[SUMMARY] Getting to 100 (#119)


Ruby Quiz

This was a popular problem and we saw many different ways to go about solving
it. The discussion on the mailing list was also very good, with many ideas
shared and explained.

I'm going to go through a few solutions today, because they are not overly long
and I think we can learn something from each of them. Let's begin with some
code by Dennis Frommknecht. Here's the entire solution:

#!/usr/bin/env ruby -W

# This solution uses 3 nested loops to divide the
# numbers into 4 groups (using regular expressions).
# Then the 3 allowed combinations of plus and minus
# are inserted between the groups.
# Finally the result is calculated using eval

NUMBERS = "123456789"
OPS = [['+', '-', '-'],
['-', '+', '-'],
['-', '-', '+']]

num_of_ops = OPS[0].length
equ_counter = 0

1.upto(NUMBERS.length - num_of_ops) do |i|
1.upto(NUMBERS.length - num_of_ops - i + 1) do |j|
1.upto(NUMBERS.length - num_of_ops - i + 1 - j + 1) do |k|
if NUMBERS.match(/(\d{#{i}})(\d{#{j}})(\d{#{k}})(\d+)/) then
OPS.each do |o|
command = "#{$1} #{o[0]} #{$2} #{o[1]} #{$3} #{o[2]} #{$4}"
res = eval command
equ_counter += 1
puts "*" * 15 if res == CORRECT_RES
puts "#{command} = #{res}"
puts "*" * 15 if res == CORRECT_RES

puts "#{equ_counter} possible equations tested"

There are really two aspects to solving the base quiz. First, you need to find
the possible permutation of the operators. There are really only three orders
they can come up in and Dennis chose to just hardcode those in the OPS constant.

The other problem is about partitioning. You need to divide the digits into
four groupings that you can insert the three operators between. Again, Dennis
pretty much hardcoded the logic for this using three nested iterators. The
interesting side of this solution though is how they work.

Basically, Dennis just counts off some number of digits from the front of the
line. Then he counts of some number of digits after that. A third count is
made for some number of digits after that, ensuring we leave at least one at the
end. This gives us three lengths, which one Regexp turns into four groups.
From there, the expression is constructed, eval()ed to get an answer, and
finally printed.

This solution does a good job of showing the process, but there are some
shortcuts to be found. For example, if you move from considering just the
ordering of the three operators to the actual positioning of the three
operators, you can drop the partitioning step altogether.

Let me use Paul's code to show you what I mean:

class String
def unique_permutations
# modified to get unique permutations from
# http://blade.nagaokaut.ac.jp/cgi-bin/scat.rb/ruby/ruby-talk/139858
# which says it was inspired by a RubyQuiz! :)
return [self] if self.length < 2
perms = Hash.new

0.upto(self.length - 1) do |n|
rest = self.split(//u) # for UTF-8 encoded strings
picked = rest.delete_at(n)
rest.join.unique_permutations.each { |x| perms[picked + x] = nil }


digits = ARGV[0]
ops = ARGV[1]
target = ARGV[2].to_i

# pad ops list with spaces to match the number of slots between the digits
ops = ops + " " * (digits.size - ops.size - 1)

# build a format string with slots between the digits
digits = digits.split("").join("%s")

operator_perms = ops.unique_permutations
operator_perms.each do |p|
# build expression by inserting the ops into the format string,
# after converting spaces to empty strings
exp = digits % p.split("").map{|x|x.chomp(" ")}
val = eval(exp)
puts "*******************" if val==target
puts exp + " = " + val.to_s
puts "*******************" if val==target
puts "%d possible equations tested" % operator_perms.size

Paul's permutation code is the first new element that jumps out at us here. It
works by recursively combining smaller and smaller permutations until it has
generated all possible combinations. The permutations are stored in Hash keys
so duplicates will automatically be eliminated.

Next we see that Paul nails pretty much all of the extra credit by importing the
digits, operators, and target from command-line parameters. This means that he
can't hardcode any logic, because he can't count on what will be passed to his

The next two lines are the interesting part. Paul counts the number of spaces
between all the digits. These are the positions in which operators need to be
placed. Paul pads the operator list to include extra spaces, so it will match
the position count. Then the digit list is transformed into a printf() style
pattern that will insert an operator, or extra space, between each number.

From there it just takes one call to the unique_permutations() method to walk
the choices. Paul's output code is very similar to the code we saw from Dennis

Going one step further, let's examine how Christian Neukirchen walked the same
position list:

(00000000..'22222222'.to_i(3)).map { |x| x.to_s(3).rjust(8, "0").
tr('012', '-+ ') }.
find_all { |x| x.count("-") == 2 and x.count("+") == 1 }.
map { |x|
t = "1" + x.split(//).zip((2..9).to_a).join.delete(" ")
[eval(t), t]
}.sort.each { |s, x|
puts "*****************" if s == 100
puts "#{x}: #{s}"
puts "*****************" if s == 100

This code does the same permutation of positions Paul's code did, using simple
counting. You see, counting from 00000000 (a pretty way to show 0 with eight
positions) to whatever number '22222222'.to_i(3) is, in base 3, will yield all
possible combinations of 0s, 1s, and 2s. Christian then culls the list for the
proper mixes of operators and transliterates those digits to operators or
spaces. The rest of the code we have seen before.

Together these solutions show quite a few options for handling permutations in
Ruby. Another common option from the submissions I didn't show was to load an
external library that handles permutations. This is a great option in practice,
because you get road-tested code, but I wanted to show how you might handle such
a thing for this summary.

My thanks to all the people who sent in permutations of Ruby code to solve this
permutations problem. They are great stuff.

Tomorrow's problem is something you can handle on your fingers...


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