Related
I am exploring how a Dynamic Programming design approach relates to the underlying combinatorial properties of problems.
For this, I am looking at the canonical instance of the coin change problem: Let S = [d_1, d_2, ..., d_m] and n > 0 be a requested amount. In how many ways can we add up to n using nothing but the elements in S?
If we follow a Dynamic Programming approach to design an algorithm for this problem that would allow for a solution with polynomial complexity, we would start by looking at the problem and how it is related to smaller and simpler sub-problems. This would yield a recursive relation describing an inductive step representing the problem in terms of the solutions to its related subproblems. We can then implement either a memoization technique or a tabulation technique to efficiently implement this recursive relation in a top-down or a bottom-up manner, respectively.
A recursive relation to solve this instance of the problem could be the following (Python 3.6 syntax and 0-based indexing):
def C(S, m, n):
if n < 0:
return 0
if n == 0:
return 1
if m <= 0:
return 0
count_wout_high_coin = C(S, m - 1, n)
count_with_high_coin = C(S, m, n - S[m - 1])
return count_wout_high_coin + count_with_high_coin
This recursive relation yields a correct amount of solutions but disregarding the order. However, this relation:
def C(S, n):
if n < 0:
return 0
if n == 0:
return 1
return sum([C(S, n - coin) for coin in S])
yields a correct amount of solutions while regarding the order.
I am interested in capturing more subtle combinatorial patterns through a recursion relation that can be further optimized via memorization/tabulation.
For example, this relation:
def C(S, m, n, p):
if n < 0:
return 0
if n == 0 and not p:
return 1
if n == 0 and p:
return 0
if m == 0:
return 0
return C(S, m - 1, n, p) + C(S, m, n - S[n - 1], not p)
yields a solution disregarding order but counting only solutions with an even number of summands. The same relation can be modified to regard order and counting number of even number of summands:
def C(S, n, p):
if n < 0:
return 0
if n == 0 and not p:
return 1
if n == 0 and p:
return 0
return sum([C(S, n - coin, not p) for coin in S])
However, what if we have more than 1 person among which we want to split the coins? Say I want to split n among 2 persons s.t. each person gets the same number of coins, regardless of the total sum each gets. From the 14 solutions, only 7 include an even number of coins so that I can split them evenly. But I want to exclude redundant assignments of coins to each person. For example, 1 + 2 + 2 + 1 and 1 + 2 + 1 + 2 are different solutions when order matters, BUT they represent the same split of coins to two persons, i.e. person B would get 1 + 2 = 2 + 1. I am having a hard time coming up with a recursion to count splits in a non-redundant manner.
(Before I elaborate on a possible answer, let me just point out that counting the splits of the coin exchange, for even n, by sum rather than coin-count would be more or less trivial since we can count the number of ways to exchange n / 2 and multiply it by itself :)
Now, if you'd like to count splits of the coin exchange according to coin count, and exclude redundant assignments of coins to each person (for example, where splitting 1 + 2 + 2 + 1 into two equal size parts is only either (1,1) | (2,2), (2,2) | (1,1) or (1,2) | (1,2) and element order in each part does not matter), we could rely on your first enumeration of partitions where order is disregarded.
However, we would need to know the multiset of elements in each partition (or an aggregate of similar ones) in order to count the possibilities of dividing them in two. For example, to count the ways to split 1 + 2 + 2 + 1, we would first count how many of each coin we have:
def partitions_with_even_number_of_parts_as_multiset(n, coins):
results = []
def C(m, n, s, p):
if n < 0 or m <= 0:
return
if n == 0:
if not p:
results.append(s)
return
C(m - 1, n, s, p)
_s = s[:]
_s[m - 1] += 1
C(m, n - coins[m - 1], _s, not p)
C(len(coins), n, [0] * len(coins), False)
return results
Output:
=> partitions_with_even_number_of_parts_as_multiset(6, [1,2,6])
=> [[6, 0, 0], [2, 2, 0]]
^ ^ ^ ^ this one represents two 1's and two 2's
Now since we are counting the ways to choose half of these, we need to find the coefficient of x^2 in the polynomial multiplication
(x^2 + x + 1) * (x^2 + x + 1) = ... 3x^2 ...
which represents the three ways to choose two from the multiset count [2,2]:
2,0 => 1,1
0,2 => 2,2
1,1 => 1,2
In Python, we can use numpy.polymul to multiply polynomial coefficients. Then we lookup the appropriate coefficient in the result.
For example:
import numpy
def count_split_partitions_by_multiset_count(multiset):
coefficients = (multiset[0] + 1) * [1]
for i in xrange(1, len(multiset)):
coefficients = numpy.polymul(coefficients, (multiset[i] + 1) * [1])
return coefficients[ sum(multiset) / 2 ]
Output:
=> count_split_partitions_by_multiset_count([2,2,0])
=> 3
Here is a table implementation and a little elaboration on algrid's beautiful answer. This produces an answer for f(500, [1, 2, 6, 12, 24, 48, 60]) in about 2 seconds.
The simple declaration of C(n, k, S) = sum(C(n - s_i, k - 1, S[i:])) means adding all the ways to get to the current sum, n using k coins. Then if we split n into all ways it can be partitioned in two, we can just add all the ways each of those parts can be made from the same number, k, of coins.
The beauty of fixing the subset of coins we choose from to a diminishing list means that any arbitrary combination of coins will only be counted once - it will be counted in the calculation where the leftmost coin in the combination is the first coin in our diminishing subset (assuming we order them in the same way). For example, the arbitrary subset [6, 24, 48], taken from [1, 2, 6, 12, 24, 48, 60], would only be counted in the summation for the subset [6, 12, 24, 48, 60] since the next subset, [12, 24, 48, 60] would not include 6 and the previous subset [2, 6, 12, 24, 48, 60] has at least one 2 coin.
Python code (see it here; confirm here):
import time
def f(n, coins):
t0 = time.time()
min_coins = min(coins)
m = [[[0] * len(coins) for k in xrange(n / min_coins + 1)] for _n in xrange(n + 1)]
# Initialize base case
for i in xrange(len(coins)):
m[0][0][i] = 1
for i in xrange(len(coins)):
for _i in xrange(i + 1):
for _n in xrange(coins[_i], n + 1):
for k in xrange(1, _n / min_coins + 1):
m[_n][k][i] += m[_n - coins[_i]][k - 1][_i]
result = 0
for a in xrange(1, n + 1):
b = n - a
for k in xrange(1, n / min_coins + 1):
result = result + m[a][k][len(coins) - 1] * m[b][k][len(coins) - 1]
total_time = time.time() - t0
return (result, total_time)
print f(500, [1, 2, 6, 12, 24, 48, 60])
I studied math, and I come up with this problem.
There are two permutations A and B and a integer M.
We say A almost equal to B if we can make from A to B doing following operations.
-1 Choose a M-length segment of the permutation A.
-2 Perform a cyclic shift on it towards the right.(So,if sub segment is "1 2 3 4 5"(m=5), then after this operation , it will be "5 1 2 3 4".)
Question : Does A almost equal to B?
I think it is typical , but I couldn't find the answer.
How to solve it?(not brute force!)
number of elements in the permutation<=10^5
example
A "1 2 3"
B "2 3 1"
m=2
answer YES
explain "1 2 3"->"2 1 3"->"2 3 1"
Here's a proof of my conjecture. Let n be the length of the permutations and m be the length of the windows that we are allowed to rotate, where 1 ≤ m ≤ n. Permutations P and Q are almost equal if there exists a sequence of window rotations that transforms P into Q. Almost equality is an equivalence relation. Here's the claimed characterization of the equivalence classes.
(1) m = 1: P almost equals Q if and only if P = Q
(2) m = n: P almost equals Q if and only if they're rotations of each other
(3) 1 < m < n, m odd: P almost equals Q if and only if they have the same parity
(4) 1 < m < n, n even: P almost equals Q
The first two claims are obvious. As for (3), the necessity of the parity condition follows from the fact that rotating a window of odd length is an even permutation.
The meat of the argument here is to find an algorithm for when n = m + 1 ≥ 4, since in general, we can use an algorithm similar to insertion sort to transform P so that all but the last m + 1 elements match Q, and specifically, the case (n, m) = (3, 2) can be solved by inspection. In case m is even, we ensure further that the transformation matches the parity of Q, by rotating the last m elements once if necessary. (For m odd, we just assume equal parities.)
We need a technique for moving fewer than m elements at once. Suppose that the state is as follows.
1, 2, 3, 4, ..., m, m + 1
Rotate the second window m - 1 times (i.e., once in reverse).
1, 3, 4, ..., m, m + 1, 2
Rotate the first window m - 1 times.
3, 4, ..., m, m + 1, 1, 2
Second, twice.
3, 2, 4, ..., m, m + 1, 1
3, 1, 2, 4, ..., m, m + 1
We've succeeded in rotating the first three elements. This suffices in combination with conjugation by rotations to "insertion sort" the first m - 1 elements of Q into place. The other two are in the right order by the parity match.
I studied math, and I come up with this problem.
There are two permutations A and B and a integer M.
We say A almost equal to B if we can make from A to B doing following operations.
-1 Choose a M-length segment of the permutation A.
-2 Perform a cyclic shift on it towards the right.(So,if sub segment is "1 2 3 4 5"(m=5), then after this operation , it will be "5 1 2 3 4".)
Question : Does A almost equal to B?
I think it is typical , but I couldn't find the answer.
How to solve it?(not brute force!)
number of elements in the permutation<=10^5
example
A "1 2 3"
B "2 3 1"
m=2
answer YES
explain "1 2 3"->"2 1 3"->"2 3 1"
Here's a proof of my conjecture. Let n be the length of the permutations and m be the length of the windows that we are allowed to rotate, where 1 ≤ m ≤ n. Permutations P and Q are almost equal if there exists a sequence of window rotations that transforms P into Q. Almost equality is an equivalence relation. Here's the claimed characterization of the equivalence classes.
(1) m = 1: P almost equals Q if and only if P = Q
(2) m = n: P almost equals Q if and only if they're rotations of each other
(3) 1 < m < n, m odd: P almost equals Q if and only if they have the same parity
(4) 1 < m < n, n even: P almost equals Q
The first two claims are obvious. As for (3), the necessity of the parity condition follows from the fact that rotating a window of odd length is an even permutation.
The meat of the argument here is to find an algorithm for when n = m + 1 ≥ 4, since in general, we can use an algorithm similar to insertion sort to transform P so that all but the last m + 1 elements match Q, and specifically, the case (n, m) = (3, 2) can be solved by inspection. In case m is even, we ensure further that the transformation matches the parity of Q, by rotating the last m elements once if necessary. (For m odd, we just assume equal parities.)
We need a technique for moving fewer than m elements at once. Suppose that the state is as follows.
1, 2, 3, 4, ..., m, m + 1
Rotate the second window m - 1 times (i.e., once in reverse).
1, 3, 4, ..., m, m + 1, 2
Rotate the first window m - 1 times.
3, 4, ..., m, m + 1, 1, 2
Second, twice.
3, 2, 4, ..., m, m + 1, 1
3, 1, 2, 4, ..., m, m + 1
We've succeeded in rotating the first three elements. This suffices in combination with conjugation by rotations to "insertion sort" the first m - 1 elements of Q into place. The other two are in the right order by the parity match.
I found this interview question, and I couldn't come up with an algorithm better than O(N^2 * P):
Given a vector of P natural numbers (1,2,3,...,P) and another vector of length N whose elements are from the first vector, find the longest subsequence in the second vector, such that all elements are uniformly distributed (have the same frequency).
Example : (1,2,3) and (1,2,1,3,2,1,3,1,2,3,1). The longest subsequence is in the interval [2,10], because it contains all the elements from the first sequence with the same frequency (1 appears three times, 2 three times, and 3 three times).
The time complexity should be O(N * P).
"Subsequence" usually means noncontiguous. I'm going to assume that you meant "sublist".
Here's an O(N P) algorithm assuming we can hash (assumption not needed; we can radix sort instead). Scan the array keeping a running total for each number. For your example,
1 2 3
--------
0 0 0
1
1 0 0
2
1 1 0
1
2 1 0
3
2 1 1
2
2 2 1
1
3 2 1
3
3 2 2
1
4 2 2
2
4 3 2
3
4 3 3
1
5 3 3
Now, normalize each row by subtracting the minimum element. The result is
0: 000
1: 100
2: 110
3: 210
4: 100
5: 110
6: 210
7: 100
8: 200
9: 210
10: 100
11: 200.
Prepare two hashes, mapping each row to the first index at which it appears and the last index at which it appears. Iterate through the keys and take the one with maximum last - first.
000: first is at 0, last is at 0
100: first is at 1, last is at 10
110: first is at 2, last is at 5
210: first is at 3, last is at 9
200: first is at 8, last is at 11
The best key is 100, since its sublist has length 9. The sublist is the (1+1)th element to the 10th.
This works because a sublist is balanced if and only if its first and last unnormalized histograms are the same up to adding a constant, which occurs if and only if the first and last normalized histograms are identical.
If the memory usage is not important, it's easy...
You can give the matrix dimensions N*p and save in column (i) the value corresponding to how many elements p is looking between (i) first element in the second vector...
After completing the matrix, you can search for column i that all of the elements in column i are not different. The maximum i is the answer.
With randomization, you can get it down to linear time. The idea is to replace each of the P values with a random integer, such that those integers sum to zero. Now look for two prefix sums that are equal. This allows some small chance of false positives, which we could remedy by checking our output.
In Python 2.7:
# input:
vec1 = [1, 2, 3]
P = len(vec1)
vec2 = [1, 2, 1, 3, 2, 1, 3, 1, 2, 3, 1]
N = len(vec2)
# Choose big enough integer B. For each k in vec1, choose
# a random mod-B remainder r[k], so their mod-B sum is 0.
# Any P-1 of these remainders are independent.
import random
B = N*N*N
r = dict((k, random.randint(0,B-1)) for k in vec1)
s = sum(r.values())%B
r[vec1[0]] = (r[vec1[0]]+B-s)%B
assert sum(r.values())%B == 0
# For 0<=i<=N, let vec3[i] be mod-B sum of r[vec2[j]], for j<i.
vec3 = [0] * (N+1)
for i in range(1,N+1):
vec3[i] = (vec3[i-1] + r[vec2[i-1]]) % B
# Find pair (i,j) so vec3[i]==vec3[j], and j-i is as large as possible.
# This is either a solution (subsequence vec2[i:j] is uniform) or a false
# positive. The expected number of false positives is < N*N/(2*B) < 1/N.
(i, j)=(0, 0)
first = {}
for k in range(N+1):
v = vec3[k]
if v in first:
if k-first[v] > j-i:
(i, j) = (first[v], k)
else:
first[v] = k
# output:
print "Found subsequence from", i, "(inclusive) to", j, "(exclusive):"
print vec2[i:j]
print "This is either uniform, or rarely, it is a false positive."
Here is an observation: you can't get a uniformly distributed sequence that is not a multiplication of P in length. This implies that you only have to check the sub-sequences of N that are P, 2P, 3P... long - (N/P)^2 such sequences.
You can get this down to O(N) time, with no dependence on P by enhancing uty's solution.
For each row, instead of storing the normalized counts of each element, store a hash of the normalized counts while only keeping the normalized counts for the current index. During each iteration, you need to first update the normalized counts, which has an amortized cost of O(1) if each decrement of a count is paid for when it is incremented. Next you recompute the hash. The key here is that the hash needs to be easily updatable following an increment or decrement of one of the elements of the tuple that is being hashed.
At least one way of doing this hashing efficiently, with good theoretical independence guarantees is shown in the answer to this question. Note that the O(lg P) cost for computing the exponential to determine the amount to add to the hash can be eliminated by precomputing the exponentials modulo the prime in with a total running time of O(P) for the precomputation, giving a total running time of O(N + P) = O(N).
I'm struggling to get some solution, but I have no idea for that.
RobotA and RobotB who choose a permutation of the N numbers to begin with. RobotA picks first, and they pick alternately. Each turn, robots only can pick any one remaining number from the permutation. When the remaining numbers form an increasing sequence, the game finishes. The robot who picked the last turn (after which the sequence becomes increasing) wins the game.
Assuming both play optimally , who wins?
Example 1:
The original sequence is 1 7 3.
RobotA wins by picking 7, after which the sequence is increasing 1 3.
Example 2:
The original sequence is 8 5 3 1 2.
RobotB wins by selecting the 2, preventing any increasing sequence.
Is there any known algorithm to solve that? Please give me any tips or ideas of where to look at would be really thankful!
Given a sequence w of distinct numbers, let N(w) be the length of w and let L(w) be the length of the longest increasing subsequence in w. For example, if
w = 3 5 8 1 4
then N(w) = 5 and L(w) = 3.
The game ends when L(w) = N(w), or, equivalently, N(w) - L(w) = 0.
Working the game backwards, if on RobotX's turn N(w) - L(w) = 1, then the optimal play is to remove the unique letter not in a longest increasing subsequence, thereby winning the game.
For example, if w = 1 7 3, then N(w) = 3 and L(w) = 2 with a longest increasing subsequence being 1 3. Removing the 7 results in an increasing sequence, ensuring that the player who removed the 7 wins.
Going back to the previous example, w = 3 5 8 1 4, if either 1 or 4 is removed, then for the resulting permutation u we have N(u) - L(u) = 1, so the player who removed the 1 or 4 will certainly lose to a competent opponent. However, any other play results in a victory since it forces the next player to move to a losing position. Here, the optimal play is to remove any of 3, 5, or 8, after which N(u) - L(u) = 2, but after the next move N(v) - L(v) = 1.
Further analysis along these lines should lead to an optimal strategy for either player.
The nearest mathematical game that I do know is the Monotone Sequence Game. In a monotonic sequence game, two players alternately choose elements of a sequence from some fixed ordered set (e.g. 1,2,...,N). The game ends when the resulting sequence contains either an ascending subsequence of length A or a descending one of length D. This game has its origins with a theorem of Erdos and Szekeres, and a nice exposition can be found in MONOTONIC SEQUENCE GAMES, and this slide presentation by Bruce Sagan is also a good reference.
If you want to know more about game theory in general, or these sorts of games in particular, then I strong recommend Winning Ways for Your Mathematical Plays by Berlekamp, Conway and Guy. Volume 3, I believe, addresses these sorts of games.
Looks like a Minimax problem.
I guess there is more fast solution for this task. I will think.
But I can give you an idea of solution with O(N! * N^2) complexity.
At first, note that picking number from N-permutation is equivalent to the following:
Pick number from N-permutation. Let's it was number X.
Reassign numbers using rule:
1 = 1
2 = 2
...
X-1 = X-1
X = Nothing, it's gone.
X+1 = X
...
N = N - 1
And you get permutation of N-1 numbers.
Example:
1 5 6 4 2 3
Pick 2
1 5 6 4 3
Reassign
1 4 5 3 2
Let's use this one as move, instead just picking. It's easy too see that games are equivalent, player A wins in this game for some permutation if and only if it wins in original.
Let's give a code to all permutations of N numbers, N-1 numbers, ... 2 numbers.
Define F(x) -> {0; 1} (where x is a permutation code) is function which is 1 when current
player wins, and 0 if current player fails. Easy to see F(1 2 .. K-1 K) = 0.
F(x) = 1 if there is at least on move which transforms x to y, and F(y) = 0.
F(x) = 0 if for any move which transforms x to y, F(y) = 1.
So you can use recursion with memorization to compute:
Boolean F(X)
{
Let K be length of permutation with code X.
if you already compute F for argument X return previously calculated result;
if X == 1 2 .. K return 0;
Boolean result = 0;
for i = 1 to K do
{
Y code of permutation get from X by picking number on position i.
if (F(y) == 0)
{
result = 1;
break;
}
}
Store result as F(X);
return result;
}
For each argument we will compute this function only once. There is 1! permutations of length 1, 2! permutations of length 2 .. N! permutations of length N. For permutation length K, we need to do O(K) operation to compute. So complexity will be O(1*1! + 2*2! + .. N*N!) =<= O(N! * N^2) = O(N! * N^2)
Here is Python code for Wisdom's Wind's algorithm. It prints out wins for RobotA.
import itertools
def moves(p):
if tuple(sorted(p)) == p:
return
for i in p:
yield tuple(j - (j > i) for j in p if j != i)
winning = set()
for n in range(6):
for p in itertools.permutations(range(n)):
if not winning.issuperset(moves(p)):
winning.add(p)
for p in sorted(winning, key=lambda q: (len(q), q)):
print(p)