Question

def different(s):
   x = len(s)
   for i in range(1, 1 << x):
      u.append([s[j] for j in range(x) if (i & (1 << j))])

It takes a list and makes different combinations

(a,b,c) = ((a,b,c),(a,b),(a,c) ...)

But what does the range do? From 1 to what. I don’t understand the "<<"

and also, if (i & (1 << j)) what does this do? It checks if i and 2 to the power of j? Doesn’t make any sense to me.

Answer 1

The range function returns a list of numbers from zero to the given number minus one. It also has two- and three-argument forms (see the doc for more info):

range(n) == [0, 1, 2, ..., n - 1]

<< is the left-shift operator, and has the effect of multiplying the left hand side by two to the power of the right hand side:

x << n == x * 2**n

Thus the above range function (range(1, 1 << x)) returns [1, 2, 3, ..., 2**x - 1].

In the seconds usage of <<, the left-shift is being used as a bit-mask. It moves the 1-bit into the j-th bit, and performs a bit-wise and with i, so the result will be non-zero (and pass the if test) if and only if the j-th bit of i is set. For example:

j = 4
1 << j = 0b1000 (binary notation)

i = 41 = 0b101001
i & (1 << j) = 0b101001
             & 0b001000
             = 0b001000 (non-zero, the if-test passes)

i = 38 = 0b100110
i & (1 << j) = 0b100110
             & 0b001000
             = 0b000000 (zero, the if-test fails)

In short, x & (1 << y) is non-zero iff the y-th bit of x is set.