I have a simple two lines of code but fail to understand what is actually happening.       s = 'We\'re' + 'Here'     print(s[4::2])   The result is just eee. I fail to see how the [4::2] is working to cause the code to print eee.
I need a good explanation (references are a plus) on Python's slice notation.   To me, this notation needs a bit of picking up.   It looks extremely powerful, but I haven't quite got my head around it.
I am beginning with Abstract Algebra and I'm trying to understand Lagrange's Theorem. The theorem reads      For any finite group $G$, the order of every subgroup $H$ of $G$ should divide the order of $G$.   It seems simple and I've used it to solve some exercices but I believe I'm missing the essence of it. Is there an example that can help me understand it better, or visualize it? Is there a geometric interpretation? Is there a difference between "Joe said" and "said Joe"? Does the subject/verb order make a difference when writing a dialog tag?  &quot;The sky is blue,&quot; Joe said. &quot;The sky is blue,&quot; said Joe.  Is one preferable over the other? Does one emphasize the speaker (or the method of speech, such as shout) more than the other? (Personally, I would use the second form to emphasize that Joe made the statement) Reading/writing an INI file Is there any class in the .NET framework that can read/write standard .ini files:  [Section] &lt;keyname&gt;=&lt;value&gt; ...   Delphi has the TIniFile component and I want to know if there is anything similar for C#? The meaning of Bit-wise NOT in Python Prove :(P → Q) ∨ (Q → P) using natural deduction Allowed inference rules: ∨-I, ∨-E, ∧-I, ∧-E, →-I, →-E, ¬-I, ¬-E  I tried to prove a contradiction by assuming $¬ ((P → Q) ∨ (Q → P))$ but got stuck, or am I doing it in the wrong way?  Edit:  My proof attempt  $1.\qquad¬ ((P → Q) ∨ (Q → P))\qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad Assum$  $2.\qquad P → Q\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\  Assum$  $3.\qquad(P → Q) ∨ (Q → P)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\; ∨-I(2)$  $4.\qquad(¬ ((P → Q) ∨ (Q → P))) ∧ ((P → Q) ∨ (Q → P))\qquad\qquad\qquad\; ∧-I(1,3)$  $5.\qquad  ¬ (P → Q)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\;\;\; ¬-I(2,4)$  $6.\qquad  Q → P\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad\qquad\quad\; Assum$  $7.\qquad  (P → Q) ∨ (Q → P)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\;\; ∨-I(6)$  $8.\qquad  (¬ ((P → Q) ∨ (Q → P))) ∧ ((P → Q) ∨ (Q → P))\qquad\qquad\qquad\;\; ∧-I(1,7)$  $9.\qquad  ¬ (Q → P)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\; ¬-I(6,8)$  $10.\qquad ¬ (P → Q) ∧ ¬ (Q → P)\qquad\qquad\qquad\qquad\qquad\qquad\;\qquad\qquad\quad\;\;\, ∧-I(5,9)$  Stuck at this line, can't spot any contradiction to line 1 When the automorphism group is trivial...... If $G$ is a trivial group , obviously $Aut(G)$  is  trivial . Does the converse  hold $?$ .  When  we  are  given  that a  group $G$  has  trivial  automorphism  group , can we  conclude  that the group  will  be  trivial $?$ .  I  guess  not . For  one  thing , we know  that  $$G/Z(G)\cong Inn(G)$$. So , $Aut(G)$ being  trivial  would  imply  that $$G\cong Z(G)$$  i.e.  $G$  is  commutative . But I  see  nothing about $G'$s being trivial .   The problem  is  that  while I cannot  prove  it , I  could  not  find  a  counter example  either , from  among  the  groups  I  am  familiar  with .   What  is  the non-trivial  commutative  group  that  has  trivial  automorphism  group  $?$ How to integrate a paper where I am third author, but did a lot of work, into my master's thesis? I am writing my thesis currently and I want to include papers that I am included on as an author.  I will have a couple papers where I am 1st author, but there is one where I am third but did a lot of the work.  How do I go about integrating it into my thesis?  Can I copy/paste most of the text still or do I need to rewrite it?  I don't want to plagiarize but don't want to rewrite it if I don't have to. CSS Selector that applies to elements with two classes I was asked to evaluate the following sum: $$\sum_{k=1}^{\infty}\frac{1}{k}\int_{\pi k}^{\infty}\frac{\sin(x)}{x}dx$$  I'm trying to use $$\int_{0}^{\infty} \frac{\sin(x)}{x}dx=\frac{\pi}{2}$$ However, it doesn't seem to work. Any help is greatly appreciated. What is the coefficient of the term $a^i b^j c^k$ in the expansion of $(a+b+c)^n$? What is the coefficient of the term $a^i b^j c^k$  in the expansion of   $(a+b+c)^n$ ?   Given solution:     Before collecting like terms, each term in the expansion is a product   of $n$ factors each  of which is $a, b,$ or $c$.  Those terms which   have exactly i a’s, j b’s and k c’s can be combined into   the single term $a^i b^j c^k$ and the number of such terms will be   equal to the number of n letter strings containing i a’s, j b’s and k   c’s, which by the Mississippi rule is $\frac{n!}{(i!\times j! \times  k!)}$.   I understand that this problem is actually about the derivation of the coefficient of the multinomial theorem,but I am not sure about the combinatoric reasoning,what I mean is that I understand the analogy problem but what I couldn't is how the analogy holds here,Could anybody explain this to me? How to center an element horizontally and vertically Subset data frame based on number of rows per group .NET obfuscation tools/strategy Evaluate $\mathop {\lim }\limits_{x \to 0} \left( {{1 \over {{{\sin }^2}x}} - {1 \over {{x^2}}}} \right)$ I tried l'Hospital but that will require a lot (and I mean A LOT!!!) of differentiating  Is there a shortcut? $$\mathop {\lim }\limits_{x \to 0} \left( {{1 \over {{{\sin }^2}x}} - {1 \over {{x^2}}}} \right)$$  Thanks in advance ACPI error when trying to make dual boot with Ubuntu 18.04 and Windows 10 How to scan an entire network using Nmap? Is there a way to scan an entire network using ?   What I want to do is scan my network for all the devices that are currently connected to it. Uncertainty modeling for samples I have instances of data that are composed of 6 attributes. Each attribute takes values between 1-10. Any instance has a utility calculated by some utility function:  Example: {1,1,2,5,5,2} , utility = 0.89  In some readings for these attribute values, undefined values may appear in the instance, I mark them as question mark '?'  Example: {1,1,?,5,?,2}  These question marks represents uncertainty in the instance.  How can I model this uncertainty in a statistical or mathematical way ?   Thanks Clearing all cookies with JavaScript Code coverage for wrapper class is 0% I have written below Wrapper class.  global class BI_VA_AutomationWrapper {      global class BI_VA_AutomationRequest {      webservice String userName;     webservice String emailId;     webservice String buinessUnit;     webservice String orgName;     webservice String country;     webservice String environment;     webservice String myShopNumber;     webservice String BidsId;//for testing     webservice String Type; //for testing  }  global class BI_VA_AutomationResponse {      webservice String responseStatus;     webservice String myShopNumber;     //webservice list&lt;Error&gt; errorMessages;     webservice String  errorMessages;//for testing   } global class BI_VA_Error {      webservice String type;     webservice String message;    }}   And test class is,  @isTest public class BI_VA_AutomationWrapperTestClass  { public static testMethod void unitTest(){      BI_VA_AutomationWrapper.BI_VA_AutomationRequest wrap= new BI_VA_AutomationWrapper.BI_VA_AutomationRequest();        wrap.userName='Test@test.com';     wrap.emailId='Test@test11.com';     wrap.buinessUnit='AH';     wrap.orgName='ORD';     wrap.country='US';     wrap.environment='QA';     wrap.myShopNumber='123456789';     wrap.BidsId='123';     wrap.Type='ResetPassword';              BI_VA_AutomationWrapper.BI_VA_AutomationResponse res= new BI_VA_AutomationWrapper.BI_VA_AutomationResponse();     res.responseStatus = 'Success';     res.myShopNumber = '987654321';     res.errorMessages = 'Test error';      BI_VA_AutomationWrapper.BI_VA_Error error = new BI_VA_AutomationWrapper.BI_VA_Error();     error.type = 'Test';     error.message = 'Test';  }}   Code coverage is 0% only. Limiting floats to two decimal points