discrete mathematics – Fermat's last sentence: How to solve (partially) through programs

First, it is known that it suffices to look at odd prime numbers $ n $

Grief has shown that when $ p $ divides the counters none of the Bernoulli numbers $ B_2, B_4, ldots, B_ {p-3} $ then Fermat's last sentence applies to $ n = p $, Such primes are called regular. Vandiver specified a criterion that deals with irregular prime numbers. The criterion is somewhat complicated and is given in Section 4 of Wagstaff, The Irregular Prime Numbers to 125000.

An integer must be specified for the criterion $ t $ that satisfies some property. Wagstaff indicates this in practice $ t = 2 $ always works, but a priori we can not guarantee that any value of $ t $ works; and even if no value of $ t $ works, but that does not mean that Fermat's last sentence applies. So the criterion is sufficient, but not known to be necessary, and also a good event $ t $ exists, it is not clear how to find it. As stated above, but in practice $ t = 2 $ always works.

The work you describe, which proves Fermat's last sentence for larger primes, probably uses the same framework. You can consult Buhler, Crandall, Ernvall and Metsänkylä, Irregular Primes and Cyclotomic Fields up to four million.

Reduction from the independent sentence to the restricted independent sentence and vice versa

Independent set

Input: undirected graph G = (V, E) and an integer k ≥ 0.

Issue: Does G have an independent set S ⊆ V of size k (yes or no)?

Limited independent set

Input: undirected graph G = (V, E), a vertex u ∈ V and an integer k ≥ 0,

Issue: Does G have an independent set S ⊆ V of size k that contains u? (Yes or no).

Is it possible to reduce the independent quantity to a limited independent quantity, and is it possible to reduce the limited independent quantity to an independent quantity?

Proofing – How to prove the function of a recursive big theta without using repeated substitution, mastering the sentence, or having the closed form?

I have defined a function: $ V (j, k) $ Where $ j, k in mathbb {N} $ and $ t> 0 in mathbb {N} $ and $ 1 leq q leq j – 1 $, Note $ mathbb {N} $ includes $ 0 $,

$ V (j, k) = begin {cases} tj & k leq 2 tk & j leq 2 tjk + V (q, k / 2) + T (j – q, k / 2) & j , k> 2 end {cases} $

I am not allowed to use repeated substitution and I want to prove this by induction. I can not use the main clause because the recursive part is not in that form. Any ideas how I can solve it with given limitations?

When I start induction: I fix $ j, q $ and introduce $ k $, Then the base case $ k = 0 $, Then $ V (j, 0) = tj $, The question indicated that the function may be $ Theta (jk) $ or maybe $ Theta (j ^ 2k ^ 2) $ (but it does not necessarily have to be either).

I choose $ Theta (j, k) $, In the base case, this would mean that I had to prove that $ tj = theta (j, k) $ when $ j = 0 $, But if I start with the big-oh, I have to show it $ km leq mn = m cdot0 = 0 $ which I currently do not think possible.

I am not sure if I did the basic case wrong or if there is another approach.

MongoDB conditional sentence

I use MongoDB + Mongoose + Node.JS

App logic: We have a game similar to the minesweeper. Save custom cells in the game field in fields: Arrayand bomb sites store in bombs: Array,

When users make a request to select a cell in the game, the server must be set fields(cell) = 1 // 1 means cell is selected, Is the bomb in the cell bombs(cell) === 1 // 1 means bomb in cell we have to do another update, like fields(cell) = 1; winning = 0, How can this be realized in a query?

I tried to do a query that just changes field(cell) = 1but do not change winning = 0 If a bomb was detected in the field.


const selection = await db.Minesweeper.findOneAndUpdate({
    winning: null,
}, {
    $set: { (`fields.${cell}`): 1, },
    // I need to set winning = 0, if (bombs.${cell}) = 1

Match details:

const MinesweeperSchema = new mongoose.Schema({
    uid: { type: Number, required: true },
    bet: { type: Number, required: true },
    winning: { type: Number, default: null },
    fields: { type: Array, default: new Array(5*5).fill(0) },
    bombs: { type: Array, required: true },
{ timestamps: true });

Again good:

I need that logic: if the selected cell contains a bomb, put it field(cell_index) = 1; winning = 0but if the cell contains no bomb, put field(cell_index) = 1

Website design – Should Emoji / Emoticon replace a punctuation mark at the end of the sentence?

If the sentence ends in the content of the website or in a personal message with emoji / emoticon, should a punctuation mark follow?

example 1

I'm glad you made it. 🙂 We had a great time. – This looks good and clean
I'm glad you made it 🙂 We had a great time. – However, this looks strange to me
I'm glad you made it. 🙂 We had a great time. – Emoji belongs to the next sentence
I'm glad you made it :] We had a great time. – This one is super funny

Example # 2

Is it really like that? I doubt it. – This is fine, since the question mark means it's a question
Is it really like that? I doubt it. – That does not look like a question anymore
Is that really so 🤔 I doubt it. – Again, Emoji belongs to the next movement
Is that really so = /? I doubt it. – And this emoticon is something else

The emoji / emoticon understanding varies according to the position. Is there a common way to use them or are there best practices?

Mining theory – how long does it take to find a UTXO in a UTXO sentence?

It's impossible to answer this accurately without knowing details like the miner's hardware, the Utxo storage method, and the validation process.

In general, however, the utxo set can be saved as a list with the utxo identifier (txid:vout). This returns O (1) or a complexity search constant for a single utxo.

However, the utxo set may not fully fit into memory on many systems, and the actual polling time may be longer than expected in an ideal scenario after considering disk access time and other implementation-specific details.

In the case of a miner, they validate incoming transactions and create a block template by acting asynchronously to the actual mining operation. In this way, they have a good chance of maximizing their revenue per block without blocking the mining process through tx and block validation systems.

How do I create an autoincrement field within a sentence in LibreOffice Writer?

I'm using LibreOffice 6.0.2 on a Mac. I need to be able to use an autoincrement that appears in sentences that occur multiple times in a document.

For example:

A sentence that does not have the number. A sentence with the number 1. Several pages without numbers. Still some pages without numbers. A sentence with the number 2. More content.

End of the example. Expected behavior:

If I add a numbered sentence between number 1 and number 2, that new sentence gets the 2, and what was previously a 2 is changed to a three. This leads to:

A sentence that does not have the number. A sentence with the number 1. Several pages without numbers. A new sentence with the number 2. Several more pages without numbers. A sentence with the number 3. More content.