Can any cookies collect my entire browsing history when they're dropped? I use CCleaner to remove all cookies after I close browser, exceptions are the sites I have accounts.
When I visit a website it sometimes puts cookies on my browser. Usually cookies keep track of browsing info from that particular site. If I don't delete those cookies and browse other websites, will the cookies be able to know which site I am browsing? Also, can any website read browsing history which is saved by browsers? Can any website know about my browsing history by help of cookies, browsing histories saved by browsers, or by any scripting language e.g. javascript?
Mankind prepares for a war with aliens by warring among itself I am looking for the title or author of a atory I read back in the 90's, no idea when it was written. The plot goes something like this: Humanity has gained access to space travel and sends 9 or 12 survey ships towards galactic center, only 1 returns with stories of an enemy so rapacious that humanity starts preparing for war. After a long time, the enemy hasn't shown up and people begin to doubt the need to keep building for war. The government engineers a plan to turn the core systems against the fringe worlds and start warring amongst ourselves so that we may be prepared for when the real enemy shows up.  Written with the focus on one spaceship captain battling in our own internal war. I can't solve this problem:  Suppose $f$ is polynomial function of even degree $n$ with always $f\geq0$.  Prove that $f+f'+f''+\cdots+f^{(n)}\geq 0$. Using divisibility theorems, prove that if $\gcd(a,b)=1$ and $a|c$ and $b|c$, then $ab|c$.   This is pretty clear by UPF, but I'm having some trouble proving it using divisibility theorems. I was looking at , and if this could be proven, then the solution would follow immediately. Add lines to cron from script Is there a way to add lines to a user's cron via script?  I usually do it using crontab -e, but I would like to automate this task with a shell script. Creating a Snapshot of a Windows 7 System Is there a way to create a snapshot of a Windows 7 system?  My goal is to be able to save the current state of my Windows 7 OS, on external storage, and be able to restore it on the same hardware in the future. Much like a snapshot on a virtual machine.   One solution would be to install two hard drives in RAID 1, then take one of them out an store it. Is there an easier solution? So basically as it stands I have a laptop which has no cd/dvd drive, and I don't have a usb drive.  The laptop has windows 7 installed with ubuntu 11.04 installed through wubi.  What I want to do is remove windows completely, and make ubuntu the only OS installed on the system.  Is there a way to do this without re-installing ubuntu? (i.e. can I take my wubi install away from windows?)  or is there a way to from inside ubuntu have it run the ubuntu iso somehow so I can just wipe the system and install it fresh? (even if it means I need to have an e.g. 2gb partition just for the image to reside in). How to set vertex weights using Blender's Python API? Looking at the API docs I found , but It cannot be called in Edit Mode. When I try to call it, I always get:     RuntimeError: VertexGroup.add(): cannot be called while object is in edit mode.   The buttons in the vertex group list use bpy.ops.objects.vertex_group_assign(), but to work with it you have to manipulate selections, this is not nice to do from a script.    Why I try to use VertexGroup.add() in Edit Mode? My script works this way:  Human weight paint a bone, "Arm_1.L", "Arm_1.L" shares vertices with "Arm_2.L" in this case they share half the vertices (4 bone arm rig, trick to have nice deform in certain movements). When script run, all selected vertices that exist in both groups (Arm_1.L and Arm_2.L) will copy the weights of Arm_1.L into Arm_2.L but inverted. This way, I only need to weight paint "Arm_1.L" and "Arm_3.L", the script will fill the weights for "Arm_2.L".  import bpy  # ensure we are in edit mode bpy.ops.object.mode_set(mode="EDIT")  # configurable parameters src_group = "Arm_03.L" dst_group = "Arm_02.L"  obj = bpy.context.edit_object src_group_index = obj.vertex_groups[src_group].index dst_group_index = obj.vertex_groups[dst_group].index  mesh = obj.data sel_verts = [v for v in mesh.vertices if v.select] #sel_verts = mesh.vertices  # SEGFAULT in this line bpy.ops.object.mode_set(mode="OBJECT")  print("Working")  for v in sel_verts:     v1in = False     v2in = False     for g in v.groups:         if g.group == src_group_index:             v1in = True         if g.group == dst_group_index:             v2in = True     if v1in and v2in:         print("True")         obj.vertex_groups[dst_group].add([v.index], 1.0 - obj.vertex_groups[src_group].weight(v.index), 'REPLACE')  print("Finished") How to remove trailing whitespaces with sed? I need a bit of help with this.      She is a photographer but that camera is her brother's-in-law     or:        She is a photographer but that camera is her brother-in-law's   I am aware that I could change the construction to make things easier, like:      She is a photographer but that camera belongs to her brother-in-law       However, I would really like to find out what's the correct way to write the sentence. I'm having some issues SSH'ing to a server when logged into a particular VM.  From my local PC, I can log in to the server just fine using the command ssh -i .ssh/[my private key] username@domain.  If I then SSH into another unrelated server which also has the same private key and run the above command it will also log in just fine in to the server.  But, when I SSH into a particular OpenStack VM (again has the same private key) on the unrelated server and then SSH in to the server from there I start getting prompted for a passphrase. But a passphrase hasn't been set for the key so inevitably I get a "Permission denied (publickey)." error.  To double check, I removed the private key from the VM and scp'd the private key from my PC into it again and tried the same command I still get prompted for a passphrase.  In other words:   PC - SSH - Server1 = Works fine  PC - SSH - Server2 - SSH - Server1 =     Works fine  PC - SSH - Server2 - SSH - Virtual Machine - SSH - Server 1 =         Asks for a passphrase   What could be the reason behind this? It's definitely the right private key I'm using and it has the appropriate permissions.  Edit - Output of ssh -v -i .ssh/[my private key] username@domain as requested  OpenSSH_7.4p1, OpenSSL 1.0.2k-fips  26 Jan 2017 debug1: Reading configuration data /etc/ssh/ssh_config debug1: /etc/ssh/ssh_config line 58: Applying options for * debug1: Connecting to xxx.xxx.xxx.xxx [xxx.xxx.xxx.xxx] port 22. debug1: Connection established. debug1: key_load_public: No such file or directory debug1: identity file /.ssh/[my private key].ppk type -1 debug1: key_load_public: No such file or directory debug1: identity file /.ssh/[my private key].ppk-cert type -1 debug1: Enabling compatibility mode for protocol 2.0 debug1: Local version string SSH-2.0-OpenSSH_7.4 debug1: Remote protocol version 2.0, remote software version OpenSSH_7.2p2 Ubuntu-4ubuntu2.4 debug1: match: OpenSSH_7.2p2 Ubuntu-4ubuntu2.4 pat OpenSSH* compat 0x04000000 debug1: Authenticating to xxx.xxx.xxx.xxx:22 as '[my username]' debug1: SSH2_MSG_KEXINIT sent debug1: SSH2_MSG_KEXINIT received debug1: kex: algorithm: curve25519-sha256@libssh.org debug1: kex: host key algorithm: ecdsa-sha2-nistp256 debug1: kex: server-&gt;client cipher: chacha20-poly1305@openssh.com MAC: &lt;implicit&gt; compression: none debug1: kex: client-&gt;server cipher: chacha20-poly1305@openssh.com MAC: &lt;implicit&gt; compression: none debug1: kex: curve25519-sha256@libssh.org need=64 dh_need=64 debug1: kex: curve25519-sha256@libssh.org need=64 dh_need=64 debug1: expecting SSH2_MSG_KEX_ECDH_REPLY debug1: Server host key: ecdsa-sha2-nistp256 SHA256:ZiugTjR5fM0E3evOwHoePFKspDQChA0Ab6L0q88KP/g debug1: Host 'xxx.xxx.xxx.xxx' is known and matches the ECDSA host key. debug1: Found key in /home/centos/.ssh/known_hosts:2 debug1: rekey after 134217728 blocks debug1: SSH2_MSG_NEWKEYS sent debug1: expecting SSH2_MSG_NEWKEYS debug1: SSH2_MSG_NEWKEYS received debug1: rekey after 134217728 blocks debug1: SSH2_MSG_EXT_INFO received debug1: kex_input_ext_info: server-sig-algs=&lt;rsa-sha2-256,rsa-sha2-512&gt; debug1: SSH2_MSG_SERVICE_ACCEPT received debug1: Authentications that can continue: publickey debug1: Next authentication method: publickey debug1: Trying private key: /.ssh/[my private key].ppk Enter passphrase for key '/.ssh/[my private key].ppk': Enter passphrase for key '/.ssh/[my private key].ppk': Enter passphrase for key '/.ssh/[my private key].ppk': debug1: No more authentication methods to try. Permission denied (publickey). How to remove session entries from LightDM? How compatible are my old Nikon lenses? I'm looking at buying a new SLR body. I have an old film Nikon entry-level SLR (about ten years old) with a kit lens. If I buy a Nikon body, how likely is it that my old lens will work on it? Would I be better off buying a new lens? What does the number next to the review link in the top bar actually mean? Show that  $$\frac{\sin x}{\cos 3x}+\frac{\sin 3x}{\cos 9x}+\frac{\sin 9x}{\cos 27x} = \frac{1}{2}\left(\tan 27x-\tan x\right)$$ This is more of a panic question after reading reviews on Istanbul Atatürk Airport. They said it's messy and hard to navigate, therefore I'm asking questions here in name of my girlfriend who will need to transit through this airport.  My girlfriend (Canadian national) traveling from Toronto (YYZ) to Istanbul (IST) for a couple hours (~4 hrs, so transiting) in order to come to Pristina (PRN). I would like a quick guideline on how to pass through this airport seamlessly without facing any trouble or delay.  Her tickets are both with Turkish Airlines. She purchased it via a third party (Flighthub.ca). It's only one booking and it includes both flights in it.   Once my girlfriend departs, she's obviously required to follow the signs and go to the transit area. Once she's there, will she be required to present a ticket? Where does she get that ticket? Does she get it in YYZ or does she need to visit a counter in IST and get it? After she's got the proper ticket, what does she need to do exactly? Wait in line to have her passport checked? Will they also control her carryon?  After this step, does she simply go to the PRN flight gate and that's it?   We're simply worried about Istanbul Atatürk Airport because of the recent things that happened in Turkey (coup attempt and ISIS attack) and she doesn't wanna somehow end up locked up there. I've read that the airport is a mess lately and a lot of people will be there waiting in line. So, another question: Are 4 hours enough?  I would appreciate a step by step guideline on how she needs to navigate and what she needs to do in that airport to have a smooth experience without facing any delays. Also some misc information on Istanbul Atatürk Airport would be appreciated. Save entire process for continuation after reboot We can say ‘fried fish’, ‘baked potato’ or ‘minced pork’ using past participles for modifiers.  However, ‘roast’ is different - either ‘roast duck’ or ‘roasted duck’ works, it seems to me.  How should we analyze this?  Is ‘roast’ a noun modifying a noun here, or is it a different form of the past participle?  (According to freedictionary.com, in Middle English the past participle was ‘roste’ – does that mean that this was once a strong verb?) Ball motion with air resistance coupled differential equation for fourth-order Runge-Kutta I've created a MATLAB function for solving coupled differential equation with the fourth-order Runge-Kutta method based on what is provided  (Simultaneous Equations of First Order). Here the function:  function [y] = rk4_c(f, g, h, x, y, z, n) % Runge-Kutta % Implementation of the fourth-order method for coupled equations % h = dt % x is the time here  for ii=1:(n-1)     k1 = h * f(x(ii), y(ii), z(ii));     l1 = h * g(x(ii), y(ii), z(ii));     k2 = h * f(x(ii) + 0.5*h, y(ii) + 0.5*k1, z(ii) + 0.5*l1);     l2 = h * g(x(ii) + 0.5*h, y(ii) + 0.5*k1, z(ii) + 0.5*l1);     k3 = h * f(x(ii) + 0.5*h, y(ii) + 0.5*k2, z(ii) + 0.5*l2);     l3 = h * g(x(ii) + 0.5*h, y(ii) + 0.5*k2, z(ii) + 0.5*l2);     k4 = h * f(x(ii) + h, y(ii) + k3, z(ii) + l3);     l4 = h * g(x(ii) + h, y(ii) + k3, z(ii) + l3);     y(ii+1) = y(ii) + (1/6)*(k1 + 2*k2 + 2*k3 + k4);     z(ii+1) = z(ii) + (1/6)*(l1 + 2*l2 + 2*l3 + l4); end   I would like to use this function for solving motion equations for a ball with air resistance. So starting with the basic equations (I hope they are correct): $$md^2x/dt^2 = F_{D_x}$$ and $$md^2y/dt^2 = F_{D_y}-mg$$ with $$F_{D_x} = -Dvvcos(\theta)$$ and $$F_{D_y} = -Dvvsin(\theta)$$ where $D$ is a drag coefficient. From my understanding, and from what I read, both second-order equation can be converted to two coupled equations that could be used with my coupled Runge-Kutta method function. Can someone help me understand how to convert my equation to a set of coupled equations that could be used with my MATLAB function?  EDIT  I was able to use my rk4_c function, however it only works for solving the velocities. I'm now trying to see how I could solve for the positions x and y. I'm kind of learning about Runge-Kutta methods at the same time, so sorry for the questions that may sound trivials.  % Runge Kutta code to solve projectile motion with quadratic drag % dVx/dt = -(D/m)*vx*sqrt(vx^2+vy^2) % dVy/dt = -(D/m)*vy*sqrt(vx^2+vy^2) - g  clc clear all  % Constant D = 0.24; % m = 2; % kg g = 9.80665; % m/s^2  % Define function handles fVx = @(t,vx,vy) -(D/m)*vx*sqrt(vx^2+vy^2); fVy = @(t,vx,vy) -(D/m)*vy*sqrt(vx^2+vy^2) - g;  % Initial conditions v0 = 200; % m/s theta = 30*pi/180; % rad t(1) = 0; vx(1) = v0*cos(theta); vy(1) = v0*sin(theta);  % Step size h = 0.01; % s tFinal = 2; N = ceil(tFinal/h);  % RK4 simultaneous coupled loop for ii = 1:N     % Update time     t(ii+1) = t(ii) + h;      % Update vx and vy     k1 = h * fVx(t(ii), vx(ii), vy(ii));     l1 = h * fVy(t(ii), vx(ii), vy(ii));     k2 = h * fVx(t(ii) + 0.5*h, vx(ii) + 0.5*k1, vy(ii) + 0.5*l1);     l2 = h * fVy(t(ii) + 0.5*h, vx(ii) + 0.5*k1, vy(ii) + 0.5*l1);     k3 = h * fVx(t(ii) + 0.5*h, vx(ii) + 0.5*k2, vy(ii) + 0.5*l2);     l3 = h * fVy(t(ii) + 0.5*h, vx(ii) + 0.5*k2, vy(ii) + 0.5*l2);     k4 = h * fVx(t(ii) + h, vx(ii) + k3, vy(ii) + l3);     l4 = h * fVy(t(ii) + h, vx(ii) + k3, vy(ii) + l3);     vx(ii+1) = vx(ii) + (1/6)*(k1 + 2*k2 + 2*k3 + k4);     vy(ii+1) = vy(ii) + (1/6)*(l1 + 2*l2 + 2*l3 + l4); end  % Plot the solution figure(1) plot(t, vx, t,vy) xlabel('Time (s)') ylabel('Velocities (m/s)') legend('Vx', 'Vy') Ahlfors "Prove the formula of Gauss" He says:     Prove the formula of Gauss:   $$ (2\pi)^\frac{n-1}{2} \Gamma(z) = n^{z - \frac{1}{2}}\Gamma(z/n)\Gamma(\frac{z+1}{n})\cdots\Gamma(\frac{z+n-1}{n}) $$   This is an exercise out of Ahlfors.  By taking the logarithmic derivative, it's easy to show the left &amp; right hand sides are the the same up to a multiplicative constant.  After that I'm lost. It's easy using another identity when $n$ is even to use induction. But when $n$ is odd I am lost.  It's obvious when $n$ is a power of 2. How to navigate to C drive in bash on WSL-Ubuntu?