Basically trying to learn Python and I was doing the hackerrank 30 day challenge.
Task
Given an integer, n , perform the following conditional actions:
If n is odd, print ok
If n is even and in the inclusive range of 2 to 5, print nok
n = 4
if n in range(2,5) and type(n/2) == int:
print ("ok")
else:
print ("nok")
it prints "nok" no matter what n is.
In Python 3 / always returns a float (isinstance(4 / 2, int) is False since 4 / 2 returns 2.0).
Since you are essentially checking for the parity of n, check it explicitly:
... and n % 2 == 0:
Because, n/2 = 2.0 not 2, i.e division returns a float not int here are some ways to check it :
n in range(2,5) and n % 2 == 0
#or
n in range(2,5) and n & 1 != 1 #because, every odd number has last bit 1
I've seen a couple posts explaning how to iterate through the digits of a number in Python, but they all turn the number into a string before iterating through it...
For example:
n=578
print d for d in str(n)
How can I do this without the conversion into a string?
10**int(log(n, 10)) is basically 10*, such that it is the same length as n. The floor division of n by that will give us the leading digit, while the modulo % gives us the rest of the number.
from math import log
def digits(n):
if n < 0:
yield '-'
n = -1 * n
elif n == 0:
yield 0
return
xp = int(log(n, 10).real)
factor = 10**xp
while n:
yield int(n/factor)
n = n % factor
try:
xp, old_xp = int(log(n, 10).real), xp
except ValueError:
for _ in range(xp):
yield 0
return
factor = 10**xp
for _ in range(1, old_xp-xp):
yield 0
for x in digits(12345):
print(x)
prints
1
2
3
4
5
Edit: I switched to this version, which is much less readable, but more robust. This version correctly handles negative and zero values, as well as trailing and internal 0 digits.
I'm trying to solve this problem in hackerrank. At some point I have to check if a number divides n(given input) or not.
This code works perfectly well except one test case(not an issue):
if __name__ == '__main__':
tc = int(input().strip())
for i_tc in range(tc):
n = int(input().strip())
while n % 2 == 0 and n is not 0:
n >>= 1
last = 0
for i in range(3, int(n ** 0.5), 2):
while n % i == 0 and n > 0:
last = n
n = n // i # Concentrate here
print(n if n > 2 else last)
Now you can see that I'm dividing the number only when i is a factor of n.For example if the numbers be i = 2 and n = 4 then n / 2 and n // 2 doesn't make any difference right.
But when I use the below code all test cases are getting failed:
if __name__ == '__main__':
tc = int(input().strip())
for i_tc in range(tc):
n = int(input().strip())
while n % 2 == 0 and n is not 0:
n >>= 1
last = 0
for i in range(3, int(n ** 0.5), 2):
while n % i == 0 and n > 0:
last = n
n = n / i # Notice this is not //
print(n if n > 2 else last)
This is not the first time.Even for this problem I faced the same thing.For this problem I have to only divide by 2 so I used right shift operator to get rid of this.But here I can't do any thing since right shift can't help me.
Why is this happening ? If the numbers are small I can't see any difference but as the number becomes larger it is somehow behaving differently.
It is not even intuitive to use // when / fails. What is the reason for this ?
The main reason of the difference between n // i and n / i given that n and i are of type int and n % i == 0 is that
the type of n // i is still int whereas the type of n / i is float and
integers in Python have unlimited precision whereas the precision of floats is limited.
Therefore, if the value of n // i is outside the range that is accurately representable by the python float type, then it will be not equal to the computed value of n / i.
Illustration:
>>> (10**16-2)/2 == (10**16-2)//2
True
>>> (10**17-2)/2 == (10**17-2)//2
False
>>> int((10**17-2)//2)
49999999999999999
>>> int((10**17-2)/2)
50000000000000000
>>>
So I was studying recursion function online. And the one question asks me to write a function to add up a number's digits together. For example (1023) -> 1 + 0 + 2 + 3 = 6. I used % and // get get rid of a digit each time. However, I don't know how to add them up together. The closest I can get is to print out each digit. Can anyone help me solve it or give me a hint please?
def digitalSum(n):
if n < 10:
sum_total = n
print(sum_total)
else:
sum_total = n % 10
digitalSum((n - (n % 10))//10)
print(sum_total)
digitalSum(1213)
Your function should return the current digit plus the sum of the rest of the digits:
def digitalSum(n):
if n < 10: return n
return n % 10 + digitalSum(n // 10)
print digitalSum(1213)
For completeness, you can also handle negative numbers:
def digitalSum(n):
if n < 0: sign = -1
else: sign = 1
n = abs(n)
if n < 10: return n
return sign * (n % 10 + digitalSum(n // 10))
print digitalSum(1213)
A correct version of your function is as follows:
from math import log10
def sum_digits(n, i=None):
if i is None:
i = int(log10(abs(n)))
e = float(10**i)
a, b = (n / e), (abs(n) % e)
if i == 0:
return int(a)
else:
return int(a) + sum_digits(b, (i - 1))
print sum_digits(1234)
print sum_digits(-1234)
Example:
$ python -i foo.py
10
8
>>>
Updated: Updated to properly (IHMO) cope with negative numbers. e.g: -1234 == -1 + 2 + 3 + 4 == 8
NB: Whilst this answer has been accepted (Thank you) I really think that perreal's answer should have been accepted for simplicity and clarity.
Also note: that whilst my solution handles negative numbers and summing their respective digits, perreal clearly points out in our comments that there are ate least three different ways to interpret the summing of digits of a negative number.
I have the following code for Project Euler Problem 12. However, it takes a very long time to execute. Does anyone have any suggestions for speeding it up?
n = input("Enter number: ")
def genfact(n):
t = []
for i in xrange(1, n+1):
if n%i == 0:
t.append(i)
return t
print "Numbers of divisors: ", len(genfact(n))
print
m = input("Enter the number of triangle numbers to check: ")
print
for i in xrange (2, m+2):
a = sum(xrange(i))
b = len(genfact(a))
if b > 500:
print a
For n, I enter an arbitrary number such as 6 just to check whether it indeed returns the length of the list of the number of factors.
For m, I enter entered 80 000 000
It works relatively quickly for small numbers. If I enter b > 50 ; it returns 28 for a, which is correct.
My answer here isn't pretty or elegant, it is still brute force. But, it simplifies the problem space a little and terminates successfully in less than 10 seconds.
Getting factors of n:
Like #usethedeathstar mentioned, it is possible to test for factors only up to n/2. However, we can do better by testing only up to the square root of n:
let n = 36
=> factors(n) : (1x36, 2x18, 3x12, 4x9, 6x6, 9x4, 12x3, 18x2, 36x1)
As you can see, it loops around after 6 (the square root of 36). We also don't need to explicitly return the factors, just find out how many there are... so just count them off with a generator inside of sum():
import math
def get_factors(n):
return sum(2 for i in range(1, round(math.sqrt(n)+1)) if not n % i)
Testing the triangular numbers
I have used a generator function to yield the triangular numbers:
def generate_triangles(limit):
l = 1
while l <= limit:
yield sum(range(l + 1))
l += 1
And finally, start testing:
def test_triangles():
triangles = generate_triangles(100000)
for i in triangles:
if get_factors(i) > 499:
return i
Running this with the profiler, it completes in less than 10 seconds:
$ python3 -m cProfile euler12.py
361986 function calls in 8.006 seconds
The BIGGEST time saving here is get_factors(n) testing only up to the square root of n - this makes it heeeaps quicker and you save heaps of memory overhead by not generating a list of factors.
As I said, it still isn't pretty - I am sure there are more elegant solutions. But, it fits the bill of being faster :)
I got my answer to run in 1.8 seconds with Python.
import time
from math import sqrt
def count_divisors(n):
d = {}
count = 1
while n % 2 == 0:
n = n / 2
try:
d[2] += 1
except KeyError:
d[2] = 1
for i in range(3, int(sqrt(n+1)), 2):
while n % i == 0 and i != n:
n = n / i
try:
d[i] += 1
except KeyError:
d[i] = 1
d[n] = 1
for _,v in d.items():
count = count * (v + 1)
return count
def tri_number(num):
next = 1 + int(sqrt(1+(8 * num)))
return num + (next/2)
def main():
i = 1
while count_divisors(i) < 500:
i = tri_number(i)
return i
start = time.time()
answer = main()
elapsed = (time.time() - start)
print("result %s returned in %s seconds." % (answer, elapsed))
Here is the output showing the timedelta and correct answer:
$ python ./project012.py
result 76576500 returned in 1.82238006592 seconds.
Factoring
For counting the divisors, I start by initializing an empty dictionary and a counter. For each factor found, I create key of d[factor] with value of 1 if it does not exist, otherwise, I increment the value d[factor].
For example, if we counted the factors 100, we would see d = {25: 1, 2: 2}
The first while loop, I factor out all 2's, dividing n by 2 each time. Next, I begin factoring at 3, skipping two each time (since we factored all even numbers already), and stopping once I get to the square root of n+1.
We stop at the square_root of n because if there's a pair of factors with one of the numbers bigger than square_root of n, the other of the pair has to be less than 10. If the smaller one doesn't exist, there is no matching larger factor.
https://math.stackexchange.com/questions/1343171/why-only-square-root-approach-to-check-number-is-prime
while n % 2 == 0:
n = n / 2
try:
d[2] += 1
except KeyError:
d[2] = 1
for i in range(3, int(sqrt(n+1)), 2):
while n % i == 0 and i != n:
n = n / i
try:
d[i] += 1
except KeyError:
d[i] = 1
d[n] = 1
Now that I have gotten each factor, and added it to the dictionary, we have to add the last factor (which is just n).
Counting Divisors
Now that the dictionary is complete, we loop through each of the items, and apply the following formula: d(n)=(a+1)(b+1)(c+1)...
https://www.wikihow.com/Determine-the-Number-of-Divisors-of-an-Integer
All this formula means is taking all of the counts of each factor, adding 1, then multiplying them together. Take 100 for example, which has factors 25, 2, and 2. We would calculate d(n)=(a+1)(b+1) = (1+1)(2+1) = (2)(3) = 6 total divisors
for _,v in d.items():
count = count * (v + 1)
return count
Calculate Triangle Numbers
Now, taking a look at tri_number(), you can see that I opted to calculate the next triangle number in a sequence without manually adding each whole number together (saving me millions of operations). Instead I used T(n) = n (n+1) / 2
http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/runsums/triNbProof.html
We are providing a whole number to the function as an argument, so we need to solve for n, which is going to be the whole number to add next. Once we have the next number (n), we simply add that single number to num and return
S=n(n+1)2
S=n2+n2
2S=n2+n
n2+n−2S=0
At this point, we use the quadratic formula for : ax2+bx+c=0.
n=−b±√b2−4ac / 2a
n=−1±√1−4(1)(−2S) / 2
n=−1±√1+8S / 2
https://socratic.org/questions/how-do-you-solve-for-n-in-s-n-n-1-2
So all tri_number() does is evaluate n=1+√1+8S / 2 (we ignore the negative equation here). The answer that is returned is the next triangle number in the sequence.
def tri_number(num):
next = 1 + int(sqrt(1+(8 * num)))
return num + (next/2)
Main Loop
Finally, we can look at main(). We start at whole number 1. We count the divisor of 1. If it is less than 500, we get the next triangle number, then try again and again until we get a number with > 500 divisors.
def main():
i = 1
while count_divisors(i) < 500:
i = tri_number(i)
return i
I am sure there are additional ways to optimize but I am not smart enough to understand those ways. If you find any better ways to optimize python, let me know! I originally solved project 12 in Golang, and that run in 25 milliseconds!
$ go run project012.go
76576500
2018/07/12 01:56:31 TIME: main() took 23.581558ms
one of the hints i can give is
def genfact(n):
t = []
for i in xrange(1, n+1):
if n%i == 0:
t.append(i)
return t
change that to
def genfact(n):
t=[]
for i in xrange(1,numpy.sqrt(n)+1):
if(n%i==0):
t.append(i)
t.apend(n/i)
since if a is a divisor than so is b=n/a, since a*b=a*n/b=n, That should help a part already (not sure if in your case a square is possible, but if so, add another case to exclude adding the same number twice)
You could devise a recursive thing too, (like if it is something like for 28, you get 1,28,2,14 and at the moment you are at knowing 14, you put in something to actually remember the divisors of 14 (memoize), than check if they are alraedy in the list, and if not, add them to the list, together with 28/d for each of the divisors of 14, and at the end just take out the duplicates
If you think my first answer is still not fast enough, ask for more, and i will check how it would be done to solve it faster with some more tricks (could probably make use of erastothenes sieve or so too, and some other tricks could be thought up as well if you would wish to really blow up the problem to huge proportions, like to check the first one with over 10k divisors or so)
while True:
c=0
n=1
m=1
for i in range(1,n+1):
if n%i==0:
c=c+1
m=m+1
n=m*(m+1)/2
if c>500:
break
print n
this is not my code but it is so optimized.
source: http://code.jasonbhill.com/sage/project-euler-problem-12/
import time
def num_divisors(n):
if n % 2 == 0: n = n / 2
divisors = 1
count = 0
while n % 2 == 0:
count += 1
n = n / 2
divisors = divisors * (count + 1)
p = 3
while n != 1:
count = 0
while n % p == 0:
count += 1
n = n / p
divisors = divisors * (count + 1)
p += 2
return divisors
def find_triangular_index(factor_limit):
n = 1
lnum, rnum = num_divisors(n), num_divisors(n + 1)
while lnum * rnum < 500:
n += 1
lnum, rnum = rnum, num_divisors(n + 1)
return n
start = time.time()
index = find_triangular_index(500)
triangle = (index * (index + 1)) / 2
elapsed = (time.time() - start)
print("result %s returned in %s seconds." % (triangle, elapsed))