## Python: Built-in function for the & # 39; sum & # 39; Unicode of a chain?

In Python, is there a built-in function to calculate the & # 39; sum & # 39; Unicode of a chain?

The function would take a string of any length, essentially execute ord () on each character, then return the sum of all ord () calls.

So something like `unicode_sum (& # 39; abc & # 39;) == unicode_sum (& # 39; cba & # 39;)` would be `True`.

I realize that this could be easily coded, but a built-in function would be nice.

## Curves and surfaces – Analytical expression for the normal vector of the 3D height map computed as sum of sin

I created a field in Matlab for a robotics project in the following way:

``````global X Y Z;

xmin = -3; xmax = 3; xstep = .01;
ymin = -3; ymax = 3; ystep = .01;
[X, Y] = meshgrid (xmin: xstep: xmax, ymin: ystep: ymax);

nsin = 30;
C = 1;
k = .5; % .03;
norm = 500 * (nsin) ^ 2;

fmin = 10; fmax = 50;
f = unifrnd (fmin, fmax, 1, nsin);

for i = 1: nsin
a (i) = unifrnd (-C / f (i), C / f (i));
finish
a = a / sum (a) * C;

o = unifrnd (0, 2 * pi, 1, nsin);

Z = zeros (size (X));
for i = 1: nsin
Z = Z + a (i) * sin (f (i) * (k * X + o (i)));
Z = Z + a (i) * sin (f (i) * (k * Y + o (i)));
finish
Z = Z / norm;
``````

Basically, given a mesh of mesh in X, Y calculated the corresponding Z as a sum of sinusoids with a group of random parameters such as amplitude, frequency and phase.

Now, the goal would be to find an analytic expression for the normal vectors given any x, and where z is defined.

Greetings to anyone who can help!

## Performance – Python program to find the subarray with maximum sum

A task at school required that I write a program for this task using the Kadane algorithm (explained here):

1. Define the function `find_max_subarray` that takes a list as an argument and two indexes, start and end. Find the maximum subarray in the
distance `[start, end – 1]`.
2. The function `find_max_subarray` returns the tuple `(l, r, m)` where `l`, `r` are the left and right indexes of the maximum subarray and `subway` that's
sum.
3. The function uses a loop to track the maximum subarray that ends in the index `I`. That is, the maximum subarray that the element has.
in the index `I` as his last element.
4. Then, the maximum subarray will simply be the maximum of all these sub-arrays.
5. The maximum subarray that ends in the index `i + 1` is given by `max (list[i]` + maximum subarray sum ending in `me, list[i])`.
6. The function tracks the maximum sum of the subarray seen so far and its indexes to the left and to the right as the loop repeats.

Here is my solution for this task, using Python:

``````def find_max_subarray (alist, start, end):

max_ending_at_i = max_seen_so_far = alist[start]
max_left_at_i = max_left_so_far = start
max_right_so_far = start + 1
for i in range (start + 1, end):
if max_ending_at_i> 0:
max_ending_at_i + = alist[i]
plus:
max_ending_at_i = alist[i]
max_left_at_i = i
if max_ending_at_i> max_seen_so_far:
max_seen_so_far = max_ending_at_i
max_left_so_far = max_left_at_i
max_right_so_far = i + 1
return max_left_so_far, max_right_so_far, max_seen_so_far

alist = input (& # 39; Enter the list of numbers: & # 39;)
alist = alist.split ()
alist = [int(x) for x in alist]
start, end, maximum = find_max_subarray (alist, 0, len (alist))
print (& # 39; The maximum subarray starts in the index {}, ends in the index {}, and has the format of sum {}. & # 39 ;. (start, end - 1, maximum))
``````

Here are some examples of outputs:

``````Enter the list of numbers: 3 -2 4 -6 7 8 -10 4
The maximum subarray starts at index 4, ends at index 5 and has the sum 15.

Enter the list of numbers: 4 5 -2 0 -5 15
The maximum subarray starts at index 0, ends at index 5 and has the sum 17.

Enter the list of numbers: 3
The maximum subarray starts at index 0, ends at index 0 and has the sum 3.
``````

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

Any help would be greatly appreciated.

## nt.number theory – Estimating a sum of algebraic numbers

Leave $$alpha_1, ldots, alpha_n$$ Be algebraic numbers and leave $$p_1, ldots, p_n$$ They will be the corresponding minimum polynomials with integer coefficients.
We denote by $$H$$ the maximum magnitude between all the coefficients of all $$p_i$$.

Assume that $$alpha_1 + ldots + alpha_n not = 0$$. It is true that
$$| alpha_1 + ldots + alpha_n | ge {CH} ^ {- q ( text {deg} (p_1) + ldots text {deg} (p_n))}$$ for some fixed polynomials $$q$$ and constant $$C$$?

## mysql – Total sum of values ​​of a hierarchical tree formed by a Foreach with IF conditions in PHP

I go to this community to see if they can guide me, I have been fighting with this function and I would like to know how I could obtain the total sum of a variable that runs through the array (hierarchical tree of 5 levels) formed by a foreach and if conditions of a query in the DB, I tried to do it but it only returns the values ​​assigned to me in the penultimate level of the tree.

Here the function:

``````function show_points (\$ id, \$ limit, \$ level) {
\$ list = array ();
global \$ pdo;

\$ sql = \$ pdo-> prepare ("SELECT (@rownum: = @ rownum + 1) AS No, users.id, users.id_pai, users.patent, users.nome, patents.nome as p_nome
FROM (SELECT @rownum: = 0) r, users
LEFT JOIN patents ON patente.id = usuarios.patente
WHERE users.id_pai =: id ");
\$ sql-> bindValue (": id", \$ id);
\$ sql-> execute ();

if (\$ sql-> rowCount ()> 0) {
\$ list = \$ sql-> fetchAll (PDO :: FETCH_ASSOC);
\$ puntosum = 0;
foreach (\$ list as \$ chave => \$ user) {
\$ list[\$chave]['filhos']    = array ();
if (\$ limit == 0) {
\$ list[\$chave]['filhos']    = show_points (\$ user['id'], \$ limit -1, \$ level-0, \$ puntosum + = \$ level);
echo \$ user['nome']. ' -Points: '. \$ Level.'';
}
elseif (\$ limit == 1) {
\$ list[\$chave]['filhos']    = show_points (\$ user['id'], \$ limit -1, \$ level - 5, \$ puntosum + = \$ level);
echo \$ user['nome']. ' -Points: '. \$ Level.'';
}
elseif (\$ limit == 2) {
\$ list[\$chave]['filhos']    = show_points (\$ user['id'], \$ limit -1, \$ level - 10, \$ puntosum + = \$ level);
echo \$ user['nome']. ' -Points: '. \$ Level.'';
}
elseif (\$ limit == 3) {
\$ list[\$chave]['filhos']    = show_points (\$ user['id'], \$ limit -1, \$ level - 25, \$ puntosum + = \$ level);
echo \$ user['nome']. ' -Points: '. \$ Level.'';
}
elseif (\$ limit == 4) {
\$ list[\$chave]['filhos']    = show_points (\$ user['id'], \$ limit -1, \$ level - 50, \$ puntosum + = \$ level);
echo \$ user['nome']. ' -Points: '. \$ Level.'';
}
elseif (\$ limit == 5) {
\$ list[\$chave]['filhos']    = show_points (\$ user['id'], \$ limit -1, \$ level - 100, \$ puntosum + = \$ level);
echo \$ user['nome']. ' -Points: '. \$ Level.'';
}
}

}
echo \$ puntosum. ' Total';
return \$ puntosum;
}
``````

As you may notice level 1 is assigned -100 pts until you reach level 6, but at the time of doing the summation only takes the values ​​of the last cent that is met, in this case when \$ limit == 5.

Beforehand thank you very much!

## c – Sum of columns and diagonals of a matrix.

I made a simple matrix program that prints some data … I want to know how to reduce the number of lines with some functions or whatever, because there are a lot of if-s, and for loops.

I tried to do functions, but almost all for loops are different, so it did not work.

``````#include
principal()
{
int i, j, m, n, sMain, sAnti, sCol, sMainUp, sMainDown, sAntiUp, sAntiDown;
int arr[30][30];

printf ("What matrix dimensions do you want? (1 number)");
scanf ("% d", & m);
n = m;

for (i = 0; i <m; i ++) {
for (j = 0; j <n; j ++) {
printf ("Elements of type:");
printf ("Element[%d,%d]: ", i, j);
scanf ("% d", & arr[i][j]);
}
}
for (i = 0; i <m; i ++) {
for (j = 0; j <n; j ++) {
printf ("% d", arr[i][j]);
}
printf (" n");
}
sMain = 0;
sAnti = 0;
sMainUp = 0;
sMainDown = 0;
sAntiUp = 0;
sAntiDown = 0;

for (i = 0; i <m; i ++) {
for (j = 0; j <n; j ++) {
yes (i == j) {
sMain + = arr[i][j];
}
yes ((i + j) == (n - 1)) {
sAnti + = arr[i][j];
}
if I < j){
sMainUp += arr[i][j];
}
if(i > j) {
sMainDown + = arr[i][j];
}
yes ((i + j) < (n - 1)){
sAntiUp += arr[i][j];
}
if((i + j) > (n - 1)) {
sAntiDown + = arr[i][j];
}
}
}
sCol = 0;
printf ("Sum of the column you want?");
scanf ("% d", & j);
for (i = 0; i <m; i ++) {
sCol + = arr[i][j];
}
printf ("Sum of the main diagonal:% d.  n", sMain);
printf ("Sum of the anti diagonal (counter):% d.  n", sAnti);
printf ("Sum over the main diagonal:% d.  n", sMainUp);
printf ("Sum in main diagonal:% d.  n", sMainDown);
printf ("Sum above anti diagonal:% d.  n", sAntiUp);
printf ("Sum under anti diagonal:% d.  n", sAntiDown);
printf ("The sum of the column% d is:% d.", j, sCol);

}
``````

## probability: how to calculate this formula \$ operatorname {E} ({ frac {2} {n}} Y_ {i} sum _ {j = 1} ^ {n} Y_ {j}) \$

I am learning the justification of the variance of the sample.

{ displaystyle { begin {aligned} operatorname {E} [sigma _{Y}^{2}]& = operatorname {E} left[{frac {1}{n}}sum _{i=1}^{n}left(Y_{i}-{frac {1}{n}}sum _{j=1}^{n}Y_{j}right)^{2}right] (1.1) \[5pt] & = { frac {1} {n}} sum _ {i = 1} ^ {n} operatorname {E} left[Y_{i}^{2}-{frac {2}{n}}Y_{i}sum _{j=1}^{n}Y_{j}+{frac {1}{n^{2}}}sum _{j=1}^{n}Y_{j}sum _{k=1}^{n}Y_{k}right ] (1.2) \[5pt] & = { frac {1} {n}} sum _ {i = 1} ^ {n} left[{frac {n-2}{n}}operatorname {E} [Y_{i}^{2}]- { frac {2} {n}} sum _ {j neq i} operatorname {E} [Y_{i}Y_{j}]+ { frac {1} {n ^ {2}}} sum _ {j = 1} ^ {n} sum _ {k neq j} ^ {n} operatorname {E} [Y_{j}Y_{k}]+ { frac {1} {n ^ {2}}} sum _ {j = 1} ^ {n} operatorname {E} [Y_{j}^{2}]Right](1.3) \[5pt] end {aligned}}}

it is easy to understand how equation (1.1) would take equation (1.2)

I am trying to understand how equation (1.2) would be carried out in equation (1.3)

{ displaystyle { begin {aligned} operatorname {E} [sigma _{Y}^{2}]& = operatorname {E} left[{frac {1}{n}}sum _{i=1}^{n}left(Y_{i}-{frac {1}{n}}sum _{j=1}^{n}Y_{j}right)^{2}right] (1.1) \[5pt] & = { frac {1} {n}} sum _ {i = 1} ^ {n} operatorname {E} left[Y_{i}^{2}-{frac {2}{n}}Y_{i}sum _{j=1}^{n}Y_{j}+{frac {1}{n^{2}}}sum _{j=1}^{n}Y_{j}sum _{k=1}^{n}Y_{k}right ] (1.2) \[5pt] & = { frac {1} {n}} sum _ {i = 1} ^ {n} left[ operatorname {E}(Y_{i}^{2})- operatorname {E} ({frac {2}{n}}Y_{i}sum _{j=1}^{n}Y_{j})+operatorname {E}({frac {1}{n^{2}}}sum _{j=1}^{n}Y_{j}sum _{k=1}^{n}Y_{k})right ] (2.2) \[5pt] end {aligned}}}

with linearity of expectation equation (1.2) would lead to equation (2.2)

The question is, how to calculate the second term within the brackets in equation (2.2)?

$$operatorname {E} ({ frac {2} {n}} Y_ {i} sum_ {j = 1} ^ {n} Y_ {j})$$

## java – Sum of two matrices

For a task, they asked me to do this problem. It worked, but it seems messy in my opinion. I want to know if this is the best way to perform this task and if improvements can be made in this code, such as the efficiency and compactness of the code. The task is explained below:

Write a method called matrixSum that accepts two 2D parameters as parameters
arrays of integers, treats the matrices as 2D arrays and adds them,
returning the result. The sum of two matrices A and B is a matrix C
where for each row i and column j, Cij = Aij + Bij. You can assume
That the matrices passed as parameters have the same dimensions.

A = {{1, 2, 3}, {4, 4, 4}}

B = {{5, 5, 6}, {0, -1, 2}}

maxtrixSum (a, b)
=> {{6, 7, 9}, {4, 3, 6}}

My solution

``````int public[][]    matrixSum (int[][]    a, int[][]    b) {
if (a.length == 0 || b.length == 0) {return new int[0][0]; }
In t[][]    sum = new int[a.length][a[to[una[a[0].length];
// You can use only `a` for the length because a and b have the same dimensions
for (int i = 0; i <a.length; i ++) {
for (int j = 0; j <a[0].length; j ++) {
sum[i][j]    = a[i][j]    + b[i][j];
}
}
sum of return
}
``````

## Sum including combination

I'm having trouble evaluating the summary:

$$sum_ {k = 0} ^ {2n} (- 1) ^ kk ^ n {2n choose k}$$

Can someone take me to a solution?

Also, is there a general or simple way to address sums that include combinations in them?

## nt.number theory – finite sum that involves the root of the unit

I have the following sum:

$$sum _ { sigma = 1, odd} ^ { frac {p} {2} -2} frac { sin frac {r sigma pi} {p}} { sin frac {q sigma pi} {p}}$$

where $$p = 2 ; text {mod} ; 4$$, $$p, q$$ they are coprime numbers and $$p-q ge1$$Y $$r = 1, …, p-1 ; text {odd}$$. I know the result is the way

$$pm frac { frac {p} {2} pm 1} {2}$$

but the signs depend on the relation of r, p, q and it is not easy to guess.

Is there any formula for this type of sums?

I tried to reduce it using exponentials like the following:

$$sum _ { sigma = 1, odd} ^ { frac {p} {2} -2} e ^ {i pi frac {(qr) sigma} {p}} frac {e ^ { 2i pi frac {r sigma} {p}} – 1} {e ^ {2i pi frac {q sigma} {p}} – 1}$$

and define $$omega = e ^ {2 i pi frac {q sigma} {p}}$$ which is a primitive root of unity, but I do not know how this can help.