[SOLUTION] Splitting the Loot (#65) (My second attempt)

I had a small chat with Manuel K., who found out that my version
can’t split all loots correctly.
For example, my old version didn’t split

ruby loot.rb 6 34 78 21 70 45 67 70 19 90 76 54 20 30 19 80 7 65 43 56
46

But it’s possible to split this loot to:
1: 7 78 80
2: 30 45 90
3: 19 70 76
4: 46 54 65
5: 21 34 43 67
6: 19 20 56 70

Here’s my second attempt. It should now split all loots correctly.
It now remembers splits that didn’t result in a complete solution and
trys again.
I also added some comments, so it’s easier to understand the code I use.

And because I had problems with Thunderbird wrapping the lines wrong I
attached my program to the email, too. Just in case

class Array
def sum
inject { |s,x| s + x }
end
def delete_one! n
(i = index(n)) ? delete_at(i) : nil
end
def count n
inject(0) { |c,x| x == n ? c+1 : c }
end
end

class Knapsack
def initialize target, numbers, avoid=Array.new
@target,@numbers,@avoid = target, numbers,avoid.compact.uniq
end
def solve
solver @numbers.map { |n| [n] }
end
def solver paths
new_paths = Array.new # New paths will be stored here
paths.uniq.each do |path| # For each path do
return path if path.sum == @target && (!@avoid.include?path) # If
it’s a valid soulution and shouldn’t be avoided return in
@numbers.uniq.each do |n| # Add each number to the path
# If we have numbers left to add to the path and the sum will
not get greater then the target we want
if (path.count(n)<@numbers.count(n)) && (path.sum+n <= @target)
# Store the new path in our new new_path array if it
shouldn’t be avoided
new_path = path.dup
new_path << n
unless @avoid.include?new_path
new_paths << new_path unless new_path.sum == @target
return new_path if new_path.sum == @target
end
end
end
end
return nil if new_paths.empty? # We walked the whole tree and no
new path has been found, return nil
solver new_paths # Launch again with the new paths
end
end

def find_split fair_split,loot,avoid=Array.new
current_loot = loot.dup # Remember the loot before a split has been
stakes = Array.new

Try splitting the loot

begin
stake = Knapsack.new(fair_split,loot,avoid).solve
stakes << stake
stake.each { |s| loot.delete_one!(s) } unless stake.nil? # Remove
from the loot what has been found
end until stake.nil? || loot.empty? # Loop until the loot is empty,
or the algorithm found no valid solution
if loot.empty? # The whole loot is empty, a fair split has been found
return stakes
else
if current_loot == loot # The algorithm splitted nothing, it’s not
possible to fairly split the loot
return nil
else # The algorithm splitted something, but it wasn’t correct. Try
again avoiding the already found solutions
return find_split(fair_split,current_loot,stakes+avoid)
end
end
end

adventures,loot = ARGV.shift.to_i,ARGV.map { |a| a.to_i }

stakes = (loot.sum%adventures).zero? ? find_split(fair_split,loot) : nil

if stakes.nil?
puts “It is not possible to fairly split this treasure #{adventures}
ways.”
else
stakes.size.times { |i| puts “#{i+1}: " + stakes[i].sort.join(” ") }
end

On 2/6/06, Patrick D. [email protected] wrote:

For example, my old version didn’t split

ruby loot.rb 6 34 78 21 70 45 67 70 19 90 76 54 20 30 19 80 7 65 43 56 46

My turn to say ‘Doh!’
I found a case where the greedy algorithms failed and mine passed, but
now Patrick has found one that mine fails. But I have a simple 2 line
fix. I just need to move gems to my ‘pile2’ one at a time, instead of
a share-at-a-time. I had considered doing this before, but managed to
convince myself it was slower and not needed. Oops.

##################################
#loot.rb
#evenly splits an array into n sets of equal value

class Array
def sum
inject(0){|s,v| s+v}
end
def subtract arr
return clear if arr==self
arr.each{|e| if (n=index(e)) then delete_at(n); end }
self
end
#fast version which misses some subsets.
#useful as a rough filter.
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
return set if sum == n
end
}
nil
end
def find_subset_with_sum n
s = quick_find_subset_with_sum n
return s if s
possibilities, seen = [self.select{|e| e<=n}],{}
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

#1: put all loot in pile 1
#2: find a share from pile 1
#3: if you can’t find one, it can’t be split
#4: find a share in the remaining gems
#5: repeat unitl you find all shares
#6: if you can’t find enough shares
#7: move one item from the first share to pile2
#8: repeat starting from step 2, include pile2 when searcing the
remainder in step 4

if all the gems are moved to pile2, there is no possible solution

def splitter n, loot
splits=[]
pile1,pile2=loot.dup.sort.reverse,[]
total = loot.sum
share = total/n
return nil if total%n != 0 || loot.size < n || loot.max > share

until pile1.empty?
splits[0] = pile1.find_subset_with_sum(share)
break if !splits[0]
remaining = pile1.subtract(splits[0])+pile2
(1…n).each do |i|
break if nil == (splits[i] =
remaining.find_subset_with_sum(share))
remaining.subtract(splits[i])
end
return splits if splits[n-1]
#~ pile2 += splits[0] ##This line changes to the following two
lines
pile2 << splits[0].shift
pile1 += splits[0]
end
return nil
end

if FILE == \$0

n = ARGV.shift.to_i
if ARGV.size < 2 || n < 1
puts “Usage: #{\$0} partners item1 item2 …”
else
shares = splitter(n, ARGV.map{|a| a.to_i })
if !shares
puts “This loot can not be evenly divided into #{n} parts!”
else
shares.sort_by{|a|[a.size,-a.max]}.each_with_index{|share,i|
puts “#{i}: #{share.sort.reverse.join(’ ')}”}
puts “everyone gets #{shares[0].sum}”
end
end
end

My turn to say ‘Doh!’

Yet again. Your new version doesn’t split

6-ways
26 77 26 77 1 39 17 10 90 89 3 20 47 37 51 9 34 15 22 22 94 52

possible solution:
1: [10, 39, 94]
2: [1, 52, 90]
3: [3, 51, 89]
4: [9, 20, 37, 77]
5: [15, 17, 34, 77]
6: [22, 22, 26, 26, 47]

Manuel K.

On 2/8/06, Manuel K. [email protected] wrote:

My turn to say ‘Doh!’

Yet again. Your new version doesn’t split

6-ways
26 77 26 77 1 39 17 10 90 89 3 20 47 37 51 9 34 15 22 22 94 52

Aargh.
into my last submission.
I added 2 more tests to detect unsplittable sets before I start
iterating, and I fixed the iterator to be sure it tests every
combination, and quits as soon as it finds a value which can not fit
into any fair share. Oh, and I added one helper to Integer:

######################
class Integer
def odd?
self % 2 != 0
end
end

def splitter n, loot
splits=[]
pile1,pile2=loot.dup.sort.reverse,[]
total = loot.sum
share = total/n
num_odd = loot.inject(0){|s,g| g.odd? ? s+1 : s}

size;

shares

with odd values

if the share size is odd, there must be an even number plus one

for every share

return nil if total%n != 0 || loot.size < n || loot.max > share
return nil if loot.find_all{|g| g>share/2}.size > n
num_odd-=n if share.odd?
return nil if num_odd < 0 || num_odd.odd?

#pile1 holds all the items we haven’t tried to make a share with.
#take a candidate from the pile.
#if you can’t make a share using that one, it is impossible to
divide the loot.
#othewise, keep trying to make shares.
#if you get stuck, move the candidate to pile2, and start again.
#if pile1 becomes empty, give up

until pile1.empty?
candidate = pile1.shift
remaining = (pile1+pile2)
splits[0] = remaining.find_subset_with_sum(share - candidate)
break if !splits[0]
splits[0].unshift candidate
(1…n).each do |i|
break if nil == (splits[i] =
remaining.find_subset_with_sum(share))
remaining.subtract(splits[i])
end
return splits if splits[n-1]
pile2 << splits[0].shift
end
nil
end

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

Aargh.