Projection matrix that sum to identity are orthogonal
Orthogonal projections with $\sum P_i =I$, proving that $i\ne j \Rightarrow  P_{j}P_{i}=0$
I'm reading . Then I'm not able to figure out the difference between hard voting and soft voting in context to ensemble based methods.  I quote descriptions of them from the book. The first two images from the top are description for hard voting, and last one is for soft voting.        In my view hard voting is majority decision, but I don't understand soft voting and the reason why soft voting is better than hard voting. Would anyone teach me these? What's the difference between Invoke() and BeginInvoke() Just wondering what the difference between BeginInvoke() and Invoke() are?   Mainly what each one would be used for.  EDIT: What is the difference between creating a threading object and calling invoke on that and just calling BeginInvoke() on a delegate? or are they the same thing? xdg-open on debian 9 fails to open browser I decided to try lxdm (was using fluxbox and xfce), and discovered that for many programs the url handler was failing, producing this error message;   Quite strange as you can see, it's prepending the user directory to the url.  The example here is from telegram, but it happens in discord, as well as when executing from the command line; xdg-open https://www.google.com produces a similar error. xdg-settings get default-web-browser output's firefox.desktop which works as a link in both xfce and lxdm.  More information; I ran bash -x on it and...  $ bash -x /usr/bin/xdg-open http://www.google.com + check_common_commands http://www.google.com + '[' 1 -gt 0 ']' + parm=http://www.google.com + shift + case "$parm" in + '[' 0 -gt 0 ']' + '[' -z '' ']' + unset XDG_UTILS_DEBUG_LEVEL + '[' 0 -lt 1 ']' + xdg_redirect_output=' &gt; /dev/null 2&gt; /dev/null' + '[' xhttp://www.google.com '!=' x ']' + url= + '[' 1 -gt 0 ']' + parm=http://www.google.com + shift + case "$parm" in + '[' -n '' ']' + url=http://www.google.com + '[' 0 -gt 0 ']' + '[' -z http://www.google.com ']' + detectDE + unset GREP_OPTIONS + '[' -n LXDE ']' + case "${XDG_CURRENT_DESKTOP}" in + DE=lxde + '[' xlxde = x ']' + '[' xlxde = x ']' + '[' xlxde = x ']' + '[' xlxde = xgnome ']' + '[' -f /run/user/1000/flatpak-info ']' + '[' xlxde = x ']' + DEBUG 2 'Selected DE lxde' + '[' -z '' ']' + return 0 + case "${BROWSER}" in + case "$DE" in + open_lxde http://www.google.com + pcmanfm --help -a is_file_url_or_path http://www.google.com ++ file_url_to_path http://www.google.com ++ local file=http://www.google.com ++ echo http://www.google.com ++ grep -q '^file:///' ++ echo http://www.google.com + local file=http://www.google.com + echo http://www.google.com + grep -q '^/' ++ pwd + file=/home/nesmerrill/.local/share/applications/http://www.google.com + pcmanfm /home/nesmerrill/.local/share/applications/http://www.google.com + '[' 0 -eq 0 ']' + exit_success + '[' 0 -gt 0 ']' + exit 0   The important part seems to be pcmanfm --help -a is_file_url_or_path http://www.google.com but, that command if that's how it was used, doesn't seem to do much of anything?  $ pcmanfm --help -a is_file_url_or_path http://www.google.com Usage:   pcmanfm [OPTION…] [FILE1, FILE2,...]    Help Options:   -h, --help                   Show help options   --help-all                   Show all help options   --help-gtk                   Show GTK+ Options  Application Options:   -p, --profile=PROFILE        Name of configuration profile   -d, --daemon-mode            Run PCManFM as a daemon   --no-desktop                 No function. Just to be compatible with nautilus   --desktop                    Launch desktop manager   --desktop-off                Turn off desktop manager if it's running   --desktop-pref               Open desktop preference dialog   --one-screen                 Use --desktop option only for one screen   -w, --set-wallpaper=FILE     Set desktop wallpaper from image FILE   --wallpaper-mode=MODE        Set mode of desktop wallpaper. MODE=(color|stretch|fit|crop|center|tile|screen)   --show-pref=N                Open Preferences dialog on the page N   -n, --new-win                Open new window   -f, --find-files             Open a Find Files window   --role=ROLE                  Window role for usage by window manager   --display=DISPLAY            X display to use A big thank you, Tim Post As you may have seen, this was ’s last week at Stack Overflow. To say that he has been a staple of this company for a very long time is an understatement. Tim has served , evangelist, thought leader, anthropologist, psychologist, and mediator for this site for almost as long as it’s been around. For me personally, Tim has been my right hand in this role. He has educated me on the history of our community, been the person I’ve gone to to ask “why is this thing this way?“, bounced ideas off of, collaborated with, and relied on as a steady voice and a thoughtful teammate. We’re all proud to send him off and support him in any way we can in his next role. Tim has agreed to meet with me once a month to talk these things through moving forward and I’m excited to continue getting his feedback on our initiatives and goals. Thank you, Tim, for being such a stalwart here for so long. We’ll miss you, and look forward to sneaking you into our “Among Us” games for as long as we can. We're also excited to see you request to rejoin the ranks of Stack Overflow moderators and continue to be a pillar in the community. Can a Mac be used by different users at the same time?  I mean to say that we have a single Mac but 3 users. Can they access the same machine remotely at the same time for developing applications on iPhone or Mac?  Does Mac OS X server allow us to do this?  If so, how do I configure it for that? Attacks of the MAC construction $\mathcal{H}(m\mathbin\|k)$ for common hashes $\mathcal{H}$? Consider a common practically-collision-resistant hash function $\mathcal{H}$ (e.g. , , , ), perhaps based on the  as are the first three. We define a  $\mathcal{C}$ $$(k,m) \mapsto \mathcal{C}(k,m)=\mathcal{H}(m\mathbin\|k)$$ where $\mathbin\|$ denotes concatenation, $k$ is a secret key (constant, or at least of fixed size), and $m$ is a message (possibly of variable length). Assume that an adversary can (iteratively) submit queries with $m_j$ and obtain $C(k,m_j)$, and wants to obtain $k$ or otherwise compute $C(k,m)$ for some $m\ne m_j$.  That MAC $\mathcal{C}$ is not trivially bad. In particular, if $\mathcal{H}$ was indistinguishable from a random function in the , $\mathcal{C}$ would be secure. And even though $\mathcal{H}$ may have the , it does not turn into a devastating attack on $\mathcal{C}$.  The less impractical generic attack that I see is that if a collision was known for $\mathcal{H}$ with the colliding messages of moderate identical length, it could be deduced countless collisions for $\mathcal{C}$. Hence security is demonstrably not better than collision-resistance of $\mathcal{H}$ (for identical-length messages). We could assume that $k$ is half the size of the result of $\mathcal{H}$, and hope that the security is about  or  or , 280, 2128, 2256 hash rounds for SHA-1, RIPEMD-160, SHA-256, SHA-512.   What are the known attacks against $\mathcal{C}$ (better than the above), and their cost, for each of these common hashes?  Is there hope for an argument that an attack against $\mathcal{C}$ would turn into an attack of similar cost against $\mathcal{H}$, or hint of the contrary?  Update: this  is of interest, but I fail to find that it really answers the present question.  Update 2: I am aware that the construction considered is weaker than , and in particular is vulnerable to collision on $\mathcal H$; I stated that, and that it is thus pointless to have the key wider hopeless to target security against some attacks better than half the hash's size. I'm asking exactly what cryptanalytic attack better than finding a collision on $\mathcal H$ there are. There is room for such an attack only by exploiting a weakness in the structure or/and the round function of a concrete $\mathcal H$. Hydrostatic pressure - doesn't density vary with depth? Our class is learning about hydrostatic water pressure and we have been told that we can calculate the force of the liquid on an object at any depth using "the density x 9.8 x the depth". However, as the depth increases, wouldn't the density of the liquid increase because of the weight of the liquid above it compressing it? So should't there be something in the equation to account for the varying density? To me, "density x 9.8 x depth" seems like it is saying that the density will be constant... I can't seem to access instance members of the surrounding class from inside an enum, as I could from inside an inner class. Does that mean enums are static? Is there any access to the scope of the surrounding class's instance, or do I have to pass the instance into the enum's method where I need it?  public class Universe {     public final int theAnswer;      public enum Planet {         // ...         EARTH(...);         // ...          // ... constructor etc.          public int deepThought() {             // -&gt; "No enclosing instance of type 'Universe' is accessible in this scope"             return Universe.this.theAnswer;         }     }      public Universe(int locallyUniversalAnswer) {         this.theAnswer = locallyUniversalAnswer;     } } This is just a curiosity. I have come across multiple proofs of the fact that there are infinitely many primes, some of them were quite trivial, but some others were really, really fancy. I'll show you what proofs I have and I'd like to know more because I think it's cool to see that something can be proved in so many different ways.  Proof 1 : Euclid's. If there are finitely many primes then $p_1 p_2 ... p_n + 1$ is coprime to all of these guys. This is the basic idea in most proofs : generate a number coprime to all previous primes.  Proof 2 : Consider the sequence $a_n = 2^{2^n} + 1$. We have that  $$ 2^{2^n}-1 = (2^{2^1} - 1) \prod_{m=1}^{n-1} (2^{2^m}+1), $$  so that for $m &lt; n$, $(2^{2^m} + 1, 2^{2^n} + 1) \, | \, (2^{2^n}-1, 2^{2^n} +1) = 1$. Since we have an infinite sequence of numbers coprime in pairs, at least one prime number must divide each one of them and they are all distinct primes, thus giving an infinity of them.  Proof 3 : (Note : I particularly like this one.) Define a topology on $\mathbb Z$ in the following way : a set $\mathscr N$ of integers is said to be open if for every $n \in \mathscr N$ there is an arithmetic progression $\mathscr A$ such that $n \in \mathscr A \subseteq \mathscr N$. This can easily be proven to define a topology on $\mathbb Z$. Note that under this topology arithmetic progressions are open and closed. Supposing there are finitely many primes, notice that this means that the set $$ \mathscr U \,\,\,\, \overset{def}{=} \,\,\, \bigcup_{p} \,\, p \mathbb Z $$ should be open and closed, but by the fundamental theorem of arithmetic, its complement in $\mathbb Z$ is the set $\{ -1, 1 \}$, which is not open, thus giving a contradiction.   Proof 4 : Let $a,b$ be coprime integers and $c &gt; 0$. There exists $x$ such that $(a+bx, c) = 1$. To see this, choose $x$ such that $a+bx \not\equiv 0 \, \mathrm{mod}$ $p_i$ for all primes $p_i$ dividing $c$. If $a \equiv 0 \, \mathrm{mod}$ $p_i$, since $a$ and $b$ are coprime, $b$ has an inverse mod $p_i$, call it $\overline{b}$. Choosing $x \equiv \overline{b} \, \mathrm{mod}$ $p_i$, you are done. If $a \not\equiv 0 \, \mathrm{mod}$ $p_i$, then choosing $x \equiv 0 \, \mathrm{mod}$ $p_i$ works fine. Find $x$ using the Chinese Remainder Theorem.   Now assuming there are finitely many primes, let $c$ be the product of all of them. Our construction generates an integer coprime to $c$, giving a contradiction to the fundamental theorem of arithmetic.  Proof 5 : Dirichlet's theorem on arithmetic progressions (just so that you not bring it up as an example...)  Do you have any other nice proofs? If $M$ isn't Noetherian, $M$ has a submodule maximal with respect to being not finitely generated. I'm answering this question:     Let $M$  be a module. Show that if  $M$ is not Noetherian then $M$ has a submodule $N$ such that $N$ is not finitely generated but $A$ is finitely generated whenever $N &lt; A ≤ M$.   The solution starts as follows (I'm following a given solution): Let $Y$ be the set of non-finitely-generated submodules of $M$. If $(N_α)$ is a chain in $Y$ then $Q= \bigcup_αN_α $ is in $Y$, else $Q$ is f.g., which implies $Q=N_α$ for some $α$, whence $N_α$  is f.g., contradiction.  I don't understand why if $Q$ is f.g., then $Q=N_α$ for some $α$? Am I missing some fact?  Thanks! Proving $1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$ using induction How can I prove that     $$1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$$    for all $n \in \mathbb{N}$? I am looking for a proof using mathematical induction.  Thanks Need hint: Determine whether the series $\sum_{n=1}^{\infty}(\frac{n}{n+1})^{n^2}$ is convergent or divergent. How to prove floor identities? Are there any series whose convergence is unknown? Are there any infinite series about which we don't know whether it converges or not? Or are the convergence tests exhaustive, so that in the hands of a competent mathematician any series will eventually be shown to converge or diverge? EDIT: People were kind enough to point out that without imposing restrictions on the terms it's trivial to find such &quot;open problem&quot; sequences. So, to clarify, what I had in mind were sequences whose terms are composed of &quot;simple&quot; functions, the kind you would find in an introductory calculus text: exponential, factorial, etc. Why does `find` not delete the files it found? I'm trying to understand why the find command is not deleting files in one particular case. I've got the following command:  find C:\\path\\to\\Pictures\\pmcctv -name cap_*.jpg -o -name cap_*.ogg -o -name cap_*.flv -o -name cap_*.mp4 -o -name cap_*.webm -mtime +7   Which returns the files I would expect:  C:\path\to/Pictures/pmcctv/cap_20160915T193251_620067800.jpg C:\path\to/Pictures/pmcctv/cap_20160915T193322_742708800.jpg C:\path\to/Pictures/pmcctv/cap_20160916T081046_394767500.jpg C:\path\to/Pictures/pmcctv/cap_20160916T081125_615129600.jpg C:\path\to/Pictures/pmcctv/cap_20160916T081129_503678200.jpg C:\path\to/Pictures/pmcctv/cap_20160916T081255_842394600.jpg C:\path\to/Pictures/pmcctv/cap_20160916T161008_693586800.jpg C:\path\to/Pictures/pmcctv/cap_20160916T161012_749396800.jpg C:\path\to/Pictures/pmcctv/cap_20160916T161043_774132200.jpg C:\path\to/Pictures/pmcctv/cap_20160916T161150_497251900.jpg C:\path\to/Pictures/pmcctv/cap_20160916T161154_437379600.jpg C:\path\to/Pictures/pmcctv/cap_20160916T161337_350955300.jpg C:\path\to/Pictures/pmcctv/cap_20160916T161341_452596000.jpg C:\path\to/Pictures/pmcctv/cap_20160916T161412_870774000.jpg C:\path\to/Pictures/pmcctv/cap_20160916T161443_969064400.jpg C:\path\to/Pictures/pmcctv/cap_20160916T161554_468109900.jpg C:\path\to/Pictures/pmcctv/cap_20160916T161558_378086000.jpg C:\path\to/Pictures/pmcctv/cap_20160916T161916_668433100.jpg   Now if I try to run the same command with -delete, the files are not deleted:  find C:\\path\\to\\Pictures\\pmcctv -name cap_*.jpg -o -name cap_*.ogg -o -name cap_*.flv -o -name cap_*.mp4 -o -name cap_*.webm -mtime +7 -delete   I've also tried with -exec rm, but they are also not deleted:  find C:\\path\\to\\Pictures\\pmcctv -name cap_*.jpg -o -name cap_*.ogg -o -name cap_*.flv -o -name cap_*.mp4 -o -name cap_*.webm -mtime +7 -exec rm {} \;   find returns no error and the files are owned by me with permissions 644.  Any idea what could be causing this?  (note: I've cut the commands to make it more readable but normally there's no line break) If the Universe is expanding, how is that galaxies run into each other? As the title says. If everything is moving outward, thrown by the Big Bang, how could two Galaxies end up with velocity vectors that run them into each other? How do I wrap text with no whitespace inside a ? Remove HTML tags from a String "Results found containing ___" too close to header I have a shape (rounded rectangle) in PowerPoint 2007 of which I would like to increase the vertical size only (to illustrate de-bottlenecking). So, the width will stay the same but the height will grow (both from the top and bottom, so the centre will remain in the same location).  Custom Animation -> Emphasis only appears to change the whole shape.  Is it possible to increase only the vertical size (stretch the shape)?  This is during animation of the shape.