How to perform orthogonal check on two circular binary strings?

Question

Say we have two circular binary strings $a = a_0a_1...a_{n-1}$ and $b = b_0b_1...b_{n-1}$ with arbitary starting point, and define a and b are orthogonal if $\sum_{i=0}^{n-1}a_ib_i = 0$. Is there a $O(nlogn)$ algorithm can tell a rotation of such circular binary string is orthogonal to another?

Answer 1

Say we have strings $a = a_0a_1...a_{n-1}$ and $b=b_0b_1...b_{n-1}$. First concatenate $a$ with itself and remove the last character. You should get a new string $a=a_0a_1...a_{n-1}a_0a_1...a_{n-2}$. Now we are interested in substrings of size $n$ in new string $a$ that are orthogonal to $b$ (this is a very classic trick to deal with cyclic strings/arrays).

Let's reinterpret them as polynomials in the following way.

For $a=a_0a_1...a_{2n-2}$ use value at position $i$ as the coefficient associated with power $i$ in a polynomial such that $P_a(x) = a_0x^0 + a_1x^1+...+a_{2n-2}x^{2n-2}$.

Now reverse $b$ and get the new string $b=b_{n-1}...b_1b_0$ and build a polynomial similarly as we did above. You get $P_b(x)=b_{n-1}x^0+...+b_1x^{n-2}+b_0x^{n-1}$.

Now let the magic happen. Multiply both polynomials into $P(x) = P_a(x) \cdot P_b(x)$ (Remember to do this fast). Wait, what ... where is the answer to the problem.

Let's see what is the $n-1$ coefficient of the new polynomial: $$P(x) = c_0x^0+c_1x^1+...+c_{3n-3}x^{3n-3}$$

Coefficient at position $n-1$ is equal to $$c_{n-1} = a_0 \cdot b_0 + a_1 \cdot b_1 + ... a_{n-1}\cdot b_{n-1}$$

Awesome, that is the dot product of original string $b$ with original string $a$ (without rotations). Keep looking at further terms of $P$. Coefficient at position $n$ is equal to $$c_{n} = a_1 \cdot b_0 + a_2 \cdot b_1 + ... + a_{n-1} \cdot b_{n-2} + a_{n} \cdot b_{n - 1}$$

But keep in mind that $a_n=a_0$ since this is just $a$ concatenated with itself. So we have the dot product of $a$ rotated one step to the left with $b$. You can see that terms of $c_{i}$ from $n - 1 \le i \le 2n - 1$ will give you the dot product of all rotations of $a$ with array $b$. Therefore, after computing $P$, you can find whether there is at least one dot product with value equals to $0$ easily.

In general polynomial multiplication is a useful representation to model convolutions of two arrays. It turns out that a lot problems can be modeled this way in areas such as: combinatoric, dynamic programming, string matching with least mismatches.

This is indeed relevant because there is a very efficient way to multiply polynomials which is via FFT in $O(n \log n)$ (or Karatsuba in $O(n^{\frac{3}{2}})$) instead of the naive $O(n^2)$ algorithm.