Develop Reference

How to add all even numbers in a given range recursively?

I have to find the sum of all even numbers between a given range x and y.
I have tried:
def sum_even(x, y):
if x == y:
return x
else:
if x % 2 == 0:
return x + sum_even(x+2, y)
If my given range is (5, 11) then my output should be 24.

Try this:
def sum_even(x, y):
if x >= y:
return 0
if x % 2 == 0:
return x + sum_even(x + 1, y)
return sum_even(x + 1, y)
Examples:
>>> sum_even(5, 11)
24
>>> sum_even(2, 3)
2
>>> sum_even(2, 4)
2
>>> sum_even(2, 5)
6
My code does not include the upper bound y in the sum (see the examples). It might or might not be the desired behavior, but that's depend on your case. If you want to include the upper bound use this code:
def sum_even(x, y):
if x > y:
return 0
if x % 2 == 0:
return x + sum_even(x + 1, y)
return sum_even(x + 1, y)
Examples:
>>> sum_even(5, 11)
24
>>> sum_even(2, 3)
2
>>> sum_even(2, 4)
6
>>> sum_even(2, 5)
6

You might even solve this problem with a O(1) solution. Indeed, you don't need to actually look at ALL the numbers in your range. All you need is a simple mathematical formula. For example, in order to sum all the numbers from 0 to 10^10, you can either do s = sum(x for x in range(10**10 + 1)) (i.e., s = sum(range(10**10 + 1))) or you can use the famous formula s = n * (n + 1) // 2 (i.e., s = 10**10 * (10**10 + 1) // 2). The second approach is much much faster, because you don't need to go through all the numbers in the range.
Similarly in this case you can do this:
def sum_even(x, y):
n1 = y // 2
n2 = (x - 1) // 2
return n1 * (n1 + 1) - n2 * (n2 + 1)
Examples:
>>> sum_even(5, 11)
24
>>> sum_even(2, 3)
2
>>> sum_even(2, 4)
6
>>> sum_even(2, 5)
6
So you don't need recursion and you don't need iteration to solve this problem. Just some math :)

This is your solution patched, I think it could be correct:
def sum_even(x, y):
if x == y or x == y-1:
return x
else:
if x % 2 == 0:
return x + sum_even(x+2, y)
else:
return sum_even(x+1, y)

I assume you want to include x and y (if they're even). For instance, sum_even(2,6) would return 2+4+6.
If you want to exclude the upper bound instead, the code snippets below need a small correction.
# Using python builtin functions `sum` and `range`:
def f1(x, y):
return sum(range(x+x%2, y+1, 2))
# Using arithmetic:
def f2(x, y):
start, end = (x+1)//2, y//2
return end*(end+1) - start*(start-1)
# Using a `for`-loop:
def f3(x, y):
result = 0
for z in range(x, y+1):
if z % 2 == 0:
result += z
return result
# Using recursion:
def f4(x, y):
if x > y:
return 0
elif x % 2 == 1:
return f4(x+1, y)
else:
return x + f4(x+2, y)
# Testing:
print(' x, y --> f1,f2,f3,f4')
for (x, y) in [(5,11), (0,10), (2,6), (12, 17)]:
results = [f(x,y) for f in (f1,f2,f3,f4)]
print('{:2d},{:2d} --> {:2d},{:2d},{:2d},{:2d}'.format(x,y,*results))
# x, y --> f1,f2,f3,f4
# 5,11 --> 24,24,24,24
# 0,10 --> 30,30,30,30
# 2, 6 --> 12,12,12,12
# 12,17 --> 42,42,42,42

The faculty of two numbers python 3

def choose (x, y):
if y > x:
print ("False")
elif y == 0 or y == x:
return 1
elif y == 1:
return x
else:
if (x-y) > y:
biggest = x-y
smallest = y
else:
biggest = y
smallest = x-y
resultatet = x * choose (x-1, biggest)
res = resultatet // smallest
return res
My function is working perfectly with whatever x input I insert but with bigger Y inputs like 8000 for example I'm getting
File "/home/nazel607/labb3b_2.py", line 20, in choose
resultatet = x * choose (x-1, biggest)
File "/home/nazel607/labb3b_2.py", line 3, in choose
if y > x:
RuntimeError: maximum recursion depth exceeded in comparison
Is there a way I can overcome this problem or is it impossible in python due to its limits? Is there another way than increasing the limit?

It seems that you can get rid of the recursion:
def choose2(x, y):
if y > x:
raise ValueError()
if y == 0 or y == x:
return 1
if y == 1:
return x
result = 1
while y != x:
big, small = max(x-y, y), min(x-y, y)
result *= x // small
x -= 1
y = big
return result
I've tested it over few examples:
for x, y in [
(4, 2),
(17, 9),
(125, 79),
(8005, 13),
(9005, 13),
# (19005, 7004) # exceeds max recursion depth on my machine
]:
assert choose(x, y) == choose2(x, y)
and seems to work fine.

You are not exiting the program ...
def choose (x, y):
if y > x:
print ("False")
return
# ...rest of your program

Need to find the wrong output in the following python code

def max3bad(x,y,z):
maximum = 0
if x >= y:
if x >= z:
maximum = x
elif y >= z:
maximum = y
else:
maximum = z
return(maximum)
wrong output for what input?
get an input for which u get wrong output

for case: x = 2, y = 1, z = 3
the code output is 0 not 3.
It's better to get max number by :
max(x, y, z)
or you fix bug in your code:
def max3bad(x,y,z):
maximum = 0
if x >= y:
if x >= z:
maximum = x
else:
maximum = z
elif y >= z:
maximum = y
else:
maximum = z
return maximum

Returning the middle value when giving the function 3 integer arguments

This code is restarting the Python shell, and I cannot work out the errors of my code.
def middle(x,y,z):
if x > y and x < y:
return x
elif y > x and y < z:
return y
elif z > x and z < y:
return z
else:
return False
#Main Routine
middle(1,11,111)

Note that Python can chain comparisons for you (see the docs), and you are missing several cases:
def middle(x, y, z):
"""Return the middle of the three input values."""
if y < x < z or z < x < y: # or min(y, z) < x < max(y, z)
return x
elif x < y < z or z < y < z:
return y
elif x < z < y or y < z < x:
return z
return False
In use:
>>> middle(1, 11, 111)
11
If you want to see results when running the script directly, you will have to be explicit about this; as Martijn suggested in the comments, you could print middle(1, 11, 111). Otherwise the result will be evaluated, but not actually shown on-screen.
You can also simplify by sorting the inputs:
def middle(x, y, z):
"""Return the middle of the three input values."""
x, y, z = sorted((x, y, z))
return y if x < y < z else False

I am not sure what you want to do with this but your function looks weird to me
For example:
if x > y and x < y:
Will never be true
Also, if this function aims at returning the median of the three value it does not do this.
it would be more like
def middle (x, y, z):
t = [x, y, z]
t.sort()
return t[1]
Hope this helped

Python checking if a point is in sphere with center x, y ,z

I'm trying to check if a point is within a sphere with a center point of (x, y, z) where (x, y, z) is not (0, 0, 0).
This code I'm using to generate the points I want to check:
def generatecoords(self, i):
x, y, z = generatepoint()
if i >= 1:
valid = False
while valid == False:
coords = self.checkpoint(x, y, z)
for b in world.starlist:
if coords == world.starlist[b].coords:
coords = self.checkpoint(x, y, z)
else:
valid = True
else:
coords = self.checkpoint(x, y, z)
return coords
def checkpoint(self, x, y, z):
d = math.sqrt(x * x + y * y + z * z)
while d >= self.radius:
x, y, z = generatepoint()
d = math.sqrt(x * x + y * y + z * z)
coords = (int(x), int(y), int(z))
return coords
def generatepoint():
x, y, z = [int(random.uniform(-self.radius, self.radius)) \
for b in range(3)]
return x, y, z
These function are called in a for loop to generate the points in a dictionary, while also checking the unlikely chance that points aren't placed on top of another(mostly because I can).
I trying to figure out what I need to add to math.sqrt(x * x + y * y + z * z) so that it accounts for a center that isn't (0, 0, 0). I do know of one way to do it, but it would require several lines of code and I'd rather do it in one. I would have asked this in the comments of the answer in another question, but I'm not allowed to comment on answers yet.

The formula is:
A point (x,y,z) is inside the sphere with center (cx,cy,cz) and radius r if
(x - cx)^2 + (y - cy)^2 + (z - cz)^2 < r^2

Here is a very short function that returns True if the point is in the sphere, and False if not.
The inputs are two numpy arrays: point = [x,y,z] and ref = [x,y,z] and the radius should be a float.
import numpy as np
def inSphere(self, point, ref, radius):
# Calculate the difference between the reference and measuring point
diff = np.subtract(point, ref)
# Calculate square length of vector (distance between ref and point)^2
dist = np.sum(np.power(diff, 2))
# If dist is less than radius^2, return True, else return False
return dist < radius ** 2