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I have written the following program to find the permutation of all the elements in an array. The values are created properly but the problem occurs when I try to assign the generated sequence into a new array. The old values will get cleared and the new values are copied as per the array size
def find_perm(nums, answer, set)
if nums.empty?
p set
answer.push(set)
p answer.object_id
p answer
return true
end
for i in (0..nums.length - 1) do
new_nums = nums.clone
new_nums.delete_at(i)
set.push(nums[i])
find_perm(new_nums, answer, set)
set.pop
end
end
def permute(nums)
answer = []
set = []
element = find_perm(nums, answer, set)
return element
end
permute([1,2,3])
This are the observations that I have found out while debugging:
[1, 2, 3]
47167191669680
[[1, 2, 3]]
[1, 3, 2]
47167191669680
[[1, 3, 2], [1, 3, 2]]
[2, 1, 3]
47167191669680
[[2, 1, 3], [2, 1, 3], [2, 1, 3]]
[2, 3, 1]
47167191669680
[[2, 3, 1], [2, 3, 1], [2, 3, 1], [2, 3, 1]]
[3, 1, 2]
47167191669680
[[3, 1, 2], [3, 1, 2], [3, 1, 2], [3, 1, 2], [3, 1, 2]]
[3, 2, 1]
47167191669680
[[3, 2, 1], [3, 2, 1], [3, 2, 1], [3, 2, 1], [3, 2, 1], [3, 2, 1]]
The problem was each time I was pushing the same set array into the answer array so each element in the answer array will have same reference (same object_id).
Solution: Is to clone the set array during the each push to the answer array so that each element have different reference.
The solution:
def fact(n)
return 1 if n == 1
n*fact(n-1)
end
def find_perm(nums, answer, set)
if nums.empty?
answer.push(set.clone)
end
for i in (0..nums.length - 1) do
new_nums = nums.clone
new_nums.delete_at(i)
set.push(nums[i])
find_perm(new_nums, answer, set)
set.pop
return answer if fact(nums.count) == answer.count
end
end
def permute(nums)
answer = []
set = []
element = find_perm(nums, answer, set)
return element
end
p permute([1,2,3])
I believe the approach you are taking is similar to the following.
Suppose we wish to obtain the permutations of the elements of the array
arr = [1, 2, 3, 4]
Begin with the array
[4]
This array has only a single perumutation:
perms3 = [[4]]
(3 in perms3 denotes the index of 4 in arr.) Now obtain the permuations of
[3, 4]
We see that is
perms2 = [[3, 4], [4, 3]]
We simply take each element of perms3 ([4] is the only one) and create two permuations by inserting 3 before 4 and then 3 after 4;
Now suppose the array were
[2, 3, 4]
Then
perms1 = [[2, 3, 4], [3, 2, 4], [3, 4, 2], [2, 4, 3], [4, 2, 3], [4, 3, 2]]
We create three 3-element arrays from [3, 4], one by inserting 2 before 3, one by inserting 2 between 3 and 4 and one by inserting 2 after 4. Simlarly, three 3-element arrays are generated from [4, 3] in a simlar way. This generates six arrays. (Indeed, 3! #=> 6).
Lastly we generating the 4! #=> 24 permuations of [1, 2, 3, 4] by inserting 1 in four locations of each element of perm1, the first four derived from perms1[0] #=> [2, 3, 4]:
[[1, 2, 3, 4], [2, 1, 3, 4], [2, 3, 1, 4], [2, 3, 4, 1]]
We can do this in code as follows.
def my_permutations(arr)
perms = [[arr.last]]
(arr.size-2).downto(0) do |i|
x = arr[i]
perms = perms.flat_map do |perm|
(0..(perm.size)).map { |i| perm.dup.insert(i, x) }
end
end
perms
end
my_permutations(arr)
#=> [[1, 2, 3, 4], [2, 1, 3, 4], [2, 3, 1, 4], [2, 3, 4, 1], [1, 3, 2, 4],
# [3, 1, 2, 4], [3, 2, 1, 4], [3, 2, 4, 1], [1, 3, 4, 2], [3, 1, 4, 2],
# [3, 4, 1, 2], [3, 4, 2, 1], [1, 2, 4, 3], [2, 1, 4, 3], [2, 4, 1, 3],
# [2, 4, 3, 1], [1, 4, 2, 3], [4, 1, 2, 3], [4, 2, 1, 3], [4, 2, 3, 1],
# [1, 4, 3, 2], [4, 1, 3, 2], [4, 3, 1, 2], [4, 3, 2, 1]]
See Enumerable#flat_map and Array#insert. Note that we need to make a copy of the array perm before invoking insert.
We could of course have gone "forward" in arr (starting with [[1]]), rather than "backward", though the elements of the array of permutations would be ordered differently.
I’m getting some weird results implementing cyclic permutation on the children of a multidimensional array.
When I manually define the array e.g.
arr = [
[1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5]
]
the output is different from when I obtain that same array by calling a method that builds it.
I’ve compared the manual array to the generated version and they’re exactly the same (class and values, etc).
I tried writing the same algorithm in JS and encountered the same issue.
Any idea what might be going on?
def Build_array(child_arr, n)
#Creates larger array with arr as element, n times over. For example Build_array([1,2,3], 3) returns [[1,2,3], [1,2,3], [1,2,3]]
parent_arr = Array.new(4)
0.upto(n) do |i|
parent_arr[i] = child_arr
end
return parent_arr
end
def Cylce_child(arr, steps_tocycle)
# example: Cylce_child([1, 2, 3, 4, 5], 2) returns [4, 5, 1, 2, 3]
0.upto(steps_tocycle - 1) do |i|
x = arr.pop()
arr.unshift(x)
end
return arr
end
def Permute_array(parent_array, x, y, z)
#x, y, z = number of steps to cycle each child array
parent_array[0] = Cylce_child(parent_array[0], x)
parent_array[1] = Cylce_child(parent_array[1], y)
parent_array[2] = Cylce_child(parent_array[2], z)
return parent_array
end
arr = Build_array([1, 2, 3, 4, 5], 4)
# arr = [[1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5]]
puts "#{Permute_array(arr, 1, 2, 3)}"
# Line 34: When arr = Build_array([1, 2, 3, 4, 5], 4)
# Result (WRONG):
# [[5, 1, 2, 3, 4], [5, 1, 2, 3, 4], [5, 1, 2, 3, 4], [5, 1, 2, 3, 4]]
#
# Line 5: When arr = [[1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, # 2, 3, 4, 5]]
# Result (CORRECT):
# [[5, 1, 2, 3, 4], [4, 5, 1, 2, 3], [3, 4, 5, 1, 2], [1, 2, 3, 4, 5]]
#
The problem is in the way you build the array.
This line:
parent_arr[i] = child_arr
does not put in parent_arr[i] a copy of child_arr but a reference to it.
This means your initial array contains four references to the same child array. Later on, when the code changes parent_arr[0], it changes the same array that child_arr was referring to in the build method. And that array is also parent_arr[1] and parrent_arr[2] and so on.
A simple solution to the problem is to put in parent_arr[i] a copy of child_arr:
parent_arr[i] = Array.new(child_arr)
I see where the bug was. Added the clone method to line 8 so that it now reads:
parent_arr[i] = child_arr.clone
#Old: parent_arr[i] = child_arr
Thanks Robin, for pointing me in the right direction.
This is a fairly common mistake to make in Ruby since arrays do not contain objects per-se, but object references, which are effectively pointers to a dynamically allocated object, not the object itself.
That means this code:
Array.new(4, [ ])
Will yield an array containing four identical references to the same object, that object being the second argument.
To see what happens:
Array.new(4, [ ]).map(&:object_id)
# => => [70127689565700, 70127689565700, 70127689565700, 70127689565700]
Notice four identical object IDs. All the more obvious if you call uniq on that.
To fix this you must supply a block that yields a different object each time:
Array.new(4) { [ ] }.map(&:object_id)
# => => [70127689538260, 70127689538240, 70127689538220, 70127689538200]
Now adding to one element does not impact the others.
That being said, there's a lot of issues in your code that can be resolved by employing Ruby as it was intended (e.g. more "idiomatic" code):
def build_array(child_arr, n)
# Duplicate the object given each time to avoid referencing the same thing
# N times. Each `dup` object is independent.
Array.new(4) do
child_arr.dup
end
end
def cycle_child(arr, steps_tocycle)
# Ruby has a rotate method built-in
arr.rotate(steps_tocycle)
end
# Using varargs (*args) you can just loop over how many positions were given dynamically
def permute_array(parent_array, *args)
# Zip is great for working with two arrays in parallel, they get "zippered" together.
# Also map is what you use for transforming one array into another in a 1:1 mapping
args.zip(parent_array).map do |a, p|
# Rotate each element the right number of positions
cycle_child(p, -a)
end
end
arr = build_array([1, 2, 3, 4, 5], 4)
# => [[1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5]]
puts "#{permute_array(arr, 1, 2, 3)}"
# => [[5, 1, 2, 3, 4], [4, 5, 1, 2, 3], [3, 4, 5, 1, 2]]
A lot of these methods boil down to some very simple Ruby so they're not especially useful now, but this adapts the code as directly as possible for educational purposes.
I have a range of numbers R = (1..n). I also have another character 'a'. I want to generate strings of length L (L > n + 2) that have all the numbers in the same order, but go through every repeated permutation of 'a' to fill the length L. For example, if n = 3, and L = 7, then some valid strings would be :
"123aaaa",
"1a23aaa",
"1aa2a3a",
"aaaa123"
while the following strings would be invalid:
"213aaaa", # invalid, because 1,2,3 are not in order
"123a", #invalid, because length < L
"1123aaa", # invalid because a number is repeated
I am currently doing this, which is way too inefficient:
n = 3
L = 7
all_terms = (1..n).to_a + Array.new(L - n, 'a')
all_terms.permutation.each do |permut|
if(valid_permut? permut) # checks if numbers are in their natural order
puts permut.join
end
end
How do I directly generate valid strings more efficiently?
The problem is equivalent to: select n elements from index 0 to L - 1, fill these with 1 to n accordingly, and fill the rest with some constant character.
In your example, it's taking 3 elements from 0..6:
(0..6).to_a.combination(3).to_a
=> [[0, 1, 2], [0, 1, 3], [0, 1, 4], [0, 1, 5], [0, 1, 6], [0, 2, 3], [0, 2, 4],
[0, 2, 5], [0, 2, 6], [0, 3, 4], [0, 3, 5], [0, 3, 6], [0, 4, 5], [0, 4, 6], [0, 5, 6],
[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 2, 6], [1, 3, 4], [1, 3, 5], [1, 3, 6], [1, 4, 5],
[1, 4, 6], [1, 5, 6], [2, 3, 4], [2, 3, 5], [2, 3, 6], [2, 4, 5], [2, 4, 6], [2, 5, 6],
[3, 4, 5], [3, 4, 6], [3, 5, 6], [4, 5, 6]]
Every subarray here represents a possible result. For example, [0, 2, 3] corresponds to '0a12aaa', [3, 5, 6] corresponds to 'aaa0a12', etc. The code for this conversion is straight-forward.
You can model this as all possible interleavings of two strings, where relative order of the input elements is preserved. Here's a recursive solution. It works by choosing an element from one list, and prepending it to all possible subproblems, then doing it again where an element is chosen from the second list instead, and combining the two solution sets at the end.
# Returns an array of all possible interleaving of two strings
# Maintains relative order of each character of the input strings
def interleave_strings_all(a1, a2)
# Handle base case where at least one input string is empty
return [a1 + a2] if a1.empty? || a2.empty?
# Place element of first string, and prepend to all subproblems
set1 = interleave_strings_all(a1[1..-1], a2).map{|x| a1[0] + x}
# Place element of second string and prepend to all subproblems
set2 = interleave_strings_all(a1, a2[1..-1]).map{|x| a2[0] + x}
# Combine solutions of subproblems into overall problem
return set1.concat(set2)
end
if __FILE__ == $0 then
l = 5
n = 3
a1 = (1..n).to_a.map{|x| x.to_s}.join()
a2 = 'a' * (l - n)
puts interleave_strings_all(a1, a2)
end
The output is:
123aa
12a3a
12aa3
1a23a
1a2a3
1aa23
a123a
a12a3
a1a23
aa123
So what Im trying to accomplish is write a (shorter) condition that makes sure each element is different from the other array. This is confusing but I hope this example clears it up.
array = [1, 2, 3]
new_array = array.shuffle
until array[0] != new_array[0] &&
array[1] != new_array[1] &&
array[2] != new_array[2]
new_array = array.shuffle
end
So what Im doing is making sure that every single element/index pair does not match in the other array.
# [1, 2, 3] => [3, 1, 2] yayyyy
# [1, 2, 3] => [3, 2, 1] not what I want because the 2 didnt move
Is there a better way to do what I want to do? Ive looked up the .any? and .none? but I cant seem to figure out how to implement them. Thanks!
I would do this:
array.zip(new_array).all? { |left, right| left != right }
Here are two approaches that do not involve repeated sampling until a valid sample is obtained:
Sample from the population of valid permutations
Construct the population from which you are sampling:
array = [1, 2, 3, 4]
population = array.permutation(array.size).reject do |a|
a.zip(array).any? { |e,f| e==f }
end
#=> [[2, 1, 4, 3], [2, 3, 4, 1], [2, 4, 1, 3], [3, 1, 4, 2], [3, 4, 1, 2],
# [3, 4, 2, 1], [4, 1, 2, 3], [4, 3, 1, 2], [4, 3, 2, 1]]
Then just choose one at random:
10.times { p population.sample }
# [4, 3, 1, 2]
# [3, 4, 1, 2]
# [3, 4, 1, 2]
# [4, 3, 1, 2]
# [2, 1, 4, 3]
# [2, 1, 4, 3]
# [4, 1, 2, 3]
# [2, 1, 4, 3]
# [4, 3, 1, 2]
# [3, 4, 1, 2]
Sequentially sample for each position in the array
def sample_no_match(array)
a = array.each_index.to_a.shuffle
last_ndx = a[-1]
a.dup.map do |i|
if a.size == 2 && a[-1] == last_ndx
select = a[-1]
else
select = (a-[i]).sample
end
a.delete(select)
array[select]
end
end
10.times.each { p sample_no_match(array) }
# [2, 4, 3, 1]
# [4, 3, 1, 2]
# [2, 1, 3, 4]
# [1, 3, 4, 2]
# [1, 3, 2, 4]
# [1, 3, 2, 4]
# [1, 4, 3, 2]
# [3, 4, 2, 1]
# [1, 3, 4, 2]
# [1, 3, 4, 2]
I have been unable to prove or disprove that the second method produces a random sample. We can, however, determine relative frequencies of outcomes:
n = 500_000
h = n.times.with_object(Hash.new(0)) { |_,h| h[sample_no_match(array)] += 1 }
h.keys.each { |k| h[k] = (h[k]/(n.to_f)).round(4) }
h #=> {[1, 2, 3, 4]=>0.0418, [2, 1, 3, 4]=>0.0414, [1, 4, 2, 3]=>0.0418,
# [3, 4, 2, 1]=>0.0417, [4, 3, 2, 1]=>0.0415, [3, 1, 4, 2]=>0.0419,
# [2, 3, 1, 4]=>0.0420, [4, 2, 3, 1]=>0.0417, [3, 2, 1, 4]=>0.0413,
# [4, 2, 1, 3]=>0.0417, [2, 1, 4, 3]=>0.0419, [1, 3, 2, 4]=>0.0415,
# [1, 2, 4, 3]=>0.0418, [1, 3, 4, 2]=>0.0417, [2, 4, 1, 3]=>0.0414,
# [3, 4, 1, 2]=>0.0412, [1, 4, 3, 2]=>0.0423, [4, 1, 3, 2]=>0.0411,
# [3, 2, 4, 1]=>0.0411, [2, 4, 3, 1]=>0.0418, [3, 1, 2, 4]=>0.0419,
# [4, 3, 1, 2]=>0.0412, [4, 1, 2, 3]=>0.0421, [2, 3, 4, 1]=>0.0421}
avg = (h.values.reduce(:+)/h.size.to_f).round(4)
#=> 0.0417
mn, mx = h.values.minmax
#=> [0.0411, 0.0423]
([avg-mn,mx-avg].max/avg).round(6)
#=> 0.014388
which means that the maximum deviation from the average was only 1.4% percent of the average.
This suggests that the second method is a reasonable way of producing pseudo-random samples.
Initially, the first line of this method was:
a = array.each_index.to_a
By looking at the frequency distribution for outcomes, however, it was clear that that method did not produce a pseudo-random sample; hence, the need to shuffle a.
Here's one possibility:
until array.zip(new_array).reject{ |x, y| x == y }.size == array.size
new_array = array.shuffle
end
Note, though, that it will break for arrays like [1] or [1, 1, 1, 2, 3], where the number of instances of 1 exceeds half the size of the array. Recommend Array#uniq or similar, along with checking for arrays of sizes 0 or 1, depending on how trustworthy your input is!
I'd like to get [[2, 1, 3], [1, 3, 2]] from [1, 2, 3] in Ruby.
For [1, 2, 3, 4], I'd like to get [[2, 1, 3, 4], [1, 3, 2, 4], [1, 2, 4, 3]]
Rule: Within two numbers, if left one is smaller then it swap the position.
I have the following codes so far but it returns [[2, 3, 1], [2, 3, 1]]
What am I doing wrong here? I appreciate any inputs.
In amidakuji.rb
class Amidakuji
def initialize(column, rung)
#column = column
#rung = rung
#myarr = []
#per_arr = []
#build_arr = []
end
def build_initial
#arr = (1..#column).to_a
end
def swap_element
i = 0
arr = build_initial
while i < #column - 1 do
#build_arr << swap(arr, i)
i += 1
end
#build_arr
end
def swap(arr, a)
if arr[a] < arr[a + 1]
arr[a], arr[a + 1] = arr[a + 1], arr[a]
end
arr
end
end
In amidakuji_spec.rb
it 'should create an array with swapped elements' do
expect(#kuji1.swap_element).to eq ([[2, 1, 3], [1, 3, 2]])
end
Results
Failures:
expected: [[2, 1, 3], [1, 3, 2]]
got: [[2, 3, 1], [2, 3, 1]]
You can do this quite compactly by using the methods Enumerable#each_cons and Enumerable#map.
Code
def doit(arr)
(0...arr.size).each_cons(2).map do |i,j|
a = arr.dup
a[i], a[j] = a[j], a[i]
a
end
end
Examples
doit([1,2,3]) #=> [[2, 1, 3], [1, 3, 2]]
doit([1,2,3,4]) #=> [[2, 1, 3, 4], [1, 3, 2, 4], [1, 2, 4, 3]]
doit([1,2,3,4,5]) #=> [[2, 1, 3, 4, 5], [1, 3, 2, 4, 5],
#=> [1, 2, 4, 3, 5], [1, 2, 3, 5, 4]]
Explanation
arr = [1,2,3,4]
b = (0...arr.size).each_cons(2)
#=> #<Enumerator: 0...4:each_cons(2)>
To view the contents of this enumerator:
b.to_a
#=> [[0, 1], [1, 2], [2, 3]]
Lastly
b.map do |i,j|
a = arr.dup
a[i], a[j] = a[j], a[i]
a
end
#=> [[2, 1, 3, 4], [1, 3, 2, 4], [1, 2, 4, 3]]
In the last step, consider the first element of b that is passed to map, which assigns the following values to the block variables:
i => 0
j => 1
We then make a copy of arr, swap the elements offsets 0 and 1, making
a => [2, 1, 3, 4]
and then enter a at the end of the block, causing map to replace [0, 1] with that array.
Given what you're trying to accomplish and the output you're getting, it looks like you're reusing the same array when you want distinct arrays instead. Specifically this line:
#build_arr << swap(arr, i)
is always passing the same 'arr' to swap.
So first time, it swaps the 1 and the 2 to give you [2, 1, 3]
Second time, it swaps the 1 and the 3 give you [2, 3, 1]
You push the same array onto #build_arr twice, which is why it repeats.