text:
Acta Phys. -Chim. Sin. ›› 2021, Vol. 37 ›› Issue (9): 2009049.

Special Issue: Fuel Cells

• ARTICLE •

### Enhanced Performance and Durability of High-Temperature Polymer Electrolyte Membrane Fuel Cell by Incorporating Covalent Organic Framework into Catalyst Layer

Liliang Tian1, Weiqi Zhang1, Zheng Xie1, Kai Peng1, Qiang Ma1, Qian Xu1, Sivakumar Pasupathi2, Huaneng Su1,*()

1. 1 Institute for Energy Research, Jiangsu University, Zhenjiang, 212013, Jiangsu Province, China
2 South African Institute for Advanced Materials Chemistry, University of the Western Cape, Bellville 7535, South Africa
• Received:2020-09-15 Accepted:2020-10-18 Published:2020-10-23
• Contact: Huaneng Su E-mail:suhuaneng@ujs.edu.cn
• About author:Huaneng Su, Email: suhuaneng@ujs.edu.cn; Tel.: +86-511-88799500
• Supported by:
the National Key Research and Development Program of China(2018YFE0121200);the National Natural Science Foundation of China(21676126);the Priority Academic Program Development (PAPD) of Jiangsu Higher Education Institutions

Abstract:

Proton exchange membrane fuel cells (PEMFCs) are considered one of the most promising technologies for efficient power generation in the 21st century. However, several challenges for the PEMFC power technology are associated with low operating temperature, such as complex water management and strict fuel purification. PEMFC operating at high temperature (HT, 100–200 ℃) has in recent years been recognized as a promising solution to meet these technical challenges. At present, HT-PEMFC based on phosphoric acid (PA)-doped polybenzimidazole (PBI) is considered to be the trend of PEMFC future development due to its good environmental tolerance and simplified water/ thermal management. In this HT-PEMFC system, the proton transfer in both catalyst layer (CL) and membrane relies on liquid PA. Thus, a proper amount of PA is required to impregnate the membrane and the CL in order to achieve good proton conductivities in an HT-PEMFC system. Therefore, reducing the loss of PA electrolyte in the membrane electrode assembly (MEA) is crucial to maintaining the good durability of HT-PEMFC. In this work, a Schiff base networks (SNW)-type covalent organic framework (COF) material is proposed as the CL additive to enhance the durability of HT-PEMFC. The well-defined porous structure and tailored functional groups endow the proposed COF network with not only excellent PA retention capacity but also good proton transfer ability, thus leading to the superior durability of the HT-PEMFC in an accelerated stress test (AST). After 100 h operation at heavy load (0.2 V) and high flow rate of air purge, the accumulative PA loss of the COF-based MEA was ~4.03 mg, which is almost an order of magnitude lower than that of the conventional MEA (~13.02 mg), consequently leading to a much lower degradation rate of current density (~0.304 mA·cm-2·h-1) than that of the conventional MEA (~1.01 mA·cm-2·h-1). Moreover, it was found that the electrode incorporating a proper amount (5%–10%, mass fraction) of the COF material possessed a higher electrochemical surface area (ECSA) and lower ohmic and charge transfer resistances, which further improved the performance of the HT-PEMFC. At the usual operating voltage of 0.6 V, the current density of the MEA containing 10% COF was up to 0.361 A·cm-2, which is ~30% higher than that of the conventional MEA at 150 ℃, H2/Air and ambient pressure. These results indicate that incorporating COF materials into CL is a promising strategy to enhance the performance and durability of HT-PEMFC.

MSC2000:

• O646


text:
This notebook contains an excerpt from the Python Programming and Numerical Methods - A Guide for Engineers and Scientists, the content is also available at Berkeley Python Numerical Methods.

The copyright of the book belongs to Elsevier. We also have this interactive book online for a better learning experience. The code is released under the MIT license. If you find this content useful, please consider supporting the work on Elsevier or Amazon!

# If-Else Statements¶

A branching statement, If-Else Statement, or If-Statement for short, is a code construct that executes blocks of code only if certain conditions are met. These conditions are represented as logical expressions. Let $$P$$, $$Q$$, and $$R$$ be some logical expressions in Python. The following shows an if-statement construction.

CONSTRUCTION: Simple If-Else Statement Syntax

if logical expression:
code block

The word “if” is a keyword. When Python sees an if-statement, it will determine if the associated logical expression is true. If it is true, then the code in code block will be executed. If it is false, then the code in the if-statement will not be executed. The way to read this is “If logical expression is true then do code block.”

When there are several conditions to consider you can include elif-statements; if you want a condition that covers any other case, then you can use an else statement.

Note! Python gives the same level of indentation to every line of code within a conditional statement.

CONSTRUCTION: Extended If-Else Statement Syntax

if logical expression P:
code block 1
elif logical expression Q:
code block 2
elif logical expression R:
code block 3
else:
code block 4

In the previous code, Python will first check if $$\textit{P}$$ is true. If $$\textit{P}$$ is true, then code block 1 will be executed, and then the $$\textit{if-statement}$$ will end. In other words, Python will not check the rest of the statements once it reaches a true statement. However, if $$\textit{P}$$ is false, then Python will check if $$\textit{Q}$$ is true. If $$\textit{Q}$$ is true, then code block 2 will be executed, and the if-statement will end. If it is false, then $$\textit{R}$$ will be executed, and so forth. If $$\textit{P}$$, $$\textit{Q}$$, and $$\textit{R}$$ are all false, then code block 4 will be executed. You can have any number of elif statements (or none) as long as there is at least one if-statement (the first statement). You do not need an else statement, but you can have at most one else statement. The logical expressions after the if and elif (i.e., such as P, Q, and R) will be referred to as conditional statements.

TRY IT! Write a function my_thermo_stat(temp, desired_temp). The return value of the function should be the string ‘Heat’ if temp is less than desired_temp minus 5 degrees, ‘AC’ if temp is more than the desired_temp plus 5, and ‘off’ otherwise.

def my_thermo_stat(temp, desired_temp):
"""
Changes the status of the thermostat based on
temperature and desired temperature
author
date
:type temp: Int
:type desiredTemp: Int
:rtype: String
"""
if temp < desired_temp - 5:
status = 'Heat'
elif temp > desired_temp + 5:
status = 'AC'
else:
status = 'off'
return status

status = my_thermo_stat(65,75)
print(status)

Heat

status = my_thermo_stat(75,65)
print(status)

AC

status = my_thermo_stat(65,63)
print(status)

off

EXAMPLE: What will be the value of y after the following script is executed?

x = 3
if x > 1:
y = 2
elif x > 2:
y = 4
else:
y = 0
print(y)

2

We can also insert more complicated conditional statements using logical operators.

EXAMPLE: What will be the value of y after the following code is executed?

x = 3
if x > 1 and x < 2:
y = 2
elif x > 2 and x < 4:
y = 4
else:
y = 0
print(y)

4

Note that if you want the logical statement a < x < b, this is two conditional statements, a < x and x < b. In Python, you can type a < x < b as well. For example:

x = 3
if 1 < x < 2:
y = 2
elif 2 < x < 4:
y = 4
else:
y = 0
print(y)

4

A statement is called nested if it is entirely contained within another statement of the same type as itself. For example, a nested if-statement is an if-statement that is entirely contained within a clause of another if-statement.

EXAMPLE: Think about what will happen when the following code is executed. What are all the possible outcomes based on the input values of x and y?

def my_nested_branching(x,y):
"""
Nested Branching Statement Example
author
date
:type x: Int
:type y: Int
:rtype: Int
"""
if x > 2:
if y < 2:
out = x + y
else:
out = x - y
else:
if y > 2:
out = x*y
else:
out = 0
return out

Note! As before, Python gives the same level of indentation to every line of code within a conditional statement. The nested if-statement have deeper indentation by increasing four white spaces. You will get an IndentationError if the indentation is not correct, as we saw in the defining of the functions.

import numpy as np

all([1, 1, 0])

False

There are many logical functions that are designed to help you build branching statements. For example, you can ask if a variable has a certain data type with function isinstance. There are also functions that can tell you information about arrays of logicals like any, which computes to true if any element in an array is true, and false otherwise, and all, which computes to true only if all the elements in an array are true.

Sometimes you may want to design your function to check the inputs of a function to ensure that your function will be used properly. For example, the function my_adder in the previous chapter expects doubles as input. If the user inputs a list or a string as one of the input variables, then the function will throw an error or have unexpected results. To prevent this, you can put a check to tell the user the function has not been used properly. This and other techniques for controlling errors are explored further in Chapter 10. For the moment, you only need to know that we could use the $$\texttt{raise}$$ statement with a $$\texttt{TypeError}$$ exception to stop a function’s execution and throw an error with a specific text.

EXAMPLE: Modify my_adder to throw out a warning if the user does not input numerical values. Try your function for non-numerical inputs to show that the check works. When a statement is too long, we can use ‘' symbol to break a line to multiple lines.

def my_adder(a, b, c):
"""
Calculate the sum of three numbers
author
date
"""

# Check for erroneous input
if not (isinstance(a, (int, float)) \
or isinstance(b, (int, float)) \
or isinstance(c, (int, float))):
raise TypeError('Inputs must be numbers.')
# Return output
return a + b + c

x = my_adder(1,2,3)
print(x)

6

x = my_adder('1','2','3')
print(x)

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-13-c3e353c636b0> in <module>
2 print(x)

10             or isinstance(b, (int, float)) \
11             or isinstance(c, (int, float))):
---> 12         raise TypeError('Inputs must be numbers.')
13     # Return output
14     return a + b + c

TypeError: Inputs must be numbers.

There is a large variety of erroneous inputs that your function may encounter from users, and it is unreasonable to expect that your function will catch them all. Therefore, unless otherwise stated, write your functions assuming the functions will be used properly.

The remainder of the section gives a few more examples of branching statements.

TRY IT! Write a function called is_odd that returns ‘odd’ if the input is odd and ‘even’ if it is even. You can assume that input will be a positive integer.

def is_odd(number):
"""
function returns 'odd' if the input is odd,
'even' otherwise
author
date
:type number: Int
:rtype: String
"""
# use modulo to check if the input is divisible by 2
if number % 2 == 0:
# if it is divisible by 2, then input is not odd
return 'even'
else:
return 'odd'

is_odd(11)

'odd'

is_odd(2)

'even'

TRY IT! Write a function called my_circ_calc that takes a numerical number, r, and a string, calc as input arguments. You may assume that r is positive, and that calc is either the string ‘area’ or ‘circumference’. The function my_circ_calc should compute the area of a circle with radius, r, if the string calc is ‘area’, and the circumference of a circle with radius, r, if calc is ‘circumference’.

np.pi

3.141592653589793

def my_circ_calc(r, calc):
"""
Calculate various circle measurements
author
date
:type r: Int or Float
:type calc: String
:rtype: Int or Float
"""
if calc == 'area':
return np.pi*r**2
elif calc == 'circumference':
return 2*np.pi*r

my_circ_calc(2.5, 'area')

19.634954084936208

my_circ_calc(3, 'circumference')

18.84955592153876

Note! The function we write not only working on single value, but on Numpy arrays as well (that is the same operation will apply on each item of the array). See the following example, that we could calculate the circumferences for radius as [2, 3, 4] using a Numpy array.

my_circ_calc(np.array([2, 3, 4]), 'circumference')

array([12.56637061, 18.84955592, 25.13274123])

format: plain


text:
# Problem: Chromium crystallizes with a body-centered cubic unit cell. The radius of a chromium atom is 125 pm. Calculate the density of solid crystalline chromium in grams per cubic centimeter.

🤓 Based on our data, we think this question is relevant for Professor Wright's class at CCNY.

###### FREE Expert Solution

A body-centered cubic (BCC) unit cell is composed of a cube with one atom at each of its corners and one atom at the center of the cube

Recall that the edge length (a) of a BCC unit cell can be calculated using the equation:

$\overline{){\mathbf{a}}{\mathbf{=}}\frac{\mathbf{4}\mathbf{r}}{\sqrt{\mathbf{3}}}}$

Step 1: Calculate the volume of 1 unit cell using density

• molar mass Cr = 51.99 g/mol
• 1 mole = 6.022x1023 entities (Avogadro' number)
entities = atoms, ions, molecules, formula units

Recall the # of atoms present per 1 BCC unit cell: corner atoms contribute 1/8 and the center atom contribute 1

# of atoms = 2 per 1 unit cell

Solving volume of 1 unit cell:

volume of unit cell = 2.406 x 10-23 cm3

###### Problem Details

Chromium crystallizes with a body-centered cubic unit cell. The radius of a chromium atom is 125 pm. Calculate the density of solid crystalline chromium in grams per cubic centimeter.


text:
# Kakutani theorem

Let $X$ be a non-empty compact subset of $\mathbb{R}^{n}$, let $X^{*}$ be the set of its subsets, and let $f: X \to X^{*}$ be an upper semi-continuous mapping such that for each $x \in X$, the set $f(x)$ is non-empty, closed and convex. The theorem then states that $f$ has a fixed point (i.e., there is a point $x \in X$ such that $x \in f(x)$). S. Kakutani showed in [1] that from his theorem, the minimax principle for finite games does follow.


text:
# Electrical conductivity of organic solvents

Do organic solvents such as pure ethanol and pure methanol conduct electricity? why is this so?

I have done some research and have not been able to find anything conclusive. I would think it would not conduct because I cannot see any medium for the passage of charge.

$\ce{2H2O <=> H3O+ + OH-}$
For an alcohol $\ce{ROH}$,
$\ce{2ROH <=> ROH2+ +~ ^{-}OR}$
For water the self-ionization constant is $K_w \approx10^{-14}$. For ethanol or methanol, I was unable to find data on the self-ionization constant, which is essentially the same as the pKa of the alcohol in a solution of itself. The pKa of ethanol in water is slightly higher than the pKa of water in water. Also, the lower dielectric constant in alcohols will further disfavor self-ionization relative to water. So for methanol and ethanol the self-dissociation into ions will be weaker than in water.
These ions will still be present, though, and will conduct electricity even in pure ethanol and methanol solutions...but the conductivities will be very low. Very pure water has a resistivity of 18.2 $\text{M}\Omega\cdot \text{cm}$. If they self-ionize to a lesser extent than water does, the resistivities of methanol and ethanol will be even higher than water's.$%edit$


text:
Generalising Gelfand's spectral theory - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T13:43:51Z http://mathoverflow.net/feeds/question/79420 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/79420/generalising-gelfands-spectral-theory Generalising Gelfand's spectral theory Salvo Tringali 2011-10-28T17:59:03Z 2011-10-29T13:25:05Z
This is primarily a request for references and advices.

Question (edited on 10/29/2011). What's known about comprehensive generalisations of Gelfand's spectral theory for unital [associative] normed algebras [over the real or complex field] (*)?

Here, a generalisation should be meant as a framework, say, with the following distinctive features (among the others):

It should be founded on somehow different bases than the classical theory - especially to the extent that the notion itself of spectrum isn't any longer defined in terms of, and cannot be reduced to, the existence of any inverse in some unital algebra.

It should recover (at least basic) notions and results from the classical theory for unital Banach algebras in some appropriate incarnation (more details on this point are given below), for which the "generalised spectrum" does reproduce the classical one.

It should be unsensitive to completeness [under suitable mild hypotheses] in any setting where completeness is a well-defined notion (**), so yielding as a particular outcome that an element in a unital normed algebra, $\mathfrak{A}$, shares the same spectrum as its image in the Banach completion of $\mathfrak{A}$.

(*) If useful to know, my absolute reference here is the (let me say) wonderful book by Charles E. Rickart: General Theory of Banach Algebras (Academic Press, 1970).

(**) At least in principle, the kind of generalisation that I've in mind is tailored on the properties of topological vector spaces, though I've worked it out only in the restricted case of normed spaces.

Historical background.

As acknowledged by Jean Dieudonné in his History of Functional Analysis, the notion of spectrum (along with the foundation of modern spectral theory) was first introduced by David Hilbert in a series of articles inspired by Fredholm's celebrated work on integral equations (the word spectrum seems to have been lent by Hilbert from an 1897 article by Wilhelm Wirtinger) in the effort of lifting properties and notions from matrix theory to the broader (and more abstract) framework of linear operators. Especially, this led Hilbert to the discovery of complete inner product spaces (what we call, today, Hilbert spaces just in his honour). In 1906, Hilbert himself extended his previous analysis and discovered the continuous spectrum (already present but not fully recognised in earlier work by George William Hill in connection with his own study of periodic Sturm-Liouville equations).

A few years later, Frigyes Riesz introduced the concept of an algebra of operators in a series of articles culminating in a 1913 book, where Riesz studied, among the other things, the algebra of bounded operators on the separable Hilbert space. In 1916 Riesz himself created the theory of what we call nowadays compact operators. Riesz's spectral theorem was the basis for the definitive discovery of the spectral theorem of self-adjoint (and more generally normal) operators, which was simultaneously accomplished by Marshall Stone and John von Neumann in 1929-1932.

The year 1932 is another important date in this story, as it saw the publication of the very first monography on operator theory, by Stefan Banach. The systematic work of Banach gave new impulse to the development of the field and almost surely influenced the later work of von Neumann on the theory of operator algebras (developed, partly with Francis Joseph Murray, in a series of articles starting from 1935). Then came the seminal work of Israil Gelfand (partly in collaboration with Georgi E. Shilov and Mark Naimark), who introduced Banach algebras (under the naming of normed rings) and elaborated the corresponding notion of spectrum starting with a 1941 article in Matematicheskii Sbornik.

Now, it is undoubtable that Gelfand's work has deeply influenced the subsequent developments of spectral theory (and, accordingly, functional analysis). Yet, as far as I can understand in my own small way, something is still missing. I mean, something which may still be done, on the one hand, to clean up some inherent "defects" (or better fragilities) of the classical theory and, on the other, to make it more abstract and, then, portable to different contexts.

Naïve stuff.

As I learned from an anonymous user on MO (here), the term spectrum, in operator theory as well as in the context of normed algebras, is seemingly derived from the Latin verb spècere ("to see"), from which the root spec- of the Latin word spectrum ("something that appears, that manifests itself, vision"). Furthermore, the suffix -trum in spec-trum may come from the Latin verb instruo (like in the English word "instrument", which follows in turn the Latin noun instrumentum). So, the classical (or, herein, Gelfand) spectrum may be really considered, even from an etymological perspective, as a tool to inspect (or get improved knowledge of) some properties. I like to think of it as a sort of magnifying glass; we can move it through the algebra, zoom in and out on its elements, and get local information about them and/or global information about the whole structure.

Now, taking in mind (some parts of) another thread on this board about "wrong" definitions in mathematics, we are likely to agree that the worth of a notion is also measured by its sharpness (let me be vague on this point for the moment). And the classical notion of spectrum is, in fact, so successful because it is sharp in an appropriate sense, to the extent that it reveals deep underlying connections, say, between the algebraic and topological structures of a complicated object such as a Banach algebra (which is definitely magic, at least in my view). On another hand, what struck my curiosity is the consideration that the same conclusion doesn't hold (not at least with the same consistency) if Banach algebras are replaced by arbitrary (i.e. possibly incomplete) normed algebras, where the spectrum of a given element, $\mathfrak{a}$, can be scattered through the whole complex plane (in the complete case, as it is well-known, the spectrum is bounded by the norm of $a$, and indeed compact). So the question is: Why does this happen? And my answer is: essentially because the classical notion of spectrum is too algebraic, though completeness can actually conceal its true nature and make us even forgetful of it, or at least convinced that it must not be really so algebraic (despite of its own definition!) if it can dialogue so well with the topological structure. Yes, any normed algebra can be isometrically embedded (as a dense subalgebra) into a Banach one, but I don't think this makes a difference in what I'm trying to say, and it does not seriously explain anything. Clearly enough, the problem stems from the general failure in the convergence of the Neumann series $\sum_{n=0}^\infty (k^{-1}\mathfrak{a})^n$ for $k$ an arbitrary scalar with modulus greater than the norm of $\mathfrak{a}$. And why this? Because the convergence of such a Neumann series follows from the cauchyness of its partial sums, which is not a sufficient condition to convergence as far as the algebra is incomplete. According to my humble opinion, this is something like a "bug" in the classical vision, but above all an opportunity for getting a better understanding of some facts.

Motivations.

In the end, my motivation for this long post is that I've seemingly developed (the basics of) something resembling a spectral theory for linear (possibly unbounded) operators between different normed spaces. To me, this stuff looks like a sharpening of the classical theory in that it removes some of its "defects" (including the one addressed above); and also as an abstraction since, on the one hand, it puts standard notions from the operator setting (such as the ones of eigenvalue, continuous spectrum, and approximate spectrum) on a somehow different ground (so possibly foreshadowing further generalisations) and, on the other, it recovers familiar results (such as the closeness, the boundedness, and the compactness of the spectrum as well as the fact that all the points in the boundary are approximate eigenvalues) as a special case (while revealing some (unexpected?) dependencies).

Then, I'd really like to know what has been already done in these respects before putting everything in an appropriate form, submitting the results to any reasonable journal, and being answered, possibly after several months, that I've just reinvented the wheel. It would be really frustrating... Yes, of course, I've already asked here around (in Paris), but I've got nothing more concrete than contrasting (i.e. negative and positive) feelings. Also, I was suggested to contact a few people, and I've done it with one of them some weeks ago (sending him something like a ten page summary after checking his availability by an earlier email), but I've got no reply so far and indeed he seems to have disappeared... Then, I resolved to come here and consult the "oracle of MO" (as I enjoy calling this astounding place). :-)

Thank you in advance for any help.
 http://mathoverflow.net/questions/79420/generalising-gelfands-spectral-theory/79450#79450 Answer by Anatoly Kochubei for Generalising Gelfand's spectral theory Anatoly Kochubei 2011-10-29T05:06:32Z 2011-10-29T05:06:32Z
There exists an approach by Berezansky and Brasche to define a kind of generalized selfadjointness and develop some eigenfunction expansions for operators between different Hilbert spaces. This is done within the theory of rigged Hilbert spaces. Their motivation was apparently different from yours - they wanted to cover the case of Schroedinger operators with distribution potentials. Nevertheless there can be some connection. See

Yu.M. Berezansky and J. Brasche, Generalized selfadjoint operators and their singular perturbations. Methods Funct. Anal. Topol. 8, No. 4, 1-14 (2002);

Yu.M. Berezansky, J. Brasche, and L. P. Nizhnik, On generalized selfadjoint operators on scales of Hilbert spaces, Methods Funct. Anal. Topol. 17, No. 3, 193-196 (2011).

The second paper is available online: http://www.imath.kiev.ua/~mfat/


text:
Blog

# Expanding data from a frequency table into case form in R and STATA

I’ve been learning a bit of R this summer. R is a pretty powerful language and environment used for statistical computing. I chose it because it’s open-source (and therefore free!). I’ve been using RStudio in conjunction with R — RStudio has a more intuitive GUI that’s a little easier for me to use as a beginner.

As I mentioned in earlier posts, part of my project involves regression analysis. This type of analysis is usually done on case data. However, my datasets yielded frequency tables, and I found myself needing a way to expand these tables into case form.

The answer is a function called expand.dft that is part of the vcdExtra package for R. Downloading vcdExtra is simple, but keep in mind that you may need to install a few dependencies to get it to work correctly.

Likewise, using vcdExtra is pretty simple. Make sure you know which column in your raw data contains the frequency numbers, and name that column Frequency for simplicity’s sake.I started with a frequency table that looked a little like this:

Gender Enrollment Frequency
Male 0 50
Male 1 25
Female 0 50
Female 1 15

Here’s the code:

>FrequencyTable  read.csv(“rawdata.csv”, header = TRUE)>Library(vcdExtra)>CaseForm  expand.dft(FrequencyTable, freq=”Frequency”)

Easy peasy! The end result is a table of 140 rows (50+25+50+15), one for each person represented in the original frequency data.

Another note: I’m working with a large dataset that contains millions of points. Expanding that data to case form, then running a regression on it can be quite a task for a computer that doesn’t have a ton of RAM. So an alternative, if you are lucky enough to have access to STATA, is to simply open your dataset and type “expand Frequency” into the command area to automatically expand the data in your frequency column into instances of case data.

To check whether your data’s been expanded properly, simply ensure that the number of rows in your new dataset are equal to the sum of the cells in your frequency column.


text:
Keywords = degree pattern
##### OD-Characterization of almost simple groups related to $L_{3}(25)$

Volume 40, Issue 3, June 2014, Pages 765-790


text:
### The average of seven numbers is 18. If one of the numbers is 17 and if it is replaced by 31, then the average becomes :

A. 21.5 B. 19.5 C. 20 D. 21 Answer: Option C

### Solution(By Apex Team)

According to the question, Actual number = 17 New number = 31 Difference = 31 – 17 = 14 Difference ’14’ effects the seven numbers $=\Large\frac{14}{7}$=2 ∵ Present average = 18 ∴ New average = 18 + 2 = 20

A. 20
B. 21
C. 28
D. 32

A. 18
B. 20
C. 24
D. 30

A. 10 years
B. 10.5 years
C. 11 years
D. 12 years


text:
Regular and Chaotic Dynamics
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 Regul. Chaotic Dyn.: Year: Volume: Issue: Page: Find

 Regul. Chaotic Dyn., 2008, Volume 13, Issue 3, Pages 141–154 (Mi rcd566)

Gibbs Ensembles, Equidistribution of the Energy of Sympathetic Oscillators and Statistical Models of Thermostat

V. V. Kozlov

V. A. Steklov Institute of Mathematics, Russian Academy of Sciences, ul. Gubkina 8, Moscow, 119991 Russia

Abstract: The paper develops an approach to the proof of the "zeroth" law of thermodynamics. The approach is based on the analysis of weak limits of solutions to the Liouville equation as time grows infinitely. A class of linear oscillating systems is indicated for which the average energy becomes eventually uniformly distributed among the degrees of freedom for any initial probability density functions. An example of such systems are sympathetic pendulums. Conditions are found for nonlinear Hamiltonian systems with finite number of degrees of freedom to converge in a weak sense to the state where the mean energies of the interacting subsystems are the same. Some issues related to statistical models of the thermostat are discussed.

Keywords: Hamiltonian system, sympathetic oscillators, weak convergence, thermostat

DOI: https://doi.org/10.1134/S1560354708030015

Bibliographic databases:

MSC: 37A60, 60K35, 70H05, 82B30, 40A9
Accepted:28.10.2007
Language:

Citation: V. V. Kozlov, “Gibbs Ensembles, Equidistribution of the Energy of Sympathetic Oscillators and Statistical Models of Thermostat”, Regul. Chaotic Dyn., 13:3 (2008), 141–154

Citation in format AMSBIB
\Bibitem{Koz08} \by V.~V.~Kozlov \paper Gibbs Ensembles, Equidistribution of the Energy of Sympathetic Oscillators and Statistical Models of Thermostat \jour Regul. Chaotic Dyn. \yr 2008 \vol 13 \issue 3 \pages 141--154 \mathnet{http://mi.mathnet.ru/rcd566} \crossref{https://doi.org/10.1134/S1560354708030015} \mathscinet{http://www.ams.org/mathscinet-getitem?mr=2415369} \zmath{https://zbmath.org/?q=an:1229.82091}

• http://mi.mathnet.ru/eng/rcd566
• http://mi.mathnet.ru/eng/rcd/v13/i3/p141

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Related articles on Google Scholar: Russian articles, English articles

This publication is cited in the following articles:
1. Andrea Carati, Luigi Galgani, Alberto Maiocchi, Fabrizio Gangemi, Roberto Gangemi, “The FPU Problem as a Statistical-mechanical Counterpart of the KAM Problem, and Its Relevance for the Foundations of Physics”, Regul. Chaotic Dyn., 23:6 (2018), 704–719
2. A. L. Kuzemsky, “Nonequilibrium statistical operator method and generalized kinetic equations”, Theoret. and Math. Phys., 194:1 (2018), 30–56
3. Carati A. Galgani L. Gangemi F. Gangemi R., “Relaxation Times and Ergodic Properties in a Realistic Ionic-Crystal Model, and the Modern Form of the Fpu Problem”, Physica A, 532 (2019), 121911


text:
# How do you translate a graph to the left or right?

Oct 25, 2014

When translating a graph to the left or to the right, it means moving the entire graph to the left or to the right of its current location, which is usually relative to $\left(0 , 0\right)$

For example, let's say you have a circle centered at $\left(0 , 0\right)$ with radius 2. Its standard equation would be

${x}^{2} + {y}^{2} = 4$

Now, let's say we translate the circle 5 units to the left. Your circle will now be centered at (-5, 0) with the radius still equal to 2. Its new standard equation would be

${\left(x + 5\right)}^{2} + {y}^{2} = 0$

This is also true for linear equations.
For example, let's say you have a line with slope 1 with the x-intercept equal to 0. Its equation will be

$y = x$

When we translate the line 3 units to the right, its slope will remain the same, but its x-intercept will now be 3. Its new equation will be

$y = x - 5$


text:
# Ganyu

Main Page:

## 📄Ganyu

The secretary at Yuehai Pavilion. The blood of the qilin, an illuminated beast, flows within her veins.

## Basic Mechanics

### Aimed Shot Damage Dropoff

By: f̸̒͂ỏ̶̂o̵͌̚s̶͊̏h̷̤̀ḯ̴̊#9920
Last tested: 2022-10-23 (v. 3.1)
Discussion

Finding:
Aimed Shot damage and GU dropoff is based on the amount of time that the arrow travels in the air.

For the first 0.7s, there is no damage dropoff.
After 0.7s, every 0.05s:

1. The damage will reduce by 10%, for a maximum of a 90% reduction.
2. The Gauge Units will reduce by 10%, for a maximum of a 90% reduction.

Evidence:

1. Imgur: arrow travels for 55 frames (13 frames of dropoff), DMG and GU reduction should be 40%; DMG was reduced from 2006 to 1204, Cryo aura lasted for 8.5s which lines up with the formula (0.6×2.5+7).
2. Imgur: arrow travels for 50 frames (8 frames of dropoff), DMG and GU reduction is expected to be 20%; DMG reduced from 2319 to 1855, aura lasted for exactly 9 seconds (2.5*0.8+7).
3. Imgur: 63 frames travel time (21 frames of dropoff), DMG reduction is as expected 70% (2350 -> 705).
4. Imgur: arrow travels for 96 frames (54 frames of dropoff), DMG was reduced from 2350 to 235, aura lasted for 7s 19 frames which is roughly in line with 0.1GU's duration (2.5+0.1+7=7.25s), proof that DMG dropoff caps at 90%.

Significance:
Better understanding of Aimed Shot dropoff.

## Weapon Mechanics

### Bloom’s Innate Travel Time for Amos’ Bow

By: SavageCabbage#1445
Last tested: 2021-01-24 (v. 1.2)

Finding:

Bloom has an innate travel time of 0.3s with Amos’ bow. Currently no record of Amos’ bow’s passive dropping below 3 stacks for FFB.

With Ganyu’s C1, the 15% Cryo RES Shred is applied to the FFB of the first shot so long as FFA hits.

Evidence:

Tested with Ganyu, no artifact, R1 Amos, c6 and level 8 talent. consistent 889 DMG with 2090 Bloom in the first video. After the initial charged shot, the resulting Bloom DMG did not change at close range meaning that the Cryo shred was applied to the first Bloom. by running maths, we found out that the Bloom has 0.3 travel time. (204.8% x 1.12 x 0.9)/(348.16% x (1.12 + 0.24) x 1.025) = 889/2090.

Significance:

Ganyu synergises well with Amos’ Bow. C1 is slightly better due to the finding.

### Viridescent Hunt’s Passive Can Activate off of Bloom

By: Kidz#8667
Last tested: 2021-01-25 (v. 1.2)

Finding:

Regardless of whether both the initial and the Bloom of Ganyu’s Charged Attack hit the enemy, Viridescent Hunt’s passive has been found to be activatable by both instances. Additionally, because both the arrow and the Charged Attack can trigger Viridescent Hunt’s passive, the 50% chance of activation is rolled twice if both the arrow and the Bloom hit the enemy.

Evidence:

User Kidz#8557 tested for Viridescent Hunt triggers by purposefully missing the initial shot during testing in order to only hit the enemies with the Bloom, and sure enough despite the initial arrow completely missing, Ganyu’s Bloom still triggered Viridescent Hunt’s passive

Viridescent hunt passive Bloom test

Significance:

Due to the fact that Viridescent Hunt can trigger on both the Bloom or the arrow this means that on average Viridescent Hunt’s passive has a 75% chance at triggering with refinement 1 when users hit a target with both the Bloom and the arrow, meaning that Viridescent Hunt is theoretically more consistent on Ganyu than other characters.

### Specific Distances Required for Each Stack of Amos’ Buff

By: Shinku#8903
Last tested: 2021-01-26 (v. 1.2)

Finding:

The distance required in order to get an individual stack of the Amos’ passive buff of 8% DMG bonus per 0.1 seconds of travel time is approximately 5 in game meters per stack with the first stack applying at a distance of 3 in game meters. For reference, to get maximum stacks with the Amos bow, the distance is approximately half of Ganyu’s Elemental Burst in length.

Evidence:

User Shinku used the training dummy daily commission along with a world quest as a marker for distance, (starting at 544m away from said marker at point blank range), and subsequently tested the distance required for each stack by measuring his total distance away from the marker, and then tracking how many meters he had to travel before the next stack of Amos proc’d. He in turn found the following values.

[video] Amos bow stack range test

544m is our 0, 1769 is our damage without stacks.

547m damage changes to 1896, first stack (3m)

552m damage changes to 2022, second stack (8m)

557m damage changes to 2149, third stack (13m)

562m damage changes to 2275, fourth stack (18m)

567m damage changes to 2409, fifth stack (23m)

580 damage falloff gives 2161 (36m)

Significance:

These distance calculations will allow for players to have a better understanding of how much distance is required to get maximum stacks of Amos’ Bow’s passive, instead of having to just guess how far they need to be from their target.

## Charged Attack Mechanics

### Bloom Damage Scales with Charged Attack DMG Bonus

By: That’s so Poyo#6921
Last tested: 2021-01-24 (v. 1.2)

Finding:

The Bloom damage from Ganyu’s Frostflake Arrow is considered to be a Charged Attack along with the initial arrow itself, meaning that both the Bloom and the arrow are affected by Charged Attack multipliers such as the ones received from 4-Piece Wanderers Troupe, Bolide, and Heart of Depth.

Evidence:

User Poyo tested Ganyu’s Frostflake Arrow with the 4-Piece Wanderers Troupe set, then swapped out a piece of the set with similar sub stats to test if there would be a significant damage difference in Bloom damage between having the 35% charge ATK Bonus, and not. The following findings show that there is indeed a significant damage difference when equipping the 4-Piece Wanderer’s Troupe set, implying that the Bloom is considered a Charged Attack, and is in turn affected by Charged Attack multipliers.

Wanderer Troupe Bloom damage test

Significance:

This finding means that set bonuses that improve Charged Attacks such as Wanderers Troupe, Bolide, and Heart of Depth, will have a significant boost to Ganyu’s Bloom damage, implying that Ganyu will have significantly more potential options for viable sets then we initially thought.

### Ganyu Cannot Perform the ARCC Tech That All Other Bow Users Can Perform With Her Frostflake Arrow (with the exception of C6)

By: SavageCabbage#1445
Last tested: 2021-01-24 (v. 1.2)

Finding:

Ganyu is unable to perform the ARCC tech with her Frostflake Arrow meaning that she cannot benefit from the frame advantages that a majority of the bow users gain when using their respective charged shots.

Evidence:

User SavageCabbage tested out trying to perform the ARCC tech with Ganyu’s Frostflake Arrow, which resulted in the arrow reverting back to a non-Elemental aimed shot, meaning that it is not possible to perform the ARCC tech with Ganyu’s charged shot.

ARCC tech test

Significance:

With this knowledge we now know that there is no potential frame advantage from attempting to perform the ARCC tech with Ganyu’s charged attack, meaning that there is significantly less room for improvement for individual players in terms of animation cancels that they would need to learn.

### Frostflake Bloom Hitbox Range

By: Kidz#8667 SavageCabbage#1445 Pinupski#9999
Last tested: 2021-01-14 (v. 1.2)

Finding:

Damage from FFB is applied in a cylinder relative to the impact point of the FFA. Regardless of vertical distance from the impact point of the FFA the damage is applied to all units inside the AoE on the same frame. Relative to the animation, the AoE cylinder extends further down than the animation suggests and also slightly above the apex of the FFB animation.

Evidence:

Same frame damage regardless of vertical, same frame can be deduced from the opaqueness and size of the damage numbers: Image 1;

Upper limit of AoE and its radius shown by upper hilichurl: Image 2;

Upper limit of AoE and its radius shown by upper hilichurl: Image 3;

Lower Limit of AoE shown by bottom hilichurl: Image 4;

FFB Radius shown by hilichurl behind the 5: Image 5.

Significance:

Plots out the AoE of the FFB and show that the FFB damage is being applied to all mobs in the AoE at the same frame.

### Charged Shot Arrow Deviation When Aiming Near Objects

By: qwertyasdf007#9101
Last tested: 2021-02-08 (v. 1.3)

Finding:

Theory: Charged Shot originates at the bow and targets the frontmost object in the reticle line of sight.

Subtheory: Collision detection hitboxes are wonky for some objects seen in the video (e.g. thorns).

Evidence:

• First clip supports the fact that the charge shot originates at the bow.
• Second and third clip support the fact that the charge shot is aimed toward some object in the reticle line of sight.
• Fourth clip makes the distinction that the charge shot is aimed toward the frontmost object.

Bow Character Charge Shot Aim Deviation

Given that if the theory is correct, then the subtheory is supported by the fact that in clips 2-4 the charge shot is targeting something in front of the "first" object in the reticle line of sight. This seems to imply that the "aim shot" hitbox is different from the actual hitbox for collision detection.

Significance:

Some objects provide strange interactions when aiming charge shots. As the bow character is to the left of the reticle, the released charge shot will be:

• to the left of the reticle when the arrow is in front of the target;
• to the right of the reticle when the arrow is behind the target.

For moving targets, as they move toward you, the target is closer to the character than where the target was originally. In this case, the arrow will be to the left of the target, and potentially miss. It's better to aim toward the right edge of the target. Arrow deviation will be increased the closer the target is.

Special thanks to SavageCabbage#1445 for founding this theory and discussing its significance.

### Airspeed Velocity Of An Unladen Frostflake Arrow

By: Rare Possum#0511
Last tested: 2022-10-13 (v. 3.1)
Discussion

Finding:
Ganyu's Fully Charged Frostflake Arrow has an Airspeed Velocity of approximately 55 m/s.

Evidence:

Significance:
Amos Stacks.

## Skill Mechanics

### C0 Ganyu: 2 Simultaneous Ice Lotuses

By: Exuma#1917 and JenjenJL#6582
Last tested: 2021-06-17 (v. 1.6)
Discussion

Finding:
Two Ice Lotuses can exist simultaneously on C0 Ganyu.

Evidence: Video

Significance:
It is possible to spawn multiple Ice Lotuses without C2 Ganyu.

### Ganyu Elemental Skill has no ICD

By: TWICE#9958
Last tested: 2022-01-10 (v. 2.4)
Discussion

Finding:
Ganyu E does not have ICD.

Evidence:
YouTube - Ganyu equipped with Sacrificial Bow. Applied 2U of Pyro to the Fatui Agent using Bennett's Skill. Used Ganyu Skill twice using Sac Bow passive, both hits melted. First damage instance appears at frame 455, second at frame 530 => 1.25 seconds between damage, which is less than the standard 2.5 seconds.

Significance:
Provides evidence for "ICD: None" on TCL for Ganyu's Skill.

### Ganyu’s Trail of Qilin can taunt every enemy in the game that is not considered a boss monster

By: TiagoESC#6773
Last tested: 2021-01-13 (v. 1.2)

Finding:

Against just about every single normal enemy in the game, Ganyu’s Trail of Qilin (E) has been shown to properly taunt the various enemies. The enemies that were included in testing are: Abyss mage, Slime, Boar King, Cryo Cicin Mage, Electro Cicin Mage, Geovishap Hatchling, Ruin Guard, Ruin Hunter, Hili/Samachurl, Lawachurl, Mitachurl, Pyro Agent, Electro/Hydro Skirmishers, Pyro Skirmisher and (Cryo) Whopperflower.

Evidence:

User Tiago thoroughly tested Ganyu’s Trail of Qilin by going to each individual enemy in the game (except for eye of the storm and bosses) and activating Ganyu’s E ability to see if the taunt would work against said enemy. The result showed that against every single enemy that was tested against, Ganyu’s Trail of Qilin was able to taunt the enemy as intended.

Ganyu taunt test

Significance:

As Ganyu’s Trail of Qilin is capable of properly taunting just about every single enemy in the game it will prove to be a highly valuable asset in crowd control for Ganyu players, especially in Melt comps due to it attracting enemies near Xiangling’s Guoba, along with being a major asset for abyss 11-2.

### Ganyu Construct Height Placement Shenanigans

By: Empyress#9451
Last tested: 2022-04-08 (v. 2.6)
Discussion

Finding:
Ganyu's Ice Lotus placement seems to have a specific height difference past which it will be placed mid-air.

Evidence:

Significance:
Knowing more about construct spawn conditions at different heights.

## Burst Mechanics

### Icicle drop frequency of Ganyu’s Q

By: TiagoESC#6773
Last tested: 2021-01-24 (v. 1.2)

Finding:

Ganyu’s Q creates 50 icicles in 15 seconds, or 1 every 0.3 seconds.

Evidence:

Shard hit the ground in a 18 frame interval (at 60 FPS) or 9 frames in 30 FPS.

Icicle drop counting

Significance:

Knowing the number of icicles during Ganyu’s Q will help with other findings like Icicle ICD.

### Icicle Internal Cooldown in Ganyu’s Burst

By: Artesians#0002
Last tested: 2021-01-12 (v. 1.2)

Finding:

Ganyu’s Q, Celestial Shower, drops an icicle on a random enemy that has not been targeted by an icicle in the last 4 icicle drops. The icicles drop randomly among "legal" targets, which are defined as enemies that have not been targeted by the last 4 icicle drops. If there are no legal targets, they will drop on the ground, where the AoE can still strike a target.

Evidence:

User Artesians#0002 walked into F12 and up to the Hypostasis, used Ganyu’s Q, and counted the frames it took for an icicle to spawn above an enemy:

Ganyu's Q against an isolated enemy in a group

In the video, the icicle spawns at frames: 603, 710, and 818.

107/60 = 1.783s

108/60 = 1.8s

It was tested against the Hypostasis, and it took again 108 frames exactly for an icicle to spawn above it; however, these results were later disproved by other clips and testing where an ICD was seemingly 1.5s.

Thus, a series of experiments were conducted on groups of 1, 5, 6, and 7 enemies within Ganyu ultimate. At around 10 seconds, a hilichurl breaks out from the pack and 4 icicles are dropped randomly amongst the 6 other targets, but he is struck again by the 5th icicle:

Ganyu's Q against several enemies

Difficult to count the icicles initially, but 4 icicles are dropped on the ground before one strikes the Fatui:

Ganyu's Q against 1 enemy

Significance:

Ganyu's Q is really bad against single-target enemies for Cryo application. Her Cryo application really shines when you can abuse the fact that the icicles have separate Cryo applications on enemies that are grouped.

### Ganyu’s Ascension 4 and Constellation 4 Effects Carry Over on Consecutive Casts of Ganyu’s Elemental Burst

By: SavageCabbage#1445 TiagoESC#6773
Last tested: 2021-01-25 (v. 1.2)

Finding:

On subsequent casts of Ganyu’s Celestial Shower, if done in quick succession (immediately after the cooldown generally), Ganyu’s ascension 4 will actually trigger for the Elemental Burst unlike in the initial cast, and furthermore; Ganyu’s constellation 4 debuff will carry over to any enemy that is currently in the Elemental Bursts aura.

Evidence:

Users SavageCabbage and Tiago did some testing of Ganyu’s Elemental Burst by casting the Burst in succession against a Cryo Regisvine, and a group of slimes respectively in order to test for both C4 and non C4 interactions. The result of which showed that after initial casting, both ascension 4 and constellation 4 were able to carry over providing additional damage for Ganyu.

Significance:

This finding implies that Ganyu’s overall Elemental Burst damage spikes significantly if done off-cooldown due to A4 being able to go into effect if the Elemental Burst is cast in quick succession. This in turn greatly improves the value of constellation 1, minorly improves the value of constellation 4, and makes Ganyu’s ascension 4 talent even better than it already is.

## Ascension Mechanics

### A1: Undivided Heart

#### Ganyu’s A1 Passive Does Not Apply to the Initial Bloom

By: SavageCabbage#1445
Last tested: 2021-01-24 (v. 1.2)

Finding:

Ganyu’s A1 Passive does not apply to itself. Meaning the Bloom from the first Frostflake Arrow arrow shown within 5 seconds will not have an increased CRIT Rate

Evidence:

Tests were run by user SavageCabbage with Ganyu at 80% CRIT Rate on their artifacts. Due to this, if Ganyu’s A1 Passive applies to itself they would never experience a non-critical Bloom during testing, which was not the case.

Ascension 1 passive test

Significance:

This knowledge allows for a clearer understanding of her talents, and implies that for shorter fights Ganyu will be marginally weaker at the start of the fight due to her first shot not having the A1 Passive.

#### Ganyu’s A1 Passive Persists After Swap

By: Nello#1928
Last tested: 2021-01-25 (v. 1.2)

Finding:

Ganyu’s A1 Passive persists after swapping characters, meaning that it is possible to maintain the 20% CRIT Rate from A1 while swapping to supports to activate their abilities and Elemental Bursts as long as the swap time is within the 5 second period. (Note: the CRIT Rate increase does not apply to others).

Additionally, Buffs applied to the character after an arrow has been released are not added to the fired arrow. The arrow's damage is “snapshot” from the state of the character when released.

With these two things in mind, as long as the FrostFlakeArrow is fired within 5 seconds of the previous arrow regardless of what you do during this timeframe the 20% additional CRIT buff is maintained.

Evidence:

The activation of the A1 Passive is visible after Ganyu shoots a Frostflake Arrow (as can be seen from the yellow visual effect around Ganyu) and the effect persists after swap:

A1 Passive swap test

After shooting an arrow, moving inside/outside Bennett’s Burst does not change the damage dealt by arrows previously released, therefore arrows do not dynamically update buffs, they keep the same state after being fired:

Arrow snapshot test

Significance:

This proves that it is possible to use other character’s Elemental Bursts and/or Skills while maintaining Ganyu’s 20% CRIT Rate buff which both improves her viability by making her significantly less of a selfish carry, and allows for potential usage (although overall probably not optimal) in swap comps with Ganyu.

### A4: Harmony Between Heaven and Earth

#### Ganyu’s Ascension 4 Passive Will Not Apply to the First Elemental Burst Cast

By: SavageCabbage#1445
Last tested: 2021-01-24 (v. 1.2)

Finding:

Ganyu’s Elemental Burst will not apply the 20% bonus Cryo damage from its initial cast to itself, regardless of if the user stands inside or outside of the Elemental Burst post casting, implying that Ganyu’s Elemental Burst is snapcast in nature.

Evidence:

User SavageCabbage tested out using Ganyu’s Elemental Burst on a group of enemies and recorded the damage values he received. After this, he took said recorded values and inputted them into the damage formula to determine if the values he observed matched the values he would theoretically be receiving with the 20% damage buff.

[Video] A4 test on Elemental Burst

Image

Significance:

These findings are important for Ganyu players to know as their Elemental Bursts will deal less damage then they might assume on paper due to the common misconception that Ganyu’s ultimate is not snapcast in nature. Along with this due to the fact that the 20% bonus Cryo damage will apply if you Burst off of cooldown further incentivises players to do so.

### Ganyu Elemental Burst and Charge Shots Snapshot

By: Karzak#7126 Nello#1928 Neptunya#8291
Last tested: 2021-01-28 (v. 1.2)

Finding:

Both Ganyu’s Frostflake Arrow (on release) as well as Ganyu’s Elemental Burst (on initial cast) are snapshot in nature, meaning that any buffs or effects applied to the character will only affect them if they are on Ganyu during the initial release, and cast respectively

Evidence:

User Neptunya tested Ganyu’s Elemental Burst by casting it along with a Bennett ult, varying between being inside the Bennett ult during cast, walking into the Bennett ult after cast, and not being in the Bennett ult whatsoever in order to determine if Ganyu’s Elemental Burst was snapshot in nature, which she found to be true. Similarly, user Nello tested Ganyu’s charge shot by shooting a hilichurl at extreme distances in order to give them enough time to walk in and outside of a Bennett ult mid shot. Through this they similarly discovered that Ganyu’s Frost Flake arrow is also snapshot in nature.

Ganyu charge shot snapshotting

Ganyu Burst alone

Ganyu Burst outside of Bennett buff

Ganyu Burst inside of Bennett buff

Significance:

By the fact that both Ganyu’s CA and Elemental Burst are snapshot in nature, we now have a better understanding of how Bennet’s Elemental Burst interacts with Ganyu’s abilities, which in particular is very important to know for Melt comp, and even to an extent can be decently important for a Ganyu Freeze comp if they decide to run Bennett for the extra damage.

### Ganyu Burst does not Overlap

By: EdisonsMathsClub#6469
Last tested: 2021-04-17 (v. 1.4)

Finding: Ganyu's ultimate cannot be duplicated or have 2 on the field at the same time

Evidence:

1. Slap on massive CDR via Chong's C2, Anemo Resonance, and a -20% CDR abyss card
2. Ult repeatedly off cooldown
3. Test whether casting inside her ultimate allows it to coexist
4. Test whether casting outside of her ultimate field, but have the 2nd one overlap the ultimate, allows it to coexist
5. Test whether casting completely outside her ultimate field allows it to coexist

Significance:
In a scenario where Ganyu has lots of cooldown reduction to the point where you can cast her Burst before it ends, it is advised to wait until it has ended to cast her Burst again.

### Ganyu Burst Frequency

By: Yukarix#6534, Greyhound#7836
Last tested: 2021-06-24 (v. 1.6)
Discussion

Finding:
Ganyu's rng ult icycles are more likely to hit a target(beyond the base 10) if they are larger in size (who'd have thought :keqewide: ) up to all 50 for enemies large enough such as Azhdaha

Evidence:
Azhdaha 50 hits: Video
Hilichurl 13/11/16/11/16/14 hits: Video
Humanoid 11/12/10/13/10 hits: Video
Fat Fatui 14/12/12/11/15/14 hits: Video
Geo Vishap 20/19/23/25/22/18 hits: Video
Primo Vishap 25/24/23/24/19/26/22/23/21 hits: Video
Geo Lawachurl 14(ran outside)/17/18/16/17 hits: Video
Mitachurl 12(ran outside)/18/17/14/15/19 hits: Video
Ruin guard 18/19/21/21/23/28 hits: Video

Significance:
Ganyu's ult is better than expected

## Constellation Mechanics

### C1: Dew-Drinker

#### Constellation 1’s Cryo Resistance Shred Will Apply to the First Shot’s Bloom

By: Karzak#7126
Last tested: 2021-01-31 (v. 1.2)

Finding:

When shooting your first Frostflake Arrow at an enemy if you hit the enemy directly with the arrow, the initial Bloom of the shot will get the bonus 15% Cryo resistance shred

Evidence:

User Karzak tested this interaction by intentionally missing the first arrow and only hitting the enemy with a Bloom, and compared that damage to when they shot the enemy with both the arrow and the Bloom, finding that the Bloom's damage increased slightly when the enemy was hit with the initial arrow first. ALong with this, they also noticed that the debuff arrows that surround the enemy started appearing after the initial hit, rather than from the Bloom hit.

Significance:

The 15% Cryo resistance shredding from Ganyu’s constellation 1 is slightly better during short encounters as the enemy will get the resistance shred applied to the first Bloom.

#### Ganyu’s Constellation 1 Generates Energy Without Dealing Damage

By: Pinupski#9999
Last tested: 2021-01-31 (v. 1.2)

Finding:

Ganyu can regenerate Energy through her C1 passive ability even when shooting immune enemies.

Evidence:

Slimes are immune to Cryo and therefore take no damage. Despite taking no damage, it can be seen in the video below that C1’s passive is functioning:

Video

Significance:

It can be useful to know that even in the event of immunity or resistance you can still generate Energy from a FFA/B.

### C6: The Clement

#### ARCC Shot Tech is Doable Only With Ganyu’s Constellation 6

By: SavageCabbage#1445
Last tested: 2021-01-26 (v. 1.2)

Finding:

You can perform an ARCC shot with Ganyu only if you have Ganyu’s constellation 6 and have an instant charged shot primed.

Evidence:

ARCC shot test with C6

Significance:

Allows for quick execution of charged shot -> e -> charged shot -> e -> charged shot combo.

## Miscellaneous

### Ganyu’s E Can Trigger Small Pressure Plates

By: TiagoESC#6773
Last tested: 2021-01-14 (v. 1.2)

Finding:

Ganyu’s Trail of Qilin is classified as a construct similar to Amber’s bunny, and all of the Geo character’s constructs. As such, her Trail of Qilin can function as a trigger for smaller pressure plates which can help in exploration

Evidence:

User Taigo tested by placing Ganyu’s Trail of Qilin on a large pressure plate where it was shown to be able to trigger half of the plate, implying that the construct’s weight is equal to that of Ambers bunny and can trigger either a small pressure plate, or half of a large one.

Ganyu E pressure plate test

Significance:

With this in mind, Ganyu is potentially a better character to use for exploration as she can fill the niche of a pressure plate activator, especially if the player has access to constellation 2 giving them a second charge of Trail of Qilin.

### Ganyu’s Bloom from Frostflake Arrow is Considered a Projectile and Can Headshot

By: Karzak#7126 SavageCabbage#1445
Last tested: 2021-01-31 (v. 1.2)

Finding:

The Frostflake arrow’s Bloom although at first glance would appear to be a separate ability unable to headshot or trigger weak points, after further investigation proved that in certain scenarios it is actually possible to trigger weak points with the Bloom, in turn activating Prototype Crescents buff.

Evidence:

User SavageCabbage noticed that they at one point accidentally managed to trigger the prototype crescent’s boost without hitting the enemy with their initial arrow. After discussing this finding with user Karzak, after further testing they discovered that the Bloom indeed was considered a projectile, and had the capability (although very unrealistic in practice) to headshot.

Ganyu Bloom weak point hit

Significance:

In practice when using the Prototype Crescent it is very possible to accidentally get a passive proc from the bow.

### Ganyu can provide an ice bridge via her charged shots

By: Cola#6861
Last tested: 2021-01-31 (v. 1.2)

Finding:

Ganyu’s Frostflake arrow when shot on water applies the ice effect for long enough to allow for a 2nd charge shot to be used, in turn allowing for Ganyu to make an icebridge.

Evidence:

User ColaSen tested shooting Frostflake Arrow arrows repeatedly to try and travel across water and found that there was just enough time to be able to build a constant bridge if you are fast enough with your Frostflake arrows.

Ice bridge building

Significance:

Ganyu is a decent character when attempting to travel across water as her Frostflake Arrows has no cooldown unlike abilities which will theoretically allow for faster overall travel across water.

### Unlimited Ganyu Works

By: Pinupski#9999
Last tested: 2021-04-05 (v. 1.4)

The culmination of all Ganyu macro-related testing and iteration since release. Now compiled into one mostly comprehensible document.

Finding: Most optimal/reliable timings for Ganyu rotations tested and verified via macro.

• Example tech macros for C6 E FFA Instacast, Quickshot and Aimcast variants.
• Example tech for standalone Quickshot.

Significance:
Mostly so close in timings that there's not a huge difference for normal players to use one over the other, preference should be given to whatever's most reliable/comfortable.

Significance:
QoL and DPS increase to fluidly execute all as one motion.

### Sacrificial Passive With Multiple E Stacks (Ganyu C2)

By: mol#3280
Last tested: 2021-05-06 (v. 1.5)
[Discussion]

Finding:
When C1 Sucrose has both stacks of E down, a sacrificial fragments proc will reset the cooldown showed by the timer on E.

Evidence:
Streamable Video lost to the sands of time
I cast one Sucrose E, waited until the first cooldown was down to 5s, and cast the second E. A reset occurred, and the time of the second cooldown was ~15s (minus 10% due to Anemo resonance in video). So, whatever timer is currently showing is the one which will be reset.

Significance:
Unlike Sacrificial-type-weapon users with single-cast Skills, any serial-cooldown, multi-cast Skill user may waste cooldown resets by causing a reset when their first cooldown is near 0s. Since the timer on stacked Skills begins with the first Skill cast out of the stack, and that Skill is the one which is reset (not the longer, still waiting cooldowns), resets can be completely wasted on characters with multiple stacks. For example, with Sucrose it is possible to reset a Skill with 0.1s of cooldown remaining, then have a stack of 15s go onto cooldown afterward. To get the most out of the cooldown benefit of sacrificial weapons, characters with multiple stacks might consider using stacks as quickly as possible.


text:
# Why is there n-1 degrees of freedom and not n-2 (student distribution)

#### Rafaelle55

##### New Member
When we want to compute a confidence interval for the mean of a sample of n persons, it is always explained that the number of degrees of freedom used should be equal to the sample size minus the number of estimated parameters.
But, in my opinion, two parameters are estimated: the mean is estimated by the sample mean and the variance is estimated by the sample variance (corrected). Then why shouldn't we use the t-distribution with n-2 degrees of freedom?

#### Dason

There is only one parameter that needs to be estimated before you calculate the sample variance

#### Rafaelle55

##### New Member
But, what's different between estimating this one parameter (sample mean) and estimating the sample variance?

#### Dason

The estimate of the sample mean is required before estimating the sample variance. So the number of parameters you need to estimate before you actually have enough information to estimate the variance is the number that is important for the degrees of freedom for the variance.

Go ahead and work it out - try and estimate the variance without first having some estimate of the mean (you won't be able to).

#### Rafaelle55

##### New Member
Yes I understand that the variance has n-1 degrees of freedom because of the estimation of the sample mean.
BUT the t-distribution uses the estimate of the sample variance, which should remove ANOTHER degree of freedom?

I hope my misunderstanding is clear enough?

#### Dason

To provide a counter point to your assertion that what you're describing should remove another degree of freedom:

why?

#### obh

##### Member
Hi Rafaelle ,

I will try to say in other words what Dason said.

The refection of the DF is Not for any parameter you estimate BUT because you estimate the Variance based on the estimation of the Mean, so there is a connection between the mean and the variance and here you loose the degree of freedom.

For example in z test you only estimate the mean, so you don't loose any degree of freedom.

#### Rafaelle55

##### New Member
Hi,

So if I understand well, this statement is wrong?
"Why do we have n-1 degrees of freedom? The reason is because to calculate the test statistic, we have used the sample mean as an estimator for the population mean µ"

#### Rafaelle55

##### New Member
To provide a counter point to your assertion that what you're describing should remove another degree of freedom:

why?
My reasoning was: you estimate 2 values so you lose 2 degrees of freedom. I know it is wrong but I'm still not convinced of why.

#### obh

##### Member
Hi Dason and Rafaelle.

My common sense says n-2 also I understand it is not correct as all know it is n-1...
if you have 5 values: x1,x2,x3,x4,x5
if you know the estimation of mean - average(x) and estimation of variance - sample var(x) and you know x1 x2 and x3 you can calculate the x4 and x5

I found the following explanation not sure if correct... He wrote:
"Also, note in a hypothesis test involving a 1 sample t-distribution the only known is the mean mu. The standard deviation/variance is estimated from the sample size so is not known. If you do know the standard deviation of the population then you should be using a Z-test not a T-test."

https://www.quora.com/Why-is-the-de...mean-and-the-standard-variance-are-both-given

#### Archidamus

##### Member
I may be able to add some clarity if you are familiar with MLE (maximum likelihood estimator) methods. In the process of finding the MLEs of a normal distribution (which the CLT [central limit theorem] tells us a sampling distribution converges to) we are looking to find the mean and variance. When finding the MLEs simultaneously you will find that the two parameters are independent (this can be verified with the fisher information matrix and its inverse the correlation matrix).

In short, the mean of a normal distribution does not rely on the variance and the variance does not rely on the mean, therefor when conducting inferences you only lose 1 degree of freedom. Also, the t-distribution is by definition has a mean of zero, therefor the only parameter unknown is the variance, again resulting in the 1 less degree of freedom.

If you still aren't convinced, write some R code and run a few million iterations. I do this sometimes, because I am skeptical at times too.

#### Rafaelle55

##### New Member
My common sense says n-2 also I understand it is not correct as all know it is n-1...
if you have 5 values: x1,x2,x3,x4,x5
if you know the estimation of mean - average(x) and estimation of variance - sample var(x) and you know x1 x2 and x3 you can calculate the x4 and x5
That is exactly my "problem" and I feel like I'm not convinced why it is n-1

#### Rafaelle55

##### New Member
I found the following explanation not sure if correct... He wrote:
"Also, note in a hypothesis test involving a 1 sample t-distribution the only known is the mean mu. The standard deviation/variance is estimated from the sample size so is not known. If you do know the standard deviation of the population then you should be using a Z-test not a T-test."

https://www.quora.com/Why-is-the-de...mean-and-the-standard-variance-are-both-given
I think this is different; my case is when the standard deviation is estimated and therefore not known. Of course when the standard deviation is known, a Z-test should be used.

#### Rafaelle55

##### New Member
In short, the mean of a normal distribution does not rely on the variance and the variance does not rely on the mean, therefor when conducting inferences you only lose 1 degree of freedom.
Because they are independent, you should lose 2 degrees of freedom (two independent pieces of information)? If they were dependant, I would understand n-1.

Sorry if I really don't understand

#### Dason

I think this is different; my case is when the standard deviation is estimated and therefore not known. Of course when the standard deviation is known, a Z-test should be used.
So you're fine with there being no mention of degrees of freedom in a Z test even though we estimate the mean in the process?

#### Archidamus

##### Member
Rafaelle, you are not having to guess the variance at the same time as guessing the mean.

Think of the data points creating the hyper-plane of n dimension covering all possibilities of data. Then the mean and variance in this case can have all possibilities in a 1 dimensional vector in this case (each having their own). So the resulting hyper plane is of the dimension of possibilities minus the maximum dimension of parameters.

If you have a distribution like a gamma function where the two a parameters are dependent, then the possibilities of the parameters is a 2 dimensional plane. Then the resulting degrees of freedom as n - 2.

I do encourage you to write some code and see for yourself. Create a random sample of normal data with known mean and variance. Then conduct the interval for the mean using degrees freedom of n-1 and n-2. Repeat this a couple thousand times and compare the reliability of the intervals against the intervals confidence. Doing a 95% CI you will see that n-1 will capture the mean 95% of the time, but n-2 will capture it more than 95% of the time.

#### Rafaelle55

##### New Member
So you're fine with there being no mention of degrees of freedom in a Z test even though we estimate the mean in the process?

#### Dason

That's actually a good thing that you're not sure anymore.

The degrees of freedom really are associated with the variance estimate. In the case of a z-test we *know* the variance so we don't need to estimate anything before we can get what we use for the variance. In the case of a t-test we need to estimate the mean before we can even estimate the variance since it is required for the variance calculation. That's why we lose one degree of freedom - we needed to estimate one parameter before we could estimate the variance. In a simple linear regression we need to estimate the intercept and the slope before we can get a prediction for each point which is required before we estimate the variance. So in that case we lose 2 degrees of freedom because we estimated 2 parameters before we could estimate the variance.

#### Rafaelle55

##### New Member
Rafaelle, you are not having to guess the variance at the same time as guessing the mean.

Think of the data points creating the hyper-plane of n dimension covering all possibilities of data. Then the mean and variance in this case can have all possibilities in a 1 dimensional vector in this case (each having their own). So the resulting hyper plane is of the dimension of possibilities minus the maximum dimension of parameters.

If you have a distribution like a gamma function where the two a parameters are dependent, then the possibilities of the parameters is a 2 dimensional plane. Then the resulting degrees of freedom as n - 2.

I do encourage you to write some code and see for yourself. Create a random sample of normal data with known mean and variance. Then conduct the interval for the mean using degrees freedom of n-1 and n-2. Repeat this a couple thousand times and compare the reliability of the intervals against the intervals confidence. Doing a 95% CI you will see that n-1 will capture the mean 95% of the time, but n-2 will capture it more than 95% of the time.
Aren't the mean and the variance dependent?

#### Rafaelle55

##### New Member
That's actually a good thing that you're not sure anymore.

The degrees of freedom really are associated with the variance estimate. In the case of a z-test we *know* the variance so we don't need to estimate anything before we can get what we use for the variance. In the case of a t-test we need to estimate the mean before we can even estimate the variance since it is required for the variance calculation. That's why we lose one degree of freedom - we needed to estimate one parameter before we could estimate the variance. In a simple linear regression we need to estimate the intercept and the slope before we can get a prediction for each point which is required before we estimate the variance. So in that case we lose 2 degrees of freedom because we estimated 2 parameters before we could estimate the variance.
That makes sense to me. I think I'm beginning to understand, thank you very much for your answers.

A quick question though: why are the degrees of freedom associated with the variance estimate? Is it linked to the definition of a "degree of freedom"?
Maybe you can use this example to explain me the definition of a degree of freedom (more accurately, where I misunderstand the definition of a degree of freedom):
if you have 5 values: x1,x2,x3,x4,x5
if you know the estimation of mean - average(x) and estimation of variance - sample var(x) and you know x1 x2 and x3 you can calculate the x4 and x5


text:
## Cryptology ePrint Archive: Report 2021/989

Stateful KEM: Towards Optimal Robust Combiner for Key Encapsulation Mechanism

Jia Xu and Yiwen Gao and Hoon Wei Lim and Hongbing Wang and Ee-Chien Chang

Abstract: A $(1,n)$-robust combiner combines $n$ cryptography primitives to construct a new primitive of the same type, and guarantees that if any of the ingredient primitive is secure, then the resulting primitive is secure. In recent two decades, robust combiners for various crypto primitives (e.g. public key encryption, oblivious transfer) have been proposed. Very recently, more works on robust combiners for post-quantum key encapsulation mechanism appear to achieve multi-layer of defence, to counter the future threat from Shor's algorithm running on powerful quantum computers. However, typically such combination of $n$ crypto primitives will sum up running times of all ingredient primitives and thus introduce linear overhead in time complexity, which may be a big burden on server side, since the server has to run key encapsulation mechanism (or key exchange protocol) with every online client.

We propose the very first robust combiner (of KEMs), with $O(1)$ \emph{amortized} complexity overhead, which not only breaks the linear boundary, but also achieves optimal complexity. Our experiments also confirm that the performance overhead of our robust combiner of $n$ KEMs is constant (i.e. $O(1)$) rather than linear (i.e. $O(n)$). Our cost is that, the resulting KEM has to maintain a secret dynamic state of fixed and linear size (i.e. $O(n)$) . We call such KEM as Stateful Key Encapsulation Mechanism (SKEM). SKEM is suitable for two users (or devices), who will have \emph{frequent} secure communications (e.g. via VPN or SSH). We also formally define the security formulation for SKEM and prove the security of our proposed SKEM scheme in standard model.

Category / Keywords: public-key cryptography / Key Exchange Protocol, Key Encapsulation Mechanism, Robust Combiner, Security and Performance, Parallel Combination, Series Combination, Post-quantum Cryptography


text:
1. ## The connected Game

There are n dots which are on a plane (flat surface). There are two players, A and B, who move alternatively; A moves first. The rules of the game are the same for both players: at each move they can connect two points, but they cannot connect points which were already directly connected to each other or connect a point with itself.

To put it simply, they build a graph (with predefined n vertices) by connecting some of the dots. The winner is the one who makes the graph connected (a graph is connected if there is a path between any two nodes of the graph, however, not every two nodes have to be connected directly). What is the winning strategy for player A, if such exists

• Clearly, if n = 2, the first player always wins.

•If n = 3, the second player always wins.

•If n = 4, the first player has a winning strategy.

•What if n =14?

• I suggest you find yourselves a partner its way more fun (unless you like playing with yourself) and play several games where n = 14. Determine which player (first or second) has a winning strategy? Provide convincing argument.

•Try then the general case (more involving)

2. ## Re: The connected Game

What comes to mind is this: A play loses on their turn iff they face two complete graphs (including a complete graph on one vertex). Thus the outcome is determined once the points have been connected into exactly two connected graphs (including an isolated vertex as counting as connected). The win/loss at that point will be 100% determined by the odd/even parity of the number of edges required to complete those two graphs. So your winning move is if you can connect three disjoint connected graphs, so as to make it into two disjoint connected graphs with an odd number of edges left to complete those two graphs. That's the condition that you want to find, but never can allow your opponent to find.
If you have 3 connected graphs, with number of verts a, b, c, and a total of k edges, and it's your move, can you win?
If move to connect the a & b graphs, then your opponent faces: (k+1) edges inside 2 connected graphs. Since that will fill out to [ (a+b) choose 2 ] + [ c choose 2 ] edges before someone is forced to lose, your opponent wins if [ (a+b) choose 2 ] + [ c choose 2 ] - (k+1) is odd, and your opponent loses if it's even.
Let f(n) = 0 if n congruent to 0 or 1 mod 4, and f(n) = 1 in the 2, 3 mod 4 case.
Then can show that parity[ n choose 2 ] = parity (f(n)).
So you win if you face 3 disjoint connected graphs, with total k edges and verticies counts a, b, and c, such that:
parity[ f(a+b) ] + parity[ f(c) ] - parity[k+1] = even, so if parity[ f(a+b) ] + parity[ f(c) ] + parity[k] = odd, so if parity[ f(a+b) ] + parity[ f(c) ] = opposite parity of k.
Then the question becomes, given a, b, c and a choosen parity, can you choose two of them so that that works? Not always.
Ex: If a=b=c mod 4, then no matter how you poermute a, b, and c, you end up with the condition parity[ f(2a) ] + parity[ f(a) ] = opposite parity of k.
If a=b=c=0 mod 4, then parity[ f(2a) ] + parity[ f(a) ] = even + even = even, so you're only able to make the winning move in that case if k were odd (if you'd gone 2nd).
If a=b=c=1 mod 4, then parity[ f(2a) ] + parity[ f(a) ] = odd + even = odd, so you're only able to make the winning move in that case if k were even (if you'd gone 1st).
If a=b=c=2 mod 4, then parity[ f(2a) ] + parity[ f(a) ] = even + odd = odd, so you're only able to make the winning move in that case if k were even (if you'd gone 1st).
If a=b=c=3 mod 4, then parity[ f(2a) ] + parity[ f(a) ] = odd + odd = even, so you're only able to make the winning move in that case if k were odd (if you'd gone 2nd).
Also, note that as soon as there are a total of 3 connected graphs with vert numbers a, b, and c, and total edges k, then if you don't have a winning move, then after your move, it's still the same situation, excpet for k+1 edges. Since you did NOT have a winning move, it must be the case that parity[ f(a+b) ] + parity[ f(c) ] = same parity as k.
Therefore, after you move, your opponent sees a {a, b, c; k+1} situation, and it's true that parity[ f(a+b) ] + parity[ f(c) ] = same parity as k = opposite parity as (k+1), and so your opponent wins.

This shows that the moment those n verticies have been connected up into exactly 3 disjoint connected graphs (including singletons as such), the winner is determined (assuming everyone plays correctly), and that winner is based only on the following data: the number of vertices of those 3 graphs, taken only mod 4, and the parity of the the number of edges placed. And obviously, from a player's point of view, when their turn comes up, the parity of the the number of edges placed will be solely determined by whether that player had gone 1st or 2nd.

So now the game is about deciding how to drive the initial setup into the situation I just described. That's starting to look like more thinking than I care to invest without a bunch of writing, so I'm going to stop with the details there. I'll make notation of state {a1, a2, ..., as; k} meaning it's broken into s connected graphs with vertcies count a1, a2, ... as respectively, and k edges.
It's certain that the game is decided at {a1, a2, a3; k} based only on the mod 4 values a1, a2, a3 and the parity of k.
A player looking at {a1, a2, a3, a4; k} gets to choose whether to send it to {a1, a2, a3, a4; k+1}, or (WLOG) {a1, a2, (a3+a4); k+1}, UNLESS all 4 graphs are already complete! (A single isolated vertex counts as complete.)
If there is NO permutation like {a1, a2, (a3+a4); k+1} that wins for that player, and not all 4 graphs are complete, then the next player, facing {a1, a2, a3, a4; k+1} will be able to find a permutation {a1, a2, (a3+a4); k+2} that wins (since those a's won't have changed, but the parity of the number of edges will have).
This shows that the moment there are exactly 4 connected components, the game is decided, based on the mod 4 values of their numbers of verticies, and the parity of the number of edges placed (i.e.
which player went first).
The same reasoning seems to continue, so that as soon as 5 connected components, it's decided, and so forth.
It seems to me that the thing that keeps this from being a simple pattern is that isolated vertices count as complete graphs, and a move connecting isolated vertices forms a new complete graph. Graphs being complete change the options, and so make it more complicated.

I'll make one more generic comment:
Fix n >= 4, the first player is always looking at the same situation on his/her first move - but that's also true for the 2nd player. It's the 2nd player who gets the first actual "choice" - whether to place an edge so that the two edges share or don't share a vertex. So in effect, the 2nd player gets to decide whether the 1st player "starts" (it's actually the 1st player's second move) by looking at:
$\left( \amalg_{1}^{n-3} K_{1} \right) \amalg \left( K_{3} - \{e\} \right)$
or
$\left( \amalg_{1}^{n-4} K_{1} \right) \amalg \left( K_{2} \right) \amalg \left( K_{2} \right).$

That's all I care to invest for now. Maybe I'll think about it more tomorrow.

3. ## Re: The connected Game

I don't understand the claim that the first player has a winning strategy: it just seems to me that the first player always wins with an even number of points and the second player wins with an odd number of points. The proof is by induction because we can replace a set of connected nodes by a single node to get a smaller case.

Edit: I've seen the point I was missing now.