String Equations (#112)

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][i] != 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[i] > 0 : left << [values[i], words[i]]
    elsif values[i] < 0 : right << [-values[i], words[i]]
    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…

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