I have a problem with loops and declaring variables. currently I am making a program about Collatz Conjecture, the program should check what is the biggest steps to reach one from certain amount of Collatz Sequence. here's my code :
start_num = int(input("insert a starting Number > "))
how_many = int(input("how many times you want to check? >"))
def even_or_odd(number):
if number % 2 == 0:
return 'isEven'
else:
return 'notEven'
def collatz(n):
z = n
counter = 0
while True:
if n != 1:
if even_or_odd(n) == 'isEven':
n = n/2
counter += 1
continue
if even_or_odd(n) == 'notEven':
n = (n*3)+1
counter += 1
continue
else:
print('number ' + str(z) + ' reached 1 with : ' + str(counter) + ' steps')
return counter
break
def check_biggest_steps(steps_before, steps_after):
if steps_before > steps_after:
return steps_before
if steps_after > steps_before:
return steps_after
if steps_after == steps_before:
return steps_after
def compute_collatz(n_times, collatz_number):
for _ in range(n_times):
before = collatz(collatz_number)
collatz_number += 1
after = collatz(collatz_number)
collatz_number += 1
biggest_steps = check_biggest_steps(before, after)
print('Biggest Steps is :' + str(biggest_steps))
compute_collatz(how_many, start_num)
this biggest_steps variable always return the last 2 steps. I know what causing this problem is that biggest_step variable located inside the loop but I can't get it working anywhere don't know what to do. Thanks
Don't read my code until you have tried it yourself.
Try to make a list that appends every change to a list, then to get the number of moves at the end, just get the length of the list.
.
def collatz(x):
while x != 1:
if x % 2 > 0:
x =((3 * x) + 1)
list_.append(x)
else:
x = (x / 2)
list_.append(x)
return list_
print('Please enter a number: ', end='')
while True:
try:
x = int(input())
list_ = [x]
break
except ValueError:
print('Invaid selection, try again: ', end='')
l = collatz(x)
print('\nList:', l, sep=' ')
print('Number of steps required:', len(l) - 1)
you didn't save your biggest_steps and compared always the last 2 only.
I would suggest following change.
def compute_collatz(n_times, collatz_number):
biggest_steps = 0
for _ in range(n_times):
steps = collatz(collatz_number)
if steps > biggest_steps:
biggest_steps = steps
collatz_number += 1
print('Biggest Steps is :' + str(biggest_steps))
Related
When I run the following code:
def collatz(number):
if number % 2 == 0:
return number // 2
result = number
print(number)
elif number % 2 == 1:
return 3 * number + 1
result = number
print(number)
n = input('Enter a number: ')
while n != 1:
n = collatz(int(n))
the code runs to get the number to 1, but it always seems to skip the lines
result = number
print(number)
because it does not show any output. But when I visualize it, it is running the number to one. Could someone help explain why that is happening? Thanks so much.
You are using a recursive loop that bypasses those two statements.
def collatz(number):
if number % 2 == 0:
return number // 2 <---- exits function here
result = number
print(number)
elif number % 2 == 1:
return 3 * number + 1 <---- exits function here
result = number
print(number)
The same concept applies if you have the following function.
def add(x,y):
return x + y
print(x)
You would never get to the print(x) statement as you are exiting the function before that line of code!
You need to remove the return statement. Return exits the collatz function immediatly.
this is because it returns the value and it doesn't continue after the return, so you could do:
def collatz(number):
if number % 2 == 0:
result = number / 2
print(result)
return result
elif number % 2 == 1:
result = 3 * number + 1
print(result)
return result
n = input('Enter a number: ')
while n != 1:
n = collatz(int(n))
When a function sees return it ends immediately and outputs the value after the word. So to get the lines to run, they need to be before the word return:
def collatz(number):
if number % 2 == 0:
result = number
print(number)
return number // 2
elif number % 2 == 1:
result = number
print(number)
return 3 * number + 1
n = input('Enter a number: ')
while n != 1:
n = collatz(int(n))
I'm making an arithmetic progression with python.
After the 10 first terms, I want a question with how many more terms the user wants to see. If it's 0, it will end the program. If it's not the program return the number of terms asked. And repeat until it's 0.
After the first loop, the program works, after that the program returns an empty.
i = int(input('Start of PA: '))
r = int(input('PA Reason: '))
t1 = i
cont = 1
terms = 1
n1 = 0
while cont <= 10:
t1 = t1 + r
cont += 1
print(f'{t1} > ', end='')
while terms != 0 :
terms = int(input('\nHow many terms? '))
if terms!= 0:
while cont <= (10 + terms):
t1 = t1 + r
cont += 1
print(f'{t1} > ', end='')
else:
print('END!')
resolution
Edit: sorry for my english.
while cont <= (10 + terms):
is only the correct condition on the second batch of results. The next batch should be less than 20 + terms, and so on.
Instead of adding to terms, you should just set cont back to 0 before each loop that prints the next batch of terms.
And rather than checking whether terms is zero twice, use while True: and break out of the loop when they enter 0.
while True:
terms = int(input('\nHow many terms? '))
if terms!= 0:
cont = 0
while cont <= terms:
t1 = t1 + r
cont += 1
print(f'{t1} > ', end='')
else:
print('END!')
break
Or instead of using while loops, use for and range()
while True:
terms = int(input('\nHow many terms? '))
if terms != 0:
for _ in range(terms)
t1 = t1 + r
print(f'{t1} > ', end='')
else:
print('END!')
break
In this code there are four test cases. Three test cases are camed successfully but 4th test case was not getting expected output. Please help me.
I got 0x^3 as output in 4th test case but expected output is 0
Add two polynomials
Given two polynomials A and B, write a program that adds the given two polynomials A and B
Output
Print the addition of polynomials A and B.
If the degree of polynomial is zero and the constant term is also zero, then just print 0 to represent the polynomial.
For term Cix^Pi, if the coefficient of the term Ci is 1, simply print x^Pi instead of 1x^Pi.Explanation
We need all test cases can be came when code was run. I want exact outputs for all test cases
Here is the question and test case inputs and outputs in the below url link
https://drive.google.com/file/d/1DhE2akiG_pX_Q_EoKaEY9EapSgz3xWhY/view?usp=sharing
def check_polinom(polinom):
checked = []
while polinom:
tmp = polinom.pop(0)
if polinom:
for i in range(len(polinom)-1, -1, -1):
if polinom[i][0] == tmp[0]:
tmp[1] += polinom[i][1]
polinom.pop(i)
checked.append(tmp)
return checked
def add_polinoms(pol_1, pol_2):
added = []
while pol_1 or pol_2:
if pol_1:
tmp = pol_1.pop(0)
for i in range(len(pol_2)):
if pol_2[i][0] == tmp[0]:
tmp[1] += pol_2[i][1]
pol_2.pop(i)
break
else:
tmp = pol_2.pop(0)
added.append(tmp)
added.sort(reverse=True)
return(added)
def print_polinom(polinom):
s = ''
if polinom:
if polinom[0][1] < 0:
s += '-'
polinom[0][1] = -polinom[0][1]
if polinom[0][1] == 1:
if polinom[0][0] == 0:
s += str(polinom[0][1])
elif polinom[0][0] == 1:
s += 'x'
else:
s += 'x^' + str(polinom[0][0])
else:
if polinom[0][0] == 1:
s += str(polinom[0][1]) + 'x'
elif polinom[0][0] == 0:
s += str(polinom[0][1])
else:
s += str(polinom[0][1]) + 'x^' + str(polinom[0][0])
polinom.pop(0)
for el in polinom:
if el[1] == 0:
continue
elif el[1] < 0:
s += ' - '
el[1] = -el[1]
else:
s += ' + '
if el[1] == 1:
if el[0] == 0:
s += str(el[1])
elif el[0] == 1:
s += 'x'
else:
s += 'x^' + str(el[0])
else:
if el[0] == 1:
s += str(el[1]) + 'x'
elif el[0] == 0:
s += str(el[1])
else:
s += str(el[1]) + 'x^' + str(el[0])
print(s)
def input_data():
while True:
try:
n = int(input())
break
except:
print('enter an integer N')
continue
a = list()
i = 1
while n > i-1:
try:
tmp = list(map(int,(input()).split()))
if len(tmp) != 2:
print('enter two space separated integers')
continue
a.append(tmp)
i += 1
except:
print('enter two space separated integers')
return a
a = check_polinom(input_data())
b = check_polinom(input_data())
c = add_polinoms(a,b)
print_polinom(c)
It seems that you code does not clean up 0 coefficients. An additional phase should be added to your processing. Once you have the final coefficients of your polinomial, you should have two adjustments. You have already described one, that turns items with coefficient 1 from the form 1x to simply x. Another one is when you turn 0x to nothing. Then you only have to be careful to output something, so you should not turn every term into nothing. That is why there is a description to represent it as 0.
The easiset way to make this work would thus be:
Add a function to check if a polinomial has all zero coefficients: is_zero_polinom
Do something like this in print_polinom:
if is_zero_polinom(polinom):
print('0')
return
<here comes the current code>
so i have some code that works perfectly without a function. But i want to change this inside a function, but it does not work properly.
For example, i have the end="". This doesn't work in a function, without using print.
I also have more than one print statements. When i turn these to return, they don't work. So could someone please help me change this code to work in a function?
Thanks!
My code
def underscore_hash_staircase(number):
if number > 0:
k = 2 * number - 2
for i in range(0, number):
for j in range(number-1, k):
print(end=" ".replace(" ", "_"))
k = k - 1
for j in range(0, i + 1):
print("#", end="")
print("")
else:
number = int(str(number).replace("-", ""))
i = number
while i >= 1:
j = number
while j > i:
print('', end=' '.replace(" ", "_"))
j -= 1
k = 1
while k <= i:
print('#', end='')
k += 1
print()
i -= 1
print(underscore_hash_staircase(8))
so the code above doesn't work properly in a function, without the print statements. Please let me know how to get this working in a function without the print statements. Using returns. It should be exact output as what is being returned in this not function-working code.
Thanks again!
Since a function can only return one value, instead of printing, you want to add to a variable to return instead of printing. Try:
def underscore_hash_staircase(number):
returnValue = "" # start as empty string
if number > 0:
k = 2 * number - 2
for i in range(0, number):
for j in range(number-1, k):
returnValue += "_"
k = k - 1
for j in range(0, i + 1):
returnValue += "#"
returnValue += "\n" # adding a new line
else:
number = int(str(number).replace("-", ""))
i = number
while i >= 1:
j = number
while j > i:
returnValue += "_"
j -= 1
k = 1
while k <= i:
returnValue += "#"
k += 1
returnValue += "\n"
i -= 1
print(underscore_hash_staircase(8))
Edit: missed a print when replacing
The function should append to a string instead of printing, and then return the string. Append \n to add a newline.
def underscore_hash_staircase(number):
result = ""
if number > 0:
k = 2 * number - 2
for i in range(0, number):
for j in range(number-1, k):
result += "_"
k = k - 1
for j in range(0, i + 1):
result += "#"
result += "\n"
else:
number = -number
i = number
while i >= 1:
j = number
while j > i:
result += "_"
j -= 1
k = 1
while k <= i:
result += "#"
k += 1
result += "\n"
i -= 1
return result
print(underscore_hash_staircase(8))
You also don't need all those inner loops. You can repeat a string by multiplying it.
def underscore_hash_staircase(number):
result = ""
if number > 0:
k = 2 * number - 2
for i in range(1, number + 1):
result += "_" * (number - i)
result += "#" * i
result += "\n"
else:
number = -number
for i in range(number, 0, -1):
result += "_" * (number - i)
result += "#" * i
result += "\n"
return result
print(underscore_hash_staircase(8))
print(underscore_hash_staircase(-8))
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