Obtain the multiplication of two values ​​in a variable in asp.net core mvc 2.0

I have a table where I have two values, the quantity of a product and the price, and I want to obtain the total of the multiplication of those two values ​​that is equal to (total = quantity * price), I have seen that it can be done of the following way in the class of the Models folder at the time of creating the table by means of migration:

            public class Invoice
{
    public decimal Quantity {get; set; }
    public decimal Price {get; set; }

    public decimal Total ()
    {
        return Amount * Price;
    }

}

Searching the internet I saw that it can be done that way, but when I do the migration and I create the driver, I do not get the total, I must do some action or consult the controller to get the result or what I should do, if there is another way to do what? I need your help!! Please!!

Algorithm of systolic matrices for the multiplication of matrices.

(I received classes with a very fresh teacher that I find very bad in pedagogical / pedagogical skills). I tried to search but I do not understand this. (It seems he's so busy, and there's no TA either, everything I want to understand and pass this course). I hope you can help.

I do not have enough right to insert the image here.

This is what he noted in his presentation:

"The systolic matrix is ​​a way to perform the matrix multiplication algorithm with $ n ^ 2 $ processors and
$ O (n) $ complexity of time, by $ (i) $ placing the $ n ^ 2 $ processors in square ($ n times n $), Y $ (ii) $
assigning the computation of $ I (i, j) $, $ A (i, j) $Y $ O (i, j) $ to the $ (i, j) $-th processor In another
words, you can think of the systolic matrix as the combination of $ (i) $ the multiplication of matrices
algorithm and $ (ii) $ a programming strategy for $ n ^ 2 $ processors. "

What I do not understand here are:

That I am here? For what? What does it look like? Any example?
What is A here, too? For what? What does it look like? Any example?

I just understood a little about the "systolic arrangements" of the CMU slides. But it is not the same as my teacher taught.

Also, what does this mean? (In I, A, O)

$[(i,j) mapsto 0]$

Linear algebra – Is the functional sum equal to the multiplication of matrices?

I have a problem in understanding the equations and I try to relate it to the matrix representation for programming purposes.

Could you explain the following equation?

$ Big ( sum _ { omega} sum_ {x_ {s}} F ^ { ast} (x_ {s}, omega) ^ {T} F ^ { ast} (x_ {s}, omega) Big) (x, x) = sum _ { omega} omega ^ {4} sum_ {x_ {s}} | u (x, x_ {s}, omega) | $ 2

where $ ( cdot) ^ {T} $ it is synonymous with transposition.

Does the functional $ F ^ { ast} (x_ {s}, omega) $ equivalent to the matrix $ F ^ { ast} $ with row of $ x_ {s} $ and column of $ omega $?

It also does $ | u (x, x_ {s}, omega) | $ 2 medium $ u ^ {T} u $?

Thank you!

performance – Python program to solve the multiplication of the matrix chain

A homework assignment in the school required that I write a program for this task:

In the problem of multiplication of the matrix chain, we are given a sequence of
matrices A (1), A (2), ..., A (n). The objective is to calculate the product.
A (1) ..., A (n) with the minimum number of scalar multiplications.
Therefore, we have to find an optimal parenthesis of the matrix.
product A (1) ..., A (n) such that the cost of computing the product is
minimized.

Here is my solution for this task (in Python):

def matrix_product (p):
""
Return m and s.

subway[i][j]    is the minimum number of scalar multiplications needed to calculate the
product of matrices A (i), A (i + 1), ..., A (j).

s[i][j]    is the index of the matrix after which the product is divided into a
Optimal parenthesis of the parent product.

P[0... n] is a list such that the matrix A (i) has dimensions p[i - 1] x p[i].
""
length = len (p) # len (p) = number of matrices + 1

# m[i][j]    is the minimum number of multiplications needed to calculate the
# product of matrices A (i), A (i + 1), ..., A (j)
# s[i][j]    It is the matrix after which the product is divided into the minimum.
# number of multiplications needed
m = [[-1]* length for _ in range (length)]s = [[-1]* length for _ in range (length)]matrix_product_helper (p, 1, length - 1, m, s)

return m, s


def matrix_product_helper (p, start, end, m, s):

""
Return the minimum number of scalar multiplications needed to calculate the
product of matrices A (start), A (start + 1), ..., A (end).

The minimum number of scalar multiplications needed to calculate the
product of matrices A (i), A (i + 1), ..., A (j) is stored in m[i][j].

The index of the matrix after which the previous product is divided into an optimal
the parenthesis is stored in s[i][j].

P[0... n] is a list such that the matrix A (i) has dimensions p[i - 1] x p[i].
""
yes m[start][end]    > = 0:
return m[start][end]

    

    

    

    if start == end:
q = 0
plus:
q = float (& # 39; inf & # 39;)
for k in range (start, end):
temp = matrix_product_helper (p, start, k, m, s) 
+ matrix_product_helper (p, k + 1, final, m, s) 
+ p[start - 1]*P[k]*P[end]
            if q> temp:
q = temperature
s[start][end]    = k

subway[start][end]    = q
come back


def print_parenthesization (s, start, end):
""
Print the optimal parenthesis of the matrix product A (start) x
A (start + 1) x ... x A (end).

s[i][j]    is the index of the matrix after which the product is divided into a
Optimal parenthesis of the parent product.
""
if start == end:
print (& # 39; A[{}]& # 39; .format (start), end = & # 39; & # 39;)
he came back

k = s[start][end]

    

    

    

    print (& # 39 ;, & end = & # 39; & # 39;)
print_parenthesization (s, start, k)
print_parenthesization (s, k + 1, final)
print (& # 39;) & # 39 ;, end = & # 39; & # 39;)


n = int (entry (& # 39; Enter the number of arrays: & # 39;))
p = []
for i in the range (n):
temp = int (entry (& # 39; Enter the number of rows in the array {}: & # 39; .format (i + 1)))
p.append (temp)
temp = int (entry (& # 39; Enter the number of columns in the array {}: & # 39; .format (n)))
p.append (temp)

m, s = matrix_product (p)
print (& # 39; The number of scalar multiplications needed: & # 39 ;, m[1][n])
print (& # 39; Optimum parenthesis: & # 39 ;, end = & # 39; & # 39;)
print_parenthesization (s, 1, n)

Here is an example of output:

Enter the number of matrices: 3
Enter the number of rows in the matrix 1: 10
Enter the number of rows in the matrix 2: 100
Enter the number of rows in the matrix 3: 5
Enter the number of columns in the matrix 3: 50
The number of scalar multiplications needed: 7500.
Optimum parenthesis: ((A[1]A[2])A[3])

NOTE – The time needed to print_parenthesization () (for this example) it is 0: 00: 15.332220 seconds.

Therefore, I would like to know if I could make this program shorter and more efficient.

Any help would be greatly appreciated.

Could all CPU instructions be faster if a better multiplication method were developed?

I was reading the article by David Harvey and Joris Van Der Hoeven called multiplication of integers in time O (n log n).

Could this discovery increase the performance of future CPUs? If so, could we determine how much preventively?

abstract algebra – Law of proof of cancellation for the multiplication of natural numbers.

The cancellation law for the multiplication of natural numbers is:

$$ forall m, n in mathbb N, forallp in mathbb N – {0 }, m cdot p = n cdot p Rightarrow m = n. $$

Is it possible to show this using induction?

I tried to define $$ X = {n in mathbb N: forall m in mathbb N, forallp in mathbb N – {0 }, m cdot p = n cdot p Rightarrow m = n }. $$

It is easy to verify that $ 0 in X $ but I can not show that if $ n in X $ so $ n + 1 in X $.

My attempt: Suppose $$ m cdot p = n cdot p Rightarrow m = n $$
for all $ m en mathbb N $ Y $ p in mathbb N – {0 } $. Supposing that

$$ m cdot p = (n + 1) cdot p $$ we should show $ m = n + 1 $. But:

$$ m cdot p = (n + 1) cdot p = n cdot p + p, $$ and I do not see how to use the hypothesis of induction.

Well, I could try this after the trichotomy. If it were the case that $ m neq n + 1 $ so $ m> n + 1 $ or $ m <n + 1 $. In the first case we would have $ m = n + 1 + r $ for some $ r in mathbb N – {0 } $. So:

$ m cdot p = (n + 1 + r) cdot p = (n + 1) cdot p + r cdot p. $As $ r, p in mathbb N – {0 } $ would follow $$ m cdot p> (n + 1) cdot p, $$ what contradicts $$ m cdot p = (n + 1) cdot p. $$ Is there another way to do this test?

Multiplication of a vector by a scalar [on hold]

They gave us a task question in vectors and I'm not sure how to solve it.
Screenshot of the problem

Thank you!

abstract algebra – Find the addition and multiplication tables for GF (7)

Find the addition and multiplication tables for GF (7).

I know that to do this, I have to express it in terms of $ Z_7[x]/ (x + 1) $, or some other irreducible polynomial of order 1. But I'm not sure how to proceed from here. Any help would be great, thanks in advance!

Functional analysis: understand a test that a compact multiplication operator is zero

This answer gives proof that if $ g in L ^ infty (0,1) $ and the multiplication operator $ T_g: L ^ 2 (0,1) rightarrow L ^ 2 (0,1) $ It is compact, then $ g = 0 $ Almost anywhere:

We show that if $ g $ is not the equivalence class of the null function, then $ M_g $ it's not compact Let $ c> 0 $ such that $ lambda ( {x, | g (x) |> c })> 0 $ (such $ c $ it exists by supposition). Leave $ S: = {x, | g (x) |> c } $, $ H_1: = L ^ 2[0,1]$, $ H_2: = {f in H_1, f = f chi_S } $. So $ T colon H_2 to H_2 $ given by $ T (f) = T_g (f) $ is in fact, if $ h in H_2 $, so $ T (h cdot chi_S cdot g ^ {- 1}) = h cdot chi_S = h $.

As $ H_2 $ it is a closed subspace of $ H_1 $, it's a Banach space. This gives, by the open mapping theorem that $ T $ It's open. It is also compact, so $ T (B (0,1)) $ It is open and has compact closure. By the Riesz theorem, $ H_2 $ It is finite dimensional.

But for each $ N $, we can find $ N + 1 $ disjoint subsets of $ S $ that they have a positive measure, and their characteristic functions will be linearly independent, which gives a contradiction.

I'm interested in the bold part. My question is, why the fact that $ T (B_1) $ It is open and its closing is compact implies that $ H_2 $ is finite-dimensional? The answer cites Riesz's theorem, but that theorem simply says that a Banach space whose closed unit ball is compact must be of finite dimension. Why the fact that the closure of the image of the ball of the open unit below $ T $ Is compact implies that the closing of the ball of the open unit is compact?

Or is there an error in this test?

Move and add to obtain multiplication values?

multiplying by 33 -> $ 2 ^ 5 + 2 ^ 0 $
is the same as (num << 5) + 1

and for 65599 as (num << 16) + (num << 6) – (num << 0)

How do you get these numbers, that is, 16.6? I know you can use the registry but is there any other faster way to do this?