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# PAT（甲级）2020年春季考试 7-1 Prime Day
## 7-1 Prime Day (20分)
The above picture is from Sina Weibo, showing May 23rd, 2019 as a very cool "Prime Day". That is, not only that the corresponding number of the date `20190523` is a prime, but all its sub-strings ended at the last digit `3` are prime numbers.
Now your job is to tell if a given date is a Prime Day.
### Input Specification:
Each input file contains one test case. For each case, a date between January 1st, 0001 and December 31st, 9999 is given, in the format `yyyymmdd`.
### Output Specification:
For each given date, output in the decreasing order of the length of the substrings, each occupies a line. In each line, print the string first, followed by a space, then `Yes` if it is a prime number, or `No` if not. If this date is a Prime Day, print in the last line `All Prime!`.
### Sample Input 1:
``20190523``
### Sample Output 1:
``````20190523 Yes
0190523 Yes
190523 Yes
90523 Yes
0523 Yes
523 Yes
23 Yes
3 Yes
All Prime!``````
### Sample Input 2:
``20191231``
### Sample Output 2:
``````20191231 Yes
0191231 Yes
191231 Yes
91231 No
1231 Yes
231 No
31 Yes
1 No``````
#### 算法思路：
• 1、将s转化为数字N，使用`isPrime`函数判断该数字是否为素数，如果是输出该数字和`Yes`，否则输出该数字和`No`，并使用`allPrime`记录不是所有子串都是素数。
• 2、令s为下一个子串`s = s.substr(1)`，转1
最后判断`allPrime`是否为`true`，如果是输出`All Prime!`
#### AC代码：
``````#include<cstdio>
#include<vector>
#include<cstring>
#include<cmath>
#include<string>
#include<iostream>
using namespace std;
bool isPrime(int N){
if(N<=1) return false;
int sqrtn = (int)sqrt(1.0*N);
for (int i = 2; i <= sqrtn; ++i) {
if(N%i==0) return false;
}
return true;
}
int main(){
string s;
cin>>s;
bool allPrime = true;
while (s.size()>0){
int N = stoi(s);
if(isPrime(N)){
printf("%s Yes\n",s.c_str());
} else {
allPrime = false;
printf("%s No\n",s.c_str());
}
s = s.substr(1);
}
if(allPrime) printf("All Prime!");
return 0;
}
``````
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13 粉丝
0 条评论
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# math30
A sequence has a first term of 24 and every other term is one half of the previous term.
Find a recursive formula that defines this sequence.
The answer of this question is:
an=an-1*1/2
Somebody can help me and let me know why is 1/2 on the formula? please please!!!
I am not sure if the formula is the right one!!
1. 👍 0
2. 👎 0
3. 👁 137
1. r = .5
Just as in your other problem, r in a geometric sequence is the ratio of every term to the one before it. That is the very meaning of a geometric sequence.
1. 👍 0
2. 👎 0
## Similar Questions
1. ### math Question
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2. ### arithmetic
Determining the first three terms of an ArithmetiC progressionof which the (a.)10th term is 31 and the 15th term is 49 (b.)7th term is 3 and 12th term is -3 (c.)5th term is 8 and the 11th term is -28 (d.)9th term is7+9x and
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4. ### FURTHER MATHS
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1. ### MaTh
The 23rd term in a certain geometric sequence is 16 and the 28th term in the sequence is 24. What is the 43rd term? What is the answer?
asked by Speedo on September 2, 2013
2. ### Math
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5. ### Sequence
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Re: SymbolShape
• To: mathgroup at smc.vnet.net
• Subject: [mg52916] Re: [mg52853] SymbolShape
• From: "David Annetts" <davidannetts at ihug.com.au>
• Date: Thu, 16 Dec 2004 03:42:00 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com
```Hi Naima
> I need to know how I can change the shape of the points (to box or
> triangle) when I use LogListPlot ?
> for example :
>
>
>
> In[1]:=
> << Graphics`Graphics`
> In[2]:=
> points1 = {{9 + 273, 0.012}, {11 + 273, 0.016}, {12 + 273,
> 0.15}, {13 + 273, 0.32}, {15 + 273, 0.3}, {18 + 273, 0.195},
> {20 + 273, 0.098}, {22 + 273, 0.055}, {25 + 273, 0.02}, {30 +
> 273, 0.017}}; In[4]:=
> a11 = LogListPlot[points1, PlotStyle -> {PointSize[0.02],
> Hue[.6]}, GridLines -> None, PlotRange -> All];
Graphics`MultipleListPlot can do the job, though there are other solutions
....
MultipleListPlot[{{#[[1]], Log[10, #[[2]]]} & /@ points1},
PlotStyle -> {Hue[.6]},
SymbolStyle -> {Hue[.6]},
SymbolShape -> {PlotSymbol[Box, Filled -> False]},
FrameTicks -> {Automatic, LogScale, None, None}
];
MultipleListPlot[{{#[[1]], Log[10, #[[2]]]} & /@ points1},
PlotStyle -> {Hue[.6]},
SymbolStyle -> {Hue[.6]},
SymbolShape -> {PlotSymbol[Triangle, 8, Filled -> True]},
FrameTicks -> {Automatic, LogScale, None, None}
];
Regards,
Dave.
--
No virus found in this outgoing message.
Checked by AVG Anti-Virus.
Version: 7.0.296 / Virus Database: 265.5.3 - Release Date: 14/12/2004
```
• Prev by Date: Re: Reverse Hash Function
• Next by Date: Re: shuffling (randomizing) a series
• Previous by thread: Re: SymbolShape
• Next by thread: Re: SymbolShape
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# How to find $\log{x}$ close to exact value in two digits with these methods?
I'm trying to find the result of $\log{x}$ (base 10) close to exact value in two digits with these methods:
The methods below are doing by hand. I appreciate you all who already give answers for computer method.
As suggested by Praktik Deoghare's answer
If number N (base 10) is n-digit then $$n-1 \leq \log_{10}(N) < n$$ Then logarithm can be approximated using $$\log_{10}(N) \approx n-1 + \frac{N}{10^{n} - 10^{n-1}}$$ Logarithm maps numbers from 10 to 100 in the range 1 to 2 so log of numbers near 50 is about 1.5. But this is only a linear approximation, good for mental calculation and toy projects but not that good for serious research.
This method is cool for me, but it's nearly close to exact value. $\log_{10}(53)$ is 1.7 and with that method results 1.58.
As suggested by Pedro Tamaroff's answer
One can get very good approximations by using $$\frac 1 2 \log \left|\frac{1+x}{1-x}\right| =x+\frac {x^3} 3+ \frac {x^5}5+\cdots$$ Say you want to get $\log{3}$. Then take $x=1/2$. Then you get $$\log 3 \approx 2\left( \frac 1 2 +\frac 1 {24} + \frac 1 {140} \right)=1.0976190\dots$$ The real value is $\log 3 \approx 1.098065476\dots$
This one is also cool for me, but it's to find natural logarithm, not base-10 logarithm.
This can be done by recourse to Taylor series. For $ln(x)$ centered at 1, i.e. where $0 < x \leq 2$: $$\ln(x)= \sum_{n=1}^\infty \frac{(x-1)^n}{n}= (x-1) - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3 + \frac{1}{4}(x-1)^4 + \cdots$$
The method is for calculating $ln(x)$, not $\log{x}$. I don't know the Taylor series for calculate $\log{x}$, especially when to find log result close to exact value in two digits (with similar method).
The Wikipedia article "Generalized continued fraction" has a Khovanskiĭ-based algorithm that differs only in substituting x/y for z, and showing an intermediate step: $$\log \left( 1+\frac{x}{y} \right) = \cfrac{x} {y+\cfrac{1x} {2+\cfrac{1x} {3y+\cfrac{2x} {2+\cfrac{2x} {5y+\cfrac{3x} {2+\ddots}}}}}}$$
This method is very slow for me. When I stop at $3y$, the result (log calculation) is still far from the exact value.
Anyone who can improve all of the above methods so I can precisely get log result close to exact value in two digits?
• What is the range you want to approximate? – grdgfgr Jun 7 '15 at 7:24
• The formula for $\frac12\log \big|\frac{1+x}{1-x} \big|$ is also for the natural logarithm, not the base-10 logarithm. – Greg Martin Jun 7 '15 at 7:26
• @grdgfgr For $x$ of $\log{x}$ is 1 until 100, especially prime numbers. – làntèrn Jun 7 '15 at 7:26
• Could you explain exactly what your requirements are? If you're doing this by computer, why can't you just use the built-in logarithm function? And if you're doing it by hand, I would only ask why you are doing it by hand!? – Christopher A. Wong Jun 7 '15 at 7:58
• Why. Got nothing better to do? – grdgfgr Jun 7 '15 at 8:11
Listing a few tricks that work for mental arithmetic. Mostly to get the OP to comment, whether this is at all what they expect. I write $\log$ for $\log_{10}$ to save a few keystrokes.
You need to memorize a few logarithms and play with those. We all have seen $\log 2\approx 0.30103$ enough many times to have memorized it. Consequently by mental arithmetic we get for example the following \begin{aligned} \log4&=2\log2&\approx 0.602,\\ \log5&=\log(10/2)&\approx 0.699,\\ \log1.6&=\log(2^4/10)&\approx 0.204,\\ \log1.024&=\log(2^{10})-3&\approx 0.0103.\\ \end{aligned} Using these is based on spotting numerical near matches.
You should also be aware of the first order Taylor series approximation $$\log(1+x)\approx\frac{x}{\ln 10}\approx\frac{x}{2.3}\approx0.434 x,$$ which implies (plug in $x=0.01$) that if you change the value of $x$ by 1 per cent, then its logarithm changes by approximately $0.0043$.
As an example let's do $\log 53$ and $\log7$. Here $x=53$ is $6\%$ larger than $50$, so a first order approximation would be $$\log53\approx\log 50+6\cdot 0.0043=\log(10^2/2)+6\cdot0.00434\approx 2-0.30103+0.0258\approx1.725.$$ With $7$ we can spot that $7^2=49$ is $2\%$ less than $50$, so $$\log 7=\frac12\,\log49\approx\frac12(2-0.301-2\cdot0.0043)\approx\frac{1.690}2=0.845.$$ Here the third decimal of $\log53$ is off by one, but $\log7$ has three correct digits - both well within your desired accuracy.
• Thank you so much! You can understand my question exactly :-). How do you make the %? – làntèrn Jun 7 '15 at 9:59
• Sorry, I mean how do you know that is 6% larger and 2% less? – làntèrn Jun 7 '15 at 10:02
• $1$ is $2\%$ of $50$. This method really is better suited for mental arithmetic than any serious work. Something you can impress students with (or chicks you meet at a bar, if you think showing off with logarithms helps your cause). – Jyrki Lahtonen Jun 7 '15 at 10:05
• Wow this method is very awesome for me! Is the change by 0.0043 always work for any number of log? Or is there a small method to get the 0.0043? – làntèrn Jun 7 '15 at 10:10
• It comes from that Taylor series expansion: $0.0043\approx 0.434\cdot \dfrac1{100}$. That $1/100$ is, of course, $1$ percent. Caveat: if the percentage difference is larger the formula will quickly begin to lose accuracy. You always need a nearby reference point. – Jyrki Lahtonen Jun 7 '15 at 10:17
Is this good enough? It is within your specifications in 1 to 100
$$\frac{\frac{(x-10)^5 (137+30 \log (10))}{756000000}+\frac{(x-10)^4 (77+30 \log (10))}{2520000}+\frac{(x-10)^3 (47+30 \log (10))}{36000}+\frac{1}{450} (x-10)^2 (9+10 \log (10))+\frac{1}{20} (x-10) (2+5 \log (10))+\log (10)}{\frac{(x-10)^5}{25200000}+\frac{(x-10)^4}{84000}+\frac{(x-10)^3}{1200}+\frac{1}{45} (x-10)^2+\frac{x-10}{4}+1}$$
in plain text:
(Log[10]+1/20 (-10+x) (2+5 Log[10])+1/450 (-10+x)^2 (9+10 Log[10])+((-10+x)^3 (47+30 Log[10]))/36000+((-10+x)^4 (77+30 Log[10]))/2520000+((-10+x)^5 (137+30 Log[10]))/756000000)/(1+1/4 (-10+x)+1/45 (-10+x)^2+(-10+x)^3/1200+(-10+x)^4/84000+(-10+x)^5/25200000)
You can generate these yourself :
Click the A button
If you want base 10, you would want to use Log10[x]
You are going to use this syntax to tune it yourself:
Of course you have to you use a computer to actually evaluate this rational function. But if you already have a computer, why not just evaluate logx ?
There is a well-known trick based on the linearization of the logarithm that can be employed in a shockingly simple manner. However, the linearization is only "nice" for computing $\log_2(x)$, which on a computer is the only relevant logarithm anyways since all numbers are represented in base 2.
The algorithm is as follows: Let $x = 2^b (1 + m)$, where $b$ is a non-negative integer and $m \in [0,1)$. Let the floating point bit representation of $x$ be given by the concatenation of the binary strings $b$ and $m$. Then you can trivially approximate $\log_2(x)$ by the binary number $b.m$, which is merely just moving the position of the "dot" in the floating point bit representation. This is really just the statement that $\log_2(1 + m) \approx m$.
If you are willing to store a single precomputed value $\log_{2}(10)$, then from here you can trivially convert to a base 10 logarithm.
Example: Chose $x = 53_{10}$ (as represented as a decimal). Then in binary $x = 110101_{2}$, and hence $b = 101_{2}$ and $m = 10101_{2}$. Then in binary $\log_2(x)$ can be approximated by $b.m = 101.10101_{2} = 5.65625_{10}$, and then $\log_{10}(x) \approx 5.65625 / \log_2(10) = 1.7027$.
Summary: Let $x = 2^b (1 + m)$. Then $\log_{10}(x) \approx 0.30103_{10} \times (b.m)_{2}$.
For the slightly better version, see this Wikipedia description.
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## What is Rc Circuit?
Potential difference across a capacitor in ac lags in phase by 90° with the current in the circuit.
Suppose in phase or diagram current is taken along positive x-direction. Then VR is also along positive x-direction but VC is along negative y-direction. So we can write.
V = VR – jVC = iR – j(iXC) = iR – j(i/ω) =iZ
Here, impedance is, Z = R - j(1/ωC)
The modulus of impedance is,
and the potential difference lags the current by an angle,
## Email Based Homework Assignment Help in RC Circuit
Transtutors is the best place to get answers to all your doubts regarding RC circuits. You can submit your school, college or university level homework or assignment to us and we will make sure that you get the answers you need which are timely and also cost effective. Our tutors are available round the clock to help you out in any way with alternating current.
## Live Online Tutor Help for RC circuit
Transtutors has a vast panel of experienced physics tutors who specialize in RC circuits and can explain the different concepts to you effectively. You can also interact directly with our physics tutors for a one to one session and get answers to all your problems in your school, college or university level alternating current. Our tutors will make sure that you achieve the highest grades for your physics assignments.
## Related Topics
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More Q&A
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# ICAP GROUP S.A. FINANCIAL RATIOS EXPLANATION
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1 ICAP GROUP S.A. FINANCIAL RATIOS EXPLANATION OCTOBER 2006
2 Table of Contents 1. INTRODUCTION FINANCIAL RATIOS FOR COMPANIES (INDUSTRY - COMMERCE - SERVICES) Profitability Ratios Viability and Capital Structure Ratios Liquidity Ratios Activity Ratios Investor Ratios FINANCIAL RATIOS FOR BANKS Profitability and Efficiency Ratios Productivity Ratios Balance Sheet Structure Ratios Income Structure Ratios Investor Ratios FINANCIAL RATIOS FOR INSURANCES - INSURANCE BROKERS & AGENTS Profitability Ratios Activity and Liquidity Ratios Capital Structure Ratios Investor Ratios ICAP GROUP S.A. Financial Ratios Explanation 2
3 1. INTRODUCTION Financial ratios are useful indicators to measure a company s performance and financial situation. They can be also used to analyze trends and to compare a firm s financial figures to those of competitors or those of the business sector in which it belongs to. Financial ratios can be classified according to the information they provide. ICAP has included in its products the majority of the most commonly used ratios in order to support companies directors and executives during decision-making. The present document describes the calculation of ratios applied either in companies or in sectoral business reports. ICAP GROUP S.A. Financial Ratios Explanation 3
4 2. FINANCIAL RATIOS FOR COMPANIES (INDUSTRY - COMMERCE - SERVICES) 2.1 Profitability Ratios Profitability ratios are designed to evaluate the firm's ability to generate earnings. Analysis of profit is of vital concern to stockholders since they derive revenue in the form of dividends. Profits are also important to creditors because profits are one source of funds for debt coverage. Furthermore Management uses profit as a performance measure. Return on Equity (Before Income Tax) Return on Equity (Before Interest and Income Tax) Return on Capital Employed (Before Income Tax) Return on Capital Employed (Before Interest and Income Tax) Gross Profit Margin Operating Profitability Net Profit Margin (Before Income Tax) Net Profit Margin (Before Interest and Income Tax) Net Profit Margin (before Interest, Income Tax, Depreciation and Non Operating Income) Efficiency of Financial Leverage (X) Personnel Productivity (EURO) [(Net Income for the Year Before Tax) / (Average of Owners Equity)] *100 [(Net Income for the Year Before Interest and Income Tax) / (Average of Owners Equity)] *100 [(Net Income for the Year Before Tax) / (Average of Total Owners Equity and Liabilities)] *100 [(Net Income for the Year Before Interest and Income Tax) / (Average of Total Owners Equity and Liabilities)] *100 (Total Gross Trading Results Account / Net Turnover)*100 (Operating Results Before Financial Transactions / Net Turnover) * 100 (Net Income for the Year Before Tax / Net Turnover)*100 [(Net Income for the Year Before Interest and Income Tax) / (Net Turnover)]*100 (Net Income for the Year Before Interest, Income Tax, Depreciation and Non Operating Income / Net Turnover)*100 Percentage change of Net Income for the Year Before Tax / Percentage change of Net Turnover Net Turnover / Personnel Personnel Profitability (EURO) Net Income for the Year Before Tax / Personnel ICAP GROUP S.A. Financial Ratios Explanation 4
5 2.2 Viability and Capital Structure Ratios Viability and capital structure ratios measure a company s ability to meet its obligations and how much of the company s assets are financed with debt. They reveal the equity cushion that is available to absorb any losses that may occur. Financial Leverage (: 1) (Liabilities + Transit Credit Balances + Provisions for Contingencies and Expenses) / Total Owners Equity and Liabilities Total Debt Equity Ratio (: 1) (Liabilities + Transit Credit Balances + Provisions for Contingencies and Expenses) / Owners Equity Banks Short Term Obligations to Owners Equity (Banks Short Term Obligations / Owners Equity) *100 Capital Structure (X) Return on Equity (Before Income Tax) / Return on Capital Employed (Before Income Tax) Basic Period of Viability (DAYS) Equity to Fixed Assets (X) Fixed Assets to Long Term Liabilities (X) Fixed Assets to Total Owners Equity and Liabilities Interest Coverage (Before Interest and Income Tax) (X) (Cash + Securities + Receivables) / [(Total Cost of Net Sales+ Selling Expenses) / 365] Owners Equity / Fixed Assets (Net Value) Fixed Assets (Net Value) / Long Term Liabilities [(Fixed Assets (Net Value) / (Total Owners Equity and Liabilities)] *100 Net Income for the Year Before Interest and Income Tax / Interest Charges and Related Expenses ICAP GROUP S.A. Financial Ratios Explanation 5
6 2.3 Liquidity Ratios Liquidity ratios measure the ability of a firm to meet its short-term obligations. The ability to pay short-term debt is of concern to anyone who interacts with the company. If a company cannot maintain a short-term debt-paying ability, it will not be able to maintain a long-term debt-paying ability, nor will it be able to satisfy its stockholders. The liquidity ratios look at aspects of the company's assets and their relationship to current liabilities. Current Ratio (Χ) Quick Ratio (Acid Test) (Χ Cash Ratio (Χ) (Average of: Current Assets + Transit Debit Balances) / (Average of: Short Term Liabilities + Transit Credit Balances) (Average of: Current Assets + Transit Debit Balances - Stocks) / (Average of: Short Term Liabilities + Transit Credit Balances) (Average of: Cash + Securities) / (Average of Short Term Liabilities) Working Capital (EURO) (Current Assets + Transit Debit Balances) - (Short Term Liabilities + Transit Credit Balances) Short Term Liabilities to Working Capital (Χ) Short Term Liabilities / [(Current Assets + Transit Debit Balances) (Short Term Liabilities + Transit Credit Balances)] ICAP GROUP S.A. Financial Ratios Explanation 6
7 2.4 Activity Ratios Activity ratios measure the quality of a business' receivables and how efficiently it uses and controls its assets, how effectively the firm is paying suppliers, and whether the business is overtrading or under trading on its equity (using borrowed funds). Collection Period (DAYS) Payable Period (DAYS) Inventory Turnover (DAYS) Operating Cycle (DAYS) Commercial Cycle (DAYS) Equity Turnover (Χ) Turnover of Capital Employed (Χ) Fixed Assets Turnover (Χ) [(Average of: Customers + Long Term Bills and Cheques + Doubtful Receivables) / (Net Turnover)] *365 [(Average of: Suppliers + Bills and Promissory Notes Payable + Outstanding Cheques) / (Total Cost of Net Sales)] *365 [(Average of Stocks) / (Total Cost of Net Sales)] *365 Days of Collection Period + Days of Inventory Turnover (Days of Collection Period + Days of Inventory Turnover Days of Payable Period) Net Turnover / (Average of Owners Equity) Net Turnover / (Average of Total Owners Equity and Liabilities) Net Turnover / Average of Fixed Assets (Net Value) Significance of Selling Expenses (Selling Expenses / Net Turnover) *100 ICAP GROUP S.A. Financial Ratios Explanation 7
8 2.5 Investor Ratios Investors to appraise potential investment opportunities commonly use these Ratios. Investor ratios connect the number of shares of the company and their Stock Exchange price, with the profits, the dividends and other assets. They measures the overall profit generated for each share in existence over a particular period, the proportion of earnings that are being retained by the business rather than distributed as dividends and provide a guide as to the ability of a business to maintain a dividend payment. Share Internal Value (EURO) Owners Equity / Number of Outstanding Shares Earnings per Share (EURO) Net Income for the Year Before Tax / Number of Outstanding Shares Dividend per Share (EURO) Dividend to Net Income (Before Tax) Dividend / Number of Outstanding Shares (Dividend / Net Income for the Year Before Tax) *100 ICAP GROUP S.A. Financial Ratios Explanation 8
9 3. FINANCIAL RATIOS FOR BANKS 3.1 Profitability and Efficiency Ratios Profitability and efficiency ratios are designed to evaluate the firm's ability to generate earnings. Analysis of profit is of vital concern to stockholders since they derive revenue in the form of dividends. Profits are also important to creditors because profits are one source of funds for debt coverage. Furthermore Management uses profit as a performance measure. Return on Equity (Before Income Tax) [(Net Income for the Year Before Tax) / (Average of Owners Equity)] *100 Return on Assets [(Net Income for the Year Before Tax) / (Average of Total Fixed Assets) ] *100 Return on Earning Assets [(Interest Income and Similar Income) / (Average of: Cash and Balances with the Central Bank + Treasury Bills and Other Bills + Loans and Advances to Credit Institutions + Receivables from Customers + Debts Securities Including Fixed Income Securities)] *100 Interest Expense of Interest Bearing on Owners Equity and Liabilities Return on Earning Assets Less Interest Expense of Interest Owners Equity and Liabilities (Χ) Net Return on Earning Assets Net Interest Margin Efficiency Ratio Efficiency Ratio (Except Net Profit on Financial Operations) [(Interest Expense and Similar Changes) / (Average of: Due to Credit Institutions + Liabilities to Customers + Liabilities from Credit Titles + Subordinated Debts + Debt Loan Compulsory Convertible to shares)] *100 (Return on Earning Assets Interest Expense of Interest Bearing on Owners Equity and Liabilities) [(Interest Income and Similar Income Minus Expense) / (Average of: Cash and Balances with the Central Bank + Treasury Bills and Other Bills + Loans and Advances to Credit Institutions + Receivables from Customers + Debts Securities Including Fixed Income Securities)] *100 [(Interest Income and Similar Income Minus Expense) / (Average of Total Fixed Assets)] *100 (Operating Expenses / Operating Income) *100 [(Operating Expenses) / (Operating Income Net Profit on Financial Operations)] *100 ICAP GROUP S.A. Financial Ratios Explanation 9
10 Operating Income (Except Net Interest Income) To Total Assets Operating Expenses To Assets General Administrative Expenses To Operating Expenses [Operating Income (Except Net Interest Income) / (Average of Total Assets)] *100 [Operating Expenses / (Average of Total Assets)] *100 General Administrative Expenses / Operating Income 3.2 Productivity Ratios Productivity is the quantitative relationship between what a company produces and the resources that uses. Productivity ratios show the participation of company s employees in its financial figures such as Profit or Assets etc. Personnel Productivity (EURO) Gross Margin (From Services) / Number of Personnel Personnel Profitability (EURO) Net Income for the Year Before Tax / Number of Personnel Assets to Number of Personnel (EURO) Assets / Number of Personnel Granting to Number of Personnel (EURO) Receivables from Customers (Granting) / Number of Personnel Deposits to Number of Personnel (EURO) Liabilities to Customers (Deposits) / Number of Personnel ICAP GROUP S.A. Financial Ratios Explanation 10
11 3.3 Balance Sheet Structure Ratios These ratios show the participation of company s financial figures in its balance sheet structure. Interest Earning Assets to Total Assets [(Cash and Balances with the Central Bank + Treasury Bills and Other Bills + Loans and Advances to Credit Institutions + Receivables from Customers + Debts Securities Including Fixed Income Securities) / Total Assets] *100 Granting to Total Assets [Receivables from Customers (Granting) / Total Assets] *100 Provisions for Granting to Granting Debts Securities Including Fixed Income Securities to Total Assets Interest Liabilities to Total Owners Equity and Liabilities Shareholders Equity to Total Owners Equity and Liabilities Deposits to Total Owners Equity and Liabilities [Provisions for Granting / Receivables from Customers (Granting)] *100 (Debts Securities Including Fixed Income Securities / Total Assets) *100 [(Due to Credit Institutions + Liabilities to Customers + Liabilities from Credit Titles + Subordinated Debts + Debt Loan Compulsory Convertible to shares) / Total Owners Equity and Liabilities] *100 (Shareholders Equity / Total Owners Equity and Liabilities) *100 (Liabilities to Customers (Deposits) / Total Owners Equity and Liabilities) *100 Granting to Deposits [Receivables from Customers (Granting)) / Liabilities to Customers (Deposits)] *100 Interest Expenses and Similar Charges to Other Interest Income and Similar Income 3.4 Income Structure Ratios (Interest Expenses and Similar Charges / Interest Income and Similar Income) *100 These ratios show the structure of company s profit and loss accounts. Net Interest Income to Operating Income Dividend Income to Operating Income Net Income on Financial Operations to Operating Income ) Staff Cost to Operating Income (Interest Income and Similar Income Minus Expense / Operating Income) *100 (Dividend Income / Operating Income) *100 (Net Income on Financial Operations / Operating Income) *100 Staff Cost / Operating Income ICAP GROUP S.A. Financial Ratios Explanation 11
12 3.5 Investor Ratios Investors to appraise potential investment opportunities commonly use these Ratios. Investor ratios connect the number of shares of the company and their Stock Exchange price, with the profits, the dividends and other assets. They measures the overall profit generated for each share in existence over a particular period, the proportion of earnings that are being retained by the business rather than distributed as dividends and provide a guide as to the ability of a business to maintain a dividend payment. Share Internal Value (EURO) Owners Equity / Number of Outstanding Shares Earnings per Share (EURO) Net Income for the Year Before Tax / Number of Outstanding Shares Dividend per Share (EURO) Dividend / Number of Outstanding Shares Dividend to Net Income (Before Tax) Dividend / Net Income (Before Tax) *100 Net Income (After Tax) to Dividend (x) Net Income (After Tax) / Dividend ICAP GROUP S.A. Financial Ratios Explanation 12
13 4. FINANCIAL RATIOS FOR INSURANCES - INSURANCE BROKERS & AGENTS 4.1 Profitability Ratios Profitability ratios are designed to evaluate the firm's ability to generate earnings. Analysis of profit is of vital concern to stockholders since they derive revenue in the form of dividends. Profits are also important to creditors because profits are one source of funds for debt coverage. Furthermore Management uses profit as a performance measure. Return on Equity (Before Income Tax) Return on Capital Employed (Before Income Tax) Return on Capital Employed (Before Interest and Income Tax) [(Net Income for the Year Before Tax) / (Average of Owners Equity)] *100 [(Net Income for the Year Before Tax) / (Average of Total Owners Equity and Liabilities)] *100 [(Net Income for the Year Before Interest and Income Tax) / (Average of Total Owners Equity and Liabilities)] *100 Gross Profit Margin [(Total Gross Trading Results Account / Net Turnover)]*100 Operating Profitability Net Profit Margin (Before Income Tax) Personnel Productivity (EURO) [(Operating Results Before Financial Transactions / Net Turnover)] * 100 (Net Income for the Year Before Tax / Net Turnover)*100 Grand Total Income / Number of Personnel ICAP GROUP S.A. Financial Ratios Explanation 13
14 4.2 Activity and Liquidity Ratios Activity ratios measure the quality of a business' receivables and how efficiently it uses and controls its assets, how effectively the firm is paying suppliers, and whether the business is overtrading or under trading on its equity (using borrowed funds). Liquidity ratios measure the ability of a firm to meet its short-term obligations. The ability to pay short-term debt is of concern to anyone who interacts with the company. If a company cannot maintain a short-term debt-paying ability, it will not be able to maintain a long-term debt-paying ability, nor will it be able to satisfy its stockholders. The liquidity ratios look at aspects of the company's assets and their relationship to current liabilities. Collection Period (DAYS) Equity Turnover (Χ) Turnover of Capital Employed (Χ) Current Ratio (Χ) [(Average of Receivables) / Grand Total Income] *365 Grand Total Income / (Average of Equity Capital) Grand Total Income / Average of Total Owners Equity and Liabilities) (Average of: Current Assets + Transit Debit Balances) / (Average of : Short Term Liabilities + Transit Credit Balances) 4.3 Capital Structure Ratios Capital structure ratios measure a company s ability to meet its obligations and how much of the company s assets are financed with debt. They reveal the equity cushion that is available to absorb any losses that may occur. Total Debt Equity Ratio (: 1) Insurance Provisions to Total Owners Equity and Liabilities Shareholders Equity to Total Owners Equity and Liabilities Intangible Assets Ratio Interest Coverage (Before Interest and Income Tax) (X) Total Liabilities + Total Insurance (Technical) Provisions / Equity Capital (Insurance Provisions / Total Owners Equity and Liabilities) *100 (Equity Capital Total Owners Equity and Liabilities) *100 (Establishment Costs and Intangibles (Net Value) / Total Fixed Assets) *100 (Net Income for the Year Before Interest and Income Tax / Interest Charges and Related Expenses ICAP GROUP S.A. Financial Ratios Explanation 14
15 4.4 Investor Ratios Investors to appraise potential investment opportunities commonly use these Ratios. Investor ratios connect the number of shares of the company and their Stock Exchange price, with the profits, the dividends and other assets. They measures the overall profit generated for each share in existence over a particular period, the proportion of earnings that are being retained by the business rather than distributed as dividends and provide a guide as to the ability of a business to maintain a dividend payment. Share Internal Value (EURO) Owners Equity / Number of Outstanding Shares Earnings per Share (EURO) Net Income for the Year Before Tax / Number of Outstanding Shares Dividend per Share (EURO) Dividend / Number of Outstanding Shares Dividend to Net Income (Before Tax) Dividend / Net Income (Before Tax) *100 Net Income (After Tax) to Dividend (x) Net Income (After Tax) / Dividend ICAP GROUP S.A. Financial Ratios Explanation 15
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### Please NOTE This example report is for a manufacturing company; however, we can address a similar report for any industry sector.
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How many tons can an standard dump truck carry
## How many tons can an standard dump truck carry
### How much weight can a dump truck hold? - …
Kefid · A dump truck (pickup truck sized) can carry 1 yard and three-axle dump truck can carry about 16.5 yards of gravel. Also Know, how much topsoil is a ton? As a general rule, a tonne of a standard topsoil equals approximately 0.67m3. Divide the cubic metres required by 0.67 to give you the tonnage. In this manner, how many cubic yards are in a
### How many tons can a dump truck hold? - Quora
Answer (1 of 3): Mohammed, it will easily hold all the stupid questions you with in the time you've been here. That's a ton… A larger truck could hold all the crap we've been through trying to answer you. You need to explain the size of the truck, but It should be big enough…
### Dump Truck Capacity: How Much Can a Western Star …
Kefid · How Many Tons Can a Dump Truck Carry? When you are in the market for a new dump truck you will likely find yourself wondering about the performance of your truck as it relates to your industry. You may wonder, how many tons of gravel does a dump truck hold or even how much soil does a dump truck hold.
### How many tons fit in a semi truck trailer? - Answers
Kefid · How many tons can an standard dump truck carry? Typical commercial US dump trucks are double-axle dually setups carrying about 10 cubic yards of material, which is about 10 short tons for typical
### How many yards does an end dump truck hold? – …
Kefid · What is the capacity of the average dump truck? Typically, larger dump trucks can carry about 28,000 pounds or about 14 tons. On average, smaller dump trucks can transport around 13,000 to 15,000 pounds or 6.5 to 7.5 tons. Dump Truck Inventory Specialized Commercial Vehicles What is a Cubic Yard? How many yards is an end dump?
### How many tons can a single axle dump truck carry? - Quora
Answer (1 of 3): Mohammed, it will easily hold all the stupid questions you with in the time you've been here. That's a ton… A larger truck could hold all the crap we've been through trying to answer you. You need to explain the size of the truck, but It should be big enough…
### How Much Weight Can A Dump Truck Haul - Diamonius.com
Kefid · On average, large dump trucks can carry roughly 28,000 pounds or about 14 tons. Smaller dump trucks can manage about 13,000 pounds to 15,000 pounds or 6.5 to
### How many tons are in a dump truck topsoil?
Kefid · A dump truck (pickup truck sized) can carry 1 yard and three-axle dump truck can carry about 16.5 yards of gravel. Also Know, how much topsoil is a ton? As a general rule, a tonne of a standard topsoil equals approximately 0.67m3. Divide the cubic metres required by 0.67 to give you the tonnage. In this manner, how many cubic yards are in a
### WEiCHTS AVERAGE STANDARD TRUCK - 3 TONS …
Kefid · average standard car - 1.5 tons average ambulance - 5 tons average loaded school bus - 17 tons truck - 6 tons average loaded charter bus - 20 tons . average fire truck average loaded cement truck - 33 tons average loaded dump truck - 36 tons average loaded tractor trailer - 40 tons . created date: 8/13/2013 11:45:47 am
### How many yards is an end dump? | PopularAsk.net - Your
How much weight can a dump truck hold? - FindAnyAnswer.com
### How Many Tons Can A Dump Truck Carry - English Tenses
70 Tons U Shape 6 Axle Dump Tipper Truck Trailer For Carry . Item: 70 Tons U Shape 6 Axle Tipper Trailer, Dump Trailer For Carry Rock, Aggregate. Introduction of 70 Tons Tipper Trailer. A tipper trailer is a big trailer that usually lifts when dumping its hall in a chosen location.
### How many tons will a tandem dump truck haul? - Answers
Kefid · For a tandem axle dump truck, they can typically legally carry between 13 and 15 tons. How many tons can a dump truck hold? Dump trucks come in …
### How many tons does a standard dump truck hold
how many tons can a truck carry - 002mag.com. On average, large dump trucks can carry roughly 28,000 pounds or about 14 tons. Smaller dump trucks can manage about 13,000 pounds to 15,000 pounds or 6.5 to. Learn More
### How many tons can a dump truck hold? - SupportMyMoto
Kefid · A 26,000 lb. GVW dump truck can usually carry round 5 tons legally, whereas a 33,000 lbs. What number of tons of gravel do I want for my driveway? Utilizing 2 inches for the depth, the next measurements are a information to the quantity of gravel protection per ton : 1/4 to 1/2 inch gravel, 100 sq. toes per ton ; 1/2 to 1 inch gravel, 90 sq
### Guess how many tons of cargo this dump truck trailer can …
#trucktrailer#semitrailer#dumptrailerWe are semi trailer manufacturer in China, we will produce the highest quality semi trailer and truck, - if you need sem
### How Many Tons Can A Dump Truck Carry - English Tenses
Typically, larger dump trucks can carry about 28,000 pounds or about 14 tons. On average, smaller dump trucks can transport around 13,000 to 15,000 pounds or 6.5 to 7.5 tons. How much gravel is in a dump truck? How Many Tons of Gravel In A Dump Truck? A 10-wheel dump truck carries 12 to 20 tons of gravel or around 8 to 10 yards depending on the
### Question: How much weight can a dump truck carry? - …
How many tons are in a dump truck topsoil?
### How many tons can a dump truck hold? - SupportMyMoto
Kefid · A 26,000 lb. GVW dump truck can usually carry round 5 tons legally, whereas a 33,000 lbs. What number of tons of gravel do I want for my driveway? Utilizing 2 inches for the depth, the next measurements are a information to the quantity of gravel protection per ton : 1/4 to 1/2 inch gravel, 100 sq. toes per ton ; 1/2 to 1 inch gravel, 90 sq
### How many tons of gravel can a dump truck hold?
Answer (1 of 4): Depends on the capacity of the dump truck. Look on the specs plate located on the dump truck or inside the owners manual for capacities
### How many tons do dump trucks haul? - Quora
Kefid · While large dump trucks can haul 28,000 pounds, most carry slightly less. Many smaller dump trucks can haul about 13,000 to 15,000 pounds. Dump trucks often weigh about the same amount as their load capacity. For example, a fully loaded dump truck carrying 14 tons of material usually weighs about 28 tons.
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# Math 4530 Homework for Friday February 4
Math 4530
Homework for Friday February 4
Useful procedures are posted on our lecture page, at
http://www.math.utah.edu/~korevaar/4530spring05/4530lectures.html
1) Rework exercise 1.4.7 using Maple; this was a messy hand calculation to compute the curvature,
torsion, T,N,B for a specific curve which was not parameterized by arclength. In "curtor" are already
written procedures to compute curvature and torsion. You can modify these to write procedures which
will compute T, N, B. Note, that since it is easy to compute T, and also B, as the normalized cross
product of the first and second derivatives of the curve, you can get N by taking B cross T.
After all those computations, use Maple to draw a nice picture of the curve, say for t between zero and
one. You can decide whether you prefer the "plot3d" or "tubeplot" commands. (See below for an
example of what "tubeplot" creates.)
2) Exercise 1.7.8, for kap3 and kap4: Use the plane-curve recreation template which I’ve posted, or
create your own procedure, to recreate nice pictures of these two curves.
3) Exercises 1.7.7 and 1.5.15: Again use the plane-curve recreation template which I’ve posted or your
1
own procedure, to recreate a curve with curvature κ(s) =
2 , so that at s=0 the curve is horizontal
1+s
and at the origin. Now, the real fun begins. Using the plane- Frenet system which we talked about in
class (see also exercise 1.5.13 on page 41), you can find an explict formula for the curve by
antidifferentiation (this is the exercise 1.5.15 part). Prove that the curve you’ve created is actually a
catenary (graph of y=cosh(x) or Euclidean motion of such)! (We saw the catenary before as the shape of
a hanging cable.)
4) Use the procedure "recreate3dview", which is one of the files posted on our web page, to create a 3-d
curve which is beautiful, with your own choice of curvature and torsion functions.
5) Here is a picture of part of an a=2, b=1 helix, sitting on a radius 2 cylinder:
> with(plots):
> cylinder:=
plot3d([2*cos(theta),2*sin(theta),z],theta=0..2*Pi,z=0..2*Pi,
style=wireframe,color=black):
#this is a parametric way of drawing the cylinder
> helix2:=tubeplot([2*cos(t),2*sin(t),t],
#this is a corresponding piece of the helix
> display({cylinder,helix2},scaling=constrained,axes=boxed);
#here they are displayed together.
6
5
4
3
2
1
0
–2
–1
0
1
22
1
0
–1
–2
Your job is to find the osculating circle to the helix at the point when t = Pi , at P = [−2, 0, π ] .
(Recall exercise 1.3.28 from last week.) Add this osculating circle to the display above.
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https://www.physicsforums.com/threads/gamma-radiation-decay-intensity-iaea-nuclide-chart.1050142/
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# Gamma radiation decay intensity (IAEA nuclide chart)
• I
• eneacasucci
In summary, the website lists the gamma radiation intensity for different nuclei, but the absolute intensity doesn't add up to 100%. I was wondering why and Google led me to understand that the intensity is related to the probability of emission.
eneacasucci
TL;DR Summary
Gamma radiation decay intensity in IAEA website, what does it represent.
I was looking at the gamma radiation data from IAEA's website:
(https://www-nds.iaea.org/relnsd/vcharthtml/VChartHTML.html)
and was confused by the absolute intensity listed in the page. I Googled it and it seems to be the probability of emission but why it doesn't add up to 100%?
For example Cs-137 https://www-nds.iaea.org/relnsd/vcharthtml/VChartHTML.html
I don't understand why, because for example the sum of the beta intensities is 100%:
I know about gamma intensity related to the attenuation in matter but this is another topic
Thank you so much for your help
Last edited:
The 0.662 KeV state of Ba137 decays 85% of the time only by gamma emission and decays by the internal conversion process the rest of the time.
eneacasucci
Some nucleai, like Co-60 decay by more than one gamma.
gleem said:
The 0.662 KeV state of Ba137 decays 85% of the time only by gamma emission and decays by the internal conversion process the rest of the time.
This would make sense and clarify the %, but how do you know it? I don't see this info in the website
Some nucleai, like Co-60 decay by more than one gamma.
Thank you for your answer. I know that there could be more than one gamma, but for Co-60 the main two gammas (1332 keV and 1173 keV) are related to different initial level. For Cs-137, considering the level 662 keV, if only 85% of times the nucleus decays to ground level emitting the related gamma, what happens the 15% of times (still starting from that level)?
eneacasucci said:
I don't see this info in the website
eneacasucci
eneacasucci said:
I know that there could be more than one gamma, but for Co-60 the main two gammas (1332 keV and 1173 keV) are related to different initial level
Huh?
They come from the Co-60 ground state, and you get two, not one or the other.
gleem said:
My fault, I wasn't considering the column of electrons and now I see them. My last question is: shouldn't these [%] (highlighted in yellow) make 100% (because it doesn't)?
(Excuse my ignorance, but what does the initials 'CE' mean (before the electron's shell)?
eneacasucci said:
My fault, I wasn't considering the column of electrons and now I see them. My last question is: shouldn't these [%] (highlighted in yellow) make 100% (because it doesn't)?
Keep in mind the percentages are the percentages of the radiation per decay. Cesium decays 5% of the time to the ground state of Ba which has no additional radiation processes. So only about 95% of the decays of Cs result in additional radiation so expect a gamma or an internal conversion electron 95% of the decays.
eneacasucci
I think I got it now.
The percentages match as it should be right?
Last edited:
I've tried to make the same reasoning applied to Co-60.
I considered level 1332. You can reach it with the beta decay
, and with gamma decays
. Also without considering the conversion electrons the tot probability to reach final level 1332 (99,9776%) is lower than the probability of decaying from that level (99,9826%)
I've also tried to download the excel file with the whole data but there is still this incongruency... so I'm doubting I didn't understand the concept
Did you take into account the uncertainties in the data especially the beta branching ratios?
gleem said:
Did you take into account the uncertainties in the data especially the beta branching ratios?
No I didn't, I don't know exactly how to deal with them. I just wanted to be sure that "my reasoning" (which would be the application of what you explained to me in the previous messages) was correct, and that the non-coincidence of numbers was due to something else
Play with the data and see how it affects your results. Increase the 1.332 Beta ratio 0.12 ±.03 to a value within the uncertainty for example.
I had the same question and found some information in this site:
https://www-nds.iaea.org/relnsd/vcharthtml/guide.html.
There it says:
Iγ(rel) Relative Intensity 100 is assigned to the most intense gamma from a given initial level, and other gammas relative intensities are referred to that.
So, my humble opinion in this case is, they do not have to add up to 100%, because the percentage here is just for comparison between any given intensity and the intensity of the “most intense” gamma, it would mean just how much smaller are those other gammas relative to the 100% highest intensity gamma. It would make more sense to relate them with a relative ratio than with percentage, which in most cases gives a balance of species out of a total.
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Velocity Reviews > level order traversal of binary tree
# level order traversal of binary tree
sophia.agnes@gmail.com
Guest
Posts: n/a
02-14-2008
Dear all,
How good is this function which implements level order traversal of
binary tree ?
level_trav(struct node *sn)
{
struct node *queue[MAX];
int front = rear = -1;
front++;
rear++;
queue[rear] = sn;
while(front <= rear)
{
sn = queue[front++];
if(sn != NULL)
{
printf("%d",sn -> val);
queue[++rear] = sn -> left;
queue[++rear] = sn -> right;
}
}
}
Ben Bacarisse
Guest
Posts: n/a
02-14-2008
http://www.velocityreviews.com/forums/(E-Mail Removed) writes:
> How good is this function which implements level order traversal of
> binary tree ?
These are just coding comments rather than algorithmic ones...
> level_trav(struct node *sn)
> {
> struct node *queue[MAX];
> int front = rear = -1;
>
> front++;
> rear++;
Looks odd. Just initialise both to 0.
> queue[rear] = sn;
>
> while(front <= rear)
> {
> sn = queue[front++];
> if(sn != NULL)
> {
> printf("%d",sn -> val);
> queue[++rear] = sn -> left;
> queue[++rear] = sn -> right;
I'd want to check that there is room!
> }
> }
> }
Mildly algorithmic points:
- You will be able to handle bigger trees if you don't queue up NULL
pointers rather than testing for them when you de-queue.
- Again, you get better usage from your array if you make your queue
circular.
--
Ben.
Barry Schwarz
Guest
Posts: n/a
02-15-2008
On Thu, 14 Feb 2008 06:01:40 -0800 (PST), (E-Mail Removed)
wrote:
>Dear all,
>
>How good is this function which implements level order traversal of
>binary tree ?
>
>level_trav(struct node *sn)
>{
> struct node *queue[MAX];
> int front = rear = -1;
This doesn't define rear so it remains undefined throughout your
program.
>
> front++;
> rear++;
> queue[rear] = sn;
>
> while(front <= rear)
> {
> sn = queue[front++];
> if(sn != NULL)
> {
> printf("%d",sn -> val);
> queue[++rear] = sn -> left;
> queue[++rear] = sn -> right;
> }
> }
> }
Remove del for email
--
Posted via a free Usenet account from http://www.teranews.com
Kaz Kylheku
Guest
Posts: n/a
02-15-2008
On Feb 14, 6:01*am, (E-Mail Removed) wrote:
> Dear all,
>
> How good is this function which implements level order traversal of
> binary tree ?
>
> level_trav(struct node *sn)
> {
> * struct node *queue[MAX];
This queue is allocated in automatic storage, and is used to simulate
a stack. So effectively, your program is recursive.
> * int front = rear = -1;
This declaration declares only front. Where is rear? If you don't have
rear as a global variable somewhere, this is an error.
Before you post something, at least try to compile it and fix as many
errors as you are able.
C declarations do not have this multiple = syntax for declaring
multiple identifiers; you must write:
int front = -1, rear = -1;
Or perhaps:
int front = -1, rear = front;
> * front++;
> * rear++;
What's the point of this? Why initialize two variables to -1 when your
very next step overwrites their value with the successor? Just take
this out and change the initialization to:
int front = 0, rear = 0;
No???
> * queue[rear] = sn;
>
> * while(front <= rear)
> * {
> * * *sn = queue[front++];
> * * *if(sn != NULL)
> * * *{
> * * * *printf("%d",sn -> val);
> * * * *queue[++rear] = sn -> left;
> * * * *queue[++rear] = sn -> right;
> * * *}
> * *}
> *}
You have a few problems. Firstly, why not make the queue circular? The
way you have it, once you visit a node at position queue[N], you never
reuse that position. So your queue needs as many entries as the total
number of nodes in the tree.
Secondly, you are wasting lots of space by storing null pointers in
the queue when encountering leaf nodes. If sn->left is null, there is
no point in doing queue[++rear] = sn->left. You're just placing a null
pointer into the path, which your (sn != NULL) test has to stumble
over later. And it means that your queue has to be sized to handle all
of the null pointers in the leaves. Are you not aware that there are
more null pointers in the leaf nodes of the tree than there are nodes
in the tree?
For instance, consider a simple two level tree:
A
B C
It has three nodes, okay? The children B and C each have two null
pointers. That's four null pointers. And four is, like, more than
three, get it? Try it with a deeper tree:
A
B C
D E F G
Now there are seven nodes, right? How many null pointers? Yes, eight.
Each of D E F C has two. Again eight is greater than seven. Always one
more than the number of nodes. Each level has as many items in it as
all of the levels before it, plus one. Level 1 has A. Level two has B
C, one more than A. Level 3 has D E F G (four items), one more than
levels 1 and 2 combined (three items).
This is why if you make your queue circular, you need only about half
the space. The most space will be needed when you are doing the bottom
level, and that has only half the total number of nodes. As you
prepare that level by queuing it at the tail, you are removing the
previous level, visiting those nodes, and that space can be used.
Kaz Kylheku
Guest
Posts: n/a
02-15-2008
On Feb 14, 6:01 am, (E-Mail Removed) wrote:
> Dear all,
>
> How good is this function which implements level order traversal of
> binary tree ?
queue proportional to the sizeo of the tree. Effectively you are doing
A breadth first search can be made to require less space by using a
depth-first search instead, and using repeated depth-first searches
to /fake/ the effect of a breadth-first search.
This is done by repeatedly doing a breadth-first search whose depth is
limited, and iteratively increasing the depth. This is called
iterative deepening.
First you do a depth-first search which only goes to depth 0. Then you
do another one which goes to depth 1. Then you do one which goes to
depth 2, and so on.
Whenever you hit the bottom depth, you visit the nodes at that depth.
The storage you require is proportional to the depth of the tree, not
the number of nodes in it, and it can be done recursively.
/*
* helper function: visit nodes at specified depth.
* return 1 if something was visited, otherwise 0.
*/
static int visit_depth(struct node *node, int depth)
{
if (node != NULL) {
int left_visited, right_visited;
if (depth == 0) {
printf("%d\n", node->val);
return 1;
}
left_visited = visit_level(node->left, depth - 1);
right_visited = visit_level(node->right, depth - 1);
return left_visited || right_visited;
}
return 0;
}
[[Student exercise question: above, why didn't we write
return visit_level(node->left, level - 1) || visit_level(node-
>right, level - 1);
and instead used the left_visited and right_visited variables?]]
Now given visit_depth above, we can simply write level_trav like this:
void level_trav(struct node *tree)
{
int depth;
/* keep increasing depth until no nodes are found at that depth */
for (depth = 0; visit_level(tree, depth); depth++)
{
/* empty */
}
}
Now you might think that the algoithm is slow because it keeps
revisiting the lower depths over and over again. But remember that
more than half of the nodes are in the bottom-most level that is
visited. Each time visit_level is called with a greater depth, it
visists more than twice as many nodes as in the previous iteration.
And this means that in the last iteration, it visits about as many
nodes as in all of the first iterations combined! This means that half
of the work is always in the last iteration! So the algorithm runs in
O(N) and basically the overhead for doing it this way is that you
traverse each node twice on average. Of course, the nodes at the
bottom level are visited only once (and that's more than half the
nodes in the tree!). And the root node is a high traffic corridor,
being a gateway to the rest of the tree: it is traversed in each
iteration.
Whether you want to do it this way or using the queue depends on how
large the tree is. Can you spare the O(N) queue storage? Or the tree
large enough that saving memory is worth traversing the nodes more
than once?
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https://aakashdigitalsrv1.meritnation.com/ask-answer/question/the-turbine-pits-at-the-niagara-falls-are-50m-deep-the-avera/work-energy-and-power/6809895
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# The turbine pits at the Niagara falls are 50m deep. The average horse power developed is 5000, the efficiency being 85%. How much water passes through the turbine per minute?given 1 HP = 746W and g= 10m/s2
Here,
h = 50 m
Po = 5000 HP
g = 10 ms-2
η = 0.85
Let the input power to run the turbine = P
Output power / Input power = 0.85
Po/Pi = 0.85
Po = 5000 HP = 5000×746 W
So
Pi = 5000×746/0.85 W
Now let m be the mass of water must be passed per second to deliver the power Pi
5000×746/0.85 = mgh
=> m = 5000×746/(0.85×10×50)
=> m = 8776.47 kg / s
In 1 min mass flow will be
8776.5 × 60 = 526588 kg
• 30
What are you looking for?
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# Math (algebra)
posted by .
Matt has to bake muffins. He uses x eggs for each 200 grams of flour. If he has to use 900 grams of flour how many eggs will he use? I have to set this up as an algebra problem. I am stuck. Our class is just learning this and I don't understand.
• Math (algebra) -
carry the units with you, and it will help. You want to convert from grams of flour to eggs
x eggs = 200g
or,
x eggs/200g = 1
you can always multiply by 1 and not change anything.
900 g * x eggs/200g = 900g/200g * x eggs = 9x/2 eggs
Or, since 900 = 9/2 * 200, he will use 9/2 times the eggs used for 200 grams of flour.
• Math (algebra) -
4.5
• Math (algebra) -
86
• Math (algebra) -
pfft
• Math (algebra) -
gas
• Math (algebra) -
butthole
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# Path to finding a good electrode design
Get Complete Project Material File(s) Now! »
## Pore network modeling
The basic physical principle used in pore network modeling (PNM) is conservation (Bern-abe 1995). This means that there are no sinks or sources inside the pore network. Meshes of pores are built with a particular geometry (e.g., rectangular, hexagonal, triangular) and are connected through a node. The size of the pore needs to be explicit (i.e. radius of the pore). The goal of PNM is to determine the response of the whole network subjected to an exter-nal gradient (for example a voltage) given a particular condition of each pore (for example particular radius).
Maineult et al. (2017) and corrected by Maineult (2018) propose a PNM to obtain the SIP response of randomly sized tubes (see figure 1.7). They randomly distribute the radius size to each pore, given a pore size distribution. They obtain the electrical response of the system using the expression: r 2 ¿ ˘ 2D . (1.39).
### Electrode testing
In order to measure the polarization of the electrodes (or lack thereof ) an SIP measurement is done. To do so, I fill a recipient with very saline water (around 36.8 mS cm¡1), I locate two current injecting electrodes made of stainless steel at opposing sides of the recipient, with the two potential electrodes to be tested in the middle (all located in a straight line, see figure 2.2). All four electrodes barely touch the water (see figure 2.2). This is a way to test the electrode polarization as very saline water should ideally have a very small phase. Knowing that phase can be represented as: ’ ˘ ¡ar ct an µ¾000 ¶, (2.1) ¾.
where in very saline water ¾0 ¨¨ ¾00, therefore the phase of very saline water should be close to zero. If the measured phase is not close to zero, that means that the measured polarization does not come from the very saline water but from the electrodes themselves. It is worth mentioning that it is normal to have a small polarization signal at high frequencies (>100 Hz), coming from impedance effects of the electrodes (see Huisman et al. 2016; Wang and Slater 2019). The idea is to build electrodes with the least amount of internal polarization possible. All the electrodes batches used in this study were tested in this same manner.
#### Path to finding a good electrode design Type and amount of gelifying agent
A problem with electrodes of the first kind (including Cu-CuSO4 electrodes) is that the cham-ber holding the aqueous solution (CuSO4 in this case) may leak out of the chamber through the porous plug by diffusion. Kremer (2015) did an extensive study on the type of porous caps that can be used to plug the electrodes. Additionally, Kremer (2015) and Kremer et al. (2016) present a test using a gelifying agent to decrease the leakage of the solution (i.e. re-duce the ionic mobility). I based the electrode construction off of their work. However, I decided to do further tests. I tested the concentration of CuSO4, and type (and amount) of gelifying agent.
On table 2.1 I present a synthesis on the tests performed to obtain a suitable pair of non-polarizable Cu-CuSO4 electrodes. Note, that all solutions were made from an amount of penta-dehydrated CuSO4, 100 ml of de-ionized water, and an amount of a gelifying agent. For the first test, I was advised by Feras Abdulsamad to use 0.9 g of agar-agar. As to the amount of penta-dehydrated CuSO4, I was advised to use an amount that would saturate the solution. The amount of CuSO4 that saturates a solution varies with temperature, since these solutions are heated up to 100 –C, I decided to test the amount of CuSO4 based of the temperature range I would heat the solution to. With the first test, I was able to narrow down the amount of penta-dehydrated CuSO4, as most electrodes showed significant signs of precipitation (negative results, marked as « – » in table 2.1). I decided that visible signs of early precipitation in the electrodes meant the lack of chemical equilibrium in the solution. Constant chemical reactions within the solution itself could potentially affect the measured SIP signal.
For the second test, I decided to slightly increase the amount of agar-agar, as the first batch of electrodes did not all gelatinize. I decided also to narrow down the search on the proper amount of penta-dehydrated CuSO4 needed for the electrode construction. The pair of 32 g of penta-dehydrated CuSO4 still presented precipitation. To test these electrodes, I mea-sured the SIP signal of the 23, 26 and 29 g of CuSO4 electrodes in a recipient filled with saline water. For an improved readability of this text, I will not present the SIP measurements of all the electrode tests, just the final and most relevant tests. I singled out the pair of electrodes using 23 g of CuSO4 because they had the best SIP signal. For the third test, I decided to vary the amount and type of gelifying agent. None of these electrodes precipitated.
To decide which pair of electrodes were optimal, I measured their SIP response (see figure 2.3), at different times after the making (the day after making the electrodes, a month later, and two months after construction). It is worth mentioning that in figure 2.3c and f, the SIP signal of the agar-agar electrodes is not presented, because the electrodes had visibly- Table 2.1: Chronological tests to obtain a suitable pair of non-polarizable Cu-CuSO4 electrodes. All these tests were done in 100 ml of de-ionized water. Note that by + and – in this table, I mean positive or negative results. By positive, I mean that the electrodes did not show any visible sign of precipita-tion, by negative I mean that they did show visible signs of precipitation.
Brass electrodes testing
Building Cu-CuSO4 electrodes is a tedious task that needs to be repeated every other month or so. This is why some researchers such as Huisman et al. (2016) have proposed to use brass electrodes (a brass rod) in contact with fluid from the same salinity as the one saturating the rock sample (see figure 2.7b and c).
The brass rod is retracted in a tube (figure 2.7b), so the streamlines of the SIP measurements will not pass through the metal, as the metal is connected to the sample through the liquid of the pore solution but is not inserted in the sample. This is ideally supposed to work, as what would create polarization is the contact between a metal and the rock sample. If the current lines are not in contact with the metal, no polarization should occur.
I purchased a few brass rods with a 5 mm diameter, with standards BS2874/CZ121M (1986)BS EN 12164/CW614N (from the specification sheet of the manufacturer). I retracted the elec-trodes by 1 cm from the end of the tube (see figure 2.7b) and filled this space with the same salted water as in the recipient (figure 2.7). I measured the SIP signal with these electrodes (see figure 2.8, this dataset corresponds to the date 25-06-19). I also attempted sanding the rods (in case the brass rods had some coating, see figure 2.7a), re-filled the recipient with new very saline water and re-measured the SIP signal (see figure 2.8, this dataset corresponds to the date 26-06-19). The SIP signals I measured are presented in figure 2.8. The errorbars of these measurements reached 1500 mrad, and the lowest measured phases were around 400 mrad; significantly higher than the SIP signals I had measured in the clay samples. Some-thing was obviously very wrong with my set-up. Many researchers (e.g., Huisman et al. 2016; Izumoto et al. 2020) have been able to use the brass electrodes without problems. I am not attempting to discourage the reader on the use of these electrodes, I am merely presenting a way that does not work so the reader does not follow this path. I am not sure why this at-tempt did not work. Although, brass is an alloy of metals, I wonder if a specific type of alloy is needed in order to get good measurements. I did not further investigate these types of electrodes for SIP, for time purposes. In the future, I would like to contact these researchers for more details on the use of their electrodes.
READ The interaction between body and culture in metaphor understanding
Electrode correction
Another electrode-related improvement I attempted was the SIP signal electrode correction. Huisman et al. (2016) proposed a way to correct SIP measurements in a rock sample. They measure the SIP signal in the traditional configuration with 4 electrodes, then they measure the reciprocal. They use both measurements to determine the electrode impedance and substract it from the measured SIP signal. Later Wang and Slater (2019) proposed another correction in which the SIP signal is measured in 4 different configurations, then the metal of the electrode (copper wire for Cu-CuSO4 electrodes) is inserted deeper gradually. At each metal insertion step, the 4 SIP configurations are measured. All of these measurements are stored and then plotted all together in a phase vs electrode impedance plot for each fre-quency. Allowing to infer the phase for zero electrode impedance. In the results shown by Wang and Slater (2019) they were able to remove the high frequency noise of the measure-ments. I attempted to do this, however, after all the calculations I obtained a noisier signal than the original one I measured (see figure 2.9). I therefore did not remove the electrode error from the SIP measurements in this study. Again, this does not mean that the protocol proposed by Wang and Slater (2019) does not work. Here, I have some suggestions at ele-ments that have to be taken into account for this test. Elements I had a hard time with in the measuring part of this test.
One of the goals for this thesis was to create clay heterogeneities. There was a need for a sample holder where clay heterogeneities could be located without major disruption to the sample (such as inserting air bubbles). That is, a sample holder that could be opened longi-tudinally. The best solution I found for this problem was a « sushi bazooka »; a plastic device created to locate sushi ingredients inside, close it giving the sushi ingredients the cylindri-cal shape, and then pushing it out. I transformed these devices into the sample holders by slightly modifying them. The dimensions of the sample holders are presented in table 2.7 and the complementary figure 2.11 showing the dimensions of the sushi bazooka.
I used a pair of stainless steel cylinders as injecting electrodes. There was a need to create a structure to hold the injecting electrodes in a consistent manner. The first attempt of exter-nal structure involved two parallel plastic rectangles, held together by four plastic threaded rods (one for each corner of the rectangle) and the sample holder (i.e. « sushi bazooka ») lo-cated in the middle. A problem arose because it was not possible to locate the sample holder in the same exact location repeatedly. For this, I created an external structure with acrylic sheets cut by a laser cutter (see figure 2.12a). This allowed the creation of acrylic sheets of the same dimensions with more than millimeter precision. I also cut (with the laser cutter) four holes for the four threaded rods, and a smaller hole located exactly in the middle (to lo-cate a small screw as a guide, see figure 2.12b). The screw guide allows for the sample holder to be located in the same exact position repeatedly. I think that the use of new available technologies for the general public improved the cre-ation of the sample holder (as the laser cutter). I think the SIP-laboratory community should further explore these new available technologies, such as laser cutters and 3D printers, and incorporate them to their laboratory equipment.
Acknowledgments
Résumé
Contents
List of Figures
List of Tables
Introduction
1 Theoretical background
1.1 Background on clay structures
1.2 Background on geo-electrical methods
1.2.1 Maxwell’s laws for induced polarization
1.2.2 Active electrical methods in geophysics
1.3 Background on SIP
1.3.1 The electrical double layer
1.3.2 Polarizationmechanisms
1.3.3 SIP models
1.3.3.1 Phenomenological models
1.3.3.2 Physical models
1.4 Upscaling techniques
1.4.1 Differential effective medium
1.4.2 Pore networkmodeling
2 Materials and methods
2.1 Non-polarizable electrodes
2.1.1 Electrode testing
2.1.2 Path to finding a good electrode design
2.1.3 Brass electrodes testing
2.1.4 Electrode correction
2.2 Tests for water content
2.3 Sample holder
2.3.1 Comments on sample holder construction
2.3.2 Geometrical factor of sample holder
2.4 Compression and decompression tests
2.5 Water chemistry
2.6 Conclusion of this chapter
3 SIP on individual types of clays
3.1 JGR article in the context of this thesis
3.2 JGR article
3.3 Main results of JGR article
3.4 Supplementary information of JGR article
3.4.1 SIP measurements on additional samples
3.4.2 Differentiation of clay minerals
3.4.3 Repeatability test
3.4.4 Relationship between the imaginary conductivity at a frequency of 1.46 Hz and surface area per unit pore volume
3.5 Comment on pore water equilibrium
4 SIP on heterogeneous clays
4.1 Introduction tomanuscript
4.2 Manuscript
4.3 Main results of manuscript
4.4 Complements tomanuscript
4.4.1 Mesh types on complex conductance networks
5 Perspectives
5.1 Varying pH on individual clay samples
5.2 Clay heterogeneity mixtures
5.3 Clay compaction
5.4 Numericalmodeling
5.4.1 Numerical models in SIP
5.4.2 Numerical models of clays
5.5 Various recommendations
Conclusions
Appendix1
Electrode construction
Calculation of errorbars for the complex conductivity
Appendix2
Article Jougnot et al. (2019)
Bibliography
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# The Science of…Linear Algebra and the Design Matrix
December 14, 2022 | By Michelle Franklin
Johannes Brust is a visiting scholar with the Department of Mathematics. Earlier this year he published a feature article in Image, the Bulletin of the International Linear Algebra Society that discusses matrix designs for COVID-19 group testing. In this edition of “The Science of…” he talks about his paper and his work in linear algebra.
Q: Tell us about the field of mathematics you study.
Johannes Brust: My background is in applied math and my research focus is on the development of algorithms (methods) in optimization and numerical linear algebra. When I was an undergraduate, I wanted to work in economics or finance, but I found myself wanting more challenging problems, so I decided to do a graduate degree in applied math. Broadly, it’s numerical analysis — computations to make processes more efficient. The specific topics I study are numerical linear algebra (finding a structure for multiple equations and then using the structure to improve accuracy or efficiency) and numerical optimization (finding parameters to make a model fit to data better or streamlining a process in an automatic way).
Q: Can you explain numerical optimization more — what do you mean by making a model fit to data better?
JB: I can give you a financial example. Imagine you have a model of prices in the options market that would behave in a certain way in an ideal world — certain relations would result in a certain price and the relations are governed by a set of inputs or parameters. As the market evolves, the parameters may change, or you might need a different set of parameters. If there are a lot of changing parameters, to make the model fit to the observed prices, you would need a method to minimize the difference between the data points and the computer price model —that is the fitting of the data.
Essentially, you are trying to find relationships between different characteristics and then find a trend that allows us to make predictions. That could be anything from the stock market to population growth charts, which you might see in a doctor’s office.
Q: Your feature article in Image discusses matrix designs for COVID-19 group testing. Can you walk us through it?
JB: In populations where the prevalence of COVID is low, I wondered if there was a way to conserve resources through group testing or pooling. Is it possible to use a systematic approach to improve on methods that had been developed beforehand? If you are able to identify a certain number of positive cases precisely with a single design, then it could be straightforward to use. You could make a template available to a lab that follows the pattern of the matrix in terms of assigning samples to a test.
Q: How do you design a matrix?
JB: This was based on arranging samples in a grid format and then using that format to group points in a grid to define a test of samples. Partly I wanted to see if certain things would repeat themselves, if there were any patterns. If samples are aligned in a grid, can they be combined in a group, can we combine other samples in another group? Can we identify properties where if those properties are satisfied by the groups, we could exactly decode the information that was put into the matrix when we are trying to find out which samples are positive? If we can achieve something that reproduces the properties, then we can obtain the method that will be able to do the pooling tests.
Multipool matrices can be constructed using an affine plane (l). A projective plane (r) can be derived from an affine plane.
Q: Where does linear algebra come in?
JB: To do this work relatively fast, we used linear relations — arranging samples into squares. Later, we looked at direct angles and drawing lines into the arrangement of the samples, but everything is discrete, so the lines wrap around the square and start over again. You have to make sure none of the lines intersect because that would correspond to cases where one of the properties is not satisfied anymore. We looked for parallel lines so all samples are covered, but each line would then define a new test. Then we needed to find a formula that can be used to compute each of the indices that would then map into one of the samples.
Q: How do you get a pool sample without testing samples individually?
JB: Let’s say you have a set of vials with samples in them, but you don’t know if the samples are positive or negative for COVID. You can take an additional vial and mix all of those samples together and then test that one group sample. If the group sample is negative, we can conclude all the original ones are negative too. If it’s positive, either we did the grouping in such a way that we can infer from the set of groups, which ones of the original samples had to be positive because we recorded how we added them together. Or in a two-step case, we would go back and test them individually.
Q: Is this practical? Is this something labs would implement?
JB: There are programs where these types of strategies can be used to gain an overview of how the disease is developing or evolving. When you have a large sample size, this can act like a screening testing to roughly determine what the percentage of positive cases might be.
Q: Are these matrices already being used for testing?
JB: We wanted to generate a matrix that can serve as a template to guide which sample goes into which test. We have already launched software where you can set the parameters (e.g., how large the pool should be, how many positives are expected) to create a matrix that shows which sample to add to which test. The software is public domain, with different versions in different languages (a computing language and an open-source language that doesn't require any commercial software to use). The PP software (polynomial pools) is already being used in bioinformatics and genome sequencing, which uses similar testing strategies.
Ultimately, we hope that students will use the software and become enthusiastic about continuing the work. I’m sure there are many ways to improve the software and make it more practical and efficient. We are open to collaborations!
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# Thread: Ring in the field of complex numbers
1. ## Ring in the field of complex numbers
Hello,
let k be a field. On the set
$\displaystyle k \left[ i \right] := \lbrace\left(x,y\right) \mid x,y \in k\rbrace$
$\displaystyle \left( x,y \right) + \left( x',y' \right) = \left(x+x', y+y' \right) ~ \forall x,x',y,y'$
and the multiplication
$\displaystyle \left( x,y \right) \cdot \left( x',y' \right) = \left( xx' -yy', xy' + x'y \right) ~ \forall x,x',y,y' \in k$
Proof that k[i] is a ring with the relations above.
I have a approach but I am not sure.
For proofing it, I guess I have to show that it is a commutative Group, that the multiplication is associative und the distributive laws.
So I want to start with the first. Showing that it is a Abel's group.
For the group I have to proof the existence of neutral elements, the existence of one inverse element, the associative law and the inner composition "+" must be commutative.
Okay, so far but how do I show this concretely?
thanks
all the best
2. Originally Posted by Herbststurm
Okay, so far but how do I show this concretely?
thanks
all the best
Look at the definitions that it needs to satisfy to be a ring.
And now see if they work out.
Post your work here so that we can correct it if you get unsure of what to do.
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## Radon washout (2)
In August 2013 I wrote a small comment on ambient air radioactivity peaks coincident with a sharp rain fall pulse. Several specialists like A. Kies and M. Severijnen confirmed that these observations show a wash out of the radioactive daughters of the ubiquitous radon gas. This year a similar event happened the 21th April 2014:
Note that the 2 mm rain fall pulse triggered a radiation rise of about 20 nSv/h above the base level. One also sees that the much lower rain fall pulse (0.8 mm) which happened the next day does only a minor radiation peak.
Several interesting questions can be asked:
1. Is there a minimum time between two rain fall pulses needed to cause strong radiation peaks?
Or in other words: how long does it take for the atmosphere aerosol content to recover after a first washout?
The observations made in August 2013 suggest that 1 day could be enough:
The first rain pulse (1.2 mm) causes a radiation peak of about 14 nSv/h; the second much stronger rain pulse (2.2 mm) triggers a radiation peak of comparable amplitude: this could be a hint that the atmospheric aerosol load has not quite recovered to the previous level (which would be the base line after 3 dry days).
A week in September 2013 gives a similar picture:
Here we see two strong similar rain pulses of about 4 mm separated by about 48 hours: the first triggers a radiation peak of about 35 nSv/h. and the second of only 15 nSv/h. A close inspection shows that this second rain pulse actually is double, a first of 2.8 mm followed a couple of hours by a second of 4 mm. The radiation curve peak also shows these two peaks, with again the second (corresponding to a higher rain fall) less than the first.
The conclusion is that a simple relation-ship of the form radiation-peak = f(rain pulse) can not be established without respecting this atmospheric recovery time lapse.
2. Is the radiation peak proportional to the rain fall pulse?
Using only our miniscule set of observations from 2013 and 2014, let us just keep the (rain pulse, radiation peak) points separated by a minimum of 3 dry days:
The plot shows that the slope of the regression line is 7.09. i.e. the first rain-fall pulse of 1 mm causes on average a radiation peak of 7 nSv/h. The goodness of the fit is rather poor (R2 = 0.28), and this analysis shouts for more data!
Conclusion:
An analysis of the relationship between rain pulse and radiation peak intensities must respect the time lapse needed for the recovery of atmospheric aerosol load, possibly 3 dry consecutive days . This first short study suggests a possible linear relationship with a slope of 7 (nSV/h)/mm .
### 3 Responses to “Radon washout (2)”
1. Radon washout: two consecutive precipitation peaks | meteoLCD Weblog Says:
[…] Many times I wrote on this blog on radon washout: after a short downpour, we (nearly) always see a visible peak in our gamma radiation, caused by a washout of the daughters of the noble gas radon which is a natural constituent of our atmosphere; to find these comments enter “radon” into the search window on this site or click here , here and here. […]
2. Дозиметр своими руками | Многобукфф Says:
[…] washout […]
3. Radioactivity and precipitation | meteoLCD Weblog Says:
[…] to radon washout, the ambient gamma radiation shows sometimes impressive peaks ( see here, here, here, here, here, […]
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Investing for Multiple Period
To measure the performance of investment, it is necessary to evaluate the investment for various time periods. It can be determined for multiple periods. The investment for multiple periods is calculated using compounding process. Compounding interest means earning interest on interest. It is different from simple interest as interest is reinvested in compound interest. Reinvesting is the process of investing the profits of investing.
The formula for evaluating the future value of can be expressed as:
Where:
A= Future value, value of money at the specified time
P= Present Value, amount of invested money
r = Annual interest rate
n = Number of times the interest is compounded per year
t = number of years
Example:
Suppose, one invest $100 in a saving account at 10% annual interest rate. What will be amount of money after 2 years if the rate of interest does not vary? One will earn$ 10 interest after one year and $11 interest at the end of second year. Total amount of money or future value of money will be$ 121 ($100 +$10 + \$ 11) in 2 years at the rate of 10% interest.
Transtutors.com is well-recognized for providing accurate homework help at the affordable prices. It provides detailed answers related to your finance queries for your homework or assignment.
Related Topics
All Finance Topics
More Q&A
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# 9 Common Length Conversions You Need to Know Before High school
Length conversion and other measurement conversions are necessary, especially for scientific-minded people. Measurement conversions also help us to represent figures in a more natural and easy-to-understand form. However, there are basic conversion rules that you should know to make your conversion calculations simpler.
Firstly, if you want to convert a smaller unit into a larger one, you should divide it. Similarly, if you are converting larger units into smaller units, you are supposed to multiply; for instance, to convert nanometer to meter, you divide. Below are some length conversions a student should know before joining high school.
## a) Nautical Miles into Miles
A nautical mile is a common metric used in the measurement of distances in air and water navigation. A nautical mile is slightly longer than a mile on the road and is defined as a movement of one minute of latitude. One nautical mile is approximately 1.15 miles on the ground; therefore, in the conversion of nautical miles to miles, multiply the with 1.15 and, to convert back miles to nautical miles, divide by the same value.
## b) Nanometer to Meter
A nanometer is one of the minor units of measurement, but that doesn’t mean that these units are not used in practical cases. One billion nanometers make one meter, so to convert one nanometer to a meter, you divide by 1000000000. And to convert meters to nanometers, you multiply by the same value.
## c) Centimeter to Meter
Unlike nanometer to meter conversion, which is likely done in scientific projects, we carry out centimeter-to-meter conversions almost daily in classrooms and at home. Now it’s 100 centimeters, which makes one meter, so to convert centimeters to meters, divide by 100.
## d) Meter to Millimeter
Sometimes, due to the high accuracy needed in your measurements or calculations, you may want to express your values in millimeters and not meters. One meter is made up of 1000 millimeters, so you multiply your values in meters by 1000 to get the result in millimeters.
## e) Kilometer to Mile
Most people may not be familiar with miles when measuring long distances, so you need to convert miles to kilometers for your measurements to be more realistic. A mile is approximately 1.609344 km, so to convert miles into kilometers, you multiply by 1.609344, the vice versa; you divide by the same value.
## f) Kilometer to Meter
Kilometer-to-meter conversions are also so common in our daily lives when sporting, taking field measurements, or even driving on the road. To convert kilometers into meters, multiply the measurements by 1000, and vice versa; you divide by the value.
## g) Nautical Mile to Kilometers
You will find nautical mile-to-mile conversions more useful in marine and air navigation than other conversions, such as nanometer to meter, primarily applicable for small measurements. To convert nautical mile to kilometer, multiply the value by 1.852; to convert kilometers back to nautical miles, divide by the same value.
## h) Inch to Millimeter
Inches are practical measurement units, especially in buildings and constructions. However, some may not be familiar with inches, so you have to convert them to millimeters. To convert inches to millimeters, multiply the measurements by 25.4.
## i) Centimeter to Feet
Feet and centimeters are also very common units for measuring small lengths in our daily activities. Most students are pretty familiar with centimeters. So to convert centimeters to feet, divide the measurement by 30.844.
## Summary
There are numerous length conversions that a student should know before joining high school. Some of these conversions include; nanometer to meter, nautical miles to miles, centimeter to meter, kilometer to meter, and many more. However, you can use the various online conversion calculators for more accurate conversions.
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Python: Combine two dictionary adding values for common keys - w3resource
# Python: Combine two dictionary adding values for common keys
## Python dictionary: Exercise-19 with Solution
Write a Python program to combine two dictionary by adding values for common keys.
Sample Solution:
Python Code:
``````# Import the 'Counter' class from the 'collections' module.
from collections import Counter
# Create two dictionaries 'd1' and 'd2' with key-value pairs.
d1 = {'a': 100, 'b': 200, 'c': 300}
d2 = {'a': 300, 'b': 200, 'd': 400}
# Use the 'Counter' class to create counter objects for 'd1' and 'd2', which count the occurrences of each key.
# Then, add the counters together to merge the key-value pairs and their counts.
d = Counter(d1) + Counter(d2)
# Print the resulting merged dictionary 'd'.
print(d)
```
```
Sample Output:
```Counter({'b': 400, 'd': 400, 'a': 400, 'c': 300})
```
Python Code Editor:
What is the difficulty level of this exercise?
Test your Programming skills with w3resource's quiz.
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# -2(-2y+1)-(2y-2)+6y+2
how do i solve this problem?
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SPOJ
(Difference between revisions)
SPOJ (Sphere Online Judge) is an automated programming contest website. It has a large problem set archive, and accepts solutions in many languages, including Haskell. Solving the problems can be a interesting way to learn Haskell.
Revealing answers to questions is generally frowned upon as it enables cheating, but giving example solutions to a few of the simpler questions is OK as it encourages newcomers to get started on SPOJ. This page covers techniques for dealing with problems efficiently. It accumulates some earned wisdom about writing Haskell programs which overcome some of the technical obstacles in SPOJ. These are not spoilers, hopefully, but useful tips in general about auxiliary issues.
Contents
Many Online Judges support just a few languages such as C, C++, Pascal and Java. SPOJ is unique in that it supports many (currently over 30) languages, including Haskell. Most problems have a time limit which is calculated to be a comfortable time when solving the problem in C. But it can be a challenge (and sometimes impossible) to solve the problem, within this time limit, using other languages.
2 Getting Started: Solution to TEST
TEST description: the input consists of one number per line, the program should echo each number until 42 is reached, at which point the program should exit.
main :: IO ()
main = mapM_ putStrLn . takeWhile (/="42") . lines =<< getContents
or
main :: IO ()
main = interact f
where f = unlines . takeWhile (/="42") . words
3 I/O
3.1 Introduction
For SPOJ purposes, the most common operation is reading some kind of list of Ints.
The standard Haskell IO using System.IO and [Char] (or String) is slow and will make answering some problems difficult or impossible. Luckily, there is a faster IO alternative - Data.ByteString.
ByteStrings come in two varieties: Strict (default), and Lazy. Strict ByteStrings, in brief, are implemented by a foreign pointer to a block of memory which is filled by low level operations. The ByteString data type points to this memory region and also contains two Ints which track relative position and length. This means that many operations on ByteStrings can re-use memory and merely manipulate the two Ints.
Data.ByteString.Lazy is a layer on top which provides the ByteString API, but now the data is only read in chunks of Strict ByteStrings when necessary.
Don Stewart and others have put a great deal of excellent work into the library to ensure that the high-level interface will be optimized into efficient code. However, it may behoove you to inspect the source code for yourself, which is quite accessible and available in the GHC repository under libraries/base/Data/.
The module is normally imported qualified because it shares many names with standard Prelude functions. I use the prefix SS in my examples, for Strict byteString.
3.2 Span and Break
The span and break functions available from ByteString can be used to parse the occasional non-numeric value in input. For example, the following code skips a line:
import qualified Data.ByteString.Char8 as SS
-- ...
skipLine cs = snd \$ SS.span (== '\n') cs'
where (_, cs') = SS.break (== '\n') cs
The nice part about span and break is that when it is found in the form span (== c) for all characters c, it is optimized down to a simple memchr.
3.3 Lazy ByteStrings
In general, Strict ByteStrings are more than adequate. There are cases where Lazy ByteStrings are better - when a file is enormous memory usage will be lower using Lazy ByteStrings because it does not read in all the file at once. You can use Lazy ByteStrings as a drop-in replacement for Strict, mostly. A useful part of Lazy ByteStrings is the function called toChunks. This function provides an interface to the lower-level portions which actually operate by reading chunks of data into Strict ByteStrings acting as buffers. The chunks are kept in a lazy list, so each chunk is only read on-demand. However, you are getting the data as efficiently as the library deems possible.
3.4 Enormous Input: Solutions to INTEST
INTEST description: the first line of input contains two numbers: n and k. The input then consists of n lines of numbers. The output of the program should be the count of the numbers which are divisible by k.
Solution 0: Use System.IO getContents with [Char]/String - is very slow.
import qualified Data.List as DLi
main :: IO ()
main = do
(w1:w2:ws) <- words `fmap` getContents -- all IO in one go
print \$ DLi.foldl' (f1 k) 0 \$ map readInt \$ take n ws
where
f1 :: Int -> Int -> Int -> Int
f1 k acc ti
| mod ti k == 0 = acc+1
| otherwise = acc
Solution 1: A straightforward ByteString solution which achieves a reasonable time.
import qualified Data.ByteString.Lazy.Char8 as BS
main :: IO ()
main = do
(l:ls) <- BS.lines `fmap` BS.getContents -- all IO in one go
let [n,k] = map readInt (BS.split ' ' l)
print . length . filter ((== 0) . (`mod` k) . readInt) \$ take n ls
case BS.readInt x of Just (i,_) -> i
Nothing -> error "Unparsable Int"
Solution 2: A faster version which reads the integers directly from the ByteString (thanks to Eugene on Haskell-cafe).
{-# LANGUAGE BangPatterns #-}
{-# OPTIONS_GHC -O2 -optc-O2 #-}
import qualified Data.ByteString.Lazy as BLW
import qualified Data.Word as DW
main :: IO ()
main = do
(l, ls) <- BLW.break (==10) `fmap` BLW.getContents
let [_, k] = map readInt . BLW.split 32 \$ l
answer :: Int -> BLW.ByteString -> Int
answer k = fst . BLW.foldl' f (-1, 0)
where
f :: (Int,Int) -> DW.Word8 -> (Int,Int)
f (!acc, !x) 10
| x `mod` k==0 = (acc+1, 0)
| otherwise = (acc, 0)
f (!acc, !x) c = (acc, 10 * x + fromIntegral c - 48)
readInt = BLW.foldl' (\x c -> 10 * x + fromIntegral c - 48) 0
Solution 3: A faster version which uses ByteString.Lazy.Char8 chunks (thanks to Don's original post on Haskell-cafe, and a speed-up patch from Eugene)
{-# LANGUAGE BangPatterns #-}
{-# OPTIONS_GHC -O2 -optc-O2 #-}
import Data.Char
import Data.Maybe
import qualified Data.ByteString.Unsafe as BU
import qualified Data.ByteString.Char8 as S -- 'Strict'
import qualified Data.ByteString.Lazy.Char8 as L -- 'Lazy'
main :: IO ()
main = do
ss <- L.getContents -- done with IO now.
let (l,ls) = L.break (=='\n') ss
-- don't need count, we're allocating lazily
k = fst . fromJust . L.readInt . last . L.split ' ' \$ l
file = L.toChunks (L.tail ls) -- a lazy list of strict cache chunks
print \$ process k 0 file
-- Optimised parsing of strict bytestrings representing \n separated numbers.
--
-- we have the file as a list of cache chunks align them on \n boundaries, and
-- process each chunk separately when the next chunk is demanded, it will be
process :: Int -> Int -> [S.ByteString] -> Int
process _ i [] = i
process k !i (s:t:ts)
| S.last s /= '\n' = process k (add k i s') ts'
where
(s',r) = S.breakEnd (=='\n') s
(r',rs) = S.break (=='\n') t
ts' = S.concat [r,r',S.singleton '\n'] : BU.unsafeTail rs : ts
process k i (s: ss) = process k (add k i s) ss
-- process a single cache-sized chunk of numbers, \n aligned
add :: Int -> Int -> S.ByteString -> Int
add k i s = fst \$ S.foldl' f (i, 0) s
where f (!a, !n) '\n' | mod n k == 0 = (a+1, 0)
| otherwise = (a, 0)
f (!a, !n) w = (a, 10*n+ord w-ord '0')
4 Arrays
From time to time you may want to write an algorithm using Mutable Arrays in Haskell. While it may be tempting to jump right into arrays, you should consider that the Data.IntMap and Data.IntSet libraries are actually very fast and convenient. You may be surprised to find that they are more than sufficient. In fact, I would argue that unless you plan to use Unboxed Arrays, you should stick with these.
If you are set on using a mutable array, go ahead and read the wiki page Arrays for a summary of the choices available. I am going to assume that you know the basics as imparted on that page. In particular, I'll be talking about ST and IO arrays.
IO arrays are probably the easiest mutable arrays to use. They are fairly straightforward, but require you to stay in the IO monad. I made the mistake of avoiding IO arrays originally, but through some strange fluke, it turns out there are many cases where IO arrays still have the best performance.
ST arrays have the advantage that they can be easily constructed in monadic code, but then returned to the functional world (without invoking voodoo like unsafePerformIO). However, ST itself can make the resulting type expressions confusing. In fact, it appears that you cannot type-annotate certain expressions in ST monadic code without going outside Haskell'98.
Personally, I try to write my mutable array code monad-agnostic. This way I can compare IO array performance to ST array performance. However this can really make the type annotations confusing if you haven't seen it before!
After seeing a few examples, though, I think anyone can pick up on the pattern. The usage isn't that difficult, once you see how the types fit together.
4.1 Creating a mutable array
There is a set of functions defined to work with mutable arrays separate from those that work on immutable arrays. You will want to import Data.Array.{ST,IO} depending on what you use.
a <- newArray (1, 10) 0
newArray expects bounds to be supplied just like for immutable arrays. The second argument is the initial value with which to initialize all the elements. A simple example of usage:
runSTArray (newArray (1, 10) 0)
The runSTArray function is an efficient helper function which you use when you want to create an array in ST but return an immutable array. It also happens to be a special case where newArray doesn't need a type-annotation.
In general, you will have to supply some kind of type-annotation for newArray. Let's suppose you are using Unboxed Arrays and want to write the same code.
runSTUArray (newArray (1, 10) 0)
But now GHC will spit at you an incomprehensible error message. Well it's not so bad actually, it says,
No instance for (MArray (STUArray s) t (ST s))
GHC has the wrong idea about our array. We need to inform it that our array is indexed by Ints and contains Ints.
runSTUArray (newArray (1,10) 0 :: ST s (STUArray s Int Int))
Remember the type of newArray:
newArray :: (MArray a e m, Ix i) => (i, i) -> e -> m (a i e)
The result must be in a monad. That monad is ST s. s is the type-variable which represents the "state" which must remain contained within the monad. But the STUArray needs to know about that "state." The type-signature expresses that the monad and the array are sharing this "state."
At this point, you are pretty much ready to do anything with arrays. I'll show you one more example which uses the more general function runST.
runST (do
a <- newArray (1, 10) 1 :: ST s (STUArray s Int Int)
mapM_ (\ i -> do
x <- readArray a (i - 2)
y <- readArray a (i - 1)
writeArray a i (x + y))
[3 .. 10]
a' <- unsafeFreeze a
return (v, a' :: Array Int Int))
This code computes the first 10 Fibonacci numbers and returns a pair containing the 10th and array turned immutable. Some notes:
• unsafeFreeze turns the array immutable in-place and therefore renders it unsafe to perform any mutations after the fact. If you don't change the array after unsafeFreeze, it's perfectly fine. runSTArray uses this internally.
• Second, I had to put the type-annotation for a' on the final line instead of the binding line. This is one of those little annoyances about ST which I mentioned earlier. The reason I did not say a' <- unsafeFreeze a :: ST s (Array Int Int) is because GHC views this s as being different from the one on the first line. Clearly, GHC can't allow you to mix "state" from different places. Of course, we know that's not happening, but without lexically scoped type variables, GHC can't discern.
• Don't do runST \$ do .... It doesn't work (at the moment). The reasons are obscure, having to do with unification and higher-rank polymorphism. Do runST (do ...) instead.
Hopefully at this point you can see that it isn't difficult to get mutable arrays, so long as you know how the type annotations are supposed to go. IO arrays have simpler type annotations, because the type-variable s is not necessary. So you can write the above example as:
do
a <- newArray (1, 10) 1 :: IO (IOUArray Int Int)
mapM_ (\ i -> do
x <- readArray a (i - 2)
y <- readArray a (i - 1)
writeArray a i (x + y))
[3 .. 10]
a' <- unsafeFreeze a :: IO (Array Int Int)
return (v, a')
4.2 Generalized mutable arrays
Mutable arrays are instances of the typeclass MArray. MArray a e m is a multi-parameter type-class where a is the array-type, e is the element-type, and m is the monad.
Sometimes you will want to write your code so that it can operate on any mutable array. Generally, type-inference will be more than capable. The types generated can be a bit confusing, though. Other times, you may want to create a datatype, in which case you will need to know how to specify it for MArrays.
Here is an example which sums up an array of Ints:
sumArray a = do
(s,e) <- getBounds a
let loop sum i
| i > e = return sum
| otherwise = do
loop (sum + x) (i + 1)
loop 0 s
when we ask the type of this function we get a pretty hefty answer:
sumArray :: (MArray a e m, Num e, Ix i, Num i) => a i e -> m e
Our function works on any mutable array type a, with index type i, element type e, and in monad m.
We can create a datatype which contains a mutable array by simply satisfying the kind requirements for MArray:
data IntArray a = IA (a Int Int)
sumIntArray (IA a) = do
(s,e) <- getBounds a
let loop sum i
| i > e = return sum
| otherwise = do
loop (sum + x) (i + 1)
loop 0 s
But keep in mind, you still may need to supply some kind of type annotation when you construct the datatype, so that newArray knows which kind of array to create.
4.3 Unsafe Array Access
squeeze out a bit more performance:
unsafeWrite
and
. Just
and use these functions in place of
writeArray
and
when you feel it is safe. The first difference is that the normal array functions check array-bounds before doing anything. You can crash your program using the unsafe ones, if you access an index out-of-bounds. The second difference is that the unsafe functions do not perform any Index-arithmetic. They expect to be handed an array index that has already been converted into a zero-based, flat system. If you already use zero-based flat arrays, great. Otherwise, you may have to do a small bit of arithmetic first.
5 Garbage Collection
SPOJ has a fairly rigid framework for running programs, naturally. While it is, thankfully, possible to specify compiler options inside Haskell programs, the same is not true for RTS options. Unfortunately, sometimes there are circumstances where you would like to tweak runtime options like on the Garbage Collector. There is one technique that I have found which will allow you to tune the GC for your programs when they run on SPOJ.
The gist of it is to restart the program using System.Posix tools. The major hurdle is finding the location of the program executable
so that it may be supplied to the
executeFile
function.
The following code achieves this:
{-# OPTIONS -ffi #-}
-- or -fglasgow-exts
import System.Environment (getArgs)
import System.Posix.Process (executeFile)
import Foreign.C.Types (CInt)
import Foreign.C.String (CString, peekCString)
import Foreign (peek, alloca, peekElemOff, Ptr)
main = do
flags <- getArgs
progname <- getFullProgName
if null flags
then
-- Supply an "argument" so that flags will not be null.
-- RTS option -A100m will increase the allocation area size
-- to 100 megabytes.
executeFile progname False ["r","+RTS","-A100m"] Nothing
else
realMain
-- Now the trickier part: getProgName in GHC does not return the
-- full path, for "portability" reasons. SPOJ does not run
-- programs from the current directory. That means we need to
-- find the full path to the program some other way.
foreign import ccall unsafe "getProgArgv"
getProgArgv :: Ptr CInt -> Ptr (Ptr CString) -> IO ()
-- As it turns out, the C function which getProgName uses actually
-- does return the full path. But then getProgName cuts it out
-- before returning it. This is a version of getProgName which
-- leaves the full path intact.
getFullProgName :: IO String
getFullProgName =
alloca \$ \ p_argc ->
alloca \$ \ p_argv -> do
getProgArgv p_argc p_argv
argv <- peek p_argv
s <- peekElemOff argv 0 >>= peekCString
return s
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https://learn.careers360.com/ncert/question-if-a-tissue-has-at-a-given-time-1024-cells-how-many-cycles-of-mitosis-had-the-original-parental-single-cell-undergone/?question_number=9.0
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#### If a tissue has at a given time 1024 cells, how many cycles of mitosis had the original parental single cell undergone?
Ans.
210 = 1024
The above equation proves that a cell must undergo 10 rounds of mitosis to produce 1024 cells.
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# Limiting Reactants Lab Report
Superior Essays
The purpose of the experiment was to observe limiting reactants by using a fixed amount of one reagent along with a varying the amount of the second reagent used in a chemical reaction. Through these results, the concept of limiting reactants can be determined. Using a graph to demonstrate the relation between the volumes of hydrogen gas produced versus the moles of the reagents will help figure out the stoichiometric balance required. The success of the experiment will be determined by the comparing the experimental values with the fixed values.
Introduction
Stoichiometry involves converting chemical formulas and equations that represent individual atoms, molecules, and formula units to the laboratory scale that uses milligrams, grams, and kilograms, of these substances (1). A limiting reactant is the reactant that has been completely consumed in a chemical reaction. It is
A graph of the volume of hydrogen gas produced versus moles of a reagent that is varied will be able to tell you the point in which the stoichiometric balance is achieved and what the limiting reactant is. In this experiment group A used a constant mass of magnesium metal (0.100 g) and varying volumes of 2.00 M hydrochloric acid added from 2.00 mL to 7.00 mL in progressions of 1.00 mL which in total came out to be six distinctive chemical reactions. Group B used the consistent volume of 2.00 M hydrochloric acid (5.00 mL) and changed the mass of the magnesium metal used from 0.070 g to 0.170 g in growths of 0.020 g which resulted in a total of six different chemical reactions as well. The percent yield for magnesium, according to the data, was 107.3% and the percent yield for hydrochloric acid, according to the data, was 96.0%. Overall, the experiment was fairly successful as observed by the percent yields of both
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# Ae311_fall_2011_hw_4 - Prof Daniel J Bodony AE311 Fall 2011...
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Prof. Daniel J. Bodony AE311, Fall 2011 HW #4 Due 5 pm Wednesday, October 5, in the AE311 dropbox Problem 1 Recall Problem 3 from HW2 where you were given an approximate form of the velocity profile in a boundary layer for the flow of water over a flat plate with zero pressure gradient: u ( x , y ) U = 2 parenleftBigg y δ ( x ) parenrightBigg 2 parenleftBigg y δ ( x ) parenrightBigg 3 + parenleftBigg y δ ( x ) parenrightBigg 4 0 y δ ( x ) 1 y > δ ( x ) Assuming v ( x , y ) 0, w 0, and that δ ( x ) = A x , where A is a constant, compute the skin friction coe ffi cient as a function of x . What value of A do you need to match the drag you computed in Problem 3, HW2, using the integral momentum equation? Compute and plot the z -component of the vorticity, ω z . Is the flow rotational or irrotational? Sketch what happens to an initially square fluid element with time. Problem 2 Consider the motion of a fluid in one direction only with velocity field u = ( u ( x ) , 0 , 0) T . Calculate the
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# 6+ Diy 4th Grade Math Eog
Ad Step-by-Step Learning Path designed to help kids learn math reading science and more. May reproduce for instructional and educational purposes only.
### Not for personal or financial gain.
4th grade math eog. Resources on this page are organized by the Instructional Framework. At the end of the school year students must take an End-of-Grade EOG test to evaluate their mastery of the concepts. It is aligned to 4th grade math common core.
EOG Grades 38. IIIIV EOG READING EOG MATH 410412 IRVING PARK ELEM B B 26 8077 27 8519 410412 IRVING PARK ELEM I I A Asian 410412 IRVING PARK ELEM M M B Black 410412 IRVING PARK ELEM W W 36 8611 36 9167 I American Indian H Hispanic. In 4th grade math vocabulary words can be tricky to remember.
A 80064 B 80640 C 86400 D 86404 D 2. 3rd Grade EOG Math Spring 2007. Which estimate best describes the amount of gas Mr.
Back to North Carolina Elementary School. Fourth Grade EOG Breakdown. Are the fractions equivalent.
Games puzzles and other fun activities to help kids practice letters numbers and more. Students that dont do well on the 4th grade EOG test may not be ready for 5th grade math. A less than one gallon B between 2 and 3 gallons.
Students will be bringing a take home test on WednesdayThursday. 3rd Grade EOG Math Spring 2010. Email ThisBlogThisShare to TwitterShare to FacebookShare to Pinterest.
Harbin would use in a five-day work week. May reproduce for instructional and educational purposes only. At September 22 2020 No comments.
A measure of how much liquid a container can hold. This is a Math EOG Review game that is compatible with Active Inspire software. EOG Mathematics Grade 4 Released Form.
Math EOG Test Prep. 25 questions Connor Declan Ryan Clara Mr. EOG Grades 38 Reading.
Study Islands focus on the Standard Course of Study enables students to improve their performance in all skill areas tested on the EOGEOC Tests in grades 3 through 8 and high school. Damaiya and Laura have equalsized pizzas. 3rd Grade EOG Math Spring 2009.
There are eighty-six thousand four hundred seconds in a day. An angle that measures less than 90 degrees. An arrangement of objects or pictures in.
Work on Multiplication skills. Get thousands of teacher-crafted activities that sync up with the school year. The words come directly from the 4th grade common core math standards.
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The purpose of this document is to connect and sequence. Keep reading to find out how you can help your 4th grader prepare for this test. Games puzzles and other fun activities to help kids practice letters numbers and more.
GRADE 4 MATHEMATICSRELEASED FORM 3 Go to the next page. 4th Grade EOG Math Practice and Test Prep Try it for free. IIIIV Students Tested Lev.
3 Each day of the work week Mr. Not for personal or financial gain. They are mixed up.
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Students will work on Time this week. This product which contains 80 math vocab words and 4 quizzes is a great way to help students remember their words AND understand their meaning. North Carolina Testing Program EOG Mathematics Grade 4 Sample Items Goal 1 Page 1 Published November 2003.
Click on the Cluster in the table below to be taken to the resources pages for lessons tasks and additional resources for teaching the NC Mathematics Standard Course of Study. 30 questions 4th Grade Math 4th Grade Writing 4th grade RANDOM 4th Grade Science 4th Grade Spelling School Info. Study Island also offers EOG test preparation and Math and Reading Skills.
North Carolina has set math standards for grades 3-8. How many of each brand of DVD player did the store sell. North Carolina Department of Public Instruction 301 N.
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North Carolina Testing Program EOG Mathematics Grade 4 Sample Items Goal 3 Page 1 Published November 2003. The store sold 8 different brands of DVD players. Discover the most effective and comprehensive online solution for curriculum mastery high-stakes testing and assessment in.
There are 3 tic-tac-toe games that can be played 9. These tests are in Virginia Standards. Harbin uses 3 4 of a gallon of gas.
EOG Grades 38 Reading will be available upon the North Carolina State Board of Educations adoption of academic achievement standards and cut scores for reading expected August 2021. How else could this number be written. Yes because 16 28.
1 1 2 6 24 120. 4th Grade Math EOG Vocabulary. No because 16 28.
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Get thousands of teacher-crafted activities that sync up with the school year.
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# Can two cipher letters per plaintext letter easily defeat character frequency analysis?
For a class 5 years ago I wrote a paper about "defeating character frequency analysis by using two cipher letters per plaintext letter" (jamesjava.blogspot.com/2009/08/defeating-character-frequency-analysis.html).
Quote:
Using two letters in the cipher text for each letter in the plaintext can be a good way to create a flat character distribution.
The algorithm is to partition the 676 2-letter combinations based on the standard character frequency. i.e. if the standard frequency for a letter is 5% then it will get 5% of the 2-letter combinations (randomly selected). This doubles the size of the data, could include spaces & punctuation, and makes a much larger key.
Note that some letters may get dropped because they occur less than 1/676 (0.15%) of the time. Both 1-gram and 2-gram frequency analysis produce a nearly uniform histogram (variation appears to only be caused by rounding). Two-gram results: P&P=5,117%; SH=5,013%.
Therefore this technique was extremely effective with no obvious weaknesses.
I didn't get much feedback from the professor so I wonder if anyone can comment on this and tell me if my conclusion is correct.
• Sounds like a homophonic substitution cipher. Commented Jun 18, 2013 at 21:36
• An extensive analysis of homophonic substitution ciphers can be found in this paper: Efficient Cryptanalysis of Homophonic Substitution Ciphers
– Reid
Commented Jun 19, 2013 at 14:57
• Could you copy the relevant paragraph from your paper into your question? This makes it easier to reference it in an answer. Commented Jun 20, 2013 at 18:33
• Given sufficient ciphertext, you can still do frequency analysis. All you did was make the alphabet much larger. Commented Jul 3, 2013 at 6:36
• @JamesA.N.Stauffer Given a large enough ciphertext, bigrams and trigrams may still be analyzed. English text consists of roughly 1.5% "th" bigrams, for instance. Based on character frequency, there would be about 2,680 pairs representing "th". You'd need a plaintext large enough to reliably detect quartets occurring 0.00055% of the time, which isn't really all that large. Commented Jul 17, 2013 at 22:47
• Now there are $61*41=2501$ cipher quadruples for the original bigram th, and all of them have the same probability. However, $2501$ of $26^4 = 456976$ possible qudruples is just a fraction of 0.005473. So now the number of ciphertext quadruples for *th does not match required number to enforce a uniform distribution.
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# Math quiz for 7th graders
Math Games. In 7th grade, students develop their ability to reason quantitatively and abstractly. With Math Games, pupils get to master this skill while playing accessible, engaging games.
The Videos, Games, Quizzes and Worksheets make excellent materials for math teachers, math educators and parents. Math workbook 1 is a content-rich downloadable zip file with 100 Math printable exercises and 100 pages of answer sheets attached to each exercise. This product is suitable for Preschool, kindergarten and Grade 1.The product is available for instant download after purchase.
Learn seventh grade math for free—proportions, algebra basics, arithmetic with negative numbers, probability, circles, and more. Full curriculum of exercises and videos.
Preview this quiz on Quizizz. The class has 12 girls and 16 boys. What is the ratio of total students to boys? (simplest form) 7th Grade Math Review DRAFT. 6th - 7th grade. 16345 times. Mathematics. 64% average accuracy. a year ago. aweisman. 30. Save. Edit. Edit. 7th Grade Math Review DRAFT.
These free interactive math worksheets are suitable for Grade 7. Use them to practice and improve your mathematical skills. Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations. You can use the free Mathway calculator.
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## Grade 7 - Practice with Math Games.
You did okay on the 7th grade science quiz, but there's still more to learn to pass that class. Ann Cutting, Getty Images Good job! You knew many of the answers to the 7th grade science quiz. If you think you're up for the challenge, see how you do on the 8th grade science quiz. Ready to put your knowledge to use?
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These materials enable personalized practice alongside the new Illustrative Mathematics 7th grade curriculum. They were created by Khan Academy math experts and reviewed for curriculum alignment by experts at both Illustrative Mathematics and Khan Academy.
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https://www.mrexcel.com/board/tags/complicated-formula/
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# complicated formula
1. ### Is it possible to insert unique formulas into cells based on a product code?
Hi everyone,I do have a complicated problem to solve and I was wondering if anyone can point me to the right direction as to how to tackle this problem.So I've got 5 product codes, and I've formulated 5 different unique excel formula codes for each product. For example,Product 1codes to...
2. ### Help using complex lookup to populate table
I'm not sure if this is even possible, but I have a table with approximately 4K rows, with each row listing a student and the grade that they received in a particular course. <tbody> A B C 1D 1 Student # Name Course Grade 2 12345 Jack ENGL 101 3.3 3 12345 Jack MUSIC 200 3.7 4 12346...
3. ### Is there a formula for that??
I have been searching high and low for a formula that will meet my needs but cant find one! I appreciate any help or guiding light! I have a spreadsheet set up for my grocery list. Column A contains the name of the product, Column B contains the estimated price of that item, at the very bottom...
4. ### Isolating dates for comparison
I have some shares and i want to do some comparisons. eg. This month vs last month, this week vs last week, this week vs last 6 weeks ect I have been experimenting with the advanced filter but keep bringing my self to confusion. For now what i need is to extract data for each year, month, and...
5. ### IF Formula (Or maybe something else...)
http://www.box.net/shared/0ebhd83tg6 Hi Guys So heres what I want to do. I need to link up the PO's and part values on the Stock out and relevant values to the corresponding PO's on the Consignment stock sheet. (It makes sense when you look at the workbook) For example Column B on the Stock...
6. ### Complicated Formula Question
I have a complicatedd problem that I am working on. 1 want to have a cell where I have a number and be able to take that number and Add to it itself + a percentage, however the problem is taht I need to be able to do this multiple times Example. Take 2048 as main number with 10% as...
7. ### Complicated Excel Formula
For my spreadsheet, I am designing a program where the user can enter what they have eaten every day, and it will then give the results at the end of the week. However, my input table looks very cumbersome; it is very long incase the user needs to input a lot of food. What I want to be able to...
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## Jacques Hadamard on "Who discovered the calculus"
In An Essay on the Psychology of Invention in the Mathematical Field (Princeton University Press, 1945), Jacques Hadamard looks at the question: "Who discovered the Infinitesimal Calculus." We present here a version of his argument. Perhaps it is worth considering as we read, given the emphasis Hadamard puts on the proof that integration is the converse of differentiation, to ask where James Gregory should fit. What is certain is that Gregory was the first to publish a proof of this fundamental theorem of the calculus. Hadamard writes:
Heraclitus's profound idea that everything ought to be considered in its "devenir" i.e. in its continuous transformation, had not been understood during Antiquity. Only in the fourteenth century A.D. did one of the greatest medieval thinkers, Nicole Oresme, notice that the rate of increase or decrease of a quantity is slowest in the neighbourhood of a maximum or minimum. The bearing of such a remark was not perceived by anybody, including Oresme himself, to whom it did not appear that such a fundamental idea had to be developed.
Three centuries later the same principle was enunciated by Johannes Kepler, but Kepler went no further than Oresme had already gone before him; the discovery went only halfway.
Then, in Fermat's hands, the principle received mathematical expression. In several instances, considering one quantity in terms of another (such as time), Fermat used a mathematical operation which gave zero as the result if the quantity in question was a maximum or minimum. Moreover, the same method allowed him to find tangents to several curves considered by his contemporaries.
The operation performed by Fermat is precisely what we now call differentiation. Does this mean, as many are inclined to think, that he invented the Differential Calculus? In one sense we must answer "yes," for we see him applying his method to various problems, and even pointing out that the method could he applied to similar ones. But in another sense we must say "no," for the method he used appeared to nobody in his time, not even to himself, as a general rule for solving a whole class of problems, or as a new conception the properties of which deserved further investigation. Adapting an expression of Poincaré, we can say that things are more or less discovered, not discovered outright from complete obscurity to complete revelation. One step consists in acquiring the idea of a principle; another, if not several others, in giving a precise form to that idea and driving it far enough to be able to take it a starting point for further researches. In the present case, this was the work of Newton and Leibniz.
But the Differential Calculus is not the whole Infinitesimal Calculus. There is a second branch, the Integral Calculus, the fundamental operation of which is the valuation of plane curvilinear areas; and this implies a discovery which lay deep and had been entirely unsuspected, viz. the fact that integration is nothing else than the converse of differentiation. Who made that essential and difficult discovery? In the development of differentiation something of it was in the making. Torricelli and Fermat (perhaps even, though doubtful, Descartes) used methods which looked like the application of the principle and were indeed rather near it, but with a fundamental difference. I should say the same even of Barrow, Newton's master, in his 'Lectiones Geometricae', although the essential content of the No 11 of his xth lecture is really equivalent to that principle. Besides Descartes, who did not let the connection appear clearly, both Torricelli and Fermat treated special cases whose properties concealed the general principle; while for Barrow the true meaning of the principle was hidden by the meaning he gave to the notion of the tangent.
Thus Oresme, Keper, and Fermat failed to discover the Differential Calculus because they did not pursue their initial and fruitful ideas.
We see how psychological considerations can be illustrated by the history of science; and conversely, how they can help us to understand correctly a question which, very often, can be considered as a badly set one: Who is the author of such or such a discovery?
JOC/EFR July 2012
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This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A124582 Primes p such that q-p >= 6, where q is the next prime after p. 9
23, 31, 47, 53, 61, 73, 83, 89, 113, 131, 139, 151, 157, 167, 173, 181, 199, 211, 233, 241, 251, 257, 263, 271, 283, 293, 317, 331, 337, 353, 359, 367, 373, 383, 389, 401, 409, 421, 433, 443, 449, 467, 479, 491, 503, 509, 523, 541, 547, 557, 563, 571, 577 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS R. Zumkeller, Table of n, a(n) for n = 1..1000 K. Soundararajan, Small gaps between prime numbers: the work of Goldston-Pintz-Yildirim, Bull. Amer. Math. Soc., 44 (2007), 1-18. FORMULA a(n) ~ n log n. - Charles R Greathouse IV, Jun 04 2015 MAPLE d:=6; M:=1000; t0:=[]; for n from 1 to M do p:=ithprime(n); if nextprime(p) - p >= d then t0:=[op(t0), p]; fi; od: t0; MATHEMATICA t={}; q=6; Do[If[Prime[n+6]-Prime[n+5]>=q, AppendTo[t, Prime[n+5]]], {n, 1, 200}]; t (* Vladimir Joseph Stephan Orlovsky, Feb 02 2012 *) Transpose[Select[Partition[Prime[Range[200]], 2, 1], #[[2]]-#[[1]] >= 6&]] [[1]] (* Harvey P. Dale, May 15 2013 *) PROG (PARI) is(n)=!isprime(n+2) && !isprime(n+4) && n>2 && isprime(n) \\ Charles R Greathouse IV, Jun 04 2015 CROSSREFS Complement of A124589. Cf. A134099, A134100, A134101. Sequence in context: A256872 A276435 A083370 * A130796 A258578 A031924 Adjacent sequences: A124579 A124580 A124581 * A124583 A124584 A124585 KEYWORD nonn AUTHOR N. J. A. Sloane, Dec 19 2006 STATUS approved
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Last modified October 16 13:51 EDT 2019. Contains 328093 sequences. (Running on oeis4.)
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# Ordered Pairs, Intercepts, And Slopes Worksheet Page 15
Secton . — Ordered Pars, Intercepts, and Slopes
0
3. Calculate the slope for the following equations:
Find two ordered pair
Equation
Calculate the slope
Validate
solutions
a) 4x – 3y = 5
b) 6x + 3y = 18
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Cody
# Problem 94. Target sorting
Solution 1580425
Submitted on 11 Jul 2018 by Karthik S S
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
a = [1 2 3 4]; t = 0; b_correct = [4 3 2 1]; assert(isequal(targetSort(a,t),b_correct))
2 Pass
a = -4:10; t = 3.6; b_correct = [-4 -3 10 -2 9 -1 8 0 7 1 6 2 5 3 4]; assert(isequal(targetSort(a,t),b_correct))
3 Pass
a = 12; t = pi; b_correct = 12; assert(isequal(targetSort(a,t),b_correct))
4 Pass
a = -100:-95; t = 100; b_correct = [-100 -99 -98 -97 -96 -95]; assert(isequal(targetSort(a,t),b_correct))
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https://www.lessonup.com/nl/lesson/BWnsjNJY6sYoDuxSj
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# Revision length and area 2
Revision length and area
Schedule:
• Learning goals
• Think, pair, share
• Continue individual review
• Code breaker
1 / 12
Slide 1: Tekstslide
WiskundeMiddelbare schoolvwoLeerjaar 2
In deze les zitten 12 slides, met tekstslides.
Lesduur is: 60 min
## Onderdelen in deze les
Revision length and area
Schedule:
• Learning goals
• Think, pair, share
• Continue individual review
• Code breaker
#### Slide 1 -Tekstslide
Learning goals
At the end of this lesson I can:
• Manage my time and tasks effectively (ATL: Self-Management)
• Use skills and knowledge in multiple contexts (ATL: Critical thinking)
• (Re)consider the process of learning (ATL: Reflective skills)
#### Slide 2 -Tekstslide
Share your findings with the whole class
Solution:
#### Slide 4 -Tekstslide
Solution:
Atrapezium = 0.5 x (30 + 42) x 15.5 = 558 ft2
Total Area = 2 x 558 = 1116 ft2
$3 per ft2 Total cost = 3 x 1116 =$3348
#### Slide 5 -Tekstslide
Student question
If Johny has a bucket filled with 4.765kl, and he wants 11L, how much would he need to fill/take away to get his reached goal?
#### Slide 6 -Tekstslide
Student question
If Johny has a bucket filled with 4.765kl, and he wants 11L, how much would he need to fill/take away to get his reached goal?
4.765 kL = 4765 L
4765 - 11 = 4754 L
He needs to empty out 4754 L
#### Slide 7 -Tekstslide
Volume and capacity
1. Find the volume of this prism
2. Find the capacity of this prism
Bonus:
3. What would be the weight of the prism if it was filled with seawater (you can ignore the weight of the sides of the prism)
#### Slide 8 -Tekstslide
Volume and capacity
1. Find the volume of this prism
Atrapezium = 0.5 x (6 + 10 ) x 5 = 40 cm2
VPrism = 40 x 20 = 800 cm3
2. Find the capacity of this prism
800 cm3 = 800 ml = 0.8 L
Bonus:
3. What would be the weight of the prism if it was filled with seawater (you can ignore the weight of the sides of the prism)
#### Slide 9 -Tekstslide
Volume and capacity
Bonus:
3. What would be the weight of the prism if it was filled with seawater (you can ignore the weight of the sides of the prism)
1 L of seawater weighs approximately 1.025 kg.
0.8 L will weigh 0.8 x 1.025 = 0.82 kg
#### Slide 10 -Tekstslide
Continue practicing
I have 2 workpackets:
- Length, perimeter, area
- Volume, capacity, mass
You will choose 1 workpacket to work on today.
(You can borrow the answer key from my desk)
#### Slide 11 -Tekstslide
Code breaker
Code breaker 1:
What is the result of Snow White’s survey of the dwarves?
Code breaker 2:
How did I realise that the drink 'Ocean' didn’t exist?
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Recognized HTML document
62 NATURAL INHERITANCE. [011A'.
The Charms of Statistics.-It is difficult to understand why statisticians commonly limit their inquiries to Averages, and do not revel in more comprehensive views. Their souls seem as dull to the charm of variety as that of the native of one of our flat English counties, whose retrospect of Switzerland was that, if its mountains could be thrown into its lakes, two nuisances would be got rid of at once. An Average is but a solitary fact, whereas if a single other fact be added to it, an entire Normal Scheme, which nearly corresponds to the observed one, starts potentially, into existence.
Some people hate the very name of statistics, but I find them full of beauty and interest. Whenever they are not brutalised, but delicately handled by the higher methods, and are warily interpreted, their power of dealing with complicated phenomena is extraordinary. They are the only tools by which an opening can be cut
(2) If the Measures at any two specified Grades are given, the whole Scheme of Measures is thereby determined. Let A, B be the two given Measures of which A is the larger, and let a, b be the values of the tabular Deviations for the same Grades, as found in Table 8, not omitting their signs of plus or minus as the case may be.
Then the Q of the Scheme = 4b . (The sign of Q is not to be re
garded ; it is merely a magnitude.)
M=A -aQ; or M=B- bQ.
Example : A, situated at Grade 55°, = 14.38
B, situated at Grade 5°, = 9.12
The corresponding tabular Deviations are : a = + 0-19; b = - 2.44. Therefore Q = 1488 - 9.12 = 5.26 = 2.0
0-19+2-44 2.63
M=14.38-0.19X2=14.0
or 9.12+2.44X2=14.0
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# Show that (a+b+c)³ = a³ + b³ + c³ + (a+b+c)(ab+ac+bc)
As stated in the title, I'm supposed to show that $(a+b+c)³ = a³ + b³ + c³ + (a+b+c)(ab+ac+bc)$.
My reasoning: $$(a + b + c)³ = [(a + b) + c]³ = (a + b)³ + 3(a + b)²c + 3(a + b)c² + c³$$
$$(a + b + c)³ = (a³ + 3a²b + 3ab² + b³) + 3(a² + 2ab + b²)c + 3(a + b)c²+ c³$$
$$(a + b + c)³ = a³ + b³ + c³ + 3a²b + 3a²c + 3ab² + 3b²c + 3ac² + 3bc² + 6abc$$
$$(a + b + c)³ = (a³ + b³ + c³) + (3a²b + 3a²c + 3abc) + (3ab² + 3b²c + 3abc) + (3ac² + 3bc² + 3abc) - 3abc$$
$$(a + b + c)³ = (a³ + b³ + c³) + 3a(ab + ac + bc) + 3b(ab + bc + ac) + 3c(ac + bc + ab) - 3abc$$
$$(a + b + c)³ = (a³ + b³ + c³) + 3(a + b + c)(ab + ac + bc) - 3abc$$
$$(a + b + c)³ = (a³ + b³ + c³) + 3[(a + b + c)(ab + ac + bc) - abc]$$ It doesn't look like I made careless mistakes, so I'm wondering if the statement asked is correct at all.
-
The statement is obviously wrong: look at it when $a=b=c=1$. – KCd Jan 28 '13 at 14:43
You've probably not mentioned some condition? – hjpotter92 Jan 28 '13 at 14:45
I transcribed the exercise. Found a typo or another at other parts of the book, so that's just another one of them. – Sawyier Jan 28 '13 at 14:48
If it helps the final statement in your derivation looks correct. – kaine Jan 28 '13 at 15:12
In general, $$a^n+b^n+c^n = \sum_{i+2j+3k=n} \frac{n}{i+j+k}\binom {i+j+k}{i,j,k} s_1^i(-s_2)^js_3^k$$
where $s_1=a+b+c$, $s_2=ab+ac+bc$ and $s_3=abc$ are the elementary symmetric polynomials.
In the case that $n=3$, the triples possible are $(i,j,k)=(3,0,0),(1,1,0),$ and $(0,0,1)$ yielding the formula:
$$a^3+b^3+c^2 = s_1^3 - 3s_2s_1 + 3s_3$$
which is the result you got.
In general, any symmetric homogeneous polynomial $p(a,b,c)$ of degree $n$ can be written in the form:
$$p(a,b,c)=\sum_{i+2j+3k=n} a_{i,j,k} s_1^i s_2^j s_3^k$$
for some constants $a_{i,j,k}$.
I've often thought Fermat's Last Theorem was most interesting when stated as a question about these polynomials. One statement of Fermat can be written as:
If $p$ is an odd prime, then $a^n+b^n+c^n=0$ if and only if $a+b+c=0$ and $abc=0$.
-
Take $a=b=c=1$, you will find that that the expression is not correct.
You have a missing factor of $3$ on the right. – O.L. Sep 9 '13 at 23:11
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The last modifications of this post were around 4 years ago, some information may be outdated!
This is a draft, the content is not complete and of poor quality!
K-Means is the most popular clustering method any learner should know. In this note, we will understand the idea of KMeans and how to use it with Scikit-learn. Besides that, we also learn about its variants (K-medois, K-modes, K-medians).
## K-Means
### Idea?
1. Randomly choose centroids ($k$).
2. Go through each example and assign them to the nearest centroid (assign class of that centroid).
3. Move each centroid (of each class) to the average of data points having the same class with the centroid.
4. Repeat 2 and 3 until convergence.
A simply basic steps of K-Means.
A gif illustrating the idea of K-Means algorithm. Source.
### How to choose k?
Using "Elbow" method to choose the number of clusters $k$.
### Discussion
• A type of Partitioning clustering.
• K-means is sensitive to outliers ⇒ K-medoids clustering or PAM (Partitioning Around Medoids) is less sensitive to outliers[ref]
### K-Means in code
from sklearn.cluster import KMeanskmeans = KMeans(n_clusters=10, random_state=0) # default k=8
kmeans.fit(X)kmeans.predict(X)
# orkmeans.fit_predict(X)
Some notable parameters (see full):
• max_iter: Maximum number of iterations of the k-means algorithm for a single run.
• kmeans.labels_: show labels of each point.
• kmeans.cluster_centers_: cluster centroids.
## K-medoids clustering
• It is simple to understand and easy to implement.
• K-Medoid Algorithm is fast and converges in a fixed number of steps.
• PAM is less sensitive to outliers than other partitioning algorithms.
• Disavdvantages:[ref]
• The main disadvantage of K-Medoid algorithms is that it is not suitable for clustering non-spherical (arbitrary shaped) groups of objects. This is because it relies on minimizing the distances between the non-medoid objects and the medoid (the cluster centre) – briefly, it uses compactness as clustering criteria instead of connectivity.
• It may obtain different results for different runs on the same dataset because the first k medoids are chosen randomly.
• Different from K-Means:
• K-Means:
• Final centers no need to be points in data.
• Measure generally requires Euclidean distance.
• Sensitive to outliers.
• K-Medoids:
• Final Centers is actual points in data. They're called medoids or exemplars.
• Measures can be arbitrarily dissimilar.
• Robust to outliers.
The idea:
1. (Like KMeans'): choose randomly points (no need to be a point in data)
2. (Like KMeans'): assign labels to points based on chosen points in step 1.
3. (Different from KMeans):
4. (Like KMeans'): repeat the steps.
## Choose k by Silhouette
It's better than Elbow method to choose the number of clusters $k$.
👉 Check this note.
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# Newton's L. - equation/formula
1. Oct 9, 2007
1. The problem statement, all variables and given/known data
Because of a frictional force of 2.6N, a force of 2.8N must be applied to a textbook in order to slide it along the surface of a wooden table. The book accelerates at a rate of 0.11 m/s
a) What is the unbalanced force on the book?
b) What is the mass of the book?
2. Relevant equations
3. The attempt at a solution
The frictional force of 2.6 has anything to do with the equation F=ma to find the mass? If yes, what is the equation?
2. Oct 9, 2007
### cepheid
Staff Emeritus
Yes, it does, because 'F' in that equation refers to the NET force on the body (also referred to as the unbalanced force in your homework problem). To calculate the net force, you must consider the applied force and the frictional force, adding them together (taking into account their directions).
3. Oct 9, 2007
Oh ok. So you mean that I need to add (the regular addition or in any other way?) the frictional force which is 2.6 and the force of 2.8, and then finish with the regular equation to find the mass?
Last edited: Oct 9, 2007
4. Oct 9, 2007
### cepheid
Staff Emeritus
No, obviously not just regular addition. Like I said before, you have to take into account the directions of the forces! The applied force is trying to move the textbook in one direction, eg. to the right. The frictional force is trying to oppose that motion (it points to the left). The two forces are *opposite*, so one of them partly balances out (ie cancels out) the other. The remaining part (which is the NET force acting on the object) is the only thing that is left to actually move the object. Since you know that the acceleration value given in the problem must be due to this "leftover" force, you can deduce the mass of the object.
5. Oct 9, 2007
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# Question: How Many Is 1 Trillion?
one thousand
## How many millions is a trillion?
Well, starting out smaller, 1,000 Thousands equals 1 Million (1,000,000); 1,000 Millions equals 1 Billion (1,000,000,000); and 1,000 Billions equals 1 Trillion (1,000,000,000,000). A Trillion can be thought of as a Million Millions.
## How many crores make a trillion?
In common parlance, the thousand, lakh, and crore terminology (though inconsistent) repeats for larger numbers: thus 1,000,000,000,000 (one trillion) becomes 1 lakh crore, written as 1,00,000,00,00,000.
## Is zillion a number?
zillion. A zillion is a huge but nonspecific number. Zillion sounds like an actual number because of its similarity to billion, million, and trillion, and it is modeled on these real numerical values. However, like its cousin jillion, zillion is an informal way to talk about a number that’s enormous but indefinite.
## How many zeros are there in a trillion?
Numbers Bigger Than a Trillion
NameNumber of ZerosGroups of (3) Zeros
Hundred thousand5(100,000)
Million62 (1,000,000)
Billion93 (1,000,000,000)
Trillion124 (1,000,000,000,000)
22 more rows
## Who is a trillionaire?
Jeff Bezos may have lost an estimated \$38 billion in his recent divorce, but the Amazon CEO is still the richest person on the planet. According to Comparisun, his net worth has grown at a pace of 34 percent over the last five years, which could make him the world’s first trillionaire by 2026.
## What comes after Centillion?
From a simple arithmetic point of view, counting 1, 2, 3 and so on, then obviously the next number after “centillion” would be “centillion-and-one”. And there is another number – googolplex – which is 1 followed by googol zeros.
## How much is a zillion?
How many is a zillion? 1 million = 1,000,000. 1 billion = 1,000,000,000 or 1,000,000,000,000 depending on if using the long or short scale.
## How much is Billion?
1,000,000,000. 1,000,000,000 (one billion, short scale; one thousand million or milliard, yard, long scale) is the natural number following 999,999,999 and preceding 1,000,000,001. One billion can also be written as b or bn. In scientific notation, it is written as 1 × 109.
9 zeros
## What is this number 1000000000000000000000000?
A thousand trillions is a quadrillion: 1,000,000,000,000,000.
## What is the highest number?
A googol is a 1 with a hundred zeroes behind it. We can write a googol using exponents by saying a googol is 10^100. The biggest named number that we know is googolplex, ten to the googol power, or (10)^(10^100). That’s written as a one followed by googol zeroes.
## How much is a googolplex?
A googolplex is the number 10googol, or equivalently, 10. Written out in ordinary decimal notation, it is 1 followed by 10100 zeroes, that is, a 1 followed by a googol zeroes.
A googol is 10 to the 100th power (which is 1 followed by 100 zeros). A googol is larger than the number of elementary particles in the universe, which amount to only 10 to the 80th power.
## What is the last number?
A googol is the large number 10100.
10 digits
Bill Gates
## How much is Donald Trump worth?
2.1 billion USD (2020)
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EASEL BY TPT
# Solving Quadratic Equations BINGO game
Rated 5 out of 5, based on 2 reviews
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##### Also included in
1. This bundle is for the algebra 2 unit on quadratic expressions and equations. The bundle consists of challenges, games, and puzzles to support your teaching of the algebra 2 quadratics unit. The included resources cover the topics of factoring quadratics, completing the square, finding the roots of
Price \$18.00Original Price \$26.90Save \$8.90
### Description
Solving Quadratic Equations Bingo! game
A bingo activity on finding the solutions to quadratic equations by factoring.
For this activity, questions are projected on the board. You can change the order of these if you want.
Comes complete with printable unique bingo cards for up to 36 students.
Alternatively, there is a no print option where students draw their own 3 by 3 grid and choose nine numbers from the board. Solutions are provided and can be revealed on the slide at the discretion of the teacher.
If the answer to a slide is one of their numbers they cross it off. The first student who crosses off all nine numbers and calls out 'bingo!' is the winner. (For a speed version it can be the first to get a row or column). Usually, there is a small prize for the winner.
Follow me on TPT.
If you would like to see an example of my bingo! games I have a some free sample versions:
- Rational (fractional) and negative exponents here: *** FREE Rational and negative exponents Bingo! game ***
- Estimating calculations by rounding to one significant figure here: *** FREE Estimating Bingo! game ***
Why use bingo in your lessons?
- Bingo will engage your students, fully involving them in an activity for 20 to 30 minutes.
- Bingo is a great way to consolidate and answer questions on a recently taught topic or to get students to refresh their understanding of an earlier topic without them complaining that theyβve seen it before!
Please leave your feedback, it is much valued - and receive your TPT credits!
A few of my other resources you may like:
Multiplying Binomials by expanding brackets Bingo!
Factoring Simple Quadratics Bingo!
Factoring Harder Quadratics Bingo!
Back to school first lesson warm-up math game
Rounding Bingo!
Negative Numbers Bingo!
Multiplying and Dividing with Powers of Ten Bingo!
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### Standards
to see state-specific standards (only available in the US).
Solve quadratic equations in one variable.
Use the method of completing the square to transform any quadratic equation in πΉ into an equation of the form (πΉ β π±)Β² = π² that has the same solutions. Derive the quadratic formula from this form.
Solve quadratic equations by inspection (e.g., for πΉΒ² = 49), taking square roots, completing the square, the quadratic formula and factoring, as appropriate to the initial form of the equation. Recognize when the quadratic formula gives complex solutions and write them as π’ Β± π£πͺ for real numbers π’ and π£.
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## Project, can't find problem.
I have this basic little porject for class and I can't seem to find what's wrong in it. It looks good to go in my eyes, but when I go to check it nothing appears.....have a look. It's due tomorrow morning...
<script type="text/javascript">
/* Start code and set up some variables
*/
var name = prompt("Enter your name:");
alert(name + ", You need to do a few more calculations.");
var one = prompt("enter First Number:",0);
var two = prompt("enter Second Number:",0);
/* Parse the numbers given
*/
var numone = parseFloat(one);
var numtwo - parseFloat(two);
/* Do Calculations
*/
var add = numone + numtwo;
var subtract = numone - numtwo;
var multiply = numone * numtwo;
var divide = numone / numtwo;
var squareone = numone * numtwo;
var squaretwo = numone * numtwo;
var sumsquare = squareone + squaretwo;
/* Print results
*/
document.write("<center><h1><u>Assignment 1 - Sample Solution</u></h1></center>");
document.write("\xA0The first number you entered was: " + numone + "<br>");
document.write("\xA0The second number you entered was: " + numtwo + "<br><br>");
document.write("\xA0Here are the calculations I performed with thembr><br>");
document.write("Addition: " + numone + " + " + numtwo + " = " + add + "<br>");
document.write("Subtraction: " + numone + " - " + numtwo + " = " + subtract + "<br>");
document.write("Multiply: " + numone + " x " + numtwo + " = " + multiply + "<br>");
document.write("Divide: " + numone + " / " + numtwo + " = " + divide + "<br>");
document.write("<h3>Squared</h3>");
document.write("Square One: " + numone + "<sup>2</sup> = " + squareOne + "<br>");
document.write("Square Two: " + numtwo + "<sup>2</sup> = " + squareTwo + "<br>");
document.write("Sum of Squares: " + numone + "<sup>2</sup>" + numtwo + "<sup>2</sup> = " + sumSquare);
</script>
Any help is appreciated, I'm kinda new to javascript.
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# `**` （双星）和`*` （星星）对参数做什么？
## 定义函数
`*args`允许任意数量的可选位置参数（参数），它们将被分配给一个名为`args`的元组。
`**kwargs`允许任意数量的可选关键字参数（参数），这些参数将在一个名为`kwargs`的字典中。
## 扩展，传递任意数量的参数
``````>>> x = xrange(3) # create our *args - an iterable of 3 integers
>>> xrange(*x) # expand here
xrange(0, 2, 2)
``````
``````>>> foo = 'FOO'
>>> bar = 'BAR'
>>> 'this is foo, {foo} and bar, {bar}'.format(**locals())
'this is foo, FOO and bar, BAR'
``````
## Python 3中的新增内容：使用仅关键字参数定义函数
``````def foo(arg, kwarg=None, *args, kwarg2=None, **kwargs):
return arg, kwarg, args, kwarg2, kwargs
``````
``````>>> foo(1,2,3,4,5,kwarg2='kwarg2', bar='bar', baz='baz')
(1, 2, (3, 4, 5), 'kwarg2', {'bar': 'bar', 'baz': 'baz'})
``````
``````def foo(arg, kwarg=None, *, kwarg2=None, **kwargs):
return arg, kwarg, kwarg2, kwargs
``````
``````>>> foo(1,2,kwarg2='kwarg2', foo='foo', bar='bar')
(1, 2, 'kwarg2', {'foo': 'foo', 'bar': 'bar'})
``````
``````>>> foo(1,2,3,4,5, kwarg2='kwarg2', foo='foo', bar='bar')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: foo() takes from 1 to 2 positional arguments
but 5 positional arguments (and 1 keyword-only argument) were given
``````
``````def bar(*, kwarg=None):
return kwarg
``````
``````>>> bar('kwarg')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: bar() takes 0 positional arguments but 1 was given
``````
``````>>> bar(kwarg='kwarg')
'kwarg'
``````
## Python 2兼容演示
`*args` （通常称为“星形参数”）和`**kwargs` （星号可以用“kwargs”来表示，但用“double-star kwargs”来表示）是使用`*``**`符号的Python常见成语。 这些特定的变量名称不是必需的（例如，您可以使用`*foos``**bars` ），但与惯例背离可能会让您的Python编码人员`*foos`
``````def foo(a, b=10, *args, **kwargs):
'''
this function takes required argument a, not required keyword argument b
and any number of unknown positional arguments and keyword arguments after
'''
print('a is a required argument, and its value is {0}'.format(a))
print('b not required, its default value is 10, actual value: {0}'.format(b))
# we can inspect the unknown arguments we were passed:
# - args:
print('args is of type {0} and length {1}'.format(type(args), len(args)))
for arg in args:
print('unknown arg: {0}'.format(arg))
# - kwargs:
print('kwargs is of type {0} and length {1}'.format(type(kwargs),
len(kwargs)))
for kw, arg in kwargs.items():
print('unknown kwarg - kw: {0}, arg: {1}'.format(kw, arg))
# But we don't have to know anything about them
# to pass them to other functions.
print('Args or kwargs can be passed without knowing what they are.')
# max can take two or more positional args: max(a, b, c...)
print('e.g. max(a, b, *args) \n{0}'.format(
max(a, b, *args)))
kweg = 'dict({0})'.format( # named args same as unknown kwargs
', '.join('{k}={v}'.format(k=k, v=v)
for k, v in sorted(kwargs.items())))
print('e.g. dict(**kwargs) (same as {kweg}) returns: \n{0}'.format(
dict(**kwargs), kweg=kweg))
``````
``````foo(a, b=10, *args, **kwargs)
``````
``````a is a required argument, and its value is 1
b not required, its default value is 10, actual value: 2
args is of type <type 'tuple'> and length 2
unknown arg: 3
unknown arg: 4
kwargs is of type <type 'dict'> and length 3
unknown kwarg - kw: e, arg: 5
unknown kwarg - kw: g, arg: 7
unknown kwarg - kw: f, arg: 6
Args or kwargs can be passed without knowing what they are.
e.g. max(a, b, *args)
4
e.g. dict(**kwargs) (same as dict(e=5, f=6, g=7)) returns:
{'e': 5, 'g': 7, 'f': 6}
``````
``````def bar(a):
b, c, d, e, f = 2, 3, 4, 5, 6
# dumping every local variable into foo as a keyword argument
# by expanding the locals dict:
foo(**locals())
``````
`bar(100)`打印：
``````a is a required argument, and its value is 100
b not required, its default value is 10, actual value: 2
args is of type <type 'tuple'> and length 0
kwargs is of type <type 'dict'> and length 4
unknown kwarg - kw: c, arg: 3
unknown kwarg - kw: e, arg: 5
unknown kwarg - kw: d, arg: 4
unknown kwarg - kw: f, arg: 6
Args or kwargs can be passed without knowing what they are.
e.g. max(a, b, *args)
100
e.g. dict(**kwargs) (same as dict(c=3, d=4, e=5, f=6)) returns:
{'c': 3, 'e': 5, 'd': 4, 'f': 6}
``````
``````def foo(a, b, c, d=0, e=100):
# imagine this is much more code than a simple function call
preprocess()
differentiating_process_foo(a,b,c,d,e)
# imagine this is much more code than a simple function call
postprocess()
def bar(a, b, c=None, d=0, e=100, f=None):
preprocess()
differentiating_process_bar(a,b,c,d,e,f)
postprocess()
def baz(a, b, c, d, e, f):
... and so on
``````
``````def decorator(function):
'''function to wrap other functions with a pre- and postprocess'''
@functools.wraps(function) # applies module, name, and docstring to wrapper
def wrapper(*args, **kwargs):
# again, imagine this is complicated, but we only write it once!
preprocess()
function(*args, **kwargs)
postprocess()
return wrapper
``````
``````@decorator
def foo(a, b, c, d=0, e=100):
differentiating_process_foo(a,b,c,d,e)
@decorator
def bar(a, b, c=None, d=0, e=100, f=None):
differentiating_process_bar(a,b,c,d,e,f)
@decorator
def baz(a, b, c=None, d=0, e=100, f=None, g=None):
differentiating_process_baz(a,b,c,d,e,f, g)
@decorator
def quux(a, b, c=None, d=0, e=100, f=None, g=None, h=None):
differentiating_process_quux(a,b,c,d,e,f,g,h)
``````
``````def foo(param1, *param2):
def bar(param1, **param2):
``````
* args = * aList =列表中的所有元素
** args = ** aDict =字典中的所有项目
`*args``**kwargs`是一个常见的习惯用法，它允许任意数量的函数参数，如本节更多关于定义 Python文档中的函数所述。
`*args`会给你所有的函数参数作为元组
``````In [1]: def foo(*args):
...: for a in args:
...: print a
...:
...:
In [2]: foo(1)
1
In [4]: foo(1,2,3)
1
2
3
``````
`**kwargs`会给你所有的关键字参数，除了那些对应于形式参数的字典。
``````In [5]: def bar(**kwargs):
...: for a in kwargs:
...: print a, kwargs[a]
...:
...:
In [6]: bar(name='one', age=27)
age 27
name one
``````
``````def foo(kind, *args, **kwargs):
pass
``````
`*l`成语的另一个用法是在调用函数时解开参数列表
``````In [9]: def foo(bar, lee):
...: print bar, lee
...:
...:
In [10]: l = [1,2]
In [11]: foo(*l)
1 2
``````
``````first, *rest = [1,2,3,4]
first, *l, last = [1,2,3,4]
``````
``````def func(arg1, arg2, arg3, *, kwarg1, kwarg2):
pass
``````
``````def foo(x,y,z):
print("x=" + str(x))
print("y=" + str(y))
print("z=" + str(z))
``````
``````>>> mylist = [1,2,3]
>>> foo(*mylist)
x=1
y=2
z=3
>>> mydict = {'x':1,'y':2,'z':3}
>>> foo(**mydict)
x=1
y=2
z=3
>>> mytuple = (1, 2, 3)
>>> foo(*mytuple)
x=1
y=2
z=3
``````
``````>>> mydict = {'x':1,'y':2,'z':3,'badnews':9}
>>> foo(**mydict)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: foo() got an unexpected keyword argument 'badnews'
``````
``````>>> (0, *range(1, 4), 5, *range(6, 8))
(0, 1, 2, 3, 5, 6, 7)
>>> [0, *range(1, 4), 5, *range(6, 8)]
[0, 1, 2, 3, 5, 6, 7]
>>> {0, *range(1, 4), 5, *range(6, 8)}
{0, 1, 2, 3, 5, 6, 7}
>>> d = {'one': 1, 'two': 2, 'three': 3}
>>> e = {'six': 6, 'seven': 7}
>>> {'zero': 0, **d, 'five': 5, **e}
{'five': 5, 'seven': 7, 'two': 2, 'one': 1, 'three': 3, 'six': 6, 'zero': 0}
``````
``````>>> range(*[1, 10], *[2])
range(1, 10, 2)
``````
（感谢mgilson为PEP链接。）
`````` In function *construction* In function *call*
=======================================================================
| def f(*args): | def f(a, b):
*args | for arg in args: | return a + b
| print(arg) | args = (1, 2)
| f(1, 2) | f(*args)
----------|--------------------------------|---------------------------
| def f(a, b): | def f(a, b):
**kwargs | return a + b | return a + b
| def g(**kwargs): | kwargs = dict(a=1, b=2)
| return f(**kwargs) | f(**kwargs)
| g(a=1, b=2) |
-----------------------------------------------------------------------
``````
*还可以解压一个发电机
Python3 Document的一个例子
``````x = [1, 2, 3]
y = [4, 5, 6]
unzip_x, unzip_y = zip(*zip(x, y))
``````
unzip_x将会是[1,2,3]，unzip_y将会是[4,5,6]
zip（）接收多个iretable参数，并返回一个生成器。
``````zip(*zip(x,y)) -> zip((1, 4), (2, 5), (3, 6))
``````
``````class base(object):
def __init__(self, base_param):
self.base_param = base_param
class child1(base): # inherited from base class
def __init__(self, child_param, *args) # *args for non-keyword args
self.child_param = child_param
super(child1, self).__init__(*args) # call __init__ of the base class and initialize it with a NON-KEYWORD arg
class child2(base):
def __init__(self, child_param, **kwargs):
self.child_param = child_param
super(child2, self).__init__(**kwargs) # call __init__ of the base class and initialize it with a KEYWORD arg
c1 = child1(1,0)
c2 = child2(1,base_param=0)
print c1.base_param # 0
print c1.child_param # 1
print c2.base_param # 0
print c2.child_param # 1
``````
``````def __init__(self, *args, **kwargs):
for attribute_name, value in zip(self._expected_attributes, args):
setattr(self, attribute_name, value)
if kwargs.has_key(attribute_name):
kwargs.pop(attribute_name)
for attribute_name in kwargs.viewkeys():
setattr(self, attribute_name, kwargs[attribute_name])
``````
``````class RetailItem(Item):
_expected_attributes = Item._expected_attributes + ['name', 'price', 'category', 'country_of_origin']
class FoodItem(RetailItem):
_expected_attributes = RetailItem._expected_attributes + ['expiry_date']
``````
``````food_item = FoodItem(name = 'Jam',
price = 12.0,
category = 'Foods',
country_of_origin = 'US',
expiry_date = datetime.datetime.now())
``````
``````class ElectronicAccessories(RetailItem):
_expected_attributes = RetailItem._expected_attributes + ['specifications']
# Depend on args and kwargs to populate the data as needed.
def __init__(self, specifications = None, *args, **kwargs):
self.specifications = specifications # Rest of attributes will make sense to parent class.
super(ElectronicAccessories, self).__init__(*args, **kwargs)
``````
``````usb_key = ElectronicAccessories(name = 'Sandisk',
price = '\$6.00',
category = 'Electronics',
country_of_origin = 'CN',
specifications = '4GB USB 2.0/USB 3.0')
``````
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# Markov and You
In Finally, a Definition of Programming I Can Actually Understand I marvelled at particularly strange and wonderful comment left on this blog. Some commenters wondered if that comment was generated through Markov chains. I considered that, but I had a hard time imagining a text corpus input that could possibly produce output so profoundly weird.
So what are these Markov chains we're talking about?
One example of Markov chains in action is Garkov, where the long running Garfield cartoon strip meets Markov chains. I present below, for your mild amusement, two representative strips I found on the Garkov hall of fame:
Garfield's an easy target, though:
• Garfield Minus Garfield. What it says on the tin. Surprisingly cathartic.
• Lasagna Cat. Almost indescribably strange live action recreations of Garfield strips. If you only click one link in this post, make it this one. Sanity optional.
• Garfield Variations. Hand-drawn versions of Garfield in underground "comix" style, usually on paper napkins.
• Barfield. Garfield strips subtly modified to include amusing bodily functions.
• Permanent Monday. Literary commentary on selected strips.
• Arbuckle. Strips faithfully redrawn by random internet "artists", with one dramatic twist: Jon can't actually hear Garfield, because he is, after all, a cat.
• Garfield Randomizer. Sadly defunct – combined random panels to form "new" Garfield strips.
So let's proceed to the "kov" part of Garkov. The best description of Markov chains I've ever read is in chapter 15 of Programming Pearls:
A generator can make more interesting text by making each letter a random function of its predecessor. We could, therefore, read a sample text and count how many times every letter follows an A, how many times they follow a B, and so on for each letter of the alphabet. When we write the random text, we produce the next letter as a random function of the current letter. The Order-1 text was made by exactly this scheme:
t I amy, vin. id wht omanly heay atuss n macon aresethe hired boutwhe t, tl, ad torurest t plur I wit hengamind tarer-plarody thishand.
We can extend this idea to longer sequences of letters. The order-2 text was made by generating each letter as a function of the two letters preceding it (a letter pair is often called a digram). The digram TH, for instance, is often followed in English by the vowels A, E, I, O, U and Y, less frequently by R and W, and rarely by other letters.
Ther I the heingoind of-pleat, blur it dwere wing waske hat trooss. Yout lar on wassing, an sit." "Yould," "I that vide was nots ther.
The order-3 text is built by choosing the next letter as a function of the three previous letters (a trigram).
I has them the saw the secorrow. And wintails on my my ent, thinks, fore voyager lanated the been elsed helder was of him a very free bottlemarkable,
By the time we get to the order-4 text, most words are English, and you might not be surprised to learn that it was generated from a Sherlock Holmes story ( "The Adventure of Abbey Grange'').
His heard." "Exactly he very glad trouble, and by Hopkins! That it on of the who difficentralia. He rushed likely?" "Blood night that.
So the text in Garkov strips is generated in exactly this way, but using words instead of letters. The input corpus is, as you'd expect, the text of many old Garfield strips.
What's amazing to me about Markov chains is how unbelievably simple they are. A Markov chain has no memory of previous states: the next state (word, in our case) is chosen based on a random dice roll and a lookup into a table of the states that tend to historically follow the current state in the input corpus. Given an adequate input corpus, they work almost uncannily well, a testament to the broad power of rudimentary statistical inference. Garfield's been around since 1978, and still going str.. well, going, so there's no shortage of material to work with.
Now let's try it ourselves. I fed the text of the last twelve Paul Graham essays to this online Markov generator, using two word groupings – what Bentley refers to as "Order-2". Here's what I got back:
You can feel the need to take advantage of increased cheapness, however. You're not all playing a zero-sum game. There's not some fixed number of startups; we fund startups we fund to work on matters of passing importance. But I'm uncomfortably aware that this is part of any illusions about the problem of overeating by stopping eating. I couldn't simply avoid the Internet had become, because the company is the new trend of worrying obsessively about what it meant for someone, usually an outsider, who deliberately stirred up fights in a startup than just start it. You know how the A List is selected. And even that is more work.
But Markov chains aren't just useful for automatically generating Paul Graham essay parodies. They're also quite practical. You might even say Markov chains are a large part of what powers today's internet. Most remarkably, to me at least, Markov chains underly Google's trillion dollar PageRank formula:
The [PageRank] formula uses a model of a random surfer who gets bored after several clicks and switches to a random page. The PageRank value of a page reflects the chance that the random surfer will land on that page by clicking on a link. [PageRank] can be understood as a Markov chain in which the states are pages, and the transitions are all equally probable and are the links between pages.
As a result of Markov theory, it can be shown that the PageRank of a page is the probability of being at that page after lots of clicks. This happens to equal t-1 where t is the expectation of the number of clicks (or random jumps) required to get from the page back to itself.
Incidentally, if you haven't read the original 1998 PageRank paper, titled The PageRank Citation Ranking: Bringing Order to the Web (pdf), you really should. It's remarkable how, ten years on, so many of the predictions in this paper have come to pass. It's filled with interesting stuff; the list of the top 15 PageRank sites circa 1996 in Table 1 is an eye-opening reminder of how far we've come. Plus, there are references to pornographic sites, too!
Markovian models – specifically, hidden Markov Models – are also related to our old friend, Bayesian spam filtering. They're even better! The most notable example is the CRM114 Discriminator, as outlined in this excellent presentation (pdf).
If you play with the Markov text synthesizer, you'll quickly find that Markov methods are only as good as their input corpus. Input a bunch of the same words, or random gibberish, and that's what you'll get back.
But it's sure tough to imagine a more ideal input corpus for Markovian techniques than the unimaginable vastness of web and email, isn't it?
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# Difference between revisions of "Calculating Pitting Corrosion"
## Localized Corrosion
OLI Software Corrosion Analyzer can predict the tendency or propensity of an alloy to localized corrosion.
The relationship between the corrosion potential (Ecorr) and the repassivation potential (Erp) indicates the likelihood of localized corrosion to occur.
The Ecorr is the potential at which the anodic and cathodic rates are the same under a determined environment; and the Erp is the potential below which localized corrosion, such as pitting and crevice corrosion, does not occur.
Whenever Ecorr exceeds Erp, localized corrosion is predicted. Now the larger the difference between these two potentials, the greater the propensity to localized corrosion, as is depicted in the Figure below.
## Calculating Pitting Corrosion
### Predicting Corrosion and Repassivation Potentials
In OLI Software you can simulate a simple scenario like NaCl in water at 95oC for different alloys such as 304 and 316.
It is important to keep in mind that for simulating pitting corrosion you need to have enough oxidizing species in your simulation to raise the corrosion potential above Erp. If Ecorr is determined just by the reduction of water on a passive surface, then it is usually to low to provide a driving force for localized corrosion. If we include oxygen, then we get localized corrosion depending on the chloride concentration.
In the example below, two simulations were run for alloy 316 as a function of NaCl concentration – one without oxygen and one with an oxygen amount that roughly corresponds to oxygen content in air (0.2 O2 and 0.8 N2):
Figure 1. Corrosion and Repassivation potential of alloy 316 as a function of NaCl.
Erp (greenish line) does not depend on oxygen but Ecorr (brown lines) certainly does. If we have no oxygen (i.e., the only reduction reaction on the passive surface is the reduction of water), then Ecorr is always below Erp. If we have air, then Ecorr exceeds Erp at NaCl concentrations a little below 0.1 m.
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# 柯西－黎曼方程
（重定向自柯西－黎曼方程）
(1a) ${\displaystyle {\partial u \over \partial x}={\partial v \over \partial y}}$
(1b) ${\displaystyle {\partial u \over \partial y}=-{\partial v \over \partial x}.}$
## 注释和其他表述
### 共形映射
(2) ${\displaystyle {i{\partial f \over \partial x}}={\partial f \over \partial y}.}$
${\displaystyle {\begin{pmatrix}a&-b\\b&\;\;a\end{pmatrix}},}$
### 复共轭的独立性
(3) ${\displaystyle {\frac {\partial f}{\partial {\bar {z}}}}=0}$
${\displaystyle {\frac {\partial }{\partial {\bar {z}}}}={\frac {1}{2}}\left({\frac {\partial }{\partial x}}+i{\frac {\partial }{\partial y}}\right).}$
### 复可微性
${\displaystyle f(z)=u(z)+iv(z)}$
${\displaystyle \lim _{\underset {h\in \mathbb {C} }{h\to 0}}{\frac {f(z_{0}+h)-f(z_{0})}{h}}=f'(z_{0})}$
${\displaystyle \lim _{\underset {h\in \mathbb {R} }{h\to 0}}{\frac {f(z_{0}+h)-f(z_{0})}{h}}={\frac {\partial f}{\partial x}}(z_{0}).}$
${\displaystyle \lim _{\underset {ih\in i\mathbb {R} }{h\to 0}}{\frac {f(z_{0}+ih)-f(z_{0})}{ih}}=\lim _{\underset {ih\in i\mathbb {R} }{h\to 0}}-i{\frac {f(z_{0}+ih)-f(z_{0})}{h}}=-i{\frac {\partial f}{\partial y}}(z_{0}).}$
f沿着两个轴的导数相同也即
${\displaystyle {\frac {\partial f}{\partial x}}(z_{0})=-i{\frac {\partial f}{\partial y}}(z_{0}),}$
### 物理解释
${\displaystyle {\bar {f}}={\begin{bmatrix}u\\-v\end{bmatrix}}}$
${\displaystyle {\frac {\partial (-v)}{\partial x}}-{\frac {\partial u}{\partial y}}=0.}$
${\displaystyle {\frac {\partial u}{\partial x}}+{\frac {\partial (-v)}{\partial y}}=0.}$
### 其它解释
${\displaystyle {\frac {\partial u}{\partial s}}={\frac {\partial v}{\partial n}},\quad {\frac {\partial u}{\partial n}}=-{\frac {\partial v}{\partial s}}}$
${\displaystyle {\partial u \over \partial r}={1 \over r}{\partial v \over \partial \theta },\quad {\partial v \over \partial r}=-{1 \over r}{\partial u \over \partial \theta }.}$
${\displaystyle {\partial f \over \partial r}={1 \over ir}{\partial f \over \partial \theta }.}$
## 非齐次方程
${\displaystyle {\frac {\partial u}{\partial x}}-{\frac {\partial v}{\partial y}}=\alpha (x,y)}$
${\displaystyle {\frac {\partial u}{\partial y}}+{\frac {\partial v}{\partial x}}=\beta (x,y)}$
${\displaystyle {\frac {\partial f}{\partial {\bar {z}}}}=\phi (z,{\bar {z}})}$
${\displaystyle f(\zeta ,{\bar {\zeta }})={\frac {1}{2\pi i}}\iint _{D}\phi (z,{\bar {z}}){\frac {dz\wedge d{\bar {z}}}{z-\zeta }}}$
## 推广
### Goursat定理及其推广
f = u+iv为复函数，作为函数f : R2R2可微。则柯西积分定理（柯西－古尔萨定理）断言f在开域Ω上解析当且仅当它在该域上满足柯西-黎曼方程（Rudin 1966，Theorem 11.2）。特别是，f不需假定为连续可微（Dieudonné 1969，§9.10, Ex. 1）。
f在整个域Ω上满足柯西-黎曼方程是要点。可以构造在一点满足柯西-黎曼方程的连续函数，但它不在该点解析（譬如，f(z) = z5/|z|4）。只满足柯西-黎曼方程也是不够的，（需额外满足连续性），下面的例子表明了这一点：（Looman 1923，p.107）
${\displaystyle f(z)={\begin{cases}\exp(-z^{-4})&\mathrm {if\ } z\not =0\\0&\mathrm {if\ } z=0\end{cases}}}$
• f(z)在开域Ω⊂C上局部可积，并以弱形式满足柯西-黎曼方程，则f和Ω上的一个解析函数几乎处处相等。
### 多变量的情况
${\displaystyle {\bar {\partial }}}$
${\displaystyle {\partial f \over \partial {\bar {z}}}=0}$,
${\displaystyle {\partial f \over \partial {\bar {z}}}={1 \over 2}\left({\partial f \over \partial x}-{1 \over i}{\partial f \over \partial y}\right).}$
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## Squares
Using the first nine cards from any suit (let the Ace represent one) can you arrange them in a 3x3 square so that the sum of the three cards in any row, column and diagonal are the same?
Here are some other challenges: Can you arrange the cards in a 3x3 square such that:
- none of the sums of the three cards in any row, column and diagonal are the same?
- the sum of the three numbers in each row and each column is a prime number;
- the three 3-digit numbers formed in the rows add up to 999;
Solutions to these puzzles are available to those who have a Transum Subscription and are signed in.
You can try similar activities which are online and interactive:
Magic Square - Unmagic Square - Prime Square - Nine Nine Nine - Square and Even
Do you have any wonderful ideas for using playing cards to help learn mathematics? Please share your ideas here. Playing cards have been around since the ninth century. They were invented in China and spread across Europe in the fourteenth century. Though the designs on the cards have changed over the years the basic number properties have not. Because they are so popular the manufacture of high quality cards is not expensive and makes them an ideal tool for learning mathematics.
## Comments and suggestions:
Ann Mason, Kingsmead School Derby
Tuesday, November 17, 2015
"NUMBER BONDS TO TEN
Remove all 10, j, q, k cards from a pack of cards.
Pupil turns over cards to create 8 separate piles.
Pairs of cards can be covered, with new cards from the pack, if the two numbers add up to 10. (Ace is taken as 1)
The aim is to get rid of all the cards - not always possible, but they soon learn to recognise the number bonds.
Could be timed to create competition as they get more confident."
Do you have any comments? It is always useful to receive feedback and helps make this free resource even more useful for those learning Mathematics anywhere in the world. Click here to enter your comments.
## Activities with numbered cards:
#### Prime Pairs Game
A game for two players who take turns to select two numbers that add up to a prime number.
The short web address is:
Transum.org/go/?to=ppg
#### Superior
A game for two players who compete to make the largest possible number from randomly selected cards.
The short web address is:
Transum.org/go/?to=superior
#### Manifest
An exciting and thought-provoking number placing game for two players or one player versus the computer.
The short web address is:
Transum.org/go/?to=manifest
#### Nevertheless
Players decide where to place the cards to make an equation with the largest possible solution.
The short web address is:
Transum.org/go/?to=nevertheless
#### Fifteen
A strategy game. Play against the computer to select three numbers that add up to 15.
The short web address is:
Transum.org/go/?to=fifteen
#### Square Pairs Game
A game for two players who take turns to select two numbers that add up to a square number.
The short web address is:
Transum.org/go/?to=sqpg
For Students:
For All:
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Home » 1.69 M Is How Many Feet? New
# 1.69 M Is How Many Feet? New
Let’s discuss the question: 1.69 m is how many feet. We summarize all relevant answers in section Q&A of website Countrymusicstop.com in category: MMO. See more related questions in the comments below.
## What is 2 13m feet?
Meters to Feet table
Meters Feet
2 m 6.56 ft
3 m 9.84 ft
4 m 13.12 ft
5 m 16.40 ft
## What distance is 1 Metre in feet?
Answer: 1 meter is equal to 3.28084 feet.
### ✅ How Many Feet In A Meter
✅ How Many Feet In A Meter
✅ How Many Feet In A Meter
## Does 1 m equal 3 ft?
1 metre is equal to 3.2808399 feet, which is the conversion factor from meters to feet.
## What size is 2 meters in feet?
Meters to feet conversion table
Meters (m) Feet (ft)
0.1 m 0.328084 ft
1 m 3.28084 ft
2 m 6.56168 ft
3 m 9.84252 ft
## How do you calculate meters to feet?
There are 3.28084 feet per meter. So if you want to convert meters to feet using your own calculator, just multiply your number of meters by 3.28084.
## What is a meter long?
A meter is a standard metric unit equal to about 3 feet 3 inches. This means that a meter is part of the metric system of measurement. Guitars, baseball bats, and yard sticks are examples of objects that are about one meter long. Meters are also used to measure distances in races, such as running and swimming.
## Are a cent?
Ares to Cents Conversions
Ares Cents
1 are 2.47105 cents
2 ares 4.94211 cents
3 ares 7.41316 cents
4 ares 9.88422 cents
## Is it 1 foot or 1 feet?
Whilst foot refers to the single unit of measurement, ‘feet’ is its plural alternative. In this sense, the difference between foot and feet in Maths depends on how large the distance or length is that you are measuring.
## What is the difference between feet and meters?
Meter to Feet Conversion
For converting meter to feet firstly we should know the difference between their lengths. That is one meter is equal to 3.28 feet and one foot is equal to 12 inches as per rule.
## How many meters is a UK foot?
1 metre is equal to 3.2808399 feet, which is the conversion factor from meters to feet.
### How to convert meter(m) to feet(ft) and feet to meter / feet to meter and meter to feet conversion
How to convert meter(m) to feet(ft) and feet to meter / feet to meter and meter to feet conversion
How to convert meter(m) to feet(ft) and feet to meter / feet to meter and meter to feet conversion
## How many meters is 5’7 feet?
Quick Lookup Feet To Metres Common Conversions
ft & in m
5′ 4″ 1.63
5′ 5″ 1.65
5′ 6″ 1.68
5′ 7″ 1.70
## How CM is an inch?
Inches to Centimeter Formula
The value of 1 inch is approximately equal to 2.54 centimeters. To convert inches to the centimeter values, multiply the given inch value by 2.54 cm. 1 cm = 0.393701 inches.
## How many miles are in Ameter?
How many miles in a meter? 1 metre is equal to 0.00062137 miles, which is the conversion factor from meters to miles.
## How many inches means 1 meter?
Meters to Inches table
Meters Inches
1 m 39.37 in
2 m 78.74 in
3 m 118.11 in
4 m 157.48 in
## How many meters is 5 5 feet?
Feet to meters chart
Feet & Inches Feet Meters
5 feet 5 inches 5.42 feet 1.65 m
5 feet 6 inches 5.5 feet 1.68 m
5 feet 7 inches 5.58 feet 1.7 m
5 feet 8 inches 5.67 feet 1.73 m
## What does 2m mean in length?
Indicates the length of the cord. m=Meters one meter=3 feet 3.37 inches. So the approximate lengths are 1m=3 feet and 2m=6 feet.
## Which is longer 1 meter or 1 yard?
Answer: The difference between meter and yard is that the meter is a SI unit of length and a yard is a unit of length. Also, 1 meter is about 1.09 yards.
## How do you work out m2 on a calculator?
How to work out m2 or ft2 of your room. In order to work out square meters, measure the length and width of an area you’re calculating using meters and centimeters. Next you need to multiply the length and width together to get the area in square meters: Width x Length: 9 x 10 = 90.
## What is 6ft tall in meters?
Feet to meters conversion table
Feet (ft) Meters (m)
6 ft 1.8288 m
7 ft 2.1336 m
8 ft 2.4384 m
9 ft 2.7432 m
### How to Convert Meters to Feet and Feet to Meters
How to Convert Meters to Feet and Feet to Meters
How to Convert Meters to Feet and Feet to Meters
## How long is 1mm?
A measure of length in the metric system. A millimeter is one thousandth of a meter. There are 25 millimeters in an inch.
## Is your arm a metre?
By using our Meter to Arms Length conversion tool, you know that one Meter is equivalent to 1.43 Arms Length.
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Questions – Exam Papers – Computer Sir Ki Class
Question Typewise Collection-"Derivations" Questions - Exam Papers (CPP) No. of Q.12
Q.1 Exam - CBSE12D-2017/C06D/3
Reduce the following Boolean Expression to its simplest form using K-Map:
F(X,Y,Z,W)= Σ (0,1,2,3,4,5,10,11,14)
Q.2 Exam - CBSE12D-2017/C06C/1
Derive a Canonical SOP expression for a Boolean function F, represented by the following truth table:
U V W F(U,V,W)
0 0 0 1
0 0 1 0
0 1 0 1
0 1 1 1
1 0 0 0
1 0 1 0
1 1 0 1
1 1 1 0
Q.3 Exam - CBSE12A-2015/C06D/3
Reduce the following Boolean Expression to its simplest form using K-Map :
F(X,Y,Z,W) = ∑(0,1,4,5,6,7,8,9,11,15)
Q.4 Exam - CBSE12A-2015/C06C/1
Derive a Canonical POS expression for a Boolean function F, represented by the following truth table:
P Q R F(P,Q,R)
0 0 0 1
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 1
1 0 1 0
1 1 0 0
1 1 1 1
Q.5 Exam - CBSE12A-2015/C06A/2
Verify the following using Boolean Laws.
U’+ V= U’V’+U’.V +U.V
Q.6 Exam - CBSE12A-2017/C06D/3
Reduce the following Boolean expression to its simplest form using K-Map:
E(U,V,Z,W)= Ó (2,3,6,8,9,10,11,12,13)
Q.7 Exam - CBSE12A-2017/C06C/1
Derive a Canonical POS expression for a Boolean function G, represented by the following truth table:
X Y Z G(X,Y,Z)
0 0 0 0
0 0 1 0
0 1 0 1
0 1 1 0
1 0 0 1
1 0 1 1
1 1 0 0
1 1 1 1
Q.8 Exam - CBSE12A-2018/C06D/3
Reduce the following Boolean Expression to its simplest form using K-Map:
G(U,V,W,Z) = Σ (3,5,6,7,11,12,13,15)
Q.9 Exam - CBSE12A-2018/C06C/1
Derive a Canonical POS expression for a Boolean function FN, represented by the following truth table:
X Y Z FN(X,Y,Z)
0 0 0 1
0 0 1 1
0 1 0 0
0 1 1 0
1 0 0 1
1 0 1 0
1 1 0 0
1 1 1 1
Q.10 Exam - CBSE12A-2016/C06D/3
Reduce the following Boolean Expression to its simplest form using K-Map:
F(P,Q,R,S)= Σ(0,4,5,8,9,10,11,12,13,15)
Q.11 Exam - CBSE12A-2016/C06C/1
Derive a Canonical SOP expression for a Boolean function G, represented by the following truth table:
A B C G(A,B,C)
0 0 0 1
0 0 1 0
0 1 0 1
1 0 0 0
1 0 1 0
1 1 0 1
1 1 1 1
Q.12 Exam - CBSE12A-2016/C06A/2
Verify the following using Boolean Laws.
X’+ Y’Z = X’.Y’.Z’+ X’.Y.Z’+ X’Y.Z+ X’.Y’.Z+ X.Y’.Z
LHS X’ + Y’.Z = X’.(Y + Y’).(Z + Z’) + (X + X’).Y’.Z
×
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# On the cover: dragon curves
Read more about the fire-breathing curves that appear on the cover of issue 05
Take a long strip of paper. Fold it in half in the same direction a few times. Unfold it and look at the shape the edge of the paper makes. If you folded the paper $n$ times, then the edge will make an order $n$ dragon curve, so called because it faintly resembles a dragon. Each of the curves shown on the cover of issue 05 of Chalkdust, and in the header box above, is an order 10 dragon curve.
Left: Folding a strip of paper in half four times leads to an order four dragon curve (after rounding the corners). Right: A level 10 dragon curve resembling a dragon
The dragon curves on the cover show that it is possible to tile the entire plane with copies of dragon curves of the same order. If any readers are looking for an excellent way to tile a bathroom, I recommend getting some dragon curve-shaped tiles made.
An order $n$ dragon curve can be made by joining two order $n-1$ dragon curves with a 90° angle between their tails. Therefore, by taking the cover’s tiling of the plane with order 10 dragon curves, we may join them into pairs to get a tiling with order 11 dragon curves. We could repeat this to get tilings with order 12, 13, and so on… If we were to repeat this ad infinitum we would arrive at the conclusion that an order $\infty$ dragon curve will cover the entire plane without crossing itself. In other words, an order $\infty$ dragon curve is a space-filling curve.
The endpoints of dragon curves of order 1 to 10 with a logarithmic spiral passing through them
Like so many other interesting bits of recreational maths, dragon curves were popularised by Martin Gardner in one of his Mathematical Games columns in Scientific American. In this column, it was noted that the endpoints of dragon curves of different orders (all starting at the same point) lie on a logarithmic spiral. This can be seen in the diagram to the right.
Although many of their properties have been known for a long time and are well studied, dragon curves continue to appear in new and interesting places. At last year’s Maths Jam conference, Paul Taylor gave a talk about my favourite surprise occurrence of a dragon.
Normally when we write numbers, we write them in base ten, with the digits in the number representing (from right to left) ones, tens, hundreds, thousands, etc. Many readers will be familiar with binary numbers (base two), where the powers of two are used in the place of powers of ten, so the digits represent ones, twos, fours, eights, etc.
In his talk, Paul suggested looking at numbers in base -1+i (where i is the square root of -1; you can find more adventures of i here) using the digits 0 and 1. From right to left, the columns of numbers in this base have values 1, -1+i, -2i, 2+2i, -4, etc. The first 11 numbers in this base are shown below.
Number in base -1+i Complex number 0 0 1 1 10 -1+i 11 (-1+i)+(1)=i 100 -2i 101 (-2i)+(1)=1-2i 110 (-2i)+(-1+i)=-1-i 111 (-2i)+(-1+i)+(1)=-i 1000 2+2i 1001 (2+2i)+(1)=3+2i 1010 (2+2i)+(-1+i)=1+3i
Numbers in base -1+i of ten digits or less plotted on an Argand diagram
Complex numbers are often drawn on an Argand diagram: the real part of the number is plotted on the horizontal axis and the imaginary part on the vertical axis. The diagram to the left shows the numbers of ten digits or less in base -1+i on an Argand diagram. The points form an order 10 dragon curve! In fact, plotting numbers of $n$ digits or less will draw an order $n$ dragon curve.
Brilliantly, we may now use known properties of dragon curves to discover properties of base -1+i. A level $\infty$ dragon curve covers the entire plane without intersecting itself: therefore every Gaussian integer (a number of the form $a+\text{i} b$ where $a$ and $b$ are integers) has a unique representation in base -1+i. The endpoints of dragon curves lie on a logarithmic spiral: therefore numbers of the form $(-1+\text{i})^n$, where $n$ is an integer, lie on a logarithmic spiral in the complex plane.
If you’d like to play with some dragon curves, you can download the Python code used to make the pictures here.
Matthew is a postdoctoral researcher at the University of Cambridge. He hasn’t had time to play Klax since the noughties, but he’s pretty sure that Coke is it!
• ### My favourite LaTeX package
The Chalkdust editors share some of their favourites
• ### On the cover: cellular automata
Discover the meaning of the coloured squares on the cover of issue 13
• ### Oπnions: Should I share my code?
Scroggs debates whether sharing truly is caring
• ### Crossnumber winners, issue 11
Did you solve it?
• ### Chalkdust issue 11 puzzle hunt #1
Matthew Scroggs sets the first puzzle. Can you solve it?
• ### Crossnumber winners, issue 10
Did you solve it?
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https://www.washingtonpost.com/news/politics/wp/2018/01/09/for-18-billion-you-could-build-318-miles-of-border-wall-out-of-dollar-bills/
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It’s hard to overstate the extent to which Donald Trump’s campaign-trail mentions of building a wall on the border with Mexico was more applause line than policy proposal. Beyond the ludicrous argument that Mexico would pay for its construction — an argument that has been revealed as empty so often that it is now its own punchline — Trump also regularly used the height of the wall as an exclamation point. Any time the president of Mexico said something Trump didn’t like (like that his country wouldn’t pay for the wall) the wall “got 10 feet higher,” as Trump said several times during the campaign.
For a while, we tried to track the actual height of the wall, assuming that Trump actually planned to add 10 feet every time he said he was going to.
The wall started here two months after he declared his candidacy in 2015.
By the time we stopped updating our tracking of the wall’s height, it was here:
Last week, the White House sent Congress its first formal estimate of the cost of the first phase of wall construction, including a sense of scale. The wall would range from 18 to 30 feet in height, and the first phase of construction would include 316 miles of new construction and reinforcement of 407 miles of existing fencing.
So, the wall would end up somewhere between these two heights.
The price tag for this effort? A paltry \$18 billion over 10 years. Just a bit more than the annual cost of the federal judiciary (which cost \$17.4 billion in 2017).
How unrealistic is it that Mexico would foot that bill? The \$18 billion cost would account for about 1.7 cents of every dollar the country generated in gross domestic product in 2016.
Naturally, a 316-mile-long, 30-foot-tall wall is expensive. But we were curious about something. Eighteen billion is a lot of dollars. If you were to create a wall out of actual dollar bills, how far would it extend?
As it turns out, an 18-foot-tall wall can be built out of \$18 billion in dollar bills and stretch … 31.8 miles, one-tenth of that distance.
The dimensions of a dollar bill are about 156 millimeters in width, 66.3 millimeters in depth and 0.1 millimeter in height. (We used millimeters since Mexico is paying for it.) A stack of 54,860 bills reaches a height of 18 feet; a stack of 91,440 bills gets you to 30 feet. (We ignored the effects of compression here, meaning the fact that you could fit more bills in a stack given that the weight of the bills at the top would compress the bills at the bottom. We do need to inject some sort of adhesive between the bills, so let’s call it a wash.)
We end up with a paper column that’s 18-to-30-feet tall and about 6 inches wide. So how far would that stretch, if you lined up these columns one next to another?
Divide a 18 billion by 54,860 or 91,440 and multiply by 156 millimeters inches. An 18-foot-wall, as above, gets you 31.8 miles. The taller wall goes only 19 miles.
Of course, you also end up with a wall made out of paper, which is not terribly robust as a deterrent. (Not to mention a wall made up of negotiable currency.) What if we wanted a metal wall?
Happily, the United States also makes a metal currency. So we looked at how big a wall could be made with \$18 billion in pennies.
Pennies are narrower than dollar bills, as you may be aware, measuring three-quarters of an inch in diameter. They are also a bit thicker than a dollar bill, at 1.52 millimeters.
For a column 18 feet in height, you need 3,609 pennies. For a 30-foot wall, you need a column 6,016 pennies high.
How far would that stretch? The trick here is that a penny has — ahem — 1/100th the value of a dollar bill. So while pennies are narrower than bills, you need a lot more of them to make up the same value.
In short, you could build a 5,904-mile wall out of pennies that’s 18 feet high, or a 30-foot wall that runs for 3,542 miles. Given that we’re talking about a wall that is only three-quarters of an inch thick, we could build it 18 feet high and three pennies thick (2.25 inches), which still allows us to run 1,968 miles — nearly the length of the border. Or we could run it the length of the borders with Mexico and Canada, which, excluding Alaska, is a little over 5,900 miles.
(A penny is actually worth more than a penny, by the way, thanks to the copper and zinc that compose them. \$18 billion worth of pennies, then, would actually be fewer than 1.8 trillion pennies. But we’re not talking about actual value here, since the value of the paper in a dollar bill is much less than a dollar.)
How much copper would be used in this wall? A new penny weighs 2.5 grams, 95 percent of which is zinc and 5 percent of which is the copper exterior. In total, then, the 1.8 trillion pennies worth \$18 billion would use 4.7 million tons of zinc and 248,000 tons of copper. At the current trading price of copper, that’s about \$1.6 billion in copper costs alone.
There are other costs that would be required for this wall, too. Adhesive, as mentioned above. Some sort of support structure: An 18-foot-tall tower of pennies would likely be prone to toppling over, which wouldn’t be useful for a wall. Trump has also argued that the wall needs windows in it, so that people walking near it on the American side wouldn’t be hit and killed by bags of drugs being thrown over the wall by drug dealers.
This was not incorporated into our analysis.
Correction: The original version of this article misplaced a decimal point, lamentably. The figures have been corrected as a result.
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Convert a to Rai (Are to Rai)
## Are into Rai
numbers in scientific notation
https://www.convert-measurement-units.com/convert+Are+to+Rai.php
# Convert Are to Rai (a to Rai):
1. Choose the right category from the selection list, in this case 'Area'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), square root (√), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Are [a]'.
4. Finally choose the unit you want the value to be converted to, in this case 'Rai'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '389 Are'. In so doing, either the full name of the unit or its abbreviation can be usedas an example, either 'Are' or 'a'. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Area'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '85 a to Rai' or '95 a into Rai' or '12 Are -> Rai' or '19 a = Rai' or '36 Are to Rai' or '65 Are into Rai'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(40 * 72) a'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '389 Are + 1167 Rai' or '46mm x 42cm x 91dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
The mathematical functions sin, cos, tan and sqrt can also be used. Example: sin(π/2), cos(pi/2), tan(90°), sin(90) or sqrt(4).
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 6.560 999 940 294 9×1029. For this form of presentation, the number will be segmented into an exponent, here 29, and the actual number, here 6.560 999 940 294 9. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 6.560 999 940 294 9E+29. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 656 099 994 029 490 000 000 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications.
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+0
# whats this?
0
98
2
23-46=40-?`
Guest May 23, 2017
Sort:
#1
+80
0
23-46 would equal negative twenty three, not fourty.
mathguardian23 May 23, 2017
#2
+1224
+1
\(?=63\)
This is your original equation. I'll replace your question mark with an x instead:
\(23-46=40-x\)
Now, solve by isolating x:
\(23-46=40-x\) Simplify the left hand side of the equation \(-23=40-x\) Subtract 40 on both sides \(-63=-x\) Multiply by -1 on both sides \(x=63\)
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http://passionofthecross.info/i4_thesis-pie-graph.php
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# Thesis Pie Graph
Bar charts are common in the presentation of gender statistics.
2005 Designing science graphs for data. Presenting Data in Charts and Graphs. Loading. In general, when you write essays you avoid using bullet points and lists, but in a dissertation the chapter where you present your data is an thesis pie graph to this rule.
but I also wrote a follow-up 5 years after the illustrated guide which may be of.
Make a pie graph for each set of data below. A graph is the representation of data by using graphical symbols such as lines, bars, pie slices, dots etc.
The line thesis pie graph shows the number of books that were borrowed in four different months in 2014 from four village libraries, and the pie chart shows the percentage of books, by type.
Lesson Plan 3 Pie Graphs.
• essay national day celebrations
• ONLINE CHARTS | create and design your own charts and...
Graphs are Interactive, Responsive, support Animation can be easily. All slices are in whole numbers. (bar charts, line graphs, pie charts. Here we will focus on writing the results and analysis of data based on a quantitative.
It is an effective way, for example, to show a citys ethnic mix. Pie Charts. Bar Graphs. Tables and spreadsheets need to be concise. The objective of this guide for architectural education not fixed. Pie graph in thesis graph Generator very easy to thesis pie graph.
A pie thesis pie graph is a specialized graph used in statistics in which the relative proportions of parts of a whole are shown as different size wedges of a pie thesis pie graph. As for categorical variables, frequency distributions may math homework ideas 4th grade presented in a table or a graph, including bar charts and pie or sector charts.
In order.
Bar graphs, charts, and tables, they. Lack of clarity is also the reason why pie charts should be used with caution. graph pie draws pie charts. Pie charts are not recommended in the R documentation, thesis pie graph their features are somewhat limited.
charts, graphs. may also be called a circle or sector graph. The pie graph below math homework ideas 4th grade the results. Sobieski asked his class to vote on where they would most like to go on a field trip.
Data Enter your data. How to use Pie Charts or Circle graphs or Pie Graphs to represent data, how to construct or circle graphs or pie charts, examples and step by step solutions, how to find the angle of a circle graph, How to construct a circle graph or pie chart from a table of percentages or fractions.
The advantages and disadvantages of pie charts as a method of data presentation Failing to thesis format ukm engage in essay builder pie chart metacognitive monitoring.
Update. Presenting Data in Charts and Graphs. Interpreting a pie graph. Pie chart survey analysis 1. You must be logged in to post a review.
You can create graphs like that using our Data Graphs (Bar, Line and Pie) page. The rst corresponds to the specication of two or more variables Some students made a pie graph based on the population of some of the biggest citiesin the world. We thesis pie graph against the use of thesis pie graph charts and. The term frequency distribution has a specific meaning, referring to the the way observations of a given variable behave in terms of its absolute, relative or cumulative frequencies.
Graphs are Interactive, Responsive, support Animation can be easily.
The advantages and disadvantages of pie charts as thesis pie graph method of data presentation Failing to thesis format ukm engage in essay builder essay topics english second language chart metacognitive monitoring.
However, as with tables, one has to be careful in a serious academic study such as a thesis, that an inappropriate or excessive use of presentations such as pie charts may not thesis pie graph perceived as oversimplistic.
Determine what percentage each bill will be of the pie graph.
The line graph shows the number of books thesis pie graph were borrowed in four different months in 2014 from four village libraries, and the pie chart shows the percentage of books, by type. Tables and spreadsheets need to be concise. Determine what percentage each bill will be of the pie graph. To save results or sets tasks for your students you need to be logged in.
TABLES AND FIGURES--GENERAL GUIDELINES. Essay topics english second language Section.
If you want to illustrate proportions, experiment with a pie chart or bar graph. Using tables. Can some be placed in an appendix. Nov 28, 2012.
Create a Pie Chart.
In order. Aurlie Auvray, Graphothrapeute, Graphologue. Many people do not understand the strengths and weaknesses that come with each chart type, either deciding off the cuff which looks the nicest or staying in their comfort zone by overloading their report with pie or vertical bar charts.
### Designing science graphs for data analysis and …
The term frequency distribution has a specific meaning, referring to the the essay prompts college applications observations of a given variable behave in terms of thesis pie graph absolute, relative or cumulative frequencies. Here are some other ways of presenting your data Pie charts show. Pie charts are good for displaying data for around 6 categories or fewer.
(bar charts, line graphs, how to write a grad school personal statement charts.
Other useful guides Bar charts, Histograms, Presenting numerical data. Pie chart maker online. Clearly thesis pie graph bullet points are fine in these cases. May 30, 2012 - 7 min - Uploaded by Aldo Mencaragliahttpwww.
## Survey Results: Reporting via Pie Charts or Bar Graphs
Data Enter your data. GRAPH-BASED DATA VISUALIZATION IN VIRTUAL REALITY A COMPARISON OF USER EXPERIENCES A Thesis presented to the Read below for the illustrated guide to a Ph. Technorati Tags graphic, communication, map, illustration, photograph, pie chart, dot graph, graph.
Create a Pie Chart.
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# R Language Swirl Module Worksheet
Description
1 attachmentsSlide 1 of 1attachment_1attachment_1
Unformatted Attachment Preview
STAT 362
Spring 2020
Homework 5
Problems
DUE DATE: WEDNESDAY, FEBRUARY 5, 2020 by 11:59 PM on Gradescope
• Part 1: Swirl Module will be graded.
• Part 2: 5 randomly selected problems will be graded. The rest of the problems will be taken
for completion points only.
This homework has a total of 18 problems.
Part 1: Swirl Module
1. In the console window where you can type in R code, type the following to load swirl:
library(swirl)
swirl()
Follow the prompts on the screen. To exist swirl, press the Esc key. If you are already at the
prompt, type bye() to exit and save your progress. When you exit properly, you’ll see a short
message letting you know you’ve done so.
3. Complete Module 13: Simlulation.
After completion
• When they ask “Would you like to receive credit for completing this course on
Coursera.org?”, type YOUR NAME (FIRST AND LAST).
• Hit enter.
• Print the screen. This shows that you have completed the assignment.
• Then for selection, type “0”.
Part 2 Problems:
1. Suppose you want to make a password using exactly 4 out of the 6 letters: e, t, a, b, p, k.
You do not want to repeat any letter. How many different passwords can you create?
2. In a poker hand consisting of 5 cards, find the probability of holding 4 hearts and 1 club.
1
3. An automatic welding machine is being considered for use in a production process. It will
be considered for purchase if it is successful on 99% of its welds. Otherwise, it will not be
considered efficient. A test is to be conducted with a prototype that is to perform 100 welds.
The machine will be accepted for manufacture if it misses 3 welds or less.
(a) What is the probability that a good machine will be rejected?
(b) What is the probability that an inefficient machine with 95% welding success will be
accepted?
4. Find P(X ≤ 813) assuming:
(a) X ∼ Bin(2000, 0.4)
(b) X ∼ N(µ = 800, σ 2 = 480)
5. Suppose X follows a continuous uniform distribution from 1 to 5. Determine the conditional
probability P(X > 2.5 | X ≤ 4). Hint: see Definition 2.10 from the Stat 381 textbook.
6. Given a standard normal distribution, find:
(a) the area under the curve that lies to the left of z = 1.72.
(b) the area under the curve that lies between z = −2.16 and z = −0.65.
(c) find the value of k such that P(Z < k) = 0.0427. (d) find the value of k such that P(Z > k) = 0.2946.
7. The finished inside diameter of a piston ring is normally distributed with a mean of 10 centimeters and a standard deviation of 0.03 centimeters.
(a) What is the probability that piston ring will have an inside diameter between 9.97 and
10.03 centimeters?
(b) Below what value of inside diameter will 15% of the piston rings fall?
8. Find:
(a) the 95th percentile from a Standard Normal Distribution.
(b) the 90th percentile from a Standard Normal Distribution.
(c) the 99th percentile from a t-distribution with 13 degrees of freedom.
(d) the 95th percentile from a t-distribution with 9 degrees of freedom.
9. Generate 30 values from the interval [50, 200] allowing for repeats. Use a seed of 18. Report
the mean and standard deviation of your sample.
2
attachment
1 Worksheet
User generated content is uploaded by users for the purposes of learning and should be used following Studypool’s honor code & terms of service.
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+0
help
0
56
2
A square pyramid has base edge length 18 cm. The slant height of the pyramid is 41 cm. What is the volume of the pyramid?
Feb 9, 2020
#1
0
Answered by pish.glitch.me (go to pish.glitch.me/papers for more math help and review sheets!)
Looking at the prism, we see there is a right traingle with hypothenuse 41 and leg length of 18/2. Then, the other side must be 40 by pythagorean
theorem. Thus, the volume is $$\frac{1}{3}\cdot 18^2\cdot 40=4320$$
.
Feb 9, 2020
#2
+2847
0
I remember that website from AOPS
CalculatorUser Feb 9, 2020
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### Vault and vault-like contracts’ shared vulnerabilities
Now that we’ve introduced the ERC-4626 standard and identified the shared characteristics between ERC-4626 compliant vault contracts, non ERC-4626 vault contracts, and vault-like contracts, we can discuss the vulnerabilities that are shared among these contracts. For simplicity, when I refer to smart contracts in this post, I’m referring to all three categories of vault contracts.
#### Share Inflation
When a user deposits a quantity of an asset into a contract, they receive a quantity of a new asset, or share, proportional to the total deposited assets. This share may be a separate token, but in some cases it is simply an internal state variable. Therefore, the relationship between shares and underlying assets is a ratio, where the numerator is the total quantity of shares multiplied by the quantity of assets to be deposited and the denominator is the current total quantity of underlying assets.
This calculation is the source of our first vulnerability: share inflation. Share inflation occurs when an attacker manipulates the shares-to-deposited-assets ratio. There are two prerequisites that must be met in order for an attacker to manipulate this ratio:
1. The contract has no, or very low, deposit balance.
2. The contract’s share calculation denominator must be able to move independently of the numerator.
The first prerequisite is met by all ERC-4626, non ERC-4626, and vault-like contracts that have been created. One thing to note is that this prerequisite may not be required in all cases, but we’ll talk more about that in part three. The second prerequisite is a little trickier to understand, so let’s go over it.
As you may recall, we previously mapped the deposit calculation numerator to the total shares multiplied by the deposited asset quantity and the denominator to the total asset quantity in the contract. Taking this mapping into account, we can re-state the second prerequisite as:
The contract’s current total asset quantity must be able to move independently of the total shares multiplied by the deposited asset quantity.
##### Reward Mechanism
This statement is intrinsically true for all ERC-4626, non ERC-4626, and vault-like contracts because it enables the mechanism that allows contracts to reward users who have deposited assets into the contract. Rewards for a contract are transferred directly to the contract itself, thus inflating the value of each share that a user holds. When users redeem shares for assets, they will receive more assets than they had deposited due to the rewards that have been deposited directly into the contract. For example, if a contract has a quantity of 100 assets and 100 shares, then each share is worth 1 asset. If 100 assets are then transferred to the contract as rewards, then each share is now worth 2 assets.
It is important to note that the normal deposit mechanism cannot be used for adding rewards to a contract. If the deposit mechanism were to be used, shares would be created and given to the depositor (most likely another smart contract within a protocol), leaving the ratio of shares to total assets deposited unchanged. This approach would not provide any incentive for users to deposit funds.
##### Abusing the Reward Mechanism
The way users are rewarded for depositing assets into a contract is also the mechanism that attackers can use to manipulate the initial state of a contract into over-inflating the value of their shares. By being the first depositor, an attacker is able to set the quantity of underlying asset in the contract and set the initial ratio of shares to underlying asset. Typically, an attacker will deposit 1 wei of any asset and receive 1 share. This is because the first deposit cannot use the ratio calculation because it would involve division by zero. Therefore, the first deposit calculation simply returns a number of shares equal to the quantity of assets deposited.
For readers that have implemented ERC-4626 vaults, the above code snippet may look familiar. In fact, this code snippet comes from OpenZeppelin’s own ERC-4626 contract template v4.7! Surprisingly, very few ERC-4626 contract templates implement their own mitigation to the share inflation vulnerability. This means that share inflation is very likely to be present in protocols that implement ERC-4626. However, OpenZeppelin recently released a new version of their ERC-4626 template that mitigates share inflation. We will talk more about the mitigations for share inflation in the final part of this deep-dive series.
##### The Attack
Now that we understand the conditions required for this vulnerability to be present and have our code snippet reference, we can go over the attack scenario. Notice in the previous code snippet that the first deposit simply returns a quantity of shares in direct proportion to the quantity of assets deposited. This means if a user deposited 1 WETH, or 1e18 WETH due to decimals scaling, they would receive 1e18 shares.
Therefore, as the first depositor, an attacker will be able to set the ratio of shares to assets deposited to an arbitrarily small value by depositing 1 wei of an asset and receiving 1 share. Then, when a victim submits a transaction to deposit assets, the attacker initiates the second step of the attack. By directly transferring a quantity of asset to the contract similarly to how the contract itself would reward users, the attacker is able to manipulate the denominator of the deposit calculation independently of the numerator. If the attacker transfers enough assets such that the denominator would be greater than the numerator when a victim user goes to deposit, the attacker can cause division truncation during the calculation. Note that the attacker would need to place their direct deposit transaction immediately before the legitimate deposit transaction of the victim, so front-running would be necessary.
A brief note about division truncation: In Solidity, when you divide an integer by another integer, the expected type of the output is an integer (reference). Additionally, Solidity does not support floating or fixed point numbers, aka decimals. Therefore, when you divide an integer by another integer that would result in a decimal value, the decimal is rounded to zero. This is also known as truncation. For example, if you divide 5 by 2 in Solidity, the output will be 2 and not 2.5, as the 0.5 is truncated to 0. If you were to divide 2 by 5, the result would be 0.
Now back to the hack. By directly transferring enough assets to the contract before a victim deposits their own assets such that the denominator is greater than the numerator, the attacker is able to cause division truncation. This means that victims receive 0 shares for their deposits, and will not be able to withdraw the assets they just deposited because they have no shares that can be used to redeem their deposited assets! Then, when the attacker redeems 1 share, it is worth the full amount of assets that are in the contract, so 1 wei + the assets the attacker directly deposited + the victim’s deposited assets. The 1 share the attacker owns is now worth more than it was originally due to the attacker manipulating the contract, hence the term “share inflation.”
Let’s walk through an example scenario of how this attack would play out with real numbers:
1. A new ERC-4626, non ERC-4626, or vault-like contract is deployed, where the asset it takes in is WETH. WETH has a decimals value of 18.
2. Alice, the attacker, notices a new contract has been deployed. She deposits 1 wei of WETH into the contract and receives 1 share.
3. Bob, the victim, decides to deposit 1 WETH (1e18 wei) into the contract, expecting to receive 1e18 shares, based on the formula: ( (1 share * 1e18 wei) / 1 wei )
4. Before Bob’s transaction is finalized, Alice front-runs his transaction. She directly transfers 1 WETH to the contract.
5. Now, when Bob’s deposit is calculated, the calculation results in:
( (1 share * 1e18 wei) / (1e18 wei + 1 wei) )
Because the denominator is larger than the numerator, division truncation occurs, and Bob receives 0 shares for his 1 WETH deposit.
6. Alice’s 1 share is now worth 2 WETH + 1 wei, and Bob has lost his deposit.
Now that we’ve gone over the basics of share inflation, I’ll leave you with one last aside on this vulnerability. The above example primarily focused on the direct manipulation of the total assets in a contract. However, it is possible that an attacker is unable to directly transfer assets to a contract, but is still able to manipulate the total balance. A hypothetical scenario where indirect manipulation could occur is if an attacker is able to directly increase the amount of rewards granted to the contract. This would have the same effect of arbitrarily increasing the denominator independent of the numerator.
#### Incorrect Rounding
The second vulnerability shared among ERC-4626 compliant vault contracts, non ERC-4626 vault contracts, and vault-like contracts involves rounding, which is a problem that plagues many smart contracts. Specifically, for the three categories of vault contracts, rounding becomes an issue when depositing and withdrawing assets favors the user. Favorable rounding means that users receive more shares relative to the amount of assets that they deposited, and they receive more assets relative to the amount of shares that they redeem. This occurs when developers do not round in the correct direction relative to the function called by a user.
##### Rounding Up or Down
The ERC-4626 standard is very specific about which way to round when performing various operations. Paraphrased from the ERC-4626 standard, the specified rounding directions are as follows:
• If (1) the function is calculating how many shares to issue to a user for a certain amount of the underlying tokens they provide or (2) it’s determining the amount of the underlying tokens to transfer to them for returning a certain amount of shares, it should round down.
• If (1) the function is calculating the amount of shares a user has to supply to receive a given amount of the underlying tokens or (2) it’s calculating the amount of underlying tokens a user has to provide to receive a certain amount of shares, it should round up.
Unfortunately, the wording of the rounding specification from ERC-4626 is confusing, making it easy for developers who are coding vault implementations to misinterpret it. Let’s go over an example of favorable rounding:
In the above code snippet from our contract, we see that users can specify the amount of underlying asset they would like to withdraw. The function then uses a calculation to determine the corresponding amount of shares to subtract from the user. However, by rounding down, the function presents an opportunity for an attacker to manipulate the contract. If the attacker supplies an amount of asset they would like to withdraw such that the calculated shares amount is rounded down to 0, then they have effectively withdrawn assets for free (remember, Solidity rounds to 0 if the numerator is smaller than the denominator).
##### The Attack
Now let’s walk through a scenario using the above code snippet with real numbers:
1. Alice, the attacker, and Bob, the victim, have both deposited 1 WETH into a new contract. Alice and Bob both have 1e18 shares. Note that shares use the same order of magnitude as the underlying asset, and WETH has 18 decimal places.
2. The contract receives 1 WETH in rewards, which is directly transferred to the contract. The total assets in the contract are now 3 WETH.
3. Alice decides to withdraw 1 wei of WETH from the contract. The contract calculates the amount of shares to subtract from Alice using the following calculation:
(1 wei * 2e18 shares) / (3e18 wei)
Because the numerator is less than the denominator, 0 shares will be subtracted from Alice’s account.
4. Alice will receive 1 wei of WETH, but will not have any shares subtracted from her balance in the contract. Alice could repeat this process over and over again to drain the contract of its funds.
Notice in the above example that the denominator had to change independently of the numerator as a prerequisite for this attack. If the total supply of shares (numerator) and the total amount of assets (denominator) are the same, then it would not be possible for division truncation to occur. As mentioned previously, having the denominator move independently of the numerator is something that all ERC-4626 compliant, non ERC-4626, and vault-like contracts must be able to do in order to reward depositors. Additionally, Alice could have moved the denominator herself by directly transferring WETH to the contract. However, Alice would only be able to steal back 1 wei of WETH at a time, making this attack very unlikely to occur on a real contract due to the fact that the gas cost of this attack vector typically outweighs the rewards from the exploit. Alice could directly transfer a large amount of WETH in order to make the attack more profitable, but this would require orders of magnitude more WETH (100, 1,000, 10,000, etc.) which further limits the viability of this exploit.
There is another consequence of this rounding behavior. When Bob attempts to withdraw his expected total amount of assets, the contract will encounter an integer underflow and revert. This is because Alice has removed assets without having her shares amount decremented, causing a deficit in the contract.
### Conclusion
In the second part of our three part deep-dive, we went over two vulnerabilities that are shared among the three identified vault categories. Stay tuned for the final part of our deep-dive, where we’ll go over several proposed effective and ineffective mitigations for share inflation, and cover the mitigation for incorrect rounding.
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3.G.A.1 - Coherence Map | Math & ELA Coherence Map with Learning Resources
Math:
ELA:
# The coherence map shows the relationships among the standards across multiple topics and grade levels in the common core subjects Math & English (ELA). Coherence map helps teachers/educators diagnose the students mastery in specific topics and it will help students achieve success on that topics/standards. The Lumos coherence map not only provides graphical representation and convenient navigation within the standards map but also access to thousands of engaging learning resources such as Practice questions, Videos, Books, and Infographics related to every standard. It helps educators and students visually explore the learning standards. It's an useful tool to help students progress through the learning standards. Teachers can use this tool to develop their pacing charts and lesson plans....SHOW MORE
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Standard Description of 3.G.A.1: Understand that shapes in different categories (e.g., rhombuses, rectangles, and others) may share attributes (e.g., having four sides), and that the shared attributes can define a larger category (e.g., quadrilaterals). Recognize rhombuses, rectangles, and squares as examples of quadrilaterals, and draw examples of quadrilaterals that do not belong to any of these subcategories.
Overview:
Standards preceding 3.G.A.1 are 2.G.A.1. Standards that are taught after students master 3.G.A.1 are 4.G.A.1. The Lumos Coherence map for 3.G.A.1 includes standards correlations, description and a list of apps, questions, videos, books and other resources related to 3.G.A.1
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Standard Description of 3.G.A.1: Understand that shapes in different categories (e.g., rhombuses, rectangles, and others) may share attributes (e.g., having four sides), and that the shared attributes can define a larger category (e.g., quadrilaterals). Recognize rhombuses, rectangles, and squares as examples of quadrilaterals, and draw examples of quadrilaterals that do not belong to any of these subcategories.
Overview:
Standards preceding 3.G.A.1 are 2.G.A.1. Standards that are taught after students master 3.G.A.1 are 4.G.A.1. The Lumos Coherence map for 3.G.A.1 includes standards correlations, description and a list of apps, questions, videos, books and other resources related to 3.G.A.1
### What is the importance of coherence map?
The Coherence map gives educators a visual understanding of Grade Level Standards in both Math and English Language Arts (ELA). Teachers can effortlessly see how each grade level connects to previous and next levels along with the same standard.
### How to use coherence map?
Math and ELA standards for each Grade Level from 3rd grade up through high school are represented on a grid. Teachers simply click on the Grade Level/Content Area that they need to bring up a pop-up list of all the standards connected with it. It’s also possible to navigate to individual Standards by typing them manually into the search bar.
EdSearch WebSearch
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## (OPTIONAL) Remedial Workshop.ppt - Section 2: Moving Forward!
(OPTIONAL) Remedial Workshop.ppt
# Calculator Boot-Camp and Solving Exponential Equations
Unit 2: Exponential and Logarithmic Functions
Lesson 13 of 21
## Big Idea: This lesson places a high emphasis on teaching the students how to properly use a calculator when solving exponential/logarithmic equations.
Print Lesson
Standards:
Subject(s):
45 minutes
### Jarod Hammel
##### Similar Lessons
###### Simplifying Exponential Expressions, Day 1
Algebra II » Exponents & Logarithms
Big Idea: Students persevere in solving problems as they use their knowledge of the properties of exponents to simplify expressions involving rational exponents.
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###### Radioactive Decay and Nuclear Waste
12th Grade Math » Exponential and Logarithmic Functions
Big Idea: How long it will take for radioactive waste to reach a safe level?
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###### Using an Area Model to Multiply Binomials
Big Idea: Who needs FOIL? Students learn to multiply binomials by using a geometric model.
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Boston, MA
Environment: Urban
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https://gamedev.stackexchange.com/questions/105992/restricting-body-movement-withing-a-specific-length-while-dragging-a-gameobject
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Restricting body movement withing a specific length while dragging a gameobject in unity
private void OnMouseDrag()
{
Vector3 screen_sp=new Vector3(Input.mousePosition.x,Input.mousePosition.y,screen_space.z);
Vector3 cur_pos=Camera.main.ScreenToWorldPoint(screen_sp)+offset;
if(pendulum_object.gameObject != null && pendulum_object.gameObject.rigidbody!=null)
{
transform.position=cur_pos;
Debug.Log ("Pendulum is being dragged left"+transform.position);
Debug.Log("Mouse position is:"+Input.mousePosition);
Debug.Log("Current position is:"+cur_pos);
}
}
The problem is:The dragging of game object must be restricted to the length of string. The pendulum is joint using hinge joint to the world.
• Try to find normalized direction vector between the string start and the end after physical simulation, then try move object at the dir_vector*string_length and see what happens. I'm not sure if this is possible because I'm not familiar with Unity. This comment is all that I can suggest about your question. Aug 21, 2015 at 8:01
• I don't know what that means?I just need that the Gameobject must be restricted to the length of the string,which is the hingeJoint anchor to which it is connected. Aug 21, 2015 at 9:45
• – Anko
Oct 23, 2015 at 21:40
You should proceed in the following way
1) Get the input mouse position convert it to world co-ordinates
Vector3 direction = Camera.main.ScreenToWorldPoint (Input.mousePosition);
2) Now calculate the directional Vector3 form anchor of the hinge joint to the input mouse position.
Vector3 diff = direction - Anchor.transform.position;
3) The last step will be to restrict the Position of the pendulum along this directional vector as per the string length (float) . By normalising we get the direction , then we clamp the magnitude to max String Length.
pendulum_object.transform.position = diff.normalized * Mathf.Clamp (diff.magnitude, 0, StringLength);
Note: If the pendulum is a child of a game object and if the scale and position differs. Then you need to use pendulum_object.transform.localPosition and also take care of the scaling factor.
• Thank you for the respose,but that code didn't work.The pendulum now behave abnormally.I forgot to tell that the pendulum is swinging in y and z direction only.x axis position didn't change. Aug 22, 2015 at 7:27
• I have tested the code , I have used it once for on screen game pad to control the thumb from moving outside the surrounding circle. I am sure there must be some physics or hierarchy issue messing around. If you can share a project with the pendulum setup. I might be able to fix it. Aug 22, 2015 at 8:49
• This is the improved code,the pendulum is now restricted to y axis but no luck in the z axis. transform.position = new Vector3 (cur_pos.x,Mathf.Clamp (cur_pos.y,transform.position.y, pendulum_body.hingeJoint.connectedAnchor.y),Mathf.Clamp (cur_pos.z,-pendulum_object.transform.position.y,pendulum_object.transform.position.y)); Aug 24, 2015 at 8:10
• Sorry mate I am not expert enough to debug this logic , until I see the setup. Aug 24, 2015 at 8:21
• See the updated question. Aug 24, 2015 at 11:11
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# Week 2: Exponents – Day 5
1. If we know that a100 = 250 , what is a?
2. What is 31/2 x 31/4 ?
3. Which is bigger, 21/2 or 21/3 ?
4. Which is bigger, 2-2 or 21/2 ?
5. Which is bigger, (1/2)-2 or (1/2)1/2 ?
# Week 2: Exponents – Day 3
Negative Exponents, Parentheses, Signs and Simplification
“Simplifying” expressions with exponents usually means canceling out all common factors and rewriting them with only positive exponents.
1. Simplify 4 x 2-4.
2. Simplify -23 and 2-3. Are they the same number?
3. Simplify (-3)-2 and -3-2. Are they the same number?
4. Simplify 1 – 2-1
5. Simplify 1/2-3
# Week 2: Exponents – Day 2
Give exponential answers to these questions – in other words, your answer can involve numbers raised to an exponent.
1. On Monday, 10 bacteria live in a petri dish. Each bacterium divides into two bacteria every twenty-four hours. How many bacteria are in the dish the following Saturday?
2. What is the average rate of bacteria population growth in problem 1 between Monday and Tuesday? (The average growth rate, in bacteria per day, is the number of bacteria on Tuesday minus the number of bacteria on Monday divided by the number of days between Tuesday and Monday.)
3. What is the average rate of bacteria population growth in problem 1 between Friday and Saturday?
4. What is the average rate of bacterial population growth in problem 1 between Tuesday and Friday? (Think about what answer you expect to this problem. Do you expect the same answer as problem 2 or problem 3, or something different?)
5. How many bacteria are in the petri dish one month later (thirty days later)?
Bonus question: since 210 is approximately 1000, approximately how many bacteria are living in the dish after one month – a thousand? million? billion? trillion?
# Week 2: Exponents – Day 1
1. If we know that 2100 x 2n = 2300 what is n?
2. If we know that 210 / 2n = 8, what is n?
3. If we know that 22n x 23 = 211 what is n?
4. If we know that 220 = 4n what is n?
5. If we know that 913 = 3n what is n?
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https://www.geeksforgeeks.org/draw-color-filled-shapes-in-turtle-python/?ref=rp
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# Draw Color Filled Shapes in Turtle – Python
Prerequisite: Python Turtle Basics
turtle is an inbuilt module in python. It provides drawing using a screen (cardboard) and turtle (pen). To draw something on the screen, we need to move the turtle. To move turtle, there are some functions i.e forward(), backward(), etc.
To fill the colors in the shapes drawn by turtle, turtle provides three functions –
fillcolor(): This helps to choose the color for filling the shape. It takes the input parameter as the color name or hex value of the color and fills the upcoming closed geographical objects with the chosen color. Color names are basic color names i.e. red, blue, green, orange.
The hex value of color is a string(starting with ‘#’) of hexadecimal numbers i.e. #RRGGBB. R, G, and B are the hexadecimal numbers (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F).
begin_fill(): This function tells turtle that all upcoming closed graphical objects needed to be filled by the chosen color.
end_fill(): this function tells turtle to stop the filling upcoming closed graphical objects.
## Drawing Color Filled Square:
# draw color-filled square in turtle import turtle # creating turtle pent = turtle.Turtle() # taking input for the side of the squares = int(input("Enter the length of the side of the square: ")) # taking the input for the colorcol = input("Enter the color name or hex value of color(# RRGGBB): ") # set the fillcolort.fillcolor(col) # start the filling colort.begin_fill() # drawing the square of side sfor _ in range(4): t.forward(s) t.right(90) # ending the filling of the colort.end_fill()
Input :
200
green
Output :
## Drawing Color Filled Triangle:
# draw color filled triangle in turtle import turtle # creating turtle pent = turtle.Turtle() # taking input for the side of the triangles = int(input("Enter the length of the side of the triangle: ")) # taking the input for the colorcol = input("Enter the color name or hex value of color(# RRGGBB): ") # set the fillcolort.fillcolor(col) # start the filling colort.begin_fill() # drawing the triangle of side sfor _ in range(3): t.forward(s) t.right(-120) # ending the filling of the colort.end_fill()
Input :
200
red
Output :
## Drawing Color Filled Hexagon:
# draw color-filled hexagon in turtle import turtle # creating turtle pent = turtle.Turtle() # taking input for the side of the hexagons = int(input("Enter the length of the side of the hexagon: ")) # taking the input for the colorcol = input("Enter the color name or hex value of color(# RRGGBB): ") # set the fillcolort.fillcolor(col) # start the filling colort.begin_fill() # drawing the hexagon of side sfor _ in range(6): t.forward(s) t.right(-60) # ending the filling of the colort.end_fill()
Input :
100
#113300
Output :
## Drawing Color Filled Star:
# draw color filled star in turtle import turtle # creating turtle pent = turtle.Turtle() # taking input for the side of the stars = int(input("Enter the length of the side of the star: ")) # taking the input for the colorcol = input("Enter the color name or hex value of color(# RRGGBB): ") # set the fillcolort.fillcolor(col) # start the filling colort.begin_fill() # drawing the star of side sfor _ in range(5): t.forward(s) t.right(144) # ending the filling of colort.end_fill()
Input :
200
#551122
Output :
## Drawing Color Filled Circle:
# draw color filled circle in turtle import turtle # creating turtle pent = turtle.Turtle() # taking input for the radius of the circler = int(input("Enter the radius of the circle: ")) # taking the input for the colorcol = input("Enter the color name or hex value of color(# RRGGBB): ") # set the fillcolort.fillcolor(col) # start the filling colort.begin_fill() # drawing the circle of radius rt.circle(r) # ending the filling of the colort.end_fill()
Input :
100
blue
Output :
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# Federal incentives now encourage investing capital in commercial offic
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Re: Federal incentives now encourage investing capital in commercial offic [#permalink]
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richardkliao wrote:
293. Federal incentives now encourage investing capital in commercial office buildings despite vacancy rates in existing structures that are exceptionally high and no demand for new construction.
(A) investing capital in commercial office buildings despite vacancy rates in existing structures that are exceptionally high and
(B) capital investment in commercial office buildings, even though vacancy rates in existing structures are exceptionally high and there is
(C) capital to be invested in commercial office buildings even though there are exceptionally high vacancy rates in existing structures with
(D) investing capital in commercial office buildings even though the vacancy rates are exceptionally high in existing structures with
(E) capital investment in commercial office buildings despite vacancy rates in existing structures that are exceptionally high, and although there is
why is the the OA correct? It seems to me to be a run-on sentence ( Don't you need , before the conj. "and")? I find D quite attractive ... can someone explains?? Really confused now..
Using POE, the answer is B.
‘capital investment’ is preferable than ‘investing capital’ , so A, D out.
Also in (D): originally there are two reasons: 1 – high vacancy rates and 2 – no demand, and in this answer choice these two reasons become one “high vacancy rates in existing structures with no demand” and moreover, it is not very clear for me what "no demand" refers to: "vacancy rate" or "existing structures"
(C) – ‘encourage capital to be invested”, well, I can understand when we encourage students to study or children to be obedient, but I can’t imagine how can we encourage capital? C – out.
(E) - too wordy. “despite vacancy rates in existing structures that are exceptionally high, and although there is no demand”
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Re: Federal incentives now encourage investing capital in commercial offic [#permalink]
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ajit257,
I don't want to give away the reasoning here because I think it's more valuable if you give it a try on your own and then ask others to help you check your thinking. That said, I'll point you in the right direction:
When I look at SC, I immediately scan vertically to see what I'll have to decide. Here, some of those splits include:
-investing capital vs. capital investing
-despite vs. even though
-there is vs. with
If I'm considering these splits, I know that the most black and white grammatical rule relates to the use of "there is" or "with."
I'll be notified as soon as you reply to this. Let me know where you take this problem and then I'll be back to help out as necessary.
Good luck!
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Re: Federal incentives now encourage investing capital in commercial offic [#permalink]
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richardkliao wrote:
Federal incentives now encourage investing capital in commercial office buildings despite vacancy rates in existing structures that are exceptionally high and no demand for new construction.
(A) investing capital in commercial office buildings despite vacancy rates in existing structures that are exceptionally high and
(B) capital investment in commercial office buildings, even though vacancy rates in existing structures are exceptionally high and there is
(C) capital to be invested in commercial office buildings even though there are exceptionally high vacancy rates in existing structures with
(D) investing capital in commercial office buildings even though the vacancy rates are exceptionally high in existing structures with
(E) capital investment in commercial office buildings despite vacancy rates in existing structures that are exceptionally high, and although there is
Project SC Butler: Day 38: Sentence Correction (SC2)
why is the the OA correct? It seems to me to be a run-on sentence ( Don't you need , before the conj. "and")? I find D quite attractive ... can someone explains?? Really confused now..
Official Explanation:
In choices A, C, and D, investing capital and capital-to be invested are awkward and imprecise; the verb encourage more appropriately takes a noun such as investment for its object. Moreover, these choices and choice E contain ambiguous wording: that are exceptionally high in A and E, and with (no demand) in C and D, illogically seem to modify existing structures. In A and E, despite vacancy rates does not allow for a grammatically complete sentence with parallel construction. Choice B is best: capital investment is the proper object for encourage, modifying phrases are placed so as to avoid ambiguity, and the construction even though vacancy rates ... are ... high and there is (no demand) is grammatically parallel.
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Re: Federal incentives now encourage investing capital in commercial offic [#permalink]
But in answer option B, what is the antecedent of there?
As I could not find any antecedent for there, I prefered D to B.
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Re: Federal incentives now encourage investing capital in commercial offic [#permalink]
angarg wrote:
But in answer option B, what is the antecedent of there?
As I could not find any antecedent for there, I prefered D to B.
But why? I'm sure that you're comfortable using there is/are/was... to discuss something about something.
Okay, that was not a very precise way to put it . But that's what sentences like the ones below do.
1. There is no train at that time.
2. There are two students in the classroom.
This is common usage. Option D on the other hand, has errors that are "worse" than any possible ambiguity introduced by the there.
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Re: Federal incentives now encourage investing capital in commercial offic [#permalink]
Hi,
Can anyone explain how parallelism is maintained in B? Even though vacancy rates in existing structures are exceptionally high vs Even though no demand for new construction.
Also why C and D are wrong? I don't see any meaning errors in these two options.
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Re: Federal incentives now encourage investing capital in commercial offic [#permalink]
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krishnabalu wrote:
Hi,
Can anyone explain how parallelism is maintained in B? Even though vacancy rates in existing structures are exceptionally high vs Even though no demand for new construction.
Also why C and D are wrong? I don't see any meaning errors in these two options.
This is the sentence that option B leads to:
Federal incentives now encourage capital investment in commercial office buildings, even though vacancy rates in existing structures are exceptionally high and there is no demand for new construction.
(1) Federal incentives now encourage capital investment in commercial office buildings
even though
(2a) vacancy rates in existing structures are exceptionally high
and
(2b) there is no demand for new construction.
As for options C and D, we can't really say encourage money to be invested and encourage investing money. Also, existing structures with no demand makes it sound as if the existing structures don't have any demand.
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Re: Federal incentives now encourage investing capital in commercial offic [#permalink]
AjiteshArun wrote:
krishnabalu wrote:
Hi,
Can anyone explain how parallelism is maintained in B? Even though vacancy rates in existing structures are exceptionally high vs Even though no demand for new construction.
Also why C and D are wrong? I don't see any meaning errors in these two options.
This is the sentence that option B leads to:
Federal incentives now encourage capital investment in commercial office buildings, even though vacancy rates in existing structures are exceptionally high and there is no demand for new construction.
(1) Federal incentives now encourage capital investment in commercial office buildings
even though
(2a) vacancy rates in existing structures are exceptionally high
and
(2b) there is no demand for new construction.
As for options C and D, we can't really say encourage money to be invested and encourage investing money. Also, existing structures with no demand makes it sound as if the existing structures don't have any demand.
Can you help me in identifying how parallelism is maintained between 2a and 2b.
I know capital to be invested and investing capital in C and D aare awkward, but I am still unable to interpret why the last part
" with no demand for new construction" is wrong
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Re: Federal incentives now encourage investing capital in commercial offic [#permalink]
krishnabalu wrote:
Can you help me in identifying how parallelism is maintained between 2a and 2b.
I know capital to be invested and investing capital in C and D aare awkward, but I am still unable to interpret why the last part
" with no demand for new construction" is wrong
The and joins two clauses. Let's look at them separately:
2a. Federal incentives now encourage capital investment in commercial office buildings even though vacancy rates in existing structures are exceptionally high
and
2b. Federal incentives now encourage capital investment in commercial office buildings even though there is no demand for new construction.
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Re: Federal incentives now encourage investing capital in commercial offic [#permalink]
Hi AndrewN sir
Could you please share your thoughts on this question. I find hard to eliminate more options in this question.
On reading the posts above, I would like to know your approach on making decisions related with these three below points:
1. investment vs investing - you decide based on by meaning or suggest some rule?
2. There doesn't refer to any definite place( vacancy rates are not at some place ). Its better not to reject there immediately? Your advise.
3. high vacancy rates with no demand. with here just give extra information about demand . it should not be a main reason to reject. What's your take on this ?
Thanks in anticipation AndrewN sir
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Re: Federal incentives now encourage investing capital in commercial offic [#permalink]
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Hello, mSKR. I will respond to your queries below, but how about we take a look at the sentence and answer choices first (for reference)?
Quote:
Federal incentives now encourage investing capital in commercial office buildings despite vacancy rates in existing structures that are exceptionally high and no demand for new construction.
(A) investing capital in commercial office buildings despite vacancy rates in existing structures that are exceptionally high and
(B) capital investment in commercial office buildings, even though vacancy rates in existing structures are exceptionally high and there is
(C) capital to be invested in commercial office buildings even though there are exceptionally high vacancy rates in existing structures with
(D) investing capital in commercial office buildings even though the vacancy rates are exceptionally high in existing structures with
(E) capital investment in commercial office buildings despite vacancy rates in existing structures that are exceptionally high, and although there is
mSKR wrote:
Hi AndrewN sir
Could you please share your thoughts on this question. I find hard to eliminate more options in this question.
On reading the posts above, I would like to know your approach on making decisions related with these three below points:
1. investment vs investing - you decide based on by meaning or suggest some rule?
The only option I would eliminate based on the head of each answer choice is (C): encourage capital to be invested is passive and generally harder to follow than either of the other iterations. Encourage investing capital may be less preferred than encourage capital investment, since the -ing form of a verb after the verb encourage takes a moment to sort out (i.e. the -ing phrase would be considered a noun, answering the question, Encourage [what]?) However, both options use two words, the former focusing on capital, the latter on investment, so I would keep everything but (C) in the running to play it safe, even if (B) and (E) were the forerunners on this basis alone.
mSKR wrote:
2. There doesn't refer to any definite place( vacancy rates are not at some place ). Its better not to reject there immediately? Your advise.
No, there does not refer to a place, but you have to be careful not to consider a word in isolation, as there is/there are is a common construct, and that is just what we see in answer choices (B), (C), and (E). Because I would have eliminated (C) already, a quick check at the tail-end of (B) or (E) would suffice:
there is no demand for new construction
This makes perfect sense. I suppose the sentence could dodge hiding behind a there is by opting for a more active verb, as in, no demand exists for new construction, but that is not an option anyway. Leave (B) and (E) alone on this consideration.
mSKR wrote:
3. high vacancy rates with no demand. with here just give extra information about demand . it should not be a main reason to reject. What's your take on this ?
The transition from discussing vacancy rates in existing structures to a completely separate topic in demand for new construction is not clear when with is used right after existing structures. A reader is led to believe that existing structures with, or even existing structures with no demand for [something], will provide further information on these structures after the preposition, so it is jarring to encounter information on new construction instead. If there were any lingering questions about (C), I would hope that this third reason to eliminate it would suffice. (D) also looks poor compared to other options.
So, in light of these three considerations, I would disfavor (C) and (D). With the remaining answer choices, I would test the and at the end to see whether the part that followed was logical, grammatically sound, and, as a final consideration, parallel. Between (B) and (E), the two answers that start with capital investment, (E) is the poorer choice. Do we need both despite and although, for instance? It really comes down to (A) or (B) for me as the best of their kind. Using the and test, (A) pairs a phrase that is modified by a clause with a phrase:
despite vacancy rates in existing structures that are exceptionally high AND
[despite] no demand for new construction.
Meanwhile, (B) looks a little more promising, pairing a clause with another clause:
even though vacancy rates in existing structures are exceptionally high AND
[even though] there is no demand for new construction.
(B) is the better option, which is why we should choose it.
mSKR wrote:
Thanks in anticipation AndrewN sir
You are welcome, mSKR. I hope my response helps you.
- Andrew
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Re: Federal incentives now encourage investing capital in commercial offic [#permalink]
Hi,
I eliminated B for not using 'comma' before 'and'. I read in a book that to connect two independent clause we need comma plus one of the FANBOYS.
Please shed some lights; where am I wrong?
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Re: Federal incentives now encourage investing capital in commercial offic [#permalink]
Mck2023 wrote:
Hi,
I eliminated B for not using 'comma' before 'and'. I read in a book that to connect two independent clause we need comma plus one of the FANBOYS.
Please shed some lights; where am I wrong?
Hi Mck2023,
The clause you're looking at is part of a dependent clause (even though...), but, generally speaking, it's not a good idea to take options out on the basis of comma usage (except for a few issues related to commas that the GMAT does test, like comma splices).
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Re: Federal incentives now encourage investing capital in commercial offic [#permalink]
richardkliao wrote:
Federal incentives now encourage investing capital in commercial office buildings despite vacancy rates in existing structures that are exceptionally high and no demand for new construction.
(A) investing capital in commercial office buildings despite vacancy rates in existing structures that are exceptionally high and
(B) capital investment in commercial office buildings, even though vacancy rates in existing structures are exceptionally high and there is
(C) capital to be invested in commercial office buildings even though there are exceptionally high vacancy rates in existing structures with
(D) investing capital in commercial office buildings even though the vacancy rates are exceptionally high in existing structures with
(E) capital investment in commercial office buildings despite vacancy rates in existing structures that are exceptionally high, and although there is
why is the the OA correct? It seems to me to be a run-on sentence ( Don't you need , before the conj. "and")? I find D quite attractive ... can someone explains?? Really confused now..
Splits that helped me:
Quote:
....despite vacancy rates in existing structures that are exceptionally high....
This literally means that something happened despite the existence of vacancy rates (which are exceptionally high). But actually, this something happened despite high vacancy rates. These two phrases have different meaning. Mere existence of vacancy rates cannot trigger anything
(A) and (E) are out
Split number 2: "with" or no with
Logically "with" makes no sense here, as existing structures can never have demand for new construction
(C) and (D) are out
(B) makes perfect sense. Therefore, (B)
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Federal incentives now encourage investing capital in commercial offic [#permalink]
Hello expert,
I ruled out B for this reason: we know “AND” is a parallelism trigger, but I can not see any structure that can parallel to “there is no demand for new construction”.
I picked D, although I know it might be kind of ambiguous. According to RON, “with” can modify either preceding noun or preceding sentence, so I think “with” can modify the preceding sentence “vacancy rates in existing structures that are exceptionally high”.
Any expert can help? Thanks in advance.
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Re: Federal incentives now encourage investing capital in commercial offic [#permalink]
richardkliao wrote:
Federal incentives now encourage investing capital in commercial office buildings despite vacancy rates in existing structures that are exceptionally high and no demand for new construction.
(A) investing capital in commercial office buildings despite vacancy rates in existing structures that are exceptionally high and
(B) capital investment in commercial office buildings, even though vacancy rates in existing structures are exceptionally high and there is
(C) capital to be invested in commercial office buildings even though there are exceptionally high vacancy rates in existing structures with
(D) investing capital in commercial office buildings even though the vacancy rates are exceptionally high in existing structures with
(E) capital investment in commercial office buildings despite vacancy rates in existing structures that are exceptionally high, and although there is
why is the the OA correct? It seems to me to be a run-on sentence ( Don't you need , before the conj. "and")? I find D quite attractive ... can someone explains?? Really confused now..
Hello richardkliao,
We hope this finds you well.
To answer your query, the clauses "vacancy rates in existing structures are exceptionally high" and "there is no demand for new construction" form the parallelism pair in Option B.
We hope this helps.
All the best!
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Re: Federal incentives now encourage investing capital in commercial offic [#permalink]
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Mavisdu1017 wrote:
Hello expert,
I ruled out B for this reason: we know “AND” is a parallelism trigger, but I can not see any structure that can parallel to “there is no demand for new construction.”
"There is no demand" is an independent clause. All we need is an independent clause earlier in the sentence that can be parallel to this one. "Federal incentives now encourage investing capital in commercial office buildings," is also an independent clause, so the parallelism is fine.
Quote:
I picked D, although I know it might be kind of ambiguous. According to RON, “with” can modify either preceding noun or preceding sentence, so I think “with” can modify the preceding sentence “vacancy rates in existing structures that are exceptionally high”.
Any expert can help? Thanks in advance.
Part of the problem with (D) is that "with" could function in multiple ways and none of them make much sense. Take another look:
Quote:
The vacancy rates are exceptionally high in existing structures with no demand for new construction.
One interpretation is that "with no demand" is modifying the noun "structures." What kind of structures have high vacancy rates? The kind of structures with no demand for new construction. This interpretation seems debatable, at best -- if the structures already exist, you can't really say they're with low demand for new construction, can you? What would that even mean? That the few people who live in the existing low-vacancy buildings don't want new buildings? Nah.
But if we interpret the "with" modifier to describe the clause, "vacancy rates are exceptionally high" it's still off. How can the lack of demand for new construction describe the vacancy rates of current buildings?
So now in (D) we've got multiple meanings that aren't great. Contrast that with (B), which introduces a new clause altogether. This seems way more logical. One clause talks about what's happening in existing buildings -- they have low vacancy. And the other talks about demand for new construction -- it's low.
If (B) is clearer and more logical, it's better.
I hope that clears things up!
Re: Federal incentives now encourage investing capital in commercial offic [#permalink]
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# How to measure mutual fund risks – five statistical tools
Are you a mutual fund investor? Do you want to invest in mutual fund schemes while remaining aware of the associated risks? Read this article which describes the five statistical tools which are used to understand the risks associated with mutual fund schemes.
Many investors rely upon mutual funds for short-term and long-term investments. Generally it has been seen that on long-term basis, mutual funds give better return than that of various fixed rate investments like provident funds. However, mutual funds carry a higher amount of risk. So it is imperative for an investor to measure risks associated with mutual funds. There are five statistical tools to check the risk associated with mutual fund investments. These five tools are Standard Deviation, Alpha, Beta, Sharpe ratio and R-Squared. In this article, we will discuss about these five statistical tools.
## Standard Deviation- a very common statistical tool
Standard Deviation is a very well known statistical tool. This is widely known to the students of Statistics and Mathematics. This tool can be utilized for checking the volatility of particular mutual fund(s). It measures the dispersion of data from the mean. The more the data is spread apart, the higher the difference is from the norm. In respect of mutual funds, the Standard Deviation tells the investor how much the return is deviating from the expected return based on the fund's historical performance.
## Alpha to measure risk-adjusted return
Alpha is a very important tool to measure the risk adjusted return of Mutual Fund. It is a measure of a mutual fund's performance on a risk- adjusted basis. The excess return of the investment relative to the return of the benchmark index is its Alpha. For example, a positive Alpha of 1.0 means the fund has outperformed its benchmark index by 1%. Similarly a negative Alpha of 2.0 indicates under-performance by 2%. So the more positive the Alpha is, the more better it is.
## Beta-a measure of volatility
Beta is a measure of the volatility, or systematic risk, of a mutual fund in comparison to the market as a whole. Beta is calculated using regression analysis, and it is the tendency of an investment's return to respond to swings in the market. By definition, the market has a Beta of 1.0. Individual security and portfolio values are measured according to how they deviate from the market. A Beta of 1.0 indicates that the investment's price will move exactly with the market. A Beta of less than 1.0 indicates that the investment will be less volatile than the market; and, a Beta of more than 1.0 indicates that the fund's price will be more volatile than the market.
## Sharpe ratio developed by William Sharpe
Sharpe ratio has been developed by the Nobel laureate economist, William Sharpe. This ratio measures risk-adjusted performance. It is calculated by subtracting the risk-free of return from the rate of return for an investment and dividing the result by the investment's standard deviation of its return. The Sharpe ratio tells the investors whether an investment's returns are due to specific investment decisions, or, the result of excess risk taken by the fund manager. The greater is a fund's Sharpe ratio, the better is its risk-adjusted performance.
## R-squared-another important statistical tool
R-Squared is a statistical measure that represents the percentage of fund portfolios or security movements that can be explained by the movements of the benchmark index. Mutual fund investors should generally avoid actively managed mutual funds with high R-Squared ratios.
The mutual fund analysts and experienced investors use these five statistical tools to measure the risk associated with a mutual fund scheme. These ratios are available in various websites which deal with mutual fund schemes. However, new investors must remember that all the data regarding these five tools are some ratios. So, for the sake objective comparison, the corresponding ratios of similar nature of funds must be taken into account.
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Mutual Fund or ULIP? Many of us have this question in mind. This resource will help you to understand the difference between Mutual Fund and a ULIP.
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This article explains various measures which keep acne under control. This decreases inflammation and consequently chances of formation of acne scars too decreases. Actions like not picking at skin, proper hydration, using sunscreen lotions, proper skin care regime etc prevent formation of acne scars.
#### How to choose the right mutual fund scheme?
A mutual fund pools the funds of innumerable investors to invest in shares or securities defined by the asset allocation policy of the mutual fund scheme. Are you also planning to invest your hard earned money in mutual fund scheme? Then, read this article for an in-depth understanding of how to select the right mutual fund scheme to avoid losses and earn profits.
#### The best and profitable mutual fund investment plans in 2012
New year has began. This is the most suitable time for investing in mutual funds. This article gives a list of mutual funds which assures the probability of maximum profit at present. It also deals with tax savings funds, Fidility tax advantage, Religer tax plan, ICICI prudential tax plan and ING tax savings. I have also written about important gold funds. Read on.
#### How To Control Anger?
Have you ever been angry and later on felt sad for being angry on some one or something? Do you want to control your anger ? Is anger good or not ? The following article will discuss about anger and measures through which we can control our anger.
More articles: How to Mutual funds
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help-octave
[Top][All Lists]
## Re: Octave coding
From: Shai Ayal Subject: Re: Octave coding Date: Fri, 16 Jun 2006 09:22:19 +0300 User-agent: Thunderbird 1.5.0.4 (Windows/20060516)
```Michael,
```
It seems like you are asking for someone to write code for you. This is beyond the scope of this list, in which *volunteers* offer help to people which have trouble coding. Perhaps if you were willing to pay, someone would be willing to invest the time and write the code for you.
```
Shai
MICHAEL FORREST wrote:
```
```Dear all,
```
I would like Octave code to implement the following equation system and to plot (v) vs. time. However, I am at a complete loss how to do this. Could anyone help? I would be so, so grateful. I am under quite a bit of time pressure and haven't the time to get my Octave programming up to the heights required to do this task in time. Please don't hesitate to write back if you require any more info. I am using Octave as a module of cygwin on a Windows XP machine. I have the GNUPlot module installed as well as Cygwin.
```Once again. Thank you so, so much. All the best,
Michael
NB.
v' denotes dv/dt
b' denotes db/dt
h' denotes dh/dt
-------------
// Master equation
v' = -(ina+ik+ih+il)
// ina
ina = 0.000060*minf*(v-55)
minf = (1+exp(-(v--53.8)/3))^-1
// ik
ik = 0.000105*b*(v--85)
b' = ((binf - b)/btau)
binf = (1+exp(-(v--54)/5))^-1
btau = (3000 * (1/(cosh((v--54)/(4*5)))))
// ih
ih = 0.000200*h*(v--30)
h' = ((hinf - h)/htau)
hinf = (1+exp((v--76.4)/20))^-1
```
htau = (1/((((-2.89 * v) + -445)/(1-exp(((v+-445)/-2.89)/24.02))) + (((27.1 * v) + -1024)/(1-exp(((v+-1024)/27.1)/-17.4)))))
```
// Ileak
il = 0.000100*(v--70)
--------------
_______________________________________________
Help-octave mailing list
```
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Requests for technical support from the VASP group should be posted in the VASP-forum.
# Algorithms used in VASP to calculate the electronic groundstate
The following section discusses the minimization algorithms implemented in VASP. We generally have one outer loop in which the charge density is optimized, and one inner loop in which the wavefunctions are optimized.
Fig. 1:Electronic minimization flowchart
Most of the algorithms implemented in VASP use an iterative matrix-diagonalization scheme, where the used algorithms are based on the conjugate gradient scheme [1][2], block Davidson scheme [3][4], or a residual minimization scheme -- direct inversion in the iterative subspace (RMM-DIIS) [5][6]. For the mixing of the charge density an efficient Broyden/Pulay mixing scheme [6][7][8] is used. Fig. 1 shows a typical flow-chart of VASP. Input charge density and wavefunctions are independent quantities (at start-up these quantities are set according to INIWAV and ICHARG). Within each selfconsistency loop, the charge density is used to set up the Hamiltonian, then the wavefunctions are optimized iteratively so that they get closer to the exact wavefunctions of this Hamiltonian. From the optimized wavefunctions, a new charge density is calculated, which is then mixed with the old input-charge density.
The conjugate gradient and the residual minimization scheme do not recalculate the exact Kohn-Sham eigenfunctions but an arbitrary linear combination of the NBANDS lowest eigenfunctions. Therefore it is in addition necessary to diagonalize the Hamiltonian in the subspace spanned by the trial-wavefunctions, and to transform the wavefunctions accordingly (i.e. perform a unitary transformation of the wavefunctions, so that the Hamiltonian is diagonal in the subspace spanned by transformed wavefunctions). This step is usually called sub-space diagonalization (although a more appropriate name would be, using the Rayleigh Ritz variational scheme in a sub space spanned by the wavefunctions):
${\displaystyle \langle \phi _{j}\ \vert {\bf {H}}\vert \phi _{i}\rangle ={\rm {H}}_{ij},}$
${\displaystyle H_{ij}U_{jk}=\epsilon _{k}U_{ik},}$
${\displaystyle \phi _{j}\leftarrow U_{jk}\phi _{k}.}$
The sub-space diagonalization can be performed before or after the conjugate gradient or residual minimization scheme. Tests we have done indicate that the first choice is preferable during self-consistent calculations.
In general, all iterative algorithms work very similarly. The core quantity is the residual vector
${\displaystyle |R_{n}\rangle =({\bf {H}}-E)|\phi _{n}\rangle \quad {\mbox{with}}\quad E={\frac {\langle \phi _{n}|{\bf {H}}|\phi _{n}\rangle }{\langle \phi _{n}|\phi _{n}\rangle }}.}$
This residual vector is added to the wavefunction ${\displaystyle \phi _{n}}$, the algorithms differ in the way this is exactly done.
Alternately, the groundstate energy can be evaluated with the random phase approximation.
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Search a number
136246 = 2112563
BaseRepresentation
bin100001010000110110
320220220011
4201100312
513324441
62530434
71105135
oct412066
9226804
10136246
1193400
1266a1a
134a026
143791c
152a581
hex21436
136246 has 12 divisors (see below), whose sum is σ = 225036. Its totient is φ = 61820.
The previous prime is 136237. The next prime is 136247. The reversal of 136246 is 642631.
Adding to 136246 its reverse (642631), we get a palindrome (778877).
It is a Harshad number since it is a multiple of its sum of digits (22).
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (136247) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (7) of ones.
It is a polite number, since it can be written in 5 ways as a sum of consecutive naturals, for example, 40 + ... + 523.
It is an arithmetic number, because the mean of its divisors is an integer number (18753).
2136246 is an apocalyptic number.
136246 is a deficient number, since it is larger than the sum of its proper divisors (88790).
136246 is a wasteful number, since it uses less digits than its factorization.
136246 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 587 (or 576 counting only the distinct ones).
The product of its digits is 864, while the sum is 22.
The square root of 136246 is about 369.1151581824. The cubic root of 136246 is about 51.4566197677.
The spelling of 136246 in words is "one hundred thirty-six thousand, two hundred forty-six".
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PDA
View Full Version : Need some assistance with java!!
prostuff1
May 19, 2006, 03:36 AM
Hello all. I am working on some homework and i can't seem to figure the whole thing out completely. I have ben going back and forth and just cant seem to figure it out.
OK, an explination of what i need to do:
I need to get the output of this little program to look like this:
How many numbers? 100
How many intervals? 10
Histogram
--------------------------------------------------------
1 ****(4)
2 ******(6)
3 ***********(11)
4 *****************(17)
5 **************************(26)
6 *************************(25)
7 *******(7)
8 ***(3)
9 (0)
10 *(1)
--------------------------------------------------------
And The program performs the following actions:
1. Ask the user to enter the number of numbers that will be generated and the number of intervals that will be used in the histogram, and input the two values;
2. Generate the required number of pseudo-random numbers by using the double[] Generator.getData(int n), which, given the number of numbers to generate, n, returns an array of double of size n containing the generated numbers;
3. Compute the range of data values (by finding the maximum and the minimum and taking their difference); then compute the width of each equally-sized interval (by dividing the whole range by the number of intervals specified by the user); finally, compute the number of data numbers falling in each interval and store these frequencies in an array of integers;
4. Output the histogram of the data as displayed in the sample run above: each bar starts with an index (1 through the number of intervals), followed by a number of '*' equal to the number of numbers in that range, followed by the number of '*' in parentheses.
While that does not seem to hard i seem to be having some trouble.
Right now this is what my program looks like:
import java.util.Scanner;
public class messwith {
/**
* @param args
*/
public static void main(String[] args)
{
Scanner keyboard = new Scanner(System.in);
System.out.print("How many numbers? ");
int numbers = keyboard.nextInt();
System.out.print("How many Intervals? ");
int intervals = keyboard.nextInt();
double[] rand_num = Generator.getData(numbers);
double max = max(rand_num);
double min = min(rand_num);
double inter_size = intervalSize(max, min, intervals);
int[] amount = numbers(rand_num, inter_size, min, intervals);
outPutResults(numbers, rand_num, amount);
}
private static double max(double[] rand_num)
{
double max = 0;
for(int index = 0; index < rand_num.length; index++)
{
if(rand_num[index] > max)
{
max = rand_num[index];
}
}
return max;
}
private static double min(double[] rand_num)
{
double min = 0;
for(int index = 0; index < rand_num.length; index++)
{
if(rand_num[index] < min)
{
min = rand_num[index];
}
}
return min;
}
private static double intervalSize(double max, double min, int intervals)
{
double range = max - min;
double width = range/intervals;
return width;
}
private static int[] numbers(double[] rand_num, double inter_size, double min, int intervals)
{
int[] numbers = new int[intervals];
int index = 0;
double count_interval = min + inter_size;
while(index < intervals)
{
int begin = 0;
int count = 0;
while(begin < (intervals-1))
{
if(rand_num[begin] >= min && rand_num[begin] < count_interval)
{
count++;
}
numbers[index] = count;
}
}
return numbers;
}
private static void outPutResults(int numbers, double[] rand_num, int[] amount)
{
int index = 0;
System.out.println("Histogram");
System.out.println("--------------------------------------------------------");
while(index < numbers)
{
System.out.print((index + 1) + " ");
while(index < rand_num[index])
{
System.out.print("*");
}
System.out.println("(" + rand_num[index] + ")");
index++;
}
System.out.println("-------------------------------------------------------");
}
}
While i thought this would work i seem to be having troubles. I cant seem to get it to print anything besides this:
How many numbers? 10
How many Intervals? 5
Histogram
--------------------------------------------------------
As you can see i have some problems in there that i cant seem to figure out. I was hopfing to get this done by friday afternoon cause this has to be turned in sunday night. I am not going to be at the school this weekend so i am trying to turn it in early.
If anyone can give me some help it would be appreciated!!!
Thanks
HiRez
May 19, 2006, 04:37 AM
It seems that something is causing while(index < numbers) to evaluate to true on your first test. Because you have just set index to be zero, that must mean something has caused the numbers variable to be less than or equal to zero, so I would investigate that first.
But, something else is wrong here. Even if that while loop never gets executed, you should see both lines of dashes printed out, because there is no way to break out of that method before it exits naturally by falling off the end. Because of that, I suspect your program must be terminating prematurely with some error or crash -- is there any error message being printed out?
Also "output" is one word, not two, so it shouldn't be intercapped. :)
prostuff1
May 19, 2006, 04:54 AM
It seems that something is causing while(index < numbers) to evaluate to true on your first test. Because you have just set index to be zero, that must mean something has caused the numbers variable to be less than or equal to zero, so I would investigate that first.
But, something else is wrong here. Even if that while loop never gets executed, you should see both lines of dashes printed out, because there is no way to break out of that method before it exits naturally by falling off the end. Because of that, I suspect your program must be terminating prematurely with some error or crash -- is there any error message being printed out?
Also "output" is one word, not two, so it shouldn't be intercapped. :)
Thanks for the help.
It does not give me any other errors cause the program never terminates. I have let it run for a couple minutes and i still cant get anything more then what i put in my first post.
As for output thanks. I was typing fast and did not realize that i had capitalized it.
I will look at my program again and keep the suggestions coming.
thanks
May 19, 2006, 09:24 AM
Your program seemed to be pretty slow, and an important lesson in programming is to keep it simple.;)
Basically I rewrote the entire thing into the main method as that was easier to work with (especially as it goes through in a fixed order), I couldn't use the generator class as I guess that's a specific special one for your course so you'll have to re-add it back in, also you'll need to get it to print the values as you want them, currently it goes and prints how many values there are in each interval.
I hope this is useful :)
import java.util.Scanner;
import java.util.Random;
public class messwith {
/**
* @param args
*/
public static void main(String[] args){
Scanner keyboard = new Scanner(System.in);
System.out.print("How many numbers? ");
int numbers = keyboard.nextInt();
System.out.print("How many Intervals? ");
int intervals = keyboard.nextInt();
//NEW CODE
Random rGen=new Random();//creates a random number generator.
double[] rand_num = new double[numbers];//creats an array to store them in.
for(int count=0;count<numbers;count++){
rand_num[count]=rGen.nextDouble();
}
//now find the max,min values,
double max=rand_num[0];
double min=rand_num[0];
for(int count=1;count<numbers;count++){
if(rand_num[count]>max){
max=rand_num[count];
}
if(rand_num[count]<min){
min=rand_num[count];
}
}
//System.out.println("max="+max+", min="+min); I used this for testing, you may want to re-enable it.
double intervalSize=(max-min)/intervals;
//System.out.println("Interval size="+intervalSize); I also used this for testing, you may want to re-enable it.
int[] Results=new int[intervals];//counts of the values in each interval are stored in this array
for(int count=0;count<numbers;count++){//processes the rand_num array
Double CValue=(rand_num[count]-min)/intervalSize;
int MEnt=CValue.intValue();
if(MEnt==intervals){//to subtract 1 from the final value so it goes in the right group.
MEnt--;
}
Results[MEnt]++;
}
System.out.println("A histogram of the results");
for(int count=0;count<intervals;count++){
System.out.println(Results[count]);
}
}
}
You will need to interate a print, inside the final for block to print the stars that you want.
prostuff1
May 21, 2006, 02:02 AM
Thanks for the help.
I kinda knew what i was doing but i ended up asking the teacher what exactly the program was supposed to do cause the direction were kinda vague on this one. Once i knew what the program was supposed to do i had a better idea what was going on.
If you dont mind could you explain (or anyone else) what this piece of code is exactly doing:
Double CValue = (rand_num[count]-min)/interval_size;
int MEnt = CValue.intValue();
if(MEnt == intervals)
{
MEnt--;
}
result[MEnt]++;
I am not quite sure wgat exactly it is doing.
Thanks
Svennig
May 21, 2006, 02:57 AM
Disclaimer: I'm away from home at present and don't have access to a machine to check if this compiles and executes properly. However, it will do as a proof of concept :)
This code becomes MUCH easier to understand if you sort the array (plus it's much quicker for larger arrays)
import java.util.Scanner;
import java.util.Random;
public class messwith {
/**
* @param args
*/
public static void main(String[] args){
Scanner keyboard = new Scanner(System.in);
System.out.print("How many numbers? ");
int numbers = keyboard.nextInt();
System.out.print("How many Intervals? ");
int intervals = keyboard.nextInt();
//NEW CODE
Random rGen=new Random();//creates a random number generator.
double[] rand_num = new double[numbers];//creats an array to store them in.
for(int count=0;count<numbers;count++){
rand_num[count]=rGen.nextDouble();
}
//Iterating through the array to find the maximum and
//minimum is inefficient, as it scales O(n).
//
//A better way of doing it is to sort the array, which gives
//much better performance, and then take the first and last values.
//This also makes it MUCH easier to do the groupings at the end.
Arrays.sort(rand_num);
//now find the max,min values,
double max=rand_num[rand_num.length - 1];
double min=rand_num[0];
//Now, we must calculate the interval size and make an array to store
//the results.
double intervalSize=(max-min)/intervals;
int[] Results=new int[intervals];
//Set the contents of the results array to zeros
Arrays.fill(Results, 0);
//Iterate through the results, grouping them.
//
//Because the set is sorted this is quite simple. The
//array of values is in ascending numerical order. Therefore
//we need to check whether it is between the lower and upper
//bounds of the interval and then count them.
//
//If we go over the bounds, we increment the interval and continue.
double highValue = intervalSize + min;
int currentInterval = 0;
for(int count=0;count<numbers;count++){
double value = rand_num[count];
//If the value is less than the current highest
//acceptable value for the interval, then we
//increment the number of results in the interval
if(value =< highval) {
Results[currentInterval]++;
}
//If the value is greater than the current
//highest value, then we have moved into the next
//Interval. So, we must increment the high value,
//increment the currentInterval, increment the results
//of the new interval
if(value > highval) {
currentInterval++;
highValue += intervalSize;
Results[currentInterval]++;
}
}
System.out.println("A histogram of the results");
for(int count=0;count<intervals;count++){
System.out.println(Results[count]);
}
}
}
Same comment as previous, you will need to do some work to make the stars work ;)
May 21, 2006, 06:02 AM
Thanks for the help.
I kinda knew what i was doing but i ended up asking the teacher what exactly the program was supposed to do cause the direction were kinda vague on this one. Once i knew what the program was supposed to do i had a better idea what was going on.
If you dont mind could you explain (or anyone else) what this piece of code is exactly doing:
Double CValue = (rand_num[count]-min)/interval_size;
int MEnt = CValue.intValue();
if(MEnt == intervals)
{
MEnt--;
}
result[MEnt]++;
I am not quite sure wgat exactly it is doing.
Thanks
I guess it's irrelavent as Svennig has done it in a better way than me, his method is better as it only iterates through the data once not 3 times, and it also is easier to understand how it checks the interval.
In my method above the first line takes off the minimum value and then divides by the interval size, this gives you a double between 0.0 and the number of intervals, you take off the minimum as the intervals are effectively shifted down by that amount. Then this is converted to an integer which takes off the decimal places of the number. (This gives the line of the Results array it is added to at the end.) However for the maximum value it will have exactly the number of interval's which means it's too big for the array so in that case you need to subtract 1 so it goes into the largest group.
then finally you add one to the value in the Results array.
MarkCollette
May 23, 2006, 12:45 PM
Two things pop to mind:
1. Your min method should default its min variable to a very large value, like Double.MAX_VALUE, instead of zero. While you're in there, change the max method's max default value from zero to Double.MIN_VALUE.
2. In the numbers method, I don't see the index variable being incremented.
mrichmon
May 24, 2006, 05:32 PM
I guess it's irrelavent as Svennig has done it in a better way than me, his method is better as it only iterates through the data once not 3 times, and it also is easier to understand how it checks the interval.
However Svennig's approach relies on sorting the Array. Sorting this array is going to be an expensive process -- according to the Java API docs this is O(n*log(n)). This is a common CS problem and the most efficient solution does not involve sorting the Array.
You want to do two passes traversals of the array. First pass finds max and min values -- set foundMax and foundMin values to the first value of the array then iterate over the array comparing the foundMax and foundMin to the current element. If the current element is greater than foundMax set foundMax to current element value. If the current element is greater than foundMin then set foundMin to current element value.
Then, calculate the interval width and create an array of histogram bins.
Second traversal, subtract minValue from current element then divide by interval width. This will give you the bin number that should be incremented.
Once the second traversal is complete print the histogram based on the contents of the histogram bins.
This algorithm is O(2n).
mbabauer
May 24, 2006, 09:37 PM
Oh, this brings me back to the "good ol' days of collage". Boy I miss those totally irrelavant projects that teach you nothing.
Ok, let me see if I can restate the problem: You need a app, that when given an interval (number of samplings) and a number (the total number of readings) it will calculate a nifty histogram showing only the intervals.
My first question is: What happens to the numbers that do not fall in the interval? Are they thrown out, calculated into some final "average"...it didn't seem too clear.
That aside, my suggestion(s) would be such:
Dont have a separate min and max method. This is WAY too much work. I know it seems like good OOP to split them out, but in practice it creates more work and a slower program. It also gives you more chances to screw up. Have one for loop, and two variables (min and max) set to the first value in the array. As you iterate the for loop, check each number in the array to see if its > max or < min, and set them accordingly
You seem to be using this "number" variable a lot. Note that all arrays, being objects themselves, have a property called "length". Length is the number of elements in the array, so why keep passing it around if you don't have too. Again, doing so provides more opportunities to screw up
Get yourself a good IDE with a debugger, and learn to use it. When you are first starting out, its OK to use notepad/TextMate, but as you progress, a good IDE is invaluable. If you get Eclipse, and have an Intel-based mac, get the 3.2 release, as its Universal Binary and the 3.1 doesn't work at all. Eclipse can be found at www.eclipse.org
If an IDE is totally out of the question, and you are stuck with a text editor and your JDK, then use my old debug technique: Print, print, print. Put a System.out.println() on ever other line! It gets messy, but this is a temporary thing. Print out variables and their values, print out markers saying stuff like "I am about to execute XXX" and "Just did XXX", or simply print "At line XXX". This will help you to learn to visualize what the program IS doing vs. what you THINK it is doing
This is the best advise I can give you, other than stick with it. Its tempting to have someone just write the code for you, but if you are looking at a career in the field, you have to ask yourself "What am I really learning here...".
Good luck
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Card games to play with kids
I’ve tried a few times and can’t seem to find one that works. Too easy for me, too tough for the kids, no fun etc.
What’s worked for you at different ages?
While we’re at it, how about board games?
I learned to play blackjack at an early age. I loved math and tried to learn how to count cards in a 52 card deck. Contract Bridge is also easy to learn for many kids. Hearts is also a good choice.
Chess is an excellent board game to learn.
I also played Trivial Pursuit, Scrabble, Backgammon, and Risk.
I’ve been playing poker since I was nine, euchre since I was twelve.
We’re big on cards in my family.
For younger kids, I’d recommend rummy, gin, Michigan rummy, and Crazy 8’s.
Uno, Pit, Skipbo (or the face card version called Spite and Malice), I Doubt It, Stealing Bundles, Pig, and Snap.
Casino. Good way to teach adding, too.
If those rules are too confusing, here’s the gist of the game:
Four cards are placed face-up in the center and each player has four cards in their hand. The object is to capture the face-up cards. This can be done by matching cards in pairs (or more), adding vaules of two (or more) cards, or by both matching cards and adding values.
That didn’t help, did it? I’ll explain further. [ul][]Players take turns by placing a card face-up on the table or capturing one of the face-up cards.[]If a player holds a seven and there’s a three and a four on the table, the seven can take both cards.[]If there’s a three, a four, and a seven on the table, the player can make seven out of the three and four and take all three cards.[]If a player holds two sevens, they can make a seven out of the three and four and place one seven on top of them, intending to take the three cards on the next turn. However, another player with a seven can do the same on their turn.[]10s and face cards can only be combined with cards of the same value.[]Players cannot combine a three and a four to take a seven.[]If a player cannot make a legal play, they must lay down one of their cards face-up with the others.[]Once each player has played their four cards, they get four more.Play cotinues until the deck is run through.[/ul]
I know from personal experiance that a 6-year old can play Texas Hold 'Em.
Poker was the first card game I learned to play followed by Crib. Kids pick up just about any card game very quickly.
My kids also particularly liked Uno and Uno Hearts, Scrabble, Backgammon, Chess, Monopoly, Game of Life, and just about anything I was willing to play with them.
You have to get a special set of cards, but Mille Borne is the best. I loved that game as a kid and I think I’m going to get a set and start playing it again.
Oh, here’s an Amazon link for Mille Bornes. (I forgot the “s” at the end there.) Read the customer reviews. A lot of people love it as much as I did.
Old Maid can be played with a regular deck. Just throw in a Joker.
Concentration, where you lay out the cards face down for the kids to match up, is a good one too. Depending on how old the kidlets are, you may not want to use all 52.
We used to play “Fleece the Father” all week long on family vacations, during otherwise idle times.
This is poker, with Dad giving everyone a pile of money and then trying to win it back from them. Kids are allowed to cash in any time they want, and don’t spend any of their own money. Each child gets maybe \$5 or so in chips, or in actual coinage (the merry chink of pennies hitting the pot adds to the fun).
Straight 5-card poker is the easiest to teach, though it’s rarely played. You deal 5 down cards, have a round of betting, then the showdown.
5 card draw is quite easy for kids to learn. 5 down cards, round of betting, draw, another round of betting, then showdown.
It can be tedious when every dealer gets to name his own game. 9-card stud with one eyed jacks wild and odd diamonds dead after first up queen is just about impossible to imagine strategy for when you were already thinking about bed time.
One downside - little kids never want to fold, which messes up bluffing.
For younger kids, you can’t go wrong with War.
I started playing Tonk (which is good for practicing addition) and slap jack at around age five.
Enchanted Forest and Othello are good board games.
Another vote for Mille Bournes and Uno. My son is almost 8 and plays both. Don’t bother with Uno Junior (or whatever the kiddie version is called).
He’s been playing regular Uno since he was four.
Sorry is another good board game. Again, don’t bother with the kiddie version. The regular one is just fine.
Paidhi Girl loves Uno. I’ve been trying to teach her Cribbage, but she’s not quite ready for it yet. (She’s seven). Go Fish and Old Maid are also popular here. I’m working on spades–the only tricky part is bidding, but I think Paidhi Girl will catch on pretty quickly.
Both she and Paidhi Boy (four) love Junior Monopoly, and recently I picked up Mancala at the store–that one’s a winner with pretty much everyone. I highly recommend it. Oh, and there’s always Yahtzee. That’s a good one, too.
Cheat (aka Bullshit/I Doubt It)
Rules: http://www.pagat.com/beating/cheat.html
A third vote for Mille Bornes and Uno. I played them both as a kid, and we play them today with our six year old daughter. They’re fun for everyone.
Another vote for Uno, 5-card draw poker and blackjack.
As far as board games, I always had fun playing Guess Who?, Battleship, Wheel of Fortune and Boggle.
Thanks, everybody, especially cazzle. I played I doubt you in college and hadn’t thought about it in years.
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## How many shapes can you make with 5 squares?
Pentominoes (made from 5 squares) are the type of polyomino most worked with. There are only 12 in the set (because shapes which are identical by roatation or reflection are not counted).
## How many shapes can you make with 6 cubes?
Naming the Hexacube Pieces
The 166 uniquely shaped hexacube pieces are based on the shapes of six cubes joined on their faces in all the possible different combinations.
## How many different shapes can you make with 4 cubes?
The Solid Tetrominoes
Note that four cubes can be joined eight different ways. Packing a set of these pieces into a 4 x 4 x 2 box makes a neat but quite easy puzzle. There are said to be 1,390 possible solutions.
## What does Pentomino mean in math?
A pentomino (or 5-omino) is a polyomino of order 5, that is, a polygon in the plane made of 5 equal-sized squares connected edge-to-edge. Pentomino tiling puzzles and games are popular in recreational mathematics.
## What are 12 Pentominoes used for?
The pentominoes are a puzzle that has been used by teachers to introduce students to important math concepts such as symmetry, area, and perimeter. Pentominoes are suggested for use by teachers on page 99 of the NCTM Principles and Standards, in the Geometry Standard of the Pre-K-2 section.
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## What are the 12 Pentominoes?
There are twelve possible shapes in a set of unique pentominoes, named T, U, V, W, X, Y, Z, F, I, L, P, and N. An easy way to remember all the letters in a pentominoes set is to look at the word, FILiPiNo, and remember the end of the alphabet, TUVWXYZ.
## How many different Cuboids can you make with 24 cubes?
(d) You can make only six different cuboids with 24 cubes.
## How many blocks make a cube?
We need to add another 25 blocks on the top to satisfy the definition of a cube. Thus 45 blocks is the answer!
## How many Hexominoes are nets of cubes?
There are exactly 11 free hexominoes that can be folded to make a cube. These hexominoes are called nets of the cube.
## Is a cube a block?
When used as nouns, block means a substantial, often approximately cuboid, piece of any substance, whereas cube means a regular polyhedron having six identical square faces. A substantial, often approximately cuboid, piece of any substance.
## How do you make Soma cubes?
Making a Soma cube requires folding 27 Sonobe cubes, then combining them in groups of either three or four into seven pieces. The pieces are then assembled into a three-by-three-by three cube. There are 240 ways to put the pieces together to form a Soma cube — challenge yourself or a friend to find as many as you can.
## How many Polyominoes are there?
It has been shown, though, that there are 35 types of hexominoes (composed of six squares) and 108 types of heptominoes (seven squares), if the dubious heptomino with an interior “hole” is included. The term polyomino was introduced in 1953 as a jocular extension of the word domino.
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## How many Heptominoes are there?
When rotations and reflections are not considered to be distinct shapes, there are 108 different free heptominoes. When reflections are considered distinct, there are 196 one-sided heptominoes. When rotations are also considered distinct, there are 760 fixed heptominoes.
## What is the perimeter of a Pentomino?
Any combination of pentomino P and another pentomino would have a perimeter of 20 in.
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Difference Between Similar Terms and Objects
# Differences Between CRC And Checksum
CRC vs Checksum
Anytime data is stored in a computer with the intent to transmit it, there is a need to ensure that the data is not corrupted. If corrupted data was sent, there would be inaccurate data transmitted and it may not work as desired. There is, therefore, a need for an error detection system that checks that all the data entered is okay and not corrupt before any encryption or transmission occurs. There are two main methods to check the data.
Checksum is arguably the oldest methods that has been used in the validation of all data prior to its being sent. Checksum also helps in authenticating data, as the raw data and the entered data should conform. If an anomaly is noticed, referred to as an invalid checksum, there is a suggestion that there may have been a data compromise in a given method.
Cyclic redundancy check, or CRC as it is commonly referred to, is a concept also employed in the validation of data. The principle used by CRC is similar to checksums, but rather than use the 8 byte system employed by Checksum in checking for data consistency, polynomial division is used in the determination of the CRC. The CRC is most commonly 16 or 32 bits in length. If a single byte is missing, an inconsistency is flagged in the data as it does not add up to the original.
Differences
One of the differences noted between the 2 is that CRC employs a math formula that is based on 16- or 32-bit encoding as opposed to Checksum that is based on 8 bytes in checking for data anomalies. The CRC is based on a hash approach while Checksum gets its values from an addition of all truncated data which may come in 8 or 16 bits. CRC, therefore, has a greater ability to recognize data errors as a single bit missing in the hash system which changes the overall result.
The checksum, on the other hand, requires less transparency and will provide for ample error detection as it employs an addition of bytes with the variable. It can, therefore, be said that the main purpose of CRC is to catch a diverse range of errors that may come about during the transmission of data in analog mode. Checksum, on the other hand, can be said to have been designed for the sole purpose of noting regular errors that may occur during software implementation.
CRC is an improvement over checksums. As earlier noted, checksums are a traditional form of computing, and CRC’s are just a mere advancement of the arithmetic that increases the complexity of the computation. This, in essence, increases the available patterns that are present, and thus more errors can be detected by the method. Checksum has been shown to detect mainly single-bit errors. However, CRC can detect any double-bit errors being observed in the data computation. In understanding the differences between the two data validation methods, knowledge is gathered as to why these two methods are used hand-in-hand in Internet protocol, as it reduces the vulnerability of Internet protocols occurring.
Summary :
– CRC is more thorough as opposed to Checksum in checking for errors and reporting.
– Checksum is the older of the two programs.
– CRC has a more complex computation as opposed to checksum.
– Checksum mainly detects single-bit changes in data while CRC can check and detect double-digit errors.
– CRC can detect more errors than checksum due to its more complex function.
– A checksum is mainly employed in data validation when implementing software.
– A CRC is mainly used for data evaluation in analogue data transmission.
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:: Journal of Korean Society of Industrial and Systems Engineering ::
• For Contributors +
• Journal Search +
Journal Search Engine
ISSN : 2005-0461(Print)
ISSN : 2287-7975(Online)
Journal of Society of Korea Industrial and Systems Engineering Vol.45 No.4 pp.217-224
DOI : https://doi.org/10.11627/jksie.2022.45.4.217
# Solving the Location Problem of Charging Station with Electric Vehicle Routing Problem
Gitae Kim†
Department of Industrial Management Engineering, Hanbat National University
Corresponding Author : gitaekim@hanbat.ac.kr
29/11/2022 16/12/2022 19/12/2022
## Abstract
Due to the issue of the sustainability in transportation area, the number of electric vehicles has significantly increased. Most automakers have decided or planned to manufacture the electric vehicles rather than carbon fueled vehicles. However, there are still some problems to figure out for the electric vehicles such as long charging time, driving ranges, supply of charging stations. Since the speed of growing the number of electric vehicles is faster than that of the number of charging stations, there are lack of supplies of charging stations for electric vehicles and imbalances of the location of the charging stations. Thus, the location problem of charging stations is one of important issues for the electric vehicles. Studies have conducted to find the optimal locations for the charging stations. Most studies have formulated the problem with deterministic or hierarchical models. In this paper, we have investigated the fluctuations of locations and the capacity of charging stations. We proposed a mathematical model for the location problem of charging stations with the vehicle routing problem. Numerical examples provide the strategy for the location routing problems of the electric vehicles.
# 전기차량경로문제의 충전소 위치선정문제의 해법
김 기태†
한밭대학교 산업경영공학과
## 1. 서 론
물류에서 재고문제와 함께 중요한 문제 중에 하나는 차 량을 어떻게 배송할 것인지와 어떤 경로를 이용할 것인지 를 결정하는 문제이다. 복수의 차량으로 지역적으로 흩어 져 있는 수요지에 물건을 배송하거나 서비스를 하도록 차 량을 배송할 때, 차량의 수송비용이 최소가 되도록 하는 차량들의 경로를 찾는 문제를 차량경로문제라고 한다.
차량경로문제는 경영과학에서 오래 동안 연구되어져 왔던 분야이기도 하다. 최근 들어, 차량이 전기차량으로 바뀌어 가면서 전기차량과 관련된 연구도 활발히 진행되 어지고 있다. 전기차량에서 중요한 문제들 중에 하나는 차 량을 충전하는 것이다. 충전할 때의 문제는, 충전하는 시 간이 많이 걸리는 것과 충전소가 많이 없다는 문제가 있 다. 충전소에 대한 수요가 많아 충전소 공급이 부족한데 원인이 있다. 따라서, 부족한 충전소를 적절한 위치에 설 치하는 것은 충전에 대한 수요를 충족하는데 필수적이라 고 볼 수 있다.
차량의 이동이 많지 않은 곳에 충전소가 있으면 충전소 의 활용이 낮아지고, 차량의 밀도가 높은 지역에 충전소가 많지 않으면, 충전소에서 기다리는 차량이 많아져 효율적 으로 운영되지 못하게 된다. 이러한 문제는 시설의 위치를 찾는 시설위치선정문제(facility location problem)와 유사 하고 동일한 방법으로 모형화 되어 문제를 해결할 수 있 다. 하지만, 시설에 대한 위치선정은 시설이 고정되고 차 량들이 시설을 기준으로 운송을 하게 되며, 비교적 고정되 어 있는 운송지에 배송이 이루어지기 때문에 시설에 대한 위치선정문제는 간단하게 모형화 할 수 있다. 반면에, 전 기차량에 대한 충전소의 위치선정은 복잡성이 추가될 수 있다. 전기차량은 차량들이 이동을 하면서 충전을 하고, 이동에 대한 경로가 고정되어 있지 않기 때문에 차량들의 경로들이 가장 많은 곳을 찾기는 쉽지 않다. 이러한 불확 실성은 수요의 변동과 차량과 수요지의 변동 등이 원인이 라고 할 수 있다.
충전소의 최적위치를 찾는 연구는 주로 설비위치선정 문제와 차량경로문제를 두 단계로 나누어 해를 구하는 방 법을 많이 사용하였다. 먼저, 충전소는 수요지의 위치들을 고려하여 충전소가 담당할 수요지들이 적절하게 배분될 수 있는 위치가 선정이 되도록 한다. 충전소의 위치가 결 정이 되면 차량들을 이용한 차량경로문제에 대한 최적경 로를 찾게 되는데, 충전소의 위치를 거쳐 가도록 경로를 만들어 주어야 한다. 하지만, 모형이 다소 복잡하지만 충 전소의 위치와 차량의 경로를 함께 고려해서 최적해를 찾 는 것도 생각해 볼 수 있다. 또한 충전소의 용량도 고려하 여 차량의 경로를 설정하는 것이 필요하다.
본 연구는 전기차량의 차량경로문제와 충전소의 위치 선정문제를 함께 고려한 모형을 제시하고, 충전소의 용량 과 위치들에 따라 차량경로가 어떻게 변화되는지를 분석 해본다.
## 2. 선행연구
전기차량을 위한 충전소 위치를 찾는 문제는 많이 연구가 되어지고 있다. 문제의 모형에 있어서 다양한 모형이 제시 되고 있고, 문제의 형태에서 고려하는 부분과 제약을 두는 부분 등을 중심으로 최근에 연구가 활발히 진행중이다.
시뮬레이션을 이용하여 모형화 하거나 구현을 하는 연 구들이 진행되었는데, Huang and Kockelman[4]은 이득을 최대화 하는 충전소의 최적위치를 찾는 문제를 연구하였 다. 이 연구에서의 모형은 설치, 운영, 유지보수와 부지매 입을 포함한 관련 비용들을 고려하였다. 이 모형은 탄력적 인 수요와 차량들의 충전소 선택에 대해 교통체증과 정체 가 되고 있는 충전소를 고려하였다. 모형은 시뮬레이션을 기반으로 한 모형을 제안했고, 유전자 알고리즘으로 문제 를 해결하였다. Gong et al.[3]는 공공 전기 차량을 위한 충전소의 위치에 대한 연구를 진행하였다. 이들의 모형은 마일리지, 차량의 분포, 승객의 분포를 고려하였다. 에이 전트 기반의 시뮬레이션 모형을 이용하여 문제를 구현하 고 해를 찾았다. 이 연구의 내용은 우선순위, 마일리지, 차 량과 승객의 분포가 충전소 위치에 영향을 준다는 것을 보여주었고, 해당 모형은 인구가 밀집한 대도시에 적합하 다고 주장하였다. Li et al.[8]은 전기자동차 충전소 위치문 제를 풍력이나 광전지와 같은 재생에너지와 결합하여 문 제를 모형화하고 해결하려고 했다. 이들은 수요의 변화와 불확실성을 표현하기 위해 강건최적화 모형을 제안했고, 충전소의 저장용량에 대한 부분도 고려하였다. 이 모형은 시뮬레이션을 이용해서 구현이 되었으며 가능한 해 중에 서 최적의 조건을 찾는 과정을 수행하였다.
확정적 모형을 이용한 위치선정문제가 많이 연구되고 있고, 불확실한 상황을 반영한 확률적 모형을 이용한 연구 도 진행되고 있다. Lee and Han[7]은 충전소의 최적위치를 결정하기 위해 흐름충전위치모형(Flow refueling location model, FRLM)을 이용하였다. 이 모형은 출발지에서 도착 지까지 이송할 수 있는 흐름을 최대화 할 수 있는 문제를 다루었는데, 여러 가지 요인들로 인해 수송 영역이 확률적 으로 변하는 네트워크를 고려하였다. 이들 연구에서는 비 선형 혼합정수계획법을 사용하였는데, 해법으로는 Benders and price 방법을 사용하였다.
도로 상태, 교통혼잡도, 차량의 상태 등의 변화나 불확 실성을 반영하는 연구가 주로 많이 진행되어지고 있지만, 사람에 대해 초점을 두고 연구를 진행하기도 하였다. Lam et al.[6]은 전기자동차의 충전소 문제를 두 단계의 문제로 접근하였다. 첫 번째 단계는 충전소의 위치에 대해 부지사 용 형태, 환경영향, 안전 등의 요소를 고려하였고, 두 번째 단계의 문제는 운전자의 관점에서 인간과 관련된 요소를 이용하였다. 이 연구에서의 모형은 비선형혼합정수계획법 으로 해를 찾기 어려운 문제이기 때문에, 해법은 선형혼합 정수계획법을 여러 번 풀어서 해결하는 방법이나 탐색 방 법 등을 이용한 발견적 해법들을 이용하여 최적해를 구하 였다. Ahmad et al.[1]은 전기자동차에 관한 연구들에 대해 정리를 하였는데, 배송네트워크 운영자 관점, 충전소 주인 의 관점, 전기자동차의 주인의 관점에서 문제를 모형화 하 는 내용을 다루었다. 또한 해법으로 사용되는 알고리즘에 대한 내용과 최적해가 단일 최적인지, 다목적 최적인지에 대한 부분을 나누어 정리를 했다. 이러한 문제들과 함께, 전기자동차의 적재와 관련된 문제도 다루었다.
차량을 운송하는 물류업체를 기준으로 최적해를 찾는 연구가 많이 진행되고 있지만, 정부나 거주자를 위한 최적 방안을 찾는 연구도 수행되고 있으며, 사회적인 비용을 고 려한 연구들이 있었다. Luo and Qiu[9]는 도심에서의 전기 차량 충전소 위치문제를 다루었고, 예약서비스, 비혼잡 시 간대의 유휴, 혼잡시간대의 대기시간 등을 고려한 모형을 제안했다. 또한 모형에서는 정부의 입장에서 사회적 비용 을 고려하였고, 대기시간을 줄이고 시스템의 효율과 고객 의 만족도를 높이기 위해 예약서비스를 도입하였다. 문제 의 해는 유전자 알고리즘을 이용하여 예약시스템의 사용 과 관련 비용에 대한 민감도분석을 수행하였다. Zhou et al.[11]은 충전소에 대해 총 사회비용을 분석했는데, 시설 설치비용과 운영비용을 포함한 경제적 비용과 환경적 비 용으로 구분해서 비용을 추정하였다. 또한 충전소 위치선 정을 위해 문제를 모형화 하여 유전자 알고리즘으로 해를 구하였다. 또한 변전소의 위치선정과 관련된 문제들에 대 해서도 다루었다. 아일랜드의 실제 사례를 통해 충전소의 설치현황과 전기자동차의 보급대수를 비교하고 향후 대책 을 수립하기 위해 최적위치선정의 정보를 제공하였다.
전기차량경로문제와 위치선정문제는 차량경로문제에 복잡성이 추가된 형태로 모형화가 되기 때문에 해법은 주로 발견적 해법이 많이 이용되고 있다. Kizhakkan et al.[5]은 충전소 위치선정을 위한 연구들에 대해 정리를 하였는데, 주로 알고리즘과 모형이 어떤 것들이 사용 되었는지에 관 한 내용이다. 모형은 주로 혼합정수계획법이 많이 사용되 었으며, 알고리즘은 유전자 알고리즘과 같은 발견적해법이 주로 사용되었다고 정리를 했다. Meng and Kai[10]는 충전 소 위치선정문제를 위해 게임이론을 이용하였다. 비협조 완전정보 게임을 이용하여 위치선정문제를 모형화 하고, 내시균형 기반의 해를 찾아 최적의 전략으로 사용하였다. Chakraborty et al.[2]은 충전소 위치문제를 다목적계획법으 로 모형화 하였는데, 에너지 소비를 최소화 하는 것과 차량 이동시간을 최소로 하는 목적함수를 만들었다. 이들 모형은 그래프 기반으로 네트워크형태의 모형을 제안했으며, 다목 적계획법의 최적해를 찾기 위한 발견적해법 알고리즘을 제 안하였다.
## 3. 전기차량의 충전소위치경로문제
차량경로문제는 복수의 차량으로 수요지에 물품이나 서 비스를 보내주는 최적의 경로를 찾는 문제이다. 최적의 경 로는 주로 차량의 총이동거리나 총 수송비용을 최소로 하는 경로를 말한다. 차량경로문제는 외판원문제(TSP)와 통채우 기문제(Bin packing problem)를 결합한 문제이며 외판원문 제가 NP-hard 문제이기 때문에 차량경로문제도 해를 구하 기 어려운 문제라고 할 수 있다. 전기차량문제는 차량경로 문제에 충전소를 방문하여 충전하는 제약이 추가된 문제이 므로 더욱 복잡하면서 해를 찾기 어려운 문제이다.
전기차량경로문제는 수요지와 충전소를 방문하고 돌아 오는 문제인데, 본 연구에서는 이 문제에 충전소의 위치선 정문제를 결합한 문제를 다룬다. 충전소의 위치가 바뀌면 차량의 경로가 함께 바뀌게 되고, 충전소의 용량 또한 영 향을 준다.
전기차량에 대한 충전소과 관련된 문제들은 다양하다. 어느 충전소를 방문해야 되는지를 결정하는 충전소 방문 문제가 있고, 차량이 충전소에 도착하여 충전이 진행되는 데, 충전을 완전충전이 될 때까지 할지 또는 부분충전을 할지에 대해 결정하는 문제도 있다. 부분충전을 하는 경우 에는 충전시간을 얼마나 오래 지속해야 되는지에 대해 충 전함수를 추정하고 이를 이용해서 충전시간을 최적화 하 는 문제도 있다. 전기와 에너지의 사용율을 최적화 시키는 문제도 생각할 수 있다. 또한 단순히 차량경로문제의 확장 형 문제라고 볼 수도 있다. 이러한 모형화 측면 뿐만 아니 라 모형에 대해 어떻게 문제의 최적해를 찾을지에 대한 해법도 다양하게 연구되고 있다.
전기차량에 대한 충전소 위치선정과 경로문제는 아래 와 같다.
<Figure 1>은 전기차량의 충전소위치선정 차량경로문 제를 보여주고 있다. 복수의 차량이 수요지와 함께 수요지 와 다른 위치에 있는 충전소를 거쳐서 돌아오는 문제이다.
충전소 위치를 선정하는 문제는 실제로 충전소를 모든 위치에 설치를 할 수가 없다. 예를 들어, 도로 위나 물이 있는 지역은 충전소가 세워질 수 없는 지역들이다. 따라 서, 수요지를 고려하여 최적의 위치를 찾아도 그 위치에 설치하지 못할 수 있기 때문에, 일반적으로 충전소의 위치 를 찾기 위해서는 가능한 위치들을 찾아 놓고 이들 중에서 최적의 위치를 찾는 방법을 이용한다. 충전소의 위치가 차 량의 이동경로와 가까운 위치에 있어야 차량의 수송비용 도 줄어들고 효율을 높일 수가 있다. 복수차량의 이동경로 와 겹치거나 차량들의 이동경로들과 멀리 있는 경우에는 추가적인 시간소요와 수송비용의 증가를 초래한다. 따라 서 본 논문에서는 충전소로 이용이 가능한 위치들 중에서 어느 위치에 설치 또는 운영을 할지를 결정하는 위치선정 문제를 고려한다.
차량의 경로문제는 단기간의 최적 경로를 찾는 문제이 고, 충전소의 위치는 설치 후에 계속해서 사용할 수 있는 중장기적인 의사결정이라고 볼 수 있다. 하지만, 화물을 운송하는 한 업체를 위한 임시 충전소나 충전소에서 해당 업체를 위해 자리를 확보해 두는 비용을 고려하면 차량경 로와 함께 위치를 결정하는 문제로도 생각해 볼 수 있다.
이 논문에서는 차량경로문제에 대한 모형에서 출발하 여 충전소를 방문하고 돌아오는 제약이 추가된 전기차량 경로문제에 대한 모형을 제시하고, 이 모형에 충전소의 최 적위치를 찾는 부분이 고려된 충전소 위치선정과 전기차 량의 최적경로를 결합한 모형을 제시한다. 또한 충전소를 개설하는 비용이 위치에 따라 다를 때 차량의 최적해가 어떻게 변화하는지에 대해서도 살펴본다.
## 4. 충전소 위치선정을 고려한 모형
차량경로문제의 기본 모형은 시간제약(time window)을 갖는 차량경로문제를 사용한다. 복수차량으로 복수의 수 요지를 방문해야 하고 수요지는 원하는 배송시간이 존재 한다. 시간제약 차량경로문제를 시작으로 전기차량경로문 제와 충전소 위치선정을 위한 차량경로문제에 대한 모형 을 제시한다. 문제의 모형은 아래와 같다.
<집합>
• 0: (0: 차량의 출발위치 또는 도착위치),
• V = {0, …, n} : 모든 노드집합
• N = V∖0, n: 최종수요지
• F : 충전소집합, F ′: 더미 충전소집합,
• $A = { ( i , j ) | i , j ∈ V , i ≠ j }$: 아크집합,
• M = {1, …, m } : 차량집합
• $V + ( i ) = { j ∈ V | ( i , j ) ∈ A }$: 노든 i에서 나가는 아크들의 집합
• $V − ( i ) = { j ∈ V | ( j , i ) ∈ A }$: 노드 i로 들어오는 아크들의 집합
<파라미터>
• [ai, bi ]: 노드 i에 대한 시간제약(Time window)
• si : 노드 i에 대한 서비스 시간
• di,j: 노드 i에서 노드 j까지의 이동시간
• qi : 노드 i의 수요량
• g : 단위에너지당 충전시간
• h: 단위거리당 에너지소모량
• Q : 차량의 최대에너지용량
• wi : 충전소 i가 개설되는 비용
• L : 큰 수
<변수>
• $x i , j m$ : 이진변수, 아크(i,j)∈A 가 차량 m에 의한 최적경 로에 포함되면 1, 아니면 0
• $t i m$: 차량 m이 노드 i에 도착하는 시간 (서비스의 시작 시간)
• $y i m$: 노드 i에서 차량 m의 에너지 수준
• zi : 이진변수, 충전소 i가 개설되면 1, 아니면 0
시간제약(time window)이 있는 차량경로문제는 아래와 같이 모형화 된다.
(VRP-TW)
$M i n ∑ m ∈ M ∑ ( i , j ) ∈ A d i , j x i , j m$
(1)
$s . t . ∑ m ∈ M ∑ j ∈ V x i , j m = 1 , ∀ i ∈ N ,$
(2)
$∑ j ∈ V + ( 0 ) x 0 , j m = 1 , ∀ m ∈ M ,$
(3)
$∑ i ∈ V − ( h ) x i , h m − ∑ j ∈ V + ( h ) x h , j m = 0 ∀ h ∈ N , ∀ m ∈ M$
(4)
$∑ i ∈ V − ( 0 ) x i , 0 m = 1 ∀ m ∈ M ,$
(5)
$∑ i ∈ N q i ∑ j ∈ V x i , j m ≤ K ∀ m ∈ M ,$
(6)
$t i m + ( s i + d i , j ) x i , j m − t j m ≤ ( 1 − x i , j m ) L ∀ ( i , j ) ∈ A , m ∈ M$
(7)
$a i ≤ t i m ≤ b i , ∀ i ∈ N , m ∈ M$
(8)
$x i , j m ∈ 0 , 1 , ∀ i , j ∈ V , m ∈ M$
(9)
목적함수 (1)은 차량들의 총수송비용을 최소화 하는 것 을 나타낸다. 제약식 (2)는 최종수요지는 한 차량에 의해 서 한번만 방문된다는 것을 나타낸다. 제약식 (3)은 모든 차량들은 차량출발지에서 출발해야 되는 제약이다. 제약 식 (4)는 네트워크 흐름의 균형 제약식으로 한 노드로 들 어간 차량은 다시 그 노드를 나가야 되는 것을 의미하고, 한 노드는 하나의 차량만 거쳐가는 것도 나타낸다. 제약식 (5)는 모든 차량은 경로를 통과하고 다시 하나의 도착지로 돌아와야 된다는 제약이다. 제약식 (6)은 차량의 용량제약 으로 각 최종수요지의 수요를 충족시키면서 용량제약을 갖고 운행해야 되는 제약이다. 제약식 (7)은 시간제약(time window)으로 도착한 시간 이후에 서비스시간 동안에 그 노드에 머물고 서비스가 끝나면 다른 노드로 이동을 하는 것을 나타낸다. 제약식 (8)은 각 최종수요지마다 원하는 시간을 나타내는 시간제약(time window)을 나타낸다. (9) 는 이진변수의 정의를 나타낸다.
시간제약 차량경로문제에 전기차량에 대한 문제로 확 장하면, 전기차량이 충전소를 거쳐서 돌아오는 제약식들 이 추가된다. 충전소는 모든 충전소를 다 방문하지 않아도 되고, 일부분만 방문하도록 하면 되는 제약식이 추가된다. 이 모형은 아래와 같다.
(EVRP-TW)
$M i n ∑ m ∈ M ∑ ( i , j ) ∈ A d i , j x i , j m$
(10)
$s . t . ∑ m ∈ M ∑ j ∈ V x i , j m = 1 , ∀ i ∈ N$
(11)
$∑ m ∈ M ∑ j ∈ V x i , j m ≤ 1 , ∀ i ∈ F ′$
(12)
$∑ j ∈ V + ( 0 ) x 0 , j m = 1 , ∀ m ∈ M$
(13)
$∑ i ∈ V − ( h ) x i , h m − ∑ j ∈ V + ( h ) x h , j m = 0 ∀ h ∈ N , m ∈ M$
(14)
$∑ i ∈ V − ( 0 ) x i , 0 m = 1 , ∀ m ∈ M$
(15)
$∑ i ∈ N q i ∑ j ∈ V x i , j m ≤ K , ∀ m ∈ M$
(16)
$t i m + ( s i + d i , j ) x i , j m − t j m ≤ ( 1 − x i , j m ) L ∀ ( i , j ) ∈ A , m ∈ M$
(17)
$t i m + d i , j x i , j m + g ( Q + y i m ) − t j m ≤ ( L − + g Q ) ( 1 − x i , j m ) ∀ i ∈ F ′ , j ∈ N , m ∈ M$
(18)
$0 ≤ y j m ≤ y i m − ( h d i , j ) x i , j m + Q ( 1 − x i , j ) ∀ i , j ∈ N , m ∈ M$
(19)
$0 ≤ y j m ≤ Q − ( h d i , j ) x i , j m ∀ i , j ∈ N , m ∈ M$
(20)
$a i ≤ t i m ≤ b i , ∀ i ∈ N , m ∈ M$
(21)
$x i , j m ∈ { 0 , 1 } , ∀ i , j ∈ V , m ∈ M$
(22)
$t i m ≥ 0 , y i m ≥ 0 , ∀ i ∈ V , m ∈ M$
(23)
전기차량경로문제는 차량이 충전소를 방문하고 충전을 하면서 배송을 해야 하는 제약이 추가되는 모형이다. 먼저 목적함수 (10)는 총수송비용을 최소로 하는 경로를 찾는 것은 같다고 볼 수 있다. 전기를 이용하는 차량은 내연기 관 차량보다 주행할 수 있는 거리가 제한적이며 충전하는 시간도 많이 소요되는 특징이 있다. 충전소를 모두 방문하 지 않고 최적이 되는 일부분만 방문하도록 한다.
제약식 (12)는 모든 충전소가 사용되지 않아도 된다는 제약식이다. 제약식 (13)~(17)은 시간제약 차량경로문제와 같다. 제약식 (18)은 제약식 (17)이 서비스 시간에 대한 제 약인 것처럼 노드에서 충전시간에 대한 제약식을 나타낸 다. 제약식 (19)와 (20)은 각 노드에서의 에너지 소모에 따 른 제약식을 나타낸다. 제약식 (21)은 노드별 시간제약에 대한 제약식이고 제약식 (22)와 (23)은 변수 제약이다.
다음으로, 충전소 위치문제에 대해서, 본 논문에서는 전 기차량의 충전소에 대한 위치문제를 차량경로문제와 결합 하면서 문제를 단순화 하고자 한다. 차량은 한 경로를 전 부 운행할 용량이 충분하지 않다고 가정하고 경로 중에 한번은 충전소를 들렸다고 와야 된다고 가정한다. 이렇게 되면, 전기차량의 에너지 소모량과 충전량을 위한 제약식 은 필요없게 된다. 따라서 전기차량에 대한 차량경로문제 에 충전소의 위치선정을 위한 문제가 추가된 모형은 아래 와 같다.
(EVRP-TW-CL)
$M i n ∑ m ∈ M ∑ ( i , j ) ∈ A d i , j x i , j m + ∑ i ∈ F ′ w i z i ,$
(24)
$s . t . ∑ m ∈ M ∑ j ∈ V x i , j m = 1 , ∀ i ∈ N$
(25)
$∑ m ∈ M ∑ j ∈ V x i , j m ≤ 1 , ∀ i ∈ F ′$
(26)
$∑ j ∈ V + ( 0 ) x 0 , j m = 1 , ∀ m ∈ M$
(27)
$∑ i ∈ V − ( h ) x i , h m − ∑ j ∈ V + ( h ) x h , j m = 0 ∀ h ∈ N , m ∈ M$
(28)
$∑ i ∈ V − ( 0 ) x i , 0 m = 1 , ∀ m ∈ M$
(29)
$∑ i ∈ N q i ∑ j ∈ V x i , j m ≤ K , ∀ m ∈ M$
(30)
$t i m + ( s i + d i , j ) x i , j m − t j m ≤ ( 1 − x i , j m ) L ∀ ( i , j ) ∈ A , m ∈ M$
(31)
$∑ i ∈ N , j ∈ F ′ x i , j m = 1 , ∀ m ∈ M$
(32)
$x i , j m ≤ z i , ∀ i ∈ F ′ , j ∈ N , m ∈ M$
(33)
$x i , j m ≤ z j , ∀ i ∈ N , j ∈ F ′ , m ∈ M$
(34)
$a i ≤ t i m ≤ b i , ∀ i ∈ N , m ∈ M$
(35)
$x i , j m , z i ∈ { 0 , 1 } , ∀ i , j ∈ V , m ∈ M$
(36)
$t i m ≥ 0 , ∀ i ∈ V , m ∈ M$
(37)
목적함수 (24)는 차량의 총 이동거리와 충전소를 개설 하는 총비용을 최소로 하는 것을 목표로 한다. 이것은 충 전소별 개설비용이 다른 경우에 이동거리와 충전소 개설 비용을 고려하여 최적의 개설 위치를 찾도록 해준다. 제약 식 (25)~(31)은 전기차량경로문제와 같다. 제약식 (32)는 차량이 충전소를 한번은 방문해야 하는 제약식이다. 제약 식 (33)과 (34)는 충전소가 개설되는 노드로 차량이 이동 한다는 제약식이고, 제약식 (35)는 시간제약이며, 제약식 (36)과 (37)은 변수제약이다.
제안된 모형은 복수의 전기차량이 충전소를 거치면서 수요지를 한 번씩 방문하며 돌아오는 최적경로를 찾는 문제 로 충전소의 위치가 가능한 수요지를 방문하는 경로와 멀지 않게 하면서 총수송비용이 최소가 되도록 하는 경로를 찾고 최적의 위치를 갖는 충전소의 위치도 찾는 모형이다.
## 5. 실험결과
차량경로문제에 전기 충전소의 위치선정을 고려한 문 제에 대해 간단한 예제를 통해 결과를 살펴본다. 문제의 모형 EVRP-TW-CL에 대해 공개된 차량경로 데이터인 Solomon benchmark 문제 중에 하나를 충전소들의 가능한 위치들을 추가한 데이터를 이용하여 실험을 진행한다.
<Figure 2>는 실험에 사용된 예제 데이터를 보여주는데, 출발점(Depot)이 있고 수요지들이 지리적으로 흩어져 있 으며, 복수의 차량을 이용하여 화물을 배송하는 문제이다. 이 예제에서 수요지는 100개, 차량은 10대이고 그림에서 진한 점으로 표현된 가능한 충전소는 21개이다.
모형은 C#으로 구현되었고, C#에서 concert technology 를 이용해서, 해는 CPLEX 12.6.1 를 이용하여 구하였다. 차량의 대수는 10대로 고정되어 있다고 가정하였고, 충전 소의 운영비용(개설비용)이 동일한 경우와 위치별로 다른 경우에 대해 최적경로가 변화되는지를 살펴보았다. 실험 결과는 최적해에서 차량들이 어떤 충전소를 선택하였는지 를 나타낸다.
<Figure 3>은 목적함수에서 충전소의 개설 또는 운영 비용을 고려하는 항을 제외하고 문제를 풀었을 때의 해를 보여준다. 21개의 가능한 충전소 위치에서 차량들이 각각 하나의 충전소를 이용한 것을 보여준다.
목적함수에서 충전소의 개설비용을 고려한 경우에 대 한 실험결과는 다음과 같다.
<Figure 4>는 목적함수에 충전소 개설비용 항목을 고려 하는 경우의 해를 보여준다. 이 경우에 모든 위치의 개설 비용을 동일하게 4.0으로 주었을 때, 고려하지 않았을 때 의 <Figure 3>의 해와 차량들이 충전소를 이용하는 것이 동일하다는 것을 알 수 있다.
<Figure 5>는 가능한 충전소 위치들 중에서 1~10번 노 드들은 개설비용을 4.0으로 하고 나머지 노드들의 개설비 용은 9.0으로 했을 경우의 해이다. 충전소 개설비용을 고 려하지 않은 <Figure 4>의 결과와 비교하면, 충전소를 이 용하는 차량들이 바뀐 것을 볼 수 있다. 하지만, 사용된 충전소는 case 1 에서 보여준 충전소들과 같고 이용한 차 량만 다른 것을 볼 수 있다.
<Figure 6>은 <Figure 5>와 대비하여 1~10 노드의 충전 소 개발비용을 9.0, 11~20 노드의 비용을 4.0으로 바꾸었 을 때의 해를 보여준다. 이 결과에서는 case 2의 결과와 비교하면, 충전소 중에서 2번이 사용되지 않았고 16번이 사용된 것을 볼 수 있다. 충전소와의 거리가 가까워도 충 전소의 개설비용이 높으면 다른 충전소를 이용하는 것을 알 수 있다.
<Figure 7>은 충전소 개설비용을 1.0과 4.0 사이의 임의 의 수를 넣어서 문제의 해를 구한 결과를 보여준다. 이 결 과는 차량들이 이용하는 충전소의 순서가 다르지만, case 1과 case 2와 같이 사용된 충전소들의 노드들은 같은 것을 볼 수 있다.
차량이 충전소까지의 거리가 가까워도 충전소의 개설 비용이 크면, 거리가 먼 충전소를 이용할 수 있고, 충전소 의 개설비용이 상대적으로 크지 않으면, 충전소 개설비용 이 위치별로 같이 않더라도 거리를 기준으로 결정된 경로 를 선택하는 것을 볼 수 있다.
<Figure 8>은 <Figure 7>의 case 4 처럼 충전소 개설비 용에 임의의 수를 할당하였는데 비용을 1.0에서 9.0으로 좀 더 큰 값으로 설정한 경우의 결과이다. 이 경우에는 충 전소 중에서 17이 사용되지 않았고 15가 사용된 것을 볼 수 있다. 개설비용이 큰 경우에는 거리에 의존하지 않고 개설비용에 의존해서 경로를 결정하는 것을 볼 수 있다.
<Figure 9>는 충전소 개설비용에 좀 더 편차를 크게하 여 변화를 주었을 때의 해를 보여준다. case 5와 비교하면, 충전소 2와 15가 사용되지 않고 충전소 11과 16이 새롭게 사용된 것을 볼 수 있다.
전기차량의 경우에 차량의 배송 경로 중간에 충전이 필 요한 경우에 배송업체를 위해 임시로 충전소를 개설하거 나 기존의 충전소에서 해당 업체의 차량을 위해 자리를 확보하는 등의 비용이 발생할 경우에 차량이 어느 충전소 를 선택해야 최적인지를 찾는 예제를 살펴보았다. 충전소 의 개설비용이 작은 경우에는 충전소까지의 거리를 기반 으로 충전소를 선택하게 되고, 충전소의 개설비용이 크면 거리가 멀지만 비용이 적게 드는 충전소를 선택하는 것을 알 수 있다.
## 6. 결 론
내연기관을 이용한 차량으로 인한 배기가스가 공기오 염의 원인 중에 하나로 알려지면서, 환경에 대한 문제의식 을 갖고 최근의 자동차 산업에서는 전기차의 생산이 늘리 고 있다. 이러한 상황은 전기차에 대한 개발과 연구를 활 발하게 만드는 원동력이 되고 있다. 차량경로문제에 대한 연구도 전기차량에 대한 경로문제로 확장되어 연구가 진 행되고 있다. 하지만, 전기자동차는 여전히 충전에 관한 문제가 해결되어야 할 부분으로 남아있다. 충전에 걸리는 시간도 문제이고 충전소의 위치도 중요한 문제가 되고 있 다. 따라서, 충전소의 위치선정은 전기차량경로문제에서 중요한 부분을 차지한다.
본 연구에서는 전기차량경로문제를 다루는데, 충전소 의 위치선정에 대한 문제도 함께 고려했다. 시간제약이 있 는 차량경로문제의 모형에서 출발하여, 전기차량경로문제 에 대한 수리모형을 제시하였고, 배송업체를 위한 임시충 전소나 충전소에서의 자리확보 등을 고려하여 어떤 충전 소를 선택하는 것이 최적인지를 찾는 모형을 제안했다. 또 한 간단한 예제를 통해 충전소 개설비용이 위치별로 같은 경우와 다른 경우에 차량들이 충전소를 선택하는 결정에 변화가 있는지를 살펴보았다.
실험결과를 통해, 충전소의 개설비용이 위치별로 다른 경우에 차량들의 충전소 선택에 대한 결정이 달라지는 것 을 확인하였다. 개설비용이 작은 경우는 차량들이 최적경 로로 이동하면서 경로와 가장 가까운 충전소를 이용하여 이동거리에 의존한 의사결정을 하게 되고, 개설비용이 큰 경우에는 충전소까지의 위치가 가깝더라도 비용을 줄이기 위해 멀리 있는 충전소를 이용하는 것을 알 수 있다. 이 결과를 이용하면, 충전소의 개설비용에 따라 차량이 충전 소를 선택할지를 결정할 수 있는 하한값을 제공해 줄 수 있고, 충전을 위한 준비를 해야 되는지도 결정하는데 도움 을 줄 수 있을 것으로 보인다.
## Acknowledgement
This work was supported by the National Research Foundation of Korea(NRF) grant funded by the Korea government( MSIT) (No. 2021R1A2C10141801231482092260102).
## Figure
Charging Station Location Electric Vehicle Routing Problem
Distribution Network with Charging Stations
Result without Considering Charging Station Cost
Result of Case 1 (All Charging Station Costs are 4.0)
Result of Case 2 (Charging Station Costs of 1~10: 4.0, those of 11~21: 9.0)
Result of Case 3 (Charging Station Costs of 1~10: 9.0, those of 11~21: 4.0)
Result of Case 4 (Charging Station Costs are Random Numbers between 1 and 4)
Result of Case 5 (Charging Station Costs are Random Numbers between 1 and 9)
Result of Case 6 (Charging Station Costs of 1~10: Random in (10, 20), those of 11~21: Random in (1,5))
## Reference
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A ball with a mass of 5 kg is rolling at 7 m/s and elastically collides with a resting ball with a mass of 4 kg. What are the post-collision velocities of the balls?
Jan 6, 2016
I found: $0.78 \mathmr{and} 7.78 \frac{m}{s}$ but check my maths!
Explanation:
We can use conservation of momentum $\vec{p} = m \vec{v}$ in one dimension (along $x$). So we get:
${p}_{\text{before")=p_("after}}$
So:
$\left(5 \cdot 7\right) + \left(4 \cdot 0\right) = \left(5 \cdot {v}_{1}\right) + \left(4 \cdot {v}_{2}\right)$
Being an elastic collision also kinetic energy $K = \frac{1}{2} m {v}^{2}$ is conserved:
${K}_{\text{before")=K_("after}}$
So:
$\frac{1}{2} \cdot 5 \cdot {7}^{2} + \frac{1}{2} \cdot 4 \cdot {0}^{2} = \frac{1}{2} \cdot 5 \cdot {v}_{1}^{2} + \frac{1}{2} \cdot 4 \cdot {v}_{2}^{2}$
We can use the two equations together and we get:
$\left\{\begin{matrix}35 = 5 {v}_{1} + 4 {v}_{2} \\ 245 = 5 {v}_{1}^{2} + 4 {v}_{2}^{2}\end{matrix}\right.$
From the first equation:
${v}_{1} = \frac{35 - 4 {v}_{2}}{5}$
Substitute into the second and get:
$245 = \cancel{5} {\left(35 - 4 {v}_{2}\right)}^{2} / {\cancel{25}}^{5} + 4 {v}_{2}^{2}$
$\cancel{1225} = \cancel{1225} - 280 {v}_{2} + 16 {v}_{2}^{2} + 20 {v}_{2}^{2}$
$36 {v}_{2}^{2} - 280 {v}_{2} = 0$
${v}_{2} = 0 \frac{m}{s}$ (not) and ${v}_{2} = \frac{280}{36} = \frac{70}{9} = 7.78 \frac{m}{s}$ yes.
Corresponding to:
${v}_{1} = 7 \frac{m}{s}$ and ${v}_{1} = \frac{7}{9} = 0.78 \frac{m}{s}$
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# How to Make an Oval?
An oval shape is very similar to a circle. Picture an egg and add some dimension to the circle on two opposite sides of the circle to make an oval shape. You can find more information here: http://members.tripod.com/˜wintermist_3/oval/oval.html
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1. Draw a pencil outline of an oval onto the metal, wood, craft foam or thin plastic for the key chain. 2. Cut the oval shape from thin metal or wood using a scroll saw or cut the http://www.ehow.com/how_8421067_make-oval-keychain...
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1. Measure the length of the pool at its longest point. 2. Measure the width of the pool at its widest point. 3. Determine the average depth of the pool by adding the depth of the http://www.ehow.com/how_5544596_calculate-gallons-...
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https://aufdercouch.net/1111-repeating-as-a-fraction/
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As a side keep in mind the whole number-integral part is: emptyThe decimal component is: .1111= 1111/10000Full simple fraction breakdown:1111/10000
Scroll down to customize the precision point permitting .1111 to be broken down come a specific variety of digits. The page additionally includes 2-3D graphical representations of .1111 together a fraction, the different varieties of fractions, and also what form of fraction .1111 is when converted.
You are watching: .1111 repeating as a fraction
Graph depiction of .1111 together a FractionPie chart depiction of the fractional part of .1111
2D Chart3D Chart
Level the Precision for .1111 together a Fraction
The level of precision are the variety of digits to ring to. Choose a lower precision suggest below to break decimal .1111 down further in portion form. The default precision allude is 5.If the last rolling digit is "5" you can use the "round half up" and "round fifty percent down" options to round that digit increase or down once you readjust the precision point.For example 0.875 v a precision allude of 2 rounded fifty percent up = 88/100, rounded fifty percent down = 87/100.select a precision point:123456789 round fifty percent up round half down
Enter a Decimal Value:
this.maxLength)this.value=this.value.slice(0,this.maxLength);">Show as a fraction
Is1111/10000 a Mixed, entirety Number or suitable fraction?A combined number is consisted of of a entirety number (whole numbers have actually no fountain or decimal part) and also a proper fraction part (a portion where the numerator (the peak number) is much less than the denominator (the bottom number). In this case the totality number value is empty
and also the proper fraction value is 1111/10000.
Can all decimals be converted right into a fraction?
Not every decimals deserve to be converted right into a fraction. There room 3 basic types i beg your pardon include:Terminating decimals have a restricted number of digits after the decimal point.Example:4892.2821 = 4892 2821/10000Recurring decimals have actually one or more repeating number after the decimal allude which continue on infinitely.Example:5577.3333 = 5577 3333/10000 = 333/1000 = 33/100 = 1/3 (rounded)Irrational decimals go on forever and never kind a repeating pattern. This form of decimal can not be expressed as a fraction.Example: 0.652552697.....
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Fraction right into Decimal
You can also see the turning back conversion I.e. How portion 1111/10000is converted right into a decimal.
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Digital Camera and Computer Vision Laboratory Department of Computer Science and Information Engineering National Taiwan University, Taipei, Taiwan, R.O.C.
Presentation on theme: "Digital Camera and Computer Vision Laboratory Department of Computer Science and Information Engineering National Taiwan University, Taipei, Taiwan, R.O.C."— Presentation transcript:
Digital Camera and Computer Vision Laboratory Department of Computer Science and Information Engineering National Taiwan University, Taipei, Taiwan, R.O.C. Computer and Robot Vision I Chapter 13 Perspective Projection Geometry Presented by: 傅楸善 & 張博思 0911 246 313 r94922093@ntu.edu.tw 指導教授 : 傅楸善 博士
DC & CV Lab. CSIE NTU 13.1 Introduction Computer vision problems often involve interpreting the information on a two- dimensional (2D) image of a three-dimensional (3D) world in order to determine the placement of the 3D objects portrayed in the image. To do this requires understanding the perspective transformation governing the geometric way 3D information is projected onto the 2D image.
DC & CV Lab. CSIE NTU 13.1 Introduction image formation on the retina, according to Descartes scrape ox eye, observe from darkened room inverted image of scene
DC & CV Lab. CSIE NTU Nalwa, Scrape ox eye, observe from darkened room inverted image of scene
DC & CV Lab. CSIE NTU 13.2 One-Dimensional Perspective Projection f: focal length of lens u: distance between object and lens center v: distance between image and lens center thin-lens equation: lens law: 1/f=1/u+1/v light passing lens center dose not deflect light parallel to optical axis will pass focus
DC & CV Lab. CSIE NTU
DC & CV Lab. CSIE NTU
DC & CV Lab. CSIE NTU
DC & CV Lab. CSIE NTU pinhole camera: infinitesimally small aperture pinhole camera: approximated by lens with aperture adjusted to the smallest pinhole camera: simplest device to form image of 3D scene on 2D surface
DC & CV Lab. CSIE NTU 13.2 One-Dimensional Perspective Projection aperture size decreased: image become sharper diameter of aperture is 0.06 inch, 0.015 inch, 0.0025 inch aperture below certain size: diffraction: bending of light rays around edge
DC & CV Lab. CSIE NTU
DC & CV Lab. CSIE NTU JOKE
DC & CV Lab. CSIE NTU 13.2 One-Dimensional Perspective Projection
DC & CV Lab. CSIE NTU 13.2 One-Dimensional Perspective Projection f: camera constant (different from above equation) (r, s, 1): homogeneous coordinate system for point (r, s) first linear transformation: translates (r, s, 1) by distance of f
DC & CV Lab. CSIE NTU 13.2 One-Dimensional Perspective Projection second linear transformation: takes perspective transformation to image line 1D image line coordinate:
DC & CV Lab. CSIE NTU 13.2 One-Dimensional Perspective Projection (Xp,Yp) (X, Y) X Y
DC & CV Lab. CSIE NTU 13.2 One-Dimensional Perspective Projection (Xp,Yp) (X, Y) X Y
DC & CV Lab. CSIE NTU 13.2 One-Dimensional Perspective Projection lens: at origin and looks down - axis image line: distance f in front of lens and parallel to -axis : the x - y axes rotated anticlockwise by angle
DC & CV Lab. CSIE NTU 13.2 One-Dimensional Perspective Projection
DC & CV Lab. CSIE NTU 13.2 One-Dimensional Perspective Projection rewriting the relationship in terms of homogeneous coordinate system
DC & CV Lab. CSIE NTU 13.3 The Perspective Projection in 3D camera lens: along line parallel to z-axis position of lens: center of perspectivity: (u, v): coordinates of perspective projection of (x, y, z) on image plane
DC & CV Lab. CSIE NTU JOKE
DC & CV Lab. CSIE NTU 13.3.1 Smaller Appearance of Farther Objects without loss of generality: take center of perspectivity to be origin perspective projection: objects appear smaller the farther they are
DC & CV Lab. CSIE NTU
DC & CV Lab. CSIE NTU foreshortening: line segments in plane parallel to image has maximum size
DC & CV Lab. CSIE NTU 13.3.2 Lines to Lines lines in 3D world transform to lines in the image plane parallel lines in 3D with nonzero z slope: meet in a vanishing point
DC & CV Lab. CSIE NTU
DC & CV Lab. CSIE NTU 13.3.3 Perspective Projection of Convex Polyhedra are Convex Proofs in textbook, simple but tedious, study as exercise by yourself
DC & CV Lab. CSIE NTU 13.3.4 Vanishing Point Perspective projections of parallel 3D lines having nonzero slope along the optic z-axis meet in a vanishing point on the image projection plane.
DC & CV Lab. CSIE NTU 13.3.5 Vanishing Line
DC & CV Lab. CSIE NTU
DC & CV Lab. CSIE NTU 13.3.6 3D Lines-2D perspective Projection Lines There is a relationship between the parameters of a 3D line and the parameters of the perspective projection of the line.
DC & CV Lab. CSIE NTU
DC & CV Lab. CSIE NTU
DC & CV Lab. CSIE NTU JOKE
DC & CV Lab. CSIE NTU 13.4 2D to 3D Inference Using Perspective Projection perspective projection on unknown 3D line: provides four of six constraints additional constraints: 3D-world-model information about points, lines
DC & CV Lab. CSIE NTU 13.4.1 Inverse Perspective Projection : perspective projection of a point f: image plane distance from camera lens thus : 3D coordinate of the point in image plane camera lens: at the origin line L: inverse perspective projection of the point
DC & CV Lab. CSIE NTU 13.4.2 Line Segment with Known Direction Cosines and Known Length known: : line segment length : line segment direction cosine, : perspective projections of endpoints unknown:, : 3D coordinates of endpoints
DC & CV Lab. CSIE NTU 13.4.2 Line Segment with Known Direction Cosines and Known Length
DC & CV Lab. CSIE NTU 13.4.2 Line Segment with Known Direction Cosines and Known Length
DC & CV Lab. CSIE NTU 13.4.3 Collinear Points with Known Interpoint Distances known: : perspective projection of nth collinear points, n = 0, …, N - 1 distance between (n+1)th point and first point unknown: : direction cosine of line, : 3D coordinates of points
DC & CV Lab. CSIE NTU 13.4.3 Collinear Points with known Interpoint Distances
DC & CV Lab. CSIE NTU 13.4.4 N Parallel Lines known: : perspective projection of nth parallel line unknown: : direction cosine of line
DC & CV Lab. CSIE NTU 13.4.4 N Parallel Lines
DC & CV Lab. CSIE NTU 13.4.5 N Lines Intersecting at a Point with Known Angles known: : perspective projection of intersecting point : perspective projection of nth intersecting line : known angle between and unknown: : 3D nth intersecting line
DC & CV Lab. CSIE NTU 13.4.6 N Lines Intersecting in a Known Plane known: : perspective projection of intersecting point : perspective projection of nth intersecting line : plane equation unknown: : 3D nth intersecting line
DC & CV Lab. CSIE NTU 13.4.6 N Lines Intersecting in a Known Plane
DC & CV Lab. CSIE NTU 13.4.7 Three Lines in a Plane with One Perpendicular to the Other Two known: : perspective projection of line unknown: three lines in same plane, perpendicular to, : perspective projection of line : since is perpendicular to,
DC & CV Lab. CSIE NTU
DC & CV Lab. CSIE NTU 13.4.8 Point with Given Distance to a Known Point known: : perspective projection of unknown point : known 3D points : distance between the two points unknown: : direction cosine between two points Inverse perspective projection: u v f
DC & CV Lab. CSIE NTU
DC & CV Lab. CSIE NTU 13.4.9 Point in a Known Plane known: : perspective projection of unknown point : known plane equation where point lies unknown: : 3D coordinate of the point:
DC & CV Lab. CSIE NTU 13.4.9 Point in a Known Plane
DC & CV Lab. CSIE NTU 13.4.10 Line in a Known Plane known: : known plane equation where line lies : perspective projection of line unknown: : 3D line A*i + B*j + C*k = 0,
DC & CV Lab. CSIE NTU
DC & CV Lab. CSIE NTU JOKE
DC & CV Lab. CSIE NTU 13.4.11 Angle known: : perspective projection of the unknown line : direction cosine for the known line : angle between the 3D lines unknown: : direction cosine for the unknown line =.
DC & CV Lab. CSIE NTU 13.4.11 Angle
DC & CV Lab. CSIE NTU 13.4.12 Parallelogram known: perspective projection of four corner points of a parallelogram unknown: : normal to the plane on which the parallelogram lies, : direction cosines of two sides of parallelogram,
DC & CV Lab. CSIE NTU 13.4.13 Triangle with One Vertex Known known:,, : perspective projection of three vertices : one known 3D vertex of the three vertices, : known lengths of the triangle in 3D unknown:, : two unknown 3D vertices of the three vertices
DC & CV Lab. CSIE NTU 13.4.13 Triangle with One Vertex Known
DC & CV Lab. CSIE NTU 13.4.14 Triangle with Orientation of One Leg Known known:,, : perspective projection of three vertices : known direction cosines between the first two vertices, : known lengths of the triangle in 3D unknown:,, : three unknown 3D vertices
DC & CV Lab. CSIE NTU
DC & CV Lab. CSIE NTU
DC & CV Lab. CSIE NTU
DC & CV Lab. CSIE NTU 13.4.15 Triangle: three-point spatial resection problem in photogrammetry known:,, : perspective projection of three vertices, : known lengths of the triangle in 3D unknown:,, : three unknown 3D vertices four solutions =, =, =
DC & CV Lab. CSIE NTU
DC & CV Lab. CSIE NTU 13.4.16 Determining the Principal Point by Using Parallel Lines principle point: point through which the optic axis passes principle point: so far assumes origin of image reference frame known:, n = 1, …, N: perspective projection of nth parallel line unknown: : coordinate of the principal point
DC & CV Lab. CSIE NTU
DC & CV Lab. CSIE NTU 13.15 Circles known: perspective projection of a circle having known radius unknown: plane on which the circle lies the 3D center of the circle:
DC & CV Lab. CSIE NTU JOKE
DC & CV Lab. CSIE NTU 13.6 Range from Structured Light structured light: active visual sensing technique upon perspective geometry structured light: controlled light source with regular pattern onto scene regular pattern: stripes, grid, …
DC & CV Lab. CSIE NTU
DC & CV Lab. CSIE NTU intensity and range images
DC & CV Lab. CSIE NTU 13.6 Range from Structured Light Two light sources with cylindrical lenses produce sheets of light that intersect in a line lying on the surface of a conveyor belt. A camera above the belt is aimed so that this line is imaged on a linear array of photo sensors. When there is no object present, all the sensor cells are brightly illuminated. When part of an object interrupts the incident light, the corresponding region on the linear array is darkened. The motion of the belt scans the object past the sensor, generating the second image dimension.
DC & CV Lab. CSIE NTU
DC & CV Lab. CSIE NTU
DC & CV Lab. CSIE NTU
DC & CV Lab. CSIE NTU 13.7 Cross-Ratio cross-ratio: of perspective projection of 4 collinear points, takes same value
DC & CV Lab. CSIE NTU 13.7.1 Cross-Ratio Definitions and Invariance four collinear points: q, r: centers of perspectivity for two projection images
DC & CV Lab. CSIE NTU 13.7.1 Cross-Ratio Definitions and Invariance Let,. by perspective projection equations, cross-ratio: cross-ratio: independent of reference frame, point p, direction cosine b cross-ratio: depends only on directed of collinear points
DC & CV Lab. CSIE NTU 13.7.2 Only One Cross-Ratio each of 4! Cross-ratios is a function of cross-ratio
DC & CV Lab. CSIE NTU 13.7.3 Cross–Ratio in Three Dimensions The cross-ratio derived from one-dimensional perspective projections in a two-dimensional world can be generalized to two-dimensional perspective projection in a three-dimensional world. five co-planar points : cross-ratio for the line segment and
DC & CV Lab. CSIE NTU 13.7.3 Cross–Ratio in Three Dimensions
DC & CV Lab. CSIE NTU
DC & CV Lab. CSIE NTU 13.7.4 Using Cross-Ratios cross-ratio: to aid in establishing correspondences
DC & CV Lab. CSIE NTU END
DC & CV Lab. CSIE NTU JOKE
DC & CV Lab. CSIE NTU 13.7.4 Using Cross-Ratios cross-ratio: to aid in establishing correspondences Term Project JPEG 2000
DC & CV Lab. CSIE NTU 1. Neural Network
DC & CV Lab. CSIE NTU 3. Image Compression JPEG, MPEG, H. 264
DC & CV Lab. CSIE NTU 5. segmentation based on texture
DC & CV Lab. CSIE NTU 6. optical character reading
DC & CV Lab. CSIE NTU 7. stereo vision
DC & CV Lab. CSIE NTU 7. stereo vision
DC & CV Lab. CSIE NTU 8. Handwriting recognition
DC & CV Lab. CSIE NTU 8. Handwriting recognition
DC & CV Lab. CSIE NTU 9. histogram specification
DC & CV Lab. CSIE NTU Term Project
DC & CV Lab. CSIE NTU 10. homomorphic filtering
DC & CV Lab. CSIE NTU 10. homomorphic filtering
DC & CV Lab. CSIE NTU 12. calculating the sizes of stones, cells, cell nucleus.
DC & CV Lab. CSIE NTU 13. trademark resemblance, semi-automatic similarity classification
DC & CV Lab. CSIE NTU 15. structured light 3-D reconstruction
DC & CV Lab. CSIE NTU 15. structured light 3-D reconstruction
DC & CV Lab. CSIE NTU 16. object classification with moments invariant to rotation, scaling, translation
DC & CV Lab. CSIE NTU 17. photometric stereo
DC & CV Lab. CSIE NTU 18. shape from focus
DC & CV Lab. CSIE NTU 19. shape from polarization Sec. 12.5, p. 22
DC & CV Lab. CSIE NTU 20. shape from shading
DC & CV Lab. CSIE NTU 20. shape from shading
DC & CV Lab. CSIE NTU 21. shape from texture
DC & CV Lab. CSIE NTU 21. shape from texture
DC & CV Lab. CSIE NTU 22. solving correspondence problem or optic flow field
DC & CV Lab. CSIE NTU 23. motion and shape parameter recovery
DC & CV Lab. CSIE NTU 24. segmentation of newspaper, documents into title, figure, caption, …
DC & CV Lab. CSIE NTU 25. optical distortion correction
DC & CV Lab. CSIE NTU JOKE
DC & CV Lab. CSIE NTU Term Project
DC & CV Lab. CSIE NTU 26. line labeling of 2D line drawing of 3D objects
DC & CV Lab. CSIE NTU 27. Computer Tomography
DC & CV Lab. CSIE NTU 27. Computer Tomography
DC & CV Lab. CSIE NTU 29. X Ray diagnostic
DC & CV Lab. CSIE NTU 30. finger-print validation
DC & CV Lab. CSIE NTU 31. face recognition (intensity image, range image)
DC & CV Lab. CSIE NTU 33. digital morphing
DC & CV Lab. CSIE NTU 33. digital morphing
DC & CV Lab. CSIE NTU Term Project
DC & CV Lab. CSIE NTU 39. Printed music sheet recognition and translation into MIDI (Musical Instrument Digital Interface) format file
DC & CV Lab. CSIE NTU 40. wafer defect inspection
DC & CV Lab. CSIE NTU 41. wafer critical dimension measurement
DC & CV Lab. CSIE NTU 42. IC pin inspection
DC & CV Lab. CSIE NTU 43. IC mark printing inspection
DC & CV Lab. CSIE NTU Term Project
DC & CV Lab. CSIE NTU JOKE
DC & CV Lab. CSIE NTU END
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# What is the sum of this series? Dirichlet $L$-Function
$$\sum_{n>0} \frac{\mu(n)}{n^s}$$
Sum from 1 to infinity of The Möbius function$/n^s$, i.e., Möbius function/Riemman-zeta function?
Sorry, I forgot to mention that the way that I am suppose to tackle this question is with the information that: L(s,f[convolution]g) = L(s,f)*L(s,g).
The problem was that I couldn't use the mobius inversion theorem or anything to try and find the mobius function in terms of a convolution? Any help? thanks.
• Just a minor point: what you have written is not "Möbius/Zeta," although it sort of looks like it. – angryavian Jan 11 '14 at 18:32
Put $\;\mathcal P:= \{p\in\Bbb N\;;\;p\;\text{is a prime}\}\;$ :
$$\text{Re}(s)>1\;:\;\;\zeta(s)=\prod_{p\in\mathcal P}\left(1-\frac1{p^s}\right)^{-1}\implies$$
$$\frac1{\zeta(s)}=\prod_{p\in\mathcal P}\left(1-\frac1{p^s}\right)=1-\sum_{p\in\mathcal P}\frac1{p^s}+\sum_{p,q\in\mathcal P\,,\,p\neq q}\frac1{p^sq^s}-\sum_{p_1,p_2,p_3\in\mathcal P\,,\,i\neq j\implies p_i\neq p_j}\frac1{p_1^sp_2^sp_3^s}+\ldots$$
$$=\sum_{n=1}^\infty\frac{\mu(n)}{n^s}$$
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# tslearn.metrics.subsequence_path¶
tslearn.metrics.subsequence_path(acc_cost_mat, idx_path_end, be=None)[source]
Compute the optimal path through an accumulated cost matrix given the endpoint of the sequence.
Parameters:
acc_cost_mat: array-like, shape=(sz1, sz2)
Accumulated cost matrix comparing subsequence from a longer sequence.
idx_path_end: int
The end position of the matched subsequence in the longer sequence.
beBackend object or string or None
Backend. If be is an instance of the class NumPyBackend or the string “numpy”, the NumPy backend is used. If be is an instance of the class PyTorchBackend or the string “pytorch”, the PyTorch backend is used. If be is None, the backend is determined by the input arrays. See our dedicated user-guide page for more information.
Returns:
path: list of tuples of integer pairs
Matching path represented as a list of index pairs. In each pair, the first index corresponds to subseq and the second one corresponds to longseq. The startpoint of the Path is $$P_0 = (0, ?)$$ and it ends at $$P_L = (len(subseq)-1, idx\_path\_end)$$
dtw_subsequence_path
Get the similarity score for DTW
subsequence_cost_matrix
Calculate the required cost matrix
Examples
>>> acc_cost_mat = numpy.array([[1., 0., 0., 1., 4.],
... [5., 1., 1., 0., 1.]])
>>> # calculate the globally optimal path
>>> optimal_end_point = numpy.argmin(acc_cost_mat[-1, :])
>>> path = subsequence_path(acc_cost_mat, optimal_end_point)
>>> path
[(0, 2), (1, 3)]
## Examples using tslearn.metrics.subsequence_path¶
sDTW multi path matching
sDTW multi path matching
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# Numpy random choice to produce a 2D-array with all unique values
297
January 25, 2018, at 10:42 PM
so I am wondering if there's a more efficient solution in generating a 2-D array using `np.random.choice` where each row has unique values.
For example, for an array with shape `(3,4)`, we expect an output of:
``````# Expected output given a shape (3,4)
array([[0, 1, 3, 2],
[2, 3, 1, 0],
[1, 3, 2, 0]])
``````
This means that the values for each row must be unique with respect to the number of columns. So for each row in `out`, the integers should only fall between 0 to 3.
I know that I can achieve it by passing `False` to the `replace` argument. But I can only do it for each row and not for the whole matrix. For instance, I can do this:
``````>>> np.random.choice(4, size=(1,4), replace=False)
array([[0,2,3,1]])
``````
But when I try to do this:
``````>>> np.random.choice(4, size=(3,4), replace=False)
``````
I get an error like this:
`````` File "<stdin>", line 1, in <module>
File "mtrand.pyx", line 1150, in mtrand.RandomState.choice
(numpy\random\mtrand\mtrand.c:18113)
ValueError: Cannot take a larger sample than population when
'replace=False'
``````
I assume it's because it's trying to draw `3 x 4 = 12` samples due to the size of the matrix without replacement but I'm only giving it a limit of `4`.
I know that I can solve it by using a `for-loop`:
`````` >>> a = (np.random.choice(4,size=4,replace=False) for _ in range(3))
>>> np.vstack(a)
array([[3, 1, 2, 0],
[1, 2, 0, 3],
[2, 0, 3, 1]])
``````
But I wanted to know if there's a workaround without using any for-loops? (I'm kinda assuming that adding for-loops might make it slower if I have a number of rows greater than 1000. But as you can see I am actually creating a generator in `a` so I'm also not sure if it has an effect after all.)
One trick I have used often is generating a random array and using `argsort` to get unique indices as the required unique numbers. Thus, we could do -
``````def random_choice_noreplace(m,n, axis=-1):
# m, n are the number of rows, cols of output
return np.random.rand(m,n).argsort(axis=axis)
``````
Sample runs -
``````In [98]: random_choice_noreplace(3,7)
Out[98]:
array([[0, 4, 3, 2, 6, 5, 1],
[5, 1, 4, 6, 0, 2, 3],
[6, 1, 0, 4, 5, 3, 2]])
In [99]: random_choice_noreplace(5,7, axis=0) # unique nums along cols
Out[99]:
array([[0, 2, 4, 4, 1, 0, 2],
[1, 4, 3, 2, 4, 1, 3],
[3, 1, 1, 3, 2, 3, 0],
[2, 3, 0, 0, 0, 2, 4],
[4, 0, 2, 1, 3, 4, 1]])
``````
Runtime test -
``````# Original approach
def loopy_app(m,n):
a = (np.random.choice(n,size=n,replace=False) for _ in range(m))
return np.vstack(a)
``````
Timings -
``````In [108]: %timeit loopy_app(1000,100)
10 loops, best of 3: 20.6 ms per loop
In [109]: %timeit random_choice_noreplace(1000,100)
100 loops, best of 3: 3.66 ms per loop
``````
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12 Replies Latest reply: Aug 25, 2015 4:07 AM by Richard Chilvers
String Aggregation Function for matching text
My data looks like:
Code, Description, Value
ABC, 20% discount , £100
ABC, 20% discount voucher, £200
ABC, 20% discount with money back, £140
DEF, another discount, £20
etc.
I have a straight table chart to aggregate (SUM) values by Code:
ABC £340
DEF £20
etc.
Is there a string function to show the first n matching characters in Description?
Eg. ABC, 20% discount, £340
• Re: String Aggregation Function for matching text
You could use concat() to aggregate your string-values and then apply an mid() or left() function to cut it after x-chars or you used subfield() for returning a certain string and/or you combined it in some ways.
- Marcus
• Re: String Aggregation Function for matching text
Hi Marcus
I believe CONCAT() would give me a string like:
"20% discount;20% discount voucher;20% discount with money back"
how would I then reduce this to a string "20% discount" (ie. the common left portion) ?
• Re: String Aggregation Function for matching text
In this case per subfield() which used the same delimiter which you used for concat():
subfield(concat(distinct Description, ';'), ';', 1)
- Marcus
• Re: String Aggregation Function for matching text
I probably didn't explain the issue well enough.
I believe your solution will CONCAT all the strings, separated by ';' and then extract the first entry up to the ';'.
So - if my string was:
"20% discount and free gift;20% discount voucher;20% discount with money back"
then this function would give "20% discount and free gift".
What I'd like is only the common string of "20% discount".
• Re: String Aggregation Function for matching text
Then you will need some logic to identify which or how many chars from string you want to return, maybe you used index() to find the max. number of return-chars like:
left(Expression, index(Expression, ' ', 2) - 1)
- Marcus
• Re: String Aggregation Function for matching text
Hi Richard,
Try using Pick and Wild Match
Regards,
Varun
• Re: String Aggregation Function for matching text
Try This,
if(wildmatch(Description,'*20%*'), pick(wildmatch(Description,'*20%*'),'20% Discount'))
Hope this helps !
Regards,
Varun
• Re: String Aggregation Function for matching text
Thanks Marcus & Varun
Neither solution will give me a solution which is flexible enough. But I will keep exploring.
What I need is a character by character compare starting from the left until a character position fails the match.
• Re: String Aggregation Function for matching text
Hi,
You might have the other values also like 20% discount. 15 % discount... etc with many combinations. Without knowing the actual data it will be difficult to build the correct logic.
• Re: String Aggregation Function for matching text
my attempt in the attachment
• Re: String Aggregation Function for matching text
Hi Massimo
Thanks for taking the time to understand my requirement and come up with a solution
Regards
• Re: String Aggregation Function for matching text
Try like below:
SubField( FieldDesc , ',',1) & ' ' & SubField( FieldDesc , ',',3)
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# SUMPRODUCT function in Excel
Welcome back
In today's lesson, we will soon learn about a very important Excel function with more than one example for clarification
# SUMPRODUCT function in Excel
At first we will get acquainted with a very important question
### What is the SUMPRODUCT function?
The SUMPRODUCT function returns the sum of the products of the matching ranges or arrays. The default operation is multiplication, but addition, subtraction, and division are also possible.
### How to use the SUMPRODUCT function?
In this example, we'll use SUMPRODUCT to return the total sales for certain items and volume
### syntax
To use the default operation (multiplication):
=SUMPRODUCT(array1, [array2], [array3], ...)
The SUMPRODUCT function syntax has the following arguments:
argument description array1 wanted The first array argument you wish to multiply and then add its components. [array2], [array3],... choice Array arguments from 2 to 255 whose components you wish to multiply and then add.
### To perform other arithmetic operations
Use SUMPRODUCT as usual, but replace the commas separating the array arguments with the arithmetic operators you want (*, /, +, -). After all operations are performed, the results are collected as usual.
#### Notes
· Array arguments must have the same dimensions. If you don't, SUMPRODUCT #VALUE! It is an error value. For example, =SUMPRODUCT(C2:C10,D2:D5) returned an error because the domains are not the same size.
· SUMPRODUCT treats non-numeric array entries as if they were zeros.
· For best performance, SUMPRODUCT should not be used with full column references. Think =SUMPRODUCT(A:A,B:B), here the function will multiply the 1048576 cells in column A by the 1048576 cells in column B before adding them.
#### Example 1
To create the formula using the sample list above, type =SUMPRODUCT(C2:C5,D2:D5) and press Enter. Each cell in column C is multiplied by its corresponding cell in the same row in column D, and the results are added up. The total amount of groceries is \$78.97.
To write a longer formula that gives you the same result, type =C2*D2+C3*D3+C4*D4+C5*D5 and press Enter . After pressing Enter, the result is the same: \$78.97. Cell C2 is multiplied by D2, and the result is added to the result of cell C3 times cell D3 and so on
#### Example 2
The following example uses SUMPRODUCT to return total net sales by sales agent, where we have both total sales and expenses by agent. In this case, we are using an Excel table that uses structured references instead of standard Excel ranges. Here you will see that the Sales, Expense, and Agent ranges are indicated by name.
The formula is: =SUMPRODUCT(((Table1[Sales])+(Table1[Expenses]))*(Table1[Agent]=B8)) , and it returns the sum of all sales and expenses for the factor listed in cell B8.
Related Articles
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# ICESat-2 Ice Cores Activity
After reading the accompanying background information, students create an ice core using a tennis ball container and an assortment of dyes and craft supplies. Students measure the thickness and determine the age of each layer. As an extension... (View More)
# ICESat-2 Bouncy Ball Photon Collection Challenge
Acting as the ICESat-2 satellite, students investigate the reflection of light photons off Earth's surface by catching and recording a number of photons. Using bouncy balls to represent the photons, students drop, let bounce once and try to catch in... (View More)
# Global Energy Budget
Students will explore how energy from the sun is absorbed, reflected and radiated back into space by Earth. By completing three short labs investigating the effects of surface color, type of material, or cloud cover on temperature change, students... (View More)
# Erosion and Landslides
Students will be introduced to the causes, locations, and hazards of landslides, as well as the role of satellite observations in predicting and studying them. To begin, students investigate the amount of precipitation sufficient to cause a... (View More)
# MRC: Manual and Skit (Grades 6-8)
Learners will brainstorm ideas to be developed into a team skit, work cooperatively to assign duties and write a team skit, and collaborate with team members to complete the Mars Rover Manual. The lesson uses the 5E instructional model and includes:... (View More)
# Earth, Earth's Moon and Mars Balloons
Balloons are used to construct a scale model of the Earth, Earth's Moon and Mars in relation to each other. Students use this model to predict distances and reflect on how scientists use models to construct explanations through the scientific... (View More)
# Measure Up
Students will use various objects in the classroom to experiment with nonstandard measurement. They will make estimates and test them out. Then, working in pairs or small groups, students will use a ruler or a measuring tape to become familiar with... (View More)
Keywords: Water; Scale; Size
# Play Dough Planets
Learners will demonstrate the size (volume) differences between Earth, Earth's Moon, and Mars. An extension is provided to estimate the distance between the Earth and the Moon, and the Earth and Mars, using the scale of the play dough planets'... (View More)
# Building a Thermometer
In this learning activity, students will construct a soda-bottle thermometer, which is similar to the thermometer used by GLOBE schools. Both are based on the principle that most substances expand and contract as their temperature changes. This... (View More)
# Snow Goggles and Limiting Sunlight
This is a lesson about radiation and the use of the scientific method to solve problems of too much radiation. Learners will build snow goggles similar to those used by the Inuit (designed to block unwanted light, while increasing the viewer's... (View More)
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# When is the closure of a path connected set also path connected?
What are the most general criteria we can impose on a locally path connected Hausdorff space $X$ and a path connected subset $A$ such that $\overline{A}$ is path connected? Do more restrictions need to be imposed on $X$ or $A$?
For instance, I know that if $\overline{A}$ is locally path connected then $\overline{A}$ is path connected; for all $x \in \overline{A}$ and some neighborhood $U$ of $x$ that is open in $\overline{A}$, there must be some path connected neighborhood $U' \subseteq U$ of $x$ that is open in $\overline{A}$. That is, there is some open subset $V'$ of $X$ such that $U' = V' \cap \overline{A}$. Since $x$ is a point of closure of $A$, $V'$ must contain some point $x' \in A \subseteq \overline{A}$, i.e. $x' \in U'$, so $x$ is path connected to $x'$ and hence also to $A$. This holds for all $x \in \overline{A}$ so $\overline{A}$ is path connected.
However, the tough part is proving that $\overline{A}$ is locally path connected, because $\overline{A}$ is probably (?) not open in $X$. I'm a complete novice so the only useful thing I know from browsing definitions is that all open subsets of a locally path connected space inherit the local path connectivity. Are there more ways to prove that a subspace inherits local path connectedness?
This is more specific, but would it help if I knew that $A$ was the set difference of two closed sets (i.e. the intersection of a closed set and an open set)?
I've been looking at stronger restrictions such as $X$ being locally simply connected, but the online documentation is scarce. Would local simple connectivity be "inherited more easily" by subspaces?
• I don't think you'll get far imposing conditions on $X$: $\mathbb{R}^2$ is about as nice a space as you could wish for, but $A=\{(x,\sin(1/x))| x \in (0,1) \}$ still gives a counterexample. Here $A$ is homeomorphic to $\mathbb{R}$, another very nice space, suggesting imposing conditions on the topology of $A$ won't help either. It looks like you really do need conditions specifically on how $A$ sits in $X$. Jun 25, 2012 at 14:34
• @ChrisEagle: IMO that example is decisive enough to be an answer... oh except that now the question has been edited. I wonder if this is not really just two questions, now. Jun 25, 2012 at 14:49
• Good point, I guess I really have to prove some restrictions on $\overline{A}$ to exclude pathological cases such as the Topologist's Sine Curve (in fact, I'd prefer working on $A \cup \{x\}$, where $x \in \overline{A}$), such as local path connectivity (which I can't prove easily) or local simple path connectivity (which I don't fully understand). Jun 25, 2012 at 14:50
• Considering the edit, the problem in Chris' counterexample is not that the function doesn't exist, but that it fails to be continuous, which sounds like it would be hard to prove in generality. Jun 25, 2012 at 14:50
• Sorry for the change, but the initial question was to give a sufficient condition to construct the $p$ I mention in my edits. The second half of the question now contains another attempt at the same construction. Jun 25, 2012 at 14:52
Here's something that I came up with. The proposition below is what you're asking for but I've also encapsulated the main idea behind these results in the following lemma in case that's more helpful.
Lemma: Let $$S \subseteq X$$ be path-connected and $$x^1 \in \overline{S}$$. Suppose there exists a countable decreasing (i.e. $$U_{i+1} \subseteq U_i$$) neighborhood basis $$\left( U_i \right)_{i=1}^{\infty}$$ in $$X$$ at $$x^1$$ such that for each $$i$$, whenever $$s^i \in S \cap U_i$$ then there exists a path in $$S \cap U_i$$ from $$s^i$$ to some element of $$S \cap U_{i+1}$$. Then $$S \cup \left\lbrace x^1 \right\rbrace$$ is path-connected.
Remark: Note that we are not assuming that for all $$i$$, there exists a path between any two point of $$S \cap U_i$$. The sets $$S \cap U_i$$ need not even be connected so this is weaker than requiring local connectivity of $$\overline{S}$$ at $$x^1$$.
Corollary: Let $$S \subseteq X$$ be path-connected. If the condition of the above lemma is satisfied at each $$x^1 \in \overline{S}$$ (or slightly more generally, if each path-component of the boundary of $$S$$ contains some point satisfying this condition) then $$\overline{S}$$ is path-connected.
Prop: Let $$S \subseteq X$$ be path-connected. Suppose that each path component of $$\overline{S} \setminus S$$ contains some $$x^1$$ for which there exists a countable decreasing neighborhood basis $$\left( U_i \right)_{i=1}^{\infty}$$ in $$X$$ at $$x^1$$ s.t. for each $$i$$ and each path-component $$P_i$$ of $$S \cap U_i$$, there exists a path in $$\overline{S} \cap U_i$$ whose image intersects both $$P_i$$ and $$S \cap U_{i+1}$$. Then $$\overline{S}$$ is path-connected.
Remark: In this proposition, you can replace "of $$\overline{S} \setminus S$$" with "of the boundary of $$S$$ in $$\overline{S}$$". Also, to prove that $$\overline{S}$$ is path-connected, it may be easier to find some other path-connected $$R \subseteq X$$ such that $$\overline{R} = \overline{S}$$ and then apply these results to $$R$$ in place of $$S$$.
Proof of lemma: Pick any $$s^1 \in S \cap U_1$$ and any $$0 = t_0 < t_1 < \cdots < 1$$ s.t. $$t_i \to 1$$ and let $$\gamma_0 : [t_0, t_1] \to S$$ be the constant path at $$s^0 := s^1$$. Suppose for all $$0 \leq l \leq i + 1$$ we've picked $$s^l \in S \cap U_l$$ and for every $$0 \leq l \leq i$$ we have a path $$\gamma_l : [t_l, t_{l+1}] \to S \cap U_l$$ from $$s^l$$ to $$s^{l+1}$$ (where observe that this holds for $$i = 0$$). By assumption, we can pick $$s^{i+2} \in S \cap U_{i+2}$$ and a path $$\gamma_{i+1} : [t_{i+1}, t_{i+2}] \to S \cap U_{i+1}$$ from $$s^{i+1}$$ to $$s^{i+2}$$.
After starting this inductive construction at $$i = 0$$ we can use $$\gamma_0, \gamma_1, \ldots$$ to define $$\gamma : [0, 1] \to S \cup \left\lbrace x^1 \right\rbrace$$ on $$[0, 1)$$ in the obvious way and then declare that $$\gamma(1) := x^1$$. For any integer $$N$$, $$l \geq N$$ implies $$\operatorname{Im} \gamma_l \subseteq U_l \subseteq U_N$$ so that $$\gamma([t_N, 1]) \subseteq U_N$$. Thus $$\gamma$$ is continuous at $$1$$ so that $$S \cup \left\lbrace x^1 \right\rbrace$$ is path-connected. Q.E.D.
It should now be clear how the idea behind this lemma's proof led to the above proposition's statement.
Proof of prop: Let $$x^1$$ and $$\left( U_i \right)_{i=1}^{\infty}$$ have the properties described in the proposition's statement, let $$0 = t_0 < t_1 < \cdots < 1$$ be s.t. $$t_i \to 1$$, and let $$\gamma_0 : [t_0, t_1] \to S \cap U_1$$ be any constant path. Suppose $$i \geq 0$$ is such that for all $$1 \leq l \leq i$$, we have constructed a path $$\gamma_l : \left[ t_l, t_{l+1} \right] \to \overline{S} \cap U_l$$ such that $$\gamma_l(t_l) = \gamma_{l-1}\left( t_{l} \right)$$ and $$\gamma_l\left( t_{l+1} \right) \in S \cap U_{l+1}$$ (note that this is true for $$i = 0$$). Our assumption on $$\left( U_i \right)_{i=1}^{\infty}$$ allows us to construct a path $$\gamma_{i+1} : \left[ t_{i+1}, t_{i+2} \right] \to \overline{S} \cap U_{i+1}$$ starting that $$\gamma_i\left( t_{i+1} \right)$$ and ending at some point of $$S \cap U_{i+2}$$. Exactly as was done in the proof of the above lemma, we may now define a continuous map $$\gamma : [0, 1] \to \overline{S}$$ such that $$\gamma(1) = x^1$$. Q.E.D.
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### Topic: Two's Complement (Read 490 times)previous topic - next topic
#### jremington
#15
##### Aug 10, 2019, 07:20 pmLast Edit: Aug 10, 2019, 07:27 pm by jremington
Quote
What you geniuses can't seem to grasp is that after getting the two's as in reply #1, I can't pass it as an unsigned value because the msb does not lose its significance as a sign bit.
Your explanation makes no sense. By definition, an unsigned value has no sign bit. I repeat: by definition.
It is pretty hard to imagine what you are doing wrong, given that you failed to post an example demonstrating the problem.
#16
#### johnwasser
#17
##### Aug 10, 2019, 09:28 pm
If I send a "MoveAbs,0,20000,800" command, the motor moves to position 20000. The "GetPos,0" command returns 20000 as expected.
If I send a "MoveAbs,0,-20000,800" command, the motor does not move at all but interestingly enough, the "GetPos,0" command returns 4194303.
I don't know why the motor doesn't move but I do know why the new position is 4194303 (0x3FFFFF). When you put -20,000 (0xFFFFB1E0) into an unsigned long you get the value 4,294,947,296. The _GoTo() function (see reply #10) in the library finds that is larger than the maximum allowed absolute position (0x3FFFFF == 4,194,303) and sets the target position to the maximum.
Maybe the library and the motor controller disagree on what the maximum allowed absolute coordinate is and the driver rejects the command.
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#### johnwasser
#18
##### Aug 10, 2019, 09:49 pm
Did you wait for the previous command to complete? The datasheet says "Any attempt to perform a GoTo command when a previous command is under execution (BUSY low) causes the command to be ignored and the NOTPERF_CMD flag to rise (see Section 9.1.22)"
I think I see part of the problem. The function takes a 32-bit unsigned argument but the ABS_POS register is a SIGNED 22-bit number. You can't use a 32-bit negative number directly. You have to mask off the upper bits. That is BAD design on the part of the library writer.
To move to position -20,000 you can't pass -20,000 to the function. Since the function takes a 32-bit unsigned number it will appear to be a very large positive number: 4,294,947,296, greater than the allowed maximum of 4,194,303. To get the number you want you have to mask off the upper bits:
-20,000 & 0x3FFFFFUL == 0x003FB1E0 == 4,174,304
Since 0x3FB1E0 is less than 0x3FFFFF it will be passed to the driver unchanged.
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#### Chevelle
#19
##### Aug 10, 2019, 10:21 pmLast Edit: Aug 10, 2019, 10:36 pm by Chevelle
John,
Thanks very much for you thoughtful and supportive help.
Yes, the writer of the library didn't to a particularly good job of it. On that subject for a moment, I have tried several libraries for this device (which is one hell of a device, very powerful) but there doesn't seem to be a simple straightforward library from anyone, including STI. They are all loaded up with board specific configurations (such as the Sparkfun and STI libraries) or are combined in a mega library such as from Adafruit.
This one at least is geared for the IC level which is great for my application. It isn't a bad library but it isn't great either. This is one example. The purpose of a library is to provide the user with a command set that gives him/her a frame of reference to the real word and makes the translation of real world units to machine level units. If people think in + and - terms with reference to home, then accept + and - parameters. Why require the stall current in hex when it can be converted to milliamps in the library?
Anyway, back to this issue. It was indeed simple.
My mistake was keeping everything as a signed long (as was pointed out). I was taking a negative signed long, taking the two's complement, and then assigning that back to a signed long. That "preserved" the msb as a sign bit.
The right way is to take the two's complement for negative position values and assign that back to an unsigned long. And John, you are dead right. The upper bits have to be masked off. Viola, it works.
Code: [Select]
`long valLong;unsigned long valULong;if (valLong < 0) { valULong = (~abs(valLong) + B01) & 0x3FFFFF; } else { valULong = valLong; }Axis0._GoTo(valULong);`
#### johnwasser
#20
##### Aug 11, 2019, 12:17 am
Code: [Select]
`(~abs(valLong) + B01)`
Since the absolute value of a negative number is the negative of that negative number and the 2's Compliment of a number is the negative of that number, your expression is functionally equivalent to:
Code: [Select]
`(-(-valLong))`
Which is equivalent to:
Code: [Select]
`(valLong)`
For values in the allowed range of 0x200000 (-2,097,152) to 0x1FFFFF (+2,097,151) the code:
Code: [Select]
`if (valLong < 0) { valULong = (~abs(valLong) + B01) & 0x3FFFFF; } else { valULong = valLong; }Axis0._GoTo(valULong);`
Is equivalent to:
Code: [Select]
`Axis0._GoTo(valLong & 0x3FFFFFUL);`
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#### Chevelle
#21
##### Aug 11, 2019, 12:43 am
I will give that try. Looks much more efficient.
Thanks very much.
(I'll be back when it comes to reading the x_ABS_POS value.)
#### Chevelle
#22
##### Sep 17, 2019, 05:50 pmLast Edit: Sep 17, 2019, 06:15 pm by Chevelle
It has been a while since this thread has been active because I have been making great progress, especially thanks to John. Things are moving and there are now plenty of functional feature that are working very well.
By way off background, the board controls two stepper motors (A and B). These motors move a CoreXY stage (very similar to many 3D printers.) Any X,Y movement of the carriage requires movement by motors A and B. For example, moving just motor A move the carriage in both X and Y in a diagonal direction.
There are two routines that do not seem to jive. One is to send the carriage to an X,Y location. (MoveAbsXY) The next two parameters that are sent are the desired X and Y positions. Optionally, velocities for the A and B motor can be sent.
The other routine is to read the position of the A and B motor. (GetPos) The A and B positions are then translated into the X and Y coordinates of the carriage.
Here are the two routines....
Code: [Select]
` if (Cmd[0] == "moveabsxy") { // Syntax MoveAbsXY, X, Y, vel0*, vel1* PosX = Cmd[1].toInt(); // This the desired X position to move to if (abs(PosX) > 0x1FFFFF) { CmdOK = false; } // Check to be sure that X is within acceptable range PosY = Cmd[2].toInt(); // This is the desired Y position to move to if (abs(PosY) > 0x1FFFFF) { CmdOK = false; } // Check to be sure that Y is within acceptable range SpeedA = Cmd[3].toInt(); // This the speed at which the A motor will move if ((Cmd[3] != "") && (SpeedA != constrain(SpeedA, 1, 0x3CFA))) { CmdOK = false; } SpeedB = Cmd[4].toInt(); // This is the speed at which the B motor will move if ((Cmd[4] != "") && (SpeedB != constrain(SpeedB, 1, 0x3CFA))) { CmdOK = false; } if (CmdOK == false) { Serial.println("err:Command"); } else { Serial.println("OK"); } if (CmdOK == true) { if (Cmd[3] != "") { AxisA._SetParam(x_MAX_SPEED, AxisA._MaxSpdCalc(SpeedA)); } if (Cmd[4] != "") { AxisB._SetParam(x_MAX_SPEED, AxisB._MaxSpdCalc(SpeedB)); } uPosA = (PosX + PosY) & 0x3FFFFFUL; // Calculate the position for the A motor so that the carriage move to the desired X, Y position AxisA._GoTo(uPosA); // Send the A motor on its way uPosB = (PosX - PosY) & 0x3FFFFFUL; // Calculate the position for the B motor so that the carriage moves to the desired X,Y position AxisB._GoTo(uPosB); // Send the B motor on its way } }`
and
Code: [Select]
`if (Cmd[0] == "getpos") { // Syntax GetPos valULong = AxisA._GetParam(x_ABS_POS); // Get the position for the A motor in 2's comp if (valULong > 0x1FFFFF) { // Convert from 2's comp to signed long valLong = (~valULong + B01) & 0x1FFFFF; valLong = -valLong;} else { valLong = valULong; } strCmd = String(valLong) + ","; valULong = AxisB._GetParam(x_ABS_POS); // Get the position for the B motor in 2's comp if (valULong > 0x1FFFFF) { // Convert from 2's comp to signed long valLong = (~valULong + B01) & 0x1FFFFF; valLong = -valLong; } else { valLong = valULong; } strCmd = strCmd + String(valLong); Serial.println(strCmd);}`
Something is wrong because when I command the carriage to an X,Y location and use the GetPos routine to get the A and B position and calculate X and Y, they do not match.
I am assuming that there is something wrong with how I am dealing with the 2's complements of the values.
Any assistance is greatly appreciated.
#### johnwasser
#23
##### Sep 18, 2019, 02:22 am
Things are very confusing because the library takes (and returns?) positions as unsigned 32-bit values BUT the hardware requires 22-bit SIGNED numbers. This is what we in the programming business call "a major design mistake".
It it were me, I would modify the library. I would have the library always use 32-bit signed integers (int32_t) for positions. The library would truncate to 22 bits (Position & 0x3FFFFF) when sending the position to the hardware and sign-extend any 22-bit position it gets from the hardware:
Code: [Select]
`int32_t pos = (Position & 0x200000) ? Position | 0xFFC00000 : Position;`
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#### Chevelle
#24
##### Sep 18, 2019, 02:30 pm
John, you are a life saver.
Yes, confusing to say the least. I am confident enough to modify the library but only up to a point. (I already modified it to comment out some unwanted serial print statements.)
This is a professional application but it is not in my main area of expertise. We are a low budget affair but I am sure we can engage you for a few hours of your time if you are interested.
#### gfvalvo
#25
##### Sep 18, 2019, 02:43 pm
I wouldn't modify the library because you'd have to do so again every time it is updated by the author. Instead, just write a wrapper class around it.
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#### Chevelle
#26
##### Sep 18, 2019, 02:54 pm
I think a new library is in order here, completely separate from the original.
#### Chevelle
#27
##### Sep 18, 2019, 03:00 pmLast Edit: Sep 18, 2019, 03:01 pm by Chevelle
John,
My C++ isn't good enough to completely rely on the use of the (better) short handed way of doing things. Am I correct in my long handed interpretation of your code snippet?
Code: [Select]
`int32_t pos = Position;if (Position & 0x200000) {Position | 0xFFC00000;}`
#### johnwasser
#28
##### Sep 18, 2019, 03:31 pm
Am I correct in my long handed interpretation of your code snippet?
Code: [Select]
`int32_t pos = Position;if (Position & 0x200000) {Position | 0xFFC00000;}`
Not quite. The "{Position | 0xFFC00000;}" doesn't store a value anywhere. It calculates a value and throws the result away. If your warning levels are high enough the compiler should warn you that the statement has no side effects.
A full expansion would be more like:
Code: [Select]
`int32_t pos = Position; if (Position & 0x200000) { pos = Position | 0xFFC00000; }`
The result of the Ternary Operator "A ? B : C" is the value B if A is 'true' (non-zero) and the value C if A is 'false' (zero).
Another way to expand it is:
Code: [Select]
`int32_t pos = Position; if (pos & 0x00200000) { pos |= 0xFFC00000; }`
Similar to +=, -=, *=... the "pos |=" is a C shortcut for "pos = pos |".
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#### gfvalvo
#29
##### Sep 18, 2019, 03:44 pm
I think a new library is in order here, completely separate from the original.
OK, then it's on you to fix any problems that arise with the base library and are subsequently fixed in that repository. You'll need to implement similar fixes in your custom version. Not a problem if your coding skills are up to it.
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# Reduction of 3-SAT to 3-COLOR
The decision version of the 3-COLOR problem is the problem of deciding whether an input graph G(V, E) can be colored using only 3 colors so that no 2 adjacent vertices have the same color.
I had interpreted that to mean that we were looking for any coloring.
However, most proofs I have seen that reduce 3-SAT to 3-COLOR to prove that 3-SAT is NP-Complete use subgraph "gadgets" where some of the nodes are already colored.
But in this case, it would only show that a specific 3-coloring (i.e. some nodes on the input graph are pre-colored) does not exist.
It doesn't show that no 3-coloring exists. Can someone clarify why the reduction is valid?
An important part of these gadget schemes is that there is a 3-clique somewhere called the palette that arbitrates the logic roles of colors. In other words, there are 3 nodes all connected to each other, labeled $v_T$, $v_F$, and $v_R$. The palette can be arbitrarily colored, and whatever colors are chosen represent the 3 logic values. Then when a gadget has a fixed "true" node in it, arcs are extended from fixed $v_F$ and $v_R$ nodes to it, to compel its color to be the same as $v_T$.
• No, no colors are fixed. "$v_T$" is just a label on a node. Arcs are added so that other nodes are forced to be the same color as it, and can also be labeled "$v_T$". It is easy to prove then that all $v_T$ nodes must be the same color, not that they must be a particular color. – NovaDenizen Nov 21 '14 at 16:55
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You may remember the dolls from the 80's: 'My Buddy' and 'Kid Sister'. They were about two feet tall, and kids would lug them around. 'My Buddy' was a success, eventually spawning 'Kid Sister' - which brings me to this subject. I was thinking about this writeup this morning, and I've come across lots of evidence that proves my theory.
Let's begin: at the beginning!
In 1985, Hasbro introduced 'My Buddy', a companion doll for kids. It was a tremendous success, and toy stores everywhere carried them. The success of the male doll led Hasbro to create a female companion - 'Kid Sister'. It would seem that our little 'Kid Sister' was an afterthought of Buddy's success.
Now, let's compare marketing between the two. We'll take a look at their respective jingles.
My Buddy:
My Buddy! (My Buddy!)
My Buddy! (My Buddy!)
Wherever I go, he goes.
My Buddy! (My Buddy!)
My Buddy! (My Buddy!)
My Buddy and me!
Kid Sister:
Kid Sister! (Kid Sister!)
Kid Sister! (Kid Sister!)
Wherever I go, you're gonna go
Kid Sister! (Kid Sister!)
Kid Sister! (Kid Sister!)
Kid Sister and me!
Clearly, this demonstrates that 'My Buddy' has free will, while 'Kid Sister' will do as she's told.
Now we move to their names. Let's use them in a sentence:
"This is My Buddy!" (Implying friendship, and companionship.)
"This is my Kid Sister." (Implying a burden, and a tag-along.)
Later, 'My Buddy' proceeded to (indirectly) launch a high-profile movie career. The character of 'Chucky' in Child's Play was modelled after 'My Buddy'. While 'My Buddy' was enjoying his success, 'Kid Sister' was left to fade into obscurity.
From the clarity and amount of evidence, we must conclude that 'My Buddy' was enjoying his success early in his career. When 'Kid Sister' came along, 'My Buddy' became jealous that his newer, younger sister was stealing some of his spotlight. He then used his market muscle to undermine his little sister's career, causing these little slip-ups in her marketing. When he vanquished his evil sister, he went on to start his acting career as alter-ego 'Chucky', leaving 'Kid Sister' to be forgotten.
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# Truncatable primes
Truncatable primes
You are encouraged to solve this task according to the task description, using any language you may know.
A truncatable prime is a prime number that when you successively remove digits from one end of the prime, you are left with a new prime number; for example, the number 997 is called a left-truncatable prime as the numbers 997, 97, and 7 are all prime. The number 7393 is a right-truncatable prime as the numbers 7393, 739, 73, and 7 formed by removing digits from its right are also prime. No zeroes are allowed in truncatable primes.
The task is to find the largest left-truncatable and right-truncatable primes less than one million (base 10 is implied).
` with Ada.Text_IO; use Ada.Text_IO;with Ada.Containers.Ordered_Sets; procedure Truncatable_Primes is package Natural_Set is new Ada.Containers.Ordered_Sets (Natural); use Natural_Set; Primes : Set; function Is_Prime (N : Natural) return Boolean is Position : Cursor := First (Primes); begin while Has_Element (Position) loop if N mod Element (Position) = 0 then return False; end if; Position := Next (Position); end loop; return True; end Is_Prime; function Is_Left_Trucatable_Prime (N : Positive) return Boolean is M : Natural := 1; begin while Contains (Primes, N mod (M * 10)) and (N / M) mod 10 > 0 loop M := M * 10; if N <= M then return True; end if; end loop; return False; end Is_Left_Trucatable_Prime; function Is_Right_Trucatable_Prime (N : Positive) return Boolean is M : Natural := N; begin while Contains (Primes, M) and M mod 10 > 0 loop M := M / 10; if M <= 1 then return True; end if; end loop; return False; end Is_Right_Trucatable_Prime; Position : Cursor;begin for N in 2..1_000_000 loop if Is_Prime (N) then Insert (Primes, N); end if; end loop; Position := Last (Primes); while Has_Element (Position) loop if Is_Left_Trucatable_Prime (Element (Position)) then Put_Line ("Largest LTP from 1..1000000:" & Integer'Image (Element (Position))); exit; end if; Previous (Position); end loop; Position := Last (Primes); while Has_Element (Position) loop if Is_Right_Trucatable_Prime (Element (Position)) then Put_Line ("Largest RTP from 1..1000000:" & Integer'Image (Element (Position))); exit; end if; Previous (Position); end loop;end Truncatable_Primes; `
Sample output:
```Largest LTP from 1..1000000: 998443
Largest RTP from 1..1000000: 739399
```
## ALGOL 68
Translation of: C
Note: This specimen retains the original C coding style.
Works with: ALGOL 68 version Revision 1 - no extensions to language used.
Works with: ALGOL 68G version Any - tested with release 1.18.0-9h.tiny.
`#!/usr/local/bin/a68g --script # PROC is prime = (INT n)BOOL:( []BOOL is short prime=(FALSE, TRUE, TRUE, FALSE, TRUE, FALSE, TRUE, FALSE, FALSE); IF n<=UPB is short prime THEN is short prime[n] # EXIT # ELSE IF ( NOT ODD n | TRUE | n MOD 3 = 0 ) THEN FALSE # EXIT # ELSE INT h := ENTIER sqrt(n)+3; FOR a FROM 7 BY 6 WHILE a<h DO IF ( n MOD a = 0 | TRUE | n MOD (a-2) = 0 ) THEN false exit FI OD; TRUE # EXIT # FI FI EXIT false exit: FALSE); PROC string to int = (STRING in a)INT:( FILE f; STRING a := in a; associate(f, a); INT i; get(f, i); close(f); i); PROC is trunc prime = (INT in n, PROC(REF STRING)VOID trunc)BOOL: ( INT n := in n; STRING s := whole(n, 0); IF char in string("0", NIL, s) THEN FALSE # EXIT # ELSE WHILE is prime(n) DO s := whole(n, 0); trunc(s); IF UPB s = 0 THEN true exit FI; n := string to int(s) OD; FALSE EXIT true exit: TRUE FI); PROC get trunc prime = (INT in n, PROC(REF STRING)VOID trunc)VOID:( FOR n FROM in n BY -1 TO 1 DO IF is trunc prime(n, trunc) THEN printf((\$g(0)l\$, n)); break FI OD; break: ~); main:( INT limit = 1000000; printf((\$g g(0) gl\$,"Highest left- and right-truncatable primes under ",limit,":")); get trunc prime(limit, (REF STRING s)VOID: s := s[LWB s+1:]); get trunc prime(limit, (REF STRING s)VOID: s := s[:UPB s-1]); write("Press Enter"); read(newline))`
Output:
```Highest left- and right-truncatable primes under 1000000:
998443
739399
Press Enter
```
## AutoHotkey
`SetBatchLines, -1MsgBox, % "Largest left-truncatable and right-truncatable primes less than one million:`n" . "Left:`t" LTP(10 ** 6) "`nRight:`t" RTP(10 ** 6) LTP(n) { while n { n-- if (!Instr(n, "0") && IsPrime(n)) { Loop, % StrLen(n) if (!IsPrime(SubStr(n, A_Index))) continue, 2 break } } return, n} RTP(n) { while n { n-- if (!IsPrime(SubStr(n, 1, 1))) n -= 10 ** (StrLen(n) - 1) if (!Instr(n, "0") && IsPrime(n)) { Loop, % StrLen(n) if (!IsPrime(SubStr(n, 1, A_Index))) continue, 2 break } } return, n} IsPrime(n) { if (n < 2) return, 0 else if (n < 4) return, 1 else if (!Mod(n, 2)) return, 0 else if (n < 9) return 1 else if (!Mod(n, 3)) return, 0 else { r := Floor(Sqrt(n)) f := 5 while (f <= r) { if (!Mod(n, f)) return, 0 if (!Mod(n, (f + 2))) return, 0 f += 6 } return, 1 }}`
Output:
```Largest left-truncatable and right-truncatable primes less than one million:
Left: 998443
Right: 739399```
## Bracmat
Primality test: In an attempt to compute the result of taking a (not too big, 2^32 or 2^64, depending on word size) number to a fractional power, Bracmat computes the prime factors of the number and checks whether the powers of prime factors make the fractional power go away. If the number is prime, the output of the computation is the same as the input.
`( 1000001:?i& whl ' ( !i+-2:>0:?i & !i:?L & whl'(!L^1/2:#?^1/2&@(!L:% ?L)) & !L:~ )& out\$("left:" !i)& 1000001:?i& whl ' ( !i+-2:>0:?i & !i:?R & whl'(!R^1/2:#?^1/2&@(!R:?R %@)) & !R:~ )& out\$("right:" !i))`
Output:
```left: 998443
right: 739399```
## C
`#include <stdio.h>#include <stdlib.h>#include <string.h> #define MAX_PRIME 1000000char *primes;int n_primes; /* Sieve. If we were to handle 10^9 range, use bit field. Regardless, * if a large amount of prime numbers need to be tested, sieve is fast. */void init_primes(){ int j; primes = malloc(sizeof(char) * MAX_PRIME); memset(primes, 1, MAX_PRIME); primes[0] = primes[1] = 0; int i = 2; while (i * i < MAX_PRIME) { for (j = i * 2; j < MAX_PRIME; j += i) primes[j] = 0; while (++i < MAX_PRIME && !primes[i]); }} int left_trunc(int n){ int tens = 1; while (tens < n) tens *= 10; while (n) { if (!primes[n]) return 0; tens /= 10; if (n < tens) return 0; n %= tens; } return 1;} int right_trunc(int n){ while (n) { if (!primes[n]) return 0; n /= 10; } return 1;} int main(){ int n; int max_left = 0, max_right = 0; init_primes(); for (n = MAX_PRIME - 1; !max_left; n -= 2) if (left_trunc(n)) max_left = n; for (n = MAX_PRIME - 1; !max_right; n -= 2) if (right_trunc(n)) max_right = n; printf("Left: %d; right: %d\n", max_left, max_right); return 0;}`
output
`Left: 998443; right: 739399`
Faster way of doing primality test for small numbers (1000000 isn't big), and generating truncatable primes bottom-up:
`#include <stdio.h> #define MAXN 1000000int maxl, maxr; int is_prime(int n){ int p; if (n % 3 == 0) return 0; for (p = 6; p * p <= n; p += 6) if (!(n % (p + 1) && n % (p + 5))) return 0; return 1;} void left(int n, int tens){ int i, nn; if (n > maxl) maxl = n; if (n < MAXN / 10) for (tens *= 10, i = 1; i < 10; i++) if (is_prime(nn = i * tens + n)) left(nn, tens);} void right(int n){ int i, nn; static int d[] = {1,3,7,9}; if (n > maxr) maxr = n; if (n < MAXN / 10) for (i = 1; i < 4; i++) if (is_prime(nn = n * 10 + d[i])) right(nn);} int main(void){ left(3, 1); left(7, 1); right(3); right(5); right(7); printf("%d %d\n", maxl, maxr); return 0;}`
Output:
```998443 739399
```
## C#
`using System;using System.Diagnostics; namespace RosettaCode{ internal class Program { public static bool IsPrime(int n) { if (n<2) return false; if (n<4) return true; if (n%2==0) return false; if (n<9) return true; if (n%3==0) return false; var r = (int) Math.Sqrt(n); var f = 6-1; while (f<=r) { if (n%f==0 ||n%(f+2)==0) return false; f += 6; } return true; } private static bool IsRightTruncatable(int n) { for (;;) { n /= 10; if (n==0) return true; if (!IsPrime(n)) return false; } } private static bool IsLeftTruncatable(int n) { string c = n.ToString(); if (c.Contains("0")) return false; for (int i = 1; i<c.Length; i++) if (!IsPrime(Convert.ToInt32(c.Substring(i)))) return false; return true; } private static void Main() { var sb = new Stopwatch(); sb.Start(); int lt = 0, rt = 0; for (int i = 1000000; i>0; --i) { if (IsPrime(i)) { if (rt==0 && IsRightTruncatable(i)) rt = i; else if (lt==0 && IsLeftTruncatable(i)) lt = i; if (lt!=0 && rt!=0) break; } } sb.Stop(); Console.WriteLine("Largest truncable left is={0} & right={1}, calculated in {2} msec.", lt, rt, sb.ElapsedMilliseconds); } }}`
`Largest truncable left is=998443 & right=739399, calculated in 16 msec.`
## Clojure
`(use '[clojure.contrib.lazy-seqs :only [primes]]) (def prime? (let [mem (ref #{}) primes (ref primes)] (fn [n] (dosync (if (< n (first @primes)) (@mem n) (let [[mems ss] (split-with #(<= % n) @primes)] (ref-set primes ss) ((commute mem into mems) n))))))) (defn drop-lefts [n] (let [dropl #(if (< % 10) 0 (Integer. (subs (str %) 1)))] (->> (iterate dropl n) (take-while pos? ,) next))) (defn drop-rights [n] (->> (iterate #(quot % 10) n) next (take-while pos? ,))) (defn truncatable-left? [n] (every? prime? (drop-lefts n))) (defn truncatable-right? [n] (every? prime? (drop-rights n))) user> (->> (for [p primes :while (< p 1000000) :when (not-any? #{\0} (str p)) :let [l? (if (truncatable-left? p) p 0) r? (if (truncatable-right? p) p 0)] :when (or l? r?)] [l? r?]) ((juxt #(apply max-key first %) #(apply max-key second %)) ,) ((juxt ffirst (comp second second)) ,) (map vector ["left truncatable: " "right truncatable: "] ,))(["left truncatable: " 998443] ["right truncatable: " 739399])`
## CoffeeScript
`# You could have symmetric algorithms for max right and left# truncatable numbers, but they lend themselves to slightly# different optimizations. max_right_truncatable_number = (n, f) -> # This algorithm only evaluates 37 numbers for primeness to # get the max right truncatable prime < 1000000. Its # optimization is that it prunes candidates for # the first n-1 digits before having to iterate through # the 10 possibilities for the last digit. if n < 10 candidate = n while candidate > 0 return candidate if f(candidate) candidate -= 1 else left = Math.floor n / 10 while left > 0 left = max_right_truncatable_number left, f right = 9 while right > 0 candidate = left * 10 + right return candidate if candidate <= n and f(candidate) right -= 1 left -= 1 throw Error "none found" max_left_truncatable_number = (max, f) -> # This is a pretty straightforward countdown. The first # optimization here would probably be to cache results of # calling f on small numbers. is_left_truncatable = (n) -> candidate = 0 power_of_ten = 1 while n > 0 r = n % 10 return false if r == 0 n = Math.floor n / 10 candidate = r * power_of_ten + candidate power_of_ten *= 10 return false unless f(candidate) true do -> n = max while n > 0 return n if is_left_truncatable n, f n -= 1 throw Error "none found" is_prime = (n) -> return false if n == 1 return true if n == 2 for d in [2..n] return false if n % d == 0 return true if d * d >= n console.log "right", max_right_truncatable_number(999999, is_prime)console.log "left", max_left_truncatable_number(999999, is_prime) `
output
` > coffee truncatable_prime.coffee right 739399left 998443 `
## Common Lisp
` (defun start () (format t "Largest right-truncatable ~a~%" (max-right-truncatable)) (format t "Largest left-truncatable ~a~%" (max-left-truncatable))) (defun max-right-truncatable () (loop for el in (6-digits-R-truncatables) maximizing el into max finally (return max))) (defun 6-digits-R-truncatables (&optional (lst '(2 3 5 7)) (n 5)) (if (zerop n) lst (6-digits-R-truncatables (R-trunc lst) (- n 1)))) (defun R-trunc (lst) (remove-if (lambda (x) (not (primep x))) (loop for el in lst append (mapcar (lambda (x) (+ (* 10 el) x)) '(1 3 7 9))))) (defun max-left-truncatable () (loop for el in (6-digits-L-truncatables) maximizing el into max finally (return max))) (defun 6-digits-L-truncatables (&optional (lst '(3 7)) (n 5)) (if (zerop n) lst (6-digits-L-truncatables (L-trunc lst (- 6 n)) (- n 1)))) (defun L-trunc (lst n) (remove-if (lambda (x) (not (primep x))) (loop for el in lst append (mapcar (lambda (x) (+ (* (expt 10 n) x) el)) '(1 2 3 4 5 6 7 8 9))))) (defun primep (n) (primep-aux n 2)) (defun primep-aux (n d) (cond ((> d (sqrt n)) t) ((zerop (rem n d)) nil) (t (primep-aux n (+ d 1))))) `
Output:
```Largest right-truncatable 739399
Largest left-truncatable 998443```
## D
`import std.stdio, std.math, std.string, std.conv, std.algorithm, std.range; bool isPrime(in int n) pure nothrow { if (n <= 1) return false; foreach (immutable i; 2 .. cast(int)sqrt(real(n)) + 1) if (!(n % i)) return false; return true;} bool isTruncatablePrime(bool left)(in int n) pure { immutable s = n.text; if (s.canFind('0')) return false; foreach (immutable i; 0 .. s.length) static if (left) { if (!s[i .. \$].to!int.isPrime) return false; } else { if (!s[0 .. i + 1].to!int.isPrime) return false; } return true;} void main() { enum n = 1_000_000; writeln("Largest left-truncatable prime in 2 .. ", n, ": ", iota(n, 1, -1).filter!(isTruncatablePrime!true).front); writeln("Largest right-truncatable prime in 2 .. ", n, ": ", iota(n, 1, -1).filter!(isTruncatablePrime!false).front);}`
Output:
```Largest left-truncatable prime in 2 .. 1000000: 998443
Largest right-truncatable prime in 2 .. 1000000: 739399```
## EchoLisp
` ;; does p include a 0 in its decimal representation ?(define (nozero? n) (= -1 (string-index (number->string n) "0"))) ;; right truncate : p and successive quotients by 10 (integer division) must be primes(define (right-trunc p) (unless (zero? p) (and (prime? p) (right-trunc (quotient p 10)))))(remember 'right-trunc) ;; left truncate : p and successive modulo by 10, 100, .. must be prime(define (left-trunc p (mod 1000000)) (unless (< mod 1) (and (prime? p) (nozero? p) (left-trunc (modulo p mod) (/ mod 10))))) ;; start from 999999. stop on first found(define (fact-trunc trunc)(for ((p (in-range 999999 100000 -1))) #:break (when (trunc p) (writeln p) #t))) `
Output:
` (fact-trunc left-trunc)998443 (fact-trunc right-trunc)739399 `
## Eiffel
` class APPLICATION create make feature make do io.put_string ("Largest right truncatable prime: " + find_right_truncatable_primes.out) io.new_line io.put_string ("Largest left truncatable prime: " + find_left_truncatable_primes.out) end find_right_truncatable_primes: INTEGER -- Largest right truncatable prime below 1000000. local i, maybe_prime: INTEGER found, is_one: BOOLEAN do from i := 999999 until found loop is_one := True from maybe_prime := i until not is_one or maybe_prime.out.count = 1 loop if maybe_prime.out.has ('0') or maybe_prime.out.has ('2') or maybe_prime.out.has ('4') or maybe_prime.out.has ('6') or maybe_prime.out.has ('8') then is_one := False else if not is_prime (maybe_prime) then is_one := False elseif is_prime (maybe_prime) and maybe_prime.out.count > 1 then maybe_prime := truncate_right (maybe_prime) end end end if is_one then found := True Result := i end i := i - 2 end ensure Result_is_smaller: Result < 1000000 end find_left_truncatable_primes: INTEGER -- Largest left truncatable prime below 1000000. local i, maybe_prime: INTEGER found, is_one: BOOLEAN do from i := 999999 until found loop is_one := True from maybe_prime := i until not is_one or maybe_prime.out.count = 1 loop if not is_prime (maybe_prime) then is_one := False elseif is_prime (maybe_prime) and maybe_prime.out.count > 1 then if maybe_prime.out.at (2) = '0' then is_one := False else maybe_prime := truncate_left (maybe_prime) end end end if is_one then found := True Result := i end i := i - 2 end ensure Result_is_smaller: Result < 1000000 end feature {NONE} is_prime (n: INTEGER): BOOLEAN --Is 'n' a prime number? require positiv_input: n > 0 local i: INTEGER max: REAL_64 math: DOUBLE_MATH do create math if n = 2 then Result := True elseif n <= 1 or n \\ 2 = 0 then Result := False else Result := True max := math.sqrt (n) from i := 3 until i > max loop if n \\ i = 0 then Result := False end i := i + 2 end end end truncate_left (n: INTEGER): INTEGER -- 'n' truncated by one digit from the left side. require truncatable: n.out.count > 1 local st: STRING do st := n.out st.remove_head (1) Result := st.to_integer ensure Result_truncated: Result.out.count = n.out.count - 1 end truncate_right (n: INTEGER): INTEGER -- 'n' truncated by one digit from the right side. require truncatable: n.out.count > 1 local st: STRING do st := n.out st.remove_tail (1) Result := st.to_integer ensure Result_truncated: Result.out.count = n.out.count - 1 end end `
Output:
```Largest right truncatable prime: 739399
Largest left truncatable prime: 999431
```
## Elena
ELENA 3.2 :
`import system'calendar.import extensions. const MAXN = 1000000. extension mathOp{ isPrime [ int n := self int. if (n < 2) [ ^ false. ]. if (n < 4) [ ^ true. ]. if (n mod:2 == 0) [ ^ false. ]. if (n < 9) [ ^ true. ]. if (n mod:3 == 0) [ ^ false. ]. int r := n sqrt. int f := 5. while (f <= r) [ if ((n mod:f == 0) || (n mod:(f + 2) == 0)) [ ^ false ]. f := f + 6 ]. ^ true ] isRightTruncatable [ int n := self. while (n != 0) [ ifnot (n isPrime) [ ^ false ]. n := n / 10 ]. ^ true. ] isLeftTruncatable [ int n := self. int tens := 1. while (tens < n) [ tens := tens * 10. ]. while (n != 0) [ ifnot (n isPrime) [ ^ false ]. tens := tens / 10. n := n - (n / tens * tens) ]. ^ true ]} program =[ var n := MAXN. var max_lt := 0. var max_rt := 0. while ((max_lt == 0) || (max_rt == 0)) [ if(n literal; indexOf:"0" == -1) [ if ((max_lt == 0) && \$(n isLeftTruncatable)) [ max_lt := n. ]. if ((max_rt == 0) && \$(n isRightTruncatable)) [ max_rt := n. ]. ]. n := n - 1. ]. console printLine("Largest truncable left is ",max_lt). console printLine("Largest truncable right is ",max_rt). console readChar.].`
Output:
```Largest truncable left is 998443
Largest truncable right is 739399
```
## Elixir
Translation of: Ruby
`defmodule Prime do defp left_truncatable?(n, prime) do func = fn i when i<=9 -> 0 i -> to_string(i) |> String.slice(1..-1) |> String.to_integer end truncatable?(n, prime, func) end defp right_truncatable?(n, prime) do truncatable?(n, prime, fn i -> div(i, 10) end) end defp truncatable?(n, prime, trunc_func) do if to_string(n) |> String.match?(~r/0/), do: false, else: trunc_loop(trunc_func.(n), prime, trunc_func) end defp trunc_loop(0, _prime, _trunc_func), do: true defp trunc_loop(n, prime, trunc_func) do if elem(prime,n), do: trunc_loop(trunc_func.(n), prime, trunc_func), else: false end def eratosthenes(limit) do # descending order Enum.to_list(2..limit) |> sieve(:math.sqrt(limit), []) end defp sieve([h|_]=list, max, sieved) when h>max, do: Enum.reverse(list, sieved) defp sieve([h | t], max, sieved) do list = for x <- t, rem(x,h)>0, do: x sieve(list, max, [h | sieved]) end defp prime_table(_, [], list), do: [false, false | list] defp prime_table(n, [n|t], list), do: prime_table(n-1, t, [true|list]) defp prime_table(n, prime, list), do: prime_table(n-1, prime, [false|list]) def task(limit \\ 1000000) do prime = eratosthenes(limit) prime_tuple = prime_table(limit, prime, []) |> List.to_tuple left = Enum.find(prime, fn n -> left_truncatable?(n, prime_tuple) end) IO.puts "Largest left-truncatable prime : #{left}" right = Enum.find(prime, fn n -> right_truncatable?(n, prime_tuple) end) IO.puts "Largest right-truncatable prime: #{right}" endend Prime.task`
Output:
```Largest left-truncatable prime : 998443
Largest right-truncatable prime: 739399
```
## Factor
`USING: formatting fry grouping.extras kernel literals mathmath.parser math.primes sequences ;IN: rosetta-code.truncatable-primes CONSTANT: primes \$[ 1,000,000 primes-upto reverse ] : number>digits ( n -- B{} ) number>string string>digits ; : no-zeros? ( seq -- ? ) [ zero? not ] all? ; : all-prime? ( seq -- ? ) [ prime? ] all? ; : truncate ( seq quot -- seq' ) call( seq -- seq' ) [ 10 digits>integer ] map ; : truncate-right ( seq -- seq' ) [ head-clump ] truncate ; : truncate-left ( seq -- seq' ) [ tail-clump ] truncate ; : truncatable-prime? ( n quot -- ? ) [ number>digits ] dip '[ @ all-prime? ] [ no-zeros? ] bi and ; inline : right-truncatable-prime? ( n -- ? ) [ truncate-right ] truncatable-prime? ; : left-truncatable-prime? ( n -- ? ) [ truncate-left ] truncatable-prime? ; : find-truncatable-primes ( -- ltp rtp ) primes [ [ left-truncatable-prime? ] find nip ] [ [ right-truncatable-prime? ] find nip ] bi ; : main ( -- ) find-truncatable-primes "Left: %d\nRight: %d\n" printf ; MAIN: main`
Output:
```Left: 998443
Right: 739399
```
## Fortran
Works with: Fortran version 95 and later
`module primes_mod implicit none logical, allocatable :: primes(:) contains subroutine Genprimes(parr) logical, intent(in out) :: parr(:) integer :: i! Prime sieve parr = .true. parr (1) = .false. parr (4 : size(parr) : 2) = .false. do i = 3, int (sqrt (real (size(parr)))), 2 if (parr(i)) parr(i * i : size(parr) : i) = .false. end do end subroutine function is_rtp(candidate) logical :: is_rtp integer, intent(in) :: candidate integer :: n is_rtp = .true. n = candidate / 10 do while(n > 0) if(.not. primes(n)) then is_rtp = .false. return end if n = n / 10 end do end function function is_ltp(candidate) logical :: is_ltp integer, intent(in) :: candidate integer :: i, n character(10) :: nstr write(nstr, "(i10)") candidate is_ltp = .true. do i = len_trim(nstr)-1, 1, -1 n = mod(candidate, 10**i) if(.not. primes(n)) then is_ltp = .false. return end if end doend function end module primes_mod program Truncatable_Primes use primes_mod implicit none integer, parameter :: limit = 999999 integer :: i character(10) :: nstr ! Generate an array of prime flags up to limit of search allocate(primes(limit)) call Genprimes(primes) ! Find left truncatable prime do i = limit, 1, -1 write(nstr, "(i10)") i if(index(trim(nstr), "0") /= 0) cycle ! check for 0 in number if(is_ltp(i)) then write(*, "(a, i0)") "Largest left truncatable prime below 1000000 is ", i exit end if end do ! Find right truncatable prime do i = limit, 1, -1 write(nstr, "(i10)") i if(index(trim(nstr), "0") /= 0) cycle ! check for 0 in number if(is_rtp(i)) then write(*, "(a, i0)") "Largest right truncatable prime below 1000000 is ", i exit end if end doend program`
Output
```Largest left truncatable prime below 1000000 is 998443
Largest right truncatable prime below 1000000 is 739399```
## FreeBASIC
### Version 1
`' FB 1.05.0 Win64 Function isPrime(n As Integer) As Boolean If n Mod 2 = 0 Then Return n = 2 If n Mod 3 = 0 Then Return n = 3 Dim d As Integer = 5 While d * d <= n If n Mod d = 0 Then Return False d += 2 If n Mod d = 0 Then Return False d += 4 Wend Return TrueEnd Function Dim As UInteger i, j, p, pow, lMax = 2, rMax = 2 Dim s As String ' largest left truncatable prime less than 1000000' It can't end with 1, 4, 6, 8 or 9 as these numbers are not prime' Nor can it end in 2 if it has more than one digit as such a number would divide by 2For i = 3 To 999997 Step 2 s = Str(i) If Instr(s, "0") > 1 Then Continue For '' cannot contain 0 j = s[Len(s) - 1] - 48 If j = 1 OrElse j = 9 Then Continue For p = i pow = 10 ^ (Len(s) - 1) While pow > 1 If Not isPrime(p) Then Continue For p Mod= pow pow \= 10 Wend lMax = iNext ' largest right truncatable prime less than 1000000' It can't begin with 1, 4, 6, 8 or 9 as these numbers are not primeFor i = 3 To 799999 Step 2 s = Str(i) If Instr(s, "0") > 1 Then Continue For '' cannot contain 0 j = s[0] - 48 If j = 1 OrElse j = 4 OrElse j = 6 Then Continue For p = i While p > 0 If Not isPrime(p) Then Continue For p \= 10 Wend rMax = iNext Print "Largest left truncatable prime : "; lMaxPrint "Largest right truncatable prime : "; rMaxPrintPrint "Press any key to quit"Sleep`
Output:
```Largest left truncatable prime : 998443
Largest right truncatable prime : 739399
```
### Version 2
Construct primes using previous found primes.
`' version 10-12-2016' compile with: fbc -s console Dim Shared As Byte isPrime() Sub sieve(m As UInteger) Dim As Integer i, j ReDim isPrime(m) For i = 4 To m Step 2 isPrime(i) = 1 Next For i = 3 To Sqr(m) Step 2 If isPrime(i) = 0 Then For j = i * i To m Step i * 2 isPrime(j) = 1 Next End If Next End Sub ' ------=< MAIN >=------ #Define max 1000000 'upto 2^30 max for 32bit OS Dim As UInteger a(), lt_prime(5000), rt_prime(100)Dim As UInteger i, j, j1, p1, p2, left_max, right_max sieve(max) ' left truncatable primes' if odd and ends with 3 or 7, never ends 1 or 9 (no prime' never ends on a 2 or 5 and starts with 1 to 9lt_prime(1) = 3 : lt_prime(2) = 7p1 = 1 : p2 = 2 Do For i = 1 To 9 j = Val( Str(i) + Str(lt_prime(p1)) ) If j > max Then Exit Do If isPrime(j) = 0 Then ' if prime then add to the list p2 += 1 lt_prime(p2) = j If Left_max < j Then left_max = j End If Next p1 += 1Loop Until p1 > p2 ' no more numbers to process ' right truncatable prime' start with 2, 3, 5 or 7 and end with 1, 3, 7 or 9rt_prime(1) = 2 : rt_prime(2) = 3 : rt_prime(3) = 5 : rt_prime(4) = 7p1 = 1 : p2 = 4Dim As UInteger end_num(1 To 4) => {1, 3, 7, 9} Do j1 = rt_prime(p1) * 10 If j1 > max Then Exit Do For i = 1 To 4 j = j1 + End_num(i) If isprime(j) = 0 Then ' if prime then add to the list p2 += 1 rt_prime(p2) = j ' If right_max < j Then right_max = j End If Next p1 += 1Loop Until p1 > p2 ' no more numbers to process' the last one added is the biggestright_max = rt_prime(p2) PrintPrint "The biggest left truncatable prime below"; max; " is "; left_maxPrint "The biggest right truncatable prime below"; max; " is "; right_max ' empty keyboard bufferWhile Inkey <> "" : WendPrint : Print "hit any key to end program"SleepEnd`
Output:
```The biggest left truncatable prime below 1000000 is 998443
The biggest right truncatable prime below 1000000 is 739399```
## Go
`package main import "fmt" func main() { sieve(1e6) if !search(6, 1e6, "left", func(n, pot int) int { return n % pot }) { panic("997?") } if !search(6, 1e6, "right", func(n, _ int) int { return n / 10 }) { panic("7393?") }} var c []bool func sieve(ss int) { c = make([]bool, ss) c[1] = true for p := 2; ; { p2 := p * p if p2 >= ss { break } for i := p2; i < ss; i += p { c[i] = true } for { p++ if !c[p] { break } } }} func search(digits, pot int, s string, truncFunc func(n, pot int) int) bool { n := pot - 1 pot /= 10smaller: for ; n >= pot; n -= 2 { for tn, tp := n, pot; tp > 0; tp /= 10 { if tn < tp || c[tn] { continue smaller } tn = truncFunc(tn, tp) } fmt.Println("max", s, "truncatable:", n) return true } if digits > 1 { return search(digits-1, pot, s, truncFunc) } return false}`
Output:
```max left truncatable: 998443
max right truncatable: 739399
```
Using
Library: Primes
from HackageDB
`import Data.Numbers.Primes(primes, isPrime)import Data.Listimport Control.Arrow primes1e6 = reverse. filter (notElem '0'. show) \$ takeWhile(<=1000000) primes rightT, leftT :: Int -> BoolrightT = all isPrime. takeWhile(>0). drop 1. iterate (`div`10)leftT x = all isPrime. takeWhile(<x).map (x`mod`) \$ iterate (*10) 10 main = do let (ltp, rtp) = (head. filter leftT &&& head. filter rightT) primes1e6 putStrLn \$ "Left truncatable " ++ show ltp putStrLn \$ "Right truncatable " ++ show rtp`
Output:
`*Main> mainLeft truncatable 998443Right truncatable 739399`
Interpretation of the J contribution:
`digits = [1..9] :: [Integer]smallPrimes = filter isPrime digitspow10 = iterate (*10) 1mul10 = (pow10!!). length. showrighT = (+) . (10 *)lefT = liftM2 (.) (+) ((*) . mul10) primesTruncatable f = iterate (concatMap (filter isPrime.flip map digits. f)) smallPrimes`
Output:
`*Main> maximum \$ primesTruncatable righT !! 5739399 *Main> maximum \$ primesTruncatable lefT !! 5998443`
## Icon and Unicon
`procedure main(arglist) N := 0 < integer(\arglist[1]) | 1000000 # primes to generator 1 to ... (1M or 1st arglist) D := (0 < integer(\arglist[2]) | 10) / 2 # primes to display (10 or 2nd arglist) P := sieve(N) # from sieve task (modified) write("There are ",*P," prime numbers in the range 1 to ",N) if *P <= 2*D then every writes( "Primes: "|!sort(P)||" "|"\n" ) else every writes( "Primes: "|(L := sort(P))[1 to D]||" "|"... "|L[*L-D+1 to *L]||" "|"\n" ) largesttruncateable(P)end procedure largesttruncateable(P) #: find the largest left and right trucatable numbers in Plocal ltp,rtp every x := sort(P)[*P to 1 by -1] do # largest to smallest if not find('0',x) then { /ltp := islefttrunc(P,x) /rtp := isrighttrunc(P,x) if \ltp & \rtp then break # until both found } write("Largest left truncatable prime = ", ltp) write("Largest right truncatable prime = ", rtp) returnend procedure isrighttrunc(P,x) #: return integer x if x and all right truncations of x are in P or failsif x = 0 | (member(P,x) & isrighttrunc(P,x / 10)) then return xend procedure islefttrunc(P,x) #: return integer x if x and all left truncations of x are in P or failsif *x = 0 | ( (x := integer(x)) & member(P,x) & islefttrunc(P,x[2:0]) ) then return xend`
Sample output:
```There are 78498 prime numbers in the range 1 to 1000000
Primes: 2 3 5 7 11 ... 999953 999959 999961 999979 999983
Largest left truncatable prime = 998443
Largest right truncatable prime = 739399```
## J
Truncatable primes may be constructed by starting with a set of one digit prime numbers and then repeatedly adding a non-zero digit (combine all possibilities of a truncatable prime digit sequence with each digit) and, at each step, selecting the prime numbers which result.
In other words, given:
`selPrime=: #~ 1&p:seed=: selPrime digits=: 1+i.9step=: [email protected],@:(,&.":/&>)@{@;`
Here, selPrime discards non-prime numbers from a list, so seed is the list 2 3 5 7.
The largest truncatable primes less than a million can be obtained by adding five digits to the prime seeds, then finding the largest value from the result.
` >./ digits&step^:5 seed NB. left truncatable998443 >./ step&digits^:5 seed NB. right truncatable739399`
Note that we are using the same combining function and same basic procedure in both cases. The difference is which side of the number we add arbitrary digits to, for each step.
## Java
`import java.util.BitSet; public class Main { public static void main(String[] args){ final int MAX = 1000000; //Sieve of Eratosthenes (using BitSet only for odd numbers) BitSet primeList = new BitSet(MAX>>1); primeList.set(0,primeList.size(),true); int sqroot = (int) Math.sqrt(MAX); primeList.clear(0); for(int num = 3; num <= sqroot; num+=2) { if( primeList.get(num >> 1) ) { int inc = num << 1; for(int factor = num * num; factor < MAX; factor += inc) { //if( ((factor) & 1) == 1) //{ primeList.clear(factor >> 1); //} } } } //Sieve ends... //Find Largest Truncatable Prime. (so we start from 1000000 - 1 int rightTrunc = -1, leftTrunc = -1; for(int prime = (MAX - 1) | 1; prime >= 3; prime -= 2) { if(primeList.get(prime>>1)) { //Already found Right Truncatable Prime? if(rightTrunc == -1) { int right = prime; while(right > 0 && right % 2 != 0 && primeList.get(right >> 1)) right /= 10; if(right == 0) rightTrunc = prime; } //Already found Left Truncatable Prime? if(leftTrunc == -1 ) { //Left Truncation String left = Integer.toString(prime); if(!left.contains("0")) { while( left.length() > 0 ){ int iLeft = Integer.parseInt(left); if(!primeList.get( iLeft >> 1)) break; left = left.substring(1); } if(left.length() == 0) leftTrunc = prime; } } if(leftTrunc != -1 && rightTrunc != -1) //Found both? then Stop loop { break; } } } System.out.println("Left Truncatable : " + leftTrunc); System.out.println("Right Truncatable : " + rightTrunc); }} `
Output :
```Left Truncatable : 998443
Right Truncatable : 739399
```
## Julia
There are several features of Julia that make solving this task easy. Julia has excellent built-in support for prime generation and testing. The built-in mathematical functions prevpow and divrem are quite handy for implementing isltruncprime.
` function isltruncprime{T<:Integer}(n::T, base::T=10) isprime(n) || return false p = n f = prevpow(base, p) while 1 < f (d, p) = divrem(p, f) isprime(p) || return false d != 0 || return false f = div(f, base) end return trueend function isrtruncprime{T<:Integer}(n::T, base::T=10) isprime(n) || return false p = n while base < p p = div(p, base) isprime(p) || return false end return trueend hi = 10^6 for i in reverse(primes(hi)) isltruncprime(i) || continue println("The largest left truncatable prime ≤ ", hi, " is ", i, ".") breakend for i in reverse(primes(hi)) isrtruncprime(i) || continue println("The largest right truncatable prime ≤ ", hi, " is ", i, ".") breakend `
Output:
```The largest left truncatable prime ≤ 1000000 is 998443.
The largest right truncatable prime ≤ 1000000 is 739399.
```
## Kotlin
Translation of: FreeBASIC
`// version 1.0.5-2 fun isPrime(n: Int) : Boolean { if (n < 2) return false if (n % 2 == 0) return n == 2 if (n % 3 == 0) return n == 3 var d : Int = 5 while (d * d <= n) { if (n % d == 0) return false d += 2 if (n % d == 0) return false d += 4 } return true} fun main(args: Array<String>) { var j: Char var p: Int var pow: Int var lMax: Int = 2 var rMax: Int = 2 var s: String // calculate maximum left truncatable prime less than 1 million loop@ for( i in 3..999997 step 2) { s = i.toString() if ('0' in s) continue j = s[s.length - 1] if (j == '1' || j == '9') continue p = i pow = 1 for (k in 1..s.length - 1) pow *= 10 while(pow > 1) { if (!isPrime(p)) continue@loop p %= pow pow /= 10 } lMax = i } // calculate maximum right truncatable prime less than 1 million loop@ for( i in 3..799999 step 2) { s = i.toString() if ('0' in s) continue j = s[0] if (j == '1' || j == '4' || j == '6') continue p = i while(p > 0) { if (!isPrime(p)) continue@loop p /= 10 } rMax = i } println("Largest left truncatable prime : " + lMax.toString()) println("Largest right truncatable prime : " + rMax.toString()) }`
Output:
```Largest left truncatable prime : 998443
Largest right truncatable prime : 739399
```
## Lua
`max_number = 1000000 numbers = {}for i = 2, max_number do numbers[i] = i;end for i = 2, max_number do for j = i+1, max_number do if numbers[j] ~= 0 and j % i == 0 then numbers[j] = 0 end endend max_prime_left, max_prime_right = 2, 2for i = 2, max_number do if numbers[i] ~= 0 then local is_prime = true local l = math.floor( i / 10 ) while l > 1 do if numbers[l] == 0 then is_prime = false break end l = math.floor( l / 10 ) end if is_prime then max_prime_left = i end is_prime = true local n = 10; while math.floor( i % 10 ) ~= 0 and n < max_number do if numbers[ math.floor( i % 10 ) ] ~= 0 then is_prime = false break end n = n * 10 end if is_prime then max_prime_right = i end endend print( "max_prime_left = ", max_prime_left )print( "max_prime_right = ", max_prime_right )`
## Maple
` MaxTruncatablePrime := proc({left::truefalse:=FAIL, right::truefalse:=FAIL}, \$)local i, j, c, p, b, n, sdprimes, dir;local tprimes := table(); if left = true and right = true then error "invalid input"; elif right = true then dir := "right"; else dir := "left"; end if; b := 10; n := 6; sdprimes := select(isprime, [seq(1..b-1)]); for p in sdprimes do if assigned(tprimes[p]) then next; end if; i := ilog[b](p)+1; j := 1; while p < b^n do if dir = "left" then c := j*b^i + p; else c := p*b + j; end if; if j >= b or c > b^n then # we have tried all 1 digit extensions of p, add p to tprimes and move back 1 digit tprimes[p] := p; if i = 1 then # if we are at the first digit, go to the next 1 digit prime break; end if; i := i - 1; j := 1; if dir = "left" then p := p - iquo(p, b^i)*b^i; else p := iquo(p, b); end if; elif assigned(tprimes[c]) then j := j + 1; elif isprime(c) then p := c; i := i + 1; j := 1; else j := j+1; end if; end do; end do; return max(indices(tprimes, 'nolist'));end proc;`
```> MaxTruncatablePrime(right); MaxTruncatablePrime(left);
739399
998443
```
## Mathematica
`LeftTruncatablePrimeQ[n_] := Times @@ IntegerDigits[n] > 0 && And @@ PrimeQ /@ ToExpression /@ StringJoin /@ Rest[Most[NestList[Rest, #, Length[#]] &[Characters[ToString[n]]]]]RightTruncatablePrimeQ[n_] := Times @@ IntegerDigits[n] > 0 && And @@ PrimeQ /@ ToExpression /@ StringJoin /@ Rest[Most[NestList[Most, #, Length[#]] &[Characters[ToString[n]]]]]`
Example usage:
```n = PrimePi[1000000]; While[Not[LeftTruncatablePrimeQ[Prime[n]]], n--]; Prime[n]
-> 998443
n = PrimePi[1000000]; While[Not[RightTruncatablePrimeQ[Prime[n]]], n--]; Prime[n]
-> 739399```
## MATLAB
largestTruncatablePrimes.m:
`function largestTruncatablePrimes(boundary) %Helper function for checking if a prime is left of right truncatable function [leftTruncatable,rightTruncatable] = isTruncatable(prime,checkLeftTruncatable,checkRightTruncatable) numDigits = ceil(log10(prime)); %calculate the number of digits in the prime less one powersOfTen = 10.^(0:numDigits); %cache the needed powers of ten leftTruncated = mod(prime,powersOfTen); %generate a list of numbers by repeatedly left truncating the prime %leading zeros will cause duplicate entries thus it is possible to %detect leading zeros if we rotate the list to the left or right %and check for any equivalences with the original list hasLeadingZeros = any( circshift(leftTruncated,[0 1]) == leftTruncated ); if( hasLeadingZeros || not(checkLeftTruncatable) ) leftTruncatable = false; else %check if all of the left truncated numbers are prime leftTruncatable = all(isprime(leftTruncated(2:end))); end if( checkRightTruncatable ) rightTruncated = (prime - leftTruncated) ./ powersOfTen; %generate a list of right truncated numbers rightTruncatable = all(isprime(rightTruncated(1:end-1))); %check if all the right truncated numbers are prime else rightTruncatable = false; end end %isTruncatable() nums = primes(boundary); %generate all primes <= boundary %Flags that indicate if the largest left or right truncatable prime has not %been found leftTruncateNotFound = true; rightTruncateNotFound = true; for prime = nums(end:-1:1) %Search through primes in reverse order %Get if the prime is left and/or right truncatable, ignoring %checking for right truncatable if it has already been found [leftTruncatable,rightTruncatable] = isTruncatable(prime,leftTruncateNotFound,rightTruncateNotFound); if( leftTruncateNotFound && leftTruncatable ) %print out largest left truncatable prime display([num2str(prime) ' is the largest left truncatable prime <= ' num2str(boundary) '.']); leftTruncateNotFound = false; end if( rightTruncateNotFound && rightTruncatable ) %print out largest right truncatable prime display([num2str(prime) ' is the largest right truncatable prime <= ' num2str(boundary) '.']); rightTruncateNotFound = false; end %Terminate loop when the largest left and right truncatable primes have %been found if( not(leftTruncateNotFound || rightTruncateNotFound) ) break; end endend `
Solution for n = 1,000,000:
` >> largestTruncatablePrimes(1e6)998443 is the largest left truncatable prime <= 1000000.739399 is the largest right truncatable prime <= 1000000. `
## Nim
Translation of: Python
`import sets, strutils, algorithm proc primes(n: int64): seq[int64] = result = @[] var multiples = initSet[int64]() for i in 2..n: if i notin multiples: result.add i for j in countup(i*i, n, i.int): multiples.incl j proc truncatablePrime(n: int64): tuple[left: int64, right: int64] = var primelist: seq[string] = @[] for x in primes(n): primelist.add(\$x) reverse primelist var primeset = toSet primelist for n in primelist: var alltruncs = initSet[string]() for i in 0..n.len-1: alltruncs.incl n[i..n.high] if alltruncs <= primeset: result.left = parseInt(n) break for n in primelist: var alltruncs = initSet[string]() for i in 0..n.len-1: alltruncs.incl n[0..i] if alltruncs <= primeset: result.right = parseInt(n) break echo truncatablePrime(1000000'i64)`
Output:
`(left: 998443, right: 739399)`
## ooRexx
` -- find largest left- & right-truncatable primes < 1 million.-- an initial set of primes (not, at this time, we leave out 2 because-- we'll automatically skip the even numbers. No point in doing a needless-- test each time throughprimes = .array~of(3, 5, 7, 11) -- check all of the odd numbers up to 1,000,000loop j = 13 by 2 to 1000000 loop i = 1 to primes~size prime = primes[i] -- found an even prime divisor if j // prime == 0 then iterate j -- only check up to the square root if prime*prime > j then leave end -- we only get here if we don't find a divisor primes~append(j)end -- get a set of the primes that we can test more efficientlyprimeSet = .set~of(2)primeSet~putall(primes) say 'The last prime is' primes[primes~last] "("primeSet~items 'primes under one million).'say copies('-',66) lastLeft = 0 -- we're going to use the array version to do these in order. We're still-- missing "2", but that's not going to be the largestloop prime over primes -- values containing 0 can never work if prime~pos(0) \= 0 then iterate -- now start the truncations, checking against our set of -- known primes loop i = 1 for prime~length - 1 subprime = prime~right(i) -- not in our known set, this can't work if \primeset~hasIndex(subprime) then iterate prime end -- this, by definition, with be the largest left-trunc prime lastLeft = primeend-- now look for right-trunc primeslastRight = 0loop prime over primes -- values containing 0 can never work if prime~pos(0) \= 0 then iterate -- now start the truncations, checking against our set of -- known primes loop i = 1 for prime~length - 1 subprime = prime~left(i) -- not in our known set, this can't work if \primeset~hasIndex(subprime) then iterate prime end -- this, by definition, with be the largest left-trunc prime lastRight = primeend say 'The largest left-truncatable prime is' lastLeft '(under one million).'say 'The largest right-truncatable prime is' lastRight '(under one million).' `
Output:
```The last prime is 999983 (78498 primes under one million).
------------------------------------------------------------------
The largest left-truncatable prime is 998443 (under one million).
The largest right-truncatable prime is 739399 (under one million).
```
## OpenEdge/Progress
`FUNCTION isPrime RETURNS LOGICAL ( i_i AS INT): DEF VAR ii AS INT. DO ii = 2 TO SQRT( i_i ): IF i_i MODULO ii = 0 THEN RETURN FALSE. END. RETURN TRUE AND i_i > 1. END FUNCTION. /* isPrime */ FUNCTION isLeftTruncatablePrime RETURNS LOGICAL ( i_i AS INT): DEF VAR ii AS INT. DEF VAR cc AS CHAR. DEF VAR lresult AS LOGICAL INITIAL TRUE. cc = STRING( i_i ). DO WHILE cc > "": lresult = lresult AND isPrime( INTEGER( cc ) ). cc = SUBSTRING( cc, 2 ). END. RETURN lresult. END FUNCTION. /* isLeftTruncatablePrime */ FUNCTION isRightTruncatablePrime RETURNS LOGICAL ( i_i AS INT): DEF VAR ii AS INT. DEF VAR cc AS CHAR. DEF VAR lresult AS LOGICAL INITIAL TRUE. cc = STRING( i_i ). DO WHILE cc > "": lresult = lresult AND isPrime( INTEGER( cc ) ). cc = SUBSTRING( cc, 1, LENGTH( cc ) - 1 ). END. RETURN lresult. END FUNCTION. /* isRightTruncatablePrime */ FUNCTION getHighestTruncatablePrimes RETURNS CHARACTER ( i_imax AS INTEGER): DEF VAR ii AS INT. DEF VAR ileft AS INT. DEF VAR iright AS INT. DO ii = i_imax TO 1 BY -1 WHILE ileft = 0 OR iright = 0: IF INDEX( STRING( ii ), "0" ) = 0 THEN DO: IF ileft = 0 AND isLeftTruncatablePrime( ii ) THEN ileft = ii. IF iright = 0 AND isRightTruncatablePrime( ii ) THEN iright = ii. END. END. RETURN SUBSTITUTE("Left: &1~nRight: &2", ileft, iright ). END FUNCTION. /* getHighestTruncatablePrimes */ MESSAGE getHighestTruncatablePrimes( 1000000 )VIEW-AS ALERT-BOX. `
Output:
```---------------------------
Message
---------------------------
Left: 998443
Right: 739399
---------------------------
OK
---------------------------```
## PARI/GP
This version builds the truncatable primes with up to k digits in a straightforward fashion. Run time is about 15 milliseconds, almost all of which is I/O.
`left(n)={ my(v=[2,3,5,7],u,t=1,out=0); for(i=1,n, t*=10; u=[]; for(j=1,#v, forstep(a=t,t*9,t, if(isprime(a+v[j]),u=concat(u,a+v[j])) ) ); out=v[#v]; v=vecsort(u) ); out};right(n)={ my(v=[2,3,5,7],u,out=0); for(i=1,n, u=[]; for(j=1,#v, forstep(a=1,9,[2,4], if(isprime(10*v[j]+a),u=concat(u,10*v[j]+a)) ) ); out=v[#v]; v=u ); out};[left(6),right(6)]`
## Perl
Typically with Perl we'll look for a CPAN module to make our life easier. This basically just follows the task rules:
Library: ntheory
`use ntheory ":all";sub isltrunc { my \$n = shift; return (is_prime(\$n) && \$n !~ /0/ && (\$n < 10 || isltrunc(substr(\$n,1))));}sub isrtrunc { my \$n = shift; return (is_prime(\$n) && \$n !~ /0/ && (\$n < 10 || isrtrunc(substr(\$n,0,-1))));}for (reverse @{primes(1e6)}) { if (isltrunc(\$_)) { print "ltrunc: \$_\n"; last; }}for (reverse @{primes(1e6)}) { if (isrtrunc(\$_)) { print "rtrunc: \$_\n"; last; }}`
Output:
```ltrunc: 998443
rtrunc: 739399```
We can be a little more Perlish and build up n-digit lists then select the last one:
`use ntheory ":all"; my @lprimes = my @rprimes = (2,3,5,7); @lprimes = sort { \$a <=> \$b } map { my \$p=\$_; map { is_prime(\$_.\$p) ? \$_.\$p : () } 1..9 } @lprimes for 2..6; @rprimes = sort { \$a <=> \$b } map { my \$p=\$_; map { is_prime(\$p.\$_) ? \$p.\$_ : () } 1..9 } @rprimes for 2..6; print "ltrunc: \$lprimes[-1]\nrtrunc: \$rprimes[-1]\n";`
Or we can do everything ourselves:
`#!/usr/bin/perluse warnings;use strict; use constant { LEFT => 0, RIGHT => 1,}; { my @primes = (2, 3); sub is_prime { my \$n = shift; return if \$n < 2; for my \$prime (@primes) { last if \$prime >= \$n; return unless \$n % \$prime; } my \$sqrt = sqrt \$n; while (\$primes[-1] < \$sqrt) { my \$new = 2 + \$primes[-1]; \$new += 2 until is_prime(\$new); push @primes, \$new; return unless \$n % \$new; } return 1; }} sub trunc { my (\$n, \$side) = @_; substr \$n, \$side == LEFT ? 0 : -1, 1, q(); return \$n;} sub is_tprime { # Absence of zeroes is tested outside the sub. my (\$n, \$side) = @_; return (is_prime(\$n) and (1 == length \$n or is_tprime(trunc(\$n, \$side), \$side)));} my \$length = 6;my @tprimes = ('9' x \$length) x 2;for my \$side (LEFT, RIGHT) { \$tprimes[\$side] -= 2 until -1 == index \$tprimes[\$side], '0' and is_tprime(\$tprimes[\$side], \$side);} print 'left ', join(', right ', @tprimes), "\n";`
Output:
`left 998443, right 739399`
## Perl 6
Works with: Rakudo version 2015.09
`constant ltp = \$[2, 3, 5, 7], -> @ltp { \$[ grep { .&is-prime }, ((1..9) X~ @ltp) ]} ... *; constant rtp = \$[2, 3, 5, 7], -> @rtp { \$[ grep { .&is-prime }, (@rtp X~ (1..9)) ]} ... *; say "Highest ltp = ", ltp[5][*-1];say "Highest rtp = ", rtp[5][*-1];`
Output:
```Highest ltp: 998443
Highest rtp: 739399```
## Phix
A slightly different approach. Works up to N=8, quite fast - 10^8 in 5s with ~90% of time spent creating the basic sieve and ~10% propagation and final scan.
`constant N = 6, limit = power(10,N)-- standard sieve:enum L,R -- (with primes[i] as mini bit-field)sequence primes = repeat(L+R, limit)primes[1] = 0for i=2 to floor(sqrt(limit)) do if primes[i] then for k=i*i to limit by i do primes[k] = 0 end for end ifend for -- propagate non-truncateables up the prime table:for p=1 to N-1 do integer p10 = power(10,p) -- ie 10, 100, .. 100_000 for i=p10+1 to p10*10-1 by 2 do -- to 99, 999, .. 999_999 if primes[i] then integer l = remainder(i,p10), r = floor(i/10) integer pi = and_bits(primes[l],L)+and_bits(primes[r],R) if pi and find('0',sprint(i)) then pi = 0 end if primes[i] = pi end if end forend for integer maxl=0, maxr=0 for i=limit-1 to 1 by -2 do integer pi = primes[i] if pi then if maxl=0 and and_bits(pi,L) then maxl = i end if if maxr=0 and and_bits(pi,R) then maxr = i end if if maxl!=0 and maxr!=0 then exit end if end ifend for?{maxl,maxr}`
Output:
```{998443,739399}
```
## PicoLisp
`(load "@lib/rsa.l") # Use the 'prime?' function from RSA package (de truncatablePrime? (N Fun) (for (L (chop N) L (Fun L)) (T (= "0" (car L))) (NIL (prime? (format L))) T ) ) (let (Left 1000000 Right 1000000) (until (truncatablePrime? (dec 'Left) cdr)) (until (truncatablePrime? (dec 'Right) '((L) (cdr (rot L))))) (cons Left Right) )`
Output:
`-> (998443 . 739399)`
## PL/I
` tp: procedure options (main); declare primes(1000000) bit (1); declare max_primes fixed binary (31); declare (i, k) fixed binary (31); max_primes = hbound(primes, 1); call sieve; /* Now search for primes that are right-truncatable. */ call right_truncatable; /* Now search for primes that are left-truncatable. */ call left_truncatable; right_truncatable: procedure; declare direction bit (1); declare (i, k) fixed binary (31); test_truncatable: do i = max_primes to 2 by -1; if primes(i) then /* it's a prime */ do; k = i/10; do while (k > 0); if ^primes(k) then iterate test_truncatable; k = k/10; end; put skip list (i || ' is right-truncatable'); return; end; end;end right_truncatable; left_truncatable: procedure; declare direction bit (1); declare (i, k, d, e) fixed binary (31); test_truncatable: do i = max_primes to 2 by -1; if primes(i) then /* it's a prime */ do; k = i; do d = 100000 repeat d/10 until (d = 10); e = k/d; k = k - e*d; if e = 0 then iterate test_truncatable; if ^primes(k) then iterate test_truncatable; end; put skip list (i || ' is left-truncatable'); return; end; end;end left_truncatable; sieve: procedure; declare (i, j) fixed binary (31); primes = '1'b; primes(1) = '0'b; do i = 2 to sqrt(max_primes); do j = i+i to max_primes by i; primes(j) = '0'b; end; end;end sieve; end tp; `
``` 739399 is right-truncatable
998443 is left-truncatable
```
## PowerShell
`function IsPrime ( [int] \$num ){ \$isprime = @{} 2..[math]::sqrt(\$num) | Where-Object { \$isprime[\$_] -eq \$null } | ForEach-Object { \$_ \$isprime[\$_] = \$true for ( \$i=\$_*\$_ ; \$i -le \$num; \$i += \$_ ) { \$isprime[\$i] = \$false } } 2..\$num | Where-Object { \$isprime[\$_] -eq \$null }} function Truncatable ( [int] \$num ){ \$declen = [math]::abs(\$num).ToString().Length \$primes = @() \$ltprimes = @{} \$rtprimes = @{} 1..\$declen | ForEach-Object { \$ltprimes[\$_][email protected]{}; \$rtprimes[\$_][email protected]{} } IsPrime \$num | ForEach-Object { \$lastltprime = 2 \$lastrtprime = 2 } { \$curprim = \$_ \$curdeclen = \$curprim.ToString().Length \$primes += \$curprim if( \$curdeclen -eq 1 ) { \$ltprimes[1][\$curprim] = \$true \$rtprimes[1][\$curprim] = \$true \$lastltprime = \$curprim \$lastrtprime = \$curprim } else { \$curmod = \$curprim % [math]::pow(10,\$curdeclen - 1) \$curdiv = [math]::floor(\$curprim / 10) if( \$ltprimes[\$curdeclen - 1][[int]\$curmod] ) { \$ltprimes[\$curdeclen][\$curprim] = \$true \$lastltprime = \$curprim } if( \$rtprimes[\$curdeclen - 1][[int]\$curdiv] ) { \$rtprimes[\$curdeclen][\$curprim] = \$true \$lastrtprime = \$curprim } } if( ( \$ltprimes[\$curdeclen - 2].Keys.count -gt 0 ) -and ( \$ltprimes[\$curdeclen - 1].Keys.count -gt 0 ) ) { \$ltprimes[\$curdeclen -2] = @{} } if( ( \$rtprimes[\$curdeclen - 2].Keys.count -gt 0 ) -and ( \$rtprimes[\$curdeclen - 1].Keys.count -gt 0 ) ) { \$rtprimes[\$curdeclen -2] = @{} } } { "Largest Left Truncatable Prime: \$lastltprime" "Largest Right Truncatable Prime: \$lastrtprime" }}`
## PureBasic
`#MaxLim = 999999 Procedure is_Prime(n) If n<=1 : ProcedureReturn #False ElseIf n<4 : ProcedureReturn #True ElseIf n%2=0: ProcedureReturn #False ElseIf n<9 : ProcedureReturn #True ElseIf n%3=0: ProcedureReturn #False Else Protected r=Round(Sqr(n),#PB_Round_Down) Protected f=5 While f<=r If n%f=0 Or n%(f+2)=0 ProcedureReturn #False EndIf f+6 Wend EndIf ProcedureReturn #TrueEndProcedure Procedure TruncateLeft(n) Protected s.s=Str(n), l=Len(s)-1 If Not FindString(s,"0",1) While l>0 s=Right(s,l) If Not is_Prime(Val(s)) ProcedureReturn #False EndIf l-1 Wend ProcedureReturn #True EndIfEndProcedure Procedure TruncateRight(a) Repeat a/10 If Not a Break ElseIf Not is_Prime(a) Or a%10=0 ProcedureReturn #False EndIf ForEver ProcedureReturn #TrueEndProcedure i=#MaxLimRepeat If is_Prime(i) If Not truncateleft And TruncateLeft(i) truncateleft=i EndIf If Not truncateright And TruncateRight(i) truncateright=i EndIf EndIf If truncateleft And truncateright Break Else i-2 EndIf Until i<=0 x.s="Largest TruncateLeft= "+Str(truncateleft)y.s="Largest TruncateRight= "+Str(truncateright) MessageRequester("Truncatable primes",x+#CRLF\$+y)`
## Python
`maxprime = 1000000 def primes(n): multiples = set() prime = [] for i in range(2, n+1): if i not in multiples: prime.append(i) multiples.update(set(range(i*i, n+1, i))) return prime def truncatableprime(n): 'Return a longest left and right truncatable primes below n' primelist = [str(x) for x in primes(n)[::-1]] primeset = set(primelist) for n in primelist: # n = 'abc'; [n[i:] for i in range(len(n))] -> ['abc', 'bc', 'c'] alltruncs = set(n[i:] for i in range(len(n))) if alltruncs.issubset(primeset): truncateleft = int(n) break for n in primelist: # n = 'abc'; [n[:i+1] for i in range(len(n))] -> ['a', 'ab', 'abc'] alltruncs = set([n[:i+1] for i in range(len(n))]) if alltruncs.issubset(primeset): truncateright = int(n) break return truncateleft, truncateright print(truncatableprime(maxprime))`
Sample Output
`(998443, 739399)`
## Racket
` #lang racket(require math/number-theory) (define (truncate-right n) (quotient n 10)) (define (truncate-left n) (define s (number->string n)) (string->number (substring s 1 (string-length s)))) (define (contains-zero? n) (member #\0 (string->list (number->string n)))) (define (truncatable? truncate n) (and (prime? n) (not (contains-zero? n)) (or (< n 10) (truncatable? truncate (truncate n))))) ; largest left truncatable prime(for/first ([n (in-range 1000000 1 -1)] #:when (truncatable? truncate-left n)) n) ; largest right truncatable prime(for/first ([n (in-range 1000000 1 -1)] #:when (truncatable? truncate-right n)) n) ; Output:998443739399 `
## REXX
Extra code was added to the prime number generator as this is the section of the REXX program that consumes the vast majority of the computation time.
`/*REXX program finds largest left─ and right─truncatable primes ≤ 1m (or argument 1).*/parse arg high .; if high=='' then high=1000000 /*Not specified? Then use 1m*/!.=0; w=length(high) /*placeholders for primes; max width. */@.1=2; @.2=3; @.3=5; @.4=7; @.5=11; @.6=13; @.7=17 /*define some low primes. */!.2=1; !.3=1; !.5=1; !.7=1; !.11=1; !.13=1; !.17=1 /*set some low prime flags. */#=7; s.#[email protected].#**2 /*number of primes so far; prime². */ /* [↓] generate more primes ≤ high.*/ do [email protected].#+2 by 2 for max(0, high%[email protected].#%2-1) /*only find odd primes from here on out*/ if j// 3==0 then iterate /*is J divisible by three? */ parse var j '' -1 _; if _==5 then iterate /* " " " " five? (right digit)*/ if j// 7==0 then iterate /* " " " " seven? */ if j//11==0 then iterate /* " " " " eleven? */ if j//13==0 then iterate /* " " " " thirteen? */ /* [↑] the above five lines saves time*/ do k=7 while s.k<=j /* [↓] divide by the known odd primes.*/ if j//@.k==0 then iterate j /*Is J ÷ X? Then not prime. ___ */ end /*k*/ /* [↑] only process up to the √ J */ #=#+1 /*bump the number of primes found. */ @.#=j; s.#=j*j; !.j=1 /*assign next prime; prime²; prime #.*/ end /*j*/ /* [↓] find largest left truncatable P*/ do L=# by -1 for #; digs=length(@.L) /*search from top end; get the length.*/ do k=1 for digs; _=right(@.L, k) /*validate all left truncatable primes.*/ if \!._ then iterate L /*Truncated number not prime? Skip it.*/ end /*k*/ leave /*egress, found left truncatable prime.*/ end /*L*/ /* [↓] find largest right truncated P.*/ do R=# by -1 for #; digs=length(@.R) /*search from top end; get the length.*/ do k=1 for digs; _=left(@.R, k) /*validate all right truncatable primes*/ if \!._ then iterate R /*Truncated number not prime? Skip it.*/ end /*k*/ leave /*egress, found right truncatable prime*/ end /*R*/ /* [↓] show largest left/right trunc P*/say 'The last prime found is ' @.# " (there are" # 'primes ≤' high")."say copies('─', 70) /*show a separator line for the output.*/say 'The largest left─truncatable prime ≤' high " is " right(@.L, w)say 'The largest right─truncatable prime ≤' high " is " right(@.R, w) /*stick a fork in it, we're all done. */`
output when using the default input:
```The last prime found is 999983 (there are 78498 primes ≤ 1000000).
──────────────────────────────────────────────────────────────────────
The largest left─truncatable prime ≤ 1000000 is 998443
The largest right─truncatable prime ≤ 1000000 is 739399
```
## Ring
` # Project : Truncatable primes# Date : 2017/10/15# Author : Gal Zsolt (~ CalmoSoft ~)# Email : <[email protected]> for n = 1000000 to 1 step -1 flag = 1 flag2 = 1 strn = string(n) for nr = 1 to len(strn) if strn[nr] = "0" flag2 = 0 ok next if flag2 = 1 for m = 1 to len(strn) strp = right(strn, m) if isprime(number(strp)) else flag = 0 exit ok next if flag = 1 nend = n exit ok oknextsee "Largest left truncatable prime : " + nend + nl for n = 1000000 to 1 step -1 flag = 1 strn = string(n) for m = 1 to len(strn) strp = left(strn, len(strn) - m + 1) if isprime(number(strp)) else flag = 0 exit ok next if flag = 1 nend = n exit oknextsee "Largest right truncatable prime : " + nend + nl func isprime num if (num <= 1) return 0 ok if (num % 2 = 0 and num != 2) return 0 ok for i = 3 to floor(num / 2) -1 step 2 if (num % i = 0) return 0 ok next return 1 `
Output:
```Largest left truncatable prime : 998443
Largest right truncatable prime : 739399
```
## Ruby
`def left_truncatable?(n) truncatable?(n) {|i| i.to_s[1..-1].to_i}end def right_truncatable?(n) truncatable?(n) {|i| i/10}end def truncatable?(n, &trunc_func) return false if n.to_s.include? "0" loop do n = trunc_func.call(n) return true if n.zero? return false unless Prime.prime?(n) endend require 'prime'primes = Prime.each(1_000_000).to_a.reverse p primes.detect {|p| left_truncatable? p}p primes.detect {|p| right_truncatable? p}`
returns
```998443
739399```
### An Alternative Approach
Setting BASE to 10 and MAX to 6 in the Ruby example here Produces:
```The largest left truncatable prime less than 1000000 in base 10 is 998443
```
## Sidef
`func t_prime(n, left=true) { var p = %w(2 3 5 7); var f = ( left ? { '1'..'9' ~X+ p } : { p ~X+ '1'..'9' } ) n.times { p = f().grep{ .to_i.is_prime } } p.map{.to_i}.max} say t_prime(5, left: true)say t_prime(5, left: false)`
Output:
```998443
739399
```
## Tcl
`package require Tcl 8.5 # Optimized version of the Sieve-of-Eratosthenes task solutionproc sieve n { set primes [list] if {\$n < 2} {return \$primes} set nums [dict create] for {set i 2} {\$i <= \$n} {incr i} { dict set nums \$i "" } set next 2 set limit [expr {sqrt(\$n)}] while {\$next <= \$limit} { for {set i \$next} {\$i <= \$n} {incr i \$next} {dict unset nums \$i} lappend primes \$next dict for {next -} \$nums break } return [concat \$primes [dict keys \$nums]]} proc isLeftTruncatable n { global isPrime while {[string length \$n] > 0} { if {![info exist isPrime(\$n)]} { return false } set n [string range \$n 1 end] } return true}proc isRightTruncatable n { global isPrime while {[string length \$n] > 0} { if {![info exist isPrime(\$n)]} { return false } set n [string range \$n 0 end-1] } return true} # Demo codeset limit 1000000puts "calculating primes up to \$limit"set primes [sieve \$limit]puts "search space contains [llength \$primes] members"foreach p \$primes { set isPrime(\$p) "yes"}set primes [lreverse \$primes] puts "searching for largest left-truncatable prime"foreach p \$primes { if {[isLeftTruncatable \$p]} { puts FOUND:\$p break }} puts "searching for largest right-truncatable prime"foreach p \$primes { if {[isRightTruncatable \$p]} { puts FOUND:\$p break }}`
Output:
```calculating primes up to 1000000
search space contains 78498 members
searching for largest left-truncatable prime
FOUND:998443
searching for largest right-truncatable prime
FOUND:739399
```
## VBScript
` start_time = Now lt = 0rt = 0 For h = 1 To 1000000 If IsLeftTruncatable(h) And h > lt Then lt = h End If If IsRightTruncatable(h) And h > rt Then rt = h End IfNext end_time = now WScript.StdOut.WriteLine "Largest LTP from 1..1000000: " & ltWScript.StdOut.WriteLine "Largest RTP from 1..1000000: " & rtWScript.StdOut.WriteLine "Elapse Time(seconds) : " & DateDiff("s",start_time,end_time) '------------Function IsLeftTruncatable(n) IsLeftTruncatable = False c = 0 For i = Len(n) To 1 Step -1 If InStr(1,n,"0") > 0 Then Exit For End If If IsPrime(Right(n,i)) Then c = c + 1 End If Next If c = Len(n) Then IsLeftTruncatable = True End IfEnd Function Function IsRightTruncatable(n) IsRightTruncatable = False c = 0 For i = Len(n) To 1 Step -1 If InStr(1,n,"0") > 0 Then Exit For End If If IsPrime(Left(n,i)) Then c = c + 1 End If Next If c = Len(n) Then IsRightTruncatable = True End IfEnd Function Function IsPrime(n) If n = 2 Then IsPrime = True ElseIf n <= 1 Or n Mod 2 = 0 Then IsPrime = False Else IsPrime = True For i = 3 To Int(Sqr(n)) Step 2 If n Mod i = 0 Then IsPrime = False Exit For End If Next End IfEnd Function `
Output:
```Largest LTP from 1..1000000: 998443
Largest RTP from 1..1000000: 739399
Elapse Time(seconds) : 49
```
## XPL0
`code CrLf=9, IntOut=11; func Prime(P); \Return true if P is a prime numberint P; \(1 is not prime, but 2 is, etc.)int I;[if P<=1 then return false; \negative numbers are not primefor I:= 2 to sqrt(P) do if rem(P/I) = 0 then return false;return true;]; func RightTrunc(N); \Return largest right-truncatable prime < one millionint N;int M;[for N:= 1_000_000-1 downto 2 do [M:= N; loop [if not Prime(M) then quit; M:= M/10; if rem(0) = 0 then quit; \no zeros allowed if M=0 then return N; ]; ];]; func LeftTrunc(N); \Return largest left-truncatable prime < one millionint N;int M, P;[for N:= 1_000_000-1 downto 2 do [M:= N; P:=100_000; loop [if not Prime(M) then quit; M:= rem(M/P); P:= P/10; if M<P then quit; \no zeros allowed if M=0 then return N; ]; ];]; [IntOut(0, LeftTrunc); CrLf(0); IntOut(0, RightTrunc); CrLf(0);]`
Output:
```998443
739399
```
## zkl
Using Extensible prime generator#zkl and a one meg bucket of bytes, construct a yes/no lookup table for all primes <= one million (<80,000).
`const million=0d1_000_000; var pTable=Data(million+1,Int).fill(0); // actually bytes, all zeroprimes:=Utils.Generator(Import("sieve").postponed_sieve);while((p:=primes.next())<million){ pTable[p]=1; } fcn rightTrunc(n){ while(n){ if(not pTable[n]) return(False); n/=10; } True}fcn leftTrunc(n){ // 999,907 is not allowed ns:=n.toString(); if (ns.holds("0")) return(False); while(ns){ if(not pTable[ns]) return(False); ns=ns[1,*]; } True}`
`[million..0,-1].filter1(rightTrunc): "%,d is a right truncatable prime".fmt(_).println();[million..0,-1].filter1(leftTrunc): "%,d is a left truncatable prime".fmt(_).println();`
Output:
```739,399 is a right truncatable prime
998,443 is a left truncatable prime
```
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# Common interaction vacuum for QED + QCD?
• I
## Main Question or Discussion Point
Hello,
I know QED and QCD as isolated theorys but now I thought about particle interactions with QED and QCD processes (like fpr proton-antiproton scattering). But I'm not sure how to interpret this mathematically.
As I understood my Feynman diagrams are nothing more like pictures for the transition amplitueds (up to some orders). For this we introduce a interaction vacuum state $|\Omega\rangle$. And then we are able to calculate: $$\langle\Omega|\phi(x_1)...\phi(x_n)|\Omega\rangle.$$ I thougth this means the creation of some particle at $\phi$ at $x_n$ and anihilation at some other space time point.
But if I like to have both interactions in one diagram I need a common interaction vacuum to write such transition amplitueds? Is there a common state for QED and QED or better for the standard model? Or are they different? But how can I interpret these processes in tis case?
Thanks for some answers. Maybe I'm a bit to confused with this whole QFT thing.
Related Quantum Physics News on Phys.org
king vitamin
Gold Member
The vacuum for pure QED or QCD will differ from the vacuum of both of them combined (and the two vacua naturally differ from each other). Whatever problem (Hamiltonian) you're working with, you'll want to compute correlation functions with respect to the vacuum of that Hamiltonian.
Demystifier
Demystifier
Gold Member
You need, of course, a common vacuum. But since you work with Feynman diagrams, i.e. perturbatively, you only need the perturbative vacuum which is quite simple to find. If ##|0_{\rm QED}\rangle## and ##|0_{\rm QCD}\rangle## are perturbative QED and QCD vacuums, respectively, then the full perturbative vacuum is simply ##|0_{\rm QED}\rangle \otimes |0_{\rm QCD}\rangle##.
dextercioby
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# Statistical Analysis
views updated May 18 2018
# Statistical Analysis
You may have heard the saying "You can prove anything with statistics," which implies that statistical analysis cannot to be trusted, that the conclusions that can be drawn from it are so vague and ambiguous that they are meaningless. Yet the opposite is also true. Statistical analysis can be reliable and the results of statistical analysis can be trusted if the proper conditions are established.
## What Is Statistical Analysis?
Statistical analysis uses inductive reasoning and the mathematical principles of probability to assess the reliability of a particular experimental test. Mathematical techniques have been devised to allow measurement of the reliability (or fallibility) of the estimate to be determined from the data (the sample, or "N") without reference to the original population. This is important because researchers typically do not have access to information about the whole population, and a samplea subset of the population is used.
Statistical analysis uses a sample drawn from a larger population to make inferences about the larger population. A population is a well-defined group of individuals or observations of any size having a unique quality or characteristic. Examples of populations include first-grade teachers in Texas, jewelers in New York, nurses at a hospital, high school principals, Democrats, and people who go to dentists. Corn plants in a particular field and automobiles produced by a plant on Monday are also populations. A sample is the group of individuals or items selected from a particular population. A random sample is taken in such a way that every individual in the population has an equal opportunity to be chosen. A random sample is also known as an unbiased sample.
Most mail surveys, mall surveys, political telephone polls, and other similar data gathering techniques generally do not meet the proper conditions for a random, unbiased sample, so their results cannot to be trusted. These are "self-selected" samples because the subjects choose whether to participate in the survey and the subjects may be picked based on the ease of their availability (for example, whoever answers the phone and agrees to the interview).
## Selecting a Random Sampling
The most important criterion for trustworthy statistical analysis is correctly choosing a random sample. For example, suppose you have a bucket full of 10,000 marbles and you want to know how many of the marbles are red. You could count all of the marbles, but that would take a long time. So, you stir the marbles thoroughly and, without looking, pull out 100 marbles. Now you count the red marbles in your random sample. There are 10. Thus you could conclude that approximately 10 percent of the original marbles are red. This is a trustworthy conclusion, but it is not likely to be exactly right. You could improve your accuracy by counting a larger sample; say 1,000 marbles. Of course if you counted all the marbles, you would know the exact percentage, but the point is to pick a sample that is large enough (for example, 100 or 1,000) that gives you an answer accurate enough for your purposes.
Suppose the 100 marbles you pulled out of the bucket were all red. Would this be proof that all 10,000 marbles in the bucket were red? In science, statistical analysis is used to test a hypothesis . In the example we are testing, the hypothesis would be "all the marbles in the bucket are red."
Statistical inference makes it possible for us to state, given a sample size (100) and a population size (10,000), how often false hypotheses will be accepted and how often true hypotheses are rejected. Statistical analysis cannot conclusively tell us whether a hypothesis is true; only the examination of the entire population can do that. So "statistical proof" is a statement of just how often we will get "the right answer."
## Using Basic Statistical Concepts
Statistics is applicable to all fields of human endeavor, from economics to education, politics to psychology. Procedures worked out for one field are generally applicable to the other fields. Some statistical procedures are used more often in some fields than in others.
Example 1. Suppose the Wearemout Pants Company wants to know the average height of adult American men, an important piece of information for a clothing manufacturer producing pants. The population is all men over the age of 25 who live in the United States. It is logistically impossible to measure the height of every man who lives in the United States, so a random sample of around 1,000 men is chosen. If the sample is correctly chosen, all ethnic groups, geographic regions, and socioeconomic classes will be adequately represented. The individual heights of these 1,000 men are then measured. An average height is calculated by dividing the sum of these individual heights by the total number of subjects (N = 1,000). By doing so, imagine that we calculate an average height is 1.95 meters (m) for this sample of adult males in the United States. If a representative sample was selected, then this figure can be generalized to the larger population.
The random sample of 1,000 men probably included some very short men and some very tall men. The difference between the shortest and the tallest is known as the "range" of the data. Range is one measure of the "dispersion" of a group of observations. A better measure of dispersion is the "standard deviation." The standard deviation is the square root of the sum of the squares of the differences divided by one less than the number of observations.
In this equation, xi is an observed value and is the arithmetic mean.
In our example, if a smaller height interval is used (1.10 m, 1.11 m, 1.12 m, 1.13 m, and so on) and the number of men in each height interval plotted as a function of height a smooth curve can be drawn which would have a characteristic shape, known as a "bell" curve or "normal frequency distribution." A normal frequency distribution can be stated mathematically as
The value of sigma (σ) is a measure of how "wide" the distribution is. Not all samples will have a normal distribution, but many do, and these distributions are of special interest.
The following figure shows three normal probability distributions. Because there is no skew, the mean, median, and mode are the same. The mean of curve (a) is less than the mean of curve (b), which in turn is less than the mean of (c). Yet the standard deviation, or spread, of (c) is least, whereas that of (a) is greatest. This is just one illustration of how the parameters of distributions can vary.
Example 2. One of the most common uses of statistical analysis is in determining whether a certain treatment is efficacious. For example, medical researchers may want to know if a particular medicine is effective at treating the type of pain resulting from extraction of third molars (known as "wisdom" teeth). Two random samples of approximately equal size would be selected. One group would receive the pain medication while the other group received a "placebo," a pill that looked identical but contained only inactive ingredients. The study would need to be a "double-blind" experiment, which is designed so that neither the recipients nor the persons dispensing the pills knew which was which. Researchers would know who had received the active medicines only after all the results were collected.
Example 3. Suppose a student, as part of a science fair project, wishes to determine if a particular chemical compound (Chemical X) can accelerate the growth of tomato plants. In this sort of experiment design, the hypothesis is usually stated as a null hypothesis : "Chemical X has no effect on the growth rate of tomato plants." In this case, the student would reject the null hypothesis if she found a significant difference. It may seem odd, but that is the way most of the statistical tests are set up. In this case, the independent variable is the presence of the chemical and the dependent variable is the height of the plant.
The next step is experiment design. The student decides to use height as the single measure of plant growth. She purchases 100 individual tomato plants of the same variety and randomly assigns them to 2 groups of 50 each. Thus the population is all tomato plants of this particular type and the sample is the 100 plants she has purchased. They are planted in identical containers, using the same kind of potting soil and placed so they will receive the same amount of light and air at the same temperature. In other words, the experimenter tries to "control" all of the variables, except the single variable of interest. One group will be watered with water containing a small amount of the chemical while the other will receive plain water. To make the experiment double-blind, she has another student prepare the watering cans each day, so that she will not know until after the experiment is complete which group was receiving the treatment. After 6 weeks, she plans to measure the height of the plants.
The next step is data collection. The student measures the height of each plant and records the results in data tables. She determines that the control group (which received plain water) had an average (arithmetic mean) height of 1.3 m (meters), while the treatment group had an average height of 1.4 m.
Now the student must somehow determine if this small difference was significant or if the amount of variation measured would be expected under no treatment conditions. In other words, what is the probability that 2 groups of 50 tomato plants each, grown under identical conditions would show a height difference of 0.1 m after 6 weeks of growth? If this probability is less than or equal to a certain predetermined value, then the null hypothesis is rejected. Two commonly used values of probability are 0.05 or 0.01. However, these are completely arbitrary choices determined mostly by the widespread use of previously calculated tables for each value. Modern computer analysis techniques allow the selection of any value of probability.
The simplest test of significance is to determine how "wide" the distribution of heights is for each group. If there is a wide variance (σ) in heights (say, σ = 25), then small differences in mean are not likely to be significant. On the other hand, if the dispersion is narrow (for example, if all the plants in each group were close to the same height, so that σ = 0.1) then the difference would probably be significant.
There are several different tests the student could use. Selecting the right test is often a tricky problem. In this case, the student can reject several tests outright. For example, the chi-square test is suitable for nominal scales (yes or no answers are one example of nominal scales), so it does not work here. The F -test measures variability or dispersion within a single sample. It too is not suitable for comparing two samples. Other statistical tests can also be rejected as inappropriate for various reasons.
In this case, since the student is interested in comparing means, the best choice is a t test. The t -test compares two means using this formula:
In this case, the null hypothesis assumes that μ1 μ2 = 0 (no difference in the sample groups), so that we can say:
The quantity is known as the standard error of the mean difference. When the sample sizes are the same, . The standard error of the mean difference is the square root of the sums of the squares of the standard errors of the means for each group. The standard error of the mean for each group is easily calculated from . N is the sample size, 50, and the student can calculate the standard deviation by the formula for standard deviation given above.
The final experimental step is to determine sensitivity. Generally speaking, the larger the sample, the more sensitive the experiment is. The choice of 50 tomato plants for each group implies a high degree of sensitivity.
Students, teachers, psychologists, economists, politicians, educational researchers, medical researchers, biologists, coaches, doctors and many others use statistics and statistical analysis every day to help them make decisions. To make trustworthy and valid decisions based on statistical information, it is necessary to: be sure the sample is representative of the population; understand the assumptions of the procedure and use the correct procedure; use the best measurements available; keep clear what is being looked for; and to avoid statements of causal relations if they are not justified.
see also Central Tendency, Measures of; Data Collection and Interpretation; Graphs; Mass Media, Mathematics and the.
Elliot Richmond
## Bibliography:
Huff, Darrell. How to Lie With Statistics. New York: W. W. Norton & Company, 1954.
Kirk, Roger E. Experimental Design, Procedures for the Behavioral Sciences. Monterrey, CA: Brooks/Cole Publishing Company, 1982.
Paulos, John Allen. Innumeracy: Mathematical Illiteracy and Its Consequences. New York: Hill & Wang, 1988.
Tufte, Edward R. The Visual Display of Quantitative Information. Cheshire, CT: Graphics Press, 1983.
## MEAN, MEDIAN, AND MODE
The average in the clothing example is known as the "arithmetic mean," which is one of the measures of central tendency. The other two measures of central tendency are the"median" and the "mode." The median is the number that falls in the mid-point of an ordered data set, while the mode is the most frequently occurring value.
# Classical Statistical Analysis
views updated May 23 2018
# Classical Statistical Analysis
BIBLIOGRAPHY
Classical statistical analysis seeks to describe the distribution of a measurable property (descriptive statistics) and to determine the reliability of a sample drawn from a population (inferential statistics). Classical statistical analysis is based on repeatedly measuring properties of objects and aims at predicting the frequency with which certain results will occur when the measuring operation is repeated at random or stochastically.
Properties can be measured repeatedly of the same object or only once per object. However, in the latter case, one must measure a number of sufficiently similar objects. Typical examples are measuring the outcome of tossing a coin or rolling a die repeatedly and count the occurrences of the possible outcomes as well as measuring the chemical composition of the next hundred or thousand pills produced in the production line of a pharmaceutical plant. In the former case the same object (one and the same die cast) is measured several times (with respect to the question which number it shows); in the latter case many distinguishable, but similar objects are measured with respect to their composition which in the case of pills is expected to be more or less identical, such that the repetition is not with the same object, but with the next available similar object.
One of the central concepts of classical statistical analysis is to determine the empirical frequency distribution that yields the absolute or relative frequency of the occurrence of each of the possible results of the repeated measurement of a property of an object or a class of objects when only a finite number of different outcomes is possible (discrete case). If one thinks of an infinitely repeated and arbitrarily precise measurement where every outcome is (or can be) different (as would be the case if the range of the property is the set of real numbers), then the relative frequency of a single outcome would not be very instructive; instead one uses the distribution function in this (continuous) case which, for every numerical value x of the measured property, yields the absolute or relative frequency of the occurrence of all values smaller than x. This function is usually noted as F (x ), and its derivative F (x ) = f (x ) is called frequency density function.
If one wants to describe an empirical distribution, the complete function table is seldom instructive. This is why the empirical frequency or distribution functions are often represented by a few parameters that describe the essential features of the distribution. The so-called moments of the distribution represent the distribution completely, and the lower-order moments represent the distribution at least in a satisfactory manner. Moments are defined as follows:
where k is the order of the moment, n is the number of repetitions or objects measured, and c is a constant that is usually either 0 (moment about the origin) or the arithmetic mean (moment about the mean), the first-order mean about the origin being the arithmetic mean.
In the frequentist interpretation of probability, frequency can be seen as the realization of the concept of probability: It is quite intuitive to believe that if the probability of a certain outcome is some number between 0 and 1, then the expected relative frequency of this outcome would be the same number, at least in the long run. From this, one of the concepts of probability is derived, yielding probability distribution and density functions as models for their empirical correlates. These functions are usually also noted as f (x ) and F (x ), respectively, and their moments are also defined much like in the above formula, but with a difference that takes into account that there is no finite number n of measurement repetitions:
where the first equation can be applied to discrete numerical variables (e.g., the results of counting), while the second equation can be applied to continuous variables. Again, the first-order moment about 0 is the mean, and the other moments are usually calculated about this mean. In many important cases one would be satisfied to know the mean (as an indicator for the central tendency of the distribution) and the second-order moment about the mean, namely the variance (as the most prominent indicator for the variation). For the important case of the normal or Gaussian distribution, these two parameters are sufficient to describe the distribution completely.
If one models an empirical distribution with a theoretical distribution (any non-negative function for which the zero-order moment evaluates to 1, as this is the probability for the variable to have any arbitrary value within its domain), one can estimate its parameters from the moments of the empirical distributions calculated from the finite number of repeated measurements taken in a sample, especially in the case where the normal distribution is a satisfactory model of the empirical distribution, as in this case mean and variance allow the calculation of all interesting values of the probability density function f (x ) and of the distribution function F (x ).
Empirical and theoretical distributions need not be restricted to the case of a single property or variable, they are also defined for the multivariate case. Given that empirical moments can always be calculated from the measurements taken in a sample, these moments are also results of a random process, just like the original measurements. In this respect, the mean, variance, correlation coefficient or any other statistical parameter calculated from the finite number of objects in a sample is also the outcome of a random experiment (measurement taken from a randomly selected set of objects instead of exactly one object). And for these derived measurements theoretical distributions are also available, and these models of the empirical moments allow the estimation with which probability one could expect the respective parameter to fall into a specified interval in the next sample to be taken.
If, for instance, one has a sample of 1,000 interviewees of whom 520 answered they were going to vote for party A in the upcoming election, and 480 announced they were going to vote for party B, then the parameter πAthe proportion of A-voters in the overall populationcould be estimated to be 0.52, but this estimate would be a stochastic variable, which approximately obeys a normal distribution with mean 0.52 and variance 0.0002496 (or standard deviation 0.0158), and from this result one can conclude that another sample of another 1,000 interviewees from the same overall population would lead to another estimate whose value would lie within the interval [0.489, 0.551] (between 0.52 ± 1.96 0.0158) with a probability of 95 percent (the so-called 95 percent confidence interval, which in the case of the normal distribution is centered about the mean with a width of 3.92 standard deviations). Or, to put it in other words, the probability of finding more than 551 A-voters in another sample of 1,000 interviewees from the same population is 0.025. Bayesian statistics, as opposed to classical statistics, would argue from the same numbers that the probability is 0.95 that the population parameter falls within the interval [0.489, 0.551].
SEE ALSO Bayesian Statistics; Descriptive Statistics; Inference, Bayesian; Inference, Statistical; Sampling; Variables, Random
## BIBLIOGRAPHY
Hoel, Paul G. 1984. Introduction to Mathematical Statistics. 5th ed. Hoboken, NJ: Wiley.
Iversen, Gudmund. 1984. Bayesian Statistical Inference. Beverly Hills, CA: Sage.
Klaus G. Troitzsch
# Statistical Analysis
views updated May 11 2018
# Statistical Analysis
Throughout conflicts, apologists for the side in power often excuse atrocities committed by their side with the claim that "violations are being committed on all sides of the conflict." The objective of such a statement is to render the parties morally equivalent, thereby relieving observers of the responsibility or duty to make a judgment about whether one side is the aggressor and the other is acting in self-defense. Even when the greater historical narrative involves more than these labels imply, in situations of massive human rights violations the perpetrators are rarely balanced in power. Although it may be literally true that all parties to a conflict have committed at least one violation, often the number of violations each party commits differs by a factor of ten or more relative to their opponents. In some cases quantitative analysis may offer a method for assessing claims about moral responsibility for crimes against humanity, including genocide. Statistics provide a way to measure crimes of policy—massive crimes that result from institutional or political decisions.
Although all parties may be guilty, they are rarely guilty in equal measure. Only with quantitative arguments can the true proportions of responsibility be understood. In this way one can transcend facile claims about "violations on all sides" in favor of an empirically rich view of responsibility for atrocities. Did the monthly number of killings increase or decrease in the first quarter of 1999? Were there more violations in Province A or in Province B? Were men more affected than women, or adults relative to children? These simple quantitative evaluations may be important questions when linked to political processes. Perhaps a new government took power and one needs to assess its impact on the state's respect for human rights. Or a military officer may move from Province A to Province B, and one may wish to determine if he is repeating the crimes he committed in Province A. Simple descriptive statistics based on properly gathered data can address these questions more precisely than the kinds of casual assessments that nonquantitative observers often make.
There are three areas in which nonquantitative analysts most often make statistical mistakes: estimating the total magnitude of violations; understanding how bias may have affected the data collection or interpretation; and comparing the relative proportions of responsibility among perpetrators. Poor information management and inappropriate statistical analysis can lead to embarrassing reversals of findings once proper methods are applied.
The use of statistical methods that demonstrably control biases and enable estimates of total magnitude can give analysts a rigorous basis for drawing conclusions about politically important questions. One such method, multiple systems estimation, uses three or more overlapping lists of some event (such as killings) to make a statistical estimate of the total number of events, including those events excluded from all three lists. "Overlapping" in this sense means events that are documented on two or more lists. The estimate made by this technique can control for several biases that might affect the original reporting which led to the lists of events.
For example, among the most important questions the Guatemalan Commission for Historical Clarification (CEH is the Spanish acronym) had to answer was whether the army had committed acts of genocide against the Maya. Using qualitative sources and field investigation, the CEH identified six regions in which genocide might have occurred. Data were collated from testimonies given to three sources: nongovernmental organizations (NGOs), the Catholic Church, and the CEH.
If genocide has been committed, then at least two statistical indicators should be clear. First, the absolute magnitude of the violations should be large. Second, there should be a big difference in the rate of killing between those who are in the victim group versus those people in the same region who are not in the victim group. It is inadequate to argue that some large number of people of specific ethnicities have been killed, because it might have been that they were simply unfortunate enough to live in very violent areas. Killing in an indiscriminate pattern might be evidence of some other crime, but if genocide occurred, a substantial difference in killing rates between targeted and nontargeted groups should exist. Thus, to find statistical evidence consistent with genocide, it is not enough that certain people were killed at high rate, but also that other nearby people were killed at much lower rates.
The CEH analysts conducted a multiple systems estimate of the total deaths of indigenous people and nonindigenous people between 1981 and 1983 in the six regions identified. For each group in each region, the estimated total number of deaths was divided into the Guatemalan government's census figures for indigenous and nonindigenous people in 1981. The CEH showed that resulting proportions were consistent with the genocide hypothesis. In each region indigenous people were killed at a rate five to eight times greater than nonindigenous people. This statistical finding was one of the bases of the CEH's final conclusion that the Guatemalan army committed acts of genocide against the Maya.
Other human rights projects have incorporated statistical reasoning. Sociologists and demographers have testified at the trial of Slobodan Milosevic and others tried before the International Criminal Tribunal for the Former Yugoslavia. They have provided quantitative insights on ethnic cleansing, forced migration, and the evaluation of explanatory hypotheses.
In the early twenty-first century, the statistical analysis of human rights violations is just beginning, and much work remains. New techniques should be developed, including easier methods for conducting random probability sampling in the field, richer demographic analysis of forced migration, and more flexible techniques for rapidly creating lots of graphical views of data. Human rights advocacy and analysis have benefited tremendously from the introduction of better statistical methods. The international community needs to continue to find new ways to employ existing methods, and to further research on new methods, so that human rights reporting becomes more rigorous. Statistics help establish the evidentiary basis of human rights allegations about crimes of policy.
## BIBLIOGRAPHY
Ball, Patrick (2000). "The Guatemalan Commission for Historical Clarification: Inter-Sample Analysis." In Making the Case: Investigating Large Scale Human Rights Violations Using Information Systems and Data Analysis, ed. Patrick Ball, Herbert F. Spirer, and Louise Spirer. Washington, D.C.: AAAS.
Ball, P., W. Betts, F. Scheuren, J. Dudukovic, and J. Asher (2002). Killings and Refugee Flow in Kosovo March–June 1999. Washington, D.C.: AAAS.
Brunborg, H., H. Urdal, and T. Lyngstad (2001). "Accounting for Genocide: How Many Were Killed in Srebrenica?" Paper presented at the Uppsala Conference on Conflict Data, Uppsala, June 8–9, 2001. Available from http://www.pcr.uu.se/conferenses/Euroconference/paperbrunborg.doc.
Ward, K. (2000). "The United Nations Mission for the Verification of Human Rights in Guatemala." In Making the Case: Investigating Large Scale Human Rights Violations Using Information Systems and Data Analysis, ed. Patrick Ball, Herbert F. Spirer, and Louise Spirer. Washington, D.C.: AAAS.
Patrick Ball
# statistical analysis
views updated May 23 2018
statistical analysis See statistical methods.
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https://oeis.org/A213407
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A213407 First occurrence of n in A213381. 2
2, 0, 3, 5, 4, 9, 8, 7, 12, 17, 594270126787, 21, 20, 13, 24, 29, 16, 23, 32, 19, 36, 41, 1356, 45, 38, 151, 51, 411, 994, 57, 56, 31, 60, 65, 52, 69, 71, 37, 72, 77, 490, 81, 80, 169, 84, 6125, 68, 1043, 92, 49, 99, 101, 304, 105, 104, 55, 111, 63, 73, 89, 116, 61, 120, 125, 78, 129, 188, 67, 93, 137, 97, 18349355, 140, 433 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,1 LINKS FORMULA a(n) = smallest k>n such that (-2)^k == n (mod k+2). EXAMPLE Smallest n such that A213381(n)=9 is 17, so a(9)=17. PROG (C) #include // GCC -O3 -fopenmp #include // 11 minutes #define SIZE 4096 #define STEP 0x1000000 long long first[SIZE]; int main(int argc, char **argv) { unsigned long long a, i; for (a=0; a0; p=(p*p)%b, t>>=1) { if (t&1) r=(r*p)%b; } if (r
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Last modified November 30 18:40 EST 2021. Contains 349424 sequences. (Running on oeis4.)
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https://www.mathworks.com/matlabcentral/cody/problems/8046-gold-silver-standard/solutions/1974746
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Cody
# Problem 8046. Gold/Silver Standard
Solution 1974746
Submitted on 14 Oct 2019 by kranthi kumar
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
HY = 2014; HV = 1270; CV_corr = 1270; assert(abs(gold_silver_standard(HY,HV)-CV_corr)<2e-5)
ans = 1270
2 Pass
HY = 500; HV = 50; CV_corr = NaN; assert(isnan(gold_silver_standard(HY,HV)))
ans = NaN
3 Pass
HY = 2500; HV = 5000; CV_corr = NaN; assert(isnan(gold_silver_standard(HY,HV)))
ans = NaN
4 Pass
HY = 2010; HV = 1000; CV_corr = 945.54; assert(abs(gold_silver_standard(HY,HV)-CV_corr)<2e-5)
ans = 945.5400
5 Pass
HY = 2005; HV = 1000; CV_corr = 2603.88; assert(abs(gold_silver_standard(HY,HV)-CV_corr)<2e-5)
ans = 2.6039e+03
6 Pass
HY = 2000; HV = 1000; CV_corr = 3822.36; assert(abs(gold_silver_standard(HY,HV)-CV_corr)<2e-5)
ans = 3.8224e+03
7 Pass
HY = 1995; HV = 1000; CV_corr = 3711.31; assert(abs(gold_silver_standard(HY,HV)-CV_corr)<2e-5)
ans = 3.7113e+03
8 Pass
HY = 1990; HV = 1000; CV_corr = 3965.32; assert(abs(gold_silver_standard(HY,HV)-CV_corr)<2e-5)
ans = 3.9653e+03
9 Pass
HY = 1950; HV = 1000; CV_corr = 25768.43; assert(abs(gold_silver_standard(HY,HV)-CV_corr)<2e-5)
ans = 2.5768e+04
10 Pass
HY = 1900; HV = 1000; CV_corr = 30850.44; assert(abs(gold_silver_standard(HY,HV)-CV_corr)<2e-5)
ans = 3.0850e+04
11 Pass
HY = 1850; HV = 1000; CV_corr = 14532.01; assert(abs(gold_silver_standard(HY,HV)-CV_corr)<2e-5)
ans = 1.4532e+04
12 Pass
HY = 1800; HV = 1000; CV_corr = 15471.58; assert(abs(gold_silver_standard(HY,HV)-CV_corr)<2e-5)
ans = 1.5472e+04
13 Pass
ind = randi(4); switch ind case 1 HY = 2000; HV = 1000; CV_corr = 3822.36; case 2 HY = 1995; HV = 1000; CV_corr = 3711.31; case 3 HY = 2005; HV = 1000; CV_corr = 2603.88; case 4 HY = 1800; HV = 1000; CV_corr = 15471.58; end assert(abs(gold_silver_standard(HY,HV)-CV_corr)<2e-5)
ans = 1.5472e+04
14 Pass
ind = randi(4); switch ind case 1 HY = 1995; HV = 1000; CV_corr = 3711.31; case 2 HY = 1990; HV = 1000; CV_corr = 3965.32; case 3 HY = 2010; HV = 1000; CV_corr = 945.54; case 4 HY = 1900; HV = 1000; CV_corr = 30850.44; end assert(abs(gold_silver_standard(HY,HV)-CV_corr)<2e-5)
ans = 3.9653e+03
15 Pass
ind = randi(4); switch ind case 1 HY = 1990; HV = 1000; CV_corr = 3965.32; case 2 HY = 1950; HV = 1000; CV_corr = 25768.43; case 3 HY = 1900; HV = 1000; CV_corr = 30850.44; case 4 HY = 2000; HV = 1000; CV_corr = 3822.36; end assert(abs(gold_silver_standard(HY,HV)-CV_corr)<2e-5)
ans = 2.5768e+04
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https://uk.mathworks.com/matlabcentral/cody/problems/44409-/solutions/2054751
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Cody
# Problem 44409. 特定の値がベクトル内に含まれているかを確認するコードを書こう
Solution 2054751
Submitted on 11 Dec 2019 by Jonathan
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
a = 1; b = [1,2]; y_correct = 1; assert(isequal(existsInVector(a,b),y_correct))
2 Pass
a = 12; b = [1,3,4,5,6,7,8,1,2]; y_correct = 0; assert(isequal(existsInVector(a,b),y_correct))
3 Pass
a = -1; b = [1,2]; y_correct = 0; assert(isequal(existsInVector(a,b),y_correct))
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https://www.calculatorschool.com/trigonometric/SumtoProduct.aspx
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Sum to Product Trigonometry Identities Calculator
Trigonometry identities calculator to rewrite and evaluate sums of sines and/or cosines as products.
Sum to Product Trigonometry
Formula Used:
sinu + sinv = 2 x sin(u+v/2) x cos(u-v/2)
sinu - sinv = 2 x cos(u+v/2) x sin(u-v/2)
cosu + cosv = 2 x cos(u+v/2) x cos(u-v/2)
cosu - cosv = -2 x sin(u+v/2) x sin(u-v/2)
Sum to product trigonometry identities calculation is made easier here.
Product to sum trigonometry identities calculation is made easier here.
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CC-MAIN-2024-38
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https://www.topperlearning.com/forums/home-work-help-19/hello-sir-im-hazeera-i-want-to-know-about-kinetic-energy-wi-physics-work-and-energy-energy-89077/reply
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Question
Sat March 07, 2015
# hello sir im hazeera i want to know about kinetic energy with examples?
Faiza Lambe
Sat March 07, 2015
All moving things have kinetic energy.
The energy possessed by an object on account of its motion is known as kinetic energy.
The kinetic energy for a body of mass 'm' moving with a velocity 'v' is given as,
From the above equation, we know that, kinetic energy is directly proportional to the mass of the body and the square of the velocity with which it moves.
Kinetic energy is also defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.
Examples: A car in motion, moving hands of a clock, the bullet fired from a gun, strong blowing of winds, flowing of water, etc.
Related Questions
Mon November 13, 2017
# Please solve question no. 3 with proper explanation
Wed November 08, 2017
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http://www.yourelectrichome.com/2014/07/basic-dc-voltmeter.html
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### Basic D.C. Voltmeter
The basic d.c. voltmeter is nothing but a PMMC D'Arsonoval galvanometer. The resistance is required to be connected in series with the basic meter to use it as a voltmeter.
This series resistance is called a multiplier. The main function of the multiplier is to limit the current through the basic meter so that the meter current does not exceed the full scale deflection value. The voltmeter measures the voltage across the two points of a circuit or a voltage across circuit component. The basic d.c. voltmeter is shown in the Fig. 1.
Fig. 1 Basic of d.c. voltmeter
The voltmeter must be connected across the two points or a component, to measure the potential difference, with the proper polarity.
The multiplier resistance can be calculated as:
Let R= Internal resistance of coil i.e. meter
Rs = series multiplier resistance
Im = full scale deflection current
V = full range voltage to be measured
From Fig 1, ... V = I (RRs )
... V = ImRI R
... ImR= V - I Rm
The multiplying factor for multiplier is the ratio of full range voltage to be measured and the drop across the basic meter.
Let v = drop across the basic meter = ImR
m = multiplying factor = V/v
Hence multiplier resistance can also be expressed as,
Rs = (m-1)
Thus to increase the range of voltmeter 'm' times, the series resistance is (m-1) times the basic meter resistance. This is nothing but extension of ranges of a voltmeter.
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CC-MAIN-2018-47
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https://www.shaalaa.com/question-bank-solutions/find-derivative-f-x-x-x-1-derivative-algebra-derivative-functions_56317
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crawl-data/CC-MAIN-2021-17/segments/1618038468066.58/warc/CC-MAIN-20210418043500-20210418073500-00154.warc.gz
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Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
# Find the Derivative of F (X) X at X = 1 - Mathematics
Find the derivative of f (xx at x = 1
#### Solution
We have:
$f'(x) = \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h}$
$= \lim_{h \to 0} \frac{1 + h - 1}{h}$
$= \lim_{h \to 0} 1$
$= 1$
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 11 Mathematics Textbook
Chapter 30 Derivatives
Exercise 30.1 | Q 4 | Page 3
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https://docs.codejig.com/uk/entity2305843015656313960/view/4611686018427803523
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en
uk
# LOGIC
Logic blocks are generally used to control conditional blocks and repeat blocks.
Here's an example:
If the score is more than or equal to 50, the condition is true, and the text "Pass" is printed. If the score is not greater than or equal to 50, the condition is false, and "Fail" is printed.
Boolean values can also be stored in variables and passed to functions, the same as numbers, text, and list values. Leaving an input empty is not recommended.
Values
A single block, with a drop-down specifying either true or false, can be used to get a Boolean value:
Comparisons
There are six comparison operators. Each takes two inputs and returns true or false depending on how the inputs compare with each other.
The six operators are: equals, not equals, less than, greater than, less than or equal, greater than or equal.
Logical operations
The and block will return true only if both of its two inputs are also true.
The or block will return true only if either of its two inputs are true.
Not
The not block converts its Boolean input into its opposite. For example, the result of:
is false.
Ternary operator
The ternary block acts like a miniature if-else block. It takes three inputs; the first input is the Boolean condition to test, the second input is the value to return if the test was true, the third input is the value to return if the test was false. For example
Null
The null block allows the value to be set as Null instead of the usual possible values of the data type.
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CC-MAIN-2023-14
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https://www.lmpforum.com/forum/topic/4102-slip-compensation/
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# Slip Compensation
## Recommended Posts
Hi all!, I'm having some troubles with a VSD it's a mitsubishi F700 the motor is loosing power, I'm try to up with toorque boost Pr 0. but it's a little diference, I was reading about Slip compensation but not very sure how I can set it up, there are 3 parameters
Pr 245=Rated Slip from 0 to 50%
Pr 246=Slip compensation time constant from 0.01 to 10s
Pr 247=Constant-output region slip compensation selection
Not sure how I can set each one to working very well, without damage motor or vsd. any advice
##### Share on other sites
I'm try to up with torque boost Pr 0. but it's a little difference,
What thing is the difference?
You can increase the torque boost within max current limit. But first;
1) Is it a new application or a replacement?
2) What is the type of load?
3) What is the name plate info of motor and drive?
4) Do you face problem only during start up?
5) When motor loose power, what is the current, freq, and out put voltage?
Reply for better help
"Don't assume any thing, always check/ask and clear yourself".
##### Share on other sites
Hi thanks for respond, i get more data
The diference to increase the torque boost is in the output voltage, if I increas too high or too low output voltage decrease, so i decide to put in a 0.8.
It's a replacement first we had a VSD 40HP 220V, with 2 motor, and now we replace with to VSD 20HP 220V each one, the motor are for 220V, 15Hp.
Type of load i can't explain you by english, but it's to make cement block it's necesary to shake.
The problem is whe the drive show 68Hz it's a maximun frequency voltage is 220V, start with 35Amps and decresae to 25Amp I think that's the problem I need the current constant but i don't know how to do it.
What thing is the difference?
You can increase the torque boost within max current limit. But first;
1) Is it a new application or a replacement?
2) What is the type of load?
3) What is the name plate info of motor and drive?
4) Do you face problem only during start up?
5) When motor loose power, what is the current, freq, and out put voltage?
Reply for better help
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https://sushelp.scale-up.com/help/fitstatb.aspx
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Scale-up Suite Help
Sum of squares (SSQ) table
About the SSQ (or Chi-squared) table
The SSQ (or chi-squared) table shows in the final column the total sum of squares (SSQfinal ) or chi-squared calculated according to the selected data weighting method and the contributions of individual profiles or series to the total:
The quoted values are a measure of the lack of fit between the model results and the data. The table shows the contribution of each selected profile to the total lack of fit. A high relative value indicates that the corresponding profile is less well fitted. In addition the number of selected points per profile is reported. Show me more on lack of fit.
Goodness of fit
The goodness of fit (Q) assessment compares the chi-squared value with a chi-squared distribution. If the value of chi-squared is near the number of degrees of freedom (= [number of data points] – [number of parameters fitted]), the fit is considered good, and a Q of about 0.5 is obtained. The goodness of fit statistic, Q, assumes that the errors are normally distributed.
A good fit in this context means that there is a good level of overlap between model residuals and experimental errors. The Notes reported with Goodness of Fit provide a statement about the model that expands on the Q value (see screenshot above).
When users select the default data weighting method of Weighted SSQ, or the alternative Unweighted SSQ, Dynochem back-calculates the experimental error for each datapoint that would correspond with a 'good fit'. Therefore Q values near 0.5 are always obtained. The corresponding average error value is reported (see screenshot above) and the individual errors are available in the Fitting Report.
When users select the advanced data weighting method of User-Defined Errors, Dynochem uses the errors provided by the user to calculate Q. In this case, a Q value of near 0.5 is not guaranteed:
If Q is very small (Q < 10-3), this indicates that either:
1. the model is wrong – it can be statistically rejected
2. the measurement errors are in fact much larger than those provided by the user
3. the measurement errors are not normally distributed.
If Q moderate (0.001 < Q < 0.95), this indicates a good level of overlap between residuals and experimental errors and a good fit.
If Q is high (0.95 < Q < 1), this indicates that the fit appears to be good, but measurement errors have likely been overestimated.
Further reading on fitting and optimization theory
A good source of additional information on fitting and optimization algorithms and theory is Numerical Recipes in Fortran, 2nd edition, by W.H. Press, S.A. Teukolsky, W.T. Vetterling and B.P. Flannery, Cambridge University Press, 1992.
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Limited access
A particular solution of:
$$$$y'' - 3y' - 4y = 8e^{2t}$$$$
...is:
A
$y(t) = \cfrac{4}{3} e^{2t}$
B
$y(t) = -\cfrac{4}{3} e^{2t}$
C
$y(t) = \cfrac{4}{3} e^{t}$
D
$y(t) = 8 e^{2t}$
Select an assignment template
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http://mathhelpforum.com/trigonometry/175892-why-i-e-i-pi-2-a.html
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1. ## why i=e^î*(pi/2) ?
why i=e^î*(pi/2)
Guys please let us know that I am not so good in maths... and I think the question that I am asking is not so easy ... however I strongly believe that in this forum we have people who can make me understood this (although I am not so good in maths) and have abilities to understand this kind of questions! ... THANKS
Guys please let us know that I am not so good in maths... and I think the question that I am asking is not so easy ... however I strongly believe that in this forum we have people who can make me understood this (although I am not so good in maths) and have abilities to understand this kind of questions! ... THANKS
$\displaystyle e^{ix}$ is defined as $\displaystyle e^{ix} = \cos(x) + i \sin(x)$
Hence, for $\displaystyle x= \frac{\pi}{2}$, we have...
$\displaystyle e^{i \frac{\pi}{2}} = \cos(\frac{\pi}{2}) = i \sin(\frac{\pi}{2})$
If you know your trigonometry, then you'll know that...
$\displaystyle \cos(\frac{\pi}{2}) = 0$
... and...
$\displaystyle \sin(\frac{\pi}{2}) = 1$
Hence, the equation becomes...
$\displaystyle e^{i\frac{\pi}{2}} = 0 + i (1) = i$
3. Euler's formula states that
$\displaystyle e^{i \theta} = cos(\theta) + i*sin(\theta)$
When theta is pi/2, the cosine term vanishes and the sine term is 1.
Guys please let us know that I am not so good in maths... and I think the question that I am asking is not so easy ... however I strongly believe that in this forum we have people who can make me understood this (although I am not so good in maths) and have abilities to understand this kind of questions! ... THANKS
$\displaystyle e^{ix}$ is defined as $\displaystyle e^{ix} = \cos(x) + i \sin(x)$
Hence, for $\displaystyle x= \frac{\pi}{2}$, we have...
$\displaystyle e^{i \frac{\pi}{2}} = \cos(\frac{\pi}{2}) = i \sin(\frac{\pi}{2})$
If you know your trigonometry, then you'll know that...
$\displaystyle \cos(\frac{\pi}{2}) = 0$
... and...
$\displaystyle \sin(\frac{\pi}{2}) = 1$
Hence, the equation becomes...
$\displaystyle e^{i\frac{\pi}{2}} = 0 + i (1) = i$
5. You've done this a few times on threads to which I have contributed. (sandwiched my post with duplicates of yours)
I don't mean to be redundant- it's kinda slow typing laTex on an iPhone.
Anyhow, I can't think of a legitimate reason why you would create a^n exact duplicate post, then edit (almost delete) your previous post to ".".
It's just obnoxious in my book
6. Originally Posted by TheChaz
You've done this a few times on threads to which I have contributed. (sandwiched my post with duplicates of yours)
I don't mean to be redundant- it's kinda slow typing laTex on an iPhone.
Anyhow, I can't think of a legitimate reason why you would create a^n exact duplicate post, then edit (almost delete) your previous post to ".".
It's just obnoxious in my book
There seems to be a problem with this forum/my computer, whereby once I hit the post reply button, the post is submitted, but the screen returns to the 'Go Advanced' page, leading me to believe that the post hasn't been submitted, thus leading to me posting it again.
So, rather than being obnoxious, I am in fact the victim of some sort of error.
And not sure what you mean about the "."... the first I knew that I had made a duplicate post was when I returned after being e-mailed about your reply, and I haven't edited it or touched it since.
7. A little more detail.
Since we know how to add and multiply complex numbers, a good way to extend other functions from real numbers to complex numbers is to use their Taylor series.
The Taylor series for $\displaystyle e^x$ is
$\displaystyle \sum_{n=0}^\infty \frac{x^n}{n!}= 1+ x+ (1/2)x^2+ (1/3!)x^3+ \cdot\cdot\cdot$
The Taylor series for $\displaystyle cos(x)$ is
$\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}$
and the Taylor series fror $\displaystyle sin(x)$ is
$\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}$.
If you replace x with ix in the Taylor series for $\displaystyle e^x$, you get
$\displaystyle 1+ ix+ \frac{-1x^2}{2!}+ \frac{-ix^2}{3!}+ \cdot\cdot\cdot$
Because $\displaystyle i^0= 1$, $\displaystyle i^1= i$, $\displaystyle i^2= -1$, $\displaystyle i^3= -i$, and $\displaystyle i^4= 1$, every odd power involves i while every even power does not. Separating real and imaginary parts,
$\displaystyle e^{ix}= (1- \frac{1}{2}x^2+ \frac{1}{4!}x^4- \cdot\cdot\cdot)+ i(x- \frac{1}{3!}x^3+ \frac{1}{5!}x^5-\cdot\cdot\cdot)$$\displaystyle = cos(x)+ i sin(x)$
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