Maximum xor of a range of numbers

Question

I am grappling with this problem Codeforces 276D. Initially I used a brute force approach which obviously failed for large inputs(It started when inputs were 10000000000 20000000000). In the tutorials Fcdkbear(turtor for the contest) talks about a dp solution where a state is d[p][fl1][fr1][fl2][fr2].

Further in tutorial

We need to know, which bits we can place into binary representation of number а in p-th position. We can place 0 if the following condition is true: p-th bit of L is equal to 0, or p-th bit of L is equal to 1 and variable fl1 shows that current value of a is strictly greater then L. Similarly, we can place 1 if the following condition is true: p-th bit of R is equal to 1, or p-th bit of R is equal to 0 and variable fr1 shows that current value of a is strictly less then R. Similarly, we can obtain, which bits we can place into binary representation of number b in p-th position.

This is going over my head as when ith bit of L is 0 then how come we can place a zero in a's ith bit. If L and R both are in same bucket(2^i'th boundary like 16 and 24) we will eventually place a 0 at 4th whereas we can place a 1 if a = 20 because i-th bit of R is 0 and a > R. I am wondering what is the use of checking if a > L or not.
In essence I do not get the logic of

1. What states are
2. How do we recur

I know that might be an overkill but could someone explain it in descriptive manner as editorial is too short to explain anything.

I have already looked in here but suggested solution is different from one given in editorial. Also I know this can be solved with binary search but I am concerned with DP solution only

Answer 1

If I got the problem right: Start to compare the bits of l and r from left (MSB) to right(LSB). As long as these bits are equal there is no freedom of choice, the same bits must appear in a and b. the first bit differing must be 1 in r and 0 in l. they must appear also in a (0) and b(1). from here you can maximise the XOR result. simply use zeros for b an ones for a. that gives a+1==b and the xor result is a+b which is always 2^n-1.

Answer 2

I'm not following the logic as written above but the basic idea is to look bit by bit.

If L and R have different values in the same bit position then we have already found candidates that would maximize the xor'd value of that position (0 xor 1 = 1 xor 0 = 1). The other case to consider is whether the span of R-L is greater than the position value of that bit. If so then there must be two different values of A and B falling between L and R where that bit position has opposite values (as well as being able to generate any combinations of values in the lower bits.)