Word Search (#107)

Ruby
1

I had a few moments this weekend and sat down to work this problem. I
was quite
surprised to find it tougher than I had expected. I fiddled around for
about an
hour trying to find a solution I felt was elegant, but never really got
there.
Luckily, we have many submitters smarter than me.

Almost all of the solutions this week took different approaches and they
are all
quite interesting. This doesn’t seem to be one of those problems we
have a
standard approach for. Some solutions did a boring search using X and Y
coordinates; others started by building a map of points in the puzzle to
letters
or even letters to points; one even works by performing board
transformations.
I’m going to show the basic X and Y search in this summary, but the
other method
were equally interesting and well worth a look.

Let’s talk about the solution from Bob S., which begins like
this:

class WordSearch

class Board < Array

 def to_s
   collect {|s| s.split(//).join(' ')}.join("\n")
 end

end

…

Here we see an Array subclass with a trivial modification. Board
objects are
just two-dimensional Arrays that know how to draw themselves in the quiz
output
format.

Let’s move on to the setup code:

…

attr_reader :board, :solution

creates a new, empty solver

def initialize
@board = Board.new
@solution = Board.new
end

resets the solution

def reset
@solution.clear
@board.each {|row| @solution << row.gsub(/./, ‘+’)}
end

…

Here we have the initialization for two Board objects. One will hold
the actual
board, or puzzle object, and the other the solution-in-progress.
reset() gives
us hints at the solution object structure, which is cleared to a board
full of
plus characters.

The board is loaded with the following code:

…

checks that the board contains only letters and that it has a

uniform

rectangular shape

def validate
@board.size > 0 or raise “Board has no rows”
@board.grep(/[^A-Z]/).empty? or raise “Board contains non-letters”
w = @board.collect {|row| row.size}.uniq
w.size == 1 or raise “Board rows are not all the same length”
w.first > 0 or raise “Board has no columns”
end

parses the board by reading lines from io until a blank line (or

EOF)

is read.

def parse(io = ARGV)
@board.clear
while line = io.gets
line = line.strip.upcase
break if line == ‘’
@board << line
end
validate
reset
end

…

validate() is just a reality check for the board. It verifies that we
have some
rows, those rows contain letters, there are columns in each row, and in
fact the
same number of columns. That check is run towards the end of parse(),
which
just reads line by line filling the board object. Note that reset() is
also
called to prepare the solution object.

The real work comes in the methods that search for words:

…

search for word. returns number of times found. solution is

updated

with all occurences.

def search(word)
found = 0
0.upto(board.size-1) do |y|
0.upto(board[y].size-1) do |x|
[-1, 0, 1].each do |dy|
[-1, 0, 1].each do |dx|
next if dx == 0 and dy == 0
found += 1 if search_for(word.strip.upcase, x, y, dx, dy)
end
end
end
end
found
end

search for word in board starting at position (x,y) and moving in

direction (dx,dy). returns true if found, false if not found.

def search_for(word, x, y, dx, dy)
return false if x < 0 or
x >= board.first.size or
y < 0 or
y >= board.size
return false if board[y][x] != word[0]
prev = solution[y][x]
solution[y][x] = board[y][x]
return true if word.length <= 1
found = search_for(word[1,word.length-1], x + dx, y + dy, dx, dy)
solution[y][x] = prev unless found
found
end

…

The search() method manages the hunt for a single term. It’s a trivial
brute-force find from every starting square in all eight directions.
The method
maintains and returns a count, for all occurrences found during the
search. The
actual letter match is handed off to search_for() though.

In search_for() a recursive search is used to find letter by letter.
The tricky
part here is that the solution object is modified as we search, assuming
we will
find the word. This means that an extra walk isn’t needed to update the
solution later, but it also means the code must revert the changes to
the
solution object if the word isn’t found. Those are the gymnastics you
see in
the second half of the method.

The next two methods wrap an interface around these calls:

…

creates a new puzzle by parsing the board from io. see

WordSearch#parse

def self.parse(io = ARGF)
obj = new
obj.parse(io)
obj
end

def to_s
solution.to_s
end

end

…

The class method parse() constructs an object and parses the board from
the
passed IO object. The instance method to_s() can then be used to show
final
results, after one or more calls to search().

The solution ends with code to kick-off the process:

…

parse the board first

p = WordSearch.parse

parse the words until a blank line is read

words = []
while line = ARGF.gets
line = line.strip.upcase
break if line == ‘’
words += line.gsub(’,’, ’ ').split
end

submit each word and show how many times it was found

for word in words.sort.uniq
n = p.search(word)
puts word + ’ was ’ + (n == 0 ? ‘not found’ :
n == 1 ? ‘found once’ :
“found #{n} times”)
end

show the solution

puts p

Here we see the WordSearch object constructed and the board parsed out
of the
input. The remainder of the input is divided into words and search() is
called
for each one. A found count is printed for each word as the search is
made,
then the final solution is displayed.

      • M + + + + + +
        • Y + + + + +
          T H A N K S + + + +
          O + + + + + + + + +
  • L L A + T H E + +
        • C + + + + +
    • S O L V E R S +
        • e + + + + +
        • V + + + + +
        • E + + + + +
        • R + + + + +

Tomorrow we have a new word game, so stay tuned…