## Ag.algebraic geometry – characteristic polynomial of a symmetrical \$ 8 times 8 \$ matrix with indefinite entries in relation to octonionic multiplication

I look at $$1, i, j, k, l, m, n, o$$ the standard base of the (complexed at will) octon ions ($$mathbb {O}$$ for the octon ions).
To let $$a = x_1.1 + ldots + x_8.o$$. $$b = x_9.1 + ldots + x_ {16} .o$$ and $$c = x_ {17} .1+ ldots + x_ {24} .o$$, Where $$x_1, ldots, x_ {24}$$ are indefinite about the basic field (pron $$mathbb {C}$$).

I denote by $$L_a$$ the $$8 times 8$$ Matrix that represents the left multiplication with $$a$$ in the $$mathbb {O} simeq mathbb {C} ^ 8$$ and $$R_a$$ the $$8 times 8$$ Matrix that represents the correct multiplication with $$a$$, Similar names for $$b$$ and $$c$$, I want to calculate the characteristic polynomial of the symmetric matrix:
$$S = R_a L_b L_c + {} ^ {t} (R_a L_b L_c),$$
Where $${} ^ {t} X$$ is the transpose of $$X$$,

I tried Macaulay2 and this calculation seems to go far beyond what my computer (which is supposed to be a fairly powerful portable workstation) offers.

A simple reformulation of the eigenvalue problem on a well-chosen basis (namely let $$mathbb {H}$$ be the quaternionic subalgebra of $$b$$ and $$c$$, Splits $$mathbb {O}$$ how $$mathbb {H} bigoplus mathbb {H} .e$$, Where $$e$$ is orthogonal to $$mathbb {H}$$ and take a base adapted to this decomposition) shows that:
$$(T – mathrm {Re} (( overline {b} c) overline {a})) ^ 4 textrm {divides} det (S-T.id),$$
Where $$mathrm {Re} (z)$$ is the real part of $$z in mathbb {O}$$,

I put $$f (T) = dfrac {det (S-Tid)} {(T – mathrm {Re} ((bc) overline {a}) ^ 4}$$, A variety of calculations over finite fields and specialization of the $$x_i$$ random values ​​suggests that $$f (T)$$ is indeed a square, we say $$f (T) = g (T) ^ 2$$, Where $$g$$ is a quadratic polynomial in $$T$$,

I would like a closed expression of $$g (T)$$, May it be a clean formula $$a, b$$ and $$c$$ or a dirty "in coordinates" polynomial. I would be very happy about any suggestion. I would also be interested in a theoretical argument that shows that $$f (T)$$ is indeed a square.

Many thanks!

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## magento2 – CSV Uploader randomly throws errors for identical entries

I am trying to upload 12000 SKUs (1000 each) using the CSV uploader.

I get some random errors for identical data.

Example: Each cell in the attribute_set_column column has a 9, but I get this error

`````` Invalid value for Attribute Set column (set doesn't exist?) in row(s): 575, 577
``````

Is there a way to fix such inconsistencies or get less vague error messages?

Thank you very much

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## views – Calculates the total number of field entries

I have a content type called "Books" and a number field called "Ratings" and an entity reference called "Author Name". I created several nodes with the book titles, the author's name and the corresponding rating.

I would like to display the author name and the total number of ratings of this author with 10, 9.5, 9 etc. in a table view. I tried using Views Conditional to indicate that & # 39; if this field = Rating & # 39; & # 39; equal to 10 & # 39; & # 39; Then output this … & # 39; & # 39; is, and my only options are {{field_rating}}. What I want to spend is the total time the author has received 10. So the table would show: Jon Doe 10 (3), 9.5 (6), 9 (5) and so on.

I have created a table view that contains the fields "Views: Views" for each possible rating (10, 9.5, 9, 8.5, 8, 7.5, 7) together with the fields "Content: Rating" and "Content: author name" is used. I thought Views Aggregation Plus might work, but I don't seem to be able to get it to work.

Maybe I should try to create a custom view ad?

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## Rename and group Google Sheets like entries

I want to create the "Nuts", "Dairy" and "Nuts and Dairy" answers without duplications. I only want the three answers, but my data currently shows about 20 with additional information that I don't actually need or want. How would I rename everything so that an observation like "nuts, peanuts, almonds" only returns "nuts"?

This has to be done in Google Sheets. ## views – Calculates the total number of field entries

I am trying to find a solution for a Drupal 8 table view I have created that outputs book ratings per author. I want to display a table that shows the author name with the total number of ratings of 10.0, 9.5, 9.0, etc. Not the sum, but the "number" of times that a given author received a 10, 9.5, 9, etc. The ideal output would be the following example:

(field_author_name) Jon Doe received 2 x 10 reviews, 3 x 9.5 reviews, 5 x 9 reviews, etc.

Hopefully I'll explain that correctly. (See attached screenshot for a visual representation) Does anyone know if this is possible with Views Conditional? Or is there a better solution out there?

Many thanks to the Drupal community! Posted on Categories Articles

## index – indexing only for the latest entries

Consider the following table:

``````|-----------------------------------------------------|
| raffle                                              |
|----|---------|----------|-----|---------------------|
| id | shuffle |  user_id | ... |           notify_at |
|----|---------|----------|-----|---------------------|
| 1  | 4D6G8Z1 |      542 | ... | 2019-12-01 14:00:00 |
| 2  | 64G264D |        6 | ... | 2019-12-28 14:00:00 |
| 3  | 4IPF93D |       58 | ... | 2020-01-01 14:00:00 |
| 4  | D25LF03 |       58 | ... | 2020-01-14 14:00:00 |
| 5  | G04LDWE |      684 | ... | 2020-03-02 13:00:00 |
``````

In this table, most inquiries are not made to the `id` Column but on the `user_id` and `created_at`This is a 64-bit timestamp (not a 2038 error):

``````SELECT * FROM (raffle) WHERE (user_id) = ? and (created_at) = ?
``````

The table grows by the minute, but that's not the problem, it's the records for them `notify_at` in the current month, most are accessed as the rest. An index of the `user_id` and `notify_at` Totals 160MB, of which only 1% are heavily accessed.

Is there any way to optimize an index (or other strategy) to get the records for the current month faster?

## oracle – queries all audit entries that are issued with the & # 39; audit & # 39; have been defined

Depending on the method used for the check, you will find this information in the following views:

• DBA_OBJ_AUDIT_OPTS
• DBA_STMT_AUDIT_OPTS
• DBA_PRIV_AUDIT_OPTS
• AUDIT_UNIFIED_POLICIES
• AUDIT_UNIFIED_ENABLED_POLICIES

The content of the audit trail, also depending on the configuration and version, can be found in the following views:

• DBA_AUDIT_TRAIL
• UNIFIED_AUDIT_TRAIL
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## r – Replace column entries starting with a specific word

I am editing a data frame in R in which one of the columns is dates, but there are text entries in the same column that prevent the date from having the format YYYY-MM-DD. Since these texts are not important, my idea was to replace them with NA so that I could enter the date correctly, but I cannot make the replacement.

The texts in this column always start with the word "Legal". So my idea was to code a function to identify these characters and replace all occurrences. I thought of the following code:

``````UHE\$Ato(UHE\$Ato == '^Ofício') <- NA
``````

But when I run this line of code, nothing changes in the data frame.
Would anyone know a way to solve this?

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## python – City Construction Solution requires a timeout for extensive entries

problem

The country of hacker land has `n` Cities connected by `m`
unidirectional streets. The cities are numbered by `1` to `n`,
The government recently decided to build new cities in the hacker land.

Your job is to simulate `q` Events. An event can be of two types
described below:

• `1 x d` : A new city `n + 1` is built in the hacker land and connected to the city `x`, If `d = 0` the direction of the new road
is of `x` to `n + 1`, If `d = 1` The direction of the new street is
from `n + 1` to `x`,

• `2 x y` : To press `Yes` If it is possible to move out of the city `x` to the city `y`, to press `No` Otherwise.

limitations

0 <n, m <= 50000

0 <q <= 10000

x, y always correspond to an existing city in the hackerland.

The total number of cities in the Hackerland will not exceed 50,000

A full description of the problem can be found here

My efforts

I tried to use one `BFS` here, however, all test cases fail except for the default due to a timeout. For these test cases `n >= 4000. m >= 20000`,

code

``````n, m = map(int, raw_input().split())
for _ in xrange(m):
u, v = map(int, raw_input().split())
else:
children = set()

q = input()
for _ in xrange(q):
c, x, d = map(int, raw_input().split())

if c == 1:
n = n + 1
y = n
if d == 0:
else:
children = set()
else:
children = set()
else:
y = d
visited = set()
current = list()
current.append(x)
while current:
parent = current.pop(0)
if parent == y:
break