Can you solve my favorite interview question? (math + cs)

[@jomakaze]

Check out my second channel: kzbin.info/door/WhwybsQg-EeBJpsqtmbscw for snackable educational content

[@Ultrajamz]

Joma Tech what laptop or screen are you using to draw? Ipad? What might you recommend to do the same thing using linux/windows machines?

[@ppcamps]

God damn, man I found your channel today, I loved it the way that you put some funny things in videos, congrats! I thxs 4 this tip

[@daxi-yw8ui]

Hey bro calm down that's called Montecarlo and it's a procedure literally everybody knows and my computer science teacher gave me this shit First day of second year High school

[@freemind.d2714]

There have onther way to do so, and generate one point only use the random function once : ) 1: Take 1 as fixed radius, Hand choose a circle angle N ( any angle value with in (0, 180] ) Loop: (Reapet 2~3, get as many point as possible) 2: Use uniformly distributed random function generate number 0~1 and multiply it by N as random angle 3: Draw the point on polar coordinate system, format: (radius, angle) 4: Calculate the distance between two adjacent points and add them up to approximate Arc length ("adjacent " can be done easily by ordering the x, y value of coordinate) 5: Because: Arc length = N/180·πr, and r = 1; so: π ≈ approximate Arc length * 180 / N More point we have, the smaller N we use, better the approximation we get, as this kind of prerequisite the only thing we can do is approximation.

[@tisonludovic6858]

num_point_total will be n 😉. You should try with 3 or more dimensions. Monte Carlo is made for bigger dimensions.

[@hylash5444]

I feel like he only made this video to flex on us, especially his Patrick drawing skills

[@aaryashah2462]

He does that in every video

[@jgfiugdiugiud9132]

@@aaryashah2462 true lol

[@ansadahmad1958]

If you have been watching his videos you would know he is a major flexer😂but the good kind

[@Phoenix-hd1tk]

Man, i am so dumb.

[@paolosalgo6987]

bruh. the boy is not humble and i love it.

[@Waiel]

My favorite interview question is *“What’s your name?”*

[@moustafadarwish3674]

and i sometimes get it wrong

[@DrRAZI99]

Thats a pretty hard question

[@КостяПак-я5я]

Such a tough question.

[@namenotavailable9989]

Too bad. My name is harder to pronounce.

[@pipedownandeatyourpudding2986]

its my favorite as well, but it can get difficult at times

[@beachego2654]

It was actually really funny when he was 'acting' to be smart saying 'oh, you didn't know the formula to the area of a circle?' and then the next shot is a view of his computer screen looking up the area of the circle. It was clever how he did that transition ha.

[@SO-fb4ef]

I didn't even realise 😂😂😂

[@TimBoundy]

It was so well done that I didn't even notice it

[@arminbond2336]

😂😂😂

[@vraycool]

What the- 😂😂😂 didn't realize that!

[@bali5890]

I was about to call you captain obvious but seems like many people didn´t realize it so the comment is fine.

[@joeface448]

"Fuckin' idiot..." made me almost cry from laughing so hard. You bastard, gotta call me out like that xD

[@nathanmagro928]

It was funny, he had the equation looked up in google on his screen

[@zilberorph9879]

Through jokes its clear who's full of shit.

[@docbrown2045]

Yeah I felt offended too, but also I laughed like never before.

[@goodguysaladshendelzare1919]

@Nathan Magro hahahahahaha wth

[@Ledeblogger]

Long time I didn’t laugh with yt video

[@vejovim]

Please do more of these 10 minute videos with interesting questions and their answers! This would be a great format to learners. As soon as you posed the question, I hopped on pycharm and coded it up! It was fun!

[@Ben-xj6su]

They asked me this at my interview for "assistant to the cashier" at McDonald's. I aced it.

[@YunisRajab]

Lmao the whole interview should be can you count and stand a lot?

[@steelfalconx2000]

Damn you're a cashier's assistant!? Jealous!

[@vancekangyishu]

@@steelfalconx2000 assistant to the cashier

[@oregano2771]

@@vancekangyishu Dwight.

[@theakatsuki2113]

He is kidding ,he actually knows it

[@chaulenguyenminh2538]

7:32 Actually 2.92 is closer to Pi than 3.6

[@enki7952]

thats called the BrAiN GaMeS

[@Luis-Torres]

Oh you didn't know? *chuckles* idiot.

[@darienrafaelmarquezvazquez7694]

@@Luis-Torres "Google: What is more close to pi, 2.92 or 3.6????????????????"

[@mathnetic7511]

I saw that too. Xd

[@Priestessfly]

damn brilliant

[@jorgevasquezang]

3:04 I needed to pause the video because I was laughing too hard

[@arthurcasillas2912]

Same, lol

[@ameenurrahmankhan6933]

Same here 😂😂😂😂😂

[@daniel-fich]

I didn't even realize at first. That's amazing

[@daniel-fich]

@Haze The Space Commie Look at his monitor at 3:04 ...

[@Tom60]

Same 🤣

[@mgancarzjr]

When you're finished, you can show them another way and name-drop Euler for extra credit. double x = 0; for(int i = 1; i < some_limit; i++) { x += 1 / (i * i) } print(sqrt(x * 6)); // fixed by Eutu Salli The other, modern, fast formulas (Ramunajan and Chudnovsky) include polynomials and factorials. This would either require using libraries or writing your own functions.

[@thaituanly8955]

how bout - Using a presentation of a circle (center (0,0) r = 1) in a cartesian coordinate system (which would be x^2 + y^2 = r^2) - Find the area that is bound by the circle in the first quarter and the x and y axis using integral, and this area would be 1/4 of a circle - Calculate Pi from that area This way (if works) will require 0 call for random(), and gives you constant runtime.

[@mgancarzjr]

@@thaituanly8955 the integral of sqrt(r^2 - x^2)dx is 0.5 * (x * sqrt(r^2 - x^2) + r^2 * arctan(x / sqrt(r^2 - x^2))) Assuming you can pull this formula out of your head or derive it (I cannot), the arctan() function becomes a problem. If you can't code an arctan function, you'll need to rely on a math library. If you can access the math library, looking up pi will be far faster. Converting to the polar coordinate system creates a far easier integration and avoids arctan, but then you're integrating in radians which require the use of pi.

[@mgancarzjr]

@@thaituanly8955 you can do a Reimann sum where you accumulated the areas of very thin rectangles from 0 to 1 whose heights are the sqrt(1 - x.position^2). The issue with that is the same issue with the Monte Carlo simulation in that you're calling a sqrt function n times which is slow.

[@etusax]

Are you sure your formula there is correct? I implemented your solution and got results pretty far from pi (around 7,7). I'm pretty sure I triple checked the implementation to be correct.

[@mgancarzjr]

@@etusax you are correct. It should be sqrt(x * 6). I posted a link to a working implementation on an online compiler/debugger, but KZbin deleted the post. I must be on a naughty list.

[@vasiliynkudryavtsev]

Fun fact is that you don't need the random function to compute with this technique. Just using points on the grid would result the same. E.g. double loop for x and y. The finer the grid - the more precise is the result.

[@bastiaan9127]

That's what I was thinking. Regular grids should even be slightly more accurate than random numbers. But be careful to exclude two of the four borders. Starting off with random numbers is probably just to confuse applicants, or test their ability to generalize and simplify.

[@fruitygranulizer540]

yea but the point of the original question is to find pi with a bunch of seemingly ranodm numbers. if the question mentioned a grid, its a bit easier to come up with the solution.

[@jeffery_tang]

you would have to deal with edge cases like what if the point was right on the line, but that's probably a trivial solution to it

[@Andreas92]

3:04 googling the formula lmao, nice touch

[@pominchuang462]

lol

[@escanorescanor9025]

"Oh U didn't know? Fucking idiot"

[@l.1244]

Pointing out the obvious makes it unfunny.

[@tekneiq]

He got the joke boys, pack it up.

[@hl7297]

@@l.1244 I didn't pay attention to that detail so his comment's still relevant

[@tyto125]

This dude is flexed so hard that youtube had no other choice but to recommend this to everyone

[@Dgsrgv]

honestly, sometimes i find it really hard to tell whether joma is actually being serious to straight up trolling. Cuz him having a serious face can easily fool someone gullible like me.

[@barryrims9098]

That's cuz ur stupid it's ok

[@Dgsrgv]

@@barryrims9098 its ok i know u aint got friends

[@Dgsrgv]

@@choiceindot its ok i know your a low energy human being trying to find the meaning of your existence ill just give it to you

[@Dgsrgv]

@@choiceindot your actually an idiot lol

[@Dgsrgv]

@@choiceindot you implicitly admitted you`re an idiot lol nice job idiot

[@IgnisZephyr]

There is another way (more computing intensive though), you can transform the numbers into integer and check if they are coprime. The probability that two integers are coprime is equal to 6/π^2. And boom, you have a solution that is way longer to run

[@kolbstar]

That's using a lot more than just a uniform random variable generator. You're using multiplication, square root, and the definition of coprime.

[@StrangerHappened]

@@kolbstar the random generator is already ridiculously overboard, unnecessary and slow for pi calculation. So the root comment is just upping the game. And I am sure there are ways to calculate pi that are even slower. There HAS to be a competition for the slowest algorithm.

[@kolbstar]

@@StrangerHappened I planted a tree in my back yard. I will wait for it to grow to maturity, cut it down, dry it, put it on a lathe to make a rod, then measure its circumference and diameter.

[@thaituanly8955]

how bout - Using a presentation of a circle (center (0,0), r = 1) in a cartesian coordinate system (which would be x^2 + y^2 = r^2) - Find the area that is bound by the circle in the first quarter and the x and y axis using integral, and this area would be 1/4 of a circle - Calculate Pi from that area This way (if works) will require 0 call for random(), and gives you constant runtime + better accuracy.

[@astronemir]

@@StrangerHappened that’s not true the random generator is a pretty damn good empirical way of calculating pi.

[@Anshuhd]

I felt like I was being humiliated this whole time

[@antibulletdodger101]

lol same here

[@sproutdood]

When you realize he actually searched up for circle area formula. 😂😂😂

[@bjorn1724]

Did you like it? "50 shades of Joma Tech"-like?

[@jhay_vine5083]

Bro 😂😂

[@TheJustinmulli]

Better solution without the random function (could suggest both solutions, although they never said use the random function anyway) Calculate the area under the first quadrant of the unit circle curve using a riemann sum and multiply it by 4 x = np.linspace(0,1,1000) pi_approx = sum(np.sqrt(1-x**2)*x[1])*4 Also for the solution using the random function just create arrays of random numbers and perform the operations once instead of looping: x = random(n) y = random(n) pi_approx = (sum(x**2+y**2

[@tiankaixiong2340]

That's like the very first example any professor would show when teaching about Monte Carlo.

[@TrioLOLGamers]

My teacher inserted that in an exercise for the test... It was cool, but later I discovered that he copied everything in the tests from forums... So he knew only few things... Not as teacher at the University

[@CamaradaArdi]

@@TrioLOLGamers at least here in Spain you have to be a doctor to be a professor

[@TrioLOLGamers]

@@CamaradaArdi really?

[@srgk26]

@@CamaradaArdi Isn't that how it is everywhere?

[@srgk26]

@@jarolrivera640 I suppose you're from the US? Don't know how it is there, I know PhD students also teach but didn't know they're called professors too. I'm in the UK, and here only a fraction of PhD holders even have a chance of possibly becoming professor. It's the highest "rank" you can have in the academic ladder, and even that's not tenured. If you don't produce results in a certain time, you'll be asked to leave. Though by that time, you're probably quite crazy about the field you would probably produce results. With regard to lecturing, at research intensive universities would normally be working for at least 5-10 years or so to be a lecturer. Not sure if this is a good setup but that's how it is here.

[@AashishMishra_Nygma]

1:40 to 1:43 can cure depression, end wars and achieve world peace

[@AashishMishra_Nygma]

don't believe me? click it back to back

[@gabrielrezcp1788]

@@AashishMishra_Nygma shure i do

[@badunius_code]

8:40 you could also calculate root mean square of M estimates of size N, and get better precision while maintaining O(n + m) complexity

[@therealdanielfournier]

My first thought was to use the randomly generated points to do numerical integration. I had to first generate points, then sort them, then integrate numerically using trapezoidal sums (average of upper and lower sums). Since i used a simple sorting algorithm that would be in n^2 time it doesn’t seem to be a very efficient solution. I like the way you did it without getting overly complicated. I’ve spent so much time doing heavy pure maths proofs so my brain must have been stuck up there and missed the easier solution. Good job!

[@Philgob]

how the fuck is this your first thought

[@Governorrr]

Joma: *gives function total number of points as n* Joma: *proceeds to calculate total number of points as num_point_total*

[@phamtuanbinh9655]

@ :v Then why won't he name the func param like that. It's simply because of the order of his thoughts.

[@icaruscorp1529]

I think it’s probably a habit from working on bigger projects, where a single letter variable isn’t really a good idea if multiple people work on it

[@tutankhamen5080]

Legit happens to me

[@soumojitchowdhury151]

@@icaruscorp1529 but its a waste of time. You are passing n to the function and also counting it in side the loop, it adds extra computation for literally no reason

[@bo17age]

you got me b*tch!

[@va9if]

I thought he wasn't serious when he said "calculate π"

[@pablogarin]

in python: use command help('modules') to see all modules installed, use command help(module) to see documentation, use command dir(module) to see available commands

[@marceloaugusto7830]

or just use google....

[@ShibayanMondal]

2:58 The greatest sense of humour I have ever seen!

[@jepoyburner]

This is the classic Monte Carlo introduction problem. Ptsd to my thesis days.

[@yogendraadhikari4275]

If someone asks me this kind of question, I would just walk away...

[@jhay_vine5083]

Facts bro 😂😂😂

[@faizanali193]

To be honest, If someone applies for data scientist positions and doesn't even know the basic Monte Carlo, then he should definitely walk away. So many ML/DL junkies wanna be a data scientist these days, can't even pass a basic statistics test...

[@stm12p]

@@faizanali193 who the hell claims themselves DS guy if they don't know Monte Carlo? i mean, you could do it without any coding skill.

[@Dew_-km7gr]

3:01 my math teacher every time I ask him something 😂😔👌

[@pukhatozhimo]

Lol, I'm a math teacher. 😂😂

[@odikuhschrank7936]

Hey! I wanted to add a little side note for those that are interested in the maths behind it. Lets imagine that we are throwing arrows at a square of side length 1 (Lets call it "A"). Then the Probability of our arrow landing in the Part of the circle that intersects the square (Lets call it "C") is: Area(C) / Area(A) = pi/4 / 1 = pi/4. Now let X_k be a random variable, with X_k : [0,1)^2 -> {0 , 1} | (x, y) -> { 1, if |(x,y)| < 1 ; 0 in any other case We therefore have: P(X_k = 1) = pi/4, P(X_k = 0) = 1 - pi/4 => Expected Value: E(X_k) = P(X_k = 1) * 1 + P(X_k = 0) * 0 = pi/4 X_k represents our k-th throw. You get 1$ if your arrow lands in the circle and 0$ if it doesnt. We now use the "Weak law of large numbers" which states that 1/N * ( sum(X_k) from 1 to N ) -> E(X_k). Here we are just taking the mean of all our payments gained from throwing arrows. This converges (in probability) towards E(X_k) = pi/4. By using the markov-inequality we can even calculate the minimum number of throws we need to have a certain probality that we are closer than a>0 to pi, for any a>0.

[@amaizel]

This is one of the first cases of Monte Carlo you learn! Very simple and elegant solution to estimate Pi. The funny thing is that you can do the same calculation using a cardboard, a baby powder and a scale!

[@Hyttelus]

This was literally one of the exam question in a programming and microcontroller course I took this semester.

[@jdmaine51084]

Dude... this is the second video of yours I've watched, and so far they're both hilarious. I genuinely thought you were calling me an idiot, and then the next cut shows the area of a circle Google results on screen hahaha.

[@Supertweeb]

5:32 that googling google to go to Google killed me 🤣🤣🤣🤣

[@jamiejackson2144]

Instead of using another variable incrementing for total points, you can directly use "n" which came as input argument.

[@ellmango]

lmao

[@javand86]

The fact that I'm doing my homework and its a very similar problem to this, so I'm using it for a reference. So glad I watched this earlier, great content man!

[@braer240]

I swear I laugh the most watching your videos. So many programming/coding/tech channels try to be funny and it’s just awkward. But joma always delivers

[@Oscarrudnas]

I was going to comment: I could learn this all day if you would teach it. And then joma academia. Nice thanks

[@stevenspearman5697]

5:27 - You went to google.com to google google and clicked on the first google link to take you to google. BIG BRAIN!

[@anubhavagrawal3963]

Teaching recursion for free. 😌

[@abeecee]

More of these pls I love them, also the coding explanations really help beginners. Love u joma

[@frankpujo8073]

2:52 don't mind me, just wanted the golden moment of the video to replay

[@luchodasavage]

I don't even like math, but can't stop watching videos of this dude

[@Glendragon]

first number: 3.6 (0,46 away) second number: 2.92 (0,22 away) Joma: "that is not more accurate"

[@francoisa3231]

You can actually just divide by n

[@jomakaze]

Yea oops

[@3lietechhack646]

How? Could you please explain?

[@francoisa3231]

nb_loop = 10000000 nb_point = 0 x = 0 y = 0 for _ in range(nb_loop): y = x x = random.uniform(0,1) if x**2 +y**2

[@Hellopiermy]

Super fast way: from numpy import sum, random as r N = 10000 (your choice) sum(r.rand(N)**2 + r.rand(N)**2 < 1) * 4.0 / N

[@felipemotorhead]

It doesn't matter, I'm not going to write any of these calcs for my applications anyways

[@lexus7154]

Omg, i love this guy!! I'm a data science major and I am so happy I stumbled across your channel. I'm gonna sign up with your services.

[@wristocrat]

Easily the funnest programming tutorial I've ever seen, great information too

[@bogdns]

num_point_total is really useful( “n” is useless)

[@jeremyshaw1]

Interesting question, thanks! This took me way longer to program than it should have (nearly 30min in C). I did learn that floats can get less precise as the numbers get more massive (which I somehow forget; it's obvious), and to use doubles in this case. My INT_MAX iterations got me to 3.141575.

[@angry_moose94]

Hey btw if anyone is trying to plot the points in the circle graph like I did, remember that the points range from 0-1. So only in the upper right quadrant. If you want to plot them in a full circle, you have to change the range from -1 to 1.

[@kxmode]

"You have a function called random that generates a number from 0 to 1 randomly, and it's uniformly distributed. Now calculate the number Pi." _Question 1A:_ var a = []; function Random(min, max, cap) { for (var i = 0; i < cap; i++) { var v = (Math.random() * (max - min) + min).toString(); v = v.substring(0,4); v = parseFloat(v); a.push(v); } return a; } // returns 50 floating point values between 0 and 1 in an array console.log(Random(0, 1, 50)); Example output: [0.95, 0.76, 0.7, 0.48, 0.89, 0.75, 0.26, 0.63, 0.28, 0.91, 0.48, 0.36, 0.56, 0.29, 0.23, 0.64, 0.5, 0.73, 0.2, 0.87, 0.81, 0.02, 0.97, 0.91, 0.4, 0.77, 0.24, 0.71, 0.62, 0.08, 0.89, 0.93, 0.94, 0.8, 0.49, 0.84, 0.91, 0.26, 0.68, 0.23, 0.98, 0.55, 0.58, 0.77, 0.01, 0.95, 0.15, 0.54, 0.96, 0.11] _Question 1B:_ 3.14159265359

[@Oleg-mv6cx]

import math n = int(input("Input: ")) pi=0 for i in range(n): if i!=0: pi += 1/float(i*i) pi = math.sqrt(6*pi) print("Output: "+str(pi))

[@omegacon4]

This was a outstanding video. Thanks for making it. I'm just starting to learn Python so appreciate the coding segment in this video as well.

[@tarunpatel8168]

Joma you are amazing, I laughed hard at 5:30 😂😂 Amazing content too

[@OpannapO]

agree with u 😂

[@nikolatesla399]

😂😂

[@bryanfeliciano4102]

I look at coding interview questions everytime I wanna feel stupid Edit: I learned python, javascript,and c++ and algos still make me feel stupid 😂.

[@modellking]

tbh most usual interview questions are easy at companies outside of FANG and where the bosses understand anything about engineering. (Had some during the last year)

[@Heroseuss]

Lmaoo

[@DrRAZI99]

Lol

[@bryanfeliciano4102]

@@DrRAZI99 glad that made you giggle 😂

[@nyb_ok]

Its ok. Noone can know everything.

[@soumojitchowdhury151]

That logic of thinking with the graph is where the main clue is... Awesome idea

[@krisztianszabo3588]

import random def calculate_pi(n): return 4*(sum([1 for _ in range(n) if random.uniform(0,1)**2+random.uniform(0,1)**2

[@calosth]

All the knowledge inside this piece of humor, I had to stop and laugh for every joke. Really well done 👏🏼

[@ameysawant2301]

Why no one is talking about how funny at the same time educational his videos are😂😂😂😂😂😂😂😂😂 !!!! You are doing a great job JOMA ✨✨✨✨🔥🔥🔥🔥🔥

[@firepro20]

I really love learning from this channel, it makes learning complicated stuff fun, with a perfect blend of humour. Also, I really like that Patrick, let the artist know it's great work, whoever it might be.

[@yoonhero3701]

Joma, How about this way? import math def getDistance(x, y): return math.sqrt(x**2 + y**2) def getPos(x, y, r, total): return r/total * x, r/total*y def getPI(n): radius = 1 num = 0 for x in range(n): for y in range(n): posX, posY = getPos(x, y, radius, n) if getDistance(posX, posY)

[@manucanedo8039]

Did it in a different way. Calculated points in a hemisphere centred at 0.5 and with those calculated half a perimeter. With the perimeter calculated PI. Get's closer the more points you generate. Here's the C++ for it: #include #include #include #include #include constexpr double RADIUS = 0.5; constexpr double CENTRE = 0.5; constexpr double ITERS = 1000000; constexpr double get_abs(const double x, const double y) { return std::sqrt(x * x + y * y); } constexpr double get_y(const double x) { constexpr double sqr_H = RADIUS * RADIUS; return std::sqrt(sqr_H - x * x); } double rand_double() { return static_cast(rand()) / static_cast(RAND_MAX); } int main() { srand(static_cast(time(0))); std::map data; for (int i = 0; i < ITERS; ++i) { double x = rand_double() - CENTRE; data.emplace(x, get_y(x)); } auto prev_it = data.cbegin(); auto it = ++data.cbegin(); long double circle_len = 0.0; for (; it != data.cend(); ++it) { circle_len += get_abs(it->first - prev_it->first, it->second - prev_it->second); prev_it = it; } printf("Calculated PI is: %Lf", circle_len / RADIUS); return 0; }

[@mamertvonn]

5:26 dis cures my depression...

[@pmoe7]

I genuinely died when he searched up the area of the circle 😂😂😂

[@saisagar5639]

Yes

[@thelastilluminati6959]

why are you still alive then

[@legohexman2858]

@@thelastilluminati6959 because he was revived by the funny

[@EmissaryOfSmeagol]

pdf of given funciton: p(x)=1, x in [0,1], 0 otherwise. integral(0,1) 4*arctan(p(x)) dx = pi

[@alexchepilko4298]

My first though was gauss integral

[@videostop7564]

I concur

[@MoritzWallis]

This is super smart. Never would have thought of this.

[@net2cn]

I've done this on my TI graphing calculator back when I was in high school and it was a ton of fun learning about the Monte Carlo method.

[@TrueZenquiorra]

Hey Joma! I really loved this content, gonna share with my friends to blow their minds.

[@muhammadwasi5460]

Dude this man is my favourite...except he shows up on my interview day lamo..i'm dead...

[@PriyanshGupta2906]

This is called the Montecarlo simulation

[@jomardomingos7886]

Your sense of humor is great!

[@jaycee6689]

I actually understood this omg. Thanks Joma!!! thanks for the vid might also add that youre “weirdly” entertaining like you know how to do it somehow

[@Martyballin]

These questions are good for boosting the ego of the interviewer, nothing more.

[@jeremyshaw1]

Supposedly, they should be near impossible questions that aren't realisitically solvable within the confines of an interview, and the purpose is to show the approach and problem solving of the interviewee. A kobayashi maru, perhaps. Obviously, it doesn't quite work that way.

[@marwanassih3582]

It's a classical problem. making an approximation of pi using proba. here we used the monte carlo method.

[@ignaciosavi7739]

@@jeremyshaw1 i solved it in 5 mins.

[@daniele4839]

@@ignaciosavi7739 I solved it in 2

[@ignaciosavi7739]

@@daniele4839 great job!

[@calvintran6544]

6:08 when he typed in x^2 first before x**2 😂 gets me every time

[@linx3519]

Joma: .. the area of the circle? oh you didn't know? 3:04 😂😂 love this one

[@l.meyers3348]

“Fuckin idiot”…both makes me laugh and cry. Genius, just genius!

[@mionszu]

Wow, I remember that. I had it once in my high school classes. Good to know it can be an interview question

[@josephajibodu2377]

I thought I've seen it all at 3:04, I never knew there was more at 5:30. 😂😂

[@callforjhed]

when he started typing "p" in google all the search results and auto texts were blurred and the titles are long and there are dashes and xs and stuffs.... 😆 just kidding. nice video by the way cheers!

[@saimyintmyat9373]

Pork 😂😂😂😂

[@jeps7688]

5:29 thats me having trust issue googling google.

[@vejymonsta3006]

In chrome

[@futbolcolombiano320]

I didn't get why he did that. Would you let me know?

[@michaeldang8189]

That is a recursive algorithm

[@Saladmints]

Based off of just drawing the quadrant with all the dots: you find the proportion of points where x^2 + y^2

[@stevesmith9447]

Tell your interviewer you'll compete against them. They write their version, you write yours. If yours is faster, you get double your salary the first year. If it's slower, you get one tenth your salary the first year. When they accept, skip the RNG. Use a uniform field. You still take a resolution, N, and you take the same approach. But for starters, you work only in the +x,+y quarter of the unit circle (the area ratio is the same). There are N^2 points in the square. How many are in the circle? Start with the point D = (sqrt(0.5), sqrt(0.5)), which is on x=y and also on the unit circle. It reveals a square of size 0.5*N^2; those points are all within the circle. That's O(1). Now we can make two triangles by drawing lines from D to (1,0) and (0,1). The area of these is also within the circle; their combined area is sqrt(0.5)*(1-sqrt(0.5))*N^2. Again, O(1). From there, we just need to iterate from the hypotenuse of the two triangles out to the circle edge in order to compute the number of points in each column that are within the circle. We only need to do it once and then double it because they're symmetrical. We can search the upper triangle by rows and binary search over N to the edge for each row. We want to search by rows because 1-sqrt(0.5) is less than sqrt(0.5), so we're minimizing the iteration. 90% of the points are computed in O(1). The remaining 10% are computed in approximately (N log N) / 4 operations. log(1M)=20 (about). 5M*0.1 = 0.5M. Twice as fast as the O(N) RNG algorithm.

[@hypno5690]

And then they say, no thanks just complete the question please. I bet you win so many shower arguments. Like using the rng is the whole point of the challenge lmao.

[@bluehornet6752]

Very cool algorithm. Messing around with this a bit, and implemented it in C++ and C# as well. The results end up being very close to the actual value of Pi, just as shown here in the video. The cool this is that the C# version executes just about as fast as the C++ version does. I guess that's to be expected though, given that the heavy-lifting is done in the loop.

[@carlavntr]

What kind of program do you use to draw like that? Thank you you're awesome

[@nilanjanmukhopadhyay8369]

Solution with time and space complexity O(1) When I have heard the problem I started to think about it in a different way than the solution he showed. Later, I attempted to find the solution in my way nonetheless. I have come up with the solution with the time and space complexity of O(1). However, I doubt if it would be considered as a solution. The algorithm not only will solve for any number at the range [0, 1] but also will work for literally any number in the range of (-infinity, infinity). The process is as follows, *f(x) = cos(x)* This function will spit out real numbers in the range of [-1, 1]. Let's just play with the function a bit so that it spits output in the range [0, 1]. _f(x) = 0.5 * cos(x) + 0.5 // domain = (-infinity, infinity), range = [0, 1]_ Now, whenever the output will be 0 the input will be pi. (I know for infinitely many inputs it will output zero but the range of the arccosine function is [0, pi], so no issue in this particular case). Using the randomly generated number the function can be modified in a way such that irrespective of the random number the *x* will always be *pi*. let the_random_number = rand _f(x) = 0.5 * cos(x) + 0.5 + rand // thus if the output of the function is _*_rand_*_ then _*_x_*_ will be _*_pi_* Here on we can solve for *x* _rand = 0.5 * cos(x) + 0.5 + rand_ or, cos(x) = (rand - rand) - 0.5 / 0.5 _or, x = arccos((rand - rand) - 0.5 / 0.5)_ Thus, the *x* will always be *pi* regardless of what actually the *rand* is. However, it turns out that the RHS of the equation always becomes -1. The equation exactly the same as saying x = arccos(-1) which is always pi. Given the process can anybody please argue why or why not it will be accepted as a solution? *#python** implementation of the program* *import math, random* *def func():* *rand = random.uniform(0,1)* *return math.acos((rand - rand) - 0.5 / 0.5)*

[@TNothingFree]

Very neat question, The moment you said PI I said to myself "Probably calculating distances" which I was close to >< Circles and Algebra are one way to challenge someone to inspect how their thinking process works.

[@ChristopherLandicho-uc5vp]

I tried implementing this in C#, and so far, it's been working fine. This is my code : using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Threading.Tasks; namespace Calculate_Pi { class Program { static void Main(string[] args) { Console.WriteLine("Enter the number of points:"); int numPoints = int.Parse(Console.ReadLine()); double pi = EstimatePi(numPoints); Console.WriteLine($"Estimated value of pi: {pi}"); Console.WriteLine("Press Enter to exit..."); while (Console.ReadKey().Key != ConsoleKey.Enter) { // Keep looping until the Enter key is pressed } } static double EstimatePi(int numPoints) { Random random = new Random(); int pointsInsideCircle = 0; int pointsInsideSquare = 0; for (int i = 0; i < numPoints; i++) { double x = random.NextDouble(); double y = random.NextDouble(); double distance = x * x + y * y; if (distance

[@ariadnamartinleon4217]

please, do more videos like this one, with actual coding, it helps a lot!

[@sohumsharma2892]

I was able to solve after you drew the graph

[@hashimalsmael3396]

3:04 I see what you did there I nearly died laughing LMFAO

[@osmiforcentrio7866]

that's like genius. Especially if the interviewers are staring at you ........judging you

[@Whiteleon13]

Bro your humor is real epic love it, not even cringe

[@franciscoramirez-lt3zc]

Took me a minute to wrap my head around this. Here it is for the rest of the smewth brains (I think): Area of a Circle: Pi x r² Pythagorean Theorem: √ (a² + b²) *Note:* A circle has 4 quadrants. Since our numbers are in the range 0-1, we only care about the 1st quadrant (all other quadrants contain a negative in either the x or y-axis). *Process:* If we scribe a square around a circle it will contain the circle's maximum x and y values (1, 1). By taking the ratio (points in circle/points outside of the circle) we can find a number that tends towards pi. So we use Pythagorean to find the location of the point and to populate some of that area in the square (in or outside the circle.)*Since this is only for one quadrant the number will _actually_ be one-quarter of pie.* The more iterations we allow, the higher accuracy we will find (more area populated in the square).

[@tristunalekzander5608]

The question doesn't even say you need to use the random function to solve it.

[@justinchun5975]

Literally no one: Joma: GOOGLES GOOGLE

[@eformance]

I can see this as being a useful way to test RNG data sources, the closer they get to pi, the more random they are.

[@dgory]

import numpy as np def estimate_pi(n): return (4*np.mean( np.linalg.norm( np.random.rand(n, 2), axis=1) < 1))

[@UmarMash]

3:04 i lit laughed so hard... after all the bragging....lol

[@toaderadrian3146]

2:58 ,my brain after exams

[@SurferFlash]

My favorite interview question to ask for engineering positions: - How many tennis balls fit in an airplane?

[@technologyondemand4538]

>asks siri in interview

[@bilbobaggins7075]

At least 1

[@Amsoniac]

Wait now I want to know the answer. Would it be something like B = ball radius A = airplane volume X = number of balls So: A = (Br^2) * X Or some other function, or are you looking to see if they react a certain way. I know that function is wrong but bathroom breaks only last so long

[@Borg8]

double sum = 0; double c = 0; Random r = new Random(); for (;;) { double x = r.nextDouble(); ++c; sum += Math.sqrt(1 - x * x); // you don't need to choose random y, it will be more accurate to compute it from given x. System.out.println("pi~" + sum * 4 / c); }

[@NStripleseven]

You could do the same thing with a smaller circle, checking the distance to 0.5, 0.5 instead of the origin and including points within 0.5 of it.