Table Formating with Equations Inside

As you can put whatever you want in a table environment, it is simpler to put inside an flalign*. I added the caption package to have a more decent spacing under the caption

1. Does altering a table and adding multiple columns rebuild the table multiple times

Will there be any difference between the above two approaches, in terms of how expensive the operation will be?No in practice. Because in the second approach the operations execution is one-by-one, not parallel. Cause of this is simple - in theory the next step can be dependent by previous one and so impossible until next step fully performed. What are the best practices when performing such operations?Create new table copy data (lock source table copy data changes in source made by another clients while first copy) drop old rename new restore access rights, FKs and triggers. What does a table rebuild, really mean?In practice this procedure is performed as described in answer 2 exclusive lock on the table modified.Is there a way to avoid a table rebuild, or make the operation faster: (ALGORITHM=INPLACE) ?Add separate table with new columns linked to original table as 1:1 instead.

2. Create HTML table choice item displaying code in the table output

Recapping our discussion: If you are using the Advanced -> Column Options -> Custom, your can set the expression for that item to point to the Value in the JSON sub-object rather than directly at the column object. For a choice column named Change Outcome, that expression would look like:User fields are a similar case of having a subject with multiple values, but the properties are different. A User field from SharePoint will look like this in your results:So, if you wanted to just show the user's email in your table, you would reference:.

3. Sideways in table

You do not need parbox as you have only one word in the first column. Further you can use the second (optional) argument of multirow to adjust the vertical placement like:Adjust it appropriately. Further, the first column can be of c type. Code:.

4. The backup GPT table is corrupt

First, what tools and procedures did you use to get to the point of the "backup GPT table is corrupt" message, and is that error message reproducible by reproducing those steps? It could be that you did something wrong in creating the partitions, and that this is causing problems, but without knowing how you created the partition table and filesystems, that's impossible to know for sure.Second, if the error is reproducible and if you are using common partitioning tools (such as Disk Utility in OS X, GParted or gdisk in Linux, etc.), then it's very unlikely that you would get a consistent and reproducible problem in the partition table. This makes me think that you may have a defective USB flash drive on your hands. I can not be sure of that, though, particularly if you used some oddball tool to create the partitions.Third, if a partition (such as /dev/sdb2) is inaccessible, you may need to use filesystem repair tools such as fsck on it. Further diagnosis would require knowledge of what filesystem it is (I would guess HFS, if you intend to put OS X on it). The output of sudo fsck /dev/sdb2 and/or sudo blkid /dev/sdb2 would be useful in further diagnosis.Fourth, if you are trying to create a bootable OS X disk, you should probably be doing this in OS X, not in Linux. Is there some reason you are trying to use Linux for this task?I recommend you edit your question to provide all this additional information; it's too much to fit in comments

5. Page table - I don't understand how this table has been made [closed]

You have four physical pages of memory and eight virtual pages. All of the physical pages start out unallocated for any particular purpose. The virtual table of eight entries starts out with all the virtual pages being invalid.That's the starting situation. One more detail may be useful here -- defining HOW new physical pages will be allocated. Given the results, I think it's clear that the allocation method starts out by allocating physical page 0, first. Then physical page 1. Etc. Once all the physical pages have been allocated at least once, this is when the LRU (least recently used) algorithm comes into play.So that sets up the process.One more thing might be useful to consider adding. This is how the LRU itself gets handled. Think of this as a stack of eaight cards. One card representing each virtual page. So let's call the cards: 0, 1, 2, 3, etc., for obvious reasons. The deck always has eight cards in it. Initially, the deck starts out as 01234567 (or any other order you like.) When allocating a new page, we always take the top card (which will be the left-most digit here) from the deck, use whatever physical page it indicates, and then place that card back at the bottom of the deck. So, for example, if we were to need a new physical page we would go to this initial deck of "0123", pull the top card (which is "0"), allocate that physical page, then place that card back into the deck at the bottom. So the new deck would be "1230" after we were done with this step. Every time a physical page is accessed, we find its card in this deck and move it to the bottom, too.To summarize the LRU deck: When we need a new physical page, we always take the top card and use whatever page is indicated by the value there and then move that card to the bottom of the deck. Whenever we access a physical page, we always find the card associated with that page and move it to the bottom of the deck.That's it. I will show the deck along with each step taken.Now all you have to do is "act like the computer" and do the work by hand, running down the list of successive addresses in order. It's just simple hand work.Each virtual address has a page and an offset. The offset value is the low-order 9 bits of the virtual address and the page is the upper 3 bits of the virtual address. The logic is as follows:If the virtual table entry is invalid, then allocate a page from a list of free physical pages. But if there are no free physical pages left, then from a search of valid virtual table entries allocate the physical page from the virtual table entry holding the least recently used page. Also, in this case, mark the found virtual table entry as invalid because the physical page it owns is being re-allocated now. Then take this physical page and place it into the current virtual table entry and mark it as valid. Finally, pull this virtual table entry to the top of the LRU list, indicating it is very recently used.If the virtual table entry is marked as valid, then pull this virtual table entry to the top of the LRU list, again indicating it is very recently used.The above algorithm is repeated over and over for each of your virtual addresses, in sequence order. It ignores the details of forming the full 11-bit physical memory address. But I do not need to worry about that part of it, because your question is not about that but instead about how that table got into the condition it was in. So I am neglecting that detail to avoid excess detail here.Here are the steps with your addresses, in order:$$ beginarraylrllll underlinetextrmStep & underlinetextrmAction & underlinetextrmFree & underlinetextrmLRU & underlinetextrmValid & underlinetextrmFrame & & 0123 & & textrmFFFFFFFF & textrmoooooooo & textbfnew address 1:265 & & & & (1) & alloc = free & 123 & & textrmFFFFFFFF & textrmoooooooo (2) & frame_1 = alloc & 123 & & textrmFFFFFFFF & textrmo0oooooo (3) & valid_1 = true & 123 & & textrmFTFFFFFF & textrmo0oooooo (4) & LRU(1) & 123 & 1 & textrmFTFFFFFF & textrmo0oooooo & textbfnew address 3:511 & & & & (1) & alloc = free & 23 & 1 & textrmFTFFFFFF & textrmo0oooooo (2) & frame_3 = alloc & 23 & 1 & textrmFTFFFFFF & textrmo0o1oooo (3) & valid_3 = true & 23 & 1 & textrmFTFTFFFF & textrmo0o1oooo (4) & LRU(3) & 23 & 31 & textrmFTFTFFFF & textrmo0o1oooo & textbfnew address 2:175 & & & & (1) & alloc = free & 3 & 31 & textrmFTFTFFFF & textrmo0o1oooo (2) & frame_2 = alloc & 3 & 31 & textrmFTFTFFFF & textrmo021oooo (3) & valid_2 = true & 3 & 31 & textrmFTTTFFFF & textrmo021oooo (4) & LRU(2) & 3 & 231 & textrmFTTTFFFF & textrmo021oooo & textbfnew address 2:76 & & & & (4) & LRU(2) & 3 & 231 & textrmFTTTFFFF & textrmo021oooo & textbfnew address 1:34 & & & & (4) & LRU(1) & 3 & 123 & textrmFTTTFFFF & textrmo021oooo & textbfnew address 0:129 & & & & (1) & alloc = free & & 123 & textrmFTTTFFFF & textrmo0o1oooo (2) & frame_0 = alloc & & 123 & textrmFTTTFFFF & textrm3021oooo (3) & valid_0 = true & & 123 & textrmTTTTFFFF & textrm3021oooo (4) & LRU(0) & & 0123 & textrmTTTTFFFF & textrm3021oooo & textbfnew address 6:129 & & & & (1) & alloc = frame_LRU & & 012 & textrmTTTFFFFF & textrm3021oooo (2) & frame_6 = alloc & & 012 & textrmTTTFFFFF & textrm3021oo1o (3) & valid_6 = true & & 012 & textrmTTTFFFTF & textrm3021oo1o (4) & LRU(6) & & 6012 & textrmTTTFFFTF & textrm3021oo1o endarray $$You should be able to match up the last entry above with your table, I think. The only trick here is to note that the LRU virtual table entry is '3' but the physical page owned by that virtual table entry is physical page '1'. The virtual table entry for '3' is marked false, but the alloc variable is set to 1, since that is the physical page that was assigned there.EDIT: To your question:We did not start with 11. That column is the physical page number. Clearly, we did not allocate the physical page 3 (11 binary) right off the bat. Instead, it just turns out that when address 0:129 came along, there was only one free physical page left -- page 3. But that goes into virtual page table row 0, of course! Because the virtual page referenced by 0:129 is row 0 in the virtual table. But the physical page (frame) is 3. So 3 is stored there (11 binary.)

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