🔒 How is this binary search code not working??

[final jungle]

Code:

n=int(input("Enter the size of ur list: "))
lis1=[]
for i in range(n):
    ele=int(input(f"Enter element {i+1}: "))
    lis1.append(ele)
print(f"Your list is: \n{lis1}")
key=int(input("Enter the element you wish to find: "))
def sort_list(lis):
    sorted_list=[]
    while lis:
        small_num=min(lis)
        sorted_list.append(small_num)
        lis.remove(small_num)
    print(sorted_list)
    return sorted_list
sorted_list=sort_list(lis1)
def binary_search(upper,lower=0):
    result=None
    while lower<=upper:
        mid=(lower+upper)//2
        if sorted_list[mid]==key:
            result=mid
        elif sorted_list[mid]<key:
            binary_search(upper,mid+1)
        else:
            binary_search(mid-1)
    return result
if binary_search(n-1)==None:
    print("Your no is not present in the list!")
else:
    print(f"{key} was found at position {result}!")

[empty sequoia]

@final jungle

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[final jungle]

Gets stuck here

[minor galleon]

Explain what this code should do

[final jungle]

binary search

algorithm

should tell if the element is there in the list and if yes, which pos

if not then print the respective statement

[minor galleon]

Ok

[final jungle]

Its a really famous algorhithm u can look it up

I just dont get why my code would not work, + its not an error it just.. gets stuck

[minor galleon]

First instead of sorlist function u created u can simply type

Lis1.sort()

This will sort out your list in ascending order

[final jungle]

Ohk

But sort list function works fine

[minor galleon]

It make your code lenthy and hard to understand

[sacred night]

u defined the binary_search function with 2 parameters yet u are only giving it 1 argument

[minor galleon]

Instead of writing 10 lines u can do the same thing in one line

[final jungle]

Yes because one has a default value

[final jungle]

Ok i have now edited it

[final jungle]

So we do not need to pass value to it

only to the upper arguement

[minor galleon]

@sacred night well i dosnt matter cause he have mention in the parameters that lower is by default = 0

[final jungle]

we now get an error @minor galleon

why is mid nonetype tho

both lower and upper are integers

and //2 should also result in integer

[minor galleon]

If key in lis1:
      Print(f'The key is at index : {lis1.index(key)}')
Else:
       Print("your key is not present in the list")

Instead of binarysearch write this

[modest smelt]

It's not mid, it's sorted_list

[final jungle]

Dude

That defeats the whole purpose

[modest smelt]

That is None

[final jungle]

Thats LINEAR SEARCH

[final jungle]

Ohh

Wait wut

how

Wait thats cus of the correction @minor galleon suggested

[modest smelt]

!e

my_list = [1, 3, 2]
other = my_list.copy()
print(my_list.sort()) # returns None
print(my_list) # is sorted
print(sorted(other)) # sorted list
print(other) # but isn't saved to the list```

[empty sequoia]

:white_check_mark: Your 3.12 eval job has completed with return code 0.

001 | None
002 | [1, 2, 3]
003 | [1, 2, 3]
004 | [1, 3, 2]

[final jungle]

sort function sorts lis1 but returns none in sorted_list

[lone rover]

i don't think binary searches are supposed to recurse...
also, when you're recursing, you're not doing anything with the return value anyway

[final jungle]

I have now corrected it:

lis1.sort()
def binary_search(upper,lower=0):
    result=None
    while lower<=upper:
        mid=(lower+upper)//2
        if lis1[mid]==key:
            result=mid
        elif lis1[mid]<key:
            binary_search(upper,mid+1)
        elif lis1[mid]>key:
            binary_search(mid-1)
    return result

But now there is this error

No error it just gets stuck

[modest smelt]

You want to set result to whatever the recursed function returns

[final jungle]

I remember making it using recrusion in school

[final jungle]

Yeah

[lone rover]

sure, you could also make a linear search recursive, it's just terribly inefficient

[final jungle]

Ohh

so this is inefficient than iterative one

Also i found a code online

[minor galleon]

Here i found this
https://www.geeksforgeeks.org/python-program-for-binary-search/

[final jungle]

if arr[mid] == x:
            return mid

        # If element is smaller than mid, then it
        # can only be present in left subarray
        elif arr[mid] > x:
            return binarySearch(arr, low, mid-1, x)

        # Else the element can only be present
        # in right subarray
        else:
            return binarySearch(arr, mid + 1, high, x)

[final jungle]

Apparently i m supposed to return

In the if-else cases as well

[final jungle]

Yea that one

Nah nvm its still the same

[lone rover]

yes, recursion inflates the call stack and has a lot of overhead, iterating does not have that

[final jungle]

For anyone new on this thread the current code is:

n=int(input("Enter the size of ur list: "))
lis1=[]
for i in range(n):
    ele=int(input(f"Enter element {i+1}: "))
    lis1.append(ele)
print(f"Your list is: \n{lis1}")
key=int(input("Enter the element you wish to find: "))
lis1.sort()
def binary_search(list,upper,lower=0):
    result=None
    while lower<=upper:
        mid=(lower+upper)//2
        if lis1[mid]==key:
            result=mid
        elif lis1[mid]<key:
            return binary_search(list,upper,mid+1)
        elif lis1[mid]>key:
            return binary_search(list,mid-1)
    return result
output=binary_search(lis1,n-1)
if output==None:
    print("Your no is not present in the list!")
else:
    print(f"{key} was found at position {result}!")

[final jungle]

I used to think it was less lines of code and hence should be preffered

I never liked recrusion

Guys i made it work!

Code:

n=int(input("Enter the size of ur list: "))
lis1=[]
for i in range(n):
    ele=int(input(f"Enter element {i+1}: "))
    lis1.append(ele)
print(f"Your list is: \n{lis1}")
key=int(input("Enter the element you wish to find: "))
lis1.sort()
def binary_search(list,upper,lower=0):
    result=None
    if lower<=upper:
        mid=(lower+upper)//2
        if lis1[mid]==key:
            result=mid
            return result
        elif lis1[mid]<key:
            return binary_search(list,upper,mid+1)
        elif lis1[mid]>key:
            return binary_search(list,mid-1)
    else:
        return result
output=binary_search(lis1,n-1,0)
if output==None:
    print("Your no is not present in the list!")
else:
    print(f"{key} was found at position {output+1}!")

Thanks all for ur help

[lone rover]

recursion can make some things cleaner or more concise at the cost of performance, i tend to avoid it on principle though - you can implement your own recursion stack with a list and a while loop, even

[final jungle]

i had to use if and not while upper>lower

[final jungle]

while is used when doing iteratively

[final jungle]

Ohh so its basically easy on the coder but not on the systen

K thx

!close

[empty sequoia]

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