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Words overlapping in a Word Search game


Words overlapping in a Word Search game

By : Abdullah Hanefi Önal
Date : November 20 2020, 09:01 AM
Hope this helps Your logic to place a word should be:
pick a word pick random coordinates and an orientation until the word can be placed at these coordinates and with that orientation. place the word
code :
reCheck4:
    if ((randPosX + strlen(dictionary[r1]) > 10)) {
        randPosX--;
        goto reCheck4;
    }  
while (randPosX + strlen(dictionary[r1]) > 10) randPosX--;
goto reRand2;
amend = true;
int incr = 0;

for (int x = 0; x < 4; x++) {
    // Do stuff with incr, but not with x
    incr++;
}
for (int incr = 0; incr < 4; incr++) {
    // Do stuff with incr, but not with x
}
if (incr == 1) {
    r1 = r2;
}
if (incr == 2) {
    r1 = r3;
}
if (incr == 3) {
    r1 = r4;
}
r1 = index[incr];
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <time.h>

int isvalid(char matrix[10][10], int x, int y, int dx, int dy, int len)
{
    for (int i = 0; i < len; i++) {
        if (matrix[y][x] != '-') return 0;
        x += dx;
        y += dy;
    }

    return 1;
}

void matrix_fill(char matrix[10][10], const char *words[4])
{
    int dx[3] = {1, 0, 1};      // horizontal strides
    int dy[3] = {0, 1, 1};      // vertical strides

    startover:

    for (int i = 0; i < 10; i++) {
        for (int j = 0; j < 10; j++) {
            matrix[j][i] = '-';
        }
    }

    for (int i = 0; i < 4; i++) {
        const char *word = words[i];
        int len = strlen(word);

        int orient, x, y;
        int tries = 0;
        int valid;

        if (len > 8) break;     // just to be sure

        do {
            orient = rand() % 3;
            x = rand() % (10 - dx[orient] * len);
            y = rand() % (10 - dy[orient] * len);

            valid = isvalid(matrix, x, y,
                            dx[orient], dy[orient], len);            
            tries++;
        } while (!valid && tries < 100);

        if (!valid) goto startover;

        for (int i = 0; i < len; i++) {
            matrix[y][x] = word[i];
            x += dx[orient];
            y += dy[orient];
        }
    }

}

void matrix_print(char matrix[10][10])
{
    printf("%4s    ", "");
    for (int i = 0; i < 10; i++) {
        printf("%4c", 'A' + i);
    }
    printf("\n\n");

    for (int j = 0; j < 10; j++) {
        printf("%4d    ", j);

        for (int i = 0; i < 10; i++) {
            printf("%4c", matrix[j][i]);
        }
        printf("\n\n");
    }
}

int main()
{
    const char *dictionary[4] = {"apple", "pear", "cherry", "lemon"};
    char matrix[10][10];

    srand(time(NULL));

    matrix_fill(matrix, dictionary);
    matrix_print(matrix);

    return 0;
}


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Creating a words grid for a search word game

Creating a words grid for a search word game


By : Evan
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further There is no one answer to your question. You could use individual UIButtons for each letter or set up your own touch coordinate tracking system corresponding to letters or do something completely different. It boils down to what you feel comfortable working with and your experience level. To get the ball rolling, I would suggest you take a look at this Ray Wenderlich tutorial which shows how to create a letter word game by dragging letter tiles. The tech in this tutorial will definitely help you with your project.
Updated: If you go the button route, you can create them programmatically and pretty much do anything you want. Change their title (in your case) letter, show or hide them, track various kind of touches, and so on. For the letters and tracking touches, there are probably a number of methods of doing this. Try putting your letters on a subview while keeping track of their center coordinates. Then add a new subview over your letters to track the user's touches.
Python word game. Last letter of first word == first letter of second word. Find longest possible sequence of words

Python word game. Last letter of first word == first letter of second word. Find longest possible sequence of words


By : Deniz Akdere
Date : March 29 2020, 07:55 AM
seems to work fine Your idea of calling a function inside of itself is a good one. We can solve this with recursion:
code :
def get_neighbors(word, choices):
    return set(x for x in choices if x[0] == word[-1])

def longest_path_from(word, choices):
    choices = choices - set([word])
    neighbors = get_neighbors(word, choices)

    if neighbors:
        paths = (longest_path_from(w, choices) for w in neighbors)
        max_path = max(paths, key=len)
    else:
        max_path = []

    return [word] + max_path

def longest_path(choices):
    return max((longest_path_from(w, choices) for w in choices), key=len)
words = ("audino bagon baltoy banette bidoof braviary bronzor carracosta "
         "charmeleon cresselia croagunk darmanitan deino emboar emolga "
         "exeggcute gabite girafarig gulpin haxorus")

words = frozenset(words.split())
>>> longest_path(words)
['girafarig', 'gabite', 'exeggcute', 'emolga', 'audino']
How to traverse a list of words and search each word and count the occurrences of a given substring within the word?

How to traverse a list of words and search each word and count the occurrences of a given substring within the word?


By : marc Katz
Date : March 29 2020, 07:55 AM
seems to work fine If I understand correctly you have to iterate over the list of words and count the number of occurrences for each word in the list, and print them.
Now printing and returning are two different tasks. The above statement however already defines a skeleton for the function you aim to implement:
code :
def freq_count(substr,list):
    for word in list:
        count = 0
        while ...: #we find an occurence
            count += 1
        print(word+str(count))
def freq_count(substr,list):
    for word in list:
        count = 0
        pos = word.find(substr)
        while pos >= 0:#we find an occurence
            count += 1
            pos += len(substr)
            pos = word.find(substr,pos)
        print(word+str(count))
>>> freq_count('aba',['','a','aba','ababa','abaaba','abababa','abacaba'])
0
a0
aba1
ababa1
abaaba2
abababa2
abacaba2
Using Google Script, how do I exclude a word when using regex to search for other words within a given word limit?

Using Google Script, how do I exclude a word when using regex to search for other words within a given word limit?


By : CreamBeta
Date : March 29 2020, 07:55 AM
I wish this helpful for you If you only want to know either a certain word is contained in a cell or not, you do not need Regex
Much easier is the use of IndexOf()
code :
if(column.indexOf("dispute")!=-1){
  Logger.log("Column contains dispute");
  }
else if (regexp.exec(column) !== null) {
  result.push(row); 
}
function regexwithsheetpop() {
  var ss = SpreadsheetApp.getActiveSpreadsheet();
  var historySheet = ss.getSheetByName('Sheet1');  
  var resultsSheet = ss.getSheetByName('Results1');     //mind the name
  var totalRowsWithData = historySheet.getDataRange().getNumRows();
  var data = historySheet.getRange(1, 1, totalRowsWithData, 3).getValues(); 
  var regexp = /\W*(account)\W*\s+(\w+\s+){0,10}(the)|(the)\s+(\w+\s+){0,10}(account)/i;
  var result = []; 
  for (var i = 0; i < data.length; i += 1) {
     var row = data[i];
     var column = row[0];
     if(column.indexOf("dispute")!=-1){
       Logger.log("Column in row "+(i+1)+" contains dispute");
     } else if (regexp.exec(column) !== null) {
        result.push(row); }
  }
  if (result.length > 0) {
     var resultsSheetDataRows = resultsSheet.getDataRange().getNumRows();
     resultsSheetDataRows = resultsSheetDataRows === 1 ? resultsSheetDataRows : resultsSheetDataRows + 1; 
     var resultsSheetRange = resultsSheet.getRange(resultsSheetDataRows, 1, result.length, 3);
     resultsSheetRange.setValues(result);
  }
}
Algorithm to choose random letters for word search game that allows many words to be spelled

Algorithm to choose random letters for word search game that allows many words to be spelled


By : kusoo
Date : March 29 2020, 07:55 AM
it helps some times Here's a simple method:
Write a fast solver for the game using the same word list that the player will use. Generate say 100 different possible boards at random (using letter frequencies is probably a good idea here, but not essential). For each board calculate all the words that can be generated and score the board based on the number of words found or the count weighted by word length (i.e. the total sum of word lengths of all words found). Then just pick the best scoring board from the 100 possibilities and give that to the player.
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