Unable to free space in /boot Running: Linux ips-svf-1 3.0.0-22-server #36-Ubuntu SMP Tue Jun 12 17:56:20 UTC 2012 x86_64 x86_64 x86_64 GNU/Linux  I'm trying to free space in /boot with apt-get -y purge, but every time I run the command I get an error about unmet dependencies:  sudo apt-get -y purge linux-headers-3.0.0-12 Reading package lists... Done Building dependency tree        Reading state information... Done You might want to run 'apt-get -f install' to correct these: The following packages have unmet dependencies:  linux-headers-3.0.0-12-server : Depends: linux-headers-3.0.0-12 but it is not going to be installed  linux-image-server : Depends: linux-image-3.0.0-26-server but it is not going to be installed E: Unmet dependencies. Try 'apt-get -f install' with no packages (or specify a solution).   However, I can't run apt-get -f install since /boot is full.  I also tried running dpkg --purge and dpkg --remove manually, but both give the same unmet dependency error.  Any suggestions on how I can successfully free space in /boot?  Is there anything I can safely delete with rm?
How do I free up more space in /boot? My /boot partition is nearly full and I get a warning every time I reboot my system. I already deleted old kernel packages (linux-headers...), actually I did that to install a newer kernel version that came with the automatic updates.  After installing that new version, the partition is nearly full again. So what else can I delete? Are there some other files associated to the old kernel images?  Here is a list of files that are on my /boot partition:  :~$ ls /boot/ abi-2.6.31-21-generic         lost+found abi-2.6.32-25-generic         memtest86+.bin abi-2.6.38-10-generic         memtest86+_multiboot.bin abi-2.6.38-11-generic         System.map-2.6.31-21-generic abi-2.6.38-12-generic         System.map-2.6.32-25-generic abi-2.6.38-8-generic          System.map-2.6.38-10-generic abi-3.0.0-12-generic          System.map-2.6.38-11-generic abi-3.0.0-13-generic          System.map-2.6.38-12-generic abi-3.0.0-14-generic          System.map-2.6.38-8-generic boot                          System.map-3.0.0-12-generic config-2.6.31-21-generic      System.map-3.0.0-13-generic config-2.6.32-25-generic      System.map-3.0.0-14-generic config-2.6.38-10-generic      vmcoreinfo-2.6.31-21-generic config-2.6.38-11-generic      vmcoreinfo-2.6.32-25-generic config-2.6.38-12-generic      vmcoreinfo-2.6.38-10-generic config-2.6.38-8-generic       vmcoreinfo-2.6.38-11-generic config-3.0.0-12-generic       vmcoreinfo-2.6.38-12-generic config-3.0.0-13-generic       vmcoreinfo-2.6.38-8-generic config-3.0.0-14-generic       vmcoreinfo-3.0.0-12-generic extlinux                      vmcoreinfo-3.0.0-13-generic grub                          vmcoreinfo-3.0.0-14-generic initrd.img-2.6.31-21-generic  vmlinuz-2.6.31-21-generic initrd.img-2.6.32-25-generic  vmlinuz-2.6.32-25-generic initrd.img-2.6.38-10-generic  vmlinuz-2.6.38-10-generic initrd.img-2.6.38-11-generic  vmlinuz-2.6.38-11-generic initrd.img-2.6.38-12-generic  vmlinuz-2.6.38-12-generic initrd.img-2.6.38-8-generic   vmlinuz-2.6.38-8-generic initrd.img-3.0.0-12-generic   vmlinuz-3.0.0-12-generic initrd.img-3.0.0-13-generic   vmlinuz-3.0.0-13-generic initrd.img-3.0.0-14-generic   vmlinuz-3.0.0-14-generic   Currently, I'm using the 3.0.0-14-generic kernel.
At 9th level, the Armorer Artificer (from Tasha's Cauldron of Everything) gains the &quot;Armor Modifications&quot; feature :  You learn how to use your artificer infusions to specially modify your Arcane Armor. That armor now counts as separate items for the purposes of your Infuse Items feature: armor (the chest piece), boots, helmet, and the armor’s special weapon. Each of those items can bear one of your infusions, and the infusions transfer over if you change your armor’s model with the Armor Model feature. In addition, the maximum number of items you can infuse at once increases by 2, but those extra items must be part of your Arcane Armor.  This special weapon is called 'Thunder Gauntlets' (a simple melee weapon) for the 'Guardian' Armor Model, and 'Lightning Launcher' (a simple ranged weapon) for the 'Infiltrator' Model. But I am wondering which Artificer Infusions are appropriate for this special weapon ? Is it only the 'Enhanced Weapon' and 'Radiant Weapon' infusions, or are there others ? Equation numbering problems in AMSMath environments with \pause multiline command chunks in bash history into multiple lines In Bash (as well as Bash in WSL), multiline written commands can call-back with up/down arrows correctly also built-in history command report them correctly (until I'm still logged-in in bash). But after exiting from bash and press up/down arrows, multiline commands appears each as separate entries and that's because cat ~/.bash_history has chucked the command on multiple separate entries and writing problem here; the history command result messed too. I have set these options in ~/.bashrc shopt -s cmdhist lithist HISTTIMEFORMAT='%F %T '  and now writing problem fixed but history command still report in multiple line and separate entries (reading problem now). Bash version is 4.3, same issue on v4.4 there is  since a long years but no solution yet. Update: it's also to mention that below multiline command as an example is correctly displaying even after exit Bash: for i in {1..10} do echo $i done  but this is failing: awk ' { print $1}' &lt;&lt;&lt;'first last'  Update 2: the issue mentioned above is fixed in Bash v5.0.0+ that changed to recognize entries between &quot;timestamp as delimiter&quot; as single entry of command (HISTTIMEFORMAT, cmdhist and lithist must be set). Order of nested random effects in lme4 I just fit a model in lme4, and I'm wondering what the heck I fit...  I have individuals id, and each is measured pass/fail on items that can be described using two factors, f1 and f2. My theory says that id and f2 can interact. So I want to compare these models:  # f2 doesn't matter m1 &lt;- lmer(pass ~ f1 + (1|id), family="binomial")  # f2 is a fixed effect, doesn't interact with individuals m2 &lt;- lmer(pass ~ f1 + f2 + (1|id), family="binomial")  # f2 is a nested random effect, interacting with individuals m3 &lt;- lmer(pass ~ f1 + (1|f2/id), family="binomial")   First, am I right that there's a random effect of each level of f2, and each individual's level of f2 is partially pooled, depending on the number of observations for that individual?  Second, does it make sense to fit f2 as a fixed effect with random individuals nested within each level of f2? How does one write that?  Third, what's this, with f2 and id reversed in the formula? I typed it by accident, but now I want to understand it.  m3weird &lt;- lmer(pass ~ f1 + (1|id/f2), family="binomial") StandardSetController with History records How can I selectively disable paste blockers What is a single word for a person who, as a habit, exaggerates a lot when relating information?   So for example, they relate something that happened but exaggerate so much that they are including clear falsehood and they are well known for doing that. How do I access a specific element in a vector of vectors clojure How to highlight an entire paragraph? What is the function of "do" in this sentence? Rarely, though, do countries introduce a single reform alone.   What function does do have in this sentence? Emission of EM radiation from an electron I have read about an electron making a transition between two energy levels and electromagnatic radiation will be emitted. The problem is how and why e.m radiation is emitted. I would like to show that ring of order $p^2$ is commutative.   Taking $G=(R, +)$ as group, we have two possible isomorphism classes $\mathbb Z /p^2\mathbb Z$ and $\mathbb Z/ p\mathbb Z \times \mathbb Z /p\mathbb Z$.   Since characterstic must divide the size of the group then we have two possibilities $p$ and $p^2$.   Now IU don't understand how can I reason to say that the multiplication is commutative and how can I conclude for the case when characterstic is $p$? Server 2012 deduplication: run on Hyper-V host or guest VM? I have a Windows Server 2012 Hyper-V host with VMs on a dedicated RAID-5 volume. One VM is also running Server 2012 and is being used to serve files. The other VMs are smaller and mostly used for temporary testing scenarios.   The main file server VM has two dynamically expanding VHDX files - one for the OS and one for storage. The VHDX being used for storage has quite a few duplicated files on it.  Should I run Server 2012 deduplication on the volume containing the storage VHDX file on the host, or within the file server VM itself? I understand that running it within the VM will not decrease the size of the VHDX file on the host, but my goal is to enable as much storage capacity within the VHDX file itself. I also understand that if I run it on the host, I will have to take the VM offline to do it. Downtime isn't a concern in this instance, just the best way to do this. Why am I getting zero divergence of function $\vec{f} = \frac{1}{r^2} \hat{r}$, where $r$ is the distance from the origin and $\hat{r}$ is the unit vector in the radial direction. The divergence of this function over a sphere of radius $R$, which includes the origin.  $$\nabla \cdot f = \frac{1}{r^2} \frac{\partial}{\partial r}(r^2 f_r) = \frac{1}{r^2} \frac{\partial}{\partial r}(r^2 \frac{1}{r^2}) = 0$$ In Java Edition, I can name a mob by using NBT tags like this: /summon zombie ~ ~ ~ {CustomName:'{&quot;text&quot;:&quot;TheUndead&quot;}'}  But in Bedrock Edition, I've heard NBT is not accessible from commands. What can I do to summon a mob with a custom name tag using the /summon command? Why does the /winsxs folder grow so large, and can it be made smaller? There is an exercise in Humphreys's An Introduction to Lie Algebras and Representation Theory: "Any nilpotent Lie algebra contains a codimension 1 ideal".  The proof I am thinking of is the following.  Suppose the Lie algebra $L$ satisfies $L\neq[L,L]$. Then $\pi\colon L\to L/[L,L]$ is a projection onto an abelian algebra. Choose a codimension 1 subspace $I\subset L/[L,L]$. Then $I$ is an ideal and its preimage $\pi^{-1}(I)$ is necessarily an ideal in $L$, because any subspace containing the derived algebra is an ideal. Moreover, $\pi^{-1}(I)$ has codimension 1. $\square$  The only condition on $L$ is that $L\neq[L,L]$, which holds for solvable Lie algebras as well. Is this proof correct? How come Humphreys didn't set the exercise with "nilpotent" replaced by "solvable"? 0 byte file created in a simple script I have a script as below:  #!/bin/bash df -k | tr -s " " "," | awk 'BEGIN {FS=","} {print $1,$5}'|sed 1d &gt; file1.txt while read partition percentUsed do if [ $percentUsed &gt; 75 ] then echo Partition: ${partition} space is ${percentUsed} else echo Pration: $partition: OK!! fi done &lt; file1.txt   The script is executing properly, however, it is creating a zero byte file by the name 75. Any ideas why this is happening?  $ sh diskUsed.sh Partition: C:/Users/Public/Documents/CYGWIN/bin space is 75% Partition: C:/Users/Public/Documents/CYGWIN/lib space is 75% Partition: C:/Users/Public/Documents/CYGWIN space is 75% Partition: C: space is 75% Partition: H: space is 91%   $ ls -lrt total 2 -rwxrwxrwx 1 diwvibh Domain Users 284 Nov 21 04:17 diskUsed.sh -rw-r--r-- 1 diwvibh Domain Users 133 Nov 21 04:29 file1.txt -rw-r--r-- 1 diwvibh Domain Users   0 Nov 21 04:29 75 LEFT OUTER JOIN in LINQ Suppose there is a message that is encrypted with AES-128-CBC.  The message is as follows, new lines are used to delimit the 16 byte boundary for each block:  Wire funds from: Alice to Bob in   the amount of $ 1   Because this message is encrypted using CBC mode,  any modification of the first block of cipher text would propagate throughout the message.   However, the amount of the fund transfer falls in the last block.  I am under the assumption that an attacker can use the property of  to alter the last block and produce a valid amount that is greater than \$1 in far fewer than $2^{128}$ operations.   Is this a correct assumption?  Is there a name for this attack or do you know of a real world example?    My instinct is that if you just need one byte of the 16 byte block to be of a specific value then it would only require $2^8$ operations because the other 120 bits of the plain text could be any value. For example lets say we wanted the last block of cipher text to decrypt to 9\0 (where \0 denotes a single null byte, and the message would be interpreted as \$9), then it would take only $2^{16}$ operations to find this cipher text block.  Is there something I am missing?