 # [SUMMARY] Happy Numbers (#93)

I posted a link to some additional information early in the quiz discussion that
expanded on the definition provided by the quiz:

According to that document, you can tell that a number is unhappy if the sum of
the squares of the digits ever reaches 0, 4, 16, 20, 37, 42, 58, 89, or 145.
That's really just a different way to find the repeat pattern mentioned in the
quiz:

0: 0
4: 16 37 58 89 145 42 20 4
16: 37 58 89 145 42 20 4 16
20: 4 16 37 58 89 145 42 20
37: 58 89 145 42 20 4 16 37
42: 20 4 16 37 58 89 145 42
58: 89 145 42 20 4 16 37 58
89: 145 42 20 4 16 37 58 89
145: 42 20 4 16 37 58 89 145

Here's the code I used to generate the above list, which just loops over the sum
of the squares until a repeat is found:

#!/usr/bin/env ruby -w

[0, 4, 16, 20, 37, 42, 58, 89, 145].each do |n|
print "#{n}: "

seen = {n => true}
loop do
sum = n.to_s.split("").inject(0) { |tot, d| tot + d.to_i ** 2 }

print "#{sum} "
if seen[sum]
puts
break
else
seen[sum] = true
n = sum
end
end
end

The advantage of using the list is that you don't need to wait for the pattern
to start repeating and thus you find answers quicker.

Let's examine a solution that uses these numbers and another couple of
optimizations. Here's my own code, strongly influenced by Simon Kroeger's
solution:

#!/usr/bin/env ruby -w

UNHAPPY = [0, 4, 16, 20, 37, 42, 58, 89, 145].freeze

happy = Hash.new do |found, num|
digits = num.to_s.split("").sort.map { |d| d.to_i }.
delete_if { |d| d.zero? }
happiness = digits.inject(0) { |sum, d| sum + d * d }
found[num] = if happiness == 1
true
elsif UNHAPPY.include? happiness
false
else
found[happiness]
end
end

(1..100_000).each { |n| p n if happy[n] }

This is the standard Hash memoization pattern Ruby Quiz regulars are probably
pretty familiar with by now. By creating a Hash and providing a block that can
calculate the values from the keys, we ensure that Ruby will only run the code
the first time it is needed. All other access is a simple Hash lookup and
generally quite fast since Ruby's Hash is written in C.

The Hash block is where you will find all the hard work for this solution. The
first step taken there is to convert the number into an Array of digits and you
will find two more optimizations in this conversion. First note the final call
to delete_if(). Zero squared is still zero and adding zero has no effect, so we
can safely strip those out of the digits. That can take a number like 1,000,000
down to just the digit list of , skipping a fair amount of busy work.

The second optimization in here is the call to sort(). This consolidates what
we need to store in the Hash a good deal. The numbers 123 and 321 both involve
the same calculations, so we normalize digit order and take advantage of the
ability to skip several calculations.

From there the block gets almost boring. A happiness rating is figured, which
is just the sum of the digit squares. That rating is then checked for a known
happy or unhappy value. If found, the Hash can set and return true or false.
Otherwise the answer is determined by recursing to find the happiness of the
sum.

This solution ends with a trivial iteration to print all happy numbers between
one and 100,000.

My code just checked whether or not a given number is happy. The quiz mentioned
other challenges and most people took them on. One such challenge involved
finding out just how happy a number really is. Here's the start of some
optimized code from Hans Fugal that does just that:

require 'set'

class Happy
def initialize
@happy_numbers = { 1 => 0 }
@unhappy_numbers = Set.new
end

# ...

You can see that Hans intends to track both happy and unhappy numbers. Happy
numbers will be stored in a Hash with the number as the key and the value being
the happiness rank for that number. Unhappy numbers will be a Set of numbers.

Note that you can't just use the keys for the happy numbers Hash to determine if
a number is unhappy. Not being in that list may just mean the number hasn't
been checked yet.

Here's the beginning of the method that does all the work:

# ...

def happy(x)
return true if @happy_numbers.has_key?(x)
return false if @unhappy_numbers.include?(x)

path = [x]
loop do
sum = 0
while x > 0
x, r = x.divmod(10)
sum += r**2
end

# ...

This method is used to check if a number is happy, but it squirrels away the
happiness rank for the number as it finds the answer. You can see that it
begins with checks that short-circuit the process when the result is already
known. If the result is not yet known, the code enters a loop() to figure it
out.

The path variable will eventually hold each step from the original number, to
the squares sum that is known to be happy or unhappy. It begins with just what
we currently know: the number itself.

The first bit of code in the loop() is a digit splitter and squares summation
all-in-one. It divides the digits out and adds them to a running sum as it
goes. This is quite a bit quicker than the multiple iterators used to do the
same in my code.

Once we have a sum, it's time to check it for happiness:

# ...

if @unhappy_numbers.include?(sum)
return false
elsif @happy_numbers.has_key?(sum)
r = @happy_numbers[sum]
path.each_with_index do |x,i|
@happy_numbers[x] = r + path.size - i
end
return true
end

path.push sum

# ...

If the sum is unhappy, we know all we need to know and the result is immediately
returned to the user.

If the sum is happy, we need to add all steps on the current path to the happy
numbers Hash. Their rank is the rank of the sum we found plus their distance
from the end of the current path. With that saved, true is returned to the
calling code.

If we didn't find the number in either place it is just another step on the path
and push() is called to reflect this.

Now we need the exit condition for the loop():

# ...

if [0, 1, 4, 16, 20, 37, 42, 58, 89, 145].include?(sum)
if sum == 1
s = path.size
path.each_with_index do |x,i|
@happy_numbers[x] = s - i - 1
end
return true
else
path.each do |x|
end
return false
end
end

x = sum
end
end

# ...

This final bit of code checks for the known happy and unhappy sums. If the
number is happy, we again place each step in the path in the happy numbers Hash
according to their distance from the end of the path. If the number is unhappy,
all steps in the path are added to the unhappy numbers Set.

If the code makes it through all of that with no results, the current number is
switched out for the squares sum and the code loop()s to find the answer.

The method we just digested saved the number's happiness rank, so we now need a
way to get it back out:

# ...

def rank(x)
raise ArgumentError, "#{x} is unhappy." unless happy(x)
return @happy_numbers[x]
end
end

# ...

This method first ensures the number has been ranked with a call to happy().
Once it is known to be in the Hash, it's a simple lookup to locate and return a
rank.

Here's the user interface code Hans included with the solution, which will give
the happiness rank for any numbers passed via STDIN:

# ...

haphap = Happy.new
ARGF.each_line do |l|
l.scan(/\d+/) do |token|
x = token.to_i
if haphap.happy(x)
puts "#{x} is happy with rank #{haphap.rank(x)}"
end
end
end

Be sure and walk the other solutions. Many nice examples were given for finding
happy bases. Daniel Martin even sent in a great NArray solution for that.

My thanks to all happy coders who got to play with happy numbers, allowing me to
write this happy summary.

Tomorrow you all get a chance to earn double-O status, if your code is small
enough and accurate...