3.5nd dnd – What is the maximum number of uses of bardic music that a level 15 bard can have?

I am working on a (somewhat) functional version of the classic Bard / Barbarian hybrid. The lynchpin uses Casting of rage(Dragon n ° 310), which requires that it be a free action, citing Quicken Spell(PHB) as a potential option. To do this, I use Metamagic song(RoS) to feed an accelerated spell and use Fast metamagic(CM) to make it acceptable for spontaneous spell casting, and for other ways to reduce metamagic costs in order to reduce costs. Therefore, I have to have so many uses of Bardic Music a day.

How many bardic musical uses is it possible for a level 15 bard? Ideally, I would like it to be at least 60-70.

Why does it turn on the event scheduler on my maximum processor?

I've been searching the web for last week trying to figure this out.
I've exported my tables and events, reinstalled the latest version of MySQL and imported.
I tried to check the individual procedures.

My situation stops me completely, here it is:
I'm installing the latest version of MySQL 8.0 from the MySQL installer on Windows 10 x 64 bit, here is the slightly modified my.ini file: my.ini

Whenever I activate event_scheduler,
I see an undefined number of "event_worker" threads opening all which are stuck in the state "opening tables" and sleeping.
My processor reaches its maximum and slows down all my PC.

In addition, I have 8 GB of RAM and an i7-3520M processor at 2.90 GHz

It does not appear, when I check in the complete list of processes, that something is frozen because of locks, which only leaves me to believe that this is part of the server configuration.

Ideas? Need additional information?

dnd 5e – What happens if the same attack that brings a Druid back to Wild Shape also reduces their maximum HP?

Simultaneous effects (XGtE 77):

Most of the effects of the game follow each other in a definite order.
by the rules or the DM. In rare cases, effects may occur at the same time
time, especially at the beginning or end of a creature's turn. If two or
more things happen at the same time on the turn of a character or a monster,
the person at the gaming table, whether it's a player or a DM, who controls that
the creature decides the order in which these things occur. For example,
if two effects occur at the end of the turn of a player character, the
The player decides which of the two effects occurs first.

The creature's controller decides the order of the effects on his turn and therefore which of the options is correct.

So, if this happens at the druid's turn, the druid chooses his option, otherwise the DM or the other controller decides to order.

Option 2 is not correct because the druid and the beast do not share an HP pool.

Compare with SA Compendium V2.3 on Tough (page 8), so that the Druid's or Beast's CV is reduced, it's a binary choice. (I'm using this label as PHB 67: "you assume the health of the beast and the dice." And "the statistics of your game are replaced by the statistics of the beast", which leaves room for a misinterpretation.)

Does the difficult gift have an effect on a druid when he is in the form of a beast? The intention is no. The difficult feat affects the hit points of a druid,
which are replaced by the health of the beast when using Wild Shape.

The simple reduction of HP can not result in a higher HP reduction unless the HP reduction is greater than the total HP amount of the Beast, in which case the beast and therefore the druid would die instead.

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linear algebra – Maximum determinant of matrix nxn with elements 0 and 1

Generalize https://math.stackexchange.com/questions/3265627/largest-value-of-a-third-order-determinant-whose-elements-are-0-or-1 I would like to propose two related issues

(a) find the maximum value of the determinant of a $ n times n $-matrix whose elements are $ 0 or $ 1 $ (Christianized a matrix 0-1 here).

(b) find all possible values ​​of the determinants of a matrix 0-1 of order $ n $.

I found the solutions for the little ones $ n = 2, …, $ 5 by "brute force" in Mathematica by simply listing the values ​​of the determinants as much as possible $ 2 ^ {n ^ 2} $ matrices.

On my PC, I could not go further $ n = $ 5 due to lack of memory.

I have two of them questions

1) Is anyone aware of an analytical solution?

2) Can you improve the Mathematica code that I will show soon in an automatic answer? (This part is intentionally delayed to allow the reader to find his own solutions).

probability – Distribution such that the maximum expected is close to the expected sum

For a distribution $ mathcal {D} $ on the non-negative reals with an expectation equal to 1, $ X_1 $, $ X_2 $, $ ldots $, $ X_n $ be $ n $ independent samples. If we want linked $ mathbb {E}[max_i X_i]$, a classic technique is to write $$ mathbb {E}[max_i X_i] mathbb {E} Big[sum_i X_i Big] = n. $$ In fact, we can also show the factor $ n $ here is tight; for example take $ mathcal {D} $ to be the distribution that is gaining value $ 1 / $ epsilon with probability $ epsilon $ and value $ 0 with probability $ 1- $ epsilon; as $ epsilon $ go to 0 we have that $ mathbb {E}[max_i X_i] $ tends to $ n $.

I'm interested to know if there is a good way to classify the possible $ mathcal {D} $ who satisfy this property up to the constants; specifically, we should have $$ mathbb {E}[max_i X_i] = Omega (n). $$ More complicated constructions for $ mathcal {D} $ would also be interesting.

javascript – Picking up a maximum of fruits in two baskets of at most 2 types of fruits


is from LeetCode

In a row of trees, the i-th tree produces fruits of type tree[i].
You start at the tree of your choice, and then make several times the
Following steps:

  1. Add a fruit from this tree to your baskets. If you can not, stop.
  2. Go to the next tree, to the right of the current tree. If there is no tree on the right, stop.

Note that you have no choice after the initial choice of
starting tree: you must execute step 1, then step 2, then go back to step
1, then step 2 and so on until you stop.

You have two baskets, and each basket can contain any amount of fruit,
but you want each basket to only carry one type of fruit. What is
the total amount of fruit that you can collect with this procedure?

Example 1

Contribution: [1,2,1]
Released: 3
// Explanation: We can collect [1,2,1].

Example 2

Contribution: [0,1,2,2]
Released: 3
// Explanation: We can collect [1,2,2].
// If we started at the first tree, we would only collect [0, 1].

Example 3

Contribution: [1,2,3,2,2]
Released: 4
// Explanation: We can collect [2,3,2,2].
// If we started at the first tree, we would only collect [1, 2].

Example 4

Contribution: [3,3,3,1,2,1,1,2,3,3,4]
Released: 5
// Explanation: We can collect [1,2,1,1,2].
// If we start from the first or the eighth tree, we would only harvest four fruits.


  1. 1 <= tree.length <= 40000
  2. 0 <= tree[i] <tree.length

My solution:

has the complexity of time and space of Sure). At first, I thought it was easy. But then, I got mixed up with the one of the test cases (i.e. I = [3,3,3,1,2,1,1,2,3,3,4];) and everything inside the other-block is a bit hacky after. Perhaps there is a more elegant solution to that.

/ **
* @param {number[]} tree
* @return {number}
* /
var totalFruit = function (tree) {
const set = new Set (tree);
if (set.size <= 2) {
return tree.length;

const fruits = new Set ();
let i = 0;
let j = 0;
leave max = 0;
let count = 0;
while (j <tree.length) {
if (fruits.size <= 2 &&! fruits.has (tree[j])) {
fruits.add (tree[j])

if (fruits.size <= 2) {
count ++;
max = Math.max (max, count);
j ++;
} other {
fruits.delete (tree[i])
const lastIndex = tree.slice (i, j - 1) .lastIndexOf (tree[i])
i + = lastIndex + 1;
count- = lastIndex + 1;
back max;

leave me = [1,2,1];
I = [0,1,2,2];
I = [3,3,3,1,2,1,1,2,3,3,4];
console.log (totalFruit (I));

Randomized algorithms – How can a maximum number of minimum cuts in a graph be exactly $ n choose $ 2?

According to my instructor, $ n choose $ 2 is the maximum number of minimum cuts that we can have on a graph. To prove it, he showed the lower limit using an n cycle chart. To prove the upper limit, he drew the argument of two facts:

  • Probability of finding $ i ^ {th} $ min cut $ geq frac {2} {n (n-1)} = frac {1} {n choose 2} $
  • Event to find $ i ^ {th} $ min cut is disjointed.

So, by adding the probabilities, he proved the upper limit of $ n choose $ 2.

Now, if we look at a tree, graphically, with $ n $ nodes, then we will be able to conclude $ (n-1) $ cuts min which is less than $ n choose $ 2 cuts ($ n geq3) $. Did I miss something?

javascript – Regex pattern for any integer between -20 and 100, including -20 and 100.Numeric 3 characters maximum and no decimal

Can any one please confirm if the regex below will work for this condition

Any integer between -20 and 100, including -20 and 100.Numeric 3 characters max. And without decimals - E.g - -20, -18, etc.

^ ((\ - ([1-9]| 1[0-9]| 20)) | ([0-9]|[1-8][0-9]| 9[0-9]| 100)) $

graphs – Resolution of the minimum edge coverage using the maximum matching algorithm

To resolve an edge coverage instance, we can use the maximum matching algorithm.

Edge cover: A chart's edge cover is a set of edges such that each vertex of the chart is incident on at least one edge of the set [from Wikipedia].

Maximum correspondence: a corresponding or independent edge defined in a graph is a vertex without common vertices [from Wikipedia].

For example, to find the minimum edge coverage of the example below, we can:

1- Find a maximum match.

2- The extension greedily so that all the summits are covered.

The image below shows this solution:

enter the description of the image here

My question is why this reduction works, is there any evidence of this result? or at least an intuition!

How can one be sure that the final solution is the minimal edge coverage of the chart and that there is no other edge coverage smaller than the calculated solution?