R
Ruby Quiz
In reading through all the interesting solutions to this quiz, I noticed two
things. Some solutions were easy to follow and have clever ways to do the work.
Others were very fast, thanks to good optimizations. Let's examine one of each.
First, here is Brian Schroeder's complete solution (minus a line I removed):
#!/usr/bin/ruby
# Break early version, checking if a number is weird
def weird_number(n)
sum = 0
subset_sums = Hash.new
subset_sums[0] = true
for d in 1...n
next unless n % d == 0
# Calculate sum of all divisors
sum += d
# Calculate sums for all subsets
subset_sums.keys.each do | s |
return false if s + d == n
subset_sums[s + d] = true
end
end
sum > n
end
def weird_numbers(range)
range.select { | n | weird_number(n) }
end
# Argument parsing
raise "Input exactly one number" unless ARGV.length == 1
max = ARGV[0].to_i
# Call it
puts weird_numbers(1..max)
This code is not blazing fast, but it's easy to follow and a unique approach.
That's worth a look, I think.
All the action is in weird_number(), which just returns true or false to
indicate if the passed argument is indeed weird. It begins by initializing an
overall sum variable and a Hash for subset_sums. Notice that 0 is added to
subset_sums here. We will look at that more in a bit. (All the values of this
Hash are set to true, but they are really unused. Brian just wanted the unique
property of Hash keys.)
The method then walks from 1 to the number, looking for divisors. When a
divisor is found, it's added to the sum and then added to each previous
subset_sum (or just 0 on the first occurrence). Each time a new subset_sum is
generated, the total is checked against the number itself. This allows the code
to return an early false, when it finds a match.
If none of the subset_sums short-circuited the process, a final check is made to
ensure that the overall sum exceeds the number. When it does, a weird number is
found.
The rest of Brian's code is just argument parsing, the search through the range
of numbers, and output. This is very simple stuff.
I might suggest one change and that would be that printing the numbers as they
are found makes the wait a little less tedious, I think. That's an easy fix.
The last line of code can be switched to:
(1..max).each { |n| puts n if weird_number n }
Brian's code needs over a minute to find the weird numbers from 1 to 10,000.
That's the downside. If you want to do it faster, you have to find some
shortcuts and some submitters found great ones.
Let's switch gears to Ryan Leavengood's code, which can do the same calculation
in under a second. It starts with a simple helper method:
class Array
def sum
inject(0) do |result, i|
result + i
end
end
end
# ...
I assume the standard Ruby idiom for summation needs little introduction. Let's
move on to the heart of the algorithm:
# ...
class Integer
def weird?
# No odd numbers are weird within reasonable limits.
return false if self % 2 == 1
# A weird number is abundant but not semi-perfect.
divisors = calc_divisors
abundance = divisors.sum - 2 * self
# First make sure the number is abundant.
if abundance > 0
# Now see if the number is semi-perfect. If it is, it isn't weird.
# First thing see if the abundance is in the divisors.
if divisors.include?(abundance)
false
else
# Now see if any combination sums of divisors yields the abundance.
# We reject any divisors greater than the abundance and reverse the
# result to try and get sums close to the abundance sooner.
to_search = divisors.reject{|i| i > abundance}.reverse
sum = to_search.sum
if sum == abundance
false
elsif sum < abundance
true
else
not abundance.sum_in_subset?(to_search)
end
end
else
false
end
end
# ...
Finding out if a number is weird requires a fair amount of processing. We need
all of the divisors (a lot of work to find on big numbers), subset sums for
those divisors, etc. However, we know some things about weird numbers that can
be tested faster. If less work rules out that process some of the time, it can
add up to a big win.
First of all, there are no known odd weird numbers, so we might as well toss out
half of the set right off the bat. (It's possible there are some very large odd
weird numbers, but we would have trouble calculating those anyway.) Going a
step further, there are simple mathematical formulas to determine if a number is
abundant or semi-perfect. We can use those to quickly eliminate many numbers,
because weird numbers are always abundant and never semi-perfect. If we make it
that far, we will still have to do the work, but that allows us to skip a good
deal of numbers that would have cost us time.
# ...
def calc_divisors
res=[1]
2.upto(Math.sqrt(self).floor) do |i|
if self % i == 0
res << i
end
end
res.reverse.each do |i|
res << self / i
end
res.uniq
end
def sum_in_subset?(a)
if self < 0
false
elsif a.include?(self)
true
else
if a.length == 1
false
else
f = a.first
remaining = a[1..-1]
(self - f).sum_in_subset?(remaining) or sum_in_subset?(remaining)
end
end
end
end
# ...
There are even shortcuts to be found in the work itself. The biggest is in
calculating divisors. You can just find those between 1 and the square root of
the number, then use those to get the rest. Also, when checking divisor sums,
work with the big numbers first, to get totals closer to the actual number that
may rule it out quickly.
Another interesting option, much debated in the quiz solution thread, is the
ability to add a cache to the program. If a weird number is added to the cache
when found, future queries can be lightning quick. Here's a nice bit of code
Ryan added just to shut me up about the merits of caching (a noble goal!):
# ...
class WeirdCache
def initialize(filename='.weirdcache')
@filename = filename
if test(?e, filename)
@numbers = IO.readlines(filename).map do |i|
i.chomp.to_i
end
else
@numbers=[]
end
@added = false
end
def each(&block)
@numbers.each(&block)
end
def <<(i)
@added = true
@numbers << i
end
def save
if @added
File.open(@filename, File::RDWR|File::CREAT|File::TRUNC) do |file|
file.puts @numbers
end
end
end
end
# ...
Nothing tricky there. We just have a container for numbers with the ability to
save it out to disk. You can see that it is reloaded upon creation.
Finally, here's the code that ties all this together:
# ...
if $0 == __FILE__
if ARGV.length < 1
puts "Usage: #$0 <upper limit>"
exit(1)
end
puts "Weird numbers up to and including #{ARGV[0]}:"
start = Time.now
cache = WeirdCache.new
at_exit {cache.save}
limit = ARGV[0].to_i
i = 69
cache.each do |i|
if i <= limit
puts i
end
end
(i+1).upto(limit) do |j|
if j.weird?
cache << j
puts j
end
end
puts "This took #{Time.now - start} seconds"
end
Notice the nice use of variable scoping there. The variable i is set to 69
before the cache is walked. (Another optimization. 70 is the first weird
number, so we can safely skip anything before that.) Numbers in the cache will
replace i, leaving it holding the highest known weird number. Then, if the
limit is higher than that, the code only needs to calculate from there up.
I've barely scratched the surface of the solutions here. There were many more
gems hidden in them. I do recommend browsing the source of the others for
picking up new tricks.
My thanks to all who solved this problem, took my abuse about caching, and
especially to Ryan for building what I pestered him for.
We have two back-to-back quizzes for all you gamers out there coming up next.
First, Kalah anyone?
things. Some solutions were easy to follow and have clever ways to do the work.
Others were very fast, thanks to good optimizations. Let's examine one of each.
First, here is Brian Schroeder's complete solution (minus a line I removed):
#!/usr/bin/ruby
# Break early version, checking if a number is weird
def weird_number(n)
sum = 0
subset_sums = Hash.new
subset_sums[0] = true
for d in 1...n
next unless n % d == 0
# Calculate sum of all divisors
sum += d
# Calculate sums for all subsets
subset_sums.keys.each do | s |
return false if s + d == n
subset_sums[s + d] = true
end
end
sum > n
end
def weird_numbers(range)
range.select { | n | weird_number(n) }
end
# Argument parsing
raise "Input exactly one number" unless ARGV.length == 1
max = ARGV[0].to_i
# Call it
puts weird_numbers(1..max)
This code is not blazing fast, but it's easy to follow and a unique approach.
That's worth a look, I think.
All the action is in weird_number(), which just returns true or false to
indicate if the passed argument is indeed weird. It begins by initializing an
overall sum variable and a Hash for subset_sums. Notice that 0 is added to
subset_sums here. We will look at that more in a bit. (All the values of this
Hash are set to true, but they are really unused. Brian just wanted the unique
property of Hash keys.)
The method then walks from 1 to the number, looking for divisors. When a
divisor is found, it's added to the sum and then added to each previous
subset_sum (or just 0 on the first occurrence). Each time a new subset_sum is
generated, the total is checked against the number itself. This allows the code
to return an early false, when it finds a match.
If none of the subset_sums short-circuited the process, a final check is made to
ensure that the overall sum exceeds the number. When it does, a weird number is
found.
The rest of Brian's code is just argument parsing, the search through the range
of numbers, and output. This is very simple stuff.
I might suggest one change and that would be that printing the numbers as they
are found makes the wait a little less tedious, I think. That's an easy fix.
The last line of code can be switched to:
(1..max).each { |n| puts n if weird_number n }
Brian's code needs over a minute to find the weird numbers from 1 to 10,000.
That's the downside. If you want to do it faster, you have to find some
shortcuts and some submitters found great ones.
Let's switch gears to Ryan Leavengood's code, which can do the same calculation
in under a second. It starts with a simple helper method:
class Array
def sum
inject(0) do |result, i|
result + i
end
end
end
# ...
I assume the standard Ruby idiom for summation needs little introduction. Let's
move on to the heart of the algorithm:
# ...
class Integer
def weird?
# No odd numbers are weird within reasonable limits.
return false if self % 2 == 1
# A weird number is abundant but not semi-perfect.
divisors = calc_divisors
abundance = divisors.sum - 2 * self
# First make sure the number is abundant.
if abundance > 0
# Now see if the number is semi-perfect. If it is, it isn't weird.
# First thing see if the abundance is in the divisors.
if divisors.include?(abundance)
false
else
# Now see if any combination sums of divisors yields the abundance.
# We reject any divisors greater than the abundance and reverse the
# result to try and get sums close to the abundance sooner.
to_search = divisors.reject{|i| i > abundance}.reverse
sum = to_search.sum
if sum == abundance
false
elsif sum < abundance
true
else
not abundance.sum_in_subset?(to_search)
end
end
else
false
end
end
# ...
Finding out if a number is weird requires a fair amount of processing. We need
all of the divisors (a lot of work to find on big numbers), subset sums for
those divisors, etc. However, we know some things about weird numbers that can
be tested faster. If less work rules out that process some of the time, it can
add up to a big win.
First of all, there are no known odd weird numbers, so we might as well toss out
half of the set right off the bat. (It's possible there are some very large odd
weird numbers, but we would have trouble calculating those anyway.) Going a
step further, there are simple mathematical formulas to determine if a number is
abundant or semi-perfect. We can use those to quickly eliminate many numbers,
because weird numbers are always abundant and never semi-perfect. If we make it
that far, we will still have to do the work, but that allows us to skip a good
deal of numbers that would have cost us time.
# ...
def calc_divisors
res=[1]
2.upto(Math.sqrt(self).floor) do |i|
if self % i == 0
res << i
end
end
res.reverse.each do |i|
res << self / i
end
res.uniq
end
def sum_in_subset?(a)
if self < 0
false
elsif a.include?(self)
true
else
if a.length == 1
false
else
f = a.first
remaining = a[1..-1]
(self - f).sum_in_subset?(remaining) or sum_in_subset?(remaining)
end
end
end
end
# ...
There are even shortcuts to be found in the work itself. The biggest is in
calculating divisors. You can just find those between 1 and the square root of
the number, then use those to get the rest. Also, when checking divisor sums,
work with the big numbers first, to get totals closer to the actual number that
may rule it out quickly.
Another interesting option, much debated in the quiz solution thread, is the
ability to add a cache to the program. If a weird number is added to the cache
when found, future queries can be lightning quick. Here's a nice bit of code
Ryan added just to shut me up about the merits of caching (a noble goal!):
# ...
class WeirdCache
def initialize(filename='.weirdcache')
@filename = filename
if test(?e, filename)
@numbers = IO.readlines(filename).map do |i|
i.chomp.to_i
end
else
@numbers=[]
end
@added = false
end
def each(&block)
@numbers.each(&block)
end
def <<(i)
@added = true
@numbers << i
end
def save
if @added
File.open(@filename, File::RDWR|File::CREAT|File::TRUNC) do |file|
file.puts @numbers
end
end
end
end
# ...
Nothing tricky there. We just have a container for numbers with the ability to
save it out to disk. You can see that it is reloaded upon creation.
Finally, here's the code that ties all this together:
# ...
if $0 == __FILE__
if ARGV.length < 1
puts "Usage: #$0 <upper limit>"
exit(1)
end
puts "Weird numbers up to and including #{ARGV[0]}:"
start = Time.now
cache = WeirdCache.new
at_exit {cache.save}
limit = ARGV[0].to_i
i = 69
cache.each do |i|
if i <= limit
puts i
end
end
(i+1).upto(limit) do |j|
if j.weird?
cache << j
puts j
end
end
puts "This took #{Time.now - start} seconds"
end
Notice the nice use of variable scoping there. The variable i is set to 69
before the cache is walked. (Another optimization. 70 is the first weird
number, so we can safely skip anything before that.) Numbers in the cache will
replace i, leaving it holding the highest known weird number. Then, if the
limit is higher than that, the code only needs to calculate from there up.
I've barely scratched the surface of the solutions here. There were many more
gems hidden in them. I do recommend browsing the source of the others for
picking up new tricks.
My thanks to all who solved this problem, took my abuse about caching, and
especially to Ryan for building what I pestered him for.
We have two back-to-back quizzes for all you gamers out there coming up next.
First, Kalah anyone?