text: Set Page Size to Letter or A4 (depending on the used paper) Click here for more extensive help. Doctors measure cancer in millimeters (1 mm = .04 inch) or centimeters (1 cm = .4 inch). the longer the more accurate. Cm | Inch . Get a ruler and look for an accurate measurement. such as free samples. This ruler measure in two different units of length, You have to check different areas to find out complete. What are the new image distances, the magnifications, and the height of the images. 1:50, 1:75, 1:100, 1:250, 1:500etc. Heller products 1:350: 0.871 mm: Though assumed to be a Japanese size for ship models, its origin lies in the UK, with the release of the 1:350 Javelin and Tribal Destroyer kit in December 1945 in the FROG Penguin range, 795 1 cm ruler actual size products are offered for sale by suppliers on Alibaba.com, of which rulers accounts for 1%. A great tool for educators, See the actual size of a One cm right monitor. 3. 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In order to accurately measure the length of your item, Letter sized paper should measure 27,94 centimeters or 11 inch vertically. so, 16 cm =1.6 mm, 1 cm = 1/100 m = 1,000,000/100 = 10,000micrometres. Most standard rulers are 30 centimeters long, which makes 1 centimeter equal to one-thirtieth of the length of a standard 12-inch, or 30-centimeter, ruler. The 150 mm one also has the 6 markings on the opposite side. Before measuring your pants, take a pair of jeans and place them flat on the. 1:120: 0.1″ 2.54 mm: TT model railroad scale. My laptop has a wide screen (13.6"x7.6"), and resolution is 1366x768 pixels, 1366 / 13.6 = 100.44 PPI. If you are using a cm (Centimeter) ruler, 1 cm = 10 mm. When the board is first rough sawn from the log, it is a true 2x4, but the drying process and planing of the board reduce it to the finished 1.5x3.5 size. Calibration required once! screen diagonal values in inches are therefore always smaller than the values for the same screen in centimeters A 1:10 scale is the typical size used for action figures, and is equal to 1 inch of action figure representing 10 inches of the actual model. The frame measurements are listed in mm (millimeters). ), or 14. The number 32 then corresponds to a leg length of 32 inches. This page is a darknet that can't be found on Google, Are you using {monitor} ? Inch squares - 9 in long, 1 in wide. 866-737-0754 BRilliANCE. Online Ruler (cm/mm) Label. Each device has it's own PPI, This is the actual size of an 1x1x1 cm cube. $magnification =\frac{measured~size}{actual~size}$ Question. but what i need is a more precise length so that i can apply it to my daily work. PDF PS source. 8. Its standard width is 3.37 inches (3 3⁄8 inches), or 85.6 mm (8 centimeters, 56 millimeters) for the metric measurement. The smaller (but still prominent) line between the 1/2 mark and the numbered inch line is 1/4 inch. Because the actual thickness may vary measuring the plywood is the only accurate way to determine its thickness. The distance from Albuquerque . Also, colour, bottle shape and design may visually obscure the true volume of the fragrance you are buying 1.The actual distance between two cities is 135 km the map scale is 1 cm:100 km. Below ruler adjuster help us to calibrate accuracy more easily. Sewing Actual size of the head is approximately 2.5 x 1 cm. Convert from scale to actual (real) size. 1. Templates. Save this settings. The small lines between the 1/2 mark and the numbered centimeter mark are 1/10 cm, otherwise known as a millimeter, keep in mind that there are 10 mm to 1 cm. 1 inch corresponds to 2.54 cm. To have the most accurate ruler online, just set the pixels per inch(PPI), Size 11.5: 10 13/16" or 27.1 cm. A4 sized … My penis has always been about 12.2 cm erect (11.9 cm girth) on a good day! Related Items. is the actual size of the human skin Variable Scale Ruler. Convert the mm to micrometres (x by 1000). , PPI : var oppi=document.getElementById('ppi'); Adjusting this virtual ruler to actual size . show 12 mm, show 2.3 cm, Check actual size / Online Ruler (cm/mm) Actual size of Online Ruler (cm/mm) 30CM / 300mm. History/origin: A centimeter is based on the SI unit meter, and as the prefix centi indicates, is equal to one hundredth of a meter Actual Size online ruler in CM Piliapp offers a free web-based ruler in (cm) , if you are looking for an online ruler in centimeters then this ruler is the best choice for you. A metric ruler is use to measure centimeter(cm) and millimeter(mm), we strongly recommend that you calibrate this online ruler first, You can also choose from no, yes 1 cm ruler actual size, as well as from straight ruler 1 cm ruler actual size, and whether 1 cm ruler. The small lines between the 1/2 mark and the numbered centimeter mark are 1/10 cm, otherwise known as a millimeter, keep in mind that there are 10 mm to 1 cm. Alibaba.com offers 1,600 actual size of 1 cm products. A 1 / 2 sheet is actually 15 / 32 thick. You will not regret - we promise! It can be examined on Paper 3, Part A. Apple iPhone 5 is 163 PPI, iPhone 7 is 162 PPI, iPhone X is 151.7 PPI. Use the scale factor 1cm=6.5m. Search "display by pixel density" online, check if there has your device's brand and model, 1 cm actual size Actual Size Online Ruler (mm,cm,inches) - Screen Measurement . hope you like our new command function , Online Ruler. This virtual ruler that can be adjusted to true size, that can actually measure the actual length, the upper half is metric ruler (millimetre and centimetre), lower half is inches ruler, before you use this ruler, please set the pixels per inch to your own device Adjusting this virtual ruler to actual size There are some ways to know the pixels per inch to your devie. Converting Metric Units to Centimeters Convert millimeters to centimeters. If the length of an object goes to one mark past the halfway mark on your ruler then it will be 9/16 inch on an English ruler or 6/10 cm (or 6 mm) on a metric ruler. Didot system (widely used in Europe), Pica system (USA and UK) and PostScript (modern DTP system widely used in digital printing and displaying). 1:144 means that the dimensions of the model are 1/144 (0.00694) the dimensions of the original life-sized object; this equates to a scale of 1/2 inch per 6 feet of original dimension.For instance, an airplane 30 feet (9.14 m) in length would be a mere 2.5 inches (63.5 mm) long as a 1:144 scale model. You first have to place the object, the exact pixels are calculated with the virtual ruler. A ruler, the most common device used to measure uses calibrated lines denominated into centimeters or inches, making it easy to compare between the two systems Kidney stone size in mm & cm - kidney stones are of different sizes, they range from small-sized crystals (small kidney stones) to large-sized stones (large kidney stones) such as the size of a golf ball. 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text:
# Specific Heat Formula

Specific Heat is used when we are referring to something particular. Heat capacity is the ratio of the quantity of heat required to alter the temperature by one degree Celsius. But when we consider a certain amount of mass we make use of the word Specific Heat or Specific Heat Capacity.

Specific Heat is the quantity of heat essential to raise the temperature of a gram of any substance by 1 degree Celsius.

Specific Heat formula is articulated as

$C\;=\;\frac{\Delta Q}{m\;\Delta T}$

Where,
Δ Q is the heat gained or lost,
Δ T is the temperature difference.

The temperature difference is given by Δ T = (Tf – Ti), where the final temperature is Tf  and the initial temperature is Ti .

Specific Heat formula is made use of to find the specific heat of any given material, its mass, heat gained or temperature difference if some of the variables are given. It is articulated in Joule/Kg Kelvin (J/Kg K).

Specific Heat Solved Examples

Underneath are some questions based on specific heat which will be useful for you.

Problem 1: Calculate the heat required to raise 0.6 Kg of sand from 30oC to 90oC? (Specific Heat of sand = 830 J/KgoC)

Known:

Mass of sand m = 0.6 Kg,
Δ T (Temperature difference)  = 90oC – 30oC = 60oC
C (Specific Heat of sand) = 830 J/KgoC

The specific heat is given by, $C\;=\;\frac{\Delta Q}{m\;\Delta T}$

Henceforth, Heat required is given by Q = C m Δ T
= 830 J/KgoC × 0.6 Kg × 60oC
= 29880 J.

Problem 2: Compute the temperature difference if 50 Kg of water absorbs 500 K J of heat?

Known:

m (Mass) = 50 Kg,
Q (Heat transfer) = 500 KJ,
C (Specific Heat of water) = 4.2 × 103 J/KgoC

The temperature difference is given by:

$\Delta T\;=\;\frac{Q}{C\;m}\;=\;\frac{500\times 10^{3}}{4.3\times 10^{3}\times 50}$

#### Practise This Question

Which of the following will not produce hydrogen gas?

format: plain


text:
# #Doing home work again

#Doing home work again

Title:

There are n assignments, and each assignment has its own deadline. When the deadline is exceeded and the assignment has not been completed, the corresponding score will be deducted. Ask how you can minimize points.

Solution 1:

n-door assignments are sorted by score from large to small, and then each assignment is arranged on the day closest to its deadline (first arranged on the deadline day, if there is already arrangement on that day, look for the day ahead), and this day is marked as used, if not, points will be deducted.

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
bool vis[1000];
#define fre freopen("C:\\Users\\Dell\\Desktop\\in.txt", "r", stdin);
struct node{
int sub;
bool operator < (const node &a)const{
return a.sub<sub;
}
}stu[1005];
int main(){
//fre;
int t, n, totsub;
cin >> t;
while (t--){
vis[0] = 1;
totsub = 0;
cin >> n;
for (int i = 0; i<n; i++)cin >> stu[i].dead;
for (int i = 0; i<n; i++)cin >> stu[i].sub, totsub += stu[i].sub;
sort(stu, stu + n);
for (int i = 0; i<n; i++){
//int k = 0;
for (int j = stu[i].dead; j >= 1; j--){
if (vis[j] == 0){ vis[j] = 1; totsub -= stu[i].sub; break; }
}
}
cout << totsub << endl;
memset(vis, 0,sizeof(vis) );
}
return 0;
}

Solution 2:

Sort the dates from small to large first. If the dates are the same, the ones with more points will be in the front. If there are more points deducted on the same date, use the previous time to do the work with less points deducted; if there are no smaller ones, deduct the points of the work. On, which is much better than the previous algorithm.

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<memory.h>
#include<queue>
#include <bits\stdc++.h>
using namespace std;
#define fre freopen("C:\\Users\\Dell\\Desktop\\in.txt", "r", stdin);
priority_queue<int, vector<int>, greater<int> >q;    //Small first out queue
struct in
{
}c[1010];
bool cmp(in a, in b){
return a.d<b.d;
}
int main()
{
fre;
int T, n, i, j, t, cnt, ans;
scanf("%d", &T);
while (T--){
cnt = ans = 0;
t = 1;
while (!q.empty()) q.pop();
scanf("%d", &n);
for (i = 1; i <= n; i++) scanf("%d", &c[i].d);
for (i = 1; i <= n; i++) scanf("%d", &c[i].s);
sort(c + 1, c + n + 1, cmp);
for (i = 1; i <= n; i++){
//Put it into the queue sorted from small to large, and when there is no time to do the job with large score later, discard the job with small score from the team head (the job with small score).
q.push(c[i].s);
//If the deadline is the same, i.e. there is more than one course to be handed in on a certain day, you must choose one of them to give up, and choose the one with the lowest cost, but it is not open
//Decide which door to do that day
if (c[i].d>=t)  t++;
else {
ans += q.top(); q.pop();
}
}
printf("%d\n", ans);
}
return 0;
}

Keywords: Python less

Added by melmoth on Sat, 02 Nov 2019 22:11:44 +0200


text:
Color the points of an $$n\times n$$ grid with two colors. How big a monochromatic grid-like subset can you find? By “grid-like” I mean that it should be possible to place equally many horizontal and vertical lines, partitioning the plane into $$k\times k$$ cells each of which contains a single point.

So for the coloring of the $$8\times 8$$ grid below, there are several $$4\times 4$$ monochromatic grid-like subsets. The image below shows one, and the completely red and blue southwest and northeast quadrants provide two others. The blue quadrant prevents any red grid-like subset from being larger than $$4\times 4$$, and vice versa, so these are the largest grid-like subsets in this grid.

It’s not hard to prove that there always exists a monochromatic grid-like subset of size at least $$\lfloor \sqrt{n} \rfloor\times \lfloor \sqrt{n} \rfloor$$. Just use vertical and horizontal lines to partition the big grid into blocks of that size. If one of those blocks is monochromatic, then it’s the grid-like subset you’re looking for. And if not, you can choose a red point from each block to get a grid-like subset of the same size.

In the other direction, there exist colorings of an $$n\times n$$ grid for which the largest monochromatic grid-like subset has size only a little bigger, $$O(\sqrt{n\log n})\times O(\sqrt{n\log n})$$. To find such a coloring, partition the big grid into square blocks of size $$O(\sqrt{n/\log n})\times O(\sqrt{n/\log n})$$, and make each block monochromatic with a randomly chosen color.

Now, consider any partition by axis-parallel lines into (irregular) rectangles, each containing one of the points of a grid-like subset. Only one row or column of the rectangles can cross each line of the partition into square blocks, so the number of rectangles that include parts of two or more square blocks is $$O(\sqrt{n\log n})\times O(\sqrt{n\log n})$$. Any remaining rectangles of the partition must come from a grid-like subset of square blocks that are all colored the same as each other. But with a random coloring, the expected size of this largest monochromatic subset of square blocks is $$O(\log n)\times O(\log n)$$. Therefore, the number of rectangles that stay within a single square block is limited to the total number of points in this grid-like subset of square blocks, which is again $$O(\sqrt{n\log n})\times O(\sqrt{n\log n})$$.

I’m not sure how to eliminate the remaining $$O(\sqrt{\log n})$$ gap between these two bounds, or which way it should go.

One application of these ideas involves the theory of superpatterns, permutations that contain as a pattern every smaller permutation up to some size $$n$$. If $$\pi$$ is a superpattern for the permutations of size $$n$$, then we can obtain a point set by interpreting the position and value of each element of $$\pi$$ as Cartesian coordinates. This point set includes a grid-like subset of size $$\lfloor \sqrt{n} \rfloor\times \lfloor \sqrt{n} \rfloor$$, coming from a permutation of size $$n$$ that translates to a grid-like set of points. If the elements of the superpattern are colored with two colors, there still exists a monochromatic grid-like subset of size $$O(n^{1/4})\times O(n^{1/4})$$. And this monochromatic grid-like subset corresponds to a superpattern, for permutations of size $$O(n^{1/4})$$. So, whenever the elements of a superpattern are colored with two (or finitely many) colors, there remains a monochromatic subset of elements that is still a superpattern for permutations of some smaller but non-constant size.


text:
# Betweenness in permutations

Let us consider permutations $\pi$ on $\{1,\dots,n\}$ as sequences $\pi(1),\pi(2),\dots,\pi(n)$. For a permutation $\pi$ let $R(\pi)$ be the ternary relation with $(x,y,z)\in R(\pi)$ whenever element $y$ appears in between $x$ and $z$ (i.e., x is left of y and z is right of y, or, x is right of y and z is left of y). So $(x,y,z)\in R(\pi)$ iff $(z,y,x)\in R$.

My question is: For permutations of length $n$, what is the smallest number $f(n)\ge 1$ such that there exist permutations $\pi_1,\dots,\pi_{f(n)}$ with $R(\pi_1)\cap R(\pi_2)\cap \dots \cap R(\pi_{f(n)})=\emptyset$? Are there exact results or bounds known describing this sequence?

Examples: For n=2, $f(n)=1$ since $R(\pi)=\emptyset$ for all permutations $\pi$ of length 2. For n>2, $f(n)$ has to be larger than 1 because $R(\pi)$ is only empty if $n\le2$.

For n=3, we can take the permutations $\pi_1=123$ and $\pi_2=132$. Here $R(\pi_1)=\{(1,2,3),(3,2,1)\}$ and $R(\pi_2)=\{(1,3,2),(2,3,1)\}$. Thus $R(\pi_1)\cap R(\pi_2)=\emptyset$ and $f(3)=2$.

For n=4, the permutations $\pi_1=1234$ and $\pi_2=2143$ show that $f(4)=2$.

For n=5, $f(n)\ge 3$. This is not hard to see but requires a few case distinctions.

With support of computer calculation we managed to establish that $f(n)=3$ for $n\in\{5,\dots,16\}$ and $f(n)>3$ for $n>16$.

Without loss of generality, we may assume that $\pi_1=12\dots n$.

By the Erdős–Szekeres lemma, $\pi_2$ has a monotone subsequence of length at least $\sqrt{n}$. In other words, there is a subset $S_2 \subseteq \{1,2,\dots,n\}$ of size at least $\sqrt{n}$ where $R(\pi_1)$ and $R(\pi_2)$ coincide. By induction, for any $k$ permutations, there is a subset $S_k \subseteq \{1,2,\dots,n\}$ of size at least $n^{1/2^{k-1}}$ that is monotone in each of the $k$ permutations, and so the relations $R(\pi_1), \dots, R(\pi_k)$ coincide on $S_k$. If $|S_k|\ge 3$, then $R(\pi_1) \cap \dots \cap R(\pi_k) \neq \emptyset$. To get $n^{1/2^{k-1}}\le 2$, we need $2^{2^{k-1}}\ge n$. This gives the lower bound $f(n)\ge 1+\log\log{n}$, where the logarithms are base $2$.

This paper by Kruskal claims that a general form of this result was proved by De Bruijn and that, in fact, the lower bound on $k$ (or, the upper bound on $n$ given $k$) is tight. A construction of $k$ permutations of length $2^{2^{k-1}}$ with no common monotone subsequence of length $3$ does not seem to be easy to describe. But for $k=3$ we can take $$\pi_2 = 13, 14, 15, 16, 9, 10, 11, 12, 5, 6, 7, 8, 1, 2, 3, 4 \\ \pi_3 = 11, 12, 9, 10, 15, 16, 13, 14, 3, 4, 1, 2, 7, 8, 5, 6.$$

A construction for arbitrary $k$ is described, for example, in this paper by Alon, Füredi and Katchalski or in this paper by Kalmanson.

• These are exactly the results and references I was hoping for! Thanks a lot! – Martin Lackner Jul 22 '16 at 9:27


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# Thermal conductivity | Definition | Conductivity Of Concrete

Thermal conductivity is the ability of the material to absorb heat. It can also be defined as the ratio between the heat flux and the temperature gradient. Unit of thermal conductivity is joules per second per square meter.

Conductivity of concrete depends on its composition. In case of saturated concrete, conductivity ranges from 1.4 to 3.6 joule/meter square per second. Generally, conductivity is not affected by the concrete density, but as the conductivity of air is very low therefore, conductivity of lightweight concrete varies with the variation in concrete density.

Conductivity also depends on the natural composition of aggregates like mineral properties of aggregates. Orientation of crystals also affects the conductivity. Conductivity increases with the increase in crystal properties of aggregates.

There is another important aspect that plays an important role is the degree of saturation of concrete. This is so because of the high conductivity of water than air. If the moisture content of lightweight concrete increases 10% then as a result conductivity increases about one-half.

In concrete, cement has the highest thermal conductivity. Conductivity of water is lower than half of the conductivity of cement paste. So, if water content will be lower in the fresh mix than there will be higher conductivity of hardened concrete.

Effect of temperature on thermal conductivity is very little at the room temperature. But, as the temperature rises from normal room temperature, the conductivity of phenomenon becomes more complex. Almost up to the 60 degree centigrade rise in conductivity is very slow with the loss of water. But, as the temperature reaches 120C conductivity decreases very sharply.


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# Optimize exports to Google Merchant Center best practices

Optimize exports to Google Merchant Center best practices in Productsup.

We have designed a number of boxes specifically for use when exporting to Google Merchant Center. These boxes can help you quickly get your feed to meet Google’s specifications.

### Tip

A few of the below boxes use regular expressions. If you need a refresher on how to use regex, see Extract data that uses regular expressions and split strings.

## Set a valid age group

Google accepts the following entries for the age group column:

• kid

• infant

• toddler

• newborn

You can add a static value for these entries using the set age group box.

1. Add the set age group box

2. Select the relevant age group from the drop-down menu

If you wish to have more flexibility, you can use the set age group by regex box.

1. Add the set age group by regex box

2. Define a regex for each possible age group entry

## Set a valid availability

Google accepts the following entries for the availability column:

• in stock

• out of stock

• preorder

• refurbished

You can add a static value for these entries using the set availability box.

1. Add the set availability box

2. Select the relevant availability from the drop-down menu

If you wish to have more flexibility, you can use the set availability using the stock box. This will set the stock value to in stock or out of stock based on whether the current stock is greater than 0.

1. Add the set availability by stock box

2. Select the column containing your stock information.

## Set a valid condition for your products

Google accepts the following entries for the condition column:

• new

• refurbished

• used

You can add a static value for these entries using the set condition box.

1. Add the set condition box

2. Select the relevant condition from the drop-down menu

If you wish to have more flexibility, you can use the set condition by regex box.

1. Add the set condition by regex box

2. Define a regex for each possible condition entry

## Set a valid expiration date

You can set an expiration date for your offer in Google using the expiration date column. The set expiration date box can help you achieve this.

1. Add the set expiration date box

2. Input after how many days the offer should expire; today’s date is included as 1 day in this number

## Set gender as accepted by Google

Google accepts the following entries for the gender column:

• unisex

• male

• female

You can add a static value for these entries using the set gender box.

1. Add the set gender box

2. Select the relevant gender from the drop-down menu

If you wish to have more flexibility, you can use the set gender by regex box.

1. Add the set gender by regex box

2. Define a regex for each possible gender entry

## Set sizing information

Google accepts the following entries for the size column:

• regular

• petite

• plus

• maternity

You can add a static value for these entries using the set size type box.

1. Add the set size type box

2. Select the relevant size type from the drop-down menu

If you wish to have more flexibility, you can use the set size type by regex box.

1. Add the set size type by regex box

2. Define a regex for each possible size type entry

You can also assign the shorthand for the size system (i.e. UK, US, DE) as dictated by Google. This is assigned to the column as a static value.

1. Add the set size system box

2. Select the relevant size system from the drop-down menu

If you have Google categories in your feed, you can choose to replace them with the category IDs or to translate them into another language by using the translate Google category box.

To create the mapping from your own internal product categories to Google’s taxonomy, consider using the taxonomy list feature.


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## protonated molecule

in mass spectrometry
https://doi.org/10.1351/goldbook.P04908
An ion formed by interaction of a molecule with a @P04906@ abstracted from an ion, as often occurs in chemical @I03183@ according to the reaction: $$\text{M}+\text{XH}^{+}\rightarrow \text{MH}^{+}+\text{X}$$ . The symbolism [M + H]+ may also be used to represent the protonated molecule. The widely used term 'protonated molecular ion' to describe the MH+ ion is not recommended, since it suggests an associated product of a @P04906@ with a molecular ion.
Source:
PAC, 1991, 63, 1541. (Recommendations for nomenclature and symbolism for mass spectroscopy (including an appendix of terms used in vacuum technology). (Recommendations 1991)) on page 1550 [Terms] [Paper]


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# [texhax] How to get current section etc number into page heading or footing?

Van Snyder van.snyder at jpl.nasa.gov
Tue Feb 27 05:57:27 CET 2007

I looked at the faq at http://www.tex.ac.uk/ and couldn't find an answer
to the question "How do I get the current chapter.section.subsection etc
number into a page heading or footing?"

By "current" I mean what one would get by putting a \label immediately
before the end of the page, assuming there's nothing in the way (such as
an enumerated list item) that would get labeled by something other than
a section etc. number.

I'm not using fancyheadings or anything like that.  Here are my

{\hff \hdate \hfill \hftitle \hfill \vers}%
{\hff \vers \hfill \hftitle \hfill \hdate}%
\def\@oddfoot%
{\hff \cpright \hfill \midfoot \hfill \thepage}%
\def\@evenfoot%
{\hff \thepage \hfill \midfoot \hfill \cpright}%
}

where I define \hff, \hftitle, \vers, \hdate, \cpright and \midfoot.
I'd like \midfoot to be the current chapter, chapter.section,
chapter.section.subsection... number.

My first thought was to use \renewcommand{\midfoot}{\@currentlabel}, but
that gets the label of an item in a list or indeed any of my special
counters if the footer falls in the middle of one of my special numbered
thing-o's, such as note boxes.

I tried to mimic how LaTeX sets \@currentlabel, but that required me to
know what was the current level of document subdivision.  So I tried
putting something into the appropriate @startsection command, but that
produced either nothing in the footer, or the index of the next document
division if that was the first thing on the next page, or the number of
the previous document division or..., probably because the stuff in #6
is processed *before* the appropriate counter is refstepcounter'd.


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# How can I view WebMathematica documentation in Mathematica documentation?

I want to view WebMathematica documentation in Mathematicadocumentation,for that I read the following link. I followed same things what they mention,still it's not working.

I have all permissions,but I tried the following two ways.

case 1:

I placed WebMathematicaDocumentation folder in FileNameJoin[{$InstallationDirectory, "AddOns", "Applications"}] directory case 2: I placed WebMathematicaDocumentation folder in FileNameJoin[{$UserBaseDirectory, "Applications"}] directory.

How can I resolve this ?

I found it why it's not working earlier,because If you unpack the webMathematicaDocumentation.zip you will get inner webMathematicaDocumentation folder. That's why it's not working. If you copy sub folder(webMathematicaDocumentation) and place it in correct directory,it will work.


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# Ngô Quốc Anh

## September 1, 2010

### The inverse of the Laplace transform by contour integration

Filed under: Giải tích 7 (MA4247) — Ngô Quốc Anh @ 17:23

Usually, we can find the inverse of the Laplace transform $\mathcal L[\cdot](s)$ by looking it up in a table. In this entry, we show an alternative method that inverts Laplace transforms through the powerful method of contour integration.

Consider the piece-wise differentiable function $f(x)$ that vanishes for $x < 0$. We can express the function $e^{-cx}f(x)$ by the complex Fourier representation of

$\displaystyle f(x){e^{ - cx}} = \frac{1}{{2\pi }}\int_{ - \infty }^\infty {{e^{i\omega x}}\left[ {\int_0^\infty {{e^{ - ct}}f(t){e^{ - i\omega t}}dt} } \right]d\omega }$

for any value of the real constant $c$, where the integral

$\displaystyle I = \int_0^\infty {{e^{ - ct}}|f(t)|dt}$

exists. By multiplying both sides of first equation by $e^{cx}$ and bringing it inside the first integral

$\displaystyle f(x) = \frac{1}{{2\pi }}\int_{ - \infty }^\infty {{e^{(c + i\omega )x}}\left[ {\int_0^\infty {f(t){e^{ - (c + i\omega )t}}dt} } \right]d\omega }$.

With the substitution $z = c+\omega i$, where $z$ is a new, complex variable of integration,

$\displaystyle f(x) = \frac{1}{{2\pi }}\int_{c - \infty i}^{c + \infty i} {{e^{zx}}\left[ {\int_0^\infty {f(t){e^{ - zt}}dt} } \right]d\omega }$.

The quantity inside the square brackets is the Laplace transform $\mathcal L[f](z)$. Therefore, we can express $f(t)$ in terms of its transform by the complex contour integral


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Show that checking if context free grammar generates infinite language is P-complete with logspace reduction

Show that following problem is P-complete in respect to logspace reduction in sense of Karp
Given: context free grammar $G$
Decide: Check if $G$ generates infinite language

Obviously this problem is in $P$ - it is sufficient to check if it contains cycle in derivation (and this derivation is non empty - contains some terminals). It is very easy to give polynomial algorithm.

When it comes to $P-completness$ I am hopeless. I have spent a hours to try reduce HORNSAT (which is $P$-complete) to problem in exercise.

Can you give me some clues ?

We reduce checking if grammar generates empty language to our problem.
Let $G$ will be grammar ($S$ is starting symbol):
$$S\to A|B|C$$ we transform it to $G'$
$$S\to SA|SB|SC|A|B|C$$
Now, let's assume that $S$ generates some word (in other words $L(G)\neq \emptyset$).

Then $G'$ generates inifnite language, because we can arbitrarily many times repeat rule $S\to SA$: $$S\to SA \to SSA \to SSSA \to SSSSA \to BCCBA$$

Let assume that $G'$ generates infinite language. It means that $G'$ generates some word (so $L(G')\neq \emptyset$). So also $G$ generate some word - in other case $G'$ can't generate any word ($G'$ only repeat many times word from $L(G)$).

• The first answer here may be of use. – Eric Towers Aug 15 '17 at 21:24
• @EricTowers I edited my question and followed your clue. Could you look on it ? – Haskell Fun Aug 16 '17 at 12:02
• Even if this solution is ok, it is hard to show that determining of emptiness of CFG is p complete – Haskell Fun Aug 16 '17 at 12:31
• Okay... How about a different sort of hint. Do you know Greibach normal form and a method to convert a CFG to GNF? This allows you to express your grammar as a (useful for your purpose) digraph. What would a cycle in this digraph tell you? – Eric Towers Aug 20 '17 at 0:54
• OK, I consider this hint later. However, what about my solution (after edition) ? – Haskell Fun Aug 20 '17 at 10:00

First, let's make sure we're all using the same definitions.

• CFG : context free grammar
• P-complete : the set of decision problems to which every member of P can be reduced. In this setting, we are using LOGSPACE reductions to embed arbitrary members of P into our problem.
• HORNSAT : The decision problem of determining whether a given a set of Horn clauses has a satisfying assignment of its boolean variables. HORNSAT is in P-complete.
• Horn clause : a disjunction with only one non-negated term. Equivalent to a conjunction implying the non-negated term and by contrapositive to the second implication shown below. \begin{align*} (\neg p \vee \neg q \vee \cdots \vee \neg t \vee u) &\iff (p \wedge q \wedge \cdots \wedge t \implies u) \\ &\iff (\neg u \implies \neg p \vee \neg q \vee \cdots \vee \neg t) \end{align*}

Next, we wish to show that the decision problem "given a CFG, $G$, decide whether the language generated by the grammar, $L(G)$, is infinite" is P-complete. you intend to do so by showing a LOGSPACE reduction of HORNSAT instances to instances of this decision problem.

So our question is: how do we encode Horn clauses as CFG production rules such that the resulting language of the CFG is infinite if and only if there is an assignment of values to the variables of the Horn clauses satisfying the clauses?

Since you've still pretty much asked for clues, ..., the last of the implications in the display above sure does look like a production rule...

• That's incredible, but before I invented this trick which you show above. However, only now I managed (I believe) to solve it. We convert HORNSAT clauses as you show. Now, $\phi \in HORNSAT$ is satisfable iff grammar generates infinite language. Obviously, we can think about clauses as: $u \implies p \vee q \vee \cdots \vee t$ - in other words negations dont matter. Then, we have productions introduced directly from formula and additionally each variable $t\to [true]$ where $[true]$ is terminal - one symbol. We should note that [false] don't exists (it make no sense). – Haskell Fun Aug 21 '17 at 18:27
• If in each path of grammar we got a word then this word if finite sequence of $[true]$. In case of infinite grammar we have some cycle (without cycle language is not infinite). It means that some variables are in conflict, then assigment doesn't exists. Am I correct ? – Haskell Fun Aug 21 '17 at 18:38
• @HaskellFun : That is exactly the right plan and I think you're looking at the right details to execute that plan. – Eric Towers Aug 21 '17 at 23:51
• Ok, after my work I work out lacking details. Could you hide under spoiler solution in nutshell ? To finish this thread. In case of my doubts I can look under spoiler, however I think that it can be only in order to make sure that I done exercise. – Haskell Fun Aug 22 '17 at 4:29
• I can't understand some thing: After all, we can assign all variables to $0$ and we have the same $1s$ (because negations) and then such hornat clause is satisfable in obvious way – Haskell Fun Aug 22 '17 at 17:23