I'm a python beginner and I'm trying to figure out the following below two python code samples. Both code looks same but printing a different result. The main function of the code is divisor summation.
CODE 1:
def divisor_sum(n):
no_div = 0
tot = int(n / 2) + 1
for i in range(1,tot):
if n % i == 0:
no_div += 1
print no_div
CODE 2:
def divisor2(m):
max_div = int(m / 2) + 1
val = 0
for x in range(1, max_div):
if m % x == 0:
val += x
print val
when calling the function?
divisor_sum(6)
divisor2(6)
Code ouput:
3
6
How it is producing two different results when both code samples are same?
Not exactly the same code:
CODE 1:
no_div += 1
CODE 2:
val += x
The first variant increments by 1 when a divisor is encountered:
no_div += 1
The second variant increments by the divisor:
val += x
This means that the first variant counts the number of divisors. The second variant sums the values of the divisors.
The divisors of 6 are 1, 2 and 3. So, there are a total of 3 divisors whose values sum to 6.
Code in first function i.e divisor_sum() will be giving you the count of divisors of a number (then name of function is misleading). On the other hand code in the second function i.e divisor2() will give you sum of all the divisors of a number.
both function has different code :
divisor_sum() has : "no_div += 1 " that means increment no_div by 1
divisor2() has :"val += x" which means increment the value of val by x i.e if x=2 then val will become val+2.
Related
I'm new to programming. While trying to solve this problem, I'm getting the wrong answer. I checked my code a number of times but was not able to figure out the mistake. Please, help me on this simple problem. The problem is as follows:
Given a positive integer N, calculate the sum of all prime numbers between 1 and N (inclusive). The first line of input contains an integer T denoting the number of test cases. T testcases follow. Each testcase contains one line of input containing N. For each testcase, in a new line, print the sum of all prime numbers between 1 and N.
And my code is:
from math import sqrt
sum = 0
test = int(input())
for i in range(test):
max = int(input())
if max==1:
sum = 0
elif max==2:
sum += 2
else:
sum = sum + 2
for x in range(3,max+1):
half = int(sqrt(max)) + 1
for y in range(2,half):
res = x%y
if res==0:
sum = sum + x
break
print(sum)
For input 5 and 10, my code is giving output 6 and 48 respectively, while the correct answer is 10 and 17 respectively. Please, figure out the mistake in my code.
Here, I implemented simple program to find the sum of all prime numbers between 1 to n.
Consider primeAddition() as a function and ip as an input parameter. It may help you to solve your problem.Try it.
Code snippet:
def primeAddition(ip):
# list to store prime numbers...
prime = [True] * (ip + 1)
p = 2
while p * p <= ip:
# If prime[p] is not changed, then it is a prime...
if prime[p] == True:
# Update all multiples of p...
i = p * 2
while i <= ip:
prime[i] = False
i += p
p += 1
# Return sum of prime numbers...
sum = 0
for i in range (2, ip + 1):
if(prime[i]):
sum += i
return sum
#The program is ready... Now, time to call the primeAddition() function with any argument... Here I pass 5 as an argument...
#Function call...
print primeAddition(5)
This is the most broken part of your code, it's doing the opposite of what you want:
res = x%y
if res==0:
sum = sum + x
break
You only increment sum if you get through the entire loop without breaking. (And don't use sum as you're redefining a Python built-in.) This can be checked using the special case of else on a for loop, aka "no break". I've made that change below as well as corrected some inefficiencies:
from math import sqrt
T = int(input())
for _ in range(T):
N = int(input())
sum_of_primes = 0
if N < 2:
pass
elif N == 2:
sum_of_primes = 2
else:
sum_of_primes = 2
for number in range(3, N + 1, 2):
for odd in range(3, int(sqrt(number)) + 1, 2):
if (number % odd) == 0:
break
else: # no break
sum_of_primes += number
print(sum_of_primes)
OUTPUT
> python3 test.py
3
5
10
10
17
23
100
>
A slight modification to what you have:
from math import sqrt
sum = 0
test = int(input())
max = int(input())
for x in range(test,max+1):
if x == 1:
pass
else:
half = int(sqrt(x)) + 1
for y in range(2,half):
res = x%y
if res==0:
break
else:
sum = sum + x
print(sum)
Your biggest error was that you were doing the sum = sum + x before the break rather than outside in an else statement.
PS: (although you can) I'd recommend not using variable names like max and sum in your code. These are special functions that are now overridden.
Because your logic is not correct.
for y in range(2,half):
res = x%y
if res==0:
sum = sum + x
break
here you check for the factors and if there is a factor then adds to sum which is opposite of the Primes. So check for the numbers where there is no factors(except 1).
from math import sqrt
test = int(input())
for i in range(test):
sum = 0
max = int(input())
if max==1:
sum = 0
elif max==2:
sum += 2
else:
sum = sum + 2
for x in range(3,max+1):
half = int(sqrt(x)) + 1
if all(x%y!=0 for y in range(2,half)):
sum = sum + x
print(sum)
First of all, declare sum to be zero at the beginning of the for i loop.
The problem lies in the if statement at almost the very end of the code, as you add x to the sum, if the res is equal to zero, meaning that the number is indeed not a prime number. You can see that this is the case, because you get an output of 6 when entering 5, as the only non-prime number in the range 1 to and including 5 is 4 and you add 2 to the sum at the beginning already.
Last but not least, you should change the
half = int(sqrt(max)) + 1
line to
half = int(sqrt(x)) + 1
Try to work with my information provided and fix the code yourself. You learn the most by not copying other people's code.
Happy coding!
I believe the mistake in your code might be coming from the following lines of code:
for x in range(3,max+1):
half = int(sqrt(max)) + 1
Since you are looping using x, you should change int(sqrt(max)) to int(sqrt(x)) like this:
for x in range(3,max+1):
half = int(sqrt(x)) + 1
Your code is trying to see if max is prime N times, where you should be seeing if every number from 1-N is prime instead.
This is my first time answering a question so if you need more help just let me know.
I'm trying to get this program to return all possible multiples of 3 and 5 below 1001 and then add them all together and print it but for some reason these lines of code only seem to be printing one number and that number is the number 2 which is obviously wrong. Can someone point me in the right direction to why this code is grossly wrong?
n = 0
x = n<1001
while (n < 1001):
s = x%3 + x%5
print s
You've got a few mistakes:
x is a boolean type
Your loop never ends
adding values to mimic lists?
Edit
Didn't see the part where you wanted sum, so you could employ a for-in loop or just a simple one like so:
sum = 0
for i in range(1001):
if(i % 3 == 0 or i % 5):
sum += i
print(sum)
(Python 3)
You need to stop while at some point by incrementing n. Here is some code:
nums = []
n = 0
while (n < 1001):
# in here you check for the multiples
# then append using nums.append()
n += 1
This creates a loop and a list that accounts for every number in 0 to 1000. In the body, check for either condition and append to the list, then print out the values in the list.
num is a list where you are going to store all the values that apply, (those numbers who are divisible by 3 and 5, which you calculate with modulo). You append to that list when a condition is met. Here is some code:
nums = []
n = 0
while (n < 1001):
if(n % 3 == 0 or n % 5 ==0):
nums.append(n)
n += 1
print(n) #or for loop for each value
Explanation: a list of numbers called num stores the numbers that are divisible by 3 or 5. The loop starts at zero and goes to 1000, and for all the values that are divisible by 3 or 5, they will be added to the list. Then it prints the list.
Of course there is a simpler approach with a range:
for i in range(1001):
if(i % 3 == 0 or i % 5 == 0):
print(i)
This will print out all the values one by one. It is 1001 because the upper limit is exclusive.
true=1
false=0
so:
x = n<1001
we have x=1 because 0<1001 is true
s = x%3 + x%5
the remainder of 1/3 is 1 and 1/5 is 1
In your code:
1. x=n<1001 - generates a boolean value; on which we can't perform a mathematical operation.
In while loop:
your variable n,x are not changing; they are constant to same value for all the iterations.
Solution 1:
Below code will help you out.
s=0
for i in range(1,1002):
if( i%3 == 0 or i%5 == 0):
s = s + i
print(s)
Solution: 2
There is one more approach you can use.
var = [i for i in range(1,1002) if i%3==0 or i%5 ==0]
print(sum(var))
So the question comes like this, I'm new to python:
def factorial_cap(num): For positive integer n, the factorial of n (denoted as n!), is the product
of all positive integers from 1 to n inclusive. Implement the function that returns the smallest
positive n such that n! is greater than or equal to argument num.
o Assumption: num will always be a positive integer.
# Examples
# factorial_cap(20) output is 4 since 3!<20 but 4!>20
# factorial_cap(24) output is 4 since 4!=24
# factorial_cap(1) output is 1 since 1!=1
# And here is what I got
def factorial_cap(num):
n = 1
for i in range (1,num+1):
n = n*i
I'm pretty sure this is the right function for factorial def. But I just couldn't figure out, instead of getting the 'total value', how can I just get the right output as I posted example above?
Btw, should I use 'return' at the end of def, or it does not matter in this case?
There needs to be a test for when the current total is greater than or equal to the requested number. So you can use the condition of a while loop to perform that check, and increment a counter, i, that keeps track of the current iteration. Then it's a matter of returning the current value of i that produced the value >= the required number:
def factorial_cap(num):
n = 1
i = 1
while n < num:
i += 1
n *= i
return i
>>> factorial_cap(20)
4
>>> factorial_cap(24)
4
>>> factorial_cap(25)
5
>>> factorial_cap(1)
1
>>> factorial_cap(3628800)
10
You want a return but that isn't n, but i
def factorial_cap(num):
n = 1
i = 0
while True:
i += 1
n = n*i
if n >= num:
break
return i
print(factorial_cap(20))
print(factorial_cap(24))
print(factorial_cap(1))
I'm working on this one and I seem to have a working solution but I have difficulty understanding its behaviour.
Here is what I have.
#!/usr/bin/python
def even_fib_sums(limit):
number = 1
last = 0
before_last = 0
total = 0
for counter in range (0,limit):
before_last = last
last = number
number = before_last + last
if not number % 2:
total += number
yield total
print sum(even_fib_sums(4000000))
I'm new to programming but it makes sense to me that this is not very effective considering I need to cycle through all 4000000 numbers in the range.
If I use the same approach in generating the Fibonacci sequence up to 5 as follows, you will see the results below.
def generate_fib(limit):
number = 1
last = 0
before_last = 0
total = 0
for counter in range (0,limit):
before_last = last
last = number
number = before_last + last
print number
generate_fib(5)
Result: 1,2,3,5,8
Of these numbers in the result, only 2 and 8 % 2 == 0.
The sum should be 10 but I am returning 12 if I am to use the first snippet above. Why so?
By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
You only need to loop until you hit a fib that is > 400000 not the 4 millionth fibonacci number which your code is trying to do, you can simplify to a using generator function with sum, only yielding even numbers and breaking the loop when you hit a fibonacci number > 4000000:
def fib(n):
a, b = 0, 1
while a <= n:
a, b = b, a + b
if not b & 1:
yield b
print(sum(fib(4000000)))
It takes a fraction of a second to compute:
In [5]: timeit sum(fib(4000000))
100000 loops, best of 3: 6 µs per loop
trying to timeit even_fib_sums(4000000) is still running after a few minutes.
by using for counter in range(0, limit) you are having 'limit' iteration in your function. for example, if your 'limit' variable is 10, you won't have the sum of even fibonachi numbers that are less than 10, but you will have the sum of the first 10 fibonachi numbers that are even.
To make your code works properly, you need to remplace for counter in range(0, limit) by while last < limit, and each time you find that last is even, you add it to total.
You can probably clean up that generating function a bit. Here is how I would write it.
def fib(x):
a = 1
b = 1
yield a
yield b
a,b = b,a+b
while b<=x:
yield b
a,b = b,a+b
This will give you a generating function which will give you all Fibonacci numbers less than or equal to x (we should be a little more careful here, as we will return the first two numbers no matter what).
Then we can just do
sum(x for x in fib(4000000) if x%2==0)
You should change your code to just yield the number, not the sum or just change yield to return, and remove the sum() keyworkd like this:
def even_fib_sums(limit):
number = 1
last = 0
before_last = 0
total = 0
for counter in range (0,limit):
before_last = last
last = number
number = before_last + last
if not number % 2:
total += number
return total
print even_fib_sums(5)
In the first code snippet, you sum the total of round numbers, instead of just yielding the number. If you expect to get 10 in your first snippet for an input of 5, you should amend the code in either of the following ways (not trying to be efficient here, just to fix the problem):
...
number = before_last + last
if not number % 2:
yield number
print sum(even_fib_sums(4000000))
or
...
number = before_last + last
if not number % 2:
total += number
return total
print even_fib_sums(4000000)
Beginner here- trying to make a simple python program that will compute/answer this problem:
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.
Currently this is what I have:
a = 0
b = 0
while a < 1000:
a = a + 3
print (a)
while b < 1000
b = b + 5
print (b)
This will print all the numbers being considered. I just need to add them together and that's my answer.
I would like one of two things to happen, instead of the code that I have written:
I would like all of this to happen internally, and therefore not have to use the "print" function. The just print the sum of all of those multiples.
I would like all of this stuff to print, but then I want to be able to print the sum of all of them too. Is there a way to make the computer take the value of everything it has printed?
Actually this problem can be solved in O(1) instead of O(N) without using any loops or lists:
The required sum is the sum of all multiples of 3 plus sum of all multiples of 5 minus the sum of multiples of (3*5=15) below the given number 1000 (LIMIT=999). The sums are calculated as a sum of arithmetic series.
It can be calculated in following way:
LIMIT=999
# Get the upper bounds for the arithmetic series
upper_for_three = LIMIT // 3
upper_for_five = LIMIT // 5
upper_for_fifteen = LIMIT // 15
# calculate sums
sum_three = 3*upper_for_three*(1 + upper_for_three) / 2
sum_five = 5*upper_for_five*(1 + upper_for_five) / 2
sum_fifteen = 15*upper_for_fifteen*(1 + upper_for_fifteen) / 2
# calculate total
total = sum_three + sum_five - sum_fifteen
# print result with Python 3
print(int(total))
The result is:
>>>
233168
It is possible to do this in one line in Python using a generator expression:
print(sum(x for x in range(1000) if x % 3 == 0 or x % 5 == 0))
The range(1000) produces all the integers from 0 to 999 inclusive. For each one of those integers, if it is divisible by 3 or divisible by 5, then it is included in the result. The sum(...) function adds up all those numbers, and finally print(...) prints the result.
I would use a for loop to iterate over each number in your selected range. Then you can check if the modulus % is equal to 0, meaning it has no remainder when divided by those values, if so, add it to the total.
total = 0
for num in range(1000):
if num % 3 == 0 or num % 5 == 0:
print(num)
total += num
>>> total
233168
While a for loop would work, you could also use a generator expression and sum:
sum(n for n in range(1000) if n % 3 == 0 or n % 5 == 0)
def sum_multiply (n):
data = []
for num in range (1, n):
if num % 3 == 0 or num % 5 == 0:
data.append(num)
return sum(data)
sum_multiply(1000)
total=0
i=0
while i<1000:
if i%3==0 or i%5==0:
print(num)
total+=i
i+=1
print("Total is: ")
**with python**
print(sum([i for i in range(1,1000) if i%3==0 or i%5==0 ]))