Updating password of user in active Directory using C#
C#: code error while changing the active directory user's password
Build a basic Python iterator How would one create an iterative function (or iterator object) in python? Auto Increment Problem in Mysql Why would a journal accept my previous paper for publication, but reject latest manuscript due to being "outside of the journal's scope?" I submitted a manuscript last year to an IEEE journal, and the paper was accepted for publication. I recently submitted a new manuscript to the same journal, but, this time, the paper was rejected, and I was told that I should submit to another journal since my paper was not within the journal's scope. The two papers both target similar applications. Furthermore, the second paper uses some work in the first paper, making substantial improvements at solving a much more difficult problem. Based on the reviewer's comments (and my own personal assessment), I do not believe that the work presented in the second paper was "incremental" in nature. Far from it. Two of the three reviewers gave positive reviews. The third reviewer, however, mentioned the "outside of journal's scope" issue and the associate editor and editor-in-chief both sided with the 3rd reviewer, and the paper was stopped dead in its tracks.  I'm scratching my head trying to come to terms with how in the world my first paper could be accepted by this journal, but the second paper focused on the same application space and achieving substantial improvements on a much more difficult problem (backed up with measured results) could be rejected. Any ideas? The additive group $\mathbb{G}_a$ over an infinite field $k$ is not a topological group An algebraic group is not necessarily a topological group, since for an algebraic group, the product map $G \times G \to G$ is continuous when $G \times G$ is given the Zariski topology, which is finer than the product topology.   So does this tell you that the additive group $\mathbb{G}_a$ over an infinite field $k$ is not a topological group? The underlying variety of $\mathbb{G}_a \times \mathbb{G}_a$ is $\mathbb{A}_k^1 \times_k \mathbb{A}_k^1 = \mathbb{A}_k^2$, whose Zariski topology is the cofinite topology, where every open set is also open in the product topology (since $\mathbb{A}_k^1$ also has the Zariski topology). Hyphenation in compound adjectives Possible Duplicates:                In the sentence "Portland is known to be one of the most bike friendly cities in the US", is a hyphen necessary in "bike friendly"? As far as I know, the hyphen is only required when leaving it out would cause ambiguity. Wikipedia uses the example of "a small appliance factory" vs "a small-appliance factory".   As far as I can tell, the sentence above is pretty clear. A quick google search for "bike friendly city" brings up instances of both hyphenated and non-hyphenated usage. Thoughts?  Please excuse the poor formatting, sent from my iPhone. Why is January month 0 in Java Calendar? In java.util.Calendar, January is defined as month 0, not month 1. Is there any specific reason to that ?  I have seen many people getting confused about that... Can I use the Google font 'Lato' for a logo design? I am creating a logo and would like to use the Google Font 'Lato', can I use this font in my logo design or are there copyright/limitations I need to worry about? I can't find anything online other than that it is an open source font that can be used in anyway you want (maybe that's my answer but I just want to make sure). Also, is it a bad idea to use a web-based font for a logo? Prove $\sum\limits_{i=2}^{n}\frac{1}{(i-1)i}$ = $\frac{(n-1)}{n}$ using induction. I need to prove $\sum\limits_{i=2}^{n}\frac{1}{(i-1)i}$ = $\frac{(n-1)}{n}$ using induction. I am getting stuck midway through the inductive step.   Here is what I have:  $\forall n\geq 2$, where $n\in\mathbb{N}, P(n)$ is the statement "$\sum\limits_{i=2}^{n}\frac{1}{(i-1)i}$ = $\frac{(n-1)}{n}$".  Base case:  $\sum\limits_{i=2}^{2}\frac{1}{(i-1)i}$ = $\frac{1}{2}$ = $\frac{(2-1)}{2}$ so $P(2)$ is true.  Inductive assumption: Assume $P(n)$ is true. Need to show $\sum\limits_{i=2}^{n+1}\frac{1}{(i-1)i}$ = $\frac{((n+1)-1)}{(n+1)}$.  $\sum\limits_{i=2}^{n}\frac{1}{(i-1)i}$ + $\frac{1}{n^{2}}$ = $\frac{(n-1)}{n}$ + $\frac{1}{n^{2}}$  = $\frac{n(n-1) + 1}{n^{2}}$ = $\frac{n^{2}-n+1}{n^{2}}$  I got stuck here and tried working backward from $\frac{((n+1)-1)}{(n+1)}$, which gives me:  $\frac{((n+1)-1)}{(n+1)}$ = $\frac{n}{(n+1)}$   Any suggestions on what to do next or where I may be going wrong? Continuing in the Visual Studio debugger after an exception occurs When I debug a C# program and I get an exception throwed (either thrown by code OR thrown by the framework), the IDE stops and get me to the corresponding line in my code.  Everything is fine for now.  I then press "F5" to continue. From this moment, it seams like I'm in an infinite loop. The IDE always get me back to the exception line. I have to Shift + F5 (stop debugging/terminate the program) to get out of his.  I talked with some co-workers here and they told me that this happens sometime to them too.  What's happening? I have a string that contains both double-quotes and backslashes that I want to set to a variable in Python. However, whenever I try to set it, the quotes or slashes are either removed or escaped. Here's an example:  &gt;&gt;&gt; foo = 'baz "\"' &gt;&gt;&gt; foo 'baz ""'   So instead of baz "\" like I want I'm getting baz "". If I then try to escape the backslash, it doesn't help either:  &gt;&gt;&gt; foo = 'baz "\\"' &gt;&gt;&gt; foo 'baz "\\"'   Which now matches what I put in but wasn't what I originally wanted. How do you get around this problem? Do you really want to create a new tag? I have detached a process from my terminal, like this:  $ process &amp;   That terminal is now long closed, but process is still running, and I want to send some commands to that process's stdin. Is that possible? Restore Python installation to default plus apt package dependencies What is DOCTYPE? How do I create a permanent Bash alias? At the end of ‘The Return of the King’, Bilbo, Frodo, Gandalf and others boarded a boat and left Middle-earth for Valinor.  Is there any canon explanation why they couldn’t just take the ring there in the first place and leave Middle-earth altogether? Would Sauron have gone after them? Couldn’t the elves living there safeguard it?  I do realize that without Frodo taking the ring to Mount Doom there would be no story at all, but I wonder whether or not there is any canon explanation for this. I have a Microsoft XBOX 360 gamepad with the wireless adapter and I was wondering what I would have to install to use them with Ubuntu I do know that when using this on Windows I have to manually install the drivers for the wireless receiver (if that is relevant.) Can X.org Load Two Different Drivers? (Triple-Head Setup with Intel and NVidia Card) We want to set up a 3 screens with an onboard intel graphics adapter and an nvidia card.  Both cards are recognized in lspci   01:00.0 VGA compatible controller: NVIDIA Corporation GT218 [GeForce 210] (rev a2) 00:02.0 Display controller: Intel Corporation 2nd Generation Core Processor Family Integrated Graphics Controller (rev 09)   And bascially work: If I switch to the console, I see output on the Intel card, X is only running on the Nvidia card.  But when I type xrandr | grep conn I get only one connected port:   DVI-I-0 disconnected (normal left inverted right x axis y axis) VGA-0 disconnected (normal left inverted right x axis y axis) DVI-I-1 connected 1920x1200+0+0 (normal left inverted right x axis y axis) 518mm x 324mm HDMI-0 disconnected (normal left inverted right x axis y axis)   I am on Ubuntu 12.10 (and XFCE) and installed the proprietary nvidia driver.  I was assuming that X maybe has a problem to autodetect both cards, so I thought the following xorg.conf could help:   Section "ServerLayout"     Identifier     "Layout0"     Screen      0  "Screen0"     Screen      1  "Screen1"     InputDevice    "Keyboard0" "CoreKeyboard"     InputDevice    "Mouse0" "CorePointer" EndSection  Section "Files" EndSection  Section "InputDevice"     # generated from default     Identifier     "Mouse0"     Driver         "mouse"     Option         "Protocol" "auto"     Option         "Device" "/dev/psaux"     Option         "Emulate3Buttons" "no"     Option         "ZAxisMapping" "4 5" EndSection  Section "InputDevice"     # generated from default     Identifier     "Keyboard0"     Driver         "kbd" EndSection  Section "Monitor"     Identifier     "Monitor0"     VendorName     "Unknown"     ModelName      "Unknown"     HorizSync       28.0 - 33.0     VertRefresh     43.0 - 72.0     Option         "DPMS" EndSection  Section "Device"     Identifier     "Device0"     Driver         "nvidia"     VendorName     "NVIDIA Corporation"     BoardName      "GeForce 210" EndSection  Section "Screen"     Identifier     "Screen0"     Device         "Device0"     Monitor        "Monitor0"     DefaultDepth    24     SubSection     "Display"         Depth       24     EndSubSection EndSection   Section "Monitor"     Identifier     "Monitor1"     VendorName     "Unknown"     ModelName      "Unknown"     HorizSync       28.0 - 33.0     VertRefresh     43.0 - 72.0     Option         "DPMS" EndSection  Section "Device"     Identifier     "Device1"     Driver         "intel"     VendorName     "Intel"     BoardName      "Internal Intel Graphics" EndSection  Section "Screen"     Identifier     "Screen1"     Device         "Device1"     Monitor        "Monitor1"     DefaultDepth    24     SubSection     "Display"         Depth       24     EndSubSection EndSection   But it is still not working.   Is this generally possible to use two different drivers in X ? Is it necessary to create an xorg.conf manually then? (because Ubuntu 12.10 normally works without the xorg.conf) Are there any errors in my xorg.conf? How do I force a column-break in a two-column document? For all $1 \leq i < j \leq k$, the subtrees $T_i$ and $T_j$ have a vertex in common. Show that $T$ has a vertex which is in all of the $T_i$. Can someone please verify my proof or offer suggestions for improvement? I am aware that there is a similar question elsewhere, but I want help with my proof in particular.     Let $T_1, \ldots, T_k$ be subtrees of a tree $T$ such that for all $1 \leq i &lt; j \leq k$, the trees $T_i$ and $T_j$ have a vertex in common. Show that $T$ has a vertex which is in all of the $T_i$.   Suppose, for the sake of contradiction, that for every vertex $v \in V(T)$, there is a subtree $T'$ in the list $T_1, T_2, \ldots, T_k$ such that $v \notin V(T')$.  Pick $v_1 \in T_1$. There is a subtree $T^{(1)}$ such that $v_1 \notin T^{(1)}$. $T_1$ and $T^{(1)}$ have a vertex $y_1$ in common. Now, there exists a tree $T^{(2)}$ such that $y_1 \notin T^{(2)}$. But now, $T^{(2)}$ and $T^{(1)}$ have a vertex $y_2$ in common. Also, $T_1$ and $T^{(2)}$ have a vertex $y_3$ in common.  There exists a path $p_1$ in $T_1$ from $y_3$ to $y_1$. Also, there is a path $p_2$ in $T^{(1)}$ from $y_1$ to $y_2$ and a path $p_3$ in $T^{(2)}$ from $y_2$ to $y_3$. Note that the path formed by the union of $p_1$ and $p_2$ passes through $y_1$, while $p_3$ does not, since $y_1 \notin T^{(2)}$. Therefore, the union of $p_1, p_2$, and $p_3$ forms a cycle, contradicting the fact that $T$ is a tree. Therefore, it must be the case that there is a vertex $x$ in all of the $T_i$.