nt.number theory – Upper limits on the length of the shortest vector in the networks associated with polynomial congruences

We consider a network $ Lambda subset mathbb {Z} ^ 2 $, and puts $ lambda_1 ( Lambda), lambda_2 ( Lambda) $ for successive network minima $ Lambda $.

By a well-known theorem of Minkowski, we have that

$$ displaystyle lambda_1 ( Lambda) leq 2 ( det ( Lambda) / pi) ^ {1/2}. $$

For me it is quite annoying that the right constant is greater than one, so I wonder if it is possible to get an upper bound with a constant less than one, at least in some special cases.

A special case where this is possible is as follows. Consider congruence

$$ displaystyle x ^ 2 + 1 equiv 0 pmod {d}. $$

The solutions to this congruence are based on a finite union of networks, defined by

$$ displaystyle u equiv omega v pmod {d} $$

for some people $ omega in mathbb {Z} $. Note that $ det ( Lambda_ omega) = d $. For such a network $ Lambda_ omega $, we must have $ lambda_1 ( Lambda_ omega) leq d / sqrt {2} $, since in each of these networks, it is possible to find a vector $ (r, s) $ such as $ r ^ 2 + s ^ 2 = d $.

Is it possible to improve the bound provided by Minkowski's theorem for $ lambda_1 ( Lambda) $ for general $ Lambda subset mathbb {Z} ^ 2 $? What about networks that result from polynomial congruences in general?

javascript – Regex for password must contain at least eight characters, at least one number and lower and upper case letters and special characters

I would answer Peter Mortensen, but I don't have enough reputation.

Its expressions are perfect for each of the minimum requirements specified. The problem with its expressions that don't require special characters is that they don't allow special characters either, so they also impose maximum requirements, which I don't believe the OP asked. Normally, you want to allow your users to make their password as strong as they want; why restrict strong passwords?

Thus, its expression "at least eight characters, at least one letter and one number":

^(?=.*(A-Za-z))(?=.*d)(A-Za-zd){8,}$

meets minimum requirement, but remaining characters may only be letters and numbers. To allow (but not require) special characters, you should use something like:

^(?=.*(A-Za-z))(?=.*d).{8,}$ to allow all characters

or

^(?=.*(A-Za-z))(?=.*d)(A-Za-zd$@$!%*#?&){8,}$ to allow specific special characters

Likewise, "at least eight characters, at least one uppercase letter, one lowercase letter and one number:"

^(?=.*(a-z))(?=.*(A-Z))(?=.*d)(a-zA-Zd){8,}$

meets this minimum requirement but only authorizes letters and numbers. Use:

^(?=.*(a-z))(?=.*(A-Z))(?=.*d).{8,}$ to allow all characters

or

^(?=.*(a-z))(?=.*(A-Z))(?=.*d)(A-Za-zd$@$!%*?&){8,} to allow specific special characters.

probability distributions – How can I use chernoff inequality to find an upper bound?

Given a series of random variables so that for everything $ i $ $ in $ $ mathbb N $:
$ Pr (X_i = -9) = 1/10 $
$ Pr (X_i = 1) = 9/10 $

And I define everything $ n $ $ in $ $ mathbb N $: $ S_n = sum ^ n_ {i = 1} X_i $.

How can i use chernoff inequality to find an upper limit for $ Pr (| S_n |> 10 sqrt {n}) $ ?

The upper area affects my right and left area

I use sharepoint 2013 sites as part of a work tool.
I can't explain anything to you as it is covered with confidential documents, but I will try to illustrate my problems.

So every site I get comes with this default zone construction
width of an upper, left and right area that looks like a part 66/34

Default configuration

The problem is
As soon as I add a Web Part to the upper area, it changes the relationship between the left and right areas
After adding a Web Part to the upper area

I do not want that :]
but as far as i can't say anything, i add a fixed width or height.
So why does adding something to the top bar change by default?

st.statistics – Clarification on insanity, standard deviation and upper limit

I have a little doubt about the inequality of the mean absolute difference (crazy) and the standard deviation.
From "Upper and lower limits of sample standard deviation"I read the following relationship:
$$ MAD leq sigma leq frac {R} {2} $$ with
$$ MAD = frac { sum_ {i = 1} ^ {n} | x_i – bar {x} |} {n} $$
and $ sigma $ standard deviation, but after an example, the relationship is not true.
In my example, I have N multidimensional vectors instead of having unique values, after that I compute the classic formula on the standard deviation by considering the vectors.
From Popoviciu's inequality, I read the following relation:
$$ V (X) leq { frac {(x_ {max} -x_ {min}) ^ 2} {4}} $$
to express the upper limit of the variance.
Since I have dimension vectors $ d $ I write:
$$ sigma leq { sqrt { frac {d * (x_ {max} -x_ {min}) ^ 2} {4}}} $$
My question is:

  1. Can I write the last expression with d?
  2. Why the MAD value is not less than the
    standard deviation?

Thank you very much in advance.

nt.number theory – Labeling of the points of the grid $ mathbb N ^ k $ by subsets according to a permutation group, "global" upper limit for the sets which could appear

Let $ G le S_n $ to be a finite permutation group with generators $ g_1, ldots, g_k $. We watch the action of $ G $ on the subsets
with $ A ^ g = { alpha ^ g: alpha in A } $ for $ A subseteq Omega $
and $ g in G $. Label the grid $ mathbb N ^ k $ with subsets $ varphi: mathbb N ^ k to mathcal P ( {1, ldots, n }) $ according to the following diagram. Together $ varphi (0, ldots, 0) = {1 } $ and
$$ begin {align}
varphi (n_1, ldots, n_k) & = varphi ( min (n_1-1,0), n_2, ldots, n_k) ^ {g_1}
\ & quad cup varphi (n_1, min (n_2-1,0), ldots, n_k) ^ {g_2}
\ & qquad qquad vdots
\ & quad cup varphi (n_1, n_2, ldots, min (n_ {k-1} -1,0), n_k) ^ {g_ {k-1}}
\ & quad cup varphi (n_1, n_2, ldots, n_ {k-1}, min (n_k-1,0)) ^ {g_k}.
end {align} $$

It means that we start with $ {1 } $ at the origin, and the label of any other point is the union of the action of the generators on the label of its predecessor points, that is to say say these points which are one unit less in a single coordinate. For example if $ G = S_3 $ with $ g_1 = (1 2) $ and $ g_2 = (1 2 3) $
then $ varphi (0,0) = {1 } $
$$ begin {align}
varphi (1,0) & = {2 } \
varphi (2,0) & = {1 } \
varphi (0,1) & = {2 } \
varphi (0,2) & = {3 } \
varphi (0.3) & = {1 } \
varphi (1,1) & = varphi (0,1) cup varphi (1,0) = {2 } \
varphi (1,2) & = {2,3 }
end {align} $$

etc.

Now a row or column is stable at a certain value $ N $, if no new set appears along this row or column of $ N $ leave, for example the $ j $-th
line would be stable at $ N $ if $ { varphi (i, j): 0 le i le N } = { varphi (i, j): i ge 0 } $. As $ mathcal P ( {1, ldots, n }) $ is finished, each row or column will be stable from a given point in time.

But we could choose a "global" $ N $ so that each row or column will be stable after this point, or say differently if we search for each row or column $ N $ points right or up, each set that appears as labels among these rows or columns will appear among these first $ N $ Labels.

This could be seen by the following argument, if we choose any line for example, then the labels have the form that the singleton sets appear first, then sets of one greater cardinality, in the worst case of sets with two elements, then with three and soon. And if the size of the subsets does not increase, it traverses subsets of the same size. By this simple observation, we see that after at most $ binom {n} {1} + binom {n} {2} + ldots + binom {n} {n} = 2 ^ n $ steps, we saw each subset. The same reasoning applies to the columns,
and $ N = 2 ^ n $ would be such a "global" $ N $.

But I somehow feel that it could be done much better than $ 2 ^ n $, incorporating the cycle structure and the cycle structure that we have for action on the subsets. But I can't find a better formula. So could we find a better upper limit for the "global" $ N $? Or at least some asymptotics?

I hope my explanation is clear, let me know if something is unclear!

dnd 3.5e – Can the evocation of shadows (upper) duplicate a contingency spell as well as the companion spell?

This is a longstanding argument in the 3.5e D&D forums. We are not going to settle this argument here.

Your case, however, is not a problem. When evocation of the shadow Acts like contingency, he acts like contingency, which includes the caster with the ability to cast another spell like contingencyFate of companion. If you wanted to, it could even be another cast of (more) evocation of the shadow.

The real problem, as HeyICanChan suggests in a comment, is whether or not a caster can believe their own illusion. You can choose to fail a save throw, but does the caster even have the chance, then?

And there is simply no answer to that. RAW, the lack of a special exception for the caster's own illusions means they are capable of falling for them, but judging the illusions has a lot of room for the DM's decision, so in in practice, it is not at all unreasonable to say that it is simply impossible. Certainly, being able to use greater evocation of shadows launch contingency when you have prohibited Evocation is a heavy nail in the coffin of this school, which is a good reason to refuse the combination, but it is not in the rules (which assume that Evocation is a school too good than the others, even if it is not).

Anyway, really, contingency should be prohibited anyway. Fate is almost certainly the strongest of the game. Its existence completely changes the nature of the game, and not in a good way.

computation – How do you choose the upper and lower functions for the compression theorem?

Find the limit of the sequence $ a_n = frac {cos ^ 2n} {2 ^ n} $

Since $ 0 le frac {cos ^ 2n} {2 ^ n} le frac {1} {2 ^ n}, $ for everyone $ n $

The compression theorem:

if $ f (n) le g (n) le h (n) $ when $ n $ is near $ a $ and
$ lim limits_ {n à a} f (n) = lim limits_ {n à a} h (n) = L $ then $ lim limits_ {n to a} g (n) = L $

Here, $ f (n) = 0, g (n) = frac {cos ^ 2n} {2 ^ n} $ and $ h (n) = frac {1} {2 ^ n} $

What I don't understand: how were the upper and lower functions of inequality $ f (n) le g (n) le h (n) $ selected?

$ lim limits_ {n to infty} frac {1} {2 ^ n} $ and $ lim limits_ {n to infty} 0 $ are fairly simple to assess. But how is it that the author was able to choose upper and lower functions that were so easy to evaluate?

Any help is greatly appreciated

dnd 3.5e – Can the evocation of shadows (upper) duplicate a contingency spell?

I guess Contention is not eligible because Contingency must have a companion spell and an example would be Contigence (Haste) which is not an existing spell.

(Maybe I should have included an evocation spell in the contingency instead, but Celerity was the first spell that came to my mind)

Related question: Can Wish give me a contingent spell?

javascript – Search in a string for the letters that appear in upper and lower case

I had a challenge recently and they asked me to return the largest letter (in alphabetical order, for example between b and the zI return the z) so that, string, appears in upper and lower case.

I did it as follows

function solution15(S) {
    const arr = S.split("");
    const lowercase = arr.filter((a)=> a === a.toLowerCase());
    const uppercase = arr.filter((a)=> a === a.toUpperCase());
    const coincidences = uppercase.filter(val => lowercase.includes(val.toLowerCase()));
    if (coincidences.length) {
        coincidences.sort();
        return coincidences(coincidences.length-1);
    }
    return "NO";
}

Which works but is not very efficient.
I can't think of another better solution but I understand that there are. It can be done with regex for example?

Thank you!