I've been trying to extend the xor-swap to more than two variables, say n
variables. But I've gotten nowhere that's better than 3*(n-1)
.
For two integer variables x1
and x2
you can swap them like this:
swap(x1,x2) {
x1 = x1 ^ x2;
x2 = x1 ^ x2;
x1 = x1 ^ x2;
}
So, assume you have x1
... xn
with values v1
... vn
. Clearly you can "rotate" the values by successively applying swap:
swap(x1,x2);
swap(x2,x3);
swap(x3,x4);
...
swap(xm,xn); // with m = n-1
You will end up with x1 = v2
, x2 = v3
, ..., xn = v1
.
Which costs n-1
swaps, each costing 3
xors, leaving us with (n-1)*3
xors.
Is a faster algorithm using xor and assignment only and no additional variables known?
As a partial result I tried a brute force search for N=3,4,5 and all of these agree with your formula.
Python code:
from collections import *
D=defaultdict(int) # Map from tuple of bitmasks to number of steps to get there
N=5
Q=deque()
Q.append( (tuple(1<<n for n in range(N)), 0) )
goal = (tuple(1<<( (n+1)%N ) for n in range(N)))
while Q:
masks,ops = Q.popleft()
if len(D)%10000==0:
print len(D),len(Q),ops
ops += 1
# Choose two to swap
for a in range(N):
for b in range(N):
if a==b:
continue
masks2 = list(masks)
masks2[a] = masks2[a]^masks2[b]
masks2 = tuple(masks2)
if masks2 in D:
continue
D[masks2] = ops
if masks2==goal:
print 'found goal in ',ops
raise ValueError
Q.append( (masks2,ops) )
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