I've 8 bytes before encryption (for example: 01 02 03 04 05 06 07 08) and 8 bytes after DES encryption: 08 07 06 06 05 04 03 02 01. Is it possible to calculate the key?
I know DES is outdated, but will is is secure in CBC mode? Can anyone help me understand why not?
How to update to the latest numpy and scipy on Ubuntu 14.04LTS On my Ubuntu 14.04LTS, for python3, I have   &gt;&gt;&gt; import numpy  &gt;&gt;&gt; import scipy  &gt;&gt;&gt; numpy.__version__  '1.8.2'  &gt;&gt;&gt; scipy.__version__  '0.13.3'   I want to update numpy and scipy to the most , I tried with the following commands  sudo pip3 install --upgrade numpy sudo pip3 install --upgrade scipy   both commands ran successfully, but the version numbers didn't change to 1.9.2 and 0.16.0 at all for both python packages (after restarting the computer). Anyone knows how to update to the most version? Thank you very much!! How to show that the set of orthogonal n x n matrices forms a group under multiplication I am studying orthogonal matrices and I am not sure if to show if a set of orthogonal $n \times n$ matrices forms a group under multiplication. We must check each of the group axioms. I found that the axioms are:  Closure Associativity Existence of identity matrix Existence of the inverse matrix.  I edited my question, since I was able to find more information about this topic. This group is called $O(n)$. To check the four axioms I did: Let $A \text{ and } B \in O(n)$, denoted as orthogonal matrices and assume that $C=AB$, then, Closure :  To prove that $C \in  O(n)$ we must prove that $C$ is a real $n \times n$ orthogonal matrix with uni-modular determinant. Since A and B are real $n \times n$ matrices, $C$ is also a real $n \times n$ matrix so, $C^TC=(AB)^T AB=B^T A^T AB = B^TB=I$  Associativity :  Matrix multiplication is associative, so the law holds for $O(n)$ group elements.  I am not sure if this is enough to prove associativity. Identity element :  The $n \times n$ identity matrix $I_{n \times n}$ represents the identity element.  In this case I am not sure if this is enough to prove the identity element. Inverse element :  Let $A^{-1}$ be the inverse of $A$, then we need to prove that  $A^{-1} \in  O(n)$ since $(A^{-1})^T=(A^T)^{-1}$. We have that: $(A^{-1})^T A^{-1}=(A^T)^{-1} A ^{-1}=(AA^T)^{-1}=I^{-1}=I$  Can anyone check if what I did is correct? I also would like to know if I can prove the associativity and the identity element in a better way. thanks Is there difference in force when the height increases? Consider a man with mass 50kg.   When he jumps from a 2 meter height, the total force is $F=ma$ ie (mass * gravity), but nothing happens to him. When he jumps from a 50 meter height, the total force is (mass * gravity ) same, but he dies.   Why is this happening? $\sum_{n=-\infty}^{\infty}\frac{1}{(u +n)^2}=\frac{\pi^2}{(\sin \pi u)^2}$ I've already see a  by Marko Riedel which I list it follows:     The standard way to treat these sums is to integrate   $$ f(z) = \frac{1}{(z+\alpha)^2} \pi \cot(\pi z)$$   along a contour consisting of a circle of radius $R$ and with $R$ going to infinity and hence being larger than $\alpha$, where the circle does not pass through the poles on the real axis.   Now along the semicircle in the upper half plane we have   $$|f(z)| \le \frac{1}{(R-|\alpha|)^2}\pi \left|\frac{e^{i\pi R\exp(i\theta)} + e^{-i\pi R\exp(i\theta)}} {e^{i\pi R\exp(i\theta)} - e^{-i\pi R\exp(i\theta)}}\right|= \frac{1}{(R-|\alpha|)^2} \pi  \left|\frac{e^{2i\pi R\exp(i\theta)}+1}{e^{2i\pi R\exp(i\theta)}-1}\right| &lt; \frac{1}{(R-|\alpha|)^2} \pi   \frac{1+e^{-2\pi R\sin(\theta)}e^{2i\pi R\cos(\theta)}} {1-e^{-2\pi R\sin(\theta)}e^{2i\pi R\cos(\theta)}}$$   This last term is clearly $O(1/R^2)$ as $R$ goes to infinity as the quotient of the two exponentials goes to one since $\exp(-R)$ vanishes and there is no singularity when $\theta = 0$ or $\theta = \pi$ as $R\cos\theta = \pm R$, which is not an integer by the assumption that the circle avoids the poles and hence cannot be one.     My question: It is clear that for each $z=Re^{i \theta}$ , $f(z)=O(1/R^2)$ ，but can this guarantee that the last term on RHS is uniform bounded ? Since $\theta$ varies on a compact interval $[0, \pi]$ , I want to show that for each $ \theta \in [0, \pi]$ , and a fixed positive number $M \gt 1 $ there exist an open ball contains $\theta$ and a fixed $R_{\theta}\gt 0$ , for every $R \ge R_{\theta}$ and $x \in $ the open ball $$\frac{1+e^{-2\pi R\sin(x)}e^{2i\pi R\cos(x)}} {1-e^{-2\pi R\sin(x)}e^{2i\pi R\cos(x)}} \le M$$  For $\theta \neq 0,\pi$ , it is easy to find the desired $R_{\theta}$ and the open ball , but if $\theta = 0$ or $ \theta = \pi$ , for every open ball containing $\theta$ , it behave erratically near $\theta$ , I have no idea how to deal with this . I have reviewed this:   but it does not answer my question. The suggested fixes in this question is to set up FB, IG, and other social media pages. We have had these for months. We have 576 Instagram followers, 293 followers on Facebook, 17,425 views on Vimeo, Youtube, Pinterest and LinkedIn presence. Here is the response in a YouTube search:  Let's be clear. If you search for my company name and product name YT changes the spelling to something that doesn't exist and suggests a video with no matching result terms (using either spelling) disparaging OTHER luxury handbags, while insinuating it is referring to my handbags. When a user searches for our business, , Google and Youtube changes the spelling to Aslan Vaughn and displays these results:  For the first few months it just seemed like it was taking some time to percolate, but over 7 months we have had 17,500 unique visitors and nearly 90 backlinks.  The only results for Aslan Vaugn is the birth announcement of the second child of a musician in New Jersey. It is a Facebook post and has 14 likes. Google is favoring a spelling that has only two results with 14+ views over dozens of links and tens of thousands of views. How is this possible? An FB post with 14 likes is that much more popular than a business with 17,500 unique visitors (despite the Google obfuscation)? Is my company on some list? On other search engines my company covers the first few pages in results. Through Google, you cannot find our company no matter how many pages in you go. Unless you search for the term in quotes, . In which case we cover the first few pages here as well: Is it possible to activate an iPhone 4 without a SIM? How to properly secure a wireless home network I'm soon going to be setting up a wireless network for me and some buddies and (hopefully) making it secure. I have some questions and concerns regarding this and figured someone here could give me a hand and steer me in the right direction.   I have a WRT55AG (version 2) loaded with the factory firmware. I've heard that it's possible to load your own firmware onto routers like these. Is it worth it? What can be achieved with it? I was wondering what kind of security I should use on my wireless. Ideally I would want all devices to be 'accepted', but I'm not sure how to accomplish this. I'm guessing it has something to do with MAC addresses, but I don't know the first thing about them. It would be great that ALONG WITH encryption (I'm told WPA2 is the best), every device would need to be added to a list of some sort before even being able to access the AP. Another feature I was thinking of would be completely hiding my wireless from showing up on other people's computer. Is this possible (like an SSID broadcast thing). Is it worth the trouble? One more option I was looking to implement would be bandwidth calculators. Ideally something in the router to measure the amount of bandwidth being used and how much has been used for that month (this has to do with bills and limits, obviously). Is it possible to throttle down speed when a certain limit is reached?   These are alot of points but I was hoping someone who has more knowledge than me in this subject could give me a hand and a few pointers. Hopefully all this is possible and I'm not dreaming :) Why Are There So Many Double Initials in Comic Books? Not being a comics expert, I can't be sure, but I think this applies mostly to Marvel comics.  Why do so many characters have double initials?  There's Pepper Potts, Reed Richards, Sue Storm, Peter Parker and I think at some times there was Bruce Banner (when it wasn't David Banner).  Is this an inside joke or is there some reasoning for this? Inbox notification and reputation score is misaligned in mobile web view The inbox notification indicator and reputation score indicator are misaligned in the mobile web view. Which site deals with career development questions for programmers? Is there any site in the network where career development questions about the programming profession can be asked ? I thought Programmer's was the site for it, but that doesn't appear to be the case. Email from IOS with attachments larger than 7.5Mb gets stuck in outbox and constantly tries to resend. Costs us hundreds in data fees Diamond moderator rollbacks can resurrect blacklisted tags I want to pose a question such as:  What is your name? Joe  How would I accomplish this using Console.WriteLine to also wait for the response on that same line instead of it being broken into:  What is your name? Joe How to clone ArrayList and also clone its contents? How can I clone an ArrayList and also clone its items in Java?  For example I have:  ArrayList&lt;Dog&gt; dogs = getDogs(); ArrayList&lt;Dog&gt; clonedList = ....something to do with dogs....   And I would expect that objects in clonedList are not the same as in dogs list. As opposed to the generic polynomial form for utilizing the Eisenstein Criterion ($a_nx^n+a_{n-1}x^{n-1}+\dots+a_0\in\mathbb{Z}[x]$ is irreducible in $\mathbb{Q}$) how do we prove that if $p$ is a prime, $x^{p-1}+x^{p-2}+\dots+x+1$ is irreducible over $\mathbb{Q}$? Linear Algebra: If $A^3 = I$, does $A$ have to be $I$? How to prove $n^5 - n$ is divisible by $30$ without reduction I have recently a purchased Lenovo Thinkpad E480 laptop, I have installed Ubuntu 18.04 on it, everything is working fine except am not able to connect to network via WiFi. Please help.  I followed this , but am unable to solve the issue.  Below are the outputs of some commands.  lsb_release -a  No LSB modules are available. Distributor ID: Ubuntu Description:    Ubuntu 18.04.1 LTS Release:    18.04 Codename:   bionic   uname -r  4.15.0-33-generic   rfkill list all  0: tpacpi_bluetooth_sw: Bluetooth     Soft blocked: no     Hard blocked: no 1: hci0: Bluetooth     Soft blocked: no     Hard blocked: no   lspci -nnk | grep 0280 -A3  05:00.0 Network controller [0280]: Realtek Semiconductor Co., Ltd. RTL8821CE 802.11ac PCIe Wireless Network Adapter [10ec:c821]     Subsystem: Lenovo RTL8821CE 802.11ac PCIe Wireless Network Adapter [17aa:c024] 06:00.0 SD Host controller [0805]: O2 Micro, Inc. SD/MMC Card Reader Controller [1217:8621] (rev 01)     Subsystem: Lenovo SD/MMC Card Reader Controller [17aa:5068]   sudo lshw -C network   *-network                         description: Ethernet interface        product: RTL8111/8168/8411 PCI Express Gigabit Ethernet Controller        vendor: Realtek Semiconductor Co., Ltd.        physical id: 0        bus info: pci@0000:03:00.0        logical name: enp3s0        version: 10        serial: 8c:16:45:a3:d9:f1        size: 100Mbit/s        capacity: 1Gbit/s        width: 64 bits        clock: 33MHz        capabilities: pm msi pciexpress msix vpd bus_master cap_list ethernet physical tp mii 10bt 10bt-fd 100bt 100bt-fd 1000bt 1000bt-fd autonegotiation        configuration: autonegotiation=on broadcast=yes driver=r8169 driverversion=2.3LK-NAPI duplex=full firmware=rtl8168g-3_0.0.1 04/23/13 ip=192.168.2.2 latency=0 link=yes multicast=yes port=MII speed=100Mbit/s        resources: irq:16 ioport:d000(size=256) memory:f1204000-f1204fff memory:f1200000-f1203fff   *-network UNCLAIMED        description: Network controller        product: Realtek Semiconductor Co., Ltd.        vendor: Realtek Semiconductor Co., Ltd.        physical id: 0        bus info: pci@0000:05:00.0        version: 00        width: 64 bits        clock: 33MHz        capabilities: pm msi pciexpress cap_list        configuration: latency=0        resources: ioport:c000(size=256) memory:f1100000-f110ffff   It seems I have to clone, compile and install the required WiFi driver, please guide me which WiFi driver to use and how to install (with steps if possible).  P.S.: Let me know if any further information is needed to solve this issue, I will add the required info (if any) in the question as per the guidance given. Are antiparticles just particle-shaped holes? If particles are simply regions of space where certain quantum fields have non-zero divergence, are anti-particles simply the corresponding regions of opposite divergence?    This seems like the intuitive answer, especially when considering the process of annihilation.  I have heard before that anti-particles are analogous to particles moving through time in reverse, which seems to indicate that an annihilation event is just a point of symmetry in that (arguably, single) particle's history.  This analogy breaks down, however, when it comes to gravitation, since there seems to be no evidence of a negative-mass particle.    So what then is the meaning of an anti-particle, when their symmetry is preserved across certain fields, but not other? How use spaces instead of underscores in file names? I'm referring to this "feature"   It was not happening before, but it is now, and it is messing up the structure of some folders I previously created. Is there a way to disable it?