I have this code that counts up -3, -2, -1. I have also written another that counts down 3, 2, 1. If the condition is true for another after an input, the program should stop. For Example, The input i enter -3 for negatives it gives -3, -2, -1, if i enter positive number like 3, the result should be 3, 2, 1 in a single combined code.
I have tried to put a function into an if statement of the first function but fails with "NameError: name 'countdown' is not defined" for count down but count up works.
print('+=+=+=+=+=+=+=+=+=+=+=+=+ F1')
#count up
def countup(n):
if (n >= 0):
print('Hello')
if n==0:
print('Blastoff!')
else:
print(n)
countdown(n-1)
else:
print(n)
nth = n + 1
countup(nth)
result = int(input('Enter Number: '))
countup(result)
Have you tried replacing the second else with elif? And fixing the indenting? The resulting code should be this:
print('+=+=+=+=+=+=+=+=+=+=+=+=+ F1')
#count up
def countup(n):
if (n >= 0):
print('Hello')
if n==0:
print('Blastoff!')
elif (condition):
print(n)
countdown(n-1)
else:
print(n)
nth = n + 1
countup(nth)
result = int(input('Enter Number: '))
countup(result)
I solve the problem by doing this code below:
def countup(n):
if (n >= 0):
print('Blastoff 1')
else:
print(n)
nth = n + 1
return countup(nth);
def zero(n):
if (n==0):
print('You have erntered 0 program exiting')
exit()
def countdown(n):
if (n <= 0):
print('explode 2')
else:
print(n)
nth = n - 1
return (countdown(nth))
num = int(input("Enter number: "));
if (num <= 0):
print(countup(num))
elif num == 0:
print(zero(num))
else:
print(countdown(num))
The output is either negative or positive, regardless of what integer you fuse in
Related
I tried to find the prime factor for a number by the following code:
def pf(n):
f=[]
while n != 1:
for i in range(2, n+1):
if n%i == 0:
f.append(i)
n //=i
return f
The output for pf(8) is [2, 4], rather than [2,2,2] as my expect.
The output for pf(16) is [2,4,2], rather than [2,2,2,2].
can anyone help me to figure out what's wrong with my code?
If you break inside the loop, you will get what you want :
def pf(n):
f=[]
while n != 1:
for i in range(2, n+1):
if n%i == 0:
f.append(i)
n //=i
break
return f
Why was it wrong ?
in you first for loop, 8 is can be divied by 2, so n is equal to 4. And when i is then equals to 4, then 4 is added as a prime factor
Other implementation:
I prefer this implementation :)
def pf(n):
f=[]
i = 2
while n != 1:
if n%i == 0:
f.append(i)
n //=i
else:
i += 1
return f
This implementation avoids you to recompute the numbers in range you already know is not dividable by.
# 8-bit-binary adder
#if you have questions about the code just ask
# arrays and funtions
Array1 = []
Array2 = []
#Input A and Validation
def vaildNumberA():
a = int(input("Enter your A value:"))
if (a < 0):
print("Please Enter A Valid Number For Input A")
elif (a > 255):
print("Please Enter A Valid Number For Input A")
else:
Array1 = [int(x) for x in list('{0:08b}'.format(a))]
#Input B and Validation
def vaildNumberB():
b = int(input("Enter your B value:"))
if (b < 0):
print("Please Enter A Valid Number For Input B")
elif (b > 255):
print("Please Enter A Valid Number For Input B")
else:
Array2 = [int(x) for x in list('{0:08b}'.format(b))]
# and gate
# AND Gate
def AND (a,b):
if (a == 1 and b == 1):
return 1
else:
return 0
# or gate
#OR Gate
def OR(a,b):
if (a == 1 or b == 1):
return 1
else:
return 0
# XOR GATEE
#XOR Gate
def XOR (a,b):
if (a == b):
return 0
else:
return 1
#carry formula
def carryformula(a,b,c,d):
return OR(AND(a,b), AND(c,d))
# this is where the calculation should be done
#formula for sum
def calculateSum(Array1,Array2):
carry = ""
sumof = []
for index, in range(len(Array1)):
list2 = Array2[index]
sec_xor_form = XOR(Array1[index],Array2[index])
sumof.append(XOR(sec_xor_form,carry))
carry = carryformula(Array1[index],Array2[index],sec_xor_form,carry)
return list(reversed(sumof))
calculateSum(Array1,Array2)
def main(Array1,Array2):
vaildNumberA()
vaildNumberB()
while True:
a = Array1
b = Array2
total = calculateSum(list(reversed(Array1)),list(reversed(Array2)))
print(total)
quit = input("if want to quit type q: ")
if quit == 'q':
break
main(Array1,Array2)
in the send it prints 0
The only problem in your code is that you need to return Array1 and Array2 from your functions and assign them inside the while true loop, once you do that the code works fine.
The updated code will be
# 8-bit-binary adder
#if you have questions about the code just ask
# arrays and funtions
Array1 = []
Array2 = []
#Input A and Validation
def vaildNumberA():
Array1 = []
a = int(input("Enter your A value:"))
if (a < 0):
print("Please Enter A Valid Number For Input A")
elif (a > 255):
print("Please Enter A Valid Number For Input A")
else:
Array1 = [int(x) for x in list('{0:08b}'.format(a))]
#Return the array
return Array1
#Input B and Validation
def vaildNumberB():
Array2 = []
b = int(input("Enter your B value:"))
if (b < 0):
print("Please Enter A Valid Number For Input B")
elif (b > 255):
print("Please Enter A Valid Number For Input B")
else:
Array2 = [int(x) for x in list('{0:08b}'.format(b))]
#Return the array
return Array2
# AND Gate
def AND (a,b):
if (a == 1 and b == 1):
return 1
else:
return 0
#OR Gate
def OR(a,b):
if (a == 1 or b == 1):
return 1
else:
return 0
#XOR Gate
def XOR (a,b):
if (a == b):
return 0
else:
return 1
#carry formula
def carryformula(a,b,c,d):
return OR(AND(a,b), AND(c,d))
# this is where the calculation should be done
#formula for sum
def calculateSum(Array1,Array2):
carry = ""
sumof = []
for index in range(len(Array1)):
list2 = Array2[index]
sec_xor_form = XOR(Array1[index],Array2[index])
sumof.append(XOR(sec_xor_form,carry))
carry = carryformula(Array1[index],Array2[index],sec_xor_form,carry)
return list(reversed(sumof))
#No need of a main function
while True:
#Call the function from within the while True loop
Array1 = vaildNumberA()
Array2 = vaildNumberB()
total = calculateSum(list(reversed(Array1)),list(reversed(Array2)))
print(total)
quit = input("if want to quit type q: ")
if quit == 'q':
break
And the output will look like
Enter your A value:5
Enter your B value:5
[0, 0, 0, 0, 1, 0, 1, 1]
if want to quit type q: 7
Enter your A value:9
Enter your B value:8
[0, 0, 0, 1, 0, 0, 0, 1]
....
I am trying to write a function that returns the number of prime numbers that exist up to and including a given number.
Initially this was my code:
def count_primes(num):
prime = [2]
x = 3
if num < 2:
return 0
while x <= num:
for y in prime:
if x%y == 0:
print('not prime')
x+=2
break
else:
prime.append(x)
x += 2
return len(prime)
How ever I realise this code will run forever because of the following line of code:
for y in prime:
if x%y == 0:
print('not prime')
x+=2
break
else:
prime.append(x)
x += 2
Can anyone help to explain to me why will this end up with an infinite loop compared to the following code?
for y in prime:
if x%y == 0:
print('not prime')
x+=2
break
else:
prime.append(x)
x += 2
I'm trying to write my own formula to find a prime number, but it does not completely work and I cannot find the flaw in my logic. Bare in mind I have taken a look around but cannot find an algorithm that I find similar to mine.
My code:
#Challenge 7
prime = []
num = 0
found = False
while found == False:
if num == 0 or num == 1:
num+=1
else:
for value in range(2, num+1):
if len(prime) == 50:
print('Found all')
found = True
break
if num % value == 0:
num+=1
else:
if num not in prime:
prime.append(num)
else:
pass
print(prime)
This code works for first few primes (3, 5, 7...)
but it also gives incorrect values like 10, and I don't understand why. If someone could explain it to me so that I can understand where the logical mistake is, I'd appreciate it.
The error comes from this part
if num % value == 0:
num+=1
else:
if num not in prime:
prime.append(num)
else:
pass
You assume that the integer is a prime as soon as we find the first occurence of a non-divisor. But the def for primes is that every integer in the interval [2..prime] is a non-divisor. How do we check if any number does not have any divisors?
def isPrime(x):
for v in range(2, x):
if (x % v == 0):
return False;
return True;
Something like this would work to check if any given number is a prime or not. And since we now have taken the isPrime part out of the main loop, we no longer need a for loop inside the while. Something like this would do
def isPrime(x):
for v in range(2, x):
if (x % v == 0):
return False;
return True;
prime = [}
num = 2
found = False
while found == False:
if len(prime) == 50:
print("found all")
found = True
break
if(isPrime(num)):
print(num)
prime.append(num)
num+=1
else:
num+=1
If you set a breakpoint for when num == 10 you will see the problem clearly.
When you start doing you division check inside of for value in range(2, num + 1): the second number is 3, so num (10) modulo value (3) is 1, which is your test for determining a prime. What your test should be is that it not divisible by any number less than it (less than half is actually sufficient since you check with 2 anyway).
So, consider instead:
else:
is_indivisible = True
# loop through all numbers less than it not including itself
# (because x % x == 0)
for value in range(2, num - 1):
# it is only indivisible if it was previously indivisible
# And the check is same as before, modulo != 0
is_indivisible = is_indivisible and (num % value != 0)
if not is_indivisible:
break
# if it is indivisible and it doesn't exist in prime list yet
if is_indivisible and num not in prime:
prime.append(num)
# move on to the next number
num += 1
Have tried searching for this, but can't find exactly what I'm looking for.
I want to make a function that will recursively find the factors of a number; for example, the factors of 12 are 1, 2, 3, 4, 6 & 12.
I can write this fairly simply using a for loop with an if statement:
#a function to find the factors of a given number
def print_factors(x):
print ("The factors of %s are:" % number)
for i in range(1, x + 1):
if number % i == 0: #if the number divided by i is zero, then i is a factor of that number
print (i)
number = int(input("Enter a number: "))
print (print_factors(number))
However, when I try to change it to a recursive function, I am getting just a loop of the "The factors of x are:" statement. This is what I currently have:
#uses recursive function to print all the letters of an integer
def print_factors(x): #function to print factors of the number with the argument n
print ("The factors of %s are:" % number)
while print_factors(x) != 0: #to break the recursion loop
for i in range(1,x + 1):
if x % i == 0:
print (i)
number = int(input("Enter a number: "))
print_factors(number)
The error must be coming in either when I am calling the function again, or to do with the while loop (as far as I understand, you need a while loop in a recursive function, in order to break it?)
There are quite many problems with your recursive approach. In fact its not recursive at all.
1) Your function doesn't return anything but your while loop has a comparision while print_factors(x) != 0:
2) Even if your function was returning a value, it would never get to the point of evaluating it and comparing due to the way you have coded.
You are constantly calling your function with the same parameter over and over which is why you are getting a loop of print statements.
In a recursive approach, you define a problem in terms of a simpler version of itself.
And you need a base case to break out of recursive function, not a while loop.
Here is a very naive recursive approach.
def factors(x,i):
if i==0:
return
if x%i == 0:
print(i)
return factors (x,i-1) #simpler version of the problem
factors(12,12)
I think we do using below method:
def findfactor(n):
factorizeDict
def factorize(acc, x):
if(n%x == 0 and n/x >= x):
if(n/x > x):
acc += [x, n//x]
return factorize(acc, x+1)
else:
acc += [x]
return acc
elif(n%x != 0):
return factorize(acc, x+1)
else:
return acc
return factorize(list(), 1)
def factors(x,i=None) :
if i is None :
print('the factors of %s are : ' %x)
print(x,end=' ')
i = int(x/2)
if i == 0 :
return
if x % i == 0 :
print(i,end=' ')
return factors(x,i-1)
num1 = int(input('enter number : '))
print(factors(num1))
Recursion is a functional heritage and so using it with functional style yields the best results. This means avoiding things like mutations, variable reassignments, and other side effects. That said, here's how I'd write factors -
def factors(n, m = 2):
if m >= n:
return
if n % m == 0:
yield m
yield from factors(n, m + 1)
print(list(factors(10))) # [2,5]
print(list(factors(24))) # [2,3,4,6,8,12]
print(list(factors(99))) # [3,9,11,33]
And here's prime_factors -
def prime_factors(n, m = 2):
if m > n:
return
elif n % m == 0:
yield m
yield from prime_factors(n // m, m)
else:
yield from prime_factors(n, m + 1)
print(list(prime_factors(10))) # [2,5]
print(list(prime_factors(24))) # [2,2,2,3]
print(list(prime_factors(99))) # [3,3,11]
def fact (n , a = 2):
if n <= a :
return n
elif n % a != 0 :
return fact(n , a + 1 )
elif n % a == 0:
return str(a) + f" * {str(fact(n / a , a ))}"
Here is another way. The 'x' is the number you want to find the factors of. The 'c = 1' is used as a counter, using it we'll divide your number by 1, then by 2, all the way up to and including your nubmer, and if the modular returns a 0, then we know that number is a factor, so we print it out.
def factors (x,c=1):
if c == x: return x
else:
if x%c == 0: print(c)
return factors(x,c+1)