Symmetry unique atom coordinates

One thing I love about Mathematica is how easily I can go from the name of a molecule to estimated coordinates of its atoms, with a command like

AtomList[Molecule[Entity["Chemical", "Toluene"]], All, {"AtomicNumber", "AtomCoordinates"}]

(although, oddly enough, "AtomCoordinates" does not appear in the "AtomList" documentation)

I can also easily get the point group:

Molecule[Entity["Chemical", "Toluene"]]["PointGroup"]

This is exciting because this is exactly the input I need to run GAMESS and do quantum chemistry calculations (starting with a geometry optimization, of course, since JM has informed me that these coordinates are heuristic guesses).

But, really, this is not exactly the input that I need: what I really need are coordinates of only the symmetry-unique atoms.

I don’t suppose there’s a way to get coordinates of symmetry-unique atoms, which I can use for GAMESS input? I know there’s some functions related to point group symmetry, but I haven’t thought of how to do it.

Algorithm for Estimating Number of Unique Monthly Visitors

Is there a way to estimate the number of unique monthly visitors to a site based on a limited sample of one week of data? I have information about when a given user visited the site. This isn’t as simple as just multiplying the number of unique visitors the first week by 4, due to the hotel problem. If 10 people visit your site the first week and the same people are the only visitors to your site the second, third, and fourth week, the total number of monthly unique visitors to your site is only 10.

I know you can use HLL to estimate the number of unique visitors to a site in O(1) space. I’m wondering if there’s a similar approach to estimate how many unique visitors there will be after a month, preferably that also works in O(1) space.

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What are the “unique magic abilities” that draconians get instead of a breath weapon?

I like dragonborn, but I don’t really like the breath weapon that much. You can’t use it very often, and it doesn’t seem to scale very well at all, so it’s hard to imagine it having much relevance to a character after the first few levels. (And since I’m looking to play a warlock, eldritch blast says “why bother?”)

But I note a sidebar on page 34 of Player’s Handbook (or this section of the basic rules) about the draconians of Dragonlance, and it includes this tempting note:

In place of their draconic breath weapons, [draconians] have unique magical abilities.

What are those abilities? Where are they detailed? Certainly doesn’t seem to be in Player’s Handbook.

Count Unique Subsequences to Destination?

I am looking at this post:

Jamie is walking along a number line that starts at point 0 and ends at point n. She can move either one step to the left or one step to the right of her current location , with the exception that she cannot move left from point 0 or right from point n.

In other words, if Jamie is standing at point i,she can move to either i-1 or i+1 as long as her destination exists in the inclusive range [0,n]. She has a string, s, of movement instruction consisting of the letters l and r, where l is an instruction to move one step left and r is an instruction to move one step right. Jamie followed the instructions in s one by one and in order. For example if s=‘rrlr’, she performs the following sequence of moves: one step right ->one step right ->one step left -> one step right. Jamie wants to move from point x to point y following some subsequence of string s instruction and wonders how many distinct possible subsequence of string s will get her from point x to point y. Recall that a subsequence of a string is obtained by deleting zero or more characters from string.

It has four parameters

  • A String , s giving a sequence of moves using the characters l( i.e. move left one unit ) and r (i.e. move right one unit)
  • An integer n, denoting the length of the number line.
  • An integer x, denoting Jamie’s starting point on the number line
  • An integer y , denoting Jamie’s ending point on the number line.

The function must return an integer denoting the total number of distinct subsequence of strings that will lead Jamie from point x to point y as this value can be quite large.

Sample Input rrlrlr




output = 7

Let’s add few more constraints to simply the questions:

  • 1 <= length of s <= 10^3
  • 0 <= x, y < n <= 2500

I wonder what the runtime is for the proposed solution there:

int distinctSequences (int n, int a, int b, const std::string& actions) {     std::unordered_map<int, int> readyForR, readyForL;     auto res{0};      if (a > 0) {         readyForL.emplace(a, 1);     }      if (a < n) {         readyForR.emplace(a, 1);     }      std::unordered_map<int, int> nextReadyForR, nextReadyForL;      for (auto c: actions) {         nextReadyForR.clear();         nextReadyForL.clear();          if (c == 'r') {             for (auto [pos, count]: readyForR) {                 auto pp1{pos+1};                                  nextReadyForR.emplace(pp1, count);                 readyForL[pp1] += count;                  if (pp1 == b) res += count;             }              nextReadyForR.erase(n);             std::swap(readyForR, nextReadyForR);         } else {             for (auto [pos, count]: readyForL) {                 auto pm1{pos-1};                                  nextReadyForL.emplace(pm1, count);                 readyForR[pm1] += count;                  if (pm1 == b) res += count;             }              nextReadyForL.erase(0);             std::swap(readyForL, nextReadyForL);         }     }      return res; } 

What’s its runtime complexity?

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Given $n$ unique items and an $m^{th}$ normalised value, compute $m^{th}$ permutation without factorial expansion

We know that the number of permutations possible for $ n$ unique items is $ n!$ . We can uniquely label each permutation with a number from $ 0$ to $ (n!-1)$ .

Suppose if $ n=4$ , the possible permutations with their labels are,

0:  1234 1:  1243 2:  1324 3:  1342 4:  1432 5:  1423 6:  2134 7:  2143 8:  2314 9:  2341 10: 2431 11: 2413 12: 3214 13: 3241 14: 3124 15: 3142 16: 3412 17: 3421 18: 4231 19: 4213 20: 4321 21: 4312 22: 4132 23: 4123 

With any well defined labelling scheme, given a number $ m, 0 \leq m < n!$ , we can get back the permutation sequence. Further, these labels can be normalised to be between $ 0$ and $ 1$ . The above labels can be transformed into,

0:       1234 0.0434:  1243 0.0869:  1324 0.1304:  1342 0.1739:  1432 0.2173:  1423 0.2608:  2134 0.3043:  2143 0.3478:  2314 0.3913:  2341 0.4347:  2431 0.4782:  2413 0.5217:  3214 0.5652:  3241 0.6086:  3124 0.6521:  3142 0.6956:  3412 0.7391:  3421 0.7826:  4231 0.8260:  4213 0.8695:  4321 0.9130:  4312 0.9565:  4132 1:       4123 

Now, given $ n$ and $ m^{th}$ normalised label, can we get the $ m^{th}$ permutation while avoiding the expansion of $ n!$ ? For example, in the above set of permutations, if we were given the $ m^{th}$ normalised label to be $ 0.9$ , is it possible to get the closest sequence 4312 as the answer without computing $ 4!$ ?