algorithms – Finding the largest sum of $ k $ elements below the threshold

You can use meet-in-the-middle to reduce the time $ O (n ^ { lceil k / 2 rceil}) $.

To simplify, I guess $ k $ is same.

The idea is as follows:

  • Partition $ A $ in two parts.
  • For each part, calculate a sorted list of sums of $ k / $ 2 parts of the room.
  • For each $ k / $ 2-sum in the first part, use binary search to find the best $ k / $ 2-sum in the second part conforms to the constraints.

As noted, the idea poses several problems:

  1. It assumes that the optimal solution contains exactly $ k / $ 2 comes out of each game.
  2. It works in time $ O (n ^ {k / 2} log n) $.
  3. He uses $ O (n ^ {k / 2}) $ Memory.

To manage the first difficulty, there are several options. We could just repeat the whole algorithm $ sqrt {k} $ time. Alternatively, there could be deterministic means to achieve the same goal. Here is a hybrid solution:

  • Randomly partition $ A $ in $ k ^ 3 $ rooms $ P_1, ldots, P_ {k ^ 3} $ (something much bigger $ k ^ 2 $ would work). With a high probability, each element of the optimal solution is in its own part.
  • Consider all possible partitions of $ A $ in two parts of the form $ P_1, ldots, P_i $ and $ P_ {i + 1}, ldots, P_ {k ^ 3} $. One of them will contain exactly $ k / $ 2 elements of the optimal solution.

To manage the second difficulty, we must be slightly more careful in the implementation. Using fusion (the familiar mergesort routine), it should be possible to calculate $ k / $ 2– sums of each part $ O (n ^ {k / 2}) $. The last step can be implemented by $ O (n ^ {k / 2}) $ using a classic technique with two pointers (the first pointer goes up on the first half, the second goes down on the second half).

There are tricks to reduce the memory required from $ O (n ^ {k / 2}) $ at $ O (n ^ {k / 4}) $: further divide each part into two sub-parts. You can easily review all $ k / $ 2– are in the first part considering all the pairs of $ k / $ 4– are in its subparts. Binary search on the second part can then be implemented using the two-point technique.

Break a large list to avoid the limit view threshold on Sharepoint Online

I find solutions to this limit view threshold problem in Sharepoint Online. I have already read the steps recommended by Microsoft (Managing a large list and library in Sharepoint) and I have seen that the former IT manager had done all the steps.

I am thinking of breaking this list down into separate libraries. Any of you able to do it? Or am I better with the retention policy for SPO?

Thank you!


machine learning – Neuron: question on the NO door and the threshold

It is a neuron of a NON gate as illustrated in the book ML by Rojas. I have a question about his behavior. I understand that the neuron produces a signal if it entered $ x_1 $ summed are $ geq 0 $.

In the following image, if $ x_1 $ is 0, then threshold 0 is reached and there is an output from this neuron. Yes $ x_1 $ is 1, then the threshold is always reached and the neuron leaves 1?

Appreciate if someone can provide advice on how this neuron works with the inhibitory signal.

Thank you

do not

distributed systems – Why is a threshold set for the Byzantine fault tolerance of an "asynchronous" network? (where he cannot tolerate even a faulty knot)

In the following response (LINK:, it has been shown that for Asynchronous Byzantine Agreement:

"we cannot tolerate that a third or more of the nodes are dishonest or we lose
either security or alertness. "

For this proof, the following conditions / requirements have been taken into account:

  1. Our system is asynchronous.
  2. Some participants may be malicious.
  3. We want security.
  4. We want liveliness.

A fundamental question is that:

Considering the well known document titled: "Impossibility of distributed consensus with faulty process" (LINK:

showing that:

no fully asynchronous consensus protocol can tolerate even a single unexpected process death,

Can we still assume that the network is asynchronous? As in this case, the network cannot tolerate even a faulty node.

co.combinatorics – Reference – Lower limit of the probability threshold for Ramsey properties

I'm looking at an article by Rodl and Rucinski: threshold function for ramsey properties, also available in school semantics. They state the following:

For all the whole $ r geq 2 $, and for each graph $ G $ which is not a star forest, there are constants $ c $ and $ C $ such as
$$ lim_ {n to infty} mathbb {P} (K (n, p) to (G) _r ^ 2) = left {
begin {array} {ll}
0 & text {if} pCn ^ {- 1 / m_G ^ {(2)}}
end {array} right. $$

Or $ K (n, p) $ is the usual binomial random graph model, and $ H to (G) _r ^ 2 $ the usual Ramsey notation, which means that everything $ r $– coloring of the edges of $ H $ contains a monochromatic $ G $.

They prove Statement 1 in this same article. For instruction 0, it is written

The proof of statement 0 of Theorem 1 & # 39; appeared in (RR93)

And the reference reads

(RR93) _, Lower Limits of Probability Thresholds for Ramsey Properties, Combinatorics, Paul Erdos is eighty years old (Volume 1) Keszthely (Hungary), Bolyai Soc. Math. studies, 1993, p. 317-346

I only found one source for this article, Rucinski's own website.

However, in this, the declaration 0 is not proven for any number of colors, but only for 2 (or I missed the generalization). The proven theorem is

For each graph $ G $ which is not a stellar forest, there is a positive constant $ c_G $ such as
$$ lim_ {n to infty} mathbb {P} (K (n, cn ^ {- 1 / m_G ^ {(2)}}) to (G) _2 ^ 2) = 0 $$

The document only includes a final note stating

Added in evidence. Very recently, the authors have proved Theorem 3 for an arbitrary number of colors.

I found a dozen articles referring to the complete result (i.e. the $ r $ colors), all with the same references, but not the actual paper.

Can someone direct me to a place where I could find it?

Edit: I should add that I was unable to access the original cited newspaper "Paul Erdos is Eighty Vol1", neither online nor in print.

List / library display threshold alert with filter on indexed column

According to the following:

  • Threesome library view is 39,000 items (configured at farm level)
  • Library contains 50,000 items
  • We have created a view with a filter on an indexed column that corresponds to a single element
  • On the site collection side, search works well

However I still have the alert despite the use of this view: the view cannot be displayed because it exceeds the list view threshold…
Do you have any idea where it could come from?

Best regards,

Get a € 300 threshold for Google ads (AdWords)

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Get a € 300 threshold for Google ads (AdWords)

I will share with you a method to get thresholds of 300 € on google ads. if you used it 2 times you can get 600 €, and 1500 € for 5 times …

Get a € 300 threshold for Google ads (AdWords)

I will share with you a method to get thresholds of 300 € on google ads. if you used it 2 times you can get 600 €, and 1500 € for 5 times …

dnd 5th – Do the mounts have an impact on the group's XP threshold in order to create combat encounters?

Yes, if the frame is a smart ally. Otherwise no.

There is no hard and fast rule for this, as the rules for building combat encounters are just guidelines to get started. That said, I think the important question here is whether the mount is just a tool the player uses to improve his movements, or a smart ally who will move, attack and make decisions for himself.

For the specific example of a mastiff horse siccation on an enemy, if you allow that to happen, the mastiff is a smart ally and should count. However, it should be noted that according to the rules, this is not something that you can normally do with a mount. Using the normal rules of combat on horseback, a non-intelligent mount can only move and perform the Dash, Disengage and Dodge actions.

Even with animal companions, who are notably smarter and obedient than an ordinary pet, they are generally not considered capable of acting alone. With that in mind, it seems that DnD generally assumes that only very intelligent (usually magical) creatures can act completely independently.