Anonymní profil Epoxi – Programujte.com

normální sudoku to resi normalne... mam za ukol udelat diagonalni sudoku. Jde o to, ze kdyz zadam same nuly tak mi to vyresi, ale ignoruje prvni diagonalu pole 00 a 88 jsou shodna. A pokud zadam nejake zadani z internetu, neni schopen to vyresit. Vyresi jenom nejake pripady, ale to je nahoda. Zvlastni je, ze pokud smazu

for (i = 0; i < MAX - 1; i++) { k = 8; m = 0; m = k - i; if ( sudoku[i][m] == 0 ) continue; for (j = i + 1; j < MAX; j++) { n = 8; l = 0; l = n - j; if (sudoku[j][l] == sudoku[i][m]) return (FALSE); } }

tak to vyresi. Potreboval bych na to neco lepsiho.

No, dopsal jsem to takto a nefunguje to
 

int volne_pole_cislo(int cislo, int radek, int sloupec)
/**********************/
{
/* Test if the number is already in
the line, or the column, or the sub-square of 3 X 3
afferent at the cell pointed to by line and column. */
	char buffer[MAX + 10];
	int i, j,  k, l, m, n;
	for (i = 0; i < MAX; i++) if (sudoku[radek][i] == cislo) return(FALSE);
	/* If the number is used in the line return FALSE */
	for (i = 0; i < MAX; i++) if (sudoku[i][sloupec] == cislo) return(FALSE);
	/* If the number is used in the column return FALSE */


	for (i = 0; i < MAX - 1; i++) {
		if ( sudoku[i][i] == 0 ) continue;
		for (j = i + 1; j < MAX; j++) {
			if (sudoku[j][j] == sudoku[i][i]) return (FALSE);
        }
    }

	for (i = 0; i < MAX - 1; i++) {
       k = 8;
       m = 0;
	   m = k - i;
		if ( sudoku[i][m] == 0 ) continue;
		for (j = i + 1; j < MAX; j++) {
		    n = 8;
		    l = 0;
		    l = n - j;
			if (sudoku[j][l] == sudoku[i][m]) return (FALSE);
        }
    }
	/* If the number is used in the sub-square of 3 X 3 return FALSE */
	transformuj_ctverecek_poli(buffer, radek, sloupec);
	for (i = 0; i < MAX; i++) if (buffer[i] == cislo) return (FALSE);
	/* whereby the number is available for the cell, return TRUE. */
	return(TRUE);
}/* end free_cell_for_number */