Signal rate, data rate and bandwidth in digital signals
Could anyone explain in more detail the meaning of the case factor c? In practice, what does it mean to say that the signal rate depends on the data pattern?


The explanation in the text isn't very clear, and this term is not used in other texts I know of. I think what it's saying is that different messages might produce different signal spectra. For example, in a an 2-level FSK  system, a message composed of all 1's or all 0's would just be single tone, and have a very narrow bandwidth; while a message composed of alternating 1's and 0's would contain both the one-level tone and the zero-level tone (as well as a spread of frequency content related to switching between them) and produce a broader spectrum if measured on a spectrum analyzer. 


  Why does the minimum bandwidth for a digital signal equal the signal rate?


This is not correct. The minimum bandwidth for a digital signal is given by the Shannon-Hartley theorem,

\$ C = B\log_2\left(1+\frac{S}{N}\right)\$

Turned around, 

\$B = \frac{C}{\log_2\left(1+{S}/{N}\right)}\$.

Approaching this bandwidth minimum depends on making engineering trade offs between encoding scheme (which would relate to the number of bits per symbol), equalization, and error correcting codes (actually sending extra symbols to include redundant information that allows recovering the  signal even if a transmission error occurs).

A typical rule of thumb used for on-off coding in my industry (fiber optics) is that the channel bandwidth in Hz should be at least 1/2 of the baud rate. For example, a 10 Gb/s on-off-keyed transmission requires at least 5 GHz of channel bandwidth. But that is specific to the very simple coding and equalization methods used in fiber optics.


  If we set c to 1/2 in the formula for the minimum bandwidth to find Nmax (the maximum data rate for a channel with bandwidth B), and consider r to be log2(L) (where L is the number of signal levels), we get Nyquist formula. Why? What is the meaning of setting c to 1/2?


Choosing between L signal levels is equivalent to a \$\log_2(L)\$-bit digital-to-analog conversion. So it's not surprising Nyquist's formula is lurking in the shadows somewhere.