CODEWITHSUNDEEP
knuthtaocp
SSteve
SSteve

Reputation: 10728

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

At the bottom of page 5 is the phrase "changes k to k ⊕ (1j+1)2". Isn't 1 to any power still 1 even in binary? I'm thinking this must be a typo. I sent an email to Dr. Knuth to report this, but I don't expect to hear back for months. In the meantime, I'm trying to figure out what this is supposed to be.

Upvotes: 1

Views: 1022

Answers (2)

mas
mas

Reputation: 1117

This can be resolved by using the convention that (...)2 represents a bitwise representation. (1j+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 = (g01a10b)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 ⊕(1j+1)2 is equated with k ⊕ (2j+1 - 1) implying that (1j+1)2 = (2j+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 = (24 - 1)

Hope that helps.

Upvotes: 4

neesh
neesh

Reputation: 5265

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.

Upvotes: 0

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