Is in unreasonable to require FizzBuzz for frontend devs? It's OMG MATH, after all

Is in unreasonable to require FizzBuzz for frontend devs? It's OMG MATH, after all.

Attached: woman-filtered-by-fizzbuzz.png (1600x6283, 1.72M)

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en.wikipedia.org/wiki/Sieve_of_Eratosthenes
cs.hmc.edu/~oneill/papers/Sieve-JFP.pdf
danluu.com/culture/
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is it unreasonable to ask "how to convert from base 10 to base 2"?

there are a million little math tricks and they're not relevant in code monkey positions.

I've never been fizzbuzzed in interviews. All my interviews require me to build an entire CRM within 2 hours. I guess they just really don't want me to work there.

I've been fizzbuzzed on a front end interview. I failed. It was retarded bc it was a large company that had actual back end and front end divisions. Bitch I don't need to fizzbuzz when you're hiring me to make CSS/JS animations you fuck face.

>can't program
>thinks he deserves a programming job

For a developer, absolutely not. For UX person/designer, yes.

Do people really apply to programming jobs and not even be capable of doing fizzbuzz?
That shit is like day 2 of a programming class.

that's easy as fuck too though.

She later got a nepotism boost from a Googler reading her blog and inviting her for an interview. She got to cut through the screening process.

She went to a JS bootcamp in preparation, still failed the interview, and was told indirectly by Goog that she sucks at coding.

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It's perfectly reasonable. All you need is a modulo operator - that isn't anything special either, it's just a fucking char operator like + and -. If you're serious about applying for a frontend job you should be able to do the prime numbers up to 100:
var nums = [];
for (var i = 2; i < 100; i++){
if (nums.every(e => i % e != 0)) {
console.log(i);
}
nums.push(i);
}

I like your 7-line implementation of the Eratosthenes sieve.

Thank you user :)

That's not the Sieve of Eratosthenes, see:
en.wikipedia.org/wiki/Sieve_of_Eratosthenes
cs.hmc.edu/~oneill/papers/Sieve-JFP.pdf

they wpuldnt need isodd if they knew fizzbuzz

Honestly yes they do. They learn enough buzzwords to get past a recruiter/resume filter. You need to weed them out early.

One company I worked for ended up actually hiring one of those retards. By lunchtime on his first day it was obvious he had no idea what he was doing.
Stupid euro employment law and retarded HR meant that it took like a month to get rid of him.

No, in fact we should be giving them NP-hard problems. That's how you really filter by IQ

Give me an np hard problem

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Write a program that will find the prime factors of any integer. It must be O(n log n) or better.

Dan Luu claims—based on real-world experience—that it's unreasonable to require FizzBuzz for people writing concurrent processor microcode.
>Since we were a relatively coding oriented hardware shop (verification engineers primarily wrote software and design engineers wrote a lot of tooling), we tried asking a simple coding question where people were required to code up a function to output Fibonacci numbers given a description of how to compute them (the naive solution was fine; a linear time or faster solution wasn't necessary). We dropped that question because no one got it without being walked through the entire thing in detail, which meant that the question had zero discriminatory power for us.
>Despite not really asking a coding question, people did things like write hairy concurrent code (internal processor microcode, which often used barriers as the concurrency control mechanism) and create tools at a higher velocity and lower bug rate than I've seen anywhere else I've worked.
danluu.com/culture/

# Python program to print prime factors

import math

# A function to print all prime factors of
# a given number n
def primeFactors(n):

# Print the number of two's that divide n
while n % 2 == 0:
print 2,
n = n / 2

# n must be odd at this point
# so a skip of 2 ( i = i + 2) can be used
for i in range(3, int(math.sqrt(n))+1, 2):

# while i divides n, print i ad divide n
while n % i == 0:
print i,
n = n / i

# Condition if n is a prime
# number greater than 2
if n > 2:
print n

# Driver Program to test above function

n = 315
primeFactors(n)

# This code is contributed by Harshit Agrawal

>python
>import math
Stopped reading there. Filtered.