# latex

This wiki has very good and detailed infos: Latex guide 3.
and this too

Latex code begins with: \( and ends with: \)
to center it u can use: \[ and \]

short Latex-test $\lfloor \approx 200\rfloor$
$\boxed{3}$ \(\boxed{3}\) to write this as text use: \\(\)(\boxed{3}\\)

Note by Num Ic
4 months, 3 weeks ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
• Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

## Comments

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Top Newest

i found some interesting TedEx video about coding: Episode10. i found the last episode first, but i assume it's better to watch them from the start ;-) think like a coder

- 1 month, 2 weeks ago

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ah, that, I've already watched the whole series before ;)

- 1 month, 2 weeks ago

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Nice & helpful discussion @num IC

- 4 months, 3 weeks ago

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;-) ty. i didnt meant to make an official post. by accident i pressed save instead of edit.
thanks to u i know how to use notes. since then i use notes to test latex before i post it.
before i discovered this, my latex posts had a lot of post-edit-post-edit-post chains.

- 4 months, 3 weeks ago

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I have deleted the chatbot discussion .Is there any problem or it is fine?

- 4 months, 3 weeks ago

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Your welcome.

- 4 months, 3 weeks ago

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$\large{\boldsymbol{ \Box \oplus} \nabla \boldsymbol{ \oslash \mathbb{O} \Diamond} \Delta \boldsymbol{ \ominus \otimes \odot ~~~ \ast \circ \triangle \Theta \Xi \Phi \Omega} }$
$\large{ \Box \oplus \nabla \oslash \mathbb{O} \Diamond \Delta \ominus \otimes \odot ~~~ \ast \circ \triangle \Theta \Xi \Phi \Omega }$

- 4 months ago

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its all fine. ty again for the idea of creating a chatbot. when i find some spare time i ll try to implement it, like it is shown in the video.

- 4 months, 3 weeks ago

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Your huge welcome!

- 4 months, 3 weeks ago

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$and ends with: 200$ lol

- 4 months, 2 weeks ago

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Very simple latex

- 4 months, 2 weeks ago

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;-) and can be copied (-;

- 4 months, 2 weeks ago

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@num IC i noticed you have been having problems with this Oren-Ben-Dov guy can you png me as a reply to one of his messages for me and i will leave a nice suprise making sure he does not do that again

- 4 months ago

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i just wrote to him. i assume he just have no idea how to have a fruitful discusion.
i dont want you to "send a surprise". i assume talking to him makes more sense.
but it would be nice if you could teach me, how to surprise someone. ;-)

- 4 months ago

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no it is solely on;y for me and percy as hamza has disgraced us and leaked it to another person specificly going against what it is for

- 4 months ago

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@mention[10644076:Toska Pi] i cannot see your post since i m no premium member, if you have a question, please feel free to post it here.

- 4 months ago

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@Num Ic – Thank you, I'll remember that. This was what my post said from the shifting supports challenge: Hi again! This was a very helpful reply and it answered my question. Thank you for helping me out (even though my question was confusing to understand) :):)

- 4 months ago

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as i remember: the question was about "fixed CoM". you are welcome, i m glad that i could help

- 4 months ago

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Latex is a bit confusing, to be honest. I don't really get it, but I don't think I need it much for Brilliant solution posting.

- 4 months ago

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@Jonathan Eshetu nice, you found the page ;-)
since you know how to code, latex will be easy for you. when u wanna try it, i offer my help.
but anyway i just refered to this page, so we can communicate.

- 4 months ago

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Thanks! :)

Can I also chat to you on here about python, instead of just latex? Just in case I have any questions.

- 4 months ago

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;-) here you can chat about everything. there is no off-topic delete like in the daylies.
feel free to ask. there are better coders here, but i will help as good as i can.
i created this page by accident. i just wanted to test the latex output to avoid a post-change-post-chain.
but now i keep it bcs its convenient.

- 4 months ago

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Heyo, someone give me a problem to solve with python.

- 4 months ago

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what about the daily challenges?

- 4 months ago

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I've already completed most of the ones that you could solve with programming...but thanks for the suggestion.

- 4 months ago

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for starters: find the greatest common divisor for two numbers, without using the python gcd-function

- 4 months ago

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  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 print("Pick two numbers to find the greatest common divisor") a = input() b = input() cd = 0 a = int(a) b = int(b) if a > b: for i in range(1, b+1): if a%i == 0 and b%i == 0: if i > cd: cd = i else: for i in range(1, a+1): if a%i == 0 and b%i == 0: if i > cd: cd = i print(cd) 

- 4 months ago

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Is that good?

- 4 months ago

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overal nice. some improvements: a = int(input())

i = min(a,b)
while a%i !=0 or b%i !=0:
i -= 1
cd = i
print(cd)


- 4 months ago

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Wow, I didn't think of that approach, I'm so used to for loops. I feel so stupid for not thinking of using min() too.

I'm assuming you meant b%i by the way.

- 4 months ago

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ty i corrected it.
its not my intention to make u feel stupid.

- 4 months ago

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what is the difference between the condition that i used and your condition?
there is an algorithm from euklid for gcd.
can you implement it as recursive function?

- 4 months ago

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Sorry, I lost you at euklid.

What does euklid and recursive mean?

- 4 months ago

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i assume u will get that in school soon too.
its about a/b = x + remainder and then divide by the remainder and repeat until the final remainder is 0.
i propose to check the web to find it. i can explain it here, but one approach to programming is to find algorithms and implement them.

- 4 months ago

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Oh ok, so I did some research. I knew about the euclidean algorithm for finding the gcd, I just didn't know that was the name. So you want me to implement the euclid approach in python, using recursion (calling a function within itself)? I'll try.

- 3 months, 4 weeks ago

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 1 2 3 4 5 6 7 def euclidGCD(a, b): if a == 0: print(b) elif b == 0: print(a) else: euclidGCD(b, a%b) 

- 3 months, 4 weeks ago

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one change:

def euclidGCD(a, b):
if a == 0:
return(b)
elif b == 0:
return(a)
else:
return(euclidGCD(b, a%b))


very good. you are fast. i would return the gcd, so it can be used instead of print, but as i said: very good and quick

- 3 months, 4 weeks ago

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I used return at first, but I accidentally put  return a  instead of  return(a)  And nothing outputted when I called the function, so I got confused.

- 3 months, 4 weeks ago

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hm, as far as i know, "return a" and "return(a)" should return the same value. how did u do the box around the return ?
return just returns a value ;-).
to print it, call print(euclidGCD(34, 4))

- 3 months, 4 weeks ago

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Ohhh, that's my mistake. I didn't add the print, I just called the function.

- 3 months, 4 weeks ago

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yeah, print is mainly for checking.
to use it, it would be like:
g = euclidGCD(8, 4)

- 3 months, 4 weeks ago

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Yeah, that makes sense. To be fair though, in an actual program you could just use the actual gcd function.

- 3 months, 4 weeks ago

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yeah, the calling of your own functions is used the same way as built-in functions.
since you already created code for the daily challenges, you can also check the coding solves by other users. its a good way to get some coding experience just to read coding done by others.
carsten kaminski writes very impressive code. it sometimes took me a while to understand his cool solutions.

- 3 months, 4 weeks ago

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Ah, I've seen some of his solutions. They're very hard to understand, but I try.

- 3 months, 4 weeks ago

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in your text, the return is in a little box. how did you do that?

- 3 months, 4 weeks ago

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just put three tick marks  before and after.  like this 

- 3 months, 4 weeks ago

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ok thank you, i didnt know that

- 3 months, 4 weeks ago

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You're most certainly welcome. I found it out on accident :)

- 3 months, 4 weeks ago

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Thank you though!

- 3 months, 4 weeks ago

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so check the other codes first. all reading of code helps to get new ideas. (i proposed carstens code just if u wanted to have a challenge.)

- 3 months, 4 weeks ago

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It does. You can learn a lot by reading others solutions on brilliant, one of my favorite features.

- 3 months, 4 weeks ago

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i did the free course on this page: coursera.org
it is in russian, but i translated the pages (i ignored the videos)

- 4 months ago

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I'll check that out, thanks!

I've just been using Brilliant's python course and self-study, but that seems better.

- 4 months ago

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we used different conditions. can you explain what the difference between those conditions is?

- 3 months, 4 weeks ago

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I might have the wrong definition of condition, but essentially you checked if it DIDN'T work (a%i != 0 or b%i != 0), from the top down (i -= 1). This allows the program to stop when it finds a value. In my case, I checked if it DID work (a%i == 0 and b%i == 0) and went from the bottom up (for i in range(1, b+1)). Since we're looking for the GREATEST common divisor, going from the top down allows you to stop when you find a value, because that value is already the greatest possible number.

- 3 months, 4 weeks ago

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very good. you got it and you also explained it very well.

- 3 months, 4 weeks ago

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Thanks! These mini-exercizes/explanations are helping a lot.

- 3 months, 4 weeks ago

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you can also use "python" in the search bar. some users post python quests.
i just found this: Using Recursion (Python loop) and this: Break the Code

- 3 months, 4 weeks ago

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The first problem I got pretty easily, but I'm still confused over the second...

Could you please explain, if you got it? I'm just scratching my head right now...

- 3 months, 4 weeks ago

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nice solution for the first problem.
for the second: its a nice example to show that it might cause problems if a function is used as argument for the function itself.
the return value is 3 + something. so i checked what happened if n=3.
_ = 4 will result in return 0 -> no problem
_ = 3 will result in return 0 -> no problem
_ = 2 will result in return 3+something -> might cause a problem
but you can also use a loop and call the function with different values to check the results. i do this sometimes to figure out if there might be an error.
have u checked the russian course?

- 3 months, 4 weeks ago

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Wait...but by that logic, don't all values not work?

If n = 5, then 4 3 and 2 all result in leaving something....right?

And no, I haven't checked the course out, sorry! I'll get to that soon.

- 3 months, 4 weeks ago

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;-) it was no "task" to check that course. just a proposal.

- 3 months, 4 weeks ago

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:) But still, wouldn't all values be correct for the problem?

- 3 months, 4 weeks ago

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you can just create the code for all 3 _ values and then check it for 1 to 100 as input.

- 3 months, 4 weeks ago

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I'll get back to you in the morning on this, because my head is hurting a little too much right now. Good night!

- 3 months, 4 weeks ago

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np. take care

- 3 months, 4 weeks ago

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Ok, so I think I get it now.

The reason it's 2 is because eventually, the program will simplify to

3 + rec(3)

Because when you keep subtracting the number, it will eventually go below 0, and then everything will simplify to the above equation. But when you calculate the return value, it becomes 3 + 3 + rec(rec(3-_), and if the blank is less than 3, it will result in rec(natural number), which causes an infinite loop. Eg.  3 + rec(rec(5-3)); 3 + rec(rec(2)); 3 + rec(3 + rec(rec(2-3))); 3 + rec(3 + rec(rec(-1))); 3 + rec(3 + 0); 3 + rec(3) (it always simplifies to this); 3 + rec(rec(3-3))) (since the blank is 3, the rec function goes down to 0); 3 + 0; 3; However, if it was say 2; 3 + rec(3); 3 + rec(rec(3-2)); 3 + rec(rec(1)); 3 + rec(3 + rec(rec(1-3))); 3 + rec(3 + rec(rec(-2))); 3 + rec(3 + 0); 3 + rec(3); Hence the infinite loop. 

- 3 months, 4 weeks ago

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very good, you got it.
u can also test it in the coding environment here.
nothing bad will happen. there will just be a message that there are too much recursions.
i did the test with _=4 and n=10000.

- 3 months, 4 weeks ago

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Oh wow. I just used IDLE to play around and see what happened. I set a variable that increased 1 for ever recursion. and then made it stop at 20 so I could review the loop.

- 3 months, 4 weeks ago

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yeah for local test, i used idle too. but here i use the brilliant coding environment.
have u seen that carsten proposed two webpages?
have u checked the community for coding problems, that are posted by the members?

- 3 months, 4 weeks ago

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I've looked around here and there for problems posted by members, but everything is either WAY too easy or WAY too hard.

A website I just found out, that I'm going to use a lot now, is edabit.com. It has a bunch of different challenges anywhere from Very Easy -Expert, with a ton of different languages.

- 3 months, 4 weeks ago

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thats a great hint ty

- 3 months, 4 weeks ago

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I also didn't see the websites carsten proposed. Could you please send a link?

- 3 months, 4 weeks ago

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@num IC this explanation was actually a bit off, as the pattern goes:

3 + rec(3)
3 + 3 + rec(rec(3-3))
3 + 3 + 0
6

or in some cases;

3 + rec(3 + rec(3))
3 + rec(3 + 3 + rec(rec(3-3)))
3 + rec(6)
3 + 3 + rec(rec(6-3)
3 + 3 + rec(rec(3))
3 + 3 + rec(3 + rec(rec(3-3)))
3 + 3 + rec(3)
3 + 3 + 3 + rec(rec(3-3))
3 + 3 + 3 + 0
9

But I'm assuming you see the pattern by now.

- 3 months, 4 weeks ago

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Sorry that it's cramped, I can't do line breaks for some reason. You don't need to read it, I just think I understand now.

- 3 months, 4 weeks ago

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i do linebreaks by two spaces and then return. 4 spaces should create a linebreak too.
i had four spaces before the 4, but it seems not to work.
btw: u can test such things by creating a note.
but u can do it here too ;-) (if the others in this chat are annoyed, they will unsubscribe this chat)

- 3 months, 4 weeks ago

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you can just try this ;-) :

def rec(n):
print(n)
if n <= 0:
return 0
return 3 + rec(rec(n - 2))

N = 5
for i in range(1, N):
print(i, rec(i))


the recursion depth is limited, so big input values will cause a problem even with _=3 and _=4.

task: create a short code to find the max recursion depth.

- 3 months, 4 weeks ago

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Hm, what do you mean by recursion depth? Sorry if that's a stupid question...

- 3 months, 4 weeks ago

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questions are rarely stupid. if the factorial function is called with the input 10, then it calls itself 10 times, if it is called with 100, then it calls itself 100 times. the recursion depth is how often a function calls itself until it finishes.
eg factorial 4:

remember the 4 and call f(3)
remember the 3 and call f(2)
remember the 2 and call f(1)
return 1
return the remembered 2 multiplied by 1 (=2)
return the remembered 3 multiplied by 2 (=6)
return the remembered 4 multiplied by 6 (=24)


-> recursion depth 4

- 3 months, 4 weeks ago

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Ohhh, ok. That makes a ton more sense now. I'll do that program tmrw morning. :)

- 3 months, 4 weeks ago

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Well uhh....I already solved it intuition wise accidentally....

- 3 months, 4 weeks ago

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Now that sounds fun. Though, I'm a bit busy with schoolwork rn, so I'll make the recursion depth program and solve that problem when I get s'more work done.

- 3 months, 4 weeks ago

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sure. schoolwork first.
this is a proposal for a coding environment: repl.it
this is for learning: realpython.com
(the 2nd seems to be similar to your edabit.com)

- 3 months, 4 weeks ago

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I need to test my programming skills.

- 4 months ago

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i can sure propose something, but i would prefer if people use "please" when they ask for something.

- 4 months ago

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Sorry about that.

I was a bit....impatient when I wrote that. Completely my fault. Could you please?

- 4 months ago

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@Mahdi Raza hello Mahdi Raza , you created a very nice animation in this daily problem: Sum These Segments. what tool can be used to create such an animation? or did you move the lines by hand, step-by-step?

- 3 months, 4 weeks ago

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Hi, I have used Keynote. I grouped the lines and then rotated. Refer this if you have any more questions

- 3 months, 4 weeks ago

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this is great ty. i checked your pentagon. there you mentioned keynote. i thought that was only a kind of a powerpoint clone. but you are creating very impressive animations. ty for that and ty for the hint.

- 3 months, 4 weeks ago

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Thank you, would like to see your animations too...

- 3 months, 4 weeks ago

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@num IC It was getting cramped earlier, so I'm just making a new post to continue our conversation.

1) I checked out repl.it, it seems pretty helpful, but realpython seems like basically pay to python.

2) Still working on those 2 problems by the way (smallest 10-digit number and recursion depth.) :)

3) I was thinking earlier, isn't there a much easier factorial function?

 1 2 3 4 5 6 7 8 9 def factorial(n): nFac = 1 for i in range(1, n+1): nFac *= i return nFac # 5 => 120 # 7 => 5040 # 10 => 3628800 # 300 => 306057512216440636035370461297268629388588804173576999416776741259476533176716867465515291422477573349939147888701726368864263907759003154226842927906974559841225476930271954604008012215776252176854255965356903506788725264321896264299365204576448830388909753943489625436053225980776521270822437639449120128678675368305712293681943649956460498166450227716500185176546469340112226034729724066333258583506870150169794168850353752137554910289126407157154830282284937952636580145235233156936482233436799254594095276820608062232812387383880817049600000000000000000000000000000000000000000000000000000000000000000000000000 ;) 

- 3 months, 4 weeks ago

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yeah. and it is faster, since a function call takes more time than a loop.
it is just used as a simple way to understand recursions.
you can also use carstens codes to understand recursions ;-)
ty, i didnt check realpython. i just saw that one did recommend it.
for recursion depth, a simple code is enough. just dont be affraid.

- 3 months, 4 weeks ago

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Here's the program I used for the 10 digit problem:

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 # Input import itertools import math nums = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] for i in itertools.permutations(nums): if i[0] == 0: continue i = int(''.join(map(str, i))) if math.sqrt(i).is_integer(): print(i) break # Output 1026753849 

- 3 months, 4 weeks ago

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nice. i knew that was too easy for you, since you already used permutations. ;-)

- 3 months, 4 weeks ago

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Thanks :D

- 3 months, 4 weeks ago

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And isn't the recursion depth infinite?

As for any number, it would keep subtracting _ until the number became <= 0, and the number could be infinitely large...

- 3 months, 4 weeks ago

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in theory it is infinite. but that would create an endless loop. there is a limit.

- 3 months, 4 weeks ago

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Ok, my mind is blown.

For starters, I edited the function a little bit, so that it measures how many times the function is called.

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 # Input a = [] # This is an array so we can print the values if needed def rec(n): a.append(n) if n <= 0: return 0 return 3 + rec(rec(n - 3)) for x in range(1, 50): print(x, "-", rec(x), len(a)) a.clear() # Output 1 - 3 3 2 - 3 3 3 - 3 3 4 - 6 7 5 - 6 7 6 - 6 7 7 - 9 15 8 - 9 15 9 - 9 15 10 - 12 31 11 - 12 31 12 - 12 31 13 - 15 63 14 - 15 63 15 - 15 63 16 - 18 127 17 - 18 127 18 - 18 127 19 - 21 255 20 - 21 255 21 - 21 255 22 - 24 511 23 - 24 511 24 - 24 511 25 - 27 1023 26 - 27 1023 27 - 27 1023 28 - 30 2047 29 - 30 2047 30 - 30 2047 31 - 33 4095 32 - 33 4095 33 - 33 4095 34 - 36 8191 35 - 36 8191 36 - 36 8191 37 - 39 16383 38 - 39 16383 39 - 39 16383 40 - 42 32767 41 - 42 32767 42 - 42 32767 43 - 45 65535 44 - 45 65535 45 - 45 65535 46 - 48 131071 47 - 48 131071 48 - 48 131071 49 - 51 262143 

- 3 months, 4 weeks ago

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After analyzing:
1) Since it's steadily increasing, isn't the depth infinite?
2) This is crazy. If you add 1 to each of the depths, you end up with a very noticeable pattern.
4, 4, 4, 8, 8, 8, 16, 16, 16, 32, 32, 32, 64, 64, 64... Why would powers of 2 appear here?

- 3 months, 4 weeks ago

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i have to think about the logic to understand why the powers of 2 occure here.
but there is a missunderstanding. my intention was not to check the depth of this particular function. (50 is much too low to test this one).
its more easy to use a simple function that just increases the given value.

- 3 months, 4 weeks ago

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Oh, you want me to create a simple function and then test the recursion depth with another program?

- 3 months, 3 weeks ago

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hm, no. just a small code, that continuously calls itself and then you run it and check the error message

- 3 months, 3 weeks ago

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 1 2 3 4 5 6 7 8 9 a = 0 def letsBreakIDLE(): global a a += 1 print(a) letsBreakIDLE() letsBreakIDLE() # Output stops at 1012 

- 3 months, 3 weeks ago

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nice. mine stopped at 997 when i tried this in brilliant:

def rec(n):
print(n)
rec(n+1)

rec(1)


- 3 months, 3 weeks ago

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Even in IDLE, the function starts to slow down at around 60.

- 3 months, 3 weeks ago

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hm yeah it was just an idea. i thought it might be interesting for you to know how to check the limits of python (or any ide).
currently i feel a bit incapable bcs i am not able to express that intention clear enough.
have you found some interesting tasks on edabit.com?

- 3 months, 3 weeks ago

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Sorry about this, I had my hands full of schoolwork last night.

Don't feel incapable, if anything my limited python vocab is stopping me from understanding.

I've just been doing easy tasks for now on edabit, but I've started switching a bit more to repl.it

- 3 months, 3 weeks ago

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ok. i have tried none of them.
do you reccomend them? do you have questions?

- 3 months, 3 weeks ago

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There was 1 problem, something like "create a stutter function that gets a parameter, and returns the first 2 letters, ..., the first 2 letters again, ..., and then the full word with a question mark.
Ex. stutter(python) -> py...py...python?
I have the idea, I just don't exactly know how to code it.

- 3 months, 3 weeks ago

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t="hello"
print(t[:2]) what did u use to get the first 2 letters?

- 3 months, 3 weeks ago

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That's the thing, I didn't know what to use. Thank you!

- 3 months, 3 weeks ago

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I thought [num:num] only worked on arrays.

- 3 months, 3 weeks ago

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this works in brilliant:

t="hello"
print(t[:2],"...",t[:2],"...",t, sep='')


a string is an array, just filled with chars

- 3 months, 3 weeks ago

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Ah, I didn't think about it that way. Completely forgot about [:] and sep. Thanks!

- 3 months, 3 weeks ago

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yw. i had to google sep bcs i usually dont use it. its also possible to use a formated print, but i dont know how that works.

- 3 months, 3 weeks ago

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this is a good idea to train using strings and dictionaries:
https://brilliant.org/problems/you-ninja-name/

- 3 months, 3 weeks ago

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sorry this is so late.

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 def ninjaToEnglish(phrase): engText = [] ninjaDict = {"a": "ka","b": "u","c": "mi","d": "te","e": "ku","f": "lu","g": "ji","h": "ri","i": "ki","j": "zu","k": "me","l": "ta","m": "rin","n": "to","o": "mo","p": "no","q": "ke","r": "shi","s": "ari","t": "chi","u": "do","v": "ru","w": "mei","x": "na","u": "fu","z": "zi"," ": " ",} phrase = phrase.split('~') for i in phrase: for key, value in ninjaDict.items(): if i == value: engText.append(key) engText = ''.join(engText) return engText print(ninjaToEnglish("chi~ri~ku~ ~ka~to~ari~mei~ku~shi~ ~chi~mo~ ~chi~ri~ki~ari~ ~no~shi~mo~u~ta~ku~rin~ ~ki~ari~ ~ari~ki~na~ ~te~ki~ru~ki~te~ku~te~ ~u~fu~ ~ri~ka~ta~lu")) #output: the answer to this problem is six divided by half (I almost put 3) 

- 3 months, 3 weeks ago

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nice

- 3 months, 3 weeks ago

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thx

- 3 months, 3 weeks ago

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what does this?: engText = ''.join(engText)
and ty: i was not aware that there is no direct access from value to key.
i assume that is why they created that reverse dict in the solution.

- 3 months, 3 weeks ago

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changes it from ['e', 'x', 'a', 'm', 'p', 'l', 'e'] to 'example'

- 3 months, 3 weeks ago

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cool trick. ty

- 3 months, 3 weeks ago

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welcome.

- 3 months, 3 weeks ago

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i would have used:
engText = ""
and then:
engText += key

but good to know how to convert an array into a string.

- 3 months, 3 weeks ago

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true, i didn't think of that.

- 3 months, 3 weeks ago

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I had to do a bit of searching up, since I wasn't familiar with dictionaries.

- 3 months, 3 weeks ago

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yeah thats where i struggled with the online course: i had no experience with python dictionaries.

- 3 months, 3 weeks ago

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Hey @num IC, I finally made a one liner!

Basically, it takes a number, returns its divisors, the sum of its divisors, and if it's a perfect number (a number where the sum of its divisors equals itself) or not.

 1 2 def perfectNumberCheck(n): print(n, "\n", "divisors:", ' '.join(map(str, [i for i in range(1, n) if n%i == 0] + [n])), "\n", "sum of divisors (aside from itself):", sum([i for i in range(1, n) if n%i == 0]), "\n", "Perfect number:", "True" if sum([i for i in range(1, n) if n%i == 0]) == n else "False", "\n", "\n") 

Try it out!

- 3 months, 2 weeks ago

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very cool.
its possible to skip long lines by using \
and i know that for some ppl it is a challenge to solve problems in few lines, but it is more easy to read code if it is split into several commands.
and i m impressed: you already know more about how to use the print command than i do.
so i should try to catch up again ;-)

- 3 months, 2 weeks ago

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I knew I could make the code a lot more compact if it was in several lines, but I wanted to see if I could do it in one, yknow?

- 3 months, 2 weeks ago

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Here it is a bit..neater, because I assigned the divisors array so I wouldn't have to spam it.

 1 def perfectNumberCheck(n): divisors = [i for i in range(1, n) if n%i == 0] + [n]; print(n, "\n", "divisors:", ' '.join(map(str, divisors)), "\n", "sum of divisors (aside from itself):", sum(divisors) - n, "\n", "Perfect number:", "True" if sum(divisors) - n == n else "False", "\n") 

- 3 months, 2 weeks ago

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this is great !! i wanted to say that repeating the same is not efficient, but this 2nd thing is very good

- 3 months, 2 weeks ago

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Thank you! I appreciate it :)

- 3 months, 2 weeks ago

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Hey Num Ic, could you please give me a coding challenge? It can be on brilliant, or anything really. Thanks!

- 3 months, 2 weeks ago

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hm, i had to find some.
i really recommend the corrected coursera page that i posted.
coursera learn python
that was one task i liked:
korovay
it has to return correct russian word for cow (korov, korova or korovy):
For a given number n <100, finish the phrase “Grazing in the meadow ...” with one of the possible continuation: “n cows”, “n cow”, “n cows”, correctly declining the word “cow”.

Input format
A natural number is entered.

Output format
The program should print the entered number n and one of the words: korov, korova or korovy. There must be exactly one space between the number and the word.

Note
If Russian is not your native language and you find it difficult to cope with this task, you can find a hint on declensions here: https://www.coursera.org/learn/python-osnovy-programmirovaniya/programming/rsYze/korovy/discussions/threads/-FJYlWcKEeiMwRLm6nkdwA .
"n коров" if 10 <n <20 or the last digit of n is one of 0, 5, 6, 7, 8, 9.
"n корова" if the last digit n == 1.
"n коровы" in all other cases.

- 3 months, 2 weeks ago

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wait, there's no difference between "n cows" and "n cows"?

- 3 months, 2 weeks ago

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oh wait nvm.

- 3 months, 2 weeks ago

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it translated to english for some reason.

- 3 months, 2 weeks ago

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;-) yeah, the same happened to me when i translated the page.
its a double challenge to code, when the task is given in a foreign language ;-)
i saved some of the tasks, but i saved them translated to german not english ;-(

- 3 months, 2 weeks ago

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@Percy Jackson you asked for sides to learn python.
i did the free course on this page: coursera.org
it is in russian, but i translated the pages (i ignored the videos)

a member here recommended edabit.com (it seems to be a free page to learn python).
other pages:
realpython.com (might be pay to use) repl.it (online coding environment)

- 3 months, 2 weeks ago

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@Percy Jackson ty for showing your javascript code for the candy problem.
it helped me a lot to understand javascript.
i found a nice solution to change the speed.
if you want to see it, i can post it here.

- 3 months, 2 weeks ago

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Ok, you can post it here. Thanks :)

- 3 months, 2 weeks ago

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Thanks :)

- 3 months, 2 weeks ago

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javascript:

var img1Load, img2Load;
var total = '';
var customColor, customWeight, customSpeed;
var linecolor, weight, speed;

function preload() {
img1Load = loadImage('PicCandy.JPG');
img2Load = loadImage('Colour.png');
}

function restar() {
image(img1Load, 220, 200, 440, 400);
lastFrame = frameCount;
waitFrame = true;
state = 0;
}

function setup() {
createCanvas(650, 450);
background('white');
imageMode(CENTER);
image(img1Load, 220, 200, 440, 400);
image(img2Load, 535, 140, 160, 20);

customColor = createSlider(0, 360, 0);
customWeight = createSlider(2, 15, 5);
customSpeed = createSlider(1, 120, 60);
cButton = createButton("Restart");

customColor.position(450, 100);
customWeight.position(450, 210);
customSpeed.position(450, 300);
cButton.position(500, 380);
cButton.mousePressed(restar);

lastFrame = 0;
waitFrame = true;
state = 0;

textAlign(CENTER);
strokeWeight(5);
noStroke();
textSize(19);
text('Path Color', 530, 100);
text('Path Width', 535, 210);
text('Animation Speed', 535, 300);

textSize(15);
text('Thin', 455, 250);
text('Thick', 605, 250);
text('Low', 605, 340);
text('High', 455, 340);
}

function draw() {
//frameCount = 0;

linecolor = customColor.value();
weight = customWeight.value();
speed = customSpeed.value();

textAlign(CENTER);
strokeWeight(weight);
noStroke();
textSize(19);
rect(0, 400, 500, 50);
text('Total : ' + total, 250, 430);
colorMode(HSB);
stroke(360 - linecolor, 100, 100);
colorMode(RGB);

if (waitFrame) {
if (frameCount >= lastFrame + speed) {
waitFrame = false;
lastFrame = frameCount;
}
} else {

if (state <= 15) {
state = state + 1;
waitFrame = true;
}

if (state === 1) {
line(200, 268, 358, 151);
total = '10';
}

if (state === 2) {
line(358, 151, 358, 268);
total = '10 + 9';
}

if (state === 3) {
line(358, 268, 200, 268);
total = '10 + 9 + 7';
}

if (state === 4) {
line(200, 268, 40, 385);
total = '10 + 9 + 7 + 4';
}

if (state === 5) {
line(40, 385, 40, 268);
total = '10 + 9 + 7 + 4 + 4';
}

if (state === 6) {
line(40, 268, 200, 268);
total = '10 + 9 + 7 + 4 + 4 + 6';
}

if (state === 7) {
line(200, 268, 200, 151);
total = '10 + 9 + 7 + 4 + 4 + 6 + 4';
}

if (state === 8) {
line(200, 151, 358, 151);
total = '10 + 9 + 7 + 4 + 4 + 6 + 4 + 8';
}

if (state === 9) {
line(358, 151, 358, 36);
total = '10 + 9 + 7 + 4 + 4 + 6 + 4 + 8 + 6';
}

if (state === 10) {
line(358, 36, 40, 36);
total = '10 + 9 + 7 + 4 + 4 + 6 + 4 + 8 + 6 + 7';
}

if (state === 11) {
line(40, 36, 40, 151);
total = '10 + 9 + 7 + 4 + 4 + 6 + 4 + 8 + 6 + 7 + 5';
}

if (state === 12) {
line(40, 151, 200, 151);
total = '10 + 9 + 7 + 4 + 4 + 6 + 4 + 8 + 6 + 7 + 5 + 7';
}

if (state === 13) {
line(200, 151, 358, 36);
total = '10 + 9 + 7 + 4 + 4 + 6 + 4 + 8 + 6 + 7 + 5 + 7 + 7';
}

if (state === 14) {
total = '84';
}

}

}


- 3 months, 2 weeks ago

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Thanks for the code, it works :)

- 3 months, 2 weeks ago

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I wish you success with your exams.

- 3 months, 2 weeks ago

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Thanks :)

- 3 months, 2 weeks ago

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@Percy Jackson @Jonathan Eshetu
i m not sure if i linked the correct coursera course.
this is the one i did: https://www.coursera.org/learn/python-osnovy-programmirovaniya/home/welcome

- 3 months, 2 weeks ago

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@Mitch Connor
I have deleted my solution for the problem, but I apologize for the incorrect equation. I got them mixed up, which is completely my bad. Thanks for telling me!

- 3 months, 1 week ago

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Hey @num IC , I have a quick question, if you're here

- 2 months, 3 weeks ago

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hi. i m here now.
but since i still have to understand some of the lines from your cool code, i m not sure if i really still can help you ;-)

- 2 months, 3 weeks ago

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haha, thanks ;-)

I can't stay for long, so sorry if i don't respond, but say I was making a text rpg in python

in this rpg, i plan to have an effects system, ex. an attack inflicting the opponent with burn, poison, freeze, etc. etc.

what would you say is the most efficient way to do this? the player has a class, so you can add attributes if necessary. here is the class code (done by me)

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 class RPGclass: def __init__(self, strength, defense, agility, magic, gold, health, MaxMP, name, wins, EXP): self.atksDict = {} self.strength = strength self.defense = defense self.agility = agility self.magic = magic self.gold = gold self.health = health self.currenthealth = health self.MaxMP = MaxMP self.currentMP = MaxMP self.name = name self.wins = wins self.EXP = 0 self.level = 0 def add_atk(self, atk, str, cooldown=0, manacost=0, effects=None): self.atksDict[atk] = [str, cooldown, manacost, effects] Rogue = RPGclass(2, 1, 2, 0, 0, 90, 40, "Rogue", 0, 0) #edit strength, defense,agility, and magic here Warrior = RPGclass(3, 1, 1, 0, 0, 110, 20, "Warrior", 0, 0) Knight = RPGclass(2, 2, 1, 0, 0, 120, 30, "Knight", 0, 0) Mage = RPGclass(1, 1, 1, 3, 0, 100, 70, "Mage", 0, 0) Rogue.add_atk("Needles", [i for i in range(round(1/3*Rogue.strength + 2/3*Rogue.agility), round(2/3*Rogue.strength+Rogue.agility)+1)], 0, 0) Rogue.add_atk("Surprise Attack!", round(Rogue.strength+2*Rogue.agility), 5,0) Warrior.add_atk("Sword Strike", [i for i in range(round(Warrior.strength*0.75), round(Warrior.strength*1.5)+1)],0,0) Warrior.add_atk("Whirlwind", Warrior.strength*2, 7,0) Knight.add_atk("Shield Bash", [i for i in range(round(1/2*Knight.strength + 1/2*Knight.defense), round(2/3*Knight.strength+Knight.defense)+1)],0,0) Knight.add_atk("The More, the Stronger!", 0, 5,0) Mage.add_atk("Staff Strike", [i for i in range(round(Mage.strength*1.5), round(Mage.strength*2.6)+1)],0,0) Mage.add_atk("Fireball", [i for i in range(round(Mage.magic*1.5), round(Mage.magic*2)+1)], 3, 15, "Burned") 

- 2 months, 3 weeks ago

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as you can see, i have effects in the attack setting, but how should i go about adding it in an actual battle? going through a bunch of if statements (if attack[2] =="Burned", if attack[2] == "Freezed", etc. etc.) making a dict to match effect name with effects? thank you in advance! I have to go now, so i'll respond tmrw morning!

- 2 months, 3 weeks ago

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if they cause the same effect then you could use if attack[2] in ....
if the effects are different then a dict sounds like a good idea.

- 2 months, 3 weeks ago

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and i would add something like: "inflictedWith" and "inflictedTimer" to the character stats.

- 2 months, 3 weeks ago

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You're obviously more advanced at this then me, lol
so this is all being done on repl.it (remember that), so if you have an account, I can invite you to the repl ;)
however, i'm not the owner, so i'll have to wait until the owner gets online.

- 2 months, 3 weeks ago

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plus, a lot of the other code was done by me and therefore isn't that clean, so not only this, any help with the rpg would be appreciated!

- 2 months, 3 weeks ago

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i have no exp in creating a rpg, but i m glad when i can contribute some help

- 2 months, 3 weeks ago

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it's not only rpg, you're just advanced in coding in general.
so would you like to join? i'd invite you some time tmrw, and we could really use the help lol. i understand if you say no ;-)

- 2 months, 3 weeks ago

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oh, i love to join. its just that seeing you young guys doing all this cool stuff makes me feel i miss a lot that i should have done when i was younger.
i ll try to find some energy to create an repl acc.

- 2 months, 3 weeks ago

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it's never too late :).
Life's sometimes that way, but who says you can't do all those things this year, or maybe more?

Plus, you don't have to worry about final exams ;-)

- 2 months, 2 weeks ago

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yeah. i wish all i had to worry about were final exams. ;-)

- 2 months, 2 weeks ago

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just respond whenever you make an account, and i'll ask the owner to add you asap.

- 2 months, 2 weeks ago

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ty i ll do.
could you do me a favour and check what brilliant has written here? https://brilliant.org/daily-problems/circles-parallelogram/?from_notification=15616976#!/dispute-comments/36816/
i cant check because i dont have premium

- 2 months, 2 weeks ago

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sorry this is late, they said: "Thank you for your feedback. I've updated the problem statement to reflect this.

I've marked this report as resolved."

- 2 months, 2 weeks ago

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ty for checking. its not late. i was just curious bcs they updated the wording already some days ago, so i was wandering if there was more then just marking it as resolved. ty again

- 2 months, 2 weeks ago

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yeah yeah, yw!
don't forget to tell me if you make a repl acc lol

- 2 months, 2 weeks ago

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i checked if your nice solutions runs in bril env.
i used that as input:

eq = "aaaa - bbb + cc - d = 8765"
leading0 = "false"


and it worked well. i tried this too:

eq = "aaaa-bbb+cc-d=8765"
leading0 = "false"


and got 0 solutions.

it does not matter for the designed purpose, but i assume u will process input data in the future too.

- 2 months, 2 weeks ago

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Yeah, I set it to .split(" + "), .split(" - "), and .split(" = ")`, so that's why it does that. If I remove the spaces, it might count them as numbers...

- 2 months, 2 weeks ago

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yeah input processing can be become very complicated. thats one reason why i'm so impressed by your code.

- 2 months, 2 weeks ago

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Thank you, that really means a lot coming from you ;-)

- 2 months, 2 weeks ago

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@Paul Romero Bello sry for the deleted msg, i tried to get your mention id, to "invite" you to this note.
the 2 links in this note are very helpful. i found some more too. and i can help whenever you are ready to try some latex.
additional: an idea to post a solution without pictures would be:
51234
23451
12345
34512
45123

lol, the numbers where just cycling. i have not recognized that before.

- 2 months ago

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You are very kind. Thank you :)

- 2 months ago

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you are very welcome.
first "challenge": copy the next line and send it as reply
\(\boxed{\LaTeX}\)

- 2 months ago

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$\boxed{\LaTeX}$

- 2 months ago

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Cooool!!!! :)

- 2 months ago

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great !!
voila, your first latex post :)
did you check how it works? or maybe you know it without reading the links.

- 2 months ago

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Yeah, I realized how it works by myself. Now, I am taking notes from your post. I hope to post a solution for tomorrows problem, if I get it right, of course!!!

- 2 months ago

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you are very good. do you know how to post a picture?

- 2 months ago

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wait, i'm a bit confused on $\boxed{\LaTeX}$

how exactly does it work?

- 2 months ago

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ohhh, wait I think I got it

so what if I...

$\textbf{\LaTeX}$

- 2 months ago

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Nice!

- 2 months ago

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$a_{b_{c_{d_{e_{f_{g_{h_{i_{j}}}}}}}}}$

- 2 months ago

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Amazing!!!

- 2 months ago

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Thank you!

- 2 months ago

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very good. can you do this?: $^{\text {up}} \text{middle} _{\text {low}}$

- 2 months ago

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hm....${up}^middle_{low}$ ?

- 2 months ago

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oh no, that's not it

- 2 months ago

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$^{up}middle_{low}$

- 2 months ago

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got it! sorry for the 4 pings, lol

- 2 months ago

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$^{ ^{ ^{highest_{higher_{high}}}}}middle_{low_{lower_{lowest}}}$

- 2 months ago

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i have to try that $^{ ^{ ^{highest_{higher_{high}}}}}middle_{low_{lower_{lowest}}}$

- 2 months ago

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very nice. ty for that idea.
i don't mind the pings. i enjoy the interaction with you two.
@Paul Romero Bello please feel free to try it too.
if you don't want to get pings from this note, then you can click "unsubscribe", but i'm not sure what kind of notifications you will still get then.

- 2 months ago

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maybe you want to use this. just an alternative to using percy's ;-)
i'm still curious. i assume a mac has a screenshot function?
else you might have something similar like snip&sketch:
when the "print" button is pressed, it offers some screen capture options.

- 1 month, 4 weeks ago

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i don't have a mac, but the web says shift-command-3 create a screenshot.
u said u use a pc-keyboard, so i assume the command key is different from the funktion key. is it?

- 1 month, 4 weeks ago

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Thanks for the link. The way I interpreted it was this: take any group of two students, there are 60/2=30 and there are 60 x 3 = 180 possible answers of which 60 x 2 =120 correct whatever the distribution. 60 answers are either all right, all wrong, or a combination of right and wrong. Person A could get Q1 right, and person B could get Q2 right. That's 2 questions right between 2 students. Or, A and B both got Q1 right but C and D got Q2 right. Or, if person A got Q1 right, B got Q2 right, and C got Q3 right, then D could have gotten Q1 or Q2 right or even Q3. I'm probably making too much out of this.

- 2 months ago

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i just copy the statement, so it's more easy to compare.
"During an examination, 60 students had to answer 3 questions. After the examination, it's shown that for any two students, there is always at least 1 question which both students could solve."

- 2 months ago

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it deprnds on the original wording but 60/2 is dividing the 60 students into groups of 2. but any two students would be 60*59/2 (combinatorics: pick 2 out of 60).
so this is much more than 30.

regarding the 1 question that both could solve. the wording indicates that both had the same question correct.

and as i said: the total number of questions is not stated. it could be the 3 same questions for all 60 students, but it could be more depending on the conditions of the problem.

the problem like you described it sounds more like:
60 students are divided into groups of 2.
each group had to answer 3 questions. both students answered individually.
every student gave at least one correct answer.

- 2 months ago

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i was used to program in basic and c, but python is very nice.
i'm still on my way to understand how to use python-goodies like range and permutation.
but please feel free to ask questions regarding coding. it would be a pleasure for me to help you.

- 2 months ago

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Thank you. Long ago (30 years) I used Fortran, Watfive, Pascal and Pascal took the lead. I got into a lot of Finite Element modeling with programming but quickly canned programs came on the scene in the form of software. So, I drifted away from programming and relied on software. But I'm impressed with the flexibility Python offers and want to learn it. Been doing a little here and there but get lost in the goodies (permutations and other libraries). I'll be sure to take you up on that with questions as they arise. Thanks again!

- 2 months ago

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nice i used pascal too. i have seen fortran but never used it. is/was it fun to work with it?
i dont know watfive. i will search that.
you said you'r a bridge engineer. that sounds very interesting.

- 2 months ago

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Watfiv had some additional features over Fortran but as Fortran indicates "formula translation" was its main feature. With advent of Pascal working with arrays took on new meaning. Bridges are fun because they encompass so many aspects of civil engineering, structural, hydraulic, highway, geotechnical and it all has to be designed around those criteria. Kind of like a puzzle with constraints.

- 2 months ago

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i could find some fortran code samples. i'm glad that i could mainly understand them.
fortran whatF iv was harder to find, they looked a bit like assembler code.

- 1 month, 3 weeks ago

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Thank you! I didn't even consider that there could be more than 180 questions. Just assumed they were the same 3 questions.

- 2 months ago

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hm i assume 180 is the maximum if all student get different questions.
(there might be more but that should be mentioned like: ... had to answer 3 questions out of a set of questions.)
my first assumption was: all students get the same 3 questions.
i still assume this is the idea. but i have no idea what the question behind that setting is.

is this the original wording or is it just a draft for a challenge that you prepare?

- 2 months ago

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It's the actual wording of a problem in the challenges community I believe under logic. The writer intended for the groups of two to get the same question right. But I thought there could be different interpretations.

- 2 months ago

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ok nice. can you post a link here?
just use this without the blanks: [ some text ] ( html )

- 2 months ago

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https://brilliant.org/problems/during-an-examination/

- 2 months ago

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Hahaha, I got that wrong actually. I took the second point literally, when I should be doing 60 x 2 / 3 = 40.

- 2 months ago

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that was a nice problem. ty for showing it.

- 2 months ago

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You're welcome.

- 2 months ago

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"there is always at least 1 question which both students could solve."
--> implies that each student had at least 1 question solved.
--> if even one of them only scored 1/3, then that particular question must be solved by 59 others to make this claim true. We can see this in just a group of three, let ABC be the students and 123 the questions. Say q1 is the easiest one, so average students A and B could only manage that one. But even when brilliant C solved the other difficult two, but made careless mistake / ran out of time on q1, the claim will be invalid (A1, B1, C23).
--> if there are no 'easy' question which is scored by ALL the students, then all 60 must have solved AT LEAST 2 questions each.

- 2 months ago

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Thanks! I knew you'd make sense of this one. You are both a logician and mathematician.

- 2 months ago

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Thank you, Joe, but that compliment is too high for me. I am still struggling when confronted with questions under category group theory. Too much lines and nodes than I can handle. And sorry I didn't think deeper than taking it as the 60 are just the usual class of hs students answering the same 3 questions. Perhaps the assumptions brought along from Three Questions problem we just had around Xmas.

- 2 months ago

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I liked that problem around Xmas but then there was more information to give a certain exactness. My approach is always to flesh out the knowns so when confronted with abstractions I'm in unchartered territory.

- 2 months ago

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if you love math then you might like this question: Minimum value of equation - Exam Edition
i got scrambled so i used wolfram alpha to solve it.

- 2 months ago

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regarding 'beating a dead horse': hint: each char is coded as 2 digit number.
meaning: to waste effort on something when there is no chance of succeeding.
(beat a dead horse)
(i assume he wants to say that there is no benefit in discussing the different ways of solving the challenges. - in my opinion the expression does not fit here very well.)
in old times some riders did beat a horse to make it run faster, but if the horse is already done then this will have no effect.

- 1 month ago

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Expression means "We've discussed this so much and in so many ways but aren't making any progress." I value different approaches and don't understand why people think they have the ultimate solution. It's like asking, "who's better - the republican or democratic nominee? Obviously it depends on what you value. So, I'm advocating that all approaches that yield the correct answer are valuable. The lengthy one that experts find redundant appeals to the person who lacks that knowledge to solve. The programmer who has solved many cryptograms and is honing his programming skills is saving time but providing knowledge to those who don't know programming. Etc.

- 1 month ago

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good point. i assumed by 'beating a dead horse' you meant that that discussion has no positive effect.
my hope is that the discussion helps the critic to understand that there is a value in presenting different solutions (even coded ;-).

- 1 month ago

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i can tell you how the first coding was done, but i assume you want to try it by yourself first.

- 1 month ago

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Thanks, num IC! For the horse proverbs explanation and the hint, but still I'm getting no progress. I don't understand 2 parts there, one is the many 7 there (though your reply had an almost balanced quantity of 6 & 7), and the other is why there are abcd included there too.

- 1 month ago

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67736c6876 2068766576 6d68206b6 66f6f767720 626c66207a 6f686c206c 7575206769 7a78702072 20687676

- 1 month ago

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