PROBLEM LINK:

Practice
Contest

DIFFICULTY:

Easy-Medium

PREREQUISITES:

Pigeonhole Principle, Sets

PROBLEM:

You're given N 64 bit integers such that any two successive numbers differ at exactly 1 bit. Your job is to find out if there are 4 integers such that their xor is equal to 0.

QUICK EXPLANATION:

Whenever N>= 130, it is always possible to find 4 integers of the given list such that their xor is 0. For smaller N, one could do an exhaustive search in O(N^3 log N).

DETAILED EXPLANATION:

Solution of this problem is simpler than many of you might've imagined. For all N>= 130, answer is always YES. Let's see why : Assume there are >= 130 numbers in input array A.

Let x_i = A[2i] xor A[2i+1] for i in [0, N/2) As any two successive values of A differ at exactly one bit, binary representation of each x_i has exactly 1 bit as 1 and all other bits are 0. Also as N>= 130, we've at least 65 values of x_i. Note all of them can be distinct as each x_i has only 1 bit set and all x_i are at maximum 64 bits long. So say x_j = x_k for some j and k

Then x_j xor x_k = 0
=> A(2j) xor A(2j+1) xor A(2k) xor A(2k+1) = 0
And so the answer is YES.

Small Case: What if N< 130? We could do an exhaustive search. One could try moving all 4 numbers and see if their xor is 0 or not. This takes O(N^4) time and is probably too costly. We can do this in O(N^3 log N) as well. Let's see how.

Say if there are four indices i1, i2, i3, i4 such that A[i1] xor A[i2] xor A[i3] xor A[i4] = 0.
Then A[i1] xor A[i2] xor A[i3] = A[i4]

So we could move over all three numbers of array, find their xor and check if it is present in set as well or not.

for i in 1 to N
  for j in i+1 to N
    for k in j+1 to N
      x = A[i] ^ A[j] ^ A[k]
      if A[k+1...N] contains x, return true

return false

Note : Actually 130 is only an upper bound. We don't have a case for N> 67 for which answer is No. Maybe such a case doesn't exist and our proof given above could be strengthened. If you have a case where N> 67 and answer is NO, please share it here.

SETTER'S SOLUTION:

Can be found here.

TESTER'S SOLUTION:

Can be found here.

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UVA 11898