algos-and-data-structs

[lusty lantern]

U didn't what was the task was
Also it wasn't index//index

https://www.codewars.com/kata/525c65e51bf619685c000059/solutions/python/

[distant dock]

so ur trying to tell me available[name] and amount are not the index but the index?

[lusty lantern]

Those value were only test cases and not permanent

[distant dock]

then if its for test cases, u should ought to change them to integers like the problem

let me take another look give me a few moments to read and get some coffee

[lusty lantern]

I didn't want to waste time on test cases because they were not the point of my question

[distant dock]

what im confused about is, how do you know how much makes one cake

[distant dock]

nvm im stupid lul

sorry let me injest coffee

[distant dock]

def cakes4():
    recipe = {"apples": [3], "flour": [300], "sugar": [150], "milk": [100], "oil": [100]}
    available = {"sugar": [500], "flour": [2000], "milk": [2000]}
    
    
    for index, keys in enumerate(recipe):
        for index2,keys2 in enumerate(available):
            KEYS1 = recipe[keys]
            KEYS2 = available[keys2]

        for stuff, stuff2 in zip(KEYS1, KEYS2):
            calc = stuff//stuff2
            return calc
        else:
            if stuff > stuff2:
                calc = stuff//stuff2
                return calc
``` im getting the right answer but its giving me a type error for the interpretor on the website but in my IDE it runs just fine

yes its sloppy

but ehh

ahh its because im using a list

eitherway that should be faster than ur code and it still gives the right answer, now what i want u to do is figure out why it wont run with a list instantiated

2.842704500000764, 2.159629199999472, 1.8395526999993308, 1.0164244000006875, last one is mine

hmmmm, i dont watch tutorials anymore, back then when I did that I was trying to understand real basic stuff like list comp and dictionaries which i still need to become more versed in. I personally like to read tutorials/documentation/sourcecode examples. Make print statements within the code to see what variables mean, what is actually going on in that code is what u gotta do.

algorithms in my opinion (just scratching in the surface to them myself) is really just learning to traverse and manipulate arrays

think about it like ur trying to sort a list from least to greatest, what algo or what condition or even conditions would that consist of?

tbh with u, that problem above is the first algo problem i solved

first one i ever worked on

https://tenor.com/view/fantastic-beard-windy-big-beard-huge-beard-good-beard-gif-20180852

[fiery cosmos]

i would use bubble sort, learnt it from a Tom Scott video haha

[fiery cosmos]

that would be really nice

so you just, decided to

solve a problem and did it? 🤔

[distant dock]

yes

yeah i guess u could say i did, didnt think of it like that but ye

not like i solved quantum entanglement though

[dapper sapphire]

Is the space and time complexity for .join() method O(n)?

llist = ["This", "is", "a", "test"]
string = " ".join(llist)

print(string)
>> "This is a test"

[fiery cosmos]

why is ds algo so hard for me 🥲

i can't understand a thing

can anyone suggest me any books

[craggy wagon]

@fiery cosmos please lemme know if u find one 😆

[fiery cosmos]

i found one

[craggy wagon]

please?

[fiery cosmos]

grokking algorithm

[craggy wagon]

Oh

[fiery cosmos]

i have been told its a good book

[craggy wagon]

I did happened to come across that

[fiery cosmos]

did that book help

[craggy wagon]

No I just looked at first few chapters, It's fine ig

[fiery cosmos]

oh okie

[craggy wagon]

What I'm thinking of is to somehow learn problem solving along with ds algo

[fiery cosmos]

yes same

if you find one please tell me also

🥺

[craggy wagon]

At present I'm trying checkio.org btw

I just started, so not much review I can provide but it looks like a game

[fiery cosmos]

what is check.io

[fiery cosmos]

noice

[craggy wagon]

You should definitely check it out then

[fiery cosmos]

yup

it looks good

nice ui they have

[craggy wagon]

@dapper sapphire Space is definitely O(n) so is time, even I had feeling of quadratic. But checking stackover flow happened to be O(n)
And I think how it's done is you find the cumulative sum of length of all the sub-strings, then allocate required space, then you start copying! 😅

@fiery cosmos plus creative solutions by others 🎉

[dapper sapphire]

Yeah thanks! I am not sure how it's implemented. But if I were to implement something like this I guess I would use bytearray and pass in the size / length and then copy each element over of each substring index by index since bytearray is mutable.

[valid cedar]

can sombody suggest me a good book or any other reference for DSA.I have just started competitive programming any kind of help would be appreciated .thank you!

[fervent saddle]

https://github.com/python/cpython/blob/39dd141a4ba68bbb38fd00a65cdcff711acdafb5/Objects/stringlib/join.h#L57

[halcyon plank]

Objects/stringlib/join.h line 57

/* Here is the general case.  Do a pre-pass to figure out the total```

[fervent saddle]

This says you’re right

[dapper sapphire]

Oh wow thanks. What does "bytes-like" mean, when it says, and see whether all arguments are * bytes-like?

Oh so these are different types of bytes, like bytearray, mmap.mmap, array.array.
https://stackoverflow.com/questions/46750606/what-is-a-bytes-like-object

[empty swift]

Hey guys

I want to say that jesus loves you❤️ ✝️ ❤️🙌

[green seal]

Hey guys for algorithm and data structure kindly visit tutorialspoint for better explaination on algorithm and data structures 🙏🏽🙏🏽

[fiery cosmos]

DSA proselytizers

[craggy wagon]

@fervent saddle thanks for the insight
I've always heard bad about goto but I see a lot of these here!

[strange coral]

maybe someone has a idea to extract corners of a jigsaw puzzle piece... i tried almost everything... hough lines... polar coords with angle velocity.... contour defects...

[fiery cosmos]

https://www.programiz.com/dsa

[rocky moss]

How do people feel about new finger(tip) and palm(up) forms of UI? https://www.youtube.com/watch?v=sbHvkL8IpOc

You know Guido is working for Microsoft now https://github.com/JIoffe/HoloFluidBall

I wonder how his IDE will look in HoloLens

[fiery cosmos]

wow, very cool

[chilly crane]

hey guys, what kind of file you would propose to hold some info and by one click can load it to the entry boxes ?

i mean for easy writing and reading

[craggy wagon]

@chilly crane r u creating such file or looking for‽

[chilly crane]

i want load from file and save to file but wonder which would be easy to manipulate

[fiery cosmos]

https://stackoverflow.com/questions/67805191/modified-knapsack-problem-gets-stuck-in-infinite-loop

can someone help pls?

[lament oriole]

i want to "reverse search" a dictionary, but id prefer to avoid for loops.

im dynamically generating lists of unknown length, and all the items in a given list should correspond to a particular string.
i can easily construct a dictionary as {"string":[generated list here]}, but that would make searching for the key via an item in the list slow.
i can also generate {"item":"string"} for each of the items, but again thats slow.

is there a better datatype for having multiple keys go to a single value without having to generate a dictionary entry for every key? or is there a method for searching a dictionary this way that doesnt need loops?

[craggy wagon]

@lament oriole did you try sets in place of [generated list here], I believe even sets have O(1) time complexity

[lament oriole]

i have cast the list to sets (its a library generating the lists) but still have to iterate over the dictionary when searching, making it O(n) at worst

[craggy wagon]

You looking for O(1) 🔥

[lament oriole]

every day

[craggy wagon]

@lament oriole
I wonder how could this be slow:
i can also generate {"item":"string"} for each of the items, but again thats slow.

[lament oriole]

reversedDict = {
        item: "string" for item in theList
    } | {
        item: "string" for item in theList
    }

for == sadness

[main flower]

Well, you won't achieve O(1) If you also need to search that list

Also

also wdym by "search"

Do u just want to know If there is an element in that list

Or u want index or sth

[craggy wagon]

btw what that theList contains

[main flower]

Also, what are those "items" in a list? Integers, strings or sth else?

What I'd do is

For each string I'd have a BST (or even Red-Black or AVL tree)

This would make searching O(logn) which is not that bad

[lament oriole]

the list is a list of strings, each of those strings is unique across both its list and all other lists, so casting to set isnt a bad idea.
for searching, i want to have some way of addressing the dict/whatever type by an item in a list, and it returns to me the key that its under.
ideally, theres a way to generate and index without iteration.

i can do generation without iteration;
{"string": myList} but i then have to iterate twice when searching, once for the dict and once for the list.
i can do searching without iteration, but then the list generation looks like

mydict = {}
for l in all_lists:
  mydict.update({item: "string" for item in l})

which uses 2 loops.

i just wanna do this without loops, and faster than loops

[lament oriole]

im very new to data structures, idk what those 3 are lol

[lament oriole]

decided to bite the bullet on generation to ensure faster searching, just used this as it's fast enough at the scale im using it

reversedDict = {
        item: "string" for item in theList
    } | {
        item: "string" for item in theList
    }

[echo goblet]

you think i could build a discreet system for breaking things apart, based on the heat equation?

like instead of heat its would be stress and if it hit a certian amount the lattice connection would snap

@mint hedge #❓｜how-to-get-help

also

!rule 8

[halcyon plank]

Rules

8. Do not help with ongoing exams. When helping with homework, help people learn how to do the assignment without doing it for them.

[mint hedge]

im sorry

[loud trail]

@lament oriole https://pypi.org/project/bidict/

but use tuples instead of lists 🙂

[lament oriole]

>>> from bidict import bidict
>>> element_by_symbol = bidict({'H': 'hydrogen'})
>>> element_by_symbol['H']
'hydrogen'
>>> element_by_symbol.inverse['hydrogen']
'H'

would

>>> element_by_symbol = bidict({'H': {'hydrogen'}}) # <- changed to singleton set
>>> element_by_symbol.inverse['hydrogen']

still work?

[loud trail]

i doubt it, also sets aren't hashable, you need frozenset

[lament oriole]

oh right, shouldve made that a tuple really

[loud trail]

element_by_symbol = bidict({'H': ('hydrogen',)})
element_by_symbol.inverse[('hydrogen',)]

this should work

for individual list elements, yeah you might need to build 2 dicts or use a fancier data structure

[lament oriole]

element_by_symbol.inverse[('hydrogen',)] would this singleton tuple work if there were multiple items in the stored tuple?

[loud trail]

no

[lament oriole]

dang

still though, this seems like a useful library. ill make note of it. thank you

[loud trail]

red-black and AVL trees are binary search trees that can be rebalanced (or are self-balancing) to provide better guarantees about time complexity when searching

things = {
  'H': ('hydrogen', 'helium'),
  'C': ('carbon', 'cobalt'),
}

things_rev = {
    val: key
    for key, vals in things.items()
    for val in vals
}

query = 'H'
lookup = things.get(query, things_rev.get(query, None))

this could work if you don't know if you're looking for a key or value, and if you know that a value can only ever appear under one key

slow to construct + double storage, but still fast to do lookups

things = {
  'H': ('hydrogen', 'helium'),
  'C': ('carbon', 'cobalt'),
  '?': ('cobalt', 'hydrogen'),
}

from collections import defaultdict
things_rev = defaultdict(list)
for key, vals in things.items():
    for val in vals:
        things_rev[key].append(val)

this works in the more general setting where you might need the reverse lookup to be multi-valued

you can also use https://pypi.org/project/multidict/ for the multi-dict

things = {
  'H': ('hydrogen', 'helium'),
  'C': ('carbon', 'cobalt'),
  '?': ('cobalt', 'hydrogen'),
}

from multidict import MultiDict
things_rev = MultiDict()
for key, vals in things.items():
    for val in vals:
        things_rev.add(key, val)

[indigo berry]

hi guys how can i write base64 in a text file using pyhton

i want to clear the file before i write it too

[loud trail]

@indigo berry just use with open in "write" mode

[indigo berry]

thanks

[craggy wagon]

@fiery cosmos I don't think it's due to recursive function, by default maximum-recursion-depth is set to 1000, if that's not modified then it should yell error at you. I believe the problem is somewhere else. Correct if I'm wrong!

[strange terrace]

umm hello can someone teach the binary search

im just learn the python basic and i want to know more about binary search

[craggy wagon]

@strange terrace know what sorting is?

[strange terrace]

no i dont

what is sorting?

[craggy wagon]

gotta learn that first

[strange terrace]

oh k thanks

can i private chat with you so i can talk more about binary

just send a request friend

[fervent saddle]

You must know what sorting is

Sorting as in putting things in order from least-to-greatest or greatest-to-least

Binary search is checking the middle spot between all the possible values that could have the value you’re looking for

It’s like if you have to guess a number between 1 and 100, and first you guess 50, then 25, then 12 or 13, etc

[winged hamlet]

Hello. I needed guidance from the experts. Is there a way to compare two numbers in the list using xor without using brute force coding?

seq = [1, 2, 3, 3, 4, 4, 4, 10, 10, 10]
def algorithm():
result = 0
for i in range(len(seq)):
for j in range(i+1, len(seq)):
findNo1 = seq[i]
findNo2 = seq[j]

        if findNo1 ^ findNo2 == 0:
            print('X={}, Y={}'.format(findNo1, findNo2))
            result += findNo1 ^ findNo2 == 0
print('Total Matches is {}'.format(result))

algorithm()

Is there a better solution to show the same output print without using brute force?

[runic tinsel]

no

that's it

you can only do better without xor

[winged hamlet]

No offense. The project want me to use xor without using brute force

[runic tinsel]

can you show the task

maybe it's about something else

[winged hamlet]

Given a sequence of positive integers, SEQ, sorted in ascending order, design and
implement an algorithm with Python to determine if there exist two integers X and Y in
the sorted sequence such that X XOR Y = 0

[runic tinsel]

that doesn't say you need to use xor

or that you need to count matches

oh, it's about adding the end points probably

no that's not it

[winged hamlet]

The project wanted me to use xor? Since using for loop twice is getting the numbers in list is treated as brute force.

[runic tinsel]

like, I would never ever think that means I need to use xor

it's useless here

I would add the length to the first number

(1+10-1) = 10

that's the minimum the end number can be if there are no repeats

you can look at the end number, if it was 9, you immediately know there's a repeat

and then you can reevaluate that as you scan the list

and what about counting matches?

[winged hamlet]

Yes it needs me to use xor

Yes xor is necessary

[runic tinsel]

proof?

[winged hamlet]

Counting matches is there but need to have better solution

A brute-force algorithm compares all possible pairs in the sorted sequence and
check if they Exclusive OR to 0. However, using this approach will only give you a
minimal points.

[runic tinsel]

what about counting matches

[winged hamlet]

if counting matches using xor operator then its fine

[runic tinsel]

alright

[winged hamlet]

Thank You Experts. Sorry I'm a newbie.

[winged hamlet]

Hmm. No other solution

?

[strange terrace]

can someone check my code

i need experts

i just learn backtracking and this is the code

[halcyon plank]

Hey @strange terrace!

Uh-oh! It looks like your message got zapped by our spam filter. We currently don't allow .txt attachments, so here are some tips to help you travel safely:

• If you attempted to send a message longer than 2000 characters, try shortening your message to fit within the character limit or use a pasting service (see below)

• If you tried to show someone your code, you can use codeblocks
(run !code-blocks in #bot-commands for more information) or use a pasting service like:

https://paste.pythondiscord.com

[strange terrace]

from typing import List, Optional, Tuple

Matrix = List[List[int]]

initial_grid: Matrix = [
[3, 0, 6, 5, 0, 8, 4, 0, 0],
[5, 2, 0, 0, 0, 0, 0, 0, 0],
[0, 8, 7, 0, 0, 0, 0, 3, 1],
[0, 0, 3, 0, 1, 0, 0, 8, 0],
[9, 0, 0, 8, 6, 3, 0, 0, 5],
[0, 5, 0, 0, 9, 0, 6, 0, 0],
[1, 3, 0, 0, 0, 0, 2, 5, 0],
[0, 0, 0, 0, 0, 0, 0, 7, 4],
[0, 0, 5, 2, 0, 6, 3, 0, 0],
]

no_solution: Matrix = [
[5, 0, 6, 5, 0, 8, 4, 0, 3],
[5, 2, 0, 0, 0, 0, 0, 0, 2],
[1, 8, 7, 0, 0, 0, 0, 3, 1],
[0, 0, 3, 0, 1, 0, 0, 8, 0],
[9, 0, 0, 8, 6, 3, 0, 0, 5],
[0, 5, 0, 0, 9, 0, 6, 0, 0],
[1, 3, 0, 0, 0, 0, 2, 5, 0],
[0, 0, 0, 0, 0, 0, 0, 7, 4],
[0, 0, 5, 2, 0, 6, 3, 0, 0],
]

def is_safe(grid: Matrix, row: int, column: int, n: int) -> bool:
for i in range(9):
if grid[row][i] == n or grid[i][column] == n:
return False

for i in range(3):
    for j in range(3):
        if grid[(row - row % 3) + i][(column - column % 3) + j] == n:
            return False

return True

def is_completed(grid: Matrix) -> bool:
return all(all(cell != 0 for cell in row) for row in grid)

def find_empty_location(grid: Matrix) -> Optional[Tuple[int, int]]:
for i in range(9):
for j in range(9):
if grid[i][j] == 0:
return i, j
return None

def sudoku(grid: Matrix) -> Optional[Matrix]:

if is_completed(grid):
    return grid

location = find_empty_location(grid)
if location is not None:
    row, column = location
else:
    return grid

for digit in range(1, 10):
    if is_safe(grid, row, column, digit):
        grid[row][column] = digit

        if sudoku(grid) is not None:
            return grid

        grid[row][column] = 0

return None

def print_solution(grid: Matrix) -> None:

its a sudoku with backtracking

is this wrong or right

im a begginer so sorry if its wrong😄

[glacial spire]

Looks about right? Does it solve your example sudoku?

[strange terrace]

i just test it and it say cannot find solution

and i dont know what is wrong with it

[glacial spire]

Wait, I think I see the problem

[strange terrace]

?

[glacial spire]

wait no

[strange terrace]

let me see the code again

hmmm what is the problem

i just check it and nothing wrong with it

[glacial spire]

When I run sudoku(initial_grid) it just gives me a solution?

In [1]: sudoku(initial_grid)
Out[1]:
[[3, 1, 6, 5, 7, 8, 4, 9, 2],
 [5, 2, 9, 1, 3, 4, 7, 6, 8],
 [4, 8, 7, 6, 2, 9, 5, 3, 1],
 [2, 6, 3, 4, 1, 5, 9, 8, 7],
 [9, 7, 4, 8, 6, 3, 1, 2, 5],
 [8, 5, 1, 7, 9, 2, 6, 4, 3],
 [1, 3, 8, 9, 4, 7, 2, 5, 6],
 [6, 9, 2, 3, 5, 1, 8, 7, 4],
 [7, 4, 5, 2, 8, 6, 3, 1, 9]]

[strange terrace]

let me check

wait.

ohh just solved the problem thanks

thank you so much😀

XD

[winged hamlet]

I need experts to be able to help me for this code

[glacial spire]

So if I understand correctly, you want to find the number of equal pairs?

[winged hamlet]

Yes

using xor operators without brute force coding

[glacial spire]

... why using xor?

[winged hamlet]

The project want me to compare using binary

so need to use xor operator

[glacial spire]

"the project"?

[winged hamlet]

yes

Are u keen to help me to debug

[glacial spire]

Well normally I'd just combinations.Counter and some combinatoric maths

But it sounds like you have some sort of homework assignment

[winged hamlet]

So is there a solution?

[glacial spire]

Not without knowing what exactly you need to do

[fallen ridge]

I'm new to having to wrangle bunches of data in bespoke ways, and I've got a question about the best way to accomplish something if anyone's got a second

[winged hamlet]

Need to print out x and y that has the same numbers

and each time the same number compare by the xor operator if it is true then result add 1

[fallen ridge]

I've got a couple of nested classes that aren't really working like I expected, I was hoping to imitate some logic I've seen from some ML repos I've dug around in, but I have no idea how they implemented what they did.

Some semi-dummy code:

class TOKEN:
    def __init__(self, address, abi):
        self.address = address
        self.abi = abi
        self.contract = w3.eth.contract(address = self.address, abi = self.abi)

    class ROUTER:
        def __init__(self, address, abi):
            self.address = address
            self.abi = abi
            self.contract = w3.eth.contract(address = self.address, abi = self.abi)
        def quote()
            self.quote = get_quote


BUSD = TOKEN(abi.busd_address, abi.erc20_abi)
PCS = BUSD.ROUTER(abi.pcs_router_address, abi.pcs_router_abi)

print(BUSD.address)
print(BUSD.contract)
print(PCS.address)
print(PCS.contract)

All of my print statements return as expected, but I'm going to have multiple PCS objects as sub-objects of different TOKEN objects

so BUSD will have a PCS and BSP class, but I'll also have another token TETH that will also have PCS and BSP classes.

How do I keep this clean?

I thought I could do print(BUSD.PCS.address) or print(BUSD.BSP.address), but apparently not.

[fiery cosmos]

ok so i am gonna drop a bomb here, and i really need some help. Like please, i dont know calculus, and i found this formula for calculating exp exponentially for a discord bot. here it is

(xp + xp_to_add) // 42) ** 0.55

xp is the amount of xp a user already has, which is pulled from my database. xp_to_add is a random int from 1 to 20, that is added every time a message is sent(if it passes the time lock.

now everything works except that i dont know how to calculate the ex needed to get to another level. Since its exponential, it needs to be gotten using a derivative(?), which i have no idea what that is since i am a middle schooler lmao. Anyway here is my final result(the rank card, and the problem is the /370 XP ~ that needs to be dynamic, calculating the xp needed to level up(and i have no idea) plzzzzz help me this is 3d day of headaches lol

[thorn bay]

What is the best collision detection algorithm?

[fiery cosmos]

i will drop pastebin if more elaboration is needed:
https://pastebin.com/QidQEweK

[woven trench]

any good free course for dsa?

[fiery cosmos]

if you need to get the rate of change in the value of the exponents as the user gradually levels up, then just use sympy to do the derivative needed for that its a calculus library

[main flower]

What you could do is traverse that list and chech If i-th element is the same as i+1-th element. If yes, then the answer is yes, otherwise no

xor is equal to 0 only when both integers are same

This will be O(n)

And that would be the fastest you can go

[fiery cosmos]

hmm ok thx

[fiery cosmos]

i get the sympy part but can you explain the rate of change plz i lost u there lol

correct me if im wrong but i could technically just get the exponential value but using the ** operator so something like this?
exp_to_lvl = int((xp + xp_to_add) ** 0.55)

ugh, doesnt seem to work

[vocal gorge]

Since its exponential, it needs to be gotten using a derivative(?), which i have no idea what that is since i am a middle schooler lmao.
nah, you just need to inverse the function.
Your level is calculated from xp as int(xp ** 0.55), if I understand it correctly. So to reach lvl, it needs to be that xp**0.55 is at least lvl. Put both sides to the power of 1/0.55, and you'll get that xp = lvl**(1/0.55)

[rustic roost]

@winged hamlet does the data come sorted or unsorted?

[fiery cosmos]

i cant thank you enough, thank you so much man, this saves my day, ive been scouring the internet for calculus formulas for way too long lol

[rustic roost]

@winged hamlet Please paste the question if you cna

[vernal urchin]

guys, how can i use subprocess to call function of another file ?
#script1.py
import subprocess
l=[3,5,6,7,8,9]
for i in l:
call fun(i) of script2

#script2.py
def fun(f):
f=f*f
print(f)
this is just an example. when i run code for 900 videos(research purpose), it works stopped after 10 videos despite of using gpu

[winged hamlet]

I have paste the question above

[winged hamlet]

Refer to this

[winged hamlet]

it is sorted

[rustic roost]

If they just want you to determine if there exists two integers such that their XOR is 0 then y are you printing x and y and total number of such pairs?

@winged hamlet

[winged hamlet]

Yes

[rustic roost]

Why are you doing that I am asking, you could break the loop on the first pair saying such pair exists

whats the output format?

seq = [1,2,3,4,4,6,6,10,10,10]
def isPair(seq):
    for i in range(len(seq)):
        if(seq[i]^seq[i+1] == 0):
            return True
        return False

This should do the if pair exists trick

@winged hamlet

[rain snow]

Hey, Im trying to write a mathmatical program to calculate pi based using newtons method of using the unit circle formula, rearanging it into a binomial formula and Using the binomial theorem to solve for pi/

########### Atempting to Calculate PI using the binomial series and Calculus
import math 

####Binomial series
iterations  = 10
#x = int(input("Attempting to Calculate PI using the binomial series and Calculus\n The equation of a unit circle is x^2 + y^2 = 1 but we can rearrange that to get route(1-x^2) and simplify this \
#to (1-x^2)^0.5. Which is the equation for the top half of a circle since x can only be positive. by integrating this between the bounds 0 and 1 we can find the area of 1/4 of the circle \
   #and therefore pi will be equal to 4 times that. "))
#n = float(input("n: "))
x= 1
n = 0.5
## First 2 terms in the binomial theorem
number = x+(n*(-x**3)/3)
print(number)
for i in range(1, iterations):
    ###Third term index
    index = i+1
    print(f"index: {index}")
    ##Working out the factorial
    factorial = 1
    
    for j in range(1, index+1):

        factorial *=j
    print( f"factorial: {factorial}")
    ####Working out the coefficient of x.
    coefficient = 1
    for k in range(0, -i-1, -1):
        coefficient *= (n+k)
    print(f"first coefficient: {coefficient} ")
    #### Coefficient before integrating
    coefficient= coefficient/factorial
    print(f"real coefficient: {coefficient} ")
    newindex = index + 1
    value = -(coefficient)* (((x)**(2*newindex)) / 2*newindex)
    print(f"value: {value}")
    print(f"current value of pi is {number*4}")
    number+=value
    print("======================")

temp = math.sqrt(3) /8
pi =  (12*number)- temp
pi = 4*number
print(f"pi = {pi}")

3.89660390218099 is the closest I can get

it should be more exact than that but for some reason I cant anyone see an error?

these are my workings if it helps

[winged hamlet]

Output format must be

X=3, Y=3
X=4, Y=4
X=4, Y=4
X=4, Y=4
X=10, Y=10
X=10, Y=10
X=10, Y=10
Total match is 7.

It must be format like this

[winged hamlet]

@rustic roost

[brave oak]

you have way too many print statements in that

it makes it hard to understand what the logic is

[rain snow]

I was using it for debugging

[brave oak]

ye

but it also makes your code hard to read

[rain snow]

alright one sec

########### Atempting to Calculate PI using the binomial series and Calculus
import math
from decimal import *
####Binomial series

iterations  = 124



#x = int(input("Attempting to Calculate PI using the binomial series and Calculus\n The equation of a unit circle is x^2 + y^2 = 1 but we can rearrange that to get route(1-x^2) and simplify this \
#to (1-x^2)^0.5. Which is the equation for the top half of a circle since x can only be positive. by integrating this between the bounds 0 and 1 we can find the area of 1/4 of the circle \
   #and therefore pi will be equal to 4 times that. "))
#n = float(input("n: "))
x= 0.5
n = 0.5

### First 2 terms in the binomial theorem

number = x+(n*(-x**3)/3)
print(number)
for i in range(1, iterations):
    ###Third term index
    index = i+1
    #print(f"index: {index}")
    ##Working out the factorial
    factorial = 1
    
    for j in range(1, index+1):

        factorial *=j
    #print( f"factorial: {factorial}")
    ####Working out the coefficient of x.
    coefficient = 1
    for k in range(0, -i-1, -1):
        coefficient *= (n+k)
    #print(f"first coefficient: {coefficient} ")
    #### Coefficient before integrating
    coefficient= coefficient/factorial
    #print(f"real coefficient: {coefficient} ")
    newindex = index 
    value = (coefficient)* (x**((2*index)+1))/ ((2*index)+1)
    #print(f"value: {value}")
    #print(f"current value of pi is {number*4}")
    number+=value
    #print("======================")
sqrt = Decimal(3).sqrt
temp = sqrt /8
print(temp)
pi =  (12*(number- temp))
#pi = 4*number
print(f"pi = {pi}")
print(cmath.sqrt(3)) 

[brave oak]

AAAAAAAAAAAAAAAA

you just commented out the prints...

[rain snow]

BECAUUSE I NEED THEM

[brave oak]

okay honestly my head hurts just looking at it

let me

take a while

one suggestion

space out your operators

in general you should have spaces around arithmetic operators

so like

value = coefficient * (x ** (2 * index + 1)) / (2 * index + 1)
instead of:
value = (coefficient)* (x**((2*index)+1))/ ((2*index)+1)

[rain snow]

kk

[brave oak]

if your code is easy to read

it will be easier for you to get help

ngl I feel tired just looking at that chunk

[rain snow]

okay

[brave oak]

also you might want to break your code up into smaller functions?

because then you can test them individually

to know they work

e.g. the factorial

[rain snow]

no it's really not worth it

[brave oak]

but anyway...

!e from math import factorial; print(factorial(5))

[halcyon plank]

@brave oak :white_check_mark: Your eval job has completed with return code 0.

120

[brave oak]

just so you know

what is newindex even for?

you don't use it anywhere

[rain snow]

yeah wanted to see if I could do it myself

it was a bug fixing attempt I was going to add 1 to the index before I integrate but then I realised u cant so that

[brave oak]

yeah, that's another thing

remove dead code

goes back to the readability thing

[rain snow]

its removed on the main

[brave oak]

main what

[rain snow]

but I copied and pasted this ages ago

on my pc

the actual.code

########### Atempting to Calculate PI using the binomial series and Calculus
import math
from decimal import *
####Binomial series

iterations  = 124



#x = int(input("Attempting to Calculate PI using the binomial series and Calculus\n The equation of a unit circle is x^2 + y^2 = 1 but we can rearrange that to get route(1-x^2) and simplify this \
#to (1-x^2)^0.5. Which is the equation for the top half of a circle since x can only be positive. by integrating this between the bounds 0 and 1 we can find the area of 1/4 of the circle \
   #and therefore pi will be equal to 4 times that. "))
#n = float(input("n: "))
x= 0.5
n = 0.5

### First 2 terms in the binomial theorem

number = x+(n*(-x**3)/3)
print(number)
for i in range(1, iterations):
    ###Third term index
    index = i+1
    #print(f"index: {index}")
    ##Working out the factorial
    factorial = 1
    
    for j in range(1, index+1):

        factorial *=j
    #print( f"factorial: {factorial}")
    ####Working out the coefficient of x.
    coefficient = 1
    for k in range(0, -i-1, -1):
        coefficient *= (n+k)
    #print(f"first coefficient: {coefficient} ")
    #### Coefficient before integrating
    coefficient= coefficient/factorial
    #print(f"real coefficient: {coefficient} ")
   
    value = (coefficient)* (x**((2*index)+1))/ ((2*index)+1)
    #print(f"value: {value}")
    #print(f"current value of pi is {number*4}")
    number+=value
    #print("======================")
sqrt = Decimal(3).sqrt
temp = sqrt /8
print(temp)
pi =  (12*(number- temp))
#pi = 4*number
print(f"pi = {pi}")

it's late I gtg

bye

[winged hamlet]

Hmm it does not show the output that they wanted

[runic tinsel]

wait no

too many matches

!e

seq = [1, 2, 3, 3, 4, 4, 4, 10, 10, 10]
def algorithm():
    result = 0
    counts = {}
    for i in seq:
       counts[i] = counts.get(i,0)+1
    for k,v in counts.items():
        p = (v*v-v)//2
        for _ in range(p):
            print('X={}, Y={}'.format(k, k))
        result += p
    print('Total Matches is {}'.format(result))

algorithm()

[halcyon plank]

@runic tinsel :white_check_mark: Your eval job has completed with return code 0.

001 | X=3, Y=3
002 | X=4, Y=4
003 | X=4, Y=4
004 | X=4, Y=4
005 | X=10, Y=10
006 | X=10, Y=10
007 | X=10, Y=10
008 | Total Matches is 7

[runic tinsel]

it doesn't use sortedness

[fiery cosmos]

How open addressing and chaining help to resolve collision resolution in hashing?

i am in my exam and this is question of Data Structure

someone help pls

[charred fractal]

i am 13

and i joined this server for fun

and i need help

its not related to python AT ALL

[fiery cosmos]

lol ok

ask

me

i guess

[fiery cosmos]

what is about

u want me to sub u?

[austere sparrow]

@charred fractal If it's not related to Python, we have off-topic channels.
(also keep in mind that we don't allow advertisement, if that's what you're hinting towards)

[prisma lily]

I was wondering how to use a list comparison to iterate over the first element in the defined list?

[jolly mortar]

could you show an example of what you mean?

[charred fractal]

i almost kept channel link ty

[fiery cosmos]

dude same i was wondering yestarday

[austere sparrow]

no @charred fractal, this is not a platform to advertise your youtube channel if that's what you're here for.

[fiery cosmos]

i mean this , i was wondering about this yestarday

[prisma lily]

!e

[halcyon plank]

Command Help

!eval [code]
Can also use: e

*Run Python code and get the results.

This command supports multiple lines of code, including code wrapped inside a formatted code
block. Code can be re-evaluated by editing the original message within 10 seconds and
clicking the reaction that subsequently appears.

We've done our best to make this sandboxed, but do let us know if you manage to find an
issue with it!*

[prisma lily]

!e for i in range(len(vectors)):
if (mask[i] != True):
near_dist = [[dissimilarity(vectors[i], vectors[i2]), [i, i2]] for i2 in range(len(vectors)) if (((dissimilarity(vectors[i], vectors[i2])) < near_dist[0]) and (i != i2) and (mask[i2] != True))]

[halcyon plank]

@prisma lily :x: Your eval job has completed with return code 1.

001 | Traceback (most recent call last):
002 |   File "<string>", line 1, in <module>
003 | NameError: name 'vectors' is not defined

[austere sparrow]

[charred fractal]

ok i just came for help

so ye

[austere sparrow]

that's a pretty complicated comprehension

[prisma lily]

:/

Yeah

Uhh

[austere sparrow]

!code could you format it with this?

[halcyon plank]

Here's how to format Python code on Discord:

```py
print('Hello world!')
```

These are backticks, not quotes. Check this out if you can't find the backtick key.

[prisma lily]

Lemme make a quick simple version

In short though instead of appending over a list I was wondering if it was possible for a list comprehension to replace the list

Alrighty

[halcyon plank]

@prisma lily :x: Your eval job has completed with return code 1.

001 | Traceback (most recent call last):
002 |   File "<string>", line 1, in <module>
003 |   File "<string>", line 1, in <listcomp>
004 | NameError: name 'i' is not defined

[prisma lily]

outlist = [i*i for i in range(10)]
print(outlist)

This list comprehension returns:

[0, 1, 4, 9, 16, 25, 36, 49, 64, 81]

I was wondering what I would need to do to iterate the first element of the list over all elements resulting in the list:

[81]

Would my only option be a for loop?

[runic tinsel]

yes

i don't know what you mean, but it's a for loop

[dusk bison]

hey everyone, i couldn't find this online but does interpolation search work for strings in a list too?

[prisma lily]

It's currently doing this:

outlist = []
    
    for i in range(10):
        outlist.append(i*i)
    print(outlist)

[dusk bison]

That's not what you want?

[prisma lily]

And I want it to do something more like this:

 for i in range(10):
    outlist[0] = i*i
print(outlist)

[jolly mortar]

why do you want a list in that case? why not modify a single variable in the loop?

also, what are you iterating over in your main code?

[prisma lily]

This is to be nested within a list comprehension

I'm storing other elements in the list that relate to the iteration with the overall function

I'm iterating over an iteratable iterating over the length of another list if that makes any sense

[spark cedar]

hello can anyone suggest me a course for DS&A ??? Are joma tech's courses any good ??

[runic tinsel]

depends on length of the list

dictionary is faster eventually

use dictionary

[vocal gorge]

It depends on the use case, but I think this "eventually" may come as soon as n=1

for this case:

def f(n):
    lst = [(i,i+1) for i in range(n)]
    %timeit next((el for el in lst if el[0]==n), None)
def g(n):
    dct = {i:i+1 for i in range(n)}
    %timeit dct.get(n,None)

(comparing a list of tuples to a dict), even for n=1 g(n) is faster; because the dict way doesn't need a genexpr with a comparison (all of which happens in Python code)

[thorn bay]

...

[vocal gorge]

what do you mean by "best"?

and for that matter, by "collision detection algorithm"? Like, do you want Continious Collision Detection or something?

[fiery cosmos]

Worth mentioning in general searching a list like that takes time proportional to the length of the list O(N), whereas a dictionary lookup is constant time O(1) (with high probability ). So dict is definitely faster unless the list is tiny.

[uneven yarrow]

you can thank hashes for that

[uneven yarrow]

I don't understand it entirely either, but it's an integral piece.

Try asking someone else (and relay it to me if you could)

[half lotus]

What do you want to know about hashing?

There's a hash function, and there may sometimes be collisions (2 keys result in the same hash). There are generally 2 solutions to deal with collisions: one is to create a linked list and store all values that have the same hash in it. The other is a probing technique, where it keeps modifying the key until it results in a hash that hasn't been used before.

[fervent saddle]

It’s basically making a list with a lot of empty space, and then doing some math formula on the value you want to put in the list to get a valid index position, and putting it at that index

A simple example for strings would be adding the numeric values of all the characters, then using the remainder of that sum divided by the length of the list

Then if you want to check if a string is in the list, you just use that formula and check at that index if it’s holding an equivalent string

That’s the basic idea of it. Then there’s also a lot of ways to deal with if multiple values get the same index position, like mark mentioned. If you don’t deal with that in some way then it won’t work right

But the idea of hashing is just using some formula to get an arbitrary non-random index position

[fiery cosmos]

The MIT 6.006 lectures 8 9 and 10 are on hashing and not bad. https://www.youtube.com/playlist?list=PLUl4u3cNGP61Oq3tWYp6V_F-5jb5L2iHb In fact the whole DSA course and coursework is online.

[craggy wagon]

@fiery cosmos 2021?

[fiery cosmos]

hmm

The data structures haven't changed much in 10 years

[lean dome]

Or 40, honestly 😄

[runic tinsel]

40 is conservative yeah

[fiery cosmos]

I was trying to make a math function that subtracts the minimum value of a list (lists are used as input). I end up getting this error:
Arr.minimum(*list1, *list2)
TypeError: minimum() takes 1 positional argument but 8 were given

My code:

class Arr():

    def __init__(self, f_list, s_list):
        self.f_list = f_list
        self.s_list = s_list

    def minimum(self):
        self.f_list = min(f_list)
        self.s_list = min(s_list)
        difference = self.f_list - self.s_list
        print(difference)

    def maximum(self):
        self.f_list = max(f_list)
        self.s_list = max(s_list)
        addes = self.f_list + self.s_list
        print(addes)

list1 = [1,3,2,6]
list2 = [5,2,5,6]

Arr.minimum(*list1, *list2)```

[fiery cosmos]

I don't know much python. But can't you just make minimum() takes 2 lists instead?

[gritty marsh]

What would a linked list implementation in Python look like especially since there's no way to store a reference to a certain location in memory?

[fiery cosmos]

If you have a Node class it can have a self.next member that is another Node instance

[marble bronze]

I'm currently taking a course about Data Structures and algos, I have this hackerrank thing that I have to do, and it is about Binary Search Trees. Can I inquire here? Very, very new at coding

[fiery cosmos]

Ask away.

[marble bronze]

you know when talking it out sometimes helps?

thanks

HAHAHA

[fiery cosmos]

Rubber duck debugging is a thing lel.

Good luck with your schoolwork.

[marble bronze]

I'm gonna be frequent here in the future, haha

[untold crown]

I fixed that part but for some reason it still says f_list and s_list are undefined, even though it clearly shows them in the variables section on the right-hand side of this image

[fiery cosmos]

Thank you, maybe the error appeared due to the fact that I did not put self.f_num in min ()?

[untold crown]

I think Ik what the problem is. In this case, f_list and s_list are objects rather than variables. The _init_ function passes these as variables, which is why they're not defined as variables, instead being defined as objects when called in the Arr class

[untold crown]

What about using *args or **kwargs instead of what I did? Maybe that'll work

[austere sparrow]

!e

reference to a certain location in memory?
Actually, in Python you can't do anything but reference a certain location in memory 🙂

xs = (1, (2, (3, (4, (5, None)))))

def iterate_over_linked_list(ll):
    while ll:
        head, ll = ll
        yield head

for i in iterate_over_linked_list(xs):
    print(i)

[halcyon plank]

@austere sparrow :white_check_mark: Your eval job has completed with return code 0.

001 | 1
002 | 2
003 | 3
004 | 4
005 | 5

[austere sparrow]

Or do something like this:

class Node:
    def __init__(self, head, rest):
        self.head = head
        self.tail = tail

which will allow you to attach methods to it

[oak spindle]

what this means?

[austere sparrow]

in the interactive console, _ will store the result of the last evaluated expression (12.5625 in this case)

where does this code come from?

[oak spindle]

but i'm getting error

this only works on console?

[austere sparrow]

Yes.

[oak spindle]

okay, thanks

[austere sparrow]

also, you don't need semicolons

[oak spindle]

yea, i just followed from c++ and it became a habit, lol

[winter rampart]

btw how many of u guys use virtual studio

[oak spindle]

vs code?

[winter rampart]

yep that one

[oak spindle]

i'm using that one

[gritty marsh]

Yup, thanks for the reminder

Sometimes you confuse different languages, ya know

[last fulcrum]

Why is an inner loop quadratic?

For example if I do

n = 4
for j in range(n):
    for i in range(j + 1):

i is not quadratic for me. Even combining the sum of i and j goes nowhere near 16 times.

[gritty marsh]

It doesn't have to be n^2 exactly to be quadratix

In that example, the complexity is n(n+1)/2

Simplifying you get (n^2+n)/2

Dropping the lower order terms, you get n^2 ;)

Big-O notation describes how an algorithm grows

[untold crown]

@fiery cosmos Is this what you were looking for? (Sorry for the late response)

[gritty marsh]

So your example grows quadratically

It doesn't necessarily have to be quadratic

Did this answer your question @last fulcrum ?

[untold crown]

and I realised I didn't even use the *args variable

[fiery cosmos]

Yes, thank you 🙂

[untold crown]

same result but without the extra *arglists in the bottom two functions

[last fulcrum]

Yes a little. I'll like to find out how you got n(n+1)/2

[untold crown]

np, glad to help

[gritty marsh]

It's the partial sum of an arithmetic series! Are you familiar with that topic in maths?

[last fulcrum]

So we would always take the highest number right?

[gritty marsh]

Highest degree term, yes

[last fulcrum]

No. Not at all. Do I need to to really understand DS & A?

[untold crown]

the global statement I did is redundant, I was testing out some things with **kwargs so that's why it's there (got errors so I gave up on it, but I guess I forgot to remove the global statement afterwards)

[winged hamlet]

Hi can i ask is it possible to get the key function to able to sort like name, title, etc if is selected?

[last fulcrum]

@gritty marsh the book I'm reading says this. Although I don't understand the maths.

[gritty marsh]

@last fulcrum ah. Are you familiar with arithmetic series?

[gritty marsh]

... would you like me to prove that formula for you right now? 😆

[last fulcrum]

Yes if you can. I'll like you to help me with maths if you can.

[last fulcrum]

Never heard of it

[gritty marsh]

An arithmetic series is simply a series of numbers produced by adding some other number to an original number repeatedly

Wow that's a bad definition

take for example the example series in that image you sent

1 + 2 + 3 + 4 + 5 + ... + (n - 2) + (n - 1)

here, the original number is 1

the common difference is 1

you get the second term of the series by adding 1 to 1 (resulting in 2)

you get the third term of the series by adding 2 to 1 (resulting in 3)

[last fulcrum]

Oh I've heard of it

I saw a video about it last month

[gritty marsh]

mmhmm

so the general formula that will be used to represent an arithmetic series is as follows:

(a_1) + (a_1 + d) + (a_1 + 2d) + ... + (a_1 + (n-2)d) + (a_1 + (n-1)d)

a_1 is the starting number

d is the common difference

n is the number of terms you want

[last fulcrum]

so how does that go and transform into n2?

[gritty marsh]

Do you understand this formula?

[last fulcrum]

Yes.

[gritty marsh]

Alright, great!

Notice how we can represent this series in a different way

[last fulcrum]

if d is 7 you always need to add 7 to the previous number to get the next

[gritty marsh]

Correct!

[gritty marsh]

Instead of starting from the first term, we start from the last term

The last term in this case would be a_n

so we would represent that as follows:

(a_n) + (a_n - d) + (a_n - 2d) + ... + (a_n - (n-2)d) + (a_n - (n-1)d)

Do you understand that this is simply the series above but in reverse?

[last fulcrum]

what is a _ n? Does_mean multiplication?

[gritty marsh]

a_n is simply the last term in this series

if a_1 is the first term, and a_2 is the second term

and we have n terms

then a_n is the last term

_ in this case is just subscript notation

[last fulcrum]

Oh that makes sense. So if we have 10 numbers we can say find a10

[last fulcrum]

yes

Makes sense now

[gritty marsh]

Alright, great!

Now, the result when you add up all of those terms is known as the "partial sum"

we call it the partial sum because of course, the series is infinite

so if we have the sum of the first 10 terms of the series it's only a part of the series

we use s_n to denote partial sum

[last fulcrum]

Ok

[gritty marsh]

if s_1 means the sum of the first 1 terms, and s_2 means the sum of the first 2 terms, then s_n is the sum of the first n terms

Are you following along with me?

[last fulcrum]

Yes. Basically like a_n So we can say get the sum of the first 30 terms, which would be s30

[gritty marsh]

Yep

Okay, we have all the bits and pieces to start our proof now

we have two equations to represent the partial sum:
s_n = (a_1) + (a_1 + d) + (a_1 + 2d) + ... + (a_1 + (n-2)d) + (a_1 + (n-1)d)
and
s_n = (a_n) + (a_n - d) + (a_n - 2d) + ... + (a_n - (n-2)d) + (a_n - (n-1)d)

Now tell me, do you notice anything interesting or useful about the two equations above?

[last fulcrum]

One is the reverse of the other.

[gritty marsh]

Yes, anything else?

Hint: look at the common differences (d)

[last fulcrum]

both have 2d, and (n-2)d

[gritty marsh]

Nearly there!

Do you see anything interesting with the signs? (+/-)

[last fulcrum]

All I see is that the signs are reversed for both

[gritty marsh]

Correct!

And that is great!

Do you have any idea why that is great?

[last fulcrum]

Give me a min

I think that means we can get the first value of a sum if we are given the last

And the reverse is true

and d

[gritty marsh]

Not quite, here's a hint:

s_n = (a_1) + (a_1 + d) + (a_1 + 2d) + ... + (a_1 + (n-2)d) + (a_1 + (n-1)d)
s_n = (a_n) + (a_n - d) + (a_n - 2d) + ... + (a_n - (n-2)d) + (a_n - (n-1)d)

what do you think will happen when we add them together?

[last fulcrum]

We can add and subtract like terms?

[gritty marsh]

Correct!

We can cancel out all the d's!

So our result is simply
2*s_n = (a_1+a_n) + (a_1+a_n) + (a_1+a_n) + ... + (a_1+a_n)

(a_1+a_n) is repeated n times

so we can simplify this as

2*s_n = n(a_1+a_n)

isolating the s_n, we get

[last fulcrum]

2*s_n = n(a_1+a_n) / s_n

2* = n(a_1+a_n) / s_n

[gritty marsh]

Sorry, internet went out

Do you know how to isolate that s_n variable @last fulcrum ?

To isolate that s_n variable we simply perform algebra magic

s_n = n(a_1+a_n)/2

if we have the example series above 1 + 2 + 3 + ...

then a_n = n

a_1 = 1

thus

s_n = n(n+1)/2

[last fulcrum]

ok so you just divided by 2 instead of s-n. 😅 I was thinking you divide by s-n

[gritty marsh]

Yep

[gritty marsh]

note this is only when a_1 = 1 and d = 1

which is of course the series you were asking about

but does not apply to all series

[last fulcrum]

I was getting confused by that

[gritty marsh]

this is the general equation that is applicable to all series

[gritty marsh]

this is simply a simplification dedicated for the special series above that exists because of simplification magic

because a_1 = 1 and a_n = n

[last fulcrum]

So how does the general or simplification go to n2?

[gritty marsh]

Ah, this is a whole different explanation

okay so we've established that the complexity is quadratic

because of the above equation

n(n+1) / 2 expands to (n^2 + n) / 2

which expands to (1/2)*(n^2) + (1/2)*n

in big-O notation, we don't care about the lower order terms

we care about the algorithm's growth

thus, we throw away the lower order terms

n is a lower order term

(1/2) is a lower order term

so we simply say O(n^2)

because it grows quadratically

it does not necessarily have to be exactly n^2

let me throw up a quick desmos graph

This is the graph for very very small n

notice there are noticeable differnces

of course, they're different equations. there would be differences

but notice what happens when we zoom out

This is for n on the orders of the hundreds

They look like the same graph, don't they?

Because they grow the same

[last fulcrum]

So basically it's growing quadratically compared to the other equations, even if it might not be n2.

[gritty marsh]

Correct

[last fulcrum]

Man you explained it so well

I totally get it

[gritty marsh]

😆 my pleasure

[last fulcrum]

I'll come to you for more help.

Thank you

[gritty marsh]

Welcome!

[winged hamlet]

for key in emp_dict:
record_list.append(key)
templist = []
for key, value in emp_dict.items():
temp = [key, value]
templist.append(temp)

Can anyone help me for the coding

Basically i want to sort by name in dictionary. The key is Employee ID, and the following will have name, title, et

[fiery cosmos]

Have you tried googling something like "sort dictionary by value"?

[potent portal]

https://leetcode.com/problems/remove-nth-node-from-end-of-list

im failing on the test case of [1] and 1 but i cant really see why?

its expected to return an empty list but this returns [1]

[fiery cosmos]

Anybody knows a video or tutorial about how to create a simulation for image processing? With photon amount and exposure time and all of that?

[fiery cosmos]

Okay so I m gonna ask here because i can t find a proper explanation. What is a B Tree and why the hell should I use a B tree? Is there any good explanations? I can't find a satisfying one (for me)

The only thing that I see in a B tree is that the B tree is some sort of BST but with arrays

[oblique citrus]

I'm new in coding with Python and this Problem showed up when I wrote this code.

I tipped this code in the anaconda probt and this error does not disappear. Pls help me.

[vocal gorge]

Pycharm by default (if you click through the project creation dialogue) creates a venv for each project.

So installing the package via anaconda will install it into your main environment, not the venv.

Instead, click that insall package prompt and it'll install it for you (or the slightly longer way - go to the Packages tab and search scikit-learn there)

[vocal gorge]

^

(also, this isn't really what this channel is for)

[oblique citrus]

thanks

[valid slate]

Hi, I wanted to ask if I wanted to visualize merge sort then how do I exactly keep track of the main array ?

[chrome field]

hey

[unborn sundial]

huh. feeling myself being an idiot
!e

def decorator_factory(argument):
    def decorator(function):
        def wrapper(delay, *args, **kwargs):
            print(f"getting argument {argument}")
            print(f"magically knowing in advance value {delay} of the functiom!")
            result = function(delay, *args, **kwargs)
            return result
        return wrapper
    return decorator

@decorator_factory(10)
def pause(delay):
    print(f"paused for {delay} seconds!")

pause(5)

I can actually use arguments of my function in decorator in a normal way... I always extracted from *args them

[sly bronze]

anyone know if I use windows development kit's sign tool on my exe converted file (from python) will it stop seeing it as threat on other computers ?

[eager hamlet]

say i had a tree

with a bunch of numbers

i have to perform 2 operations:

sum of numbers on the path between any two vertices

increment or decrement all numbers on the path between any two vertices by a set amount

how would i do this (preferably in log (size of tree))

[whole plover]

vertices?

the space between the vertices are the space

between Documents and Downloads?

[lunar jacinth]

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def mergeTrees(self, root1: TreeNode, root2: TreeNode) -> TreeNode:
        
        if root1 and root2:
            root = TreeNode(root1.val+root2.val)
            root.left = self.mergeTrees(root1.left, root2.left)
            root.right = self.mergeTrees(root1.right, root2.right)
            print(root)
            return root
        else:
            return root1 or root2

can someone help me understand this?

[stable pecan]

some weird meld function that creates a tree with nodes whose values are the sum of the corresponding nodes in root1 and root2

[lunar jacinth]

its a leetcode problem

but like im not sure how the algo works here...

like how does it know which is left or right?

[stable pecan]

when binary trees are densely packed into a list you can tell a left node from a right node by whether it's in an odd position or even position in the list

actually, because they're providing the trees as a list, you can massively simplify the solution

[keen hearth]

Hey 🙂 Well any two vertices share a common root, so the path between them will be going up to that root from the first vertex, then down to the second vertex. If the tree is balanced, then the length of this path is O(log(num nodes in tree)).

[ashen jay]

Instead of using 'last' from the 2nd line, if I use 'list1' in its place on lines 5 and 6 the code breaks, why is that? 'list2' is used without setting it to a variable: https://paste.pythondiscord.com/nefuhoneye.rb

[hollow kayak]

please anybody that know how to get free rdp help me pls

[eager hamlet]

i mean this is for arbitrary trees

it could be a straight line, or just a bunch of nodes connected to a central one

[tall aurora]

What's a good source to learn algo and ds?

[sullen dock]

hey, anyone know how to implement multilevel queue cpu scheduling algorithm in python

[supple hemlock]

IDK i just started learning python I know nothing

[last fulcrum]

def factorial(n):
    if n == 1:
        return 1
    else:
        return n * factorial(n - 1)

print(factorial(5))

Can someone help me with the return in the else block? How does it decrease itself?

[last fulcrum]

https://www.amazon.com/Structures-Algorithms-Python-Michael-Goodrich/dp/1118290275/ref=sr_1_1?dchild=1&keywords=Data+Structures+and+Algorithms+in+Python&qid=1622009584&s=books&sr=1-1

[onyx root]

isn't n-1 less than n?

[full portal]

What's the mathematical way to find the average speed with miles and speed limits

or is it the same as finding the average with any list of numbers

[last fulcrum]

I don't really understand your question. I used 1 because n == 0 is the same as n == 1

[onyx root]

you asked how the else-return line decreases itself. n-1 is decreasing n.

[last fulcrum]

On the first iteration it's 5 - 4. But then n is still 5, that's why I'm confused. The value of 5 I gave does not change

So on the second it should still be 5 -4

[onyx root]

your call to factorial has n=5, but then it calls factorial with n=4, which calls factorial with n=3

each call keeps its own value of n, but the function calls nest like russian dolls

[last fulcrum]

Still confusing me. You said it calls factorial with n = 4, but the value of n I gave to the code never changed. I guess my problem is why is n changing. In a normal python code that would not be possible

I'll go watch some YouTube videos to help

[fiery cosmos]

I want good resource to learn algorithms

and data structures

using python

[austere sparrow]

The n will be different across function calls.
If you call factorial(2), it itself will call factorial(1) (in the else branch).

[fiery cosmos]

Also it could be better if it resource will be documented using BIG O

noatation

[last fulcrum]

https://www.amazon.com/Structures-Algorithms-Python-Michael-Goodrich/dp/1118290275/ref=sr_1_1?dchild=1&keywords=Data+Structures+and+Algorithms+in+Python&qid=1622009584&s=books&sr=1-1

[last fulcrum]

Why is the n different? Is it like using a for loop where the value of i changes on every iteration?

[austere sparrow]

!e

def print_numbers_down_from(n):
    print(n)
    if n > 0:
        print_numbers_down_from(n - 1)

print_numbers_down_from(4)

[halcyon plank]

@austere sparrow :white_check_mark: Your eval job has completed with return code 0.

001 | 4
002 | 3
003 | 2
004 | 1
005 | 0

[fiery cosmos]

Thank you but why do you recommend this book

[austere sparrow]

When you call print_numbers_down_from(4), it creates a 'frame' where all the local variables and such live inside the function. Then, inside that frame print_numbers_down_from(3) is called, and the computation for print_numbers_down_from(3) will happen separately

In other words, n isn't a global variable.

[jagged oasis]

Ya n in scope of function only

[last fulcrum]

It covers everything, including Big O notation

[onyx root]

I think the thing you are missing is that there are 5 different function calls happening, each of which has its own n.

[fiery cosmos]

Have you read it?

[last fulcrum]

Yes that is what I'm using right now.

[onyx root]

@fiery cosmos if you want to learn the basics of Big-O, this is a good less-mathematical introduction: https://bit.ly/bigopy

[austere sparrow]

Perhaps this visualization will help:
http://pythontutor.com/visualize.html#code=def print_numbers_down_from(n)%3A print(n) if n > 0%3A print_numbers_down_from(n - 1) print_numbers_down_from(4)&cumulative=false&curInstr=0&heapPrimitives=nevernest&mode=display&origin=opt-frontend.js&py=3&rawInputLstJSON=[]&textReferences=false
Or this one, for the factorial: http://pythontutor.com/visualize.html#code=def factorial(n)%3A if n %3D%3D 1%3A return 1 else%3A return n * factorial(n - 1) print(factorial(5))&cumulative=false&curInstr=0&heapPrimitives=nevernest&mode=display&origin=opt-frontend.js&py=3&rawInputLstJSON=[]&textReferences=false

[fiery cosmos]

Ok, thank you again

[last fulcrum]

Ok thank you @austere sparrow @onyx root I think what is confusing me is that I've never seen 5 different function calls happening.

[fiery cosmos]

Thank you

[fiery cosmos]

Is that you in the video?

[onyx root]

yes 🙂

[last fulcrum]

Here is the table of contents. Plus it teaches Python and OOP if you've never learnt them.

[austere sparrow]

smh rule 6 @onyx root||/s||

[onyx root]

that's the second time today i saw /s. what does that mean?

[fiery cosmos]

Wow, cool. I like the way how you explain

[austere sparrow]

it means that the preceding phrase was sarcasm, not serious

[last fulcrum]

@austere sparrow I really understood the print_numbers function. Not sure why it's not clicking for the factorial haha

[austere sparrow]

do you have Thonny?

it has a really good visual debugger

I'll show a screenshot, but I have 2gb of updates to download

[last fulcrum]

Nope. But I just understood the factorial also. I'll explain so you tell me if it's right

[onyx root]

thanks, i gathered it meant that, but didn't know why "s" 🙂

[last fulcrum]

So it's 5 right, then it calls the function on 5 saying factorial (5-1)
So now it's 4. So it calls the function on itself saying again 4-1, so it's basically like it stores the number subtracted and then on the next call uses that number.

[austere sparrow]

It's not really "changing n", it just spawns an entirely separate function call, which will have its own, completely separate n variable.

[onyx root]

@last fulcrum Look at this example:

def foo(n):
    print(n)

n = 5
foo(10)
print(n)

The call to foo doesn't change the n on the outside, only the n on the inside

[last fulcrum]

!e
def foo(n):
print(n)

n = 5
foo(10)
print(n)

[halcyon plank]

@last fulcrum :white_check_mark: Your eval job has completed with return code 0.

001 | 10
002 | 5

[last fulcrum]

So basically we give number to the function, that number then becomes a local variable, and then because the function is calling itself, each number becomes a separate local variable?

[onyx root]

exactly

[austere sparrow]

Imagine a shady Amazon seller. When you buy a book from him for $20, he goes on to buy it from another seller for $16. That seller in turn buys from another seller for $14, who just had it laying around.

Each of those transactions will have its own price, order ID etc.

[last fulcrum]

Ok. Go on. And that's a great example

So I have one more question.

So then if it's doing separate calls how does it combine them all when it reaches the base case?

[austere sparrow]

def factorial(n):
    if n == 1:
        return 1
    else:
        return n * factorial(n - 1)

If you call factorial(1), it's going to return 1.
If you call factorial(2), it's going to return 2 * factorial(1), which will in turn call factorial(1), which will return 1, so the result is 2
If you call factorial(3), it's going to return 3 * factorial(2), which will in turn call factorial(2), which will return 2 (as shown above), so the result is 6
If you call factorial(4), it's going to return 4 * factorial(3), which will in turn call factorial(3), which will return 6 (as shown above), so the result is 24

You could change the function like this:

def factorial(n):
    if n == 1:
        return 1
    else:
        previous = factorial(n - 1)
        return n * previous
```, maybe that will help.

[last fulcrum]

I like this better. Thank you and @onyx root for explaining to me

[austere sparrow]

Maybe you'll understand it better if you implement this function:

def flatten(a_list):
    ...

>>> flatten([1, 2, 3])
[1, 2, 3]

>>> flatten([1, 2, [3, 4], 5, [6, 7]])
[1, 2, 3, 4, 5, 6, 7]

>>> flatten([1, 2, [3, [4, 5]], [6, 7], [[[8, 9], 10]]])
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

[last fulcrum]

I'll work on it. Going to bed now. And thanks once again

[austere sparrow]

👍

[sick dagger]

Hi! I was browsing the internet and found this golden ratio formula. I tried to code it, unfortunately it seems endless and I have trouble conceptualizing the code. Someone would have any idea ? Thank you.

[austere sparrow]

Have you tried some implementation?

[topaz swallow]

I think it will be hard

Can you explain what you mean by implementation ?

[fiery cosmos]

@onyx root Could you please recommend an algorithms and data structure book?

[onyx root]

I don't have a recommendation.

[fiery cosmos]

I just want to have sevral choices

[fiery cosmos]

What's your source by the way

A favorite source

[keen hearth]

Check the pinned messages for some resource recommendations.

[onyx root]

I don't know what to tell you, i don't have a source for data structures etc, I guess I google when i need things.

[fiery cosmos]

Okey, thank you

[fiery cosmos]

I saw but i'm searching for a good book which helps me to prepare for the coding interviews using Python

[keen hearth]

A book called Cracking the Coding Interview is often recommended for interview prep. I own a copy but I've not read any of it, so can't vouch for its quality 😄 A very popular undergraduate algorithms textbook is _Introduction to Algorithms, by Cormen, Leiserson, Rivest, and Stein. It's a good textbook, but quite mathy.

[onyx root]

On behalf of our industry, i would like to apologize in advance for coding interviews

[fiery cosmos]

Heavy artillery math? 😄

[fiery cosmos]

But why? There is no problem i think

[onyx root]

they expect you to write code on a whiteboard, while someone is watching, with no tools. You will never have to do that on the job.

they expect you to implement red-black trees, or something, which you will never have to do on the job.

if you can study for the interview, then what is the interview assessing?

[fiery cosmos]

Interesting

[austere sparrow]

The trick to computing infinite fractions is usually to take some initial value and substitute it for the ....
For example, if you pick 1 for ... and some random depth, you end up with

phi = 1 + (1/(1 + (1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + ...))))))))
phi = 1 + (1/(1 + (1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1))))))))

...which is 1.619047619047619.

The way you could get up to this point is, for example:

1. take some value as your result, like 1
2. compute 1 + 1 / (1 + result). It's going to be a slightly better approximation.
3. store the new approximation
4. go to step 1

However, this gets a bit tricky.
Let's try to find an exact solution for phi:

φ = 1 + 1/φ
φ^2 = φ + 1
φ^2 - φ - 1 = 0

φ = (1 + √5)/2
φ = (1 - √5)/2

So there are two solutions to this equation.
If you try to substitute some number (except -1) using the algorithm above, you'll be slowly approaching (1 + √5)/2 (approx. 1.618). However, if you choose (1 - √5)/2 (approx. -0.618) as your starting point, you'll get:

φ0 = (1 - √5)/2

φ1 = 1 + 1/φ0 = 1 + 2/(1 - √5) = ((1 - √5) + 2)/(1 - √5) = (3 - √5)/(1 - √5)

Using some transformations (e.g. using the proportion rule, (3 - √5)/(1 - √5) = (1 - √5)/2 <=> (6 - 2√5) = (1 - √5)(1 - √5) <=> 6 - 2√5 = 6 - 2√5), you can see that you end up at the same point, and result is going to stick to (1 - √5)/2.
(with floating point numbers, of course, it's going to eventually drift, but it's still going to do that very slowly).

oops sorry for the text wall

[fiery cosmos]

But I think it's good idea to check problem solving skills which is very important to get fast and optimized solutions. But i'm not sure. What do you think about this

[austere sparrow]

if you implement your own optimized red-black tree for a simple CRUD application, you'll probably get fired 👀

[onyx root]

problem solving is a good thing to assess. I just don't see how studying data structures will help with an interview that assesses problem solving.

[fiery cosmos]

So again what do you think which is the most reliable way to do that?

[onyx root]

Practice solving problems.

have you built anything interesting?

[wide lake]

hey guys just completed the basic of python if i want to start dsa where do i start like any youtube channels,books,websites

any recommend

[oak spindle]

i'm also in the same path, if you get anything share with me too

[stable pecan]

job security

[austere sparrow]

well, hopefully the company has code reviews 👀 👀

[stable pecan]

i did avl trees recently, not red-black trees

i'm slacking

[fiery cosmos]

I think yes but not many

I guess it wasn't about competitive problem solving, right?

[indigo radish]

Geeeks4geeks for concepts.
Leetcode best for practicing learned concepts, filter by tags.
Elements of programming interviews Python edition, nice variety of questions, but solutions are little difficult.

But the theory of book, the approach is nice.

[fiery cosmos]

golden ratio is irrational.

Infinite and no pattern

you have formula to get closer, but never exact

Like pi

[last fulcrum]

For binary search, is it better if your item has not been found to make the new upperbound the mid - 1 or is just mid?

Same thing for lowerbound if the item has not been found (mid + 1 or mid)

And is it better to use iteration or recursion for binary search?

[runic tinsel]

performance-wise iteration is always better, you use recursion for aesthetic reasons like readability

well not always, often it's the same, depends on the compiler

for python you want iteration

[last fulcrum]

Anyone knows what I'm doing wrong?

def binary_search(sequence, item_to_find):
    lowerbound = 0
    upperbound = len(sequence) - 1

    while lowerbound <= upperbound:
        midpoint = (lowerbound + upperbound) // 2

        if sequence[midpoint] == item_to_find:
            return sequence[midpoint]

        elif sequence[midpoint] < item_to_find:
            upperbound = midpoint

        else:
            lowerbound = midpoint
    return None

[runic tinsel]

do -1 and +1

and it's > item_to_find

[last fulcrum]

Why is it >?

And why can't you do midpoint = lower, but you have to add one

[runic tinsel]

[1,2] and you look for 2

midpoint is 0

you change lower bound to 0, nothing changed

just how numbers work out

[last fulcrum]

It would be 1. 0 + 2 // 2 == 1

[runic tinsel]

0+1

[1,2] is entire array

so 0 and 1

[last fulcrum]

Ok makes sense.

[runic tinsel]

maybe -1 is not necessary

can't hurt

[fierce wolf]

hi there, I feel like I have been given a question in this practice paper which is not totally accurate and was hoping someone could help me understand. I need to apply branch-and-bound technique to a knapsack problem - but I need to do this by using jumptracking and not backtracking

I was under the impression knapsack problems were solved using backtracking and machine learning type problems were solved with jumptracking

so i'm at a loss when trying to solve a knapsack problem with jumptracking, any help is appreciated thank you

[eager hamlet]

https://codeforces.com/edu/course/2/lesson/9/3/practice/contest/307094/problem/H

would a greedy method work for this problem?

[shut phoenix]

why does this doesnt work? (Ive a file were is the data I need, and its called codigos.py), appears an error in there

from Codigos import English

objQuestionnaire = English()

objQuestionnaire.empezar()
objQuestionnaire.respuestas()
objQuestionnaire.imprimir()

[vocal gorge]

You probably want a help channel (see #❓｜how-to-get-help), this doesn't appear to be DSA-related.
(also, please post the error you get if you do)

[shut phoenix]

okok, thanks

[dapper sapphire]

If we modify the current array/list, we are not using any extra space, then is it considered O(1)?

[vocal gorge]

Yes, inplace doesn't count as extra memory. For example, many inplace sorts are O(1) extra memory.

(timsort among them, I think)

[dapper sapphire]

ok got it thanks!!

[fiery cosmos]

Hi, Is this the right place to ask about resources related to algorithms and ds in python?

[vocal gorge]

Yup

[fiery cosmos]

So I've been looking for resources lately, I tried CLRS, Princeton Uni's course on Algorithms, elements of programming interview in python but none of them were good for me atleast. I was looking for something which would just explain me the basic algorithms, ds and maybe the implementation from scratch w/o importing anything, I'm ok with video lectures but I prefer articles/books.

[jagged oasis]

I recommend implementing data structures by hand in another language.

[fiery cosmos]

like by myself?

I'll use C++

I'll look at the pinned resources first.

[marble bronze]

Is there anyone here knowledgeable about BSTs and heaps?

[drifting drum]

Came across this question in some test:
You have an array of integers.
You're supposed to remove those elements A[i] which satisfy A[i] > A[i-1]
Do this until you cannot find any more pairs. How many rounds does it take till you cannot find any more pairs?
For example

[3, 6, 2, 7, 5]
After 1 round, it becomes [3, 2, 5]
After 2 rounds, it becomes [3, 2]
And then it stays that way forever so the answer would be 2

Another example is:
[6, 5, 8, 4, 7, 10, 9]
After 1 round, [6, 5, 4, 9]
After 2 rounds, [6, 5, 4]
Then it stays that way

Constraints were len(arr) <= 10^5
Initially thought it had to do with next smaller element and stacks, but then got stuck
thoughts?

I also tried simulating it, but that timed out ofcourse

[vocal gorge]

can't you do it in one pass? Like, just extract a descending subsequence from the list

iterate over the list, remembering the last accepted element. The first element is accepted, then accept only elements <= the remembered one.

[drifting drum]

well you're supposed to extract subarrays, not subsequences right cuz it states A[i] < A[i-1]

or maybe I don't understand clearly what you're trying to say

erm, okay you mean get rid of the elements that satisfy that condition

okay but then?

[vocal gorge]

You need to remove all elements that's higher than the previous one. So for an element to remain, it has to be <= the previous one, and therefore <= than all previous ones

[drifting drum]

<= ?

right

then?

[vocal gorge]

therefore, you just need to accept only elements that are lower or equal to the last accepted element

and that's pretty much it

a single pass over the list, O(n) complexity

[drifting drum]

how does that give us the number of rounds?

getting the numbers that remain in the end is easy, O(N) with a stack, yes.
However, the # of rounds is what's required.

[vocal gorge]

oooh, I totally missed this. Hmm...

perhaps the number of turns for a number to get removed is something like "how many positions away the previous lower number is"

[drifting drum]

so did I, when I was solving it initially. Implemented the same idea before realising it -.-

haha, yeah I thought so too but then it's easy to come up with counterexamples for that

1, 10, 11, 12, 13, 4
vs
1, 10, 11, 12, 13, 14, 15, 16, 17, 18, 4
Both cases require 2 days

and I tried handling that but then it got tricky because there can be examples like above where you need 3 days, 4 days, more

[vocal gorge]

A number that gets removed but doesn't get removed right away must be a part of a nonincreasing subsequence that starts with a higher number than the last accepted number before it, so, hmm

[drifting drum]

must be -> may be

5, 10, 11, 12, 13, 4
vs
1, 10, 11, 12, 13, 4

[vocal gorge]

hmm, indeed it's not obvious

[drifting drum]

thank you for saying that, makes me feel less bad that I couldn't solve it 😄

[vocal gorge]

I do feel like it's the right way to dig, though - for each nonincreasing subarray, determine how many numbers need to be chipped off of it

[drifting drum]

yeah essentially you could simulate it till you realise there's no increasing subarrays left anymore

but that's gonna time out

[vocal gorge]

in these cases, the subsequence is 13,4, and it needs to lose 1 number in the first (because then 5>4) and 2 in the second case (because 1 is lower than both of them)

[drifting drum]

yeah, I think getting the numbers isn't the hard part

[vocal gorge]

I feel like it might be, for each nonincreasing subarray, the number of numbers in it that's higher than the last accepted number

and then you take the max over all these counts to get the value for the entire list

[drifting drum]

I mean, getting the final list of numbers that survive

i was thinking of cases where you have like
p , .....q, ..... , r, ......, s,..... w

the dots are the elements that are going to get removed

so if q > p, then q needs 2 days.

and then if r < q and r > p, it will need 3 days
but if r > q, it needs 2 days

but if q < p

then, it's like q becomes the new p

and now if r > q, r needs 2 days.

and if r < q, then.. r becomes the new q? and in turn the new p

and that's like

O(N^2) again because

it really makes no difference whether our array was
p , .....q, ..... , r, ......, s,..... w
or
p, q, r, s , w

because in the worst case

we could have p > q > r > s
and for w, we might have to wound back all the way upto p to check whether it gets removed

Agh, now that I think of it

TreeSet sounds like such a good idea

Oh god -.-

that would give us the position of w in the array [p, q, r, s] and essentially the number of days, in log(n) time

hm, I'm still fuzzy about it, maybe I'm just wrong

[vocal gorge]

it still sounds to me like you don't need any ds here. Like...

def rounds(nums):
    accepted = last = nums[0]
    max_rounds = 0
    noninc_count = 0
    for num in nums[1:]:
        #print(num,accepted,max_rounds,noninc_count)
        if num<=last:
            # we're in a nonincreasing subsequence
            noninc_count += 1
            if num <= accepted:
                max_rounds = max(max_rounds, noninc_count+1)
                accepted = num
                noninc_count = 0
        else:
            noninc_count = 0
        last = num
    max_rounds = max(max_rounds, noninc_count+1)
    return max_rounds

this seems to pass all examples you posted

[drifting drum]

The examples I posted initially? Those are the easy ones

This for example

fails nums = [5, 10, 11, 12, 4]

well, probably this is an easy one too, but what I meant to say is there were many more cases yeah

I failed on like the last 2 amongst all their cases

and ofc, the cases weren't revealed

now the answer for this case btw, should be 1, the code outputs 2

[vocal gorge]

def rounds(nums):
    accepted = last = nums[0]
    max_rounds = 0
    noninc_count = 1
    for num in nums[1:]:
        if num<=last:
            # we're in a nonincreasing subsequence
            noninc_count += 1
            if num <= accepted:
                #print(num,accepted,max_rounds,noninc_count)
                max_rounds = max(max_rounds, noninc_count-1)
                accepted = num
                noninc_count = 0
        else:
            noninc_count = 1
            #print(num,accepted,max_rounds,noninc_count)
        last = num
    max_rounds = max(max_rounds, noninc_count)
    return max_rounds

passes it. Actually, lemme build a proper automatic tester...

[drifting drum]

well this one would fail on
nums = [5, 10, 11, 12, 6, 7]

[drifting drum]

it's very similar to what I'm talking about here

[vocal gorge]

ah, I see

I generated an example [0, 1, 1, 2, 1]

which... yeah, it shows why my approach is flawed

my approach would predict that both subsequences here 1,1 and 2,1 dissolve in 2 turns

but in reality, after one turn, 0,1,1 is left, and the second 1 doesn't dissolve on that turn

[drifting drum]

also, let's just assume numbers are unique right now

(not that it changes anything ofcourse)

and yes, correct

[vocal gorge]

here's the automated tester I used, using hypothesis to generate examples:

from hypothesis import given, strategies as st

def rounds_exact(nums):
    """Inefficiently (but, hopefully, correctly) calculates rounds needed by simulating"""
    nums = list(nums)#copy
    rounds = 0
    while True:
        new = [nums[0]]
        skips = 0
        for i in range(1,len(nums)):
            if nums[i] > nums[i-1]:
                skips += 1
                continue
            new.append(nums[i])
        if skips == 0:
            break
        rounds += 1
        nums = new
    return rounds

# make hypothesis disprove this.
# Limiting the list size to 1000, otherwise rounds_exact would be too slow.
@given(st.lists(st.integers(0,1000),min_size=2,max_size=1000))
def test_rounds(nums):
    true = rounds_exact(nums)
    calc = rounds(nums)
    assert true==calc,(true,calc)

test_rounds()

[drifting drum]

ah, quite cool

what's the st.lists() and given decorator doing
@given(st.lists(st.integers(0,1000),min_size=2,max_size=1000))

oh, building a list of ints from 0-1000 of 2 <= size <= 1000

hm

[vocal gorge]

basically, hypothesis can automatically generate disproving test examples (mostly randomly, though once it finds one, it does smart stuff to reduce it to a hopefully-minimal one)

the decorator here describes the parameters - that nums should be a list of size 2 to 1000 of integers from 0 to 1000

[gleaming kite]

wow, cool

[drifting drum]

I see

interesting stuff, thank you for sharing that

[gleaming kite]

how it works? by trying to fuzz whole AST?

[vocal gorge]

you need to provide parameters for what input strategy to use (like, here, a list of ints)

[austere sparrow]

it can also use type hints to infer the data to generate, which is cool

(if your data doesn't have additional constraints)

[vocal gorge]

as for how exactly it works, uhh... I'm trying to read the source but it's unclear. I think it generates examples randomly starting from "simple" ones, but a lot of the niceness comes from it being able to shrink a disproving example to a simpler one

[gleaming kite]

https://hypothesis.readthedocs.io/en/latest/details.html#use-with-external-fuzzers
looks like it support multiple fuzzers, this can be very useful for many use case, cool

[vocal gorge]

this is the function used to decide whether to stop adding elements to a list, say:
https://github.com/HypothesisWorks/hypothesis/blob/2b7d74e5621057d0c9f0fcfbae8de1c80fbacfc9/hypothesis-python/src/hypothesis/internal/conjecture/utils.py#L164

[halcyon plank]

hypothesis-python/src/hypothesis/internal/conjecture/utils.py line 164

def biased_coin(data, p, *, forced=None):```

[vocal gorge]

it has... quite a bit of comments

[fallen breach]

Hi ! I have a problem ! I have a data file with an array of 7221721721171, I would like to split whenever it meets the number 7
the result I want is : [7..],[7..],[7..] , I tried np.split but I have some difficulties . thx you !

[past escarp]

is there a library that has max_heap implemented? i only found minheap

[fiery cosmos]

Negate all the numbers

[past escarp]

k thanks

[loud trail]

is the data in a text file? or is it already loaded as a numpy array?

are there any performance constraints? is the file really big?

[fallen breach]

hey the datas are loaded in text file

[fallen breach]

nope , I can reduce it

[halcyon plank]

Hey @fallen breach!

Uh-oh! It looks like your message got zapped by our spam filter. We currently don't allow .txt attachments, so here are some tips to help you travel safely:

• If you attempted to send a message longer than 2000 characters, try shortening your message to fit within the character limit or use a pasting service (see below)

• If you tried to show someone your code, you can use codeblocks
(run !code-blocks in #bot-commands for more information) or use a pasting service like:

https://paste.pythondiscord.com

[fallen breach]

so basically the problem is I have 2 list

[scenic goblet]

how to implement dfs in edge list vector<pair<int,int>>edges

[loud trail]

you can do it the "stupid way" like this

text = '7221721721171'
sections = []
section = []
for c in text:
    if c == '7':
        sections.append(section)
        section = [c]
    else:
        section.append(c)
print(sections)

!e ```python
text = '7221721721171'
sections = []
section = []
for c in text:
if c == '7':
sections.append(section)
section = [c]
else:
section.append(c)
print(sections)

[halcyon plank]

@loud trail :white_check_mark: Your eval job has completed with return code 0.

[[], ['7', '2', '2', '1'], ['7', '2', '1'], ['7', '2', '1', '1']]

[fallen breach]

I have another question , well basically this list is associated with time , how can I apply it on it ?

[fallen breach]

thx you a lot !

[dull yoke]

hello guys, I have a deque with maxlen of 50 items that belongs to [0, 640].

Most of the items fluctuate minimally with the range of about 50 or 60. But suddenly, some items emerge from the graph, rise above or go lower than the fluctuation range. How do I detect that change?

the reason it is maxlen of 50 is because the deque is updated per frame.

[loud trail]

can you clarify this further? this might also be more suited for a help channel as per #❓｜how-to-get-help .

[fallen breach]

may I talk to you in DM ?

[loud trail]

i'd rather not, sorry

[fallen breach]

okay so basically I have two lists one with a serie of numbers and the second with the time associated

I asked how to split for the numbers and now I have to split for the time and then substract

for example : [7112],[7212] and the time associated [0,1,2,3],[4,5,6,7] and then I have to substract the second list ( there's a lot of number in the file ) [0,1,2,3],[4-4,5-46-4,7-4] @loud trail

[loud trail]

you want to subtract the first number in a list from every number in the list?

are these lists, or numpy arrays?

[tranquil vigil]

Is there anyone who can mentor me?

I'm beginners rn

[ashen forge]

In which topic you want us to mentor you?

[ashen forge]

in DSA??

[lean spruce]

https://paste.pythondiscord.com/novesuwohi.lua How to replay game (code) ?

After typing yes in input?

[loud trail]

@lean spruce you'd have to put the entire thing in another while loop

[lean spruce]

while yes:

[loud trail]

sure, something like that

maybe it's a good idea to put your entire game into a function

def run_game():
    ...

def ask_yn(prompt):
    ask_repeat = True
    while ask_repeat:
        user_val = input(prompt).strip().lower()
        if user_val == 'n':
            return False
        elif user_val == 'y':
            return True
        else:
            print('That is not a valid response.')

def main():
    play_game = True
    while play_game:
        run_game()
        play_game = ask_yn("Play again? y/n: ")
        if play_game:
            print('Again!')
    print('Goodbye!')

something like this

[lean spruce]

oke

thanks

[grizzled hemlock]

Does anyone have a good article where I can learn about algorthims in terms of N+1 , N(0), etc...?

please ping me if you do! 🙂

[austere sparrow]

Maybe you mean O-notation? 🙂

Try this article: https://nedbatchelder.com/text/slowsgrows.html

[ashen forge]

geeksforgeeks has pretty good articles about time complexities

[grizzled hemlock]

Thanks for your recommendations 🙂

[grizzled hemlock]

That's probably what I meant haha, would N+1 be considered O-notation?

[austere sparrow]

No, that's not correct O-notation

[grizzled hemlock]

What term am I looking for that would explain N+1 complexities?

[austere sparrow]

I don't understand what you're referring to.

Did you find that somewhere?

Maybe you are referring to the 'N+1 problem'?

[grizzled hemlock]

Algorithms that suffer from N+1 , sorry I'm a bit of a noobie with time complexity.

[grizzled hemlock]

Yes!

Is that just how it's referred to, N+1?

[austere sparrow]

That's not really related to algorithms in general. It's about making N+1 queries when you really need 1.
https://stackoverflow.com/questions/97197/what-is-the-n1-selects-problem-in-orm-object-relational-mapping

[grizzled hemlock]

Gotcha understood

[ashen forge]

What study material are you referring to

[grizzled hemlock]

Just trying to prepare for a Python technical interview. So if they ask for time-complexity questions related to my algo's, I have it nailed down.

[ashen forge]

Okay

Time complexity is integral part of DSA

I learnt it from geeksforgeeks

[grizzled hemlock]

Yeah I just visited the many articles they have about time complexity

super useful stuff, thank you 👍

[fervent saddle]

It’s what type of line you’d see if you graphed how long something took vs how much data you have

So finding the length of a list is a flat horizontal line, it’s O(1), since lists keep track of their length

And finding the index position of something in an unsorted list would be a flat diagonal line, it would be O(n)

Like if the x axis was how much data you have, and the y axis was how long it took to do it

That’s why constant factors don’t matter for finding something’s time complexity

Because it doesn’t change what kind of line you see

[grizzled hemlock]

I see

[grizzled hemlock]

What would happen if the list is sorted? @fervent saddle

[fervent saddle]

Then you could use binary search

I mentioned it being unsorted so it’s clear that I’m talking about linear search and not binary search

[grizzled hemlock]

what notation would binary searches be?

[fervent saddle]

O(log(n))

[fiery cosmos]

Hey anyone who can code discord bots. I am willin to pay or do what ever you want. I just want my own discord bot for my discord server..

[sturdy crest]

why do we have gaming flag at the background?

[sturdy crest]

ey you can learn it...there is a bunch of good tutorial in yt

[cinder wigeon]

you can find an open source bot and just host it

[inland forge]

Check #discord-bots

[stiff oyster]

Hello there, I'm not sure were to ask. I try to build a data line graph with dash and plotly. The CSV sadly seperats the date in two rows: Date and Time. Is there an easy way to combine this rows with panda? So i can display the datetime on my x scale?

[gilded kernel]

hi

py

import random
word_list = ["aardvark", "baboon", "camel"]

chosen_word = random.choice(word_list)

print (chosen_word)
guess = input("chose a word ").lower()

display = ['-']
for i in chosen_word:
    if guess==i:
        print(guess)
    else:
        print(display)

how can i print my output vertically

please teach me am new to python

[cerulean elk]

wdym vertically?

[last fulcrum]

Each cell of an array must use the same number of bytes. This is true in statically typed languages like C, since I have to do
int data[100]; so everything in memory would be the same since they are all int's. In Python, the language is dynamically typed, so I can do list_1 = ['a', 'b', 4, 4, 4.0]. An int, a character and a float would all take up different spaces in memory, so how does the statement work in Python?

[west hollow]

Hello!

[austere sparrow]

Are you familiar with C?

In C, int data[100] creates a contiguous chunk of 400 bytes, and all the ints are stored in that chunk. However, you can make an array of pointers: char* strings[100]. Now you have an array of 100 strings, but the array itself only stores the memory addresses of where each string begins.

[jaunty cave]

U have to add const to make it works

[fiery cosmos]

Can someone suggest me a good python data structures algorithm book or video tutorials?

[austere sparrow]

hm?

No, you don't, const just means that you can't change something.

[last fulcrum]

Yes

[austere sparrow]

does my explanation make sense?

[last fulcrum]

Click on my profile. And yes it makes sense

[unborn sundial]

!e

import contextlib

@contextlib.contextmanager
def withandler(var):
    try:
        print("start stuff")
        yield var
    finally:
        print("finish stuff")


with withandler("123") as var:
    print(f"do stuff = {var}")

[halcyon plank]

@unborn sundial :white_check_mark: Your eval job has completed with return code 0.

001 | start stuff
002 | do stuff = 123
003 | finish stuff

[unborn sundial]

more decorators to the throne of decorators

[loud trail]

!e ```python
import contextlib

@contextlib.contextmanager
def withandler(var):
print("start stuff")
try:
yield var
finally:
print("finish stuff")

with withandler("123") as var:
print(f"do stuff = {var}")

i'd do it like this

[halcyon plank]

@loud trail :white_check_mark: Your eval job has completed with return code 0.

001 | start stuff
002 | do stuff = 123
003 | finish stuff

[hollow skiff]

[last fulcrum]

@austere sparrow can you help me explain this. The part of the referential structure.

Most significantly, the overall memory usage will be much lower for a compact structure because there is no overhead devoted to the explicit storage of the sequence of memory references (in addition to the primary data). That is, a referential structure will typically use 64-bits for the memory address stored in the array, on top of whatever number of bits are used to represent the object that is considered the element.

[jaunty cave]

Yes thats the only way to make array of strings from array of char pointer without warning after compiling. This is happening when g++ is used as compiler. To make array of string that is editable, u have to make 2 dimensional array of char, not array of char pointer.

const char* strings1[100];     //this is non-editable
char strings2[100][101];       //this is editable

[austere sparrow]

If you make a char* strings1[100], the strings will be perfectly editable, it's just that you can't put read-only strings into it.
A string literal like "Hello!" will point into a read-only part of memory, and not create a new string on the heap or the stack.
This: ```c
#include <stdlib.h>
#include <stdio.h>
#include <string.h>

int main() {
char* strings[2];

char* hello = malloc(strlen("Hello!") + 1);
strcpy(hello, "Hello!");
char* world = malloc(strlen("world") + 1);
strcpy(world, "world");

strings[0] = hello;
strings[1] = world;
printf("%s %s\n", strings[0], strings[1]);

free(hello);
free(world);

}

(also, g++ is for C++, not C)

[last fulcrum]

I miss malloc and pointers. Now python does everything

[austere sparrow]

Yeah, the sky was bluer and the grass was greSegmentation fault (core dumped)

[last fulcrum]

😂 Still, I think every new programmer should learn C before python. Makes understanding low lvl things far easier.

[austere sparrow]

I would probably have given up programming if I started with C.

[last fulcrum]

C is hard, no doubt. But the knowledge you get is so much. Most people would never learn C after python because python is far easier. So start with hard, then go to easy. But I digress.