 # Calculating single-digit summands

I have tried to make an algorithm that finds all possible combinations
of single-digit summands for a number. After an afternoon of hard and
desperate work I got something inefficient (although it works). Maybe
someone has a better idea. The link to the algorithm:
http://secam.blogspot.com/2005/12/solving-kakuro-part-i.html

On 11 Dec 2005 13:53:28 -0800, “draq” [email protected] wrote:

I have tried to make an algorithm that finds all possible combinations
of single-digit summands for a number. After an afternoon of hard and
desperate work I got something inefficient (although it works). Maybe
someone has a better idea. The link to the algorithm:
http://secam.blogspot.com/2005/12/solving-kakuro-part-i.html

I don’t understand the rules of the game from your writeup.

On Dec 11, 2005, at 3:57 PM, draq wrote:

I have tried to make an algorithm that finds all possible combinations
of single-digit summands for a number. After an afternoon of hard and
desperate work I got something inefficient (although it works). Maybe
someone has a better idea. The link to the algorithm:
http://secam.blogspot.com/2005/12/solving-kakuro-part-i.html

One thing that your code could really use is better names. Variables
like “arr”, “t”, and even “calc” tell me very little, as I’m trying

Next, there’s a lot more iterators than just each() and using them
can make your code more expressive. Some examples:

def sum (arr)
i = 0
arr.each do |k| i += k end
i
end

# … or …

def sum( enum )
inject { |sum, n| sum + n }
end

arrj.each do |a|
arri.delete(a) if sum(a) != number
end

# … or …

arrj.delete_if { |a| sum(a) != number }

You can also simplify lines like the following by using destructive
method calls:

arri = arri.uniq

# … or …

arri.uniq!

Hope that gives you some ideas.

James Edward G. II

Would you like to add some asserts, so I can figure out the
specification?

How is the data entered?

Christer

Since only single digit summands are under consideration, I don’t think
optimizations are important for this. Here’s a simple brute force
approach that performs reasonably well. It considers all possible
subsets of [1,2,3,4,5,6,7,8,9] and rejects any with the wrong depth or
sum. For this solution I reused the powerset method I wrote for a recent
Ruby Q…

class Array

def sum
inject { |sum,x| sum += x }
end

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

def calc(number, depth)
puts “number = #{number}, depth = #{depth}”
candidates = (1…9).inject([]) { |a,x| a << x }
candidates.powerset { |subset|
next unless subset.length == depth
next unless subset.sum == number
p subset
}
end

# examples

(3…5).each { |depth|
(10…20).each { |target| calc(target, depth) }
}

draq,

Array mixes in Enumerable, so inject works.

I added some asserts to be able to understand what your code does.
As I can see, this solves only a partial problem: generating all
combinations, given a sum and number of squares. This is calc(sum,
count).
Next step would probably be intersection(sum1, count1, sum2, count2)
between a row and a column, listing the possible combinations.

Curiousity: The list for arr(2) below, was too long. So I decided to cut
it by writing a method for Array. Then I found it’s already defined!
First accepts zero or one argument.

Defining a kakuro, so a program can solve it, seems to be a lot more
hassle than defining a sudoku.

Christer

def sum (arr) arr.inject { |sum,i| sum += i } end

def arr (depth, min=1, max=10-depth,t=[], arr=[])
(min…max).each do |i|
t[depth-1] = i if depth > 0
arr(depth-1, i+1, max+1, t, arr) if depth > 1
arr << t.reverse.clone if depth == 1
end
arr
end

def calc (number, depth)
arri = arr(depth)
arri.each do |a|
arri.delete_if { |a| sum(a) != number }
end
arri
end

require ‘test/unit’
class TestKakuro < Test::Unit::TestCase
def test_all
assert_equal 12, sum([3,4,5])
assert_equal [[1, 2], [1, 3], [1, 4], [1, 5], [1, 6],
[1, 7], [1, 8], [1, 9], [2, 3], [2, 4]],
arr(2).first(10)
assert_equal 9, arr(1).size
assert_equal 36, arr(2).size
assert_equal 84, arr(3).size
assert_equal 126, arr(4).size
assert_equal 126, arr(5).size
assert_equal 84, arr(6).size
assert_equal 36, arr(7).size
assert_equal 9, arr(8).size
assert_equal 1, arr(9).size

``````assert_equal [[7, 8, 9]], calc(24,3)
assert_equal [[1, 6, 8, 9],[2, 5, 8, 9],
[2, 6, 7, 9],[3, 4, 8, 9],
[3, 5, 7, 9],[3, 6, 7, 8],
[4, 5, 6, 9],[4, 5, 7, 8]], calc(24, 4)
assert_equal [[1, 2, 3, 4, 5, 6, 7, 8, 9]], calc(45, 9)
``````

end
end

I see, there is much to learn. ;-))

draq

draq wrote:

Hello Christer,

I don’t see the problem. The list of arr(2) seems to be right. The
numbers are [1,2] - [1,3] - [1,4] - [1,5] - [1,6] - [1,7] - [1,8] -
[1,9] - [2,3] - [2,4] - [2,5] - [2,6] - [2,7] - [2,8] - [2,9] - [3,4] -
[3,5] - [3,6] - [3,7] - [3,8] - [3,9] - [4,5] - [4,6] - [4,7] - [4,8] -
[4,9] - [5,6] - [5,7] - [5,8] - [5,9] - [6,7] - [6,8] - [6,9] - [7,8] -
[7,9] - [8,9].

I beg pardon for my lousy comments. I’ve tried to improve. You’ll find
http://secam.blogspot.com/2005/12/solving-kakuro-part-iii.html
Hopefully it’s now better understandable.

draq,

I was using first(10) to keep the output short.

christer

# … or …

def sum( enum )
inject { |sum, n| sum + n }
end

Don’t you need enum. ?

def sum(enum)
enum.inject {|sum,n| sum+n}
end

This is a newer algorithm which works much more faster.

def sum (arr)

# sorry James Edward G. II, but I’m not using an enum.

i = 0
arr.each do |k| i += k end
i
end

def arr (depth, min=1, max=10-depth,t=[], arr=[])
(min…max).each do |i|
t[depth-1] = i if depth > 0
arr(depth-1, i+1, max+1, t, arr) if depth > 1
arr << t.reverse.clone if depth == 1
end

arr
end

def calc (number, depth)
arri = arr(depth)

arri.each do |a|
arri.delete_if { |a| sum(a) != number }
end

arri
end

# examples

calc(24, 3).each do |a| print "#{a} - " end puts
calc(24, 4).each do |a| print "#{a} - " end puts

# even summands of 45 can be calculated now. It was impossible with the

older algorithm.
calc(45, 9).each do |a| print "#{a} - " end puts

Hello Christer,

I don’t see the problem. The list of arr(2) seems to be right. The
numbers are [1,2] - [1,3] - [1,4] - [1,5] - [1,6] - [1,7] - [1,8] -
[1,9] - [2,3] - [2,4] - [2,5] - [2,6] - [2,7] - [2,8] - [2,9] - [3,4] -
[3,5] - [3,6] - [3,7] - [3,8] - [3,9] - [4,5] - [4,6] - [4,7] - [4,8] -
[4,9] - [5,6] - [5,7] - [5,8] - [5,9] - [6,7] - [6,8] - [6,9] - [7,8] -
[7,9] - [8,9].

I beg pardon for my lousy comments. I’ve tried to improve. You’ll find
http://secam.blogspot.com/2005/12/solving-kakuro-part-iii.html
Hopefully it’s now better understandable.

On Dec 12, 2005, at 11:02 AM, draq wrote:

This is a newer algorithm which works much more faster.

def sum (arr)

# sorry James Edward G. II, but I’m not using an enum.

You don’t think so? Let’s ask Ruby…

Array.ancestors.find { |par| par.to_s =~ /enum/i }
=> Enumerable

arr = Array.new
=> []

arr.is_a? Enumerable
=> true

arr.respond_to? :inject
=> true

Ruby thinks so. Let’s try a sum:

arr = (1…10).to_a
=> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

arr.inject { |sum, n| sum + n }
=> 55

Looks good to me.

arri.each do |a|
arri.delete_if { |a| sum(a) != number }
end

The whole point of delete_if() is that you don’t need the each().
Try taking it out. James Edward G. II

Hi –

On Tue, 13 Dec 2005, draq wrote:

arr << t.reverse.clone if depth == 1

I haven’t been following this thread closely but just an observation:
Array#reverse gives you a new array, so you don’t need to clone it.

David

David A. Black
[email protected]

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

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