For a compact disk, audio is converted to digital with 16-bit samples, and is treated a stream of 8-bit bytes for storage. One simple scheme for storing this data, called direct recording, would be to represent a 1 by a land and a 0 by a pit. Instead, each byte is expanded into a 14-bit binary number. It turns out that exactly 256 (28) of the total of 16,134 (214) 14-bit numbers have at least two 0s between every pair of 1s, and these are the numbers selected for the expansion from 8 to 14 bits. The optical system detects the presence of 1s by detecting a transition for pit to land or land to pit. It detects 0s by measuring the distances between intensity changes. This scheme requires that there are no 1s in succession; hence the use of the 8-to-14 code.
The advantage of this scheme is as follows. For a given laser beam diameter, there is a minimum-pit size, regardless of how the bits are represented. With this scheme, this minimum-pit size stores 3 bits, because at least two 0s follow every 1. With direct recording, the same pit would be able to store only one bit. Considering both the number of bits stored per pit and the 8-to-14 bit expansion, which scheme stores the most bits and by what factor?