[SUMMARY] String Equations (#112)

Discussion in 'Ruby' started by Ruby Quiz, Feb 8, 2007.

  1. Ruby Quiz

    Ruby Quiz Guest

    This quiz is much more difficult than it looks. There are an infinite number of
    combinations, even for small word sets, because each term can be repeated.

    To handle this, most brave souls who solved the quiz used some matrix
    transformations from linear algebra for solving a system of linear equations.
    This makes it possible to find solutions in reasonable time, but I had to drag
    out the math textbooks to decode the solutions.

    Let's take a look into one such solution by Eric I.:

    require 'mathn'

    CompactOutput = false

    # calculate the least common multiple of one or more numbers
    def lcm(first, *rest)
    rest.inject(first) { |l, n| l.lcm(n) }
    end

    # ...

    This should be pretty easy to digest. The mathn library is pulled in here to
    get more accurate results in the calculations the code will be doing, a constant
    selects the desired output mode, and a shortcut is defined for applying lcm()
    over an Array of numbers.

    The next method is where all the action is, so let's take that one slowly:

    # ...

    # Returns nil if there is no solution or an array containing two
    # elements, one for the left side of the equation and one for the
    # right side. Each of those elements is itself an array containing
    # pairs, where each pair is an array in which the first element is the
    # number of times that word appears and the second element is the
    # word.
    def solve_to_array(words)
    # clean up word list by eliminating non-letters, converting to lower
    # case, and removing duplicate words
    words.map! { |word| word.downcase.gsub(/[^a-z]/, '') }.uniq!

    # ...

    The first comment does a good job of describing the result this method will
    eventually produce, so you may want to glance back to it when we get that far.

    The first set of operations is the word normalization process right out of the
    quiz. This code shouldn't scare anybody yet. (Just a quick side note though,
    it is possible to use delete("^a-z") here instead of the gsub() call.)

    One more easy bit of code, then we will ramp things up:

    # ...

    # calculate the letters used in the set of words
    letters = Hash.new
    words.each do |word|
    word.split('').each { |letter| letters[letter] = true }
    end

    # ...

    This code just makes a list of all letters used in the word list. (Only the
    keys() of the Hash are used.) To see what that's for, we need to dive into the
    math:

    # ...

    # create a matrix to represent a set of linear equations.
    column_count = words.size
    row_count = letters.size
    equations = []
    letters.keys.each do |letter|
    letter_counts = []
    words.each { |word| letter_counts << word.count(letter) }
    equations << letter_counts
    end

    # ...

    This code build the matrix we are going to work with to find answers. Each
    column in the matrix represents a word and each row a letter. The numbers in
    the matrix then are just a count of the letter in that word. For example, using
    the quiz equation this code produces the following matrix:

    v
    o
    l m
    d a r
    e r i
    l m v d
    o o t o d
    a r r o l l
    i m d t m o e
    +--------------
    v | 0 0 0 1 0 1 0
    l | 0 0 1 1 0 1 1
    a | 0 1 0 0 0 1 0
    m | 0 1 0 1 1 1 0
    d | 0 0 1 1 0 0 2
    o | 0 0 1 2 1 2 0
    e | 0 0 0 1 0 0 1
    r | 0 0 1 1 0 1 1
    t | 0 0 0 1 1 0 0
    i | 1 0 0 0 0 0 1

    If you glance back at the code now, it should be pretty clear how it builds the
    matrix as an Array of Arrays.

    Now we're ready to manipulate the matrix and this is the first chunk of code
    that does that:

    # ...

    # transform matrix into row echelon form
    equations.size.times do |row|
    # re-order the rows, so the row with a value in then next column
    # to process is above those that contain zeroes
    equations.sort! do |row1, row2|
    column = 0
    column += 1 until column == column_count ||
    row2[column].abs != row1[column].abs
    if column == column_count : 0
    else row2[column].abs <=> row1[column].abs
    end
    end

    # figure out which column to work on
    column = (0...column_count).detect { |i| equations[row] != 0 }
    break unless column

    # transform rows below the current row so that there is a zero in
    # the column being worked on
    ((row + 1)...equations.size).each do |row2|
    factor = -equations[row2][column] / equations[row][column]
    (column...column_count).each do |c|
    equations[row2][c] += factor * equations[row][c]
    end
    end
    end

    # ...

    Now you really don't want me to describe that line by line. Trust me. Instead,
    let me sum up what it does.

    This code transforms the matrix into row echelon form, which says that higher
    rows in the matrix have entries in further left columns and that the first
    significant entry in a row is preceded only by zeros. That sounds scarier than
    it is. Here's the transformed matrix (without the labels this time):

    1 0 0 0 0 0 1
    0 1 0 1 1 1 0
    0 0 1 2 1 2 0
    0 0 0 -1 -1 -2 2
    0 0 0 0 -1 -2 3
    0 0 0 0 0 -2 2
    0 0 0 0 0 0 0
    0 0 0 0 0 0 0
    0 0 0 0 0 0 0
    0 0 0 0 0 0 0

    The why behind this transformation is that its the first step in solving for our
    equation.

    On to the next bit of code:

    # ...

    # only one of the free variables chosen randomly will get a 1, the
    # rest 0
    rank = equations.select { |row| row.any? { |v| v != 0 }}.size
    free = equations[0].size - rank
    free_values = Array.new(free, 0)
    free_values[rand(free)] = 2 * rand(2) - 1

    # ...

    This bit of math uses the rank of the matrix to determine the free variables it
    will solve for. Free variables are just placeholders for substitutions in our
    system of equations. More concretely, they are where one or more words will be
    inserted in our string equations.

    Setting a single value to one, as the comment mentions, is basically preparing
    to to work with one word at a time. That's why the others are zeroed out.

    One more variable is prepared:

    # ...

    values = Array.new(equations[0].size) # holds the word_counts

    # ...

    As the comment explains, this will eventually be the counts for each word.

    OK, here's the last big bit of math:

    # ...

    # use backward elimination to find values for the variables; process
    # each row in reverse order
    equations.reverse_each do |row|
    # determine number of free variables for the given row
    free_variables = (0...column_count).inject(0) do |sum, index|
    row[index] != 0 && values[index].nil? ? sum + 1 : sum
    end

    # on this row, 1 free variable will be calculated, the others will
    # get the predetermined free values; the one being calculated is
    # marked with nil
    free_values.insert(rand(free_variables), nil) if free_variables > 0

    # assign values to the variables
    sum = 0
    calc_index = nil
    row.each_index do |index|
    if row[index] != 0
    if values[index].nil?
    values[index] = free_values.shift

    # determine if this is a calculated or given free value
    if values[index] : sum += values[index] * row[index]
    else calc_index = index
    end
    else
    sum += values[index] * row[index]
    end
    end
    end
    # calculate the remaining value on the row
    values[calc_index] = -sum / row[calc_index] if calc_index
    end

    # ...

    This elimination is the second and final matrix transform leading to a solution.
    The code works through each row or equation of the matrix, determining values
    for the free variables.

    Again this process is much more linear algebra than Ruby, so I won't bother to
    break it down line by line. Just know that the end result of this process is
    that values now holds the counts of the words needed to solve quiz. Positive
    counts belong on one side of the equation, negative counts on the other.

    This is the code that breaks down those counts:

    # ...

    if values.all? { |v| v } && values.any? { |v| v != 0 }
    # in case we ended up with any non-integer values, multiply all
    # values by their collective least common multiple of the
    # denominators
    multiplier =
    lcm(*values.map { |v| v.kind_of?(Rational) ? v.denominator : 1 })
    values.map! { |v| v * multiplier }

    # deivide the terms into each side of the equation depending on
    # whether the value is positive or negative
    left, right = [], []
    values.each_index do |i|
    if values > 0 : left << [values, words]
    elsif values < 0 : right << [-values, words]
    end
    end

    [left, right] # return found equation
    else
    nil # return no found equation
    end
    end

    # ...

    Assuming we found a solution, this code divides the words to be used into two
    groups, one for each side of the equation. It divides based on the positive and
    negative counts I just explained in values and the end result was described in
    that first comment at the top of this long method.

    With the math behind us, the rest of the code is easy:

    # ...

    # Returns a string containing a solution if one exists; otherwise
    # returns nil. The returned string can be in either compact or
    # non-compact form depending on the CompactOutput boolean constant.
    def solve_to_string(words)
    result = solve_to_array(words)
    if result
    if CompactOutput
    result.map do |side|
    side.map { |term| "#{term[0]}*\"#{term[1]}\"" }.join(' + ')
    end.join(" == ")
    else
    result.map do |side|
    side.map { |term| (["\"#{term[1]}\""] * term[0]).join(' + ') }.
    join(' + ')
    end.join(" == ")
    end
    else
    nil
    end
    end

    # ...

    This method just wraps the previous solver and transforms the resulting Arrays
    into the quiz equation format. Two different output options are controlled by
    the constant we saw at the beginning of the program.

    Here's the final piece of the puzzle:

    # ...

    if __FILE__ == $0 # if run from the command line...
    # collect words from STDIN
    words = []
    while line = gets
    words << line.chomp
    end

    result = solve_to_string(words)

    if result : puts result
    else exit 1
    end
    end

    This code just brings in the word list, taps the solver to do the hard work, and
    sends back the results. This turns the code into a complete solution.

    My thanks to all of you who know math so much better than me. I had to use math
    books and my pet math nerd just to breakdown how these solutions worked.

    Tomorrow we return to easier problems, pop quiz style...
     
    Ruby Quiz, Feb 8, 2007
    #1
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