On the edges of the hypercube Your job will be to write a function or a program, that will take an integer n&gt;0 as input and output a list of the edges of the n-dimensional hypercube. In graph theory an edge is defined as a 2-tuple of vertices (or corners, if you prefer), that are connected.

Example 1

A 1-dimensional hypercube is a line and features two vertices, which we will call a and b.



Therefore, the output will be:

[[a, b]]


Example 2

The 4-dimensional hypercube (or tesseract) consists of 32 edges and its graph looks like this



and the output could look like this

[[a, b], [a, c], [a, e], [a, i], [b, d], [b, f], [b, j], [c, d], [c, g], [c, k], [d, h], [d, l], [e, f], [e, g], [e, m], [f, h], [f, n], [g, h], [g, o], [h, p], [i, j], [i, k], [i, m], [j, l], [j, n], [k, l], [k, o], [l, p], [m, n], [m, o], [n, p], [o, p]]


Rules


You may name the vertices any way you like, as long as the name is unique.
The edges are undirected, i.e. [a, b] and [b, a] are considered the same edge.
Your output must not contain duplicate edges.
The output may be in any sensible format.
Standard loopholes are forbidden.


Scoring

Shortest code wins.
Python 2, 54 56 62 bytes

lambda n:{tuple({k/n,k/n^1&lt;&lt;k%n})for k in range(n&lt;&lt;n)}

Duplicate edges are removed by making a set of sets, except Python requires that set elements are hashable, so they are converted to a tuples. Note that the sets {a,b} and {b,a} are equal and convert to the same tuple. xsot saved 2 bytes with n&lt;&lt;n.
This can be cut down to 49 bytes if strings of sets are an OK output format
lambda n:{`{k/n,k/n^1&lt;&lt;k%n}`for k in range(n&lt;&lt;n)}

which gives output like
set(['set([1, 3])', 'set([2, 3])', 'set([0, 2])', 'set([0, 1])'])


lambda n:[(k/n,k/n^1&lt;&lt;k%n)for k in range(n*2**n)if k/n&amp;1&lt;&lt;k%n]

First, let's look at an older version of the solution.
lambda n:[(i,i^2**j)for i in range(2**n)for j in range(n)if i&amp;2**j]

Each a number in the interval [0,2^n) corresponds to a vertex with coordinates given by its n-bit binary strings. To vertices are adjacent if they differ in a single bit, i.e. if one is obtained from the other by xor-ing a power of 2.
This anonymous function generates all possible edges by taking every vertex and every bit position to flip. To avoid duplicating an edge in both directions, only 1's are flipped to 0's.
In the more golfed solution, k is used to encode both i and j via k=n*i+j, from which (i,j) can be extracted as (k/n,k%n). This saves a loop in the comprehension. The powers of 2 are done as 1&lt;&lt; to have the right operator precedence.
An alternative approach of generating each pair of vertices and checking if they are a bit flip apart seems longer (70 bytes):
lambda n:[(i,x)for i in range(2**n)for x in range(i)if(i^x)&amp;(i^x)-1&lt;1]