I posted a link to some additional information early in the quiz
discussion that
expanded on the definition provided by the quiz:
http://mathworld.wolfram.com/HappyNumber.html
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 Q. 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 [1], 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 F. 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|
@unhappy_numbers.add 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 M. 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…