Is there an efficient algorithm to generate a permutation from one index provided? The permutations do not need to have any specific ordering and it just needs to return every permutation once per every possible index. The set I wish to permute is all integers from 0~255.
If I understand the question correctly, the problem is as follows: You are given two integers n and k, and you want to find the kth permutation of n integers. You don't care about it being the kth lexicographical permutation, but it's just easier to be lexicographical so let's stick with that.
This is not too bad to compute. The base permutation is 1,2,3,4...n. This is the k=0 case. Consider what happens if you were to swap the 1 and 2: by moving the 1, you are passing up every single permutation where 1 goes first, and there are (n-1)! of those (since you could have permuted 2,3,4..n if you fixed the 1 in place). Thus, the algorithm is as follows:
for i from 1 to n:
j = k / (n-i)! // integer division, so rounded down
k -= j * (n-i)!
place down the jth unplaced number
This will iteratively produce the kth lexicographical permutation, since it repeatedly solves a sub-problem with a smaller set of numbers to place, and decrementing k along the way.
There is an implementation in python in module more-itertools: nth_permutation.
Here is an implementation, adapted from the code of more_itertools.nth_permutation:
from sympy import factorial
def nth_permutation(iterable, index):
pool = list(iterable)
n = len(pool)
c = factorial(n)
index = index % c
result = [0] * n
q = index
for d in range(1, n + 1):
q, i = divmod(q, d)
if 0 <= n - d < n:
result[n - d] = i
if q == 0:
break
return tuple(map(pool.pop, result))
print( nth_permutation(range(6), 360) )
# (3, 0, 1, 2, 4, 5)
Related
I would like to generate a sequence of n random integers in the interval [1,n] without duplicates, i.e. a permutation of the sequence [1,2,...,n] with O(log(n)) space complexity (or a polynomial function of log(n)).
One hint is that I can assume that I have a family of l-wise uniform hash functions h : [n] -> [k] (with l<=n) such that for any y_1, y_2,..., y_l and any distinct x_1, x_2,..., x_l :
P(h(x_1) = y_1 and h(x_2) = y_2 and ... and h(x_l) = y_l) = 1/(k^l)
My first idea was to use the hash function to generate the i-th element of the sequence, i.e. x_i = h(i) , check if x_i is already used (has already been returned by the hash function for some 0<j<i) and if it's the case increment x_i by 1 and check again until x_i is a new number. My problem is I can not have a vector of booleans of size n to check if the value x_i is already used. And if I do a recursive function to get the j-th value I will need at some point O(n log2(n)) bits...
I also found here that pseudorandom generator like Linear congruential generator can be used for this kind of problem with something like x_i+1 = (a*x_i + c)%n + 1 but I am not sure to understand how to choose a for any value of n to have a period of length n. In that case the hint is not really useful except for generating the first number of the sequence thus I don't think it's the right way.
Here's a fun super simple solution with constant space; when N is a power of 2 and your definition of "random" is incredibly loose (the resulting sequence will alternate between even and odd numbers).
N = power of 2
P = prime number larger than N.
S = random starting number between 0 and N-1
For i = 1 TO N
// add our prime to the starting random number
S += P
// S Modulus N
// Bitwise And N-1 works because N is a pow of 2
T = S & (N - 1)
//T is [0, (N-1)] => we want [1, N]
PRINT (T + 1)
Next I
JS
for(let N = 64, P = 73, S = N * Math.random(), i = 1; i <= N; i++) { S += P; console.log((S & (N - 1)) + 1); }
Another answer would probably be to consider all of the numbers [1, N] as leaf nodes in a tree and your Log(N) space is the size of a the path through the tree. Your solution would be a function that permutes all N paths through the tree. The way you permute the paths in a pseudo random way would basically be a Linear Feedback Shift Register type generator that has a period grater than N.
https://www.maximintegrated.com/en/design/technical-documents/app-notes/4/4400.html
Say that there are two sorted lists: A and B.
The number of entries in A and B can vary. (They can be very small/huge. They can be similar to each other/significantly different).
What is the known to be the fastest algorithm for this functionality?
Can any one give me an idea or reference?
Assume that A has m elements and B has n elements, with m ≥ n. Information theoretically, the best we can do is
(m + n)!
lg -------- = n lg (m/n) + O(n)
m! n!
comparisons, since in order to verify an empty intersection, we essentially have to perform a sorted merge. We can get within a constant factor of this bound by iterating through B and keeping a "cursor" in A indicating the position at which the most recent element of B should be inserted to maintain sorted order. We use exponential search to advance the cursor, for a total cost that is on the order of
lg x_1 + lg x_2 + ... + lg x_n,
where x_1 + x_2 + ... + x_n = m + n is some integer partition of m. This sum is O(n lg (m/n)) by the concavity of lg.
I don't know if this is the fastest option but here's one that runs in O(n+m) where n and m are the sizes of your lists:
Loop over both lists until one of them is empty in the following way:
Advance by one on one list.
Advance on the other list until you find a value that is either equal or greater than the current value of the other list.
If it is equal, the element belongs to the intersection and you can append it to another list
If it is greater that the other element, advance on the other list until you find a value equal or greater than this value
as said, repeat this until one of the lists is empty
Here is a simple and tested Python implementation that uses bisect search to advance pointers of both lists.
It assumes both input lists are sorted and contain no duplicates.
import bisect
def compute_intersection_list(l1, l2):
# A is the smaller list
A, B = (l1, l2) if len(l1) < len(l2) else (l2, l1)
i = 0
j = 0
intersection_list = []
while i < len(A) and j < len(B):
if A[i] == B[j]:
intersection_list.append(A[i])
i += 1
j += 1
elif A[i] < B[j]:
i = bisect.bisect_left(A, B[j], lo=i+1)
else:
j = bisect.bisect_left(B, A[i], lo=j+1)
return intersection_list
# test on many random cases
import random
MM = 100 # max value
for _ in range(10000):
M1 = random.randint(0, MM) # random max value
N1 = random.randint(0, M1) # random number of values
M2 = random.randint(0, MM) # random max value
N2 = random.randint(0, M2) # random number of values
a = sorted(random.sample(range(M1), N1)) # sampling without replacement to have no duplicates
b = sorted(random.sample(range(M2), N2))
assert compute_intersection_list(a, b) == sorted(set(a).intersection(b))
Consider a binary sequence b of length N. Initially, all the bits are set to 0. We define a flip operation with 2 arguments, flip(L,R), such that:
All bits with indices between L and R are "flipped", meaning a bit with value 1 becomes a bit with value 0 and vice-versa. More exactly, for all i in range [L,R]: b[i] = !b[i].
Nothing happens to bits outside the specified range.
You are asked to determine the number of possible different sequences that can be obtained using exactly K flip operations modulo an arbitrary given number, let's call it MOD.
More specifically, each test contains on the first line a number T, the number of queries to be given. Then there are T queries, each one being of the form N, K, MOD with the meaning from above.
1 ≤ N, K ≤ 300 000
T ≤ 250
2 ≤ MOD ≤ 1 000 000 007
Sum of all N-s in a test is ≤ 600 000
time limit: 2 seconds
memory limit: 65536 kbytes
Example :
Input :
1
2 1 1000
Output :
3
Explanation :
There is a single query. The initial sequence is 00. We can do the following operations :
flip(1,1) ⇒ 10
flip(2,2) ⇒ 01
flip(1,2) ⇒ 11
So there are 3 possible sequences that can be generated using exactly 1 flip.
Some quick observations that I've made, although I'm not sure they are totally correct :
If K is big enough, that is if we have a big enough number of flips at our disposal, we should be able to obtain 2n sequences.
If K=1, then the result we're looking for is N(N+1)/2. It's also C(n,1)+C(n,2), where C is the binomial coefficient.
Currently trying a brute force approach to see if I can spot a rule of some kind. I think this is a sum of some binomial coefficients, but I'm not sure.
I've also come across a somewhat simpler variant of this problem, where the flip operation only flips a single specified bit. In that case, the result is
C(n,k)+C(n,k-2)+C(n,k-4)+...+C(n,(1 or 0)). Of course, there's the special case where k > n, but it's not a huge difference. Anyway, it's pretty easy to understand why that happens.I guess it's worth noting.
Here are a few ideas:
We may assume that no flip operation occurs twice (otherwise, we can assume that it did not happen). It does affect the number of operations, but I'll talk about it later.
We may assume that no two segments intersect. Indeed, if L1 < L2 < R1 < R2, we can just do the (L1, L2 - 1) and (R1 + 1, R2) flips instead. The case when one segment is inside the other is handled similarly.
We may also assume that no two segments touch each other. Otherwise, we can glue them together and reduce the number of operations.
These observations give the following formula for the number of different sequences one can obtain by flipping exactly k segments without "redundant" flips: C(n + 1, 2 * k) (we choose 2 * k ends of segments. They are always different. The left end is exclusive).
If we had perform no more than K flips, the answer would be sum for k = 0...K of C(n + 1, 2 * k)
Intuitively, it seems that its possible to transform the sequence of no more than K flips into a sequence of exactly K flips (for instance, we can flip the same segment two more times and add 2 operations. We can also split a segment of more than two elements into two segments and add one operation).
By running the brute force search (I know that it's not a real proof, but looks correct combined with the observations mentioned above) that the answer this sum minus 1 if n or k is equal to 1 and exactly the sum otherwise.
That is, the result is C(n + 1, 0) + C(n + 1, 2) + ... + C(n + 1, 2 * K) - d, where d = 1 if n = 1 or k = 1 and 0 otherwise.
Here is code I used to look for patterns running a brute force search and to verify that the formula is correct for small n and k:
reachable = set()
was = set()
def other(c):
"""
returns '1' if c == '0' and '0' otherwise
"""
return '0' if c == '1' else '1'
def flipped(s, l, r):
"""
Flips the [l, r] segment of the string s and returns the result
"""
res = s[:l]
for i in range(l, r + 1):
res += other(s[i])
res += s[r + 1:]
return res
def go(xs, k):
"""
Exhaustive search. was is used to speed up the search to avoid checking the
same string with the same number of remaining operations twice.
"""
p = (xs, k)
if p in was:
return
was.add(p)
if k == 0:
reachable.add(xs)
return
for l in range(len(xs)):
for r in range(l, len(xs)):
go(flipped(xs, l, r), k - 1)
def calc_naive(n, k):
"""
Counts the number of reachable sequences by running an exhaustive search
"""
xs = '0' * n
global reachable
global was
was = set()
reachable = set()
go(xs, k)
return len(reachable)
def fact(n):
return 1 if n == 0 else n * fact(n - 1)
def cnk(n, k):
if k > n:
return 0
return fact(n) // fact(k) // fact(n - k)
def solve(n, k):
"""
Uses the formula shown above to compute the answer
"""
res = 0
for i in range(k + 1):
res += cnk(n + 1, 2 * i)
if k == 1 or n == 1:
res -= 1
return res
if __name__ == '__main__':
# Checks that the formula gives the right answer for small values of n and k
for n in range(1, 11):
for k in range(1, 11):
assert calc_naive(n, k) == solve(n, k)
This solution is much better than the exhaustive search. For instance, it can run in O(N * K) time per test case if we compute the coefficients using Pascal's triangle. Unfortunately, it is not fast enough. I know how to solve it more efficiently for prime MOD (using Lucas' theorem), but O do not have a solution in general case.
Multiplicative modular inverses can't solve this problem immediately as k! or (n - k)! may not have an inverse modulo MOD.
Note: I assumed that C(n, m) is defined for all non-negative n and m and is equal to 0 if n < m.
I think I know how to solve it for an arbitrary MOD now.
Let's factorize the MOD into prime factors p1^a1 * p2^a2 * ... * pn^an. Now can solve this problem for each prime factor independently and combine the result using the Chinese remainder theorem.
Let's fix a prime p. Let's assume that p^a|MOD (that is, we need to get the result modulo p^a). We can precompute all p-free parts of the factorial and the maximum power of p that divides the factorial for all 0 <= n <= N in linear time using something like this:
powers = [0] * (N + 1)
p_free = [i for i in range(N + 1)]
p_free[0] = 1
for cur_p in powers of p <= N:
i = cur_p
while i < N:
powers[i] += 1
p_free[i] /= p
i += cur_p
Now the p-free part of the factorial is the product of p_free[i] for all i <= n and the power of p that divides n! is the prefix sum of the powers.
Now we can divide two factorials: the p-free part is coprime with p^a so it always has an inverse. The powers of p are just subtracted.
We're almost there. One more observation: we can precompute the inverses of p-free parts in linear time. Let's compute the inverse for the p-free part of N! using Euclid's algorithm. Now we can iterate over all i from N to 0. The inverse of the p-free part of i! is the inverse for i + 1 times p_free[i] (it's easy to prove it if we rewrite the inverse of the p-free part as a product using the fact that elements coprime with p^a form an abelian group under multiplication).
This algorithm runs in O(N * number_of_prime_factors + the time to solve the system using the Chinese remainder theorem + sqrt(MOD)) time per test case. Now it looks good enough.
You're on a good path with binomial-coefficients already. There are several factors to consider:
Think of your number as a binary-string of length n. Now we can create another array counting the number of times a bit will be flipped:
[0, 1, 0, 0, 1] number
[a, b, c, d, e] number of flips.
But even numbers of flips all lead to the same result and so do all odd numbers of flips. So basically the relevant part of the distribution can be represented %2
Logical next question: How many different combinations of even and odd values are available. We'll take care of the ordering later on, for now just assume the flipping-array is ordered descending for simplicity. We start of with k as the only flipping-number in the array. Now we want to add a flip. Since the whole flipping-array is used %2, we need to remove two from the value of k to achieve this and insert them into the array separately. E.g.:
[5, 0, 0, 0] mod 2 [1, 0, 0, 0]
[3, 1, 1, 0] [1, 1, 1, 0]
[4, 1, 0, 0] [0, 1, 0, 0]
As the last example shows (remember we're operating modulo 2 in the final result), moving a single 1 doesn't change the number of flips in the final outcome. Thus we always have to flip an even number bits in the flipping-array. If k is even, so will the number of flipped bits be and same applies vice versa, no matter what the value of n is.
So now the question is of course how many different ways of filling the array are available? For simplicity we'll start with mod 2 right away.
Obviously we start with 1 flipped bit, if k is odd, otherwise with 1. And we always add 2 flipped bits. We can continue with this until we either have flipped all n bits (or at least as many as we can flip)
v = (k % 2 == n % 2) ? n : n - 1
or we can't spread k further over the array.
v = k
Putting this together:
noOfAvailableFlips:
if k < n:
return k
else:
return (k % 2 == n % 2) ? n : n - 1
So far so well, there are always v / 2 flipping-arrays (mod 2) that differ by the number of flipped bits. Now we come to the next part permuting these arrays. This is just a simple permutation-function (permutation with repetition to be precise):
flipArrayNo(flippedbits):
return factorial(n) / (factorial(flippedbits) * factorial(n - flippedbits)
Putting it all together:
solutionsByFlipping(n, k):
res = 0
for i in [k % 2, noOfAvailableFlips(), step=2]:
res += flipArrayNo(i)
return res
This also shows that for sufficiently large numbers we can't obtain 2^n sequences for the simply reason that we can not arrange operations as we please. The number of flips that actually affect the outcome will always be either even or odd depending upon k. There's no way around this. The best result one can get is 2^(n-1) sequences.
For completeness, here's a dynamic program. It can deal easily with arbitrary modulo since it is based on sums, but unfortunately I haven't found a way to speed it beyond O(n * k).
Let a[n][k] be the number of binary strings of length n with k non-adjacent blocks of contiguous 1s that end in 1. Let b[n][k] be the number of binary strings of length n with k non-adjacent blocks of contiguous 1s that end in 0.
Then:
# we can append 1 to any arrangement of k non-adjacent blocks of contiguous 1's
# that ends in 1, or to any arrangement of (k-1) non-adjacent blocks of contiguous
# 1's that ends in 0:
a[n][k] = a[n - 1][k] + b[n - 1][k - 1]
# we can append 0 to any arrangement of k non-adjacent blocks of contiguous 1's
# that ends in either 0 or 1:
b[n][k] = b[n - 1][k] + a[n - 1][k]
# complete answer would be sum (a[n][i] + b[n][i]) for i = 0 to k
I wonder if the following observations might be useful: (1) a[n][k] and b[n][k] are zero when n < 2*k - 1, and (2) on the flip side, for values of k greater than ⌊(n + 1) / 2⌋ the overall answer seems to be identical.
Python code (full matrices are defined for simplicity, but I think only one row of each would actually be needed, space-wise, for a bottom-up method):
a = [[0] * 11 for i in range(0,11)]
b = [([1] + [0] * 10) for i in range(0,11)]
def f(n,k):
return fa(n,k) + fb(n,k)
def fa(n,k):
global a
if a[n][k] or n == 0 or k == 0:
return a[n][k]
elif n == 2*k - 1:
a[n][k] = 1
return 1
else:
a[n][k] = fb(n-1,k-1) + fa(n-1,k)
return a[n][k]
def fb(n,k):
global b
if b[n][k] or n == 0 or n == 2*k - 1:
return b[n][k]
else:
b[n][k] = fb(n-1,k) + fa(n-1,k)
return b[n][k]
def g(n,k):
return sum([f(n,i) for i in range(0,k+1)])
# example
print(g(10,10))
for i in range(0,11):
print(a[i])
print()
for i in range(0,11):
print(b[i])
Given an array A with N elements, I want to find the sum of minimum elements in all the possible contiguous sub-sequences of A. I know if N is small we can look for all possible sub sequences but as N is upto 10^5 what can be best way to find this sum?
Example: Let N=3 and A[1,2,3] then ans is 10 as Possible contiguous sub sequences {(1),(2),(3),(1,2),(1,2,3),(2,3)} so Sum of minimum elements = 1 + 2 + 3 + 1 + 1 + 2 = 10
Let's fix one element(a[i]). We want to know the position of the rightmost element smaller than this one located to the left from i(L). We also need to know the position of the leftmost element smaller than this one located to the right from i(R).
If we know L and R, we should add (i - L) * (R - i) * a[i] to the answer.
It is possible to precompute L and R for all i in linear time using a stack. Pseudo code:
s = new Stack
L = new int[n]
fill(L, -1)
for i <- 0 ... n - 1:
while !s.isEmpty() && s.top().first > a[i]:
s.pop()
if !s.isEmpty():
L[i] = s.top().second
s.push(pair(a[i], i))
We can reverse the array and run the same algorithm to find R.
How to deal with equal elements? Let's assume that a[i] is a pair <a[i], i>. All elements are distinct now.
The time complexity is O(n).
Here is a full pseudo code(I assume that int can hold any integer value here, you should
choose a feasible type to avoid an overflow in a real code. I also assume that all elements are distinct):
int[] getLeftSmallerElementPositions(int[] a):
s = new Stack
L = new int[n]
fill(L, -1)
for i <- 0 ... n - 1:
while !s.isEmpty() && s.top().first > a[i]:
s.pop()
if !s.isEmpty():
L[i] = s.top().second
s.push(pair(a[i], i))
return L
int[] getRightSmallerElementPositions(int[] a):
R = getLeftSmallerElementPositions(reversed(a))
for i <- 0 ... n - 1:
R[i] = n - 1 - R[i]
return reversed(R)
int findSum(int[] a):
L = getLeftSmallerElementPositions(a)
R = getRightSmallerElementPositions(a)
int res = 0
for i <- 0 ... n - 1:
res += (i - L[i]) * (R[i] - i) * a[i]
return res
If the list is sorted, you can consider all subsets for size 1, then 2, then 3, to N. The algorithm is initially somewhat inefficient, but an optimized version is below. Here's some pseudocode.
let A = {1, 2, 3}
let total_sum = 0
for set_size <- 1 to N
total_sum += sum(A[1:N-(set_size-1)])
First, sets with one element:{{1}, {2}, {3}}: sum each of the elements.
Then, sets of two element {{1, 2}, {2, 3}}: sum each element but the last.
Then, sets of three elements {{1, 2, 3}}: sum each element but the last two.
But this algorithm is inefficient. To optimize to O(n), multiply each ith element by N-i and sum (indexing from zero here). The intuition is that the first element is the minimum of N sets, the second element is the minimum of N-1 sets, etc.
I know it's not a python question, but sometimes code helps:
A = [1, 2, 3]
# This is [3, 2, 1]
scale = range(len(A), 0, -1)
# Take the element-wise product of the vectors, and sum
sum(a*b for (a,b) in zip(A, scale))
# Or just use the dot product
np.dot(A, scale)
I have a list of N positive numbers sorted in ascending order, L[0] to L[N-1].
I want to iterate over subsets of M distinct list elements (without replacement, order not important), 1 <= M <= N, sorted according to their partial sum. M is not fixed, the final result should consider all possible subsets.
I only want the K smallest subsets efficiently (ideally polynomial in K). The obvious algorithm of enumerating all subsets with M <= K is O(K!).
I can reduce the problem to subsets of fixed size M, by placing K iterators (1 <= M <= K) in a min-heap and having the master iterator operate on the heap root.
Essentially I need the Python function call:
sorted(itertools.combinations(L, M), key=sum)[:K]
... but efficient (N ~ 200, K ~ 30), should run in less than 1sec.
Example:
L = [1, 2, 5, 10, 11]
K = 8
answer = [(1,), (2,), (1,2), (5,), (1,5), (2,5), (1,2,5), (10,)]
Answer:
As David's answer shows, the important trick is that for a subset S to be outputted, all subsets of S must have been previously outputted, in particular the subsets where only 1 element has been removed. Thus, every time you output a subset, you can add all 1-element extensions of this subset for consideration (a maximum of K), and still be sure that the next outputted subset will be in the list of all considered subsets up to this point.
Fully working, more efficient Python function:
def sorted_subsets(L, K):
candidates = [(L[i], (i,)) for i in xrange(min(len(L), K))]
for j in xrange(K):
new = candidates.pop(0)
yield tuple(L[i] for i in new[1])
new_candidates = [(L[i] + new[0], (i,) + new[1]) for i in xrange(new[1][0])]
candidates = sorted(candidates + new_candidates)[:K-j-1]
UPDATE, found an O(K log K) algorithm.
This is similar to the trick above, but instead of adding all 1-element extensions with the elements added greater than the max of the subset, you consider only 2 extensions: one that adds max(S)+1, and the other one that shifts max(S) to max(S) + 1 (that would eventually generate all 1-element extensions to the right).
import heapq
def sorted_subsets_faster(L, K):
candidates = [(L[0], (0,))]
for j in xrange(K):
new = heapq.heappop(candidates)
yield tuple(L[i] for i in new[1])
i = new[1][-1]
if i+1 < len(L):
heapq.heappush(candidates, (new[0] + L[i+1], new[1] + (i+1,)))
heapq.heappush(candidates, (new[0] - L[i] + L[i+1], new[1][:-1] + (i+1,)))
From my benchmarks, it is faster for ALL values of K.
Also, it is not necessary to supply in advance the value of K, we can just iterate and stop whenever, without changing the efficiency of the algorithm. Also note that the number of candidates is bounded by K+1.
It might be possible to improve even further by using a priority deque (min-max heap) instead of a priority queue, but frankly I'm satisfied with this solution. I'd be interested in a linear algorithm though, or a proof that it's impossible.
Here's some rough Python-ish pseudo-code:
final = []
L = L[:K] # Anything after the first K is too big already
sorted_candidates = L[]
while len( final ) < K:
final.append( sorted_candidates[0] ) # We keep it sorted so the first option
# is always the smallest sum not
# already included
# If you just added a subset of size A, make a bunch of subsets of size A+1
expansion = [sorted_candidates[0].add( x )
for x in L and x not already included in sorted_candidates[0]]
# We're done with the first element, so remove it
sorted_candidates = sorted_candidates[1:]
# Now go through and build a new set of sorted candidates by getting the
# smallest possible ones from sorted_candidates and expansion
new_candidates = []
for i in range(K - len( final )):
if sum( expansion[0] ) < sum( sorted_candidates[0] ):
new_candidates.append( expansion[0] )
expansion = expansion[1:]
else:
new_candidates.append( sorted_candidates[0] )
sorted_candidates = sorted_candidates[1:]
sorted_candidates = new_candidates
We'll assume that you will do things like removing the first element of an array in an efficient way, so the only real work in the loop is in building expansion and in rebuilding sorted_candidates. Both of these have fewer than K steps, so as an upper bound, you're looking at a loop that is O(K) and that is run K times, so O(K^2) for the algorithm.