def makebin(num):
every_squares = []
for i in range(0, num):
power = 2**i
if power > num:
break
every_squares.append(power)
reminder = None
nums = []
while reminder != 0:
print(reminder)
sq = 0
for i in range(0, num):
x = 2**i
if x > num:
print("breaked")
break
sq = 2**i
print(sq, "sq", num)
reminder = num - sq
num = reminder
nums.append(sq)
print(nums, every_squares)
binary = ""
for n in every_squares[::-1]:
if n in nums:
binary += "1"
else:
binary += "0"
return binary
x = makebin(75)
print(x)
#🔒 Trying to make a function that convert decimal to binary
faint badge
heavy obsidianBOT
```py
def makebin(num):
every_squares = []
for i in range(0, num):
...
@faint badge
Python help channel opened
faint badge
i did something like this to get the square number most close to the num py while reminder != 0: print(reminder) sq = 0 for i in range(0, num): x = 2**i if x > num: print("breaked") break sq = 2**i print(sq, "sq", num) reminder = num - sq this make sense like for a number 75 i wanted to get the square 64 is my method ok?
rotund vortex
What you want to look at is the floor division operator, the modulo operator and the divmod function.
Also the format and bin functions.
!d divmod
heavy obsidianBOT
rotund vortex
The divmod function pairs well with unpacking.
faint badge
oh.. i know floor division and modulo not divmod.. i was lacking logic this was my only logic i came up with
faint badge
What you want to look at is the floor division operator, the modulo operator and...
i actually saw a video on how to convert decimal to binary with subtraction method i thought of implementing it... does the subtraction method need this floor division or modulo isnt the division method the one that requires this.? i dont know much about binary numbers too
rotund vortex
There is subtraction involved, so I assume that is the technique I'm thinking of.
faint badge
yes num - sq
faint badge
i did something like this to get the square number most close to the num ```py
...
my actual question was is this code messy i neeed to find the square number most close to the num (75 here on 1st iteration) so i can do num -sq
rotund vortex
Squaring is not involved in the way I was thinking.
faint badge
ye i dont know that method
rotund vortex
!e py print(bin(123)) #or print(format(123, 'b'))For the easy way out.
heavy obsidianBOT
!e ```py
print(bin(123))
#or
print(format(123, 'b'))```For the easy way out.
:white_check_mark: Your 3.12 eval job has completed with return code 0.
001 | 0b1111011
002 | 1111011
the value of sq will be of its last iteration like 64 in first iteration of while
faint badge
!e ```py
print(bin(123))
#or
print(format(123, 'b'))```For the easy way out.
ye i did that first but was curios to implement this
rotund vortex
I leave to dinner.
faint badge
oh ok nvm thanks for time
muted fog
2**i are not squares 🥴
muted fog
powers of 2, yes
faint badge
okk
muted fog
but yeah, there are way easier ways of doing this process
if you have say 11 (1011 in base 2), you can extract the rightmost bit easily
by just checking if 11 is odd/even
faint badge
oh i have no idea of binary system have to study in uni
muted fog
by just checking if 11 is odd/even
then you can do something to get rid of the rightmost bit, leaving you with 101
and you can do an odd/even check again
muted fog
oh i have no idea of binary system have to study in uni
it's not really much different to how our decimal system works
faint badge
then you can do something to get rid of the rightmost bit, leaving you with 101
hmm
yeh ill learn it someday... actually this was programming question on exercism.org i spent almost 1.5hours on it lol
i need to improve my logic.. like i spend lot of time behind flawed logic.. maybe its because am still new programming (1year)
muted fog
you need to read up on the basics of the thing you are trying to do 
muted fog
!e here is the core of it
n = 11
while True:
print(n%2)
n //= 2
if n == 0:
break
:white_check_mark: Your 3.12 eval job has completed with return code 0.
001 | 1
002 | 1
003 | 0
004 | 1
Python help channel closed