The Art of Computer Programming, Vol 4, Fascicle 2 typo?

Answer 1

A list of known typos can be found on the errata page:

http://www-cs-faculty.stanford.edu/~knuth/taocp.html

Your reported typo is not there. If it really is a typo, you might be eligible for a cash reward from Knuth himself.

Answer 2

This can be resolved by using the convention that (...)₂ represents a bitwise representation. (1^j+1)₂ then consists solely of j+1 ones, rather than referring to an exponentiation. You can see this convention explained more explicitly in TAOCP Volume 4 Fascicle 1 at page 8, for example:

If x is almost any nonzero 2-adic integer, we can write its bits in the form

x = (g01^a10^b)₂

in other words, x consists of some arbitrary (but infinite) binary string g, followed by a 0, which is followed by a+1 ones and followed by b zeros, for some a >= 0 and b >= 0.

[I have substituted the symbol alpha by g to save encoding problems]

Going back to your original query; k ⊕(1^j+1)₂ is equated with k ⊕ (2^j+1 - 1) implying that (1^j+1)₂ = (2^j+1 - 1): this holds because the left-hand side is the integer whose significant bits are j+1 (contiguous) ones; the right-hand side is an exponentiation. For example, with j =3:

(1⁴)₂ = (1111)₂ = (2³ - 1)

Hope that helps.