Lately am trying to do some Python programming, so am doing some mathematical exercices in some website.
I came across this example and at first I didn't understand the exercise. So I checked the solution to at least understand the question. As a consequence I found myself learning some coding tricks (like the while True loop).
The exercise is simple:
Write a Python program to find the smallest multiple of the first n numbers. Also, display the factors.
Below is the code:
def smallest_multiple(n):
if (n<=2):
return n
i = n * 2
factors = [number for number in range(n, 1, -1) if number * 2 > n]
print(factors)
while True:
for a in factors:
if i % a != 0:
i += n
break
if (a == factors[-1] and i % a == 0):
return i
My questions are:
Why does he create a list of numbers that are superior to the input by a factor of 2?
And then the while loop is just difficult to undestand. Could someone please explain it to me (I mean the content of the loop)?
I think we have here a tiny mixture of mathematic question and programming.
first of all, please note that the spaces are important in coding. see how the code should look like. (with the spaces)
def smallest_multiple(n):
if (n<=2):
return n
i = n * 2
factors = [number for number in range(n, 1, -1) if number * 2 > n]
print(factors)
while True:
for a in factors:
if i % a != 0:
i += n
break
if (a == factors[-1] and i % a == 0):
return i
1- Why does he create a list of numbers that are superior to the input by a factor of 2 ?
Answer : because the numbers that have their double smaller than the highest number will not affect / change the result. (This is a maths question) you can check that by removing the condition and you will see that you will get the same result (same smallest multiple)
2-And then the while loop is just difficult to understand. Could someone please explain it to me (I mean the content of the loop)? Thanks for your response?
Answer : The loop is using the boolean True as the code will only stops until it finds the smallest multiple for the first n numbers. the reason of the coder doing this because he has used the keyword return that will help him to exit the function, ultimately exiting the while loop.
The loop uses the value i which is the first multiple of the highest number of the first n numbers, meaning the double of the value n. Then will check if this first multiple (meaning i) is not dividable (i % a != 0) by the numbers in the list starting from the highest.
if i % a != 0:
i += n
break
this condition is there to increase the value of i as soon as i is not dividable by any number of the first n numbers meaning that the code keeps searching through the multiples of n (n being the highest number of the list) until i is dividable by all the numbers in the list
once the i value is able to satisfy the condition below
if (a == factors[-1] and i % a == 0):
then the while loop is exited through the keyword return which exits the function and the i value is sent as a response to the function call through the line return i
also note that factors[-1] is the last item (number) of the list.
I hope the above make sense and is clear to you.
Related
I am trying to make a simple function to sum the prime numbers of a given input. I am wondering why '7' isn't coming up in my appended list of prime numberS:
def sum_primes(n):
empty = []
for i in range(2,n):
if n % i == 0:
empty.append(i)
print(empty)
sum_primes(10)
Your method for determining if numbers are prime is a bit off. Your function seems to determine if the input num n is prime but not sum all prime numbers to it.
You could make an isprime function and then loop over that for all numbers less than n to find all primes.
Actually your program have logical error. First you have to understand the prime number logic.
Step 1: We must iterate through each number up to the specified number in order to get the sum of prime numbers up to N.
Step 2: Next, we determine whether the given integer is a prime or not. If it is a prime number, we can add it and keep it in a temporary variable.
Step 3: Now that the outer loop has finished, we can print the temporary variable to obtain the total of primes.
sum = 0
for number in range(2, Last_number + 1):
i = 2
for i in range(2, number):
if (int(number % i) == 0):
i = number
break;
if i is not number:
sum = sum + number
print(sum)
def sum_primes(y):
sum = 0
# for each number in the range to the given number
for i in range(2, y):
# check each number in the sub range
for j in range(2, int(i/2)+1):
# and if its divisble by any number between
# 2 and i/2
if i % j == 0:
# then it is not a prime number
break
else:
# print current prime
print(i)
# add prime to grand total
sum += i
return sum
print(sum_primes(10))
I think you have some problems with the way you are checking if your number is prime or not. A much simpler way would be to use a library that does this for you. It takes less time, is usually a better implementation than you or I could come up with, and it takes fewer lines, therefore, looking cleaner.
I would recommend primePy. Its whole point is to work with prime numbers.
I don't have my computer with me atm so I can't make any sample code for you and test it. But something like this should work.
from primePy import primes
prime_list = primes.upto(n)
print(prime_list)
I believe that function returns a list of all prime numbers up to n. And if you wanted to get the sum of all of those numbers, you can go about whatever method you want to do that.
Afterward, slap it into a def and you should be good to go.
If it doesn't work let me know, and I will try to test and fix it when I am in front of a computer next.
help me please.
Write a function: User enters natural numbers A and B (A <B). The function prints all prime numbers separated by a space on the segment [A, B]
This is my code but it doesn't work, wrong output, I don't understand why?
def prime_numbers(a, b):
for i in range(a, b + 1):
dividers = 0
for j in range (2, i):
if i % j == 0:
dividers += 1
if dividers == 1:
return i
a = int (input ("Enter first number: ")
b = int (input ("Enter second number: ")
print (prime_numbers (a, b))
When this code is e.g. ran with a=5 and b=11 I expect the output to be 5 7 11. Instead the output is 6.
Problems with Logic
The function you wrote is meant to calculate how many dividers are in the numbers from a-b, which is different than what the prompt is asking.
You're on the right track with the loops and the % (mod) if statement, but you're off by a bit. In order to calculate the number of prime numbers, there are through a-b, you're going to have to do something different because right now I think your code is calculating the number of divisors there are for numbers a-b.
Problems with Code
When you use the keyword return, you return a value for the entire function and considering you're calculating for multiple prime numbers, I don't think that'd work out well.
Also, I'd be sure to watch out for proper tabbing, your tabbing is inconsistent, and sometimes you've used spaces, sometimes you've used tabs. Python doesn't use brackets (languages such as C, C++, Java, JavaScript aren't reliant on tabs) and is reliant on consistent spacing/tabbing.
Don't forget to close the brackets for the int() function, which is meant to convert a string to an integer.
I think what you meant to do is this:
def prime_numbers(a, b):
primes = [] # Make a list of primes
for i in range(a, b + 1): # Iterate from a-b
if is_prime(i): # Check if the number is a prime or not
primes.append(i) # If it is a prime, then add it to the list
return primes # Return the prime list
def is_prime(v): # Function to check for prime numbers
for x in range(2, v): # Iterate from 2 to the input value
if v % x == 0: # Check if v is divisible by x
return False # If it is divisible by x, return False becaues it's not a prime number
return True # Haven't found any divisors for the number, so return True.
a = int (input ("Enter first number: "))
b = int (input ("Enter second number: "))
print (prime_numbers (a, b))
I'd suggest you use websites such as YouTube and W3Schools to get a better understanding of how to code and remember, GOOGLE IS YOUR FRIEND.
Edit: My apologies, the tabbing issue was caused due to my IDE (Sublime), but it's still a good thing to keep in mind when working with Python.
I am trying to create a program that prints out a list of numbers starting at 0 and leading up to a number the user inputs (represented by the variable "number"). I am required to use "while" loops to solve this problem (I already have a functional "for" loop version of the assignment). The program should mark anything in that list divisible by 3 with the word "Fizz," divisible by 5 with the word "Buzz," and anything divisible by both with "FizzBuzz" while also including unlabeled numbers outside of those specifications.
Every time I run this program, it ignores the conditions and just prints the word "FizzBuzz" however many times is represented by the number inputted. (I typically use 15 because it has at least one example of each condition, so that means I get 15 "FizzBuzz"s in a row).
To find out why it was doing that, I used print(i) instead of the rest of the program under the first conditional and it gave me 15 counts of the number 0, so there is reason to believe the program is completely ignoring the range I gave it and just outputting copies of i based on the user number input.
Any help would be appreciated!
number = int(input("Enter a Number"))
i = 0
while(i < number + 1):
if number % 3 == 0 and number % 5 == 0:
print("Fizzbuzz")
elif number % 5 == 0:
print("Buzz")
elif number % 3 == 0:
print("Fizz")
else:
print(number)
i += 1
print ("Done!")
You meant to check the divisibility of i, which increments every loop, not of number which doesn't change.
You also meant to print(i) in the else clause.
I am new to programming and trying to write a program that prints primes less than 100. When I run my program I get an output that has most of the prime numbers, but I also have even numbers and random odd numbers. Here is my code:
a=1
while a<100:
if a%2 == 0:
a+=1
else:
for i in range(2,int(a**.5)+1):
if a%i != 0:
print a
a+=1
else:
a+=1
The first part of my code is meant to eliminate all even numbers (doesn't seem to fully work). I also don't fully understand the part of my code (for i in).
What exactly does the "for i in" part of the code do? What is 'i'?
How can I modify my code that it does accurately find all of the primes from 1-100?
Thanks in advance.
See the comments on your post to find out why your code doesn't work. I'm just here to fix your code.
First of all, make use of functions. They make code so much more readable. In this case, a function to check if a number is a prime would be a good idea.
from math import sqrt
def is_prime(number):
return all(number%i for i in range(2, int(sqrt(number))+1))
There isn't much left to do now - all you need is to count how many primes you've found, and a number that keeps growing:
primes= 0
number= 1
while primes<100:
if is_prime(number):
print number
primes+= 1
number+= 1 # or += 2 for more speed.
Suddenly it's very easy to read and debug, is it not?
I'm going to try to guide you without giving you any code to help you learn, you should probably start from scratch I think and make sure you know how each part of your code works.
Some things about primes, 1 isn't prime and 2 is so you should start with a=3 and print 2 immediately if you want to eliminate even numbers.
By doing that, you will then be able to just increment a by 2 rather than 1 as you're doing now and just skip over even numbers.
As you loop a to be less than 100, you need to check if each value of a is prime by making another loop with another variable that loops downward through all numbers that could possibly divide a (less than or equal to a's square root). If that variable divides a, then exit the inner loop and increment a, but if it makes it all the way to 1, then a is prime and you should print it, then increment a.
There's several things wrong with what you're doing here:
You don't find the first 100 primes, you find the primes less than 100 with while a < 100
You print a every time a is evenly divisble by i, and then increment a. But you've gotten ahead of yourself! a is prime only if it is not divisble by any value i, but you fail to keep track of whether it's failed for any i.
You add +1 to the range up to the square root of the number - but that's not necessary. The square root of the number is the largest number that could possibly be the smaller of two factors. You don't need to add 1 because that number is _bigger than the sqare root - in other words, if a is evenly divisble by sqrt(a)+1 , then a / (sqrt(a)+1) will be smaller than sqrt(a)+1 and therfore would have already been found.
You never print the list of primes you found; you print a every time you find a number that is not a factor ( primes have no factors other than one and themselves, so you're not printing prime numbers at all!)
This might get you thinking in the right direction:
a=0
primes = []
while len(primes) < 100:
a = a + 1
prime=True
maxfactor = int(a ** .5)
for i in primes:
if i == 1:
continue
if i > maxfactor:
break
if a % i == 0:
print a, " is not prime because it is evenly divisble by ", i
prime=False
break
if prime:
print "Prime number: ", a
primes.append(a)
for p in primes:
print p
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Simple Prime Generator in Python
First I will prompt user to input any number. Then my code will check whether does the number input by the user is a Prime number not.
Here is my codes:
num = int(raw_input("Input any of the number you like:"))
for x in range(2, int(num**0.5)+1):
if num % x == 0:
print "It is not a prime number"
else:
print "It is a prime number"
But question is I cant seem to get the output for 2 and 3. And when I randomly input any numbers like 134245, the system will output alot of sentences. And I do not know why?
Appreciate any kind souls to help me up :)
import urllib
tmpl = 'http://www.wolframalpha.com/input/?i=is+%d+a+prime+number'
def is_prime(n):
return ('is a prime number' in urllib.urlopen(tmpl % (n,)).read())
you should stop once num % x == 0 is true (no need for further testing) and print 'it is a prime number' only if the loop completed without anything printed before.
A number is prime if it only divides by 1 and itself. A pseudocode follows:
boolean prime = true;
for (int i = 2; i * i <= num; i++)
if (num % i == 0) {
prime = false;
break;
}
if (prime)
println("It is prime!");
else
println("It is not prime!");
Look at your code like this:
num = ...
for x in range(2, int(num**0.5)+1):
print something
The body of the loop is executed at every iteration. That means you're printing something at every iteration, i.e. for each x that you check to see if it's a factor of num, you print. That's not what you should be doing; in order to determine whether a number is prime, you check all possible factors first, then print your result. So you shouldn't be printing anything until after the loop.
But question is I cant seem to get the output for 2 and 3.
You're looping from 2 to ceil(sqrt(n)). For 2 and 3, this is an empty range, hence no iteration happens. Either special-case it or rewrite the code such that it assumes that n is prime and tries to disprove it in the loop.
the system will output alot of sentences.
You're printing on every iteration. Instead, use a boolean flag (or a premature return, if you factor it out into a function) to determine prime-ness and print once, after the loop, based on that prime.
Your code is not structured well-- the algorithm continues to loop all the way up to the top of your range, even if you already know that the number is composite, and it also prints some result on each iteration when it should not.
You could put the logic into a function and return True or False for prime-ness. Then you could just check the result of the function in your if statement.
def is_prime(num):
for x in range(2, int(num**0.5)+1):
if num % x == 0:
return False
return True
num = int(raw_input("Input any of the number you like:"))
if not is_prime(num):
print "It is not a prime number"
else:
print "It is a prime number"
Here are two Python routines for calculating primes. (Hint: Google for Sieve of Eratosthenese):
def pythonicSieve(maxValue):
"""
A Pythonic Sieve of Eratosthenes - this one seems to run slower than the other.
see http://love-python.blogspot.com/2008/02/find-prime-number-upto-100-nums-range2.html
"""
return [x for x in range(2,maxValue) if not [t for t in range(2,x) if not x%t]]
def sieveOfEratosthenes(maxValue):
"""
see http://hobershort.wordpress.com/2008/04/15/sieve-of-eratosthenes-in-python/
"""
primes = range(2, maxValue+1)
for n in primes:
p = 2
while n*p <= primes[-1]:
if n*p in primes:
primes.remove(n*p)
p += 1
return primes