## magento2 – Api Magento 2 – How to get the layer navigation filters available in idle api?

I have a problem with api magento 2. I need the navigation data in the layers to be available in the product collection (current category, product search), but I do not know if api magento 2 is available or not. I have created a new custom API using Magento Catalog Template Layer Resolver but it does not work. Can someone help me?

thanks guys

## beginner – Morpion Version 2

The game has become much more user friendly than the previous version – with an example of a grid to illustrate the entries, the X's and Bones are now used.

The game can now detect the draw.

The algorithm has been modified to detect the winner, it is overall less buggy, although there are still some combinations towards the end of a game in which it incorrectly determines a winner or a loser – no idea how to solve this problem or determine its cause.

I would appreciate some tips to improve the structure of the game and the code, and to avoid the mentioned bugs.

``````#understand
#understand
#understand

// Struct with all the game state variables.
struct game_data {
int win; // 0 or 1.
int turns; // Range from 1 to 9 (end of the game).
int tour; // 0 or 1 where 0 is a human player
tank grill[3][3];
};

// Initialization of game state variables
struct game_data game = {
0
1
0
{{}, & # 39;; & # 39;},
(& # 39; & # 39;) & # 39;
{& # 39; & # 39;, & # 39;}
};

void intro (void) {

printf ("Welcome to NOUGHTS AND CROSSES  n  n");
printf ("The grid on which you will play is 3x3 and your entry will be determined by the coordinates you entered, in the form" row column ".  n  n");
printf ("For example, an entry of" 1 1 "will put a" Z "on the first line of the first column, as follows:  n  n");

printf (
"+ --- + --- + --- +  n"
"| Z | | |  n"
"+ --- + --- + --- +  n"
"| | | |  n"
"--- + --- + --- |  n"
"| | | |  n"
"+ --- + --- + --- +  n"
" not");
}

void player_one_move (game struct game_data *)
{
int y_val, x_val;
printf ("You are & # 39; Cross & # 39;. Please enter the coordinates in the form & # 39; column line & # 39; for 3x3 grid:  n");
scanf ("% d% d", & y_val, & x_val);

if (game-> grid[y_val - 1][x_val - 1]    == & # 39;) {
game-> grid[y_val - 1][x_val - 1]    = & # 39; X & # 39;
printf (" nYour turn:  n  n");
}
other {
player_one_move (game);
}
}

void computer_move (struct game game_data *)
{
int x_val = rand ()% 3;
int y_val = rand ()% 3;

if (game-> grid[y_val][x_val]    == & # 39;) {
game-> grid[y_val][x_val]    = & # 39 ;;
printf (" nUnumber of computer:  n  n");
}
other {
computer_move (game);
}
}

void update (struct game game_data *)
{

printf (

"+ --- + --- + --- +  n"
"|% c |% c |% c |  n"
"+ --- + --- + --- +  n"
"|% c |% c |% c |  n"
"---- + --- + --- |  n"
"|% c |% c |% c |  n"
"+ --- + --- + --- +  n"
" not",
game-> grid[0][0], game-> grid[0][1], game-> grid[0][2],
game-> grid[1][0], game-> grid[1][1], game-> grid[1][2],
game-> grid[2][0], game-> grid[2][1], game-> grid[2][2])
}

void game_event_won (game struct game_data *)
{

int account;
// char current_mark;

// BUGGY
/ * char current_mark;

if (game-> turn == 0) {
current_mark = & # 39 ;;
other{
current_mark = & # 39;
} * /

for (int y_val = 0; y_val <3; y_val ++) {
for (int x_val = 0; x_val < 3; x_val++) {
count = 0;
while (game->Grid[y_val][x_val]    == X & # 39;) {
x_val ++;
count ++;

// BUGGY
/ * if (count == 3 && current_mark == & nbsp; X)
game-> victory = 1;
printf ("You won  n");
}
if (count == 3 && current_mark == & nbsp; & nbsp;)
game-> victory = 1;
printf ("You have lost  n");
} * /

if (count == 3) {
game-> victory = 1;
printf ("You won  n");
}
}
}
}
for (int x_val = 0; x_val <3; x_val ++) {
for (int y_val = 0; y_val < 3; y_val++) {
count = 0;
while (game->Grid[y_val][x_val]    == X & # 39;) {
y_val ++;
count ++;

if (count == 3) {
game-> victory = 1;
printf ("You won  n");
}
}
}
}
for (int y_val = 0; y_val < 3; y_val++) {
count = 0;
while (game->Grid[y_val][y_val]    == X & # 39;) {
count ++;
y_val ++;

if (count == 3) {
game-> victory = 1;
printf ("You won  n");
}
}
}
for (int y_val = 0; y_val < 3; y_val++) {
count = 0;
while (game->Grid[y_val][2 - y_val]    == X & # 39;) {
count ++;
y_val ++;

if (count == 3) {
game-> victory = 1;
printf ("You won  n");
}
}
}
}

// Repeat the previous function but for & # 39; O & # 39; s. Less concise but less buggy than the previous implementation.
void game_event_lost (struct game game_data *)
{

int account;

for (int y_val = 0; y_val <3; y_val ++) {
for (int x_val = 0; x_val < 3; x_val++) {
count = 0;
while (game->Grid[y_val][x_val]    == & # 39;)
x_val ++;
count ++;

if (count == 3) {
game-> victory = 1;
printf ("You have lost  n");
}
}
}
}
for (int x_val = 0; x_val <3; x_val ++) {
for (int y_val = 0; y_val < 3; y_val++) {
count = 0;
while (game->Grid[y_val][x_val]    == & # 39;)
y_val ++;
count ++;

if (count == 3) {
game-> victory = 1;
printf ("You have lost  n");
}
}
}
}
for (int y_val = 0; y_val < 3; y_val++) {
count = 0;
while (game->Grid[y_val][y_val]    == & # 39;)
count ++;
y_val ++;

if (count == 3) {
game-> victory = 1;
printf ("You have lost  n");
}
}
}
for (int y_val = 0; y_val < 3; y_val++) {
count = 0;
while (game->Grid[y_val][2 - y_val]    == & # 39;)
count ++;
y_val ++;

if (count == 3) {
game-> victory = 1;
printf ("You have lost  n");
}
}
}
}

int main (void)
{

srand (unsigned time (0));

intro ();

while (game.win == 0) {
if (game.turn == 0) {
player_one_move (& game);
game.turns ++;
game.turn = 1;
}

other {
game.turn = 0;
computer_move (& game);
game.turns ++;
}
if (game.turns == 9 && game.win == 0) {
game.win = 1;
printf ("You have drawn  n");
Pause;
}
update (& game);

game_event_won (& game);
game_event_lost (& game);

}

returns 0;
}
``````

## [ Insurance & Registration ] Open question: I had 2 accidents in 1 year, so I lost my loss discount (full nb for 15 years ago) is it possible to buy car insurance?

[ Insurance & Registration ] Open question: I had 2 accidents in 1 year, so I lost my loss discount (full nb for 15 years ago) is it possible to buy car insurance? .

## How to download my module in github (via composer.json) in magento 2

I want to download my module in the github repository and then I want to install it in my local directory with composer.json help. How is it possible? Please provide me with a solution, to install my module using composer.json?

## Number Theory – Is it true that \$ sum_ {k = 1} ^ infty frac { binom {2k} k ^ 2} {k16 ^ k} (H_ {2k} -H_k) = frac23 sum_ {k = 0} ^ infty frac { binom {2k} k ^ 2H_ {2k}} {(2k + 1) 16 ^ k} \$?

Six years ago, I conjectured the identity
$$sum_ {k = 1} ^ infty frac { binom {2k} k ^ 2} {k16 ^ k} (H_ {2k} -H_k) = frac23 sum_ {k = 0} ^ infty frac { binom {2k} k ^ 2H_ {2k}} {(2k + 1) 16 ^ k}, tag { * }$$
or $$H_n$$ denotes the harmonic number $$sum_ {k = 1} ^ n frac1k$$. As the two series converge slowly, I lack compelling numerical data to support $$(*)$$.

Question. Is the identity $$(*)$$ true? Can we check further? If that is true, how to prove it?

## magento2 – How to get the order collection filter with customer_is_guest and email (Magento 2)

How to get the order ID of the collection with filter `customer_is_guest = 1` and `customer email`

``````\$ order_collection = \$ objectManager-> create ('Magento  Sales  Model  Order') -> getCollection () -> addAttributeToFilter ('customer_email', \$ customer_email) -> addAttributeToFilter (& # 39; # 39; customer_is_guest # 1)
``````

The code above will give me `collection of orders`. How to get only order IDs

## algorithms – Does the conversion of the representation of the adjacency matrix of a graph of size n x n into an adjacency list always require a time O (n ^ 2)?

"Nothing is always."

Without any prior knowledge, the worst complexity in time of converting an adjacency matrix representation of a graph into 0.1 values ​​into a corresponding adjacency list representation is $$Theta (n ^ 2)$$.

However, if we know that the graph is a sparse matrix with a number of edges $$m = o (n ^ 2)$$ and its adjacency matrix is ​​stored in a compressed format like the Yale format, so we can expect that the conversion will take $$o (n ^ 2)$$ time.

## calculus – Show that \$ int ^ {R + delta} _ { delta / 2} sqrt {R ^ 2- (x- delta) ^ 2} , dx- int ^ {R} _ { delta / 2} sqrt {R ^ 2-x ^ 2} , dx leq R delta \$, for \$ 0

I need to show that if $$0 < delta then

$$int ^ {R + delta} _ { delta / 2} sqrt {R ^ 2- (x- delta) ^ 2} , dx- int ^ {R} _ { delta / 2 } sqrt {R ^ 2-x ^ 2} , dx leq R delta.$$

The result of the evaluation of the integrals is awful and seems quite difficult to manage. Is there a simpler way to get the result using a binding argument?

All clues are highly appreciated. Thank you for your time.

## 5th dnd – Can I draw 2 weapons with the same hand (by throwing one, then pulling another) using Dual Wielder's exploit?

I had an idea for a Level 5 fighter with the feat of Dual Wielder who used a melee weapon in the left hand (whip) and a throwing weapon (javelin) in the main hand.

My question is this: with the empty main hand and a whip in the left hand, can my fighter draw a javelin, throw it, throw another, end it with a whipping hand?

The main point being: Can you draw 2 weapons with the same hand using the Dual Wielder exploit?

## Complexity Theory – Why Every DNF Formula for \$ (x_ {1} vee y_ {1}) wedge (x_ {2} vee y_ {2}) wedge ldots wedge (x_ {n} vee y_ {n}) \$ have at least \$ 2 ^ {n} \$ terms?

Why each DNF formula for $$(x_ {1} vee y_ {1}) wedge (x_ {2} vee y_ {2}) wedge dots wedge (x_ {n} vee y_ {n})$$ have at least $$2 ^ {n}$$ terms?

This statement can be found on the Wikipedia page of the DNF form here: https://en.wikipedia.org/wiki/Disjunctive_normal_form

I can not understand why that's true though. I saw an explanation saying that we can distribute the ET and OR, but I was wondering if there was a better way to understand that. I do not know too much logic, and I would appreciate it if anyone could help me, please.