{"id":3179,"date":"2018-07-08T00:01:42","date_gmt":"2018-07-08T07:01:42","guid":{"rendered":"https:\/\/c-for-dummies.com\/blog\/?p=3179"},"modified":"2018-06-30T13:21:44","modified_gmt":"2018-06-30T20:21:44","slug":"fill-my-grid-solution","status":"publish","type":"post","link":"https:\/\/c-for-dummies.com\/blog\/?p=3179","title":{"rendered":"Fill My Grid – Solution"},"content":{"rendered":"

The puzzle for this month’s Exercise<\/a> was to create and fill a 20-by-20 character grid with 20 asterisks placed at random positions, no two asterisks in the same spot. I hope you found this Exercise interesting, but not too easy.
\n
\nHere is my solution for Part A:<\/p>\n

\r\n#include <stdio.h>\r\n#include <stdlib.h>\r\n#include <time.h>\r\n\r\n#define SIZE 20\r\n\r\nint main()\r\n{\r\n    char grid[SIZE][SIZE];\r\n    int column,row,count;\r\n\r\n    \/* seed the randomizer *\/<\/span>\r\n    srand( (unsigned)time(NULL) );\r\n\r\n    \/* initialize the grid *\/<\/span>\r\n    for(column=0;column<SIZE;column++)\r\n        for(row=0;row<SIZE;row++)\r\n            grid[column][row] = '.';\r\n\r\n    \/* place the random asterisks *\/<\/span>\r\n    count = 0;\r\n    while(count < SIZE)\r\n    {\r\n        column = rand() % SIZE;\r\n        row = rand() % SIZE;\r\n        if( grid[column][row] != '*' )\r\n        {\r\n            grid[column][row] = '*';\r\n            count++;\r\n        }\r\n    }\r\n\r\n    \/* display the grid *\/<\/span>\r\n    for(column=0;column<SIZE;column++)\r\n    {\r\n        for(row=0;row<SIZE;row++)\r\n            printf(\" %c \",grid[column][row]);\r\n        putchar('\\n');\r\n    }\r\n\r\n    return(0);\r\n}<\/pre>\n
The randomizer is initialized at Line 13. The 20-by-20 grid<\/code>, with the value 20 set as the SIZE<\/code> constant, is initialized in Lines 16 through 18.<\/p>\n
Variable count<\/code> tracks the number of asterisks placed, initialized at Line 21. Then a while<\/em> loop spins as random locations are obtained in Lines 24 and 25. The grid<\/code> location is tested at Line 26 to see whether an asterisk is already set at that location. If not, it’s set at Line 28 and variable count<\/code> is increased at Line 29.<\/p>\n
Lines 34 through 39 display the results.<\/p>\n
I trust you coded a similar solution. The asterisk itself acts as a flag to determine whether a specific grid location is set. And the while<\/em> loop keeps spinning as random locations are fetched until that last spot is found.<\/p>\n
To complicate matters, the Part B challenge to this Exercise required that the random location’s row and column be free of asterisks as well. You can use the code from the Part A solution to start your Part B solution, which is what I did. Extra overhead is required to scan not only the current row\/column position but all the same row<\/code> and column<\/code> positions for any asterisks.<\/p>\n
To perform the extra tests, I modified the while<\/em> loop by adding two for<\/em> loops. One scans the current row<\/code> and the other scans the current column<\/code>. If an asterisk is found in either row or column, a found<\/code> flag is set. If the found<\/code> flag remains zero, meaning both the row and column are devoid of asterisks, the asterisk is set at the random row\/column intersection. Here’s the modified while<\/em> loop:<\/p>\n
\r\n    while(count < SIZE)\r\n    {\r\n        column = rand() % SIZE;\r\n        row = rand() % SIZE;\r\n        found = 0;\r\n        if( grid[column][row] != '*' )\r\n        {\r\n            for(c=0;c<SIZE;c++)\r\n                if( grid[c][row] == '*' )\r\n                    found++;\r\n            for(r=0;r<SIZE;r++)\r\n                if( grid[column][r] == '*' )\r\n                    found++;\r\n            if( !found )\r\n            {\r\n                grid[column][row] = '*';\r\n                count++;\r\n            }\r\n        }\r\n    }<\/pre>\n
Variable found<\/code> is initialized to zero just after the random row<\/code> and column<\/code> variables are set.<\/p>\n
The first if<\/em> test checks the specific row<\/code> and column<\/code> position for an asterisk. After all, if it’s already set, why scan the rows and columns?<\/p>\n
Next, the row<\/code> is scanned for asterisks. If one is present, the found<\/code> variable is incremented. I could have stopped the loop once an asterisk is found. That’s because the code’s logic insists that no two asterisks will ever dwell in a single row or column. Still, time isn’t an issue in this code, so the loop keeps spinning.<\/p>\n
The column<\/code> is scanned in the same manner as the row<\/code>. This for<\/em> loop could be avoided when an asterisk was found in the current row. Again, time isn’t an issue in the code so I didn’t add the necessary statements to skip this loop.<\/p>\n
Finally, the found<\/code> variable is tested with “if not found.” When true, the rows and columns are clear, meaning the asterisk can be set.<\/p>\n
Click here<\/a> to view my entire solution for Part B.<\/p>\n
I hope that you enjoyed this Exercise and devised a solution that not only worked, but offered a clever way to solve the challenge.<\/p>\n","protected":false},"excerpt":{"rendered":"
The puzzle for this month’s Exercise was to create and fill a 20-by-20 character grid with 20 asterisks placed at random positions, no two asterisks in the same spot. I hope you found this Exercise interesting, but not too easy.<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"class_list":["post-3179","post","type-post","status-publish","format-standard","hentry","category-solution"],"_links":{"self":[{"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=\/wp\/v2\/posts\/3179","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=3179"}],"version-history":[{"count":2,"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=\/wp\/v2\/posts\/3179\/revisions"}],"predecessor-version":[{"id":3190,"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=\/wp\/v2\/posts\/3179\/revisions\/3190"}],"wp:attachment":[{"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=3179"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=3179"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=3179"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}