I have some code which gives me the answer I want, but I'm having trouble stopping it once I get the
answer I want.
Here's my code:
multiples = range(1,10)
n = 1
while n>0:
for i in multiples:
if n%i!=0:
n = n + 1
continue
elif n%10==0:
print(n)
This is an attempt at solving problem 5 of Project Euler. Essentially, I'm supposed to get the smallest multiple of all the digits within a given range.
Now, when i run the above code using the example given (1-10 should yield 2520 as the smallest multiple), i get the answer right. However, the code continues to run infinitely and print the answer without breaking. Also, the moment I add the break statement to the end like so:
multiples = range(1,10)
n = 1
while n>0:
for i in multiples:
if n%i!=0:
n = n + 1
continue
elif n%10==0:
print(n)
break
The code just keeps spamming the number 30. Any ideas why this is happening. For the record, I'm not really looking for an alternative solution to this question (the goal is to learn after all), but those are welcome. What I want to know most of all is where I went wrong.
You never break out of your while loop. The for is the entire while body. break interrupts only the innermost loop; you have no mechanism to leave the while loop. Note that your continue doesn't do anything; it applies to the for loop, which is about to continue, anyway, since that's the last statement in the loop (in that control flow).
I can't really suggest a repair for this, since it's not clear how you expect this to solve the stated problem. In general, though, I think that you're a little confused: you use one loop to control n and the other to step through divisors, but you haven't properly tracked your algorithm to your code.
One way to deal with this is to have an exception. At best a custom one.
multiples = range(1,10)
n = 1
class MyBreak(Exception):
pass
while n>0:
try:
for i in multiples:
if n%i!=0:
n = n + 1
continue
elif n%10==0:
print(n)
raise MyBreak()
except MyBreak:
# now you are free :)
break
With this brake you stop only for loop, to exit whole cycle you should create trigger variable, for example:
multiples = range(1,10)
n = 1
tg = 0
while n>0:
for i in multiples:
if n%i!=0:
n = n + 1
continue
elif n%10==0:
print(n)
tg = 1
break
if tg != 0:
break
Or it'll be better to use a function and stop a cycle by return:
def func():
multiples = range(1,10)
n = 1
while n>0:
for i in multiples:
if n%i!=0:
n = n + 1
continue
elif n%10==0:
print(n)
return n
Related
I need to create a program that takes in an array of numbers, check if each number is greater than the sum of all previous numbers, and output true if the condition is met, and false, if not. My trial is presented below:
import fileinput
a0 = [int(i) for i in fileinput.input()]
a = a0[1:]
b=[]
for i in range(1, a0[0]+1):
b.append(a[i])
if (a[i+1] > sum(b)):
print("true")
break
else:
print ("false")
break
My program works for half of the test cases but fails for the other test. Could you please help me figure out what i am doing wrongly. Many thanks for your assistance.
You are breaking too early in the true case. Only because the first element checks, doesn't mean all others will, too:
for i in range(1, a0[0]+1):
b.append(a[i])
if (a[i+1] <= sum(b)):
print ("false")
break
else: # for-else: else is executed only if loop isn't `break`ed out of
print("true")
Only once the loop has finished without finding a counter example, you can be sure that it holds for the entire list.
A more concise of writing this, would be:
import fileinput
_, *a = (int(i) for i in fileinput.input())
s = 0 # just keep track of total
for num in a:
if (num <= s):
print("false")
break
s += num
else:
print("true")
I'm relatively new to Python and I decided to try and code a relatively simple collatz conjecture where the user enters a number (integer). The code is just a simple function that calls itself. i is a list that should have every number that the function calculates appended to it. I'm new to executing Python scripts and I have tried using the IDLE shell to run the code. It asks me what number I want but when I enter a number nothing is printed? I'm sure I just need to edit a small bit of this code (or maybe it's all wrong yikes) but does anybody have any idea why my script returns nothing? Sorry about this and thanks.
Here's the code:
l = input("Enter a number: ")
l = int(l)
i = []
def collatz(n):
if n==1:
return i
if n%2 == 0:
n = n/2
i.append(n)
return collatz(n)
else:
n = ((n*3) + 1) / 2
i.append(n)
return collatz(n)
print(i)
collatz(l)
There are three returns before your print and one of them is inside an else statement, which means that at least one of them will be executed, so your print won't even be reached to be executed, you should move it right after the function definition to see something:
def collatz(n):
print(i) # <= print here
if n==1:
....
See more about the return statement. A snippet:
return leaves the current function call with the expression list (or None) as return value.
As others have mentioned, all of the execution paths in your function end in a return statement, so that print call is unreachable. So if you want each value of n or i to be printed you need to move the call to somewhere that it will be reachable. ;)
Also, there's a little bit of redundancy in that code. You don't need
i.append(n)
return collatz(n)
in both the if and else branches, you can move them outside the if...else block.
Here's a modified version of your code. I've also changed the / operators to // so that the results of the divisions will be integers.
i = []
def collatz(n):
print(n)
if n==1:
return i
if n%2 == 0:
n = n // 2
else:
n = ((n*3) + 1) // 2
i.append(n)
return collatz(n)
# Test
print(collatz(12))
output
12
6
3
5
8
4
2
1
[6, 3, 5, 8, 4, 2, 1]
I'm new to python and I am trying to make a code to print all the square numbers until the square of the desired value entered by the user.
n = raw_input("Enter number")
a = 1
while a < n:
a = 1
print(a*a)
a += 1
if a > n:
break
When I run this code it infinitely prints "1" ... I'm guessing that the value of a does not increase by += so it's a=1 forever. How do I fix this?
There are some problems. First, your input (what raw_input() returns) is a string, so you must convert it to integer:
n = int(raw_input(...))
Second, you are setting a = 1 each iteration, so, since the loop condition is a < n, the loop will run forever ( if n > 1). You should delete the line
a = 1
Finally, it's not necesary to check if a > n, because the loop condition will handle it:
while a < n:
print a * a
a += 1
# 'if' is not necessary
There is a small error in your code:
while a < n:
a=1 # `a` is always 1 :)
print a*a
a += 1
if a > n:
break
You're setting the value of a back to 1 on every iteration of the loop, so every time it checks against n, the value is 2. Remove the a=1 line.
As others have noted, your specific problem is resetting a each time you loop. A much more Pythonic approach to this is the for loop:
for a in range(1, n):
print(a ** 2)
This means you don't have to manually increment a or decide when to break or otherwise exit a while loop, and is generally less prone to mistakes like resetting a.
Also, note that raw_input returns a string, you need to make it into an int:
n = int(raw_input("Enter number: "))
an even better idea is to make a simple function
def do_square(x):
return x*x
then just run a list comprehension on it
n = int(raw_input("Enter #:")) #this is your problem with the original code
#notice we made it into an integer
squares = [do_square(i) for i in range(1,n+1)]
this is a more pythonic way to do what you are trying to do
you really want to use functions to define functional blocks that are easy to digest and potentially can be reused
you can extend this concept and create a function to get input from the user and do some validation on it
def get_user_int():
#a function to make sure the user enters an integer > 0
while True:
try:
n = int(raw_input("Enter a number greater than zero:"))
except TypeError:
print "Error!! expecting a number!"
continue;
if n > 0:
return n
print "Error: Expecting a number greater than zero!"
and then you can build your input right into your list
squares = [do_square(i) for i in range(1,get_user_int()+1)]
and really do_square is such a simple function we could easily just do it in our loop
squares = [x*x for x in range(1,get_user_int())]
The first line in your loop sets's a to one on every iteration.
You assign a=1 inside the loop. That means it's overwriting the a+=1.
try this:
n = eval(raw_input("Enter number"))
a=1
while a < n:
print a*a
a += 1
The issue here is that the value of a gets overridden every time you enter in the loop
Problem is in the condition of the while loop, to print squares of numbers upto a limiting value, try this..
def powers(x):
n=1
while((n**2)<=x):
print(n**2, end =' ')
n +=1
So I'm trying to find the 10,001st prime #. Here's my code -
counter = 3
primes = [1]
while len(primes) < 10002:
for i in range(2, counter):
if counter % i == 0:
counter += 1
else:
primes.append(counter)
counter += 1
print counter
So what I get as an output in primes is a list of numbers, the first few numbers are 1, 3, 5, 7, 11... so far, so good... 13, 17, 19, 23, 27... wait, 27? So at that point it breaks down and starts returning mostly primes but not all primes. And it takes forever.
I'm new to programming, made it through CodeAcademy's Python course and now trying to figure out how to get past what was essentially just an introduction to the grammar. I don't come from a math background, so while I know what a prime is, I know there are far better ways to go about this. If there's anyone in a similar boat who wants to "partner up" and work together on learning Py2.7, I'm more than happy to.
I won't implement anything for you, since that's why you're doing Project Euler, but I will point you strongly in the direction of The Sieve of Eratosthenes. It will calculate in seconds what your code will do in hours.
It works as such: (in pseudocode)
for known_prime in a huge list of numbers:
k=2
while known_prime*k < the biggest number:
known_prime*k is not prime
k += 1
Once you've made it through sqrt of the list, you've found every prime number within the list.
Since you're doing a project Euler puzzle, you obviously want comments on your code rather than the solution.
Your code:
counter = 3
primes = [1]
while len(primes) < 10002:
for i in range(2, counter):
if counter % i == 0:
counter += 1
else: # Mis-aligned else (assuming it's intended for the if)
primes.append(counter)
counter += 1
print counter
Your else is mis-aligned with the if
Your for loop goes all the way to counter, and testing divisibility against itself is bound to find remainder 0.
Your else clause will hit for each non-factor of counter. Most numbers will not be a factor, so you don't want to take action in that case. Instead break out of the loop when the if part triggers.
After exiting the for loop, you'll want to check if a factor was found and if not, add counter to your list of primes.
Looks like you are trying to implement a naive un-optimized brute force search, and that is fine.
Maybe try and write out the algorithm in words or pseudo code before coding it.
As an exercise for you instead of writing for you a simple step by step code I will write a complex line of code with a solution to a similar problem.
Try to figure out what is going on here in this line as an exercise to understand python.
This code will find the prime numbers until n number not the first n prime numbers that you want
def primes(n):
return sorted(set(range(2, n+1)) - set([p*i for p in range(2, n+1) for i in range(p, n+1)]))
If you want to write an efficient and smart code, you may use the Sieve of Eratosthenes, a good algorithm for finding prime numbers in a given range.
For more information, read this:
http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
import time
start_time = time.time()
def is_prime(limit):
aval = True
for j in range(2,int(limit**0.5)+1):
if limit == 2:
aval = True
break
elif limit % j == 0:
aval = False
break
return aval
counter = 0
for i in range(2,1000000):
if is_prime(i):
counter += 1
if counter == 10001:
print("the 10001th prime number is: ",i)
break
print("seconds: ", time.time() - start_time)
Here is my solution to this problem. It may not be so quick but it is not precisely for problem 7.
import time
start = time.time()
n = int(input('Please enter a number: '))
def nth_prime(n):
primes = [2]
x = 3
max_amount = int(input('How much would you like to go for?: '))
while True:
try:
while x < max_amount:
for i in primes:
if x % i == 0:
break
else:
primes.append(x)
x += 2
print(primes[n])
except IndexError:
max_amount = int(input("Please enter a greater number: "))
continue
else:
print(f('Good job. You\'ve found the {n+1}st prime number :) ')
break
nth_prime(n)
end = time.time()
print(str(float(end - start)) + " seconds")
I am new to python and i am still strugling to understand how the sytnax works, how you need to allign your If and else to make it work correctly. How do i really know which else goes with which if? especially when using nested code blocks.
In the code below for the else followed by the comment Prime! from what i understand that else goes with the statement if (n % div == 0): but then why is it alligned with the FOR statement instead?
the last else statement i think goes with if n == 2: but the else is not alligned with it, instead it is after. For the same statement if n == 2: why is n += 1 alligned before pime_count +=1 and not after it.
I understand that the placement of the Else and if is very important because if i decided to move any of them the code stops working. What i can't seem to understand is how does python know which else goes with which if, if the allignment doesnt seem to be consistent.
#!/usr/bin/env python
#
# Problem Set 1a
#
# A program that computes and prints the 1000th prime number.
# Finds primes using trial division (least efficient method)
#------------------------------------------------------------
prime_count = 0
n = 2
while (prime_count <= 1000):
#if even, check for 2, the only even prime
if (n % 2 == 0):
if n == 2:
prime_count += 1
n += 1
else:
# number is odd, possible prime
for div in range(3, n, 2):
if (n % div == 0):
# not a prime
n += 1
break
else:
# prime!
prime_count += 1
if prime_count == 1000:
print "The 1000 prime is", n
else:
n += 1
The rule is very simple: the else clause must have the same indentation as the statement it refers to (most commonly, an if statement).
Now, here:
for div in range(3, n, 2):
if (n % div == 0):
# not a prime
n += 1
break
else:
...
you are not using if-else, you are using for-else!
This construct means "execute the else block unless the loop has terminated through a break".
See for ... else in Python for a discussion.
An if goes with an else at the same indentation, so long as there are no other things at lower indentation between them. They have to be in the same "block". However, in your case, the else that's followed by # prime! is not actually joined to an if at all, but rather to the for div in range(3, n, 2): loop before it!
An else attached to a for loop means "execute this code if the for loop completed without hitting a break statement". It can be useful sometimes, but it is often confusing for people who haven't encountered it before!
I think this can help you to understand how python indentation works http://psung.blogspot.com/2007/12/for-else-in-python.html
In a construct like this one:
for i in foo:
if bar(i):
break
else:
baz()
the else suite is executed after the for, but only if the for terminates normally (not by a break).
In other situations else goes after if
There are 2 rules which are fairly simple,
The indent of the if and else have to be the same
for x in range(15):
if x > 10:
print 'x is more than 10'
else:
print 'x is less than or equal to 10'
Nothing with an indent lower than or equal to that of if and elseshould come in between them
So, this is invalid/ will raise a SyntaxError.
for x in range(15):
if x > 10:
print 'x is more than 10'
print x
else:
print 'x is less than or equal to 10'
Also, As stated in PEP 8
Use 4 spaces per indentation level.
for div in range(3, n, 2):
if (n % div == 0):
# not a prime
n += 1
break
else: # at same indent as for
# prime!
Also, your indent above means for...else loop is made (here else clause is executed if the for loop is exited using break), not if..else.