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+# Why FFT produces result in reversed bit order?
+
+## 1. What are natural order and reversed bit order?
+
+Let's look at the following table:
+
+| indices | bits |
+| - | - |
+| 0 | 000 |
+| 1 | 001 |
+| 2 | 010 |
+| 3 | 011 |
+| 4 | 100 |
+| 5 | 101 |
+| 6 | 110 |
+| 7 | 111 |
+
+When we increase the index, the last bit is the most modified bit,
+
+$$
+\quad \downarrow \\
+101
+$$
+
+while the first bit is the least modified bit.
+
+$$
+\downarrow \quad \\
+101
+$$
+
+The bit in the middle is modified faster than the first bit, but slower than the last bit.
+
+However, in reversed bit order, the first bit is the most modified bit, and the last bit is the least modified bit.
+
+| indices | bits |
+| - | - |
+| 0 | 000 |
+| 1 | 100 |
+| 2 | 010 |
+| 3 | 110 |
+| 4 | 001 |
+| 5 | 101 |
+| 6 | 011 |
+| 7 | 111 |
+
+This is the most modified bit.
+
+$$
+\downarrow \quad \\
+101
+$$
+
+This is the least modified bit.
+
+$$
+\quad \downarrow \\
+101
+$$
+
+Since reversed bit order is reversed by bits compared to natural order, the frequency of the bit flapping is also reversed.
+
+## 2. Why FFT produces result in reversed bit order?
+
+First, we know what to do in the first step of FFT.
+
+$$
+f(x) = C_0 + C_1 x + C_2 x^2 + \cdots + C_{n-1} x^{n-1}
+$$
+
+We divide the polynomial into two parts depending on whether the index of the term is even or odd.
+
+$$
+f_e(x) = C_0 + C_2 x + C_4 x^2 + \cdots + C_{n-2} x^{n/2-1}
+$$
+
+$$
+f_o(x) = C_1 + C_3 x + C_5 x^2 + \cdots + C_{n-1} x^{n/2-1}
+$$
+
+These two polynomials satisfy the following equation:
+
+$$
+f(x) = f_e(x^2) + x f_o(x^2)
+$$
+
+We divide the index of terms simultaneously.
+
+| indices | bits |
+| - | - |
+| 0 | 000 |
+| 1 | 001 |
+| 2 | 010 |
+| 3 | 011 |
+| 4 | 100 |
+| 5 | 101 |
+| 6 | 110 |
+| 7 | 111 |
+
+into
+
+| indices | bits |
+| - | - |
+| 0 | 000 |
+| 2 | 010 |
+| 4 | 001 |
+| 6 | 011 |
+| --- | --- |
+| 1 | 100 |
+| 3 | 110 |
+| 5 | 101 |
+| 7 | 111 |
+
+We can see that we sort the terms by the first bit of their indices.
+
+In the next step, we do the same thing for the second bit. (We only focus on the even terms for simplicity.)
+
+$$
+f_{ee}(x) = C_0 + C_4 x^4 + \cdots + C_{n-4} x^{n-4}
+$$
+
+$$
+f_{eo}(x) = C_2 x^2 + C_6 x^6 + \cdots + C_{n-2} x^{n-2}
+$$
+
+Again, we divide the index of terms simultaneously.
+
+| indices | bits |
+| - | - |
+| 0 | 000 |
+| 4 | 001 |
+| --- | --- |
+| 2 | 010 |
+| 6 | 011 |
+| --- | --- |
+| 1 | 100 |
+| 5 | 101 |
+| --- | --- |
+| 3 | 110 |
+| 7 | 111 |
+
+The last step is the same.
+
+$$
+f_{eee}(x) = C_0 + C_8 x^8 + \cdots + C_{n-8} x^{n-8}
+$$
+
+$$
+f_{eeo}(x) = C_4 x^4 + C_{12} x^{12} + \cdots + C_{n-4} x^{n-4}
+$$
+
+So as indices.
+
+| indices | bits |
+| - | - |
+| 0 | 000 |
+| --- | --- |
+| 4 | 001 |
+| --- | --- |
+| 2 | 010 |
+| --- | --- |
+| 6 | 011 |
+| --- | --- |
+| 1 | 100 |
+| --- | --- |
+| 5 | 101 |
+| --- | --- |
+| 3 | 110 |
+| --- | --- |
+| 7 | 111 |
+
+And then, as we do in the end of FFT, we simply combine the results.
+
+| indices | bits |
+| - | - |
+| 0 | 000 |
+| 4 | 001 |
+| 2 | 010 |
+| 6 | 011 |
+| 1 | 100 |
+| 5 | 101 |
+| 3 | 110 |
+| 7 | 111 |
+
+This is what we got, the result in reversed bit order.
+
+In a word, we sort the terms into reversed bit order step by step unconsciously when we divide the polynomial into two parts, since we simply combine the results of the two parts without sort them back, we get the final result in reversed bit order.