list manipulation – How do you delete sublists of a desired permutation?

I have a list given as:

d = {{{A1,A2,T3}, {A4,T1,A2}, {T5,A1,A3}}, {{T1,T2,T3}, {A5,A1,A2},{A1,A2,T3}}}

I’m trying to delete sublists of the type:

{A,A,A},{A,T,T},{T,T,T}

that way I’m left with sublists consisting of only:

{A,A,T},{T,A,A},{A,T,A}

I’m also trying to keep the numbers in list “d” so my desired outcome would look something like:

dd={{{A1,A2,T3},{A4, T1, A2}, {T5, A1, A3}},{{A1,A2,T3}}}

I have tried the following:

DeleteCases[d,{A,A,A}|{T,T,T}|{A,T,T},{2}] but nothing seems to change. I think there’s an issue with my patterns in DeleteCases where I need to include additional information to exclude the numbers in list “d” and just delete the sublists based upon the criteria of the arrangement of the A’s and T’s.

coalgebras – This is a problem about infinite order permutation groups

That’s the problem :let A be an infinite set.let D be the subgroup of Permutation group of A which move only a finite number of elements of A and let A be the set of all elements x∈D such that acts as an even permutation on the finite set of points it moves.prove that A is an infinite simple group.
My confusion :Prompted me to prove that each pair of elements of D is in a finite simple group of D.I think this means that every element in the D is in a finite single group of D and I think the finite single group will be an alternating group.But isn’t there a odd permutation in D

number theory – Euler Factors in Permutation Representation of Galois Group

Let $$k$$ be a number field and $$K / mathbb Q$$ a Galois extension containing $$k$$, with Galois group $$G=operatorname{Gal}(K/mathbb Q)$$ and let $$G_k:=operatorname{Gal}(K/k)$$. Let $$chi$$ denote the character of the permutation representation $$rho$$ of $$G$$ in $$G/G_k$$. How do I show the identity:
$$det(I-p^{-s}rho(sigma_p)|V_{p, rho}) = prod_{mathfrak{p}|p} (1-N(mathfrak{p})^{-s})$$
for every integer prime $$p$$? Here the determinant on the left is the Euler factor at prime $$p$$ appearing in the definition of the Artin $$L$$-function of the character $$chi$$ and the product on the right is over all primes $$mathfrak{p}$$ of $$k$$ which lie over $$p$$.

Algorithm – Leetcode Time and Space Complexity Issue # 31. Next Permutation

The question is as follows: given a collection of different integers, it returns all possible permutations. Example:

Input: (1,2,3)
Output:
(
(1,2,3),
(1,3,2),
(2,1,3),
(2,3,1),
(3,1,2),
(3,2,1)
)

Here is my rollback solution for the problem, where I added the num_calls variable to track the number of times the backtrack The function is called recursively.

class Solution:
def permute(self, nums):
num_calls = ()

def backtrack(combo, rem):
if len(rem) == 0:
for i in range(len(rem)):
num_calls.append(1)
backtrack(combo + (rem(i)), rem(:i) + rem(i + 1:))

if len(nums) == 0:
return None
backtrack((), nums)
print(len(num_calls))

I can't understand any of the answers I've seen so far for the time and space complexity of this solution.

Some people say their worst case is O (n * n!), But looking at the language of num_calls does not verify this claim.

For example, for:

test = Solution()
print(test.permute((1, 2, 3)))

length of num_calls = 15, what! = n * n! = 3 * (3 * 2 * 1) = 18

, for:

test = Solution()
print(test.permute((1, 2, 3, 4)))

length of num_calls = 64, which! = n * n! = 4 * (4 * 3 * 2 * 1) = 96.

, and to:

test = Solution()
print(test.permute((1, 2, 3, 4, 5)))

length of num_calls = 325, which! = n * n! = 5 * (5 * 4 * 3 * 2 * 1) = 600

Can anyone explain this in a simplified and easily understandable way?

python – How to efficiently calculate the transposition of a * permutation matrix * in Python3?

The permutation matrix is ​​represented as a list of positive integers, plus zero. My first attempt is as follows, along with a print function to help evaluate the result.
I find it too complicated.
Note that a permutation matrix multiplication is simpler than an ordinary matrix multiplication.

def transpose(m):
"""Transpose a permutation matrix.
m: list of positive integers and zero."""
c = {}
idx = 0
for row in reversed(m):
c(-row) = idx
idx += 1

return list(map(
itemgetter(1), sorted(c.items(), reverse=False, key=itemgetter(0))
))

def print_binmatrix(m):
"""Print a permutation matrix 7x7.
m: list of positive integers and zero."""

for row in m:
print(format(2 ** row, '07b'), row)

set theory: a permutation group that induces a topologically transitive action without dense orbits in \$ omega ^ * \$

Leave $$G$$ be a subgroup of the permutation group $$S_ omega$$ of the infinite accounting set $$omega$$. Each bijection $$g in G$$ supports a unique extension of a homeomorphism $$bar g$$ of the Stone-Cech compaction $$beta omega$$ from $$omega$$. Homeomorphism $$bar g$$ induces a homeomorphism of the rest $$omega ^ * = beta omega setminus omega$$ of the Stone-Cech compaction. So we get continuous action from the group $$G$$ in the compact Hausdorff space $$omega ^ *$$. I am interested in the properties of the obtained dynamic system $$( omega ^ *, G)$$. Namely, I would like to know the answer to the following

Issue. Is there a subgroup $$G subseteq S_ omega$$ such that the dynamic system $$( omega ^ *, G)$$ it is topologically transitive (= each non-empty open set has a dense orbit) but it does not have a dense orbit.

An example of such a subgroup $$G$$ exists under the assumption $$mathrm {non} ( mathcal M) < mathfrak c$$. So the question really is about the situation at ZFC.

Observation. If a group $$G subseteq S_ omega$$ induces a topologically transitive action on $$omega ^ *$$, then $$G$$ it has great cardinality, namely $$| G | ge mathsf Sigma ge max { mathfrak b, mathfrak s, mathrm {cov} ( mathcal M) }$$. More information on the cardinal $$mathsf Sigma$$ can be found in this preprint.

nt.number theory – Labeling of points of the grid \$ mathbb N ^ k \$ by subsets according to some permutation group, upper limit "global" for the sets that could appear

Leave $$G le S_n$$ be a finite permutation group with generators $$g_1, ldots, g_k$$. We look at the action of $$G$$ in subsets
with $$A ^ g = { alpha ^ g: alpha in A }$$ for $$A subseteq Omega$$
Y $$g in G$$. Label the grid $$mathbb N ^ k$$ with subsets $$varphi: mathbb N ^ k to mathcal P ( {1, ldots, n })$$ according to the following scheme. Set $$varphi (0, ldots, 0) = {1 }$$ Y
begin {align} varphi (n_1, ldots, n_k) & = varphi ( min (n_1-1,0), n_2, ldots, n_k) ^ {g_1} \ & quad cup varphi (n_1, min (n_2-1,0), ldots, n_k) ^ {g_2} \ & qquad qquad vdots \ & quad cup varphi (n_1, n_2, ldots, min (n_ {k-1} -1,0), n_k) ^ {g_ {k-1}} \ & quad cup varphi (n_1, n_2, ldots, n_ {k-1}, min (n_k-1,0)) ^ {g_k}. end {align}
This means that we start with $${1 }$$ at the origin, and the label of any other point is the union of the action of the generators in the label of their predecessor points, that is, those points that are one unit less in a single coordinate. For example, yes $$G = S_3$$ with $$g_1 = (1 2)$$ Y $$g_2 = (1 2 3)$$
then $$varphi (0,0) = {1 }$$
begin {align} varphi (1,0) & = {2 } \ varphi (2,0) & = {1 } \ varphi (0,1) & = {2 } \ varphi (0.2) & = {3 } \ varphi (0,3) & = {1 } \ varphi (1,1) & = varphi (0,1) cup varphi (1,0) = {2 } \ varphi (1,2) & = {2,3 } end {align}
and so.

Now a row or column is stable at some value $$N$$, if new sets do not appear along this row or column of $$N$$ onwards, for example the $$j$$-th
the row would be stable in $$N$$ Yes $${ varphi (i, j): 0 le i le N } = { varphi (i, j): i ge 0 }$$. How $$mathcal P ( {1, ldots, n })$$ It is finite, each row or column will be stable from some point forward.

But we could choose a "global" $$N$$ so that each row or column is stable after this point, or put another way if we look for each row or column $$N$$ point to the right or up, each set that appears as labels between these rows or columns will appear between these first $$N$$ labels

This could be seen in the following argument, if we choose any row, for example, then the tags have the form that first singleton sets appear, then sets of some larger cardinality, in the worst case with two elements, then with three and soon. And if the size of the subsets does not increase, then it scrolls through some subsets of the same size. By this simple observation we see that after a lot $$binom {n} {1} + binom {n} {2} + ldots + binom {n} {n} = 2 ^ n$$ Steps we have seen each subset. The same reasoning applies to the columns,
Y $$N = 2 ^ n$$ it would be so "global" $$N$$.

But somehow I have the feeling that it could be done much better than $$2 ^ n$$, incorporating the structure of the cycle and the structure of the cycle that we have for action in the subsets. But somehow I can't find a better formula. So, could we find a better upper limit for the "global" $$N$$? Or at least some asymptotic?

I hope my explanation is clear, let me know if something is not clear!

Need solution for the permutation merge problem in Java

If a child is climbing a ladder with "n" steps, find the maximum number of possibilities (how many different ways the ladder can climb)

permutation sum

$$f = sum_ {i = 1} ^ nia_i space space n in N space and space a_i in (1,2, …, n) space and space a_i neq a_j space yes space i neq j$$
$$let space f_k space and f_l space are space the space values ​​ space of space two space permutations space p_k space and space p_l space on space a_i, can space we space try space that space f_k neq f_l?$$
Thanks a lot!!!

set theory: permutation numbers \$ mathfrak j_1 \$, \$ mathfrak j_2 \$, are equal to \$ mathrm {non} ( mathcal M) \$

Consider the following cardinal characteristics of the continuum:

$$mathfrak j_1: = min {| H |: H subset S_ omega ; wedge ; forall A in ( omega) ^ omega ; h exists in H$$ such that $$h (A) cap A$$ It's infinite$$}$$;

$$mathfrak j_2: = min {| H |: H subset S_ omega ; wedge ; A, B in ( omega) ^ omega ; h exists in H$$ such that $$h (A) cap B$$ It's infinite$$}$$.

here $$S_ omega$$ is the group permutation group $$omega$$Y $$( omega) ^ omega$$ is the family of infinite subsets of $$omega$$.

It can be shown that $$max { mathfrak b, mathfrak s } le mathfrak j_1 le mathfrak j_2 le mathrm {non} ( mathcal M)$$.

Problem 1 It is $$mathfrak j_1 = mathfrak j_2$$ in ZFC?

Problem 2 It is $$mathfrak j_2 = mathrm {non} ( mathcal M)$$ is it ZFC?

Observation. It can be shown that the cardinals $$mathfrak j_ {1}, mathfrak j_2$$ they are equal respectively to the cardinals $$mathfrak j_ {2: 1}, mathfrak j_ {2: 2}$$ defined in this problem MO). Therefore, problems 1 and 2 are equivalent to problems 0 and 1 of this MO publication. But perhaps the equivalent descriptions of these cardinals help in some way.