R
Ruby Quiz
This is a classic algorithmic challenge and despite its seeming simplicity,
there's quite a bit you can learn from it. I'm pretty sure it was this exact
problem that finally got Big O Notation through my thick skull, so I'll take
that approach with this summary.
The obvious code that tends to come to mind for solving this problem is a
brute-force search through the subarrays. That's not a bad thing. It's very
easy to code up and may work for you if the inputs are small enough. It
certainly works for the quiz example.
Here's a solution of that by Drew Olson:
class Array
# sum the integer values of array contents
def int_sum
self.inject(0){|sum,i| sum+i.to_i}
end
# find the maximum sub array in an array
def max_sub_array
(0...self.size).inject([self.first]) do |max_sub,i|
(i...self.size).each do |j|
if max_sub.int_sum < self[i..j].int_sum
max_sub = self[i..j]
end
end
max_sub
end
end
end
# test example
if __FILE__ == $0
my_arr = [-1, 2, 5, -1, 3, -2, 1]
puts "array: #{my_arr.inspect}"
puts "maximum sub-array: #{my_arr.max_sub_array.inspect}"
end
I removed a little of Drew's printing code in the above so we could focus on the
algorithm, but the results are unchanged.
We can see that Drew's code works by walking through all of the indices, with
all possible lengths, to check each subarray. Each subarray is tested against
the current best sum and the end result is that the highest total found will be
returned.
The question we want to ask though, is how long does this take for various
inputs? It's quite zippy for the quiz example:
$ time ruby -r max_sub_array -e 'p [-1, 2, 5, -1, 3, -2, 1].max_sub_array'
[2, 5, -1, 3]
real 0m0.013s
user 0m0.007s
sys 0m0.005s
It slows down pretty quick though, with bigger inputs:
$ time ruby -r max_sub_array
-e 'p Array.new(100) { rand(11) - 5 }.max_sub_array'
[4, 0, 1, 4, -1, 1, 4, 0, 4, 3, 1, 0, 3, -4, 1, 4, -1, 0, 4, -3, 1, -3, 4, 2]
real 0m0.307s
user 0m0.301s
sys 0m0.006s
$ time ruby -r max_sub_array
-e 'p Array.new(1_000) { rand(11) - 5 }.max_sub_array'
[3, 1, -3, -1, 2, 5, 4, 3, -5, -2, 3, 1, 1, -2, -3, 4, 5, 4, 4, -3, -1, …]
real 3m39.856s
user 3m38.455s
sys 0m0.343s
The issue here is that those nested loops just execute many, many times. In
fact, that inner each() is called 500,500 times for an Array with 1,000 entries.
If we want to tackle those bigger lists we need to lower that count.
One way to find the algorithms with lower iteration counts is to randomly spot
check the solutions on an Array of similar length. In doing so, I stumbled
across this solution from Justin Either:
$ time ruby -r max_sub_array
-e 'p MaxSubArray.new.find(Array.new(1_000) { rand(11) - 5 })'
[1, 4, 1, 2, -5, -3, 4, 2, 3, 1, -2, 4, 5, 1, 3, 0, 5, -1, 4, 4, 2, 4, …]
real 0m0.016s
user 0m0.009s
sys 0m0.006s
$ time ruby -r max_sub_array
-e 'p MaxSubArray.new.find(Array.new(10_000) { rand(11) - 5 })'
[3, 1, -2, 5, 4, 5, 0, -3, 0, 3, 5, -3, -4, -3, 5, -3, -1, 4, 5, -3, 3, …]
real 0m0.047s
user 0m0.030s
sys 0m0.006s
$ time ruby -r max_sub_array
-e 'p MaxSubArray.new.find(Array.new(100_000) { rand(11) - 5 })'
[4, 1, -2, 3, 4, -4, -4, 4, 5, 1, -3, 4, -5, 5, -5, 1, -1, 0, -5, 1, -1, …]
real 0m0.286s
user 0m0.267s
sys 0m0.011s
As you can see, that's scaling to much higher counts much quicker. That's a
sure sign of a more clever algorithm, so let's take a peek at the code:
# Object defining a sub-array of integer values
# The sub-array contains a start and end index
# defining a region of the master array
class SubArray
def initialize
@start = 0
@end = 0
@sum = 0
end
# Set boundaries of the sub-array
def set_bounds(list_start, list_end)
@start, @end = list_start, list_end
end
# Provide get/set accessors
attr_reader :start, :end, :sum
attr_writer :sum
end
class MaxSubArray
# Finds the sub-array with the largest sum
# Input: a list of integers
def find(list)
max = SubArray.new
cur = SubArray.new
for i in 0...list.size
cur.sum = cur.sum + list
if (cur.sum > max.sum)
max.sum = cur.sum
cur.set_bounds(cur.start, i)
max.set_bounds(cur.start, i)
elsif (cur.sum < 0)
# If sum goes negative, this region cannot have the max sum
cur.sum = 0
cur.set_bounds(i + 1, i + 1)
end
end
list.slice(max.start, max.end - max.start + 1)
end
end
First, you need to take a look at the SubArray class. This is just a
bookkeeping tool to keep track of the bounds and sum of any given subarray.
There is a minor bug here that hinders this solution on all-negative Arrays like
[-3, -1], but it can be fixed by initializing sum to -1.0/0.0 instead of 0.
The second class holds the actual algorithm. The trick here is pretty simple
once you've seen it before. Basically, we start from the beginning of the Array
and expand the subarray to following indices. We keep track of the best total
seen thus far and replace that when we find better totals.
The trick is that we hop our subarray indices forward whenever the running total
dips into the negatives. A negative total is effectively starting over, so
skipping over all of those numbers costs us nothing. The run is broken.
This algorithm in linear, so that for iterator only executes 1,000 times for an
Array of that length. That's where your big speed gain comes from in this case
and the reason algorithms are important when dealing with larger inputs.
My thanks to all the algorists that showed off the variety of solutions that can
be applied here.
Ruby Quiz will now take a one week break to allow everyone the chance to compete
in the ICFP Contest (http://www.icfpcontest.org/). If you are a fan of
programming contests, I strongly encourage you to give this yearly competition a
shot. It's always challenging and rewarding. Best of luck!
there's quite a bit you can learn from it. I'm pretty sure it was this exact
problem that finally got Big O Notation through my thick skull, so I'll take
that approach with this summary.
The obvious code that tends to come to mind for solving this problem is a
brute-force search through the subarrays. That's not a bad thing. It's very
easy to code up and may work for you if the inputs are small enough. It
certainly works for the quiz example.
Here's a solution of that by Drew Olson:
class Array
# sum the integer values of array contents
def int_sum
self.inject(0){|sum,i| sum+i.to_i}
end
# find the maximum sub array in an array
def max_sub_array
(0...self.size).inject([self.first]) do |max_sub,i|
(i...self.size).each do |j|
if max_sub.int_sum < self[i..j].int_sum
max_sub = self[i..j]
end
end
max_sub
end
end
end
# test example
if __FILE__ == $0
my_arr = [-1, 2, 5, -1, 3, -2, 1]
puts "array: #{my_arr.inspect}"
puts "maximum sub-array: #{my_arr.max_sub_array.inspect}"
end
I removed a little of Drew's printing code in the above so we could focus on the
algorithm, but the results are unchanged.
We can see that Drew's code works by walking through all of the indices, with
all possible lengths, to check each subarray. Each subarray is tested against
the current best sum and the end result is that the highest total found will be
returned.
The question we want to ask though, is how long does this take for various
inputs? It's quite zippy for the quiz example:
$ time ruby -r max_sub_array -e 'p [-1, 2, 5, -1, 3, -2, 1].max_sub_array'
[2, 5, -1, 3]
real 0m0.013s
user 0m0.007s
sys 0m0.005s
It slows down pretty quick though, with bigger inputs:
$ time ruby -r max_sub_array
-e 'p Array.new(100) { rand(11) - 5 }.max_sub_array'
[4, 0, 1, 4, -1, 1, 4, 0, 4, 3, 1, 0, 3, -4, 1, 4, -1, 0, 4, -3, 1, -3, 4, 2]
real 0m0.307s
user 0m0.301s
sys 0m0.006s
$ time ruby -r max_sub_array
-e 'p Array.new(1_000) { rand(11) - 5 }.max_sub_array'
[3, 1, -3, -1, 2, 5, 4, 3, -5, -2, 3, 1, 1, -2, -3, 4, 5, 4, 4, -3, -1, …]
real 3m39.856s
user 3m38.455s
sys 0m0.343s
The issue here is that those nested loops just execute many, many times. In
fact, that inner each() is called 500,500 times for an Array with 1,000 entries.
If we want to tackle those bigger lists we need to lower that count.
One way to find the algorithms with lower iteration counts is to randomly spot
check the solutions on an Array of similar length. In doing so, I stumbled
across this solution from Justin Either:
$ time ruby -r max_sub_array
-e 'p MaxSubArray.new.find(Array.new(1_000) { rand(11) - 5 })'
[1, 4, 1, 2, -5, -3, 4, 2, 3, 1, -2, 4, 5, 1, 3, 0, 5, -1, 4, 4, 2, 4, …]
real 0m0.016s
user 0m0.009s
sys 0m0.006s
$ time ruby -r max_sub_array
-e 'p MaxSubArray.new.find(Array.new(10_000) { rand(11) - 5 })'
[3, 1, -2, 5, 4, 5, 0, -3, 0, 3, 5, -3, -4, -3, 5, -3, -1, 4, 5, -3, 3, …]
real 0m0.047s
user 0m0.030s
sys 0m0.006s
$ time ruby -r max_sub_array
-e 'p MaxSubArray.new.find(Array.new(100_000) { rand(11) - 5 })'
[4, 1, -2, 3, 4, -4, -4, 4, 5, 1, -3, 4, -5, 5, -5, 1, -1, 0, -5, 1, -1, …]
real 0m0.286s
user 0m0.267s
sys 0m0.011s
As you can see, that's scaling to much higher counts much quicker. That's a
sure sign of a more clever algorithm, so let's take a peek at the code:
# Object defining a sub-array of integer values
# The sub-array contains a start and end index
# defining a region of the master array
class SubArray
def initialize
@start = 0
@end = 0
@sum = 0
end
# Set boundaries of the sub-array
def set_bounds(list_start, list_end)
@start, @end = list_start, list_end
end
# Provide get/set accessors
attr_reader :start, :end, :sum
attr_writer :sum
end
class MaxSubArray
# Finds the sub-array with the largest sum
# Input: a list of integers
def find(list)
max = SubArray.new
cur = SubArray.new
for i in 0...list.size
cur.sum = cur.sum + list
if (cur.sum > max.sum)
max.sum = cur.sum
cur.set_bounds(cur.start, i)
max.set_bounds(cur.start, i)
elsif (cur.sum < 0)
# If sum goes negative, this region cannot have the max sum
cur.sum = 0
cur.set_bounds(i + 1, i + 1)
end
end
list.slice(max.start, max.end - max.start + 1)
end
end
First, you need to take a look at the SubArray class. This is just a
bookkeeping tool to keep track of the bounds and sum of any given subarray.
There is a minor bug here that hinders this solution on all-negative Arrays like
[-3, -1], but it can be fixed by initializing sum to -1.0/0.0 instead of 0.
The second class holds the actual algorithm. The trick here is pretty simple
once you've seen it before. Basically, we start from the beginning of the Array
and expand the subarray to following indices. We keep track of the best total
seen thus far and replace that when we find better totals.
The trick is that we hop our subarray indices forward whenever the running total
dips into the negatives. A negative total is effectively starting over, so
skipping over all of those numbers costs us nothing. The run is broken.
This algorithm in linear, so that for iterator only executes 1,000 times for an
Array of that length. That's where your big speed gain comes from in this case
and the reason algorithms are important when dealing with larger inputs.
My thanks to all the algorists that showed off the variety of solutions that can
be applied here.
Ruby Quiz will now take a one week break to allow everyone the chance to compete
in the ICFP Contest (http://www.icfpcontest.org/). If you are a fan of
programming contests, I strongly encourage you to give this yearly competition a
shot. It's always challenging and rewarding. Best of luck!