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Master Felix:
To get the Stone of Commune, you must kill the Archon of Halisha.
Find him by entering the Sepulchers or by hunting down about 700 monsters at the Shrine of the Loyal.


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Here at Zeal West studios we take creativity to the next level! Our studios facilitate band rehearsals, recording sessions and creative programs where our young people learn the fundamentals of audio engineering, songwriting and a fun place for them to jam after schools.
We have a small team of casual AV Technicians who help look after our studios for band rehearsals. They also operate sound and lights for any events we have at Zeal. This is a great opportunity for our young people to learn on the job and to help develop their skills to take with them in the industry.
Our rehearsal rooms in West Auckland are purpose-built and cheap to hire! All rooms are on the ground floor with easy access for loading gear in through our side entrance from the carpark. (No stairs!)
The Zeal studios are open for Rehearsal bookings Monday to Friday evenings 6-10pm.
If you would like to book rehearsal rooms outside our open hours or recording sessions the times are available upon request.
We have 3 fully sound proofed rehearsal rooms all named after Kiwi bands or Artists.
“Six60” room! our largest room that fits up to 10 people.
“Crowded house” room! Great for a 4-5 piece bands.
“Lorde” room! Great for a 4-5 piece bands.
All rehearsal rooms have…
2 x Electro-Voice ZLX-15” 2-way powered Loudspeakers
Behringer 24 bit Multi FX Processor- 12 channel audio mixing Desk
1x Full Drum Kit “Mapex brand”- BYO Cymbals/Hi Hats
1 x SM58 Microphone
1 x Bass Amp
1 x Guitar Amp
Plus all necessary cables included in the room.
Additional Microphones, Cymbals/hi hat sets, drum sticks and Amps are available to hire.
We have a small team of qualified Audio Engineers who are starting out in the industry and keen as to record your Music and help you share it with the world!
We have a purpose built studio control room with vocal booth. The control room is connected to the rehearsal rooms so you can record utilising multiple live rooms at the same time.
Check out our sweet recording equipment!
Genelec 8050 APM MDE Active Loudspeakers (studio Monitors)
Logic Keyboard Pro Tools ALU Advanced USB
Hear Technologies Hear back system ( 4 pack)
1x Keyboard M- Audio Keystation 61 MK2
Avid Pro Tools 9 Software
Avid- Artist mix- Control unit
2 x Apollo 8p - Audio Interface
Our Recording Microphone Selection includes…
1 x Beez Neez Arbella Tube Microphone
1 x Heil Sound HDK7 Drum Microphone Kit
3 x Sennheiser MD-421
3 x Shure SM- 7B


There are a lot of everyday situations where people make decisions one after the other, and what is decided earlier affects the choices later on. Most people probably go by some gut feeling about when it’s time to stop and select the next best option and settle down.
In this article we will examine the “Optimal Stopping Problem,” a mathematical puzzle that can manifest in many different real world examples. For instance:
- When should a hiring manager stop interviewing people for a new job position?
This is the classical “secretary problem” within the field of study of mathematics: probability-optimization. This problem in it’s simplest form has following features:- There is one secretarial position available.
- The number of applicants is $n$.
- The applicants are interviewed sequentially in random order, each order being equally likely.
- After each interview the hiring manager must decide to reject or hire the applicant.
- An applicant once rejected cannot later be recalled.
- Once the hiring manager decides to hire an applicant his/her search is over and there is no need to interview any more candidates.
This basic problem has a remarkably simple solution and was outlined in the basic paper by Gilbert and Mosteller ( 1966 ), with elegant derivations and extensions in a number of important directions.
-
We have a house and we wish to sell it. Each day we get a new offer for the house, which could be higher or lower than the previous offer; however, each day the house is on the market we must pay for advertising. We wish to maximise the amount you earn by choosing a stopping rule. When should we stop looking at potential buyers and finally settle the deal?
-
At some point in a lot of peoples lives, we wonder when we should stop dating and finally settle down with the right partner. Most people probably go by some gut feeling about when it’s time to stop. Editor’s Note: Of course, for mathematicians this is a purely theoretical example.
- Suppose we are a doctor and patients arrive sequentially at our clinic and must be treated immediately by one of the treatments. It is assumed that response from treatment is immediate so that the effectiveness of the treatment that the present patient receives is known before the next patient is treated. It is not known precisely which one of the treatments is best, but we must decide which treatment to give each patient, keeping in mind that our goal is to cure as many patients as possible. This may require us to give a patient a treatment which is not the one that looks best at the present time in order to gain information that may be of use to future patients. When should we stop evaluating more treatments and finally select the best one?
There are many different solutions to the problems listed above, as the strategy we choose will depend on our objective. Do we want to ensure that the option we select is better than average? What if we just want to be sure that we do not pick the worst option? In this article our goal will be to “minimize the risk” of missing the best option. In other words, we want to “maximize the chances” (i.e. probability) of selecting the best option.
The Prelude
The general form of our strategy will be to “gather information” when we are presented with our first options. We know that we will not choose any of these; however, we will examine them in order to know “how good” we should expect our options to be. After a certain point, once we are done gathering information, we will choose the first option that was better than all the ones we have seen before. After some thought, we can see that this strategy makes some intuitive sense. We don’t want to choose an option too early since there would be a high likelihood a better option would come later. But when should we stop gathering information? While we don’t want to choose an option too early, we certainly don’t want to gather information for too long and risk the best option passing us by.
This type of problem is called the “Optimal Stopping Problem”, which mathematicians and computer scientists have researched for many years.
A Real-life Example
Suppose we want to sell our house. We have a pool of n potential buyers (we number them from $1$ to $n$), who we meet one after the other. After a potential buyer has made their offer, we must decline or accept their offer. If we decide to accept their offer, we can no longer sell the house to anyone else and if we reject an offer the buyer will take their business elsewhere, and we will not be able to change our minds later on. For simplicity’s sake, we will assume there is no advertising cost (i.e. we do not have to pay to keep our house on the market).
Among our pool of $n$ people, there’s at least one who gives us the highest value for our property. We will call that person $m$ (where $m$ ranges from $1$ to $n$) – it’s who we’d ideally want to end up making deal with.
Our strategy is to discuss the deal with $m-1$ of the $n$ people and then settle with the next best person who is better. (The reason for doing the calculations with $m-1$ is that m is the minimum number of potential buyers that we will encounter; there are $m-1$ who give us the data we need before we start searching, and the $m^{th}$ is the first one who can be accepted.)
Given a pool of $n$ potential buyers, let $P(m; n)$ represent the probability that we will choose the best offer if we use the strategy discussed earlier. I.e. If there are $n$ buyers and we “gather information” during the first $m-1$ offers and pick the first offer that is better than the current best, the probability that we will select the best possible offer is $P(m; n)$.
Notice immediately that $P(1; n) = \dfrac{1}{n}$ and $P(n; n) = \dfrac{1}{n}$. These two cases essentially give no choice, meaning, it is just like the probability of picking one person at random out of total number of people ($n$) and hence the probability is $\dfrac{1}{n}$. But interesting things happen ‘in between’ these two extremes. As we see more people , we already have a notion of the “best so far,” and the next person we see that beats the current “best so far” is more likely to be the best person in the group for us. Unfortunately, it works the other way around as well. As we continue to see more people, the likelihood of having already passed the best so far increases, thereby decreasing our chances at finding the optimal person.
We want to find the optimal place to stop – a sort of middle ground that maximizes our chances of choosing the optimal person.
To do this, we want to maximize $P(m; n)$. In the process of doing this, we want to make sure our solution generalizes. A good way to generalize is to find the right ratio $\dfrac{m}{n}$, the percentage of our population that we should analyze before making a decision.
Mathematical Formalism
Our first step here is to get an equation for $P(m; n)$.
Let’s say we have collected information about $m - 1$ potential buyers and now are looking at the $k^{th}$ person in the sequence, where $m-1 \leq k \leq n$.
Now let’s break down some things that are going on here:
- We know that this $k^{th}$ person has a $\dfrac{1}{n}$ probability of being the best buyer.
- We also know that in order to be considering the $k^{th}$ buyer in this sequence, the highest ranking person thus far must be in the group of $m - 1$ people out of the $k - 1$ people we had previously inspected and rejected (otherwise we would have stopped at someone before the $k^{th}$ person).
And the probability of the highest ranked person thus far being in that group of $m - 1$ people out of the $k - 1$ people inspected is: \[ \dfrac{m - 1}{k - 1}. \]
And therefore, the overall probability of finding the best potential partner this time is: \[ (\dfrac{m - 1}{k - 1}).(\dfrac{1}{n}) = \dfrac{m - 1}{n(k - 1)}\]
Note: In the following section we will use summation notation ($ \sum $ symbol), which works as follows: Suppose we have a sequence of numbers $x_1, x_2, …, x_n$, then, by definition, $\sum_{i=1}^{n} x_i = x_1 + x_2 + x_3 + … + x_n$. Thus $\sum_{i=1}^{n} 1 = n$ and $\sum_{i=1}^{3} x_i = x_1 + x_2 + x_3$. We use this notation because it is shorter than writing out each term in the sum.
Now to determine $ P(m; n) $, we must remember that $ k $ can take on any value from $ m $ to $ n $. As a consequence, we must sum these probabilities for any value of $ k $ (note the range of $ k $ above). A compact way to express this is as follows: \[P(m; n) =\sum_{k=m}^{n} \dfrac{m-1}{n(k - 1)}= \dfrac{m-1}{n}\sum_{k=m}^{n}\dfrac{1}{k-1}. \tag{1} \label{1}\]
At this point, it is possible to determine the maximum value using methods of optimization one learns in calculus, but this can get nasty pretty quickly. We will lay out a more elegant (albeit more mathematically involved) approach in the following section.
To Get More Technical
Given a number $n$, we want to maximize the value of $P(m;n)$. Hence, the optimal value for $ m $ will be such that: \[P(m - 1; n) < P(m; n) \text{ and } P(m + 1; n) < P(m;n) \tag{2} \label{2}\]
Author’s Note: If you want to be very pedantic, you could ask what happens if there are two “best” values of $ m $, with one of those strict inequality signs replaced by a partial inequality. It turns out that the only time when equality is possible is when $ n=2 $, which is not very interesting anyway.
Let’s break this down part by part. From the first inequality in (\ref{2}) and using (\ref{1}): \[P(m - 1; n) < P (m; n)\]
\[\dfrac{m-2}{n}\sum_{k=m-1}^{n} \dfrac{1}{k - 1}<\dfrac{m-1}{n} \sum_{k=m}^{n}\dfrac{1}{k-1}\]
\[(m-2)\sum_{k=m-1}^{n} \dfrac{1}{k - 1}<(m-1) \sum_{k=m}^{n}\dfrac{1}{k-1}.\]
Rewriting the left hand side:
\[(m-2)(\dfrac{1}{m-2})+\sum_{k=m}^{n} \dfrac{1}{k - 1}<(m-1) \sum_{k=m}^{n}\dfrac{1}{k-1}.\]
After algebraic simplification, we get: \[1<\sum_{k=m}^{n}\dfrac{1}{k-1}\tag{3} \label{3}\]
Let’s now do the same thing for the other inequality from (\ref{2}):
\[P (m; n) > P (m + 1; n)\]
\[(m-1)\sum_{k=m}^{n}\dfrac{1}{k-1}>m\sum_{k=m+1}^{n}\dfrac{1}{k-1}\]
Performing a similar calculations to what we did with the last inequality, we get:
\[(m-1)(\dfrac{1}{m-1}+\sum_{k=m+1}^{n}\dfrac{1}{k-1})>m\sum_{k=m+1}^{n}\dfrac{1}{k-1}\]
\[(1+(m-1)\sum_{k=m+1}^{n}\dfrac{1}{k-1})>m\sum_{k=m+1}^{n}\dfrac{1}{k-1}\]
\[1>\sum_{k=m+1}^{n}\dfrac{1}{k-1}\tag{4}\label{4}\]
Let’s remember that our goal here is to find an appropriate approximation for the ratio $ \dfrac{m}{n} $ for the best $ m $.
One fact that is going to help us immensely in this calculation is the approximation of the partial sum of a harmonic series. Mathematically, the approximation is:
\[\sum_{k=1}^{x}\dfrac{1}{k}\approx\ln(x)+c \]
where $ c $ is the Euler-Mascheroni constant.
Now this approximation greatly helps us simplify inequalities (\ref{3}) and (\ref{4}).
Using this approximation for (\ref{3}), we derive the following:
\[1<\sum_{k=m}^{n}\dfrac{1}{k-1}=\sum_{k=m-1}^{n-1}\dfrac{1}{k}=\sum_{k=1}^{n-1}\dfrac{1}{k}-\sum_{k=1}^{m-2}\dfrac{1}{k}\approx\ln(\dfrac{n-1}{m-2})\]
Similarly, for (\ref{4}), we obtain:
\[1>\sum_{k=m+1}^{n}\dfrac{1}{k-1}=\sum_{k=m}^{n-1}\dfrac{1}{k}=\sum_{k=1}^{n-1}\dfrac{1}{k}-\sum_{k=1}^{m-1}\dfrac{1}{k}\approx\ln(\dfrac{n-1}{m-1})\]
Alright, almost done! Notice that we can make another approximation of these results for large $ n $.
For sufficiently large $ n $, the constants in our expressions will make negligible difference in our final result, allowing us to make the following claim:
\[1\approx\ln(\dfrac{n}{m})\]
Now, using the properties of natural logarithm, value of $e$ and then taking it’s inverse, we get:
\[\boxed{(\dfrac{m}{n})\approx(\dfrac{1}{e})\approx 0.37}\]
Q.E.D.
This result was proven through a lot of (good) approximations, but other more rigorous methods including those used in optimization come to the same result.
So here we have it! A proof of why you should keep $37\%$ on your mind when it comes to matters of making choices.
distill


At the Daiko Shibu Dojo, we study study the traditional Eighth Light Style of jūjutsu—Hakkō Denshin Ryū. Under the guidance of Shihan Stephen Tack, students learn the self-defense techniques of the "Gentle Art."
If you wish to learn more about the Daiko Shibu Dojo or Hakkō Denshin Ryū, you can check out our about page, send us a message, or purchase your own copy of the Hakkō Denshin Ryū Shodan Manual.


[ school(1), school(2), school(3), school(4) ]
[ "4th grade", "5th grade", "6th grade", "Total" ]
randRange( 0, COLUMNS.length - 1 )
randRange( 0, SCHOOLS.length - 1 )
(function() {
var enrollments = [];
var classMin = 60;
var classMax = 80;
for ( var i = 0; i < SCHOOLS.length; i++ ) {
var enroll = randRange( classMin, classMax, COLUMNS.length - 1 );
var total = 0;
jQuery.each( enroll, function( i, e ) {
total += e;
});
enroll.push( total );
enrollments.push( enroll );
}
var answer = enrollments[ SCHOOL_INDEX ][ COL_INDEX ];
enrollments[ SCHOOL_INDEX ][ COL_INDEX ] = " ";
return [ enrollments, answer ];
})()
The table below shows the enrollment at four different elementary schools that have 4rd through 6th grade students, except one entry is missing.
What number should go in the empty cell?
school enrollment
ANSWER students
The table is missing the total number ofthe number of COLUMNS[ COL_INDEX ] students at SCHOOLS[ SCHOOL_INDEX ] Elementary School.
The table accounts for ENROLLMENTS[ SCHOOL_INDEX ].slice( 0, COL_INDEX ).join( "+" ) students.
jQuery.each( COLUMNS, function( i, c ) {
if ( i !== 0 ) {
jQuery( ".fake_row" ).eq( SCHOOL_INDEX ).find( "span" ).eq( i )
.css( "color", KhanUtil.ORANGE );
}
});
The total number of students at SCHOOLS[ SCHOOL_INDEX ] Elementary School is simply the sum, or ANSWER.
fillInCorrectAnswer( SCHOOL_INDEX, COL_INDEX, ANSWER, KhanUtil.PINK );
The total number of students at SCHOOLS[ SCHOOL_INDEX ] Elementary School is ENROLLMENTS[ SCHOOL_INDEX ][ COLUMNS.length - 1 ].
jQuery( ".fake_row" ).eq( SCHOOL_INDEX ).find( "span" ).eq( COLUMNS.length )
.css( "color", KhanUtil.BLUE );
The table already accounts for ENROLLMENTS[ SCHOOL_INDEX ].slice( 0, COL_INDEX ).concat( ENROLLMENTS[ SCHOOL_INDEX ].slice(COL_INDEX + 1, COLUMNS.length - 1 ) ).join( "+" ) students.
jQuery.each( COLUMNS, function( i, c ) {
if ( i !== 0 ) {
jQuery( ".fake_row" ).eq( SCHOOL_INDEX ).find( "span" ).eq( i )
.css( "color", KhanUtil.ORANGE );
}
});
The missing number must be the difference of the blue and orange numbers, or ANSWER.
fillInCorrectAnswer( SCHOOL_INDEX, COL_INDEX, ANSWER, KhanUtil.PINK );


---
layout: about
permalink: "/cv"
author: Chang
weight: 1
menutitle: CV
cover: /assets/page/cover.jpg
title: Curriculum Vitae
---
{% assign cv = site.data.cv[page.author] %}
Education
{% for entry in cv.education %}
- {{entry.time}}
-
{{entry.location}}
{{entry.description}}
{% endfor %}
Research Experience
{% for employment in cv.employment %}
{{employment.time}}
{{employment.time}}
{{employment.position}}
{{employment.company}}
{% for responsibility in employment.responsibilities %}
- {{responsibility}}
{% endfor %}
{% endfor %}
PDF Version


jargon
- LiPo - Lithium-ion Polymer
- NiMH - Nickel-Metal Hydrid
- Discharge Rating
Capacity
Capacity is denonted in milliampers hours (mAh) on the battery. It is the amount of charge stored in the battery. If you have a battery with a capaciy 1 mAh and a motor is drawing 1 mAh then your motor will run for 1 h. In general the \textit{approximate} duration of a battery can be computed by:
\[\mathrm{duration} (h) = \frac{\mathrm{capaciy(mAh)}}{\mathrm{load (mA)} }\]If you have a 4700 mAh battery and it is powering a moter with a 100 mA load, it will keep runing for approximatly 47 hours. reference
Discharge
Battery bank
Combination of multiple batteries reference
- joined in series: voltage increases, capacity remains the same.
- joined in parallel: voltage remains the same, capacity increases.
- series and parallel: both voltage and capacity increase.
A guide to understanding battery specifications
Pixhawk 2.1 battery stuff
- Max continuous current sensing: 30A
- Max burst current: 100A
- Connectors: XT-60
Electricity
-
Ampere: A - base unit. Attractive/repulsive force of \(2 \times 10^{-7}\) Newtons between two wires 1m appart is the definition of \(1\,\mathrm{A}\). \(1\,\mathrm{A} = 1\,\mathrm{C} / 1\,\mathrm{s}\)
-
Coulomb: C - charge transported by a constant current of 1 A in on second. \(1\,\mathrm{C} = 1\,\mathrm{A} \cdot 1\,\mathrm{s}\)
-
Electrical work is the work done on a charged particle by an electric field.
-
Watt : W - power. \(1\,\mathrm{W} = \frac{1\,\mathrm{N} \cdot \mathrm{m}}{\mathrm{s}}\)
-
Volt: V - pressure
-
Ohm: \(\Omega\) - electrical resistance
voltage
\[V = \frac{\mathrm{potential energy}}{\mathrm{charge}} = \frac{N\cdot m}{C}\]Ohm
- Ah - ampere hour
- mAh - milli ampere hour


NASA has selected SpaceX to build a lander to take humans to the moon
SpaceX is going to the moon. NASA has selected the company, headed by Elon Musk, to build a lunar lander to take humans to the moon’s surface in 2024. SpaceX beat out two other US-based firms for the $2.9 billion contract: defense contractor Dynetics, and private spaceflight company Blue Origin, which had partnered with several other aerospace industry titans. The three contenders were announced in April 2020, and while NASA was expected to select two finalists – as it did with its commercial crew programme – it has only awarded SpaceX a continuing contract. SpaceX’s intended lunar lander is a modified version of Starship, the spacecraft that the company has already begun to build and test in Texas. So far, all four of the test flights have ended in the prototype craft dramatically exploding.
SpaceX is going to the moon. NASA has selected the company, headed by Elon Musk, to build a lunar lander to take humans to the moon’s surface in 2024. SpaceX beat out two other US-based firms for the $2.9 billion contract: defense contractor Dynetics, and private spaceflight company Blue Origin, which had partnered with several other aerospace industry titans. The three contenders were announced in April 2020, and while NASA was expected to select two finalists – as it did with its commercial crew programme – it has only awarded SpaceX a continuing contract. SpaceX’s intended lunar lander is a modified version of Starship, the spacecraft that the company has already begun to build and test in Texas. So far, all four of the test flights have ended in the prototype craft dramatically exploding.
NASA will not be putting astronauts aboard Starship until SpaceX has proven that it can land the spacecraft without it blowing up, not only on Earth but also with a test landing on the lunar surface. “We are not going to launch humans until we have a successful test,” said Lisa Watson-Morgan, the leader of NASA’s Human Landing System programme, during NASA’s announcement. This contract is part of NASA’s Artemis programme, intended to land humans – including the first woman and the first person of colour – on the moon for the first time since the final Apollo mission in 1972. If all goes well, NASA representatives say this could occur as soon as 2024.


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Mum sets up food bank in her own home
A mum has set up a food bank in her own home to help people vulnerable people in Leicester.
Emma Marsh, 37, began the heartwarming project in her Earl Shilton home several years ago, after working with numerous charities such as Help the Homeless.
Through her work, the mum became concerned that many young families and working parents were struggling to get hold of vouchers to use at food banks.
To get hold of a voucher, a person often needs to be referred from a GP or the Citizens Advice Bureau, meaning many can’t access banks and continue to struggle.
Emma, who has always kept sleeping bags and other essentials in the boot of her car, decided to set up her own food bank at her home so families can drop by instead of being referred.
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‘I know that there are loads of food banks in Leicester but for a lot of them people need referrals and vouchers so I decided to do it from home. It’s heartbreaking seeing people struggle,’ she said.
Thanks to help from her children and generous donations from locals, Emma managed to fill hundreds of shoe boxes to give back to the community over Christmas last year.
After last December’s positive response, the self-employed mum wants to encourage even more to donate toys, clothes, and food in the run-up to this festive season.
Emma said she wants to ‘reach people who will struggle over the Christmas period, especially those whose families are on low incomes or children who would usually get free school meals’.
Through word-of-mouth and social media, Emma has been able to reach hundreds of people and receives regular calls from people in need.
Emma often spends her own money to donate products that others haven’t, such as nappies and sanitary products, adding: ‘Even if it just saves someone twenty pounds it’s worth it’.
She added that her six-year-old daughter, Amelie, helps to organise things along with 16-year-old son, Jack, who now wants to join the Red Cross.
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The mother, who accepts donations to her home or picks them up herself, added: ‘Even if someone just wants to donate one jar of pasta sauce I will happily go to pick it up because that person has made an effort to help’.
As part of the donations this year, she plans to include Christmas crackers and other festive treats and is now calling for anyone who can help to donate.
She thanked everyone that has donated so far, adding: ‘It just shows that sometimes a little bit of kindness really goes far’.
Got a story for Metro.co.uk?
Get in touch with our news team by emailing us at<EMAIL_ADDRESS>For more stories like this, check our news page.


In a couple of sentences, how would you describe the command line in plain English? Can you think of an analogy for it?
The command line is quite simply just another method to access and get an overview of the file directory on your computer. It can also be used to extract (clone) information from e.g. GitHub, create new folders and files and upload (push) these files back onto GitHub along with many other functions!


Sentinel Rayen:
The orcs of Fog Mountains are becoming bolder and bolder lately. Before, they wouldn't dare set foot into the Elven Forest, and now they dare to come into the village.
Pretty soon, we plan to recruit brave brothers to help us defeat these fearless invaders. I hope you will take part in this when the time comes.
(Quest for Elven characters level 6 and above.)