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    <p>First of all the position of the letter in the grid is easy to compute:</p> <pre><code>function get_pos( c : char) begin int col ← (c-'A') modulo 4 int row ← (c-'A') div 4 return (row,col) end </code></pre> <h3>Assuming positions (6,2) and (6,3) can be used</h3> <p>We can simply define a subtraction between cell coordinates as:</p> <pre><code>function subtract( pos1, pos2 : position ) begin return (pos2.row-pos1.row, pos2.col-pos1.col) end </code></pre> <p>If the subtraction yields a pair (x,y) then the path is simply x times the chararacter 'R' if x is positive or 'L' if x is negative possibliy nothing if x is zero, and similarly y times the character 'D' if y is positive or 'U' if y is negative possibly nothing if y is zero, then we end with the character 'E'.</p> <pre><code>function print_row_path( pos1, pos2 : position ) begin path ← subtract(pos1,pos2) if path.row &gt; 0 then print_times_char(path.row,'R') else if path.row &lt; 0 print_times_char(-path.row,'L') end if end function print_col_path( pos1, pos2 : position ) begin path ← subtract(pos1,pos2) if path.col &gt; 0 then print_times_char(path.col,'D') else if path.col &lt; 0 print_times_char(-path.col,'U') end if end function print_path_direction( pos1, pos2 : position ; first_direction : direction ) begin if (first_direction = FIRST_MOVE_ROW) then print_row_path(pos1,pos2) print_col_path(pos1,pos2) else print_col_path(pos1,pos2) print_row_path(pos1,pos2) end if print 'E' end function print_path(start, end : char) begin position pos1 ← get_pos(start) position pos2 ← get_pos(end) print_path_direction(pos1,pos2, FIRST_MOVE_ROW) end </code></pre> <p>Where print_times_char(t,c) is a function which prints t times the character c. I defined two "flavours" of path printing, one printing the row movements first and another printing column movements first.</p> <h3>Assuming positions (6,2) and (6,3) are forbiden</h3> <p>If we're not allowed to use positions (6,2) and (6,3) then:</p> <ul> <li>if 'A' ≤ start,end ≤ 'X' or 'Y' ≤ start,end ≤ 'Z' : (6,2) or (6.3) will never be used in the path</li> <li>if 'A' ≤ start ≤ 'X' and 'Y' ≤ end ≤ 'Z' : to ensure to not use forbidden cells print columns movements first</li> <li>if 'A' ≤ end ≤ 'X' and 'Y' ≤ start ≤ 'Z' : this time we have to print row movements first</li> </ul> <p>In pseudo code:</p> <pre><code>function print_path_wo_forbidden(start, end : char) begin position pos1 ← get_pos(start) position pos2 ← get_pos(end) if if 'A' ≤ start ≤ 'X' and 'Y' ≤ end ≤ 'Z' then print_path_direction(pos1,pos2, FIRST_MOVE_COLUMN) else print_path_direction(pos1,pos2, FIRST_MOVE_COLUMN) end if end </code></pre> <h3>Complexity</h3> <p>Printing the path between two positions is clearly in O(1), so for a string of length n we can build an O(n) algorithm.</p>
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