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How to only show the final result after adding or multiplying all natural numbers

It has been a week since I started to self-study python and I tried making a program that adds or multiplies all natural numbers and the problem is I want to only show the final result of all the sum or product of all natural numbers. How do I do it?
repeat = 'y'
a=0
while repeat.lower() == 'y':
result = 0
choice = 0
i=0
product = 1
num = int(input("Enter the value of n: "))
if num < 1 or num > 100 :
print('must be from 1-100 only')
repeat = input("\nDo you want to try again?Y/N\n>>> ")
continue
print('1. Sum of all natural numbers')
print('2. Product of all numbers')
choice = int(input("Enter choice: "))
if choice == 1:
while(num > 0):
result += num
num -= 1
print(' ',result)
if choice ==2:
while i<num:
i=i+1
product=product*i
print(' ', product)
repeat = input("\nDo you want to try again Y/N? \n>>> ")
while repeat.lower() == 'n':
print('\nthank you')
break

The program prints you all the numbers because the print statement is in a while loop, so it gets executed with each run of the loop. Just move the print function out of the while.
if choice == 1:
while(num > 0):
result += num
num -= 1
print(' ',result)
if choice ==2:
while i<num:
i=i+1
product=product*i
print(' ', product)

You have two problems. First, your print statements that print the results need to be un-indented by one step, so they are not PART of loop, but execute AFTER the loop. Second, you need to initialize product = 1 after the if choice == 2:. As a side note, you don't need that final while loop. After you have exited the loop, just print('Thanks') and leave it at that.
So the end of the code is:
if choice == 1:
while num > 0 :
result += num
num -= 1
print(' ',result)
if choice == 2:
product = 1
while i<num:
i=i+1
product=product*i
print(' ', product)
repeat = input("\nDo you want to try again Y/N? \n>>> ")
print('thank you\n')
I presume you'll learn pretty quickly how to do those with a for loop instead of a while loop.

Lucky Number counting

I'm trying to make a lucky number counter where if there is a number containing either a 6 or an 8 between the two inputs the number is lucky, but if there's both a 6 and an 8 the number is unlucky. I'm having an issue where it's double-counting numbers, like 66, 88, etc., and not calling unlucky numbers like 68, 86, etc. Please help fast :\
l, h = [int(x) for x in input().split()]
count = 0
for i in range(l,h+1):
i = str(i)
for j in range(len(i)):
if i[j] == '6' or i[j] == '8':
count += 1
print(count)

Try this:
import sys
temp1 = ""
while len(temp1) != 2:
temp1 = input("Enter 2 digits: ")
try:
temp2 = int(temp1)
except ValueError:
print("You did not enter valid input")
sys.exit(1)
lucky = False
if "6" in temp1:
lucky = True
if "8" in temp1:
lucky = True
if ("6" in temp1) and ("8" in temp1):
lucky = False
print("Lucky number: "+str(lucky))

Something like this, probably:
number = ''
while len(number) != 2 or any(ch not in '0123456789' for ch in number):
number = input('Enter a 2-digit positive integer:')
print(f'{number} is {"lucky" if ("6" in number) != ("8" in number) else "not lucky"}')
# or, if you meant to say "exactly one 6 or 8":
print(f'{number} is {"lucky" if len([ch for ch in number if ch in "68"]) == 1 else "not lucky"}')

You could try using the count() method for Strings:
numbers = input().split(' ')
count = 0
for num in numbers:
if num.count('6') > 0:
if num.count('8') > 0:
continue
else:
count += 1
elif num.count('8') > 0:
count += 1
print(count)

this, while not as elegant or efficient, seems to be the correct code for the implied question
l, h = [int(x) for x in input().split()]
count = 0
for i in range(l,h+1):
i = str(i)
if ('6' in i and '8' in i):
continue
if i.count('6') > 0 or i.count('8') > 0:
count += 1
print(count)

Python: Cow And Bull Game

*Q.Randomly generate a 4-digit number. Ask the user to guess a 4-digit number. For every digit that the user guessed correctly in the correct place, they have a “cow”. For every digit the user guessed the number in the wrong place is a “bull.”
Every time the user makes a guess, tell them how many “cows” and “bulls” they have. Once the user guesses the correct number, the game is over. Keep track of the number of guesses the user makes throughout the game and tell the user at the end.*
**Now, the problem is that I've made the program but it could generate any 4 - digit number, and that's when the problem arises. For example:
The generated number is 3568.
The user types: 3266
Then user gets 2 Cows And 2 Bulls.
But the user has no way of knowing which are the correct numbers in the number that he typed.
I want a function that can tell the user the numbers that he guessed right.
In the example, the program should tell the user that 3 and 6 are correct in the following places.**
import random
def compare_number(number, user_guess):
cowbull = [0, 0]
for i in range(len(number)):
if number[i] == user_guess[I]:
cowbull[1] += 1
else:
cowbull[0] += 1
return cowbull
if __name__ == "__main__":
playing = True
number = str(random.randint(1000, 10000))
guesses = 0
print("Let's Play A Game Of Cows And Bulls!")
print("I Will Generate A 4 Digit Number, And You Have To Guess The Numbers One Digit At A Time.")
print("For Every Number I The Wrong Place, You Get A Bull. For Every Number In The Right Place,
You Get A Cow.")
print("The Game Will End When You Get 4 Bulls.")
print("Type Exit At Any Prompt To Exit!")
while playing:
user_guess = input("Give Me The Best You Got!: ")
if user_guess.lower() == "exit":
break
cowbull_count = compare_number(number, user_guess)
guesses += 1
print(f"You Have {cowbull_count[1]} Cows, And {cowbull_count[0]} Bulls.")
if cowbull_count[1] == 4:
playing = False
print(f"You Win The Game After {guesses} Guess(es)!. The Number Was {number}.")
break
else:
print(f"Your Guess Isn't Quite Right, Tyr Again!.")

You can do something like this:
import random
def compare_number(number, user_guess):
cowbull = [0, 0, 0, 0]
for i in range(len(number)):
if number[i] == user_guess[i]:
cowbull[i] += 1
return cowbull
if __name__ == "__main__":
playing = True
number = str(random.randint(1000, 10000))
guesses = 0
print("Let's Play A Game Of Cows And Bulls!")
print("I Will Generate A 4 Digit Number, And You Have To Guess The Numbers One Digit At A Time.")
print("For Every Number I The Wrong Place, You Get A Bull. For Every Number In The Right Place, You Get A Cow.")
print("The Game Will End When You Get 4 Bulls.")
print("Type Exit At Any Prompt To Exit!")
while playing:
user_guess = input("Give Me The Best You Got!: ")
if user_guess.lower() == "exit":
break
cowbull_count = compare_number(number, user_guess)
guesses += 1
correct = sum(cowbull_count)
wrong = len(number) - correct
print(f"You Have {correct} Cows, And {wrong} Bulls.")
if correct == 4:
playing = False
print(f"You Win The Game After {guesses} Guess(es)!. The Number Was {number}.")
break
else:
print(f"Your Guess Isn't Quite Right, Try Again!.")
if correct >= 1:
print(str([user_guess[i] for i, x in enumerate(cowbull_count) if x == 1]) + " was correct!")
Changes made to your original code:
Instead of returning [numOfCorrect,numOfWrong], i returned [is 1 correct?, is 2 correct?, is 3 correct? is 4 correct?] // you need this to know which was right and which was wrong
the number of cows is = the number of correct which is equal to sum of 1's in cowbull_count //changed because of different return of compare_number
the number of bulls is = the number of wrong which is equal to number of digits - number of wrongs = len(numbers) - correct //changed because of different return of compare_number
if not all 4 digits were correct, show them which number they got correct // this is what you wanted
Sample run

You can replace your compare number function to print the index and value of the correct number.
def compare_number(number, user_guess):
cowbull = [0, 0]
for i in range(len(number)):
if number[i] == user_guess[I]:
cowbull[1] += 1
print("The number " + number[i] + " at index " + i " is correct")
else:
cowbull[0] += 1
print("The number " + number[i] + " at index " + i " is incorrect")
return cowbull

Add another method that return a list of positions: 4 element list, 0 if the user didn't guess a digit, 1 if he did. You can use it as you want in your function.
def digit_position(number, user_guess):
right_guesses = [0, 0, 0, 0]
for i in range(len(number)):
if number[i] == user_guess[i]:
right_guesses[i] = 1
return right_guesses

# Cow and Bull Game is a game in which User
# tries to guess the Secret code chosen by computer.
# We have 2 use cases i.e
# If Value in index of both User's and Computer's number are same than it is Cow.
# If Value Exists but not on same index as computer's than ita a Bull.
import random
# Following function generate a unique 4-digit number
def checkDuplication():
r = str(random.randint(1000, 9999))
for i in r:
if r.count(i) > 1:
return checkDuplication()
return r
# Following function check both number and returns Cow and Bull COUNTS.
def cowBullGame(human):
cow_count = bull_count = 0
for i in human:
if i in computer:
if human.count(i) > 1:
print('No Repeatative Numbers Allowed!')
return 0
if human.index(i) == computer.index(i): # Checking if both the value in index i are same or not
cow_count += 1
else:
bull_count += 1
print(str(cow_count)+' Cows, '+str(bull_count)+' Bulls')
return cow_count # Returning Cow_Count to check All Numbers are on right place.
computer = checkDuplication()
print(computer)
guesses = 1
# Infinite Loop till user gets 4 Cow_Counts
while True:
human = str(int(input('Guess a Number :')))
if cowBullGame(human) == 4:
print('Game Over. You made '+str(guesses)+' guesses')
break
guesses += 1

Python multiple number guessing game

I am trying to create a number guessing game with multiple numbers. The computer generates 4 random numbers between 1 and 9 and then the user has 10 chances to guess the correct numbers. I need the feedback to display as YYYY for 4 correct numbers guessed, YNNY for first and last number guessed etc. (you get the point). the code below keeps coming back saying IndexError: list index out of range.
from random import randint
guessesTaken = 0
randomNumber = []
for x in range(4):
tempNumber = randint(1, 9)
randomNumber.append(tempNumber)
Guess = []
Guess.append(list(input("Guess Number: ")))
print(randomNumber)
print(Guess)
if randomNumber[0] == Guess[0]:
print("Y")
elif randomNumber[1] == Guess[1]:
print("Y")
elif randomNumber[2] == Guess[2]:
print("Y")
elif randomNumber[3] == Guess[3]:
print("Y")
elif randomNumber[0] != Guess[0]:
print("N")
elif randomNumber[1] != Guess[1]:
print("N")
elif randomNumber[2] != Guess[2]:
print("N")
elif randomNumber[3] != Guess[3]:
print("N")

You need four guesses to match for random numbers, you can also shorted your code using a list comp:
from random import randint
guessesTaken = 0
randomNumber = []
Guess = []
for x in range(4):
tempNumber = str(randint(1, 9)) # compare string to string
randomNumber.append(tempNumber)
Guess.append(input("Guess Number: "))
print("".join(["Y" if a==b else "N" for a,b in zip(Guess,randomNumber)]))
You can also use enumerate to check elements at matching indexes:
print("".join(["Y" if randomNumber[ind]==ele else "N" for ind, ele in enumerate(Guess)]))
To give the user guesses in a loop:
from random import randint
guessesTaken = 0
randomNumber = [str(randint(1, 9)) for _ in range(4)] # create list of random nums
while guessesTaken < 10:
guesses = list(raw_input("Guess Number: ")) # create list of four digits
check = "".join(["Y" if a==b else "N" for a,b in zip(guesses,randomNumber)])
if check == "YYYY": # if check has four Y's we have a correct guess
print("Congratulations, you are correct")
break
else:
guessesTaken += 1 # else increment guess count and ask again
print(check)

Right now you're only asking the user for one guess, and appending the guess to the Guess list. So the Guess list has one element, but you're using Guess[1], Guess[2], etc., which of course results in the IndexError

I'll rearrange your code a bit, so it doesn't stray too far from what you've done.
from random import randint
guessesTaken = 0
randomNumbers = []
Guess = [] # Combine your guesses with your loop
for x in range(4):
tempNumber = randint(1, 9)
randomNumbers.append(tempNumber)
# This should be done four times too
# In Python 2, instead of this:
# Guess.append(input("Guess Number: "))
# do this:
Guess.append(int(raw_input("Guess Number: "))) # raw_input and pass to int
# in python 3, raw_input becomes input, so do this instead:
# Guess.append(int(input("Guess Number: ")))
print(randomNumbers)
print(Guess)
You can combine these in a loop to avoid the repetitive code:
if randomNumbers[0] == Guess[0]:
print("Y")
else:
print("N")
if randomNumbers[1] == Guess[1]:
print("Y")
else:
print("N")
if randomNumbers[2] == Guess[2]:
print("Y")
else:
print("N")
if randomNumbers[3] == Guess[3]:
print("Y")
else:
print("N")
Perhaps, to print your desired result e.g. YNNY, like this:
result = []
for index in range(4):
if randomNumbers[index] == Guess[index]:
result.append("Y")
else:
result.append("N")
print(''.join(result))
If you want terser code use Python's ternary operation:
result = []
for index in range(4):
result.append("Y" if randomNumbers[index] == Guess[index] else "N")
print(''.join(result))
Or use the fact that True == 1 and False == 0 as indexes:
result = []
for index in range(4):
result.append("NY"[randomNumbers[index] == Guess[index]])
print(''.join(result))

The Next Palindrome number

I am beginner in programming, So can you please tell me what's wrong with my code?
I want to print next palindrome number if the number entered by the user (n) is not palindrome
n = int(input("Enter any number :- "))
reverse = 0
temp = n
while (n!=0):
reverse = reverse * 10
reverse = reverse + n%10
n=n//10
if(temp==reverse):
print ("Already palindrome:: ")
if(temp != reverse):
new_temp = temp
new_reverse = 0
for i in range(new_temp,new_temp+10):
while(temp != 0):
new_reverse = new_reverse * 10
new_reverse = new_reverse + temp%10
temp = temp//10
if(new_temp==new_reverse):
print ("Next pallindrome is :- ",new_temp)
break
if(new_temp != new_reverse):
temp = new_temp+1

There are two problems with your code.
1) Your "for i in range" loop calculates the reverse of the temp variable, but you don't change the temp variable's value.
You do
new_temp = temp
for i in range(new_temp,new_temp+10):
[SNIP]
if(new_temp != new_reverse):
temp = new_temp+1 #this value never changes.
So you're making 10 iterations with one and the same value.
2) Ten iterations might not be enough to find a palindrome. Keep going until you find a palindrome.
Working code:
def reverse(num):
reverse= 0
while num:
reverse= reverse*10 + num%10
num= num//10
return reverse
num= int(input("Enter any number :- "))
if num==reverse(num):
print ("Already palindrome.")
else:
while True:
num+= 1
if num==reverse(num):
print ("Next palindrome is : %s"%num)
break

To check if a number is a palindrome, you don't need to convert it to a number. In fact, its a lot simpler if you just check the string equivalent of your number.
>>> i = '212'
>>> i == i[::-1]
True
>>> i = '210'
>>> i == i[::-1]
False
Use this to your advantage, and create a function:
def is_palindrome(foo):
return str(foo) == str(foo)[::-1]
Next, to find the next palindrome, simply increment the number till your palindrome check is true.
Combine all that, and you have:
def is_palindrome(n):
return str(n) == str(n)[::-1]
n = raw_input('Enter a number: ')
if is_palindrome(n):
print('Congratulations! {0} is a palindrome.'.format(n))
else:
n1 = n
while not is_palindrome(n1):
n1 = int(n1)+1
print('You entered {0}, but the next palindrome is {1}'.format(n, n1))
Here is how it works:
$ python t.py
Enter a number: 123
You entered 123, but the next palindrome is 131
$ python t.py
Enter a number: 121
Congratulations! 121 is a palindrome.

If it helps, I believe it's possible to solve this problem with n/2 iterations where n is the length of the input number. Here's my solution in Python:
def next_palin_number(number):
number+=1
# Convert the number to a list of its digits.
number = list(str(number))
# Initialize two indices for comparing symmetric digits.
i = 0
j = len(number) - 1
while i < j:
# If the digits are different:
if number[i] != number[j]:
# If the lower-power digit is greater than the higher-power digit:
if int(number[j]) > int(number[i]):
if number[j-1]!='9':
number[j - 1] = str(int(number[j - 1]) + 1)
number[j] = number[i]
else:
number = list(str(int(''.join(number[:j]))+1))+number[j:]
else:
number[j] = number[i]
i += 1
j -= 1
# Concatenate and return the result.
return "".join(number)

This problem has a wonderful number of ways to solve them.
One of them is
def nearest_palindrome(number):
#start writitng your code here
while True:
number+=1
if str(number) == str(number)[::-1]:
return number
number=12300
print(nearest_palindrome(number))
Thanks for your time to read my answer : )

I have written this for finding next pallindrome number given a pallindrome number.
def palindrome(num):
bol=False
#x=len(str(num))
num=num+1
while(bol==False):
if(check_palindrome(num)):
bol=True
else:
num=num+1
return num
def check_palindrome(n):
temp=n
rev=0
while(n>0):
dig=n%10
rev=rev*10+dig
n=n//10
if(temp==rev):
return True
b=palindrome(8)
print(b)

def next_palin_drome(n):
while True:
n+=1
if str(n) == str(n)[::-1]:
return n
n=12231
print(next_palin_drome(n))
output:12321

def nearest_palindrome(number):
for i in range(1,number):
number=number+1
tem=str(number)
tem1=tem[-1::-1]
if(tem==tem1):
return number
else:
continue
number=12997979797979797
print(nearest_palindrome(number))

def nearest_palindrome(number):
n = len(str(number))//2
if(len(str(number)) % 2 == 0):
#number like 1221
number_1 = int((str(number))[:n]) #12
number_2 = int((str(number))[n:]) #21
if(number_1 < number_2):
number_1 += 1
number_2 = int(str(number_1)[::-1])
else:
number_2 = int(str(number_1)[::-1])
# if last half part is zero then just reverse the first number
if number_2 == 0:
number_2 = str(number_1)[::-1]
#combining the both parts
ans = int(str(number_1) + str(number_2))
return ans
else:
#numer like 12510 n=2
nu = int((str(number))[:n+1]) #add in this number
number_1 = int((str(number))[:n]) # 12
number_2 = int((str(number))[n+1:]) # 21
if (number_1 < number_2):
nu += 1
number_2 = int((str(nu))[::-1][1:])
else:
number_2 = int((str(nu))[::-1][1:])
#if last half part is zero then just reverse the first number
if number_2 == 0:
number_2 = str(nu)[::-1]
number_2 = number_2[1:]
#combinning both parts
ans = int(str(nu) + str(number_2))
return ans
number=12331
print(nearest_palindrome(number))

If a definite range is given:
# function to check if the number is a palindrome
def palin(x):
s=str(x)
if s==s[::-1]:
return True
else:
return False
n=int(input("Enter the number"))
# Putting up range from the next number till 15 digits
for i in range(n+1,int(10e14)):
if palin(i) is True:
print(i)
break

A brute force method:
def math(n):
while not var:
n += 1
if str(n) == str(n)[::-1] : f = 'but next is : '+str(n); return f
n = int(input()); t = math(n); print('Yes',t) if str(n) == str(n)[::-1] else print('No',t); global var; var = False

This is a good fast solution. I saw that the other solutions were iterating and checking through every +1 they did, but this is really slow for big numbers.
This solution has O(n) time if you look at the length of the number
beginNumber = 123456789101112131415161718 #insert number here for next palidrome
string = str(beginNumber + 1)
length = len(string)
number= [int(x) for x in list(string)]
for i in range(length//2):
if (number[i] != number[length-1-i]):
if (number[i]<number[length-1-i]):
number[length-2-i] += 1
number[length-1-i] = number[i]
print("".join([str(x) for x in number]))

I have written this for finding next pallindrome number given a pallindrome number..
#given a pallindrome number ..find next pallindrome number
input=999
inputstr=str(input)
inputstr=inputstr
#append 0 in beginning and end of string ..in case like 99 or 9999
inputstr='0'+inputstr+'0'
length=len(inputstr)
halflength=length/2;
#if even length
if(length%2==0):
#take left part and reverse it(which is equal as the right part )
temp=inputstr[:length/2]
temp=temp[::-1]
#take right part of the string ,move towards lsb from msb..If msb is 9 turn it to zero and move ahead
for j,i in enumerate(temp):
#if number is not 9 then increment it and end loop
if(i!="9"):
substi=int(i)+1
temp=temp[:j]+str(substi)+temp[j+1:]
break;
else:
temp=temp[:j]+"0"+temp[j+1:]
#now you have right hand side...mirror it and append left and right part
output=temp[::-1]+temp
#if the length is odd
if(length%2!=0 ):
#take the left part with the mid number(if length is 5 take 3 digits
temp=inputstr[:halflength+1]
#reverse it
temp=temp[::-1]
#apply same algoritm as in above
#if 9 then make it 0 and move on
#else increment number and break the loop
for j,i in enumerate(temp):
if(i!="9"):
substi=int(i)+1
temp=temp[:j]+str(substi)+temp[j+1:]
break;
else:
temp=temp[:j]+"0"+temp[j+1:]
#now the msb is the middle element so skip it and copy the rest
temp2=temp[1:]
#this is the right part mirror it to get left part then left+middle+right isoutput
temp2=temp2[::-1]
output=temp2+temp
print(output)
similarly for this problem take the left part of given number ...reverse it..store it in temp
inputstr=str(number)
if(inputstr==inputstr[::-1])
print("Pallindrome")
else:
temp=inputstr[:length/2]
temp=temp[::-1]
for j,i in enumerate(temp):
if(i!="9"):
substi=int(i)+1
temp=temp[:j]+str(substi)+temp[j+1:]
break;
else:
temp=temp[:j]+"0"+temp[j+1:]
now depending on length of your number odd or even generate the output..as in the code
if even then output=temp[::-1]+temp
if odd then temp2=temp1[1:]
output=temp2[::-1]+temp
I am not sure about this solution..but hope it helps