Related
I have a list with certain combinations between two numbers:
[1 2] [1 4] [1 6] [3 4] [5 6] [3 6] [2 3] [4 5] [2 5]
Now I want to make groups of 3 combinations, where each group contains all six digits once, e.g.:
[1 2] [3 6] [4 5] is valid
[1 4] [2 3] [5 6] is valid
[1 2] [2 3] [5 6] is invalid
Order is not important.
How can I arrive upon a list of all possible groups, without employing a brute forcing algorithm?
The language it is implemented in doesn't matter. Description of an algorithm that could achieve this is enough.
One thing to notice is that there are only finitely many possible pairs of elements you can pick from the set {1,2,3,4,5,6}. Specifically, there are (6P2) = 30 of them if you consider order relevant and (6 choose 2) = 15 if you don't. Even the simple "try all triples" algorithm that runs in cubic time in this case will only have to look at at most (30 choose 3) = 4,060 triples, which is a pretty small number. I doubt that you'd have any problems in practice just doing this.
Here's a recursive function in Python that picks a pair of numbers from a list, and then calls itself with the remaining list:
def pairs(l, picked, ok_pairs):
n = len(l)
for a in range(n-1):
for b in range(a+1,n):
pair = (l[a],l[b])
if pair not in ok_pairs:
continue
if picked and picked[-1][0] > pair[0]:
continue
p = picked+[pair]
if len(l) > 2:
pairs([m for i,m in enumerate(l) if i not in [a, b]], p, ok_pairs)
else:
print p
ok_pairs = set([(1, 2), (1, 4), (1, 6), (3, 4), (5, 6), (3, 6), (2, 3), (4, 5), (2, 5)])
pairs([1,2,3,4,5,6], [], ok_pairs)
The output (of 6 triplets) is:
[(1, 2), (3, 4), (5, 6)]
[(1, 2), (3, 6), (4, 5)]
[(1, 4), (2, 3), (5, 6)]
[(1, 4), (2, 5), (3, 6)]
[(1, 6), (2, 3), (4, 5)]
[(1, 6), (2, 5), (3, 4)]
Here's a version using Python set arithmetic:
pairs = [(1, 2), (1, 4), (1, 6), (3, 4), (5, 6), (3, 6), (2, 3), (4, 5), (2, 5)]
n = len(pairs)
for i in range(n-2):
set1 = set(pairs[i])
for j in range(i+1,n-1):
set2 = set(pairs[j])
if set1 & set2:
continue
for k in range(j+1,n):
set3 = set(pairs[k])
if set1 & set3 or set2 & set3:
continue
print pairs[i], pairs[j], pairs[k]
The output is:
(1, 2) (3, 4) (5, 6)
(1, 2) (3, 6) (4, 5)
(1, 4) (5, 6) (2, 3)
(1, 4) (3, 6) (2, 5)
(1, 6) (3, 4) (2, 5)
(1, 6) (2, 3) (4, 5)
Say I have the following ranges, in some list:
{ (1, 4), (6, 8), (2, 5), (1, 3) }
(1, 4) represents days 1, 2, 3, 4. (6, 8) represents days 6, 7, 8, and so on.
The goal is to find the total number of days that are listed in the collection of ranges -- for instance, in the above example, the answer would be 8, because days 1, 2, 3, 4, 6, 7, 8, and 5 are contained within the ranges.
This problem can be solved trivially by iterating through the days in each range and putting them in a HashSet, then returning the size of the HashSet. But is there any way to do it in O(n) time with respect to the number of range pairs? How about in O(n) time and with constant space? Thanks.
Sort the ranges in ascending order by their lower limits. You can probably do this in linear time since you're dealing with integers.
The rest is easy. Loop through the ranges once keeping track of numDays (initialized to zero) and largestDay (initialized to -INF). On reaching each interval (a, b):
if b > largestDay then
numDays <- numDays + b-max(a - 1, largestDay)
largestDay <- max(largestDay, b)
else nothing.
So, after sorting we have (1,4), (1,3), (2,5), (6,8)
(1,4): numDays <- 0 + (4 - max(1 - 1, -INF)) = 4, largestDay <- max(-INF, 4) = 4
(1,3): b < largestDay, so no change.
(2,5): numDays <- 4 + (5 - max(2 - 1, 4)) = 5, largestDay <- 5
(6,8): numDays <- 5 + (8 - max(6-1, 5)) = 8, largestDay <- 8
The complexity of the following algorithm is O(n log n) where n is the number of ranges.
Sort the ranges (a, b) lexicographically by increasing a then by decreasing b.
Before: { (1, 4), (6, 8), (2, 5), (1, 3) }
After: { (1, 4), (1, 3), (2, 5), (6, 8) }
Collapse the sorted sequence of ranges into a potentially-shorter sequence of ranges, repeatedly merging consecutive (a, b) and (c, d) into (a, max(b, d)) if b >= c.
Before: { (1, 4), (1, 3), (2, 5), (6, 8) }
{ (1, 4), (2, 5), (6, 8) }
After: { (1, 5), (6, 8) }
Map the sequence of ranges to their sizes.
Before: { (1, 5), (6, 8) }
After: { 5, 3 }
Sum the sizes to arrive at the total number of days.
8
So if I had the numbers [1,2,2,3] and I want k=2 partitions I'd have [1][2,2,3], [1,2][2,3], [2,2][1,3], [2][1,2,3], [3][1,2,2], etc.
See an answer in Python at Code Review.
user3569's solution at Code Review produces five 2-tuples for the test case below, instead of exclusively 3-tuples. However, removing the frozenset() call for the returned tuples leads to the code returning exclusively 3-tuples. The revised code is as follows:
from itertools import chain, combinations
def subsets(arr):
""" Note this only returns non empty subsets of arr"""
return chain(*[combinations(arr,i + 1) for i,a in enumerate(arr)])
def k_subset(arr, k):
s_arr = sorted(arr)
return set([i for i in combinations(subsets(arr),k)
if sorted(chain(*i)) == s_arr])
s = k_subset([2,2,2,2,3,3,5],3)
for ss in sorted(s):
print(len(ss)," - ",ss)
As user3569 says "it runs pretty slow, but is fairly concise".
(EDIT: see below for Knuth's solution)
The output is:
3 - ((2,), (2,), (2, 2, 3, 3, 5))
3 - ((2,), (2, 2), (2, 3, 3, 5))
3 - ((2,), (2, 2, 2), (3, 3, 5))
3 - ((2,), (2, 2, 3), (2, 3, 5))
3 - ((2,), (2, 2, 5), (2, 3, 3))
3 - ((2,), (2, 3), (2, 2, 3, 5))
3 - ((2,), (2, 3, 3), (2, 2, 5))
3 - ((2,), (2, 3, 5), (2, 2, 3))
3 - ((2,), (2, 5), (2, 2, 3, 3))
3 - ((2,), (3,), (2, 2, 2, 3, 5))
3 - ((2,), (3, 3), (2, 2, 2, 5))
3 - ((2,), (3, 5), (2, 2, 2, 3))
3 - ((2,), (5,), (2, 2, 2, 3, 3))
3 - ((2, 2), (2, 2), (3, 3, 5))
3 - ((2, 2), (2, 3), (2, 3, 5))
3 - ((2, 2), (2, 5), (2, 3, 3))
3 - ((2, 2), (3, 3), (2, 2, 5))
3 - ((2, 2), (3, 5), (2, 2, 3))
3 - ((2, 3), (2, 2), (2, 3, 5))
3 - ((2, 3), (2, 3), (2, 2, 5))
3 - ((2, 3), (2, 5), (2, 2, 3))
3 - ((2, 3), (3, 5), (2, 2, 2))
3 - ((2, 5), (2, 2), (2, 3, 3))
3 - ((2, 5), (2, 3), (2, 2, 3))
3 - ((2, 5), (3, 3), (2, 2, 2))
3 - ((3,), (2, 2), (2, 2, 3, 5))
3 - ((3,), (2, 2, 2), (2, 3, 5))
3 - ((3,), (2, 2, 3), (2, 2, 5))
3 - ((3,), (2, 2, 5), (2, 2, 3))
3 - ((3,), (2, 3), (2, 2, 2, 5))
3 - ((3,), (2, 3, 5), (2, 2, 2))
3 - ((3,), (2, 5), (2, 2, 2, 3))
3 - ((3,), (3,), (2, 2, 2, 2, 5))
3 - ((3,), (3, 5), (2, 2, 2, 2))
3 - ((3,), (5,), (2, 2, 2, 2, 3))
3 - ((5,), (2, 2), (2, 2, 3, 3))
3 - ((5,), (2, 2, 2), (2, 3, 3))
3 - ((5,), (2, 2, 3), (2, 2, 3))
3 - ((5,), (2, 3), (2, 2, 2, 3))
3 - ((5,), (2, 3, 3), (2, 2, 2))
3 - ((5,), (3, 3), (2, 2, 2, 2))
Knuth's solution, as implemented by Adeel Zafar Soomro on the same Code Review page can be called as follows if no duplicates are desired:
s = algorithm_u([2,2,2,2,3,3,5],3)
ss = set(tuple(sorted(tuple(tuple(y) for y in x) for x in s)))
I haven't timed it, but Knuth's solution is visibly faster, even for this test case.
However, it returns 63 tuples rather than the 41 returned by user3569's solution. I haven't yet gone through the output closely enough to establish which output is correct.
Here's a version in Haskell:
import Data.List (nub, sort, permutations)
parts 0 = []
parts n = nub $ map sort $ [n] : [x:xs | x <- [1..n`div`2], xs <- parts(n - x)]
partition [] ys result = sort $ map sort result
partition (x:xs) ys result =
partition xs (drop x ys) (result ++ [take x ys])
partitions xs k =
let variations = filter (\x -> length x == k) $ parts (length xs)
in nub $ concat $ map (\x -> mapVariation x (nub $ permutations xs)) variations
where mapVariation variation = map (\x -> partition variation x [])
OUTPUT:
*Main> partitions [1,2,2,3] 2
[[[1],[2,2,3]],[[1,2,3],[2]],[[1,2,2],[3]],[[1,2],[2,3]],[[1,3],[2,2]]]
Python solution:
pip install PartitionSets
Then:
import partitionsets.partition
filter(lambda x: len(x) == k, partitionsets.partition.Partition(arr))
The PartitionSets implementation seems to be pretty fast however it's a pity you can't pass number of partitions as an argument, so you need to filter your k-set partitions from all subset partitions.
You may also want to look at:
similar topic on researchgate.
I encountered and solved this problem as part of a larger algorithm, but my solution seems inelegant and I would appreciate any insights.
I have a list of pairs which can be viewed as points on a Cartesian plane. I need to generate three lists: the sorted x values, the sorted y values, and a list which maps an index in the sorted x values with the index in the sorted y values corresponding to the y value with which it was originally paired.
A concrete example might help explain. Given the following list of points:
((3, 7), (15, 4), (7, 11), (5, 0), (4, 7), (9, 12))
The sorted list of x values would be (3, 4, 5, 7, 9, 15), and the sorted list of y values would be (0, 4, 7, 7, 11, 12).
Assuming a zero based indexing scheme, the list that maps the x list index to the index of its paired y list index would be (2, 3, 0, 4, 5, 1).
For example the value 7 appears as index 3 in the x list. The value in the mapping list at index 3 is 4, and the value at index 4 in the y list is 11, corresponding to the original pairing (7, 11).
What is the simplest way of generating this mapping list?
Here's a simple O(nlog n) method:
Sort the pairs by their x value: ((3, 7), (4, 7), (5, 0), (7, 11), (9, 12), (15, 4))
Produce a list of pairs in which the first component is the y value from the same position in the previous list and the second increases from 0: ((7, 0), (7, 1), (0, 2), (11, 3), (12, 4), (4, 5))
Sort this list by its first component (y value): ((0, 2), (4, 5), (7, 0), (7, 1), (11, 3), (12, 4))
Iterate through this list. For the ith such pair (y, k), set yFor[k] = i. yFor[] is your list (well, array) mapping indices in the sorted x list to indices in the sorted y list.
Create the sorted x list simply by removing the 2nd element from the list produced in step 1.
Create the sorted y list by doing the same with the list produced in step 3.
I propose the following.
Generate the unsorted x and y lists.
xs = [3, 15, 7, 5, 4, 9 ]
ys = [7, 4, 11, 0, 7, 12]
Transform each element into a tuple - the first of the pair being the coordinate, the second being the original index.
xs = [(3, 0), (15, 1), ( 7, 2), (5, 3), (4, 4), ( 9, 5)]
ys = [(7, 0), ( 4, 1), (11, 2), (0, 3), (7, 4), (12, 5)]
Sort both lists.
xs = [(3, 0), (4, 4), (5, 3), (7, 2), ( 9, 5), (15, 1)]
ys = [(0, 3), (4, 1), (7, 0), (7, 4), (11, 2), (12, 5)]
Create an array, y_positions. The nth element of the array contains the current index of the y element that was originally at index n.
Create an empty index_list.
For each element of xs, get the original_index, the second pair of the tuple.
Use y_positions to retrieve the current index of the y element with the given original_index. Add the current index to index_list.
Finally, remove the index values from xs and ys.
Here's a sample Python implementation.
points = ((3, 7), (15, 4), (7, 11), (5, 0), (4, 7), (9, 12))
#generate unsorted lists
xs, ys = zip(*points)
#pair each element with its index
xs = zip(xs, range(len(xs)))
ys = zip(ys, range(len(xs)))
#sort
xs.sort()
ys.sort()
#generate the y positions list.
y_positions = [None] * len(ys)
for i in range(len(ys)):
original_index = ys[i][1]
y_positions[original_index] = i
#generate `index_list`
index_list = []
for x, original_index in xs:
index_list.append(y_positions[original_index])
#remove tuples from x and y lists
xs = zip(*xs)[0]
ys = zip(*ys)[0]
print "xs:", xs
print "ys:", ys
print "index list:", index_list
Output:
xs: (3, 4, 5, 7, 9, 15)
ys: (0, 4, 7, 7, 11, 12)
index list: [2, 3, 0, 4, 5, 1]
Generation of y_positions and index_list is O(n) time, so the complexity of the algorithm as a whole is dominated by the sorting step.
Thank you for the answers. For what it's worth, the solution I had was pretty similar to those outlined, but as j_random_hacker pointed out, there's no need for a map. It just struck me that this little problem seems more complicated than it appears at first glance and I was wondering if I was missing something obvious. I've rehashed my solution into Python for comparison.
points = ((3, 7), (15, 4), (7, 11), (5, 0), (4, 7), (9, 12))
N = len(points)
# Separate the points into their x and y components, tag the values with
# their index into the points list.
# Sort both resulting (value, tag) lists and then unzip them into lists of
# sorted x and y values and the tag information.
xs, s = zip(*sorted(zip([x for (x, y) in points], range(N))))
ys, r = zip(*sorted(zip([y for (x, y) in points], range(N))))
# Generate the mapping list.
t = N * [0]
for i in range(N):
t[r[i]] = i
index_list = [t[j] for j in s]
print "xs:", xs
print "ys:", ys
print "index_list:", index_list
Output:
xs: (3, 4, 5, 7, 9, 15)
ys: (0, 4, 7, 7, 11, 12)
index_list: [2, 3, 0, 4, 5, 1]
I've just understood what j_random_hacker meant by removing a level of indirection by sorting the points in x initially. That allows things to be tidied up nicely. Thanks.
points = ((3, 7), (15, 4), (7, 11), (5, 0), (4, 7), (9, 12))
N = len(points)
ordered_by_x = sorted(points)
ordered_by_y = sorted(zip([y for (x, y) in ordered_by_x], range(N)))
index_list = N * [0]
for i, (y, k) in enumerate(ordered_by_y):
index_list[k] = i
xs = [x for (x, y) in ordered_by_x]
ys = [y for (y, k) in ordered_by_y]
print "xs:", xs
print "ys:", ys
print "index_list:", index_list
I'm working on a program for class that involves solving the Chinese Postman problem. Our assignment only requires us to write a program to solve it for a hard-coded graph but I'm attempting to solve it for the general case on my own.
The part that is giving me trouble is generating the partitions of pairings for the odd vertices.
For example, if I had the following labeled odd verticies in a graph:
1 2 3 4 5 6
I need to find all the possible pairings / partitions I can make with these vertices.
I've figured out I'll have i paritions given:
n = num of odd verticies
k = n / 2
i = ((2k)(2k-1)(2k-2)...(k+1))/2^n
So, given the 6 odd verticies above, we will know that we need to generate i = 15 partitions.
The 15 partions would look like:
1 2 3 4 5 6
1 2 3 5 4 6
1 2 3 6 4 5
...
1 6 ...
Then, for each partition, I take each pair and find the shortest distance between them and sum them for that partition. The partition with the total smallest distance between its pairs is selected, and I then double all the edges between the shortest path between the odd vertices (found in the selected partition).
These represent the edges the postman will have to walk twice.
At first I thought I had worked out an appropriate algorithm for generating these partitions:
Start with all the odd verticies sorted in increasing order
12 34 56
Select the pair behind the pair that currently has the max vertice
12 [34] 56
Increase the second digit in this pair by 1. Leave everything to the
left of the selected pair the same,
and make everything to the right of
the selected pair the remaining
numbers in the set, sorted in
increasing order.
12 35 46
Repeat
However, this is flawed. For example, I realized that when I reach to the end and the select pair is at the left most position (ie):
[16] .. ..
The algorithm I worked out will stop in this case, and not generate the rest of the pairs that begin [16], because there is no pair to the left of it to alter.
So, it is back to the drawing board.
Does anyone who has studied this problem before have any tips that can help point me in the right direction for generating these partitions?
You can construct the partitions using a recursive algorithm.
Take the lowest node, in this case node 1. This must be paired with one of the other unpaired nodes (2 to 6). For each of these nodes, create with match 1, then find all of the pairs of the remaining 4 elements using the same algorithm on the remaining four elements.
In Python:
def get_pairs(s):
if not s: yield []
else:
i = min(s)
for j in s - set([i]):
for r in get_pairs(s - set([i, j])):
yield [(i, j)] + r
for x in get_pairs(set([1,2,3,4,5,6])):
print x
This generates the following solutions:
[(1, 2), (3, 4), (5, 6)]
[(1, 2), (3, 5), (4, 6)]
[(1, 2), (3, 6), (4, 5)]
[(1, 3), (2, 4), (5, 6)]
[(1, 3), (2, 5), (4, 6)]
[(1, 3), (2, 6), (4, 5)]
[(1, 4), (2, 3), (5, 6)]
[(1, 4), (2, 5), (3, 6)]
[(1, 4), (2, 6), (3, 5)]
[(1, 5), (2, 3), (4, 6)]
[(1, 5), (2, 4), (3, 6)]
[(1, 5), (2, 6), (3, 4)]
[(1, 6), (2, 3), (4, 5)]
[(1, 6), (2, 4), (3, 5)]
[(1, 6), (2, 5), (3, 4)]