\,\{u\}\, 4\, =\, 9$ what is u? it's the act of rearranging an equation to solve for u. It might even seem intuitively impossible. I've only found one solution to this problem... and that's by using logarithmic semi operators, though I is only defined over domain [0, 2] Does anyone know if anybody else has ever looked into this inverse function? tommy1729 Ultimate Fellow Posts: 1,454 Threads: 350 Joined: Feb 2009 04/14/2011, 10:57 PM When there is a (local) unique inverse , you might use the so-called " series reversion " for its taylor series at a fixpoint. Or you might try to express ackermann in terms of sexp slog etc but im not sure its possible. maybe this has occured in older threads. JmsNxn Long Time Fellow Posts: 481 Threads: 86 Joined: Dec 2010 04/14/2011, 11:04 PM (This post was last modified: 04/14/2011, 11:06 PM by JmsNxn.) (04/14/2011, 10:57 PM)tommy1729 Wrote: When there is a (local) unique inverse , you might use the so-called " series reversion " for its taylor series at a fixpoint. Or you might try to express ackermann in terms of sexp slog etc but im not sure its possible. maybe this has occured in older threads. you can express the domain [0, 2] using sexp slog algos, it's piece wise though. but I'll show you what happens; 0 <= q <= 1 $x\, \{q\}\, y\, =\, -q:ln(q:ln(x)\, +\, q:ln(y))\, = sexp(slog(sexp(slog(x)-q)+sexp(slog(y)-q))+q)$ solve for q, it's practically impossible; same deal for [1, 2] Xorter Fellow Posts: 93 Threads: 30 Joined: Aug 2016 09/18/2016, 11:02 AM (04/14/2011, 08:48 PM)JmsNxn Wrote: $3\,\{u\}\, 4\, =\, 9$ what is u? Well, James, according to PARI/gp u is about 1.5737, because 3[1.5737]4 = ~9 But I do not know why. I need a taylor-series formula or a recursive formula to get know why. Xorter Unizo « Next Oldest | Next Newest »

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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeJul 17th 2012

I have edited and expanded wall crossing a little

One question to Zoran:

you have designed the entry to cover the notion in great generality. But most of the references that you already had, and now also all that I have added, concern wall crossing of BPS states. Eventually we need to do something to make the entry more systematic on this point. Should we split off an “wall crossing of BPS states”, maybe?

• CommentRowNumber2.
• CommentAuthorzskoda
• CommentTimeJul 17th 2012
• (edited Jul 17th 2012)

I prefer not to split. Most of existing references are anyway about BPS states, so if one thinks of wall crossing one usually thinks of this case. On the other hand, it is beneficial for that case, to attempt the meaning which is more general. It is like Poincare duality. It is mainly about manifolds. Though it is useful sometimes to think more generally. We did not make a special entry about Poincare duality for manifolds. Do you think we should ?

Another reason why I would not like to split is that wall crossing is sometimes about BPS states essentially but it is phrased in a different language.

Of course I agree that the ’general section’ is so far vague (I originally wrote the entry on wall crossing AND stability, and the text much reflects the second aspect), but I will need more thought to make it more direct and closer to the BPS case. So if you excuse me until mid August I’d repair that (I’d appreciate some reminder if it slips my mind).

• CommentRowNumber3.
• CommentAuthorzskoda
• CommentTimeMar 18th 2013

I added a new preprint of Kontsevich and Soibleman from few days ago at wall crossing and Donaldson-Thomas invariant. The web page of Aarhus lectures cited as the main website at wall crossing in Aarhus. Can somebody correct the link ?

• CommentRowNumber4.
• CommentAuthorTodd_Trimble
• CommentTimeMar 19th 2013

I remember seeing wall crossing before and not finding what I was expecting to find: intuitively, a codimension 1 subset of a space of functions or configurations (typically infinite-dimensional) which separates chambers = connected components of the subspace consisting of nondegenerate functions or configurations. These are studied in Morse-Cerf theory; see for example here. Is there a connection between this and the material discussed in the nLab article?

• CommentRowNumber5.
• CommentAuthorzskoda
• CommentTimeMar 19th 2013

Well, walls are indeed codimension one, but the wall crossing research is not concerned with how the walls may look like in general geometric situations, but when they represent the abrupt/discontinuous change in parameters and provided the space has to do with solutions of something, in the cases the entry mainly describes, these are the space of stability conditions. Thus you have something what depends on stability conditions, like observables of QFT, and you look their dependence.

• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeMar 19th 2013

I’d think there should be room in the $n$Lab entry to mention the use of “wall crossing” as in the article that Todd is pointing to and to make the discussion of the various uses of the term more specific.

• CommentRowNumber7.
• CommentAuthorzskoda
• CommentTimeMar 20th 2013
• (edited Mar 20th 2013)

I am not sure. It may be a completely different concept and should be a separate entry (with a link) Cerf wall crossing. If I understand, Cerf crossing is a geometric path which crosses a codimension one wall in certain setup of Morse theory, so it is about a shape of something crossing the wall in the usual manifold. On the other hand, wall crossing in the sense of our entry is a phenomenon of crossing in the more complicated moduli stack of parameters. Wikipedia also calls it “phenomenon in string theory” here. I am however not an expert to judge. Having an entry about Cerf theory does not hurt, and having a link to there from wall crossing neither. But trying to explain that these are cases of a same concept may be overstretching and giving a wrong information. It would be nice if somebody can of course trace if it may be so.

• CommentRowNumber8.
• CommentAuthorzskoda
• CommentTimeMar 20th 2013
• (edited Mar 20th 2013)

Cerf wall crossing stub. Please substantiate.

• CommentRowNumber9.
• CommentAuthorUrs
• CommentTimeMar 20th 2013

Thanks. I have further cross-linked with wall crossing by adding disambiguation lines at the top of the entries.

• CommentRowNumber10.
• CommentAuthorzskoda
• CommentTimeMar 20th 2013

I am yet not sure where to put the wall crossing in representation theory, so I will also separate it: wall crossing functors with redirect wall crossing in representation theory.

• CommentRowNumber11.
• CommentAuthorUrs
• CommentTimeApr 30th 2015

did some minimal editing and added further references at wall crossing.


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# Homework Help: Optics - Double Slit Irradiance of Fringes

1. Apr 8, 2009

### peppies

[Solved] Optics - Double Slit Irradiance of Fringes

1. The problem statement, all variables and given/known data
A double slit diffraction pattern is formed using mercury green light at 546.1nm. Each slit has a width of 0.100mm (= b), slit separation is 0.400mm (=a). The pattern reveals that the fourth-order interference maxima are missing from pattern.

What is the irradiance of the first three orders of interference fringes, relative to the zeroth-order maximum?

2. Relevant equations
I = 4Io (sin($$\beta$$)/$$\beta$$))^2 * (cos($$\alpha$$))^2
$$\beta$$ = 0.5kbsin($$\theta$$)
$$\alpha$$ = 0.5kasin($$\theta$$)
k = (2(pi)/$$\lambda$$)

3. The attempt at a solution
I've been trying to figure this one out for days and the book never gave any practice problems or simple explanation on what to do here.

First I tried to solve for the first order interference fringe (m=1):

sin($$\theta$$) = (m$$\lambda$$/a), then use this result to solve for alpha and beta in the two equations above and then finally solve for the sinc and cos^2 functions in the irradiance to find the ratio to the zeroth order fringe.

The answers in the back of the book for the first three orders are (0.8106, 0.4053, 0.09006) but I am not getting these.

The book I'm using is "Introduction to Opitics 3rd edition by Pedrotti"

Thanks
Best

Last edited: Apr 8, 2009
2. Apr 8, 2009

### peppies

Nevermind, I solved it by using the RAD mode on my calculator rather than DEG.


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mersenneforum.org solving modular constraints
 Register FAQ Search Today's Posts Mark Forums Read

 2010-02-17, 20:07 #1 Joshua2     Sep 2004 13×41 Posts solving modular constraints x = 1 mod 3 x = 2 mod 4 x = 3 mod 5 so I changed the numbers from my problem, but its similar. I'm trying to find what x could be. I'm pretty sure I can do LCM of 3,4,5 and know the answer is 60N + something or something mod 60. What's the best way to find out; for small problems it would seem try 8 then 13, and so on up to 60, is there a faster way for even small problems like this? I know there's fast ways using chinese remainder theorem; maybe someone could demonstrate a little algorithm on these numbers? Thanks!
2010-02-17, 20:51   #2
R.D. Silverman

Nov 2003

22·5·373 Posts

Quote:
 Originally Posted by Joshua2 x = 1 mod 3 x = 2 mod 4 x = 3 mod 5 so I changed the numbers from my problem, but its similar. I'm trying to find what x could be. I know there's fast ways using chinese remainder theorem; !
??

What else are you looking for? There are plenty of examples.

 2010-02-17, 20:55 #3 wblipp     "William" May 2003 New Haven 1001001101102 Posts (40 + 30 + 48) mod 3 = (1+0+0) (40 + 30 + 48) mod 4 = (0+2+0) (40 + 30 + 48) mod 5 = (0+0+3) so 40 + 30 + 48 = 118 works. As you have observed, any equivalent answer mod 60 will work. Depending on your needs, the smallest answer is either 58 or -2. As you seemed to already know, this is the Chinese Remainder Theorem. The trick is to make a sum so that all but one addend is zero for each modulo. I have difficulty imagining anything simpler. It's also more help than we ought to give in homework. I justify it on the grounds of your claim to have changed the numbers and the usefulness of a pedagogical example.
 2010-02-17, 20:56 #4 Wacky     Jun 2003 The Texas Hill Country 32×112 Posts Rather than "trying" 3, 8, 13, etc., do it algebraically. If x = 3 mod 5, then x= 5A+3 for some integer A. Substituting in the other constraints 5A+3 = 1 mod 3 5A+3 = 2 mod 4 But this is the same as 2A = 1 mod 3 A = 3 mod 4 Thus, we have reduced the number of constraints. When we find permissible values for A, we can substitute and we will have the permissible values for x
2010-02-18, 07:56   #5
Joshua2

Sep 2004

53310 Posts

Quote:
 Originally Posted by Wacky Rather than "trying" 3, 8, 13, etc., do it algebraically. If x = 3 mod 5, then x= 5A+3 for some integer A. Substituting in the other constraints 5A+3 = 1 mod 3 5A+3 = 2 mod 4 But this is the same as 2A = 1 mod 3 A = 3 mod 4 Thus, we have reduced the number of constraints. When we find permissible values for A, we can substitute and we will have the permissible values for x
This way makes a ton of sense. Is this the CRT as well? Its seems we can't continue with 2A = 1 + 3A and A = 3 + 4A? I think I did that wrong. How about A = 3 + 4B and 2A = 1 + 3 B with two equations and two unknowns? I'll look at the other posts more later.

2010-02-18, 13:04   #6
Wacky

Jun 2003
The Texas Hill Country

32×112 Posts

Quote:
 Originally Posted by Joshua2 Is this the CRT as well?
Why don't you tell us? Is it, or is it not? Why?
Quote:
 Its seems we can't continue with 2A = 1 + 3A and A = 3 + 4A? I think I did that wrong.
A correct observation.
Quote:
 How about A = 3 + 4B and 2A = 1 + 3 B
That's the spirit
Quote:
 with two equations and two unknowns?
Is that what we did before?

 2010-02-19, 06:53 #7 Joshua2     Sep 2004 13×41 Posts 1. I don't think it is CRT, because I read that you use euclid's extended algorithm, and we didn't use it. 2. I think that is what we did before, reduce by one. A = -2 - B = 3 + 4B so -2 - 3 = 5B or -5 = 5B so B = -1. So is that the right idea?
2010-02-19, 06:56   #8
Joshua2

Sep 2004

21516 Posts

Quote:
 Originally Posted by wblipp (40 + 30 + 48) mod 3 = (1+0+0) (40 + 30 + 48) mod 4 = (0+2+0) (40 + 30 + 48) mod 5 = (0+0+3) so 40 + 30 + 48 = 118 works.
Where did 40, 30 and 48 come from? I assume the 1+0+0 thing is prime factored?

2010-02-19, 08:04   #9
wblipp

"William"
May 2003
New Haven

2·32·131 Posts

Quote:
 Originally Posted by Joshua2 Where did 40, 30 and 48 come from?
CRT.
The first number needs to be a multiple of 4*5 that is 1 mod 3.
The second number needs to be a multiple of 3*5 that is 2 mod 4
The third number needs to be a multiple of 3*4 that is 3 mod 5

Quote:
 Originally Posted by Joshua2 I assume the 1+0+0 thing is prime factored?
(40 + 30 + 48) mod 3
=(40 mod 3) + (30 mod 3) + (48 mod 3)
= 1 + 0 + 0
=1

2010-02-19, 15:24   #10
Wacky

Jun 2003
The Texas Hill Country

32×112 Posts

Quote:
 Originally Posted by Joshua2 1. 2. I think that is what we did before, reduce by one. A = -2 - B = 3 + 4B so -2 - 3 = 5B or -5 = 5B so B = -1. So is that the right idea?
No. I think that you may be confusing the "algebraic equality" (as in X = 5 A + 3) with the "modular equality" (as in X = 3 mod 5).
They are somewhat different concepts. Therefore, for clarification, I will use "==" in the latter case.

We are reducing the number of constraints by one by making a substitution that causes one of the constraints to be met for all values of the unknown.

We started with:

Code:
Find the set of integers "X" such that

X == 1 mod 3
and
X == 2 mod 4
and
X == 3 mod 5

X = 5 A + 3

This transformed the last constraint into

Code:
5 A + 3 == 3 mod 5
which is true for all integers A

That left us with the equivalent problem:

Code:
Find the set of integers "A" such that

5 A + 3 == 1 mod 3
and
5 A + 3 == 2 mod 4
or, equivalently:

Code:
Find the set of integers "A" such that

2 A == 1 mod 3
and
A == 3 mod 4
So, continuing this procedure:
We let A = 4 B + 3 which will meet the last constraint for all integers B, leaving only one constraint (mod 3).
Then we let B = 3 C + … (something), which will be true for all integers C.

Finally, we combine all of the substitutions to get one substitution
X = f(C), which meets all of the constraints for any integer C.

As you already know, from other posts, this will be
Code:
X = 60 C + 58
or equivalently,
Code:
X == 58 mod 60
However, you should work out the details to derive the answer yourself.
There are a couple of aspects of the derivation that might catch you unaware. You should also look to see similar coefficients in the CRT solution. Observing these similarities might give you a greater insight into "why the methods work"

Last fiddled with by Wacky on 2010-02-19 at 15:52 Reason: Formatting to clarify the presentation

2010-02-19, 21:10   #11
Joshua2

Sep 2004

13×41 Posts

Quote:
 Originally Posted by wblipp CRT. The first number needs to be a multiple of 4*5 that is 1 mod 3. The second number needs to be a multiple of 3*5 that is 2 mod 4 The third number needs to be a multiple of 3*4 that is 3 mod 5 (40 + 30 + 48) mod 3=(40 mod 3) + (30 mod 3) + (48 mod 3) = 1 + 0 + 0 =1
I think I understand CRT now. Now I just need to figure out modulo inverses, like what is inverse of 2 mod 5...

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# Linear space (geometry)

A linear space is a basic structure in incidence geometry. A linear space consists of a set of elements called points, and a set of elements called lines. Each line is a distinct subset of the points. The points in a line are said to be incident with the line. Any two lines may have no more than one point in common. Intuitively, this rule can be visualized as two straight lines, which never intersect more than once.

(Finite) linear spaces can be seen as a generalization of projective and affine planes, and more broadly, of 2-${\displaystyle (v,k,1)}$ block designs, where the requirement that every block contains the same number of points is dropped and the essential structural characteristic is that 2 points are incident with exactly 1 line.

The term linear space was coined by Libois in 1964, though many results about linear spaces are much older.

## Definition

Let L = (P, G, I) be an incidence structure, for which the elements of P are called points and the elements of G are called lines. L is a linear space if the following three axioms hold:

• (L1) two points are incident with exactly one line.
• (L2) every line is incident to at least two points.
• (L3) L contains at least two lines.

Some authors drop (L3) when defining linear spaces. In such a situation the linear spaces complying to (L3) are considered as nontrivial and those who don't as trivial.

## Examples

The regular Euclidean plane with its points and lines constitutes a linear space, moreover all affine and projective spaces are linear spaces as well.

The table below shows all possible nontrivial linear spaces of five points. Because any two points are always incident with one line, the lines being incident with only two points are not drawn, by convention. The trivial case is simply a line through five points.

In the first illustration, the ten lines connecting the ten pairs of points are not drawn. In the second illustration, seven lines connecting seven pairs of points are not drawn.

 10 lines 8 lines 6 lines 5 lines

A linear space of n points containing a line being incident with n − 1 points is called a near pencil. (See pencil)

 near pencil with 10 points


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# What is the axis of symmetry and vertex for the graph y= -4x^2?

Jan 10, 2018

Axis of symmetry is $x = 0$ and vertex is $\left(0 , 0\right)$

#### Explanation:

When an equation $y = a {x}^{2} + b x + c$ is converted in form

$y = a {\left(x - h\right)}^{2} + k$

axis of symmetry is $x - h = 0$ and vertex is $\left(h , k\right)$

As we can write $y = - 4 {x}^{2}$

as $y = - 4 {\left(x - 0\right)}^{2} + 0$

axis of symmetry is $x - 0 = 0$ i.e. $x = 0$ i.e. $y$-axis and vertex is $\left(0 , 0\right)$

graph{-4x^2 [-5.146, 4.854, -3.54, 1.46]}


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# Scientist Study

Your source for the latest research news

## Saturday, 5 June 2021

What were the first moments of the Universe like? It's a mystery that scientists have been trying to unravel for decades. The ALICE collaboration at CERN is a specialist in the subject: this detector ( A Large Ion Collider Experiment ) was designed to study quark-gluon plasma, a phase of matter that would have existed just after the Big Bang. And the team recently succeeded in recreating and characterizing this very first hypothetical material, using the Large Hadron Collider (LHC).

What are the properties of matter at the extreme densities and temperatures of the beginnings of the Universe? The scientists of the ALICE collaboration finally have some answers. The resulting material only persisted for a fraction of a second, but long enough that scientists could study its characteristics for the very first time.

It turns out that this plasma is of the liquid type and this discovery could provide a better understanding of how the early Universe evolved during the first microsecond after the Big Bang. To reproduce this primitive material, the team initiated high energy (5 TeV) heavy (lead) ion collisions within the LHC.

## Plasma that flows like water

As a reminder, quarks are elementary particles that combine to form protons and neutrons (among others). Quarks are linked together via a strong interaction, mediated by particles called gluons. The collisions that occur in the LHC generate temperatures more than 100,000 times higher than those in the center of the Sun. Under these extreme conditions, the protons and neutrons decompose, releasing the quarks and gluons that constitute them in passing: the famous quarks-gluon plasma is thus obtained.

The objective of the ALICE collaboration is to study this plasma to understand how it was able to give birth to the particles that today make up the matter of our universe. To do this, a gigantic detector was installed 56 meters underground to receive the particle beams from the LHC. Lead particles, launched at a speed close to that of light, thus made it possible to recreate the very first material that appeared after the Big Bang. The experiment had been performed successfully in the past, but this time the scientists had time to probe the characteristics of the plasma in detail.

The quark-gluon plasma exhibited characteristics typical of a perfect liquid, exhibiting almost no resistance to flow. The flow of a fluid is determined by the ratio of its viscosity to its density. While the viscosity and density of quark-gluon plasma are about 16 orders of magnitude greater than that of water, the researchers found that the relationship between the viscosity and density of the two types of fluids was the same. In other words, the very first state of matter would flow in the same way as water!

## A surprising similarity

Shortly after the Big Bang, the early Universe consisted of a dense, hot "soup" of quarks and gluons. A few microseconds later, this mixture cooled down to form the first building blocks of matter that makes up the entire universe. Thus, the material which surrounds us today has in theory very different properties from those of this primitive soup. Fluids such as water, for example, rely on assemblies of atoms and molecules much larger than primitive particles, held together by much weaker forces. But these recent experiments show that despite these differences, the kinematic viscosity - or the ability of a fluid to flow out - of primitive plasma is very similar to that of ordinary liquids.

The viscosity of a liquid can vary significantly depending on the temperature. However, there is a lower limit to this almost universal viscosity, which depends on fundamental physical constants (such as Planck's constant). However, the results of the study suggest that the viscosity of the quark-gluon plasma is very close to this universal lower limit of viscosity. "This study constitutes a rather rare and pleasing example of the possibility of establishing quantitative comparisons between extremely disparate systems", underlines in a press release Professor Matteo Baggioli, one of the members of the team.

These results are also a new illustration of the power of physics, which can translate general principles into specific predictions of complex properties, such as liquid flow in exotic types of matter such as quark-gluon plasma, adds the professor.

Quantum chromodynamics is a theory which makes it possible to describe the powerful forces of interactions between quarks and gluons (and therefore the cohesion of the atomic nucleus); but it is not sufficient to fully understand the properties of the original plasma. This astonishing similarity with fluid dynamics is therefore a further step forward in this field of research. Scientists hope to discover even more details about this plasma as CERN's accelerator is upgraded. Further studies will also provide a better understanding of how quarks and gluons organize themselves into protons and neutrons, a step that may have conditioned the extremely rapid expansion (cosmic inflation) of the Universe.

## Reference:

S. Acharya et al. Measurements of mixed harmonic cumulants in Pb–Pb collisions at


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Page content

# Making a new mode:

A mode defines the specific reaction your game should have to various events that occur. If you want the player to receive 1000 points every time a specific shot is made, you will define that code within a mode.

The Mode is the basic building block of coding your own game; Modes can be as simple as defining a response to a single switch in your machine or they can handle every switch in your machine as the complexities of the sequences in which those switches might be hit. Modes might show something specific on the display or play background music while the mode is active, or they might be “invisible” as far as the player is concerned.

### Modes in PyProcGame may:

• Respond to switch events
• Register/respond to timers
• Enable/disable/schedule lamps
• Enable/disable/schedule coils
• Play sounds
• Show animation/text on the display

When a mode is active it means it is present in the game’s ModeQueue (self.game.modes). The ModeQueue is a priority-sorted list of the Modes in the game. Inactive modes do not receive notification of switch (or other event) notifications.

In SkeletonGame, each Mode is defined by the game programmer as a subclass of “AdvancedMode” –let’s explore the creation of a mode that will serve as a skillshot.

### Defining a new Mode

To create a new SkillshotMode class, we create a file called my_modes/SkillshotMode.py with the following contents:

class SkillshotMode(procgame.game.AdvancedMode):

def __init__(self):

This file doesn’t define much functionality for the mode, but provides several key portions of the definition. At line 1 above, we see the name of the new class (SkillshotMode) and its parent class (AdvancedMode –from which our new class inheriets a lot of functionality).

Line 3 above defines the __init__() function –the constructor method that is called whenever this class is created. This function must be included in your class definition, and line 4 includes the call to super() which gives the AdvancedMode class information about how your class should be treated.

The priority is the numeric priority of this mode relative to others. Being higher priority means your mode will receive switch events before other events do. The type indicates when the mode will be activated.

mode_type = which means…
AdvancedMode.Ball The mode is activated when a ball starts and deactivated when the ball ends. In a typical 3-ball game, that means your mode is activated/started and deactivated/stopped 3-times per player.
AdvancedMode.Game The mode is activated when a player starts a game and is deactivated when the (last) player’s game ends. In a typical 3-ball game, that means your mode is activated/started exactly once per game and deactivated/stopped exactly once per game, regardless of the number of players who are added into that game.
AdvancedMode.System The mode is activated when the game code is is initially launched and deactivated when the game code is quit/shutting down. The mode will remain active across multiple games, across multiple players.
AdvancedMode.Manual The mode is not auto-added or auto-removed. The programmer will take responsibility for activating the mode and deactivating the mode whenever it is appropriate to do so. NOTE: If a mode doesn’t specify a mode_type in the call to super(), this is the default/assumed type.

Because the mode is created with a type of Ball, the mode will be added/activated when each new balls starts. We will manually deactivate the mode early, once the skillshot phase of the ball is over.

### Adding the mode to the Game class

We need to modify the T2Game.py file to import the new mode. We will modify the modes import line as:

from my_modes import BaseGameMode, ExBlankMode, SkillshotMode

and futher on in the __init__ method of the T2Game we need to add an initializer for the SkillshotMode (right after the other mode initializers). This line will be added:

       self.skillshot_mode = SkillshotMode(game=self)

If we were to test the game code, start a game, and watch the console output, we would see the mode queue will contain our new SkillshotMode. Unfortunately, the mode’s lack of interactivity with the player would make it otherwise impossible to verify its existence.

### Enhancing the SkillshotMode functionality

Modes receive events as long as they are active. Specific function names within the Mode definition will automatically be called when these events occur. The following methods can be provided:

#### Switch events

A mode that defines methods matching a speciifc naming convention indicates it wants the framework to call this method to “handle” the switch event (i.e., a switch handler method). When a Mode is created, its methods are scanned for any methods that match the naming pattern: sw_<switch name>_<switch state>(self,sw) where <switch name> is a valid named switch and <switch state> is either active or inactive.

For example, the method:

   def sw_gripTrigger_active(self, sw):

would be invoked by the framework when the switch corresponding with label gripTrigger is activated (note that the binding of the identifier gripTrigger and the specific machine switch location is defined in the machine.yaml).

Similarly, a method named sw_startButton_inactive() would be called when the startButton switch changes to an inactive state.

An optional state duration can be specified, in the case that you wish to respond to an event only after the switch has been in that state for a certain amount of time. Adding a for_<time period> suffix to the normal switch handler naming convention enables this behavior:

Examples: sw_switchName_active_for_500ms(self, sw) | called once switchName is active for 500 milliseconds sw_switchName_inactive_for_3s() | called once switchName is inactive for 3 seconds sw_switchName_inactive_for_20ms() | called once switchName is inactive for 20 milliseconds

#### Mode life-cycle events

There are two standard method signatures which will be automatically invoked when a mode is activated (added to game.modes) and a mode is deactivated (removed from game.modes), respectively.

  def mode_started(self):
# called when the mode has been actived
pass

def mode_stopped(self):
# called when the mode is no longer active
pass

Because these method names are tied to activation and deactivation, you should combine this information with the mode_type (or your knowledge of when you manually add/remove this mode) to know when these methods will be called. A different approach would be to use SkeletonGame event handlers, as described below.

Additionally, PyProcGame provides a method signature that is called every “tick” for an active mode, called mode_tick(). A tick is a single pass through run_loop() as it completes one cycle of reading events from the P-ROC and processing them (by informing modes and running their responses). A game may have a tick rate anywhere from 300Hz to 7000Hz so the tick method would potentially be called 300 to 7000 times per second; in order to keep tick rates high, any mode that needs a mode_tick handler should have that code be extremely brief in order to keep the run loop running quickly. If the tick rate drops too significantly, the run_loop() will not tickle the watchdog circuit in the P-ROC hardware, and your machine will go dark. Don’t call sleep() in code.

#### SkeletonGame events

These are regular gameplay events that the SkeletonGame system tracks. Your mode will be informed of the game event occuring so you can run code to respond to the event. The following page details all the built-in events supported by SkeletonGame:

http://skeletongame.com/the-skeletongame-event-system/

If your code defines the methods described above, the framework will call the method when that event occurs (provided your mode is active when the event occurs). Modes can also request to postpone the event propagation (i.e., the calling of the next event handler) by returning a number of seconds that the mode needs to ‘finish up’ (e.g., play a special animation or sound) or can delay and stop further propegation by returning a tuple of the delay in seconds and the second entry is True

#### Timers

Any method can be automatically called after a specified amount of time passes (i.e., delay) using the pyprocgame provided method procgame.game.Mode.delay(). For example, suppose there is a mode that should only be active for 10.5 seconds. A mode can be “deactivated” (removed from the game’s mode queue) with the line self.game.modes.remove(self), so we define a new method to contain that line, and use a delay to call that method later.

def deactivate(self):
self.game.modes.remove(self)

def mode_started(self):
self.delay(delay=10.5, handler=self.deactivate)

Delay also takes two optional arguments, a parameter to be passed to the delayed method (param) and a name (name) which is useful for canceling the delay before it fires. If a delay is not explicitly named, a name will be assigned and will be provided as the return value from the call to delay().

The following is an example of canceling a delay before it fires, by storing the return value from delay(). This example would give the player a bonus for hitting target2 within 4 seconds of hitting target1

def evt_player_added(self, player):
player.setState('t1_hit', False)

def sw_target1_active(self, sw):
self.game.setPlayerState('t1_hit', True)
self.delayed_name = self.delay(delay=4.0, handler=self.reset_var)

def sw_target2_active(self, sw):
# cancel the delay:
self.cancel_delayed(self.delayed_name)
if( self.game.getPlayerState('t1_hit')==True ):
# player gets a special bonus for hitting both
self.game.score(1000)
self.game.setPlayerState('t1_hit', False) # reset now, we've awarded
else:
# the player only hit this one target
self.game.score(10)

def reset_var(self):
# too long has passed; reset the variable
self.game.setPlayerState('t1_hit', False)

def mode_stopped(self):
# cancel the delay, the mode is over
self.cancel_delayed(self.delayed_name)

## Lighting Lights, Flashing Flashers

While the above example illustrates some advanced logic using timers, it’s not very exciting for the player, who receives zero feedback while playing the game. Flashing lights are a huge part of the pinball experience.

Lights are part of the physical machine. From Mode code, the identifier self.game.lamps leads to the dictionary of lamps in this machine. The specific lamp to be controlled is an identifier defined in your machine yaml file. A lamp can be enable()’d (turned on), disable()’d (turned off), or schedule()’d to flash with a specific pattern. Examples of all three are given below, assuming the lamps named target1, target2, and startButton are defined in the PRLamps: section of the machine yaml, and that someButton is a switch defined in the PRSwitches: section of the machine yaml:

def sw_someButton_active(self, sw):
# flash the start button ON/OFF, forever (until set otherwise):
self.game.lamps.startButton.schedule(schedule=0xff00ff00,
cycle_seconds=0, now=True)

# turn on the lamp at target1
self.game.lamps.target1.enable()

# turn off the lamp at target2
self.game.lamps.target2.disable()

## Displaying content on the HD Display

SkeletonGame provides additional display functionality on top of PyProcGame’s existing Layer metaphor.

A mode with something to show will set its self.layer to a specific instance of a Layer object (Note: Layer is actually the base-class, there are lots of specific sub-classes of Layer that provide additional functionality. The many Layer subclasses represent the various display types for content that might be shown on the DMD. Choose the right specific type of Layer based on what you want to display!)

Layers are like the Layer metaphor in Photoshop, etc. Layers are optional – if self.layer is not set, nothing is displayed by this Mode. The Z-Ordering of the Layers is dictated by the priority of the Modes highest toward the top of the display stack, lower priority mode’s appear below higher priority modes.

Because the Layer tools can be a little daunting at first, SkeletonGame provides a very powerful display helper method called displayText(), which will get text on the display without the programmer needing to worry about layers.

The complete usage of displayText will be documented in another entry, however the basic form is:

  self.game.displayText("Hello World")

which would display “Message” in the default font on the HD display for the player to see. To provide a multiline message, use the python list syntax ([,]) and each list entry will appear on its own line.

  self.game.displayText(["Hello", "World"])

To display an animation (or single frame, etc) on the display, provide a second argument to displayText, which matches a valid key in the animations section of the asset_list.yaml file. This example would display “Free” “Points!” on separate lines superimposed on a sequence of animated fire:

  def sw_target2_active(self, sw):
self.game.score(10)
self.game.displayText(["Free","Points!"], "flames")


text:
# Math

posted by Carson

The odds are 3 to 5 in favor of drawing a blue marble in one random drawing from an urn containing only red and blue marbles. There are more than 100 marbles in the urn. what is the minimum number of red marbles that there could be in the urn?

1. Reiny

prob of blue = 3/8

r/100 = 5/8 ---red = 62.5 or 63
b/100 = 3/8 --- blue = 37.5 or 38

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