I have an n-tuple, x = (x[0], .., x[n-1]) where each member of the tuple comes from a distinct, ordered set S[i] such that x[i] \in S[i]. The sets S[i] all have different cardinalities N[i]. I want to know how to generate the next tuple in lexical order given the sets S[i].
Example:
S[0] = {0,1,2}
S[1] = {1,2,3,4}
S[2] = {8,9,7}
x = {2,2,7}
xnext = {2,3,8}
xnextnext = {2,3,9}
etc
This doesn't have to be very efficient, just closed form in terms of the current tuple elements and the sets. If it's easier, it would be equivalent to think of the n-tuple as indices in the sets.
For a given tuple, you can map the elements of the tuple to their respective indices in each set of S, then try to "increment" the mixed-radix number represented by this tuple of indices. Then, take the incremented "number" and map it back to a tuple of elements. Here's a proof-of-concept in Python:
def next_tuple(x, S):
assert len(x) == len(S)
assert all(element in set_ for element, set_ in zip(x, S))
# compute the bases for our mixed-radix system
lengths = [len(set_) for set_ in S]
# convert tuple `x` to a mixed-radix number
indices = [set_.index(element) for element, set_ in zip(x, S)]
# try to increment, starting from the right
for k in reversed(range(len(indices))):
indices[k] += 1
if indices[k] == lengths[k]:
# case 1: we have a carry, rollover this index and continue
indices[k] = 0
else:
# case 2: no carry, map indices back to actual elements and return
return [set_[index] for index, set_ in zip(indices, S)]
# we have gone through each position and still have a carry.
# this means the "increment" operation overflowed, and there
# is no "next" tuple.
return None
S = [[0, 1, 2], [1, 2, 3, 4], [8, 9, 7]]
print("next tuple after {} is {}".format([2, 2, 7], next_tuple([2, 2, 7], S)))
print("all tuples in order:")
x = [0, 1, 8]
while x is not None:
print(x)
x = next_tuple(x, S)
As a final note, if you need to enumerate the entire cartesian product in order, it's simpler to use a direct algorithm rather than repeatedly using next_tuple which has to recompute indices every time.
I got it to work using this pseudocode:
# x = [2,2,7]
sets = [[0,1,2], [1,2,3,4], [8,9,7]]
def next_tuple(x):
for i, el in enumerate(x):
if(i < len(sets[i]) - 1):
x[i] = sets[i, sets[i].index(x[i])+1] // requires lists to have unique elements
return x
else :
x[i] = sets[i,0]
Basically you scan in a character from the tuple, and if it can be incremented, increment it. If not, set it to 0 and go to the next character.
Related
I am doing some stuff on leetcode and came up with solution it works fine but some cases.
Here is the problem itself:
But in case like this it doesn't:
It doesn't make sense how can I rotate elements if k is bigger than length of array.
If you have any idea how to improve this solution I would be grateful
class Solution:
def rotate(self, nums: List[int], k: int) -> None:
"""
Do not return anything, modify nums in-place instead.
"""
if len(nums) > k:
self.swap(nums, 0, len(nums)-1)
self.swap(nums, 0,k-1)
self.swap(nums, k, len(nums)-1)
def swap(self, nums, start, end):
while start < end:
nums[start], nums[end] = nums[end], nums[start]
start+=1
end-=1
In order to understand why this doesn't work for the cases where k is larger than the array length, let me try to explain some of the logic behind rotating by such values of k.
The modulo operator, % will be useful. For example, if an array is 5 long, and you want to rotate by 5, you end up with the same array. So technically, you'd optimally want to rotate by 0. This is where the % operator comes into play. 5 % 5 = 0. If we want to rotate an array length 5 by 7 spots, we would end up with the same thing as rotating the array by 2, and it turns out that 7 % 5 = 2. Do you see where I am going with this?
This also holds true if the value of k is less than the length of the array. Say we want to rotate an array length 5 by 3, we do 3 % 5 = 3.
So for any rotation of amount k and array length L, optimization rotation amount n is equivalent to n = k % L.
You should modify your code at the beginning of your rotate method to adjust the rotation amount:
k = k % L
and use this value to rotate the correct amount.
The fastest and cleanest solution by far and large is:
def rotate_right(items, shift):
shift = -shift % len(items)
return items[shift:] + items[:shift]
ll = [i + 1 for i in range(7)]
# [1, 2, 3, 4, 5, 6, 7]
rotate_right(ll, 3)
# [5, 6, 7, 1, 2, 3, 4]
rotate_right([1, 2], 3)
# [2, 1]
of course, short of using numpy.roll() or itertools.cycle().
I have a square matrix of indeterminate row & column length (assume rows and columns are equal as befits a square).
I've plotted out an example matrix as follows:
matrix = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9]
]
My goal is to get a sum from top-left to bottom-right of the diagonal values.
Obviously in this example, this is all i'll need:
diagsum = matrix[0][0]+matrix[1][1]+matrix[2][2]
#=> 15
I see the pattern where it's a +1 incremental for each row & column argument in the matrix, so the code i've developed for my matrix of indeterminate length (supplied as the argument to my method diagsum would preferably need to implement some sort of row_count method on my matrix argument.
If
arr = [[1,2,3],
[4,5,6],
[7,8,9]]
then:
require 'matrix'
Matrix[*arr].trace
#=> 15
This will sum the diagonal values.
matrix = []
matrix[0] = [1,2,3]
matrix[1] = [4,5,6]
matrix[2] = [7,8,9]
def diagsum(mat)
sum = 0
mat.each_with_index { |row,i| sum += row[i] }
sum
end
puts (diagsum matrix) # 15
Not clear what x is.
But assuming that it is the number of columns/rows, you have 0..x, while the index only goes up to x - 1. You should change it to 0...x.
You are assigning to variable i, whose scope is only in the block.
You are only using i once, perhaps intended to correspond to either row or column, but not both.
You are adding the indices instead of the corresponding elements.
each will return the receiver regardless of whatever you get in the blocks.
puts will return nil regardless of whatever you get.
Given the list of numbers
1 15 2 5 10
I need to obtain
1 2 5 10 15
The only operation I can do is "move the number X at position Y".
In the above example I only need to do "move the number 15 at position 5".
I would like to minimize the number of operations but I can't find/remember a classical algorithm for that, given the operation available.
Some background :
I'm interacting with an API for a kanban-like service.
I have about 600 cards and some actions on our bug-tracker can imply a reordering of these 600 cards in the kanban (multiple cards can move at the same time if the priority of a project is changed)
I can do it in 600 calls to the API but I'm trying to reduce that number as much as possible.
Lemma: The minimum number of (delete element, insert element) pairs you can perform to sort a list L (in increasing order) is:
Smin(L) = |L| - |LIC(L)|
Where LIC(L) is the Longest Increasing Subsequence.
Thus, you have to:
Establish the LIC of your list.
Remove the elements not in it and insert them back at the appropriate position (using binary search).
Proof:
By induction.
For a list of size 1, the longest increasing subsequence is of length... 1! The list is already sorted so the number of (del,ins) pairs required is
|L| - |LIC(L)| = 1 - 1 = 0
Now let Ln be a list of length n, 1 ≤ n. Let Ln+1 be the list obtained by adding an element en+1 to the left of Ln.
This element may or may not influence the Longest Increasing Subsequence. Let's try to see how...
Let in,1 and in,2 be the two first elements of LIC(Ln) (*):
If en+1 > in,2, then LIC(Ln+1) = LIC(Ln)
If en+1 ≤ in,1, then LIC(Ln+1) = en+1 || LIC(Ln)
Else, LIC(Ln+1) = LIC(Ln) - in,1 + en+1. We keep the LIC with the highest first element. This is done by removing in,1 from the LIC and replacing it with en+1.
In the first case, we delete en+1, we thus get to sort Ln. By the induction hypothesis, this require n (deletion, insertion) pairs. We then have to insert en+1 at the appropriate position. Thus:
S(Ln+1)min = 1 + S(Ln)min
S(Ln+1)min = 1 + n - |LIC(Ln)|
S(Ln+1)min = |Ln+1| - |LIC(Ln+1|
In the second case, we ignore en+1. We begin by deleting elements not in LIC(Ln). These elements have to be inserted again! There are
S(Ln)min = |Ln| - |LIC(Ln)|
such elements.
Now, we just have to take care and insert them in the right order (relatively to en+1). In the end, it requires:
S(Ln+1)min = |Ln| - |LIC(Ln)|
S(Ln+1)min = |Ln| + 1 - (|LIC(Ln)| + 1)
Since we have |LIC(Ln+1)| = |LIC(Ln)| + 1 and |Ln+1| = |Ln| + 1, we have in the end:
S(Ln+1)min = |Ln+1| - |LIC(Ln+1)|
The last case can be proved by considering the list L'n obtained by removing in,1 from Ln+1. In that case LIC(L'n) = LIC(Ln+1) and thus:
|LIC(L'n)| = |LIC(Ln)| (1)
From there, we can sort L'n (which takes |L'n| - |LIC(L'n| by the induction hypothesis. The previous equality (1) leads to the result.
(*): If LIC(Ln) < 2, then in,2 doesn't exist. Just ignore the comparisons with it. In that case, only case 2 and case 3 apply... The result is still valid
One possible solution is to find the longest increasing subsequence and move only elements that aren't inside it.
I can't prove it's optimal, but it is easy to prove it is correct and better than N swaps.
Here is a proof-of-concept in Python 2. I implemented it as a O(n2) algorithm, but I'm pretty sure it can be reduced to O(n log n).
from operator import itemgetter
def LIS(V):
T = [1]*(len(V))
P = [-1]*(len(V))
for i, v in enumerate(V):
for j in xrange(i-1, -1, -1):
if T[j]+1 > T[i] and V[j] <= V[i]:
T[i] = T[j] + 1
P[i] = j
i, _ = max(enumerate(T), key=itemgetter(1))
while i != -1:
yield i
i = P[i]
def complement(L, n):
for a, b in zip(L, L[1:]+[n]):
for i in range(a+1, b):
yield i
def find_moves(V):
n = len(V)
L = list(LIS(V))[::-1]
SV = sorted(range(n), key=lambda i:V[i])
moves = [(x, SV.index(x)) for x in complement(L, n)]
while len(moves):
a, b = moves.pop()
yield a, b
moves = [(x-(x>a)+(x>b), y) for x, y in moves]
def make_and_print_moves(V):
print 'Initial array:', V
for a, b in find_moves(V):
x = V.pop(a)
V.insert(b, x)
print 'Move {} to {}. Result: {}'.format(a, b, V)
print '***'
make_and_print_moves([1, 15, 2, 5, 10])
make_and_print_moves([4, 3, 2, 1])
make_and_print_moves([1, 2, 4, 3])
It outputs something like:
Initial array: [1, 15, 2, 5, 10]
Move 1 to 4. Result: [1, 2, 5, 10, 15]
***
Initial array: [4, 3, 2, 1]
Move 3 to 0. Result: [1, 4, 3, 2]
Move 3 to 1. Result: [1, 2, 4, 3]
Move 3 to 2. Result: [1, 2, 3, 4]
***
Initial array: [1, 2, 4, 3]
Move 3 to 2. Result: [1, 2, 3, 4]
***
I have an array of non-negative values. I want to build an array of values who's sum is 20 so that they are proportional to the first array.
This would be an easy problem, except that I want the proportional array to sum to exactly
20, compensating for any rounding error.
For example, the array
input = [400, 400, 0, 0, 100, 50, 50]
would yield
output = [8, 8, 0, 0, 2, 1, 1]
sum(output) = 20
However, most cases are going to have a lot of rounding errors, like
input = [3, 3, 3, 3, 3, 3, 18]
naively yields
output = [1, 1, 1, 1, 1, 1, 10]
sum(output) = 16 (ouch)
Is there a good way to apportion the output array so that it adds up to 20 every time?
There's a very simple answer to this question: I've done it many times. After each assignment into the new array, you reduce the values you're working with as follows:
Call the first array A, and the new, proportional array B (which starts out empty).
Call the sum of A elements T
Call the desired sum S.
For each element of the array (i) do the following:
a. B[i] = round(A[i] / T * S). (rounding to nearest integer, penny or whatever is required)
b. T = T - A[i]
c. S = S - B[i]
That's it! Easy to implement in any programming language or in a spreadsheet.
The solution is optimal in that the resulting array's elements will never be more than 1 away from their ideal, non-rounded values. Let's demonstrate with your example:
T = 36, S = 20. B[1] = round(A[1] / T * S) = 2. (ideally, 1.666....)
T = 33, S = 18. B[2] = round(A[2] / T * S) = 2. (ideally, 1.666....)
T = 30, S = 16. B[3] = round(A[3] / T * S) = 2. (ideally, 1.666....)
T = 27, S = 14. B[4] = round(A[4] / T * S) = 2. (ideally, 1.666....)
T = 24, S = 12. B[5] = round(A[5] / T * S) = 2. (ideally, 1.666....)
T = 21, S = 10. B[6] = round(A[6] / T * S) = 1. (ideally, 1.666....)
T = 18, S = 9. B[7] = round(A[7] / T * S) = 9. (ideally, 10)
Notice that comparing every value in B with it's ideal value in parentheses, the difference is never more than 1.
It's also interesting to note that rearranging the elements in the array can result in different corresponding values in the resulting array. I've found that arranging the elements in ascending order is best, because it results in the smallest average percentage difference between actual and ideal.
Your problem is similar to a proportional representation where you want to share N seats (in your case 20) among parties proportionnaly to the votes they obtain, in your case [3, 3, 3, 3, 3, 3, 18]
There are several methods used in different countries to handle the rounding problem. My code below uses the Hagenbach-Bischoff quota method used in Switzerland, which basically allocates the seats remaining after an integer division by (N+1) to parties which have the highest remainder:
def proportional(nseats,votes):
"""assign n seats proportionaly to votes using Hagenbach-Bischoff quota
:param nseats: int number of seats to assign
:param votes: iterable of int or float weighting each party
:result: list of ints seats allocated to each party
"""
quota=sum(votes)/(1.+nseats) #force float
frac=[vote/quota for vote in votes]
res=[int(f) for f in frac]
n=nseats-sum(res) #number of seats remaining to allocate
if n==0: return res #done
if n<0: return [min(x,nseats) for x in res] # see siamii's comment
#give the remaining seats to the n parties with the largest remainder
remainders=[ai-bi for ai,bi in zip(frac,res)]
limit=sorted(remainders,reverse=True)[n-1]
#n parties with remainter larger than limit get an extra seat
for i,r in enumerate(remainders):
if r>=limit:
res[i]+=1
n-=1 # attempt to handle perfect equality
if n==0: return res #done
raise #should never happen
However this method doesn't always give the same number of seats to parties with perfect equality as in your case:
proportional(20,[3, 3, 3, 3, 3, 3, 18])
[2,2,2,2,1,1,10]
You have set 3 incompatible requirements. An integer-valued array proportional to [1,1,1] cannot be made to sum to exactly 20. You must choose to break one of the "sum to exactly 20", "proportional to input", and "integer values" requirements.
If you choose to break the requirement for integer values, then use floating point or rational numbers. If you choose to break the exact sum requirement, then you've already solved the problem. Choosing to break proportionality is a little trickier. One approach you might take is to figure out how far off your sum is, and then distribute corrections randomly through the output array. For example, if your input is:
[1, 1, 1]
then you could first make it sum as well as possible while still being proportional:
[7, 7, 7]
and since 20 - (7+7+7) = -1, choose one element to decrement at random:
[7, 6, 7]
If the error was 4, you would choose four elements to increment.
A naïve solution that doesn't perform well, but will provide the right result...
Write an iterator that given an array with eight integers (candidate) and the input array, output the index of the element that is farthest away from being proportional to the others (pseudocode):
function next_index(candidate, input)
// Calculate weights
for i in 1 .. 8
w[i] = candidate[i] / input[i]
end for
// find the smallest weight
min = 0
min_index = 0
for i in 1 .. 8
if w[i] < min then
min = w[i]
min_index = i
end if
end for
return min_index
end function
Then just do this
result = [0, 0, 0, 0, 0, 0, 0, 0]
result[next_index(result, input)]++ for 1 .. 20
If there is no optimal solution, it'll skew towards the beginning of the array.
Using the approach above, you can reduce the number of iterations by rounding down (as you did in your example) and then just use the approach above to add what has been left out due to rounding errors:
result = <<approach using rounding down>>
while sum(result) < 20
result[next_index(result, input)]++
So the answers and comments above were helpful... particularly the decreasing sum comment from #Frederik.
The solution I came up with takes advantage of the fact that for an input array v, sum(v_i * 20) is divisible by sum(v). So for each value in v, I mulitply by 20 and divide by the sum. I keep the quotient, and accumulate the remainder. Whenever the accumulator is greater than sum(v), I add one to the value. That way I'm guaranteed that all the remainders get rolled into the results.
Is that legible? Here's the implementation in Python:
def proportion(values, total):
# set up by getting the sum of the values and starting
# with an empty result list and accumulator
sum_values = sum(values)
new_values = []
acc = 0
for v in values:
# for each value, find quotient and remainder
q, r = divmod(v * total, sum_values)
if acc + r < sum_values:
# if the accumlator plus remainder is too small, just add and move on
acc += r
else:
# we've accumulated enough to go over sum(values), so add 1 to result
if acc > r:
# add to previous
new_values[-1] += 1
else:
# add to current
q += 1
acc -= sum_values - r
# save the new value
new_values.append(q)
# accumulator is guaranteed to be zero at the end
print new_values, sum_values, acc
return new_values
(I added an enhancement that if the accumulator > remainder, I increment the previous value instead of the current value)
I need to make a random list of permutations. The elements can be anything but assume that they are the integers 0 through x-1. I want to make y lists, each containing z elements. The rules are that no list may contain the same element twice and that over all the lists, the number of times each elements is used is the same (or as close as possible). For instance, if my elements are 0,1,2,3, y is 6, and z is 2, then one possible solution is:
0,3
1,2
3,0
2,1
0,1
2,3
Each row has only unique elements and no element has been used more than 3 times. If y were 7, then 2 elements would be used 4 times, the rest 3.
This could be improved, but it seems to do the job (Python):
import math, random
def get_pool(items, y, z):
slots = y*z
use_each_times = slots/len(items)
exceptions = slots - use_each_times*len(items)
if (use_each_times > y or
exceptions > 0 and use_each_times+1 > y):
raise Exception("Impossible.")
pool = {}
for n in items:
pool[n] = use_each_times
for n in random.sample(items, exceptions):
pool[n] += 1
return pool
def rebalance(ret, pool, z):
max_item = None
max_times = None
for item, times in pool.items():
if times > max_times:
max_item = item
max_times = times
next, times = max_item, max_times
candidates = []
for i in range(len(ret)):
item = ret[i]
if next not in item:
candidates.append( (item, i) )
swap, swap_index = random.choice(candidates)
swapi = []
for i in range(len(swap)):
if swap[i] not in pool:
swapi.append( (swap[i], i) )
which, i = random.choice(swapi)
pool[next] -= 1
pool[swap[i]] = 1
swap[i] = next
ret[swap_index] = swap
def plist(items, y, z):
pool = get_pool(items, y, z)
ret = []
while len(pool.keys()) > 0:
while len(pool.keys()) < z:
rebalance(ret, pool, z)
selections = random.sample(pool.keys(), z)
for i in selections:
pool[i] -= 1
if pool[i] == 0:
del pool[i]
ret.append( selections )
return ret
print plist([0,1,2,3], 6, 2)
Ok, one way to approximate that:
1 - shuffle your list
2 - take the y first elements to form the next row
4 - repeat (2) as long as you have numbers in the list
5 - if you don't have enough numbers to finish the list, reshuffle the original list and take the missing elements, making sure you don't retake numbers.
6 - Start over at step (2) as long as you need rows
I think this should be as random as you can make it and will for sure follow your criteria. Plus, you have very little tests for duplicate elements.
First, you can always randomly sort the list in the end, so let's not worry about making "random permutations" (hard); and just worry about 1) making permutations (easy) and 2) randomizing them (easy).
If you want "truly" random groups, you have to accept that randomization by nature doesn't really allow for the constraint of "even distribution" of results -- you may get that or you may get a run of similar-looking ones. If you really want even distribution, first make the sets evenly distributed, and then randomize them as a group.
Do you have to use each element in the set x evenly? It's not clear from the rules that I couldn't just make the following interpretation:
Note the following: "over all the lists, the number of times each elements is used is the same (or as close as possible)"
Based on this criteria, and the rule that z < x*, I postulate that you can simply enumerate all the items over all the lists. So you automatically make y list of the items enumerated to position z. Your example doesn't fulfill the rule above as closely as my version will. Using your example of x={0,1,2,3} y=6 and z=2, I get:
0,1 0,1 0,1 0,1 0,1 0,1
Now I didn't use 2 or 3, but you didn't say I had to use them all. If I had to use them all and I don't care to be able to prove that I am "as close as possible" to even usage, I would just enumerate across all the items through the lists, like this:
0,1 2,3 0,1 2,3 0,1 2,3
Finally, suppose I really do have to use all the elements. To calculate how many times each element can repeat, I just take (y*z)/(count of x). That way, I don't have to sit and worry about how to divide up the items in the list. If there is a remainder, or the result is less than 1, then I know that I will not get an exact number of repeats, so in those cases, it doesn't much matter to try to waste computational energy to make it perfect. I contend that the fastest result is still to just enumerate as above, and use the calculation here to show why either a perfect result was or wasn't achieved. A fancy algorithm to extract from this calculation how many positions will be duplicates could be achieved, but "it's too long to fit here in the margin".
*Each list has the same z number of elements, so it will be impossible to make lists where z is greater than x and still fulfill the rule that no list may contain the same element twice. Therefore, this rule demands that z cannot be greater than x.
Based on new details in the comments, the solution may simply be an implementation of a standard random permutation generation algorithm. There is a lengthy discussion of random permutation generation algorithms here:
http://www.techuser.net/randpermgen.html
(From Google search: random permutation generation)
This works in Ruby:
# list is the elements to be permuted
# y is the number of results desired
# z is the number of elements per result
# equalizer keeps track of who got used how many times
def constrained_permutations list, y, z
list.uniq! # Never trust the user. We want no repetitions.
equalizer = {}
list.each { |element| equalizer[element] = 0 }
results = []
# Do this until we get as many results as desired
while results.size < y
pool = []
puts pool
least_used = equalizer.each_value.min
# Find how used the least used element was
while pool.size < z
# Do this until we have enough elements in this resultset
element = nil
while element.nil?
# If we run out of "least used elements", then we need to increment
# our definition of "least used" by 1 and keep going.
element = list.shuffle.find do |x|
!pool.include?(x) && equalizer[x] == least_used
end
least_used += 1 if element.nil?
end
equalizer[element] += 1
# This element has now been used one more time.
pool << element
end
results << pool
end
return results
end
Sample usage:
constrained_permutations [0,1,2,3,4,5,6], 6, 2
=> [[4, 0], [1, 3], [2, 5], [6, 0], [2, 5], [3, 6]]
constrained_permutations [0,1,2,3,4,5,6], 6, 2
=> [[4, 5], [6, 3], [0, 2], [1, 6], [5, 4], [3, 0]]
enter code here
http://en.wikipedia.org/wiki/Fisher-Yates_shuffle