RFC 70
Note on Padding
A Note on Padding ✨
okay bestie, this is RFC 70, and it's a whole vibe 🌸 It's titled A Note on Padding and it's from the Network Working Group, written by S. Crocker at UCLA and dated 15 October 70 🎀 The title block doesn't list anything it obsoletes or updates, and it doesn't name a category or status. Iconic start, no cap 💅
The padding on a message is a string of the form 10*. For Hosts with word lengths 16, 32, 48, etc., bits long, this string is necessarily in the last word received from the Imp. For Hosts with word lengths which are not a multiple of 16 (but which are at least 16 bits long), the 1 bit will be in either the last word or the next to last word. Of course if the 1 bit is in the next to last word, the last word is all zero 🧚
Isolating the Low-Order Bit 🦋
okay so the hard part is finding where the 1 bit sits inside its word, and that's an unpleasant coding task, ngl 😭 The obvious move is to keep testing the low-order bit and shifting the word right one position whenever that bit is zero. The techniques below are way more pleasant 💖
Let W be a non-zero word, where the word length is n. Then W is of the form
x....x10....0
\__ __/\__ __/
V V
n-k-1 kwhere 0<=k<n
and the x's are arbitrary bits.
Assuming two's complement arithmetic,
W-1 = x....x01....1
_ _
-W = x....x10....0
_ _ _
W = x....x01....1By using AND, OR and exclusive OR with various pairs of these quantities, useful new forms are obtained. It's giving math glow-up 🔥
For example,
W AND W-1 xx...x00....0
\__ __/\__ __/
V V
n-k-1 kthus removing the low-order 1 bit ✨
also W AND -W =
0....010....0
__ __/__ __/
V V
n-k-1 kthus isolating the low-order bit 💅
Below, we will focus solely on this last result; however, in a particular application it may be advantageous to use a variation 🌈
Determining the Position of an Isolated Bit 🔍
so now we know how to isolate the bit, but how do we figure out its position? 🤔 The two obvious techniques are to shift repetitively with tests, like we just talked about, or to use floating normalization hardware. On the PDP-10, in particular, the JFFO instruction is made to order*. On machines with hexadecimal normalization, e.g. IBM 360's and XDS Sigma 7's, the normalization hardware may not be very convenient. So a different approach uses division and table look-up 🫶
A word with a single bit on has an unsigned integer value of 2^k for 0<=k<n. If we choose a p such that mod(2^k,p) is distinct for each 0<=k<n, we can make a table of length p which gives the correspondence between mod(2^k,p) and k. The remainder of this paper is concerned with the selection of an appropriate divisor p for each word length n 💜
*Some of the CDC machines have a "population count" instruction which gives the number of bits in a word. Note the 2^k -1 has exactly k bits on.
Example 🍓
Let n = 8 and p = 11
Then
0
mod(2, 11) = 1
1
mod(2, 11) = 2
2
mod(2, 11) = 4
3
mod(2, 11) = 8
4
mod(2, 11) = 5
5
mod(2, 11) = 10
6
mod(2, 11) = 9
7
mod(2, 11) = 7This yields a table of the form
remainder bit position
0 --
1 0
2 1
3 --
4 2
5 4
6 --
7 7
8 3
9 6
10 5Good Divisors 👑
The divisor p should be as small as possible in order to minimize the length of the table. Since the divisor must generate n distinct remainders, the divisor will certainly need to be at least n. A remainder of zero, however, can occur only if the divisor is a power of 2. If the divisor is a small power of 2, say 2^j for j < n-1, it will not generate n distinct remainders; if the divisor is a larger power of 2, the correspondence table is either 2^(n-1) or 2^n in length. We can thus rule out zero as a remainder value, so the divisor must be at least one more than the word length. This bound is in fact achieved for some word lengths 🎀
Let R(p) be the number of distinct remainders p generates when divided into successively higher powers of 2. The distinct remainders all occur for the R(p) lowest powers of 2. Only odd p are interesting and the following table gives R(p) for odd p between 1 and 21.
p R(p) p R(p)
1 1 13 12
3 2 15 4
5 4 17 8
7 3 19 18
9 6 21 6
11 10This table shows that 7, 15, 17 and 21 are useless divisors because there are smaller divisors which generate a larger number of distinct remainders. If we limit our attention to p such that p > p' => R(p) > R(p'), we obtain the following table of useful divisors for p < 100 🌸
p R(p) p R(p)
1 1 29 28
3 2 37 36
5 4 53 52
9 6 59 58
11 10 61 60
13 12 67 66
19 18 83 82
25 20Notice that 9 and 25 are useful divisors even though they generate only 6 and 20 remainders, respectively 🙈
Determination of R(p) 🧮
okay this part is lowkey a math proof and I'm obsessed 😭 If p is odd, the remainders
0
mod(2 ,p)
1
mod(2 ,p)
.
.
.will be between 1 and p-1 inclusive. At some power of 2, say 2^t, there will be a repeated remainder, so that for some k < t, 2^t = 2^k mod p. Since 2^(t+1) = 2^(k+1) mod p and 2^(t+2) = 2^(k+2) mod p, and so on, all of the distinct remainders occur for 2^0 ... 2^(t-1). Therefore, R(p)=t 💖
Next we show that
R(p)
2 = 1 mod pWe already know that 2^R(p) = 2^k mod p for some 0<=k<R(p). Let j=R(p)-k so 0<j<=R(p). Then
k+j k
2 = 2 mod p
j k k
or 2 *2 = 2 mod p
j k
or (2 -1)*2 = 0 mod pNow p does not divide 2^k because p is odd, so p must divide 2^j -1. Thus
j
2 -1 = 0 mod p
j
2 = 1 mod pSince j is greater than 0 by hypothesis and since ther is no k other than 0 less than R(p) such that
k 0
2 = 2 mod p,we must have j=R(p), or 2^R(p) = 1 mod p.
We have thus shown that for odd p, the remainders mod(2^k,p) are unique for k = 0, 1,..., R(p)-1 and then repeat exactly, beginning with
R(p)
2 = 1 mod p.We now consider even p. Let
q
p = p'*2 ,where p' is odd. For k<q, mod(2^k,p) is clearly just 2^k because 2^k <p.
For k>=q,
k q k-q
mod(2 ,p) = 2 *mod(2 ,p').From this we can see that the sequence of remainders will have an initial segment of 1, 2, ...2^(q-1) of length q, and repeating segments of length R(p'). Therefore, R(p) = q+R(p'). Since we normally expect
R(p) ~ p,even p generally will not be useful 🥲
I don't know of a direct way of choosing a p for a given n, but the previous table was generated from the following Fortran program run under the SEX system at UCLA 💀
0
CALL IASSGN('OC ',56)
1 FORMAT(I3,I5)
M=0
DO 100 K=1,100,2
K=1
L=0
20 L=L+1
N=MOD(2*N,K)
IF(N.GT.1) GO TO 20
IF(L.LE.M) GO TO 100
M=L
WRITE(56,1)K,L
100 CONTINUE
STOP
ENDFortran program to computer useful divisors
In the program, K takes on trial values of p, N takes on the values of the successive remainders, L counts up to R(p), and M remembers the previous largest R(p). Execution is quite speedy ⚡
Results from Number Theory 🎓
The quantity referred to above as R(p) is usually written Ord_p 2 and is read "the order of 2 mod p". The maximum value of Ord_p 2 is given by Euler's phi-function, sometimes called the totient. The totient of a positive integer p is the number of integers less than p which are relatively prime to p. The totient is easy to compute from a representation of p as a product of primes:
Let p = p_1^(n_1) * p_2^(n_2) ... p_k^(n_k)where the p_i are distinct primes. Then
phi(p) = (p_1 - 1) * p_1^(n_1-1) * (p_2 - 1) * p_2^(n_2-1) ... (p_k - 1) * p_k^(n_k-1)If p is prime, the totient of p is simply
phi(p) = p-1.If p is not prime, the totient is smaller 🫧
If a is relatively prime to p, then Euler's generalization of Fermat's theorem states
a^phi(m) = 1 mod p.It is this theorem which places an upper bound on Ord_p 2, because Ord_p 2 is the smallest value such that
2^(Ord_p 2) = 1 mod pMoreover it is always true that phi(p) is divisible by Ord_p 2 ✨
Acknowledgements 🫶
Bob Kahn read an early draft and made many comments which improved the exposition. Alex Hurwitz assured me that a search technique is necessary to compute R(p), and supplied the names for the quantities and theorems I uncovered. Iconic teamwork 👑
[ This RFC was put into machine readable form for entry ]
[ into the online RFC archives by Guillaume Lahaye and ]
[ John Hewes 6/97 ]📦 where this page comes from
- ietf: rfc70 1970-10
- 💖 girlified by claude-haiku-5-5 on : a language model's rewrite of the original, under its terms