PO (mostly) of a project dragging on; doubting team's decisions

Question: TLDR:
PO-like position (+ a little dev) in a project that drags on.  Team members won't consider a less complex solution; after deadlines are missed and postponed, I suggest and proceed to implement simple solution.
How could I have handled this better, how to deal with fallout?
Details:
I've got an uncomfortable mix of roles in a joint project my team has with three other teams: I'm something between a PO, a PM, and a team representative, and also I'm a developer in my team.  I was involved in planning the project and I go to the project stand-ups.
When I sat with our architect to estimate how long the project would take, he was confident that six weeks should be enough.  Our project partners agreed, but we added another two weeks to the plan for good measure.
We're in week 15 now, and and our partners are still waiting for us.  We've been thinking for a long time now that we were close to finishing but somehow never did.
We're three people on the team: the architect, a data engineer, and myself.  Between us, we have different skills and responsibilities; I'm a decent programmer, but not a data engineer or a system architect.  My part of the technical work was therefore limited and I finished it weeks ago.  Since then, I've been waiting, supporting where I could.  I can't help thinking that the architecture is too complex and often I thought estimates were too optimistic.  However, lacking the experience and not being the one actually implementing the solution, I expressed doubts but didn't argue the point and otherwise left the decisions to the experts and implementers.
To our partners, I always presented solutions and estimates as team decisions.  This has become difficult and very stressful, since I understand our partners' frustration with our lack of progress and I can't really believe what I'm saying anymore myself.
After many missed and postponed deadlines, I've come to the conclusion that we're riding a dead horse.  I think the task is relatively simple and it should require only a simple solution; at least until we have more time to come up with possibly a more solid and scalable one.  I called a team meeting and, knowing that they were convinced that the technology was right and the architecture was not too complicated, I suggested that we implement something simple that does the job for now.  I offered they could keep working on their approach while I did the quick fix.
As I expected, they were totally against it, saying that we were close to finishing anyway, and that this was a waste of everybody's time.  They also accused me of heroism, trying to save the day all on my own.  We fought.  In the end, we ended the meeting without coming to an agreement and everyone did what they thought they should do.  It took me a day to implement most of my solution.  I haven't declared success to anyone else, yet, and when I do, I will not mention that it is my work as an individual.
Question:
I have a few ideas what I did wrong.  I should have been more vocal about my doubts early on, regardless of my role and lack of experience in their field.  I should probably also have started to convince them to implement a temporary solution a while ago, and tried to get them to do it with me.  Probably, I should not have accepted or kept my PO-like position when it became clear that I fundamentally couldn't represent the team's decisions any more.
My questions are: is there more?  And what can I do now?  I want out of this position and I want to move on to projects where I can actually contribute.  At the same time, I don't want to spoil our working relationship.
Answer: The way you describe this, this project has huge potential of damaging careers, includng yours. That's the first thing to think about, and you don't seem to have given this a thought of all.
First, go to your manager and find out clearly what your responsibilities are, and what your authority is. If you are "something between a PM and a PO and a team representative", your manager might assume that you have the responsibility for the failure of the project, so this might count very badly against you. And you need to find out what authority you have. With authority, your meeting wouldn't have ended without coming to an agreement.
And make sure that you have in writing what happened during your meeting. That the data architect, and the engineer, insisted that the task was almost finished, and rejected a simpler solution. "Almost finished" when you are far behind is a very bad estimate. You should have insisted at this point on a timeline, like "two weeks". ("Almost finished" is Ok if I have a task that I thought should take four days, and you ask me after three days about progress).
So since you talked with your manager (which you need to do), if you have nothing written yet, you can have a very quick meeting where you tell them that your manager is quite unhappy and very worried about the project, and you need them to confirm - in writing - that their solution is the right one, together with an actual estimate how long the job will take.
Comment: Actually, I'm acutely aware of the potential consequences for my career.  I just didn't mention it in order not to make my question even longer, but thank you very much for confirming.  We've made several attempts at timelines over the project, as you describe, and we never made good on them.  You're right, I need to talk to the manager, but it won't be easy as he wasn't around (our boss, yet), when we set up the project.   I don't have anything written, but we've told him a couple of times were were "almost done", so he knows.


Need help with dance studio

Question: I am doing AV for a new dance studio thats being built. They do all types of dance in there. There are three studios, two are about 1300sqft and one is 2200sqft. I believe i am going to need some form of sub so the dancers can feel the music. The ceilings are open and are about 12-13 feet high. I cant figure out how much wattage or what speakers are going to be right for the space. Any help would be more then awesome! Thanks
Edit: needs to be in the air(other then an sub)
Answer: Without seeing the space, we can only guess.
my guess would be 2 pairs of 2way speakers with a 15" woofer. Targeit dance floor SPL should be no more than 110dba/115dbc. You can figure this out by taking into account the speakers sensitivity and max spl. If a speaker is rated at 90 dbspl @1w/1m 110db at 5m would be about 200w. As a rule always spec amps at 2.5x the power you need to account for headroom, 450w. Remember this is for a 90dbspl speaker and everything changes with your speaker choice. +3db for every doubling of power and -3db for every doubling of distance. The reference is usually 1w/1m.
This is per speaker depending on impedance of the speaker and how the amp behaves with different impedances. If you have two 8 ohm speakers linked to create 1 4ohm speaker you gain some efficiency, provided your amp can handle it.
It is also necessary to provide an interface and wasy to use volume control for the operator of the system, most likely a choreographer, so make sure you take that into account. Amps and graphic eqs and anything else in the system should be behind tamper proof plates.
Comment: defiantly keep the amps and controls away from users, a nice way to do it is a "zoner" such as made by Cloud, this allows you to give the instructor a nice easy interface, for you to keep the amps locked away remotely and also to provide an emergency muting interface for the fire-alarm.

Need help with dance studio


What is the term used for logic statements which are non reversible?

Question: Something along the lines of
If A then B, if B then not. A has to be true before B can be true but if B is true first then A isn't true. Is there a term describing this in logic or philosophy?
Comment: I'm not entirely clear on what you're asking, but it sounds like you're describing the difference between sufficiency and necessity. Is that what you're looking for?
Comment: No Take-Backsies maybe?
Comment: @virmaior I was thinking along the lines of if A then B but not A if B
Comment: When you say "true first" do you mean temporally (which would make sense but require us to look at tensed logic) or logically (which would not make much sense)?
Answer: The property you're looking for is called commutativity. We say a relation R is commutative if and only if aRb is logically equivalent to bRa. in a commutative relation order doesn't matter. In arithmetic addition is commutative because 2+3=3+2. In sentential Logic conjunction and disjunction and the biconditional are both commutative.
A relation is non-commutative if it isn't commutative. The material conditional is the only non-commutative binary relation in sentential logic. Division in arithmetic is noncommutative: 2/3 != 3/2
Comment: Thank you this is the answer I was looking for, I guess I phrased the question poorly
Answer: You have it switched, existence of a limit does not imply continuity (because it may not coincide with the value), but continuity does imply existence of a limit. Conditional statement with premise and conclusion switched is called the converse of the original. There is no special name either in mathematics or in logic for conditionals that are true without their converses, this is assumed by default, unless otherwise stated. On the contrary, conditionals that are true along with their converses are distinguished by terms "biconditional" or "logical equivalence". "Irreversible" or "non-reversible" may be used informally, but they are not terms.
Comment: Thaaaaamk you, and yes I was trying to use a math which would make sense but I messed up the wording.
Answer: As Conifold pointed out, there is no specific name for conditionals that are true without their converses in logic proper. However, in the theory of computation, there is a "reversible logic", where the term logic is used in the sense of logic gates.
The key consideration is that the relationship between the inputs and the outputs has to be injective (i.e. a one to one mapping), for this to be possible.


Submit Credit Card Membership (or Event Registration) Not showing in Admin

Question: Not exactly sure what happened, but now the Submit Credit Card Membership option is not available in the Admin side of a Joomla install (v3.5.1). All seemed to be working fine, but then it disappeared. I tried recreating the Payment Processor and selecting it in the Contribution Page, but no luck. Then I opted to update to CiviCRM to 4.7.8 to see if that was the issues, still no luck. I am hoping it is just some sort of reset, otherwise our site is dead in the water with a conference coming up.
Answer: Looking at your issue: See CRM-15394
David Greenberg added a comment - 30/Sep/14 12:03 PM
The back-office 'Submit Credit Contribution' feature is only available when the site has at least one configured payment processor of the type that allows credit card data to be entered within CiviCRM forms. This includes Paypal Pro, Auth.net and iATS. The link is hidden on the demo because the configured 'Test Processor' does not support this.
Comment: I think that may be out of date - I see the link on the Drupal Demo but not on Joomla Demo.
Comment: NOTE: This is for Joomla, specifically, as per Scott's inquiry and as you noted below. We have tested here at the Sprint. It is confirmed that Dave Greenberg's explanation is correct.
Comment: @Scott - note that to process credit card payments inline in the backend you'll need PayPal Pro (not Standard)
Comment: Thanks Bruce - I added an authorize.net (dev sandbox) account and the link showed up on the Joomla demo. Not sure why the Drupal demo works with just a test processor though
Comment: Now I am really concerned, as I said using PayPal Standard the link and button where there and usable then it disappeared. So given that it was not suppose to be there to begin with, how can we process payments in Admin given our situation? Thanks for sny direction.
Answer: I am seeing the same thing in the Joomla Demo - I think you have found a bug. Please report it at https://issues.civicrm.org
Note this is only in Joomla, Drupal demo shows the link.
Update: Adding an authorize.net account on demo made the link appear. You could test your payment processor account on another instance to see if that's the problem.


What is the difference between symbolic and hard links?

Question: When would you use one over the other?
Comment: I don't feel like copying my server fault answer: http://serverfault.com/questions/10543/what-is-the-difference-between-a-soft-symbolic-link-and-a-hard-link/10550#10550
Answer: The different semantics between hard and soft links make them suitable for different things.
Hard links:

indistinguishable from other directory entries, because every directory entry is hard link
"original" can be moved or deleted without breaking other hard links to the same inode
only possible within the same filesystem
permissions must be the same as those on the "original" (permissions are stored in the inode, not the directory entry)
can only be made to files, not directories

Symbolic links (soft links)

simply records that point to another file path. (ls -l will show what path a symlink points to)
will break if original is moved or deleted. (In some cases it is actually desirable for a link to point to whatever file currently occupies a particular location)
can point to a file in a different filesystem
can point to a directory
on some file system formats, it is possible for the symlink to have different permissions than the file it points to (this is uncommon)
Comment: "[E]very directory entry is hard link." That's an excellent point that I've never seen expressed before, but I worry that someone just beginning to wrap his or head around links won't get it. For those in this situation, here's a hint: The layout of files and directories that you see when running the ls command isn't exactly the same thing as the storage system it represents. Hard links are references to an individual file on the storage system. A file is stored once. Read up on "inodes."
Comment: Nice list.  Just wanted to add that you can also break a relative path symlink by moving the symlink itself.
Comment: You could think of a symbolic link as a text file with a name in it.  It is interpreted as a symbolic link because of a special flag to the file.  Hard link examples you know are `..` and `.`.
Comment: [here is another answer that explains why hard links can't be made to directories](http://askubuntu.com/a/525129/78103). I find this answer helpful because it is more concise and easier to read.
Comment: @Mario: yup.  Every directory entry links a name to an inode.  The system call for deleting a filename is even called `unlink(2)`.  "normal" files (with a link count of 1) are just a special case.  If it helps, you can think of inodes as objects, and names as ref-counted pointers (the inode's link count is the reference-count).
Answer: The point of both types of links is to provide a way to make a file appear in two locations at the same time.  This has a lot of uses.  9 times out of 10 you want to use symbolic links.
Symbolic links, or "symlinks" work a little like Windows shortcuts.  The contents of a symlink are a pointer to the real location of the file/directory.  If you delete the real file, the symlink will become "dangling," and won't work.  Deleting the symlink does not delete the real file.  You can have as many symlinks to a single file (or even other symlinks) as you like.
Unlike Windows though, they work on the filesystem level, not shell or application level, so pretty much any application will "follow" symlinks as expected.  ls -al can be used as a quick way to see where symlinks "point" to.
Hardlinks work even on a lower level.  A hardlink is an actual, physical on-the-filesystem-level directory entry of the file.  Technically, a directory entry is a hardlink, thus each file has at least one hardlink in a directory somewhere.  Hardlinks are not separate from the file they point to; if a file has multiple hardlinks in different directories, deleting the hardlink with utilities like rm won't truly delete the file, until all hardlinks are gone.
I can't think of situation where use of hardlinks is common, or even needed, unless you intentionally want to prevent the files from getting deleted or are doing some weird low-level work with partitions or other filesystem related things.  EDIT: There's great ideas in the other answers to this question, though!
Comment: Windows shortcuts are actually quite different from symlinks: they follow their target, and they are also regular files. (Windows also has symlinks, but they're not used much.) Symlinks are purely textual, the target text is read whenever you access the file. Whether symlink permissions matter depends on the OS and filesystem.
Comment: Is it really very bad? What will happen? The most excitement I'm able to recreate is "Too many levels of symbolic links" error messages.
Comment: `ls -l` is enough to see what is being linked by a symlink, the `a` stands for `--all`, see manpage. And even if symlinks work at the file system, there are alternative functions to use symbolic links as files instead of follow.
Comment: Also, symlinks have permissions like normal files, but the operating system doesn't consult them, it consults the permissions targeted file instead to decide behavior.  And, don't do circular chains of symlinks.  Very bad.
Comment: AFAIK, the content of a symlink file is the path the symlink points to, which can be seen when looking at the size of the symlink file: `ln -s /home 1; ls -l 1` shows that the symlink 1 is 5 bytes long, whereas `ln -s /usr/share/ 2; ls -l 2` showas that 2 is 11 bytes long.
Answer: Hard links are very useful for disk-based backup mechanisms, because you can have a full directory tree for each backup while sharing the space for files that haven't changed — and the filesystem keeps track of reference counting, so when the last reference to a given version goes away because the backup was expired/removed for space reasons, the space it used is automatically reclaimed.  Some mail clients also use it for messages filed to multiple folders, for the same reason.
Comment: Maybe disk-based version control mechanisms? If you hardlink something, then it's not a backup. If the original file gets corrupted, every hardlink to it gets corrupted too.
Comment: Think of incremental backup systems like Apple's Time Machine.  (It should be obvious that these aren't disaster recovery type backups, but "oops, I deleted that file by accident" backups.)  All unchanged files in an incremental backup are hardlinked together; when the file is changed, the next incremental copies it instead of linking to the previous version.
Comment: Thanks, then incremental backup systems are pretty similar to version control systems this way =D
Comment: But how does the incremental backup mechanism preserve the "old" version of a file? 1) Backup A created, it hardlinked file F; 2) File F modified; 3) next day Backup B created... Looks like I don't get something
Answer: A soft link points to another pathname.  That pathname may or may not actually exist.  The path isn't looked for until you access the symlink.  If the path doesn't exist when you try to access it, you have a broken symlink.
With a hard link, you have one file with multiple names.  You can't say that one of those is the "real" file and the others are just a link to it.  They are all equal.  There's no such thing as a broken hard link the way there are broken symlinks.
Hard links work only within a single filesystem.  If you want to link to a file on a different filesystem (e.g. a different partition or a network share), you must use  a soft link.
Another big difference is what happens when you delete a linked file.  If you delete one of a pair of hardlinked files, then create a new file with the same name, you'll have two separate files (the link is gone).  If you delete the target of a symlink and create a new file with the same name, the link will point to the new file.
Answer: Hard links are just references to the same disk spaces, thath the 'why' you cannot hardlink something in other filesystem.
Symlinks are files linking other files (as Windows shortcuts), maybe in the same filesystem, maybe not.
EDIT: I will explain something more. Every file that exists has a minimum of 1 hard link. Hard links are the way to access the content of an inode of the filesystem. You can obtain the inode number of a file with ls -i, and get the number of hardlinks with stat as follows in this example:
$ stat plantilla-disenos.odt
  File: «plantilla-disenos.odt»
  Size: 12367       Blocks: 32         IO Block: 4096   fichero regular
Device: 803h/2051d  Inode: 319875      Links: 1
Access: (0644/-rw-r--r--)  Uid: ( 1000/   d4rio)   Gid: ( 1000/   d4rio)
Access: 2011-02-11 21:36:19.000000000 -0300
Modify: 2010-03-02 23:27:28.000000000 -0300
Change: 2010-04-10 17:46:27.000000000 -0300

Thanks @geekosaur for this reference:

The kernel has to restart
  pathname-to-inode translation
  (traversing the directory tree) to
  expand symlinks, whereas hard links
  all use the same inode. (You'll often
  see this referred to as namei, from
  the name of the kernel function that
  did this in traditional Unix.)

and this (edited):

Hard links are very useful for
  disk-based incremental backup
  mechanisms like Apple's Time
  Machine, because you can have a full
  directory tree for each backup while
  sharing the space for files that
  haven't changed — and the filesystem
  keeps track of reference counting, so
  when the last reference to a given
  version goes away because the backup
  was expired/removed for space reasons,
  the space it used is automatically
  reclaimed. Some mail clients also use
  it for messages filed to multiple
  folders, for the same reason.

Cheers
Comment: Are the performance benefits to using hard links? Or, why would you ever use a hard link instead of a symlink?
Comment: The kernel has to restart pathname-to-inode translation (traversing the directory tree) to expand symlinks, whereas hard links all use the same inode.  (You'll often see this referred to as `namei`, from the name of the kernel function that did this in traditional Unix.)
Comment: @ripper234: Hardlinks are diskspace-saving solutions.

You don't need to know about the filesystem to make a symlink, but you need to think before creating symlinks because you can create a loop or a long resolution path, so functions like `stat` will fail.
Comment: @geekosaur: I'm adding your answer to mine since it's very useful
Comment: No problem.  I actually started to write it as a comment to yours, but comments are too short.
Comment: A file can have 0 link. Normally, the file is deleted when its link count reaches 0, but it will remain in place (with no name) until no process has it open. An open unlinked file still takes space and prevents its containing directory from being removed.
Comment: @Gilles: Just Linux or every unix-like implementation does that?
Comment: @D4RIO: Every "mainstream" unix, certainly (e.g. Solaris, *BSD, OSX, Linux). More exotic unix-like OSes might not (e.g. I don't think you can delete an open file in Cygwin, because Windows doesn't work this way).
Answer: "hard" links share the same inode
$ touch foo
$ ln foo foolink # Creates a hard  link
$ ls -li foo foolink
54996 -rw-r--r-- 2 bsd users 0 2011-12-11 09:06 foo
54996 -rw-r--r-- 2 bsd users 0 2011-12-11 09:06 foolink

If I edit either foo or foolink there is only one file and it will be updated.
If I remove only one of the filenames, the inode and data will persist, foolink will survive.
$ rm foo
$ ls -li foo foolink
ls: cannot access foo: No such file or directory
54996 -rw-r--r-- 1 bsd users 0 2011-12-11 09:06 foolink

If I were to create the same, but with a "soft" or symbolic link, then
There's one file, one inode, and a new file with its own inode pointing to the first.
$ touch foo
$ ln -s foo foolink # Create symlink
$ ls -li foo foolink
55029 -rw-r--r-- 1 bsd users 0 2011-12-11 09:11 foo
55033 lrwxrwxrwx 1 bsd users 3 2011-12-11 09:11 foolink -> foo

If I edit either foo or foolink there is still only one file and it will be updated.
If I remove only the symlink, the inode and data will persist.
If I remove foo, the data will be gone, the symlink will persist but point to a non-existent file.
$ rm foo
removed `foo'
$ ls -l foo foolink
ls: cannot access foo: No such file or directory
lrwxrwxrwx 1 bsd bsd 3 2011-12-11 09:11 foolink -> foo

Comment: But what's a practical use-case for this?
Comment: One use, same as a "short cut"
Another use, having multiple versions of an application on a system allows one to install, test new version, specifying app by full path, while symlink in bin points to production. After testing is complete change symlink to new version, leave old version in place for any users that have version dependent code.
Think of perl, python, etc.
Comment: practical use-case for hard links.
At present on my filesystem I found a large number of hardlinks in /usr/share/zoneinfo
Think of all the named files representing timezones, which are all identical to EST. We save filesystem space by not having redundant copies and allow for easier package management without having the management overhead of symlinks
install/delete as packages are installed/deleted.
Even if one is removed the original data is preserved.
Sorry I didn't have time for a more pedantic explanation.
Answer: HARD LINK (Only Files) vs SOFT LINK (Files or Directories) vs BIND (HARD LINK for Directories)

(source: freesoftwareservers.com)
While daxelrod's answer explains the question well, I thought that the picture in this case made a big difference, especially to beginners who don't understand inodes and complicated Linux jargon quite yet.
Think of this, if you "deleted" everything from your drive, you could run software to restore the data, because the 1's and 0's are still there, you just deleted all the Hard Links. Recovery Software's purpose is to rebuild the Hard Links to make sense of the 0's and 1's
I read a great "one liner" that made this all make sense and I wanted to share!
All files in Linux are "Hard Links" to the 0's and 1's on the disk. When you create a data (0's & 1's) the OS creates a Hard Link in the File Tree to reference that spot on the hard disk.
Create HARD LINK 2 and delete HARD LINK 1 Original File:
You may create another hard link and delete the original file, and you still have access to the newly created hard link.
Delete FILE(HARD LINK 1) that is SOFT LINKed to:
If you deleted the HARD LINK 1, do you think the SOFT LINK would work? No, The OS will report back that the HARD LINK 1 does not exist.
Delete SOFT LINK to HARD LINK:
In reverse, if you delete the SOFT LINK, will either HARD LINK work? Yes. As long as the OS has one HARD LINK File it will report that the fill has not been deleted.
-- Also worth researching/noting is BIND, a way to BIND two directories like symlinking two directories, but it is transparent to the OS (OS's can tell when you Symlink and some have rules regarding weather they can follow Symlinks). It uses Mount, not LS and can be configured via FSTAB.
What is a BIND mount
Comment: That's quite an ambitious effort, especially for a first post.  Unfortunately, I believe that adding the material on "bind" (which wasn't asked for) just confuses matters; especially since you don't seem to have made much of an effort to *explain* "bind" mounting.  Also, I understand hard and soft/symbolic links very well, and I barely understand your picture.  I would be very surprised if a beginner could learn anything from it.
Comment: (1) Actually, on at least some versions of Linux, you *can* bind mount a file.  (2) While bind mounts look very similar to hard links, saying "Bind is simply the same as hard links (except you can't hard link a directory)" is simply wrong.
Comment: While you can symlink directories, it shows in the file system as a symlink, if you BIND, it is transparent to the OS. Shows up just like a file.
Comment: @G-Man, agreed and removed, with just a note mentioning BIND
Comment: @Free actually the soft link points to a file *name* (hard link 1) ; the schematics, should make this obvious.
Answer: Hard links are additional directory entries for the same file. That means

All hard links to a file must be on the same file system (because a directory entry cannot point to a file on a different file system), but not necessarily in the same directory.
There's no difference between the original directory entry and the new hard link; from the operating system's view, they are just two directory entries to the same file. A file is only deleted if all of the hard links are deleted (and additionally, there's no process left that has that file still open).
If you move/rename the "original", as long as you don't move it to another file system, the other hard links are not affected; they still point to the same file.
Many editors don't write the new content into the same file when saving, but rather do the following procedure:

Write the new content to a new file.
Rename the old file to a backup name (or, if not keeping backups of the previous file version, simply delete it).
Rename the newly written file to the name of the previous file.

This scheme means that any other hard links to the same file will now no longer point to the current file, but to the previous version (this is true even in case the editor deletes the old file, because under Unix, "deleting" a file just means deleting its link; only if the deleted link is the only link the actual file gets deleted).
Since the hard link goes directly to the file, you can access that file even if you don't have access to the original location of that file (for example because you don't have any permissions on the directory the original entry is in). The only rights determining your access are the access rights of the file itself (which are associated with the file, not with the link; you cannot make hard links with different permissions for the same file) and the access rights for the path the hard link is contained in (basically, the execute rights the directory the link is in, and any direct and indirect parent directories).

Symbolic links, on the other hand, store the pathname (the name to the file — or rather its directory entry — potentially including its path, like /bin/sh or subdir/foo.bar) — of another file. If the pathname is relative, it's always interpreted relative to the directory the link is contained in. That means:

A symbolic link may refer to files on a different file system (even to a file system which does not itself support hard or soft links, like FAT).
If the original file is deleted, the symbolic link doesn't preserve the file content. Unless there are other hard links to the same file, the file content will be gone. The symbolic link will then be left dangling (that is, referring to a pathname that doesn't correspond to a directory entry). On the other hand, deleting the symbolic link doesn't affect the original file, since it only refers to its pathname.
If the original file is moved or renamed, the symbolic link is not updated, but left dangling. If you move the symbolic link, it only breaks if it contains a relative path, and the path is no longer valid from the new position.
If the original file is replaced by a new file with the same name (as in the editor scenario described above), the link refers to the new file.

Most uses of hard links are basically a way to have a copy of a file without having to store the file content twice. This works best if the files are never changed again, as otherwise it's easy to accidentally break the link (see the editor scenario above). There are, of course, cases where you want the link to be broken, as in the case of keeping several backups: For files that have changed in newer backups, you don't want the copy in older backups to change as well.
Normally if you want a link, you'll use a symbolic link. One example is when you move a directory to another partition (because the one it is on gets full), you can set a soft link from the old position to the new one, so any programs trying to access the directory at the old place will access it at the new place instead. This would not be possible with hard links. However be aware that symbolic links in the moved directory can break if they contain relative paths that lead out of the moved directory.
Answer: A hard link will keep a file on disk until all hard links to it, even the first (a "filename" is technically a hard link), have been deleted. A soft link can be left "dangling" until the file it point(s/ed) to is replaced.
Answer: This is a very old question but I have a use case that is requiring me to use hard links.
I'm a musician and so I have lots and lots and lots of audio files of various sorts on several hard drives attached to my Mac.  Terabytes worth.  I have them mostly organized very nicely with symlink directories so that I can find them by content publisher, style/sound, and other criteria based on how I'm thinking at the time.  Unfortunately one program I use, Ableton Live, is completely incapable of viewing aliases or symlinks from its file browser.  The only solution I've found is to create hard links of the directories I want it to be able to see into, and then everything works great.
So, this is another case of when you might need to use hard links, that might not have occurred to others.
Comment: I would file a bug report for Ableton Live. Maybe they're able to fix this.
Comment: Yeah there is already a lot of complaining about this issue on abletons forums for years… They don't seem to be making any attempts to address it though I can't figure out why not.


What is non-decomposable and/or non-differentiable loss function?

Question: I have been reading some deep learning literature and came up with these concepts of non-decomposable and non-differentiable loss functions. My question is are these same thing? if not how are they different? How do we know if a loss function is non-decomposable/non-differentiable?
Answer: Non-differentiable (loss) functions can be labelled according to the stardard definitions of differentiation: that the function has a derivative at all points in its domain. For example, the ReLU loss function is technically non-differentiable, because its gradient is not defined at zero (or wherever the two lines meet, if you're using a variant of the standard ReLU).
Non-decomposable functions must be looked at slightly differently. A loss function that is not decomposable is usually one that is composed of several statistics across the training metrics. Take the F1 score, for example:
$$
F1 = 2 \cdot \frac{precision \cdot recall}{precision + recall}
$$
Using this as a loss function means you would have to expand the terms for precision and recall in terms of your predictions over a batch, then work out the gradients of each function, finally combining them. It can be done (but it gets complicated), but I would imagine there are trade-offs when optimising at the level of such a metric, which really summarises your model's performance as an average over samples. Using a traditional loss function focusses on improving the model at the more granular sample level. This means the gradients are concentrated at the purest level and I would imagine better optimisation is possible. It could be that optimising at the more abstract level of the F1 score leads to less overfitting or even better generalisation. This is all just an idea though; I haven't seen any work making such comparisons directly.

Here is a short reddit thread with some interesting points and examples about when you might even consider trying to optimise such a non-decompasable function directly (i.e. using it as a loss function).
Have a look at this paper, which speak about non-decomposable loss functions and offers a way to deal with them. Reading through that and comparing the functions that the authors talk about should help you gain a better understanding of the key differences.
Comment: as you mentioned, F1 would be helpful for less overfitting, is it not  better than being more granular and overfit?


Pushing LEGO city train

Question: Are you able to push the LEGO city train around the tracks without damaging it or do you always have to use the batteries to make it go?
Comment: Possibly. But that answer didnt come up in google when I was searching. I think its two different questions with the same answer.
Comment: that's why it says your question already has an _answer_ somewhere else. :-)
Answer: You can definitely use the trains without batteries and push them on the tracks. In fact, a lot of LEGO trains come without the battery option, and you can add it if you want.
Most people will get the battery option as it is fun to watch LEGO trains go on their own, but it is not necessary.
You can even build raised tracks and have the train go down on them by gravity...(c:
One thing I would also add is that it is best to either pull the train from the front, or push it from behind. Don't put pressure on it from the top; like how little kids tend to push trains putting their weight on the train as they push it from the top. That could damage the axles of the train.

Pushing LEGO city train


How to present a polynomial interpolation if the first order information of function f(x) is given?

Question: Suppose $f'(x_1),\ f'(x_2),\ f'(x_3)$ are given, how to give a polynomial interpolation $p(x)$ such that $p'(x_1)= f'(x_1),\ p'(x_2)=f'(x_2),\ p'(x_3)=f'(x_3)$? And how to give an error analysis?
Comment: This is called Hermite interpolation, and should be discussed in most textbooks on numerical analysis (e.g., Gautschi's). Note that the polynomial is not uniquely specified unless you also have some function values $f(x_i)$.
Comment: You didn't say in your questions that you set $p(x_1)=f(x_1)$.
Comment: Yes it will be uniquely determined if we set $p(x_1)=f(x_1)$. But it is not straightfoward to apply the Hermite interpolation since we do not know the informatio about $f(x_2)$ and $f(x_3)$?
Answer: The problem boils down to the solution of a linear system a equations. You get one equation for the function value at $f(x_1)$, and three more equations from your derivative equations. Then, a polynomial of degree three is suitable.
A more brute-force, not so clever way is to use computer algebra such as SymPy for a symbolical solution:
<code>from sympy import *

N = 3 # degree of polynomial - for N + 1 'pieces of information', use a polynomial of degree N

known_x = symbols(["x_%i"%i for i in range(N)])  # create symbols for supporting points
known_derivs = symbols(["d_%i"%i for i in range(N)])  # create symbols for supporting points
y1 = Symbol("y_1")  # the single zeroth derivative we know

# create symbols for the independent variable and the coefficients:
x = Symbol("x")
a = [ Symbol("a_%i"%i) for i in range(N+1) ]

# generate polynomial and the derivative
p = sum([ coeff*x**i for i, coeff in enumerate(a) ])
p_prime = diff(p, x, 1)

# define the equations to solve:
eqns = [ p_prime.subs(x, xi) - d for xi, d in zip(known_x, known_derivs) ]  # equation for the first derivative
eqns.append(p.subs(x, known_x[0]) - y1)

# solve for the coefficients and print solution:
solution = solve(eqns, a)
pretty_sol = [ str(coeff) + " = " + str(value) + "\n"
               for coeff, value in zip(solution.keys(), solution.values()) ]
print("The coefficients are:\n" + "".join(pretty_sol))
</code>
This not very elegant, but seems to work even you added another equation for the derivative (so that could use a polynomial of degree 'N = 4' for the five equations).
Comment: I think you mean "A polynomial of degree *3* is suitable"?
Comment: Yes, of course. Thank you. Edited in my reply.
Comment: Thanks, can you be more clear about your comment "The problem boils down to the solution of a linear system a equations"? It seems interesting, can it be possible to give a error analysis?


Did the USSR invite other astronauts of Apollo 11 besides Armstrong?

Question: Sometime after the return of Apollo 11, Neil Armstrong visited the USSR. I wonder why he was alone? Weren't the other astronauts invited or they refused to visit? Was it because Armstrong was an atheist while the others were devoted Christians?
Comment: Wikipedia's article on Neil Armstrong lists him as a **"Deist", and I quote**, "a person whose belief in God was founded on reason rather than on revelation, and on an understanding of God's natural laws rather than on the authority of any particular creed or church doctrine." **The article has a citation for this.** If he later became an atheist, please find a cite for that and add it to the wiki while you're at it.
Comment: @DrZ214 Does "deist" mean non-religious?
Comment: It may or may not. According to the definition given, you could be a deist and a bible-believing christian (religious). The deist definition seems to concern more about **why** you believe in a god/supernatural being, rather than what dogma surrounds it. But it seems pretty clear that deist and atheist are not the same thing.
Comment: @DrZ214 it seems to me that a deist cannot be bible-believing because bible's authority is based on revelation and church doctrine. So deist cannot be a member of a church. May be it would be better to change the phasing from atheist to non-religious.
Answer: Religion as a motivation for Armstrong's solo visit is very unlikely.  Armstrong was actually the 2nd U.S. Astronaut that travelled to the Soviet Union - the first was Frank Borman, who arrived in Moscow on July 2nd, 1969 after being invited by the U.S.S.R. Institute for Soviet-American Relations.1  Borman was among the more religious of the Apollo astronauts, and the Soviets certainly can't have missed his reading of Genesis when Apollo 8 was orbiting the Moon on December 24, 1968.
Like Borman, Armstrong was not invited by the Soviet government - his visit was arranged through the International Council of Scientific Unions' Committee on Space Research (COSPAR). Armstrong was part of a delegation of 32 American scientists and NASA officials2 and was ostensibly there to present a paper titled "Lunar Surface Exploration" at the 13th COSPAR conference in Leningrad May 20th through May 29th, 1970 (which he also subsequently presented at a conference in Berlin the next year).3 The official Soviet government response to his visit was a luke-warm, and it took 5 days after his arrival to get permission to visit the Kremlin.4
Most likely, Armstrong's visit was a natural result of the attempts of the heads of both the U.S. and Soviet scientific communities to develop a more cooperative relationship than had characterized the early space race. Several months prior to Armstrong's trip to Leningrad, Thomas O. Paine (Administer of NASA at the time) had sent his counterpart at the Soviet Academy of Sciences Mstislav Keldysh a copy of two reports on the possibility of creating a more cooperative relationship between the two agencies.

In keeping with the spirit of the Space Task Group's report, Paine
  transmitted copies of it, together with NASA's more detailed report
  America's Next Decades in Space, to the Soviet Academy of Sciences. In
  his cover letter of 10 October 1969, Paine told Keldysh that these
  documents might "suggest to you as they do to me, possibilities for
  moving beyond our present very limited cooperation to space
  undertakings in which the Soviet Union and the United States could
  undertake major complementary tasks to the benefit of both our
  countries."5

Their correspondence eventually led to a meeting in New York between Paine and Anatoli Blagonravov, chair of the Soviet Academy's Commission on Exploration and Use of Space, where Paine mentioned that Armstrong would be visiting the COSPAR meeting.

The amiable conversation touched on many subjects. Paine mentioned to
  his guest that Neil Armstrong planned to deliver a paper at the COSPAR
  meetings scheduled for 20-29 May 1970 in Leningrad, and Paine said he
  hoped that Armstrong would have an opportunity to visit some of the
  Soviet scientific facilities. Blagonravov responded that the
  cosmonauts would be pleased to show their American counterpart their
  facilities and some of the other space-related institutes.6

This ultimately formed the basis for the Apollo-Soyuz program, so it is entirely possible that Armstrong's visit was engineered specifically toward the end of fostering goodwill during the tenuous early phases of establishing a scientific relationship.
1 National Aeronautics and Space Administration. Astronautics and Aeronautics, 1969 - Chronology on Science, Technology, and Policy, p 195
2 National Aeronautics and Space Administration. Astronautics and Aeronautics, 1970 - Chronology on Science, Technology, and Policy, p 176
3 Armstrong, Neil. University of Cincinnati Résumé, September 1975, p 2
4 Hansen, James R. First Man: The Life of Neil A. Armstrong, p 582
5 Ezell, Edward and Linda. The Partnership: A History of the Apollo-Soyuz Test Project, p online
6 Ibid.
Comment: +1. In-depth answer. But were any other astronauts offered invites for that delegation or otherwise?
Comment: @LateralFractal - Buzz Aldrin and Michael Collins weren't invited to the COSPAR conference, but Armstrong's trip to Russia wasn't directly related to the Apollo 11 flight.  The official celebratory tour was much earlier - a [37 day goodwill tour](http://www.wingnet.org/rtw/RTW005JJ.HTM) around the world that included all 3 of the crew members.


What are primal Zerg?

Question: Heart of the Swarm introduces Primal Zerg.  What are they?
Answer: Primal Zergs are also referred as "original zergs". They are zerg that never left their original home planet, Zerus, and thus never evolved/mutated
Source, SC lore (and the wikis, of course!)

 As for the spoiler: Kerrigan's back to her human form, retaining her boosted psyonic powers and her zerg hairs. She was mercyless in WoL but here in HoTS she moves just by pure revenge. She is dfferent than the old Queen of Blades: she probably still has a human core on her inner inside, but she's not going to be human again. So, the answer is: 99% corrupted, 1% still Sarah.

BIG BIG BIG SPOILER (not really, you can find it in the wiki):

 In the final scene, the camera pans around space and galaxy while Kerrigan narrates, stating that she must face her true enemy after renouncing everything she has and embraces her role as the leader of the Swarm.
Comment: @PeterPAD she is just too powerful as a terran-psy to be completely controlled. Also, the Overmind exists no more and Kerrigan was "created" as a hybrid, not as a fully zerg. She'll always be part zerg and part human.
Comment: @PeterPAD The proof is every second NPC in-game telling Kerrigan she's different from the old Queen. You didn't play the game, right?
Comment: edited to answer also the spoiled question
Comment: which proofs make you said new QoBS different from older? I thought    that it may be the same and can be restored to human form.
Comment: The 2 Kerrigans are essentially the same, searching for Xel'Naga artifacts to get her back to her human form (altough se never says her objective is that in WOL, she just fights for the crystals) but yeah, in HoTS pretty much everyone tells Kerrigan she's a different queen now... :-/ Also, the cinematics when she's with Raynor. Her human side is different for sure.
Comment: @OrcJMR I've done already, yet still misunderstand
Answer: The primal zergs are the zergs that did not get corrupted by Amon (or whatever how we write the name of Dark Voice).
They are the original zergs that were made by the Xel Naga. The zergs we know that can be part of the Swarm were corrupted to be made into weapons. All the zergs in the swarm share a psyonic link with each others and that makes them dependent on their leader. A leaderless brood isn't more dangerous than wild animals (as can be seen in Flashpoint or the second book of the dark templar saga).
The primal zergs don't have that psychic link. Packs are made like wolf packs where there is an Alpha male (Brack in the Zeras mission is one of the alpha). From what I can see in the mission, most current Zerg units seems to have evolved from the primal zergs in some way. The primal zergs are also independent and only respect power. Also, each primal zerg have the ability to indivually evolve by absorbing essence while the swarm has to go through more steps to evolve.
tldr; They are the original zergs. Most zerg units have evolved from their primal zerg counsins.

What are primal Zerg?


PC not running for more than 15 minutes after cleaning CPU fan

Question: I thought that my CPU fan was gathering too much dust. So I removed it (just the fan), cleaned it and re-attached it.
Now my PC only runs for approximately 15 minutes, then the CPU shuts off. It happens again and again.
What can I do?
Comment: Perhaps the heatsink isn't seated correctly causing heat not being able to dissipate.
Comment: It is likely overheating and you need to figure out why.  Is the fan working now?
Comment: Watch in the "hardware monitor" section of your bios and write here the temperature of both CPU and motherboard.
Comment: yeah, the fan working like before, what happened?
Comment: Did you plug the fan into the proper port on the motherboard? Some mother boards have CPU and CPU aux 1 and CPU aux 2 4/3 pin connectors right next to each other.
Answer: Most CPU fans are attached to a heat sink. The heat sink must make good contact with the CPU, using a good quality heat sink compound. If you only removed the fan without touching the heat sink - the bond should be ok. But it you removed it, without cleaning and applying new heat sink compound, the heat transfer might be poor. A great heat sink compound is Arctic Silver.
Other situations to check are that all heat sink attachment legs are properly clicked in place, so that there is no gap between the cpu and heat shield.
Usually if the CPU is overheating, the fan speed will increase. But if the bond is majorly off, the fan speed might start max immediately and you wouldn't notice the difference.
There are several free utilities to monitor your CPU temp and Fan speed. Watching these things before it shuts off, should confirm these ideas.
Comment: [SpeedFan](http://www.almico.com/sfdownload.php) is one of them.
Answer: Let me ask the two obvious questions:

Did you reconnect the wires of the fan?
Did you loosen or move the heat sink while unscrewing the fan?

Both of these can cause a lack of cooling, and modern processors do an emergency shutdown when they overheat.
Finally a bit less obvious: Does the fan still work? Did you clean it with a vacuum cleaner while it was still plugged in? (If you spin it the fan engine become a generator, supplying power to the motherboard which was never intended to receive power on those pins.)
Answer: When the PC shuts itself off, this could be because of a temp over the maximum set in the BIOS for your computer, which thereby causes it to shut off to prevent a fire or damage to the CPU, or other components of the PC.
The Southbridge section of the motherboard is also sensitive to overheating.
The computer power supply can also be the cause of this problem. If it is overheating, it may also shut down the whole PC. Power supplies are closed units, and manufacturers do not recommend disassembling them, because they contain parts which hold a charge and can shock you. If you remove the power supply from the computer, and look inside below the fan, you can often see if the boards are gunked up with pet hair, dust, etc. I have used the pointy end of a small paint brush to dig out the globs of pet hair that build up in my power supply over time.
Also, if you DO disassemble your power supply, it will void the warranty most likely.
It may be a worthwhile experience to completely disassemble your PC, and completely blow it out with a low PSI air compressor. Do this outside, in your garage, or another low dust, dry area to avoid damage to your computer.
As other people mentioned, you need to be sure that you have fresh thermal compound on your heat sink, where the sink contacts the cpu itself. If you apply this compound yourself, it ONLY TAKES A DROP for a CPU. I am referring to a normal sized drop, no more! It is supposed to form a THIN layer between the cpu and the heat sink, just enough to fill the gap present.
Finally, be sure that the pins holding the sink to the motherboard are properly placed in their fully inserted position. Some people opt for screw-down mounting for CPU heat sinks, and avoiding the cheap manufacturer push-pin type connectors, for the simple fact that the push-pin system is junk & doesn't work. As you press one corner into place, it pops the opposite corner off, so be sure to push them down into place two at a time, on opposite corners.
Because of excessive amounts of pet hair in my PC, I changed from the heatsink & fan to a closed-loop liquid cooling system. That means that it is sealed, and you never add liquid to the setup. There is a metal block which attaches to the CPU (with the thermal paste, ONE DROP ONLY!), and usually has a small pump inside it. There will be two tubes which feed liquid up to a small radiator, which has a fan attached to it. When it is time to clean out the computer, you just vacuum off the radiator with a normal vacuum hose. This will eliminate the issue of having to clean as often.
Tests have shown that pushing hot air out of the computer & through the radiator actually cools the liquid in the system sufficiently. Whereas pulling room-temperature air into the PC & through the radiator produces less than optimal results, from my personal experience. I think this is because removing warm air from the computer is better than just dumping more air into the case. More air isn't always equal to better cooling, whereas instead removing heat from the computer, will get you the best results. It makes sense if you think about it, less hot air in the computer = a cooler computer.
Also, as previously mentioned there are programs which will tell you the temperature of your CPU, and other components in your computer. One such program is called Speedfan. It is a free application. You NEED TO configure it for your specific motherboard, so do that before expecting useful results.
In closing, it might also be worthwhile to look at the rest of the computers vents. Your CPU fan might be spinning, but not at full speed. Also it might be spinning proper speed, but without any air coming in the front of the machine (because the vents there are clogged with dust, dirt, hair, etc.). If completely removing the side of the computer while you turn it on gets you no better results, then you probably are NOT getting the CPU fan up to a proper speed, or a problem with the contact between the cpu and the heat sink is to blame.
If the PC is on the floor, try putting it in another place, where it gets better airflow. Likewise, if it sits in a corner, it might be getting poor airflow. Is there a heating vent under your desk somewhere? If so, this could be heating your computer up beyond what it is intended to operate as normal temp's. Last but not least, the temperature sensors in the motherboard might be failing also, which could be reporting higher than actual temps, and therefore tripping the thermal protection built into the bios. A thorough cleaning of the computer (as part of a full disassembly process) will uncover a great amount of dust, pet hair, dirt, etc. After each component is removed, and individually cleaned, placed temporarily in a dust-free area, then the case itself can also be cleaned out. I use soft nylon paint brushes to clean fan blades off, and I also use fans with removable blades (enlobal fans by enermax, specifically) and special magnetic bearings for exceptionally long life. Normal fans only lasted me a year or two, these are at 5+ years now, no joke!
I completely disassemble my whole PC annually. I vacuum out the thing monthly. I blow the thing out about 3 times a year. When I blow it out, I use a Kirby vacuum, with the hose attached to the exhaust, and a restrictor placed onto the hose. It's a neat vacuum, but is rather expensive. You can likely find a used one on craigslist or ebay for cheap. Accessories from your local kirby vacuum shop, like new hoses, and a shaft cover, will enable you to use it like I do, as a blower. Be careful using ANYTHING METAL in or around your computer. Things like Kirby's produce a high amount of static electricity, so be sure you have your computer PLUGGED IN using the three pin plug attached to your power supply, and connected to the wall using a 3-pin socket that has an actual ground wire that IS ATTACHED. An electrician can test the third pin on your sockets, to be sure they're in working order. This way, if you do shock your computer, the PSU might take the shock to ground & save your computer.
Just because your fan spins on your CPU cooler, doesn't mean it is spinning at the speed it should be! Your local computer shop can show you a NEW cooler operating at proper speed, which may be spinning much faster than yours, if yours is worn out, old, etc which happens over time & compounded by a dusty or pat-hair laden environment. Happy computing, and I hope this helps.
Comment: Answer SERIOUSLY needs some punctuation and paragraphing
Comment: Like that @JourneymanGeek? Todd, please include a few paragraphs next time – otherwise people will just skip reading your answer since it's a big wall of text.
Comment: Or at the very least add a tl;dr summary to get the solution across quickly until you fix it. Also, when the question/answer is that long, consider breaking it up into sections with headers and horizontal rules, and highlight important parts with italics or bold. Basically, if you *must* write a novel, use markup to make it readable. `;-)` (For the record, it already had punctuation, it was just one massive paragraph; and at least it wasn't all/no caps.)
Answer: I had this problem, but the CPU and GPU temperatures seemed reasonable. I'd recently changed the motherboard, so I wondered if there was a bad connection somewhere. Turns out that one of the "arms" that holds the CPU cooler in place wasn't 100% pressed down, so the CPU must have been operating at too high a temperature when gaming (though the temps didn't seem extreme to me). The computer was fine when I wasn't gaming.


Why is BCl3 a monomer whereas AlCl3 exists as a dimer?

Question: What makes dimerization possible in $\ce{AlCl3}$? Are there 3c-2e bonds in $\ce{Al2Cl6}$ as there are in $\ce{B2H6}$?
Comment: I think the question is related, but I think it is distinctively different also. On the other hand, the answer at the other question does not really explain why the boron compound does not dimerise.
Comment: Possible duplicate of [Why does AlCl3 dimerise to Al2Cl6 at low temperatures?](http://chemistry.stackexchange.com/questions/19715/why-does-alcl3-dimerise-to-al2cl6-at-low-temperatures)
Comment: Extra links in decreasing order of relevance: [1](https://www.researchgate.net/publication/253965076_Electronic_Structures_of_Molecules_XIII_Diborane_and_Related_Molecules), [2](http://pubs.acs.org/doi/abs/10.1021/acs.inorgchem.6b00118?journalCode=inocaj&#/doi/full/10.1021/acs.inorgchem.6b00118), [3](http://pubs.rsc.org/en/Content/ArticleLanding/1976/F2/F29767200539), [4](http://repositorio.ul.pt/bitstream/10451/8874/1/ulfc104237_tm_Nuno_Figueiredo.pdf), [5](http://www.sciencedirect.com/science/article/pii/0009261473802179), [6](http://www.tandfonline.com/doi/abs/10.1080/00268977100102081).
Answer: Introduction
The bonding situation in $\ce{(AlCl3)2}$ and $\ce{(BCl3)2}$ is nothing trivial and the reason why aluminium chloride forms dimers, while boron trichloride does not, cannot only be attributed to size.
In order to understand this phenomenon we need to look at both, the monomers and the dimers, and compare them to each other.
Understanding the respective bonding situation of the monomers, is key to understand which deficiencies lead to dimerisations.
Computational details
Since I was unable to find any compelling literature on the subject, I ran some calculations of my own. I used the DF-M06L/def2-TZVPP for geometry optimisations. Each structure has been optimised to a local minimum in their respective symmetry restrictions, i.e. $D_\mathrm{3h}$ for the monomers and $C_\mathrm{2v}$ for the dimers.
Analyses with the Natural Bond Orbital model (NBO6 program) and the Quantum Theory of Atoms in Molecules (QTAIM, MultiWFN) have been run on single point energy calculations at the M06/def2-QZVPP//DF-M06-L/def2-TZVPP level of theory.
A rudimentary energy decomposition analysis has been done on that level, too.
Energy decomposition analysis
The dissociation energy of the dimers $\ce{(XY3)2}$ to the monomers $\ce{XY3}$ is defined as the difference of the energy of the dimer $E_\mathrm{opt}[\ce{(XY3)2}]$ and double the energy of the monomer $E_\mathrm{opt}[\ce{XY3}]$ at their optimised (relaxed) geometries $\eqref{e-diss-def}$.
The interaction energy is defined as the difference of energy of the relaxed dimer and double the energy of the monomers in the geometry of the dimer $E_\mathrm{frag}[\ce{(XY3)^{\neq}}]$ $\eqref{e-int-def}$. That basically means breaking the molecule in two parts, but keeping these fragments in the same geometry.
The deformation energy (or preparation energy) is defined as the difference of the energy of the optimised and the non-optimised monomer $\eqref{e-def-def}$. This is the energy required to distort the monomer (in its ground state) to the configuration it will have in the dimer.
$$\begin{align}
E_\mathrm{diss} &=
  E_\mathrm{opt}[\ce{(XY3)2}] - 2E_\mathrm{opt}[\ce{XY3}]
  \tag1\label{e-diss-def}\\
E_\mathrm{int} &=
  E_\mathrm{opt}[\ce{(XY3)2}] - 2E_\mathrm{frag}[\ce{(XY3)^{\neq}}]
  %\ddag not implemented
  \tag2\label{e-int-def}\\
E_\mathrm{def} &=
  E_\mathrm{frag}[\ce{(XY3)^{\neq}}] - E_\mathrm{opt}[\ce{XY3}]
  \tag3\label{e-def-def}\\
E_\mathrm{diss} &=
  E_\mathrm{int} + 2E_\mathrm{def}\tag{1'}
\end{align}$$
Results & Discussion
The Monomers $\ce{XCl3; X{=}\{B,Al\}}$.
Let's just get the obvious out of the way: Boron is (vdW-radius 205 pm) smaller than aluminium (vdW-radius 240 pm). For comparison chlorine has a vdW-radius of 205 pm, too. That is pretty much reflected in the bond lengths and the chlorine-chlorine distance.
\begin{array}{llrrr}\hline
&\ce{X{=}}& \ce{Al} &\ce{B} &\ce{Cl}\\\hline
\mathbf{d}(\ce{X-Cl})&/\pu{pm} & 206.0 &173.6&--\\
\mathbf{d}(\ce{Cl\bond{~}Cl'})&/\pu{pm} & 356.8 & 300.6 & --\\\hline
\mathbf{r}_\mathrm{vdW}&/\pu{pm} & 240   & 205   & 205\\
\mathbf{r}_\mathrm{sing}&/\pu{pm} & 126   & 85   & 99\\
\mathbf{r}_\mathrm{doub}&/\pu{pm} & 113   & 78   & 95\\\hline
\end{array}
From this data we can draw certain conclusions without further looking. The boron monomer is much more compact than the aluminium monomer. When we compare the bond lengths to the covalent radii (Pyykkö and Atsumi) we find that the boron chloride bond is about the length that we would expect from a double bond ($\mathbf{r}_\mathrm{doub}(\ce{B}) + \mathbf{r}_\mathrm{doub}(\ce{Cl}) = 173~\pu{pm}$). While the aluminium chloride bond is still significantly shorter than a single bond ($\mathbf{r}_\mathrm{sing}(\ce{Al}) + \mathbf{r}_\mathrm{sing}(\ce{Cl}) = 225~\pu{pm}$), it is still also much longer than a double bond ($\mathbf{r}_\mathrm{doub}(\ce{Al}) + \mathbf{r}_\mathrm{doub}(\ce{Cl}) = 191~\pu{pm}$).
This itself offers compelling evidence, that there is more π-backbonding in $\ce{BCl3}$ than in $\ce{AlCl3}$. Molecular orbital theory offers more evidence for this. In both compounds is a doubly occupied π orbital. The following pictures are for a contour value of 0.05; aluminium (left/top) and boron (right/bottom)

In numbers, the main contributions are as follows (this is just a representation, not the actual formula):
$$\begin{align}
\pi(\ce{BCl3}) &=
   21\%~\ce{p_{$z$}-B} + \sum_{i=1}^3 26\%~\ce{p_{$z$}-Cl^{$(i)$}}\\
\pi(\ce{AlCl3}) &=
   13\%~\ce{p_{$z$}-Al} + \sum_{i=1}^3 29\%~\ce{p_{$z$}-Cl^{$(i)$}}
\end{align}$$
There is still some more evidence. The natural atomic charges (NPA of NBO6) fairly well agree with that assesment; aluminium is far more positive than boron.
$$\begin{array}{lrr}
                    & \ce{AlCl3} & \ce{BCl3}\\\hline
\mathbf{q}(\ce{X})~\text{[NPA]}  & +1.4       & +0.3     \\
\mathbf{q}(\ce{Cl})~\text{[NPA]} & -0.5       & -0.1     \\\hline
%\mathbf{q}(\ce{X})~\text{[QTAIM]}  & +2.4       & +2.0    \\
%\mathbf{q}(\ce{Cl})~\text{[QTAIM]} & -0.8       & -0.7     \\\hline
\end{array}$$
The analysis in terms of QTAIM also shows that the bonds in $\ce{AlCl3}$ they are predominantly ionic (left/top) while in $\ce{BCl3}$ are predominantly covalent (right/bottom).

One final thought on the bonding can be supplied with a natural resonance theory analysis (NBO6). I have chosen the following starting configurations and let the program calculate their contribution.

The overall structures in terms of resonance are the same for both cases, that is if you force resonance treatment of the aluminium monomer. Structure A does not contribute, while the others contribute to about 31%. However, when not forced into resonance, structure A is the best approximation of the bonding situation for $\ce{AlCl3}$. In the case of $\ce{BCl3}$ the algorithm finds a hyperbond between the chlorine atoms, a strongly delocalised bond between multiple centres. In this case these are 3-centre-4-electron bonds between the chlorine atoms, resulting from the higher lying degenerated π orbitals.

This all is quite good evidence that the monomer of boron chloride should be more stable towards dimerisation than the monomer of aluminium.
The Dimers $\ce{(XCl3)2; X{=}\{B,Al\}}$.
The obvious change is that the co-ordination of the central elements goes from trigonal planar to distorted tertrahedral. A look at the geometries will give us something to talk about.
\begin{array}{llrrr}\hline
&\ce{X{=}}& \ce{Al} &\ce{B} &\ce{Cl}\\\hline
\mathbf{d}(\ce{X-Cl})&/\pu{pm} & 206.7 &175.9&--\\
\mathbf{d}(\ce{X-{\mu}Cl})&/\pu{pm} & 226.1 &198.7&--\\
\mathbf{d}(\ce{Cl\bond{~}{\mu}Cl})&/\pu{pm} & 354.1 & 308.0 & --\\
\mathbf{d}(\ce{{\mu}Cl\bond{~}{\mu}Cl'})&/\pu{pm} & 323.6 & 287.3 & --\\
\mathbf{d}(\ce{B\bond{~}B'})&/\pu{pm} & 315.7 & 274.7 & --\\\hline
\mathbf{r}_\mathrm{vdW}&/\pu{pm} & 240   & 205   & 205\\
\mathbf{r}_\mathrm{sing}&/\pu{pm} & 126   & 85   & 99\\
\mathbf{r}_\mathrm{doub}&/\pu{pm} & 113   & 78   & 95\\\hline
\end{array}
In principle nothing much changes other than the expected elongation of the bonds that are now bridging. In case of aluminium the stretch is just below 10% and for boron it is slightly above 14%, having a bit more impact. In the boron dimer also the terminal bonds are slightly (> +1%) affected, while for aluminium there is almost no change.
The charges are not really a reliable tool, especially when they are that close to zero as they are for boron. In both cases one can see that charge density is transferred from the bridging chlorine to the central $\ce{X}$.
$$\begin{array}{lrr}
                    & \ce{(AlCl3)2} & \ce{(BCl3)2}\\\hline
\mathbf{q}(\ce{X})~\text{[NPA]}  & +1.3       & +0.2     \\
\mathbf{q}(\ce{Cl})~\text{[NPA]} & -0.5       & -0.1     \\\hline
\mathbf{q}(\ce{{\mu}Cl})~\text{[NPA]} & -0.4       & +0.1     \\\hline
\end{array}$$
A look at the central four-membered-ring of in terms of QTAIM offers that the overall bonding does not change. In aluminium they get a little more ionic, while in boron they stay largely covalent.

The NBO analysis offers a maybe quite surprising result. There are no hyperbonds in any of the dimers. While a description in these terms is certainly possible, after all it is just an interpretation tool, it is completely unnecessary. So after all we have two kinds of bonds in the dimers four terminal $\ce{X-Cl}$ and four bridging $\ce{X-{\mu}Cl}$ bonds. Therefore the most accurate description is with formal charges (also the simplest). The notation with the arrows is not wrong, but it does not represent the fact that the bonds are equal for symmetry reasons alone.

To make this straight: There are no hyperbonds in $\ce{(XCl3)2; X{=}\{B,Al\}}$; this includes three-centre-two-electron bonds, and three-centre-four-electron bonds. And deeper insight to those will be offered on another day.
The differentiation between a dative bond and some other for of bond does not make sense, as the bonds are equal and only introduced by a deficiency of the used description model.
A natural resonance theory for $\ce{(BCl3)2}$ gives us a overall contribution of the main (all single bonds) structure of 46%; while all other structure do contribute, there are too many and their contribution is too little (> 5%). I did not run this analysis for the aluminium case as I did not expect any more insight and I did not want to waste calculation time.
Dimerisation - yes or no
The energies offer us a clear trend. Aluminium likes to dimerise, boron not. However, there are still some things to discuss. I am going to argue for the reaction
$$\begin{align}
\ce{2XCl3 &-> (XCl3)2}&
\Delta E-\mathrm{diss}/E_\mathrm{o}/H/G&.
\end{align};$$
therefore if reaction energies are negative the dimerisation is favoured.
The following table includes all calculated energies, including the energy decomposition analysis mentioned at the beginning. All energies are given in $\pu{kJ mol^-1}$.
\begin{array}{lrcrcrcrr}
 \Delta & E_\mathrm{diss} &(& E_\mathrm{int} &+2\times& E_\mathrm{def}&)& E_\mathrm{o} &H &G\\\hline
\ce{Al} & -113.5 &(& -224.2 &+2\times& 55.4&)& -114.7 & -60.4 & -230.4\\
\ce{B}  &   76.4 &(& -111.2 &+2\times& 93.8&)&   82.6 & -47.1 &  152.5\\\hline
\end{array}
The result is fairly obvious at first. The association for aluminium is strongly exergonic, while for boron it is strongly endergonic. While both reactions should be exothermic, stronger for aluminium, the trend for the observed electronic energies ($E_\mathrm{o}$ including the zero-point energy correction) and the (electronic) dissociation energies reflect the overall trend for the Gibbs enthalpies.
While it is fairly surprising how strongly entropy favours association of $\ce{AlCl3}$, it is also surprising how it strongly disfavours it for $\ce{BCl3}$.
A look at the decomposed electronic energy offers great insight into the reasons why one dimer is stable and the other not (at room temperature).
The interaction energy of the fragments is double for aluminium than it is for boron. This can be traced back to the very large difference in the atomic partial charges. One could expect that the electrostatic energy is a lot more attractive for aluminium than it is for boron.
The deformation energy on the other hand clearly reflects the changes in the geometry discussed above. For aluminium there is a smaller penalty resulting from the elongation of the $\ce{Al-Cl}$ bond and pyramidalisation. For boron on the other hand this has a 1.5 times larger effect. The distortion also weakens the π-backbonding, which the additional bonding would need to compensate.
The four-membered-ring is certainly not an ideal geometry and the bridging chlorine atoms come dangerously close.
Conclusion, Summary and TL;DR:
The distortion of the geometry of the monomer $\ce{BCl3}$ cannot be compensated by the additional bonding between the two fragments. Therefore the monomers are more stable than the dimer. Additionally entropy considerations at room temperature favour the monomer, too.
On the other hand, the distortion of the molecular geometry in $\ce{AlCl3}$ is less severe. The gain in interaction energy of the two fragments well overcompensates for the change. Entropy also favours the dimerisation.
While size of the central atom is certainly the distinguishing factor, its impact is only severe on  the electronic structure. Steric crowding would not be a problem when the interaction energy would compensate for that. This is quite evident because $\ce{BCl3}$ is still a very good Lewis acid and forms stable compounds with much larger moieties than itself.
References
The used van der Waals radii were taken from S. S. Batsanov Inorg. Mat. 2001, 37 (9), 871-885. And the covalent radii have been taken from P. Pyykkö and M. Atsumi Chem. Eur. J. 2009, 15, 12770-12779.
Computations have been carried out using Gaussian 09 rev D.01 with NBO 6.0. Additional analyses have been performed with MultiWFN 3.3.8. Orbital pictures were generated with the incredible ChemCraft.
Comment: Martin, It would be good to perform similar analisys for boron trifluoride. It also is monomeric, by fluorine is oh so smaller than chlorine, so other factors should dominate.
Comment: @orthocresol entropy is hugely influenced by vibrational zero-energy and it falls rapidly with lowered mass of the atoms involved.
Comment: @ortho (D2h is correct, don't know where my brains went wrong - it's a long post.) There seems to be a slight numerical error in the calculation which causes (AlCl3)2 to have a tiny imaginary mode. While this usually has no effect, it can lead to wrong corrections and this is more predominant in such small systems. [Gaussian](http://www.gaussian.com/g_whitepap/thermo.htm) has a paper on how they do thermochemistry. I'll look into this further when I get the chance next week, maybe lifting the symmetry restriction will run into the minimum.
Comment: Nice. Although the entropy does intrigue me - that $-T\Delta S$ contribution is huge. I find it counter-intuitive that the entropy change can be so drastically different for two reactions that are fundamentally the same: $\ce{2A <=> A2}$. (Also would have expected the dimers to have $D_\mathrm{2h}$ symmetry.)
Comment: @permeakra This is indeed an interesting question, and when I get the chance I'll have a look. I think the reasoning will be very similar as the pi bonding should also increase with a smaller bond length. But we'll see.
Answer: TL;DR

$\ce{BCl3}$ does not dimerize to $\ce{B2Cl6}$ due to a conflict between the short $\ce{B-Cl}$ bond length in $\ce{BCl3}$ $(1.74~\mathring{\mathrm{A}}$ in my calculations$)$ and the long $\ce{B-Cl}$ bond lengths $(\sim 2~\mathring{\mathrm{A}})$ that would be required for the $\ce{B2Cl2}$ core of a $\ce{B2Cl6}$ dimer.
$\ce{B2Cl6}$ does appear to have a positive (albeit small) dissociation energy, however, forming a van der Waals complex at a $\ce{B-B}$ distance of $3.63~\mathring{\mathrm{A}}$. Thus, this weakly-bound dimeric form may have the potential to form at very low temperatures.
$\ce{Al2Cl6}$ does not possess 2e-3c $\ce{Al-Cl-Al}$ bonds, in contrast to $\ce{B2H6}$. Instead, each of the four $\ce{Al-Cl}$ bonds in the $\ce{Al2Cl2}$ core is moderately 'sub-valent,' carrying $\sim1.5~e^–$.
The valence orbital localization and orbital shapes of an optimized $\ce{B2Cl6}$ system with one $\ce{B-Cl}$ bond on each boron constrained to $2.08~\mathring{\mathrm{A}}$ are very similar to those of the fully optimized $\ce{Al2Cl6}$, in agreement with Martin's results.

These conclusions are based upon ORCA v3.0.3 DFT calculations and subsequent MultiWFN 3.3.7 electron density and ELF topology analysis on the three dimeric systems: $\ce{B2H6}$, $\ce{B2Cl6}$, and $\ce{Al2Cl6}$. Where used, "$\mathrm{E_h}$" indicates the Hartree energy unit.

As a first entry into the question, I performed relaxed surface scans on each dimer system, adjusting the $\ce{B-B}$/$\ce{Al-Al}$ distance from $9.0~\mathring{\mathrm{A}}$ down to $2.0~\mathring{\mathrm{A}}$ ($1.5~\mathring{\mathrm{A}}$ for $\ce{B2H6}$). A representative ORCA input file for these scans is:
<code>! RKS PBE0 def2-SVP def2-SVP/J RIJCOSX D3BJ
! GRID4 GRIDX5 TIGHTOPT
! PRINTBASIS

%output  PrintLevel Small  end

%geom
   Scan
      B 0 4 = 9.0, 1.5, 16
      end
   end

* xyz 0 1
  B -4.91130   1.95714  -6.10613
  H -3.80330   1.95714  -6.10613
  H -5.46530   2.39076  -5.25014
  H -5.46530   1.52351  -6.96212
  B -2.06337  -4.02491   0.00000
  H -0.95537  -4.02491   0.00000
  H -2.61737  -4.79680  -0.57003
  H -2.61737  -3.25303   0.57003
*
</code>
Note that unlike in Martin's calculations, I did not impose any symmetry constraints.
The figure below shows the potential energy curves obtained for these scans (scale increased in the right panel to show $\ce{B2Cl6}$ more clearly; click to enlarge):

The solid curves are the best-fit Morse potentials, determined by minimizing the sum-squared deviation of the fit from the data:
$$
\begin{array}{ccccc}
\hline
 & D_e~(\mathrm{E_h}) & D_e~\left(\mathrm{kcal\over mol}\right)
   & a~\left(\mathring{\mathrm{A}}^{-1}\right)
   & r_e~\left(\mathring{\mathrm{A}}\right) \\
\hline
\ce{B2H6}   &  0.0867  &  54.41  &  2.3304  &  1.7025 \\
\ce{B2Cl6}  &  0.00599 &  3.76   &  0.79832 & 3.9632 \\
\ce{Al2Cl6} &  0.0532  &  33.36  &  0.97947 & 3.0259 \\
\hline
\end{array}
$$
To note, these data likely do not run along any particular internal reaction coordinate. The images below are of the optimized geometries (left: $\ce{B2H6}$; center: $\ce{B2Cl6}$; right: $\ce{Al2Cl6}$); click each to display an animated GIF of the optimized geometries at each step of the respective scan, which show the irregularities in the progression of the scans (all GIFs pause at the step closest to the optimized geometry):

Despite these irregularities, the energy profiles seem qualitatively reasonable.  Of particular interest is the fact that $\ce{B2Cl6}$ may actually form a van der Waals complex, enabling it to be bound at a sufficiently low temperature (plotted at the $0.5$ isosurface of the reduced density gradient):

So, these computations agree with the experimental observation that $\ce{BCl3}$ does not dimerize in the way that $\ce{BH3}$ and $\ce{AlCl3}$ do.

In the course of exploring the $\ce{B2Cl6}$ system, I performed a constrained optimization where I locked one $\ce{B-Cl}$ bond on each boron atom to be considerably longer than its equilibrium value, starting from the above optimized geometry. I arbitrarily chose $2.08~\mathring{\mathrm{A}}$, which is the length of the terminal $\ce{Al-Cl}$ bonds in $\ce{Al2Cl6}$.  Once the energetic cost is paid to stretch the two $\ce{B-Cl}$ bonds, an appreciable minimum in the energy profile is exhibited:

The data in red is the same $\ce{B2Cl6}$ data from above; that in yellow is with the two $\ce{B-Cl}$ bonds constrained at $2.08~\mathring{\mathrm{A}}$. All energies are relative to the relaxed system at a $9~\mathring{\mathrm{A}}$ $\ce{B-B}$ separation.
The resulting optimized structure under this constraint is below; as above, click the image to view an animated GIF of the progression of the constrained optimization:

I suspect that the symmetry-constrained $\ce{B2Cl6}$ species studied by Martin is very similar to this constrained-optimized structure, and if he had relaxed the symmetry constraint he would have obtained the van der Waals complex above.  Based on his data, though, it appears my choice of $2.08~\mathring{\mathrm{A}}$ for the stretched $\ce{B-Cl}$ bonds was too large by $0.1~\mathring{\mathrm{A}}$ or so.
In any event, the fictional "pseudo-dissociation curve" for this "extended $\ce{B-Cl}$" system follows much the same form as that for $\ce{Al2Cl6}$:

The '$\ce{B2Cl6}$-extended' data in the above chart was shifted vertically so that $E[9~\mathring{\mathrm{A}}] = 0$, in order to illustrate its qualitative similarity to the $\ce{Al2Cl6}$ data.
So, simply stretching one $\ce{B-Cl}$ bond at each boron atom is sufficient to induce $\ce{B2Cl6}$ to behave in substantially the same fashion as $\ce{Al2Cl6}$.

$\ce{B2H6}$ does show the expected 2e-3c bonding behavior.
ORCA localization of the $\ce{B2H6}$ valence orbitals (omitting the $\ce{B}$ core orbitals) indicates their presence (the bridging hydrogens are <code>1H</code> and <code>5H</code>):
<code>----------------------------------------
LOCALIZED MOLECULAR ORBITAL COMPOSITIONS
----------------------------------------
The Mulliken populations for each LMO on each atom are computed
The LMO`s will be ordered according to atom index and type
  (A) Strongly localized MO`s have populations of >=0.950 on one atom
  (B) Two center bond orbitals have populations of >=0.850 on two atoms
  (C) Other MO`s are considered to be `delocalized`
FOUND  -   0 strongly local MO`s
       -   4 two center bond MO`s
       -   2 significantly delocalized MO`s
Bond-like localized orbitals:
MO   5:   7H  -   0.494574  and   4B  -   0.525955
MO   4:   6H  -   0.494586  and   4B  -   0.525936
MO   3:   3H  -   0.494588  and   0B  -   0.525939
MO   2:   2H  -   0.494573  and   0B  -   0.525953
More delocalized orbitals:
MO   7:  0B - 0.269 1H - 0.478 4B - 0.269
MO   6:  0B - 0.269 4B - 0.269 5H - 0.478
</code>
Both the localized orbitals and the QTAIM LCP paths are consistent with the 2e-3c bond paradigm, and ELF basin analysis finds attractors at the bridging hydrogen atoms, whose basins are calculated to contain $1.94~e^-$ each, very close to the expected value of $2.0$.

Orbital plots here and below are at the $\phi = \pm 0.02$ isosurfaces.
On the other hand, $\ce{Al2Cl6}$ does not show multicenter bonding upon localization of the valence orbitals (the bridging atoms here are <code>7Cl</code> and <code>1Cl</code>):
<code>FOUND  -   8 strongly local MO`s
       -  16 two center bond MO`s
       -   0 significantly delocalized MO`s
Rather strongly localized orbitals:
MO  47:   7Cl -   0.955366
MO  46:   7Cl -   1.001974
MO  45:   6Cl -   1.007096
MO  44:   5Cl -   1.007111
MO  43:   3Cl -   1.007113
MO  42:   2Cl -   1.007094
MO  41:   1Cl -   0.955367
MO  40:   1Cl -   1.001973
Bond-like localized orbitals:
MO  63:   7Cl -   0.844289  and   4Al -   0.149451
MO  62:   7Cl -   0.843809  and   0Al -   0.150019
MO  61:   6Cl -   0.932028  and   4Al -   0.074001
MO  60:   6Cl -   0.938027  and   4Al -   0.063923
MO  59:   6Cl -   0.746625  and   4Al -   0.250592
MO  58:   5Cl -   0.931694  and   4Al -   0.074421
MO  57:   5Cl -   0.938034  and   4Al -   0.063923
MO  56:   5Cl -   0.746451  and   4Al -   0.250727
MO  55:   3Cl -   0.931694  and   0Al -   0.074422
MO  54:   3Cl -   0.938034  and   0Al -   0.063923
MO  53:   3Cl -   0.746450  and   0Al -   0.250728
MO  52:   2Cl -   0.938028  and   0Al -   0.063922
MO  51:   2Cl -   0.932027  and   0Al -   0.074001
MO  50:   2Cl -   0.746627  and   0Al -   0.250589
MO  49:   1Cl -   0.843808  and   0Al -   0.006200
MO  48:   1Cl -   0.844290  and   0Al -   0.149450
</code>
The shape of the bond orbitals in the $\ce{Al2Cl2}$ core is consistent with strictly two-center bonding:

Interestingly, the valence orbital localization pattern and core bond orbital structures in the 'extended $\ce{B-Cl}$' system above are very similar to $\ce{Al2Cl6}$:
<code>FOUND  -   8 strongly local MO`s
       -   16 two center bond MO`s
       -   0 significantly delocalized MO`s
Rather strongly localized orbitals:
MO  39:   7Cl -   0.967482
MO  38:   7Cl -   1.006838
MO  37:   6Cl -   1.008398
MO  36:   5Cl -   1.008336
MO  35:   3Cl -   0.967481
MO  34:   3Cl -   1.006837
MO  33:   2Cl -   1.008334
MO  32:   1Cl -   1.008398
Bond-like localized orbitals:
MO  55:   7Cl -   0.830908  and   4B  -   0.171645
MO  54:   7Cl -   0.769240  and   0B  -   0.232367
MO  53:   6Cl -   0.941563  and   4B  -   0.066433
MO  52:   6Cl -   0.931032  and   4B  -   0.073365
MO  51:   6Cl -   0.658532  and   4B  -   0.343433
MO  50:   5Cl -   0.941702  and   4B  -   0.066259
MO  49:   5Cl -   0.930994  and   4B  -   0.073395
MO  48:   5Cl -   0.658437  and   4B  -   0.343525
MO  47:   4B  -   0.232364  and   3Cl -   0.769241
MO  46:   3Cl -   0.830907  and   0B  -   0.171646
MO  45:   2Cl -   0.941703  and   0B  -   0.066256
MO  44:   2Cl -   0.930995  and   0B  -   0.073394
MO  43:   2Cl -   0.658441  and   0B  -   0.343521
MO  42:   1Cl -   0.931031  and   0B  -   0.073367
MO  41:   1Cl -   0.941565  and   0B  -   0.066432
MO  40:   1Cl -   0.658533  and   0B  -   0.343432
</code>

Of particular interest, in both chlorine systems, all of the ELF basins with attractors lying along the $\ce{B-Cl}$ / $\ce{Al-Cl}$ bonds each have an integrated electron density of $\sim 1.5~e^-$. Thus, taken together, the four bonds in each rectangular core comprise six electrons in total.  While I agree with Martin that there are no multicenter bonds here, there does appear to be an interesting sort of resonance phenomenon at play.
In sum: $\ce{B2H6}$ does display 2e-3c bonding; $\ce{Al2Cl6}$ does not.
Comment: I get what you're saying, that there is a _minimum_ along that coordinate, but to call it bound would be wrong. Fixing the wording still doesn't correct that either the first plot or second plot is wrong in some way (look at the energy scale, then look at the "extended" curves).
Comment: "Once the energetic cost is paid to stretch the two B−Cl bonds, an appreciable binding energy is exhibited." At no point in that plot is the extended $\ce{(BCl3)2}$ bound (it has a positive energy at all points), yet in the next plot you show it as being bound.
Comment: @pentavalentcarbon Mm, did I not emphasize enough that the dissociation energy curve for the "extended" system is fictional? I can add that....
Answer:
What makes dimerization possible in AlCl3?

Aluminium trichloride is an electron deficient molecule and it tends to get stability by accepting a lone pair of electron from Cl atom of another AlCl3 molecule so that it now has an octet in outer shell.

Are there 3c-2e bonds in Al2Cl6 as there are in B2H6?

No, the Cl forms dative bond.

Why is BCl3 a monomer whereas AlCl3 exists as a dimer?

STABILITY, As it is seen in the following figure

The bond length and atomic size of boron in BCl3 are smaller than in AlCl3 so this leads to interatomic repulsion so BCl3 could not accept another BCl3 via a dative bond because it leads to more repulsion and finally breaking of dimer. Also some degree of pi backbonding is present which may make Cl unable to form dative bond.
One may argue that BCl4- exists but in this cases size of 4th entity which binds to Boron will be smaller than a BCl3 monomer.
Comment: I find your last sentence hard to believe. The immediate surrounding of the boron atom is identical in $\ce{BCl4-}$ and $\ce{B2Cl6}$.  The extra boron atom on the far side doesn't interact with the first.
Comment: @Jan but that would prevent dimerisation by straining the bonds.
Comment: I think your answers to the first two questions are spot on, from Cotton and Wilkinson "formation of dimers is attributable to the tendency of the metal atoms to complete their octets". AlCl3 also exists as a monomer in the gas phase near the b.p., the dH for disassociation is between 46-63 kJ/mol. For the third question, I think the MO diagram of B2H4 vs AlCl3 would help explain why the MO levels of B and H allow 3C-2E bonds while Al and Cl do not.
Answer: It is said that $\ce{AlCl_3}$ exist as tetra-coordinate aluminium when as dimer, meaning the aluminium has an $sp^3$ like structure with a coordinate bond provided by a $\ce{Cl}$ atom.
The $\ce{BX3}$ halides attain an octet by dative $\pi_{p-p}$ interaction, forming a four-centre $\pi$ molecular orbital covering all four atoms which contains two bonding electrons. This is not feasible in case of other elements in the group. They have larger atoms and cannot get effective $\pi$ overlap, so they polymerize to remedy the electron deficiency.
In case of $\ce{B2H6}$, 3 centred-2 electron bonds are formed because boron as well as hydrogen do not have enough electrons to form a classic single bond. But here, chlorine does. So there is no 3-c 2-e bonds and instead there are coordinate covalent bonds.
As dimer, they exist as 3 centred-4 electron non-hypervalent bond, forming a delocalized dative bond as shown:

Also note that similar structure is shown by $\ce{(BeCl2)_n}$ polymer, with beryllium atom having $sp^3$ hybridisation.
This type of 3c - 4e bonds are not possible in case of boron because boron is too small and the chlorine atoms will not fit in between them to form the bridge. Thus boron handles it through back bonding.
Comment: @Efram The Al-µCl-Al are 3c-4e-bonds, because symmetry dictates so. Cotton and Wilkerson are oversimplifying and the concept of the dative bond is just a book keeping tool.
Comment: I dont think the AlCl3 dimer is considered a 3C-4E bond? I think itbis more dative or backbonding of non bonding Cl electrons to fill an Al oxtet, the multi center bond requires a MO orbital with equal distribution? Do you have a source for that?
Comment: Bond length of $\ce{Al-Cl}$ bridges are equal to $2.3a$ with $2e^-$ in each bond. They all have the same bond orders (Mulliken), $\approx 0.66$. Visit [Schupf Computational Chemistry Lab](http://www.colby.edu/chemistry/webmo/Al2Cl6.html)
Comment: The definition of a multicenter multielectron bond is 3-center orbitals one bonding, one non bonding, and antibonding. An open 3 center bond is called 3 center 4 electron, where in addition to uniform electron distribution in the bonding orbital, there are a pair of electrons in the non bonding orbitals, ENTIRELY concentrated on the end atoms, giving a relatively polar bond. I do not think Al2Cl3 meets those requirements. (From Cotton&Wilkerson verbatim)
Comment: In Al2Cl6 you have 2 electrons in 2 Al-Cl bridges for a total of 4 electrons, but those 4 electrons are not distributed relative to eachother in a 3 centered orbital with 2 electrons in a bonding, and the other two removed from the center orbital in non bonding orbitals.
Comment: I found a reference for the dH association for borane to diborane giving a value of -142 kJ/mol vs dH for AlCl3 dimer formation is ~ 52 kJ/mol. The first value for diborane makes sense as a covalent bond is being formed (multicenter multielectron) and the smaller positive dH for AlCl3 is consistent with a lewis-acid lewis base dative bond interaction.
Comment: Sorry, the -142 kJ/mol is for association, and the +52 kJ/mol is for disassociation, so the bond enthalpy for diborane is +142 kJ/mol.
Comment: @EframGoldberg The dative bond is continuously being switched. That is why I believe they are distributed. So you are saying they exist as Structures 1 and 2 only and not 3, in the above picture?
Comment: @EframGoldberg, what do you say in case of $\ce{(BeCl2)_n}$? I'm pretty sure that it has a 3c-4e bond.
Comment: So are the only requirements for multicenter multielectron bonding just symmetry and equal distribution? Wouldnt all dibridged dinuclear complexes be considered the same?
Comment: @EframGoldberg According to my text book, they are considered same. But notice one thing, the structures we both refer to are in fact the same. Only difference lies in how we define multi centred bonds. I refer to Inorganic Chemistry by J D LEE for these structures.
Answer: In $\ce{AlCl3}$, $\ce{Al}$ has 6 valence $\ce{e-}$ now to complete its octet it needs 2 more $\ce{e-}$. So it accepts a pair of electron from the $\ce{Cl}$ of the neighboring $\ce{AlCl3}$. This donation of electron results in the formation of dimer.

But in $\ce{BCl3}$, due to the small size and the extent of π-bonding in $\ce{B}$ results in the short bond length, which also prevents $\ce{Cl}$ to form coordinate bond. And due to this short bond length when they dimerises (like in the above structure) chlorine atoms are closer than that in $\ce{AlCl3}$ and it breaks so it exist as monomer
Comment: $\ce{BCl4^-}$ does exist, so the size argument is void.
Comment: The second boron atom in the ring would strain the bonds, so a direct comparison to BCl4- cannot really be made.
Answer: A planar four membered ring with equal sides should form in both cases. The optimum bond angles would be 90 degrees within the ring. (This is independent of the bond lengths.)
Both the B and the Al atoms are four coordinate and thus would prefer a tetrahedral arrangement (109.5 degrees).
90 degrees is not too bad for the Cl-Al-Cl angle, (see H2S).
However the ring is too strained in the case of boron, the Cl-B-Cl angle is too strained.
It is another question why B and C atom show much higher strain compared to heavier counterparts when distorted from the tetrahedral arrangement.
Comment: Actual bond angles in Al2Cl6 are 93, 87, 110, and 122 far from tetrahedral.
Comment: 110 is not far actually from tetrahedral, but the argument was about the angles in the ring. These are close to 90 in Al2Cl6, which is tolerable for elements not in the 2nd period, but the same would cause too much strain in B2Cl6.
Answer: BH3 and AlCl3 each form "dimers" but by different processes.

AlCl3 is trigonal planar but in vapor phase at certain temperatures, it can dimerize to Al2Cl6 with a dH of ~ 46-63 kJ/mol. A few degrees above the boiling point, and the monomer is predominant. This happens by change of geometry from trigonal planar to tetrahedral and backbonding of Cl pi electrons to fill the octet of Al giving a dimer. This is possible due to the low relative energy difference between Cl and Al valence orbitals in the MO diagram. See http://pages.swcp.com/~jmw-mcw/Parsing%20the%20Bonding%20in%20the%20Aluminum%20Trichloride%20Dimer.htm for a good overview of bond angles and lengths.
BH3 is tetrahedral and forms a dimer  B2H6 but through a different process. The bonding is described by Cotton&Wilkerson as: "each boron atom uses two electrons and two roughly sp3 orbitals to form 2 center 2 electron (covalent) bonds to two hydrogen atoms. The boron atom in each BH2 group still has one electron and two hybrid orbitals for use in further bonding. The plane of the two remaining orbitals is perpendicular to the BH2 plane. Thus when two BH2 radicals approach eachother with the with the hydrogen atoms also in the plane, a B-H-B 3 center 2 electron bond is formed." The MO orbital diagram for the 3C2E bond requires close energy of the three atomic orbitals which is met between boron and hydrogen.
BCl3 does not form dimers as others have mentioned above, the difference in atomic radii between B and Cl explain the instability of the potential dimer. Additionally, larger difference in orbital energy would make such bonding unfavorable.
Comment: The website you cited is borderline pseudoscience. The author published that web page (which cites - amongst other articles by himself - Wikipedia) in a non-peer-reviewed online "journal" and is most "famously" known for arguing against the spdf model of atomic orbitals in favour of his own construction. One downvote doesn't do justice to any answer that uses him as a reference. cc @Martin
Comment: This is as wrong as most of the textbooks on this topic. Your cited source in 1 calls three-centre bonds an aberration, clearly an indicator that they have not understood the first thing about bonding. They try to explain their reasoning by using the hybridisation model completely and utterly wrong. The dimerisation process (the bonding?) is pretty much the same for BH3 and AlCl3. The size argument is only a small part of a more complicated matter. All in all, this answer just repeats what most of the others already stated.
Comment: @orthocresol I was only citing the website for the bond angles, I did not read the rest of it. I actually use Cotton & Wilkerson's definition for multicenter multielectron bonding. I think my answer adds some thermodynamic values to help show the difference and I think is the only answer to mention MO theory. Either way, if the other responses are adequate, then should the question be considered answered?


Apache: show files in a directory

Question: I have a private directory with some files, I'm able to do an HTTP authentication but when I try to show the files in the directory I get a Forbidden Error. How i can edit my .htaccess file in a way to show the directory content?
This is my .htaccess now:
AuthUserFile /private/.htpasswd
AuthName "Private"
AuthType Basic
require user member

Answer: You have to turn on directory listings. Add this:
Options +Indexes

Apache: show files in a directory


Inverse Fourier Transforms in 3D

Question: How do you calculate the inverse Fourier transform
http://www.fuw.edu.pl/~mklis/publications/Hydro/oseen.pdf
Answer: Here is one way to do it. Multiplication by $k_ik_j$ on the Fourier transform side corresponds to taking the second derivative $\dfrac{\partial^2 }{\partial x_i \partial x_j}$. Therefore, we can find the inverse Fourier transform of $1/k^4$ and then take the Hessian of that. This is also preferable (to me) because it reduces the task to transforming a scalar function.
Following the computations here, one can see that the transform of a radially symmetric function is radially symmetric. Also, the same computation implies that the transform of a function that is homogeneous of degree $d$ is homogeneous of degree $-(d+n)$ where $n$ is the dimension ($n=3$ here). Therefore, the IFT of $1/k^4$ is of the form $f(x)=c|x|$. Of course, the value of $c$ can be found once we decided on the normalization of the transform*.
The Hessian of $|x|$ is not hard to compute: $D |x| = x|x|^{-1}$, and then by the product rule $$D^2(|x|)=I|x|^{-1}-(x\otimes x)|x|^{-3}\tag1$$ This is exactly what they get for $\mathcal{F}^{-1}(k_ik_j/k^4)$, save for the factor $1/(8\pi)$.

 Naturally, I'm ignoring all convergence issues.
Comment: Thank you thank you that way is A LOT easier than what i was trying to do!

Inverse Fourier Transforms in 3D