magento2 – How come the order number in ShipWorks is displayed incorrectly?

I followed this QA method to adjust the order number on my recent Magento 2.3.1 migration. When ShipWorks downloads the orders, they still start at 1,

The ShipWorks support team has called you back and they are feeling this order number Field for order numbers and not increment_id, which is known to be the order number in human language. Once I test the effectiveness of updating the sales_order I'll post a response to the table's auto_increment primary key.

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dnd 5e – What is the highest number of stealth attacks a rogue can make in a round?

Shenanigans: Fighter (Cavalier) 18 / Rogue (Any) 2 can make a casual attack on each turn without using his reaction, and an infinite number of creatures (Arcane Gate or other mass translocation spells provide enough space for creatures to begin their moves so closely that they can come in and go out and provoke who want to provoke occasional attacks from him.

It makes sense: A Rogue (Thief) 17 can make 4 stealth attacks in one turn, as he is allowed to make two moves in the first Initiative Turn.
In the first round, he attacks as usual, and then a friendly Combat Master uses Commander's Strike to allow his reaction to a second stalking attack.
During the Rogue's second turn (in the same round), he attacks again as usual, and then a second friendly Combat Champion uses Commander's Strike to allow the fourth stalk attack with his reaction (which was triggered again because the Rogue had one Train did).

Number Theory – Does Multiplication Increase Entropy?

Does multiplication increase entropy?

The Shannon entropy of a number $ k $ in binary digits is defined as
$$ H = – log ( frac {a} {l}) cdot frac {a} {l} – log (1- frac {a} {l}) cdot (1- frac { a} {l}) $$
from where $ l = text {floor} ( frac { log (k)} { log (2)}) $ is the number of binary digits of $ k $ and $ a $ is the number of $ 1 $-s in the binary extension of $ k $,
So let's take a look at the number $ k $ as a "random variable".

Suppose that $ n, m $ are selected evenly at random in the interval $ 1 le a $,

Hypothesis 1):

$ H_ {m cdot n} $ is then "significantly" larger $ H_n $,

Hypothesis 2):

$ H_ {m + n} $ is then not "significantly" larger $ H_n $,

Here is an empirical statistical test that indicates that multiplication increases entropy, but not addition:

def entropyOfCounter (c):
S = 0
for k in c.keys ():
S + = c[k]
    prob = []
    for k in c.keys ():
prob.append (c[k]/ S)
H = sum ([ p*log(p,2) for p in prob]) .N ()
Return H

def HH (l):
return entropyOfCounter (counter (l))
N = 10 ^ 4
MX = []
MP = []
for k in the range (N):
n = Randint (1.2 ^ 500)
m = Randint (1,2 ^ 500)
Hn = HH (integer (n). Digits (2))
Hm = HH (integer (m). Digits (2))
c + = 1
M = max (Hn, Hm)
MX.append (HH (integer (n * m). Digits (2)) - Hn)
MP.append (HH (integer (n + m). Digits (2)) - Hn)

tX = mean (MX) / (sqrt (variance (MX)) / sqrt (N)). N ()
tP = average (MP) / (sqrt (variance (MP)) / sqrt (N)). N ()
prints tX, tP

31.1839027855549 0.266357305397406

The first case (multiplication) significantly increases the entropy. The second case (addition) not.

Is there any way to give a heuristic explanation of why this is generally so (if so) or is this empirical observation generally available? $ 1 le a $ not correct?


turing machines – Is the function `number TM, which ends on a blank word` calculable?

Let f: N → N be a function in which f (n) = number of Turing machines over the input alphabet {0, 1}, says: {0, 1, ..., n} and the working alphabet {_ (empty), 0, 1, ... n} ends at an empty word,

is f predictable?

Let's call a TM that solves this problem K,

It feels like this problem is not predictable. If we can not even determine if a particular TM is stopping on an empty word (the problem of stopping is reduced to the problem of stopping the empty tape), it should be even more difficult to intuitively find the number of TMs with that property. But I can not imagine a reduction, because this problem does not use a specific TM in the input. The input is just a number (in the question I've linked above – BLANK, TM takes a machine as input, so we can deploy a specific machine – anything we want – for example, machine from the instance of the HALT problem but this is not a case in this problem).

First, I wanted to reduce the BLANK problem: I wanted to measure the number of states, the alphabet, the runtime K TM with the number of states that the BLANKHALT instance has. While counting the number of TMs finished with a blank word, check that it is a BLANK machine that I have just asked. BUT I can not do this because I can not assume K lists the TMs at all, right? If K There could be a magical way to solve the question without checking all possible TMs individually, right?

theming – How to target the second field using "allowed number of values" = 2 in the branch

I have created a field for an image and set the allowed number of values ​​to 2 in the field setting (see attached screenshot)

Enter image description here)

My question is, how can I insert both pictures in another div tag into my page template?

or how can I align a first and a second image in my page template?

thank you in advance

sg.symplectic geometry – Is the minimum Chern number of a toric distributor at least 2?

I want to show that the minimum chern number $ N_M $ from a toric diverse $ M $ is at least $ 2 $, from where
N_M: = underset {l> 0} { min} lbrace exists A in H_2 (M; mathbb {Z}) : \ langle c, A rangle = l rbrace,

$ c $ denotes the first class of Chern $ (M, omega) $ (for every choice of $ omega $-compatible complex structure) and $ langle.,. rangle $ is the natural mating between cohomology and homology groups.

I do not know how to prove it, but the following interpretation of Chern's first class could help.

To let $ (M ^ {2d}, omega, mathbb {T}) $ to be a toric manifold, where $ omega $ is the symplectic form and $ mathbb {T} $ is a $ d $-dimensional torus that works effectively and in Hamiltonian fashion $ (M, omega) $, Show $ M $ as symplectic reduction of $ mathbb {C} ^ n $ through the action of a $ k $-dimensional subtorus $ mathbb {K} subset (S ^ 1) ^ n $ (therefore identification $ mathbb {T} simeq (S ^ 1) ^ n / mathbb {K} $) one can show that there is a natural isomorphism
H_2 (M; mathbb {Z}) simeq text {lie} ( mathbb {K}) _ { mathbb {Z}},

where the integral grid $ text {Lie} ( mathbb {K}) _ { mathbb {Z}} $ is the core of the exponential map $ exp: text {Lie} ( mathbb {K}) to mathbb {K} $, For every choice $ omega $compatible almost complex structure $ M $, the first class of Chern $ c in H ^ 2 (M; mathbb {Z}) simeq text {lie} ( mathbb {K}) _ { mathbb {Z}} ^ * $ writes:
c (m) = underset {j = 1} { overset {n} sum} m_j quad m in text {lie} ( mathbb {K}) _ { mathbb {Z}} quad iota (m) = (m_1, …, m_n),

from where $ iota: text {Lie} ( mathbb {K}) hookrightarrow mathbb {R} ^ n $ is the inclusion of Lie algebras induced by inclusion $ mathbb {K} subset (S ^ 1) ^ n $,

Of course in general (if $ M $ is not toric), $ N_M $ can be the same $ 1 $and you can even have that $ langle c, H_2 (M; mathbb {Z}) = 0 $ (In this case you often write $ N_M = infty $). However, as every toric manifold disintegrates in complex cells, this seems to be the case $ N_M $ should be at least $ 2 $,

Every help is appreciated. Thank you in advance.

Eigenvalued eigenvectors – Number of triangles in $ (n, n / 2,2 sqrt {n}) $ – Expandergraph.

To let $ G $ Bean $ (n, n / 2,2 sqrt {n}) $ Expander graph. That is, the second largest eigenvalue of the adjacency matrix $ 2 sqrt {n} $, To let $ M $ denote the number of triangles in $ G $I want to show that as $ n $ gets big (actually $ G $ is a family of expander graphs), we have
begin {equation}
M – frac {n ^ 3} {48} = o (n ^ 3).
end {equation}

Here is my previous work:

As usual we call with $ lambda (G) $ the second largest eigenvalue of the adjacency matrix $ A (G) $ of a graph $ G $, To let $ G $ be a $ (n, n / 2,2 sqrt {n}) $–Graph. Now the $ a ^ 3_ {ij} $ Element of $ A ^ 3 $ counts the number of paths of length $ 3 $ from $ i $ to $ j $, There are no long walks $ 3 $ From a vertex to itself with repeated vertices, we have that $ a_ {ii} ^ 3 $ counts the number of paths from $ i $ back to yourself, that is, $ a_ {ii} ^ 3 $ counts the number of triangles that contain the vertex $ i $, Thus
begin {equation}
6M = operatorname {Tr} (A ^ 3) = sum limits_ {i = 1} ^ n { lambda_i ^ 3} = d ^ 3 + lambda_2 ^ 3 + sum limits_ {i = 3} ^ n lambda_i ^ 3 = frac {n ^ 3} {8} + 8n ^ {3/2} + sum limits_ {i = 3} ^ n lambda_i ^ 3.
end {equation}

begin {equation}
left | M – frac {n ^ 3} {48} right | = left | frac {1} {6} sum limits_ {i = 2} ^ n lambda_i ^ 3 right | leq frac {1} {6} sum limits_ {i = 2} ^ n | lambda_i | ^ 3.
end {equation}

If we can show that the right side of the above is smaller than $ epsilon n ^ 3 $ for big enough $ n $, then we are done. However … I can not. I know that the second largest eigenvalue $ lambda = lambda_2 $ is $ 2 sqrt {n} $ That's fine, but there may be a negative eigenvalue that's big, for example $ d $,

Note We have
begin {equation}
left ({ sum limits_ {i = 2} ^ n | lambda_i | ^ 3} right) ^ {1/3} leq left ({ sum limits_ {i = 2} ^ n | lambda_i | ^ 2} right) ^ frac {1} {2} < sqrt {2 | E (G) |} = sqrt {nd},
end {equation}

as $ G $ is a $ d $-regular graph. Now, $ nd = n ^ 2/2 $So that shows that $ M = mathcal {O} (n ^ 3) $ but does not give the coefficient of $ n ^ 3 $ as requested.

Any tips, ideas? I hope we can somehow estimate the size of the smallest eigenvalue …