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https://projecteuclid.org/euclid.die/1356019034 | 1,558,532,680,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256812.65/warc/CC-MAIN-20190522123236-20190522145236-00105.warc.gz | 610,719,816 | 8,947 | ## Differential and Integral Equations
### Positive solutions for $n\times n$ elliptic systems with combined nonlinear effects
#### Abstract
We study the existence and multiplicity of positive solutions to $n\times n$ systems of the form \begin{align*} -\Delta u_1&=\lambda f_1(u_2)& \mbox{ in }\Omega\\ -\Delta u_2&=\lambda f_2(u_3)&\mbox{ in }\Omega\\ \vdots \quad &= \quad\vdots&\\ -\Delta u_{n-1}&=\lambda f_{n-1}(u_n)&\mbox{ in }\Omega\ \ -\Delta u_n&=\lambda f_n(u_1)&\mbox{ in }\Omega\\ u_1&=u_2=...=u_n=0 &\mbox{ on }\partial\Omega. \end{align*} Here $\Delta$ is the Laplacian operator, $\lambda$ is a non-negative parameter, $\Omega$ is a bounded domain in $\mathbb R^N$ with smooth boundary $\partial\Omega$ and $f_i\in C^1([0,\infty)),$ $i\in\{1,2,\dots,n\},$ belongs to a class of strictly increasing functions that have a combined sublinear effect at $\infty$. We establish results for positone systems ($f_i(0)\geq0,$ $i\in\{1,\dots,l-1,l+1,\dots,n\}$ and $f_l(0)>0$ for some $l\in\{1,\dots,n\}$), semipositone systems (no sign conditions on $f_i(0)$) and for systems with $f_i(0)=0,$ $i\in\{1,2,\dots,n\}.$ We establish our results by the method of sub and supersolutions.
#### Article information
Source
Differential Integral Equations, Volume 24, Number 3/4 (2011), 307-324.
Dates
First available in Project Euclid: 20 December 2012
https://projecteuclid.org/euclid.die/1356019034
Mathematical Reviews number (MathSciNet)
MR2757462
Zentralblatt MATH identifier
1240.35152
Subjects
Primary: 35J55 35J70: Degenerate elliptic equations
#### Citation
Ali, Jaffar; Brown, Ken; Shivaji, R. Positive solutions for $n\times n$ elliptic systems with combined nonlinear effects. Differential Integral Equations 24 (2011), no. 3/4, 307--324. https://projecteuclid.org/euclid.die/1356019034 | 596 | 1,807 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-22 | latest | en | 0.643521 |
http://forums.codeguru.com/printthread.php?t=455660&pp=15&page=1 | 1,591,469,818,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348517506.81/warc/CC-MAIN-20200606155701-20200606185701-00496.warc.gz | 43,808,515 | 6,871 | # counting occurences of combinations
• June 20th, 2008, 03:40 PM
gary_88
counting occurences of combinations
for the following input file
11 22 33 44
11 22
im supposed to get
11,22 2
11,33 1
11,44 1
11,22,33 1
11,22,44 1
11,33,44 1
22,33,44 1
11,22,33,44 1
there is no repetition of elements on each line ..
i used following code to count "pairs" ie {11,22} and all....How do u do it for triplets and quadruplets and so on..??
--------------------------------
#define REP(i,n) for(int i=0;i<n;i++)
#define FOR(i,a,b) for(int i=a;i<b;i++)
ifstream fin("input.txt");
ofstream fout("output.txt");
string a;
map <pair<int,int>,int> b;
vector <int> x;
set <int> ss;
int m;
int main()
{
ifstream fin("input.txt");
while(getline(fin,a))
{
stringstream s;
s<<a;
x.clear();
while(s>>m) { x.push_back(m); ss.insert(m);}
REP(i,x.size())
FOR(j,i+1,x.size())
if(x[i]<x[j])
b[make_pair(x[i],x[j])]++;
else
b[make_pair(x[j],x[i])]++;
}
for(set <int>::iterator i=ss.begin();i!=ss.end();i++)
for(set <int>::iterator j=i;j!=ss.end();j++)
if(b[make_pair(*i,*j)])
fout<<"{"<<*i<<","<<*j<<"}"<<"\t"<<b[make_pair(*i,*j)]<<endl;
}
• June 20th, 2008, 08:02 PM
Duoas
Re: counting occurences of combinations
Don't use maps or pairs. Use a set.
Hope this helps.
• June 21st, 2008, 01:25 AM
gary_88
Re: counting occurences of combinations
But after inserting elements into the set how do u do the counting part??
Can u demonstrate the code?
• June 21st, 2008, 06:57 AM
Cheezewizz
Re: counting occurences of combinations
Would a multiset not be better? A set would just ignore any duplicates so his 11,22 2 would be become 11,22 1 ....
• June 21st, 2008, 11:28 AM
mysterd429
Re: counting occurences of combinations
So, mathematically, you have several sets, A[1] = {11, 22, 33, 44}, A[2] = {11, 22}, etc. You then take the union of all of these to get B = {11, 22, 33, 44}. You then want to count the number of times that each subset of B is a subset of an A. This is a naive implementation, using a set of sets to represent A. I'm sure it's a slow as possible, but I think it does what you want.
I omitted some utility code.
Code:
```/** * Build one-tuple sets, the double sets, ..., n-tuple sets. Terminates when * tupleSize is greater than the number of elements in bee. * @param bee The set from which elements are taken. * @param aye The set of sets being generated. * @param tupleSize The number of elements in each set for this iteration. * @pre bee has all of the elements. * @pre aye is empty. * @post aye is populated with all subsets of size tupleSize. */ void BuildTupleSets(set<int> bee, set<set<int> > &aye, int tupleSize = 1) { // cerr << "called with tupleSize = " << tupleSize << endl; // first iteration. if(tupleSize == 1) { for(set<int>::iterator i = bee.begin(); i != bee.end(); i++) { set<int> ayeSubN; ayeSubN.insert(*i); // cerr << "Adding " << ayeSubN << endl; aye.insert(ayeSubN); } BuildTupleSets(bee, aye, tupleSize + 1); } // iterations where we are building sets bigger than 1 else if(tupleSize <= bee.size()) { for(set<set<int> >::iterator i = aye.begin(); i != aye.end(); i++) { // cerr << "aye[i] is " << *i << endl; // If the current size is one less than the desired size, add to it. if(i->size() + 1 == tupleSize) { for(set<int>::iterator j = bee.begin(); j != bee.end(); j++) { set<int> ayeSubN(*i); ayeSubN.insert(*j); // cerr << "Adding " << ayeSubN << endl; aye.insert(ayeSubN); } } } BuildTupleSets(bee, aye, tupleSize + 1); } } /** * Checks to see if ess is a superset ot tee. * @param ess The potential superset. * @param tee The potential subset. * @return True if ess is a superset, false otherwise. */ bool IsSubset(const set<int> &ess, const set<int> &tee) { for(set<int>::iterator j = tee.begin(); j != tee.end(); j++) { set<int>::iterator found = ess.find(*j); if(found == ess.end()) { return false; } } return true; }```
I use the function like this:
Code:
```int main() { map<set<int>, int> results; // list<set<int> > results; set<int> bee; set<set<int> > aye; vector<set<int> > input; ifstream fin("data.txt"); int i; while(fin) { char lineBuffer[65535]; set<int> inputSubN; fin.getline(lineBuffer, 65535); stringstream lin(lineBuffer, stringstream::in); while(lin) { lin >> i; // cerr << "Adding " << i << " to bee." << endl; bee.insert(i); // cerr << "Adding " << i << " to inputSubN." << endl; inputSubN.insert(i); } input.push_back(inputSubN); } // cerr << "bee is " << bee << endl; BuildTupleSets(bee, aye); // cerr << "aye is " << aye << endl; for(vector<set<int> >::iterator super = input.begin(); super != input.end(); super++) { for(set<set<int> >::iterator sub = aye.begin(); sub != aye.end(); sub++) { if(IsSubset(*super, *sub)) { // cerr << *sub << " is a subset of " << *super << endl; results[*sub] += 1; } } } for(set<set<int> >::iterator sub = aye.begin(); sub != aye.end(); sub++) { cout << *sub << " : " << results[*sub] << endl; } return 0; }```
The output is then
Code:
```{11} : 2 {11, 22} : 2 {11, 22, 33} : 1 {11, 22, 33, 44} : 1 {11, 22, 44} : 1 {11, 33} : 1 {11, 33, 44} : 1 {11, 44} : 1 {22} : 3 {22, 33} : 1 {22, 33, 44} : 1 {22, 44} : 1 {33} : 1 {33, 44} : 1 {44} : 1```
Let me know if that does what you want.
Cheers!
• June 21st, 2008, 11:37 AM
TheCPUWizard
Re: counting occurences of combinations
A nice solution :thumb: :wave:
Performance has not been given as a requirement [ie number be ables to determine all of the subset counts for an original set of size n in under x seconds]
The only things I might do would be do create some trivial derived classes to make if a bit more readable....
Code:
```class SingleSet : public std::set<int> {} class : public std::set<SingleSet> {} void BuildTupleSets(SingleSet bee, MultiSet &aye, int tupleSize = 1) { // etc.... }```
• June 21st, 2008, 02:37 PM
gary_88
Re: counting occurences of combinations
Thanks a lot mysterd429.... Although there are some small problems..
1. In the main function..this code
for(set<set<int> >::iterator sub = aye.begin(); sub != aye.end(); sub++)
{
cout << *sub << " : " << results[*sub] << endl;
}
Here cout<<*sub gives error "no match for operator<<"....the expression results[*sub] gives the right answer like u said..
2. I noticed you used char lineBuffer[65535];.. Since this doesnt work for bigger files( 1or 2mb txt files)..i tried using string linebuffer although that didnt work..Wat is the change to be made here??
3. Just wanted to know if the answer could be printed in the format like
{11} 2
{22} 2
{11,22} 1
{11,22,33} 1
{11,22,33,44} 1
something like this where lower sets could be printed first followed by the remaining higher ones.
Thanks again!! Cheers!
• June 21st, 2008, 04:16 PM
mysterd429
Re: counting occurences of combinations
Quote:
Originally Posted by gary_88
Thanks a lot mysterd429.... Although there are some small problems..
1. In the main function..this code
for(set<set<int> >::iterator sub = aye.begin(); sub != aye.end(); sub++)
{
cout << *sub << " : " << results[*sub] << endl;
}
Here cout<<*sub gives error "no match for operator<<"....the expression results[*sub] gives the right answer like u said..
I omited some of the code I used for output purposes. I've pasted the whole source file below.
Quote:
Originally Posted by gary_88
2. I noticed you used char lineBuffer[65535];.. Since this doesnt work for bigger files( 1or 2mb txt files)..i tried using string linebuffer although that didnt work..Wat is the change to be made here??
As long as no one line exceeds that 65535 characters, you should be okay. Do you have more than 65535 characters per line? If lines are longer than 65535 characters, you could either increase the buffer size, or work out some other scheme.
Quote:
Originally Posted by gary_88
]3. Just wanted to know if the answer could be printed in the format like
{11} 2
{22} 2
{11,22} 1
{11,22,33} 1
{11,22,33,44} 1
something like this where lower sets could be printed first followed by the remaining higher ones.
Thanks again!! Cheers!
You could specify a different function for the third template parameter. I don't have a moment to create a good example right now, but this page might have your answer.
Cheers!
Don
Code:
```#include <set> #include <map> #include <fstream> #include <iostream> #include <string> #include <vector> #include <sstream> #include <list> using namespace std; ostream & operator<<(ostream & lhs, set<int> rhs) { set<int>::iterator i = rhs.begin(); lhs << "{"; if(i != rhs.end()) { lhs << *i; i++; for(; i != rhs.end(); i++) { lhs << ", " << *i; } } lhs << "}"; return lhs; } ostream & operator<<(ostream & lhs, set<set<int> > rhs) { set<set<int> >::iterator i = rhs.begin(); lhs << "{"; if(i != rhs.end()) { lhs << *i; i++; for(; i != rhs.end(); i++) { lhs << ", " << *i; } } lhs << "}"; return lhs; } /** * Build one-tuple sets, the double sets, ..., n-tuple sets. Terminates when * tupleSize is greater than the number of elements in bee. * @param bee The set from which elements are taken. * @param aye The set of sets being generated. * @param tupleSize The number of elements in each set for this iteration. * @pre bee has all of the elements. * @pre aye is empty. * @post aye is populated with all subsets of size tupleSize. */ void BuildTupleSets(set<int> bee, set<set<int> > &aye, int tupleSize = 1) { // cerr << "called with tupleSize = " << tupleSize << endl; // first iteration. if(tupleSize == 1) { for(set<int>::iterator i = bee.begin(); i != bee.end(); i++) { set<int> ayeSubN; ayeSubN.insert(*i); // cerr << "Adding " << ayeSubN << endl; aye.insert(ayeSubN); } BuildTupleSets(bee, aye, tupleSize + 1); } // iterations where we are building sets bigger than 1 else if(tupleSize <= bee.size()) { for(set<set<int> >::iterator i = aye.begin(); i != aye.end(); i++) { // cerr << "aye[i] is " << *i << endl; // If the current size is one less than the desired size, add to it. if(i->size() + 1 == tupleSize) { for(set<int>::iterator j = bee.begin(); j != bee.end(); j++) { set<int> ayeSubN(*i); ayeSubN.insert(*j); // cerr << "Adding " << ayeSubN << endl; aye.insert(ayeSubN); } } } BuildTupleSets(bee, aye, tupleSize + 1); } } /** * Checks to see if ess is a superset ot tee. * @param ess The potential superset. * @param tee The potential subset. * @return True if ess is a superset, false otherwise. */ bool IsSubset(const set<int> &ess, const set<int> &tee) { for(set<int>::iterator j = tee.begin(); j != tee.end(); j++) { set<int>::iterator found = ess.find(*j); if(found == ess.end()) { return false; } } return true; } int main() { map<set<int>, int> results; // list<set<int> > results; set<int> bee; set<set<int> > aye; vector<set<int> > input; ifstream fin("data.txt"); int i; while(fin) { char lineBuffer[65535]; set<int> inputSubN; fin.getline(lineBuffer, 65535); stringstream lin(lineBuffer, stringstream::in); while(lin) { lin >> i; // cerr << "Adding " << i << " to bee." << endl; bee.insert(i); // cerr << "Adding " << i << " to inputSubN." << endl; inputSubN.insert(i); } input.push_back(inputSubN); } // cerr << "bee is " << bee << endl; BuildTupleSets(bee, aye); // cerr << "aye is " << aye << endl; for(vector<set<int> >::iterator super = input.begin(); super != input.end(); super++) { for(set<set<int> >::iterator sub = aye.begin(); sub != aye.end(); sub++) { if(IsSubset(*super, *sub)) { // cerr << *sub << " is a subset of " << *super << endl; results[*sub] += 1; } } } for(set<set<int> >::iterator sub = aye.begin(); sub != aye.end(); sub++) { cout << *sub << " : " << results[*sub] << endl; } return 0; }```
• June 21st, 2008, 04:58 PM
Duoas
Re: counting occurences of combinations
I'm confused. Are you counting permutations or combinations? They are different things...
• June 21st, 2008, 05:42 PM
gary_88
Re: counting occurences of combinations
Thanks..code was able to compile but still it isnt working for big files even though a single line is not more than 65535 characters
30 31 32
33 34 35
36 37 38 39 40 41 42 43 44 45 46
38 39 47 48
38 39 48 49 50 51 52 53 54 55 56 57 58
32 41 59 60 61 62
3 39 48
63 64 65 66 67 68
32 69
48 70 71 72
119 120 121 122 123 124 125 126 127 128 129 130 131 132 133
this a small example of how my input file is...the o/p is not coming for this..
• June 22nd, 2008, 10:44 AM
mysterd429
Re: counting occurences of combinations
Hi Gary,
There was a small bug in my input code. I was not correctly reading in from lin before I checked to see if it had failed. That section of code should look like the one included below.
The new code may or may not fix the problem that you were seeing with large files. Try making the changing and seeing if it still fails. When the program fails for large files, what happens? What error do you get? Can you use an interactive debugger to find the exact line of code where it fails?
Cheers!
P.S. - Thanks for the feedback, CPUWizard.
Code:
``` ifstream fin("data.txt"); while(fin) { char lineBuffer[65535]; set<int> inputSubN; fin.getline(lineBuffer, 65535); int i; // cerr << "Read \"" << lineBuffer << "\"." << endl; stringstream lin(lineBuffer, stringstream::in); lin >> i; while(lin) { // cerr << "Adding " << i << " to bee." << endl; bee.insert(i); // cerr << "Adding " << i << " to inputSubN." << endl; inputSubN.insert(i); lin >> i; } input.push_back(inputSubN); }```
edit: If you have 14 possible values, (using 11, 22, 33, 44 you have 4), then the total number of subsets is
2 * (14! / (1! * 13!) + 14! / (2! * 12!) + 14! / (3! * 11!) + 14! / (4! * 10!) + 14! / (5! * 9!) + 14! / (6! * 8!)) + 14! / (7! * 7!)
That's a relatively large number, 16383 combinations. If you have a 2 MiB file, that's something like 200,000 lines. That gets you 1.683e4 * 2e5 iterations of comparing set to set, which is about 3.2e9, or 3,200,000,000 set-to-set comparisons. That may take a very long time, depending on your computer. My dual processor G4 (at 1.0 GHz) has been chugging away for an hour or so. Like I said, this ain't an efficient algorithm ;-)
• June 22nd, 2008, 12:01 PM
gary_88
Re: counting occurences of combinations
hi don
Firstly i just checked up on the following input
11 22 33 44 55
66 77 88
it gave me 8 element set long answer.. I'm really sorry but that isnt wat i wanted as o/p.. Perhaps tthis wud be a bette example..
i/p
I1 I2 I5
I2 I4
I2 I3
I1 I2 I4
I1 I3
I2 I3
I1 I3
I1 I2 I3 I5
I1 I2 I3
-----------------------------------------------
o/p
6 I1
7 I2
2 I5
2 I4
6 I3
I1,I2 4
I1,I5 2
I1,I4 1
I1,I3 4
I2,I5 2
I2,I4 2
I2,I3 4
I5,I4 0
I5,I3 1
I4,I3 0
{I1,I2,I5} 2
{I1,I2,I4} 1
{I1,I2,I3} 2
{I1,I5,I4} 0
{I1,I5,I3} 1
{I1,I4,I3} 0
{I2,I5,I4} 0
{I2,I5,I3} 1
{I2,I4,I3} 0
{I5,I4,I3} 0
{I1,I2,I5,I4} 0
{I1,I2,I5,I3} 1
{I1,I2,I4,I3} 0
{I2,I5,I4,I3} 0
im really sorry for troubling you..
thanks
Cheers!
• June 24th, 2008, 02:00 AM
gary_88
Re: counting occurences of combinations | 5,138 | 16,227 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-24 | latest | en | 0.740087 |
https://www.sarthaks.com/109617/a-b-and-c-are-partners-sharing-profit-in-the-ratio-of-5-3-2-c-retires-and-his-share-is-entirely | 1,621,121,001,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991488.53/warc/CC-MAIN-20210515223209-20210516013209-00215.warc.gz | 986,146,589 | 13,515 | # A, B and C are partners sharing profit in the ratio of 5:3:2 C retires and his share is entirely
233 views
in Accounts
A, B and C are partners sharing profit in the ratio of 5:3:2 C retires and his share is entirely taken by A. calculate new profitsharing ratio of A and B.
by (106k points)
selected by
New Ratio = Old Ratio + share acquired from C | 96 | 354 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2021-21 | latest | en | 0.985083 |
http://freetofindtruth.blogspot.com/2016/08/6-13-31-52-91-allyson-felix-wins-here.html | 1,529,410,218,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267862929.10/warc/CC-MAIN-20180619115101-20180619135101-00214.warc.gz | 127,526,396 | 22,199 | ## Saturday, August 20, 2016
### 6 13 31 52 91 | Allyson Felix wins her record 6th gold medal, August 20, 2016
'Six' connects to 'prophecy', the ongoing theme for the U.S. team that sent '106' returning medalists.
The date of her record win connects with her name.
8/20/2016 = 8+20+20+16 = 64 (Allyson Felix)
Below you'll see that Allyson has a life lesson number of '52' as well.
Allyson = 1+3+3+7+1+6+5 = 26/34
Michelle = 4+9+3+8+5+3+3+5 = 40
Felix = 6+5+3+9+6 = 29
Allyson Michelle Felix = 95/104 (Allyson Felix = 55/64)
Allyson = 1+12+12+25+19+15+14 = 98
Michelle = 13+9+3+8+5+12+12+5 = 67
Felix = 6+5+12+9+24 = 56
Allyson Michelle Felix = 221 (The Bavarian Illuminati = 221)
Allyson Felix = 154
I notice write away she was born in '85, like the 8/5 start to the Rio Games, and that she was born on the 322nd day of the year, November 18.
11/18/1985 = 11+18+19+85 = 133 (Government)
11/18/1985 = 11+18+(1+9+8+5) = 52 (Prophecy) (Six)
11/18/1985 = 1+1+1+8+1+9+8+5 = 34
11/18/85 = 11+18+85 = 114
This big win comes in a span of 13-weeks or 91-days before her 31st birthday. Again, this is the 31st Summer Olympics of the modern era. Also, if you sum the numbers 1-13, it totals 91; 13 the reflection of 31.
Below are the other big headlines from today. Notice the U.S. Women's basketball team won gold for the 6th time as well. | 531 | 1,343 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2018-26 | latest | en | 0.907988 |
https://mathoverflow.net/questions/303676/counting-partitions-avoiding-some-blocks | 1,660,786,490,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573145.32/warc/CC-MAIN-20220818003501-20220818033501-00657.warc.gz | 357,448,655 | 25,593 | # Counting partitions avoiding some blocks
I am interested to find the number of non-crossing pair partitions on the set $\{1,2,...,2k\}$ such that none of these pair partitions has a pair of the form $(2r-1,2r)$ and there are exactly $t$ pairs of the form $(2i-1,2j)$ where $2i-1<2j$.
Let's go through this slowly. I think we understand what non-crossing pair partitions are, but briefly, a partition whose any block is a pair and if $i<j<k<l$ then $(i,k)$ and $(j,l)$ cannot be blocks, because if we join the elements in the same block by lines, we see that $i,k$ will be joined by a line, and $j,l$ will be joined by a line, which intersect. It is known that the number of non-crossing pair partitions on a set of size $2k$ is equal to $C_k$, the $k$-th Catalan number defined by $C_k=\dfrac{1}{k+1}{2k\choose k}$.
Clearly, since I am talking about pair partitions, my set has to be of even cardinality, hence the $2k$.
I am demanding that my non-crossing pair partitions will not have a pair of the form $(2r-1,2r)$ i.e. $(1,2),(3,4),...,(2k-1,2k)$ cannot be blocks in my partition. I can count the number of such non-crossing pair partitions, because it is easy to count the number of non-crossing pair partitions having at least one of these pairs as a block. That's because, say, we fix $(1,2)$ to be a block, then we are really counting the number of non-crossing pair partitions on a set of size $2k-2$ So this part is fine and can be tackled.
I am also demanding that my non-crossing pair partitions will have exactly $t$ pairs of the form $(2i-1,2j)$ where $2i-1$ is to the left of $2j$, that is, a line joining $2i-1$ and $2j$ begins at $2i-1$. It is well known that the number of such non-crossing pair partitions is equal to a Narayana number.
But can we compute the number of non-crossing pair partitions such that BOTH the above happen? Any help would be appreciated. All search results were giving me how to count 132 avoiding permutations, which I believe is unrelated to this problem.
• It would be helpful for you to list the partitions involved for small n. Even though you make it explicit otherwise, my mind says "avoid (2,3),(4,5)...". Maybe you can embed the list for 2n into the list for (2n+2)? Gerhard "Sometimes Likes It Very Slowly" Paseman, 2018.06.26 Jun 26, 2018 at 15:49
• Would a bijection to permutation tableaux yield anything interesting? Or, perhaps, tree-like tableaux would work even better. Jun 26, 2018 at 20:53
• Don't you necessarily have that all pairs are of the form $(2i-1,2j)$ for some $i \leq j$? Or do you allow some points to be not connected to others?
– Dirk
Jun 28, 2018 at 8:19
• Can you give an example of a non-crossing pair partition that does “not have a pair of the form $(2r-1,2r)$? I may misunderstand the definitions, but it seems like none exist. If $(1,k)$ is a pair, $k>2$, and then the partition induces a non-crossing pair partition of $\{2,\dots,k-1\}$, where $k-1>3$, which induces one of $\{3,\dots,k-2\}$, where $k-2>4$, etc. Jun 29, 2018 at 23:42
So, the question is how many non-crossing partitions have exactly $t$ pairs of the form $(2i-1,2j)$, $2i-1 < 2j$ (i.e., $i\leq j$), none of which have $i=j$. By inclusion-exclusion, the number of such partitions equals $$\sum_{i=0}^t (-1)^i \binom{k}{i} N(k-i,t-i) = \frac{1}{k+1}\binom{k+1}{t}\binom{k-t-1}{k-2t},$$ where $N(,)$ are Narayana numbers. | 1,020 | 3,381 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2022-33 | latest | en | 0.932228 |
http://mathhelpforum.com/advanced-applied-math/214027-linear-algebra-proof-regarding-linear-transformations-print.html | 1,508,221,064,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187820927.48/warc/CC-MAIN-20171017052945-20171017072945-00259.warc.gz | 269,470,201 | 3,080 | # Linear Algebra Proof Regarding Linear Transformations
• Feb 28th 2013, 11:51 PM
zachoon
Linear Algebra Proof Regarding Linear Transformations
Given the following vectors in R^2:
a(sub 1) = (1; -1), a(sub 2) = (2; -1), a(sub 3) = (-3;2),
b(sub 1) = (1;0), b(sub 2) = (0;1), b(sub 3) = (1;1),
determine whether there is a linear transformation T from R^2 into R^2 such that Ta(sub i) = b(sub i) for i = 1, 2 and 3.
Basically, I don't quite understand the process of figuring out linear transformations, so I'd like at least a few helpful pointers in figuring this out. Thank you.
• Mar 1st 2013, 12:53 AM
jakncoke
Re: Linear Algebra Proof Regarding Linear Transformations
so assume we have a linear transformation. L which takes $a_i \to b_i$
$A = \begin{bmatrix} 1 & 2 & -3 \\ -1 & -1 & 2 \end{bmatrix}$
$B = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix}$
For the matrix A, you can write the 3rd column as the linear combination of the first and second columns.
$A = \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix} \begin{bmatrix} -1 \\ -1 \end{bmatrix} = \begin{bmatrix} -3 \\ 2 \end{bmatrix}$
since a linear transformation has the property L( $c_1 v_1 + c_2 v_2$)= $c_1 L(v_1) + c_2 L(v_2)$ then
if the first column of A went to the first column of B, and the second column of A went to the second column of B then
$L(-1 * \begin{bmatrix} 1 \\ -1 \end{bmatrix} + -1 * \begin{bmatrix} 2 \\ -1 \end{bmatrix}) = -1*L(\begin{bmatrix} 1 \\ -1 \end{bmatrix}) + -1 * L(\begin{bmatrix} 2 \\ -1 \end{bmatrix}) = -1 * \begin{bmatrix} 1 \\ 0 \end{bmatrix} + -1 * \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} -1 \\ -1 \end{bmatrix}$
since this is not equal to the 3rd column of B, there is no Linear transformation which takes the first column of A to the first column of B, second .... | 643 | 1,796 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2017-43 | longest | en | 0.689805 |
https://libraryguides.centennialcollege.ca/c.php?g=717032&p=5175936&preview=ea2be4e065345451b2f08e19a906fe04 | 1,723,549,868,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641076695.81/warc/CC-MAIN-20240813110333-20240813140333-00328.warc.gz | 289,472,042 | 8,304 | # Calculus
## Derivatives of Inverse Sine and Cosine Functions
Let's consider the inverse trigonometric function $$y=\sin ^{-1}u$$. We can isolate for $$u$$ and get $$\sin y=u$$ for the values $$-\frac{\pi}{2}\leq y \leq \frac{\pi}{2}$$ for it to qualify as a function.
Now let's take the derivative of $$\sin y=u$$ in respect to $$x$$.
$\frac{du}{dx}=\cos y\frac{dy}{dx}$
Solving for $$\frac{dy}{dx}$$ we get,
$\frac{dy}{dx}=\frac{1}{\cos y}\frac{du}{dx}$
Now from the identify $$\cos ^2y+\sin ^2y=1$$, isolating from $$\cos y$$, we get $$\cos y=\sqrt{1-\sin ^2y}$$. Thus, the above equation becomes
$\frac{dy}{dx}=\frac{1}{\sqrt{1-\sin ^2y}}\frac{du}{dx}$
From our original equation $$u=\sin y$$, we now have an equation with $$u$$
$\frac{dy}{dx}=\frac{1}{\sqrt{1-u^2}}\frac{du}{dx}$
Also, $$y=\sin ^{-1}u$$. Thus, the derivative of the inverse sine function is
$\frac{d(\sin ^{-1}u)}{dx}=\frac{1}{\sqrt{1-u^2}}\frac{du}{dx}$
Through a similar process, you can show that for $$0\leq y\leq \pi$$
$\frac{d(\cos ^{-1}u)}{dx}=-\frac{1}{\sqrt{1-u^2}}\frac{du}{dx}$
If y=\tan ^{-1} u, then \begin{align} u&=\tan y \\ \frac{du}{dx} &= \sec ^2y\frac{dy}{dx} \\ \frac{dy}{dx}&=\frac{1}{ \sec ^2y}\frac{du}{dx} \\ \frac{dy}{dx}&=\frac{1}{1+ \tan ^2y}\frac{du}{dx} \end{align} Thus, the derivative is $\frac{d(\tan ^{-1}u)}{dx}=\frac{1}{1+ u^2}\frac{du}{dx}$ ## Examples Example 1: Find the derivative of $V=8\tan ^{-1}\sqrt{s}$ Solution: \begin{align} V'&=8\frac{1}{1+(\sqrt{s})^2}\left(\frac{1}{2}s^{-\frac{1}{2}}\right) \\ &=4\frac{1}{(1+s)\sqrt{s}} \\&=\frac{4\sqrt{s}}{s(1+s)}\end{align} Example 2: Find the derivative of $p=\frac{3}{\cos ^{-1}2w}$ Solution: We are going to derive the equation in the form \(p=3\left(\cos ^{-1}2w\right)^{-1} and use the chain rule
\begin{align} p'&= -3\left(\cos ^{-1}2w\right)^{-2}\left(-\frac{1}{\sqrt{1-(2w)^2}}\right)(2) \\ &= \frac{6}{\left(\cos ^{-1}2w\right)^2\left(\sqrt{1-(2w)^2} \right)}\end{align}
Example 3: Find the second derivative of $y=\tan ^{-1}2x$
Solution: The first derivative is
$y'=\frac{2}{1+2x}$
Now find another derivative
$y''=-4\left(1+2x\right)^{-2}$ | 875 | 2,132 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 3, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2024-33 | latest | en | 0.514898 |
https://testbook.com/question-answer/the-area-of-jet-and-velocity-of-jet-are-0-02-m2-an--602611aa1968cf0fb6f8daf0 | 1,632,695,948,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057973.90/warc/CC-MAIN-20210926205414-20210926235414-00591.warc.gz | 586,485,372 | 31,165 | # The area of jet and velocity of jet are 0.02 m2 and 75 m/s, respectively and the total discharge through a pelton turbine is 3 m3/s, What are the number of jets required ?
This question was previously asked in
SSC JE CE Previous Year Paper 10 (Held on 30 Oct 2020 Evening)
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1. 1
2. 4
3. 3
4. 2
Option 4 : 2
Free
ST 1: Building Material and Concrete Technology
17154
20 Questions 20 Marks 12 Mins
## Detailed Solution
Explanation:
Given;
Total discharge through pelton turbine = 3 m3/s
Area of jet = 0.02 m2
Velocity of jet = 75 m/s
Calculation:
Discharge through jet = 0.02 × 75 = 1.5 m3/s
$$Number\;of\;jet = \frac{{Total\;discharge}}{{Discharge\;through\;on\;jet}}$$ = $$\frac{3}{{1.5}}$$ = 2 | 251 | 734 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2021-39 | latest | en | 0.768258 |
https://uk.style.yahoo.com/d-drip-feed-417-month-114443918.html | 1,680,075,417,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948951.4/warc/CC-MAIN-20230329054547-20230329084547-00697.warc.gz | 678,826,505 | 128,591 | # I’d drip-feed £417 a month into a Stocks and Shares ISA to try for a million
As of 2021, there were over 2,000 ISA millionaires in the UK. So all those people have built up £1,000,000 via an account that has a maximum deposit of £20,0000 per year.
If it was just down to saving that cash, it means a 20-year-old would need to deposit the full £20,000 every single year until they turned 70. Seems utterly impossible, at first glance.
But if I work out the maths, I could get all the way to the million pound mark with only £5,000 a year (£417 a month) and in much less time than 50 years. But it would only be possible if I use one particular type of ISA.
## What type of ISA?
I’d put good money on most of those 2,000 ISA millionaires using a Stocks and Shares ISA to build their wealth. The reason is that this type of ISA offers the highest potential with 8-10% being typical annual stock market returns over long periods of time in the past. Investing in stocks does come with risks however, and no returns are guaranteed.
Another popular type of ISA is the Cash ISA. This one is very safe but with low returns that are linked to interest rates, which are 4% at present. Let’s run the numbers and see how they compare.
## Drip-feeding £417
Let’s say I’m starting from scratch and want to hit a million in my Stocks and Shares ISA. I’m going to aim to save £5,000 a year by drip-feeding £417 a month.
What kind of percentage return could I expect? Well, the FTSE 100 – the collection of the 100 largest companies listed on the London Stock Exchange – has historical annual returns of around 8%. The FTSE 250 – the 101st to 350th largest companies – has returns of around 10%. Let’s assume a 9% annual return.
Cash ISA (3%) Stocks and Shares ISA (9%) 5 years £27,597 £31,164 10 years £61,172 £79,113 20 years £151,722 £266,401 30 years £285,758 £709,781 34 years £355,932 £1,025,727
It’s clear from the table just how much investing in stocks can improve returns, especially over the long term as compound interest works its magic. And it looks like I’d potentially join the ISA millionaire club after 34 years. Now, what if I wanted to reduce that amount of time?
## How to increase returns
One way to improve returns is to invest in individual stocks. For example, if I invested in Amazon in 2005 I’d have netted 5,000% returns up to now – a 50 times return on my stake. Compare this to the FTSE 100, which is up 60% in the same timeframe. That could get me to a million a lot quicker.
Of course, companies can go bankrupt or have a lean period. I think the best philosophy is a balance of carefully chosen stocks while remaining diversified across different companies and sectors.
## Try for a million?
In reality, I’m not fixated on getting to a million just for the sake of it. But I do think this is an illuminating exercise to show why investing in stocks and shares is a powerful way to build long-term wealth.
And with the tax advantages of a Stocks and Shares ISA? I’m very happy to drip-feed regular savings to do that.
Please note that tax treatment depends on the individual circumstances of each client and may be subject to change in future. The content in this article is provided for information purposes only. It is not intended to be, neither does it constitute, any form of tax advice. Readers are responsible for carrying out their own due diligence and for obtaining professional advice before making any investment decisions.
The post I’d drip-feed £417 a month into a Stocks and Shares ISA to try for a million appeared first on The Motley Fool UK. | 855 | 3,597 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2023-14 | latest | en | 0.957332 |
http://assessment.aaas.org/topics/0/EG/373 | 1,563,237,485,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524290.60/warc/CC-MAIN-20190715235156-20190716021156-00480.warc.gz | 12,539,049 | 6,720 | Key Idea: Transformations and transfers of energy within a system usually result in some energy being released into its surrounding environment causing an increase in the thermal energy of the environment.
Students should know that:
1. When objects interact with each other or with the surrounding environment, some amount of energy is transformed into thermal energy that is transferred to the surrounding environment.
2. Because thermal energy always results from these interactions, it is impossible to convert 100% of one form of energy into another form of energy.
3. Processes that involve an interaction between objects or between an object and the surrounding environment will eventually stop unless additional energy is added to keep them running because the amount of energy available to keep the process running decreases as energy is transferred to the surrounding environment.
4. Some interactions between objects or between objects and the surrounding environment transfer more energy to their environment than others. For example, increasing the amount of friction, including air resistance, increases the amount of energy transfer to the surrounding environment.
5. There are ways to reduce the amount of energy transferred to the environment, for example, by using insulation to reduce the amount of energy transferred by conduction or using reflective materials to reduce the amount of energy transferred by radiation.
Boundaries:
1. Items do not ask students to make any calculations about how much energy is transferred to the surrounding environment. For example, students are not asked to use equations like KE = 1/2mv2 of PE = mgh.
Percent of students answering correctly (click on the item ID number to view the item and additional data)
Item ID
Number
6–8
9–12
Select This Item for My Item Bank
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?>
Frequency of selecting a misconception
Misconception
ID Number
Student Misconception
6–8
9–12
RGM010
Energy can be changed completely from one form to another (no energy losses) (Hapkiewicz, 1992).
See the ASPECt Project
RGM006
Energy only degrades when it is not conserved (Pinto, et al., 2005). In other words, energy does not degrade in systems where energy cannot enter or leave.
See the ASPECt Project
NGM060
Energy can be destroyed (Kruger, 1990; Trumper, 1998).
See the ASPECt Project
EGM048
Living things give inanimate objects energy by carrying or pushing them. For example, a person gives a bike energy by riding it or a bird give a stick energy by carrying it (Herrmann-Abell & DeBoer, 2010).
See the ASPECt Project
NGM040
The total amount of energy an object has cannot change (AAAS Project 2061, n.d.).
See the ASPECt Project
NGM037
An object always gains energy as it moves. For example, the height that a pendulum reaches after it is released is greater than its starting height because it gains energy as it swings (Loverude, 2004).
See the ASPECt Project
NGM010
Energy can be created (Kruger, 1990; Lovrude, 2004; Papadouris et al., 2008).
See the ASPECt Project
NGM009
An object has energy within it that is used up as the object moves (Brook & Driver, 1984; Kesidou & Duit, 1993; Loverude, 2004; Stead, 1980).
See the ASPECt Project
Frequency of selecting a misconception was calculated by dividing the total number of times a misconception was chosen by the number of times it could have been chosen, averaged over the number of students answering the questions within this particular idea.
NGSS Statements
Code
Statement
Energy is present whenever there are moving objects, sound, light, or heat. When objects collide, energy can be transferred from one object to another, thereby changing their motion. In such collisions, some energy is typically also transferred to the surrounding air; as a result, the air gets heated and sound is produced.
PS3.D HS
Although energy cannot be destroyed, it can be converted to less useful forms--for example, to thermal energy in the surrounding environment.
PS3.B MS
When the motion energy of an object changes, there is inevitably some other change in energy at the same time.
Energy is present whenever there are moving objects, sound, light, or heat. When objects collide, energy can be transferred from one object to another, thereby changing their motion. In such collisions, some energy is typically also transferred to the surrounding air; as a result, the air gets heated and sound is produced.
PS3.D HS
Although energy cannot be destroyed, it can be converted to less useful forms--for example, to thermal energy in the surrounding environment.
PS3.B HS
Uncontrolled systems always evolve toward more stable states--that is, toward more uniform energy distribution (e.g., water flows downhill, objects hotter than their surrounding environment cool down).
PS3.B HS
Uncontrolled systems always evolve toward more stable states--that is, toward more uniform energy distribution (e.g., water flows downhill, objects hotter than their surrounding environment cool down). | 1,137 | 5,224 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2019-30 | latest | en | 0.92628 |
https://collegephysicsanswers.com/openstax-solutions/american-football-played-100-yd-long-field-excluding-end-zones-how-long-field | 1,611,580,877,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703581888.64/warc/CC-MAIN-20210125123120-20210125153120-00576.warc.gz | 288,877,221 | 12,273 | Change the chapter
Question
American football is played on a 100-yd-long field, excluding the end zones. How long is the field in meters? (Assume that 1 meter equals 3.281 feet.)
$91.4 \textrm{ m}$
Solution Video | 61 | 212 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-04 | latest | en | 0.864402 |
https://socratic.org/questions/which-is-the-solution-to-the-equation-3-5-2h-4-5-57-75 | 1,601,460,679,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402123173.74/warc/CC-MAIN-20200930075754-20200930105754-00510.warc.gz | 613,816,935 | 6,070 | # Which is the solution to the equation 3.5(2h+4.5)=57.75?
Dec 20, 2016
$h = 6$
#### Explanation:
First, expand the terms within parenthesis:
$\left(3.5 \times 2 h\right) + \left(3.5 \times 4.5\right) = 57.75$
$7 h + 15.75 = 57.75$
Next, isolate the $h$ term on one side of the equation and the constants on the other side of the equation while keeping the equation balanced:
$7 h + 15.75 - \textcolor{red}{15.75} = 57.75 - \textcolor{red}{15.75}$
$7 h + 0 = 42$
$7 h = 42$
Now, solve for $h$ while keeping the equation balanced:
$\frac{7 h}{\textcolor{red}{7}} = \frac{42}{\textcolor{red}{7}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}} h}{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}}} = 6$
$h = 6$ | 278 | 734 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2020-40 | latest | en | 0.71043 |
http://mathhelpforum.com/differential-equations/180209-laplace-transform-calculate-integral-print.html | 1,516,349,040,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887832.51/warc/CC-MAIN-20180119065719-20180119085719-00692.warc.gz | 219,701,723 | 3,502 | # Laplace transform to calculate an integral.
• May 11th 2011, 06:45 AM
sublim25
Laplace transform to calculate an integral.
Use Laplace transform (with respect to t)to calculate the integral
I=\int([\cos(tx)/(x^2+a^2)]dx t\geqslant 0
[IMG]file:///C:/Users/Jamie/AppData/Local/Temp/msohtmlclip1/01/clip_image002.jpg[/IMG]
• May 11th 2011, 06:54 AM
TheEmptySet
Quote:
Originally Posted by sublim25
Use Laplace transform (with respect to t)to calculate the integral
I=\int([\cos(tx)/(x^2+a^2)]dx t\geqslant 0
[IMG]file:///C:/Users/Jamie/AppData/Local/Temp/msohtmlclip1/01/clip_image002.jpg[/IMG]
So you have
$I =\int\frac{\cos(tx)}{x^2+a^2}dx$
but what are the limits of integration?
• May 11th 2011, 07:02 AM
sublim25
Limits are from 0 to infinity.
• May 11th 2011, 07:41 AM
TheEmptySet
Let
$I(t) =\int_{0}^{\infty}\frac{\cos(tx)}{x^2+a^2}dx$
Then
$\mathcal{L}\{I\} =\int_{0}^{\infty}\frac{\mathcal{L}\{\cos(tx)\}}{x ^2+a^2}dx=\int_{0}^{\infty}\frac{s}{s^2+x^2}\cdot \frac{1}{x^2+a^2}dx$
Now by partial factions we get
$\mathcal{L}\{I\} =\frac{s}{s^2-a^2}\int_{0}^{\infty}\left( \frac{1}{x^2+a^2}-\frac{1}{x^2+s^2}\right)dx=\frac{\pi}{2}\left( \frac{s}{a(s^2-a^2)}-\frac{1}{s^2-a^2}\right)$
Now if you take the inverse Laplace transform we get
$I=\frac{\pi}{2a}\cosh(at)-\frac{\pi}{2}\sinh(at)$
• May 11th 2011, 07:45 AM
sublim25
Can you please explain how you get the partial fractions to come out to this?
• May 11th 2011, 07:54 AM
TheEmptySet
Quote:
Originally Posted by sublim25
Can you please explain how you get the partial fractions to come out to this?
$\frac{s}{(x^2+a^2)(x^2+s^2)}=\frac{Ax+B}{x^2+a^2}+ \frac{Cx+D}{x^2+s^2}$
$s =(Ax+B)(x^2+s^2)+(Cx+D)(x^2+a^2)$
Now expand all of this out to get
$s=Ax^3+Bx^2+s^2Ax+s^2B + Cx^3+Dx^2+a^2Cx+a^2D$
$s=(A+C)x^3+(B+D)x^2+(s^2A+a^2C)x+(s^2B+a^2D)$
So this gives 4 equations in the 4 unknowns A,B,C,D
$A+C=0 \quad B+D=0 \quad s^2A+a^2C=0 \quad s^2B+a^2D=s$
dont forget that s and a are constants. Now just solve this system.
• May 11th 2011, 08:24 AM
sublim25
I am not sure what I am doing wrong, but when I solve the system, everything is cancelling out.
• May 11th 2011, 08:38 AM
TheEmptySet
Quote:
Originally Posted by sublim25
I am not sure what I am doing wrong, but when I solve the system, everything is cancelling out.
I don't know either. Please post what you have done. Note that both A and C are equal to 0. | 951 | 2,396 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2018-05 | longest | en | 0.635271 |
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# Help with Verbal and AWA
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Help with Verbal and AWA [#permalink]
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20 Dec 2011, 22:56
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Hi,
I am Krishna from India. I am planning to take GMAT in July 2012. I am pretty good at Quant . I bought the Kaplan Premier and solved the Quant section in about 4 hours.(Had only 5 wrong answers in all). I am a little weak in the Verbal and AWA part. I bought the MGMAT Sentence Correction guide. What other books do i require? I really need a 720+ score for my internship.
Thanks,
Krishna.
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Re: Help with Verbal and AWA [#permalink]
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21 Dec 2011, 10:09
Hi ralphspencer,
A July 2012 test date - note that the new GMAT with the integrated reasoning and 1 fewer AWA section will be effective in June 2012.
Seems like your quant is pretty solid. You're using the Official Guide, right? For additional verbal, you can use the OG Verbal book. And of course, make sure you practice with GMATPrep software so you have practice doing questions in front of a computer.
Those are about the only books & resources we recommend - those and, of course, GMAT Pill
Re: Help with Verbal and AWA [#permalink] 21 Dec 2011, 10:09
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Addition of Complex Numbers
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Last updated date: 11th Aug 2024
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Introduction to Complex Number
Complex numbers are helpful in finding the square root of negative numbers. The concept of complex numbers was first referred to in the 1st century by a greek mathematician, Hero of Alexandria, when he tried to find the square root of a negative number. Complex numbers have applications in much scientific research, signal processing, electromagnetism, fluid dynamics, quantum mechanics, and vibration analysis. So in this article, we will learn what complex numbers are and how to add complex numbers.
What are Complex Numbers?
A complex number is the sum of a real number and an imaginary number. A complex number is of the form a + ib and is usually represented by z. Here both a and b are real numbers. The value 'a' is called the real part denoted by Re(z), and 'b' is called the imaginary part Im(z). Also, ib is called an imaginary number.
Some of the examples of complex numbers are $2+3 i,-2-5 i, \dfrac{1}{2}+i \dfrac{3}{2}$, etc.
How to Add Complex Numbers?
The addition of complex numbers is similar to the addition of natural numbers. In complex numbers, the real part is added to the real part, and the imaginary part is added to the imaginary part.
For two complex numbers of the form $z_1=a+i d$ and $z_2=c+i d$, the sum of complex numbers $z_1+z_2=(a+c)+i(b+d)$.
The complex numbers follow all the following properties of addition.
• Closure Law: The sum of two complex numbers is also a complex number. For two complex numbers, $z_1$ and $z_2$, the sum of $z_1+z_2$ is also a complex number.
• Commutative Law: For two complex numbers $z_1, z_2$ we have $z_1+z_2=z_2+z_1$
• Associative Law: For the given three complex numbers $z_1, z_2, z_3$ we have $z_1+\left(z_2+z_3\right)=\left(z_1+z_2\right)+z_3$.
• Additive Identity: For a complex number $z=a+i b$, there exists $0=0$ $+i 0$, such that $z+0=0+z=0$.
• Additive Inverse of Complex numbers: For the complex number $z=a+i b$, there exists a complex number $-z=-a-i b$ such that $z+(-z)=(-z)+z=0$. Here $-z$ is the additive inverse.
Solved Examples in Addition Of Complex Numbers
Let us see some solved examples in addition to complex numbers.
Q 1. Find the addition of two complex numbers $(2+3 i)$ and $(-9-2 i)$.
Ans: $(2+3 \mathrm{i})+(-9-2 \mathrm{i})$
$=2+3 \mathrm{i}-9-2 \mathrm{i}$
$=2-9+3 \mathrm{i}-2 \mathrm{i}$
$=-7+\mathrm{i}$
Q 2. Evaluate: $(2 \sqrt{ } 3+5 i)+(\sqrt{ } 3-7 i)$
Ans: $2 \sqrt{ } 3+5 i+\sqrt{3}-7 i$
$= 2 \sqrt{3}+\sqrt{ } 3+5 i-7 i$
$=3 \sqrt{ } 3-2 i$
Q 3. Express the complex number $(1-i)+(-1+6 i)$ in the standard form a $+$ ib.
Ans: $(1-i)+(-1+6 i)$
$=1-i-1+6 i$
$=1-1-i+6 i$
$=0+5 \mathrm{i}$, which is the required form.
Practice Question
Based on the explanation given above in this article, try to solve the following questions.
Q 1. Solve (4i-7)+(1-i)
Ans: 3i -6.
Q 2. Solve it. (i-1) + (1+i)
Ans: 2i
Q 3. (1+i)+ (1-i) is?
Ans: 2
Q 4. Find z+z, if z = 6i
Ans: 12i.
Summary
The' Complex' theme does not mean it is 'complicated'.
It means the two types of numbers, real and imaginary, together form a complex number, just like building a complex game on mobile or running programs on a computer. Thus we made the addition operation on complex numbers simple for you in this article. By studying the solved examples and practising the Practice Set, students will master the concept of Addition on Complex Numbers.
FAQs on Addition of Complex Numbers
1. Is the addition of complex numbers associative?
All complex numbers are Commutative and Associative under addition and multiplication, and Multiplication is distributive under addition.
2. For what complex numbers are used?
The complex number is used to easily find the square root of a negative number. Here we use the value of $i^2=-1$ to represent the negative sign of a number, which is helpful to find the square root easily. Here we have $\sqrt{ }-4=\sqrt4{i}^2 =\pm 2 i$. Further, to find the negative roots of the quadratic equation, we used complex numbers.
3. Why is it called a complex number?
Eventually, the modern terminology emerged in the 19th century: "complex numbers", meaning that they consist of two parts, real and imaginary | 1,263 | 4,378 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-33 | latest | en | 0.891232 |
http://www.cse.chalmers.se/~nad//publications/danielsson-bag-equivalence/Prelude.html | 1,481,324,857,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542851.96/warc/CC-MAIN-20161202170902-00103-ip-10-31-129-80.ec2.internal.warc.gz | 408,296,204 | 22,088 | ```------------------------------------------------------------------------
-- A small prelude
------------------------------------------------------------------------
{-# OPTIONS --without-K #-}
-- Note that parts of Agda's standard library make use of the K rule.
module Prelude where
------------------------------------------------------------------------
-- Support for universe polymorphism
-- Universe levels.
infixl 6 _⊔_
postulate
Level : Set
lzero : Level
lsuc : Level → Level
_⊔_ : Level → Level → Level
{-# BUILTIN LEVEL Level #-}
{-# BUILTIN LEVELZERO lzero #-}
{-# BUILTIN LEVELSUC lsuc #-}
{-# BUILTIN LEVELMAX _⊔_ #-}
-- Lifting.
record ↑ {a} ℓ (A : Set a) : Set (a ⊔ ℓ) where
constructor lift
field lower : A
open ↑ public
------------------------------------------------------------------------
-- Some finite types
-- The empty type.
data ⊥ {ℓ} : Set ℓ where
⊥-elim : ∀ {w ℓ} {Whatever : Set w} → ⊥ {ℓ = ℓ} → Whatever
⊥-elim ()
-- A version of the empty type which is not universe-polymorphic.
⊥₀ : Set
⊥₀ = ⊥
-- Negation.
infix 3 ¬_
¬_ : ∀ {ℓ} → Set ℓ → Set ℓ
¬ P = P → ⊥ {ℓ = lzero}
-- The unit type.
record ⊤ : Set where
constructor tt
-- Booleans.
data Bool : Set where
true false : Bool
-- Conditional.
if_then_else_ : ∀ {a} {A : Set a} → Bool → A → A → A
if true then t else f = t
if false then t else f = f
-- Not.
not : Bool → Bool
not b = if b then false else true
-- And.
infixr 6 _∧_
_∧_ : Bool → Bool → Bool
b₁ ∧ b₂ = if b₁ then b₂ else false
-- Or.
infixr 5 _∨_
_∨_ : Bool → Bool → Bool
b₁ ∨ b₂ = if b₁ then true else b₂
-- The truth predicate T is only inhabited when its argument is true.
T : Bool → Set
T b = if b then ⊤ else ⊥
------------------------------------------------------------------------
-- Natural numbers
data ℕ : Set where
zero : ℕ
suc : (n : ℕ) → ℕ
-- Support for natural number literals.
{-# BUILTIN NATURAL ℕ #-}
{-# BUILTIN ZERO zero #-}
{-# BUILTIN SUC suc #-}
-- Dependent eliminator.
ℕ-rec : ∀ {p} {P : ℕ → Set p} →
P 0 → (∀ n → P n → P (suc n)) → ∀ n → P n
ℕ-rec z s zero = z
ℕ-rec z s (suc n) = s n (ℕ-rec z s n)
infixl 6 _+_
_+_ : ℕ → ℕ → ℕ
zero + n = n
suc m + n = suc (m + n)
-- The usual ordering of the natural numbers.
infix 4 _≤_
data _≤_ (m : ℕ) : ℕ → Set where
≤-refl : m ≤ m
≤-step : ∀ {n} (m≤n : m ≤ n) → m ≤ suc n
abstract
-- Some lemmas.
zero≤ : ∀ n → zero ≤ n
zero≤ zero = ≤-refl
zero≤ (suc n) = ≤-step (zero≤ n)
suc≤suc : ∀ {m n} → m ≤ n → suc m ≤ suc n
suc≤suc ≤-refl = ≤-refl
suc≤suc (≤-step m≤n) = ≤-step (suc≤suc m≤n)
m≤m+n : ∀ m n → m ≤ m + n
m≤m+n zero n = zero≤ n
m≤m+n (suc m) n = suc≤suc (m≤m+n m n)
------------------------------------------------------------------------
-- Simple combinators working solely on and with functions
infixr 9 _∘_
infixr 0 _\$_
-- The identity function.
id : ∀ {a} {A : Set a} → A → A
id x = x
-- Composition.
_∘_ : ∀ {a b c}
{A : Set a} {B : A → Set b} {C : {x : A} → B x → Set c} →
(∀ {x} (y : B x) → C y) → (g : (x : A) → B x) →
((x : A) → C (g x))
f ∘ g = λ x → f (g x)
-- Application.
_\$_ : ∀ {a b} {A : Set a} {B : A → Set b} →
((x : A) → B x) → ((x : A) → B x)
f \$ x = f x
-- Constant functions.
const : ∀ {a b} {A : Set a} {B : Set b} → A → (B → A)
const x = λ _ → x
-- Flips the first two arguments.
flip : ∀ {a b c} {A : Set a} {B : Set b} {C : A → B → Set c} →
((x : A) (y : B) → C x y) → ((y : B) (x : A) → C x y)
flip f = λ x y → f y x
------------------------------------------------------------------------
-- Σ-types
infixr 4 _,_
infixr 2 _×_
record Σ {a b} (A : Set a) (B : A → Set b) : Set (a ⊔ b) where
constructor _,_
field
proj₁ : A
proj₂ : B proj₁
open Σ public
-- A variant where the first argument is implicit.
∃ : ∀ {a b} {A : Set a} → (A → Set b) → Set (a ⊔ b)
∃ = Σ _
-- Binary products.
_×_ : ∀ {a b} (A : Set a) (B : Set b) → Set (a ⊔ b)
A × B = Σ A (const B)
-- A map function.
Σ-map : ∀ {a b p q}
{A : Set a} {B : Set b} {P : A → Set p} {Q : B → Set q} →
(f : A → B) → (∀ {x} → P x → Q (f x)) →
Σ A P → Σ B Q
Σ-map f g = λ p → (f (proj₁ p) , g (proj₂ p))
-- Curry and uncurry.
curry : ∀ {a b c} {A : Set a} {B : A → Set b} {C : Σ A B → Set c} →
((p : Σ A B) → C p) →
((x : A) (y : B x) → C (x , y))
curry f x y = f (x , y)
uncurry : ∀ {a b c} {A : Set a} {B : A → Set b} {C : Σ A B → Set c} →
((x : A) (y : B x) → C (x , y)) →
((p : Σ A B) → C p)
uncurry f (x , y) = f x y
------------------------------------------------------------------------
-- W-types
data W {a b} (A : Set a) (B : A → Set b) : Set (a ⊔ b) where
sup : (x : A) (f : B x → W A B) → W A B
-- Projections.
head : ∀ {a b} {A : Set a} {B : A → Set b} →
W A B → A
head (sup x f) = x
tail : ∀ {a b} {A : Set a} {B : A → Set b} →
(x : W A B) → B (head x) → W A B
tail (sup x f) = f
-- If B is always inhabited, then W A B is empty.
abstract
inhabited⇒W-empty : ∀ {a b} {A : Set a} {B : A → Set b} →
(∀ x → B x) → ¬ W A B
inhabited⇒W-empty b (sup x f) = inhabited⇒W-empty b (f (b x))
------------------------------------------------------------------------
-- Support for coinduction
infix 1000 ♯_
postulate
∞ : ∀ {a} (A : Set a) → Set a
♯_ : ∀ {a} {A : Set a} → A → ∞ A
♭ : ∀ {a} {A : Set a} → ∞ A → A
{-# BUILTIN INFINITY ∞ #-}
{-# BUILTIN SHARP ♯_ #-}
{-# BUILTIN FLAT ♭ #-}
------------------------------------------------------------------------
-- M-types
data M {a b} (A : Set a) (B : A → Set b) : Set (a ⊔ b) where
dns : (x : A) (f : B x → ∞ (M A B)) → M A B
-- Projections.
pɐǝɥ : ∀ {a b} {A : Set a} {B : A → Set b} →
M A B → A
pɐǝɥ (dns x f) = x
lıɐʇ : ∀ {a b} {A : Set a} {B : A → Set b} →
(x : M A B) → B (pɐǝɥ x) → M A B
lıɐʇ (dns x f) y = ♭ (f y)
------------------------------------------------------------------------
-- Binary sums
infixr 1 _⊎_
data _⊎_ {a b} (A : Set a) (B : Set b) : Set (a ⊔ b) where
inj₁ : (x : A) → A ⊎ B
inj₂ : (y : B) → A ⊎ B
-- Eliminator for binary sums.
[_,_] : ∀ {a b c} {A : Set a} {B : Set b} {C : A ⊎ B → Set c} →
((x : A) → C (inj₁ x)) → ((x : B) → C (inj₂ x)) →
((x : A ⊎ B) → C x)
[ f , g ] (inj₁ x) = f x
[ f , g ] (inj₂ y) = g y
-- A map function.
⊎-map : ∀ {a₁ a₂ b₁ b₂}
{A₁ : Set a₁} {A₂ : Set a₂} {B₁ : Set b₁} {B₂ : Set b₂} →
(A₁ → A₂) → (B₁ → B₂) → A₁ ⊎ B₁ → A₂ ⊎ B₂
⊎-map f g = [ inj₁ ∘ f , inj₂ ∘ g ]
-- A special case of binary sums: decided predicates.
Dec : ∀ {p} → Set p → Set p
Dec P = P ⊎ ¬ P
-- Decidable relations.
Decidable : ∀ {a b ℓ} {A : Set a} {B : Set b} →
(A → B → Set ℓ) → Set (a ⊔ b ⊔ ℓ)
Decidable _∼_ = ∀ x y → Dec (x ∼ y)
------------------------------------------------------------------------
-- Lists
infixr 5 _∷_
data List {a} (A : Set a) : Set a where
[] : List A
_∷_ : (x : A) (xs : List A) → List A
-- Right fold.
foldr : ∀ {a b} {A : Set a} {B : Set b} →
(A → B → B) → B → List A → B
foldr _⊕_ ε [] = ε
foldr _⊕_ ε (x ∷ xs) = x ⊕ foldr _⊕_ ε xs
-- The length of a list.
length : ∀ {a} {A : Set a} → List A → ℕ
length = foldr (const suc) 0
-- Appends two lists.
infixr 5 _++_
_++_ : ∀ {a} {A : Set a} → List A → List A → List A
xs ++ ys = foldr _∷_ ys xs
-- Maps a function over a list.
map : ∀ {a b} {A : Set a} {B : Set b} → (A → B) → List A → List B
map f = foldr (λ x ys → f x ∷ ys) []
-- Concatenates a list of lists.
concat : ∀ {a} {A : Set a} → List (List A) → List A
concat = foldr _++_ []
-- The list monad's bind operation.
infixl 5 _>>=_
_>>=_ : ∀ {a b} {A : Set a} {B : Set b} →
List A → (A → List B) → List B
xs >>= f = concat (map f xs)
-- A filter function.
filter : ∀ {a} {A : Set a} → (A → Bool) → List A → List A
filter p = foldr (λ x xs → if p x then x ∷ xs else xs) []
------------------------------------------------------------------------
-- Finite sets
Fin : ℕ → Set
Fin zero = ⊥
Fin (suc n) = ⊤ ⊎ Fin n
-- A lookup function.
lookup : ∀ {a} {A : Set a} (xs : List A) → Fin (length xs) → A
lookup [] ()
lookup (x ∷ xs) (inj₁ tt) = x
lookup (x ∷ xs) (inj₂ i) = lookup xs i
``` | 3,105 | 8,014 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2016-50 | latest | en | 0.331637 |
https://lorensen.github.io/VTKExamples/site/Cxx/Plotting/ChartMatrix/ | 1,579,330,688,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250592261.1/warc/CC-MAIN-20200118052321-20200118080321-00541.warc.gz | 547,164,413 | 40,125 | # ChartMatrix
VTKExamples/Cxx/Plotting/ChartMatrix
### Description¶
This example illustrates the use vtkChartMatrix, a container that holds a matrix of charts. The example creates a 2 x 2 matrix of plots. The matrix elements are:
(0,0): a vtkPlotPoints (0,1): a vtkPlotPoints (1,0): a vtkPlotLine (1,1): a vtkPlotBar and a vtkPlotPoints
Question
### Code¶
ChartMatrix.cxx
#include <vtkNew.h>
#include <vtkChartMatrix.h>
#include <vtkChartXY.h>
#include <vtkNamedColors.h>
#include <vtkRenderWindow.h>
#include <vtkRenderer.h>
#include <vtkPlot.h>
#include <vtkPlotPoints.h>
#include <vtkAxis.h>
#include <vtkPen.h>
#include <vtkTable.h>
#include <vtkFloatArray.h>
#include <vtkContextView.h>
#include <vtkContextScene.h>
#include <vtkRenderWindowInteractor.h>
//----------------------------------------------------------------------------
int main( int, char * [] )
{
vtkNew<vtkNamedColors> colors;
// Set up a 2D scene, add an XY chart to it
vtkNew<vtkContextView> view;
view->GetRenderWindow()->SetSize(1280, 1024);
vtkNew<vtkChartMatrix> matrix;
matrix->SetSize(vtkVector2i(2, 2));
matrix->SetGutter(vtkVector2f(30.0, 30.0));
// Create a table with some points in it...
vtkNew<vtkTable> table;
vtkNew<vtkFloatArray> arrX;
arrX->SetName("X Axis");
vtkNew<vtkFloatArray> arrC;
arrC->SetName("Cosine");
vtkNew<vtkFloatArray> arrS;
arrS->SetName("Sine");
vtkNew<vtkFloatArray> arrS2;
arrS2->SetName("Sine2");
vtkNew<vtkFloatArray> tangent;
tangent->SetName("Tangent");
int numPoints = 42;
float inc = 7.5 / (numPoints-1);
table->SetNumberOfRows(numPoints);
for (int i = 0; i < numPoints; ++i)
{
table->SetValue(i, 0, i * inc);
table->SetValue(i, 1, cos(i * inc));
table->SetValue(i, 2, sin(i * inc));
table->SetValue(i, 3, sin(i * inc) + 0.5);
table->SetValue(i, 4, tan(i * inc));
}
// Add multiple line plots, setting the colors etc
// lower left plot, a point chart
vtkChart *chart = matrix->GetChart(vtkVector2i(0, 0));
plot->SetInputData(table, 0, 1);
dynamic_cast<vtkPlotPoints*>(plot)->SetMarkerStyle(vtkPlotPoints::DIAMOND);
plot->GetXAxis()->GetGridPen()->SetColorF(colors->GetColor3d("warm_grey").GetData());
plot->GetYAxis()->GetGridPen()->SetColorF(colors->GetColor3d("warm_grey").GetData());
plot->SetColor(
colors->GetColor3ub("sea_green").GetRed(),
colors->GetColor3ub("sea_green").GetGreen(),
colors->GetColor3ub("sea_green").GetBlue(),
255);
// upper left plot, a point chart
chart = matrix->GetChart(vtkVector2i(0, 1));
plot->SetInputData(table, 0, 2);
plot->GetXAxis()->GetGridPen()->SetColorF(colors->GetColor3d("warm_grey").GetData());
plot->GetYAxis()->GetGridPen()->SetColorF(colors->GetColor3d("warm_grey").GetData());
plot->SetColor(
255);
//
chart = matrix->GetChart(vtkVector2i(1, 0));
plot->SetInputData(table, 0, 3);
plot->GetXAxis()->GetGridPen()->SetColorF(colors->GetColor3d("warm_grey").GetData());
plot->GetYAxis()->GetGridPen()->SetColorF(colors->GetColor3d("warm_grey").GetData());
plot->SetColor(
colors->GetColor3ub("dark_orange").GetRed(),
colors->GetColor3ub("dark_orange").GetGreen(),
colors->GetColor3ub("dark_orange").GetBlue(),
255);
// upper right plot, a bar and point chart
chart = matrix->GetChart(vtkVector2i(1, 1));
plot->SetInputData(table, 0, 4);
plot->GetXAxis()->GetGridPen()->SetColorF(colors->GetColor3d("warm_grey").GetData());
plot->GetYAxis()->GetGridPen()->SetColorF(colors->GetColor3d("warm_grey").GetData());
plot->SetColor(
colors->GetColor3ub("burnt_sienna").GetRed(),
colors->GetColor3ub("burnt_sienna").GetGreen(),
colors->GetColor3ub("burnt_sienna").GetBlue(),
255);
plot->SetInputData(table, 0, 1);
dynamic_cast<vtkPlotPoints*>(plot)->SetMarkerStyle(vtkPlotPoints::CROSS);
plot->GetXAxis()->GetGridPen()->SetColorF(colors->GetColor3d("warm_grey").GetData());
plot->GetYAxis()->GetGridPen()->SetColorF(colors->GetColor3d("warm_grey").GetData());
plot->SetColor(
255);
// Lower right plot, a line chart
chart = matrix->GetChart(vtkVector2i(1, 0));
plot->SetInputData(table, 0, 3);
plot->GetXAxis()->GetGridPen()->SetColorF(colors->GetColor3d("warm_grey").GetData());
plot->GetYAxis()->GetGridPen()->SetColorF(colors->GetColor3d("warm_grey").GetData());
plot->SetColor(
colors->GetColor3ub("dark_orange").GetRed(),
colors->GetColor3ub("dark_orange").GetGreen(),
colors->GetColor3ub("dark_orange").GetBlue(),
255);
plot->SetInputData(table, 0, 3);
plot->GetXAxis()->GetGridPen()->SetColorF(colors->GetColor3d("warm_grey").GetData());
plot->GetYAxis()->GetGridPen()->SetColorF(colors->GetColor3d("warm_grey").GetData());
plot->SetColor(
colors->GetColor3ub("royal_blue").GetRed(),
colors->GetColor3ub("royal_blue").GetGreen(),
colors->GetColor3ub("royal_blue").GetBlue(),
255);
// Finally render the scene and compare the image to a reference image
view->GetRenderer()->SetBackground(colors->GetColor3d("navajo_white").GetData());
view->GetRenderWindow()->Render();
view->GetInteractor()->Initialize();
view->GetInteractor()->Start();
return EXIT_SUCCESS;
}
### CMakeLists.txt¶
cmake_minimum_required(VERSION 3.3 FATAL_ERROR)
project(ChartMatrix)
find_package(VTK COMPONENTS
vtkChartsCore
vtkCommonColor
vtkCommonCore
vtkCommonDataModel
vtkInteractionStyle
vtkRenderingContext2D
vtkRenderingContextOpenGL2
vtkRenderingCore
vtkRenderingFreeType
vtkRenderingGL2PSOpenGL2
vtkRenderingOpenGL2
vtkViewsContext2D QUIET)
if (NOT VTK_FOUND)
message("Skipping ChartMatrix: ${VTK_NOT_FOUND_MESSAGE}") return () endif() message (STATUS "VTK_VERSION:${VTK_VERSION}")
if (VTK_VERSION VERSION_LESS "8.90.0")
# old system
include(${VTK_USE_FILE}) add_executable(ChartMatrix MACOSX_BUNDLE ChartMatrix.cxx ) target_link_libraries(ChartMatrix PRIVATE${VTK_LIBRARIES})
else ()
# include all components
target_link_libraries(ChartMatrix PRIVATE ${VTK_LIBRARIES}) # vtk_module_autoinit is needed vtk_module_autoinit( TARGETS ChartMatrix MODULES${VTK_LIBRARIES}
)
endif ()
cd ChartMatrix/build
If VTK is installed:
cmake ..
If VTK is not installed but compiled on your system, you will need to specify the path to your VTK build:
cmake -DVTK_DIR:PATH=/home/me/vtk_build ..
Build the project:
make
and run it:
./ChartMatrix
WINDOWS USERS
Be sure to add the VTK bin directory to your path. This will resolve the VTK dll's at run time. | 1,795 | 6,221 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-05 | latest | en | 0.287984 |
https://cs.stackexchange.com/questions/127687/finding-a-good-estimate-for-amount-of-time-computers-spend-sorting-lists-of-what?noredirect=1 | 1,606,822,035,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141674082.61/warc/CC-MAIN-20201201104718-20201201134718-00530.warc.gz | 259,563,121 | 35,635 | # Finding a Good Estimate for Amount of Time Computers Spend Sorting Lists of What Lengths?
I have an assignment to envision and calculate the possible effects the implementation of a general sorting algorithm that is O(n) time and O(1) space ( assuming general case ) would have on computation and society.
In order to do that, I am trying to estimate how much computation time computers in server farms spend on sorting, and what general orders of magnitude the lengths of lists being sorted fall in. Server farms seem to be the most impactful area to analyze for pertinent effects, but if there are some other areas that people would expect to see an outsize effect, I would appreciate them being called out.
I can then present sets of assumptions with the "new" algorithm being faster at sorting lists at such a length and larger and how much faster than the current state of the art. eg. "If the new algorithm is even with the state of the art for lists of 500 billion items, and increases in time with "n" while the soa increases in time with "log n", and 1/2 trillion item and larger sorts take X% of server-farm computation, you could expect to see computation time for the same tasks reduced by order of magnitude Y%". Once I can get some firm numbers ( not the question here ) on how much electricity server farms use, then I can also electricity usage cost comparisons.
When I researched on the web, I found that an O(n)/O(1) sorting algorithm is considered theoretically possible, but undiscovered to this point. I suspect that there might not be much impact on computation, because there seems to be no incentives put out for someone to come up with such an algorithm ( or a proof that it cannot exist ).
I have presumed that sorting algorithms in general use, at least for large lists like web search results, are O(n log n) time and O(1) space as a general matter of course. If this is in error, please point it out!
My apologies if this forum is not appropriate for this kind of question. In that case, I would ask for pointers on where I should go with this question.
Thank you.
• researched on the web, I found that the web knows everything, right and wrong. Please specify or at least quote&attribute your sources. Note that $O(n)$ ordering requires an $O(1)$ operation to establish order between a limited number of items. Think about string comparison. – greybeard Jun 27 at 3:58
• 1) cs.stackexchange.com/questions/18536/… 2) en.wikipedia.org/wiki/Sorting_algorithm 3) staff.ustc.edu.cn/~csli/graduate/algorithms/book6/…. Citing sources is something I should have done in the original post; my apologies. – Kyle Jun 29 at 21:29
• Thank you for supplying the references - improving what begs improvement is the way to go. Note that you can(should!) edit your post(s). – greybeard Jun 30 at 3:02
• (Actual time of sorting used to be an issue when CPU was rented by the minute and tape drives needed operators requiring wages…) – greybeard Jun 30 at 3:04
• You should specify the sorting model you are using. It is NOT possible to have a comparison-based sorting algorithm with a running time smaller than $\Omega(n \log n)$ (i.e., any algorithm that requires $o(n \log n)$ time cannot be correct for sufficiently large inputs since it cannot possibly output all $n!$ permutations of an input sequence with $n$ elements). Linear-time sorting algorithms for other models/assumptions on the collection of elements to sort are known (see, e.g., counting sort and radix sort in the case of polynomially bounded integers). – Steven Jul 24 at 1:32 | 804 | 3,568 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-50 | latest | en | 0.956482 |
https://hextobinary.com/unit/speed | 1,708,591,415,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473735.7/warc/CC-MAIN-20240222061937-20240222091937-00257.warc.gz | 310,032,117 | 16,407 | # Speed Converter
Speed
Centimeter/Second
Centimeter/Minute
## How many Centimeter/Minute are in a Centimeter/Second?
The answer is one Centimeter/Second is equal to 60 Centimeter/Minute and that means we can also write it as 1 Centimeter/Second = 60 Centimeter/Minute. Feel free to use our online unit conversion calculator to convert the unit from Centimeter/Second to Centimeter/Minute. Just simply enter value 1 in Centimeter/Second and see the result in Centimeter/Minute.
## How to Convert Centimeter/Second to Centimeter/Minute (cm/s to cm/min)
By using our Centimeter/Second to Centimeter/Minute conversion tool, you know that one Centimeter/Second is equivalent to 60 Centimeter/Minute. Hence, to convert Centimeter/Second to Centimeter/Minute, we just need to multiply the number by 60. We are going to use very simple Centimeter/Second to Centimeter/Minute conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Centimeter/Second} = 1 \times 60 = \text{60 Centimeter/Minute}$$
## What is Centimeter/Second Unit of Measure?
Centimeters per second is a unit of measurement for speed. Centimeter per second is also spelled as centimetre per second. It indicates the amount of distance that could be covered in centimeters within the duration of one second. Centimeters per second is abbreviated as cm/s or cm/sec. For example, you can write 1 centimeter per second as 1 cm/s or 1 cm/sec.
## What is the symbol of Centimeter/Second?
The symbol of Centimeter/Second is cm/s. This means you can also write one Centimeter/Second as 1 cm/s.
## What is Centimeter/Minute Unit of Measure?
Centimeters per minute is a unit of measurement for speed. Centimeter per minute is also spelled as centimetre per minute. It indicates the amount of distance that could be covered in centimeters within the duration of one minute. Centimeters per minute is abbreviated as cm/min. For example, you can write 1 centimeter per minute as 1 cm/min.
## What is the symbol of Centimeter/Minute?
The symbol of Centimeter/Minute is cm/min. This means you can also write one Centimeter/Minute as 1 cm/min.
## Centimeter/Second to Centimeter/Minute Conversion Table
Centimeter/Second [cm/s]Centimeter/Minute [cm/min]
160
2120
3180
4240
5300
6360
7420
8480
9540
10600
1006000
100060000
## Centimeter/Second to Other Units Conversion Table
Centimeter/Second [cm/s]Output
1 centimeter/second in centimeter/minute is equal to60
1 centimeter/second in centimeter/hour is equal to3600
1 centimeter/second in feet/second is equal to0.032808398950131
1 centimeter/second in feet/minute is equal to1.97
1 centimeter/second in foot/hour is equal to118.11
1 centimeter/second in inch/second is equal to0.39370078740157
1 centimeter/second in inch/minute is equal to23.62
1 centimeter/second in inch/hour is equal to1417.32
1 centimeter/second in kilometer/second is equal to0.00001
1 centimeter/second in kilometer/minute is equal to0.0006
1 centimeter/second in kilometer/hour is equal to0.036
1 centimeter/second in knot is equal to0.019438445
1 centimeter/second in mach is equal to0.000029387
1 centimeter/second in meter/second is equal to0.01
1 centimeter/second in meter/minute is equal to0.6
1 centimeter/second in meter/hour is equal to36
1 centimeter/second in mile/second is equal to0.0000062137119223733
1 centimeter/second in mile/minute is equal to0.0003728227153424
1 centimeter/second in mile/hour is equal to0.022369362920544
1 centimeter/second in millimeter/second is equal to10
1 centimeter/second in millimeter/minute is equal to600
1 centimeter/second in millimeter/hour is equal to36000
1 centimeter/second in yard/second is equal to0.010936132983377
1 centimeter/second in yard/minute is equal to0.65616797900262
1 centimeter/second in yard/hour is equal to39.37
1 centimeter/second in centimeter/day is equal to86400
1 centimeter/second in dekameter/day is equal to86.4
1 centimeter/second in dekameter/hour is equal to3.6
1 centimeter/second in dekameter/minute is equal to0.06
1 centimeter/second in dekameter/second is equal to0.001
1 centimeter/second in feet/day is equal to2834.65
1 centimeter/second in furlong/day is equal to4.29
1 centimeter/second in furlong/hour is equal to0.17895490336435
1 centimeter/second in furlong/minute is equal to0.0029825817227392
1 centimeter/second in furlong/second is equal to0.000049709695378987
1 centimeter/second in furlong/fortnight is equal to60.13
1 centimeter/second in hectometer/day is equal to8.64
1 centimeter/second in hectometer/hour is equal to0.36
1 centimeter/second in hectometer/minute is equal to0.006
1 centimeter/second in hectometer/second is equal to0.0001
1 centimeter/second in inch/day is equal to34015.75
1 centimeter/second in kilometer/day is equal to0.864
1 centimeter/second in megameter/day is equal to0.000864
1 centimeter/second in megameter/hour is equal to0.000036
1 centimeter/second in megameter/minute is equal to6e-7
1 centimeter/second in megameter/second is equal to1e-8
1 centimeter/second in meter/day is equal to864
1 centimeter/second in mile/day is equal to0.53686471009306
1 centimeter/second in millimeter/day is equal to864000
1 centimeter/second in millimeter/microsecond is equal to0.00001
1 centimeter/second in nautical mile/day is equal to0.46652267818575
1 centimeter/second in nautical mile/hour is equal to0.019438444924406
1 centimeter/second in nautical mile/minute is equal to0.00032397408207343
1 centimeter/second in nautical mile/second is equal to0.0000053995680345572
1 centimeter/second in speed of light [air] is equal to3.3366416468926e-11
1 centimeter/second in speed of light [glass] is equal to5.003461444662e-11
1 centimeter/second in speed of light [ice] is equal to4.3696896581733e-11
1 centimeter/second in speed of light [vaccum] is equal to3.3356409519815e-11
1 centimeter/second in speed of light [water] is equal to4.4364024692431e-11
1 centimeter/second in speed of sound [air] is equal to0.00002938669957977
1 centimeter/second in speed of sound [metal] is equal to0.000002
1 centimeter/second in speed of sound [water] is equal to0.0000066666666666667
1 centimeter/second in yard/day is equal to944.88
Disclaimer:We make a great effort in making sure that conversion is as accurate as possible, but we cannot guarantee that. Before using any of the conversion tools or data, you must validate its correctness with an authority. | 1,823 | 6,397 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2024-10 | latest | en | 0.921132 |
https://brainly.com/question/147740 | 1,485,249,931,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560284376.37/warc/CC-MAIN-20170116095124-00193-ip-10-171-10-70.ec2.internal.warc.gz | 795,675,840 | 8,358 | # Lynn has a watering can that holds 32 cups of water, and she fills it half full. Then she waters her 27 plants so that each plant gets the same amount of water. How many cups of water will each plant get?
1
by uhfiufuikhsdvh
## Answers
2014-10-10T15:17:59-04:00
16+16=32 or 32-16=16
27-16=11 | 100 | 298 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2017-04 | latest | en | 0.908822 |
https://book.gateoverflow.in/tag/nielit2016mar-scientistc | 1,620,524,076,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988953.13/warc/CC-MAIN-20210509002206-20210509032206-00317.warc.gz | 165,467,422 | 16,112 | # Recent questions tagged nielit2016mar-scientistc
1
The ratio between the speeds of two trains is $7:8$. If the second train runs $440$kms in $4$hours, then the speed of the first train is: $47.4$ km/hr $57.19$ km/hr $48.13$ km/hr $96.25$ km/hr
2
A man walks at $5$ kmph for $6$ hr and at $4$ km/h for $12$ hr. His average speed is $4\frac{1}{3}$ km/h $9\frac{2}{3}$ km/h $9½$ km/h $8$ km/h
3
In the following question, a part of the sentence is printed in bold. Below are given alternatives to the bold part at (A),(B) and (C) which may improve the sentence. Choose the correct alternative. In case no improvement is needed, your ... Master's thesis was highly estimated and is now being prepared for publication. was highly discussed was highly commended is highly appraised No improvement
4
In the following question, a part of the sentence is printed in bold. Below are given alternatives to the bold part at (A),(B) and (C) which may improve the sentence. Choose the correct alternative. In case no improvement is needed, your answer is (D). By studying AIDS has engaged many researchers in the last decade. Important study Now that the study The study of No improvement
5
In the first $10$ overs of a cricket game, the run rate was only $3.2$. What should be the run rate in the remaining $40$ overs to reach the target of $282$ runs? $6.25$ $6.5$ $6.75$ $7$
6
A grocer has a sale of $Rs.6435,Rs.6927,Rs.6855,Rs.7230$ and $Rs.6562$ for $5$ consecutive months. How much sale must he have in the sixth month so that he gets an average sale of $Rs.6500$? $\text{Rs.4991}$ $\text{Rs.5991}$ $\text{Rs.6001}$ $\text{Rs.6991}$
7
The total electricity generation in a country is $97\:GW$. The contribution of various energy sources is indicated in percentage terms in the pie chart given below: What is the contribution of wind and solar power in absolute terms in the electricity generation? $6.79\:GW$ $9.4\:GW$ $19.7\:GW$ $29.1\: GW$
8
The total electricity generation in a country is $97\:GW$. The contribution of various energy sources is indicated in percentage terms in the pie chart given below: What is the contribution of renewable energy sources in absolute terms in the electricity generation? $\text{09.1 GW}$ $\text{26.19 GW}$ $\text{67.9 GW}$ $\text{97 GW}$
9
Worker $A$ takes $8$ hours to do a job. Worker $B$ takes $10$ hours to do the same job. How long it take both $\text{A&B}$, working together but independently, to do the same job? $40/9$ days $40/7$ days $7.5$ days $8.5$ days
10
If $34$ men completed $2/5$th of a work in $8$ days working $9$ hours a day. How many more man should be engaged to finish the rest of the work in $6$ days working $9$ hours a day? $189$ $198$ $102$ $142$
11
In the following question, a part of the sentence is printed in bold. Below are given alternatives to the bold part at (A), (B) and (C) which may improve the sentence. Choose the correct alternative. In case no improvement is needed, your answer is (D). Sordid and sensational books tend to vitiate the public taste. divide distract distort No improvement
12
$0.04\times 0.0162$ is equal to: $6.48\times 10^{-8}$ $6.48\times 10^{-4}$ $6.48\times 10^{-9}$ $6.48\times 10^{-7}$
The value of $\large\frac{(0.96)^3-(0.1)^3}{(0.96)^2+0.096+(0.1)^2}$ is : $0.86$ $0.95$ $0.97$ $1.06$
A car owner buys petrol at $Rs.7.50$, $Rs.8$ and $Rs.8.50$ per litre for three successive years. What approximately is the average cost per litre of petrol if he spends $Rs.4000$ each year? $\text{Rs.7.98}$ $\text{Rs.6}$ $\text{Rs.9.50}$ $\text{Rs.9.5}$
The expression $(11.98\times 11.98 + 11.98 \times x +0.02 \times 0.02)$ will be a perfect square for $x$ equal to: $2.02$ $0.17$ $0.04$ $1.4$ | 1,146 | 3,685 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2021-21 | latest | en | 0.931544 |
https://www.euclideanspace.com/maths/discrete/category/concrete/set/ordered/index.htm | 1,709,471,177,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476374.40/warc/CC-MAIN-20240303111005-20240303141005-00016.warc.gz | 755,624,120 | 5,869 | # Maths - Ordered Sets
On these pages we are looking at set related categories. The category theory of a set is discussed on the page here. We then go on to look at sets with some additional structure. On this page we start to look at sets with an order relation added.
First lets look at this structure from a set theoretic point of view. That is we look inside the set, at its elements, rather than looking at its external category theoretic properties.
### Strictly Ordered
First lets assume the relation is total and strictly ordered. That is for every pair of elements 'a' and 'b' then either: a<b or a>b.
Examples of strictly ordered sets are: The natural numbers. The natural numbers modulo n. strictly ordered sets are: transitive: if a
If S is a set, then the order relation on S can be considered as a subset of S×S or a map S×S-> π where π is the truth object (in this case Boolean).
More about order relation on page here
So how is the category defined?
### Objects
The objects are {}, {0}, {0,1}, {0,1,2}, {0,1,2,3} and so on.
Not necessarily labeled but it helps understanding to label with numerals.
As with general sets the objects are a 0 element set, a 1 element set, a 2 element set and so on but here
### Arrows
Arrows are order preseving functors (or cofunctors - see below) between these objects.
That is:
`if a<b then F a < F b `
### Weakly Ordered
Here the relation is total and strictly ordered. That is for every pair of elements 'a' and 'b' then then it could be that: a<=b or a>=b or both a=b.
Weakly ordered sets are: reflexive: a<=a transitive: if a<=b and b<=c then a<=c But they don't have the following properties: symmetric: not if a<=b then b<=a Although for some cases its possible that a<=b and b<=a.
If S is a set, then the order relation on S can be considered as a subset of S×S or a map S×S-> π where π is the truth object (in this case Boolean).
More about order relation on page here
So how is the category defined?
#### Objects
Not necessarily labeled but it helps understanding to label with numerals.
As with general sets the objects are a 0 element set, a 1 element set, a 2 element set and so on but here the objects are {}, {0}, {0,1}, {0,1,2}, {0,1,2,3} and so on. Also objects where the elements are repeated such as {0,0}, {0,0,1}, {0,1,1} and so on. These are known as degenerate sets.
#### Arrows
The arrows preserve this weak order.
That is:
`if a<=b then F a <= F b`
### Decomposing Arrows
Order preserving arrows can be decomposed into a sequence of arrows that insert or merge single elements one at a time.
We will call these elemental mappings di and si di is an injective mapping which goes from a set of size n to a set of size n+1 si is a surjective mapping which goes from a set of size n to a set of size n-1
So we can construct any order preserving mapping from a sequence of di and si.
### Decomposing Arrows in Setop
Setop is a category with the same objects as set but where the arrows are reversed as explained on the page here. In this case order preserving arrows can again be decomposed into a sequence of arrows although , in this case, we don't have the concept of injective and surjective mappings (or we have the reverse of these concepts).
So the elemental mappings di and si become: face map: di is a mapping (reverse of injective) which removes element 'i' and goes from a set of size n to a set of size n-1. degeneracy map: si is a mapping (reverse of surjective) which duplicates an element and goes from a set of size n to a set of size n+1
### Subobjects
We can create any subobject using a sequence of face maps.
### Equalities
Degeneracy maps create equal elements.
### Identities
Note: when combining maps below the map on the right is done first. For example, di dj means di after dj (do dj then do di).
Identity diagram
di dj = dj-1 di if i < j
di sj = s j-1 di if i < j
dj sj = id = dj+1 sj
di sj = sj di-1 if i > j+1
si sj = s j+1 s i if i ≤ j
### Simplicial Sets
See page here about simplicial sets. | 1,055 | 4,040 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-10 | latest | en | 0.915499 |
https://www.mikkymixx.com/how-to-win-the-lottery-study-the-successful-techniques/ | 1,627,251,627,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151866.98/warc/CC-MAIN-20210725205752-20210725235752-00197.warc.gz | 947,270,608 | 9,130 | # How to Win the Lottery – Study the Successful Techniques
Lottery is the game wherein the player selects six lotto numbers from a larger set of lottery numbers. In this game, exactly where a dollar stake can win million of cash, the chances against winning this game have to be exorbitant. In order to enhance your probabilities and make it easier to win, study on to understand about three approaches on how to win the lottery. These would be: Lottery Quantity Selection, Lottery Game Choice, and Lottery Balanced Wheels.
* Lottery Quantity Selection.
The very first method on how to win the lottery is referred to as the Lottery Number Selection. This includes picking the most winning lotto numbers which possess the greatest possibility of winning. A lot of folks wouldn’t gamble a lot on a horse without the need of studying its performance history beforehand. This is known as handicapping, which implies understanding the history in an effort to plan the future. Wall Street analysts practice the same strategy. They chart bonds, stocks and commodities, examining value action in the history to conclude value trends in the future. In https://ruay-ruay.com/ , we examine the past actions of the frequent winning lotto numbers to help us in resolving which numbers have the highest possibility of being drawn. Winning lotto numbers are aimlessly drawn. Nevertheless, aimlessly drawn numbers from prototypes that are to a certain extent anticipated and mastering to make use of these prototypes is the technique on how to win the lottery. The simple but incredible rule is
* Lottery Game Selection
Another method on how to win the lottery is known as the Lottery game Selection. This is performed by just picking to play the lottery game with the smallest odds, which denotes the lottery game with the lowest number fields. Majority of the states have at least two lottery games, 1 with nig lottery prizes and pretty much matchless odds, and 1 with a reduced lottery quantity field and smaller prizes for players who want to win jackpots more regularly. In retort to vast player demand for a lottery game that is simpler to win, majority of the states conformed by presenting the choose-five game, wherein just five numbers are scored on a game panel.
* Lottery Balanced Wheels
The final strategy on how to win the lottery is named the Lottery Balanced Wheels. This approach supplies your funds a lot more manage and radically improves your possibility of winning lottery jackpots. They are the most vital tools a lotto player can use to get instant odds increase. These lotto systems permit you to pick a big set of lottery numbers which are set in a scientifically resolved lottery pattern on your stake slips to present you an exact win assurance. If you entrap the six (five or four winning lotto numbers) in the massive set of lottery numbers you have chosen, you are assured to win at least 1 prize. On the other hand you can win a lot of lotto jackpots, or even the 1st prize jackpot. Winning a quantity of lottery prizes all at after is what makes these lottery systems profitable, exciting and exciting to use. | 623 | 3,132 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-31 | latest | en | 0.963985 |
https://www.hpacademy.com/courses/professional-motorsport-data-analysis/supporting-concepts-conventions/ | 1,726,334,932,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651580.73/warc/CC-MAIN-20240914161327-20240914191327-00223.warc.gz | 751,919,519 | 51,740 | ×
Sale ends todayGet 30% off any course (excluding packages)
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# Professional Motorsport Data Analysis: Conventions
-OR-
## Conventions
### 05.03
00:00 - Dealing with large amounts of data can become unnecessarily complicated and cumbersome which is the last thing we need when we're on a time crunch during a race weekend. 00:10 To help with this, conventions are used to keep things consistent, speeding up our work and allowing us to always speak the same language. 00:18 Simply put, we can think of conventions as small rules we follow when we're building up and working on our data analysis project. 00:27 At first, some of these conventions may seem a little over the top or unnecessary but I promise you that if you make an effort to stick to them now, in time you'll be thankful you did. 00:38 Especially once the analysis becomes more sophisticated. 00:42 A lot of these conventions will already be set up before we even get to analysing the data but others you'll need to set up as you go. 00:50 Ultimately as a data engineer, you're free to use whatever conventions you like but the real key here is that they just need to be consistent. 01:00 Without establishing that consistency from the outset, you're going to find yourself with a problem sooner rather than later. 01:07 In this module we're going to go over some of the most widely followed conventions in the motorsport industry. 01:13 Which is going to give you a good understanding of the basics. 01:16 Before we start though, let's take a look at the conventions cheat sheet which you can download from the related resources area underneath this video. 01:25 This is a handy summary sheet that you can keep nearby as a quick reference as you build up your analysis project. 01:31 First let's take a look at the convention we'll be using for the coordinate system of the car. 01:36 Positive directions are indicated by the direction of the arrows so X is in the direction of the car travel, Y is lateral to the direction of travel and Z is the vertical direction. 01:48 In line with our coordinate system convention, when the car is turning left, we define that as positive. 01:55 This means that when we're looking at the steering trace, positive values are for when the car is turning left and negative values of the steering trace show the car turning right. 02:04 Again referring to our coordinate system, when dealing with lateral G forces, positive is for the G force we have when the car is turning left and negative is for the G force we have when the car is turning right. 02:17 For longitudinal G force, positive shows acceleration and negative represents braking. 02:24 Next let's take a look at the use of colour. 02:26 Colour can be an incredibly useful tool in helping communicate features in the data. 02:31 One of the basic ways to use this is to show which part of the car the data relates to. 02:37 Once you're used to working with these colours and have stuck to some conventions the entire time, it means you don't have to look at the display legend constantly to remind yourself what colour represents what variable. 02:50 In this example, we see how when you're dealing with a quantity that has one channel for the front and one for the rear, like front and rear brake pressure, then we assign red to the front and blue to the rear. 03:02 We can apply the same convention to any other data source that has a front and a rear. 03:07 In the case where we have a quantity that relates to one corner of the car, for example tire pressure, then this is how we would assign the colours. 03:16 Here we're simply assigning red to the front left, green to the front right, blue to the rear left and yellow to the rear right. 03:25 You should use these colour conventions when you first set up your data analysis project or build a new math channel. 03:31 You're free to use whatever conventions make the most sense to you, whether that's to do with which values are positive or negative or if it's to do with the colours you want to use. 03:41 I just encourage you to use a system, apply it throughout your data analysis project and stick to it. 03:47 This should give you a basic understanding of some common conventions and how they're used. 03:52 But before we move on, I want to bring up just how crucial it is to be consistent when naming your logs and math channels. 04:00 It's hugely helpful for readability and organisation as things become more and more complex and of course this applies to both setting up your logger as well as your data analysis software. 04:12 Personally the way I like to do this is simply start with the element of the car, then the measurement type and then the position on the car. 04:21 So for example if I was setting up the logger, I would use channel names like damper position, front left or brake pressure rear. 04:30 To keep your channel names short it also helps to come up with some standard abbreviations for commonly used terms. 04:37 Here are some abbreviations for position on the car, pressure, position of the measurement and temperature that will help keep your channel names short. 04:47 Again, all of these examples show just one way to do it and there's nothing to say you need to follow it exactly if it doesn't work for you. 04:55 Use whatever is the most logical and readable for you, just be consistent.
We usually reply within 12hrs (often sooner)
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Experiencing website difficulties? | 1,231 | 5,499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-38 | latest | en | 0.960357 |
https://devsolus.com/2023/02/08/r-ggplot-multiple-boxplots/ | 1,679,897,756,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948609.41/warc/CC-MAIN-20230327060940-20230327090940-00549.warc.gz | 251,887,421 | 19,076 | # R: GGPlot Multiple Boxplots
I would like to plot four boxplots in one chart, however i want to add the median and range myself.
My code
``````ggplot(data = df_subset[df, aes(y = target_fed_funds_rate_median, x = Date, group = Date)) +
geom_boxplot(aes(ymin = target_fed_funds_rate_lr, lower = target_fed_funds_rate_cl, middle = target_fed_funds_rate_median,
upper = target_fed_funds_rate_ch, ymax = target_fed_funds_rate_hr))
``````
generates the following chart: What my code is generating
However, I am looking to create a chart similar to this one: How my chart should look like
What my Dataframe looks like: my dataframe df:
Any ideas how to fix this? Many thanks
Below a reproducable example:
``````time = c(1, 2, 3, 4, 5)
median = c(1, 2, 3, 4, 5)
low = c(0.5, 1.5, 2.5, 3.5, 4.5)
high = c(1.5, 2.5, 3.5, 4.5, 5.5)
center_high = c(1.25, 2.25, 3.25, 4.25, 5.25)
center_low = c(0.75, 1.75, 2.75, 3.75, 4.75)
df = data.frame(median, low, high, center_high, center_low)
ggplot(data = df, aes(y = median, x = time, group = time)) +
geom_boxplot(aes(ymin = low, lower = center_low, middle = median,
upper = center_high, ymax = high))
``````
Any ideas how to fix this? Many thanks
### >Solution :
As you want to "manually" create a boxplot by providing the boxplot stats you have to set `stat="identity"` in `geom_boxplot`:
``````library(ggplot2)
ggplot(data = df, aes(y = median, x = time, group = time)) +
geom_boxplot(aes(
ymin = low, lower = center_low, middle = median,
upper = center_high, ymax = high
), stat = "identity")
`````` | 525 | 1,554 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-14 | latest | en | 0.794651 |
http://mail.scipy.org/pipermail/scipy-dev/2009-March/011574.html | 1,411,099,055,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657130067.43/warc/CC-MAIN-20140914011210-00338-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 171,259,026 | 2,566 | [SciPy-dev] Possible Error in Kendall's Tau (scipy.stats.stats.kendalltau)
Sturla Molden sturla@molden...
Wed Mar 18 10:55:38 CDT 2009
```On 3/18/2009 4:12 PM, josef.pktd@gmail.com wrote:
> the sas reference by Bruce hat a reference to a book by Agresti, where
> I finally found a clear formal definition and it excludes matching
> ties in the in the denominator:
>
This is basically the same as in Numerical Receipes.
Hollander & Wolfe has a discussion of tie handling on page 374 and 375:
"We have recommended dealing with tied X observation and/or tied Y
observations by counting a zero in Q (8.17) counts leading to the
computation of K (8.6). This approach is statisfactory as long as the
number of (X,Y) pairs containing a tied X and/or tied Y observation does
not represent a sizable percentage of the total number (n) of sample pairs.
We should, however, point out that methods other than this zero
assignment to Q counts have been considered for dealing with tied C
and/or tied Y observations [...]"
They then suggest counting +1 or -1 by coin toss for ties, or a strategy
to be conservative about rejecting H0. They also suggest using Efron's
bootstrap for confidency intervals on tau (page 388). I don't see any
mention of tau-a, tau-b or tau-c in H&W, nor contigency tables.
I don't quite understand Hollander & Wolfe's argument. They are
basically saying that their recommended method of dealing with ties only
works when ties are so few in numbers that they don't matter anyway.
> I don't know about kendall's tau-c because m seems to be specific to
> contingency tables, while all other measures have a more general
> interpretation. Similar in view of the general definitions, I don't
> understand the talk about square or rectangular tables. But I don't
> have a good intuition for contingency tables.
The ide is that Kendall's tau works on ordinal scale, not rank scale as
Spearman's r. You can use the number of categories for X and Y you like,
but the categories have to be ordered. You thus get a table of counts.
If you for example use two categories (small or big) in X and four
categories (tiny, small, big, huge) in Y, the table is 2 x 4. If you go
all the way up to rank scale, you get a very sparse table with a lot 0
counts. With few categories, ties will be quite common, and that is the
justification for tau-b instead of gamma.
Sturla Molden
``` | 601 | 2,387 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2014-41 | longest | en | 0.92976 |
https://jurnal.unej.ac.id/index.php/prosiding/article/view/33514 | 1,670,031,812,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710918.58/warc/CC-MAIN-20221203011523-20221203041523-00484.warc.gz | 368,982,490 | 4,676 | # KLASIFIKASI DATA DIAGNOSIS COVID-19 MENGGUNAKAN METODE SUPPORT VECTOR MACHINE (SVM) DAN GENERALIZED LINEAR MODEL (GLM)
## (Classification of Covid-19 Diagnosis Data Using Support Vector Machine (SVM) and Generalized Linear Model (GLM) Methods)
• Yeni Rismawati Jurusan Matematika, Fakultas Matematika dan Ilmu Pengetahuan Alam, Universitas Jember
• I Made Tirta Jurusan Matematika, Fakultas Matematika dan Ilmu Pengetahuan Alam, Universitas Jember
• Yuliani Setia Dewi Jurusan Matematika, Fakultas Matematika dan Ilmu Pengetahuan Alam, Universitas Jember
### Abstract
Covid-19 is still a global concern. From the first time, this virus was detected, on December 31, 2019. As of March 20, 2022, there were 460 million positive cases of Covid-19, with 6.06 million deaths worldwide. The high number of Covid-19 cases is due to the rapid spread of this virus. One way to prevent the spread of this virus is by early detection of the disease and mapping the influence factors .The classification method with the support vector machine (SVM) method in machine learning can predict individuals diagnosed as positive for Covid-19 and who do not use the factors considered influential. Traditionally this can also be done with a generalized linear model (GLM). This study aims to compare two methods (SVM and GLM) in predicting individuals diagnosed as positive for Covid-19. In addition, this study also conducted an ensemble between SVM and GLM to determine whether the ensemble performed could produce better accuracy than the single classifier (SVM and GLM). The results showed that the accuracy with SVM and GLM was relatively high. However, SVM is slightly superior with 98.91% accuracy, and GLM with 95.64% accuracy. Meanwhile, the ensemble of both models achieved 98.91% accuracy, as high as SVM.
Keywords: Covid-19, Klasifikasi, Machine Learning SVM, GLM
Published
2022-08-14
How to Cite
RISMAWATI, Yeni; TIRTA, I Made; DEWI, Yuliani Setia. KLASIFIKASI DATA DIAGNOSIS COVID-19 MENGGUNAKAN METODE SUPPORT VECTOR MACHINE (SVM) DAN GENERALIZED LINEAR MODEL (GLM). UNEJ e-Proceeding, [S.l.], p. 246 - 252, aug. 2022. Available at: <https://jurnal.unej.ac.id/index.php/prosiding/article/view/33514>. Date accessed: 03 dec. 2022.
Citation Formats
Section
General | 600 | 2,265 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-49 | latest | en | 0.855671 |
https://mathematica.stackexchange.com/questions/59401/how-to-control-the-positions-of-contour-labels/59407#59407 | 1,653,482,108,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662587158.57/warc/CC-MAIN-20220525120449-20220525150449-00555.warc.gz | 458,706,444 | 68,678 | # How to control the positions of contour labels?
ContourPlot[x^2 + 5 y^2, {x, -5, 5}, {y, -2, 2},
Contours -> {2, 4, 6, 8, 10, 12, 14, 16, 20},
ContourLabels -> True,
AspectRatio -> Automatic,
gives result
But I want result like this
So I hope this code will work
ContourPlot[x^2 + 5 y^2, {x, -5, 5}, {y, -2, 2},
Contours -> {2, 4, 6, 8, 10, 12, 14, 16, 20},
ContourLabels -> (Text[#3, {#1, 0}, Background -> White] &),
AspectRatio -> Automatic,
But it didn't, it gives result
So how to control the position of ContourLabels?
• None of the workarounds offered in the answers (so far) explains why ContourPlot is giving the wrong coordinate for 14. What am I missing? Sep 11, 2014 at 9:20
• @rhermans - Unlike for the other labels, the default location for the label 14 is far from the x=0 (semi-major) axis so that when its y coordinate is just changed to zero without also correcting the x coordinate it ends up in the wrong position. The problem is not with ContourPlot but rather with the manual positioning of that label. Sep 11, 2014 at 12:32
• @Bob_Hanlon, thanks for the explanation. Sep 11, 2014 at 12:34
This can done with Epilog.
f[x_, y_] := x^2 + 5 y^2;
contours = {2, 4, 6, 8, 10, 12, 14, 16, 20};
lblXY = {#, 0} & /@ (Solve[f[x, 0] == #, x][[2, 1, 2]] & /@ contours // N);
ContourPlot[f[x, y], {x, -5, 5}, {y, -2, 2},
Contours -> contours,
AspectRatio -> Automatic,
Epilog -> {Thread[Text[contours, lblXY, Background -> White]]}]
Kind of a hack but this is what I would do.
f[x_, y_] := x^2 + 5 y^2;
c = {2, 4, 6, 8, 10, 12, 14, 16, 20};
labelPos = Solve[f[x, 0] == #, x][[2, 1, 2]] & /@ c;
Show[
ContourPlot[f[x, y], {x, -5, 5}, {y, -2, 2}, Contours -> c,
AspectRatio -> Automatic, ContourShading -> None],
Graphics[Text[#[[2]], {#[[1]], 0}, {0, 0}, Background -> White]] & /@
Transpose[{labelPos, c}]
]
Yes another workaround
Show@{ContourPlot[x^2 + 5 y^2, {x, -5, 5}, {y, -2, 2},
Contours -> Range[2, 20, 2], AspectRatio -> Automatic,
ContourShading -> None, ImageSize -> 640],
ContourPlot[x^2 + 5 y^2, {x, 0, 5}, {y, -0.0001, 0.0001},
Contours -> Range[2, 20, 2],
ContourLabels -> (Text[#3, {#1, #2}, Background -> White] &),
I decrease the possible region for labels by the second ContourPlot.
f = x^2 + 5 y^2;
ContourPlot[f, {x, -5, 5}, {y, -2, 2},
Contours -> {2, 4, 6, 8, 10, 12, 14, 16, 20},
ContourLabels -> (Text[#3, {Max[x /. Solve[f == #3 /. y -> 0, x]], 0},
Background -> White] &),
AspectRatio -> Automatic, ContourShading -> None]
Use
ContourLabels -> (Text[#3, Max[x /. Solve[(f /. y -> x/3) == #3, x]] {1, 1/3},
Background -> White] &)
to get
Update: you can also post-process the ContourPlot output to modify the location of the labels:
cp = ContourPlot[x^2 + 5 y^2, {x, -5, 5}, {y, -2, 2},
Contours -> {2, 4, 6, 8, 10, 12, 14, 16, 20}, PlotPoints -> 200,
ContourLabels -> Automatic, AspectRatio -> Automatic,
labelfunction := (Shift = RandomReal[{-30, 30}]; | 1,109 | 2,924 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2022-21 | latest | en | 0.761242 |
https://www.onlinemath4all.com/law-of-cosines-and-sines-worksheet.html | 1,713,448,720,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817206.54/warc/CC-MAIN-20240418124808-20240418154808-00637.warc.gz | 830,824,257 | 6,789 | # LAW OF COSINES AND SINES WORKSHEET
(1) In a triangle ABC, if sin A/sin C = sin(A − B)/sin(B − C), prove that a2, b2, c2 are in Arithmetic Progression.
Solution
(2) The angles of a triangle ABC, are in Arithmetic Progression and if b : c = √3 : √2, find ∠A.
Solution
(3) In a triangle ABC, if cos C = sin A / 2 sin B, show that the triangle is isosceles. Solution
(4) In a triangle ABC, prove that sin B/sinC = (c − a cosB)/(b − a cosC) Solution
(5) In a triangle ABC, prove that a cosA + b cosB + c cosC = 2a sinB sinC. Solution
(6) In a triangle ABC, ∠A = 60°. Prove that b + c = 2a cos (B − C)/2 Solution
In a triangle ABC, prove the following
(i) a sin (A/2 + B) = (b + c) sin A/2 Solution
(ii) a(cos B + cos C) = 2(b + c) sin2 A/2 Solution
(iii) (a2 − c2) / b2 = sin(A − C) / sin(A + C) Solution
(iv) a sin(B − C)/(b2 − c2) = b sin(C − A)/c2 − a2 = c sin(A − B)/(a2 − b2
(v) (a + b)/(a − b) = tan (A + B)/2 cot (A − B)/2 Solution
(8) In a triangle ABC, prove that (a2 − b2 + c2) tanB = (a2 + b2 − c2) tanC. Solution
(9) An Engineer has to develop a triangular shaped park with a perimeter 120 m in a village. The park to be developed must be of maximum area. Find out the dimensions of the park. Solution
(10) A rope of length 12 m is given. Find the largest area of the triangle formed by this rope and find the dimensions of the triangle so formed Solution
(11) Derive Projection formula from (i) Law of sines, (ii) Law of cosines. Solution
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Polar Form of a Complex Number | 671 | 1,954 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-18 | latest | en | 0.566294 |
https://www.convertunits.com/from/inch+of+air+%5B0+%C2%B0C%5D/to/zeptopascal | 1,604,132,590,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107916776.80/warc/CC-MAIN-20201031062721-20201031092721-00666.warc.gz | 636,288,872 | 8,642 | ## ››Convert inch of air [0 °C] to zeptopascal
inch of air [0 °C] zeptopascal
Did you mean to convert inch of air [0 °C] inch of air [15 °C] to zeptopascal
## ››More information from the unit converter
How many inch of air [0 °C] in 1 zeptopascal? The answer is 3.1055186930487E-21.
We assume you are converting between inch of air [0 °C] and zeptopascal.
You can view more details on each measurement unit:
inch of air [0 °C] or zeptopascal
The SI derived unit for pressure is the pascal.
1 pascal is equal to 3.1055186930487 inch of air [0 °C], or 1.0E+21 zeptopascal.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between inches of air and zeptopascals.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of inch of air [0 °C] to zeptopascal
1 inch of air [0 °C] to zeptopascal = 3.220074E+20 zeptopascal
2 inch of air [0 °C] to zeptopascal = 6.440148E+20 zeptopascal
3 inch of air [0 °C] to zeptopascal = 9.660222E+20 zeptopascal
4 inch of air [0 °C] to zeptopascal = 1.2880296E+21 zeptopascal
5 inch of air [0 °C] to zeptopascal = 1.610037E+21 zeptopascal
6 inch of air [0 °C] to zeptopascal = 1.9320444E+21 zeptopascal
7 inch of air [0 °C] to zeptopascal = 2.2540518E+21 zeptopascal
8 inch of air [0 °C] to zeptopascal = 2.5760592E+21 zeptopascal
9 inch of air [0 °C] to zeptopascal = 2.8980666E+21 zeptopascal
10 inch of air [0 °C] to zeptopascal = 3.220074E+21 zeptopascal
## ››Want other units?
You can do the reverse unit conversion from zeptopascal to inch of air [0 °C], or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Zeptopascal
The SI prefix "zepto" represents a factor of 10-21, or in exponential notation, 1E-21.
So 1 zeptopascal = 10-21 pascals.
The definition of a pascal is as follows:
The pascal (symbol Pa) is the SI unit of pressure.It is equivalent to one newton per square metre. The unit is named after Blaise Pascal, the eminent French mathematician, physicist and philosopher.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 810 | 2,549 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-45 | latest | en | 0.685338 |
http://slideplayer.com/slide/2773898/ | 1,527,280,677,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867217.1/warc/CC-MAIN-20180525200131-20180525220131-00210.warc.gz | 265,139,161 | 21,801 | # 1 Yago Diez, J. Antoni Sellarès and Universitat de Girona Noisy Road Network Matching Mario A. López University of Denver.
## Presentation on theme: "1 Yago Diez, J. Antoni Sellarès and Universitat de Girona Noisy Road Network Matching Mario A. López University of Denver."— Presentation transcript:
1 Yago Diez, J. Antoni Sellarès and Universitat de Girona Noisy Road Network Matching Mario A. López University of Denver
2 “Road Network Matching” Motivation Known scale, unknown reference system (maps may appear rotated). Find R’ In R
3 Problem Formalization -We describe maps using road crossings - Adjacency degrees act as color cathegories.
4 Given two sets of road points A and B, |A| < |B|, find all the subsets B’ of B that can be expressed as rigid motions of A. We want: the points to approximately match (fuzzy nature of real data). the adjacency degrees to coincide. One-to-one matching! (*) Rigid motion: composition of a translation and a rotation. Problem Formalization
5 Let A, B be two road point sets of the same cardinality. An adjacency-degree preserving bijective mapping f : S S’ maps each Road point P(a, r) to a distinct and unique road point f(P(a,r))= P(b,s) so that r = s. Let F be the set of all adjacency-degree preserving bijective mappings between S and S’. The Bottleneck Distance between S and S’ is is defined as: d b (S, S’ ) = min f F max P(a,r) S d(P(a,r), f(P(a,r))). Problem Formalization
6 Given two road points sets A and B, n=|A|, m=|B|, n < m, and a real positive number ε, determine all the rigid motions τ for which there exists a subset B’ of B, |B’|=|A|, such that: d b (τ(A),B’) ε (Bottleneck distance) Problem Formalization “Final Formulation”
7 Example Consider: A B Find:
8 Previous Work On Road Network Matching Previous Work Chen et Al.(STDBM’06): Similar problem with some differences: -Motions considered: - Chen et Al.: Translation + Scaling - Us: Translation + Rotation - Distance used: - Chen et Al.: Hausdorff - Us: Bottleneck
9 Previous Work On Point Set Matching Algorithms Previous Work - Alt / Mehlhorn / Wagener / Welzl (Discrete & Computational Geometry 88) - Efrat / Itai / Katz. (Comput. Geom. Theory Appl. 02) - Eppstein / Goodrich / Sun (SoCG 05) : Skip Quadtrees. - Diez / Sellarés (ICCSA 07)
10 Matching Algorithm - Tackle the problem from the COMPUTATIONAL GEOMETRY point of view. -Adapt the ideas in our paper at ICCSA 07 to the RNM problem. -Matching Algorithm: -Two main parts: Enumeration Testing OUR APPROACH:
11 Matching Algorithm Generate all possible motions τ that may bring set A near some B’. Enumeration We rule out all those pairs of points whose degrees do not coincide.
12 Matching Algorithm For every motion τ representative of an equivalence class, find a matching of cardinality n between τ(A) and S. Testing A set of calls to Neighbor operation corresponds to one range search operation in a skip quadtree Neighbor ( D(T), q ) Delete ( D(T), s ) Corresponds to a deletion operation in a skip quadtree. Amortized cost of Neighbor, Delete: log n (Under adequate assumptions)
13 Improving Running time Our main goal is to transform the problem into a series of smaller instances. We will use a conservative strategy to discard, cheaply and at an early stage, those subsets of B where no match may happen. Our process consists on two main stages: 1. Losless Filtering Algorithm 2. Matching Algorithm (already presented!)
14 Lossless Filtering Algorithm What geometric parameters, do we consider ? (rigid motion invariant ) - number of Road Points, - histogram of degrees, - max. and min. distance between points of the same degree, - CFCC codes. There cannot be any subset B‘ of B that approximately matches A fully contained in the four top-left quadrants, because A contains six points and the squares only five.
15 Initial step 1. Determine an adequate square bounding box of A. 2 s (size s) 2. Calculate associated geometric information. Lossless Filtering Algorithm
16 Calculate quadtree of B with geometric parameters............. Lossless Filtering Algorithm
17............ Points = 550 Points = 173 Points = 113 Points = 131Points = 133 23 57 56 37 20 6 53 34 54 12 14 51 49 46 34 4 0 6 1 16 1 3 22 31 3 11 1 22 20 19 6 11 Example with geometric parameter: number of points Lossless Filtering Algorithm
18 Search Algorithm a b b c Three search functions needed for every type of zone according to the current node: -Search type a zones. -Search type b zones. -Search type c zones. The search begins at the root and continues until nodes of size s are reached. Early discards will rule out of the search bigger subsets of B than later ones. Lossless Filtering Algorithm
19 - Search’s first step: Search Algorithm............ points = 550 points = 173 points = 113 points = 131 points = 133 23 57 56 37 20 6 53 34 54 12 14 51 49 46 34 4 0 6 1 16 1 3 22 31 3 11 1 22 20 19 6 11 -Target number of points = 25 - Launch search1? yes (in four sons) - Launch search2? yes (all possible couples) - Launch search3? yes (possible quartet) Lossless Filtering Algorithm
20 Search Algorithm............ points = 550 points = 173 points = 113 points = 131 points = 133 23 57 56 37 20 6 53 34 54 12 14 51 49 46 34 4 0 6 1 16 1 3 22 31 3 11 1 22 20 19 6 11 -Target number of points = 25 - Launch search1? yes (in three sons) - Launch search2? yes (all possible couples) - Launch search3? yes (possible quartet) Lossless Filtering Algorithm
21 Lossless Filtering Algorithm
22 Search Algorithm............ points= 550 points = 173 points = 113 points = 131 points = 133 23 57 56 37 19 5 54 35 54 12 14 51 49 46 34 4 0 6 1 16 1 3 22 31 3 11 1 22 20 19 6 11 -Target number of points = 25 - Launch search1? yes (in two sons) - Launch search2? yes (three possible couples) - Launch search3? yes (possible quartet) Lossless Filtering Algorithm
23 Lossless Filtering Algorithm
24 Algorithm complexity: O(m 2 ) Lossless Filtering Algorithm
25 Matching Algorithm Efrat, Itai, Katz: O( n 4 m 3 log m ) Our approach : Σ Cand.Zon O( n 4 n’ 3 log n’ ) Computational Cost
26 Implementation and Results Data used, Tiger/lines file from Arapahoe, Adams and Denver Counties:
27 Experiments Experiment 1: Does the lossless filtering step help?
28 Experiments Experiment 2: Filtering parameters comparison.
29 Experiments Experiment 3: Computational Performance
30 Experiments Experiment 3: Computational Performance
31 Conclusions - First formalization of the NRNM problem in terms of the bottleneck distance. - Fast running times in light of the inherent complexity of the problem. - Experiments show how using the lossless filtering algorithm helps reduce the running time. - We have only used information that should be evident to all observers. -We have also provided some examples on how the degree of noise in data influences the performance of the algorithm.
32 Future Work - Other values of ε (for example, those that arise directly from the precision of measuring devices). - Maps with different levels of detail.
33 Yago Diez, J. Antoni Sellarès and Universitat de Girona Noisy Road Network Matching Mario A. López University of Denver
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# Yutao Li
Last seen: etwa 2 Monate ago Active since 2022
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This topic is 3848 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
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I've come upon a very annoying problem while doing division in c++.. I'll demonstrate the problem with an example: double a = double(500000000)/9; then the value of a = 55555555.555555552 but when I do: int b = 500000000; int c = 9; double a = double(b)/c; then the value of a = 55555556.000000000 Why is there a difference when I'm doing the exact same operation? For my game I want to be able to divide variables b and c without losing precision but I don't know how.
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Floating Point (especially the section "Accuracy problems")
double a = double(b)/c;
Here you divide a double with an integer, therefore the compiler decides for itself whichever of the two to convert back and forth.
In your case, it must have chosen to convert the result to an integer and then assign it to a double.
This should yield the value 55555555.555555552 (or something close to that in contrast to the FP semantics)
double a = double(b)/(double)c;
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There is no difference. The following program gives identical output for both values of a:
#include <iostream>using namespace std;int main(){ double a = double(500000000)/9; cout << fixed << a << endl; int b = 500000000; int c = 9; a = double(b)/c; cout << fixed << a << endl;}
Output is:
55555555.55555655555555.555556
Post your exact code and we might be able to see what's wrong.
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it is still truncating you to an int. You could actually need to have
double a = double(b) / double(c);
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If I'm not mistaken, only one side of / is required to be a double to make the result double (the compiler shouldn't be free to choose the result type).
From the standard:
Quote:
--Otherwise, if either operand is double, the other shall be converted to double.
It is surprising to see that the OP's "wrong" result has been truncated "upwards".
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Yes it's true I've already tried converting both operands but I'm still getting the same result.
the exact code of my program is:
int b = 500000000;
int c = 9;
double a = double(b)/double(c);
a = double(500000000)/9;
And I check the value of variable 'a' in the debug mode of visual c++.
on line 3 , the value of a becomes 55555556.000000000
on line 4 , the value of a becomes 55555555.555555552
Does anyone have a clue?
even when I do:
double b = 500000000;
double c = 9;
double a = b/c;
the value of a is still 55555556.000000000
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That exact code gives equal output for both operations (55555555.555556) when compiled with g++. The discrepancy you see might be due to a debugger quirk. Try compiling and printing the result of the calculations through cout or printf.
By the way, what version of MSVC are you using? If it's a very old version (MSVC 6), there might be issues with nonstandard behaviour.
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Quote:
Original post by TaxCollectorAnd I check the value of variable 'a' in the debug mode of visual c++.on line 3 , the value of a becomes 55555556.000000000on line 4 , the value of a becomes 55555555.555555552
Can't replicate with MVS2008. All debug outputs show correct value.
What does debug mode mean anyway, as tooltip, as watch, compiled as debug configuration?
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After experimenting with the code I've found that at the very start of the program the calculation is done as expected.
It is only after calling the function
D3DDEVTYPE_HAL,
hWnd,
D3DCREATE_SOFTWARE_VERTEXPROCESSING,
&d3dpp,
&d3ddev);
(where d3ddev is an LPDIRECT3DDEVICE9;
and d3dpp is a D3DPRESENT_PARAMETERS;)
that the problem arises. So maybe this problem should've been posted in the DirectX part of the forum but I did not know the problem was caused by directX when I first posted the question. Because I'm a novice in DirectX, I do not know how to solve the issue.
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I believe DirectX modifies the way the floating point processor works. There is a flag "D3DCREATE_FPU_PRESERVE" that can be used (somewhere; I don't use DirectX [smile]) to keep the original behaviour. See if this is the source of your problem.
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https://www.nagwa.com/en/videos/158139808561/ | 1,675,910,677,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764501066.53/warc/CC-MAIN-20230209014102-20230209044102-00118.warc.gz | 902,160,097 | 23,305 | # Question Video: Identifying the Radii of a Given Circle Mathematics • 11th Grade
Identify all the radii of circle ๐.
01:31
### Video Transcript
Identify all the radii of circle ๐.
Thereโs a few things we should think about here. The first thing is that when we name a circle circle ๐, itโs indicating to us that the point ๐ is the center of the circle. We also need to think about what is meant by radii. Radii is the plural form of the radius of a circle, and a radius is the distance from the center to the circumference of a circle. And the circumference is the outside edge of the circle. This means weโre looking for any line that starts at the center and goes out to the outside edge of the circle.
We see that here from ๐ to ๐บ, which makes line segment ๐๐บ a radius, we can start at the center ๐ and move out to the point ๐น, which makes the line segment ๐๐น a radius. The same thing is true if we start at the center ๐ and move out to the point ๐ธ, which makes line segment ๐๐ธ a radius. ๐น๐บ cannot be a radius of the circle as it does not start or end at the center. Itโs also worth noting that the line segment ๐น๐ธ goes through the point ๐ and we call that a diameter of the circle. However, the two segments that make up the diameter are both radii. For this circle ๐, the three radii we see are ๐๐บ, ๐๐น, and ๐๐ธ. | 433 | 1,415 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2023-06 | latest | en | 0.895955 |
https://www.jiskha.com/questions/1067147/help-me-with-this-question-please-how-will-the-total-voltage-of-a-string-of-light-bulbs | 1,586,473,591,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371880945.85/warc/CC-MAIN-20200409220932-20200410011432-00138.warc.gz | 968,502,138 | 5,458 | # Physics
Help me with this question, please
How will the total voltage of a string of light bulbs in series change as you add more bulbs to the string?
A. The total voltage will decrease
B. The total voltage will stay the same
C. The total voltage will increase
d. You cannot predict what will happen to the voltage
1. 👍 0
2. 👎 0
3. 👁 229
1. it stays the same. Why would it change?
1. 👍 0
2. 👎 0
2. okay, thx for the help :)
1. 👍 0
2. 👎 0
posted by Me and I
3. If it is a series, V_T should be the same.
1. 👍 0
2. 👎 0
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asked by JORDAN on October 28, 2014
4. ### Java
NEED HELP! I would like to create a class for finding the voltages of 1, 2 or even 4 batteries. The Class can be called BatteryVoltage with a default voltage value of 1.5 (AA Battery) Must allow the user to find the total voltage
asked by Jordan on October 26, 2014
5. ### I NEED HELP WITH THIS
Three identical light bulbs are connected in series, then are disconnected and arranged in parallel. For each of the scenarios below indicate what changes (if any) take place A. Total resistance of the circuit B. Total current of
asked by Ciara on April 7, 2015
1. ### Physics !
Three identical light bulbs are connected in series, then are disconnected and arranged in parallel. For each of the scenarios below indicate what changes (if any) take place A. Total resistance of the circuit B. Total current of
asked by Ciara on April 7, 2015
2. ### Science
So I need help with a lab, I'm not looking for answers, I want to know if my answers make sense, so this is the pre lab introduction: In a series circuit, electrons travel along a single pathway. The movement of the electrons
asked by ItsGottaGetEasier on May 24, 2018
3. ### math
Solve by setting up the proper equation to describe the facts given and then carrying out the mathematical calculations to solve for the unknown variable(s). Three batteries are connected in series so that the total voltage is 54
asked by kenny on July 6, 2012
4. ### physics e&m
This question concerns your ability to manipulate combinations of capacitors. Devise a network of capacitors whose total equivalent capacitance is 2μF while the total voltage difference from one end to the other end is 400 V.
asked by njood on October 26, 2013
5. ### Engineering Science
Given the circuit below with unknown voltage source and a voltage drop of 2.233 v Volts on resistor R3, with (R1 = 2KΩ, R2 = 1KΩ, R3 = 3.3KΩ, R4 = 4.7KΩ, R5 = 4.7KΩ). Calculate and complete the table below. I need help
asked by Mike on October 14, 2018
More Similar Questions | 940 | 3,402 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2020-16 | latest | en | 0.914041 |
https://www.physicsforums.com/threads/why-does-strong-magnetic-field-destroy-spherical-symmetry-of-atom.283679/ | 1,485,146,452,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560282110.46/warc/CC-MAIN-20170116095122-00190-ip-10-171-10-70.ec2.internal.warc.gz | 967,463,503 | 14,855 | # Why does strong magnetic field destroy spherical symmetry of atom?
1. Jan 8, 2009
### xfshi2000
In most of quantum textbook, they say strong magnetic field destroy the spherical symmetry of an atom in zeeman effect. Total angular moment is not conserved. Can anyone explain what physical meaning is? I cannot picture it in my mind. How does strong magnetic field cause spherical symmetry loss? thanks
2. Jan 8, 2009
### daschaich
The Zeeman effect involves the application of a static external magnetic field. This magnetic field singles out a particular direction, thus breaking spherical symmetry.
Symmetry will be broken no matter what the strength of the field, but at sufficiently weak field it may be possible to neglect the effects of symmetry breaking without introducing too serious errors.
If there's a particular reference that's confusing you, please share it so we can try to check it out.
3. Jan 8, 2009
### xfshi2000
Thank you for your kind reply. When I read quantum textbooks, I found a lot of stuff confuse me. For example:
(1) Photon has spin 1. But it only has two projections along the propagation direction. ms=(+/-)h bar. ms=0 is not existed because of transverse property of light. Dot product of vector k and polarization direction of photon epsilon should be zero. in textbook, they derive it in mathematical calculation. Does anyone give a physical picture about this case? Simply, a photon can be described by a plane wave. But we need modify the plane wave and multiply this plane wave by polarization direction of photon to construct photon wave function.
(2) This question also is related to zeeman effect. As we know, for ground state of hydrogen atom, S-state is splitted by external magnetic field. The transition between these two splitting energys generate a photon with different polarization radiation. Anyone can explain this radiation field of this transition? Such as the transition between different angular moment quantum number with same magnetic quantum number and the transition between same angular moment quantum number with different magnetic quantum number. Sometime a circularly polarized photon or linearly polarized photon from different view plane is generated. I don't know why? Anyone can comment that? thanks a lot. | 468 | 2,286 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2017-04 | longest | en | 0.906237 |
https://www.r-bloggers.com/rcpp-example-using-na/ | 1,553,406,620,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203326.34/warc/CC-MAIN-20190324043400-20190324065400-00205.warc.gz | 882,563,989 | 16,109 | # Rcpp example using NA
August 25, 2016
By
(This article was first published on r - Brandon Bertelsen, and kindly contributed to R-bloggers)
Dealing with `NA` in `Rcpp` can be a bit tricky to sort out, but the speed boosts are incredible and if you’re dealing with large data – worth the time to figure it out. But usage cases are important to consider in the example below you can see that vectorized operations are still very competitive and very concise. `Rcpp` doesn’t shine so much in this example, but it can when you have a problem that requires a waterfall of results (one result relies on the next and so on).
Here’s a basic function that takes `NA` values into account and basically converts anything < 0 to -1 and anything > 0 to +1.
``````library(Rcpp)
library(microbenchmark)
signR <- function(x) {
for(i in 1:length(x)) {
if(is.na(x[i])) {
next
}
if (x[i] > 0) {
x[i] <- 1
} else {
x[i] <- -1
}
}
return(x)
}
signR2 <- function(x) {
x[x > 0] <- 1
x[x < 0] <- -1
x
}
cppFunction('
NumericVector signC(NumericVector x) {
int n = x.size();
NumericVector out(n);
for(int i = 0; i 0) {
out[i] = 1;
} else {
out[i] = -1;
}
}
return out;
}')
test <- sample(c(rnorm(25),NA),100000, replace = TRUE)
microbenchmark(
signR(test),
signR2(test),
signC(test)
)
# Unit: milliseconds
# expr min lq mean median uq max neval
# signR(test) 197.676873 201.344711 205.951881 202.437384 204.712847 317.973634 100
# signR2(test) 4.267238 4.335184 4.424379 4.412983 4.467380 4.869512 100
# signC(test) 2.095438 2.120071 2.185537 2.160100 2.219835 2.676364 100
all.equal(signR(test),signR2(test),signC(test))
# TRUE
``````
R-bloggers.com offers daily e-mail updates about R news and tutorials on topics such as: Data science, Big Data, R jobs, visualization (ggplot2, Boxplots, maps, animation), programming (RStudio, Sweave, LaTeX, SQL, Eclipse, git, hadoop, Web Scraping) statistics (regression, PCA, time series, trading) and more... | 633 | 2,019 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-13 | latest | en | 0.71043 |
https://bowmandickson.com/2012/05/08/ | 1,652,930,269,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522741.25/warc/CC-MAIN-20220519010618-20220519040618-00246.warc.gz | 209,220,958 | 30,704 | # Daily Archives: May 8, 2012
## Honing My Skills Instruction
Skills instruction was something that I was not good at when I first started teaching. I found it boring compared to big ideas and didn’t really understand why kids didn’t get things from me doing problems on the board and then them doing practice problems with each other or individually or for homework. For me, it was like “What is there to understand that simple practice wont solve?” Well, I have grown a lot since then. This year, I have implemented a lot of new tools (many whiteboard based) that have really helped out with my skills instruction. I feel like I really had a great sequence this year when doing the skills part of integrating with my regular class – I wanted to share and reflect, especially so it’s written down somewhere for me to use in the future, because some of the ideas will be useful when I teach other topics.
### PREAMBLE:
Just so you know how I lead up to this seminal topic in Calculus… First, I spent a considerable time (about five 45 minute class periods) exploring everything to do with Riemann Sums, both in terms of pure area and what area means in applied situations. I think that feels like a lot of time, but we tackled the conceptual side of integration very throughly and used that to motivate the idea of an antiderivative/integral. Once we motivated the integral, we focused on learning how to find antiderivatives, which is the part I want to talk about.
### 1. Guess and Check With a Partner
With inspiration from a great worksheet from Sam, I wanted students to rely on their intuition at first to find andiderivatives, instead of relying on formulae. I’ve tried things like this previously, but it really helped this time to explicitly explain that this is what we were doing – that maybe eventually we can rely on a rule, but we are going to discover the math first. I paired them up with whiteboards and set them out with the list of functions from Sam’s worksheet. Their goal: find the antiderivative of all the functions. The method: each person had a marker. One person would write down a guess for an antiderivative, and the other person would simply take the derivative of this to see if it went back to the original function. They would keep doing this until they got something correct, then write that answer down on the sheet. Then, after one person has been the “guesser’ four or five times, they switch. Example:
For the kids that actually did what I asked (others just kind of started solving them on their own, which is okay I guess), it was a really nice exercise. They worked together really well, and were so excited to tell me the rule that they had made up for integrating power functions. I had them even doing simple substitution, per great suggestion from Sam. They got good at just getting themselves to try something, and getting in the habit of checking all of their answers. One kid at the end of class told me “My brain hurts from thinking so much.” Then, after the students were done, the next class we started by collecting rules they had noticed, and it made a nice little automatic cheat sheet for them. –> SHEET WITH FUNCTIONS HERE
### 2. Power Rule Folding Game
Next was to tackle more complicated functions with which we could use our rules, mainly negative and fractional powers. I did this same exercise in the fall when learning how to differentiate these functions to much success, and then tried it again with differentiating power functions to much confusion (so I guess the activity has a specific niche). The idea is that everyone starts with a problem, does one step and folds over the sheet so that only their work is visible. Then everyone rotates their problems around. The next person does the next step, and then folds the paper so only their work is visible. The group keeps rotating the papers until they are all done, then they open them up and look for mistakes (if there are any). Example:
This was good for helping them drill some algebraic manipulation and develop the skill of checking their own work for mistakes… all while working very closely collaboratively. –> FOLDABLES HERE
### 4. Flip-Up Answers for Initial Conditions
After learning basic integration skills, we began to talk about how functions have more than one antiderivative, and how sometimes it is useful to find a specific one. After only one or two examples together, we immediately just started practicing this idea with an activity that I stole from Mimi where I placed problems around the room with the answers on the back, the idea being that students would go solve whatever problems they felt like they needed to. Example:
I enjoyed this for many of the same reasons that Mimi cited in her original post. Students could work at their own pace without feeling like they were falling behind, could pick their own problems, and could move around the room to interact with many different people (which are all huge advantages over just doing a worksheet). (though the formatting is a bit screwy)
### 5. Mistake Game
After two days of a little bit more traditional style instruction just to make the connection between the definite integral and area (a lesson that I need to make more discovery based next year), we then did the Mistake Game, an idea from Kelly, which I have described a few times now. Basically students work out problems on whiteboards and hide a mistake in their solution. They then present their work like as if they didn’t make a mistake and the other students have a discussion to try to find their error. The problems I chose for the mistake game where all functions for which you had to do some sort of simplifying before integrating (like distributing or dividing), which ended up being a great way of pushing them a little bit forward while giving them plenty of opportunity to really go in depth discussing this new mechanical process of a definite integral.
### 6. Substitution Marker
Then the last skills activity I did with integration was a few days later when we started doing substitution. I had them first try a bunch of substitution problems intuitively, and then showed them how to use a u-substitution. Then, we pulled out the whiteboards and I gave them all a sheet of problems and two markers each. They were to do all u-related work in red, and all original-integral related work in blue. What I wanted them to get comfortable with was envisioning the transition between the variables and helping see how the skeleton of the integral becomes the “outside function” of the backwards chain rule. Example: (actual student work)
This was, again, one of those activities where a bunch of students totally ignored my directions and just solved the problems (and again, not the worst thing), but I think some of the students that did it like this really benefited from using the different colors.
So why did I just ramble about all those activities? I guess what I loved about this whole sequence is how ridiculously much of the instruction for a good week and a half or so was collaborative and engaging, and forced them to think about what could have been routine material in different ways instead of just plowing through worksheets and drills. I feel like I never would have been able to pull something like this off even last year, so I am so grateful (especially to the online community) that I now have a toolbox full of sweet teaching methods. My goal is to try to mix these types of activities more often into various units, since most have skills based components. I would love any other modes of instruction that you use in your classroom to add to my toolbox!!
________________________________________________________
Side note for the Calculus people: There are a few antiderivative/integral related traps that my students fall into… any ideas on how I can stop these problems before they happen?
1. I always start with the word “antiderivative” to emphasis that it’s the opposite process of a derivative, and then try to transition to “integral” as soon as possible, but it’s really tough for them to keep the vocabulary straight. I always correct them in class (mostly just trying to replace antiderivative with integral). How do you approach that vocabulary? I even had a hard time writing this post with the correct vocabulary.
2. Many of my students had a strange barrier this year (that I have never seen before) when finding the area under a curve because they kept thinking of the function you integrate as “the derivative” and the function that you get out as “the original function.” So when we had a function they wanted to find the area underneath, they would take its derivative and then integrate, or some other strange thing like that. How do you introduce the integral as being the opposite of the derivative without getting that misconception (or rather, what did I do in my sequence to imply that)?
3. I always, always, always have so much trouble convincing some students that u-substitution is only used for specific functions that are “backward chain rules.” But after we learn how to integrate normally, we spend a ton of time on u-substitution, and then some students try to solve EVERYTHING with u-substitutions (like 1/x^6 for example). I spent a lot of time doing activities where we pick out the functions that can be integrated with substitution and those that can’t, but for a lot of students, this obviously did not sink in. Any tips?
4. I cannot for the life of me get students to remember to add a “dx” when differentiating a u to find a du. So if u = 2x, then du=2xdx. Granted we didn’t do differentials, but I still don’t understand why this was so difficult! I need some sort of conceptual trigger so they can understand why it’s so important… | 2,032 | 9,788 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2022-21 | latest | en | 0.971435 |
https://livemcqs.com/2021/10/24/cartesian-tree-mcqs/ | 1,685,671,804,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648245.63/warc/CC-MAIN-20230602003804-20230602033804-00404.warc.gz | 417,064,719 | 32,533 | # 50+ Cartesian Tree MCQs with FREE PDF
We have the best collection of Cartesian Tree MCQs and answer with FREE PDF. These Cartesian Tree MCQs will help you to prepare for any competitive exams like: BCA, MCA, GATE, GRE, IES, PSC, UGC NET, DOEACC Exams at all levels – you just have to practice regularly.
## Cartesian Tree MCQs
#### 1. What is the speciality of cartesian sorting?
a) it sorts partially sorted set of data quickly
b) it considers cartesian product of elements
c) it sorts elements in less than O(logn)
d) it is a self balancing tree
Answer: it sorts partially sorted set of data quickly
a) true
b) false
#### 3. What is a Cartesian tree?
a) a skip list in the form of tree
b) a tree which obeys cartesian product
c) a tree which obeys heap property and whose inorder traversal yields the given sequence
d) a tree which obeys heap property only
Answer: a tree which obeys heap property and whose inorder traversal yields the given sequence
#### 4. Consider a sequence of numbers to have repetitions, how a cartesian tree can be constructed in such situations without violating any rules?
a) use any tie-breaking rule between repeated elements
b) cartesian tree is impossible when repetitions are present
c) construct a max heap in such cases
d) construct a min heap in such cases
Answer: use any tie-breaking rule between repeated elements
#### 5. What happens if we apply the below operations on an input sequence?
i. construct a cartesian tree for input sequence
ii. put the root element of above tree in a priority queue
iii. if( priority queue is not empty) then
iv. search and delete minimum value in priority queue
v. add that to output
vi. add cartesian tree children of above node to priority queue
a) constructs a cartesian tree
b) sorts the input sequence
c) does nothing
d) produces some random output
Answer: sorts the input sequence
#### 6. Which of the below statements are true?
i. Cartesian tree is not a height balanced tree
ii. Cartesian tree of a sequence of unique numbers can be unique generated
a) both statements are true
b) only i. is true
c) only ii. is true
d) both are false
Answer: both statements are true
#### 7. Cartesian trees are most suitable for?
a) searching
b) finding nth element
c) minimum range query and lowest common ancestors
d) self balancing a tree
Answer: minimum range query and lowest common ancestors
#### 8. A treap is a cartesian tree with ___________
a) additional value, which is a priority value to the key generated randomly
b) additional value, which is a priority value to the key generated sequentially
c) additional heap rule
d) additional operations like remove a range of elements
Answer: additional value, which is a priority value to the key generated randomly | 642 | 2,793 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2023-23 | latest | en | 0.858431 |
https://mathoverflow.net/questions/149128/how-do-i-solve-this-nonlinear-ode-with-a-fractional-order-term | 1,582,347,165,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145648.56/warc/CC-MAIN-20200222023815-20200222053815-00462.warc.gz | 466,327,388 | 26,829 | How do I solve this nonlinear ODE with a fractional order term
Problem:
Let $p$ and $q$ be two integers, and $q > p>0$. Does the following ODE have a general solution on some finite time interval $[0,T]$? If yes, how can I obtain the solution?
$$\dot x(t) = (x(t))^{p/q} + u(t),$$ where $x(0)=x_0>0$, and $0<u(t)<b<+\infty$ for all $t$.
Based on the problem formulation, $x(t)$ for $t\in[0, T]$ should be bounded. If it is not possible to find a general solution, can we find a tight bound of $|x(t)|$ in terms of $|u(t)|$ and $x_0$?
Here is my thinking:
As the right hand side of ODE satisfy the global Lipschitz condition ($x_0 > 0$) in $x$, so the ODE has a unique solution.
Denote $f(x)=x^{p/q}$. For any constant $\epsilon>0$, $|f(x)|\leq L(\epsilon)|x|+\epsilon$, where $L$ is a constant which depends on $\epsilon$. Therefore, $$|\dot x(t)| \leq L|x(t)| + \epsilon + b,$$ which implies that $|x(t)|\leq e^{Lt}x_0+\frac{1}{L}(e^{Lt}-1)(\epsilon+b)$. In this way, the bound of $x$ depends on $\epsilon$ and $L$, which are not related to the problem. But I wish to find a more accurate bound that only depends on $x_0$ and $u$. I am not sure if it is possible to do so. Anyone can help?
• Your $L$ also depends on $x_0$ (try $x_0:=0$). – Loïc Teyssier Nov 17 '13 at 13:14
• @LoïcTeyssier Thanks for pointing out this mistake in my analysis. Indeed, $L$ depends on the initial condition $x_0$. – Zhang Changhe Nov 18 '13 at 2:22
I understand that your ODE is one-dimensional: in that case you can avoid Lipschitz continuity and replace it by transversality: The autonomous equation $$\dot x =f(x),\quad x(0)=x_0,$$ has a unique local solution provided $f$ is continuous and $f(x_0)\not=0$. Peano's theorem provides existence whereas uniqueness follows follows from separability of the equation, which can be written as $$\frac{dx}{ f(x)} =dt.$$ I know that your problem is not autonomous, but the same direct integration could be used: existence and uniqueness of a positive $C^1$ solution on $[0,T)$ follows from the classical Cauchy-Lipschitz theorem. We have $$x(t)-x_0=\int_0^t (x(s)^{p/q}+u(s))ds=R(t), \tag 1$$ so that $\dot R=x^{p/q}+u\le \bigl(R+x_0\bigr)^{p/q}+b.$ The Gronwall reasoning shows that the solutions of this differential inequality are smaller that the unique solution of the ODE $$\dot R=\bigl(R+x_0\bigr)^{p/q}+b, R(0)=0.\tag 2$$ (2) has a unique solution since $x_0>0$ and is actually separable so can be integrated explicitly. As a result, a bound for $R$ provides a bound for $x$, thanks to (1).
• Thanks so much for your clear and detailed explanation. I get your point. Your method renders a closed bound for $x(t)$. – Zhang Changhe Nov 18 '13 at 2:24 | 857 | 2,692 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2020-10 | latest | en | 0.852052 |
https://www.jiskha.com/questions/103726/Im-not-really-sure-how-i-can-write-the-possible-mole-ratios-of-these-equations-2Hgo-2Hg-o2 | 1,566,594,077,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027318986.84/warc/CC-MAIN-20190823192831-20190823214831-00455.warc.gz | 853,773,532 | 5,199 | # Chemistry 1
Im not really sure how i can write the possible mole ratios of these equations.
2Hgo-->2Hg+o2
and
4NH3+6NO-->5N2+6H20
1. 👍 0
2. 👎 0
3. 👁 225
1. 2 mols HgO to 1 mol O2
1 mol HgO to 1 mol Hg.
It's all done with the coefficients.
1. 👍 0
2. 👎 0
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asked by Masey on November 20, 2014
9. ### chemistry
2Hg+O2=2HgO if 50.0 grams of mercury reacts with 50.0 grams of oxygen gas, how many grams of mercury(ii)oxide are produced?
asked by mom on February 14, 2014
10. ### chemistry
Mercury reacts with oxygen to produce mercury(II) oxide. Hg(l) + O2(g) → HgO(s) a. Balance the reaction equation. Include physical states. 2Hg(l) + O2(g) → 2HgO(s) b. What is the theoretical yield (in grams) of HgO if 255 g of
asked by krystal on February 14, 2012
More Similar Questions | 874 | 2,558 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2019-35 | latest | en | 0.891638 |
https://forum.schizophrenia.com/t/please-convert-to-the-metric-system/151146?page=3 | 1,550,940,648,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249508792.98/warc/CC-MAIN-20190223162938-20190223184938-00427.warc.gz | 566,961,694 | 6,014 | # Please convert to the metric system
#41
#42
this should be in unusual beliefs, tbh i really dont give a damn lol, i just wanted to be heard
next you’ll be saying numbers arent really numbers, switch to the algeraic system
#43
Plus, math is so much more interesting when you use standard.
#44
when i worked in the garage, some of the measurements on the spanners were even older like witworth, it basically means that you can be screwed imperially lol
#45
But when you think about it, clocks are always imperial… and it works because there are so many ways to group and multiply (2, 3,6,12 etc)
Years wouldn’t work on a metric scale (365.25 days)
Moon cycles and months are imperial
And numbers themselves are metric (DECimal)
It’s a crazy world
#46
Wife: “How many feet in your mouth?”
Me: “Both of them.”
#47
The US tried to convert to the metric system back in the 1970’s when Jimmy Carter was president. People fought it tooth and nail. I remember buying gas in liters at one station.
#48
Yeah I remember us learning about the metric system in math class during the 70s for a while.
#49
Both my home country (South Korea) and Canada use the metric system… so I didn’t have to consider the imperial system when I went to Canada. But my cousin and her family are using it because they’re in United States.
#50
I checked and i’m 5 feet and something
I think i’m only 2 feet
As a joke lol not meant to offend.
#51
I could try. But what’s in it for me?
#52
Yes, film footage should become film meterage.
#53
The glory of being the only reasonable US American.
#54
No, that’s the British and their former colonies. Plus a few more cray cray countries. You drive on the correct side.
#55
#56
But @Treebeard, the imperial system is 2.54 times better than the metric system.
I use the metric system at work and the imperial system at home, except when I purchase a 2L bottle of soda.
#57
Oh god those freaking inches. Whenever I talk with someone about TV sizes I have no idea what they’re saying. Like 50 inches? What even is that? What do I have? 12? 90?
Yes, we use inches for screen size. And it sucks and everyone is always confused.
#58
My mom’s car can switch between miles and kilometers. Does that count?
#59
I’m for converting to the metric. Anything else is mentally lazy of us. Just forces us to work our minds until we learn it.
#60
There was a band when I was a kid called “Here Comes The Metric System”
They later changed their name to “Waving At Strangers”
I liked them so much that I bought all their CDs, and still listen to them from time to time. | 648 | 2,608 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-09 | latest | en | 0.944556 |
http://www.mywordsolution.com/question/what-is-the-probability-that-they-are-the-same/943538 | 1,480,708,571,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540563.83/warc/CC-MAIN-20161202170900-00453-ip-10-31-129-80.ec2.internal.warc.gz | 609,107,248 | 8,459 | +1-415-315-9853
info@mywordsolution.com
Algebra Math Calculus Physics Chemistry Biology Earth Science Physiology History Humanities English Sociology Nursing Science
In problem 1, we had equal quantities of the 4 nucleotides, so the random probability of identical nucleotides in two aligned sequences is 0.25 (i.e., p = 0.25). Consider a DNA database in which the ratio of A:C:G:T in the sequences is 0.35:0.15:0.15:0.35 (as would be the case for a genome sequence from an organism has a 30% G+C content), and searching this database with a query sequence with A:C:G:T equal to 0.15:0.35:0.35:0.15. If I take a random nucleotide from the database, and a random nucleotide from the query sequence, what is the probability that they are the same nucleotide? For full credit, you must describe your approach, or show enough work to make clear how you solved the problem.
• Category:- Biology
• Reference No.:- M943538
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Effective HRM Question How can an effective HRM system help facilitate the achievement of an organization's strate | 1,077 | 4,971 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2016-50 | longest | en | 0.910975 |
http://metamath.tirix.org/mpeascii/rmspecpos | 1,718,298,215,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861480.87/warc/CC-MAIN-20240613154645-20240613184645-00819.warc.gz | 18,524,231 | 2,050 | # Metamath Proof Explorer
## Theorem rmspecpos
Description: The discriminant used to define the X and Y sequences is a positive real. (Contributed by Stefan O'Rear, 22-Sep-2014)
Ref Expression
Assertion rmspecpos
`|- ( A e. ( ZZ>= ` 2 ) -> ( ( A ^ 2 ) - 1 ) e. RR+ )`
### Proof
Step Hyp Ref Expression
1 eluzelre
` |- ( A e. ( ZZ>= ` 2 ) -> A e. RR )`
2 1 resqcld
` |- ( A e. ( ZZ>= ` 2 ) -> ( A ^ 2 ) e. RR )`
3 1red
` |- ( A e. ( ZZ>= ` 2 ) -> 1 e. RR )`
4 2 3 resubcld
` |- ( A e. ( ZZ>= ` 2 ) -> ( ( A ^ 2 ) - 1 ) e. RR )`
5 sq1
` |- ( 1 ^ 2 ) = 1`
6 eluz2b1
` |- ( A e. ( ZZ>= ` 2 ) <-> ( A e. ZZ /\ 1 < A ) )`
7 6 simprbi
` |- ( A e. ( ZZ>= ` 2 ) -> 1 < A )`
8 0le1
` |- 0 <_ 1`
9 8 a1i
` |- ( A e. ( ZZ>= ` 2 ) -> 0 <_ 1 )`
10 eluzge2nn0
` |- ( A e. ( ZZ>= ` 2 ) -> A e. NN0 )`
11 10 nn0ge0d
` |- ( A e. ( ZZ>= ` 2 ) -> 0 <_ A )`
12 3 1 9 11 lt2sqd
` |- ( A e. ( ZZ>= ` 2 ) -> ( 1 < A <-> ( 1 ^ 2 ) < ( A ^ 2 ) ) )`
13 7 12 mpbid
` |- ( A e. ( ZZ>= ` 2 ) -> ( 1 ^ 2 ) < ( A ^ 2 ) )`
14 5 13 eqbrtrrid
` |- ( A e. ( ZZ>= ` 2 ) -> 1 < ( A ^ 2 ) )`
15 3 2 posdifd
` |- ( A e. ( ZZ>= ` 2 ) -> ( 1 < ( A ^ 2 ) <-> 0 < ( ( A ^ 2 ) - 1 ) ) )`
16 14 15 mpbid
` |- ( A e. ( ZZ>= ` 2 ) -> 0 < ( ( A ^ 2 ) - 1 ) )`
17 4 16 elrpd
` |- ( A e. ( ZZ>= ` 2 ) -> ( ( A ^ 2 ) - 1 ) e. RR+ )` | 671 | 1,302 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-26 | latest | en | 0.327321 |
https://brendonward.org/newfoundland-and-labrador/monte-carlo-simulation-excel-example-download.php | 1,585,829,679,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506959.34/warc/CC-MAIN-20200402111815-20200402141815-00362.warc.gz | 375,911,470 | 12,259 | ##### Project Management Final Report Example
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21.11.2019 |
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Read the previous post: weisbord six box model example | 1,196 | 5,762 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-16 | latest | en | 0.759343 |
https://math.stackexchange.com/questions/1590691/adjoint-action-on-lie-algebra-su2-a-in-su2-x-in-mathfraksu2-right | 1,718,290,775,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861451.34/warc/CC-MAIN-20240613123217-20240613153217-00578.warc.gz | 348,219,381 | 36,303 | # Adjoint action on Lie algebra su(2) ($A \in SU(2), X \in \mathfrak{su(2)} \Rightarrow AXA^{-1}\in \mathfrak{su(2)}$)
I am trying to understand ho $SU(2)/\{\pm I\} \cong SO(3)$ (see: how to show $SU(2)/\mathbb{Z}_2\cong SO(3)$) but i am not sure about the adjoint action. In especially, as I understand, the adjoint action operates on $\mathfrak{su(2)}$ therefore the following statement should hold: $A \in SU(2), X \in \mathfrak{su(2)} \Rightarrow AXA^{-1}\in \mathfrak{su(2)}$
We know that $SU(2) = \{ \left( \begin{matrix}x & y\\ -\bar y & \bar x \end{matrix}\right): |x|+|y| = 1\}$ furthermore $\mathfrak{su(2)} = span\left( \left( \begin{matrix}i & 0\\ 0 & -i \end{matrix} \right), \left( \begin{matrix}0 & -1\\ 1 & 0 \end{matrix} \right), \left( \begin{matrix}0 & i\\ i & 0 \end{matrix} \right) \right)$.
So i wanted to proof the statement on the basis of $\mathfrak{su(2)}$ (since $A\in SU(2) \Rightarrow A^{-1} = \bar{A}^t$): $AXA^{-1} = \left( \begin{matrix}a+ib & c+id\\ -c+id & a-ib \end{matrix} \right) \left( \begin{matrix}i & 0\\ 0 & -i \end{matrix} \right) \left( \begin{matrix}a-ib & -c-id\\ c-id & a+ib \end{matrix} \right) =$ $\left( \begin{matrix}((-a-i b) (c-i d)+(a-i b) (c+i d) & (-a-i b) (a+i b)+(-c-i d) (c+i d)\\ (a-i b)^2+(c-i d)^2 & (a-i b) (-c-i d)+(a+i b) (c-i d))\\ \end{matrix} \right)$
but i can't find a way to write this in the basis of $\mathfrak{su(2)}$.
Am i doing something completely wrong or was my conclusion about the adjoint action wrong?
• You can write this in the basis. Try expanding out these terms. (Sorry I can't show you the steps right now- I'm on my mobile.)
– Max
Commented Dec 28, 2015 at 1:38
An element $A$ of $\mathfrak{su}(2)$ must satisfy $A=-A^*$ and $tr(A)=0$, because, respectively, of the conditions $AA^*=I$ and $\det(A)=1$ (see Lee's Book proposition 5.38, pg 117, to see how to calculate the tangent space. hint: use $\Phi:SL(2,\mathbb{C})\to SL(2,\mathbb{C})$, given by $\Phi(X)=XX^*$, and so $SU(2)=\Phi^{-1}(I)$.
Now, if $A\in SU(2)$ and $X\in\mathfrak{su}(2)$ you have $(AXA^{-1})^*=(A^{-1})^*X^*A^*=-AXA^{-1}$, because $AA^*=I$ and $X=-X^*$. That is, $AXA^{-1}\in\mathfrak{su}(2)$.
With your calculations (you messed up the product $AXA^{-1}$, when you correct) you'll get \begin{aligned} AXA^{-1} &= \left( \begin{matrix}a+ib & c+id\\ -c+id & a-ib \end{matrix} \right) \left( \begin{matrix}i & 0\\ 0 & -i \end{matrix} \right) \left( \begin{matrix}a-ib & -c-id\\ c-id & a+ib \end{matrix} \right)\\ &=(a^2+b^2-c^2-d^2)\left( \begin{matrix}i & 0\\ 0 & -i \end{matrix} \right)+(-2ad-2bc)\left( \begin{matrix}0 & -1\\ 1 & 0 \end{matrix} \right)+(-2ac+2bd)\left( \begin{matrix}0 & i\\ i & 0 \end{matrix} \right) \end{aligned} | 1,099 | 2,701 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-26 | latest | en | 0.60198 |
https://www.cs.odu.edu/~mweigle/CS312-F11/Quiz2 | 1,537,545,474,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267157216.50/warc/CC-MAIN-20180921151328-20180921171728-00206.warc.gz | 716,514,290 | 3,729 | CS 312 - Internet Concepts Fall 2011: Tues/Thurs 3-4:15pm, Dragas 1117
Staff
# Quiz 2 - HTTP Connections, RTTs, and Delays
Thursday, Oct 13, 2011
Question 2 is an actual question from the 2010 mid-term exam.
• http://www.gomonarchs.com/banner.jpg
• http://www.odu-football.com/photos/coach.jpg
This question will consider the time it takes to download the entire main web page for www.gomonarchs.com from a remote web browser. The round trip time from the browser to any server can be represented by RTT. Transmission and queuing delays are ignored. You may assume that the IP addresses for all of the hostnames are already cached.
1. If persistent HTTP connections are used, how many TCP connections are needed to download the entire web page (base page, plus each embedded object)?
2. If persistent HTTP connections are used, how many RTTs would it take to download the entire web page? Explain the purpose of each RTT. Show your work.
3. If each object (including the base page) has a size of 1000 bytes, the access link is 4 Mbps, and the RTT to each server is 50 ms, what is the total download time for the entire web page if using persistent HTTP connections? For this question, you must calculate transmission and propagation delays. Show your work.
### Solutions
1) 3 TCP connections:
1. gomonarchs.com
2. odu-football.com
2) 7 RTTs:
1. handshake gomonarchs.com
2. request/response (req/rsp) base page
3. req/rsp banner
4. handshake odu-football.com
5. req/rsp coach
7. req/rsp tickets
3) 358 ms
``` dprop = number of RTTs * RTT
dprop = 7 * 50 ms = 350 ms
dtrans for one object = L/R
L = 1000 B = 8,000 b
8,000 b * 1000 ms 8
---------- = --- ms = 2 ms
4,000,000 b 4
dtrans = number of objects * dtrans for one object
dtrans = 4 * 2 ms = 8 ms
de2e = dtrans + dprop
de2e = 8 ms + 350 ms = 358 ms
``` | 529 | 1,843 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2018-39 | latest | en | 0.779115 |
https://learn.careers360.com/school/question-explain-solution-r-d-sharma-class-12-chapter-21-differential-equations-exercise-21-point-8-question-11-maths-textbook-solution/?question_number=11.0 | 1,716,609,851,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058773.28/warc/CC-MAIN-20240525035213-20240525065213-00012.warc.gz | 291,549,719 | 41,780 | #### Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise 21.8 Question 11 Maths Textbook Solution.
Answer: $-e^{-\left ( x+y \right )}=x+c$
Given: $\frac{dy}{dx}+1=e^{x+y}$
Hint: Differential equation of the form $\frac{dy}{dx}=\int \left ( ax+by+c \right )$ can be reduced to variable separable form by substitution $ax+by+c=v$
Solution: We have,
$\frac{dy}{dx}+1=e^{x+y}$ .......(i)
Let $x+y=v$
Differentiating with respect to x, we get,
$\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}+\mathrm{y})=\frac{\mathrm{d} \mathrm{v}}{\mathrm{d} \mathrm{x}} \\$
$\Rightarrow 1+\frac{d y}{d x}=\frac{\mathrm{d} \mathrm{v}}{\mathrm{d} \mathrm{x}} \quad \cdots \text { (ii) } \\$
Substituting (ii) in equation (i), we get,
$\frac{\mathrm{d} \mathrm{v}}{\mathrm{d} \mathrm{x}}=\mathrm{e}^{\mathrm{x}+\mathrm{y}} \\$
$\Rightarrow \frac{\mathrm{d} \mathrm{v}}{\mathrm{d} x}=\mathrm{e}^{\mathrm{v}} \quad[\because \mathrm{x}+\mathrm{y}=\mathrm{v}]$
Taking like variables on the same side, we get,
$\Rightarrow \frac{\mathrm{d} v}{\mathrm{e}^{\mathrm{v}}}=\mathrm{d} \mathrm{x} \\$
$\Rightarrow \mathrm{e}^{-\mathrm{v}} \mathrm{d} \mathrm{v}=\mathrm{d} \mathrm{x} \\$
Integrating on both sides, we get,
$\Rightarrow \int \mathrm{e}^{-\mathrm{v}} \mathrm{d} \mathrm{v}=\int \mathrm{d} \mathrm{x} \\$
$\Rightarrow-\mathrm{e}^{-\mathrm{v}}=\mathrm{x}+\mathrm{c} \\$
Putting $v=x+y$, we get,
$\Rightarrow-\mathrm{e}^{-(x+y)}=\mathrm{x}+\mathrm{c}$
(This is the required solution). | 536 | 1,527 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 16, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2024-22 | latest | en | 0.269179 |
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Actions of $Z_n$ and actions of $Z_{n-1}$
I am playing with some questions concerning connections between certain poset partitions and their linear extensions. This is not my usual playground, I just happened to stumble upon something.
When I was playing with these things, I came up with a very simple construction that I cannot find anywhere.I suspect that this construction is a minor particular case of something well known and much more general. It reminds of an equivariant map, but the group is not fixed here.
Let $n>2$, let $X$ be a finite set of cardinality $n$. Let $\oplus_n$ be an action of $Z_n$ on $X$ such that the action of $1$ is a cycle of length $n$. Pick $x\in X$ and let $\pi$ be a partition of the set $X$ such that $\{x,x\oplus_n 1\}$ is the only nonsingleton block of $\pi$.
In a very simple way, this gives rise to an action $\oplus_{n-1}$ of $Z_{n-1}$ on $\pi$ by ,,squeezing the action'' at $x$:
Put
• $\{x,x\oplus_n 1\}\oplus_{n-1} 1=\{x\oplus_n 2\}$
• $\{x\oplus_n(n-1)\}\oplus_{n-1} 1=\{x,x\oplus_n 1\}$
• $\{y\}\oplus_{n-1} 1=\{y\oplus_n 1\}$ for any $y\neq x\oplus_n(n-1)$.
It can be visualized in a simple way by a digraph construction: if we identify the action of $1$ on $X$ with an oriented cycle, this construction corresponds to a contraction of an edge.
Has anyone seen this construction before? Is there any name for it?
-
Seems to me this could fail if n is not prime power. A faithful action corresponds to a set of cycles with LCM equal to n, not necessarily a single cycle. – Omer Apr 9 2011 at 14:55 You are right, of course. I have mixed up "regular" and "faithful". I have edited the question. – Gejza Jenča Apr 9 2011 at 21:05
I might be missing something, but it seems to me that there is not much going on in your construction. In fact, your original action of $Z_n$ on $X$ does nothing more than putting a cyclic ordering on your set $X$, and your complicated-looking construction creates a new cyclically ordered set $Y$ obtained from $X$ by removing one element (and retaining the ordering). | 639 | 2,210 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2013-20 | latest | en | 0.902116 |
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# How to create a boxplot from a PDF?
Asked by Janett Göhring on 22 Apr 2013
Hello!
I have a somewhat embarrassing question, but me and my colleagues cannot figure it out since several days. Thinking block ^^ So I would appreciate help!
I have a pdf of my data called pdfxcor (598x1), which resembles a normal distribution when I plot it along a x-axis resembling the molecular weight of my data (called pixelweight (598x1)).
```plot(pixelweight,pdfxcor)
```
This is the plot: http://imageshack.us/a/img27/1694/ploti.png
```boxplot(pdfxcor)
```
The same data as boxplot: http://imageshack.us/a/img812/7528/boxplot.png
I want to display the distribution as boxplot according to the correct molecular weight.
Jette
## Products
No products are associated with this question.
Answer by Teja Muppirala on 23 Apr 2013
How about something like this. Generate the CDF from your data as Tom suggested, invert it, use the inverted CDF to generate a bunch of samples that follow your distribution exactly, and send those to BOXPLOT:
```%% Just making some data that resembles yours
x = linspace(1000,12000,598);
P = normpdf(x,5800,1800);
figure, plot(x,P), title('PDF');
```
```%% Generate the CDF
C = cumsum(P);
C = C/C(end);
figure, plot(x,C); title('CDF');
```
```%% Sample linearly along the inverse-CDF to get a bunch of points
% that have your same distribution
BigNumber = 100000;
p = interp1(C,x,linspace(C(1),C(end),BigNumber));
figure, hist(p,100); % Confirm p indeed has your distribution
figure ,h = boxplot(p);
delete(findobj(h,'tag','Outliers')) % Hide the outliers
```
Janett Göhring on 23 Apr 2013
Hi Teja and Tom,
Both are really nice solutions, but I still run into one problem with my data.
This is the cdf of my data: http://imageshack.us/a/img835/5130/cdfc.png
The histogram: http://imageshack.us/a/img41/1253/histc.png
And the resulting boxplot: http://imageshack.us/a/img138/1534/newboxplot.png
So strangly, the histogram doesn't resemble the pdf plotted against the pixelweigth. I get the same result with the inverted distribution.
Tom Lane on 23 Apr 2013
It looks like your distribution is not symmetric. The normal distribution is symmetric, so it would not resemble the histogram in that respect.
Janett Göhring on 23 Apr 2013
Hi Tom,
the curve was calculated via a Gaussian fit and is symmetric. The x-axis though is based on data, which was fitted with nlinfit and looks like a logarithmic decay. So, after correction the x-axis is not linear anymore. That's why it is so important to plot the pixelweigth against the pdf, otherwise the distribution is not symmetric anymore.
```modelFun = @(p,x) p(1)*exp(p(2)*x);
```
In between, I calculate start parameters for the fit, which is not important for the example.
Next, I fit the pixel position and the Molecular weight of the DNA standard.
```p = nlinfit(positionOfStandard, MWOfStandard, modelFun, paramEstsLin(:,1));
```
The pixelrange is just the y-length of my image in pixel. Here 1:598
```pixelweigth = p(1)*exp(p(2)*pixelrange);
```
After lots of corrections of the original data I fit a Gauss fit through it and calculate the curve, mean and sigma.
```cf3 = fit(pixelweigth',data','gauss1');
pdfxcor = cf3(pixelweigth)
```
After that I need a representation of the normal distributed data along this specialized x-axis (pixelweigth). But not as a curve ... I was asked to display it as a boxplot. And since it is a normal distribution, I thought it must be possible. But Matlab doesn't give an option in "boxplot" to specify a different axis.
thanks for the help! much appreciated :)
Answer by Tom Lane on 22 Apr 2013
The boxplot shows the median, lower quartile, and upper quartile. You may be able to calculate these for your pdf. For example, if you have the pdf as a numeric vector, you might compute cumsum on the vector, then divide by the last value to impose the correct probability normalization, then interpolate.
The boxplot also shows a notion of the range of the data, and sometimes outliers. These are harder to extend to a pdf. You could decide that you want to compute the 1% and 99% points as in the previous paragraph, and use those to represent the end points of the range. You could decide not to show outliers.
Plotting these as lines or points will be relatively simple. It would be more of a challenge to plot them in exactly the way that the boxplot function does.
## 1 Comment
Janett Göhring on 23 Apr 2013
Hello Tom, thanks for your answer! Can you explain how to interpolate in this case?
For my problem I created two solutions, but I don't like both.
a) I gauss fit my original data to create the pdf, mean and sigma. Then, I sample with randn (1Mio) & the mean and sigma as parameters. This creates a normal distribution based on my fit which can be plotted via boxplot. Since I already fit my original data with a gaussfit, I am not very interested in the outliers. I just was asked to represent the normal distribution as boxplot for easier comparison of mean and range of data. So, I would feel much better when I wouldn't have to sample a new distribution and of course it takes ages to calculate.
b) I calculate mean and the quartiles of the pdf and extract the respective position from the pixelweigth. Then I draw a barplot(colored for the upper quartile and white for the lower quartile) with error bars. I couldn't make this work, since the pdf is only normally distributed when it is plotted against the pixelweigth.
Bit stuck there ^^ Thanks for your help! Jette | 1,384 | 5,594 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2014-15 | latest | en | 0.860695 |
https://promisekit.org/2022/12/18/how-much-weight-will-i-gain-eating-3000-calories-a-day/ | 1,716,543,844,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058709.9/warc/CC-MAIN-20240524091115-20240524121115-00871.warc.gz | 415,008,155 | 12,564 | ## How much weight will I gain eating 3000 calories a day?
For some people, a 3,000-calorie may help you gain weight. An acceptable, safe rate of weight gain is 0.5–2 pounds (0.2–0.9 kg) per week.
How much protein do I need for 3000 calories?
10 percent of 3000 calories is equivalent to 300 calories. Since every four calories are equivalent to 1 gram of proteins, 300 calories are equivalent to 75 grams of proteins.
What does a 3000 calorie diet consist of?
To fit in all the calories, you need to eat three meals and three snacks a day on your 3000-calorie diet. A healthy and balanced 3000-calorie diet includes 10 ounces of grains, 4 cups of vegetables, 2 1/2 cups of fruit, 3 cups of dairy and 7 ounces of protein foods.
### Is 3300 calories enough to bulk?
The lean bulk Start by eating 5-10% more than your maintenance calories. This will equate to a few hundred calories over maintenance levels. Example: Your maintenance calories are 3,000. To lean bulk, you should start by eating between 3,150 and 3,300 calories per day.
Is 3000 calories a day enough to build muscle?
4. So a 200 lbs bodybuilder, working out hard needs to eat 3500 calories a day to gain muscle mass.
How much sugar should you have a day on a 3000 calorie diet?
For a 2,000-calorie diet, the limit works out to 12.5 teaspoons, or 50 grams, of added sugar. And for the 3,000-calorie diet, you’re looking at 18.75 teaspoons, or 75 grams.
## Can you build muscle on 3000 calories?
4. So a 200 lbs bodybuilder, working out hard needs to eat 3500 calories a day to gain muscle mass. The other thing I want to stress is healthy foods.
Is eating 3000 calories a day healthy?
According to the 2020-2025 Dietary Guidelines for Americans, people assigned male at birth who are between the ages of 15 and 35 and are active need at least 3,000 calories a day to maintain a healthy weight.
Is 3000 calories burned in a day good?
Extreme endurance athletes, such as Ironman competitors, ultramarathon runners and Olympic athletes, can easily burn 3,000 calories a day from exercise, but these are people who are able to devote several hours each day to training. For the average exerciser, a goal of 3,000 calories a day may be unrealistic.
### How much exercise does it take to burn 3000 calories?
Running: A very effective activity, running burns 850 calories an hour. However, you would need to run for 4 hours to burn 3000 calories. Cross Country Skiing: An intense one hour session burns 1100 calories. Swimming: Swimming intensely for an hour helps you burn 700 calories.
What food is highest in calories?
Calories per Cup Calories per 100g
597 calories 489 calories
How many steps burns 3000 calories?
Height 6 Feet and Above
2,000 Steps per Mile (Height 6 Feet and Above) Calories Burned by Step Count and Weight
Steps 45 kg 100 kg
1,000 28 cal. 60
2,000 55 120
3,000 83 180
## Is it possible to burn 3500 calories in a day?
It is possible to burn 3,500 calories a day (or lose a pound in one day) by creating an energy deficit. To do so, you must eat less and work out more or at a higher intensity. However, that weight loss would be from both diet and exercise, not just one 3,500-calorie-burning workout.
Is 4k calories a day good?
According to the U.S. Department of Agriculture’s 2010 Dietary Guidelines for Americans, average daily calorie needs are 1,600 to 2,000 for women and 2,000 to 3,000 for men. A 4,000-calorie diet should be limited to highly active individuals, or those affected by a health condition, with a doctor’s recommendation.
How can I get 3500 calories a day?
3500-Calorie Diet
1. Eating More Frequently.
2. Eating Foods To Help You Gain Weight Quickly. Milk. Rice. Red Meat. Whole Grain Cereals. Salmon.
3. Taking Smoothies.
4. Developing A Meal Plan.
6. Eating More Fruits.
### How much is 3500 calories in kg?
A pound of fat corresponds to roughly 3,500 calories, so losing a kilogram of weight per week means cutting 7,700 calories from your diet each week – or 1,100 each day.
Can you gain 5 pounds in a week?
“It is unlikely you gained five pounds of fat in one week,” he says. “But if it is justified, own it.” Whatever the culprit is, there are easy things you can do to fix it.
What does a 3000 calorie diet look like?
Here’s what 5 days on a 3,000-calorie diet may look like. Dinner: turkey chili made with a 4-ounce (114-gram) turkey breast, chopped onions, garlic, celery, and sweet peppers, 1/2 cup (123 grams) of canned, diced tomatoes, and 1/2 cup (120 grams) of cannellini beans, topped with 1/4 cup (28 grams) of shredded cheese.
## How fast do you gain weight on a 3000 calorie diet?
How fast you gain weight depends on how many calories you need to maintain your weight. If you maintain your weight on 2,000 calories per day, you will gain weight much quicker on a 3,000-calorie diet than someone who maintains their weight on 2,500 calories per day.
How many macronutrients do you need to eat 3000 calories a day?
The recommended percentage breakdown of macronutrients for adults is: 10 to 35 percent of a daily calories from protein, 45 to 65 percent from carbohydrates and 20 to 35 percent from fats. The amount of different macronutrients required when consuming 3000 calories a day will vary, based on the macronutrient in question.
Do nuts make you gain weight on a 3000 calorie diet?
While simply eating more calories sounds easy, it’s also important to choose the right foods so you gain healthfully. A 3,000-calorie diet is an eating plan that can help most men and women gain weight. High in calories and rich in nutrients, nuts make a healthy snack on your high-calorie diet. | 1,440 | 5,650 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-22 | latest | en | 0.927819 |
http://mrhonner.com/archives/8396 | 1,511,306,370,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806438.11/warc/CC-MAIN-20171121223707-20171122003707-00580.warc.gz | 201,198,000 | 13,772 | ## Math and Weaving — Belt Weaves
Here are some examples of what I call “belt weaving” (I’m sure there’s a better term). The basic idea is to begin with long strips of construction paper, oriented perpendicularly, and then weave and fold your way down.
Here are two examples of 2×2 belt weaves. In both cases, the same kinds of strips are used, but in a different initial configuration.
The 3×3 belt weaves offer more initial configurations, and show more complexity.
There is a rich and interesting structure to explore in these “belt weaves”. For example, these two weaves look similar, but are indeed different.
My students and I had fun exploring the mathematical relationships between the various belt weaves. I will share some of our ideas and results in my series on Weaving in Math Class. | 187 | 806 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2017-47 | latest | en | 0.939924 |
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# For systems with memory also known as dynamic systems
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Unformatted text preview: h old, t he signals are functions of space a nd time; such a system is a d istributed-parameter system. 6. S ystems w hose inputs a nd o utputs a re continuous-time signals are continuoustime systems; systems whose inputs and o utputs are discrete-time signals are discrete-time systems. I f a continuous-time signal is sampled, t he resulting signal is a d iscrete-time signal. We can process a continuous-time signal by processing t he samples of this signal with a discrete-time system. 7. S ystems w hose inputs a nd o utputs a re analog signals are analog systems; those whose i nputs a nd o utputs are digital signals are digital systems. 8. I f we c an o btain t he i nput / (t) back from t he o utput y (t) of a system S by some o peration, t he system S is s aid to be invertible. Otherwise t he system is n oninvertible. T he s ystem model derived from a knowledge of t he i nternal s tructure of the system is its i nternal description. In contrast, a n external description of a system is i ts description as seen from t he s ystem's i nput a nd o utput terminals; it can be obtained by a pplying a known input a nd measuring the resulting o utput. I n t he m ajority of p ractical systems, a n e xternal description of a system so obtained is equivalent t o i ts i nternal description. In some cases, however, t he e xternal description fails t o give adequate information a bout t he system. Such is t he case with t he so-called uncontrollable or unobservable systems. References 1. 2. 3. 4. P apoulis, A ., T he Fourier Integral and Its Applications, McGraw-Hill, New York, 1962. K ailath, T ., L inear Systems, Prentice-Hall, Englewood Cliffs, New Jersey, 1980. L athi, B .P., Signals, Systems, and Communication, Wiley, New York, 1965. Lathi, B .P., Signals and Systems, Berkeley-Cambridge Press, Carmichael, California, 1987. Problems Find the energies of the signals illustrated in Fig. P I.I-I. Comment on the effect on energy of sign change, time shifting, or doubling of the signal. What is the effect on the energy if the signal is multiplied by k? 1 .1-2 Repeat Prob. 1.1-1 for the signals in Fig. P1.1-2. 1 .1-3 ( a) Find t he energies of the pair of signals x (t) and y(t) depicted in Figs. P l.l-3a and b. Sketch and find the energies of signals x (t) + y(t) and x (t) - y(t). Can you make any observation from these results? 97 Problems (b) (a) ( d) (c) sintr\ F ig. P l.l-l f (t) 0 O t- ~ o 2 t- f 3 ( t) 1 o 2 t- 0 -1 F ig. P l.l-2 x (t) ~ o 0 2 (a) x (t) 0 1 1- -1 1- y (l) t- 0 I- (b) x (t) y (t) 1 .1-1 I- (c) 0 It I- Fig. P l.l-3 ( b) Repeat part (a) for the signal pair illustrated in Fig. P1.1-3e. Is your observation in p art (a) still valid? 98 1 .1-4 1 I ntroduction t o S ignals a nd S ystems 99 P roblems F ind t he p ower o f t he p eriodic signal f (t) s hown in Fig. P1.1-4. F ind also t he powers a nd t he r ms values of: ( a) - f(t) ( b) 2 f(t) ( c) c f(t). C omment. f (1) 0.5 o 1- 24 -I ..... t- F ig. P 1.3-2 Fig. P l.l-4 1 .1-5 S how t hat t he p ower o f a signal 1- n f (t) = L Dke jwkt is Fig. P 1....
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# Area between 2D cartesian curves, fill between two ordinates, for calculus
Posted 11 years ago
I have not been successful in finding Help information for shading the area between two 2D cartesian curves, from one ordinate x-value to another. For example, displaying graphically the filled area between the graph of y=sinx and y=cosx from x=?/4 to x= 5?/4. Or the filled area between the graph of y=sinx and the x-axis from x=?/4 to x= 5?/4.
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Posted 11 years ago
On the help page for Filling scroll down and click on Scope and then on FillingStyle and it gives the first examplePlot[{Sin[x], Cos[x]}, {x, Pi/4, 5 Pi/4}, Filling->{1->{{2}, Yellow}}]
Posted 11 years ago
Thank you Bill. What I would like to do is shade just the area from ?/4 to 5?/4, for domain 0 to 2?. My code so far as follows:Plot[{Sin[x], Cos[x]}, {x, 0, 2 Pi}, Filling -> {1 -> {2}}, FillingStyle -> Yellow, AxesLabel -> {"x", "y"}, Ticks -> {{Pi/4, 5 Pi/4, 2 Pi}, {-1, 1}}, GridLines -> {{{Pi/4, {Red, Dashed, Thickness[0.003]}}, {5 Pi/4, {Red, Dashed, Thickness[0.003]}}, 2 Pi}, {-1, -Sqrt[2]/2, Sqrt[2]/2, 1}}]
Posted 11 years ago
There certainly might be a better way of doing this, but does this give you enough of a hint that you can modify this to get what you are looking for? Show[{ Plot[{Sin[x], Cos[x]}, {x, 0, Pi/4}, PlotRange -> {{0, 2 Pi}, {-1.01, 1}}], Plot[{Sin[x], Cos[x]}, {x, Pi/4, 5 Pi/4}, Filling -> {1 -> {2}}, FillingStyle -> Yellow, AxesLabel -> {"x", "y"}, Ticks -> {{Pi/4, 5 Pi/4, 2 Pi}, {-1, 1}}, GridLines -> {{{Pi/4, {Red, Dashed, Thickness[0.003]}}, {5 Pi/4, {Red, Dashed, Thickness[0.003]}}, 2 Pi}, {-1, -Sqrt[2]/2, Sqrt[2]/2, 1}}], Plot[{Sin[x], Cos[x]}, {x, 5 Pi/4, 2 Pi}]}]
Posted 11 years ago
Thank you, very valuable, will continue my efforts.
Posted 11 years ago
You can find the dynamic sweeping area with the following code. It basically lay the filled area above the plot with curves only Manipulate[ Show[ Plot[{Sin[x], Cos[x]}, {x, Pi/4, k}, Filling -> {1 -> {{2}, Yellow}}], Plot[{Sin[x], Cos[x]}, {x, Pi/4, 5 Pi/4}], PlotRange -> {{Pi/4, 5 Pi/4}, {-1, 1}}, Axes -> None, Epilog -> Inset[Row[{"Area is:", NIntegrate[Sin[x] - Cos[x], {x, Pi/4, k}]}], {2.1, 0.1}] ], {k, Pi/4 + 0.001, 5*Pi/4}]
Posted 11 years ago
Thank you Shenghui, beautiful. | 865 | 2,430 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-33 | latest | en | 0.763987 |
https://www.profumiesaporidellamiacucina.it/equipment2/1050-how-to-find-fineness-modulus.html | 1,618,418,099,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038077843.17/warc/CC-MAIN-20210414155517-20210414185517-00025.warc.gz | 1,070,405,532 | 9,400 | # how to find fineness modulus
#### What is Fineness Modulus of Sand Fine Aggregate and Calculation
Therefore, fineness modulus of aggregate = cumulative % retained / 100 = 275/100 = 2.75. Fineness modulus of fine aggregate is 2.75. It means the average value of aggregate is in between the 2 nd sieve and 3 rd sieve. It means the average aggregate size is in between 0.3mm to 0.6mm as shown in below figure.
#### What is Fineness Modulus and how to calculate it
How to calculate Fineness Modulus Sieve the aggregate using the appropriate sieves 80 mm, 40 mm, 20 mm, 10 mm, 4.75 mm, 2.36 mm, 1.18 mm, 600 micron, 300... Note down the weight of aggregate retained on each sieve. Calculate the cumulative weight of aggregate retained on each sieve. Find out the ...
#### Fineness modulus - Wikipedia
The Fineness Modulus FM is an empirical figure obtained by adding the total percentage of the sample of an aggregate retained on each of a specified series of sieves, and dividing the sum by 100.
#### What is Fineness Modulus and How to Calculate It
How to calculate Fineness Modulus 1. Sieve the aggregate using the appropriate sieves 80 mm, 40 mm, 20 mm, 10 mm, 4.75 mm, 2.36 mm, 1.18 mm, 600 micron,... 2. Note down the weight of aggregate retained on each sieve. 3. Calculate the cumulative weight of aggregate retained on each sieve. 4. Find ...
#### Fineness Modulus - Islamic University of Gaza
To find fineness modulus of coarse aggregate we need sieve sizes of 75mm, 37.5mm, 19mm, 9.5mm, 4.75mm, 2.36mm, 1.18mm, 0.6mm, 0.3mm and 0.15mm. Fineness modulus is the number at which the average size of particle is known when we counted from lower order sieve size to higher order sieve.
#### Fineness modulus and its calculation – ProBCGuide
How to calculate Fineness Modulus. Procure to determine Fineness Modulus. Sieve the aggregate using the appropriate sieves 80 mm, 40 mm, 20 mm, 10 mm, 4.75 mm, 2.36 mm, 1.18 mm, 600 micron, 300 micron and 150 micron Note down the weight of aggregate retained on each sieve. Calculate the cumulative weight of aggregate retained on each sieve.
#### Determination of Fineness Modulus of a Sample of Fine ...
The fineness modulus of a sample of aggregate is an index number which is roughly proportional to the average size of particle in the aggregate, that is, coarser the aggregate the higher the fineness modulus. It is computed by adding the cumulative percentages coarser that each of the following IS test sieve 10 mm, 4.75 mm, 2.63 mm, 1.18 mm, 600 Micron, 300 Micron, 150 Micron and dividing the ...
#### Fineness modulus - Wikipedia
The same value of fineness modulus may therefore be obtained from several different particle size distributions. In general, however, a smaller value indi es a finer aggregate. Fine aggregates range from a FM of 2.00 to 4.00, and coarse aggregates smaller than 38.1 mm range from 6.75 to 8.00. Combinations of fine and coarse aggregates have intermediate values.
#### CRD-C104-80 Method of Calculation of the Fineness Modulus of ...
in order to calculate fineness modulus, the following procedure can be used to calculate fineness modulus using percentages passing. The example is for the same gradings as in para 4.1 but when they are expressed as percentages passing. Sieve Sizes, U. S. ...
#### Fineness Modulus - Islamic University of Gaza
modulus hence fineness modulus of coarse aggregate is higher than fine aggregate. To find fineness modulus of coarse aggregate we need sieve sizes of 75mm, 37.5mm, 19mm, 9.5mm, 4.75mm, 2.36mm, 1.18mm, 0.6mm, 0.3mm and 0.15mm. Fineness modulus is the number at which the average size of particle is known when
#### Grain Fineness Number Calculator Fineness Modulus of Sand ...
Fineness modulus of sand fine aggregate is an index number which represents the mean size of the particles in sand. It is calculated by performing sieve analysis with standard sieves. The cumulative percentage retained on each sieve is added and subtracted by 100 gives the value of fineness modulus.
#### Fineness Modulus of Aggregates: The Inside Scoop - Gilson Co.
Fineness Modulus of Sand. To calculate the fineness modulus of sand, the sum of the cumulative percentages retained on the following sieves are divided by 100: 150μm No. 100 , 300μm No. 50 , 600-μm No. 30 , 1.18mm No. 16 , 2.36mm No. 8 , and 4.75mm No. 4 for fine aggregates. If FM for coarse aggregates is desired, then sieves of 9.5mm ...
#### PDF METHOD OF CALCULATION OF THE FINENESS MODULUS OF ...
METHOD OF CALCULATION OF THE FINENESS MODULUS OF AGGREGATE
#### Fineness modulus - Wikipedia
The same value of fineness modulus may therefore be obtained from several different particle size distributions. In general, however, a smaller value indi es a finer aggregate. Fine aggregates range from a FM of 2.00 to 4.00, and coarse aggregates smaller than 38.1 mm range from 6.75 to 8.00. Combinations of fine and coarse aggregates have intermediate values.
#### Excel Sheet For Sieve Analysis Of Aggregate And To Calculate ...
For any type of mix proportioning sieve analysis results and fineness modulus is required. here is the excel sheet which includes calculations of fineness modulus. Hope it will consume less time in case of repetitive calculations. Excel Sheet For Sieve Analysis Of Aggregate And To Calculate Fineness Modulus- engineeringcivil.com
#### How to find the fineness modules of an aggregate - Quora
fineness modulus gives the average size of aggregate fineness modulus gives the number of sieve counted from the bottom. Sieve size of that is the size of aggregate Fineness modulus = sum of cumulative percentage retained / 100 Suppose if the fine...
#### What is fineness modulus? Civil4M
fineness modulus in size in a ranted in 4.75 Coarse aggregate means the aggregate which is retained on 4.75mm sieve when it is sieved through 4.75mm. To find fineness modulus of coarse aggregate we need sieve sizes of 80mm, 40mm, 20mm, 10mm, 4.75mm, 2.36mm, 1.18mm, 0.6mm, 0.3mm and
#### The Importance of Fineness Modulus Concrete Construction ...
The premise: "aggregate of the same fineness modulus will require the same quantity of water to produce a mix of the same consistnecy and give a concrete of the same strength." Before calculating FM, lab technicians perform a sieve analysis to determine the particle size distribution, or grading, of the aggregate sample. FM is the sum of the total percentages retained on each specified sieve ...
#### CHAPTER 2 SIEVE ANALYSIS AND FINENESS MODULUS Sampling
SIEVE ANALYSIS AND FINENESS MODULUS Sampling Since the reason for sampling aggregates is to determine the gradation particle size of the aggregate, it is necessary that they be sampled correctly. The results of testing will re" ect the condition and characteristics of the aggregate from which the sample is obtained. Therefore, when sampling ...
#### What is the Fineness of Cement It& 39;s Test and Effects ...
What is the fineness of the cement? A Fineness of Cement is Measure of the size of Cement Particle, and it is described in terms of the specific surface area of Cement particle. The Fineness of cement can be Calculated by particle size analysis, by using the air permeability method, by using sedimentation method.
#### PDF METHOD OF CALCULATION OF THE FINENESS MODULUS OF ...
METHOD OF CALCULATION OF THE FINENESS MODULUS OF AGGREGATE
#### Concrete Mix Ratio, Types, Proportioning of ... - Civil Lead
Pc = Fineness modulus of coarse aggregate P f = Fineness modulus of fine aggregate Pm = Fineness modulus of mix aggregate. Minimum voids Method. The minimum void method is based on the principle that the concrete with the minimum voids is the densest and most durable. The amount of fine aggregate should be such as to fill the voids. Conclusion
#### Solved: Calculate the fineness modulus of aggregate A in ...
Calculate the fineness modulus of aggregate A in problem 5.23. Is your answer within the typical range for fineness modu Is your answer within the typical range for fineness modu 60 % 500 Review
#### What is fineness modulus? Civil4M
fineness modulus in size in a ranted in 4.75 Coarse aggregate means the aggregate which is retained on 4.75mm sieve when it is sieved through 4.75mm. To find fineness modulus of coarse aggregate we need sieve sizes of 80mm, 40mm, 20mm, 10mm, 4.75mm, 2.36mm, 1.18mm, 0.6mm, 0.3mm and
#### Procedure of calcultion of fineness modulus of coarse aggregate
Procedure of calcultion of fineness modulus of coarse aggregate Products. As a leading global manufacturer of crushing, grinding and mining equipments, we offer advanced, reasonable solutions for any size-reduction requirements including, Procedure of calcultion of fineness modulus of coarse aggregate, quarry, aggregate, and different kinds of minerals.
#### Solved: Calculate The Fineness Modulus For The Following F ...
Question: Calculate The Fineness Modulus For The Following Fine Aggregate. Is This Aggregate Finer Or Coarser Than The One Shown In Table 5.7 In Your Text? Sieve Size Amount Retained g 3/8 In. 0.0 No. 4 11.5 No. 8 38.3 No. 16 95.2 No. 30 156.8 No. 50 135.3 No. 100 98.1 Pan 12.0
#### What is the Fineness of Cement It& 39;s Test and Effects ...
What is the fineness of the cement? A Fineness of Cement is Measure of the size of Cement Particle, and it is described in terms of the specific surface area of Cement particle. The Fineness of cement can be Calculated by particle size analysis, by using the air permeability method, by using sedimentation method.
#### Fineness test of Cement and its Significance Tests on Cement
Fineness test of Cement:-As per IS: 4031 Part 1 – 1996. The cement of good quality should have less than 10% of wt of cement particles larger than 90 µm. micron To determine the number of cement particles larger than 90 µm. or Fineness test of cement. The following apparatus is used.
#### A Detailed Guide on Grading of Aggregates.
Cumulative percentage of weight retained on all the six sieves = 292 Fineness Modulus of Fine Aggregates = 292/100 =2.92 kg. Now, when it is desired to obtain a combined aggregate of a definite required fineness modulus, F, the amount of fine aggregate x to be mixed with 1 volume of the coarse aggregate of a given fineness modulus can be easily determined from the relationship:
#### Fineness Modulus - Pavement Interactive
For aggregates used in PCC, another common gradation description for fine aggregate is the fineness modulus. It is described in ASTM C 125 and is a single number used to describe a gradation curve. It is defined as: The larger the fineness modulus, the more coarse the aggregate. A typical fineness modulus for fine aggregate used in PCC is ...
#### Engineers Head-Quarter: FM Fineness Modulus of Aggragate
Following procedure is adopted to calculate fineness modulus of aggregate. PROCEDURE. Sieve the aggregate using the appropriate sieves 80 mm, 40 mm, 20 mm, 10 mm, 4.75 mm, 2.36 mm, 1.18 mm, 600 micron, 300 micron and 150 micron Record the weight of aggreg ...
#### Excel Sheet For Sieve Analysis Of Aggregate And To Calculate ...
Excel Sheet For Sieve Analysis Of Aggregate And To Calculate Fineness Modulus. The sieve analysis, commonly known as the gradation test, is a basic essential test for all aggregate technicians. The sieve analysis determines the gradation the distribution ...
#### DETERMINATION OF FINENESS MODULUS OF COARSE AGGREGATES AND ...
To find fineness modulus of coarse aggregate we need sieve sizes of 80mm, 40mm, 20mm, 10mm, 4.75mm, 2.36mm, 1.18mm, 0.6mm, 0.3mm and 0.15mm. Fineness modulus is the number at which the average ...
#### Solved: Calculate the fineness modulus of aggregate B in ...
Calculate the fineness modulus of aggregate B in problem 5.25. Is your answer within the typical range for fineness modu Is your answer within the typical range for fineness modu 76 % 749 Review
#### How to Assess the Quality of Fine Aggregate at Site?
4.2. Based on Fineness Modulus: The fineness modulus is also considered as the indi ion of quality of aggregate. According to IS recommendations the aggregate should posses the F.M to the range of 2.6 to 3.2. Greater than 3.2 should be rejected. The Classifi ion of sand based on Fineness modulus is as follows,
#### Concrete Mix Design- A modifi ion of the Fineness Modulus ...
While trial batches are still recommended for previously unused aggregates the modifi ion proposed for the old fineness modulus method permits the design of equally workable batches of varied cement contents from data on trial mixes of a single cement factor. This materially reduces the number of test batches where a range of cement contents is to be used with a set of aggregates. The method ... | 3,131 | 12,743 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2021-17 | latest | en | 0.840805 |
https://inches-to-meters.appspot.com/738-inches-to-meters.html | 1,620,427,329,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988828.76/warc/CC-MAIN-20210507211141-20210508001141-00017.warc.gz | 338,792,704 | 6,289 | Inches To Meters
# 738 in to m738 Inches to Meters
in
=
m
## How to convert 738 inches to meters?
738 in * 0.0254 m = 18.7452 m 1 in
A common question is How many inch in 738 meter? And the answer is 29055.1181102 in in 738 m. Likewise the question how many meter in 738 inch has the answer of 18.7452 m in 738 in.
## How much are 738 inches in meters?
738 inches equal 18.7452 meters (738in = 18.7452m). Converting 738 in to m is easy. Simply use our calculator above, or apply the formula to change the length 738 in to m.
## Convert 738 in to common lengths
UnitLengths
Nanometer18745200000.0 nm
Micrometer18745200.0 µm
Millimeter18745.2 mm
Centimeter1874.52 cm
Inch738.0 in
Foot61.5 ft
Yard20.5 yd
Meter18.7452 m
Kilometer0.0187452 km
Mile0.0116477273 mi
Nautical mile0.0101215983 nmi
## What is 738 inches in m?
To convert 738 in to m multiply the length in inches by 0.0254. The 738 in in m formula is [m] = 738 * 0.0254. Thus, for 738 inches in meter we get 18.7452 m.
## Alternative spelling
738 Inch to Meters, 738 Inch in Meters, 738 Inch to m, 738 Inch in m, 738 in to Meter, 738 in in Meter, 738 in to m, 738 in in m, 738 Inches to Meter, 738 Inches in Meter, 738 Inches to Meters, 738 Inches in Meters, 738 in to Meters, 738 in in Meters | 439 | 1,263 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2021-21 | latest | en | 0.769388 |
https://web2.0calc.com/questions/need-some-help-please_31 | 1,621,074,184,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991801.49/warc/CC-MAIN-20210515100825-20210515130825-00290.warc.gz | 626,278,594 | 5,423 | +0
0
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+508
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Apr 1, 2019
#1
+31792
0
16.4 ft - 14.2 ft = 2.2 ft greater than the mean .....
This is 2.2 / 2.1 = 1.0476 Standar deviations above the mean
Look at 1.04 (rounded to 1.0 ) in the chart to find that .8413 fraction of sunflowers are LESS than that 16.4 ft the rest are taller
1-.8413 = .1587 is the fraction that are TALLER than 16.4 ft .1587 is equal to 15.87%
Apr 1, 2019 | 165 | 417 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2021-21 | latest | en | 0.808053 |
https://tellmeanumber.hostcoder.com/463308 | 1,642,483,720,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300722.91/warc/CC-MAIN-20220118032342-20220118062342-00534.warc.gz | 626,823,195 | 4,745 | # Question Is 463,308 a prime number?
The number 463,308 is NOT a PRIME number.
#### How to check if the number 463,308 is a prime number
A prime number can be divided, without a remainder, only by itself and by 1. For example, 13 can be divided only by 13 and by 1. In this case, the number 463,308 that you looked for, is NOT a PRIME number, so it devides by 1,2, 3, 4, 6, 12, and of course 463,308.
# Question Where is the number 463,308 located in π (PI) decimals?
The number 463,308 is at position 4334748 in π decimals.
Search was acomplished in the first 100 milions decimals of PI.
# Question What is the roman representation of number 463,308?
The roman representation of number 463,308 is CDLXMMMCCCVIII.
#### Large numbers to roman numbers
3,999 is the largest number you can write in Roman numerals. There is a convencion that you can represent numbers larger than 3,999 in Roman numerals using an overline. Matematically speaking, this means means you are multiplying that Roman numeral by 1,000. For example if you would like to write 70,000 in Roman numerals you would use the Roman numeral LXX. This moves the limit to write roman numerals to 3,999,999.
# Question How many digits are in the number 463,308?
The number 463,308 has 6 digits.
#### How to get the lenght of the number 463,308
To find out the lenght of 463,308 we simply count the digits inside it.
# Question What is the sum of all digits of the number 463,308?
The sum of all digits of number 463,308 is 24.
#### How to calculate the sum of all digits of number 463,308
To calculate the sum of all digits of number 463,308 you will have to sum them all like fallows:
# Question What is the hash of number 463,308?
There is not one, but many hash function. some of the most popular are md5 and sha-1
#### Here are some of the most common cryptographic hashes for the number 463,308
Criptographic function Hash for number 463,308
md5 9645e325b3e2cca64e2a6488d7904663
sha1 0615c0ce42fc7a0369b3d0c9ed562a75341a6acd
sha256 7f07c7d261b3fa66327c6c603442d42ebc007a744e5ffec807f97be6b18af921
sha512 e1d020a42c6e07f7f88924bc9c4f33cf68c4a2e0bb903260b54e12f9173fa8793c45225795ac765e5223dcf6c45c11db5da18256c21a234d08fe90c7e4b05158
# Question How to write number 463,308 in English text?
In English the number 463,308 is writed as four hundred sixty-three thousand, three hundred eight.
#### How to write numbers in words
While writing short numbers using words makes your writing look clean, writing longer numbers as words isn't as useful. On the other hand writing big numbers it's a good practice while you're learning.
Here are some simple tips about when to wright numbers using letters.
Numbers less than ten should always be written in text. On the other hand numbers that are less then 100 and multiple of 10, should also be written using letters not numbers. Example: Number 463,308 should NOT be writed as four hundred sixty-three thousand, three hundred eight, in a sentence Big numbers should be written as the numeral followed by the word thousands, million, billions, trillions, etc. If the number is that big it might be a good idea to round up some digits so that your rider remembers it. Example: Number 463,308 could also be writed as 463.3 thousands, in a sentence, since it is considered to be a big number
#### What numbers are before and after 463,308
Previous number is: 463,307
Next number is: 463,309 | 960 | 3,426 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2022-05 | latest | en | 0.880838 |
https://www.transtutors.com/questions/zeus-inc-produces-a-product-that-has-a-variable-cost-of-9-50-per-unit-the-company-s--2577905.htm | 1,548,159,513,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583850393.61/warc/CC-MAIN-20190122120040-20190122142040-00117.warc.gz | 961,914,926 | 17,842 | # Zeus, Inc. produces a product that has a variable cost of $9.50 per unit. The company's fixed cos... 1 answer below » Zeus, Inc. produces a product that has a variable cost of$9.50 per unit. The company's fixed costs are $40,000. The product sells for$12.00 a unit and the company desires to earn a \$20,000 profit. What is the volume of sales in units required to achieve the target profit? (Do not round intermediate calculations.)
Nasrin B
Solution:
Reuired sales volune will be calculated with the help of following formulas
Target Profit
Target Profit = Requied Sales * P/V Ratio - Fixed Cost
Here, P/V Ratio = | 156 | 623 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2019-04 | latest | en | 0.858322 |
https://www.scribd.com/document/374600812/1-s2-0-S0895717707000982-main | 1,542,535,668,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039744348.50/warc/CC-MAIN-20181118093845-20181118115845-00422.warc.gz | 998,768,820 | 66,810 | You are on page 1of 32
# Mathematical and Computer Modelling 46 (2007) 860–891
www.elsevier.com/locate/mcm
Time dependent decision-making; dynamic priorities in the
AHP/ANP: Generalizing from points to functions and from real to
complex variables
Thomas L. Saaty
University of Pittsburgh, Pittsburgh, PA 15260, United States
Received 28 November 2006; accepted 14 March 2007
Abstract
Because good decisions depend on the conditions of the future, and because conditions vary over time, to make a good decision
requires judgments of what is more likely or more preferred over different time periods. There are at least three ways to deal with
dynamic decisions. One is to include in the structure itself different factors that indicate change in time such as scenarios and
different time periods and then carry out paired comparisons with respect to the time periods using the fundamental scale of the
AHP. The second is to do paired comparisons as rates of relative change with respect to time. This is done at different points of
time as representatives of intervals to which they belong. These intervals can have different lengths. For each representative point
one needs to provide a pairwise judgment about the relative rate of change of one alternative over another and derive mathematical
functions for that matrix of comparisons for one of the time periods. The third is to do what I proposed in my first book on the
AHP and that is to use functions for the paired comparisons and derive functions from them. It is usually difficult to know what
functions to use and what they mean. Ideas and examples are presented towards the development of a theory for dynamic decision
making.
Keywords: Dynamic priorities; Time dependent AHP; AHP; ANP; Priority functions; Time horizons; Neural firing
1. Introduction
The Analytic Hierarchy Process (AHP) for decision-making is a theory of relative measurement based on paired
comparisons used to derive normalized absolute scales of numbers whose elements are then used as priorities [1,2].
Matrices of pairwise comparisons are formed either by providing judgments to estimate dominance using absolute
numbers from the 1 to 9 fundamental scale of the AHP, or by directly constructing the pairwise dominance ratios
using actual measurements. The AHP can be applied to both tangible and intangible criteria based on the judgments
of knowledgeable and expert people, although how to get measures for intangibles is its main concern. The weighting
and adding synthesis process applied in the hierarchical structure of the AHP combines multidimensional scales of
measurement into a single “uni-dimensional” scale of priorities. In the end we must fit our entire world experience
into our system of priorities if we are going to understand it.
0895-7177/\$ - see front matter
doi:10.1016/j.mcm.2007.03.028
T.L. Saaty / Mathematical and Computer Modelling 46 (2007) 860–891 861
Sound decision making not only requires that we look ahead as a process of thinking and planning, but also that
our structures be flexible to include change so that we can constantly revise our decision because the system itself is in
imbalance resulting in a non-stationary optimum. Thus optimizing inefficiently by selecting a best decision at a given
time can only be a sub-optimum over a long range time horizon because of the fundamental influence of change in any
survivable system. There are three ways to cope with this problem. The first is to make the best choice according to
short-, mid- and long-term merits with a feedback loop or a holarchy used to prioritize them according to their benefits,
opportunities, costs and risks evaluated in terms of strategic criteria used in general to guide all our decisions [3]. The
other is to make the judgments mathematically depend on time and express them as functions, discussed in this paper.
The third and less practical way is to revise a decision every once in while. Although many decisions can be tested in
advance through sensitivity analysis to determine their long-term stability, revision after implementation can be both
controversial and costly. One may have to abandon an already finished resource intensive alternative for another such
costly alternative. In politics, we circumvent doing it by electing new leaders with the perspective we want, so they
can focus on making better decisions for pressing needs in society.
So far most of us have had no way to combine dollars with yards or pounds to trade them off. We would be truly
multi-dimensional if we could combine the different dimensions into a single dimension that represents our priority
of importance. That is precisely what the AHP and ANP help us do in a more or less precise way, depending on the
level of experience that we bring to bear on a decision problem. Until recently, the AHP and ANP have been static
in that they have used numbers and derived numbers to represent priorities. What we need is to make them dynamic
by using numbers or functions and then deriving either numbers that represent functions like expected values, or
deriving functions directly to represent priorities over time. My aim here is to extend the AHP/ANP to deal with time
dependent priorities; we may refer to them as DHP/DNP (Dynamic Hierarchy Process/Dynamic Network Process). At
this point we may not know enough to develop the necessary fundamental scale of functions to use in making paired
comparisons of intangibles. But if nothing else DHP and DNP work with tangibles now and we need to weight and
trade off tangibles as functions of time.
Time dependent decision-making that we call dynamic decision-making is a subject that we need today. So far we
have thought of our decisions as known alternatives to choose from. But these alternatives may evolve over time along
with our preferences for them like stocks in the stock market whose prices constantly change over time. Our actions
need to vary over time like a medicine capsule that releases different amounts of chemical at different times. Time
dependent decisions are a reality and not a complicated idea that we can ignore. At a minimum they are needed in
technical design problems in which the influences of several tangible design factors change over time and tradeoffs
must be made among them to enable the system to respond differently and continuously over the time of its operation.
But the power and potential of the subject lie in its use of judgment to make comparisons to derive relative real
valued functions for intangibles from paired comparisons. Because we can do that for real numbers we can also do
it for complex numbers. They have a modulus (magnitude) and an argument (direction), each of which is real. That
is where we need to go later to derive relative complex functions from paired comparison expert judgments. The
modulus is estimated in the usual way of comparing magnitudes and the argument or angle is estimated along the
lines of rate of change comparison given below. In this way the two parts of a complex number are derived from
paired comparisons.
There are essentially two analytic ways to study dynamic decisions: structural, by including scenarios and time
periods as elements in the structure that represents a decision, and functional by explicitly involving time in the
judgment process. A possible third way would be a hybrid of these two.
The structural method is most familiar today and it involves using scenarios or time periods as factors in the
hierarchic or network structure of a decision, and then making appropriate judgments. Generally contrast scenarios
such as optimistic, status quo and pessimistic, or more specific ones such as different values for the economy or the
stock market are put near or at the top of a hierarchy. The likelihood of the scenarios is first determined in terms of
higher level criteria under the goal such as economic, political, social and technological, that are themselves prioritized
according to their prevailing influences over a certain time horizon. Judgments are provided for the behavior of the
alternatives with respect to the factors encountered under each scenario [4]. Synthesis reveals the best alternative to
where contrast scenarios are discussed. The other structural method is to put actual time periods at the “bottom” of the
structure, prioritize them and finally combine their priorities by, for example, using the idea of expected value. This
method was used in estimating when the US economy would recover and is illustrated in the next section.
862 T.L. Saaty / Mathematical and Computer Modelling 46 (2007) 860–891
The second approach where the judgments themselves change with time is functional in the literal sense. Whatever
the structure may be, time dependent judgments are provided using functions from which priorities are then obtained
and synthesized as one generally does with the AHP/ANP. We have two problems to solve when using dynamic
judgments in the AHP/ANP. The first is what scale to use to represent dynamic judgments and how in this case it can
be made to satisfy the axiom of homogeneity. The second is how to generate the principal eigenvector of a matrix
whose order is more than four. Because 7 is an upper bound on producing results with reliable consistency, it is
sufficient to find ways to derive the principal eigenvector of a matrix of order seven or less. This task can also be done
in two ways, analytically and numerically. This paper is about both these ways but particularly about the mathematics
of the functional way. In making pairwise comparisons we estimate how many times one element is more important
than another with respect to a common property by forming their ratios if we know their measurements and get an
expert to tell us when we do not have their exact measurements. Each judgment is expressed with a number.
Suppose now that the measurements are a function of time. If we know the functions, we can form their ratios, but
if we do not, what kind of typical functions can we use to represent ratios in a standard way just as we use numbers
in a standard way. If humans have an intrinsic ability to look at the future of two stocks whose values appreciate over
time and can say that on the whole one would be a better investment than the other, then one would like to capture this
intuitive understanding by some standard functions to apply in all situations where we have to deal with intangibles
over time. We have to do this kind of thinking even if we do not know much about the exact future. Like the US
building a 60 billion dollar anti-missile system based on imagined threats over a future time horizon that may not
materialize. The decision would be a function of time going forward or pulling back as the need may be. What are
typical functions to use in describing in an uncertain way the ratio of anticipated variations of two functions over
time? I will discuss and illustrate my proposed approach to this problem.
2. A structural dynamics example
Let us consider the problem of the turn around of the US economy and introduce 3, 6, 12, 24 month time periods
at the bottom [3]. Decomposing the problem hierarchically, the top level consists of the primary factors that represent
the forces or major influences driving the economy: “Aggregate Demand” factors, “Aggregate Supply” factors, and
“Geopolitical Context.” Each of these primary categories was then decomposed into subfactors represented in the
second level. Under Aggregate Demand, we identified consumer spending, exports, business capital investment, shifts
in consumer and business investment confidence, fiscal policy, monetary policy, and expectations with regard to such
questions as the future course of inflation, monetary policy and fiscal policy. (We make a distinction between consumer
and business investment confidence shifts and the formation of expectations regarding future economic developments.)
Under Aggregate Supply, we identified labor costs (driven by changes in such underlying factors as labor
productivity and real wages), natural resource costs (e.g., energy costs), and expectations regarding such costs in
the future. With regard to Geopolitical Context, we identified the likelihood of changes in major international political
relationships and major international economic relationships as the principal subfactors. With regard to the subfactors
under Aggregate Demand and Aggregate Supply, we recognized that they are, in some instances, interdependent.
For example, a lowering of interest rates as the result of a monetary policy decision by the Federal Reserve should
induce portfolio rebalancing throughout the economy. In turn, this should reduce the cost of capital to firms and
stimulate investment, and simultaneously reduce financial costs to households and increase their disposable incomes.
Any resulting increase in disposable income stimulates consumption and, at the margin, has a positive impact on
employment and GNP. This assumes that the linkages of the economy are in place and are well understood. This is
what the conventional macroeconomic conceptual models are designed to convey.
The third level of the hierarchy consists of the alternate time periods in which the resurgence might occur as of April
7, 2001: within three months, within six months, within twelve months, and within twenty-four months. Because the
primary factors and associated subfactors are time dependent, their relative importance had to be established in terms
of each of the four alternative time periods. Thus, instead of establishing a single goal as one does for a conventional
hierarchy, we used the bottom level time periods to compare the two factors at the top. This entailed creation of a
feedback hierarchy known as a “holarchy” in which the priorities of the elements at the top level are determined in
terms of the elements at the bottom level, thus creating an interactive loop. Fig. 1 provides a schematic representation
of the hierarchy we used to forecast the timing of the economic resurgence.
the supermatrix would have to be solved numerically. particularly in this case. there is no major theoretical difficulty in making decisions with dynamic judgments. Saaty / Mathematical and Computer Modelling 46 (2007) 860–891 863 Fig. . 1. Because of its size. given an eigenvector.20583 0.0 0.30581 0.51808 TOTAL 8. It may even turn out in the long run that this is the more efficient way to obtain priorities from the supermatrix even when we have analytic expressions to represent priorities introduced in the usual way in the supermatrix to raise it to powers to obtain its limit for various values of time.63629 Twenty-four months 12 + (24 − 12)/2 = 18. Functional dynamics: Numerical solution of the principal eigenvalue problem by raising the matrix to powers — a basic 3 × 3 example Because priorities are obtained in the form of the principal eigenvector and because this vector is obtained by raising the paired comparisons matrix to powers.5 0.45871 Six months 3 + (6 − 3)/2 = 4.0 0. to test for consistency one obtains its principal eigenvalue by forming the scalar product of the principal eigenvector and the vector of column sums of the matrix.” 3.53932 We interpret this to mean that the recovery would occur 8. with the current month as 0. So.92623 Twelve months 6 + (12 − 6)/2 = 9. Generally. “The National Bureau of Economic Research said the U. this is also the way to obtain its corresponding eigenvalue. In the ANP the problem is to obtain the limiting result of powers of the supermatrix with dynamic priorities. To obtain our forecast.18181 1. economic recession that began in March 2001 ended eight months later. not long after the Sept. we subsequently multiplied each priority by the mid-point of its corresponding time interval and added the results (as one does when evaluating expected values): Time period Mid-point of time period Priority of time period Mid-point x priority (Expressed in months from present. 11 terrorist attacks.30656 5. we do not need to solve equations. We remind the reader that if one has the principal eigenvector of a matrix.L. 2003.5 0. T. said.) Three months 0 + (3 − 0)/2 = 1. in principle. Overall view of the “2001” model. We are fortunate in this special and rare case to do that. In the AHP one simply composes by multiplying and adding functions. for the foreseeable future. The Wall Street Journal July 18.S.54 months from the time of the forecasting exercise.
108 1.214339 0.199235 0.6 0. when A(t) is consistent.652457 0.642013 0.224991 0.4 0.225 1.587395 0.630604 0.206412 0.224166 0.101 1.569422 0.. a ji (t) = ai−1 j (t).359141 1 0..603858 0.3 0.211119 0.45 0.605 1.55 0.217721 0.911059 0.957375 2.489692 0.143922 0..100125 1.125 2.1 1 1..85 0.75 0.905 ln t 1 z2 w2 (t) ≈ √ e− 2 dz.246511 0.98 2.552585 0.390876 0.18 1.171317 0.168789 0.045 1.7414125 2.5 1.35 0. 2 The following expressions are best least-squares fits in some technical sense of each set of numerical data for the components shown in Table 1 and plotted in Fig.194667 0.. .25 0.4291 0.95777 0.45 0.5 0.530484 0. .227527 0. an1 (t) an2 (t) · · · ann (t) ai j > 0..510183 0.0585 0.95 0.399583 0. It is sufficient to illustrate this entire procedure for one matrix.650441 0.373278 0.1 3 2. One expresses the judgments functionally but then derives the eigenvector from the judgments for a fixed instant of time.15 0.2 0. .7 0. Let us consider the 3 by 3 matrix with dynamic coefficients shown below. we have ai j (t) = wi (t)/w j (t).151 ln t 1 z2 w1 (t) ≈ √ e− 2 dz. Saaty / Mathematical and Computer Modelling 46 (2007) 860–891 Table 1 Eigenvectors for different values of time for matrix above t a(t) b(t) c(t) t w1 w2 w3 0 0.292855 0.805 2.5 0 0.612 2. substitutes the numerical values of the eigenvectors obtained for that instant in a supermatrix.6 0.11277 0.191125 1.709534 0.525636 0. solves the supermatrix problem and derives the priorities for the alternatives.866627 0.521875 2. They are then plotted in the diagram below that and an algorithm is used to find the analytic expressions for the best fitting curves for the three components of the eigenvector.674929 0.65207 0.189507 0.135356 0. This entire process can be made automatic in software for dynamic judgments.4 0.8 0.12202 0.316 1.005 1.08 1.15 0.169823 0.053+1.745912 0.451256 0.72 1. 2 Z −0.122939 0.32 1.05 0.157868 The typical form of a judgment matrix in dynamic form is: a11 (t) a12 (t) · · · a1n (t) a21 (t) a22 (t) · · · a2n (t) A(t) = .103375 1.469231 0.9 0.185024 0.164 1.215909 0.05 0.65 0.142875 1.4097 0.55 0. 1 a(t) a(t) = 0.219878 0. Repeating the process for different values of time one generates a curve for the priorities of the alternatives and then approximates these values by curves with a functional form for each component of the eigenvector.829 2.445 2. .25 0.580917 0. t > 0. The rows of the two tables below the matrix give the principal eigenvector for the indicated values of t.85 0.1 + t 3 b(t) A(t) = 1/a(t) 1 b(t) = 1 + 2t 2 c(t) 1 1/b(t) 1/c(t) 1c(t) = 1 + et .1 1.2 0.864 T.122032 0..443 1.344−0.425637 0.270281 0.203156 0.266375 1.35 0.29515 0.4 2π −∞ .5 0.824361 0.9 0.65 0.183833 0. .125 1.448984 0.646922 0.129205 0.229651 0.405 1.245 1.164664 1 1.203569 0.1 0.L.95 0.845 1.28 2..7 0.650917 0. . ..3 0.320802 0.640169 0.209036 0.618394 0.125189 0.02 1..006876 0.127 1.4 2π −∞ Z −0.212817 0.784156 0.62 2.374625 1.115625 1.648329 0.8 0.229802 0.222354 0.177738 0. As in the discrete case.610701 0. t > 0. The basic idea with the numerical approach is to obtain the time dependent principal eigenvector by simulation.550334 0.155057 0.75 0.346935 0.
A plot of the numerical estimate of the components of the principal right eigenvector. Then M M 0 = det(M)I . Here is an example of the foregoing ideas.979t .418539 5.06808 1 1 0. The same P P type of argument applies when a matrix is column stochastic. 2.00498879 + 0. But for the supermatrix we already know that λmax (T ) = 1 which follows from: n n X X wj max ai j ≥ ai j = λmax for max wi j=1 j=1 wi n n X X wj min ai j ≤ ai j = λmax for min wi .00498879 − 0. 4. By Perron. its principal right eigenvector and its eigenvalues are: 1 2 3 4 5 1 1 2 3 4 2 0. all of the columns of M 0 must be proportional. t > 0.582755i 3 2 1 1 1 0. What we now show is only useful when we know λmax (T ).061767 −0. thus λmax = 1.097253 −0.262518 0. We note that in our situation. They are also nonzero vectors because the rank of M is n − 1 that is “some” n − 1by n − 1 minor is nonsingular. Since the kernel of M is one-dimensional.4. the rank of M cannot be more than n − 1. Functional dynamics: Analytic solution of the principal eigenvalue problem by solving algebraic equations of degree n Because we need to solve for the principal eigenvector. because its characteristic polynomial is of degree n and vanishes at λmax (T ). On page 20 of Reference [5] there is a brief discussion of the classical adjoint. This gives us another way of getting the principal right eigenvector of A when we know λmax (T ). Perron theory gives us a way to obtain it more simply than by solving a high order algebraic equation or even by raising the matrix to arbitrarily large powers. Let M = A − λmax (T )I be an n by n matrix.068286i 1 2 4 3 2 0. Saaty / Mathematical and Computer Modelling 46 (2007) 860–891 865 Fig.582755i 1 2 3 0. The matrix A. Note the row/column transposition.L.068286i 1 1 1 1 1 5 4 3 2 . j=1 j=1 wi Thus for a row stochastic matrix we have 1 = min nj=1 ai j ≤ λmax ≤ max nj=1 ai j = 1. Let M 0 be the matrix whose entry in the ith row and jth column is (−1)i+ j times the (n − 1 by n − 1) determinant of the matrix obtained by removing the ith column and jth row of M.0390289 + 0. T. w3 (t) ≈ e−0.159923 0. Thus M M 0 = 0 and every column of M 0 is an eigenvector of M (belongs to the kernel of M).956−0. where A is the pairwise comparisons matrix and I is the identity matrix. Now M has rank n − 1. λmax (T ) is unique and cannot again be the root of a determinant of an n − 1 minor of M and the rank of M is precisely n − 1.0390289 − 0.
6099 134.69 × 10−12 4.96 × 10−14 −3. The matrix M = A − λmax (A)I is given by: −4.657 363. The reason is that in these cases one must solve a polynomial equation to obtain the principal eigenvalue and derive the corresponding principal eigenvector and we have expressions given below for both of these for n ≤ 4. solving an entire decision problem numerically.796 221.116 135.105 924.1. 3 2 1 1 1 −4.06808 2 4 3 2 1 1 1 1 −4.61 × 10−12 2. Quadratic case [1] To obtain the eigenvalue and eigenvectors of a 2 by 2 matrix.393 580.58 × 10−14 −3.35 × 10−14 3. entering these numerical values of time.58 × 10−14 1. .6099 136.35 × 10−14 3.1428 31. we must solve the problem a(t) w1 (t) w1 (t) 1 = λmax 1/a(t) 1 w2 (t) w2 (t) for which we know because the matrix is consistent that λmax (t) = 2.L.09 × 10−13 −2. −3.877 85.9371 82. But for n > 4 the story is complicated and has a long history in our time. A technical problem that arises in this approach is that because of the time dependence of the coefficients of the matrix.69 × 10−12 −9.10 × 10 −2.05 × 10−14 −3.62 × 10−12 8.393 .1526 85.20 × 10−14 1. Mathematica is a good program for providing the eigenvalues and the eigenvectors of matrices of large order.06808. In the end the several decision-outcome values can be approximated by time dependent curves.866 T. it is difficult to generate the eigenvector of priorities in symbolic form if the order of the matrix is more than four.93 × 10−14 −3.06808 2 3 .7155 49.44 × 10−14 1.855 580.657 353. 31. In this paper we give the analytic solution to derive the eigenvector for up to a 4 × 4 matrices and also give the solution of the quintic equation for the eigenvalues in closed form.63 × 10−13 1.1526 82.64 × 10−14 −2.66 × 10−12 Any column of M 0 such as the first gives the principal right eigenvector given above in normalized form.97 × 10−14 1.35 × 10−14 4. and repeating the process. 4.796 214. For the eigenvector we need to solve the system of equations: w1 (t) + a(t)w2 (t) = 2w1 (t) w1 (t)/a(t) + w2 (t) = 2w2 (t). sextic and septic equations. The reader interested in this problem can contact this author for more general information on the solution of the quintic.06808 2 3 4 5 1 −4.83 × 10−13 1.7155 52.06808 2 3 4 2 1 1 −4.06808 5 4 3 2 Its adjoint matrix M 0 is: 136.49 214.031 221.908 353.116 134.49 Note that M M 0 = 0. Saaty / Mathematical and Computer Modelling 46 (2007) 860–891 λmax (A) = 5.98 × 10−14 M M0 .67 × 10−14 1.908 20.105 M 0 = 52. It would also be useful if one were to use it for time dependent judgments by taking different times.36 × 10−14 2. −14 2.04 × 10−14 2.
w3 = .3. 4. Quartic case [1] If we define a23 a34 a24 a12 a24 a14 a12 a23 a13 a13 a34 a14 B= + + + + + + + a24 a23 a34 a14 a12 a24 a13 a12 a23 a14 a13 a34 a12 a23 a34 a14 a12 a24 a13 a34 a14 a23 a13 a24 C =3− + − + − + a14 a12 a23 a34 a13 a34 a12 a24 a13 a24 a14 a23 then 2 1/3 s 3 B2 B2 4 λmax = −8 + + 8C + − (C + 3) + 8 − − 8C 2 3 2 2 1/3 s 3 B2 B2 4 + −8 + + 8C − − (C + 3) + 8 − − 8C 2 3 2 and w1 w2 w3 w4 w1 = . w4 = Q Q Q Q where Q = (λmax − 1) + (a14 + a34 + a24 )(λ − 1) + (a12 a24 − 3) + (a13 + a23 )a34 3 2 1 1 a24 + + a14 + (λmax − 1) a12 a13 a23 a13 a24 a13 a34 a14 a32 + a12 a24 a14 − a13 + (a12 a23 a34 − a12 − a14 − a24 − a34 ) + + + + a23 a12 a13 a12 a23 a13 a24 w1 = a14 (λmax − 1)2 + (a12 a24 + a13 a34 )(λmax − 1) + a12 a23 a34 + − a14 a23 a14 a13 a34 a14 a23 w2 = a24 (λmax − 1) + a23 a34 + 2 (λmax − 1) + + − a24 a12 a12 a13 . we have for our solution w1 (t) = a(t)/[1 + a(t)]. Saaty / Mathematical and Computer Modelling 46 (2007) 860–891 867 The first equation yields w1 (t) = a(t)w2 (t). w2 (t) = 1/[1 + a(t)].2. w2 = . and because w1 (t) + w2 (t) = 1. we have ∆ w1 = D (λmax − 1)a23 + (a13 /a12 ) w2 = D −1 + (1 − λmax )2 w3 = . If we define ∆ = a12 a23 + a13 (λmax − 1) and D = a12 a23 + a13 (λmax − 1) + (λmax − 1)a23 + (a13 /a12 ) − 1 + (1 − λmax )2 . T. Cubic case [1] λmax = (a13 /a12 a23 )1/3 + (a12 a23 /a13 )1/3 + 1. D 4.L.
Let f (x) = a0 x n + a1 x n−1 + · · · + an = 0. This property holds for a reciprocal matrix.868 T. We have: a0 a1 a0 a2 a1 X 1. pp.1. 3. Saaty / Mathematical and Computer Modelling 46 (2007) 860–891 a24 a14 a14 a12 a24 w3 = a34 (λmax − 1) +2 + (λmax − 1) + + − a34 a23 a13 a12 a23 a13 a12 a23 a13 w4 = (λmax − 1)3 − 3(λmax − 1) − + .−1 = − + . a1 a2 a1 a3 a2 a3 a2 a4 a3 X 4. and a child. The quintic and higher order cases The resolution of algebraic equations [6]. a4 a3 a5 a4 Each bracket represents a power series with a monomial in the bracket as its first term: a 2 a2 a 3 a3 a3a2 a 4 a4 a 4 a2 a3 a 5 a5 a0 a0 = + 0 3 − 0 4 + 2 0 52 + 0 5 − 5 0 6 − 0 6 + · · · a1 a1 a1 a1 a1 a1 a1 a1 2 3 2 3 3 4 a14 a3 a4 a1 a1 a a3 a a4 a0 a1 a5 a1 a3 a1 a5 = + 13 − 14 − 3 + 2 + − 5 + ··· a2 a2 a2 a2 a24 a25 a25 a26 a22 a4 a23 a5 a23 a42 a2 a2 a0 a5 a1 a4 a1 a2 a5 = − 2 − 2 +2 + − + 2 + ··· a3 a3 a3 a3 a33 a33 a34 a35 a 2 a5 a1 a 2 a2 a3 a52 a0 a53 a1 a3 a53 a3 a3 a2 a5 = − 2 + 3 3 + 35 − 3 − + 4 + ··· a4 a4 a4 a4 a4 a44 a44 a45 a4 a4 = . a mother. 4.−1 = − . Family spending time at home Let us consider the simple case of a family with a father.−1 = − + . a5 a5 5. X 5. An infant will spend the same amount of time at home as the mother . X 3. Obviously the amount of time the child spends at home will depend on his age. [7] does not require the use of Tschirnhausen transformations. a0 6= 0 be an algebraic equation irreducible over a subfield of the complex numbers. Bernd Sturmfels [8] gives the solution of the quintic equation in terms that use the finest of the 2n−1 triangulations of the hypergeometric differential equations and corresponding 2n−1 series solutions. The roots of the quintic a5 x 5 + a4 x 4 + a3 x 3 + a2 x 2 + a1 x + a0 = 0 one of which is the principal eigenvalue may also be obtained in other ways given by an infinite series and the question about such expansions is whether the series always converge.−1 = − + .L. It is easy to see from this solution that if any coefficient is increased (decreased) in a given row of the pairwise comparison matrix the value of the eigenvector component corresponding to that row is increased (decreased) relative to the remaining components.−1 = − + .261–3. X 2. 5.4. Value of pairwise comparisons — an example The following example shows that by making pairwise comparisons we learn much more about the detail of a dynamic priority function than if we were to simply guess at each component of the eigenvector components individually. a13 a12 a23 Remark. A root of this equation can be expressed in terms of theta functions of zero argument involving the period matrix derived from one of two types of hyperelliptic integrals.272.
If we were to make a pairwise comparison of the different lengths of time spent at home by the different members of the family. Consider the time period corresponding to the child’s age 0–4 years. = 0. 4. as he grows older. we would get a sequence of comparison matrices each corresponding to a particular period of time. 3.5 M 2. Relative time at home — child to father..5 1/2. he will spend progressively less time at home compared to the mother. The mother and child would of course spend the same amount of time. If we were to compare the length of time spent at home by mother and child and plot this relation as a function of time (i. Comparison of father to child times yields a relationship that is a mirror image of the above—reflected about a horizontal axis halfway up the curve. This would give rise to the following matrix: F M C F 1 1/2. we would get the type of curve shown in Fig.0. 3.L. C. = 0. This yields the following eigenvector for their relative times at home: . Fig. We assume that the mother does not go out to work. This is illustrated in Fig.R.I. If we were to exclude. as the child grows older). T. then the ratio of mother’s to child’s time increases until it levels off by the time the child is in her/his mid-teens. say. The relative length of time spent by father and mother would not vary too much and could be expected to be fairly constant. we would expect the mother and child to spend about two to three times the length of time the father spends at home.5 1 1 C 2. Thus the curve begins with the home maker mother and child spending the same amount of time. C.0.5 1 1 λmax = 3.e. Saaty / Mathematical and Computer Modelling 46 (2007) 860–891 869 Fig.0. and then. Relative time at home — mother to child. 4. eight hours of the night.
Mother’s age relative to father: during child’s age from 0–16 years. along with the previous one.417.4 + ln t/2) 2 λmax =1+ + (3 − ln t/2)(0. so there is a sudden change in the relative proportions of time spent at home by mother and child and by father and child. 6. The solution of the maximum eigenvalue problem corresponding to these pairwise comparison curves for (4 ≤ t ≤ 16) is given by 1/3 1/3 (3 − ln t/2)(0.870 T. C: 0. (F: 0.L.4 + ln t/2) + /D. This matrix. 6 and 7 that depict the corresponding pairwise comparisons as time varies from zero to 16 years. gives rise to the curves in Figs. Mother’s age relative to child: from ages 0–16 years.4 + ln t/2) 1 The variable t denotes the child’s age ranging between 4 and 16 years.417) that would seem to be a reasonable estimate of the proportions of time each spends at home. Fig. M: 0. w2 = (λmax − 1)(0. Around the age of four the child begins school.167.4 + ln t/2 C 3 − ln t/2 1/(0.4 + ln t/2) 2 and the corresponding eigenvector is given by 2 w1 = ∆/D. we can express the varying proportions in a single matrix as shown below: F M C F 1 1/2 1/(3 − ln t/2) M 2 1 0. Saaty / Mathematical and Computer Modelling 46 (2007) 860–891 Fig. 5. w3 = [−1 + (−2)2 ]/D 3 − ln t/2 . Moving to the time dependent situation. 5.
0. with respect to all others.8 0.5 C 0. he begins spending even less time at home than the father. 7. The proportions once again become fairly constant and are reflected in the following (consistent) pairwise comparison matrix: F M C F 1 0. .4 + ln t/2) + 3 − ln t/2 λmax + 1 D = (λmax − 0.5)(0. 4 ≤ t ≤ 16. 8. where λmax − 1 ∆ = 0. C.L.211. The eigenvector solution is given by: F : 0.526 C : 0. Saaty / Mathematical and Computer Modelling 46 (2007) 860–891 871 Fig.25 M 2 1 2. which each spends at home (see Fig.5(0.4 1 λmax = 3.0. 3 − ln t/2 As the child finishes school.5 1. Relative proportions of time spent at home. Child’s age relative to father: during child’s age 0–16 years.0. and 16 ≤ t gives a realistic representation of the relative time. = 0.I. Fig.R. = 0. Plotting these results together for 0 ≤ t ≤ 4. 8). T.4 + ln t/2) + − 1 + (1 − λmax )2 . C.263 M : 0.
L. They assumed that the preferences for alternatives at the bottom with respect to the criteria in the fourth hierarchy level of Fig. exponential. What typical functions can we use that would for example approximate well the functions of the family example above? I do not have a conclusive answer to that question. hopefully meeting the homogeneity requirement of the 1–9 scale used in the discrete case as a limit on the range of values. Forming the functions was produced experimentally with the help of the software developed by the authors. linear. There were (n − 1) cells in a matrix selected. Thus there was no problem of inconsistency during solution of the dynamic problem. The filling of pairwise comparison matrices as the dynamic task was made as follows. Pairwise comparison judgments and scale — general discussion Recall that in making paired comparisons of intangibles. 1 ≤ a ≤ 9 an No change in relative standing integer a1 t + a2 Linear relation in t. b1 log(t + 1) + b2 Logarithmic growth up to a Rapid increase (decrease) followed by slow increase (decrease) certain point and constant thereafter c1 ec2 t + c3 Exponential growth (or decay if Slow increase (decrease) followed by rapid increase (decrease) c2 is negative) to a certain point and constant thereafter (not reciprocal of case c2 is negative is the logistic S-curve) d1 t 2 + d2 t + d3 A parabola giving a maximum or Increase (decrease) to maximum (minimum) and then decrease (increase) minimum (depending on d1 being negative or positive) with a constant value thereafter. we use the smaller or lesser element as a unit and estimate the larger one as a multiple of it. oscillatory. The relative importance of the criteria for each actor and for the alternatives shown in Tables 3 and 4 also would be subject to modification in the future. polynomial. logarithmic. and combinations of them. They. . 9 would remain constant in the future. we need to enter basic functions to represent judgments in the dynamic case. We can also do that in dynamic judgments by simply introducing a function to estimate how many times more the larger element is than the unit. reflect our intuitive feeling about relative change in trend: constant. n > 0 (n ≤ 0) with decreasing (increasing) amplitude Catastrophes Discontinuities indicated Sudden changes in intensity Polynomials are known in general to approximate to continuous functions on a closed interval arbitrarily closely.872 T. and finally discrete that allows for sudden change like a Dirac delta function. 6. Saaty / Mathematical and Computer Modelling 46 (2007) 860–891 Table 2 This author’s idea of a mathematician’s formulation of a dynamic judgment scale Time dependent Description Explanation importance intensity a Constant for all t. At the following instants the values for the other (n 2 − 2n + 1) preferences were calculated on the basis of the (n − 1) functions given (Auto). The preferences at the time coincided with preferences in the static task. Just as we used absolute numbers to enter into the comparison matrix in the static case. where the functions describing changes of appropriate preferences were made. Note that the reciprocal is a hyperbola. increasing or Steady increase in value of one activity over another decreasing to a point and then a constant value thereafter. (May be modified for skewness to the right or left) e1 t n sin(t + e2 ) + e3 Oscillatory Oscillates depending on n. However. These functions have been left in parametric form so that the parameter may be set for the particular comparison. For the pairwise comparison judgments one may attempt to fit one of the functions given in Table 2 to the dynamic judgments. if indeed there is an answer to it. they assumed that preferences for the factors located at the second hierarchy level would vary. In this case there may be a change of unit of the comparison if the larger element becomes the smaller and must serve as the unit. The following is an example given by Andreichicov and Andreichicova [9].
Saaty / Mathematical and Computer Modelling 46 (2007) 860–891 873 Fig.4t + 0.333 The dynamic paired comparison matrices were the following: FOCUS Macroeconomic Innovations Investments Buyers Macroeconomic 1 1/0.380 0.258 0.25e0.637 Investments 0.333 + 0.15t + 0.05t + 0. Table 3 Priority vectors for criteria in the static problem FOCUS Macroeconomic Innovations Investments Buyers 0.167 0.43t 1/0.069 VIS producers Vehicles producers Technologies Macroeconomic 0. The buyer’s priority will also increase slightly.2 − 0.6t Auto 1/(0.05t + 0.085 0.43t Growth 0.740 0.097 0.15t Cost 1/(0.333 0. Hierarchy for dynamic evaluation. T.094 Buyers 0.466 0.333e0.167 0.532 0.04t 2 ) 1 Auto Auto Reliability 1/(3 − 1. 10.18t 2 0. The .08t 2 ) Innovations 0.05t + 0.333e0.333 0.18t + 0.109 0.18t 2 ) Auto 1 Auto Layout 1/(0.25e0.43t Auto 1 Auto Markets 0.186 0.43t Auto Auto 1 Vehicles producers Quality Cost Reliability Layout Quality 1 0.L.025t 2 ) 1/0.402 Competitiveness Cost Simplicity Technologies 0.15t 2 ) 1 Auto Buyers 0.143 0.05t + 0.105 0. 9.186 Quality Cost Reliability Layout Vehicles producers 0.333 + 0.025t 2 1 Auto Auto Competitiveness 0.2 − 0.15t + 0.15t) Auto Auto 1 The other paired comparison matrices were the same as in the static problem.333e0.2 + 0.333e0.333 + 0.402 0.582 0.08t 2 Auto Auto 1 VIS producers Profit Growth Competitiveness Markets Profit 1 1/(0. The investor’s importance and macroeconomic influence will decrease.309 Innovations 0.15t 2 Auto Investments 1/(0.2 + 0. These modifications and the changes in the importance of the criteria for producers will result in a change in the order established for the alternatives.4t + 0.333 + 0.6t 1 0.2 − 0.054 0.2 − 0.04t 2 3 − 1. The priority of innovation in Russia is expected to increase in the future. The results are shown in Fig.18t + 0.833 Profit Growth Competitiveness Markets VIS producers 0.
The best alternative in the future will be VIS controlled.178 0.751 0. 7.178 0. As in the discrete case. Two kinds of data need to be generated. The low priority of the pneumatic VIS in both cases might be explained by its relatively high cost and low reliability. 10.075 0.258 Layout 0. Fig. spring coil.319 0. Fundamental scale of relative change — on the rate of change of intangibles! Let us now turn to a more practical way of creating a fundamental scale that is in harmony with any person’s way of looking ahead to make comparisons. but we can compare them to derive scales of relative change over time.066 0.L.648 0.637 0.109 0.592 Markets 0. Saaty / Mathematical and Computer Modelling 46 (2007) 860–891 Table 4 Priority vectors for alternatives in the static problem Coil spring Pneumatic Controlled Profit 0. moves from first place to last place.105 0. because we have intangibles we have no functions to describe their behavior absolutely. that is simple and cheap.122 Growth 0. Here we ask the same question as in the discrete case: which of the two dominates the other with regard to a common attribute they share? The other question has to do with their perceived relative change: which of the two is perceived to change more than the other with respect to a common attribute they share? This second question is answered with perhaps different estimates for different time periods.582 0.105 0. Change in factor priorities over time. The question is how to generalize this idea to the dynamic case.070 Reliability 0. One is the initial point x(0) of the curve of relative change. medium and low.751 0.333 0.429 0.874 T. an individual has the ability to recognize distinctions among high. In the case of static judgments that deal with numbers we had the nine values of the fundamental scale 1–9. 11.615 Cost 0. As in the discrete case.429 Competitiveness 0.637 Simplicity 0.070 Fig. 11. We have no problem to deal with tangibles whose behavior over time is known though a function-of-time expression for each. medium and low and for each of these also three distinctions of high.230 0.258 0.309 Quality 0. We assume for the moment that when we compare two (tangible) . which has high vibro-isolation quality and high cost as shown in Fig.143 0. Change in each alternative’s priorities over time.
668 0. Saaty / Mathematical and Computer Modelling 46 (2007) 860–891 875 Table 5 Formulation of a dynamic judgment scale Comparison of A over B Slope of line Slope of line Slope of line (0 ≤ θ ≤ 90◦ ) (0 ≤ θ ≤ 45◦ ) (0 ≤ θ ≤ 22. Homogeneity needs to be maintained not only for magnitudes but also for the angle of relative change.414 activities each described by a curve as a function of time that we can make these nine distinctions in their rates of change relative to one another.414 −0.414 0. Perhaps it should be closer to a radian 360/2π = 57. This author’s idea of a practical-minded person’s representation of a dynamic judgment scale of relative changes is shown in Table 5.199 Very strong decrease −2. time horizons may be divided into 9 periods starting with the present time and relative increases and decreases further refined.296◦ and its division into ranges of just noticeable increases in slope or angle.199 −0.5◦ ) Extreme increase ∞ 1 0. one may select the mid-point of a time horizon to draw each line segment and take their points of intersection as the positions where the composite curve connects. There needs to be a theoretical way to justify the bound on the homogeneity of relative change. Because we do not wish to use upward or downward arrows for an entire judgment over a time horizon interval. When exceeded. How can each of these curves vary so that we would be able to divide the relative variation between them into recognizable nine categories over time? We note that in the simplest case where both functions are linear in time and their relative variation expressed by their ratio is a hyperbola over time. or the soft way as an intangible divided according to our feelings about it such as short-term.414 −0. |t − a|−1 and (t − a)−2 .098 Equal 0 0 0 Moderate decrease −0. one may have to use clustering as in the comparison of different magnitudes.199 0. Here a function representing judgments may descend below x(t) = 1 and thus the coefficient may include a description of both dominance and being dominated. The angle span for homogeneity may be less than 90◦ . A better alternative is to divide each of short- term. Its transpose is the reciprocal function.668 −0. mid-term and long-term.199 Moderate increase 0.L. An application of it over time is shown in Table 6. T. the hard way as a tangible measured on a clock and make judgments over known perhaps uniform time intervals.414 0. mid-term and long-term into three intervals and provide judgments of relative change by the above scale . instead of connecting their end points to make a continuous curve.414 −0.414 0. As in the 1–9 scale.414 Very strong increase 2.303 Extreme decrease −∞ −1 −0. Note that because the time horizons are not well specified over time. What are different possible interpretations of variations (ratios) between two functions? To deal with such concepts effectively we need to think of time in two ways.303 Strong increase 1 0.098 Strong decrease −1 −0. Examples are (t − a)−1 . we can replace such a “Dirac” type function by a curve that rises or falls sharply and continuously over the interval. A hyperbola has a simple but rather sophisticated interpretation.
Combined over the Functional form horizons term term three horizons x1 (0) 0 ≤ t ≤ t1 x1 (0) + 0.0079t AT& T 1 Medium Term IBM MSFT AT& T IBM 1 1.and long-term time horizons.L.414(t2 − t1 ) t2 < t ≤ t3 of A over B −2. where a = 30. one can use the criterion ai j w j /wi 1 of the AHP or the gradient approach (more complex in the dynamic case) to improve overall consistency. 13. The example has to deal with the earnings and projected electricity use by three companies (taken from the Internet) as shown in Fig. As time lapses this ratio converges asymptotically to the line x(t) = 1.414t t2 < t ≤ t3 x (0)t1 (t1 + t)−1 0 ≤ t ≤ t1 13 x (0) t1 < t ≤ t2 Comparison a23 (t) = 2 3 of B over C 1 t +t x3 (0) 3 t2 < t ≤ t3 2 t3 + t2 for each. 8.414(t − t1 ) t1 < t ≤ t2 Comparison a12 (t) = x1 (0) + 0. The pairwise comparison matrix formed by taking the ratios of the functions of age of the two people is .797 + 0. However.499 − 0. We now show by an example that real life situations reveal themselves along similar lines to the line segments we combined to put together decisions made over short-.0104t MSFT 1 1. the ratio of their ages in the (1. medium. 14 as time lapses the difference between their ages remains constant.0103t 2. Judgments fitted with linear approximations: Short Term IBM MSFT AT& T IBM 1 1. 2) position in the paired comparison matrix below is a hyperbola given by (a + t)/t = 1 + a/t as shown in Fig.414t t1 < t ≤ t2 of A over C x2 (0) + 1.476 + 0. Saaty / Mathematical and Computer Modelling 46 (2007) 860–891 Table 6 Different possible formulations and variations using the basic scale in Table 5 Time Short.969 + 0.899 + 0.733 − 0. In that case one uses the geometric mean of all these judgments leading to a product integral. As seen in Fig. Initially the ratio is infinite when t = 0. One would possibly be dealing with nonlinear inequalities. AT& T 1 Plots of estimated dynamic priorities versus actual are shown in Fig.876 T. We have two people.668 + 0. This is time consuming but more accurate even if one were to repeat the same judgments over success intervals.0056t 2. 15. 12. Finally.089t MSFT 1 1.414t1 − 4. Mid-term Long.414(t − t2 ) x2 (0) − t 0 ≤ t ≤ t1 Comparison a13 (t) = x2 (0) + 1.414t1 − 2. A complete example with paired comparisons Let us consider the case of tangibles.828t2 + 2.0089t MSFT 1 1.943 + 0. the first is a years older than the second.1125t AT& T 1 Long Term IBM MSFT AT& T IBM 1 1.0175t . It may be possible to aggregate a judgment function into a number by assuming that each discrete judgment on a curve arises from different observations about a future determined not by a point but by an interval of time.955 − 0.089t 2.
Fig. The ideal mode solution is 1. 16.L. 18. Assume that Fig. 12. w2 (t) = t/(a + 2t). with a graph comprised of a horizontal line and a curve as shown in Fig. Let us assume that strength increases to a certain value #1 and then decreases with age eventually to zero at death as in Fig. t > 0. T. and that is strength. We continue our illustration of a multicriteria problem by considering another attribute besides age that is measurable over time. 17 is a plot of the normalized eigenvector components. Saaty / Mathematical and Computer Modelling 46 (2007) 860–891 877 Fig. a ≥ 0. Relative earnings of software companies. t/t + a. given by: (a + t)/t 1 t/(a + t) 1 and this matrix has the normalized principal right eigenvector : w1 (t) = (a + t)/(a + 2t). 18 describes the strengths of the two .
2500 2500 Forming the ratios of the functions that represent the relative strengths of the individuals we have the paired comparison matrix below: y1 (t) 1 y2 (t) . individuals and that the corresponding equations are: 1 1 y1 (t) = 1 − (t − 20)2 . Saaty / Mathematical and Computer Modelling 46 (2007) 860–891 Fig. y2 (t) 1 y1 (t) . y2 (t) = 1 − (t − 50)2 .L. 13.878 T. Estimated versus actual relative values for the three companies.
by combining the two vectors obtained for age and strength. Graph of the ratio of the functions x2 (t)/x1 (t) = t/(t + 30). 14. The above matrix has a normalized eigenvector of [y1 (t). age and strength. Hyperbola x1 (t)/x2 (t) = 1 + 30/t. 15. 16. 2 2500 2500 2500 1 1 1 u 2 (t) = 1 − (t − 50) 2 1− (t − 20) + 1 − 2 (t − 50) . Fig. 19. Saaty / Mathematical and Computer Modelling 46 (2007) 860–891 879 Fig. y2 (t)] which gives the solution vector [u 1 (t). We obtain the composite vector of the two alternatives.L. Fig. u 2 (t)] with: 1 1 1 u 1 (t) = 1 − (t − 20) 2 1− (t − 20) + 1 − 2 (t − 50) . 2 2500 2500 2500 The graph of this time dependent eigenvector solution of strength is shown in Fig. T. Age as a function of time. .
12] How do we maintain an ongoing record of the signals transmitted in the brain that is updated. 2 2 2500 2500 2500 9. We might also have given the criteria time dependent weights to do the weighting and adding. Graph of normalized time dependent eigenvector w1 (t). Finally let us assume that the two criteria age and strength are equally important so we can weight and add the two vectors w1 (t). Fig. y2 (t)]. u 2 (t)]. We have the following solution for our final composite vector: [z 1 (t). 1 1 1 1 1 z 1 (t) = (30 + t) (30 + 2t) + 1− (t − 20) 2 1− (t − 20) + 1 − 2 (t − 50) . Graph of the two strength functions [y1 (t). 18.880 T. Saaty / Mathematical and Computer Modelling 46 (2007) 860–891 Fig. Fig. 17. w2 (t) with a = 30. and synthesized. revised. 19. u 2 (t)]. z 2 (t)].L. Time dependent eigenvector solution of strength graph [u 1 (t). Hypermatrix of the Brain [11. 2 2 2 2500 2500 2500 1 1 1 1 1 z 2 (t) = t/(30 + 2t) + 1− (t − 50)2 1− (t − 20)2 + 1 − (t − 50)2 . to capture the transient information that is communicated through neural firings? Synthesis in the brain . 20. Its graph is shown in Fig. w2 (t) with a = 30 and [u 1 (t).
Saaty / Mathematical and Computer Modelling 46 (2007) 860–891 881 Fig. The approach we follow to represent interactions in the brain is a result of modeling the network of neurons by a graph. It all hangs together in one piece. Electric potentials are measured across the arcs. z 2 (t)]. In that case each column of the matrix would represent the number of synapses of all neurons with a given neuron represented at the top of the matrix. In the opposite direction we assign them a zero value. Graph of synthesized solution [z 1 (t). We assume that for convenience. which correspond to subcomponents or layers and these in turn according to the components or modules to which they belong. because every element is connected to some other element. 20. 21) and listing them again in a row above the columns of the matrix. whose nodes are the neurons themselves. In neural terminology.L. Such a representation in a matrix can be modified by multiplying its nonnegative integer entries by the damped periodic oscillations of period one corresponding to the neuron with which the synapses are associated. instead of the number one. In fact. The different components of the hypermatrix are represented as block supermatrices (Table 7). One can then enter a zero or a one in each column of the matrix to indicate whether a neuron in a row synapses with a neuron in a column. summing the elements in a column corresponds to spatial summation at an instant of time. Table 7 The hypermatrix is the most time dependent operation we know of intimately. We assign the potentials direction so that they are read in a positive direction. That would be the most elementary way to represent the connections of the brain. we have arranged the neurons into components. T. The brain creates instantaneous impressions that are synthesized from one instant to another. and whose synaptic connections with other neurons are arcs or line segments that connect the nodes. Thus we represent the neurons of the brain conceptually by listing them in a column on the left of the hypermatrix (Table 7 and Fig. The control subsystems . a positive integer can be used to indicate the number of synapses that a neuron on the left has with a neuron at the top of the matrix.
at great distances from the small piece. which consists of many individual neurons. The uniqueness of analytic continuation has the striking consequence that something happening on a very small piece of a connected open set completely determines what is happening in the entire set. The hypermatrix of the brain.1. The i. The size of the hypermatrix to describe the brain would be of the order of 100 billion by 100 billion (we have not consulted the Cray Research people about whether the development of their supercomputers comes close to satisfying the dicta of brain computing). and also to a higher level of control components. is that it is a synthesizer of the firings of individual neurons into clusters of information and these in turn into larger clusters and so on. j block of this matrix is a supermatrix whose entries are matrices whose columns are principal right eigenfunctions. Synthesis The most significant observation about the brain. We offer the reader in Figs. 21 and 22 a hard-worked-on but rudimentary application of the hypermatrix to modules and submodules of the brain to illustrate what we have in mind.882 T. “The nervous system appears constructed as if each neuron had built into it an awareness of its proper place in the system”. which they control. leading to an integrated whole. 21. or know enough about the physical reality of the brain and its synapses to create the entries in the matrix. Is there a field in the brain and where is it? We believe that the process of analytic continuation in the theory of functions of a complex variable provides insight into how neurons seem to know one another. On page 373. It is far beyond our capability to handle the size of such a matrix. and among themselves. They are obtained as follows. That is what analytic continuation does. It conditions neurons to fall on a unique path to continue information that connects with information processed by adjacent neurons with which it is connected. Fig. . 9. We warn the reader that we are simply using imagination to brave the complexity. Due to their sequential nature. Saaty / Mathematical and Computer Modelling 46 (2007) 860–891 Fig. Kuffler and Nicholls [10] in their often-cited book say. 20 shows the supermatrix of vision as part of the hypermatrix.L. are connected to the supermatrices. the firings of a neuron that precede other neurons would be lost unless there is something like a field in which all the firings fit together to form a cohesive entity which carries information.
All such two-step interactions are obtained by squaring the matrix.to post-synaptic neurons to accumulate all the transitive influences from one neuron to another and allow for feedback. the influence of each neuron on all the neurons with which one can trace a connection. Saaty / Mathematical and Computer Modelling 46 (2007) 860–891 883 Fig. A solution w(s) of this equation is a right eigenfunction. ai j > 0 generalizes to the continuous case through Fredholm’s integral equation of the second kind (which we can also derive directly from first principles to describe the response of a neuron to stimuli): Z b K (s. 22. t) > 0. yields the transient influence of neurons in the original hypermatrix. T. This means that a neuron influences another neuron through intermediate neurons. t)w(t)dt = λmax w(s) (3) a where instead of the matrix A we have a positive kernel. K (s. By raising the matrix to sufficiently large powers. including consciousness related to the Fourier transform of the single-valued sensory functions.L. . The Fourier transform that takes place as a result of firing and the density of the resulting firings gives us the desired synthesis. The supermatrix of vision: technical discussion. Three step interactions are obtained by cubing the matrix and so on. Different physical and behavioral attributes are observed to take place in different parts of the brain. The foregoing discrete formulation is given in my book about the brain [12]. The system of equations: n X ai j w j = λmax wi (1) j=1 n X wi = 1 (2) i=1 with a ji = 1/ai j or ai j a ji = 1 (the reciprocal property). Multiplying the hypermatrix by itself allows for combining the functions that represent the influence from pre. By raising the hypermatrix to powers one obtains transitive interactions.
sound. is received as a similarity transformation as. As in the finite case. t. Generalizing this result we have: K (as. If K (s. In the discrete case. It is a stretching if a > 1. A stimulus S of magnitude s. u) = K (s. we see by substituting in the equation that Cw(t) is also an eigenfunction corresponding to the same λ. A necessary and sufficient condition for w(s) to be an eigenfunction solution of Fredholm’s equation of the second kind. t) is not only positive. s) = 1 (6) so that K (s.L. t)w(t)dt = w(s) (4) a with the normalization condition: Z b w(s)ds = 1. K (s. The most important part of what follows is the derivation of the fundamental equation. t) is consistent if it satisfies the relation K (s. Theorem 1. s) = 1 for all s which is analogous to having ones down the diagonal of the matrix in the discrete case. t)K (t. and u. Theorem 3. at each instant. the normalized eigenvector was independent of whether all the elements of the pairwise comparison matrix A are multiplied by the same constant a or not. Here also. The value λ = 0 is not a characteristic value because we have the corresponding solution w(t) = 0 for every value of t. sight. the solution of (4) is given by k(s) w(s) = R . heat and cold. the kernel K (s. at) = a K (s. and a contraction if a < 1. (9) s k(s)ds We note that this formulation is general and applies to all situations where a continuous ratio scale is needed. It is clear that whatever aspect of the real world we consider. t) is consistent. (5) a An eigenfunction of this equation is determined to within a multiplicative constant. smell. t) is consistent if and only if it is separable of the form: K (s. their corresponding stimuli impact our senses numerous times. If w(t) is an eigenfunction corresponding to the characteristic value λ and if C is an arbitrary constant. for all s. t) = k(s)/k(t). (7) It follows by putting s = t = u. t) = es−t = es /et . t)K (t. t) = k(as)/k(at) = ak(s)/k(t) (10) which means that K is a homogeneous function of order one. which is the trivial case. but also reciprocal. u). An example of this type of kernel is K (s. Saaty / Mathematical and Computer Modelling 46 (2007) 860–891 The standard way in which (3) is written is to move the eigenvalue to the left hand side which gives it the reciprocal form: Z b λ K (s. with a consistent kernel that is homogeneous of order one is that it satisfy the functional equation w(as) = bw(s) (11) where b = αa. We now determine the form of k(s) and also of w(s). that K (s. . taste. touch. a functional equation whose solution is an eigenfunction of our basic Fredholm equation. we have the reciprocal property K (s. (8) Theorem 2.884 T. It applies equally to the derivation or justification of ratio scales in the study of scientific phenomena. excluded in our discussion. and thus we can replace A by aA and obtain the same eigenvector. a > 0 referred to as a dilation of s.
Even without the multivaluedness of P. but could not if the second weight is only 20. log α Note in (12) that the periodic function P(u) is bounded and the negative exponential leads to an alternating series. they could between 40 and 42 g. if p(0) < 0. Thus.L. we have the proportionality relation we just wrote down: w(as) = b. while they could not distinguish between 40 and 41 g. to a first order approximation one obtains the Weber–Fechner law for response to a stimulus s : A log s + B. This solution leads to the Weber–Fechner law (log sn − log s0 ) n= . we may call it: The Fundamental Equation of Proportionality and Order. then w(z) is a singlevalued or finitely multivalued function. In 1846 Weber found. If P is single valued and ln b/ ln a turns out to be an integer or a rational number. the function w(z) could be multivalued because ln b/ ln a is generally a complex number. that people while holding in their hand different weights.2.5 g. C < 0. Saaty / Mathematical and Computer Modelling 46 (2007) 860–891 885 When relating response to a dilated stimulus of magnitude as to response to an unaltered stimulus whose magnitude is s. then so is u which may be negative even if a and s are both assumed to be positive. w(s) We refer to (11) as: The Functional Equation of Ratio Scales. (13) Here P(u) with u = ln z/ ln a. Again if we write. is an arbitrary multivalued periodic function in u of period 1. On the other hand. Because s is the magnitude of a stimulus and cannot be negative. We assume that B = 0. T. and so on at higher levels. We need to increase a stimulus s by a minimum amount ∆s to reach a point where our senses can first discriminate between s and s + ∆s. w(a u ) = bu p(u) we get: p(u + 1) − p(u) = 0 which is a periodic function of period one in the variable u (such as cos u/2B). If in the last equation p(0) is not equal to 0. Note that C > 0 only if p(0) is positive. we can introduce C = p(0) and P(u) = p(u)/C. the exponential factor which is equal to s log b/ log a . Because of its wider implications in science. respectively. could distinguish between a weight of 20 g and a weight of 21 g. we do not have a problem with complex variables here so long as both a and b are real and both positive. ∆s is called the just noticeable difference (jnd). we have for the general response function w(s): log s log b log log s w(s) = Ce a P (12) log a where P is also periodic of period 1 and P(0) = 1. for example. Near zero. If we substitute s = a u in (11) we have: w(a u−1 ) − bw(a u ) = 0. The ratio r = ∆s/s does not depend on s. and hence the response belongs to a ratio scale. Our solution in the complex domain has the form: w(z) = z ln b/ ln a P(ln z/ ln a). Weber’s law states that change in sensation is noticed when the stimulus is increased by a constant percentage of the . Note that if a and s are real. This generally multivalued solution is obtained in a way analogous to the real case. “slims” w(s) if log b/ log a > 0 and “spreads” w(s) if log b/ log a < 0. Otherwise. 9.
By dividing each Mi by M1 we obtain the sequence of absolute numbers 1. . Paired comparisons are made by identifying the less dominant of two elements and using it as the unit of measurement. 2.). someone might think that other scale values would be better. which is possible by calibrating a unit stimulus. Here the unit stimulus is s0 . a 6= 0. . The third noticeable stimulus is s2 = s0 α 2 which yields a response of 2a log α. Saaty / Mathematical and Computer Modelling 46 (2007) 860–891 stimulus itself. . . But the latter are obtained when we solve for n. He denotes the first one by s0 . the psychophysical law of Weber–Fechner is given by: M = a log s + b. Similarly s2 = s1 + ∆s1 = s1 (1 + r ) = s0 (1 + r )2 ≡ s0 α 2 . for example using 1. 3. . Imagine comparing the magnitude of two people with respect to the magnitude of one person and using 1. Mn = na log α. . from which we must have log s0 = 0 or s0 = 1. The next noticeable stimulus is s1 = s0 α = α which yields the second noticeable response a log α. This law holds in ranges where ∆s is small when compared with s. . In general sn = sn−1 α = s0 α n (n = 0. Aggregating or decomposing stimuli as needed into clusters or into levels in a hierarchy is an effective way to extend the uses of this law. using the scale 1–9 or its verbal equivalent. Fechner noted that the corresponding sensations should follow each other in an arithmetic sequence at the discrete points at which just noticeable differences occur. Despite the foregoing derivation of the scale in the form of integers. . While the noticeable ratio stimulus increases geometrically. Thus b = 0. how many times more the dominant member of the pair is than this unit. . Thus if M denotes the sensation and s the stimulus.3 for how many there are instead of 2. and hence in practice it fails to hold when s is either too small or too large. We now need to take its inverse to develop expressions for the response. of the fundamental 1–9 scale. relying on the insensitivity of the eigenvector to small perturbations (discussed below). Thus we have for the different responses: M0 = a log s0 . We are interested in responses whose numerical values are in the form of ratios. In making paired comparisons. The reciprocal value is then automatically used for the comparison of the less dominant element with the more dominant one.886 T. What we have been doing so far is concerned with the inverse Fourier transform. In 1860 Fechner considered a sequence of just noticeable increasing stimuli. we use the nearest integer approximation from the scale. the response to that stimulus increases arithmetically. This is transformed to awareness of nature by converting the electrical synthesis (vibrations caused by a pattern) to a space–time representation. The way the brain goes back and forth from a pattern of stimuli to its electrochemical synthesis and then to a representation of its response to that spatio-temporal pattern is by applying the Fourier transform to the stimulus and the inverse Fourier transform to form its response. The next just noticeable stimulus is given by ∆s0 s1 = s1 + ∆s0 = s0 + s0 = s0 (1 + r ) s0 based on Weber’s law. 2.3 in the place of 2. . Note that M0 = 0 and there is no response. . M2 = 2a log α. Stimuli received by the brain from nature are transformed to chemical and electrical neural activities that result in summation and synthesis. We assume that the stimuli arise in making pairwise comparisons of relatively comparable activities.L. One then determines. We have: (log sn − log s0 ) n= log α and sensation is a linear function of the logarithm of the stimulus. Thus stimuli of noticeable differences follow sequentially in a geometric progression. M1 = a log α. 1.
It is known in the case of vision that the eyes do not scan pictures symmetrically if they are not symmetric. α. T. or to some sort of deformation. The reason is that complex functions contain an imaginary part. The density of neural firing is not completely analogous to the density of the rational numbers in the real number system. and on approximations to them without regard to the coefficients in (15). Our representation focuses on the real part of the magnitude rather than the phase of the Fourier transform. . a bird. 9. Saaty / Mathematical and Computer Modelling 46 (2007) 860–891 887 We now show that the space–time Fourier transform of (13) is a combination of Dirac distributions. t)K(t. If limξ →0 K (ξ s. The functions result from modeling the neural firing as a pairwise comparison process in time. This invariance would allow the network to recognize images even when they are not identical to the ones from which it recorded a given concept. on 0 ≤ s. one can make a plot of the modulus or absolute value of such a function. We use the functions: {t α e−βt . The response function w(s) of the neuron in spontaneous activity results from solving the homogeneous equation (4). The next step would be to use the network representation given here with additional conditions to uniquely represent patterns from images. In speaking of density here we may think of making a sufficiently close approximation (within some prescribed bound rather than arbitrarily close). For example. Thus. The rationals are countably infinite. t ≤ b is Lebesgue square integrable and continuously differentiable. Tests have been made to see the effect of phase and of magnitude on the outcome of a representation of a complex valued function. Nevertheless. for all s and t.g. α. b] we can approximate t α eg(t) by linear combinations of t α e−βt and hence we substitute g(t) = −βt. t) > 0. where K is a compact integral operator defined on the space L2 [0. and each stimulus value is represented by the firing of a neuron in each layer. A reciprocal kernel K is an integral operator that satisfies the condition K(s. Our solution of Fredholm’s equation here is given as the Fourier transform. β ≥ 0} to represent images and sounds. Thus we focus on representing responses in terms of Dirac functions.3. It is assumed that a neuron compares neurotransmitter-generated charges in increments of time.L. There is much more blurring due to change in magnitude than there is to change in phase. Because finite linear combinations of the functions {t α e−βt .. If the reciprocal kernel K(s. e. Taking into account this principle would allow us to represent images independently of the form in which stimuli are input into the network. the number of neurons is finite but large. Z +∞ f (ω) = F(x)e−2πiωx dx = Ceβω P(ω) (14) −∞ with its inverse transform given by: ∞ (2π n + θ (b) − x) X (1/2π ) log a 0 an i δ(2πn + θ (b) − x) (15) −∞ (log a|b| + (2πn + θ (b) − x)) where δ(2π n + θ (b) − x) is the Dirac delta function. β ≥ 0} are dense in the space of bounded continuous functions C[0. sounds and perhaps other sources of stimuli such as smell. we recognize an image even if it is subjected to a rotation. b] of Lebesgue square integrable functions. sums of such functions. the invariance principle must include affine and similarity transformations. then . The formation of images and sounds with Dirac distributions Complex valued functions cannot be drawn as one does ordinary functions of three real variables. in the eigenfunction w(t). This is supporting evidence in favor of our ratio scale model.Z b α g(t) w(t) = t e t α eg(t) dt 0 satisfies (4) for some choice of g(t). β ≥ 0. This leads to the continuous counterpart of a reciprocal matrix known as a reciprocal kernel. ξ t) exists. s) = 1. and hence our representation must satisfy some order invariant principle. The basic assumption we made to represent the response to a sequence of individual stimuli is that all the layers in a network of neurons are identical. A shortcoming of this representation is that it is not invariant with respect to the order in which the stimuli are fed into the network.
4. the x-coordinate was used to represent time and the y-coordinate to represent response to a stimulus.. R. Thus. the stimuli list consists of 124 values. k = 1. y) coordinates of the points were obtained. . We also assumed that the firing threshold of each neuron had the same width. The numerical values associated with the drawings in Figs. For illustrative purposes. It follows that the spontaneous activity of a neuron during a very short period of time in which the neuron fires is given by: R γk (t − τk )α e−β(t−τk ) X w(t) = k=1 if the neuron fires at the random times τk . . Under the assumption that each neuron fires in response to one stimulus. . and each layer of the network has n neurons. Saaty / Mathematical and Computer Modelling 46 (2007) 860–891 Before we describe how the network can be used to represent images and sounds.888 T. respectively.e.5. Empirical findings support the assumption that R and the times τk . Sound experiment In the sound experiment we first recorded with the aid of Mathematica the first few seconds of Haydn’s symphony no. whereas the sound experiment required 1000 times more data points. . The result is a set of numerical amplitudes between −1 and 1. we consider an inhomogeneous equation to represent stimuli acting on the neuron in addition to existing spontaneous activity. . However. if the perceptual range of a stimulus varies between two values θ1 and θ2 . Picture experiment In the graphics experiment the bird and rose pictures required 124 and 248 data points. Thus. t)w(t)dt = f (s). 102 in B-flat major and Mozart’s symphony no. k = 1. we created n layers with a specific number of neurons in each layer. For example. arranged in 124 layers of 124 neurons each The network and the data sampled to form the picture given in Fig. Neural responses are impulsive and hence the brain is a discrete firing system. 23. each layer of the network must also consist of n neurons with thresholds varying between the largest and the smallest values of the list of stimuli. Under the assumption that each numerical stimulus is represented by the firing of one and only one neuron.L. . but are constant for the firings of each neuron. . for the bird picture. 117. 40 in G minor. i.532 in Mozart’s symphony. Each of these amplitudes was used to make neurons fire when the amplitude falls within a prescribed threshold range. To derive the response function when neurons are stimulated from external sources. we would need the same number of neurons as the sample size. we summarize the mathematical model on which the neural density representation is based. We created a two-dimensional network of neurons consisting of layers. and we would need 1242 = 15 376 neurons. R are probabilistic. the parameters α and β vary from neuron to neuron. . This task is computationally demanding even for such simple geometric figures as the bird and the rose. 0 This equation has a solution in the Sobolev space: W pk (Ω ) of distributions (in the sense of Schwartz) in L p (W ) whose derivatives of order k also belong to the space L p (W ). we assume that there is one layer of neurons corresponding to each of the stimulus values. if the list of stimuli consists of n numerical values. and then use the resulting . where W is an open subset of Rn . Thus. were used to create a 124H124 network of neurons consisting of 124 layers with 124 neurons in each layer. Each dot in the figures is generated by the firing of a neuron in response to a stimulus falling within the neuron’s lower and upper thresholds. we solve the inhomogeneous Fredholm equation of the 2nd kind given by: Z b w(s) − λ0 K (s. Our objective was to approximate the amplitude using one neuron for each amplitude value. 23 and 24 were tabulated and the numbers provided the input to the neurons in the networks built to represent the bird and the rose. 2. Once the (x. 9. then a neuron in the ith position of the layer will fire if the stimulus value falls between θ2 − θ1 θ2 − θ1 θ1 + (i − 1) and θ1 + i n−1 n−1 9.247 in Haydn’s symphony and 144. 2. Non-spontaneous activity can be characterized as a perturbation of background activity.
Nearly 250 years later it became known form the works of Abel and Ruffini and from Galois theory that a quartic is the largest degree equation for which one can obtain the roots (eigenvalues in our problem) in the form of radicals.L. one of the oldest of the medieval universities and a school with a strong mathematical tradition. Rose. quintic and higher order equations Solution of the cubic and the quartic became common knowledge through the publication of the Ars Magna of Geronimo Cardano (1501–1576). But Cardano was not the original discoverer of the solution of either the cubic or the quartic. Fig. The names of Niccolo Fontana (1500–1557) the stammerer (Tartaglia) is also associated with the solution of the cubic because he elaborated on Del Ferro’s solution and became addicted to solving cubic equations. T. values in Mathematica to play back the music. The solution of the quartic is attributed to Ludovico Ferrari (1522–1565). Bird. It was from Ferrari in 1547–48 that the history of this spectacular discovery became public knowledge. Saaty / Mathematical and Computer Modelling 46 (2007) 860–891 889 Fig. Brief history of solving cubic. 40. quartic. A small sample of the numerical data for Mozart’s symphony is displayed in Fig. 25. 24. 10. Fig. Generally the solution of the cubic is attributed to the Italian professor Scipione del Ferro (1465–1526). Mozart’s symphony no. Ferrari reduced the solution of the general bi-quadratic equation to that of a cubic equation. from Bologna. 23. It was Cardano who considered negative numbers and called them “fictitious”. This means that the roots can be expressed as a finite formula involving only the four arithmetic operations and . 25.
Umemura [6] does not require the use of Tschirnhausen transformations. nor is PC computer technology necessarily up to the task. To solve a quintic equation in symbolic form in order to get the principal eigenvalue we are told by the chart distributed in 1994 by Wolfram Research on the quintic equation that it requires a trillion bytes. für Math. 1878) using theta series P(see below. q) = 2 n=0 ∞ X 2 θ3 (z. . Once we have the outcome. One then approximates these values by curves with a functional form for each coefficient. power series. a0 6= 0 is reduced to z n + b1 z n−4 + · · · + bn−1 x + bn = 0 with three fewer terms. Hermite. Abel (1802–1829) proved that the general equation of degree higher than four could not be solved using radicals. it is a relatively complicated process to weight and synthesize time dependent priorities for the alternatives. A root of this equation can be expressed in terms of theta functions of zero argument involving the period matrix derived from one of two types of hyperelliptic integrals. ∞ n=0 nc t n . q) = 1 + 2 q n cos 2nz. Let f (x) = a0 x n + a1 x n−1 + · · · + an = 0. One wonders what solving higher order equations would require.L. P 114–133. ∞ q (n+1/2) cos(2n + 1)z X 2 θ2 (z. P∞ and Puiseux series often encountered in this connection the form n=−∞ cn t n/k k. Hyperradicals is the term used for such functions. z)4 Camille Jordan proved in 1870 that an algebraic equation of any degree can be solved in terms of modular functions. solve it in the reduced form and then apply the inverse transformation to obtain the solution of the general equation. put the outcome in a supermatrix. showed that elliptic modular functions provide solutions to the quintic equation. The mathematics to do this in general symbolic form is not here yet. z) 4 ϕ(z) = . Jacobi and Abel in 1827 studied these functions for the first time. the equation a0 x n + a1 x n−1 + · · · + an−1 x + an = 0. and thus the quintic equation takes the form z 5 + b4 z + b5 = 0. for example. All algebraic equations of degree 4 and less can be solved using only square and cubic roots. we can analyze its rate of change with respect to time by taking its derivative. Laurent series the form ∞ n n=−∞ cn t . 87. a0 6= 0 be an algebraic equation irreducible over a subfield of the complex numbers. Repeating the process for different values of time generates a curve for the priorities. The upshot is that we can do time dependent AHP/ANP numerically by simulation. Saaty / Mathematical and Computer Modelling 46 (2007) 860–891 the extraction of roots. The general quintic equation can be solved by Kiepert’s algorithm (Auflösung der Gleichungen vom Fünften Grades. Galois (1811–1832) provided a method to characterize equations that are solvable by radicals. We have already given symbolic expressions for the principal eigenvalue and eigenvector in symbolic form for n#4 in my first book on the AHP.1. n=0 The elliptic modular function ϕ(z) is given by: s 8 θ2 (0. Theta functions are periodic functions that can be represented by series whose convergence is extraordinarily rapid. solve the problem and derive the outcome for the alternatives. J.890 T. One would express the judgments functionally but then derive the eigenvector from the judgments for a fixed moment of time. Actually. His method involved the development of group theory to determine the effect of permuting the roots of the equation on functions of the roots. The following theorem of H. or one gigabyte of storage. 10. Still even in these cases. Solutions of algebraic equations using radicals are called radical solutions. The Quintic and higher order cases The poster distributed by Wolfram Research in 1994 gives a nice introduction to the classical methods for symbolic computation of the roots of an algebraic equation. We have. The fact that radicals cannot be used in all cases of the quintic and higher order equations leads to the question as to what type of functions are needed to solve equations of higher degree than 4. sometimes also called Taylor series have the form.. The solution of the quartic equation can be related to the 4! = 24 symmetries of the tetrahedron. Tschirnhausen transformations are often used to simplify the form of an algebraic equation. θ3 (0. and those of the quintic equation to the 5! = 120 symmetries of the icosahedron. a fixed natural number).
Saaty / Mathematical and Computer Modelling 46 (2007) 860–891 891 11. RWS Publications. Management Science 26 (7) (1980). RWS Publications. in: ISAHP Proceedings 2001. Perhaps one of the advantages of dynamic judgments may be that the requirement of homogeneity is given up to the mathematics of the functions one uses. Fundamentals of Decision Making with the Analytic Hierarchy Process. Portuguese. The Czech Republic. Intuitively I think this is a considerable worthy challenge of our effort to extend the AHP/ANP. Pittsburgh. Forecasting the resurgence of the US economy in 2001: An expert judgment approach.R. J. A choice of a prospective system for vibration isolation in conditions of varying environment. [7] R. 1985. [9] A. [6] D. Birkhaüser. vol. In that case homogeneity is not needed as a requirement on the fundamental scale. [5] R. [11] T. Vargas for his dexterous and caring help with several tables and figures and for some material from our joint paper. there would always be technological problems with mathematical engineering design that require the use of AHP/ANP in dynamic form. PA 15213. Johnson. Saaty.L.L. [12] T.L. [4] Y. 2001. It seems to me that while this may be reasonable for technical but not behavioral problems. Kuffler. Hypermatrix of the brain. II. Solving algebraic equations in terms of A-hypergeometric series. pp. in: Invited Lecture. One may be able to compare the relation between two machines constructed in a systematic way so that their relative behavior is predetermined. 2000. Acknowledgment My appreciation goes to my friend and colleague Luis G. O.G. Springer-Verlag.). . Cambridge University Press. 1996). paperback. Bern. Whitaker. Blair. C. Unraveling the Mystery of How it Works: The Neural Network Process. PA 15213-2807. Saaty. From Neuron to Brain. 13–24. Horn. [2] T. Pittsburgh. The Analytic Hierarchy Process.A. Andreichicov. T. R. One thing we have to learn to do is to create a scale that is truly associated with our intuition so that we can use it in the natural and spontaneous way people have learned to use the scale 1–9 that is associated with verbally expressed judgments. Discrete Mathematics 210 (1–3) (2000) 171–181. 4922 Ellsworth Avenue. R. Saaty.L. Prague. Saaty. RWS Publications. Academy of Sciences. 4922 Ellsworth Avenue. 1980 (Later translated to Russian. and Chinese. References [1] T. but it is certain that soon after they will and I expect that extensions of this paper will make that appear even more plausible. later included in my work on the brain [12].L. in: Tata Lectures on Theta. Matrix Analysis. Socio-Economic Planning Sciences 36 (2002) 77–91. T. Saaty. 1984. New York. Pittsburgh. 2001. [3] A. USA. Nichols. T. [10] S. Boston. Mumford (Ed. Switzerland. Wind. 1994. Beyond the Quartic Equation. 1990. Sunderland. Marketing applications of the analytic hierarchy process.L. McGraw-Hill International. Resolution of algebraic equations by theta constants.L. Saaty. original edition 1994. 1976. Sturmfels. [8] B. MA. Nachtmann. Sinauer Associates. The first applications may not be taxing to our ingenuity to deal with functions instead of numbers.R. revised 2000.B. The Brain. Birkhaüser. Revised edition published in paperback. Conclusions It appears that independently from whether people like to do (or can do) dynamic comparisons. King. Andreichicova. | 22,778 | 84,391 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2018-47 | latest | en | 0.926611 |
https://www.reasonablefaith.org/forums/index.php?topic=6033815.0 | 1,611,560,429,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703565376.63/warc/CC-MAIN-20210125061144-20210125091144-00242.warc.gz | 931,648,628 | 23,421 | # Ontological Argument
#### cnearing
• 2677 Posts
##### The Modal Ontological Argument Begs the Question
« on: March 23, 2016, 09:02:02 pm »
Let G be the proposition "a maximally great being exists."
By definition, then, if G exists, G exists necessarily. Hence:
1.) G iff []G
2.) <>G iff <>[]G (from 1)
3.) <>[] G iff []G (s5)
4.) G iff <>G (from 1 - 3)
It is fairly trivial, as this argument shows, that G and <>G entail each other. This means, definitionally, that they have identical entailments and therefore identical extensions--literally, identical literal meanings.
And, of course, we can continue:
5.) If a premise in an argument and the argument's conclusion have the same literal meaning, the argument begs the question.
6.) <>G is a premise in the MOA and G is the conclusion of the MOA.
7.) <>G and G have the same literal meaning. (From 4)
C.) Therefore, the MOA begs the question.
P((A => B), A) = P(A => B) + P(A) - 1
1
#### alex1212
• 761 Posts
##### Re: The Modal Ontological Argument Begs the Question
« Reply #1 on: March 27, 2016, 01:20:37 pm »
Let G be the proposition "a maximally great being exists."
By definition, then, if G exists, G exists necessarily. Hence:
1.) G iff []G
2.) <>G iff <>[]G (from 1)
3.) <>[] G iff []G (s5)
4.) G iff <>G (from 1 - 3)
It is fairly trivial, as this argument shows, that G and <>G entail each other. This means, definitionally, that they have identical entailments and therefore identical extensions--literally, identical literal meanings.
And, of course, we can continue:
5.) If a premise in an argument and the argument's conclusion have the same literal meaning, the argument begs the question.
6.) <>G is a premise in the MOA and G is the conclusion of the MOA.
7.) <>G and G have the same literal meaning. (From 4)
C.) Therefore, the MOA begs the question.
No.
1. If it is possible that God exists, then God exists.
2. It is possible that God exists.
3. Therefore, God exists.
2 and 3 are not the same. The evidence for premise 2 can not include the conclusion, 3. Proposed evidence for premise 2 includes things like modal intuitions, religious experience, and other arguments for God's existence.
2
#### cnearing
• 2677 Posts
##### Re: The Modal Ontological Argument Begs the Question
« Reply #2 on: March 27, 2016, 02:26:32 pm »
Given that I have proven your claim trivially false, already, and you responded with nothing more than a flat denial and no response to the actual logic of my proof, you'll forgive me for assuming that you are the troll you certainly appear to be.
If you want to have an actual discussion, feel free to address my *formal logical proof* in some substantive manner.
P((A => B), A) = P(A => B) + P(A) - 1
3
#### Atheist in Louisiana
• 2631 Posts
• I ain't afraid of no ghost!
##### Re: The Modal Ontological Argument Begs the Question
« Reply #3 on: March 28, 2016, 12:03:58 am »
To answer your original question, yes the ontological argument begs the question in modal form. It does it in the others too.
Had the magazine not published these cartoons, they would not have been specifically targeted.
Consequences, AiL, consequences. - Jenna Black
Hey, if you want to, I'm more than ok with it. I love the attention. - Questions11
4
#### alex1212
• 761 Posts
##### Re: The Modal Ontological Argument Begs the Question
« Reply #4 on: March 28, 2016, 02:26:33 pm »
Given that I have proven your claim trivially false, already, and you responded with nothing more than a flat denial and no response to the actual logic of my proof, you'll forgive me for assuming that you are the troll you certainly appear to be.
If you want to have an actual discussion, feel free to address my *formal logical proof* in some substantive manner.
I don't see the problem, and I'm not trolling. It seems to me that all you've done is just described the nature of a deductive argument. I don't see, "Possibly P" and "P" as identical. I don't think the argument is a good argument but not because I think it begs the question. I did respond to the OP and you didn't reply to what I said.
5
#### alex1212
• 761 Posts
##### Re: The Modal Ontological Argument Begs the Question
« Reply #5 on: March 28, 2016, 02:33:57 pm »
But for those who think there's a problem, Craig responds by noting the de dicto and de re distinction. (of the thing vs. of the word)
6
#### alex1212
• 761 Posts
##### Re: The Modal Ontological Argument Begs the Question
« Reply #6 on: March 28, 2016, 02:39:30 pm »
7
#### cnearing
• 2677 Posts
##### Re: The Modal Ontological Argument Begs the Question
« Reply #7 on: March 28, 2016, 02:41:44 pm »
P and possibly P are not the same in all cases, but hey are trivially the same in the case of P = "a maximally great being exists" or "maximal greatness is instantiated."
As I have proven.
Craig's response regarding de dicto and de res does not address this argument at all.
My proof holds in both the de dicto mode and the de res mode.
P((A => B), A) = P(A => B) + P(A) - 1
8
#### cnearing
• 2677 Posts
##### Re: The Modal Ontological Argument Begs the Question
« Reply #8 on: March 28, 2016, 02:52:37 pm »
Also, note that *by definition* you cannot actually offer the modal ontological argument in the de res mode.
Specifically, possibility and necessity are, by definition, features of propositions, since possible worlds themselves are explicitly sets of propositions.
P((A => B), A) = P(A => B) + P(A) - 1
9
#### alex1212
• 761 Posts
##### Re: The Modal Ontological Argument Begs the Question
« Reply #9 on: March 28, 2016, 02:58:43 pm »
P and possibly P are not the same in all cases, but hey are trivially the same in the case of P = "a maximally great being exists" or "maximal greatness is instantiated."
As I have proven.
Craig's response regarding de dicto and de res does not address this argument at all.
My proof holds in both the de dicto mode and the de res mode.
What I mean is that Possibly P (the premise) entails P (conclusion). In other words, you've just described the nature of a deductive argument. The support for God possibly existing is not that God exists. If P is possibly necessary, then it's necessarily possible that P.
10
#### cnearing
• 2677 Posts
##### Re: The Modal Ontological Argument Begs the Question
« Reply #10 on: March 28, 2016, 03:38:21 pm »
The support for the possibility of Gid existing cannot--by definition--be anything *less* than the existence of God.
As I proved, the existence of God is a *necessary condition* for the possible existence of God.
You literally cannot support a proposition without supporting all of its necessary conditions at least as well. That's what necessary conditions are.
P((A => B), A) = P(A => B) + P(A) - 1
11
#### alex1212
• 761 Posts
##### Re: The Modal Ontological Argument Begs the Question
« Reply #11 on: March 28, 2016, 04:42:38 pm »
The support for the possibility of Gid existing cannot--by definition--be anything *less* than the existence of God.
As I proved, the existence of God is a *necessary condition* for the possible existence of God.
You literally cannot support a proposition without supporting all of its necessary conditions at least as well. That's what necessary conditions are.
That's totally backwards. It has to be possible for God to exist in order for God to exist. If it's impossible for God to exist, then God does not exist.
What points do you disagree with in the video?
12
#### alex1212
• 761 Posts
##### Re: The Modal Ontological Argument Begs the Question
« Reply #12 on: March 28, 2016, 04:46:27 pm »
The only thing that I will grant that you have "proved" is that you can describe the nature of deductive arguments. The nature of a deductive argument is that the conclusion is already hidden in the premises waiting to be revealed by the rules of logic.
13
#### ParaclitosLogos
• 4902 Posts
##### Re: The Modal Ontological Argument Begs the Question
« Reply #13 on: March 28, 2016, 06:29:02 pm »
1. Clark Kent is with louise at niagara falls
2. Clark is superman (Clark Kent exists iff Superman exists.Clark is with Louis iff Superman is with Louise)
3. Superman is with louise at niagara falls
To beg the question Louise needs to believe 1 or 2 because she believes 3.
Support for 1. Louis Lane is with Clark at niagara falls on an assignment.
Sport for 2. when Louis went at Niagara Falls with Clark he felt onto fire and did not burn, and his eyeglasses fell(his wonderful disguise), at all, and, after all he looks like superman.
She does not believes 1 because she believes 3, nor believes 2 because she believes 3
Sadly, afterwards Louise got a defeater for 2 since Clark Kent did not give up his facade and tried to save her when she jumped into the falls.
Arguments do not beg the question. People beg the question.
14
#### aleph naught
• 7392 Posts
• For the glory of the Canadian empire.
##### Re: The Modal Ontological Argument Begs the Question
« Reply #14 on: March 28, 2016, 06:30:54 pm »
This means, definitionally, that they have identical entailments and therefore identical extensions--literally, identical literal meanings.
That's false. Just because P and Q are logically equivalent, it doesn't follow that they mean the same thing. <>T is only logically equivalent with T on S5, it's not logically equivalent on S4. But obviously it's not a matter of semantics whether or not S5 or S4 are the proper modal logic system to do metaphysics in. | 2,482 | 9,472 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2021-04 | latest | en | 0.851821 |
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## Is chart overlay always a line graph or can it be changed it to something else?
Motivator
I'm looking to reverse the axis on this graph, but keep the visualization the same. I can't find a way to do this, because chart overlay seems to only use a line. Is it possible to force chart overlay to use something besides a line?
###
If this reply helps you, an upvote would be appreciated.
Tags (5)
1 Solution
Motivator
Hope this helps:
1) Overlay seems to be line always
2) How far away from the edge an overlay will be dependent on:
• What visualization has been chosen, for example same data when represented as line graph or area chart will overlay the line to the edge.
• How many items are being graphed for each x axis element in case of bar chart and how many x axis points are there in total. For example if you had Volume, Volume(squared) and Volume(cubed) bars for each Sunday (and you were plotting only two Sundays) then the line might be even further away from the edge as now one x axis element (first Sunday) need to plot three bars and the line point which shall represent overlay shall be the middle point of "distance" from the edge of axis till first grouping plot end. If you were to plot these for say 20 Sundays now, the line will shift closer, given the bars will be slimmer and the "distance" used to plot the three bars per Sunday will be slimmer.
What is the expected output when you mention I'm looking to reverse the axis on this graph, but keep the visualization the same.
Motivator
try partial=f on your timechart like this
your search here |timechart partial=f count by whatever
Motivator
Thank you. Result was no different.
###
If this reply helps you, an upvote would be appreciated.
Motivator
Hmmm, I guessing gokadroid is correct. It appears to be the width of the columns that is shorting the lines in the overlay. What happens if you select other types of graphs for the primary (column)?
Motivator
Hope this helps:
1) Overlay seems to be line always
2) How far away from the edge an overlay will be dependent on:
• What visualization has been chosen, for example same data when represented as line graph or area chart will overlay the line to the edge.
• How many items are being graphed for each x axis element in case of bar chart and how many x axis points are there in total. For example if you had Volume, Volume(squared) and Volume(cubed) bars for each Sunday (and you were plotting only two Sundays) then the line might be even further away from the edge as now one x axis element (first Sunday) need to plot three bars and the line point which shall represent overlay shall be the middle point of "distance" from the edge of axis till first grouping plot end. If you were to plot these for say 20 Sundays now, the line will shift closer, given the bars will be slimmer and the "distance" used to plot the three bars per Sunday will be slimmer.
What is the expected output when you mention I'm looking to reverse the axis on this graph, but keep the visualization the same.
Motivator
Point 2 does not help, but does makes sense. Thank you. To answer your question, the duration axis would be on the left, but still represented by lines, and the Volume axis would be on the right, and still represented by columns.
###
If this reply helps you, an upvote would be appreciated.
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There are two approaches to undergraduate and graduate courses in linear statistical models and experimental design in applied statistics. One is a two-term sequence focusing on regression followed by ANOVA/Experimental design. Applied Linear Statistical Models serves that market. It is offered in business, economics, statistics, industrial engineering, public health, medicine, and psychology departments in four-year colleges and universities, and graduate schools. Applied Linear Statistical Models is the leading text in the market. It is noted for its quality and clarity, and its authorship is first-rate. The approach used in the text is an applied one, with an emphasis on understanding of concepts and exposition by means of examples. Sufficient theoretical foundations are provided so that applications of regression analysis can be carried out comfortably. The fourth edition has been updated to keep it current with important new developments in regression analysis.
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Variable 3 Inferences in Regression Analysis 46 Regression Analysis 69 Measurement 138 Simultaneous Inferences and Other 152 Matrix Approach to Simple Linear 176 Part II 214 Matrix Terms 225
Part VI 793 Analysis of Factor Effects 849 TwoFactor StudiesOne Case 875 Multifactor Studies 924 Random and Mixed Effects 958 Analysis of Covariance 1010 Study Designs 1043 Randomized Block Designs 1072
Multiple RegressionII 260 Building the Regression Model 327 Part VII 329 Building the Regression Model 354 Diagnostics 361 Qualitative Predictor Variables 455 Autocorrelation in Time Series 497 Nonlinear Regression 529 Logistic Regression Poisson 567 Normal Correlation Models 631 PartV 661 Analysis of Factor Level Effects 710 ANOVA Diagnostics and Remedial 756
Nested Designs Subsampling 1121 Repeated Measures and Related 1164 Latin Square and Related 1207 Exploratory ExperimentsTwoLevel 1234 Response Surface Methodology 1280 Appendix A Some Basic Results in Probability 1313 Appendix B Tables 1336 Data Sets 1365 Rules for Developing ANOVA Models 1373 Appendix E Selected Bibliography 1392 Index 1400 Copyright | 473 | 2,301 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-22 | longest | en | 0.854333 |
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# What was as remarkable as the development of the Compact
Author Message
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What was as remarkable as the development of the Compact [#permalink]
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26 Jan 2009, 20:09
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What was as remarkable as the development of the Compact Disc has been the use of the new technology to revitalize, in better sound than was ever before possible, some of the classic recorded performances of the pre-LP era.
A) What was as remarkable as the development of the Compact Disc
B) The thing that was as remarkable as the developing the Compact Disc
C) No less remarkable than the development of the Compact Disc
D) Developing the Compact Disc has been none the less remarkable as the development than
E) Development of the Compact Disc has been no less remarkable as
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Senior Manager
Joined: 26 May 2008
Posts: 412
Schools: Kellogg Class of 2012
### Show Tags
26 Jan 2009, 20:44
1
x2suresh wrote:
What was as remarkable as the development of the Compact Disc has been the use of the new technology to revitalize, in better sound than was ever before possible, some of the classic recorded performances of the pre-LP era.
A) What was as remarkable as the development of the Compact Disc
B) The thing that was as remarkable as the developing the Compact Disc
C) No less remarkable than the development of the Compact Disc
D) Developing the Compact Disc has been none the less remarkable as the development than
E) Development of the Compact Disc has been no less remarkable as
I'll go with C
B is wrong because "The thing" is wordy and "developing" is not in sync with " the use" in the later part of the sentence.
D is wrong because it is too wordy and unidiomatic
E is wrong because "no less remarkable" requires "than" not "as"
We're left with A and C. I'll choose C because of 2 reasons
1) In A, becuase of the presence of "has been" in the later part of the sentence, the first part of the sentence should be in present tense
2) C is a lot more concise than A is
Cheers,
Unplugged
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Joined: 07 Nov 2007
Posts: 1738
Location: New York
### Show Tags
26 Jan 2009, 21:52
unplugged wrote:
x2suresh wrote:
What was as remarkable as the development of the Compact Disc has been the use of the new technology to revitalize, in better sound than was ever before possible, some of the classic recorded performances of the pre-LP era.
A) What was as remarkable as the development of the Compact Disc
B) The thing that was as remarkable as the developing the Compact Disc
C) No less remarkable than the development of the Compact Disc
D) Developing the Compact Disc has been none the less remarkable as the development than
E) Development of the Compact Disc has been no less remarkable as
I'll go with C
B is wrong because "The thing" is wordy and "developing" is not in sync with " the use" in the later part of the sentence.
D is wrong because it is too wordy and unidiomatic
E is wrong because "no less remarkable" requires "than" not "as"
We're left with A and C. I'll choose C because of 2 reasons
1) In A, becuase of the presence of "has been" in the later part of the sentence, the first part of the sentence should be in present tense
2) C is a lot more concise than A is
Cheers,
Unplugged
I didn't understand below statement and structure of the statement..
Can you elaborate what exactly it trying to say.
What is as remarkable as the development of the Compact Disc has been the use of the new technology to revitalize
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Posts: 412
Schools: Kellogg Class of 2012
### Show Tags
26 Jan 2009, 22:15
First, it says that the development of compact disc is remarkable. Second, it compares the use of new technology to reviatlize some of the old performances to the development of CD.
In other words, it says that both " the reviatlization of old performances" and " the development of CDs" are remarkable
If you remove all the so called "middle men" the sentence will be something like this... "Development of CD is as remarkable as the use of technology to X"..
Hope it is clear
Cheers,
Unplugged
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Joined: 17 Jun 2008
Posts: 1479
### Show Tags
27 Jan 2009, 01:43
Really tough one. I selected A for the reason that I found the structure of C wrong. But, it is clear now why A is wrong for the user of past and present perfect continuous tense simultaneously.
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Re: SC -remarkable [#permalink] 27 Jan 2009, 01:43
Display posts from previous: Sort by | 1,482 | 6,141 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2018-26 | latest | en | 0.949519 |
http://altrushare.com/iphone-5-cable-wiring-diagram-schematics/vW5c-12882/ | 1,532,261,766,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593223.90/warc/CC-MAIN-20180722120017-20180722140017-00032.warc.gz | 17,003,104 | 10,432 | # Lightning Detector Wiring Iphone 5 Cable Diagram Schematics
Friday, January 05th 2018, 09:15:16 AM. Diagram. Posted by Cyrielle Marjolaine
# Iphone 5 Cable Wiring Diagram Schematics
In a digital circuit, power supply voltage levels are constrained to two distinct values – “logic high voltage” (called LHV or Vdd) and “logic low voltage” (called LLV or GND). The GND node in any circuit is the universal reference voltage against which all other voltages are measured (in modern digital circuits, GND is typically the lowest voltage in the circuit). In a schematic, it is often difficult to show lines connecting all GND nodes; rather, any nodes labeled GND are assumed to be connected into the same node. Often, a downward pointing triangle symbol, is attached to a GND node in addition to (or instead of) the GND label. The Vdd node in a digital circuit is typically the highest voltage, and all nodes labeled Vdd are tied together into the same node. Vdd may be thought of as the “source” of positive charges in a circuit, and GND may be thought of as the “source” of negative charges in a circuit. In modern digital systems, Vdd and GND are separated by anywhere from 1 to 5 volts. Older or inexpensive circuits typically use 5 volts, while newer circuits use 1-3 volts.
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Copyright © 2003 - 2018 Domain Media. All sponsored products, company names, brand names, trademarks and logos arethe property of their respective owners. | 554 | 2,377 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-30 | longest | en | 0.800106 |
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UPD 7.24:留一道UVa212……溜了溜了
UVa12096
//将s1,s2合并到s中
set_union(s1.begin(), s1.end(), s2.begin(), s2.end(), inserter(s, s.begin()));
//将s1,s2取交赋给s
set_intersection(s1.begin(), s1.end(), s2.begin(), s2.end(), inserter(s, s.begin()));
//TEMPLATE BEGIN
//#define ONLINE_JUDGE
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#include <climits>
#include <cassert>
//#include <ctime>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <functional>
#include <algorithm>
#include <complex>
#include <bitset>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
using namespace std;
#define FOR(i, a, b) for (int (i) = (a); (i) <= (b); ++ (i))
#define ROF(i, a, b) for (int (i) = (a); (i) >= (b); -- (i))
#define FOREDGE(i, u, head, next) for (int (i) = (head[(u)]); (i); (i) = (next[(i)]))
#define FORSTL(it, A) for (__typeof((A).begin()) it = (A).begin(); it != (A).end(); ++ it)
#define CLR(a, b) memset((a), (b), sizeof (a))
#define CPY(a, b) memcpy((a), (b), sizeof (a))
#define DEBUG cerr << "debug" << endl
#define STAR cerr << "**********" << endl
#define RT return
#define BK break
#define CT continue
#define ALL(x) (x).begin(), (x).end()
#define HEAP priority_queue
#define st first
#define nd second
#define pf push_front
#define pb push_back
#define ppf pop_front
#define ppb pop_back
#define mp make_pair
#define re real()
#define im imag()
#define LIST0(A, n) { \
FOR (i, 0, n - 1) printf("%d ", A[i]); \
printf("\n"); \
} \
#define LIST1(A, n) { \
FOR (i, 1, n) printf("%d ", A[i]); \
printf("\n"); \
} \
#define SHOW0(A, n, m) { \
FOR (i, 0, n - 1) { \
FOR (j, 0, m - 1) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
#define SHOW1(A, n, m) { \
FOR (i, 1, n) { \
FOR (j, 1, m) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
typedef long long LL;
typedef double DB;
typedef long double LDB;
typedef unsigned int UINT;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef vector<DB> VB;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef map<int, int> MII;
typedef map<LL, LL> MLL;
typedef map<int, string> MIS;
typedef map<string, int> MSI;
const int INF(0x3f3f3f3f);
const LL LINF(0x3f3f3f3f3f3f3f3fLL);
const DB DINF(1e20);
const DB EPS(1e-12); //adjustable
const DB PI(acos(-1.0));
inline DB cot(DB x) { RT 1.0 / tan(x); }
inline DB sec(DB x) { RT 1.0 / cos(x); }
inline DB csc(DB x) { RT 1.0 / sin(x); }
template<class T> inline T max(T a, T b, T c) { RT max(max(a, b), c); }
template<class T> inline T max(T a, T b, T c, T d) { RT max(max(a, b), max(c, d)); }
template<class T> inline T min(T a, T b, T c) { RT min(min(a, b), c); }
template<class T> inline T min(T a, T b, T c, T d) { RT min(min(a, b), min(c, d)); }
template<class T> inline void Scan(T& x) { //int, LL
char c;
for (c = getchar(); c <= ' '; c = getchar()) ;
bool ngt(c == '-');
if (ngt) c = getchar();
for (x = 0; c > ' '; c = getchar()) x = (x << 3) + (x << 1) + c - '0';
if (ngt) x = -x;
}
template<class T> inline void Scan(T& x, T& y) { Scan(x), Scan(y); }
template<class T> inline void Scan(T& x, T& y, T& z) { Scan(x), Scan(y), Scan(z); }
//TEMPLATE END
stack<int> S;
map<set<int>, int> ID;
vector<set<int> > Set;
void Init() {
while (!S.empty()) S.pop();
ID.clear();
Set.clear();
}
int GetID(set<int> A) {
if (ID.count(A)) RT ID[A];
Set.pb(A);
RT ID[A] = Set.size() - 1;
}
void Main() {
int T;
cin >> T;
while (T --) {
Init();
int n;
cin >> n;
while (n --) {
string opt;
cin >> opt;
if (opt[0] == 'P') S.push(GetID(set<int>()));
else if (opt[0] == 'D') S.push(S.top());
else {
set<int> A, B, C;
A = Set[S.top()];
S.pop();
B = Set[S.top()];
S.pop();
switch (opt[0]) {
case 'U':
set_union(ALL(A), ALL(B), inserter(C, C.begin()));
BK;
case 'I':
set_intersection(ALL(A), ALL(B), inserter(C, C.begin()));
BK;
case 'A':
C = B;
C.insert(GetID(A));
BK;
}
S.push(GetID(C));
}
cout << Set[S.top()].size() << endl;
}
cout << "***" << endl;
}
}
int main() {
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
Main();
RT 0;
}
//imagasaikou!
UVa1592
//TEMPLATE BEGIN
//#define ONLINE_JUDGE
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#include <climits>
#include <cassert>
//#include <ctime>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <functional>
#include <algorithm>
#include <complex>
#include <bitset>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
using namespace std;
#define FOR(i, a, b) for (int (i) = (a); (i) <= (b); ++ (i))
#define ROF(i, a, b) for (int (i) = (a); (i) >= (b); -- (i))
#define FOREDGE(i, u, head, next) for (int (i) = (head[(u)]); (i); (i) = (next[(i)]))
#define FORSTL(it, A) for (__typeof((A).begin()) it = (A).begin(); it != (A).end(); ++ it)
#define CLR(a, b) memset((a), (b), sizeof (a))
#define CPY(a, b) memcpy((a), (b), sizeof (a))
#define DEBUG cerr << "debug" << endl
#define STAR cerr << "**********" << endl
#define RT return
#define BK break
#define CT continue
#define ALL(x) (x).begin(), (x).end()
#define HEAP priority_queue
#define st first
#define nd second
#define pf push_front
#define pb push_back
#define ppf pop_front
#define ppb pop_back
#define mp make_pair
#define re real()
#define im imag()
#define LIST0(A, n) { \
FOR (i, 0, n - 1) printf("%d ", A[i]); \
printf("\n"); \
} \
#define LIST1(A, n) { \
FOR (i, 1, n) printf("%d ", A[i]); \
printf("\n"); \
} \
#define SHOW0(A, n, m) { \
FOR (i, 0, n - 1) { \
FOR (j, 0, m - 1) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
#define SHOW1(A, n, m) { \
FOR (i, 1, n) { \
FOR (j, 1, m) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
typedef long long LL;
typedef double DB;
typedef long double LDB;
typedef unsigned int UINT;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef vector<DB> VB;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef map<int, int> MII;
typedef map<LL, LL> MLL;
typedef map<int, string> MIS;
typedef map<string, int> MSI;
const int INF(0x3f3f3f3f);
const LL LINF(0x3f3f3f3f3f3f3f3fLL);
const DB DINF(1e20);
const DB EPS(1e-12); //adjustable
const DB PI(acos(-1.0));
inline DB cot(DB x) { RT 1.0 / tan(x); }
inline DB sec(DB x) { RT 1.0 / cos(x); }
inline DB csc(DB x) { RT 1.0 / sin(x); }
template<class T> inline T max(T a, T b, T c) { RT max(max(a, b), c); }
template<class T> inline T max(T a, T b, T c, T d) { RT max(max(a, b), max(c, d)); }
template<class T> inline T min(T a, T b, T c) { RT min(min(a, b), c); }
template<class T> inline T min(T a, T b, T c, T d) { RT min(min(a, b), min(c, d)); }
template<class T> inline void Scan(T& x) { //int, LL
char c;
for (c = getchar(); c <= ' '; c = getchar()) ;
bool ngt(c == '-');
if (ngt) c = getchar();
for (x = 0; c > ' '; c = getchar()) x = (x << 3) + (x << 1) + c - '0';
if (ngt) x = -x;
}
template<class T> inline void Scan(T& x, T& y) { Scan(x), Scan(y); }
template<class T> inline void Scan(T& x, T& y, T& z) { Scan(x), Scan(y), Scan(z); }
//TEMPLATE END
MSI id;
int cnt;
int n, m;
string dat[10005][11];
void Read(int r) {
string str;
getline(cin, str);
int s = 0, t = 0;
FOR (i, 1, m) { //c
while (t < str.size()) {
if (str[t] != ',') ++ t; else BK;
}
dat[r][i] = str.substr(s, t - s);
s = (++ t);
}
}
void Main() {
while (cin >> n >> m) {
getchar();
FOR (i, 1, n) Read(i);
/*
FOR (i, 1, n) {
FOR (j, 1, m) {
cout << dat[i][j] << '*';
}
cout << endl;
}
*/
bool flag = true;
FOR (c1, 1, m - 1) FOR (c2, c1 + 1, m) {
if (!flag) BK;
cnt = 0;
id.clear();
FOR (r, 1, n) {
string t = dat[r][c1] + "," + dat[r][c2];
//cout << t << endl;
if (!id.count(t)) id[t] = r;
else {
flag = false;
cout << "NO" << endl;
cout << id[t] << ' ' << r << endl;
cout << c1 << ' ' << c2 << endl;
BK;
}
}
}
if (flag) cout << "YES" << endl;
}
}
int main() {
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
Main();
RT 0;
}
//imagasaikou!
UVa207
1）所有选手先进行前两轮，除非犯规（DQ）
2）所有犯规的选手当场出局，前两轮犯规的选手不得晋级，后两轮犯规的选手不能赢得奖金
3）前两轮结束的时候，选出70个得分相加最低的选手晋级，如果第70名有并列也一并选入，没有晋级的选手就不能继续后两轮和赢得奖金
4）晋级的选手再打两轮比赛，并获得总奖金的一定比例
5）选手获得总奖金的比例由排名决定
6）本场巡回赛中只有一个冠军
7）如果第k名有n个人并列，则第k~n-k+1名的奖金相加平均分给n个人，奖金精确到小数点后两位
8）如果犯规使得选手少于70人了，为剩下的名次准备的奖金不予分配
9）业余选手可以参赛但无法获得奖金，奖金分配跳过业余选手按照名次顺延
10）只有前70名职业选手（包括并列）能够获得奖金
1）输入输出格式题目中规定精确到了占位长度，所以注意使用相应的格式符
2）带*的选手是业余选手
3）输出时名次后带的T代表并列，但是业余选手不该有此标记！
4）在前两轮出局的选手不会被输出
5）并列选手按照字典序输出
//TEMPLATE BEGIN
//#define ONLINE_JUDGE
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#include <climits>
#include <cassert>
//#include <ctime>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <functional>
#include <algorithm>
#include <complex>
#include <bitset>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
using namespace std;
#define FOR(i, a, b) for (int (i) = (a); (i) <= (b); ++ (i))
#define ROF(i, a, b) for (int (i) = (a); (i) >= (b); -- (i))
#define FOREDGE(i, u, head, next) for (int (i) = (head[(u)]); (i); (i) = (next[(i)]))
#define FORSTL(it, A) for (__typeof((A).begin()) it = (A).begin(); it != (A).end(); ++ it)
#define CLR(a, b) memset((a), (b), sizeof (a))
#define CPY(a, b) memcpy((a), (b), sizeof (a))
#define DEBUG cerr << "debug" << endl
#define STAR cerr << "**********" << endl
#define RT return
#define BK break
#define CT continue
#define ALL(x) (x).begin(), (x).end()
#define HEAP priority_queue
#define st first
#define nd second
#define pf push_front
#define pb push_back
#define ppf pop_front
#define ppb pop_back
#define mp make_pair
#define re real()
#define im imag()
#define LIST0(A, n) { \
FOR (i, 0, n - 1) printf("%d ", A[i]); \
printf("\n"); \
} \
#define LIST1(A, n) { \
FOR (i, 1, n) printf("%d ", A[i]); \
printf("\n"); \
} \
#define SHOW0(A, n, m) { \
FOR (i, 0, n - 1) { \
FOR (j, 0, m - 1) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
#define SHOW1(A, n, m) { \
FOR (i, 1, n) { \
FOR (j, 1, m) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
typedef long long LL;
typedef double DB;
typedef long double LDB;
typedef unsigned int UINT;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef vector<DB> VB;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef map<int, int> MII;
typedef map<LL, LL> MLL;
typedef map<int, string> MIS;
typedef map<string, int> MSI;
const int INF(0x3f3f3f3f);
const LL LINF(0x3f3f3f3f3f3f3f3fLL);
const DB DINF(1e20);
const DB EPS(1e-12); //adjustable
const DB PI(acos(-1.0));
inline DB cot(DB x) { RT 1.0 / tan(x); }
inline DB sec(DB x) { RT 1.0 / cos(x); }
inline DB csc(DB x) { RT 1.0 / sin(x); }
template<class T> inline T max(T a, T b, T c) { RT max(max(a, b), c); }
template<class T> inline T max(T a, T b, T c, T d) { RT max(max(a, b), max(c, d)); }
template<class T> inline T min(T a, T b, T c) { RT min(min(a, b), c); }
template<class T> inline T min(T a, T b, T c, T d) { RT min(min(a, b), min(c, d)); }
template<class T> inline void Scan(T& x) { //int, LL
char c;
for (c = getchar(); c <= ' '; c = getchar()) ;
bool ngt(c == '-');
if (ngt) c = getchar();
for (x = 0; c > ' '; c = getchar()) x = (x << 3) + (x << 1) + c - '0';
if (ngt) x = -x;
}
template<class T> inline void Scan(T& x, T& y) { Scan(x), Scan(y); }
template<class T> inline void Scan(T& x, T& y, T& z) { Scan(x), Scan(y), Scan(z); }
//TEMPLATE END
const int DQ = 500000000;
struct Player {
string name, place;
int rd[5];
int total2, total4;
DB money;
bool amateur;
};
vector<Player> p;
bool Cmp2(Player A, Player B) {
RT A.total2 == B.total2 ? A.name < B.name : A.total2 < B.total2;
}
bool Cmp4(Player A, Player B) {
RT A.total4 == B.total4 ? A.name < B.name : A.total4 < B.total4;
}
int n;
DB money, won[100];
int Strtonum(string str) {
int res = 0;
FOR (i, 0, str.size() - 1) if (isdigit(str[i])) res = res * 10 + str[i] - '0';
RT res;
}
string Numtostr(int num) {
ostringstream ss;
ss << num;
RT ss.str();
}
void Scan() {
//prize
cin >> money;
FOR (i, 1, 70) {
cin >> won[i];
won[i] = won[i] / 100.0 * money;
}
(cin >> n).get();
p.resize(n);
//player
FOR (i, 0, n - 1) {
string str;
getline(cin, str);
//name
p[i].name = str.substr(0, 20);
//amateur
p[i].amateur = false;
ROF (j, 19, 0)
if (p[i].name[j] == '*') {
p[i].amateur = true;
BK;
}
istringstream ss(str.substr(20, str.size() - 20));
//round
FOR (j, 1, 4) p[i].rd[j] = DQ;
FOR (j, 1, 4) {
ss >> str;
if (str == "DQ") BK;
p[i].rd[j] = Strtonum(str);
}
//total score
p[i].total2 = p[i].total4 = 0;
FOR (j, 1, 2) p[i].total2 += p[i].rd[j];
FOR (j, 1, 4) p[i].total4 += p[i].rd[j];
}
}
void Solve() {
//round2
sort(p.begin(), p.end(), Cmp2);
FOR (i, 69, n - 1) if (i == n - 1 || p[i].total2 != p[i + 1].total2) { n = i + 1; BK; }
//round4
p.resize(n);
sort(p.begin(), p.end(), Cmp4);
printf("Player Name Place RD1 RD2 RD3 RD4 TOTAL Money Won\n");
printf("-----------------------------------------------------------------------\n");
int pos = 1;
FOR (i, 0, n - 1) {
if (p[i].total4 >= DQ) { //DQ!
printf("%s ", p[i].name.c_str());
FOR (j, 1, 4)
if (p[i].rd[j] < DQ) printf("%-5d", p[i].rd[j]);
else printf(" ");
puts("DQ");
CT;
}
int j = i;
int m = 0;
DB sum = 0.0;
bool flag = false;
while (j < n && p[i].total4 == p[j].total4) {
if (!p[j].amateur) {
++ m;
if (pos <= 70) {
flag = true;
sum += won[pos ++];
}
}
++ j;
}
//print [i,j)
int rank = i + 1;
while (i < j) {
printf("%s ", p[i].name.c_str());
char tmp[5];
sprintf(tmp, "%d%c", rank, m > 1 && flag && !p[i].amateur ? 'T' : ' ');
printf("%-10s", tmp);
FOR (t, 1, 4) printf("%-5d", p[i].rd[t]);
if (!p[i].amateur && flag) printf("%-10d\$%9.2lf\n", p[i].total4, sum / m);
else printf("%d\n", p[i].total4);
++ i;
}
-- i;
}
}
void Main() {
ios :: sync_with_stdio(false);
int T;
cin >> T;
while (T --) {
p.clear();
Scan();
Solve();
if (T) puts("");
}
}
int main() {
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
Main();
RT 0;
}
//imagasaikou!
UVa1593
//TEMPLATE BEGIN
//#define ONLINE_JUDGE
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#include <climits>
#include <cassert>
//#include <ctime>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <functional>
#include <algorithm>
#include <complex>
#include <bitset>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
using namespace std;
#define FOR(i, a, b) for (int (i) = (a); (i) <= (b); ++ (i))
#define ROF(i, a, b) for (int (i) = (a); (i) >= (b); -- (i))
#define FOREDGE(i, u, head, next) for (int (i) = (head[(u)]); (i); (i) = (next[(i)]))
#define FORSTL(it, A) for (__typeof((A).begin()) it = (A).begin(); it != (A).end(); ++ it)
#define CLR(a, b) memset((a), (b), sizeof (a))
#define CPY(a, b) memcpy((a), (b), sizeof (a))
#define DEBUG cerr << "debug" << endl
#define STAR cerr << "**********" << endl
#define RT return
#define BK break
#define CT continue
#define ALL(x) (x).begin(), (x).end()
#define HEAP priority_queue
#define st first
#define nd second
#define pf push_front
#define pb push_back
#define ppf pop_front
#define ppb pop_back
#define mp make_pair
#define re real()
#define im imag()
#define LIST0(A, n) { \
FOR (i, 0, n - 1) printf("%d ", A[i]); \
printf("\n"); \
} \
#define LIST1(A, n) { \
FOR (i, 1, n) printf("%d ", A[i]); \
printf("\n"); \
} \
#define SHOW0(A, n, m) { \
FOR (i, 0, n - 1) { \
FOR (j, 0, m - 1) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
#define SHOW1(A, n, m) { \
FOR (i, 1, n) { \
FOR (j, 1, m) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
typedef long long LL;
typedef double DB;
typedef long double LDB;
typedef unsigned int UINT;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef vector<DB> VB;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef map<int, int> MII;
typedef map<LL, LL> MLL;
typedef map<int, string> MIS;
typedef map<string, int> MSI;
const int INF(0x3f3f3f3f);
const LL LINF(0x3f3f3f3f3f3f3f3fLL);
const DB DINF(1e20);
const DB EPS(1e-12); //adjustable
const DB PI(acos(-1.0));
inline DB cot(DB x) { RT 1.0 / tan(x); }
inline DB sec(DB x) { RT 1.0 / cos(x); }
inline DB csc(DB x) { RT 1.0 / sin(x); }
template<class T> inline T max(T a, T b, T c) { RT max(max(a, b), c); }
template<class T> inline T max(T a, T b, T c, T d) { RT max(max(a, b), max(c, d)); }
template<class T> inline T min(T a, T b, T c) { RT min(min(a, b), c); }
template<class T> inline T min(T a, T b, T c, T d) { RT min(min(a, b), min(c, d)); }
template<class T> inline void Scan(T& x) { //int, LL
char c;
for (c = getchar(); c <= ' '; c = getchar()) ;
bool ngt(c == '-');
if (ngt) c = getchar();
for (x = 0; c > ' '; c = getchar()) x = (x << 3) + (x << 1) + c - '0';
if (ngt) x = -x;
}
template<class T> inline void Scan(T& x, T& y) { Scan(x), Scan(y); }
template<class T> inline void Scan(T& x, T& y, T& z) { Scan(x), Scan(y), Scan(z); }
//TEMPLATE END
vector<string> words[1005];
int len[1005][200], maxlen[200];
void Main() {
int lines = 0;
string str;
while (getline(cin, str)) {
++ lines;
istringstream ss(str);
while (ss >> str) {
words[lines].pb(str);
}
}
/*
FOR (i, 1, lines) {
cout << "line #" << i << endl;
FORSTL (it, words[i]) cout << *it << endl;
cout << endl;
}
cout << endl;
*/
int maxcnt = 0;
FOR (i, 1, lines) {
int cnt = 0;
FORSTL (it, words[i]) {
++ cnt;
len[i][cnt] = (int)words[i][cnt - 1].length();
}
maxcnt = max(maxcnt, cnt);
}
/*
SHOW1(length, lines, maxcnt);
*/
FOR (j, 1, maxcnt) {
FOR (i, 1, lines) {
maxlen[j] = max(maxlen[j], len[i][j]);
}
}
FOR (i, 1, lines) {
FOR (j, 1, maxcnt) {
if (!len[i][j]) BK;
if (j > 1) cout << " ";
cout << words[i][j - 1];
if (!len[i][j + 1]) BK;
FOR (k, len[i][j] + 1, maxlen[j]) {
cout << " ";
}
}
cout << endl;
}
}
int main() {
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
Main();
RT 0;
}
//imagasaikou!
UVa1594
//TEMPLATE BEGIN
//#define ONLINE_JUDGE
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#include <climits>
#include <cassert>
//#include <ctime>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <functional>
#include <algorithm>
#include <complex>
#include <bitset>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
using namespace std;
#define FOR(i, a, b) for (int (i) = (a); (i) <= (b); ++ (i))
#define ROF(i, a, b) for (int (i) = (a); (i) >= (b); -- (i))
#define FOREDGE(i, u, head, next) for (int (i) = (head[(u)]); (i); (i) = (next[(i)]))
#define FORSTL(it, A) for (__typeof((A).begin()) it = (A).begin(); it != (A).end(); ++ it)
#define CLR(a, b) memset((a), (b), sizeof (a))
#define CPY(a, b) memcpy((a), (b), sizeof (a))
#define DEBUG cerr << "debug" << endl
#define STAR cerr << "**********" << endl
#define RT return
#define BK break
#define CT continue
#define ALL(x) (x).begin(), (x).end()
#define HEAP priority_queue
#define st first
#define nd second
#define pf push_front
#define pb push_back
#define ppf pop_front
#define ppb pop_back
#define mp make_pair
#define re real()
#define im imag()
#define LIST0(A, n) { \
FOR (i, 0, n - 1) printf("%d ", A[i]); \
printf("\n"); \
} \
#define LIST1(A, n) { \
FOR (i, 1, n) printf("%d ", A[i]); \
printf("\n"); \
} \
#define SHOW0(A, n, m) { \
FOR (i, 0, n - 1) { \
FOR (j, 0, m - 1) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
#define SHOW1(A, n, m) { \
FOR (i, 1, n) { \
FOR (j, 1, m) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
typedef long long LL;
typedef double DB;
typedef long double LDB;
typedef unsigned int UINT;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef vector<DB> VB;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef map<int, int> MII;
typedef map<LL, LL> MLL;
typedef map<int, string> MIS;
typedef map<string, int> MSI;
const int INF(0x3f3f3f3f);
const LL LINF(0x3f3f3f3f3f3f3f3fLL);
const DB DINF(1e20);
const DB EPS(1e-12); //adjustable
const DB PI(acos(-1.0));
inline DB cot(DB x) { RT 1.0 / tan(x); }
inline DB sec(DB x) { RT 1.0 / cos(x); }
inline DB csc(DB x) { RT 1.0 / sin(x); }
template<class T> inline T max(T a, T b, T c) { RT max(max(a, b), c); }
template<class T> inline T max(T a, T b, T c, T d) { RT max(max(a, b), max(c, d)); }
template<class T> inline T min(T a, T b, T c) { RT min(min(a, b), c); }
template<class T> inline T min(T a, T b, T c, T d) { RT min(min(a, b), min(c, d)); }
template<class T> inline void Scan(T& x) { //int, LL
char c;
for (c = getchar(); c <= ' '; c = getchar()) ;
bool ngt(c == '-');
if (ngt) c = getchar();
for (x = 0; c > ' '; c = getchar()) x = (x << 3) + (x << 1) + c - '0';
if (ngt) x = -x;
}
template<class T> inline void Scan(T& x, T& y) { Scan(x), Scan(y); }
template<class T> inline void Scan(T& x, T& y, T& z) { Scan(x), Scan(y), Scan(z); }
//TEMPLATE END
int n;
void Trans(int* a) {
int b[20];
FOR (i, 1, n - 1) b[i] = abs(a[i] - a[i + 1]);
b[n] = abs(a[1] - a[n]);
FOR (i, 1, n) a[i] = b[i];
}
bool Check0(int a[]) {
FOR (i, 1, n) if (a[i]) RT false;
RT true;
}
int a[20];
void Main() {
int T;
cin >> T;
while (T --) {
cin >> n;
FOR (i, 1, n) cin >> a[i];
if (Check0(a)) { cout << "ZERO" << endl; CT;}
bool flag = false;
FOR (i, 1, 1000) {
Trans(a);
if (Check0(a)) { flag = true; BK; }
}
cout << (flag ? "ZERO" : "LOOP") << endl;
}
}
int main() {
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
Main();
RT 0;
}
//imagasaikou!
UVa10935
//TEMPLATE BEGIN
//#define ONLINE_JUDGE
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#include <climits>
#include <cassert>
//#include <ctime>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <functional>
#include <algorithm>
#include <complex>
#include <bitset>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
using namespace std;
#define FOR(i, a, b) for (int (i) = (a); (i) <= (b); ++ (i))
#define ROF(i, a, b) for (int (i) = (a); (i) >= (b); -- (i))
#define FOREDGE(i, u, head, next) for (int (i) = (head[(u)]); (i); (i) = (next[(i)]))
#define FORSTL(it, A) for (__typeof((A).begin()) it = (A).begin(); it != (A).end(); ++ it)
#define CLR(a, b) memset((a), (b), sizeof (a))
#define CPY(a, b) memcpy((a), (b), sizeof (a))
#define DEBUG cerr << "debug" << endl
#define STAR cerr << "**********" << endl
#define RT return
#define BK break
#define CT continue
#define ALL(x) (x).begin(), (x).end()
#define HEAP priority_queue
#define st first
#define nd second
#define pf push_front
#define pb push_back
#define ppf pop_front
#define ppb pop_back
#define mp make_pair
#define re real()
#define im imag()
#define LIST0(A, n) { \
FOR (i, 0, n - 1) printf("%d ", A[i]); \
printf("\n"); \
} \
#define LIST1(A, n) { \
FOR (i, 1, n) printf("%d ", A[i]); \
printf("\n"); \
} \
#define SHOW0(A, n, m) { \
FOR (i, 0, n - 1) { \
FOR (j, 0, m - 1) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
#define SHOW1(A, n, m) { \
FOR (i, 1, n) { \
FOR (j, 1, m) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
typedef long long LL;
typedef double DB;
typedef long double LDB;
typedef unsigned int UINT;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef vector<DB> VB;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef map<int, int> MII;
typedef map<LL, LL> MLL;
typedef map<int, string> MIS;
typedef map<string, int> MSI;
const int INF(0x3f3f3f3f);
const LL LINF(0x3f3f3f3f3f3f3f3fLL);
const DB DINF(1e20);
const DB EPS(1e-12); //adjustable
const DB PI(acos(-1.0));
inline DB cot(DB x) { RT 1.0 / tan(x); }
inline DB sec(DB x) { RT 1.0 / cos(x); }
inline DB csc(DB x) { RT 1.0 / sin(x); }
template<class T> inline T max(T a, T b, T c) { RT max(max(a, b), c); }
template<class T> inline T max(T a, T b, T c, T d) { RT max(max(a, b), max(c, d)); }
template<class T> inline T min(T a, T b, T c) { RT min(min(a, b), c); }
template<class T> inline T min(T a, T b, T c, T d) { RT min(min(a, b), min(c, d)); }
template<class T> inline void Scan(T& x) { //int, LL
char c;
for (c = getchar(); c <= ' '; c = getchar()) ;
bool ngt(c == '-');
if (ngt) c = getchar();
for (x = 0; c > ' '; c = getchar()) x = (x << 3) + (x << 1) + c - '0';
if (ngt) x = -x;
}
template<class T> inline void Scan(T& x, T& y) { Scan(x), Scan(y); }
template<class T> inline void Scan(T& x, T& y, T& z) { Scan(x), Scan(y), Scan(z); }
//TEMPLATE END
int n;
queue<int> Q;
void Main() {
while (cin >> n && n) {
FOR (i, 1, n) Q.push(i);
int cnt = 0;
cout << "Discarded cards:";
while (cnt < n - 1) {
++ cnt;
int ans = Q.front();
Q.pop();
if (cnt > 1) cout << ",";
cout << " " << ans;
ans = Q.front();
Q.pop();
Q.push(ans);
}
cout << endl << "Remaining card: " << Q.front() << endl;
Q.pop();
}
}
int main() {
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
Main();
RT 0;
}
//imagasaikou!
UVa10763
//TEMPLATE BEGIN
//#define ONLINE_JUDGE
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#include <climits>
#include <cassert>
//#include <ctime>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <functional>
#include <algorithm>
#include <complex>
#include <bitset>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
using namespace std;
#define FOR(i, a, b) for (int (i) = (a); (i) <= (b); ++ (i))
#define ROF(i, a, b) for (int (i) = (a); (i) >= (b); -- (i))
#define FOREDGE(i, u, head, next) for (int (i) = (head[(u)]); (i); (i) = (next[(i)]))
#define FORSTL(it, A) for (__typeof((A).begin()) it = (A).begin(); it != (A).end(); ++ it)
#define CLR(a, b) memset((a), (b), sizeof (a))
#define CPY(a, b) memcpy((a), (b), sizeof (a))
#define DEBUG cerr << "debug" << endl
#define STAR cerr << "**********" << endl
#define RT return
#define BK break
#define CT continue
#define ALL(x) (x).begin(), (x).end()
#define HEAP priority_queue
#define st first
#define nd second
#define pf push_front
#define pb push_back
#define ppf pop_front
#define ppb pop_back
#define mp make_pair
#define re real()
#define im imag()
#define LIST0(A, n) { \
FOR (i, 0, n - 1) printf("%d ", A[i]); \
printf("\n"); \
} \
#define LIST1(A, n) { \
FOR (i, 1, n) printf("%d ", A[i]); \
printf("\n"); \
} \
#define SHOW0(A, n, m) { \
FOR (i, 0, n - 1) { \
FOR (j, 0, m - 1) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
#define SHOW1(A, n, m) { \
FOR (i, 1, n) { \
FOR (j, 1, m) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
typedef long long LL;
typedef double DB;
typedef long double LDB;
typedef unsigned int UINT;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef vector<DB> VB;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef map<int, int> MII;
typedef map<LL, LL> MLL;
typedef map<int, string> MIS;
typedef map<string, int> MSI;
const int INF(0x3f3f3f3f);
const LL LINF(0x3f3f3f3f3f3f3f3fLL);
const DB DINF(1e20);
const DB EPS(1e-12); //adjustable
const DB PI(acos(-1.0));
inline DB cot(DB x) { RT 1.0 / tan(x); }
inline DB sec(DB x) { RT 1.0 / cos(x); }
inline DB csc(DB x) { RT 1.0 / sin(x); }
template<class T> inline T max(T a, T b, T c) { RT max(max(a, b), c); }
template<class T> inline T max(T a, T b, T c, T d) { RT max(max(a, b), max(c, d)); }
template<class T> inline T min(T a, T b, T c) { RT min(min(a, b), c); }
template<class T> inline T min(T a, T b, T c, T d) { RT min(min(a, b), min(c, d)); }
template<class T> inline void Scan(T& x) { //int, LL
char c;
for (c = getchar(); c <= ' '; c = getchar()) ;
bool ngt(c == '-');
if (ngt) c = getchar();
for (x = 0; c > ' '; c = getchar()) x = (x << 3) + (x << 1) + c - '0';
if (ngt) x = -x;
}
template<class T> inline void Scan(T& x, T& y) { Scan(x), Scan(y); }
template<class T> inline void Scan(T& x, T& y, T& z) { Scan(x), Scan(y), Scan(z); }
//TEMPLATE END
int n;
map<PII, int> m;
vector<PII> v;
void Main() {
while (scanf("%d", &n) == 1 && n) {
m.clear();
FOR (i, 1, n) {
int a, b;
Scan(a, b);
if (a > b) swap(a, b);
PII p = mp(a, b);
if (!m.count(p)) {
m[p] = 1;
v.pb(p);
}
else m[p] ^= 1;
}
bool flag = true;
FORSTL (it, v) {
PII p = *it;
if (m[p] == 1) { flag = false; BK; }
}
puts(flag ? "YES" : "NO");
}
}
int main() {
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
Main();
RT 0;
}
//imagasaikou!
UVa10391
//TEMPLATE BEGIN
//#define ONLINE_JUDGE
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#include <climits>
#include <cassert>
//#include <ctime>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <functional>
#include <algorithm>
#include <complex>
#include <bitset>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
using namespace std;
#define FOR(i, a, b) for (int (i) = (a); (i) <= (b); ++ (i))
#define ROF(i, a, b) for (int (i) = (a); (i) >= (b); -- (i))
#define FOREDGE(i, u, head, next) for (int (i) = (head[(u)]); (i); (i) = (next[(i)]))
#define FORSTL(it, A) for (__typeof((A).begin()) it = (A).begin(); it != (A).end(); ++ it)
#define CLR(a, b) memset((a), (b), sizeof (a))
#define CPY(a, b) memcpy((a), (b), sizeof (a))
#define DEBUG cerr << "debug" << endl
#define STAR cerr << "**********" << endl
#define RT return
#define BK break
#define CT continue
#define ALL(x) (x).begin(), (x).end()
#define HEAP priority_queue
#define st first
#define nd second
#define pf push_front
#define pb push_back
#define ppf pop_front
#define ppb pop_back
#define mp make_pair
#define re real()
#define im imag()
#define LIST0(A, n) { \
FOR (i, 0, n - 1) printf("%d ", A[i]); \
printf("\n"); \
} \
#define LIST1(A, n) { \
FOR (i, 1, n) printf("%d ", A[i]); \
printf("\n"); \
} \
#define SHOW0(A, n, m) { \
FOR (i, 0, n - 1) { \
FOR (j, 0, m - 1) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
#define SHOW1(A, n, m) { \
FOR (i, 1, n) { \
FOR (j, 1, m) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
typedef long long LL;
typedef double DB;
typedef long double LDB;
typedef unsigned int UINT;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef vector<DB> VB;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef map<int, int> MII;
typedef map<LL, LL> MLL;
typedef map<int, string> MIS;
typedef map<string, int> MSI;
const int INF(0x3f3f3f3f);
const LL LINF(0x3f3f3f3f3f3f3f3fLL);
const DB DINF(1e20);
const DB EPS(1e-12); //adjustable
const DB PI(acos(-1.0));
inline DB cot(DB x) { RT 1.0 / tan(x); }
inline DB sec(DB x) { RT 1.0 / cos(x); }
inline DB csc(DB x) { RT 1.0 / sin(x); }
template<class T> inline T max(T a, T b, T c) { RT max(max(a, b), c); }
template<class T> inline T max(T a, T b, T c, T d) { RT max(max(a, b), max(c, d)); }
template<class T> inline T min(T a, T b, T c) { RT min(min(a, b), c); }
template<class T> inline T min(T a, T b, T c, T d) { RT min(min(a, b), min(c, d)); }
template<class T> inline void Scan(T& x) { //int, LL
char c;
for (c = getchar(); c <= ' '; c = getchar()) ;
bool ngt(c == '-');
if (ngt) c = getchar();
for (x = 0; c > ' '; c = getchar()) x = (x << 3) + (x << 1) + c - '0';
if (ngt) x = -x;
}
template<class T> inline void Scan(T& x, T& y) { Scan(x), Scan(y); }
template<class T> inline void Scan(T& x, T& y, T& z) { Scan(x), Scan(y), Scan(z); }
//TEMPLATE END
string str;
set<string> words;
void Main() {
while (cin >> str) {
words.insert(str);
}
FORSTL (it, words) {
str = *it;
FOR (i, 0, (int)str.size() - 1) {
string str1 = str.substr(0, i + 1), str2 = str.substr(i + 1);
//cout << str1 << ' ' << str2 << endl;
if (words.count(str1) && words.count(str2)) {
cout << str << endl;
BK;
}
}
}
}
int main() {
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
Main();
RT 0;
}
//imagasaikou!
UVa1595
//TEMPLATE BEGIN
//#define ONLINE_JUDGE
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#include <climits>
#include <cassert>
//#include <ctime>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <functional>
#include <algorithm>
#include <complex>
#include <bitset>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
using namespace std;
#define FOR(i, a, b) for (int (i) = (a); (i) <= (b); ++ (i))
#define ROF(i, a, b) for (int (i) = (a); (i) >= (b); -- (i))
#define FOREDGE(i, u, head, next) for (int (i) = (head[(u)]); (i); (i) = (next[(i)]))
#define FORSTL(it, A) for (__typeof((A).begin()) it = (A).begin(); it != (A).end(); ++ it)
#define CLR(a, b) memset((a), (b), sizeof (a))
#define CPY(a, b) memcpy((a), (b), sizeof (a))
#define DEBUG cerr << "debug" << endl
#define STAR cerr << "**********" << endl
#define RT return
#define BK break
#define CT continue
#define ALL(x) (x).begin(), (x).end()
#define HEAP priority_queue
#define st first
#define nd second
#define pf push_front
#define pb push_back
#define ppf pop_front
#define ppb pop_back
#define mp make_pair
#define re real()
#define im imag()
#define LIST0(A, n) { \
FOR (i, 0, n - 1) printf("%d ", A[i]); \
printf("\n"); \
} \
#define LIST1(A, n) { \
FOR (i, 1, n) printf("%d ", A[i]); \
printf("\n"); \
} \
#define SHOW0(A, n, m) { \
FOR (i, 0, n - 1) { \
FOR (j, 0, m - 1) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
#define SHOW1(A, n, m) { \
FOR (i, 1, n) { \
FOR (j, 1, m) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
typedef long long LL;
typedef double DB;
typedef long double LDB;
typedef unsigned int UINT;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef vector<DB> VB;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef map<int, int> MII;
typedef map<LL, LL> MLL;
typedef map<int, string> MIS;
typedef map<string, int> MSI;
const int INF(0x3f3f3f3f);
const LL LINF(0x3f3f3f3f3f3f3f3fLL);
const DB DINF(1e20);
const DB EPS(1e-12); //adjustable
const DB PI(acos(-1.0));
inline DB cot(DB x) { RT 1.0 / tan(x); }
inline DB sec(DB x) { RT 1.0 / cos(x); }
inline DB csc(DB x) { RT 1.0 / sin(x); }
template<class T> inline T max(T a, T b, T c) { RT max(max(a, b), c); }
template<class T> inline T max(T a, T b, T c, T d) { RT max(max(a, b), max(c, d)); }
template<class T> inline T min(T a, T b, T c) { RT min(min(a, b), c); }
template<class T> inline T min(T a, T b, T c, T d) { RT min(min(a, b), min(c, d)); }
template<class T> inline void Scan(T& x) { //int, LL
char c;
for (c = getchar(); c <= ' '; c = getchar()) ;
bool ngt(c == '-');
if (ngt) c = getchar();
for (x = 0; c > ' '; c = getchar()) x = (x << 3) + (x << 1) + c - '0';
if (ngt) x = -x;
}
template<class T> inline void Scan(T& x, T& y) { Scan(x), Scan(y); }
template<class T> inline void Scan(T& x, T& y, T& z) { Scan(x), Scan(y), Scan(z); }
//TEMPLATE END
const int maxn = 1005;
int n, x[maxn];
map<int, set<int> > p;
PII points[maxn];
void Main() {
int T;
cin >> T;
while (T --) {
p.clear();
cin >> n;
FOR (i, 1, n) {
int xx, yy;
cin >> xx >> yy;
points[i] = mp(xx, yy);
x[i] = xx;
if (!p.count(xx)) p[xx] = set<int>();
p[xx].insert(yy);
}
sort(x + 1, x + n + 1);
int m = unique(x + 1, x + n + 1) - (x + 1);
//LIST1(x, m);
int mid;
bool flag = true;
if (m & 1) {
mid = (m + 1) >> 1;
FORSTL (it, p) {
int xx = it -> st;
FORSTL (it2, it -> nd) { //it2: set<int> :: iterator
int yy = *it2;
if (!p[x[mid] * 2 - xx].count(yy)) { flag = false; BK; }
}
}
}
else {
mid = m >> 1;
FORSTL (it, p) {
int xx = it -> st;
FORSTL (it2, it -> nd) {
int yy = *it2;
if (!p[x[mid] + x[mid + 1] - xx].count(yy)) { flag = false; BK; }
}
}
}
cout << (flag ? "YES" : "NO") << endl;
}
}
int main() {
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
Main();
RT 0;
}
//imagasaikou!
UVa12100
//TEMPLATE BEGIN
//#define ONLINE_JUDGE
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#include <climits>
#include <cassert>
//#include <ctime>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <functional>
#include <algorithm>
#include <complex>
#include <bitset>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
using namespace std;
#define FOR(i, a, b) for (int (i) = (a); (i) <= (b); ++ (i))
#define ROF(i, a, b) for (int (i) = (a); (i) >= (b); -- (i))
#define FOREDGE(i, u, head, next) for (int (i) = (head[(u)]); (i); (i) = (next[(i)]))
#define FORSTL(it, A) for (__typeof((A).begin()) it = (A).begin(); it != (A).end(); ++ it)
#define CLR(a, b) memset((a), (b), sizeof (a))
#define CPY(a, b) memcpy((a), (b), sizeof (a))
#define DEBUG cerr << "debug" << endl
#define STAR cerr << "**********" << endl
#define RT return
#define BK break
#define CT continue
#define ALL(x) (x).begin(), (x).end()
#define HEAP priority_queue
#define st first
#define nd second
#define pf push_front
#define pb push_back
#define ppf pop_front
#define ppb pop_back
#define mp make_pair
#define re real()
#define im imag()
#define LIST0(A, n) { \
FOR (i, 0, n - 1) printf("%d ", A[i]); \
printf("\n"); \
} \
#define LIST1(A, n) { \
FOR (i, 1, n) printf("%d ", A[i]); \
printf("\n"); \
} \
#define SHOW0(A, n, m) { \
FOR (i, 0, n - 1) { \
FOR (j, 0, m - 1) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
#define SHOW1(A, n, m) { \
FOR (i, 1, n) { \
FOR (j, 1, m) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
typedef long long LL;
typedef double DB;
typedef long double LDB;
typedef unsigned int UINT;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef vector<DB> VB;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef map<int, int> MII;
typedef map<LL, LL> MLL;
typedef map<int, string> MIS;
typedef map<string, int> MSI;
const int INF(0x3f3f3f3f);
const LL LINF(0x3f3f3f3f3f3f3f3fLL);
const DB DINF(1e20);
const DB EPS(1e-12); //adjustable
const DB PI(acos(-1.0));
inline DB cot(DB x) { RT 1.0 / tan(x); }
inline DB sec(DB x) { RT 1.0 / cos(x); }
inline DB csc(DB x) { RT 1.0 / sin(x); }
template<class T> inline T max(T a, T b, T c) { RT max(max(a, b), c); }
template<class T> inline T max(T a, T b, T c, T d) { RT max(max(a, b), max(c, d)); }
template<class T> inline T min(T a, T b, T c) { RT min(min(a, b), c); }
template<class T> inline T min(T a, T b, T c, T d) { RT min(min(a, b), min(c, d)); }
template<class T> inline void Scan(T& x) { //int, LL
char c;
for (c = getchar(); c <= ' '; c = getchar()) ;
bool ngt(c == '-');
if (ngt) c = getchar();
for (x = 0; c > ' '; c = getchar()) x = (x << 3) + (x << 1) + c - '0';
if (ngt) x = -x;
}
template<class T> inline void Scan(T& x, T& y) { Scan(x), Scan(y); }
template<class T> inline void Scan(T& x, T& y, T& z) { Scan(x), Scan(y), Scan(z); }
//TEMPLATE END
int n, m;
queue<pair<int, bool> > Q;
multiset<int> S;
void Main() {
int T;
cin >> T;
while (T --) {
while (!Q.empty()) Q.pop();
S.clear();
cin >> n >> m;
FOR (i, 0, n - 1) {
int a;
cin >> a;
Q.push(mp(a, i == m));
S.insert(a);
}
//FORSTL (it, S) cout << *it << ' ';
//cout << endl;
int ans = 0;
while (true) {
pair<int, bool> p = Q.front();
Q.pop();
bool flag = true;
FOR (i, p.st + 1, 9) {
if (S.count(i)) { flag = false; BK; }
}
if (flag) {
++ ans;
S.erase(S.find(p.st));
if (p.nd) { cout << ans << endl; BK; }
}
else Q.push(p);
}
}
}
int main() {
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
Main();
RT 0;
}
//imagasaikou!
UVa230
//TEMPLATE BEGIN
//#define ONLINE_JUDGE
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#include <climits>
#include <cassert>
//#include <ctime>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <functional>
#include <algorithm>
#include <complex>
#include <bitset>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
using namespace std;
#define FOR(i, a, b) for (int (i) = (a); (i) <= (b); ++ (i))
#define ROF(i, a, b) for (int (i) = (a); (i) >= (b); -- (i))
#define FOREDGE(i, u, head, next) for (int (i) = (head[(u)]); (i); (i) = (next[(i)]))
#define FORSTL(it, A) for (__typeof((A).begin()) it = (A).begin(); it != (A).end(); ++ it)
#define CLR(a, b) memset((a), (b), sizeof (a))
#define CPY(a, b) memcpy((a), (b), sizeof (a))
#define DEBUG cerr << "debug" << endl
#define STAR cerr << "**********" << endl
#define RT return
#define BK break
#define CT continue
#define ALL(x) (x).begin(), (x).end()
#define HEAP priority_queue
#define st first
#define nd second
#define pf push_front
#define pb push_back
#define ppf pop_front
#define ppb pop_back
#define mp make_pair
#define re real()
#define im imag()
#define LIST0(A, n) { \
FOR (i, 0, n - 1) printf("%d ", A[i]); \
printf("\n"); \
} \
#define LIST1(A, n) { \
FOR (i, 1, n) printf("%d ", A[i]); \
printf("\n"); \
} \
#define SHOW0(A, n, m) { \
FOR (i, 0, n - 1) { \
FOR (j, 0, m - 1) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
#define SHOW1(A, n, m) { \
FOR (i, 1, n) { \
FOR (j, 1, m) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
typedef long long LL;
typedef double DB;
typedef long double LDB;
typedef unsigned int UINT;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef vector<DB> VB;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef map<int, int> MII;
typedef map<LL, LL> MLL;
typedef map<int, string> MIS;
typedef map<string, int> MSI;
const int INF(0x3f3f3f3f);
const LL LINF(0x3f3f3f3f3f3f3f3fLL);
const DB DINF(1e20);
const DB EPS(1e-12); //adjustable
const DB PI(acos(-1.0));
inline DB cot(DB x) { RT 1.0 / tan(x); }
inline DB sec(DB x) { RT 1.0 / cos(x); }
inline DB csc(DB x) { RT 1.0 / sin(x); }
template<class T> inline T max(T a, T b, T c) { RT max(max(a, b), c); }
template<class T> inline T max(T a, T b, T c, T d) { RT max(max(a, b), max(c, d)); }
template<class T> inline T min(T a, T b, T c) { RT min(min(a, b), c); }
template<class T> inline T min(T a, T b, T c, T d) { RT min(min(a, b), min(c, d)); }
template<class T> inline void Scan(T& x) { //int, LL
char c;
for (c = getchar(); c <= ' '; c = getchar()) ;
bool ngt(c == '-');
if (ngt) c = getchar();
for (x = 0; c > ' '; c = getchar()) x = (x << 3) + (x << 1) + c - '0';
if (ngt) x = -x;
}
template<class T> inline void Scan(T& x, T& y) { Scan(x), Scan(y); }
template<class T> inline void Scan(T& x, T& y, T& z) { Scan(x), Scan(y), Scan(z); }
//TEMPLATE END
string str, opt;
map<string, string> m;
set<pair<string, string> > S;
vector<pair<string, string> > ret;
void Main() {
while (getline(cin, str) && str != "END") {
string title, author;
istringstream ss(str);
bool flag = false;
while (ss >> str && str != "by") {
if (!flag) flag = true; else title += " ";
title += str;
}
flag = false;
while (ss >> str) {
if (!flag) flag = true; else author += " ";
author += str;
}
m[title] = author;
S.insert(mp(author, title));
//cout << title << endl;
//cout << author << endl;
}
//exit(0);
//FORSTL (it, S) cout << it -> st << ' ' << it -> nd << endl; cout << endl;
//exit(0);
while (cin >> opt && opt != "END") {
string title;
if (opt == "BORROW") {
char ch;
cin.get(ch);
getline(cin, title);
S.erase(mp(m[title], title));
//cout << title << endl;
//cout << m[title] << endl;
}
else if (opt == "RETURN") {
char ch;
cin.get(ch);
getline(cin, title);
ret.pb(mp(m[title], title));
//cout << title << endl;
//cout << m[title] << endl;
}
else { //shelve
//FORSTL (it, S) cout << "test: " << it -> nd << endl;
sort(ALL(ret));
FORSTL (it, ret) {
title = it -> nd;
//cout << title << endl;
//cout << m[title] << endl;
S.insert(mp(m[title], title));
set<pair<string, string> > :: iterator it2 = S.find(mp(m[title], title));
if (it2 != S.begin()) cout << "Put " << title << " after " << (-- it2) -> nd << endl;
else cout << "Put " << title << " first" << endl;
}
ret.clear();
cout << "END" << endl;
}
}
}
int main() {
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
Main();
RT 0;
}
//imagasaikou!
UVa1596
//TEMPLATE BEGIN
//#define ONLINE_JUDGE
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#include <climits>
#include <cassert>
//#include <ctime>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <functional>
#include <algorithm>
#include <complex>
#include <bitset>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
using namespace std;
#define FOR(i, a, b) for (int (i) = (a); (i) <= (b); ++ (i))
#define ROF(i, a, b) for (int (i) = (a); (i) >= (b); -- (i))
#define FOREDGE(i, u, head, next) for (int (i) = (head[(u)]); (i); (i) = (next[(i)]))
#define FORSTL(it, A) for (__typeof((A).begin()) it = (A).begin(); it != (A).end(); ++ it)
#define CLR(a, b) memset((a), (b), sizeof (a))
#define CPY(a, b) memcpy((a), (b), sizeof (a))
#define DEBUG cerr << "debug" << endl
#define STAR cerr << "**********" << endl
#define RT return
#define BK break
#define CT continue
#define ALL(x) (x).begin(), (x).end()
#define HEAP priority_queue
#define st first
#define nd second
#define pf push_front
#define pb push_back
#define ppf pop_front
#define ppb pop_back
#define mp make_pair
#define re real()
#define im imag()
#define LIST0(A, n) { \
FOR (i, 0, n - 1) printf("%d ", A[i]); \
printf("\n"); \
} \
#define LIST1(A, n) { \
FOR (i, 1, n) printf("%d ", A[i]); \
printf("\n"); \
} \
#define SHOW0(A, n, m) { \
FOR (i, 0, n - 1) { \
FOR (j, 0, m - 1) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
#define SHOW1(A, n, m) { \
FOR (i, 1, n) { \
FOR (j, 1, m) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
typedef long long LL;
typedef double DB;
typedef long double LDB;
typedef unsigned int UINT;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef vector<DB> VB;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef map<int, int> MII;
typedef map<LL, LL> MLL;
typedef map<int, string> MIS;
typedef map<string, int> MSI;
const int INF(0x3f3f3f3f);
const LL LINF(0x3f3f3f3f3f3f3f3fLL);
const DB DINF(1e20);
const DB EPS(1e-12); //adjustable
const DB PI(acos(-1.0));
inline DB cot(DB x) { RT 1.0 / tan(x); }
inline DB sec(DB x) { RT 1.0 / cos(x); }
inline DB csc(DB x) { RT 1.0 / sin(x); }
template<class T> inline T max(T a, T b, T c) { RT max(max(a, b), c); }
template<class T> inline T max(T a, T b, T c, T d) { RT max(max(a, b), max(c, d)); }
template<class T> inline T min(T a, T b, T c) { RT min(min(a, b), c); }
template<class T> inline T min(T a, T b, T c, T d) { RT min(min(a, b), min(c, d)); }
template<class T> inline void Scan(T& x) { //int, LL
char c;
for (c = getchar(); c <= ' '; c = getchar()) ;
bool ngt(c == '-');
if (ngt) c = getchar();
for (x = 0; c > ' '; c = getchar()) x = (x << 3) + (x << 1) + c - '0';
if (ngt) x = -x;
}
template<class T> inline void Scan(T& x, T& y) { Scan(x), Scan(y); }
template<class T> inline void Scan(T& x, T& y, T& z) { Scan(x), Scan(y), Scan(z); }
//TEMPLATE END
set<char> arr;
map<char, int> sz;
set<pair<char, int> > def;
map<pair<char, int>, int> val;
int ans;
bool getans;
void Init() {
ans = 0;
getans = false;
arr.clear();
sz.clear();
def.clear();
val.clear();
}
int Getval(string str) {
bool flag = true;
FORSTL (it, str) if (!isdigit(*it)) { flag = false; BK; }
if (flag) {
int res;
istringstream ss(str);
ss >> res;
RT res;
}
int k = Getval(str.substr(2, str.size() - 3));
if (k == -1) RT -1;
if (!def.count(mp(str[0], k))) RT -1;
RT val[mp(str[0], k)];
}
void Main() {
bool flag = false;
string str;
while (getline(cin, str)) {
if (str == ".") {
if (!flag) flag = true; else BK;
if (!getans) cout << 0 << endl;
Init();
CT;
}
else flag = false;
if (getans) CT;
++ ans;
int k = str.find('=');
if (k != string :: npos) { //assignment
//cout << str.substr(0, k) << endl;
//cout << str.substr(2, k - 3) << endl;
int pos = Getval(str.substr(2, k - 3));
//cout << str << ' ' << t << endl;
int v = Getval(str.substr(k + 1));
//cout << str << ' ' << l << endl;
if (pos != -1 && v != -1 && pos < sz[str[0]]) {
def.insert(mp(str[0], pos));
val[mp(str[0], pos)] = v;
}
else { //get the ans
cout << ans << endl;
getans = true;
CT;
}
}
else { //declaration
istringstream ss(str);
char nam, t;
int s;
ss >> nam;
arr.insert(nam);
ss >> t >> s;
sz[nam] = s;
//cout << nam << ' ' << s << endl;
}
}
}
int main() {
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
Main();
RT 0;
}
//imagasaikou!
UVa1597
//TEMPLATE BEGIN
//#define ONLINE_JUDGE
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#include <climits>
#include <cassert>
//#include <ctime>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <functional>
#include <algorithm>
#include <complex>
#include <bitset>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
using namespace std;
#define FOR(i, a, b) for (int (i) = (a); (i) <= (b); ++ (i))
#define ROF(i, a, b) for (int (i) = (a); (i) >= (b); -- (i))
#define FOREDGE(i, u, head, next) for (int (i) = (head[(u)]); (i); (i) = (next[(i)]))
#define FORSTL(it, A) for (__typeof((A).begin()) it = (A).begin(); it != (A).end(); ++ it)
#define CLR(a, b) memset((a), (b), sizeof (a))
#define CPY(a, b) memcpy((a), (b), sizeof (a))
#define DEBUG cerr << "debug" << endl
#define STAR cerr << "**********" << endl
#define RT return
#define BK break
#define CT continue
#define ALL(x) (x).begin(), (x).end()
#define HEAP priority_queue
#define st first
#define nd second
#define pf push_front
#define pb push_back
#define ppf pop_front
#define ppb pop_back
#define mp make_pair
#define re real()
#define im imag()
#define LIST0(A, n) { \
FOR (i, 0, n - 1) printf("%d ", A[i]); \
printf("\n"); \
} \
#define LIST1(A, n) { \
FOR (i, 1, n) printf("%d ", A[i]); \
printf("\n"); \
} \
#define SHOW0(A, n, m) { \
FOR (i, 0, n - 1) { \
FOR (j, 0, m - 1) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
#define SHOW1(A, n, m) { \
FOR (i, 1, n) { \
FOR (j, 1, m) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
typedef long long LL;
typedef double DB;
typedef long double LDB;
typedef unsigned int UINT;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef vector<DB> VB;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef map<int, int> MII;
typedef map<LL, LL> MLL;
typedef map<int, string> MIS;
typedef map<string, int> MSI;
const int INF(0x3f3f3f3f);
const LL LINF(0x3f3f3f3f3f3f3f3fLL);
const DB DINF(1e20);
const DB EPS(1e-12); //adjustable
const DB PI(acos(-1.0));
inline DB cot(DB x) { RT 1.0 / tan(x); }
inline DB sec(DB x) { RT 1.0 / cos(x); }
inline DB csc(DB x) { RT 1.0 / sin(x); }
template<class T> inline T max(T a, T b, T c) { RT max(max(a, b), c); }
template<class T> inline T max(T a, T b, T c, T d) { RT max(max(a, b), max(c, d)); }
template<class T> inline T min(T a, T b, T c) { RT min(min(a, b), c); }
template<class T> inline T min(T a, T b, T c, T d) { RT min(min(a, b), min(c, d)); }
template<class T> inline void Scan(T& x) { //int, LL
char c;
for (c = getchar(); c <= ' '; c = getchar()) ;
bool ngt(c == '-');
if (ngt) c = getchar();
for (x = 0; c > ' '; c = getchar()) x = (x << 3) + (x << 1) + c - '0';
if (ngt) x = -x;
}
template<class T> inline void Scan(T& x, T& y) { Scan(x), Scan(y); }
template<class T> inline void Scan(T& x, T& y, T& z) { Scan(x), Scan(y), Scan(z); }
//TEMPLATE END
int n, m, sta[1505];
string pas[1505]; //original
bool print[1505]; //should be print as answer
map<string, set<int> > key;
map<string, map<int, bool> > apr;
void Main() {
cin >> n;
getchar();
int lines = 0;
FOR (i, 1, n) {
string str;
sta[i] = lines + 1;
while (getline(cin, str) && str != "**********") {
pas[++ lines] = str;
string tstr = str;
FORSTL (it, tstr) *it = isalpha(*it) ? tolower(*it) : ' ';
istringstream ss(tstr);
while (ss >> str) {
if (!key[str].count(i)) key[str].insert(i);
apr[str][lines] = true;
}
}
}
sta[n + 1] = lines + 1; //[sta[i], sta[i + 1])
//FOR (i, 1, lines) cout << pas[i] << endl;
//FOR (i, 1, n) cout << sta[i] << ' ' << sta[i + 1] - 1 << endl;
//exit(0);
cin >> m;
getchar();
while (m --) {
string str, str1, str2;
getline(cin, str);
istringstream ss(str);
CLR (print, 0);
if (str.find("AND") != string :: npos) {
ss >> str1 >> str2 >> str2;
FOR (i, 1, n) {
if (!key[str1].count(i) || !key[str2].count(i)) CT;
FOR (j, sta[i], sta[i + 1] - 1) {
if (apr[str1][j] || apr[str2][j]) print[j] = true;
}
}
}
else if (str.find("OR") != string :: npos) {
ss >> str1 >> str2 >> str2;
FOR (i, 1, n) {
if (!key[str1].count(i) && !key[str2].count(i)) CT;
FOR (j, sta[i], sta[i + 1] - 1) {
if (apr[str1][j] || apr[str2][j]) print[j] = true;
}
}
}
else if (str.find("NOT") != string :: npos) {
ss >> str1 >> str1;
FOR (i, 1, n) {
if (key[str1].count(i)) CT;
FOR (j, sta[i], sta[i + 1] - 1) {
print[j] = true;
}
}
}
else {
FOR (i, 1, n) {
if (!key[str].count(i)) CT;
FOR (j, sta[i], sta[i + 1] - 1) {
if (apr[str][j]) print[j] = true;
}
}
}
bool getans = false;
FOR (i, 1, n) {
FOR (j, sta[i], sta[i + 1] - 1) {
if (print[j]) {
if (getans) cout << "----------" << endl;
getans = true;
BK;
}
}
FOR (j, sta[i], sta[i + 1] - 1) {
if (print[j]) {
cout << pas[j] << endl;
}
}
}
if (!getans) cout << "Sorry, I found nothing." << endl;
cout << "==========" << endl;
}
}
int main() {
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
Main();
RT 0;
}
//imagasaikou!
UVa12504
//TEMPLATE BEGIN
//#define ONLINE_JUDGE
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#include <climits>
#include <cassert>
//#include <ctime>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <functional>
#include <algorithm>
#include <complex>
#include <bitset>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
using namespace std;
#define FOR(i, a, b) for (int (i) = (a); (i) <= (b); ++ (i))
#define ROF(i, a, b) for (int (i) = (a); (i) >= (b); -- (i))
#define FOREDGE(i, u, head, next) for (int (i) = (head[(u)]); (i); (i) = (next[(i)]))
#define FORSTL(it, A) for (__typeof((A).begin()) it = (A).begin(); it != (A).end(); ++ it)
#define CLR(a, b) memset((a), (b), sizeof (a))
#define CPY(a, b) memcpy((a), (b), sizeof (a))
#define DEBUG cerr << "debug" << endl
#define STAR cerr << "**********" << endl
#define RT return
#define BK break
#define CT continue
#define ALL(x) (x).begin(), (x).end()
#define HEAP priority_queue
#define st first
#define nd second
#define pf push_front
#define pb push_back
#define ppf pop_front
#define ppb pop_back
#define mp make_pair
#define re real()
#define im imag()
#define LIST0(A, n) { \
FOR (i, 0, n - 1) printf("%d ", A[i]); \
printf("\n"); \
} \
#define LIST1(A, n) { \
FOR (i, 1, n) printf("%d ", A[i]); \
printf("\n"); \
} \
#define SHOW0(A, n, m) { \
FOR (i, 0, n - 1) { \
FOR (j, 0, m - 1) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
#define SHOW1(A, n, m) { \
FOR (i, 1, n) { \
FOR (j, 1, m) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
typedef long long LL;
typedef double DB;
typedef long double LDB;
typedef unsigned int UINT;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef vector<DB> VB;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef map<int, int> MII;
typedef map<LL, LL> MLL;
typedef map<int, string> MIS;
typedef map<string, int> MSI;
const int INF(0x3f3f3f3f);
const LL LINF(0x3f3f3f3f3f3f3f3fLL);
const DB DINF(1e20);
const DB EPS(1e-12); //adjustable
const DB PI(acos(-1.0));
inline DB cot(DB x) { RT 1.0 / tan(x); }
inline DB sec(DB x) { RT 1.0 / cos(x); }
inline DB csc(DB x) { RT 1.0 / sin(x); }
template<class T> inline T max(T a, T b, T c) { RT max(max(a, b), c); }
template<class T> inline T max(T a, T b, T c, T d) { RT max(max(a, b), max(c, d)); }
template<class T> inline T min(T a, T b, T c) { RT min(min(a, b), c); }
template<class T> inline T min(T a, T b, T c, T d) { RT min(min(a, b), min(c, d)); }
template<class T> inline void Scan(T& x) { //int, LL
char c;
for (c = getchar(); c <= ' '; c = getchar()) ;
bool ngt(c == '-');
if (ngt) c = getchar();
for (x = 0; c > ' '; c = getchar()) x = (x << 3) + (x << 1) + c - '0';
if (ngt) x = -x;
}
template<class T> inline void Scan(T& x, T& y) { Scan(x), Scan(y); }
template<class T> inline void Scan(T& x, T& y, T& z) { Scan(x), Scan(y), Scan(z); }
//TEMPLATE END
string str1, str2;
map<string, string> dic1, dic2; //(name, value)
VS ans1, ans2, ans3; //+, -, *
void Init() {
dic1.clear();
dic2.clear();
ans1.clear();
ans2.clear();
ans3.clear();
}
void Work(map<string, string>& dic, string str) {
FORSTL (it, str) if (!isalpha(*it) && !isdigit(*it)) *it = ' ';
istringstream ss(str);
string name, value;
while (ss >> name >> value) dic[name] = value;
}
void Main() {
int T;
cin >> T;
while (T --) {
Init();
cin >> str1 >> str2;
Work(dic1, str1); Work(dic2, str2);
//FORSTL (it, dic1) cout << it -> st << ' '; cout << endl;
//FORSTL (it, dic2) cout << it -> st << ' '; cout << endl;
bool flag = true;
FORSTL (it, dic1) {
pair<string, string> p = *it;
string name = p.st, value = p.nd;
if (!dic2.count(name)) { //ans2
ans2.pb(name);
}
else if (dic2[name] != value) { //ans3
ans3.pb(name);
}
}
FORSTL (it, dic2) { //ans1
pair<string, string> p = *it;
string name = p.st, value = p.nd;
if (!dic1.count(name)) ans1.pb(name);
}
sort(ALL(ans1)); sort(ALL(ans2)); sort(ALL(ans3));
if (ans1.size()) { //ans1
flag = false;
cout << "+" << ans1[0];
FOR (i, 1, (int)ans1.size() - 1) cout << "," << ans1[i];
cout << endl;
}
if (ans2.size()) { //ans2
flag = false;
cout << "-" << ans2[0];
FOR (i, 1, (int)ans2.size() - 1) cout << "," << ans2[i];
cout << endl;
}
if (ans3.size()) { //ans3
flag = false;
cout << "*" << ans3[0];
FOR (i, 1, (int)ans3.size() - 1) cout << "," << ans3[i];
cout << endl;
}
if (flag) cout << "No changes" << endl;
cout << endl;
}
}
int main() {
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
Main();
RT 0;
}
//imagasaikou!
UVa511
//TEMPLATE BEGIN
//#define ONLINE_JUDGE
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#include <climits>
#include <cassert>
//#include <ctime>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <functional>
#include <algorithm>
#include <complex>
#include <bitset>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
using namespace std;
#define FOR(i, a, b) for (int (i) = (a); (i) <= (b); ++ (i))
#define ROF(i, a, b) for (int (i) = (a); (i) >= (b); -- (i))
#define FOREDGE(i, u, head, next) for (int (i) = (head[(u)]); (i); (i) = (next[(i)]))
#define FORSTL(it, A) for (__typeof((A).begin()) it = (A).begin(); it != (A).end(); ++ it)
#define CLR(a, b) memset((a), (b), sizeof (a))
#define CPY(a, b) memcpy((a), (b), sizeof (a))
#define DEBUG cerr << "debug" << endl
#define STAR cerr << "**********" << endl
#define RT return
#define BK break
#define CT continue
#define ALL(x) (x).begin(), (x).end()
#define HEAP priority_queue
#define st first
#define nd second
#define pf push_front
#define pb push_back
#define ppf pop_front
#define ppb pop_back
#define mp make_pair
#define re real()
#define im imag()
#define LIST0(A, n) { \
FOR (i, 0, n - 1) printf("%d ", A[i]); \
printf("\n"); \
} \
#define LIST1(A, n) { \
FOR (i, 1, n) printf("%d ", A[i]); \
printf("\n"); \
} \
#define SHOW0(A, n, m) { \
FOR (i, 0, n - 1) { \
FOR (j, 0, m - 1) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
#define SHOW1(A, n, m) { \
FOR (i, 1, n) { \
FOR (j, 1, m) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
typedef long long LL;
typedef double DB;
typedef long double LDB;
typedef unsigned int UINT;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef vector<DB> VB;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef map<int, int> MII;
typedef map<LL, LL> MLL;
typedef map<int, string> MIS;
typedef map<string, int> MSI;
const int INF(0x3f3f3f3f);
const LL LINF(0x3f3f3f3f3f3f3f3fLL);
const DB DINF(1e20);
const DB EPS(1e-12); //adjustable
const DB PI(acos(-1.0));
inline DB cot(DB x) { RT 1.0 / tan(x); }
inline DB sec(DB x) { RT 1.0 / cos(x); }
inline DB csc(DB x) { RT 1.0 / sin(x); }
template<class T> inline T max(T a, T b, T c) { RT max(max(a, b), c); }
template<class T> inline T max(T a, T b, T c, T d) { RT max(max(a, b), max(c, d)); }
template<class T> inline T min(T a, T b, T c) { RT min(min(a, b), c); }
template<class T> inline T min(T a, T b, T c, T d) { RT min(min(a, b), min(c, d)); }
template<class T> inline void Scan(T& x) { //int, LL
char c;
for (c = getchar(); c <= ' '; c = getchar()) ;
bool ngt(c == '-');
if (ngt) c = getchar();
for (x = 0; c > ' '; c = getchar()) x = (x << 3) + (x << 1) + c - '0';
if (ngt) x = -x;
}
template<class T> inline void Scan(T& x, T& y) { Scan(x), Scan(y); }
template<class T> inline void Scan(T& x, T& y, T& z) { Scan(x), Scan(y), Scan(z); }
//TEMPLATE END
struct Map {
string name;
DB x1, x2, y1, y2, area, ratio;
};
vector<Map> maps;
struct Map2 {
string name;
DB area, ratio, mid, se, minx;
};
struct Cmp {
bool operator () (const Map2 A, const Map2 B) {
if (A.area != B.area) RT A.area < B.area;
if (A.mid != B.mid) RT A.mid > B.mid;
if (A.ratio != B.ratio) RT A.ratio > B.ratio;
if (A.se != B.se) RT A.se < B.se;
RT A.minx > B.minx;
}
};
map<string, int> name;
DB Sqr(DB x) { RT x * x; }
vector<Map2> ans[10010];
void Main() {
string str;
cin >> str;
while (cin >> str && str != "LOCATIONS") {
DB x1, x2, y1, y2;
cin >> x1 >> y1 >> x2 >> y2;
Map t;
t.name = str;
t.x1 = min(x1, x2);
t.x2 = max(x1, x2);
t.y1 = min(y1, y2);
t.y2 = max(y1, y2);
t.area = (t.y2 - t.y1) * (t.x2 - t.x1);
t.ratio = fabs(0.75 - (t.y2 - t.y1) / (t.x2 - t.x1));
maps.pb(t);
}
int m = 0;
while (cin >> str && str != "REQUESTS") {
DB x, y;
cin >> x >> y;
name[str] = ++ m;
HEAP<Map2, vector<Map2>, Cmp> H;
FORSTL (it, maps) {
if (x >= it -> x1 && x <= it -> x2 && y >= it -> y1 && y <= it -> y2) {
Map2 t;
t.name = it -> name;
t.area = it -> area;
t.ratio = it -> ratio;
t.mid = sqrt(Sqr((it -> x1 + it -> x2) / 2.0 - x) + Sqr((it -> y1 + it -> y2) / 2.0 - y));
t.se = sqrt(Sqr(it -> x2 - x) + Sqr(it -> y1 - y));
t.minx = it -> x1;
H.push(t);
}
}
Map2 t;
if (!H.empty()) {
ans[m].pb(t = H.top());
H.pop();
}
while (!H.empty()) {
Map2 t2 = H.top();
H.pop();
if (t.area != t2.area) ans[m].pb(t = t2);
}
}
while (cin >> str && str != "END") {
int k;
cin >> k;
cout << str << " at detail level " << k;
int pos = name[str];
if (!pos) cout << " unknown location" << endl;
else if (!ans[pos].size()) cout << " no map contains that location" << endl;
else if ((int)ans[pos].size() >= k) cout << " using " << ans[pos][k - 1].name << endl;
else cout << " no map at that detail level; using " << ans[pos][(int)ans[pos].size() - 1].name << endl;
}
}
int main() {
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
Main();
RT 0;
}
//imagasaikou!
UVa822
//TEMPLATE BEGIN
//#define ONLINE_JUDGE
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#include <climits>
#include <cassert>
//#include <ctime>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <functional>
#include <algorithm>
#include <complex>
#include <bitset>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
using namespace std;
#define FOR(i, a, b) for (int (i) = (a); (i) <= (b); ++ (i))
#define ROF(i, a, b) for (int (i) = (a); (i) >= (b); -- (i))
#define FOREDGE(i, u, head, next) for (int (i) = (head[(u)]); (i); (i) = (next[(i)]))
#define FORSTL(it, A) for (__typeof((A).begin()) it = (A).begin(); it != (A).end(); ++ it)
#define CLR(a, b) memset((a), (b), sizeof (a))
#define CPY(a, b) memcpy((a), (b), sizeof (a))
#define DEBUG cerr << "debug" << endl
#define STAR cerr << "**********" << endl
#define RT return
#define BK break
#define CT continue
#define ALL(x) (x).begin(), (x).end()
#define HEAP priority_queue
#define st first
#define nd second
#define pf push_front
#define pb push_back
#define ppf pop_front
#define ppb pop_back
#define mp make_pair
#define re real()
#define im imag()
#define LIST0(A, n) { \
FOR (i, 0, n - 1) printf("%d ", A[i]); \
printf("\n"); \
} \
#define LIST1(A, n) { \
FOR (i, 1, n) printf("%d ", A[i]); \
printf("\n"); \
} \
#define SHOW0(A, n, m) { \
FOR (i, 0, n - 1) { \
FOR (j, 0, m - 1) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
#define SHOW1(A, n, m) { \
FOR (i, 1, n) { \
FOR (j, 1, m) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
typedef long long LL;
typedef double DB;
typedef long double LDB;
typedef unsigned int UINT;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef vector<DB> VB;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef map<int, int> MII;
typedef map<LL, LL> MLL;
typedef map<int, string> MIS;
typedef map<string, int> MSI;
const int INF(0x3f3f3f3f);
const LL LINF(0x3f3f3f3f3f3f3f3fLL);
const DB DINF(1e20);
const DB EPS(1e-12); //adjustable
const DB PI(acos(-1.0));
inline DB cot(DB x) { RT 1.0 / tan(x); }
inline DB sec(DB x) { RT 1.0 / cos(x); }
inline DB csc(DB x) { RT 1.0 / sin(x); }
template<class T> inline T max(T a, T b, T c) { RT max(max(a, b), c); }
template<class T> inline T max(T a, T b, T c, T d) { RT max(max(a, b), max(c, d)); }
template<class T> inline T min(T a, T b, T c) { RT min(min(a, b), c); }
template<class T> inline T min(T a, T b, T c, T d) { RT min(min(a, b), min(c, d)); }
template<class T> inline void Scan(T& x) { //int, LL
char c;
for (c = getchar(); c <= ' '; c = getchar()) ;
bool ngt(c == '-');
if (ngt) c = getchar();
for (x = 0; c > ' '; c = getchar()) x = (x << 3) + (x << 1) + c - '0';
if (ngt) x = -x;
}
template<class T> inline void Scan(T& x, T& y) { Scan(x), Scan(y); }
template<class T> inline void Scan(T& x, T& y, T& z) { Scan(x), Scan(y), Scan(z); }
//TEMPLATE END
struct Topic {
int tid, num, t0, t, dt, cnt;
};
vector<Topic> top;
struct Personnel {
int id, k;
VI tid;
int st, ed;
bool operator < (const Personnel B) const {
if (st == B.st) RT id < B.id;
RT st < B.st;
}
};
vector<Personnel> per;
int n, m, kase;
MII ha;
void Main() {
while (cin >> n && n) {
top.clear();
per.clear();
FOR (i, 0, n - 1) {
Topic t;
t.cnt = 0;
cin >> t.tid >> t.num >> t.t0 >> t.t >> t.dt;
ha[t.tid] = i;
top.pb(t);
}
cin >> m;
FOR (i, 0, m - 1) {
Personnel t;
int pid;
t.st = t.ed = 0;
t.id = i;
t.tid.clear();
cin >> pid >> t.k;
FOR (j, 1, t.k) {
int x;
cin >> x;
t.tid.pb(ha[x]);
}
per.pb(t);
}
int tim = 0;
while (true) {
bool getans = true;
FORSTL (it, top) if (it -> cnt < it -> num) { getans = false; BK; }
if (getans) BK;
sort(ALL(per));
FORSTL (it1, per) {
if (tim < it1 -> ed) CT;
FORSTL (it2, it1 -> tid) {
if (top[*it2].cnt >= top[*it2].num) CT;
int nexttime = top[*it2].t0 + top[*it2].dt * top[*it2].cnt;
if (tim < nexttime) CT;
++ top[*it2].cnt;
it1 -> ed = tim + top[*it2].t;
it1 -> st = tim;
BK;
}
}
++ tim;
}
int ans = 0;
FORSTL (it, per) ans = max(ans, it -> ed);
cout << "Scenario " << ++ kase << ": All requests are serviced within " << ans << " minutes." << endl;
}
}
int main() {
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
Main();
RT 0;
}
//imagasaikou!
UVa1598
//TEMPLATE BEGIN
//#define ONLINE_JUDGE
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#include <climits>
#include <cassert>
//#include <ctime>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <functional>
#include <algorithm>
#include <complex>
#include <bitset>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
using namespace std;
#define FOR(i, a, b) for (int (i) = (a); (i) <= (b); ++ (i))
#define ROF(i, a, b) for (int (i) = (a); (i) >= (b); -- (i))
#define FOREDGE(i, u, head, next) for (int (i) = (head[(u)]); (i); (i) = (next[(i)]))
#define FORSTL(it, A) for (__typeof((A).begin()) it = (A).begin(); it != (A).end(); ++ it)
#define CLR(a, b) memset((a), (b), sizeof (a))
#define CPY(a, b) memcpy((a), (b), sizeof (a))
#define DEBUG cerr << "debug" << endl
#define STAR cerr << "**********" << endl
#define RT return
#define BK break
#define CT continue
#define ALL(x) (x).begin(), (x).end()
#define HEAP priority_queue
#define st first
#define nd second
#define pf push_front
#define pb push_back
#define ppf pop_front
#define ppb pop_back
#define mp make_pair
#define re real()
#define im imag()
#define LIST0(A, n) { \
FOR (i, 0, n - 1) printf("%d ", A[i]); \
printf("\n"); \
} \
#define LIST1(A, n) { \
FOR (i, 1, n) printf("%d ", A[i]); \
printf("\n"); \
} \
#define SHOW0(A, n, m) { \
FOR (i, 0, n - 1) { \
FOR (j, 0, m - 1) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
#define SHOW1(A, n, m) { \
FOR (i, 1, n) { \
FOR (j, 1, m) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
typedef long long LL;
typedef double DB;
typedef long double LDB;
typedef unsigned int UINT;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef vector<DB> VB;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef map<int, int> MII;
typedef map<LL, LL> MLL;
typedef map<int, string> MIS;
typedef map<string, int> MSI;
const int INF(0x3f3f3f3f);
const LL LINF(0x3f3f3f3f3f3f3f3fLL);
const DB DINF(1e20);
const DB EPS(1e-12); //adjustable
const DB PI(acos(-1.0));
inline DB cot(DB x) { RT 1.0 / tan(x); }
inline DB sec(DB x) { RT 1.0 / cos(x); }
inline DB csc(DB x) { RT 1.0 / sin(x); }
template<class T> inline T max(T a, T b, T c) { RT max(max(a, b), c); }
template<class T> inline T max(T a, T b, T c, T d) { RT max(max(a, b), max(c, d)); }
template<class T> inline T min(T a, T b, T c) { RT min(min(a, b), c); }
template<class T> inline T min(T a, T b, T c, T d) { RT min(min(a, b), min(c, d)); }
template<class T> inline void Scan(T& x) { //int, LL
char c;
for (c = getchar(); c <= ' '; c = getchar()) ;
bool ngt(c == '-');
if (ngt) c = getchar();
for (x = 0; c > ' '; c = getchar()) x = (x << 3) + (x << 1) + c - '0';
if (ngt) x = -x;
}
template<class T> inline void Scan(T& x, T& y) { Scan(x), Scan(y); }
template<class T> inline void Scan(T& x, T& y, T& z) { Scan(x), Scan(y), Scan(z); }
//TEMPLATE END
struct Command {
string opt;
int s, p;
}cmd[10010];
int s, p, id;
Buy() { s = p = 0; }
Buy(int size, int price, int i): s(size), p(price), id(i) {}
bool operator < (const Buy B) const {
if (p == B.p) RT id > B.id;
RT p < B.p;
}
};
struct Sell {
int s, p, id;
Sell() { s = 0, p = 99999; }
Sell(int size, int price, int i): s(size), p(price), id(i) {}
bool operator < (const Sell B) const {
if (p == B.p) RT id > B.id;
RT p > B.p;
}
};
int n;
HEAP<Sell> sell;
set<int> cancel;
int bnum[100010], snum[100010];
void Init() {
while (!sell.empty()) sell.pop();
cancel.clear();
CLR (bnum, 0);
CLR (snum, 0);
}
void Main() {
int kase = 0;
while (cin >> n) {
if (kase ++) cout << endl;
Init();
FOR (i, 1, n) {
string opt;
cin >> opt;
int size, price;
if (opt == "BUY") {
cin >> size >> price;
cmd[i] = (Command){"BUY", size, price};
bnum[price] += size;
}
else if (opt == "SELL") {
cin >> size >> price;
cmd[i] = (Command){"SELL", size, price};
sell.push(Sell(size, price, i));
snum[price] += size;
}
else { //CANCEL
int id;
cin >> id;
cancel.insert(id);
if (cmd[id].opt == "BUY") bnum[cmd[id].p] -= cmd[id].s;
else snum[cmd[id].p] -= cmd[id].s;
cmd[id].s = 0;
}
while (!buy.empty() && !sell.empty()) {
while (!sell.empty() && cancel.count(sell.top().id)) sell.pop();
Sell s = sell.top();
if (b.p >= s.p) {
int ts = min(cmd[b.id].s, cmd[s.id].s);
cout << "TRADE " << ts << " " << (opt == "BUY" ? s.p : b.p) << endl;
cmd[b.id].s -= ts, cmd[s.id].s -= ts;
bnum[b.p] -= ts, snum[s.p] -= ts;
if (!cmd[s.id].s) sell.pop();
}
else BK;
}
//QUOTE
while (!sell.empty() && cancel.count(sell.top().id)) sell.pop();
Sell s;
int tn1 = 0, tn2 = 0;
if (!buy.empty()) b = buy.top(), tn1 = bnum[cmd[b.id].p];
if (!sell.empty()) s = sell.top(), tn2 = snum[cmd[s.id].p];
cout << "QUOTE " << tn1 << " " << b.p << " - " << tn2 << " " << s.p << endl;
}
}
}
int main() {
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
Main();
RT 0;
}
//imagasaikou!
UVa12333
//TEMPLATE BEGIN
//#define ONLINE_JUDGE
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#include <climits>
#include <cassert>
//#include <ctime>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <functional>
#include <algorithm>
#include <complex>
#include <bitset>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
using namespace std;
#define FOR(i, a, b) for (int (i) = (a); (i) <= (b); ++ (i))
#define ROF(i, a, b) for (int (i) = (a); (i) >= (b); -- (i))
#define FOREDGE(i, u, head, next) for (int (i) = (head[(u)]); (i); (i) = (next[(i)]))
#define FORSTL(it, A) for (__typeof((A).begin()) it = (A).begin(); it != (A).end(); ++ it)
#define CLR(a, b) memset((a), (b), sizeof (a))
#define CPY(a, b) memcpy((a), (b), sizeof (a))
#define DEBUG cerr << "debug" << endl
#define STAR cerr << "**********" << endl
#define RT return
#define BK break
#define CT continue
#define ALL(x) (x).begin(), (x).end()
#define HEAP priority_queue
#define st first
#define nd second
#define pf push_front
#define pb push_back
#define ppf pop_front
#define ppb pop_back
#define mp make_pair
#define re real()
#define im imag()
#define LIST0(A, n) { \
FOR (i, 0, n - 1) printf("%d ", A[i]); \
printf("\n"); \
} \
#define LIST1(A, n) { \
FOR (i, 1, n) printf("%d ", A[i]); \
printf("\n"); \
} \
#define SHOW0(A, n, m) { \
FOR (i, 0, n - 1) { \
FOR (j, 0, m - 1) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
#define SHOW1(A, n, m) { \
FOR (i, 1, n) { \
FOR (j, 1, m) \
printf("%d ", A[i][j]); \
printf("\n"); \
} \
} \
typedef long long LL;
typedef double DB;
typedef long double LDB;
typedef unsigned int UINT;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef vector<DB> VB;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef map<int, int> MII;
typedef map<LL, LL> MLL;
typedef map<int, string> MIS;
typedef map<string, int> MSI;
const int INF(0x3f3f3f3f);
const LL LINF(0x3f3f3f3f3f3f3f3fLL);
const DB DINF(1e20);
const DB EPS(1e-12); //adjustable
const DB PI(acos(-1.0));
inline DB cot(DB x) { RT 1.0 / tan(x); }
inline DB sec(DB x) { RT 1.0 / cos(x); }
inline DB csc(DB x) { RT 1.0 / sin(x); }
template<class T> inline T max(T a, T b, T c) { RT max(max(a, b), c); }
template<class T> inline T max(T a, T b, T c, T d) { RT max(max(a, b), max(c, d)); }
template<class T> inline T min(T a, T b, T c) { RT min(min(a, b), c); }
template<class T> inline T min(T a, T b, T c, T d) { RT min(min(a, b), min(c, d)); }
template<class T> inline void Scan(T& x) { //int, LL
char c;
for (c = getchar(); c <= ' '; c = getchar()) ;
bool ngt(c == '-');
if (ngt) c = getchar();
for (x = 0; c > ' '; c = getchar()) x = (x << 3) + (x << 1) + c - '0';
if (ngt) x = -x;
}
template<class T> inline void Scan(T& x, T& y) { Scan(x), Scan(y); }
template<class T> inline void Scan(T& x, T& y, T& z) { Scan(x), Scan(y), Scan(z); }
//TEMPLATE END
const int BASE = 10;
struct Bignum {
int dig, num[50010];
Bignum () { dig = 1, num[1] = 0;}
};
struct Node {
int id;
bool flag;
Node* ch[10];
}*root, tree[5000010];
int T;
char tstr[50010];
int front, back, cntn;
Node* NewNode() { RT &tree[++ cntn]; }
void Insert(int id, char *str, int len) {
//if (!(id % 5000)) cerr << id << endl;
//cerr << tstr + 1 << endl;
Node *x = root;
FOR (i, 1, len) {
if (!x -> ch[str[i] - '0']) x -> ch[str[i] - '0'] = NewNode();
if (!x -> id) x -> id = id;
x = x -> ch[str[i] - '0'];
}
if (!x -> id) x -> id = id;
}
int Query(string str) {
int len = (int)str.size();
Node *x = root;
FOR (i, 0, len - 1) {
if (!x -> ch[str[i] - '0']) RT -1;
x = x -> ch[str[i] - '0'];
}
RT x -> id;
}
void Print(Bignum a) {
ROF (i, back, 1) cout << a.num[i];
cout << endl;
}
Bignum a, b, c;
void Init() {
root = NewNode();
a.dig = b.dig = a.num[1] = b.num[1] = 1;
//cerr << "start" << endl;
front = back = 1;
FOR (i, 2, 99999) {
//if (!(i % 1000)) cerr << i << endl;
//c = a + b;
FOR (j, front, back) {
c.num[j] += a.num[j] + b.num[j];
if (c.num[j] >= 10) {
++ c.num[j + 1];
c.num[j] -= 10;
}
}
if (c.num[back + 1]) ++ back;
if (back - front > 50) ++ front;
//Print(c);
a = b;
b = c;
int cnt = 0, pos = back;
while (cnt <= 40 && pos > 0) tstr[++ cnt] = c.num[pos --] + '0';
Insert(i, tstr, cnt);
FOR (j, front, back + 5) c.num[j] = 0;
}
//cerr << "end" << endl;
}
void Main() {
Init();
//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);
cin >> T;
FOR (kase, 1, T) {
string str;
cin >> str;
cout << "Case #" << kase << ": ";
//cerr << "Case #" << kase << endl;
if (str != "1") cout << Query(str) << endl;
else cout << 0 << endl;
}
}
int main() {
Main();
RT 0;
}
//imagasaikou!
UVa212
To be continued...
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01-10 168 | 28,833 | 80,384 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-30 | latest | en | 0.265556 |
https://books.google.no/books?id=va82AAAAMAAJ&dq=editions:OXFORD600050087&hl=no&output=html_text | 1,643,383,831,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306301.52/warc/CC-MAIN-20220128152530-20220128182530-00666.warc.gz | 194,147,874 | 10,015 | Euclid's Elements of Geometry: From the Latin Translation of Commandine. To which is Added, a Treatise of the Nature and Arithmetic of Logarithms; Likewise Another of the Elements of Plane and Spherical Trigonometry; with a Preface, Shewing the Usefulness and Excellency of this Work
W. Strahan, 1772 - 399 sider
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Side 195 - If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz.
Side 165 - IF two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals : the triangles shall be equiangular, and shall have those angles equal which are opposite to the homologous sides.
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Side 22 - ... sides equal to them of the other. Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DF, each to each, viz. AB...
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Side 161 - And because HE is parallel to KC, one of the sides of the triangle DKC, as CE to ED, so is...
Side 207 - A plane is perpendicular to a plane, when the straight lines drawn in one of the planes perpendicular to the common section of the two planes, are perpendicular to the other plane. V. The inclination of a straight line to a plane... | 569 | 2,294 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2022-05 | latest | en | 0.89188 |
https://www.weegy.com/?ConversationId=1DN22BVT | 1,537,879,129,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267161638.66/warc/CC-MAIN-20180925123211-20180925143611-00479.warc.gz | 927,885,257 | 8,362 | If the equation of a circle is (x - 2)2 (y - 6)2 = 4, it passes through point (5, 6). True or False?
Question
Updated 7/1/2014 8:44:13 PM
Flagged by yeswey [7/1/2014 8:43:10 PM]
Original conversation
User: If the equation of a circle is (x - 2)2 (y - 6)2 = 4, it passes through point (5, 6). True or False?
Question
Updated 7/1/2014 8:44:13 PM
Flagged by yeswey [7/1/2014 8:43:10 PM]
Rating
3
If the equation of a circle is (x - 2)^2 + (y - 6)^2 = 4, it passes through point (5, 6). FALSE.
When x = 5, y = 6,
(x - 2)^2 + (y - 6)^2 = (5 - 2)^2 + (6 - 6)^2 = 9 + 0 = 9, not 4.
Confirmed by jeifunk [7/1/2014 9:19:03 PM]
27,373,405
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 761 | 2,187 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-39 | longest | en | 0.857907 |
http://au.metamath.org/mpegif/rexrsb.html | 1,529,603,069,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864256.26/warc/CC-MAIN-20180621172638-20180621192638-00435.warc.gz | 27,246,027 | 5,546 | Mathbox for Alexander van der Vekens < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > rexrsb Structured version Unicode version
Theorem rexrsb 27914
Description: An equivalent expression for restricted existence, analogous to exsb 2206. (Contributed by Alexander van der Vekens, 1-Jul-2017.)
Assertion
Ref Expression
rexrsb
Distinct variable groups: ,, ,
Allowed substitution hint: ()
Proof of Theorem rexrsb
StepHypRef Expression
1 rexsb 27913 . 2
2 alral 2756 . . . 4
3 df-ral 2702 . . . . . 6
4 19.27v 1917 . . . . . . . 8
5 pm2.04 78 . . . . . . . . . . 11
6 eleq1 2495 . . . . . . . . . . . . 13
76biimprd 215 . . . . . . . . . . . 12
8 pm2.83 73 . . . . . . . . . . . 12
97, 8ax-mp 8 . . . . . . . . . . 11
10 pm2.04 78 . . . . . . . . . . 11
115, 9, 103syl 19 . . . . . . . . . 10
1211imp 419 . . . . . . . . 9
1312alimi 1568 . . . . . . . 8
144, 13sylbir 205 . . . . . . 7
1514ex 424 . . . . . 6
163, 15sylbi 188 . . . . 5
1716com12 29 . . . 4
182, 17impbid2 196 . . 3
1918rexbiia 2730 . 2
201, 19bitri 241 1
Colors of variables: wff set class Syntax hints: wi 4 wb 177 wa 359 wal 1549 wcel 1725 wral 2697 wrex 2698 This theorem is referenced by: 2rexrsb 27916 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-3 7 ax-mp 8 ax-gen 1555 ax-5 1566 ax-17 1626 ax-9 1666 ax-8 1687 ax-6 1744 ax-7 1749 ax-11 1761 ax-12 1950 ax-ext 2416 This theorem depends on definitions: df-bi 178 df-or 360 df-an 361 df-tru 1328 df-ex 1551 df-nf 1554 df-sb 1659 df-cleq 2428 df-clel 2431 df-nfc 2560 df-ral 2702 df-rex 2703
Copyright terms: Public domain W3C validator | 767 | 1,661 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-26 | latest | en | 0.113785 |
https://www.quizzes.cc/metric/percentof.php?percent=913&of=400 | 1,582,344,337,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145648.56/warc/CC-MAIN-20200222023815-20200222053815-00060.warc.gz | 864,222,945 | 3,282 | #### What is 913 percent of 400?
How much is 913 percent of 400? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 913% of 400 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
## 913% of 400 = 3652
Calculate another percentage below. Type into inputs
Find number based on percentage
percent of
Find percentage based on 2 numbers
divided by
Calculating nine hundred and thirteen of four hundred How to calculate 913% of 400? Simply divide the percent by 100 and multiply by the number. For example, 913 /100 x 400 = 3652 or 9.13 x 400 = 3652
#### How much is 913 percent of the following numbers?
913 percent of 400.01 = 365209.13 913 percent of 400.02 = 365218.26 913 percent of 400.03 = 365227.39 913 percent of 400.04 = 365236.52 913 percent of 400.05 = 365245.65 913 percent of 400.06 = 365254.78 913 percent of 400.07 = 365263.91 913 percent of 400.08 = 365273.04 913 percent of 400.09 = 365282.17 913 percent of 400.1 = 365291.3 913 percent of 400.11 = 365300.43 913 percent of 400.12 = 365309.56 913 percent of 400.13 = 365318.69 913 percent of 400.14 = 365327.82 913 percent of 400.15 = 365336.95 913 percent of 400.16 = 365346.08 913 percent of 400.17 = 365355.21 913 percent of 400.18 = 365364.34 913 percent of 400.19 = 365373.47 913 percent of 400.2 = 365382.6 913 percent of 400.21 = 365391.73 913 percent of 400.22 = 365400.86 913 percent of 400.23 = 365409.99 913 percent of 400.24 = 365419.12 913 percent of 400.25 = 365428.25
913 percent of 400.26 = 365437.38 913 percent of 400.27 = 365446.51 913 percent of 400.28 = 365455.64 913 percent of 400.29 = 365464.77 913 percent of 400.3 = 365473.9 913 percent of 400.31 = 365483.03 913 percent of 400.32 = 365492.16 913 percent of 400.33 = 365501.29 913 percent of 400.34 = 365510.42 913 percent of 400.35 = 365519.55 913 percent of 400.36 = 365528.68 913 percent of 400.37 = 365537.81 913 percent of 400.38 = 365546.94 913 percent of 400.39 = 365556.07 913 percent of 400.4 = 365565.2 913 percent of 400.41 = 365574.33 913 percent of 400.42 = 365583.46 913 percent of 400.43 = 365592.59 913 percent of 400.44 = 365601.72 913 percent of 400.45 = 365610.85 913 percent of 400.46 = 365619.98 913 percent of 400.47 = 365629.11 913 percent of 400.48 = 365638.24 913 percent of 400.49 = 365647.37 913 percent of 400.5 = 365656.5
913 percent of 400.51 = 365665.63 913 percent of 400.52 = 365674.76 913 percent of 400.53 = 365683.89 913 percent of 400.54 = 365693.02 913 percent of 400.55 = 365702.15 913 percent of 400.56 = 365711.28 913 percent of 400.57 = 365720.41 913 percent of 400.58 = 365729.54 913 percent of 400.59 = 365738.67 913 percent of 400.6 = 365747.8 913 percent of 400.61 = 365756.93 913 percent of 400.62 = 365766.06 913 percent of 400.63 = 365775.19 913 percent of 400.64 = 365784.32 913 percent of 400.65 = 365793.45 913 percent of 400.66 = 365802.58 913 percent of 400.67 = 365811.71 913 percent of 400.68 = 365820.84 913 percent of 400.69 = 365829.97 913 percent of 400.7 = 365839.1 913 percent of 400.71 = 365848.23 913 percent of 400.72 = 365857.36 913 percent of 400.73 = 365866.49 913 percent of 400.74 = 365875.62 913 percent of 400.75 = 365884.75
913 percent of 400.76 = 365893.88 913 percent of 400.77 = 365903.01 913 percent of 400.78 = 365912.14 913 percent of 400.79 = 365921.27 913 percent of 400.8 = 365930.4 913 percent of 400.81 = 365939.53 913 percent of 400.82 = 365948.66 913 percent of 400.83 = 365957.79 913 percent of 400.84 = 365966.92 913 percent of 400.85 = 365976.05 913 percent of 400.86 = 365985.18 913 percent of 400.87 = 365994.31 913 percent of 400.88 = 366003.44 913 percent of 400.89 = 366012.57 913 percent of 400.9 = 366021.7 913 percent of 400.91 = 366030.83 913 percent of 400.92 = 366039.96 913 percent of 400.93 = 366049.09 913 percent of 400.94 = 366058.22 913 percent of 400.95 = 366067.35 913 percent of 400.96 = 366076.48 913 percent of 400.97 = 366085.61 913 percent of 400.98 = 366094.74 913 percent of 400.99 = 366103.87 913 percent of 401 = 366113 | 1,586 | 4,121 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2020-10 | latest | en | 0.79674 |
https://www.steema.com/support/viewtopic.php?f=1&p=77479 | 1,600,684,270,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400201601.26/warc/CC-MAIN-20200921081428-20200921111428-00434.warc.gz | 1,129,593,786 | 8,627 | ## Drawing Polygon on Canvas in OnAfterDraw
TeeChart for ActiveX, COM and ASP
LAL
Newbie
Posts: 40
Joined: Fri Jun 20, 2008 12:00 am
### Drawing Polygon on Canvas in OnAfterDraw
In C++ ActiveX v8, I want to draw a direction arrow on my line series that points along the line.
I've calculated the points of the arrow (below) - but I'm not sure how to put them into the VARIANT for the Polygon?
Is it supposed to be an array of CPoints? Or an integer list of coordinates?
Code: Select all
``````//------------------------------------------------------------------------------
void MyDialog::OnAfterDraw()
//------------------------------------------------------------------------------
{
int nValues = m_Chart.Series(1).GetXValues().GetCount();
int nHalf = nValues / 2;
CPoint to(m_Chart.Series(0).CalcXPos(nHalf ), m_Chart.Series(0).CalcYPos(nHalf));
CPoint from(m_Chart.Series(0).CalcXPos(nHalf - 1), m_Chart.Series(0).CalcYPos(nHalf - 1));
int x1 = to.x;
int y1 = to.y;
int x2 = from.x;
int y2 = from.y;
CPoint p1, p2, p3;
int xres = 8; // Size of arrow
double theta;
double theta2 = atan(0.5);
double hypot = sqrt(double(xres * xres) * 5/4);
double l;
p1 = CPoint(0, 0);
p2 = CPoint(0, 0);
p3 = CPoint(0, 0);
// Calculate angle of line
if (y2 == y1)
{
theta = ((x2 > x1) ? 0.0 : PI);
}
else
{
theta = atan((double) (y2 - y1) / (double) (x2 - x1));
if ((y2 > y1) && (x2 < x1))
{
theta = PI + theta;
}
if ((y1 > y2) && (x1 > x2))
{
theta = PI + theta;
}
}
// If arrow is in the middle, adjust P1 to line centre
if (TRUE)
{
l = 0.6 * hypot * cos(theta2);
p1.x = long(x2 + l * cos(theta) + 0.5);
p1.y = long(y2 + l * sin(theta) + 0.5);
}
else
{
p1.x = x2;
p1.y = y2;
}
p2.x = long((p1.x - hypot * cos (theta + theta2)) + 0.5);
p2.y = long((p1.y - hypot * sin (theta + theta2)) + 0.5);
p3.x = long((p1.x - hypot * cos (theta - theta2)) + 0.5);
p3.y = long((p1.y - hypot * sin (theta - theta2)) + 0.5);
m_Chart.GetCanvas().GetPen().SetColor(RGB(0, 0, 0));
m_Chart.GetCanvas().GetPen().SetWidth(1);
m_Chart.GetCanvas().GetBrush().SetColor(RGB(0, 0, 255));
m_Chart.GetCanvas().GetBrush().SetBackColor(RGB(0, 0, 255));
VARIANT pts; // <<<<<<------- How to construct this?
m_Chart.GetCanvas().Polygon(3, pts);
}
``````
Kyudos
Newbie
Posts: 2
Joined: Wed Sep 09, 2020 12:00 am
### Re: Drawing Polygon on Canvas in OnAfterDraw
After some trial and error I found this seems to work...is it correct?
Code: Select all
`````` VARIANT pts;
VariantInit(&pts);
pts.vt = VT_ARRAY|VT_I4;
SAFEARRAY* psa = SafeArrayCreateVector(VT_I4, 0, 6);
if (psa != NULL)
{
LONG i;
i = 0; SafeArrayPutElement(psa, &i, &p1.x);
i = 1; SafeArrayPutElement(psa, &i, &p1.y);
i = 2; SafeArrayPutElement(psa, &i, &p2.x);
i = 3; SafeArrayPutElement(psa, &i, &p2.y);
i = 4; SafeArrayPutElement(psa, &i, &p3.x);
i = 5; SafeArrayPutElement(psa, &i, &p3.y);
}
pts.parray = psa;
m_Chart.GetCanvas().Polygon(3, pts);
SafeArrayDestroy(psa);
``````
Yeray
Posts: 8910
Joined: Tue Dec 05, 2006 12:00 am
Location: Girona, Catalonia
Contact:
### Re: Drawing Polygon on Canvas in OnAfterDraw
Hello,
In the AddArrays example here, we populate arrays in a similar way than what you are doing. So it looks good to me.
- .h:
Code: Select all
``````class CVCTeeChart5Dlg : public CDialog
{
public:
//...
COleSafeArray XValues;
COleSafeArray YValues;
//...``````
- .cpp:
Code: Select all
``````BOOL CVCTeeChart5Dlg::OnInitDialog()
{
//...
DWORD numElements[] = {200000};
// Create the safe-arrays...
XValues.Create(VT_R8, 1, numElements);
YValues.Create(VT_R8, 1, numElements);
// Initialize them with values...
long index;
for(index=0; index<200000; index++) {
double val = index+((rand()*5)/RAND_MAX);
XValues.PutElement(&index, &val);
};
srand(100);
for(index=0; index<200000; index++) {
double val = index+rand();
YValues.PutElement(&index, &val);
};
//...``````
Best Regards,
Yeray Alonso Development & Support Steema Software Av. Montilivi 33, 17003 Girona, Catalonia (SP) Please read our Bug Fixing Policy | 1,357 | 3,996 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-40 | latest | en | 0.528965 |
http://math-in-the-middle.com/tag/factoring-trinomials/ | 1,493,514,466,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123635.74/warc/CC-MAIN-20170423031203-00351-ip-10-145-167-34.ec2.internal.warc.gz | 232,302,995 | 9,091 | # 5 Ways to Use Factor Trees in Middle School Math
I think factor trees are one of the coolest “tools” for middle school students! Students tend to like them and they have several different uses in middle school math.
Here are the 5 ways I use factor trees with my students:
1. Prime Factorization
This is the most obvious use of factor trees and it’s typically the first way students learn to use them.
When I teach prime factorization I introduce the idea of the Fundamental Theorem of Arithmetic (no matter how many different factor trees students make for a number, they will always get the same prime factorization). I also emphasize the two things they can/should do to check their answer:
– make sure all numbers in their prime factorization are PRIME numbers
– make sure the product of the numbers in their prime factorization is the starting number
2 & 3. GCF and LCM
Using factor trees is my second favorite way to teach the greatest common factor and least common multiple of numbers and/or monomials. (The cake method is my favorite method).
Factor trees make simplifying radicals so easy for my pre-algebra and algebra students! Students understand that finding the square root of a number means figuring out what number times itself equals the starting number, so in simplifying radicals they are simply pulling out one of each prime factor that is listed twice in the factor tree.
5. Factoring Trinomials
The idea behind factoring trinomials is pretty simple. Students need to come up with two numbers that have a certain product and sum. Students are typically pretty good at factoring simple trinomials like this, since coming up with the two numbers is a breeze:
But, many students struggle with problems like the one below, not because they don’t understand the factoring process, but because they can’t come up with the two numbers with the given sum and product. That’s where factor trees come in handy!
For students who have a tough time coming up with the numbers, I have them make a factor tree for 126. Once they have it broken into its prime factors, they just need to break up the prime factors into 2 numbers every possible way until they find the ones with a sum of 23.
While this method does take some trial and error, it gives “stuck” students a starting place and a set number of possibilities to try.
Do you use factor trees in any other ways in your math classes? If so, please share! | 528 | 2,442 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2017-17 | longest | en | 0.938573 |
https://www.physicsforums.com/threads/trigonometric-function-of-x.389594/ | 1,708,684,207,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474377.60/warc/CC-MAIN-20240223085439-20240223115439-00771.warc.gz | 966,373,063 | 19,179 | # Trigonometric function of x
In summary: For example, when x=0, what is d?In summary, the angle of elevation from the top of a house to a jet flying 2 miles above the house is represented by x radians. The horizontal distance, d, of the jet from the house can be expressed as d=2/tan(x). By plotting points on a graph, the function can be visualized for 0<x<pi.
The angle of elevation from the top of a house to a jet flying 2 miles above the house is x radians. If d represents the horizontal distance, in miles, of the jet from the house, express d in terms of a trigonometric function of x. The graph the function for 0< x <pi.
Did you draw a right triangle involving x, d and 2 miles? That's a good place to start.
yes I did, now what?
From your triangle with d, x, and 2 labeled can you think of a trig function that involves them?
no, that's why I am asking...
ok, this is what I need to figure out. tan x=opp/adj = ?/?
ok, this is what I need to figure out. tan x=opp/adj = ?/?
Very much so. Skip the last = ?/?. Just substitute the values for opp and adj.
well if I knew what they were I could..lol
maybe this graph will help..
#### Attachments
• Wk2Picture.JPG
14.2 KB · Views: 286
well if I knew what they were I could..lol
I'm starting to feel helpless here. You drew the picture, right? It should be right in front of you. Which is the adjacent side and which is the opposite side to the angle x?
maybe this graph will help..
How can you not know with such a gorgeous picture to help? Which leg is opposite the angle x and which is adjacent?
the 2 mile side would be the opposite and the other would be the adjacent side.
I have never done this before, so that is why I am lost...No I didnt draw the picture. It is supposed to help me.
the 2 mile side would be the opposite and the other would be the adjacent side.
Yes, so tan(x)=... just fill it in.
tan(x)=2/? Grrrrrr
the 2 mile side would be the opposite and the other would be the adjacent side.
No offense, but this is getting to be like pulling teeth. The 'other' i.e. the 'adjacent side' is clearly labelled in your picture. It has a letter on it.
tax(x)=2/d
I do understand your frustration with me, just think how I feel.. :(
I was wondering why do I need to forget the ?/?...
I was wondering why do I need to forget the ?/?...
I just meant you should fill in the ??'s. tan(x)=2/d is exactly what you want. You are basically done. Now just solve for d.
I truly don't know how to solve for d. I must sound pretty stupid, huh?
ok wait, is it d=2cot(x) ?
oops, I mean 2/tan(x)
ok wait, is it d=2cot(x) ?
Sure. That's more than brilliant. I was hoping for just d=2/tan(x). But that's the same if not better.
oops, I mean 2/tan(x)
2/tan(x)=2*cot(x).
ok, now I need to use this information to fill out that graph. How would I go about that?
ok, now I need to use this information to fill out that graph. How would I go about that?
You could start by plotting some points.
## What is a "Trigonometric function of x"?
A trigonometric function of x is a mathematical function that relates the angles of a right triangle to the lengths of its sides. It is commonly used in the fields of mathematics, physics, and engineering.
## What are the six main trigonometric functions?
The six main trigonometric functions are sine, cosine, tangent, cotangent, secant, and cosecant. They are abbreviated as sin, cos, tan, cot, sec, and csc respectively. Each function represents a ratio of two sides of a right triangle.
## How are trigonometric functions used in real life?
Trigonometric functions are used in a variety of real-life situations, including navigation, surveying, and engineering. They are also used in physics to describe the motion of waves and in astronomy to calculate the positions of celestial objects.
## What is the unit circle and how is it related to trigonometric functions?
The unit circle is a circle with a radius of 1 centered at the origin of a Cartesian coordinate system. It is used to define and visualize trigonometric functions. The coordinates of a point on the unit circle correspond to the sine and cosine values of a particular angle.
## What is the relationship between trigonometric functions and right triangles?
The values of trigonometric functions are based on the ratios of the sides of a right triangle. The sine function is the ratio of the opposite side to the hypotenuse, the cosine function is the ratio of the adjacent side to the hypotenuse, and the tangent function is the ratio of the opposite side to the adjacent side. This relationship is known as the "SOH-CAH-TOA" rule. | 1,139 | 4,631 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-10 | latest | en | 0.954723 |
https://opencourser.com/course/bsnj3f/mathematical-thinking-in-computer-science | 1,603,726,573,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107891428.74/warc/CC-MAIN-20201026145305-20201026175305-00664.warc.gz | 461,366,806 | 13,039 | # Mathematical Thinking in Computer Science
Introduction to Discrete Mathematics for Computer Science,
Mathematical thinking is crucial in all areas of computer science: algorithms, bioinformatics, computer graphics, data science, machine learning, etc. In this course, we will learn the most important tools used in discrete mathematics: induction, recursion, logic, invariants, examples, optimality. We will use these tools to answer typical programming questions like: How can we be certain a solution exists? Am I sure my program computes the optimal answer? Do each of these objects meet the given requirements? In the course, we use a try-this-before-we-explain-everything approach: you will be solving many interactive (and mobile friendly) puzzles that were carefully designed to allow you to invent many of the important ideas and concepts yourself. Prerequisites: 1. We assume only basic math (e.g., we expect you to know what is a square or how to add fractions), common sense and curiosity. 2. Basic programming knowledge is necessary as some quizzes require programming in Python.
OpenCourser is an affiliate partner of Coursera.
Rating 4.1★ based on 107 ratings 7 weeks 6 weeks, 2–5 hours/week Oct 12 (last week) \$79 University of California San Diego, National Research University Higher School of Economics via Coursera Alexander S. Kulikov, Michael Levin, Vladimir Podolskii On all desktop and mobile devices English Programming Mathematics Computer Science Algorithms Math And Logic
## What people are saying
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discrete mathematics
A very interesting introduction to discrete mathematics.
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Thank you UC San Diego I took this course as I am from a non-computer science background.
I would recommend everyone who is interested in Computer Science and is from a non-computer science background take this course.
course for beginners
Awesome course for beginners.
Good course for beginners of the subject discrete mathematics.
highly recommend
I haven't finished the bonus track yet, but it sounds like an interesting exercise for 15 puzzle.I learned a lot, I've taken discrete math before but definitely gained some new insights this time through, highly recommend these instructors!
I highly recommend it to anyone who wants to learn data structures and algorithms thoroughly.
mathematical thinking
I really liked this course, it's a good introduction to mathematical thinking, with plenty of examples and exercises, I also liked the use of other external graphical tools as exercises.
Great introduction to mathematical thinking and how to apply it to computational problems.
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## Write a review
Your opinion matters. Tell us what you think. | 705 | 3,517 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-45 | latest | en | 0.901699 |
https://tukarguling77.xyz/sparsely-connected-matrix-completion-for-large-graph-streams/ | 1,653,738,750,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663016373.86/warc/CC-MAIN-20220528093113-20220528123113-00005.warc.gz | 644,563,107 | 15,771 | # Sparsely Connected Matrix Completion for Large Graph Streams
Sparsely Connected Matrix Completion for Large Graph Streams – The goal of this chapter is to present a new dataset of multi-dimensional binary data sets. The dataset is composed of 16k points, each with a point-separable partition, i.e., the cluster’s membership matrix. These positions correspond to the cluster’s nodes. A typical multi-dimensional binary dataset consists of 15k points, each with a point-separable partition, i.e., the cluster’s membership matrix. Each node of the cluster (the parent nodes of the cluster) is represented by a fixed set of points, and its rank is defined as a weighted sum of its values of rank. The cluster’s membership matrix is a matrix of different lengths, i.e., its membership matrices cannot be more than the set of its positions. The clustering algorithm (LASSo) is an algorithm for finding the nearest neighbor. The goal of the paper is to define a set of rules for clustering binary data sets as the probability distributions are defined. In an extensive experimental evaluation on a number of datasets, the clustering algorithm is found to be robust to outliers and noise.
We propose an online clustering technique for clustering data with multiple dimensions. Different datasets are often represented using a set of nodes (for example, an MRI image) and a set of labels. The dataset may contain multiple dimensions such as the dimension of noise, or it may be a set of images. The clustering algorithm, which we call Online Clustering Challenge, requires a set of parameters which are determined by our algorithms. We then learn the optimal solutions to each of these parameters and use them as the parameters of the clustering model. We validate this approach on several data clustering datasets. We present the results of our algorithms for each dataset that we evaluate on two datasets. The results show that our model is competitive with existing algorithms and we show that our algorithm is more flexible and accurate. Moreover, the algorithms we evaluate show that the algorithm does not take too many parameters and can be used to estimate the parameters of multiple datasets simultaneously.
The Anatomy of True English and Baloney English: An Analysis of Lexical Features
An Unsupervised Method for Multi-Person Visual Localization
# Sparsely Connected Matrix Completion for Large Graph Streams
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• BblUq0THrUaoArtPgc4QtQfZwl3aTZ
• ghujshbjAefMKtmGTx2IE5IF2G1fXj
• rW2J0gK0JIW3ut2xttHS6uCsieF50u
• 8qHUXnDIVVLEvc6nIkm5wVWHGmXDIj
• c7K9R6G8G9WESVyI8JtX2Heas7yjPh
• GMIAVDJaXyLcP896JfkHRLrGpaagnj
• Video Game Performance Improves Supervised Particle Swarm Optimization
An Online Clustering Approach to Optimal RegressionWe propose an online clustering technique for clustering data with multiple dimensions. Different datasets are often represented using a set of nodes (for example, an MRI image) and a set of labels. The dataset may contain multiple dimensions such as the dimension of noise, or it may be a set of images. The clustering algorithm, which we call Online Clustering Challenge, requires a set of parameters which are determined by our algorithms. We then learn the optimal solutions to each of these parameters and use them as the parameters of the clustering model. We validate this approach on several data clustering datasets. We present the results of our algorithms for each dataset that we evaluate on two datasets. The results show that our model is competitive with existing algorithms and we show that our algorithm is more flexible and accurate. Moreover, the algorithms we evaluate show that the algorithm does not take too many parameters and can be used to estimate the parameters of multiple datasets simultaneously. | 1,620 | 4,892 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-21 | latest | en | 0.918514 |
https://mathematica.stackexchange.com/questions/42796/how-can-i-selectively-trigger-computations-in-manipulate | 1,582,527,453,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145897.19/warc/CC-MAIN-20200224040929-20200224070929-00462.warc.gz | 454,559,905 | 34,102 | # How can I selectively trigger computations in Manipulate?
I have a Manipulate that uses RandomVariate and Random that I would like to have redraw and recompute random values selectively, depending on what Manipulate parameters have been changed.
For example, I would like to be able to control redraws and recomputations so that
• new data and honestMeans are only computed if n has changed;
• new punishMeans are computed only when n or punish have changed;
• a redraw is triggered when zoom changes, but without triggering an update to data, honestMeans or punishMeans;
• a redraw triggered when any of data, honestMeans or punishMeans or zoom changes (without triggering a recomputation of data, honestMeans or punishMeans);
• everything updates if an explicit update is requested (somehow)
as indicated in the comments below:
scale = 5;
dist = HypergeometricDistribution[scale, Round[0.7*2*scale], 2*scale];
genVotes[punish_, n_] := Module[{},
For[i = 2, i <= n, i++,
If[Random[] < punish,
If[votes[[i - 1]] > data[[i]], vote = 0, vote = scale],
vote = data[[i]]
];
AppendTo[votes, Mean[Append[data[[1 ;; i - 1]], vote]]];
AppendTo[votes, Mean[Append[data[[1 ;; i - 1]], data[[i]]]]];
];
Manipulate[
data = RandomVariate[dist, n];(* Should only change when n changes *)
punishMeans = genVotes[punish, n]; (* Should change when punish or n change *)
honestMeans = genVotes[0, n]; (* Should only change when n changes *)
GraphicsColumn[{ (* Should redraw when any of the above change, or if zoom changes, without triggering recompilation of data, punishMeans or honestMeans *)
Show[
ListPlot[punishMeans,
Joined -> False, PlotStyle -> Gray, PlotMarkers -> Automatic,
PlotRange -> {{0, n}, {(1 - zoom)*#, (1 + zoom)*#} &[Mean[data]]},
Frame -> True, Axes -> None,
Epilog -> {Blue, Line[{{0, Mean[data]}, {n, Mean[data]}}]}],
ListPlot[honestMeans,
Joined -> True, PlotStyle -> Black, PlotMarkers -> None]
],
Show[
Histogram[data, scale, PlotRange -> {{0, scale + 1}, All},
Frame -> True,
FrameTicks -> {{Automatic,
Automatic}, {{# + .5, #, 0} & /@ Range[0, scale], None}},
Axes -> None],
DiscretePlot[n*PDF[dist, x], {x, 0, scale + 1}, Joined -> True,
PlotMarkers -> None, Filling -> None]
]
}],
{{n, 100}, 10, 5000},
{{punish, 0.5}, 0, 1},
Delimiter,
{{zoom, 0.2}, .1, .5},
TrackedSymbols :> {n, punish, zoom}]
Is there a way to do this kind of selective update of elements of a Manipulate?
• Please don't post your full-blown code/problem here but just a minimal example for other users to work with. – Dr. belisarius Feb 23 '14 at 21:45
• @belisarius: That's my MWE. (All interdependencies matter, as near as I can tell.) Don't hesitate to simplify the example if you spot an opportunity to do so. – orome Feb 23 '14 at 21:48
• fyi, your code above does not run, it has errors in it, on V 9.01 , here is screen shot !Mathematica graphics may be if you fix these first it will help. – Nasser Feb 23 '14 at 23:14
• @Nasser: Thanks. Fixed (I think). – orome Feb 23 '14 at 23:29
This uses the second argument of dynamics, which acts like an event call back. In there, you do the specific action needed when that dynamic changes. This localizes the logic with its own control variable. Makes it easier to manage. If you like more information about this method, see this question
But you really need to fix/improve the way the function genVotes works. It is very inefficient and slow (that is why I made ContinuousAction -> False below for now), and it also uses global data which is not a good way to implement it. I'll leave this for you to fix.
scale = 5;
dist = HypergeometricDistribution[scale, Round[0.7*2*scale], 2*scale];
genVotes[punish_, n_] := Module[{},
For[i = 2, i <= n, i++,
If[Random[] < punish,
If[votes[[i - 1]] > data[[i]],
vote = 0,
vote = scale
],
vote = data[[i]]
];
AppendTo[votes, Mean[Append[data[[1 ;; i - 1]], vote]]];
AppendTo[votes, Mean[Append[data[[1 ;; i - 1]], data[[i]]]]]
];
];
Manipulate[tick;
GraphicsColumn[{
Show[ListPlot[punishMeans, Joined -> False,
PlotStyle -> Gray, PlotMarkers -> Automatic,
PlotRange -> {{0, n}, {(1 - zoom)*#, (1 + zoom)*#} &[Mean[data]]},
Frame -> True, Axes -> None,
Epilog -> {Blue, Line[{{0, Mean[data]}, {n, Mean[data]}}]}],
ListPlot[honestMeans, Joined -> True, PlotStyle -> Black, PlotMarkers -> None]
],
Histogram[data, scale, PlotRange -> {{0, scale + 1}, All}, Frame -> True,
FrameTicks -> {{Automatic, Automatic},
{{# + .5, #, 0} & /@ Range[0, scale], None}}, Axes -> None],
DiscretePlot[n*PDF[dist, x], {x, 0, scale + 1}, Joined -> True,
PlotMarkers -> None, Filling -> None]
}],
Grid[{
{"n",
Manipulator[Dynamic[
n, {n = #; data = RandomVariate[dist, n]; punishMeans = genVotes[punish, n];
honestMeans = genVotes[0, n];
tick = Not[tick]} &], {10, 200, 1}, ImageSize -> Tiny], Dynamic[n]
},
{"punish",
Manipulator[Dynamic[punish, {punish = #; punishMeans = genVotes[punish, n];
tick = Not[tick]} &], {0, 1, 0.1}, ImageSize -> Tiny], Dynamic[punish]
},
{"zoom",
Manipulator[Dynamic[zoom, {zoom = #; tick = Not[tick]} &], {.1, .5, 0.1},
ImageSize -> Tiny, ContinuousAction -> True], Dynamic[zoom]
}
}
],
{{zoom, .2}, None},
{{n, 100}, None},
{{punish, .5}, None},
{{tick, False}, None},
SynchronousUpdating -> False,
SynchronousInitialization -> False,
ContinuousAction -> False,
TrackedSymbols :> {tick}
]
• Yeah, sorry: genVotes and the use of data are pretty terrible. I'll fix that. – orome Feb 24 '14 at 1:05 | 1,587 | 5,406 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-10 | latest | en | 0.70559 |
https://www.physics.uoguelph.ca/biophysics-problem-32 | 1,726,502,326,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.45/warc/CC-MAIN-20240916144317-20240916174317-00847.warc.gz | 877,694,426 | 19,307 | # Biophysics Problem 32
If the leg bones described in Question 31 are $90 \;cm$ long, what is the amount of compression in each leg at the fracture point? Assume that the leg bones remain elastic under compression until fracture, which occurs at a stress of $1.4 \times 10^8\;N\;m^{-2}.$
#### First Step
Hooke's Law applies in compression (as in this problem) as well as in tension.
From the table on page 103 of your text, $Y = 2 \times 10^{10} \;N\; m^2.$
You should be able to substitute and calculate the change of length now. (In the above formula, $L$ is the leg length).
#### Calculations
The stress was given in the problem, and the leg length is $0.9 \;m.$
You should have found that $L = 0.0063 \;m.$ | 207 | 718 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-38 | latest | en | 0.917115 |
https://electronics.stackexchange.com/questions/71280/making-an-obstruction-avoiding-device-without-microcontroller | 1,585,579,790,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370497042.33/warc/CC-MAIN-20200330120036-20200330150036-00448.warc.gz | 467,426,973 | 34,121 | Making an obstruction avoiding device without microcontroller
I want to make a toy device which spins to the left whenever there is an obstruction in front of it, without any microcontroller. Can anybody please check whether my approach in building the device is correct or not [let's assume we are driving our device in a bright room with uniform lighting]? Any tips or suggestions for improvement (if it is correct) will also be appreciated.
My approach is to use two DC motors as the rear wheels of the device. Attach an LDR to the front of the device and connect it to the left motor. Now both the motors are connected to a power supply. When the power supply is switched on and no obstruction is in front of the toy, both the motors will be moving at approximately same speed since the LDR will be in exposure to light and its resistance will be negligible, because of which the toy will run straight. Now when there is some obstruction in front of the device, the amount of light falling on the LDR will decrease, which will imply a rise in resistance of the LDR. Thus the velocity of the left motor will decrease, while the velocity of the right motor will be moreorless constant. So the device will spin to the left.
The circuit diagram corresponding to my argument is provided in the link below. I will post the mechanical simulation of the device shortly (within 2 hours from now).
• In principle it will do what you expect but you'll need to use the LDR into an amplifier so that the motor is controlled because its resistance is unlikely to be able to power much even when totally illuminated – Andy aka Jun 1 '13 at 16:02
• I am the one who asked this question [sorry for asking through the fake account, I was in a hurry actually]. Since I can't comment I am posting it as an answer. @Andy aka, The LDR is not used to power the motor, instead it is used to reduce the velocity of the motor. Since the velocity of the other motor remains constant as it is not connected to the LDR, the vehicle should turn about the direction of the motor connected to the LDR. – abcd Jun 1 '13 at 17:15
• What is the resistance of the LDR when it is lit? Is it zero ohms? Even if it is 10 ohms, you'll need the equivalent of that resistance in the uncontrolled motor line to balance things to keep it in a straight line. How much current do the motors take? 500mA? If 500mA, at half motor speed from 6V battery, the power disippated in the LDR will be likely enough to torch it. – Andy aka Jun 1 '13 at 17:24
• The motors take 500 mA current each. The LDR has minimum resistance nearly equal to 1 Ohms. The 1 Ohms resistor can be placed in the motor line to balance the speed of the motors. But do you think it will work? I am currently designing the mechanical simulation of the device and will upload it soon. – abcd Jun 1 '13 at 17:31
• I think you may be reinventing a BEAM popper or photovore – RedGrittyBrick Jun 1 '13 at 23:42
No, this won't really work. Your idea sortof makes sense in theory, but you are missing some down to earth practicality issues.
First, consider the kind of currents the motors in even a small toy would require. Then consider how low a resistance the LDR would need to have so that the voltage drop to the right motor is negligible. It would probably need to be less than 1 Ω, which is well below the practical range for LDRs.
Even if this did work, it would not be very efficient and the LDR would dissipate significant power when it is in the half-on state. Wasting power is not a good thing when batteries are in use. LDRs are much more suitalble as sensors instead of resistive pass elements for significant power.
You have given no valid reason for avoiding a microcontroller, probably because there is none. If your reasons are religious, then your whole question is off topic here. This is a engineering site where we discuss solutions to real problems.
A microcontroller reading light intensity from a LDR used as sensor, then driving each motor with PWM to modulate its drive level is the obvious and efficient way of doing this. With that setup, it would be simple to have two LDRs, one on each side, and have the device always drive towards the strongest light.
I see in some comments you made (now deleted) that you think avoiding a microcontroller reduces cost. This is not true. Using a microcontroller will be the lowest cost solution that actually works. You mentioned the price of a arduino, but only a tiny fraction of that is the cost of the microcontroller, so is not relevant. The motors, mechanical drive train assembly, and battery will all cost more than a reasonable microcontroller for this job. Many micros are available for under $1. I didn't look all that carefully, but from a first glance a PIC 12F1501 looks like it can do this job. It is listed as$.49 in 5k quantity.
Since there is some objection to providing only the 5k quantity price here for some reason, I looked up a example version of this part on microchipdirect, just like anyone else can do that wants detailed pricing. You could also look on distributor web sites, like Mouser. There are lots of places to buy PICs in various quantities.
Specifically, as a example, the PIC 12F1501-I/SN, which is the commerical temperature grade in SOIC package is available on microchipdirect in single pieces with the price for up to 25 units being $.67 each. Of course there is shipping added to that, so buying a single unit is silly. Generally you'd buy a bunch of parts for a whole project at once, preferably with a few spares from Mouser or wherever. All I'm trying to point out is that these things are cheap and to give a rough idea of price. The details are the job of anyone actually doing the design to chase down, as always. • If OP is a developer creating something for mass production, a$2500 investment in bulk chips might be appropriate. If OP is a hobbyist building a one off, 4,999 left over chips might not be the way to go. My guess is OP is a hobbyist, but I don't think we have enough information to know for sure. – mikeY Jun 2 '13 at 19:00
• @mike: Your points are invalid. 1: If the OP is a hobbyist, then the price of a single micro is pretty much irrelevant and his worry about the cost pointless. It doesn't matter if it's $.50 or$5 if he's only buying one. Still cheap compared to the motors, drive train, and batteries. 2: You are confusing bulk price with minimum order quantity. Of course you can buy single PIC 12F1501 for maybe 1$to 2$. I showed the 5k price because that's more relevant when you're worrying about a few cents per micro. I never said you could only buy them 5k at a time. Read more carefully. – Olin Lathrop Jun 2 '13 at 19:08
• My points are perfectly valid. I said we didn't have enough information to know. We don't. I think quoting things at the 5k price and "worrying about a few cents per micro" is making assumptions about the OP. I think the OP's schematic suggests he is a hobbyist. But again, we don't know. I can tell you with certainty that as a hobbyist, someone quoting a 5K price to me, even when meant as an illustration, is not helpful. – mikeY Jun 3 '13 at 0:51
• @mikey: As you say, we don't know if he is a hobbyist or not. In any case, I gave one example price point. Getting bent out of shape because that single example may not apply to the OP is a gross overreaction. Anyone can spend a minute on the internet and determine the price for whatever quantity they actually want. In the mean time, I gave a particular model number to look into further. It seems I could have left the one example price off and you would have thought the answer better. Dinging the whole answer because one example might not be relevant rather a stretch. – Olin Lathrop Jun 3 '13 at 12:25
• I didn't downvote or ding your answer- in fact I have upvoted it. You provided an actual answer and addressed OP's questions. I didn't do that. You get credit for that in my book. I'm not bent out of shape either. I made my comments because I didn't/don't like some of what you said. I think your orientation and assumptions are wrong. But it is perfectly fine if we disagree. – mikeY Jun 3 '13 at 15:09
Instead of a continuous current variation against the light falling on the sensor, if you can use an on/off approach, this would be workable, with minimal wasted power. In other words, one of the motors would simply turn off if light falling on the LDR is obstructed. The solution would cost much less than the Arduino, 10 such devices could be made for under $10. simulate this circuit – Schematic created using CircuitLab • The preset R1 shown in the schematic allows the light level tripping threshold to be set. • R2 simply prevents the base of Q1 from being shorted to Vcc if R1 is set to 0 Ohms. • Q1 is used for gain, to provide a sharper turning on or off of the MOSFET. A useful change would be to connect an LED as a light source, right next to the LED and facing the same direction. Any obstruction directly in front of the device would reflect the LED's light into the LDR, causing the light level to rise instead of fall, hence causing the LDR's resistance to fall instead of rise. Simply swap positions of the LDR and R1+R2, to invert the motor's operating mode accordingly. Common infrared reflective sensors such as the TCRT5000 are available for less than$1 for 10 pieces, possibly cheaper than the LDR mentioned in the question. This can be used as such an active obstruction sensor, eliminating the need for the LDR and ensuring that dependency on ambient room lighting is removed. The IR LED part of the TCTR5000 is wired up through a suitable current limiting resistor, and the sensor part is used to sense the reflected IR from obstructions. If the reflected intensity goes beyond a threshold you set, the motor stops.
This page, from which the above image is taken, also provides a schematic of how the IR obstruction sensor may be wired up:
This schematic shows the (photo)transistor in a common collector configuration, but it can just as well be used as common emitter. Hence. the phototransistor output can be used in place of the 2n2222 in the first schematic above, to directly drive the MOSFET. | 2,382 | 10,198 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-16 | latest | en | 0.953911 |
https://mathvis.academic.wlu.edu/tag/cylindrical-shell-method/ | 1,713,278,400,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817095.3/warc/CC-MAIN-20240416124708-20240416154708-00881.warc.gz | 367,019,240 | 13,036 | # Solid of Revolution – Comparing Methods #2
To get my solid, I rotated the area between the curves $$y=x$$ and $$y=x^2$$ around the line $$y=1.25$$. I then approximated it by cylindrical shells. The width of each of my shells was 1/16 and the height was $$x^2-x$$, where the $$x$$-value was chosen from the right endpoint of each interval.
One of the issues with the print comes from the fact aht the top shell does not connect smoothly to the shell below it. Initially I thought that this resulted from an error in the printer. When I looked at the Cinema 4D file, however, I discovered that the top shell does not make contact with the shell below it because its height is less than 1/16 (the bowl shape is very flat there). Additionally, the shell second from the top does not make contact with the shell below it, but the height difference is not noticable. Another issue with the print was that the leftover supports look bad. The inside of the object was filled with them, and even after clearing out most of the supports, it still looks terrible. We are thinking about printing the object on a different printer, such as the Form Labs printer (which uses stereolithography) or the ProJet 260 (which uses powder), in order to get the inside part of the object to look clean.
# Solid of Revolution – Comparing Methods #1
We decided to find a solid of revolution for which both the washer method and cylindrical shell method worked and to model it with both methods. Laura Taalman has done similar designs for shell approximations which can be found here and here
We decided to rotated the area between $$y=x$$ and $$y=x^2$$ about the line $$y=1.25$$. This created a bowl-like object which is curved on the outside, straight on the inside and with a hole at the bottom of the bowl. We split this project up and I created a model of the object using the washer method, while Ryan used the cylindrical shell method. We planned to each use 16 slices/shells so we could compare our two models later.
Creating this object was a very similar process to that of creating the sphere with washers. Since I was using 16 washers and the object has a height of 1, each washer was 1/16 thick. In Cinema4D I used the preprogramed tube object and made each of them to have a height of 1/16. In order to calculate the inner and outer radius of each washer I used the value of middle of each washer, 1/32, 3/32, 5/32, 7/32, etc. I plugged these values into the equation $$f(x)=1.25-x$$ for the inner radius value and $$g(x)=1.24-x^2$$ for the outer radius value. The only other thing I had to change for each was was its height coordinate. My washers in Cinema4D were parallel to the $$xz$$-plane so I adjusted each washer’s $$y$$-coordinate so they would be spread out to the correct height. I increased the $$y$$-coordinate of each was by 1/16 (the height of each washer) from the previous.
Since I used the preprogramed tube object I was unable to use Magic Merge easily to connect these object and instead just used the ‘Connect + Delete’ command under the ‘Mesh’ tab. This means that there are inner walls in my object, but since each washer is relatively small this is not hugely problematic and adds extra support.
I ran into one issue when I was finishing up with my object. The smallest washer (and the last one I created) did not touch the previous washer. All the other washers had been able to lie on top of the previous one since their outer radius was larger than the inner radius of the previous washer. The 15th washer had an inner radius of 0.34375 cm and the 16th washer had an outer radius of 0.31152cm and thus there was a gap. This is due to the fact the bowl is very flat at the bottom.
We decided to remedy this we would just delete the 16th washer and make a note of it here as well as in the description of the object on Thingiverse.
We printed Ryan’s object with cylindrical shells using the Makerbot and the supports on the inside were difficult to remove and didn’t look good. We are currently discussing printing on another printer for these objects and I will update later on how that goes!
# Volume by Cylindrical Shells
On Monday, June 15, I modeled a volume by cylindrical shells from Calculus II. I used Example 1 in 7.3 of Stewart’s Essential Calculus, which is a volume of revolution of the curve $$y=2x^2-x^3$$ about the y-axis. This is shaped a bit like a stadium. The plan is to approximate this volume using 16 cylindrical shells. I first sketched out the curve in 2-dimensions to get a feel for the profile of the shape. Next, I wrote out each point on the curve from $$x=0$$ to $$x=2$$ in intervals of length 1/8, namely $$(0,0), … , (1.875,0.439), (2,0)$$.
Each cylindrical shell is determined by its height, the thickness of the shell and either the inner or outer radius. By construction, each shell has thickness 1/8 and for each $$k=0,1,2, \dots, 15$$, the inner radius of a shell is $$k/8$$, while the outer radius is $$(k+1)/8$$. I chose the height of each shell to be $$f(k/8)$$, which is the the $$x$$-value closer to the origin.
To summarize, if $$r=$$inner radius of shell, $$R=$$outer radius of shell, and $$h=$$height of tube, then for $$k = 0, 1, 2, \dots , 15$$,
$$r = k/8, \quad R = (k+1)/8, \quad h=2r^2 – r^3.$$
Since the height function is increasing between $$x=0$$ and $$x=4/3$$, the cylindrical shells lie inside the volume of revolution. Between $$x=4/3$$ and $$x=2$$, the function is decreasing and the shells lie outside the volume.
I then made a model of this Riemann approximation for the solid in Cinema 4D by inserting tubes (Cinema 4D’s name for “cylindrical shells”) of inner radius $$r$$, outer radius $$R$$, and height $$h$$. When Cinema 4D inserts a tube, it places half of the tube above the $$xy$$-plane (the $$xz$$-plane in Cinema 4D) and half of it below. Therefore, I needed to add half of each tube’s height to put each shell onto the $$xy$$-plane.
The object was first printed with the supports setting on just in case, although I thought that they did not need supports. The result was that a couple superfluous strands of plastic running along the shells, which needed to be removed. Given the geometry of the shape, I would advise against using supports in building this object.
I would advise that you use a raft to build this object, however, because it was very difficult to remove the object from the build plate. (It took 5 minutes of very careful tugging by David Pfaff.)
You can find this objects on Thingiverse here. | 1,663 | 6,528 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-18 | latest | en | 0.939734 |
http://boards.fool.com/wow-you-must-be-able-to-get-some-cheap-regular-30287986.aspx?sort=username | 1,524,374,872,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945493.69/warc/CC-MAIN-20180422041610-20180422061610-00209.warc.gz | 47,551,240 | 23,259 | Message Font: Serif | Sans-Serif
No. of Recommendations: 0
Wow, you must be able to get some cheap regular yogurt!
I'm in Germany, so I guess the subsidies are all different. Anyway I just discovered I was way off in my calculation. Actually I'm saving 62% by making it myself!
Greek yogurt €4.79 per liter (cheapest I could find it)
Regular full fat yogurt €0.92 per liter; assume two liters needed to make one liter Greek yogurt = €1.84
(4.79 -1.84)/4.79 x 100 = 62%
If I made my own yogurt, milk here costs €0.55 per liter, so my yogurt cost would fall from €1.84 to €1.10 then the savings would be 77% ! | 173 | 613 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-17 | latest | en | 0.951373 |
https://studylib.net/doc/25777061/13.1-quiz-ak | 1,685,656,536,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648209.30/warc/CC-MAIN-20230601211701-20230602001701-00007.warc.gz | 589,987,121 | 11,448 | # 13.1 quiz AK
```Name: _____________________________ Class: __________________ Date: __________________
Assessment
Light and Reflection
13 Light and Reflection
CHARACTERISTICS OF LIGHT
1.
2.
3.
4.
5.
6.
7.
8.
9.
10. 4.6 1011 Hz
Given
d
a
c
a
b
a
c
b
no sharp division between one kind of wave
and the next.
−6
−4
= 650 μm= 650 10 m = 6.510 m
8
c =3.00 10 m/s
Solution
Rearrange the wave speed equation and
substitute values.
c=f
f =c/ =
(3.00 10 8 m/s)
(6.510 −4 m)
=
4.61011 Hz
Original content Copyright © by Holt, Rinehart and Winston. Additions and changes to the original content are the responsibility of the instructor.
Holt Physics
1
Section Quizzes
Name: _____________________________ Class: __________________ Date: __________________
Assessment
Light and Reflection
Section Quiz: Characteristics of Light
Write the letter of the correct answer in the space provided.
_____ 1. Which of the following is not a component of the electromagnetic
spectrum?
a. light waves
c. microwaves
d. sound waves
_____ 2. All of the following statements about electromagnetic waves are true
except which one?
a. Electromagnetic waves are distinguished by their different shapes.
b. Electromagnetic waves are composed of oscillating electric and
magnetic fields.
c. Electromagnetic waves are transverse waves.
d. Electromagnetic waves move in a direction that is perpendicular to
the electric and magnetic fields.
_____ 3. Given the wave speed equation c = f , what is the relationship between
frequency (f) and wavelength ()?
a. direct
b. exponential
c. inverse
d. none of the above
_____ 4. Which variable in the wave speed equation is constant in a vacuum?
a. c
b. f
c.
d. all of the above
_____ 5. Which of the following is approximately equal to the currently
accepted value for the speed of light?
a. 3.00 106 m/s
b. 3.00 108 m/s
c. 30.0 108 m/s
d. 3.00 109 m/s
_____ 6. The speed of light is incredibly fast. Physicists consider it to be
a. finite.
b. infinite.
c. immeasurable.
d. variable.
Original content Copyright © by Holt, Rinehart and Winston. Additions and changes to the original content are the responsibility of the instructor.
Holt Physics
2
Section Quizzes
Name: _____________________________ Class: __________________ Date: __________________
Light and Reflection continued
_____ 7. The intensity of light depends on the
a. speed of the light wave and the amount of light energy emitted from
a source.
b. distance from the light source and the type of surface the light
strikes.
c. amount of light energy emitted from a source and the distance from
the light source.
d. size of the surface area which the light strikes and the distance from
the light source.
_____ 8. If the distance from a light source is increased by a factor of 5, the
illuminance
a. also increases by a factor of 5.
b. decreases by a factor of 25.
c. increases by a factor of 25.
d. decreases by a factor of 5.
9. Since each type of electromagnetic wave has its own specific range, explain
why the electromagnetic spectrum is continuous rather than divided into
distinct sections.
_________________________________________________________________
_________________________________________________________________
_________________________________________________________________
10. What is the frequency of an infrared wave whose wavelength is 650 µm?
Original content Copyright © by Holt, Rinehart and Winston. Additions and changes to the original content are the responsibility of the instructor.
Holt Physics
3
Section Quizzes
``` | 914 | 3,568 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2023-23 | latest | en | 0.783869 |
http://parasys.net/how-to/error-residual-excel.php | 1,537,774,945,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267160233.82/warc/CC-MAIN-20180924070508-20180924090908-00193.warc.gz | 184,344,111 | 5,161 | # parasys.net
Home > How To > Error Residual Excel
# Error Residual Excel
## Contents
This gives the following Dialog, click on Regression and then click OK. For example: R2 = 1 - Residual SS / Total SS (general formula for R2) = 1 - 0.3950 / 1.6050 (from data in the ANOVA table) = We wish to estimate the regression line: y = b1 + b2 x2 + b3 x3 We do this using the Data analysis Add-in and Regression. The standard diagnostic graphs are not available in Excel, e.g. get redirected here
Using the critical value approach We computed t = -1.569 The critical value is t_.025(2) = TINV(0.05,2) = 4.303. [Here n=5 and k=3 so n-k=2]. This equation calculates the expected value of the second variable based on the actual value of the first variable. Photo Credits Stockbyte/Stockbyte/Getty Images Suggest an Article Correction Related Searches More Articles What Is Residual Cash Flow? "How to Use Excel to Find the Mean, Median & Mode Ranges" How to He received a Master of Science degree in wildlife biology from Clemson University and a Bachelor of Arts in biological sciences at College of Charleston.
## How To Calculate Residual In Excel
Hochgeladen am 08.12.2011 Kategorie Bildung Lizenz Standard-YouTube-Lizenz Wird geladen... Wird geladen... The predicted anxiety score is the score that was predicted from the regression equation. Step 10Hold the "Ctrl" key and highlight the data in column A.
The example we'll use is from pages 27-30 of Avery and Burkhart, 5th Edition. From this I standardize the residuals by saying (x-u)/(uRSD) where x = the observed value and u = the predicted value, so x-u = the residual. When \$n-p\$ is large, this correlation is of little import, so the Excel calculation yields almost the same results as a correct calculation would. Excel Residual Formula the normality plot of the residuals, the scatterplot or residuals against predicted values.
So I position myself so that I am looking down the line of fit, if that makes sense, so that I've reduced a dimension from the system and now I have How To Find Residual On Excel In multiple regression, it's not what we want. Detect if runtime is device or desktop (ARM or x86/x64) D&D 5e: Portent and Legendary Resistance Is it "eĉ ne" or "ne eĉ"? http://www.real-statistics.com/multiple-regression/residuals/ For example, to find 99% confidence intervals: in the Regression dialog box (in the Data Analysis Add-in), check the Confidence Level box and set the level to 99%.
Here, we tell Excel about our outcome variable. Excel Residual Sum Of Squares Wird verarbeitet... The two regressions, if both are sound from a residual analysis perspective, can be compared using the Standard Error statistic. Wähle deine Sprache aus.
## How To Find Residual On Excel
Nächstes Video Making a Residual Plot - Dauer: 4:57 Ms. In linear regression, this would be fine. How To Calculate Residual In Excel For older versions of Excel, see here. Valor Residual Excel If the dots tightly adhere to the zero baseline, the regression equation is reasonably accurate.
Du kannst diese Einstellung unten ändern. If you want to run a slightly different analysis, it is hard work, because you have to move your data around, a process which is prone to errors. This graph is useful, as it shows us the non-linear nature of the relationship. The the dataset we are using, the dependent variable is Anx, which is the column which goes from cell D1 to Cell D41. Excel Residual Analysis
Wird geladen... Honestly I don't know. Also some of this might be due to the presence of outliers (see Outliers and Influencers). more stack exchange communities company blog Stack Exchange Inbox Reputation and Badges sign up log in tour help Tour Start here for a quick overview of the site Help Center Detailed
Anmelden Transkript Statistik 83.535 Aufrufe 192 Dieses Video gefällt dir? How To Get Regression Statistics In Excel Kind regards Declan Reply Charles says: December 31, 2013 at 8:51 am Declan, I expect to add Durbin Watson test shortly. ANOVA ANOVA df SS MS F Significance F Regression 3 3017.745 1005.915 22.50151 2.22E-08 Residual 36 1609.355 44.70432 Total 39 4627.1 The ANOVA table comes next.
## Explanation Multiple R 0.895828 R = square root of R2 R Square 0.802508 R2 Adjusted R Square 0.605016 Adjusted R2 used if more than one x variable Standard Error 0.444401 This
It is not to be confused with the standard error of y itself (from descriptive statistics) or with the standard errors of the regression coefficients given below. Columns "Lower 95%" and "Upper 95%" values define a 95% confidence interval for βj. Generated Thu, 13 Oct 2016 05:07:34 GMT by s_ac4 (squid/3.5.20) Residual Plot Excel 2013 This is a version of Appendix 2 of the book "Applying regression and correlation: A guide for students and researchers." This is a version of Appendix 2 of the book "Applied
But you could have done the regression on your own, if you really wanted to. By default the x-axis labels appear in the middle of a residual graph (below a Y value of 0). Confidence intervals for the slope parameters. With multiple independent variables, then the plot of the residual against each independent variable will be necessary, and even then multi-dimensional issues may not be captured.
There is no hierarchical regression, no weighting cases, etc, etc, etc... Conclude that the parameters are jointly statistically insignificant at significance level 0.05. Excel limitations. To plot "BA_GRO" versus "CRWN_VOL" select the data (including labels - cells A1:B63) and choose Chart...
TEST HYPOTHESIS OF ZERO SLOPE COEFFICIENT ("TEST OF STATISTICAL SIGNIFICANCE") The coefficient of HH SIZE has estimated standard error of 0.4227, t-statistic of 0.7960 and p-value of 0.5095. For example, for HH SIZE p = =TDIST(0.796,2,2) = 0.5095. The results of the analysis are "dead" and will not change. Step 2Click and drag your mouse across both data sets to highlight all the values.
The data in column D should remain highlighted. Taylor embarked on a professional writing career in 2009 and frequently writes about technology, science, business, finance, martial arts and the great outdoors. Make sure that "Analysis ToolPak" is selected. | 1,528 | 6,207 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2018-39 | latest | en | 0.82688 |
https://testbook.com/question-answer/the-value-offracsqrt-768-times-sqr--6220a68e95606bac9441ac13 | 1,718,360,354,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861545.42/warc/CC-MAIN-20240614075213-20240614105213-00276.warc.gz | 521,804,315 | 45,674 | # The value of $$\frac{{\sqrt {768} \times \sqrt {3267} }}{{\sqrt {144} }}$$ is
This question was previously asked in
CTET Paper 2 Maths & Science 17th Jan 2022 (English-Hindi-Sanskrit)
View all CTET Papers >
1. 198
2. 128
3. 132
4. 134
Option 3 : 132
Free
CTET CT 1: CDP (Concept of Development) Mock Test
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10 Questions 10 Marks 10 Mins
## Detailed Solution
Given:
The given expression is (√768 × √3267)/√144
Calculation:
Prime factors of 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
∴ √768 = 16√3
Prime factors of 3267 = 3 × 3 × 3 × 11 × 11
∴ √3267 = 33√3
Prime factors of 144 = 2 × 2 × 2 × 2 × 3 × 3
∴ √144 = 12
Then, (√768 × √3267)/√144
= (16√3 × 33√3)/12
= 1584/12
= 132
∴ The value of (√768 × √3267)/√144 is 132.
Last updated on Apr 8, 2024
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# Counting
0
58
1
How many two-digit numbers are there such that the two digits differ by exactly \(3\)?
Jan 31, 2022
#1
+1374
+1
The possible pairs of numbers are \((0,3), (1,4), (2,5), (3,6) ,(4, 7),(5,8),(6,9)\)
Each of the \(7\) pairs can be ordered in \(2\) ways, making for \(14 \) numbers.
However, we need to account for the fact that the number \(03\) is not a \(2\)-digit number, so we have to subtract \(1\), making our total \(\color{brown}\boxed{13}\)
Jan 31, 2022 | 185 | 489 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2022-21 | longest | en | 0.70859 |
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# help
0
78
1
A gym instructor has a total of 60 hand weight in her studio. Some of the hand weights are 3 pound weights and the rest are 5 pound weights. If there are 10 more 5-lb weights than 3-lb weights, then how many of each kind does she have?
Jan 24, 2020
#1
+18830
0
Let x = number of 3-lb weights
Then x + 10 = number of 5-lb weights
Since there are a total of 60: x + (x + 10) = 60
x + x + 10 = 60
2x + 10 = 60
2x = 50
x = 25 (number of 3-lb weights)
x + 10 = 35 (number of 5-lb weights)
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http://www.jiskha.com/search/index.cgi?query=ana | 1,462,292,559,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860121618.46/warc/CC-MAIN-20160428161521-00002-ip-10-239-7-51.ec2.internal.warc.gz | 616,289,252 | 16,554 | Tuesday
May 3, 2016
# Search: ana
Number of results: 100
arithmetic
Ana sees a computer that costs \$1600. The store manager says next week will be on sale \$350 less. What percent will Ana save
April 2, 2014 by Anonymous
ALGEBRA
I know the answers, but am having trouble writing the equation. Their ages are 17 and 8.5, and 14.5 Can I get help with the equaiton? The sum of the ages of Sonia, Collin, and Ana is 40. Sonia is twice as old as Ana. Ana is 6 years younger than Collin. How old are they? Write ...
November 22, 2009 by BELLA
Spanish #1 (URGENT)
Can someone please check my work? Look at the words in the vocabulary list and write them in the appropriate column. Items in word bank may be used more than once. las cortinas el armario el espejo el televisor el comedor la cómoda la cama el sofá la lámpara EL COMEDOR el ...
September 17, 2009 by Larry
data management (gr.12)
ana finds 11 shirts in his size at a clearence sale. How many different purchases could ana make? the answer is 2047, any idea how to answer this? thanks!!
February 23, 2009 by joanna
Spanish!
Write the correct form of the verb tener for the following dialogue in the corresponding numbers below: Ana: Hola Isabel: ¿Tú _____ galletas cookies? Isabel: Yo _____ hamburguesas hamburgers. Ana: Yo no _____ ganas de comer hamburguesas. ¿Sabes si do you know if Amanda ____ ...
May 14, 2009 by James
Spanish for John
SINGLE OBJECT PRONOUNS = UN COMPLEMENTO SIMPLE DIRECT-OBJECT PRONOUNS: Ana no me visita. = Ann does not visit me. Ana no te visita. = Ann does not visit you (fam.) Ana no le visita = Ann does not visit him, you (ONLY IN SPAIN), otherwise: Ana no lo visita. = Ann does not visit...
May 10, 2008 by SraJMcGin
Trigonometry
A jet plane flying 6000 m above the ground passed directly above Ana. one minute later, Ana noticed that the plane's angle of elevation was 18 degrees 34 minutes. What was the horizontal distance covered by the plane?
November 2, 2013 by Sheena
math
alma is 6 years older than her sister ana. four years ago, ana's age is more than half of alma's age. what are their present ages
October 8, 2015 by abby
spanish
home work check please d1. Eduardo es _____ Chile. A al; B en C a D de E del F all of the above b2. "Romeo and Juliet" es _____ Shakespeare. A al B de C en D a E del F all of the above. c3. Voy _____ los Estados Unidos. A de B en C al D del E dal F a e4. Mi casa está _____ los...
October 4, 2010 by whitney
english
ana wants to write a descriptive essay about a place tha is important to her 1-which of these would best help ana prepare to write aboute her plkace a-reading stories about places she has not visited b-closing her eyes, picturing her place and listing details c-Listing places ...
April 14, 2013 by jake plzzz
SPANISH
This question tells us to answer if the children bahaved good or bad from the paragraph below. Todos los niños le regalaron a Ana muchas cosas fantásticas. Había muñecas, juguetes, animales de peluche … y más cosas. Susana le regaló un oso de peluche enorme. Ana estaba ...
February 3, 2015 by Carl
Spanish
Complete this conversation between two friends, Ana Margarita and Fernando. They are planning to go to a party with a group of students from the United States. Use ir, voy, vas, va, vamos, and van. 1.Ana Margarita: ¡Hola, Fernando! ¿Adónde * ahora? 2. Fernando: (Yo) * al cine...
October 9, 2010 by Christopher
Spanish
8. ____ ayudan mucho en casa. a. Tú y yo b. Ana y yo c. Ana Maria y él d. Tú Can't decide between A and B 9. "¿Qué bebes en la cena?" "________ leche." a. Bebo b. Bebes c. Beben d. Bebemos C? 10. No me gustan las uvas pero me ______ las manzanas. a. encanta b. gusta c. ...
November 18, 2010 by mysterychicken
science
Ana exerts a force of 20 N on a volleyball that has a mass of 0.15 kg. What is the force that the volleyball exerts on Ana?
November 2, 2015 by john
Spanish
Read the description of Ana’s eighth birthday celebration. Write bien or mal to describe how the guest behaved. Then write a sentence that supports your answer. ¿Cómo se portaba Susana? ¿Por qué crees eso? 1. Todos los niños le regalaron a Ana muchas cosas fantásticas. Había ...
October 23, 2015 by Alex
Geometr
Determin the distance b/w a(4,3)and(8,12)b(2,1)and(2,-9)c(-1,-5)ana(2,-3)
February 23, 2012 by Mathes
Geometr
Determin the distance b/w a(4,3)and(8,12)b(2,1)and(2,-9)c(-1,-5)ana(2,-3)
February 23, 2012 by Mathes
Spanish
What is Me llamo Ana.?Y tu'? ?Co'mo te llamas?
September 25, 2013 by Chase S.
help with homework
¿De dónde _________ Ana y Claudia?
September 8, 2015 by Anonymous
Spanish
2. Nunca como allí porque la comida es ______. a. buena b. horrible c. sabrosa d. malo D? 7. Me gustan los sandwiches de jamón y queso. Son muy _________. a. buenas b. malos c. sabrosos d. horribles C 8. ____ ayudan mucho en casa. a. Tú y yo b. Ana y yo c. Ana Maria y él d. T...
November 18, 2010 by mysterychicken
spanish
what does Por Que esta triste ana mean?
June 1, 2008 by Ashley
spanish
correct verb form of leer when used as que __ ana?
November 30, 2009 by Anonymous
English Literature
Compare my Papa's walta with ana. they are both poems.
March 3, 2010 by Devaughn
spanish
Ayer mis amigos Esteban, Ana, Olivia y yo fuimos al cine a ver la nueva película de Javier Bardem. A mí me (1) ____ la película porque Javier es uno de mis actores favoritos. Pero a (2)____ le (3)______ ; él empezó a dormirse al final. A Ana y (4)_____ Olivia les molestaron ...
October 7, 2013 by Kayee
Spanish II
9. Which word belongs in the blank? ________ compro el libro a Ana. A. La B. Le C. Lo D. Ella I think it's D. Can anyone help?
December 22, 2014 by Sandy East Ward
English Literature
Each of the poems Ana and My papa's Waltz, write an essay to describe the relationship presented.
March 3, 2010 by Devaughn
math
i have a garden 22meters long and 14 meters wide how do i find the area ana the perimeter
February 13, 2013 by adriana
To Ana
I saw you post 4-5 pages down about K for the disproportionation problem. I calculated K by RTlnK = nFEo
May 9, 2014 by DrBob222
statistics
suppose x is a uniform random variable with c=40 ana d= 70 find the probability that a randomly selected observation is between 43 and 65
November 7, 2011 by Anonymous
lit
"Maybe we should wait to begin until everyone is seated," Ana said to her band mates. did I punctuate this correctly?
May 10, 2011 by joel
Chemistry - Science (Dr. Bob222)
Posted by Ana on Saturday, May 3, 2014 at 11:23am. A zinc-copper battery is constructed as follows (standard reduction potentials given below): Zn | Zn2+ (0.10 M) || Cu2+ (2.50 M) | Cu Zn2+ + 2 e- → Zn(s) Eº = -0.76 V Cu2+ + 2 e- → Cu(s) Eº = 0.34 V The mass of ...
May 5, 2014 by Ana
Chemistry
We say that most bases are metal oxides and hydroxides, so an example of a base that is not ana oxide or hydroxide would be aqueous ammonia NH3(aq)?
April 10, 2009 by janice
world history
what page can i find "Hoe Ana on in our world by macmillan/mcgraw-hill
January 31, 2013 by lori
Spanish
These are the instructions. Write the correct conjugation of the verbos pronominales in parenthesis to correctly fill the blanks in the following sentences: Example: Yo me levanto a las ocho. (levantarse) 1. Nuria y Ana despiertan a las nueve. (despertarse) 2. Nosotros no ...
May 22, 2010 by Rosa
Math
Ana had some money. she used 4/7 to buy shoes which cost \$96. if she spent another 2/7 to buy a dress how much would she spend altogether?
April 22, 2015 by Liz
spanish
4. In the sentence "Paula nos dice la verdad." (Paula tells the truth to us.) Is the "nos" a direct or indirect object pronoun? direct 8. Ya lo ___veo___ (ver), (I can see that) Please re-write the following sentences (making them grammatically correct) adding an indirect ...
November 6, 2010 by whitney
spanish
please explain the meaning of the quesiton and each of the answers "¿Les mostraste a Ana y a Paco tu tocacintas nuevo?" "Sí, y ellos me _____ el suyo también." a. muestran b. muestra c. mostraron d. mostró thanks
May 30, 2009 by y912f
spanish
I have to fill in the blanks using ser, estar and tener and Im having trouble with a few can somebody help me please. Ana y yo____bajas. Vosotros____buenos. Uds._____simpaticos Donde____mis cosas? De donde_____usted?
September 19, 2009 by Jasmine
nursing
The ANA'S ethical theories and principles and how it incorporates in its operations. Provide an example of how it applies these theories and principles
December 17, 2011 by maria
Math
Ana bought a van that holds 12.5 gallons of gas and gets an average of 15.5 miles per gallon. How many miles can she expect to go on a full tank of gas?
September 23, 2015 by Amy
Science
Of the three local winds, which do you think could be easily harnessed for the generation of electricity by a huge wind farm and why? Land and Sea Breezes,Mountain and Valley Breezes or Chinook and Santa Ana Winds.
June 15, 2011 by Niki
Math
In 5 years, Ana's age will be twice the age of her friend Jun. 5 years ago, she was 3 times as old as her friend. Find their present ages.
November 17, 2015 by Marie
math
The sum of the digits of a two-digit number is 14. If the numbers are reversed, the new number is 18 less than the original number. Find the original number. I know Ana asked this question, but i dont understand how to get the equations.
January 6, 2008 by Astrid
physics
A child wants to throw a rock into a puddle which is 20m away and 20m down a hill. If he can throw the rock at ana ngle of 60, what velocity does he need to release the rock at?
November 17, 2010 by kristen
Spainsh
Check? Choose the correct option to complete the sentence Choose the correct option to complete the sentence 1. Marco y Ana ______________ en Acapulco por 3 semanas A. estuvo B. estuvieron C. estuvimos D. estuvisteis My answer: A Yo no _____________ comprar ropa muy cara en la...
December 13, 2012 by Unknown
Spanish
11.) El profesor___, y el estudiante___. A-termina las clase, empieza la clase B-ensena, estudia C-ensenar mucho, estudiar D-empieza la clase, terminar 18.)___. tocas la guitarra? A- Miguel B-Juan y Carlos C- Felipe, Ana y Barbara D-Senora Carreras
August 10, 2011 by Sarah
PHYSICS
Ana run in a circular path. The dimension of the circular path is r = 10m , she finished it in 20 seconds. Find his speed, distance, displacement and velocity. 10m east (starting point) and 10 m south (finish point)
July 23, 2014 by SUZI
Spanish
Completa las oraciones con los verbos entre paréntesis en el pretérito perfecto. OJO: Ana Sofía es española y usa la forma de vosotros cuando habla con Estefanía y Franklin. pues amigos, que _________ (hacer) esta semana? I am unsure? I thought it was han hacido
January 13, 2016 by Spanish
English... Urgent
Look at the following puzzles and try to decipher the well known sayings 1. Ban ana 2.nafish nafish 3.japmadean 4.issue issue issue issue issue issue issue issue issue issue 5. Lookuleap Note that I have not typed any of these words wrong
October 26, 2013 by Tim
MATH
Bob drove to the store( a distance fo 10.5) miles at ana average speed of 35 mile per hours. On the way home, he got stuck in traffic, and he only averaged 15 miles per hour along the same route. What was his average speed for the round trip ?
September 20, 2013 by Anonymous
ana g mendez
A sample of fifteen 2-liter bottles of a soft drink delivers a mean weight of 2009 milliliters with a sample standard deviation of 11 milliliters. The quality assurance department wants to determine if there is a true fill problem, so they determine the confidence interval at ...
March 27, 2016 by carol
spanish
can someone please explain this to me? its the irregular verb estar. 1. Yo_____ en mi casa (in my house). 2. Jose, María, y yo _________ alegres. 3. Mario, tú, y yo __________ en el cine (at the movies). 4. El perro ________ en el parque (at the park.) 5. Usted _________ cerca...
September 10, 2010 by whitney
Algebra
Use a system of equations to solve problem Emily emptied her bank ana totaled the nickels, dimes, and quarters that she had saved. The total amount was 14.50 The number of dimes was 2 more than 3 times the number of nickels. How many of each coin were there Can you show the ...
December 10, 2008 by Chris
Spanish
Can someone please check my work? Answer the following questions in complete sentences. 1. ¿Adónde vas los lunes a las 8:30 de la mañana? Yo voy a la clase de historia los lunes a las ocho y media de la mañana. 2. ¿Adónde vas los lunes a las 3:30 de la tarde? Yo voy a la mi ...
February 26, 2009 by Larry
Spanish for John
Spanish for John. Direct-object pronouns: SINGLE OBJECT PRONOUNS: (what you learn first) and Heaven Help me if it won't cut & paste because then I'll have to retype everything! l7. UN COMPLEMENTO SIMPLE DIRECT OBJECT PRONOUNS Ana no me visita. Ann does not visit me. Ana no te ...
May 10, 2008 by SraJMcGin
I need HELP!!!!
Use a system of equations to solve problem Emily emptied her bank ana totaled the nickels, dimes, and quarters that she had saved. The total amount was 14.50 The number of dimes was 2 more than 3 times the number of nickels. How many of each coin were there Can you show the ...
December 10, 2008 by Chris
math
ana is 5 times as old as beth and 2 2/3 times as old as cathy. What fraction of cathy's age is Beth's?
November 13, 2011 by kathleen catacutan
spanish
what would the possesive pronoun be? Se me quedaron en casa los libros. ¿Trajo Ana ____ (hers)? Uso mi teléfono celular cuando viajo. ¿Cuándo usa Ramón _____ (his)? Tu profesora es muy simpática. _____ (Ours) es simpática también. Me encantan mis clases. ¿Les gustan a tus ...
December 15, 2012 by lauren
health
if rheumatoid factor 1gM under ELISA process shows around 320.6 RU/mL and thyroid stimulating hormone TSH at 13.37 and ESR at 60mm/hour ra factor : non reactive 1:2 anti nuclear anti bodies ANA 0.39 (n) pl tell me what would happen to the patient and what precautions are ...
November 23, 2012 by R S Telagathoty
math
Can you help me with this: Paul's budget for food is \$10 more than 1/3 of his total budget. If \$100 was alloted for food, what was the total budget? Ana if you have to change the 1/3 to a decimal, can you help me with that too? Thank you!! 100-10= 1/3 B I wouldn't change it to...
January 10, 2007 by Brianna
spanish
Write the correct conjugation of tener for the following sentences: 9. ¿Por qué _________ (Jose) sueño por la mañana? 10. Mi amiga _________ tres hijos. 11. ¿Cómo se dice "Patricio tiene ganas de comer" en inglés? 12. ¿Cómo se dice "Ana has to study (estudiar)" en español? 13...
May 14, 2009 by luke
spanish
please check my homework. 1. Yo__estoy___ en mi casa (in my house). 2. Jose, María, y yo _____estoy____ alegres. 3. Mario, tú, y yo ____estoy______ en el cine (at the movies). 4. El perro ___estamos_____ en el parque (at the park.) 5. Usted ____esta_____ cerca de la casa. 6. ...
September 13, 2010 by whitney
earth science
of the three "local wind"(land and sea breezes, mountain and valley breezes, and chinook and santa ana winds)which do you think could be most easily harnessed for the generation of electricity by a huge wind farm? some important things to consider:strength and constistency of ...
June 14, 2011 by Staci
spanish sra jmcguin
1. ana tenia una casa preciosa en el centro, pero era/fue muy viejay no tenia calefaccion. Al final se mudo. we must choose era or fue -- i think it is fue. 2. A los anos descubrieron que marquitos era/fue milope y le pusieron gafas, claro. chose era or fue - i think it is era
May 14, 2009 by sam
spanish sra jmcguin
1. ana tenia una casa preciosa en el centro, pero era/fue muy viejay no tenia calefaccion. Al final se mudo. we must choose era or fue -- i think it is fue. 2. A los anos descubrieron que marquitos era/fue milope y le pusieron gafas, claro. chose era or fue - i think it is era
May 14, 2009 by sam
Spanish Sra J Mcguin Help
Answer questions positively using complete sentences 1. Me afeite esta manana antes de ir al trabajo? 2. Nos ponemos los vestidos en la habitacion del hotel? 3. Te duermes en el cine cuando ves peliculas aburridas? 4. Se sienta Ana delante de Frederico en clase? I am sorry ...
February 6, 2008 by sam
Spanish Check
Can you tell me if any of my answers are wrong, and if so which/why? Instructions: Choose por or para to fill in the following phrases; 1. Compré las flores __________ mi novia. Compré las flores para mi novia. 2. Hice la tarea __________ la tarde. Hice la tarea para la tarde...
October 12, 2014 by Joshua
spanish Homework
The directions say... Complete the following sentences with the correct preterit form of the logical verb in parentheses. Here I have listen 3 out of the 15 questions. I am not really sure what the directions are asking me to do. Mis padres me ____ (llegar / llamar) por telé...
March 10, 2015 by Michael
Physics Ana
http://www.jiskha.com/display.cgi?id=1258760143
November 20, 2009 by bobpursley
Spanish #2
Can someone please check my work? Write complete sentences to find out about the following people in Spanish class. 1. Claudia y yo/estar/cerca de la puerta Claudia y yo están cerca de la puerta. 2. La maestra/estar/contenta La maestra está contenta. 3. Los estudiantes/estar/...
February 26, 2009 by Larry
Spanish #2
Can someone please check my work? Write complete sentences to find out about the following people in Spanish class. 1. Claudia y yo/estar/cerca de la puerta Claudia y yo están cerca de la puerta. 2. La maestra/estar/contenta La maestra está contenta. 3. Los estudiantes/estar/...
February 26, 2009 by Larry
Ana
A group of hikers on a mountain began at an elevation of 3040 feet above sea level and stopped at elevation of 2319 feet above sea level. What was their change in elevation between these points? How can you tell from the change in elevation whether the hikers were going up or ...
September 9, 2012 by Math
spanish
1. The "tú"form of you is used: A all the time B as a sign of respect C with strangers D with friends and peers E only in Spain F only on Mondays Which subjects correspond to the following combinations? 2. tú y yo A yo B nosotros C ustedes D ellos E ellas F tú 3. Ana y yo A t...
September 9, 2010 by nicole
Math
A flagpole casts a shadow of 23.5 feet long.at the same time of day, Ana, who is 5.5 feet tall, casts a shadow that is 7.5 feet long. How tall in feet is the flagpole?
January 13, 2015 by Anonymous
spanish
what does yo soy tuja and dame un beso and yo te amo mucho and no me quiero ir yo soy tuja means i'm yours (girl verison) dame me un beso means give me a kiss yo te amo mucho means i love you alot no me quiero ir means i dont want to leave Thank you for using the Jiskha ...
March 12, 2007 by jose
Spanish
18. ________ ¿tocas la guitarra? a. Miguel b. Juan y Carlos c. Felipe, Ana y Barbara d. Señora Carreras A? 19. Me gusta ir a las cuatro y _______ a. cuarto b. cuarta c. quinta d. séptima I don't really understand this one.. 22. ¿Qué quiere hacer ____ sábado por la noche? a. en...
January 27, 2011 by mysterychicken
algebra 1
ana bought 12 pieces of gum consisting of only red gum and white gums. The total cost was 1.29 the red gums each cost .03 more than each white gum and she bought fewer red gums than white gums. How many white gums did she buy?
February 14, 2008 by angleica
Math (Probability)
Mrs. Ana is holding a logic contest. The 13 students who are participating randomly draw cards that are numbered with consecutive integers from 1 to 13. >The student who draws number 1 will be the host. >The students who draw the other odd numbers will be on the red team...
February 13, 2014 by Nikko
medical language
B. Define the following words or word parts. -ad -ior -ics Ana- Prone Supine viscera- transverse Hypo- Re- C. In this activity, break the medical word into its word parts. Using the word parts, create a definition of the word. Write a sentence using the word in a medical ...
May 28, 2008 by sue
(19) Chemistry - Science (Dr. Bob222)
Use the following information to answer this question: Cu+(aq) + e- → Cu(s) E° = 0.521V Cu2+ (aq) + e- → Cu+ (aq) E° = 0.153 V. Given these half cell reactions, an aqueous solution of Cu+ ion in the absence of O2(g) : A. will be thermodynamically stable B. will ...
May 8, 2014 by Ana
bio stats
After a winter ice storm, 25 parrots are found sheltering in a tool shed. Ana traps the parrots, weighs them (average weight = 36 grams) and then feeds them. The next morning she find that 6 of the parrots have died. She weighs the dead ones (average weight = 20 grams) and the...
April 26, 2009 by Frank Minimi
Sra ramired is asking her family if they did everything they were supposed to do today write the excuses with the correct preterite form like i know what that is of HACER,IR,PODER,SER or TENER paquito fuiste a la biblioteca verdad? ay no mama no ________ 1 ir ____________ 2 ...
June 29, 2010 by Marie
Pre-Calc
Ana Colon asks her broker to divide her 401K investment of \$2000 among the International Fund, the Fixed Assets Fund, and the company stock. She decides that her investment in the International Fund should be twice her investment in company stock. During the first quarter, the...
September 24, 2011 by Pauline
Finance
Anna has been saving \$450 in her retirement account each month for the last 20 years and plans to continue contributing \$450 each month for the next 20 years. Her account has been earning a 9 percent annual interest rate and she expects to earn the same rate for the next 20 ...
October 13, 2014 by Stovall
Finance
Anna has been saving \$450 in her retirement account each month for the last 20 years and plans to continue contributing \$450 each month for the next 20 years. Her account has been earning a 9 percent annual interest rate and she expects to earn the same rate for the next 20 ...
October 13, 2014 by Dashawn
Finance
Anna has been saving \$450 in her retirement account each month for the last 20 years and plans to continue contributing \$450 each month for the next 20 years. Her account has been earning a 9 percent annual interest rate and she expects to earn the same rate for the next 20 ...
October 13, 2014 by Dashawn
Spanish Check??
Conjugate the following verbs to future simple. Juan Y Ana (viajar) ______viajarán___________ a Europa el próximo año. Mi horóscopo dice que (tener) ___tendré______________un dia muy bonito. En las noticias dijeron que hoy (llover, 3rd person) ___ lloverá______________. ...
January 18, 2013 by Victoria
#1. Who was the African American who called the war with Mexico "disgraceful" and "cruel"? a) Santa Ana b) Frederick Douglass ****** c) Davy Crockett d) Sam Huston #2. When Mexico won its independence in 1821, it inherited the New Mexico province from Spain. New Mexico ...
February 28, 2014 by Mindy
Can someone help me I really don't understand this and this is how far I got. 1. ¿ De dónde es la paella? Paella es más popular en el sur de España. 2. ¿Con qué se hace el flan? El flan se hace al horno 3. ¿Cuál (which) tipo de plátano frito es dulce? Maduro son los pl...
March 5, 2013 by Alexus
English
I just want to know if i have these correct for in-text citation. January 2005. Globalization: “A portrait of exploitation, inequality, and limits” (Dunklin) Prospect, 27. Retrieved January 15, 2008 September 1977. “Body Politics: Power, Sex” (Nancy) Non-Verbal Communication ...
February 25, 2015 by Denise
Spanish
Multiple Choice Escoge el mandato correcto para las siguientes oraciones. Nota el tono, formal o informal. 6. Chicas por favor ____ (venir) conmigo a la pelicula. No quiero ir sola. A) venid B) vengais C) vengan D) vengamos 7. Estimados senores y senoras, les agradecemos su ...
July 9, 2011 by Lindsey
spanish
11. Todos los domingos Soledad _______ (correr) una mila. A corrieron B corrió C corrías D corrían E corría F corriste 12. Susana y Javier siempre _______ (tener) sueño los lunes por la mañana. A tuvieron B tenía Dtenían E tuviste F tenemos 13. Mientras yo _______ (dormir), mi...
January 19, 2016 by imane
Spanish help
1. ¿Se divirtieron Uds. En la fiesta de fin de año? Sí, pero comimos demasiado. La madre de Pablo nos ______ mucha comida. A. sirve B. saludó C. besó D. sirvió I said D 2. Carla, el tocacintas mío no funciona. ¿Puedo usar _____? ¡Claro que sí! Aquí lo tienes. A. tuyo B. ...
March 15, 2014 by Eri Uchi
Chemistry - Science (Dr. Bob222)
Liquid chloroform (CHCl3) is placed in a closed container and allowed to equilibrate with chloroform vapor. With the temperature held constant, additional chloroform is introduced into the system, and the vapor-liquid equilibrium is reestablished. Following this, the ...
May 10, 2014 by Ana
Spanish
Please see if the translation is correct, or if there are better ways of saying certain phrases, especially toward the latter of the paragraph. Please look for grammatical errors. Spanish author Ana Maria Matute was born on July 26, 1926. She was one of the strongest voices ...
April 27, 2010 by Anonymous
1. ¿Sabes quién ganó el partido de fútbol? Sé que _______. A. soportaron B. empataron C. prestaron D. fueron I said B 2. ¿Cómo te _______ cuando eras pequeño? Muy mal. Siempre me peleaba con mi hermana menor. A. llorabas B. portabas C. preferías D. eras I said D 3. Cuando era ...
March 12, 2014 by Eri Uchi
spanish
Ok so I just need someone to check my answers and tell me if they are right or wrong :) (if wrong why) And to help me with the unanswered questions (couldnt find the right answer for them) Choose the correct form of the préterito tense: 1. Susana dormía cuando ________ (sonar...
January 19, 2016 by Luna
Spanish
Sorry for the backwards accents I just got a new keyboard, so I don't know how to do the spanish accents yet. I hope a got more right this time 1. Camarero, ¿me trae ________? Quiero pedir un postre. A. El mantel B. el menú C. la servilleta D. la merienda I said B 2. El sòtano...
July 9, 2013 by Eri Uchi
Spanish for John
l8. LOS COMPLEMENTOS DOBLES When a verb has two object pronouns, the indirect object (usually the object pronoun referring to the person) precedes the direct object pronoun (usually the object pronoun referring to a thing). Juen me lo (la) da. John gives it to me. Juan me los...
May 10, 2008 by SraJMcGin
spanish
Complete each of the following sentences by writing the appropriate preterit or imperfect form of the indicated verb. BELOW THESE 43 QUESTIONS ARE MY ATTEMPT AT THE CORREECT ANSWERS. I JUST NEED THESE CORRECTED. PLEASE MARK ANY THAT ARE WRONG AND PROVIDE THE CORRECT ANSWER. 1...
February 2, 2010 by Lane
1. Pages:
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You are here: Home » ICSE Physics Standard 10 Index » ICSE Physics Class 10: Refraction of Light
# Refraction of Light Notes: Flash Cards Set 2
Refraction of Light is a chapter of ICSE class 10 Physics textbook. Here, we present Set 2 of flash card notes on Refraction of Light. These notes are totally based on latest Physics textbook of ICSE Board. This learning module contains five questions in flash cards format. You can even download these flash cards for free. Learn this Physics module on Refraction of Light to score full marks in ICSE class 10 Board exams. Quizcrazy.in and team is proud to present Refraction of Light flash card notes to you.
Refraction of Light Flash Card Notes - Set 2
Refraction of Light Flash Card 4: STATE SNELL’S LAW OF REFRACTION OF LIGHT The ratio of the sine of the angle of incidence in the first medium to the sine of the angle of refraction in the second medium is constant for a given pair of media. This constant is called refractive index of the second medium with respect to the first medium. This is Snell's Law of Refraction. Refraction of Light - Snell's law of refraction
Refraction of Light Flash Card 5: WHAT IS REFRACTIVE INDEX OF A MEDIUM? The refractive index of a medium is define as the speed of light in vacuum to the speed of light in that medium. Refraction of Light - Refractive index of a medium
Refraction of Light Flash Card 6: DEFINE PRISM A prism is a transparent refracting medium bounded by five plane surfaces inclined at some angle. Refraction of Light - Definition Prism
Flash card notes of Refraction of Light: | 384 | 1,664 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2020-45 | latest | en | 0.854188 |
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# Stat4Ci A sensitive statistical test for smooth classification images.
## Présentation au sujet: "Stat4Ci A sensitive statistical test for smooth classification images."— Transcription de la présentation:
Stat4Ci A sensitive statistical test for smooth classification images
Aspects théorique Image de classification Multiple comparaisons –Aucune correction –Bonferroni Random field theory –Pixel –Cluster –Scale Space
.* = Original imageMask of bubblesStimulus Stimuli creation Bubbles stimulus
Bubble task: gender discrimination Men or women?
Creation of the Ci +…+++++ Expected proportion =.75. ++++…, …,,,,, ==./ =
(Gibson, Gosselin, Schyns & Wasserman, Psychonomics, 2002) (Gosselin & Schyns, Vision Research, 2001) Bubble classification images
Visage à lendroit Visage à lenvers
Image de classification Régression multiple i=1..n, n nombre dessais u est une variable despace Pour n suffisamment grand
Test t p = [ 96,127,127,137,114,119,109,109,143,109,116,114,143,109,117,127,112,112,98, 37,112,109,119,106,109 ] Moyenne = 117.2, 2 = 160 v = [ 114,88,102,127,104,104,91,96,104,106,91,102,104,100,114,109,109,119, 91,81,114,119,102,111,80,119,119,123,119,114,132 ] Moyenne = 106.71, 2 = 167 t=1.97
Test t~Z score T(12)=3.56 t=1.97 Z=1.64, p=.05Z=2.35, p=.01
Comparaison multiple Aucune correction –p = % de chance de faux positif Bonferroni –p = p/N N est le nombre de comparaison –Soit 5 faux positif pour 100 images 1 test 2 tests 16*16 tests
Comparaison multiple p=.05, z=1.65 p=.05, z=1.92 p=.05, z=3.54 p=.05, z=3.89 P=0.00005
? p=.05, z=4.80 P=8.10 -8
Solutions Bonf t = 4.5228 Bonf t = 3.5463RFT t = 4.06
Random Field theory
Expected Euler Characteristic
Gaussian Random field
Seuil non corrigé
Bonferroni Correction
Pixel test
t z = 3.30
tsize resels Zmax x y ----------------------------------------- C[2.70] 970 0.44 4.17 122 129 [2.70] 917 0.41 3.95 162 129 ----------------------------------------- P3.30- p-value = [0.05] FWHM = [47.1] Minimum cluster size = 861.7 Seulement 2 paramètres –FWHM = taille du filtre de lissage –p = seuil de confiance Comment choisir FWHM ? –Pour détecter un signal donné, le meilleur filtre est un filtre de taille comparable Problèmes –Si le signal est diffus, le pic est faible Solution –Prendre en compte la taille et le Z score. Cluster test
tsize resels Zmax x y ----------------------------------------- C[2.70] 970 0.44 4.17 122 129 [2.70] 917 0.41 3.95 162 129 ----------------------------------------- P3.30- p-value = [0.05] FWHM = [47.1] Minimum cluster size = 861.7 t size resels Zmax x y ----------------------------------------- C[2.70] 1787 0.81 5.2 133208 ----------------------------------------- P3.30- p-value = [0.05] FWHM = [47.1] Minimum cluster size = 861.7
Cluster test t z = 2.5 k = 350 pixels
Scale space r=w 1 /w 2
Scale space t z = 4
Scale space t z = 2.5 k = 350 pixels
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# Week3.5Syllabus - Week 3.5 Syllabus The Metric Structure of...
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Week 3.5 Syllabus: The Metric Structure of R n , Applications, Matrix Operations 1 The Metric Structure R n R n (ro vectors for ease of typography, similarly for column vectors R n ) has a dot product on vectors defined as ~u ~v = n i =1 u i v i R , i.e., multiply the coefficients term by term but then add the results to form a single scalar . It satisfies the following basic rules for all vectors ~u, ~u i ,~v V and scalars c R : (1) ( symmetry ) ~u ~v = ~v ~u (2) ( additivity ) ( ~u 1 + ~u 2 ) ~v = u 1 ~v + u 2 ~v (3) ( homogeneity ) ( c~u ) ~v = c ( ~u ~v ) (4) ( positive-definiteness ) ~u ~u > 0 for all ~u 6 = ~ 0 . These 4 rules will become the definition for an abstract inner product on any vector space. 1.1 The Norm From the dot product we can derive the the concepts of norm of a vector k ~v k := ~v ~v = p x 2 1 + · · · x 2 n 0 satisfying k ~u k = 0 iff ~u = ~ 0 and k c~u k = | c | k ~u k . Any nonzero vector ~v can be normalized to give a (normal) unit vector ~u with k ~u k = 1 pointing a length 1 in the same direction via ~u = 1 k ~v k ~v (sloppily but conveniently written ~v k ~v k ). Remember that because the norm involves a square root, itis usually more convenient for calculations to deal with k ~v k 2 = ~v ~v, since dot products are bilinear but square roots are not. 1.2 The Cauchy-Schwarz Inequality Expanding out the positive-definiteness ~x ~x 0 for x := k ~u k ~v -k ~v k ~u yields 2 k ~u k k ~v k ( k ~u k k ~v k- ~u ~v ) 0 , establishing (applying this to both ~u and - ~u ) the Cauchy-Schwarz Inequality | ~u ~v | 6 k ~u k k ~v k . This guarantees that the ratio ~u ~v k ~u k k ~v k falls in the interval [ - 1 , 1], hence is uniquely cos( θ ) for some θ in the interval [0 , π ]; the angle θ between two vectors ~u,~v is θ = arccos ~u ~v k ~u k k ~v k · [0 , π ] . The most important angle is pi 2 (a.k.a 90 ): orthogonality (a.k.a. perpendicularity) ~u ~v of two vectors means ~u ~v = 0. The CSI implies the triangle inequality k ~u + ~v k ≤ k ~u k + k ~v k (by squaring both sides). This will be very important in measuring distances between vectors (especially functions) 1.3 Distance We have a metric concept of distance between two vectors d ( ~u,~v ) := k ~u - ~v k , satisfying d ( x, x ) = 0 ⇐⇒ x = 0 , d ( x, y ) = d ( y, x ), and the triangle inequality d ( x, z ) 6 d ( x, y ) + 1
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d ( y, z ) [this follows from the triangle inequality for vectors with u = x - y, v = y - z, u + v = x - z ]. The Pythagorean Theorem says that if ~u,~v are orthogonal then k ~u + ~v k 2 = k ~u k 2 + k ~v k 2 . From these we can (in some other course) study calculus in R n . 2 Applications The text introduces several instances where practical problems lead to systems of linear equations. There is no one set of applications of interest to all students, and the applications given in the book are extremely elementary. Many applications involve ”network analysis” where several systems are interconnected in a linear way, and the problem is to analyze the network as a whole.
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Jill Tulane University ‘16, Course Hero Intern | 1,244 | 4,355 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2018-22 | latest | en | 0.851637 |
https://www.exceldashboardtemplates.com/multi-column-stacked-chart-how-would-you-do-it-how-i-did-it/ | 1,722,680,101,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640365107.3/warc/CC-MAIN-20240803091113-20240803121113-00133.warc.gz | 605,514,759 | 33,496 | # Multi-Column Stacked Chart – How would you do it? – How I did it.
First, I want to say thank you, because you are an Excel fan. Then again, maybe you are an Excel Geek? I know I am . So I read blogs and try to learn about everything Excel. Thank you for joining me in a quest to explore and unlock all Excel puzzles. In the meantime, I wanted to show you my recommendations for the “How would you do it?” segment.
How did you do? Let me know in the comments below! Also, let me know if you were able to use this technique in your last company dashboard. Here is the Excel User question that was posed:
“I have data in this format
A B C
1 Project Name Provider Spend
2 APPLE FRUIT \$ 73,460.68
3 BURGER FAST FOOD \$ 226,331.56
4 HOTDOG FAST FOOD \$ 328,968.64
5 CHOCOLATE BAR CONFECTIONARY \$ 11,611.08
6 CARROT VEGETABLE \$ 74,283.04
7 CABBAGE VEGETABLE \$ 93,143.96
8 ONION VEGETABLE \$ 74,283.04
9 PEPPER VEGETABLE \$ 70,799.56
10 GARLIC VEGETABLE \$ 48,601.80
12 ROSEMARY HERB \$ 159,088.80
I want to present it so that it looks like this (although obviously in real life would be in proportion to the numbers!):
I’m really struggling – can anyone help? I would be very grateful.”
Well sadly, I never heard back from this forum poster . Maybe they had moved on by the time I saw their post and was able to answer their question. No worries mate, others can benefit from this question and maybe the users used it at a later date.
I got a few great comments on alternate suggestions. I feel that everyone gravitates to a first answer and mine was different then theirs. Here is what I came up with:
My final chart looks like this:
I think the Excel user may have gotten the scales a bit wrong . Look at the difference between my fast food column and theirs. Regardless, he said that the columns weren’t accurate. I think my chart looks good. So how can we do this type of chart. Here is one way. See the bottom of the post for an alternate solution.
The Breakdown
1) Setup your data in Rows
2) Create Chart and Switch Row/Column
4) Resize and Clean up Chart Junk
5) FAN ALTERNATE SOLUTIONS Check them out!
Step-by-Step
1) Setup your data in Rows
So you say the way the user had their data set up. It a great table for the data, but Excel won’t chart it the way we want. So we need to change the way we position the chart data so that Excel will make the graph we want. Here is how I positioned the Excel stacked column chart data table:
A B C D E F G H I J K L
15 Provider APPLE BURGER HOTDOG CHOCOLATE BAR CARROT CABBAGE ONION PEPPER GARLIC BREAD ROSEMARY
16 FRUIT \$ 73,461
17 FAST FOOD \$ 226,332 \$ 328,969
18 CONFECTIONARY \$ 11,611
19 VEGETABLE \$ 74,283 \$ 93,144 \$ 74,283 \$ 70,800 \$ 48,602
20 CARBOHYDRATE \$ 23,221
21 HERB \$ 159,089
Each row of data will represent a stacked column for the dashboard chart. This is why the Excel user had such issues trying to create their chart. So if you have problems creating a chart, try new and exciting ways to combine your data and Excel may then give you the graph you are looking to do.
2) Create Chart and Switch Row/Column
Now that we have our data setup correctly, creating the chart will be a piece of cake. So highlight the entire chart data table from the Provider label to the Rosemary value. Then go to the Insert Ribbon and choose the Column Chart button and the Stacked Column choice.
If you don’t know why you need to do this step, check out this post:
#### Why Does Excel Switch Rows/Columns in My Chart?
Your chart should look like this:
Now make sure you still have the chart selected and then go to the Design Ribbon and choose the Switch Row/Column button from the Data Group:
Your resulting chart should now look like this:
This chart now looks pretty close. All we need to do is to add the data labels and clean up the chart junk.
Here is where we need to do a lot of individual work. I haven’t found another way to do this any faster. If you know of a way to do this faster, please let me know in the comments below so that I can share it with everyone.
To add data labels, select the chart and then go to your Layout Ribbon and choose the Data Labels button and choose Center option. Your chart should now look like this:
Not quite the labels we are looking to display, but we can fix it. First select any of the data labels that you see and then press CTRL+1 to bring up the Format Data Labels dialog box and change the Label Contains choice to Series Name instead of Value.
Now don’t close the Format Data Labels dialog box and just keep selecting the data labels and changing the Label Contains choice for each. Do this for every data point in the chart and your chart will now look like this:
Almost done. Just a little chart junk cleanup.
4) Resize and Clean up Chart Junk
Excel is a hoarder when it comes to chart junk. So we need to delete it. So click on the legend and press the delete key. Then resize your chart to be a larger chart or decrease the size of the horizontal axis data label font size so that the horizontal axis shows the categories horizontally instead of at an angle. Your chart should now look like this:
All I did was make the chart a lot larger in size so that the horizontal categories were displaying in a horizontal fashion. That is it. A simple change to the data and you can make this type of chart. However, I had a few fans that sent me their charts. The one below was very cool, so check it out.
5) FAN ALTERNATE SOLUTIONS
I don’t usually think about using a pivot table to create a chart as my first choice. However, a fan “Jake” sent me this chart. Jake set up his data so quickly by using a pivot table, that I think I will do this in the future. Will save me lots of key strokes. Below he created a pivot table of the data and then simply charted it. I think he still had to do the same label creation, but the setup of the data was much quicker.
Here was Jake’s chart:
Thanks Jake!
Thanks for being a fan and thanks to all the fans out there. More to come, as you can do almost anything in Excel!!
Video Tutorial
Here is the video demonstration of both techniques:
Steve=True | 1,540 | 6,186 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-33 | latest | en | 0.948814 |
https://byjus.com/maths/254-in-words/ | 1,702,111,929,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100873.6/warc/CC-MAIN-20231209071722-20231209101722-00399.warc.gz | 182,245,399 | 110,071 | # 254 in Words
254 in words can be written as Two Hundred Fifty-Four. Learn the concept of count or counting which is important for the students to learn right from their primary education. If you buy a packet of biscuits for Rs. 254, then you can say that “I bought a packet of biscuits for Two Hundred Fifty-Four Rupees”. Use the numbers in words article to learn the conversion of numbers in words. Hence, the number 254 can be read as “Two Hundred Fifty-Four” in English.
254 in words Two Hundred Fifty-Four Two Hundred Fifty-Four in Numbers 254
## How to Write 254 in Words?
Learn the place value and the expanded form of 254 with the help of the table given below. There are three digits in 254 and its place value can be found here.
Hundreds Tens Ones 2 5 4
254 in expanded form is shown below:
2 × Hundred + 5 × Ten + 4 × One
= 2 x 100 + 5 x 10 + 4 x 1
= 200 + 50 + 4
= 254
= Two Hundred Fifty-Four
Therefore, 254 in words is written as Two Hundred Fifty-Four.
254 is a natural number that precedes 255 and succeeds 253.
254 in words – Two Hundred Fifty-Four
Is 254 an odd number? – No
Is 254 an even number? – Yes
Is 254 a perfect square number? – No
Is 254 a perfect cube number? – No
Is 254 a prime number? – No
Is 254 a composite number? – Yes
## Frequently Asked Questions on 254 in Words
Q1
### Write 254 in words.
254 can be written in words as “Two Hundred Fifty-Four”.
Q2
### Write Two Hundred Fifty-Four in numbers.
Two Hundred Fifty-Four can be written in numbers as 254.
Q3
### Is 254 a perfect cube number?
No, 254 is not a perfect cube number as it is not the product of three similar numbers. | 443 | 1,645 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2023-50 | latest | en | 0.906774 |
https://meters-to-feet.appspot.com/89.4-meters-to-feet.html | 1,675,401,277,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500042.8/warc/CC-MAIN-20230203024018-20230203054018-00256.warc.gz | 401,589,332 | 5,793 | Meters To Feet
# 89.4 m to ft89.4 Meters to Feet
m
=
ft
## How to convert 89.4 meters to feet?
89.4 m * 3.280839895 ft = 293.307086614 ft 1 m
A common question isHow many meter in 89.4 foot?And the answer is 27.24912 m in 89.4 ft. Likewise the question how many foot in 89.4 meter has the answer of 293.307086614 ft in 89.4 m.
## How much are 89.4 meters in feet?
89.4 meters equal 293.307086614 feet (89.4m = 293.307086614ft). Converting 89.4 m to ft is easy. Simply use our calculator above, or apply the formula to change the weight 89.4 m to ft.
## Convert 89.4 m to common lengths
UnitUnit of length
Nanometer89400000000.0 nm
Micrometer89400000.0 µm
Millimeter89400.0 mm
Centimeter8940.0 cm
Inch3519.68503937 in
Foot293.307086614 ft
Yard97.7690288714 yd
Meter89.4 m
Kilometer0.0894 km
Mile0.0555505846 mi
Nautical mile0.0482721382 nmi
## Alternative spelling
89.4 Meter to ft, 89.4 Meter in ft, 89.4 m to Foot, 89.4 m in Foot, 89.4 Meter to Feet, 89.4 Meter in Feet, 89.4 m to Feet, 89.4 m in Feet, 89.4 Meters to ft, 89.4 Meters in ft, 89.4 Meters to Foot, 89.4 Meters in Foot, 89.4 m to ft, 89.4 m in ft | 436 | 1,121 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2023-06 | longest | en | 0.76759 |
https://www.scribd.com/document/393500431/IJATES-Paper-Anil-Lor | 1,571,089,708,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986655310.17/warc/CC-MAIN-20191014200522-20191014224022-00065.warc.gz | 1,079,676,997 | 71,481 | You are on page 1of 8
# A REVIEW OF VARIOUS TURBULENCE MODELS
## USED TO SIMULATE THE FLOW PAST AN ELLIPTIC
CYLINDER
Anil Kumar Lor1, N. K. Singh2
1
Department of Mechanical Engineering, NIT Kurukshetra, India
2
Associate Professor, Department of Mechanical Engineering, NIT Kurukshetra, India
ABSTRACT
Flow around bluff bodies such as elliptical, square and circular cylinders is part of the fundamental fluid
mechanics. There are extensive literatures on the flow around a circular cylinder because of its practical
importance in hydrodynamics and aerodynamics applications and the fundamental significance in the flow
physics. Building on the understanding of the basic concepts of stationary circular cylinder, many researchers
have studied multiple stationary or oscillating cylinders, making it the synonym for the external separated flows.
Many engineering applications such as nuclear rods, heat exchangers, condensers, off shore structures, and
bridges can be modelled as a circular cylinder but on the other hand, some applications involve flow past
complex bodies like aero plane wings, rotor blades, submarines and missiles which can’t be investigated as the
flow over a circular cylinder. In such cases, other parameters such as angle-of-attack and the aspect ratio can
significantly influence the boundary layer separation, forces such as drag and lift and characteristics of the
wake region, the nature of the vortex street and the flow separation phenomenon. Flow with these phenomenon
can be modelled assuming bluff body as elliptic cylinder. On the other hand, elliptic cylinders have the
of various studies focused on modelling the flow past an elliptic cylinder. Most of the scientists and researchers
uses CFD codes to analyse the models which is subjected to various conditions and compared these results with
the experiments.
## Keywords: Cross flow, Elliptical cylinder, Numerical Simulation, Turbulence modelling
I. INTRODUCTION
When some placed are placed in a fluid stream, they generate separated flow over a considerable proportion of
their surface and hence can be classified as bluff. On bluff bodies with sharp edges, separation is fixed at the
trailing edges, whereas on bluff bodies having continuous surface curvature, the location of separation depends
both on the nature of the boundary layer and the shape of the bluff body. For low Reynolds numbers when
boundary layer separation begins, the flow around the body remains stable, but as the Reynolds number is
increased and reached to a critical value, instabilities start occurring. These instabilities can lead to an unsteady
wake motion which is organized, disorganized or a combination of both.
620 | P a g e
Numerous experimental and numerical studies has been done on the laminar and turbulent flow behind bluff
bodies. Turbulent flow contains extremely complex physics which involves high adverse pressure gradient, flow
separation, and unsteady and aperiodic vortex shedding. Due to these problems, numerical analysis is very
difficult. The main objective of the present study is to find out some numerical schemes and concepts to model
the flow around an elliptical cylinder which is in the supercritical and upper transition flow regimes. Due to
absence of sufficient data in these regimes for an elliptical cylinder, investigation data of flow over a circular
cylinder has been included in this study.
## II. DIFFERENT MODELING APPROACHES
There are several questions due to which the researchers have taken interest in the turbulent flow around
cylinders. The distribution of pressure and location of maximum pressure (force) on the surface of the bodies is
required in the design process with regard to failure studies while values of the drag coefficients and lift
coefficients on the cylinder is necessary for the assessment of fluctuating forces and heat generated on the
surface. Research on this fluid dynamics problem is done by means of experiments and by numerical simulation.
For numerical simulation, the three predominant approaches are the
1. Reynolds-averaged Navier-Stokes equations (RANS) models.
2. Large eddy simulation (LES)
3. Direct Numerical Simulation (DNS).
In RANS and LES, the averaged Navier–Stokes equations are solved, where the average is defined as a
statistical or temporal average in RANS and as a spatial average over a small volume in LES. Due to the non-
linearity of the Navier–Stokes equations, after averaging some additional terms are there for which modelling is
needed. At present there are various advanced turbulence models which can be used to simulate such flows and
researchers investigate these models to know their suitable applications. Both the above-discussed methods use
turbulence models to simulate the flow; but in DNS, Navier-Stokes equations are solved without using any
turbulence model. This means that no spatial and temporal averaging is done but their whole range of these
scales are resolved which makes this method very costly.
## III. REVIEW OF FLOW PAST AN ELLIPTIC CYLINDER
H J Lugt et al. [1] numerically investigated the 3D unbounded laminar flow past an elliptic cylinder with
infinite span at angle of attack by using various finite difference schemes. They found that there was no
periodicity in the wake for Reynolds number up to 15, and at Re=200, vortex shedding starts thus developing a
Karman vortex street. For , there were no inertial forces acting on the cylinder. Calculations were
made using the Oberbeck‟s integrals for elliptic cylinders with different focal to length ratios and also showed
streamlines and vortex patterns for these configurations.
H J Lugt [2] numerically studied the autorotation of an elliptic cylinder with its major axis perpendicular to the
flow, for . He found that due to the synchronisation between the vortex shedding and the rate of
621 | P a g e
rotation of the cylinder, autorotation phenomenon take place. Lock-in happens over a range of the vortex
shedding frequency. He gave different conditions for auto rotations to take place.
Jackson [3] by using the finite element simulations investigated the onset of periodicity in the flow which leads
to oscillatory forces on the body, in 2D, steady state, laminar flow past an elliptic cylinder, and found that the
transition corresponds to a „Hopf bifurcation‟ which was located using extended systems of equations. He found
the critical Reynolds number and critical Strouhal number for various configurations of the elliptic cylinder, for
ex , elliptic cylinder oriented perpendicular the flow. Also
found that the use of extended system techniques makes the computational cost much cheaper than time-
dependent calculations for getting these critical quantities.
J K Park et al. [4] numerically investigated the effect of the angle of attack on the unsteady laminar flow past a
thin elliptic cylinder by using the DuFort-Frankel scheme and the Buneman algorithm. They found five different
regimes of flow for low Reynolds numbers ( ). The first two regimes were steady flow regimes.
Regime one, was without any separation bubble whereas the regime two was with a separation bubble. The
critical Reynolds number for these two regimes in case of circular cylinder was found to be 8.5. The remaining
three regimes were unsteady ones where transition to the Karman vortex street took place. Also found that the
vortex shedding frequency in regime three was found to be twice as that of regime five.
Yoshihiro Mochimaru [5] simulated a two dimensional (2D), steady and incompressible flow using the Fourier
spectral method for high Reynolds number regime ( ). For different minor to major axis ratios (1/12,
1/6, 1/3, 1/2), he evaluated coefficient of drag, streamlines and isobars. Notable reduction in the drag coefficient
was observed for higher Reynolds numbers. He concluded that the steady state flow past an elliptic cylinder can
be simulated using Fourier spectral method through double exponential transformation in the space.
Mittal et al. [6] investigated the effects of three dimensional effects on the flow at Reynolds numbers for which
the flow becomes three-dimensional but was simulated as 2D. They compared the results from 2D and 3D
simulations of flow over elliptic and circular cylinders and found that the drag coefficient computed from 2D
simulations was notably higher than that from 3D simulations and experimental results. The over prediction of
the mean drag prediction was due to the higher Reynolds stresses in the wake region and is more pronounced in
bluffer bodies.
Again in 1996, Mittal et al. [7] studied the 3D flow past an elliptic cylinder by developing a spectral collocation
technique to study the concept of unsteady flow separation and the wake region. They also discussed the inflow
and boundary conditions and used discretization techniques. The effect of axis ratio and the angle-of-attack on
flow field was investigated by direct numerical simulation for Reynolds numbers up to 1000. To validate their
results, they compared the mean drag coefficients and the Strouhal numbers with the existing results from the
flow past circular cylinders. They found that the three dimensionality affects the flow phenomenon and there
was over prediction in the mean drag and lift amplitude.
M T Nair [8] numerically studied the 2D unsteady laminar flow past an elliptic cylinder with aspect ratio of 0.1.
For a combination of two Reynolds numbers (200 and 10000) with two non-dimensional rotation rates (1 and 4),
622 | P a g e
they studied the lift and drag coefficients. They concluded that the wake region was smaller for higher Reynolds
number and the shedding was also increased.
Johnson et al [9] carried out a numerical investigation of the flow past an elliptic cylinder with different axis
ratio and low Reynolds number by using the spectral-element method. The axis ratio was varied from 0.01 to 1
and Reynolds number from 30 to 200. They found that there was a peak in the Strouhal number for the transition
in wake region from normal vortex street to a secondary vortex shedding street and a minima in the drag
coefficient for an axis ratio. Also, the critical Reynolds number for which the vortex shedding starts, was
decreasing with decrement in the axis ratio.
Zhihua Li et al. [10] numerically simulated the flow over an elliptic cylinder using
turbulent model with different axis ratio and Reynolds number up to . Coefficient of pressure, rms lift
coefficient, mean drag and the local and surface averaged Nusselt number were predicted for different axis
ratios. As slenderness of the cylinders is increased, a reduction in the pressure drop and the drag forces is
reported when compared to the circular cylinder. But Nusselt number also follows the same trend. They also
compared the results of lenticular and elliptic cylinders which were found to be more or less comparable to each
other.
M S Kim et al. [11] studied the unsteady flow past an elliptic cylinder using SIMPLER algorithm of Patanker
(1980) along with Crank-Nicolson temporal integration method for various thickness-to-cord ratios (t/c=0.6, 0.8,
1.0 and 1.2) and different Reynolds numbers (Re=200, 400 and 1000). They found that the pressure drag was
dominating in total drag and increases with increment in the t/c. Also the mean friction drag depends upon the
Reynolds number. The amplitude of drag and lift forces were found to be increasing with the Reynolds number
and t/c but rate of increment is more with t/c.
Hu Ruifeng [12] performed a computational study on the 3D flow past a rotating elliptic cylinder with different
aspect ratios and rotational speeds. For Re=200, they found that the periodic vortex shedding get suppressed or
clustered with increasing the rotational speeds. He also presented iso-surfaces of wake structures for different
AR and rotating speeds. At low axis ratio, the streamwise vortices are found to be comparable to the spanwise
vortices, while increasing the AR, the strength of spanwise vortices were reported to be more.
S N Naik et al. [13] presented numerical solutions for the uniform flow over an elliptic cylinder for different
rotational speed using the immersed boundary method (IBM). They studied the periodic flow for a Re=100 and
a fixed AR=0.1. Vortex shedding was reported for all the rotation rates and „Hovering Vortex‟ were also
reported for and also studied the factors affecting this phenomenon.
With increasing the rotation rate, the lift amplitude was also found to be increasing.
## IV. VARIOUS TURBULENT MODELS
Spalart-Allmaras turbulence model is one of the most popular one-equation model which is based on eddy-
viscosity turbulence models. An additional transport equation needs to be solved for a term which is the
modified form of the turbulent kinetic viscosity for using this model. The model is extensively used in the
aerodynamics as it gives good results for these applications. In its original form, the production term is
623 | P a g e
dependent upon rotation tensor while in the modified version, the measures of both rotation and strain tensors
for the production term are combined. It gives good results for low-Reynolds-numbers. It includes damping
functions for turbulent viscosity, turbulent production and dissipation. To fulfil these requirements, the meshing
has to be very fine in the near-wall region.
K-epsilon model is most common turbulence model which is used in CFD, focuses on the mechanism that
affects the turbulent kinetic energy. It is a two equations turbulence model, one equation is for turbulent kinetic
energy and other one is for dissipation rate. The two equations are solved to calculate the turbulent viscosity,
hence Reynolds stress by using Boussinesq relationship. The standard model along with the Boussinesq
equation, works well for a broad range of engineering problems. But, when the problem include intensive
isotopes of flow or unbalanced effects, this model finally reaches to responses which are over-diffused, i.e., the
values predicted by this model will be large.
The Realizable turbulence model is comparatively a recent development and an improved version of the
standard model. It contains a new formulation and a new transport equation for the turbulent viscosity and
the dissipation rate respectively. The constant in RKE model, which is present at the turbulence viscosity
term as a closure coefficient, arranged in such a manner that it includes the effect of the strain and rotation
tensors. Furthermore, for the turbulence dissipation with functional expression, a new transport equation was
developed. This model is a high-Reynolds-number model which requires very fine mesh near the wall and is
suitable for fully turbulent flows only.
The RNG turbulence model was developed using a mathematical technique called “renormalization
group" (RNG) in which renormalization of the instantaneous Navier-Stokes equations is done to take account
for the effects of smaller eddies. The analytical derivation results in a model with constants different from those
in the standard k-ε model. Model also have some additional terms and functions in the transport equations. For
flows with separation and recirculating regions, this model along with the modified coefficients provides results
which are accurate and less diffusive than standard model.
Wilcox turbulence model (WKO) is an improved version of the standard model. Like the standard
version, two additional transport equations for the turbulence kinetic energy, k and the dissipation per unit
kinetic energy, are required to be solved. The constants which are present in both k and ω transport equations,
are expressed as functions in the WKO model that improves the performance of the standard model for
the free shear flows without affecting the boundary layer flow.
The SST turbulence model comes under the two-equation eddy-viscosity model which is more improved version
of the standard . Due to the known weakness of the standard model, which is its sensitivity to free
624 | P a g e
stream boundary conditions for free shear flows, a cross-diffusion term in the equation along with a blending
function is included. It uses the formulation in the near-wall region and for the outer free stream
conditions. The SST model is considered good in prediction capability for the flows with adverse
## pressure gradients and flow separations.
Large eddy simulation (LES) is a space filtering method in CFD. Large eddy simulation (LES) falls between
DNS and RANS in terms of the fraction of the resolved scales. LES directly computes the large-scale turbulent
structures which are responsible for the transfer of energy and momentum in a flow while modeling the smaller
scale of dissipative and more isotropic structures. A filter function is used in the LES to differentiate between
the small and large scale eddies. A filter function dictates which eddies are large by introducing a length scale,
usually denoted as in LES, the characteristic filter cut-off width of the simulation. All eddies which are larger
than the defined length scale are resolved directly, while the smaller one are approximated.
Detached eddy simulation is a 3D unsteady numerical solution, which acts as a sub-grid-scale model in regions
where the grid density is fine enough, and as a RANS model in regions of coarse grid density. The model senses
the grid density and adjusts itself to a lower level of mixing, in order to unlock the larger-scale instabilities of
the flow and to let the energy cascade extend to length scales close to the grid spacing. Therefore, the meshing
need not be that fine as in case of Large-Eddy Simulation (LES) and thus reduces the computation cost. The new
length scale in DES depends upon grid spacing. It is selected as the minimum of RANS length scale and ,
## where is the maximum dimension of the grid cell.
Dynamic LES model is one in which the Smagorinsky model constant , is dynamically computed based on the
information provided by the resolved scales of motion making the model self- tuning. The dynamic procedure
thus obviates the need for users to specify the model constant in advance which is a function of space and
time. The dynamic model provides a systematic way of adjusting or allowing it to be a function of
position which is desirable for inhomogeneous flows. The dynamic method allows adapting the filter size in the
spatial direction. It shows improvement in complex flows as for example drag crisis flow around a cylinder.
Partially averaged Navier-Stokes (PANS) model in which unresolved kinetic energy parameter varies with grid
spacing and the turbulence length scale. The two-equation turbulence model is used for the unresolved kinetic
energy and the dissipation where a constant is replaced by unresolved kinetic energy parameter ( ). The
parameter varies between zero and one and has the magnitude equal to one in the viscous sub layer, and when
the RANS turbulent viscosity becomes smaller than the LES viscosity.
625 | P a g e
In the Mesh free vortex method, the vorticity field is modeled using a cloud of point‟s vortices. The total
velocity field can be constructed by adding up the contribution of the vortices in the cloud and their image, the
uniform flow and the dipole term that models the circular cylinder. Complex velocity field induced at a point
can be obtained by using the circle theorem. The vortex method uses discrete points to model the vortices,
whose transport at each time step is carried out in a sequence.
VMS-LES hybrid model consist in splitting between the large resolved scales (LRS) i.e. those resolved on a
virtual coarser grid, and the small resolved ones (SRS). The VMS-LES method does not compute the SGS
component of the solution, but modelizes its effect on the small resolved scales which corresponds to the highest
level of discretization, and preserves the Navier-Stokes model for the large resolved scales. The SGS model is
used to modelizes the dissipation effect of the unresolved scales on the resolved scales. To distinguish the
resolved scales in to the largest and smallest ones is the main idea of this method.
In the Hybrid RANS/VMS-LES model, VMS-LES and RANS approaches are combined by the use of NLDE
(Non-Linear Disturbance Equations) technique in which the solution of Navier- Stokes equations is
decomposed into three parts. First part is mean (RANS), the second one is that takes into account the turbulent
large-scale fluctuations and a third part consist of the unresolved (SGS) fluctuations. The basic idea is to solve
the RANS equations and to correct the obtained averaged flow field by adding the resolved fluctuations in a
hybrid mode. A blending function is introduced to the suitability of the grid used for the LES. The blending
function varies between 0 and 1. When Blending function is less than one, additional resolve fluctuations are
computed while blending function is 1 when the RANS approach is recovered. In this model, RANS approach is
used in the viscous sub layer and in the separating shear layers, while for the wake region of the bluff body, LES
model is used making it a hybrid model.
In the LES with RGF algorithm, random flow generation algorithm is incorporated into a finite element code
with which the LES code is applied to generate a realistic inflow field. The algorithm takes correlation tensor of
the original flow field, length and time scales of turbulence as the input. These quantities can be obtained from a
steady state RANS simulations or experimental data. The outcome of the procedure is a time dependent flow
field and it is divergence free inhomogeneous anisotropic flow field.
Direct numerical simulation (DNS) is the most accurate approach to simulate the turbulenc. This approach
involves the numerical solution of the Navier-Stokes equations that govern fluid flow without modeling with its
accuracy is only bounded by the accuracy of numerical scheme adopted, boundary conditions and discretizations.
The main problem is that, the large number of grid points and the small size of time steps required to capture the
so small time and space scale of turbulent motion make the advanced turbulent computations cumbersome.
V. CONCLUSIONS
Significant advances were made in the last decade in understanding the fundamentals of fluid flow over a
circular cylinder with improved computational mesh generation techniques and efficiency of solvers but in the
case of an elliptical cylinder, absence of comprehensive data is observed in the supercritical and upper transition
regimes. Many engineering application, which operates at high speed or in a highly turbulent regime, are very
difficult to model. So some numerical simulation in these regimes should be conducted to avoid such
circumstances.
626 | P a g e
The circular cylinder that is a special and ideal case of an elliptical cylinder (minor to major axis ratio = 1) can
be used for the validation of different approaches to turbulence modeling. A wide number of research papers are
available in the literature. LES and different hybrid models are able to predict more accurately in supercritical
and transcritical regimes but still, most researchers use the two-equation models, mainly the SST to
simulate the flow past a circular cylinder as this model is relatively simple and reasonably accurate. If higher
accuracy is the main requirement, one should adopt models other than the two-equation models.
Overall the present study helps someone in finding the suitable method to model an elliptical cylinder in high
Reynolds number regimes.
REFERENCES
[1.] Lugt, H.J. and Haussling, H.J., 1974. Laminar flow past an abruptly accelerated elliptic cylinder at 45
incidence. Journal of Fluid Mechanics, 65(04), pp.711-734.
[2.] Lugt, H.J., 1980. Autorotation of an elliptic cylinder about an axis perpendicular to the flow. Journal of
Fluid Mechanics, 99(04), pp.817-840.
[3.] Jackson, C.P., 1987. A finite-element study of the onset of vortex shedding in flow past variously shaped
bodies. Journal of Fluid Mechanics, 182, pp.23-45.
[4.] Park, J.K., Park, S.O. and Hyun, J.M., 1989. Flow regimes of unsteady laminar flow past a slender elliptic
cylinder at incidence. International Journal of Heat and Fluid Flow, 10(4), pp.311-317.
[5.] Mochimaru, Y., “Numerical simulation of flow past an elliptical cylinder at moderate and high Reynolds
number using spectral method,” 11th Australian fluid mechanics conference, 1992.
[6.] Mittal, R. and Balachandar, S., 1995. Effect of three‐ dimensionality on the lift and drag of nominally
two‐ dimensional cylinders. Physics of Fluids, 7(8), pp.1841-1865.
[7.] Mittal, R. and Balachandar, S., 1996. Direct numerical simulation of flow past elliptic cylinders. Journal of
Computational Physics, 124(2), pp.351-367.
[8.] Sengupta, M.T.N.T.K., FLOW PAST ROTATING ELLIPTIC CYLINDERS.
[9.] Johnson, S.A., Thompson, M.C. and Hourigan, K., 2001, December. Flow past elliptical cylinders at low
Reynolds numbers. In Proc. 14th Australasian Fluid Mechanics Conference, Adelaide University, South
Australia, Dec (pp. 9-14).
[10.] Li, Z., Davidson, J.H. and Mantell, S.C., 2006. Numerical simulation of flow field and heat transfer of
streamlined cylinders in cross flow. Journal of heat transfer, 128(6), pp.564-570.
[11.] Kim, M.S. and Sengupta, A., 2005. Unsteady viscous flow over elliptic cylinders at various thickness with
different reynolds numbers. Journal of mechanical science and technology, 19(3), pp.877-886.
[12.] Ruifeng, H., 2015. Three-dimensional flow past rotating wing at low Reynolds number: a computational
study. Fluid Dynamics Research, 47(4), p.045503.
[13.] Naik, S.N., Vengadesan, S. and Arul, K., 2016. Flow past rotating low axis ratio elliptic cylinder. In 46th
AIAA Fluid Dynamics Conference (p. 4348).
627 | P a g e | 5,626 | 25,880 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-43 | latest | en | 0.90015 |
https://www.randelshofer.ch/rubik/virtual_cubes/rubik/super_cubes/triangular_super_cube.html | 1,623,510,294,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487584018.1/warc/CC-MAIN-20210612132637-20210612162637-00054.warc.gz | 871,395,815 | 2,871 | # Virtual Cubes
Activate JavaScript to see the Virtual Cubes!
### Triangular Super Cube
A Super Cube is a design variation of the Rubik's Cube which visualizes the orientation of side parts. Because of this, a Super Cube has 2048 times more positions than a regular cube (46/2 = 2048).
Compared to a Rubik's Cube, the stickers are divided into two triangles where one of them is brighter then the other.
The layout of the Triangular Super Cube was created in 2010 by Ortwin Schenker. The design was inspired by Georges Helm's Triangle Cube.
#### Algorithms to twist the centers
Algorithm
(U R U R')5 (20 ltm, 20 ftm, 20 qtm)
Permutation (++u)
Algorithm M. B. Thistlethwaite
(U · R L · U2 · R' L')2 (12* ltm, 12* ftm, 14 qtm)
Permutation (++u)
Algorithm
(U' R2 U2 R' U2 R2)3 (18 ltm, 18 ftm, 30 qtm)
Permutation (+u) (+r)
Algorithm Herbert Kociemba 2008
D' R · U D · R2 U2 MF MD B MD' MF' U2 R2 U' (14* ltm, 18 ftm, 22 qtm)
Permutation (+u) (+r)
Algorithm
(MR' MD' MR) U' (MR' MD MR) U (8* ltm, 14* ftm, 14* qtm)
Permutation (+u) (-r)
Algorithm
(MR' MD' MR) U2 (MR' MD MR) U2 (8* ltm, 14 ftm, 16* qtm)
Permutation (++u) (++r)
Algorithm
(MR' MD2 MR) U' (MR' MD2 MR) U (8* ltm, 14* ftm, 18* qtm)
Permutation (+u) (-d)
Algorithm
(MR' MD2 MR U2)2 (8* ltm, 14 ftm, 20 qtm)
Permutation (++u) (++d)
Algorithm Herbert Kociemba 2008
CD' · MR MD MR' D MR MF MR' F' MD MF' MD' F U' (13* ltm, 22 ftm, 22* qtm)
Permutation (+u) (+r) (+f) (+d) (-l) (+b)
Notation | 546 | 1,463 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-25 | longest | en | 0.645331 |
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Question
# Hence coordinates of ex-centre is Do yourself - 2 : Ans. (D) 1. The coordinates of the vertices of a triangle are and :
1. Find centroid of the triangle.
2. Find circumcentre \& the circumradius.
3. Find orthocentre of the triangle.
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Question Text Hence coordinates of ex-centre is Do yourself - 2 : Ans. (D) 1. The coordinates of the vertices of a triangle are and : Updated On Sep 3, 2023 Topic Coordinate Geometry Subject Mathematics Class Class 11 Answer Type Video solution: 1 Upvotes 78 Avg. Video Duration 6 min | 302 | 1,265 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-26 | latest | en | 0.783445 |
https://www.enotes.com/homework-help/there-an-infinite-number-solution-system-what-418686 | 1,669,623,086,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710488.2/warc/CC-MAIN-20221128070816-20221128100816-00288.warc.gz | 810,086,613 | 17,989 | # If there are an infinite number of solution to the system, what are the values of h and k in the following system? {(5x-9y=h),(8x+ky=-3):}
You need to consider the next two conditions for the system to have an infinite number of solutions, such that:
`Delta = [(5,-9),(8,k)] = 0` (`Delta` represents determinant of matrix of coefficients of variables x and y)
`Delta =5k - (-9*8) => {(Delta = 5k + 72),(Delta = 0):} => 5k + 72 = 0 => 5k = -72 => k = -72/5`
`Delta_1 = [(5,h),(8,-3)] ` = 0 (`Delta_1` represents characteristic determinant)
`Delta_1 = -15 - 8h => Delta_1 = {(Delta_1 = -15 - 8h),(Delta_1 = 0):} => -15 - 8h = 0 => 8h = -15 => h = -15/8`
Hence, evaluating k and h for the system to have an infinite number of solutions yields `k = -72/5` and ` h = -15/8.`
Approved by eNotes Editorial Team | 278 | 813 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2022-49 | latest | en | 0.846392 |
https://www.get-digital-help.com/2010/05/10/extract-unique-distinct-numbers-from-closed-workbook-in-excel-formula/ | 1,537,354,763,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156192.24/warc/CC-MAIN-20180919102700-20180919122700-00102.warc.gz | 763,636,149 | 29,483 | Author: Oscar Cronquist Article last updated on December 02, 2010
### Introduction
In this post I will show you how to extract unique distinct numbers from a closed workbook. There can´t be any blank cells in the cell range. You can only extract numbers with this method.
Unique distinct numbers are all numbers but duplicates are merged into a single number.
### Extract unique distinct numbers from closed workbook
Here is a picture of the numbers in the closed workbook.
Here is how to extract unique distinct numbers from closed workbook.
Array formula in cell A1:
=SMALL(IF(FREQUENCY('C:\temp\[closed workbook.xls]Sheet1'!\$A\$1:\$A\$6, 'C:\temp\[closed workbook.xls]Sheet1'!\$A\$1:\$A\$6)>0, 'C:\temp\[closed workbook.xls]Sheet1'!\$A\$1:\$A\$6, ""), ROW(A1)) + CTRL + SHIFT + ENTER.
Copy cell and paste it down as far as needed. | 216 | 843 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2018-39 | latest | en | 0.892069 |
https://math.answers.com/questions/Can_a_chord_also_be_a_tangent | 1,713,662,061,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817699.6/warc/CC-MAIN-20240421005612-20240421035612-00752.warc.gz | 342,440,683 | 48,593 | 0
# Can a chord also be a tangent?
Updated: 12/20/2022
Wiki User
11y ago
In the limit, a chord approaches a tangent, but is never actually a tangent. (In much the same way as 1/x approaches 0 as x increases, but is never actually 0.)
In the limit, a chord approaches a tangent, but is never actually a tangent. (In much the same way as 1/x approaches 0 as x increases, but is never actually 0.)
In the limit, a chord approaches a tangent, but is never actually a tangent. (In much the same way as 1/x approaches 0 as x increases, but is never actually 0.)
In the limit, a chord approaches a tangent, but is never actually a tangent. (In much the same way as 1/x approaches 0 as x increases, but is never actually 0.)
Wiki User
11y ago
Wiki User
11y ago
In the limit, a chord approaches a tangent, but is never actually a tangent. (In much the same way as 1/x approaches 0 as x increases, but is never actually 0.) | 248 | 926 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-18 | latest | en | 0.968446 |
https://math.stackexchange.com/questions/4250583/the-boundary-of-a-region-to-define-double-integrals | 1,632,727,392,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058373.45/warc/CC-MAIN-20210927060117-20210927090117-00584.warc.gz | 415,428,603 | 36,394 | The boundary of a region to define double integrals
Given a function $$f:\mathbb{R}^2\to\mathbb{R}$$, the calculus textbook I'm referencing defines the double integral over a region $$D$$ by forming a rectangle $$R$$ that contains $$D$$, and constructing a function F(x,y)=\left\{\begin{align*}f(x,y)\ \ \ \text{ for } (x,y)& \in D\\ 0 \ \ \ \ \ \ \ \ \text{ for } (x,y)& \in R\setminus D\end{align*}\right. and setting $$\int_D f = \int_R F.$$ The book says this holds as long as $$D$$ has a "nice enough" boundary with "nice enough" being outside the scope of the book.
I'm wondering what it is, and if it can be stated concisely? Is it some topological condition, like locally euclidean, or oriented? Is it easier to state in $$\mathbb{R}^2$$ then in general?
• This is just a guess, but I'd think that the boundary being measure 0 would be nice enough, which non-pathological boundaries will b.
– Alan
Sep 14 at 22:51 | 273 | 924 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-39 | latest | en | 0.892141 |
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