Which of the following vim operations can help me explain?

Who can help me explain the following vim operations-general Linux technology-Linux technology and application information? The following is a detailed description. The following content about vim deletion is taken from the network:

: % S = * $ = Delete the blank space at the end of the row
: G/^ s * $/d delete all empty rows
: G/str1/,/str2/d delete all the rows from the first entry containing str1 to the first entry containing str2.
: V/./,/./-1 join compressed empty rows
: G/^ $/,/./-j compressed empty rows
In the first line, I checked vim's help. = is the meaning of the last line of the line number, and there is no "/" after s. I didn't see it clearly.

The reason for deleting all empty rows in the second row is my search. It confirmed it was feasible, but I didn't understand it. I checked it: [range] g/{pattern}/[cmd], execute [cmd] (default: print) for the rows that match {pattern} in the [range ). The problem lies in this {pattern}. I think it should be a regular expression, but does ^ s * $ indicate empty rows? I tried to change it to ^ $. I just don't understand what the original expression means.

The third line is just a bit more understandable. I don't understand the four or five lines. It seems that joinspace is involved, and I don't understand it either. I hope you can help me explain it.