# Forum: Ruby Weird Numbers (#57) Solution

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on 2005-12-04 14:28
```Here is my solution. Its not the most beautiful thing in the world, but
its effective.

-----------------------------------------

def weirdo_exhaustive(from, max)
puts "none found" and return 0 if max < 70
(from..max).each do |num|
if num % 2 == 0
list, sum= [], 0
(1...num).each do |div|
list << div and sum += div if num % div == 0
end
puts "==" + num.to_s if sum > num and has_sum(num, list) == false
end
end
end

def has_sum(num, array)
if array.is_a? Integer then return false end
sum = 0
array.each do |val| sum += val end
if sum === num then return true end
#this next line saves a BUNCH of checks.
if sum < num then return false end
array.each do |removeme|
copy = array.dup
copy.delete(removeme)
if copy.size > 1 and has_sum(num, copy) == true then return true
end
end
return false
end

----------------------------------------

It took a good number of optimizations that make it less pretty, but
much faster. The first time I ran it it took 4 hours to reach 1,000.
Its faster than that now, but its still somewhere around O(3^n).

SO... I thought of a way to speed up the algorithm a lot! So, with a
*few* more optimizations, here's the weirdo_fast algorithm.

-------------------------------------------------------

def weirdo_fast(max)
list = [ 70,836,4030,5830,7192,7912,9272,10792,17272,45356,73616, #
83312,91388,113072,243892,254012,338572,343876,388076,  #
519712,539744,555616,682592,786208,1188256,1229152,1713592, #
1901728,2081824,2189024,3963968 ]
list.each do |num|
puts num if num <= max
end
if max > list[list.size-1] then weirdo_exhaustive(list[list.size-1],
max) end
end

-----------------------------------------------------

A little faster, eh? It will start exhaustively searching if your max
is out of its range.... but I warn you, my computer has been
calculating 3963970 for the last 24 hours. And, not the range up to it,
just that number. Remember, its 3^3963970 = 62_286_091_158_062_773_000
which takes a minute to calculate. Well, with my other optimizations,

I'm interested to see if anyone else found a much better way to test
for has_sum(num, array). It seems to me like there must be a clever
mathematical trick for getting the job done. Or perhaps the only clever
way is my "fast" algorithm. But, that's not even that clever.

-Hampton Catlin.
hcatlin@gmail.com```
on 2005-12-04 15:29
```On 12/4/05, Hampton <hcatlin@gmail.com> wrote:
> Here is my solution. Its not the most beautiful thing in the world, but
> its effective.
[snip]

I could not find any fast solution..

--
Simon Strandgaard

# Weird Numbers
# Simon Strandgaard  <neoneye@gmail.com>

def divisors(value)
ary = []
(value/2).times do |i|
div = i + 1
ary << div if value % div == 0
end
ary
end

\$bits = []
32.times do |bit|
\$bits << 2 ** bit
end

def has_subset_equal_to(divs, value)
pairs = divs.zip(\$bits)
1.upto(2 ** divs.size - 1) do |i|
sum = 0
pairs.each{|div,b| sum+=div if (i&b)>0 }
return true if sum == value
end
false
end

def find_weird_numbers(range_min, range_max)
ary = []
range_min.upto(range_max) do |value|
divs = divisors(value)
sum = divs.inject(0){|a,b|a+b}
ary << [value, divs] if sum > value
end
res = []
ary.each do |value, divs|
if !has_subset_equal_to(divs, value)
puts "##{value} is a WEIRD NUMBER"
res << value
else
puts "##{value} is nothing"
end
end
p res
end

p find_weird_numbers(1, 100)```
on 2005-12-04 16:47
```On Dec 4, 2005, at 7:27 AM, Hampton wrote:

>   list.each do |num|
>     puts num if num <= max
>   end
>   if max > list[list.size-1] then weirdo_exhaustive(list[list.size-1],
> max) end
> end
>
> -----------------------------------------------------
>
> A little faster, eh?

Bingo.

I was actually thinking of screen scraping them from on of the
encyclopedias mentioned in the quiz thread, but I like this even better.

Nice job.

James Edward Gray II```
on 2005-12-04 17:03
```James,

I was starting to do that... I was building my HTTP object, and I
noticed how hard it was going to be to parse...... so I said screw it,
I'll make it easier.

And FASTER!

It would even run with 6k of ram!

-h.```
on 2005-12-04 17:32
```On 12/4/05, Simon Strandgaard <neoneye@gmail.com> wrote:

Here is my solution.
<http://tinyurl.com/9qjfg> (needs data-url support in your browser)

I basically used the definition of a "weird number" to implement the
solution.  This makes the code pretty, but also really, really slow.

---

#!/usr/bin/ruby

# Rubyquiz #57 -- Weird Numbers
# -----------------------------
# Levin Alexander <levin@grundeis.net>

module Enumerable
def sum
inject(0) {|s,elem| s+elem}
end
end

class Array
def subsets
(0...2**length).collect {|i|
values_at(*(0...length).find_all {|j| i&(1<<j)>0 })
}
end
end

class Integer
def abundant?
divisors.sum > self
end

def semiperfect?
divisors.subsets.any? { |set| set.sum == self }
end

def weird?
abundant? and not semiperfect?
end

def divisors
(1...self).select { |i| self % i == 0 }
end
end

if __FILE__ == \$0
0.upto(ARGV[0].to_i) { |i|
if i.weird?
puts i
else
warn "#{i} is not weird" if \$DEBUG
end
}
end```
on 2005-12-04 17:56
```#!ruby
=begin

The Quiz was real fun and I spent quite a lot of time on it.  Below
source is rather easy, but believe me I tried lots of different ways
(such as dynamic binary knapsacks etc.), most of them suck because they
need to many method calls.

The code takes about 30 seconds to find all Weird Numbers up to 10_000:

70, 836, 4030, 5830, 7192, 7912, 9272.

The code doesn't scale rather well, the caches would need to be
optimized for that.

=end

class Integer
# 70, 836, 4030, 5830, 7192, 7912, 9272, 10430
def weird?
!has_semiperfect? && !weird2?
end

SEMIPERFECT = {nil => 'shortcut'}

def weird2?(divisors=proper_divisors)
return true  if divisors.any? { |x| SEMIPERFECT[x] }
if brute(self, divisors)
SEMIPERFECT[self] = true
else
false
end
end

SMALL_SEMIPERFECT = [6, 20, 28]    # + [88, 104, 272]

def has_semiperfect?
SMALL_SEMIPERFECT.any? { |v| self % v == 0 }
end

def proper_divisors
d = []
sum = 0
2.upto(Math.sqrt(self)) { |i|
if self % i == 0
d << i << (self / i)
sum += i + (self / i)
end
}

return [nil]  unless sum > self
d << 1
end

def brute(max, values)
values.sort!

values.delete max / 2
max = max / 2

s = values.size
(2**s).downto(0) { |n|
sum = 0
s.times { |i| sum += values[i] * n[i] }
return true  if sum == max
}
false
end
end

if ARGV[0]
n = Integer(ARGV[0])
else
n = 10_000
end

2.step(n, 2) { |i|
p i  if i.weird?
}

__END__```
on 2005-12-04 18:13
```Here is my solution. It is pretty fast, but not quite as fast as
Rob's. For example his takes 6.797 seconds to calculate all weird
numbers up to 50000, whereas mine takes 7.375. Our solutions are quite
similar. Some of the other solutions are REALLY SLOW though. Probably
one of the biggest optimizations was using the square root of a given
number to calculate half the divisors, then use those to get the other
half. For example if 2 is a divisor of 14, then so is 14/2 = 7. For
big numbers this can be a massive speed-up.

Ryan

class Array
def sum
inject(0) do |result, i|
result + i
end
end
end

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

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
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

if \$0 == __FILE__
if ARGV.length < 1
puts "Usage: #\$0 <upper limit>"
exit(1)
end

puts "Weird numbers up to and including #{ARGV[0]}:"
70.upto(ARGV[0].to_i) do |i|
puts i if i.weird?
end
end```
on 2005-12-04 18:29
```Mine's *really* fast up to 3,963,968.... then it slows down just a

For instance, mine can do up to 50,000 in 0.0012 seconds on a 300mhz
processor.
And it can do 3 million in 0.014338 seconds.

Actually, I think I can speed up my exhausted.... {walks away to code}```
on 2005-12-04 18:34
```On 12/4/05, Hampton <hcatlin@gmail.com> wrote:
> Mine's *really* fast up to 3,963,968.... then it slows down just a
>
> For instance, mine can do up to 50,000 in 0.0012 seconds on a 300mhz
> processor.
> And it can do 3 million in 0.014338 seconds.

Except your solution is missing tons and tons of valid weird numbers.

Ryan```
on 2005-12-04 18:38
```Here's my solution. It looks pretty close to others. Not too fast, but
it gets the job done. I think after I get a chance to look at other
solutions I can write a some of this better.

class Integer
def divisors
divs = []
1.upto(Math.sqrt(self).to_i) do |i|
divs += [i ,self/i].uniq if (self%i == 0)
end
divs.sort.reverse #reverse speeds things up a bit
end

def weird?
divs = self.divisors - [self]
return false if divs.sum < self
divs.each_combination do |comb|
return false if comb.sum == self
end
return true
end
end

class Array
def each_combination
(2**self.length).times do |comb|
curr = []
self.length.times do |index|
curr << self[index] if(comb[index] == 1)
end
yield curr
end
end

def sum
inject(0) { |sum, i| sum + i }
end
end

max = (ARGV[0] || 10000).to_i

max.times do |i|
puts i if i.weird?
end

-----HornDude77```
on 2005-12-04 18:46
```On 12/4/05, Ryan Leavengood <leavengood@gmail.com> wrote:
[snip]
>         remaining = a[1..-1]
>         (self - f).sum_in_subset?(remaining) or sum_in_subset?(remaining)
>       end
>     end
>   end
> end

nice and speedy.. mine is awful and slow.```
on 2005-12-04 19:10
```On 12/4/05, James Edward Gray II <james@grayproductions.net> wrote:
> >         519712,539744,555616,682592,786208,1188256,1229152,1713592, #
> > A little faster, eh?
>
> Bingo.
>
> I was actually thinking of screen scraping them from on of the
> encyclopedias mentioned in the quiz thread, but I like this even better.
>
> Nice job.

I don't know, I think this is cheating. Plus he is missing a bunch of
weird numbers in his list.

Ryan```
on 2005-12-04 19:22
```Here is my solution (it's optimized for statistics not for speed).
I don't use things like someone tried all odd numbers up to 10**18
because doesn't change the algorithm.. I'll use that in my speed
optimized version..

#############

=begin

References:

http://en.wikipedia.org/wiki/Divisor_function
http://en.wikipedia.org/wiki/Weird_number
Eric W. Weisstein. "Weird Number." From MathWorld--A Wolfram Web
Resource. http://mathworld.wolfram.com/WeirdNumber.html
Eric W. Weisstein. "Semiperfect Number." From MathWorld--A Wolfram
Web Resource. http://mathworld.wolfram.com/SemiperfectNumber.html
Eric W. Weisstein. "Perfect Number." From MathWorld--A Wolfram Web
Resource. http://mathworld.wolfram.com/PerfectNumber.html
Eric W. Weisstein. "Divisor Function." From MathWorld--A Wolfram Web
Resource. http://mathworld.wolfram.com/DivisorFunction.html
Eric W. Weisstein. "Mersenne Prime." From MathWorld--A Wolfram Web
Resource. http://mathworld.wolfram.com/MersennePrime.html
Eric W. Weisstein. "Abundance." From MathWorld--A Wolfram Web
Resource. http://mathworld.wolfram.com/Abundance.html

=end

class Integer
\$prime_factors = Hash.new # we cache prime factors...
def prime_factors position = -1
if cached = \$prime_factors[self] # cached?
return cached # yes
end

if self == 1 # we have 1 we are done
return \$prime_factors[self]=[] # return no factors
elsif position<0 # we havn't reached factor 5 yet
if self&1 == 0 # test factor 2
return \$prime_factors[self]=[2,*(self>>1).prime_factors]
elsif self%3 == 0 # and factor 3
return \$prime_factors[self]=[3,*(self/3).prime_factors]
end
end

loop do
position+=6 # increment position by 6
if position*position > self # we have a prime number return it
return \$prime_factors[self]=[self]
elsif (quo,rem = divmod(position))   and rem.zero? # test 6n-1
return \$prime_factors[self]=[position,*quo.prime_factors
(position-6)]
elsif (quo,rem = divmod(position+2)) and rem.zero? # and 6n+1
return \$prime_factors[self]=[position+2,*quo.prime_factors
(position-6)]
end
end
end

def distinct_prime_factors # rle encode the prime factors ;)
distinct_prime_fac = Hash.new{0} # setup the hash
prime_factors.each do |prime_factor| # fill it
distinct_prime_fac[prime_factor]+=1
end
distinct_prime_fac.to_a.sort_by{|(fac,count)|fac} # and return
it as sorted array
end

def divisors # get the divisors (not needed for divisor sum)
divs = [] # store divisors here
loop do
break if n*n > self # done
if (qua,rem = divmod(n)) and rem.zero? # test for division
divs << qua # add divisors
divs << n
end
n+=1
end
divs.uniq.sort[0..-2] # we don't want self
end

def semi_perfect? deficient=false # final test
cached_abundance = abundance
return deficient if cached_abundance < 0 # deficient return the
argument
return true if cached_abundance == 0 # perfect => semi perfect too

possible_values = {0=>true} # store all possible values in a hash
divs = self.divisors # get the divisors

div_sum_left = divs.inject(0){|a,b|a+b} # get the divisor sum

pos_2 = div_sum_left - self # this is a possibility too

divs.reverse.each do |div| # for each divisor
possible_values.keys.each do |value| # and each possible value
if value+div_sum_left < self # check wether it can reach the
number with the divisors left
possible_values.delete(value) # if not delete the number
(huge speedup)
end

new_value = value+div # we create a new possible value
including the divisor

if new_value == self or new_value == pos_2 # if it is the
number it's semi perfect
return true
elsif new_value < self # if it's less than the number it
could be semi perfect
possible_values[new_value]=true # add it to the possiblities
end # if it's more than the value we can ignore it
end
div_sum_left-=div # the current divisor isn't left anymore
end
false # we found no way to compose the number using the divisors
end

def restricted_divisor_sum # uses the formular from wikipedia
distinct_prime_factors.map do |(fac,count)|
comm_sum = 1
comm_mul = 1
count.times do
comm_sum += (comm_mul*=fac)
end
comm_sum
end.inject(1){|a,b|a*b}-self
end

def perfect? # perfect numbers have the form 2**(n-1)*(2**n-1)
where n is prime (and small ;) )
return false if self&1 == 1 # it isn't known weather there are
odd perfect numbers.. but it's irrelevant for my algorithm
doubled = self * 2 # the perfect number is a triangular number
of the form (n*(n+1))/2
doubled_root = Math.sqrt(doubled).floor # if we double it and
take the floored square root we get n
return false unless doubled == doubled_root*(doubled_root+1) #
if we don't get n it isn't perfect
doubled_root_string = (doubled_root+1).to_s(2) # from the first
line we know n+1 has to be of the form 2**n
return false unless doubled_root_string.count('1')==1 # we check
that here
return false unless
(doubled_root_string.size-1).prime_factors.size == 1 # and n ha to be
a prime
return false unless self.abundance == 0 # if it passes all the
earlier test we check it using the abundance
true # and if it passes it's perfect
end

def abundance
self.restricted_divisor_sum-self
end
end

require 'benchmark'

max_num = Integer(ARGV.shift||'1000') rescue 1000 # read a number
from the command line

new_semi_perfects = [] # store semi perfect numbers that can't be
constructed using other semi perfect numbers

STDOUT.sync = true
puts "Weird Numbers Up To #{max_num}:"

#begin_stat
perfect_count = 0
composed_semi_perfect_count = 0
new_semi_perfect_count = 0
weird_count = 0
deficient_count = 0

record_nums = (1..80).map{|n|max_num*n/80}

record_nums_left = record_nums.dup
next_record = record_nums_left.shift

recorded_times = []

init_time = [Benchmark.times,Time.now]

min_odd = 10**17

#end_stat

(1..max_num).each do |test_num| # counting loop

if test_num == next_record #stat
recorded_times << [Benchmark.times,Time.now] #stat
next_record = record_nums_left.shift #stat
end #stat

if test_num.perfect? # it's perfect
new_semi_perfects << test_num
perfect_count += 1 #stat
next
end

do_next = false
new_semi_perfects.each do |semi_per|
if test_num % semi_per == 0 # is it possible to compose the
current number using a semi-perfect number?
do_next = true # yes
composed_semi_perfect_count += 1 #stat
break
end
end
next if do_next
# no

case test_num.semi_perfect? nil # we don't care about deficient
numbers
when true # but we care about semi perfects
new_semi_perfects << test_num
new_semi_perfect_count += 1 #stat
when false # and even more about abundand non semi perfects
puts test_num
weird_count += 1 #stat
else #stat
deficient_count += 1 #stat
end

end

#end

#begin_stat

final_time = [Benchmark.times,Time.now]

digit_length = max_num.to_s.size

def form_float(num)
"%12s" % ("%im%#{7}ss" % [(num/60).floor,("%.4f" % (num %60))]).tr
(" ","0")
end

def rel(x,y)
"%#{y.to_s.size}i (%6s%%)" % [x,"%3.2f" % (x/y.to_f*100)]
end

puts "Stats"
puts "Time:"
puts "- User:   "+form_float(ut = final_time.first.utime -
init_time.first.utime)
puts "- System: "+form_float(st = final_time.first.stime -
init_time.first.stime)
puts "- U+S:    "+form_float(ut+st)
puts "- Real:   "+form_float(final_time.last - init_time.last)
puts
puts "Numbers:"
puts "- Total        "+rel(max_num,max_num)
puts "- Weird        "+rel(weird_count,max_num)
puts "- Perfect      "+rel(perfect_count,max_num)
puts "- Deficient    "+rel(deficient_count,max_num)
puts "- Abundand     "+rel(abundand_count = max_num-perfect_count-
deficient_count,max_num)
puts "- Semi-Perfect "+rel(abundand_count-weird_count,max_num)
puts "   (w/o perfects)"
puts ""
puts "- Passed 1st   "+rel(max_num-perfect_count,max_num)
puts "   (perfect test)"
puts "- Passed 2nd   "+rel(new_semi_perfect_count+weird_count
+deficient_count,max_num)
puts "   (composed test)"
puts "- Passed 3rd   "+rel(new_semi_perfect_count+weird_count,max_num)
puts "   (deficient test)"
puts "- Uncomposed   "+rel(new_semi_perfects.size,max_num)
puts "   (semi-perfects that arn't a multiply of another semi-perfect)"
puts

if recorded_times.size >= 80
puts "Graphs:"
puts

process_plot = []
real_plot = []

first_ustime = init_time.first.utime+init_time.first.stime
first_realtime = init_time.last

recorded_times.each do |(process_time,realtime)|
process_plot << process_time.utime+process_time.stime -
first_ustime
real_plot << realtime - first_realtime
end

max_process = process_plot.last
step_process = max_process / 22

max_real = real_plot.last
step_real = max_real / 22

def plot(plot_data,max,step)
22.downto(0) do |k|
res = ""
res = form_float(max) if k == 22
while res.size != plot_data.size
val = plot_data[res.size]
lower_range = k*step
res << ( val >= lower_range ? (val < lower_range+step ?
'#' : ':') : ' ' )
end
puts res
end
end

puts "Y: Realtime X: Number"
plot(real_plot,max_real,step_real)
puts "%-40s%40s"% [1,max_num]
puts "Y: System+Usertime X: Number"
plot(process_plot,max_process,step_process)
puts "%-40s%40s"% [1,max_num]
puts
end
#end_stat

__END__```
on 2005-12-04 20:19
```I've got an optimization I haven't seen anyone else use yet.  My
solution is roughly 20% faster than Ryan's.

The trick is that any number of the form k*(2^m)*p where m > 1 and p
is a prime such that 2 < p < 2^(m+1) is semiperfect (according to
wikipedia).  This rules out many numbers, and its cheaper than a
thorough search of subsets.

#!/usr/bin/ruby

def weird(max)
primes = sieve(max*2)
70.step(max,2){|n|
puts n if weird?(n,primes)
}
end

def weird?(n,primes)
divs = divisors(n)
abund = divs.inject(0){|a,b| a+b} - n
return false if abund <= 0
return false if spfilter(n,primes)
return false if divs.include? abund
smalldivs = divs.reverse.select{|i| i < abund}
not sum_in_subset?(smalldivs,abund)
end

def sum_in_subset?(lst,n)
#p [lst,n]
return false if n < 0
return true if lst.include? n
return false if lst.size == 1
first = lst.first
rest = lst[1..-1]
sum_in_subset?(rest, n-first) or sum_in_subset?(rest,n)
end

def divisors(n)
result = []
sr = Math.sqrt(n).to_i
(2 .. sr).each {|d|
if n.modulo(d) == 0
result << d
end
}
return [1] if result.empty?
hidivs = result.map {|d| n / d }.reverse
if hidivs[0] == result[-1]
[1] + result + hidivs[1..-1]
else
[1] + result + hidivs
end
end

def spfilter(n,primes)
m = 0
save_n = n
while n[0]==0
m += 1
n >>= 1
end
return false if m == 0
low = 2
high = 1 << (m+1)
primes.each {|p|
return false if p > high
if p > low
return true if n%p == 0
end
}
raise "not enough primes while checking #{save_n}"
end

# Sieve of Eratosthenes
def sieve(max_prime)
candidates = Array.new(max_prime,true)
candidates[0] = candidates[1] = false
2.upto(Math.sqrt(max_prime)) {|i|
if candidates[i]
(i+i).step(max_prime,i) {|j| candidates[j] = nil}
end
}
result = []
candidates.each_with_index {|prime, i| result << i if prime }
result
end

weird(ARGV[0].to_i)

regards,
Ed```
on 2005-12-04 20:40
```On Dec 4, 2005, at 11:29 AM, Ryan Leavengood wrote:

> I don't know, I think this is cheating.

It's cheating to get correct answers quickly?  ;)

> Plus he is missing a bunch of weird numbers in his list.

I haven't compared the Array in question with the posted sequences.
Perhaps it's not very complete.  Let me ask you this though, since
list or a broken calculation solution?

Didn't the solution also brute force answers when it left its list of
known numbers?

One last question:  How well does your presumably fast solution do
when it goes beyond the listed sequence?  Does it still find lots of

I'm really not trying to be mean here and I apologize if I'm coming
off that way.

I think what Hampton hit on is a simple cache optimization.  It's
fast and very effective.  I don't think we should be so quick to toss
it out as a viable approach...

James Edward Gray II```
on 2005-12-04 21:01
```#!/usr/bin/env ruby
#
# Ruby Quiz Weird Numbers solution.
#
# I found only two significant optimizations:
# 1. Use continuations when generating the powerset of
#    divisors to sum. Evaluate each sum immediately
#    and break upon determining the number is semiperfect.
# 2. Sort the divisors in descending order so that
#    subsets involving the larger divisors are considered
#    first.
#
# On my machine, generating results for numbers up to 5000
# took less than a minute. Generating results up to 12000
# took almost twenty minutes.
#
# This powerset implementation was inspired by a post to
# Ruby-Talk by Robert Klemme on May 14, 2004:
#
#
#

module Enumerable

def powerset
for element_map in 0...(1 << self.length) do
subset = []
each_with_index do |element, index|
subset << element if element_map[index] == 1
end
yield subset
end
end

end

class Array

def sum
return self.inject { |total,x| total += x }
end

end

class Integer

def weird?
divs = self.divisors.reverse
return false unless divs.sum > self
divs.powerset { |subset| return false if subset.sum == self }
return true
end

def divisors
list = []
(1..Math.sqrt(self).to_i).each do |x|
if (self / x) * x == self
list << x
list << (self / x) unless x == 1 or (self / x) == x
end
end
return list.sort
end

end

####################

(1..5000).each { |n| puts n if n.weird? }```
on 2005-12-04 21:41
```On 12/4/05, James Edward Gray II <james@grayproductions.net> wrote:
>
> It's cheating to get correct answers quickly?  ;)

Well something had to generate those answers in the first place, and
for that I'd take my, Rob, or Edward's solution any day.

> I haven't compared the Array in question with the posted sequences.
> Perhaps it's not very complete.  Let me ask you this though, since
> list or a broken calculation solution?

Well I certainly debugged mine by looking at the pre-existing list,
which was useful. But the fact that his list was missing things really
caused me pause, because if someone depended on that list to be
accurate, they would be in trouble.

> Didn't the solution also brute force answers when it left its list of
> known numbers?

Yes but his brute-force algorithm is extremely slow, and his list is
wrong.

> One last question:  How well does your presumably fast solution do
> when it goes beyond the listed sequence?  Does it still find lots of

Of the three fastest solutions so far (Edward's, Rob's and mine), all
continue to generate answers at a similar rate, consistently. For
example, to generate up to 100000, the time was 16.687, 17.281 and
18.469 seconds (in the order listed above, i.e. mine is slowest.)
Since there are 204 weird numbers between 0 and 100000, each algorithm

> I'm really not trying to be mean here and I apologize if I'm coming
> off that way.

Well I've probably come off as mean to Hampton, and I don't really
want to come off that way either. But it does rub me the wrong way to
have someone call their solution fast when the solution is WRONG.

> I think what Hampton hit on is a simple cache optimization.  It's
> fast and very effective.  I don't think we should be so quick to toss
> it out as a viable approach...

His caching optimization is certainly a good idea (and fairly
obvious), though it would be better if it actually generated the cache
and reused it so that subsequent runs got faster. I may code up
something like this.

Ryan```
on 2005-12-04 21:45
```Which ones are missing?

My set is....  [
70,836,4030,5830,7192,7912,9272,10792,17272,45356,73616, #
83312,91388,113072,243892,254012,338572,343876,388076,  #
519712,539744,555616,682592,786208,1188256,1229152,1713592, #
1901728,2081824,2189024,3963968 ]```
on 2005-12-04 21:53
```I did mine with a good dose of humor.

I certainly know that my weirdo_exhaustive is slow... hell, I called it
both "weirdo" and "exhaustive". However, I thought of the fast solution
as a bit of humor for everyone to enjoy. I do think it has merit in
*some* ways only because its damn hard to make anything go faster than
that.

I did not mean any previous comments to be "i'm better than you" in any
real way. They were all made tounge in cheek because of course just
accessing an array is going to be orders of magnatude faster.

These quizes are not for points and there is no "winner" (at least not
that i've seen assigned in previous iterations of the quiz. In fact,
there is never any qualifications for what makes the best one!

And yes, I did use the wrong number set. I found the number set for
primitive weird numbers, not exhaustive.

I used this:
http://www.research.att.com/cgi-bin/access.cgi/as/...
As opposed to this:
http://www.research.att.com/cgi-bin/access.cgi/as/...

My mistake....

But, lets all be friends! How about a beer?

-hampton c.```
on 2005-12-04 21:57
```On Dec 4, 2005, at 2:40 PM, Ryan Leavengood wrote:

> On 12/4/05, James Edward Gray II <james@grayproductions.net> wrote:
>>
>> It's cheating to get correct answers quickly?  ;)
>
> Well something had to generate those answers in the first place, and
> for that I'd take my, Rob, or Edward's solution any day.

[snip]

> His caching optimization is certainly a good idea (and fairly
> obvious), though it would be better if it actually generated the cache
> and reused it so that subsequent runs got faster. I may code up
> something like this.

NOW you're talking!  You could just calculate them once and that's
the best of both worlds...  ;)

James Edward Gray II```
on 2005-12-04 22:01
```Ryan-

It was supposed to be a funny statement. I even used a ;)

And of course calling my "exhaustive" (hint in the name) a *tad* slower
is absolutely, 100% absurdist.

I used the wrong AO list as mentioned in my other response. I apologize
to all those who my not paying attention to "primitive weird numbers"
and "weird numbers" hurt in the process.

This quiz is for *fun* not competition.

-hampton.

PS: For further clarification, about 50% of this message is also
supposed to be somewhat amusing.```
on 2005-12-04 22:01
```On Dec 4, 2005, at 2:52 PM, Hampton wrote:

> However, I thought of the fast solution
> as a bit of humor for everyone to enjoy. I do think it has merit in
> *some* ways only because its damn hard to make anything go faster than
> that.

It is a valid technique.  It can probably be used to make the fastest
posted solutions even faster.  You're example maybe wasn't the ideal
representation, but I think Ryan hit on a great one in his last message.

That's all I was saying.  Don't overlook what can help!  ;)

> But, lets all be friends! How about a beer?

<laughs>  Seriously, I didn't mean to upset a single soul.  My truest
apologies if I did.

James Edward Gray II```
on 2005-12-04 22:01
```Just up to 100,000 you are missing 191 weird numbers. Run my, Rob or
Edward's solution to see exactly which ones.

Up to 3963968, I don't even want to guess how many are missing. Even
Edward's solution would take some time to find out. They aren't quite
as rare as you think.

Ryan```
on 2005-12-04 22:10
```I realized it was a list of primitive weird numbers instead of weird
numbers.

Down below I have links to the two lists of numbers.

-hampton.```
on 2005-12-04 22:14
```On 12/4/05, Hampton <hcatlin@gmail.com> wrote:
> I did mine with a good dose of humor.

Hehehe, I could see that a bit.

> I certainly know that my weirdo_exhaustive is slow... hell, I called it
> both "weirdo" and "exhaustive". However, I thought of the fast solution
> as a bit of humor for everyone to enjoy. I do think it has merit in
> *some* ways only because its damn hard to make anything go faster than
> that.

Well there is no doubt that in a real application there would be a
cache for weird numbers, should that application need them frequently.

> I did not mean any previous comments to be "i'm better than you" in any
> real way. They were all made tounge in cheek because of course just
> accessing an array is going to be orders of magnatude faster.

Fair enough, and I would have appreciated the solution more had it been
correct.

> These quizes are not for points and there is no "winner" (at least not
> that i've seen assigned in previous iterations of the quiz. In fact,
> there is never any qualifications for what makes the best one!

Of course, but quizzes like this benefit from analyzing which
algorithms are the fastest. This one in particular had a HUGE margin
between the fast and slow solutions. Other quizzes are more about the
structure of the code and solving the problem elegantly, and you won't
see me or anyone else getting out the benchmarking tools.

> And yes, I did use the wrong number set. I found the number set for
> primitive weird numbers, not exhaustive.
>
> I used this:
> http://www.research.att.com/cgi-bin/access.cgi/as/...
> As opposed to this:
> http://www.research.att.com/cgi-bin/access.cgi/as/...
>
> My mistake....

I see, easy enough mistake to make I suppose.

> But, lets all be friends! How about a beer?

Don't take my criticism as being hostile. In fact if you read this
list you'll find me most helpful when newbies have problems. But I
don't hesitate to tell people when they are wrong. A beer would be
fine if you happen to live where I am, in South Florida ;)

Ryan```
on 2005-12-04 22:30
```I'm from Jacksonville originally, but now live in the cold tundra of
Toronto.

However, I will be there for christmas!

Yes, it was hostile. And I will beat up both your *and* your mom for
what you said about my misfunctioning algorithm! ;)

-hampton.

PS: However, I will say that on the next one, I'm going to be debugging
on 2005-12-04 22:51
```On 12/4/05, James Edward Gray II <james@grayproductions.net> wrote:
>
> NOW you're talking!  You could just calculate them once and that's
> the best of both worlds...  ;)

Well I coded this up. Instead of pasting my whole solution I'll just
paste the cache class and the updated "main" block for my solution.
I've also added some timing code so one can see the speed
improvements. To try this out, replace my if \$0 == __FILE__ code from
my original solution with all of the following:

class WeirdCache
def initialize(filename='.weirdcache')
@filename = filename
if test(?e, filename)
i.chomp.to_i
end
else
@numbers=[]
end
end

def each(&block)
@numbers.each(&block)
end

def <<(i)
@numbers << i
end

def save
File.open(@filename, File::RDWR|File::CREAT|File::TRUNC) do |file|
file.puts @numbers
end
end
end
end

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```
on 2005-12-04 22:59
```> His caching optimization is certainly a good idea (and fairly
> obvious), though it would be better if it actually generated the cache
> and reused it so that subsequent runs got faster. I may code up
> something like this.

Why not take the Seti-At-Home approach and parcel out ranges of
numbers that can be checkout by home users with spare cycles on their
machines, searched for WeirdNumbers and reported back to the mothership.```
on 2005-12-04 23:28
```On Dec 4, 2005, at 3:47 PM, Ryan Leavengood wrote:

> On 12/4/05, James Edward Gray II <james@grayproductions.net> wrote:
>>
>> NOW you're talking!  You could just calculate them once and that's
>> the best of both worlds...  ;)
>
> Well I coded this up.

Just so others see the effect of this, here are some timings from
runs.  The first is Ryan's original code, the second is the new
version's first (cache-less) run, and the third is the code taking

Neo:~/Documents/Ruby/Ruby Quiz\$ time ruby solutions/Ryan\ Leavengood/
weird_numbers.rb 50000
Weird numbers up to and including 50000:
...

real    0m6.258s
user    0m6.170s
sys     0m0.046s
Neo:~/Documents/Ruby/Ruby Quiz\$ time ruby solutions/Ryan\ Leavengood/
weird_numbers.rb 50000
Weird numbers up to and including 50000:
...
This took 6.231214 seconds

real    0m6.253s
user    0m6.169s
sys     0m0.044s
Neo:~/Documents/Ruby/Ruby Quiz\$ time ruby solutions/Ryan\ Leavengood/
weird_numbers.rb 50000
Weird numbers up to and including 50000:
...
This took 0.096576 seconds

real    0m0.116s
user    0m0.102s
sys     0m0.011s

James Edward Gray II```
on 2005-12-04 23:50
```This was really a good and enjoyable exercise. I learnt a lot. One of
the best solutions is made by Leavengood. I would like to point out a
small error. 36.calc_divisors returns [1,2,3,4,6,6,9,12,18,36]. 6 is
reckoned twice. This seems to have no implication on the result though.

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
end

The part I appreciate most is the beautiful and fast handling of the
binary set:

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

I also learnt that Ruby is really quick once you leave the debugger!

Here is my solution. It is short but slow.

def divisors(n) (1..n.div(2)).collect {|i| i if n.modulo(i)==0}.compact
end
def sum(arr) arr.inject(0) {|sum,element| sum+element} end

def subset?(n, divisors)
arr=[]
divisors.each do |i|
arr.concat arr.collect {|j| i+j}
arr << i
arr.uniq!
return true if arr.member?(n)
end
false
end

def weird(n)
return if n.modulo(2)==1
coll = divisors(n)
diff = sum(coll)-n
return if diff <= 0
return n unless subset?(diff,coll)
end

def all_weird(n) (1..n).collect {|i| weird(i)}.compact end

assert_equal [1,2,3,4,6], divisors(12)
assert_equal [1,2,5,7,10,14,35], divisors(70)
assert_equal 16, sum(divisors(12))
assert_equal 74, sum(divisors(70))
assert_equal true, subset?(12,divisors(12))
assert_equal false, subset?(70,divisors(70))
assert_equal false, subset?(4,divisors(70))
assert_equal nil, weird(2)
assert_equal nil, weird(12)
assert_equal nil, weird(20)
assert_equal 70, weird(70)
assert_equal [70], all_weird(70)
assert_equal [70,836], all_weird(836)

Christer```
on 2005-12-05 00:25
```On 12/4/05, Christer Nilsson <janchrister.nilsson@gmail.com> wrote:
> This was really a good and enjoyable exercise. I learnt a lot. One of
> the best solutions is made by Leavengood. I would like to point out a
> small error. 36.calc_divisors returns [1,2,3,4,6,6,9,12,18,36]. 6 is
> reckoned twice. This seems to have no implication on the result though.

Thank you for the compliment and for catching the error. It seems
calc_divisors will always have two divisors for the square root of any
perfect square. I suppose it was just luck that this didn't mess up
the algorithm. But there might be some number where this could cause a
problem, so res.uniq should probably be called at the end.

Ryan```
on 2005-12-05 01:21
```Christian Neukirchen <chneukirchen@gmail.com> wrote:
> #!ruby
> =begin
>
> The Quiz was real fun and I spent quite a lot of time on it.  Below
> source is rather easy, but believe me I tried lots of different ways
> (such as dynamic binary knapsacks etc.), most of them suck because they
> need to many method calls.

Same here - I tried a bunch of stuff, but the code ended up being rather
slow, so I discarded it. It's still rather slow :(

N = ARGV[0].to_i

# precalculate the list of primes

def primes_to(n)
sieve = (0..n).to_a
2.upto(n) {|i|
next unless sieve[i]
(i*i).step(n, i) {|j| sieve[j] = nil}
}
sieve[2..-1].compact
end

PRIMES = primes_to(N)

# helper method
class Array
def bsearch(n)
i = 0
j = size - 1
k = (i+j)/2
while i < k
if at(k) > n
j = k
elsif at(k) < n
i = k
else
return k
end
k = (i+j)/2
end
return i
end
end

# factorisation routines - find the prime factors, then combine them to
get a
# list of all factors

def prime_factors(x)
pf = Hash.new {|h, k| h[k] = 0}
PRIMES.each {|p|
break if p > x
while x % p == 0
pf[p] += 1
x /= p
end
}
pf
end

def expand_factors(f, pf)
return f if pf.empty?
p, n = pf.shift
powers = [p]
(n-1).times { powers << p * powers[-1] }
g = f.dup
powers.each {|i| f.each {|j| g << i*j } }
expand_factors(g, pf)
end

def factors(n)
a = expand_factors([1], prime_factors(n)).sort
a.pop
a
end

# and finally, the weirdness test

def weird?(n)
fact = factors(n)
#
# test for abundance (sum(factors(n)) > n)
sum = fact.inject {|a, i| a+i}
return false if sum < n # weird numbers are abundant

# now the hard part
partials = [0]

fact.each {|f|
if sum < n
# discard those partials that are lower than the sum of all
remaining
# factors
i = partials.bsearch(n-sum)
return false if partials[i] == (n-sum)
partials = partials[(i+1)..-1]
end

sum -= f # sum of all remaining factors
temp = []

partials.each {|p|
j = f + p
break if j > n
l = n - j
next if l > sum
return false if (j == n) or (l == sum)
temp << j
}

# handwriting a merge sort didn't help :-/
partials = partials.concat(temp).sort.uniq
}

return true
end

def all_weird(n)
weird = []
# odd numbers are not weird (unproven but true for all n < 10^17)
2.step(n, 2) {|i| weird << i if weird?(i) }
weird
end

require 'benchmark'

Benchmark.bm(10) {|x|
[1000,10000,20000].each {|n|
x.report("#{n}") {p all_weird(n)}
}
}

martin```
on 2005-12-05 01:30
```On 12/4/05, Christer Nilsson <janchrister.nilsson@gmail.com> wrote:

>         false
>       else
>         f = a.first
>         remaining = a[1..-1]
>         (self - f).sum_in_subset?(remaining) or
>                    sum_in_subset?(remaining)
>       end
>     end
>   end

Yes, this is quite nice.  I rewrote it a bit so that it does not need
the nested ifs.  What do you think?

def sum_in_subset?(div = self.divisors)
return false if self < 0
return false if div.empty?
return true if div.include?(self)

f, remaining = div[0], div[1..-1]
(self - f).sum_in_subset?(remaining) or sum_in_subset?(remaining)
end

Viele GrÃ¼Ã?e,
Levin```
on 2005-12-05 01:57
```On 12/4/05, Levin Alexander <levin@grundeis.net> wrote:
>     (self - f).sum_in_subset?(remaining) or sum_in_subset?(remaining)
>   end

I don't really like multiple returns, which is why I wrote it like
that. But the above is still pretty nice, and certainly more succinct.
Here you can also see how similar this is to Rob Leslie's method of
finding the sums of the subsets. He posted before me though and
deserves credit as well for a nice and fast solution. His solution is

Ryan```
on 2005-12-05 03:30
```I have a solution for the "effectively useless" basket. It's a naive
algorithm, and I haven't yet received a number above 70 from it.

The advantages are that it's straightforward idiomatic Ruby, and core of
the
algorithm is expressed about as tersely as Martin's definition. And I
like
my ArraySubsetList class.

class Fixnum
def divisors
(1..self).select {|i| self % i == 0 }
end
end
module Enumerable
def sum
inject(0) {|m, o| m + o }
end
end
class Array
def subsets
ArraySubsetList.new(self)
end
end
class ArraySubsetList
def initialize(array)
@array = array
end
def [](index)
return nil unless (0...size) === index
ret = []
@array.size.times {|i| ret << @array[i] if index[i] == 1 }
ret
end
def each
size.times {|bits| yield self[bits] }
end
include Enumerable
def size
1 << @array.size
end
alias length size
end
def wierd_numbers_up_to(max)
ret = []
for n in 1..max
# A weird number is defined as a number, n, such that the sum of
all
its divisors
# (excluding n itself) is greater than n, but no subset of its
divisors sums up to
# exactly n.
divs = n.divisors
divs.delete i
if divs.sum > i &&
divs.subsets.select {|subset| subset.sum == i }.empty?
ret << i
yield i if block_given?
end
end
ret
end
if \$0 == __FILE__
if ARGV.size == 1 && (ARGV[0].to_i rescue 0) > 0
wierd_numbers_up_to(ARGV[0].to_i) {|n| puts n }
else
puts "usage: #\$0 n\n  Find all weird numbers less than n"
end
end

Cheers,
Dave```
on 2005-12-05 05:29
```Wow, I got a lot from looking at the other solutions. Thanks especially
to Ed, Ryan, Jannis and Christian. I see that the single biggest
performance gain comes from using a recursive sum_in_subset? method such
as that found in Ed's solution. I've borrowed from his code for this
fairly simple and straightforward resubmission. It doesn't perform as
well as the faster solutions, but it's much, much faster than my first
submission.

#!/usr/bin/env ruby
#
# Ruby Quiz Weird Numbers solution.
# Uses recursive sum_in_subset? approach borrowed from
# other solutions.
#

class Integer

def weird?
divisors = self.divisor_list
abundancy = divisors.inject { |total,x| total += x } - self
return false unless abundancy > 0
smalldivisors = divisors.reverse.select { |j| j <= abundancy }
return false if sum_in_subset?(smalldivisors, abundancy)
return true
end

def sum_in_subset?(list, target)
return false if target < 0
return true if list.include?(target)
return false if list.length == 1
first = list.first
rest = list[1..-1]
sum_in_subset?(rest, target-first) or sum_in_subset?(rest, target)
end

def divisor_list
list = []
(1..Math.sqrt(self).to_i).each do |x|
if self % x == 0
list << x
list << (self / x) unless x == 1 or (self / x) == x
end
end
return list.sort
end

end

####################

unless ARGV.length == 1
puts "Usage: #{\$0} <max_value>"
end
max_value = ARGV.shift.to_i
(1..max_value).each { |n| puts n if n.weird? }```
on 2005-12-05 07:18
```Here's mine.
I started out with very straightforward programming, but ended up
making it muddier as I optimized.  I also implemented a bunch of
functions that I figured I could probably find alread written if I
looked, but I figured they were good exercises :  Integer#factors,
Integer#divisors, Array#all_combinations....

I thought it was pretty fast, especially compared to my initial
implementation.  Then I timed Ryan's on my machine...  Oh well.  I
enjoyed the challenge.

---
class Integer
def weird?
(d = divisors).pop   #remove self (which is always last)
d.sum > self &&
!d.quick_find_subset_with_sum(self) &&  #weed out most
!d.find_subset_with_sum(self)           #confirm the rest
end

def divisors
factors.all_combinations.uniq.inject([]){|result,combo|
result << combo.product
}.sort.uniq
end

def factors
value, candidate = self, 3
factors = [1]
while value % 2 == 0
factors << 2
value /= 2
end
while candidate <= Math.sqrt(value)
while value % candidate == 0
factors << candidate
value /= candidate
end
candidate += 2
end
factors << value if value != 1
factors
end
end

class Array
def product
inject(1){|p,v| p*v}
end
def sum
inject(0){|s,v| s+v}
end
def all_combinations
ComboIndexGenerator.new(self.size).inject([]) {|result, index_set|
result << values_at(*index_set)
}
end

#this was my first attempt, which was straightforward,
# but slow as heck for large sets
def slow_find_subset_with_sum n
return nil if sum < n
all_combinations.each {|set|
return set if set.sum == n
}
nil
end

#this is my second attempt which is fast but misses some subsets.
#but it is useful for quickly rejecting many non-weird numbers.
def quick_find_subset_with_sum n
a = self.sort.reverse
sum,set = 0,[]
a.each {|e|
if (sum+e <= n)
sum+=e
set<<e
end
return set if sum == n
}
nil
end

#this one works pretty quickly...
#it never tests subsets which are less than the sum,
#and keeps track of sets it has already calculated
def find_subset_with_sum n
possibilities, seen  = [self],{}
until possibilities.empty?
candidate = possibilities.pop
diff = candidate.sum - n
return candidate if diff == 0
break if diff < 0
candidate.each_with_index{|e,i|
break if e > diff
new_cand = (candidate.dup)
new_cand.delete_at(i)
return new_cand if e == diff
possibilities << new_cand if !seen[new_cand]
seen[new_cand]=true
}
end
nil
end
end

#this class generates an all the possible combinations of n items
#it returns an array with the next combination every time you call #next
class ComboIndexGenerator
include Enumerable
def initialize nitems
@n = nitems
@max = 2**@n
@val=0
end
def to_a
return nil if @val==@max
(0..@n).inject([]){|a,bit| a<<bit if @val[bit]==1; a}
end
def next
@val+=1 if @val<@max
to_a
end
def each &b
yield to_a
while (n=self.next)
yield n
end
end
end

if \$0 == __FILE__

if ARGV.length < 1
puts "Usage: #\$0 <upper limit>"
exit(1)
end

puts "Weird numbers up to and including #{ARGV[0]}:"
70.upto(ARGV[0].to_i) do |i|
puts i if i.weird?
end
end```
on 2005-12-05 11:57
```Here is my solution, I don't think it is as fast as the others, but it
does never calculate a list of divisors. It scales quite badly

max   time
----------------
1000  0m0.714s
2000  0m2.756s
3000  0m6.351s
4000  0m13.404s
5000  0m16.806s
6000  0m27.031s
7000  0m33.482s
8000  0m44.111s
9000  0m54.781s
10000 1m6.179s

bschroed@black:~/svn/projekte/weird-numbers\$ cat weird-numbers-be.rb
#!/usr/bin/ruby

# Break early version, checking if a number is weird
def weird_number(n)
d  = r = s = nil
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)

cheers,

Brian```
on 2005-12-08 09:48
```I was impressed when I started going through 1..1000 in under 15
seconds, so this is much slower than others report. I'm just going to
tell myself you all have much faster computers, and go to bed.

require 'set'

class Subsets
def initialize(set, start)
@set = set.to_a.uniq.sort
@num_elements = start - 1
@map = {}
@set.each_with_index {|k, v| @map[k] = v+1}
end

# returns each subset, in turn. Returns nil when there are no more
def succ
if @combo == nil or @combo == @set[-@num_elements..-1]
return nil if (@num_elements +=1) > @set.length
@combo = @set[0,@num_elements]
else
index = (1..@num_elements).find {|i| @combo[-i] < @set[-i]}
@combo[-index, index] = @set[@map[@combo[-index]], index]
end
@combo
end

def find
while(x = succ)
break if yield x
end
x
end
end

class Integer

def proper_divisors
return [] if self < 2
div = Set.new [1]
2.upto(Math.sqrt(Float.induced_from(self)).to_i) {|i|
quotient, modulus = self.divmod(i)
div.merge([i,quotient]) if modulus.zero?
}
div.to_a.sort
end

def abundant?
self > 11 and [0].concat(proper_divisors).inject {|sum,n| sum += n}
> self
end

def semiperfect?
return nil if self < 6
subsets = Subsets.new(proper_divisors, 2)
subsets.find {|subset| [0].concat(subset).inject {|sum,n| sum += n}
== self }
end

def weird?
self > 69 and abundant? and not semiperfect?
end
end

n = gets.strip
exit if n =~ /\D/ or n !~ /[^0]/
p (1..n.to_i).find_all {|i| i.weird? }```
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