CODEWITHSUNDEEP
pythonperformancealgorithmpermutation
Jernej
Jernej

Reputation: 352

efficiently converting a bijection to cycle notation

Consider the following problem. Given is an array P of length n representing a bijection. That is element 0 <= i < n is mapped to P[i].

Given that P is a permutation, it can be written in cycle notation as described for example here.

For example if

P = [16, 3, 10, 6, 5, 9, 1, 19, 13, 14, 7, 0, 2, 8, 12, 4, 17, 15, 11, 18]

then the result in cycle notation would be 

[(16, 17, 15, 4, 5, 9, 14, 12, 2, 10, 7, 19, 18, 11, 0), (3, 6, 1), (13, 8)]

Following is a Python method accomplishing this

def toCycle(p):
    covered = cur = 0 
    perm = []    
    n = len(p)
    done = [0]*n
    while covered < n:
        while cur < n and done[cur] == -1: 
            cur+=1
        cycle = [p[cur]]
        sec = p[p[cur]]
        done[p[cur]] = -1
        done[cur] = -1
        covered+=1
        while sec != cycle[0]:
            cycle.append(sec)
            done[sec] = -1
            sec = p[sec]
            covered+=1
        perm+=[tuple(cycle)]
    return perm

The algorithm clearly runs in linear time (each element of done/p is accessed a constant number of times) and hence there is not much that can be done asymptotically.

As I have to use this method on a large number of large permutations I was wondering

Can you make it faster? Do you have any suggestions for performance improvement?

Upvotes: 2

Views: 364

Answers (1)

BurningKarl
BurningKarl

Reputation: 1196

def cycling(p, start, done):
    while not done[e]:
        done[e] = True
        e = p[e]
        yield e

def toCycle(p):
    done = [False]*len(p)
    cycles = [tuple(cycling(p, 0, done))]
    while not all(done):
        start = done.index(False)
        cycles.append(tuple(cycling(p, start, done)))
    return cycles

With your example my code runs about 30% faster than yours.

Upvotes: 1

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