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Given a number of digits n = 1, 2, 3, 4, 5, 6.
I wanted to generate a nested array S that will contain all possible 4 digit permutations of n.
since 6^4 = 1296, there will be 1296 possible permutations.
Example:
S = [[1,1,1,1],[1,1,1,2],[1,1,2,2]...[6,6,6,6]]
I started the nested loop with the first index with value of [1,1,1,1]
Then used a for in loop with range 0..1295 and tried to carry over the value of S[i] to S[i+1]
then increment the value of S[i+1][x], where x always starts at 3 then is decremented until it reaches 0 then it becomes 3 again. The problem with my procedure is when i try to increment the S[i+1][x], S[i] also increments its S[i][x].
In the code below S is instead called 'all_possible_combinations'
all_possible_combinations = Array.new(1296) {Array.new(4)}
all_possible_combinations[0] = [1, 1 ,1 ,1]
x = 3
for i in 0..1295
if i + 1 == 1296
break
else
all_possible_combinations[i+1] = all_possible_combinations[i]
all_possible_combinations[i+1][x] += 1
x -= 1
if x == 0
x = 3
end
end
end
[Attached image shows debugging process where Si][x] also gets incremented
You may compute that array as follows.
a = [1, 2, 3, 4, 5, 6]
b = a.repeated_permutation(4).to_a
#=> [[1, 1, 1, 1], [1, 1, 1, 2], [1, 1, 1, 3], [1, 1, 1, 4], [1, 1, 1, 5],
# [1, 1, 1, 6], [1, 1, 2, 1], [1, 1, 2, 2], [1, 1, 2, 3], [1, 1, 2, 4],
# ...
# [6, 6, 5, 3], [6, 6, 5, 4], [6, 6, 5, 5], [6, 6, 5, 6], [6, 6, 6, 1],
# [6, 6, 6, 2], [6, 6, 6, 3], [6, 6, 6, 4], [6, 6, 6, 5], [6, 6, 6, 6]]
b.size
#=> 1296
See Array#repeated_permutation
If the array a may contain duplicates and you wish to remove duplicate permutations you may wish to tack on Array#uniq.
a = [1, 1, 3, 1, 1, 6]
b = a.repeated_permutation(4).to_a.uniq
#=> [[1, 1, 1, 1], [1, 1, 1, 3], [1, 1, 1, 6], [1, 1, 3, 1],
# [1, 1, 3, 3], [1, 1, 3, 6], [1, 1, 6, 1], [1, 1, 6, 3],
# [1, 1, 6, 6], [1, 3, 1, 1], [1, 3, 1, 3], [1, 3, 1, 6],
# [1, 3, 3, 1], [1, 3, 3, 3], [1, 3, 3, 6], [1, 3, 6, 1],
# [1, 3, 6, 3], [1, 3, 6, 6], [1, 6, 1, 1], [1, 6, 1, 3],
# [1, 6, 1, 6], [1, 6, 3, 1], [1, 6, 3, 3], [1, 6, 3, 6],
# [1, 6, 6, 1], [1, 6, 6, 3], [1, 6, 6, 6], [3, 1, 1, 1],
# [3, 1, 1, 3], [3, 1, 1, 6], [3, 1, 3, 1], [3, 1, 3, 3],
# [3, 1, 3, 6], [3, 1, 6, 1], [3, 1, 6, 3], [3, 1, 6, 6],
# [3, 3, 1, 1], [3, 3, 1, 3], [3, 3, 1, 6], [3, 3, 3, 1],
# [3, 3, 3, 3], [3, 3, 3, 6], [3, 3, 6, 1], [3, 3, 6, 3],
# [3, 3, 6, 6], [3, 6, 1, 1], [3, 6, 1, 3], [3, 6, 1, 6],
# [3, 6, 3, 1], [3, 6, 3, 3], [3, 6, 3, 6], [3, 6, 6, 1],
# [3, 6, 6, 3], [3, 6, 6, 6], [6, 1, 1, 1], [6, 1, 1, 3],
# [6, 1, 1, 6], [6, 1, 3, 1], [6, 1, 3, 3], [6, 1, 3, 6],
# [6, 1, 6, 1], [6, 1, 6, 3], [6, 1, 6, 6], [6, 3, 1, 1],
# [6, 3, 1, 3], [6, 3, 1, 6], [6, 3, 3, 1], [6, 3, 3, 3],
# [6, 3, 3, 6], [6, 3, 6, 1], [6, 3, 6, 3], [6, 3, 6, 6],
# [6, 6, 1, 1], [6, 6, 1, 3], [6, 6, 1, 6], [6, 6, 3, 1],
# [6, 6, 3, 3], [6, 6, 3, 6], [6, 6, 6, 1], [6, 6, 6, 3],
# [6, 6, 6, 6]]
b.size
#=> 81
To create a sequence where each element is generated based on the previous one, there's Enumerator.produce, e.g.:
enum = Enumerator.produce([1, 1, 1, 1]) do |a, b, c, d|
d += 1 # ^^^^^^^^^^^^
# initial value
if d > 6
d = 1
c += 1
end
if c > 6
c = 1
b += 1
end
if b > 6
b = 1
a += 1
end
if a > 6
raise StopIteration # <- stops enumeration
end
[a, b, c, d] # <- return value = next value
end
I've kept the example intentionally simple, using an explicit variable for each of the four digits. You could of course also have an array and use a little loop to handle the increment / carry.
The above gives you:
enum.count #=> 1296
enum.first(3) #=> [[1, 1, 1, 1], [1, 1, 1, 2], [1, 1, 1, 3]]
enum.to_a.last(3) #=> [[6, 6, 6, 4], [6, 6, 6, 5], [6, 6, 6, 6]]
I have a list of lists of 4 integers. All integers in the same list are distinct, e.g.,
data = [[8, 9, 3, 0], [3, 8, 4, 9], [7, 9, 6, 4], [3, 6, 4, 8], [0, 5, 3, 7], [0, 9, 4, 2], [9, 0, 1, 5], [3, 2, 8, 6], [3, 5, 4, 0], [1, 2, 5, 9], [1, 3, 6, 5], [2, 4, 5, 7], [7, 8, 6, 3], [6, 2, 9, 8], [8, 7, 5, 4], [8, 5, 1, 3]]
Currently I have a function that receives the list above and a list of distinct integers certain. The function returns a list with the probabilities of each integer (from 0 to 9) being in a list knowing that it contains the integers in certain.
def probability(data, certain):
probs = [0] * 10
counter_total = 0
set_certain = set(certain)
for d in data:
if set_certain.issubset(d):
counter_total += 1
for i in range(10):
if i in d and i not in set_certain:
probs[i] += 1
probs = [x / counter_total for x in probs]
return probs
Initially, the list certain is empty and values are added later. Is there a data structure I can use in the start of the program so that I don't have to go through all the data again every time I append a new value to certain? The list data can be very big.
I am trying to implement the Yen's K Shortest Path Algorihtm based on the pseudo-code from https://en.wikipedia.org/wiki/Yen%27s_algorithm. Here is the code.
import numpy as np
import networkx as nx
edge_list = [[0, 1], [0, 2], [0, 7], [1, 2], [1, 9], [2, 5], [2, 7], [2, 9], [3, 4], [3, 5], [3, 6], [3, 8], [4, 5], [4, 6], [4, 7], [4, 8], [5, 6], [5, 7], [5, 8], [6, 8], [7, 8]]
graph = nx.Graph()
graph.add_edges_from(edge_list)
nx.draw(graph, with_labels = True)
source_node = 8
destination_node = 9
def yen_ksp(graph, source, sink, K):
A, B = [], []
A.append(nx.shortest_path(graph, source=source, target=sink))
for k in range(1, 1+K):
for i in range(len(A[k - 1]) - 1):
spurNode = A[k-1][i]
rootPath = A[k-1][0:i+1]
removed_edges, removed_nodes = [], []
for p in A:
if rootPath == p[0:i+1] and p[i:i+2] not in removed_edges:
removed_edges.append(p[i:i+2])
for edge in removed_edges:
graph.remove_edge(edge[0], edge[1])
try:
spurPath = nx.shortest_path(graph, source=spurNode, target=sink)
except:
for edge in removed_edges:
graph.add_edge(edge[0], edge[1])
continue
totalPath = rootPath + spurPath[1:]
B.append(totalPath)
for edge in removed_edges:
graph.add_edge(edge[0], edge[1])
if B == []:
# This handles the case of there being no spur paths, or no spur paths left.
# This could happen if the spur paths have already been exhausted (added to A),
# or there are no spur paths at all - such as when both the source and sink vertices
# lie along a "dead end".
break
B.sort()
A.append(B[-1])
B.pop(-1)
return A
print(yen_ksp(graph.copy(), source_node, destination_node, 10))
This is supposed to be an undirected, unweighted graph generated from the above code.
And this is the output of the code.
[[8, 5, 2, 9],
[8, 7, 2, 9],
[8, 7, 2, 1, 9],
[8, 7, 2, 1, 2, 9],
[8, 7, 2, 1, 2, 1, 9],
[8, 7, 2, 1, 2, 1, 2, 9],
[8, 7, 2, 1, 2, 1, 2, 1, 9],
[8, 7, 2, 1, 2, 1, 2, 1, 2, 9],
[8, 7, 2, 1, 2, 1, 2, 1, 2, 1, 9],
[8, 7, 2, 1, 2, 1, 2, 1, 2, 1, 2, 9],
[8, 7, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 9]]
Obviously there are shorter paths that the algorithm missed. And, the results contain paths that have loops. I want only the ones without.
Also, in other cases, the results were in the wrong order, some longer paths appear before other paths that are shorter. In the KSP problem, the order of results is obviously important because if I stop at some k, I want to be sure that there is no shorter path that I have missed.
I am open to other algorithms that can correctly and effectively solve this problem of KSP without loops on undirected-unweighted graphs.
Please help.
Networkx provides a function for generating a list of all simple paths in a graph from source to target, starting from shortest ones: shortest_simple_paths. This procedure is based exactly on Yen's algorithm, as you can read in the documentation.
Using it is very simple:
paths = list(nx.shortest_simple_paths(graph, source_node, target_node))
If you want only the first K shortest paths you can make use of islice:
from itertools import islice
paths = list(islice(nx.shortest_simple_paths(graph, source_node, target_node), K))
Example:
from itertools import islice
K = 10
source_node = 8
target_node = 9
graph = nx.Graph()
edge_list = [[0, 1], [0, 2], [0, 7], [1, 2], [1, 9], [2, 5], [2, 7],
[2, 9], [3, 4], [3, 5], [3, 6], [3, 8], [4, 5], [4, 6],
[4, 7], [4, 8], [5, 6], [5, 7], [5, 8], [6, 8], [7, 8]]
graph.add_edges_from(edge_list)
for path in islice(nx.shortest_simple_paths(graph, source_node, target_node), K):
print(path)
Output:
[8, 5, 2, 9]
[8, 7, 2, 9]
[8, 5, 7, 2, 9]
[8, 5, 2, 1, 9]
[8, 3, 5, 2, 9]
[8, 7, 0, 1, 9]
[8, 7, 2, 1, 9]
[8, 4, 5, 2, 9]
[8, 7, 5, 2, 9]
[8, 7, 0, 2, 9]
If you want to understand how shortest_simple_path is implemented you can check out its source code: it's well written and very easy to understand!
Let i have lazy sequences: s1, s2, s3, ..., sN, with non-descending numbers, for example:
s1 = [1, 1, 2, 3, 3, 3, 4, .....]
s2 = [1, 2, 2, 2, 2, 2, 3, 3, 4, ....]
s3 = [1, 2, 3, 3, 3, 3, 4, 4, 4, ....]
what I'd like to do - is to merge it, grouping by similar items and processing it with some function, for example generate list of tuples (number, count)
for my case:
merge(s1, s2, s3) should generate [ [1, 4], [2, 6], [3, 9], [4, 5], .... ]
Are any gems, etc., to process such sequences
If you want to do it lazily, here is some code which would do that:
def merge(*args)
args.map!(&:lazy)
Enumerator.new do |yielder|
while num = args.map(&:peek).min
count = 0
while list = args.find { |l| l.peek == num }
list.next
list.peek rescue args.delete list
count += 1
end
yielder.yield [num, count]
end
end
end
s1 = [1, 1, 2, 3, 3, 3, 4]
s2 = [1, 2, 2, 2, 2, 2, 3, 3, 4]
s3 = [1, 2, 3, 3, 3, 3, 4, 4, 4]
s4 = (0..1.0/0)
merge(s1, s2, s3, s4).take(20)
# => [[0, 1], [1, 5], [2, 8], [3, 10], [4, 6], [5, 1], [6, 1], [7, 1], [8, 1], [9, 1], [10, 1], [11, 1], [12, 1], [13, 1], [14, 1], [15, 1], [16, 1], [17, 1], [18, 1], [19, 1]]
I'm new to ruby and working on such a problem:
There are n numbered letters and n numbered envelopes. The letter x can't be
put into the envelope x.(OP only wants values where no value of x is at index x-1) What I want is to print out all the possible
cases.
The index of Array + 1 ---> the number of the envelop
The element of Array ---> the number of the letter
Input: n = 3.
Output: [2, 3, 1], [3, 1, 2]
Input: n = 4.
Output: [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]
Here is my code:
$nums = []
def f( already, n, times )
if n > times
$nums << already.dup
return
else
1.upto(times) do |i|
next if ((already.include? i) || n == i)
already << i
f( already, n+1, times )
already.pop
end
end
end
I'm looking for more elegant solutions.
Make use of the permutation enumerator, rejecting all those where the value at index x-1 matches x:
def f(n, x)
(1..n).to_a.permutation.reject{|p| p[x-1] == x}
end
> f 3, 3
=> [[1, 3, 2], [2, 3, 1], [3, 1, 2], [3, 2, 1]]
> f 4, 4
=> [[1, 2, 4, 3], [1, 3, 4, 2], [1, 4, 2, 3], [1, 4, 3, 2], [2, 1, 4, 3], [2, 3, 4, 1], [2, 4, 1, 3], [2, 4, 3, 1], [3, 1, 4, 2], [3, 2, 4, 1], [3, 4, 1, 2], [3, 4, 2, 1], [4, 1, 2, 3], [4, 1, 3, 2], [4, 2, 1, 3], [4, 2, 3, 1], [4, 3, 1, 2], [4, 3, 2, 1]]
UPDATE
Looking at your question again, it's unclear if you want to use a specific x, or just that the logic should hold true for any value of x. If the second guess is what you want, then use this instead:
def f(n)
(1..n).to_a.permutation.reject{|p| p.any?{|x| p[x-1] == x}}
end
> f 3
=> [[2, 3, 1], [3, 1, 2]]
> f 4
=> [[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]]