derived algebraic geometry – DAG applied to homotopy theory: How to reach the research level?

It is my dream to one day explore applications of spectral algebraic geometry in homotopy theory. In particular, a more uniform treatment of the results was shown by uncanny calculations (of which I think there are many). There was a post about MO on how to be well versed in the DAG so that you can do your own research, but to some extent it reflected the taste of the respondent (who is a good mathematician but seems more focused on motives) ,

Can someone give a similar overview, but consider applications for (non-motivic) homotopy theory? Since we are homotopists, the local objects are binding for the DAG we use (or $ k $-Periodic for some $ k $, could be?) $ E _ { infty} $Ring spectra.

dnd 3.5e – Is it possible to reach the magic level faster than clerics or wizards?

In terms of mixing Theurge classes with fast-moving classes such as the Peace Apostle, Blighter, or Primal Priest, this response mentions that "Each time you gain a magic level before a cleric or wizard would do so, you cause problems"That made me ask – is there a way to do that? with consistency (ie not just at a certain level)? Without obvious cheese like wordplay, I do not know any tricks that would allow such progress. Even among the classes I have listed, I only know isolated cases, such as the Peace Apostle, who takes the ninth grade one stage earlier than the cleric. Does it have something to do with exploiting early exploits?

doublespend – What happens if 2 transactions with a duplicate of the others reach the same block?

I'll interpret this question from a different angle: if a node received a block with two transactions issuing the same input, what transaction would it return to mempool and which one would it discard?


When a node receives a block containing two or more transactions for which the same inputs are used, the block is marked invalid and completely discarded. None of the transactions in this block are considered confirmed, and all transactions already in the mempool are retained there. A transaction was not in memory but was not added to memory.

If the node receives this invalid block and acknowledges that it has been found to be invalid, it throws the block off and does nothing. It does not change the UTXO group and the storage pool. This means that if one of the conflicting transactions was in the memory area when the block was received, it remains there. If not, none of these transactions will be added to the mempool.

So, if one of the conflicting transactions paid a higher fee but the lower fee transaction was already in the mempool of the node, the lower fee transaction would remain in the mempool and the higher fee would be discarded with the block.

The code for this behavior is here. Connect block takes one CCoinsViewCache This is a cache for the UTXO set. This cache only writes out the changes made to it Do the washing up() Function is called. Connect block will return false because the block is invalid. Therefore, this function is already returned here Do the washing up() can be accessed here. Therefore, no changes are made to the UTXO set.

In addition, this premature return due to an invalid block means that the mempool update will also fail. The call for the mempool update here is not reached because of the early return. Therefore, the transactions that were in the mempool at the time of blocking are kept there, and those that were not in the mempool are thrown along with the block.

Runtime Analysis – Big Oh-Time Complexity with "within reach for me"

So you have the right logic if you have a loop $ O (n) $ Iterations will be the complexity of the entire loop $ O (n * text {complexity of loop operations}) $, In this case, you have the right complexity of your loop again $ O (n) $ for every iteration.

Your last bullet shows that you understand this as well, since the overall loop complexity is then $ O (n ^ 2) $, However, this trivially leads to a total time complexity of $ O (n ^ 2) $Not $ O (n) $ how you conclude.

I can imagine what you really asked for, the complexity of Area (s), It is true that the first call is to the area $ O (1) $That's because it generates a generator that needs it $ O (1) $ to return the next value, d. H. He does not immediately produce the entire range, but on demand. Remember, however, that the generator has progressed $ O (n) $ times (once for each repetition of the loop).

SharePoint Personnel Picking Error "Sorry we could not reach the server"

My users face errors in SharePoint 2016 when trying to add more than 15 users to share a file / folder. However, this error occurs intermittently and has different results on different network connections. I have extended the timeout from standard 25s to 60s. After that, the problem occurs again when a user adds names to the People Picker, left the computer inactive for a while, and when he comes back to add more names, he encounters the error message.

Does the network connection have anything to do with the People Picker extracting names from AD? Is it normal for the people picker to time out if it is not used for a long time? How can users prevent this error from occurring again?

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