This quiz was adapted from an ACM programming challenge at the suggestion of Gavin Kistner.

Let's define two operators for a simple set of string equations.

First, we will use the plus (+) operator and keep Ruby's meaning of concatenation for it. Therefore, "james" + "gray" is "jamesgray".

The other operator we will support is equality (==). Two strings will be considered equal if they are anagrams (they contain the same letters in a possibly different ordering). In other words, "cinema" == "iceman".

Using these two operators we can build equations:

"i" + "am" + "lord" + "voldemort" == "tom" + "marvolo" + "riddle"

This week's quiz is to write a program that accepts a list of words on STDIN and outputs any one string equation using those words, assuming there is one. Each word in the list may be used zero or more times, but only on one side of the equation. The program should exit(0) on success and exit(1), without printing any output, if a solution cannot be found.

Let's sidestep case sensitivity issues by lower casing all incoming words. You can also remove all non-alphanumeric characters.

Posting word lists and/or equations is not spoiler material.

Quiz Summary

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.:

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...