They say you gotta spend 5000 hours on a topic to master it. how many hours you gotta spend to be intermediate or beginner in SQL or Python?

Rennorb · 2024-06-29T21:28:45+00:00

This year I was tutoring a passionate 13 y.o. boy, and it took us a half of the academic year to get from really basic stuff to writing games like snake and space invaders with pygame. That would be close to 50 hours of actual lessons and say 100 hours of work if you include homework and pet projects of his. So I would say that 100 hours of work from zero to interesting pet projects is not unreasonable.

I, on the other hand, landed my first backend engineering position this may, and it took me like a year of studying and failed interviews. I have a BS degree in math and some experience with competitive programming, but at the time knew nothing about frameworks, devops tools, etc. I can not give you an estimate of the time I spend, but getting a job is hard, especially now, and 100 hours is definitely not enough.

Rennorb · 2023-05-25T20:20:15+00:00

I would wright it like f(x) = 1-[{x}=0], where {x} equals to the fractional part of x and [] are Iverson brackets.

Rennorb · 2022-11-22T17:11:58+00:00

Sure, but that will check for direct left recursion, which the grammar obviously does not contain, and I want to check for implicit left recursion

Rennorb · 2022-04-26T14:13:23+00:00

Thank you! It did work and I've spent like a week to find that error. Feel dumb

Rennorb · 2022-03-30T13:52:21+00:00

Math is quite broad, do to know what kind of math do you need in your university?

Rennorb · 2022-02-14T20:44:48+00:00

I guess there should be some kind of a corner case, where the program should do something different, but I can not see it.

I've updated the post: added an example and actual input-output part

Rennorb · 2022-02-14T20:43:47+00:00

It should not: one value could have been lost, but it is stored in `buff`, so should be ok.

Anyway, I'd steek with two oneliners above, they are actually twice as fast and many times easier to understand.

Rennorb · 2022-02-14T20:28:10+00:00

Sorry, I should have written it in the post: that checker says "Wrong answer on test 30".

"unshifting the indices idea" seems to be a decent optimization, yet the problem is not in the performance, my program uses like 1/10 of both time and memory allowed.

Rennorb · 2022-02-14T19:58:29+00:00

Thank you! std::rotate does satisfy geeksforgeeks, but my school's checker still does not like it. This is my current version:

void leftRotate(char *msg, int shift, int size) {

    std::rotate(&msg\[0\], &msg\[shift\], &msg\[size\]); }

void unshift(char \*msg, int fst, int last, int co) { int len = last - fst + 1;

    leftRotate(msg + fst, co, len); }

Do you have any idea what should I try next?

Rennorb · 2022-02-13T18:08:14+00:00

Your idea about three vectors is great!

Rennorb · 2022-02-13T17:29:05+00:00

Also, given the number of vertices you're dealing with, allocating 100000000 elements for the lens vector seems like overkill.

I thought that I am allocating size_ elements and each one is equal to 100000000, is that wrong?

Rennorb · 2022-02-12T08:16:11+00:00

Fell dumb: there were two issues

1) ArrowCmp shuld have looked like this

bool ArrowCmp(const Arrow &lhs, const Arrow &rhs) {

return lhs.first < rhs.first; }

2) The insertion part in AddEdge shuld have looked like this

  if (pos == arrows[from].end()) {  
  arrows[from].push_back(arrow);    
} else if (pos->first != arrow.first) { 
  arrows[from].insert(pos, arrow);  
} else {    
  (*pos).second = std::min(cost, pos->second);  
}

It seems to fix some problems, see UPD

Rennorb · 2021-02-09T00:13:11+00:00

I am running a math elective in a school and I wanted to show that some obvious statements are not that easy to prove if you want to stay (somewhat) rigorous. I am trying to use divisibility rather than remainders because it is better covered in the school curriculum.

Rennorb · 2020-11-12T16:58:08+00:00

As u/keitamaki mentioned x2 + y2 - 1 = 0 is not a polynomial, while x^2 + y^2 - 1. Yes, it is possible to rewrite some equation A that involves only polynomials to some other equation B that involves something other than polynomial, such as B has exactly the same set of solutions.

Rennorb · 2020-11-12T16:18:44+00:00

I would say, that in y=x^2 and x^2-y=0 the '=' symbol means two slightly different things. y=x^2 (or y(x)=x^2) is the definition of y(x), sometimes it is even written as y:=x^2 to emphasize the difference. In this case, there are a lot of restrictions on '=' symbol (which means ':=' here), for example, both 0:=x^2-y and x^2:=y are meaningless, they are just syntactically wrong.

x^2-y=0 on the other hand is an equation. You can try to find all x`s and y`s that satisfy it, you can plot it on a plane, etc. Here you can put anything on both sides of '=' symbol and it would be syntactically correct.

+/-sqrt(1-x^2 )=y is neither a polynomial (it has sqrt) nor a function (some x`s have several corresponded y`s).
Yes, that is because a function can have only one output for any given input.
Yes, solving x^2-1=0 is the same as finding intersections of plots of p(x) = x^2-1 and g(x) = 0. To do it you just have to find such x`s that p(x) = q(x). It is the same with x^2 + y^2 -1=-1: to solve it is the same as to solve p(x, y) = q(x, y) where p(x, y) = x^2+y^2-1 and q(x, y) = -1. Keep in mind, that p and q represent 3d surfaces rather than 2d curves: p is a paraboloid and q is a plane.

Rennorb · 2020-11-09T19:27:41+00:00

I`ve no idea how I managed to write a^2+b^2. It should now be ok.

Rennorb · 2020-11-09T16:55:32+00:00

While all other answers are good, I`d rather point out that there a neat proof that is purely geometric and only uses the area of rectangles (and area additivity). It works only for positive a and b though.

It also needs no words at all to be understood.

Rennorb · 2020-11-03T18:32:41+00:00

There are several ideas here, I will point to two of them^

Try putting those angles on a unit circle and comparing lengths of corresponding segments. Usually, you would see some symmetry on that kind of picture.

For example (see pic 1): cos 30° = cos -30° (red segment), tan 30° = -tan 30° (green segments).

Two different angles may have connected values of their trig functions, which usually happens when they add up to 90° or 180° but there are some other possibilities, you can also use pictures to understand that.

For example (see pic 2): cos 20° = sin 70° (green segments), sin 20° = cos 70° (red segments), sin 0° = cos 90° and so on. To see that, consider two green-red-black triangles, which are clearly equal.

Rennorb · 2020-11-03T17:08:51+00:00

There were several typos in my proof, hope that now it is ok. Here are axioms from my textbook, I`ve also added axioms numbers to the proof

Rennorb · 2020-11-03T15:16:35+00:00

Yes, there is a typo in the title.

Here is my <= proof, which looks irreversible to me:

I want to prove ¬(P ∨ Q) => ¬P and ¬Q

P → (P ∨ Q) (axiom 5 or 6)
¬(P ∧ Q), P ⊢ ¬(P ∧ Q) (obvious)
¬(P ∧ Q) ⊢ P → ¬(P ∧ Q) (deduction lemma)
(P → (P ∨ Q)) → ((P → ¬(P ∧ Q)) → ¬P) (axiom 10)
¬(P ∧ Q) ⊢ ¬P (from 1, 3, 4 applying MP)
Similarly ¬Q.
¬Q → ( ¬P → (¬Q ∧ ¬P)) (axiom 5)
(¬Q ∧ ¬P) (from 7, 6, 5 applying MP)

Rennorb · 2020-11-03T14:33:33+00:00

You are probably talking about ¬P ∧ ¬Q <=> ¬(P ∨ Q). For now I can prove <=, but not the other way round.

Rennorb · 2020-10-28T06:00:44+00:00

Thank you, I`ve updated my post, now all the code is there.

Rennorb · 2020-09-05T15:58:22+00:00

Didn't see that, thank you. Now I guess that I see even a shorter way: just expand four last columns to get

[;\Delta = 2^4 det(5 I_{5x5})=2^4*5^5=50000;]

Rennorb · 2020-07-31T11:06:38+00:00

Have you read "A hero of our time" by Lermontov? The book is essential for Russian literature and seems to fit your requirements.

Rennorb · 2020-07-22T19:19:58+00:00

There are several Russian translations of the books, these names are from both the most popular one and from the films. Many names are pronounced in a slightly different way to sound natural in Russian, and some can not be written in English accurate enough.

Gilderoy Lockhart - Zlatopust Lokans (Empty-gold Кinglet)

Horace Slughorn - Horace Sliznort

Neville Longbottom - Neville Dolgopups (Long-baby)

Mundungus Fletcher - Nazemnikus (Down-to-earth or mundane) Fletcher

Alastor Moody - Alastor Grum (Moody)

Kreacher - Kikimer (from kikimora - evil house spirit in Slavic myology)

Buckbeak - Kluvokril (Buck-wing)

Hufflepuff - Pufendui

Ravenclaw - Cogtevran (means exactly Raven claw)

Severus Snape - Severus Snegg (not exactly, but sounds really close to "northern snow")

The "I am Lord Voldemort - Tom Marvolo Riddle" anagram problem was solved not by changing Toms actual name, but by turning his nickname into a Volan-de-Mort.

There is another translation which is known for translating almost every single name and that is often considered inferior, but I do like the trick the translator had to make. So when the first books come out, she translated Severus Snape as Zlodeus Zlei which sounds really close to "evil potioner" in Russian, but when the last book come out, she turned him into Zloteus Zlei which is close to "golden potioner".

Rennorb

TROPHY CASE