Speed Golf - Remove Early Dups


#1

SUMMARY
What’s the fastest and/or shortest you can turn this:
in = [“i”, “v”, “w”, “e”, “l”, “d”, “u”, “f”, “e”, “v”, “f”, “e”,
“d”, “e”, “w”, “d”]

into this:
[“i”, “l”, “u”, “v”, “f”, “e”, “w”, “d”]
?

DETAILS
The goal is, given a stream of items, to produce the same ordered
stream with no duplicates, where the relative position of the last item
wins. The shortest (and fastest?) for me turns out to be simply:
in.reverse.uniq.reverse

I have the question mark because the in-place version is sometimes
faster, sometimes slower in my tests:
out = in.dup; out.reverse!; out.uniq!; out.reverse!

I thought it might be faster to loop through the array in one pass
myself, but that turns out to be about 4x slower:
seen = {}
out = in.dup
out.each_with_index{ |v,i|
if n=h[v]
out[n] = nil
end
h[v] = i
}
out.compact!

The input for me isn’t REALLY an array, but rather a series of items
that I receive one at a time from a depth-first traversal of a graph.
If this influences your answer, so be it :slight_smile:

BACKGROUND
Why is this problem mildly interesting? See
http://phrogz.net/nodes/traversingdirectedgraph.asp

I was reading that page for nostalgia the other day and thought about
porting the calculation library to Ruby for fun, and to see how much
easier solving the problem would be in Ruby.

Here’s the code I put together for fun (the cells and formulae in the
below correspond to the graphic on the above url), before adding the
‘minimal update chain’ optimization.

require ‘set’
module ClassContainer
def initialize( *args )
self.class[ self.name ] = self
super
end

def self.included( klass )
klass.class_eval{
def self.inherited( subklass )
( @subclasses ||= [] ) << subklass
end

  def self.all
    @subclasses ||= []
    @instances ||= {}
    all = @instances.dup
    @subclasses.each{ |sub| all.merge!( sub.all ) }
    all
  end

  def self.each
    all.each_value{ |i| yield i }
  end

  def self.[]( name )
    all[ name ]
  end

  def self.[]=( name, instance )
    ( @instances ||= {} )[ name ] = instance
  end

  def self.sorted
    all.values.sort_by{ |n| n.name.to_s }
  end
}

end
end

module GraphNode
attr_reader :outgoing_edges, :incoming_edges

def initialize( *args )
@outgoing_edges = Set.new
@incoming_edges = Set.new
end

def edges
@outgoing_edges + @incoming_edges
end

def add_edge_to( other_node )
@outgoing_edges << other_node
other_node.incoming_edges << self
self
end

def clear_incoming
@incoming_edges.each{ |n|
n.outgoing_edges.delete self
}
@incoming_edges = Set.new
self
end

def traverse( seen_nodes = Set.new, &block )
new_seen = seen_nodes.dup << self
@outgoing_edges.each{ |destination|
unless seen_nodes.include?( destination )
yield destination
destination.traverse( new_seen, &block )
end
}
end
end

class Cell
include GraphNode
include ClassContainer

attr_reader :name
attr_accessor :value
def initialize( name, value=0 )
raise “Duplicate ID” if Cell[ name ]
@name = name
@value = value * 1.0

super

end

def value=( new_value )
# Naive update approach
return if @value == new_value
@value = new_value
update_dependents
new_value
end

def update_dependents
traverse{ |formula|
formula.evaluate( true )
}
end

def to_s
‘<%s “%s” value=%.2f>’ % [ self.class.name, @name, @value ]
end
end

class Formula < Cell
def initialize( *args )
super
@formula, @value = @value.to_s, 0.0
update_dependencies
end

def update_dependencies
clear_incoming
@formula.scan( /@(\w+)/ ).flatten.each{ |source_name|
if incoming = Cell[ source_name.to_sym ]
incoming.add_edge_to( self )
end
}
end

def evaluate( skip_dependents = false )
scope = ValueSpace.new
Cell.each{ |n| scope.set( n.name, n.value ) }
@value = eval( @formula, scope.get_binding )
puts “Updated #{self}”
update_dependents unless skip_dependents
end

def to_s
‘<%s “%s” formula="%s" value=%.2f>’ % [ self.class.name, @name,
@formula, @value ]
end
end

class ValueSpace
def set( name, value )
instance_variable_set :"@#{name}", value
end
def get_binding
binding
end
end

if $0 == FILE
Cell.new( :a, 5 )
Cell.new( :d, 3 )
Cell.new( :e, 32 )
Cell.new( :i, 10 )
Cell.new( :l, 12 )

Formula.new( :b, “@d + @e + @a” )
Formula.new( :g, “@h * @i” ).value = 20
Formula.new( :h, “@g / @i” ).value = 2
Formula.new( :c, “@a + 3” )
Formula.new( :f, “@b + @c” )
Formula.new( :j, “@c + 8” )
Formula.new( :k, “@l + @j” )
Formula.new( :m, “@l + 10” )

Need to call here to handle circular dependencies

Formula.each{ |f| f.update_dependencies }

Naive approach…

puts “Naive initial evaluation…”
Formula.each{ |f| f.evaluate }

puts “*” * 40
puts “Initial values…”
puts Cell.sorted

puts “*” * 40
puts “Changing ‘a’ to 7…”
Cell[ :a ].value = 7

puts “*” * 40
puts “Changing ‘i’ to 36…”
Cell[ :i ].value = 36

puts ‘-’ * 40
puts “Per cell dependency…”
Cell.each{ |cell|
puts “#{cell.name} -> #{cell.outgoing_edges.map{|c| c.name}.join(’,
')}”
}

end

(which outputs)

Naive initial evaluation…
Updated <Formula “c” formula="@a + 3" value=8.00>
Updated <Formula “f” formula="@b + @c" value=8.00>
Updated <Formula “j” formula="@c + 8" value=16.00>
Updated <Formula “k” formula="@l + @j" value=28.00>
Updated <Formula “f” formula="@b + @c" value=8.00>
Updated <Formula “j” formula="@c + 8" value=16.00>
Updated <Formula “k” formula="@l + @j" value=28.00>
Updated <Formula “b” formula="@d + @e + @a" value=40.00>
Updated <Formula “f” formula="@b + @c" value=48.00>
Updated <Formula “k” formula="@l + @j" value=28.00>
Updated <Formula “g” formula="@h * @i" value=20.00>
Updated <Formula “h” formula="@g / @i" value=2.00>
Updated <Formula “m” formula="@l + 10" value=22.00>
Updated <Formula “h” formula="@g / @i" value=2.00>
Updated <Formula “g” formula="@h * @i" value=20.00>


Initial values…
<Cell “a” value=5.00>
<Formula “b” formula="@d + @e + @a" value=40.00>
<Formula “c” formula="@a + 3" value=8.00>
<Cell “d” value=3.00>
<Cell “e” value=32.00>
<Formula “f” formula="@b + @c" value=48.00>
<Formula “g” formula="@h * @i" value=20.00>
<Formula “h” formula="@g / @i" value=2.00>
<Cell “i” value=10.00>
<Formula “j” formula="@c + 8" value=16.00>
<Formula “k” formula="@l + @j" value=28.00>
<Cell “l” value=12.00>
<Formula “m” formula="@l + 10" value=22.00>


Changing ‘a’ to 7…
Updated <Formula “b” formula="@d + @e + @a" value=42.00>
Updated <Formula “f” formula="@b + @c" value=50.00>
Updated <Formula “c” formula="@a + 3" value=10.00>
Updated <Formula “f” formula="@b + @c" value=52.00>
Updated <Formula “j” formula="@c + 8" value=18.00>
Updated <Formula “k” formula="@l + @j" value=30.00>


Changing ‘i’ to 36…
Updated <Formula “h” formula="@g / @i" value=0.56>
Updated <Formula “g” formula="@h * @i" value=20.00>
Updated <Formula “g” formula="@h * @i" value=20.00>
Updated <Formula “h” formula="@g / @i" value=0.56>

Per cell dependency…
c -> f, j
e -> b
f ->
i -> h, g
l -> k, m
j -> k
b -> f
k ->
g -> h
m ->
a -> b, c
h -> g
d -> b


#2

perhaps in.reverse!.uniq!.reverse! ?


#3

Phrogz wrote:

DETAILS
The goal is, given a stream of items, to produce the same ordered
stream with no duplicates, where the relative position of the last item
wins. The shortest (and fastest?) for me turns out to be simply:
in.reverse.uniq.reverse

The input for me isn’t REALLY an array, but rather a series of items
that I receive one at a time from a depth-first traversal of a graph.
If this influences your answer, so be it :slight_smile:

In = [“i”, “v”, “w”, “e”, “l”, “d”, “u”, “f”, “e”, “v”, “f”, “e”,
“d”, “e”, “w”, “d”]

out = []

In.each{ |x| out.delete( x ); out << x }

p out


#4

does this have a quadratic time complexity? doing this by sorting might
be faster… or am i mistaken?

konstantin


#5

On Tue, 06 Dec 2005 02:27:44 +0100, ako… removed_email_address@domain.invalid wrote:

i meant in.reverse.uniq!.reverse! (the first reverse is non-destructive)

Don’t chain bang methods:

[“i”, “l”, “u”, “v”, “f”, “e”, “w”, “d”].reverse.uniq!.reverse!
NoMethodError: undefined method `reverse!’ for nil:NilClass
from (irb):3
from :0

They return nil if nothing changed…


#6

i meant in.reverse.uniq!.reverse! (the first reverse is non-destructive)


#7

Hi –

On Tue, 6 Dec 2005, Phrogz wrote:

DETAILS
The goal is, given a stream of items, to produce the same ordered
stream with no duplicates, where the relative position of the last item
wins. The shortest (and fastest?) for me turns out to be simply:
in.reverse.uniq.reverse

This entry is not in contention on shortness or speed (don’t even bother
to benchmark it – it’s completely off the charts), but I just thought
it
looked really cool :slight_smile:

a.inject([]){|r,*b|r-b+b}

David
__
David A. Black
removed_email_address@domain.invalid

“Ruby for Rails”, forthcoming from Manning Publications, April 2006!


#8

On Dec 5, 2005, at 6:47 PM, ako… wrote:

does this have a quadratic time complexity? doing this by sorting
might
be faster… or am i mistaken?

Quadratic? No. The suggested algorithms are O(2n) or O(3n).


#9

i do not see why not. this is almost exactly a bubble sort. which is in
worst case quadratic.


#10

On Dec 5, 2005, at 6:27 PM, ako… wrote:

i meant in.reverse.uniq!.reverse! (the first reverse is non-
destructive)

Array#uniq! can return nil if no changes were made. This will not
occur given the above input, but can for the general case.

Using reverse rather than dup does shave a bit of time off, though,
which puts it as the winner on my machine. So far:

Rehearsal --------------------------------------------------------
William J. 4.230000 0.060000 4.290000 ( 4.369104)
Simple Copies 1.180000 0.010000 1.190000 ( 1.227496)
Dup and In-Place 1.210000 0.010000 1.220000 ( 1.240726)
Reverse and In-Place 1.130000 0.010000 1.140000 ( 1.158873)
Hash and Compact 7.220000 0.090000 7.310000 ( 7.552673)
---------------------------------------------- total: 15.150000sec

                        user     system      total        real

William J. 4.230000 0.040000 4.270000 ( 4.465252)
Simple Copies 1.190000 0.010000 1.200000 ( 1.218064)
Dup and In-Place 1.220000 0.010000 1.230000 ( 1.268994)
Reverse and In-Place 1.120000 0.010000 1.130000 ( 1.152538)
Hash and Compact 7.220000 0.060000 7.280000 ( 7.511923)

input = [“i”, “v”, “w”, “e”, “l”, “d”, “u”, “f”, “e”, “v”, “f”, “e”,
“d”, “e”, “w”, “d”]

require ‘benchmark’
Benchmark.bmbm( 20 ) do |r|
n = 100_000

r.report( ‘William J.’ ){
n.times{
out = []
input.each{ |x| out.delete( x ); out << x }
}
}

r.report( ‘Simple Copies’ ){
n.times{
input.reverse.uniq.reverse
}
}

r.report( ‘Dup and In-Place’ ){
n.times{
out = input.dup
out.reverse!
out.uniq!
out.reverse!
out
}
}

r.report( ‘Reverse and In-Place’ ){
n.times{
out = input.reverse
out.uniq!
out.reverse!
out
}
}

r.report( “Hash and Compact” ){
n.times{
seen = {}
out = input.dup
out.each_with_index{ |val,idx|
if old_idx = seen[ val ]
out[ old_idx ] = nil
end
seen[ val ] = idx
}
}
}

end


#11

my reasoning was that the code loops over the array and on every
iteration sequentially searches another array of the same length. this
is similar to bubble sort where on every iteration the current element
is moved to the beginning of the array until its place in the order is
found. but this is not really important, so let us not argue over this
; -)
konstantin


#12

On Dec 6, 2005, at 12:17 AM, ako… wrote:

i do not see why not. this is almost exactly a bubble sort. which
is in
worst case quadratic.

It’s not an arbitrary re-sorting of all the elements involved. The
elements are all there, in the right order - you just have to weed
out the few bad ones.