How to get math equation of Python algorithm?

Question

ok, so I feel a little stupid because I don’t know this, but a colleague asked, so I ask here: I wrote a python algorithm that solves his problem. if x> 0, add all numbers together from 1 to x.

def intsum(x):
  if x > 0:
    return x + intsum(x - 1)
  else:
    return 0

intsum(10)
55

first, what is the type of this equation, and what is the correct way to get this answer, since it is clearly easier to use any other method?

Answer 1

This is recursion, although for some reason you are marked as factorial.

In any case, the sum from 1 to n is also simple:

n * ( n + 1 ) / 2

(If you want, you can use a special case for negative values.)

Answer 2

, , - Concrete Mathematics: Foundation for Computer Science, , (., , wikipedia ).

, , fac(x) = fac(x - 1) + x, , , - 1 100 , 5050 : " , , 1 , 100, 101, , 2 -, 99, 101, , , 50 , 50 101, 5050". , , 6-; -).

( ) , , , (N * (N+1)) / 2 ( , ; ).

Answer 3

S  = 1 +   2   + ... + (n-1) + n
S  = n + (n-1) + ... +   2   + 1
2S = (n+1) + (n+1) + ... + (n+1) + (n+1)
     ^ you'll note that there are n terms there.
2S = n(n+1)
S = n(n+1)/2

Answer 4

, n.

Python, , , . .

• sum()

  >>> sum(range(11))
  55
  >>> sum([2,4,6])
  12
  

• , reduce()

  >>> import operator
  >>> reduce(operator.add, range(11))
  55
  

Answer 5

, , range(), 0 base. (n + 1), .

...

sum ( (10))!= 55

sum ( (11)) == 55

Answer 6

OP .

. , , , , 1 100. , .

, . .

Answer 7

Note that N + 1, N-1 + 2, N-2 + 3, etc. all add up with the same number, and about the same N / 2 copies (exactly N / 2, if N even).

Answer 8

What you have there is called an arithmetic sequence, and as suggested, you can calculate it directly without the overhead that may result from a recursion.

And I would say that this is homework, despite what you say.