Network Working Group
Request for Comments #70
S. Crocker
UCLA
15 October 70

# A Note on Padding

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.

An unpleasant coding task is discovering the bit position of the 1 bit within its word. One obvious technique is to repeatedly test the low-order bit, shifting the word right one bit position if the low-order bit is zero. The following techniques are more pleasant.

## form

```            x....x10....0
\__ __/\__ __/

V      V
n-k-1    k
```

## Assuming two's complement arithmetic,

```            W-1    =       x....x01....1
_    _
-W    =       x....x10....0
_            _    _
W    =       x....x01....1
```

## For example,

```            W AND W-1       xx...x00....0
\__ __/\__ __/
V      V
n-k-1    k
```

## thus removing the low-order 1 bit;

```also W AND -W =           0....010....0

__ __/__ __/
V      V
n-k-1    k
```

## Determining the Position of an Isolated Bit

```The two obvious techniques for finding the bit position of an isolated
bit are to shift repetitively with tests, as above, and 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.  A different approach uses
division and table look-up.
k
A word with a single bit on has an unsigned integer value of 2  for
k
0<=k<n.  If we choose a p such that mod(2 ,p) is distinct for each

0<=k<n, we can make a table of length p which gives the correspondence
k
between mod(2 ,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
k
gives the number of bits in a word.  Note the 2 -1 has exactly k bits
```

## 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)     =    7
```

This 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                        5
```

## 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
j
2.  If the divisor is a small power of 2, say 2  for j < n-1, it will

not generate n distinct remainders; if the divisor is a larger power of
n-1     n
2, the correspondence table is either 2    or 2  in length.  We can
```

## 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     10
```

This 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     20
```

## If p is odd, the remainders

```           0
mod(2 ,p)
1
mod(2 ,p)

.
.
.
t
will be between 1 and p-1 inclusive.  At some power of 2, say 2 , there
k    t
will be a repeated remainder, so that for some k < t, 2  = 2  mod p.
t+1    k+1
Since 2    = 2    mod p
t+2    k+2
and   2    = 2    mod p

.
.
.
etc.
0    t-1
all of the distinct remainders occur for 2 ...2   .  Therefore, R(p)=t.
```

## Next we show that

```      R(p)
2     = 1 mod p
R(p)    k
We already know that 2     = 2  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 p
k                                     j
Now p does not divide 2  because p is odd, so p must divide 2 -1.  Thus

j
2 -1        =  0 mod p
j
2           =  1 mod p
```

## than 0 less than R(p) such that

```       k    0
2  = 2  mod p,

R(p)
we must have j=R(p), or 2     = 1 mod p.
k
We have thus shown that for odd p, the remainders mod(2 ,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 ,
k                     k          k
where p' is odd.  For k<q, mod(2 ,p) is clearly just 2  because 2 <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
q-1
initial segment of 1, 2, ...2    of length q, and repeating segments of
```

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
END
```

Fortran 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 2 and is
p
read "the order of 2 mod p".  The maximum value of Ord 2 is given by
p
Euler's phi-function, sometimes called the totient.  The totient of a
```

## representation of p as a product of primes:

```               n      n        n
Let p = p  1 * p  2 ... p  k
1      2        k

where the p  are distinct primes.  Then
i
k -1               k -1                 k -1
phi(p) = (p - 1) * p  1   * (p - 1) * p  2   ... (p - 1) * p  k
1        1         2        2           k        k
```

phi(p) = p-1.

## theorem states

```      phi(m)
a       = 1 mod p.

It is this theorem which places an upper bound Ord 2, because Ord 2 is
p              p
the smallest value such that

Ord 2
2   p  = 1 mod p
```

## 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.

```       [ This RFC was put into machine readable form for entry ]
[ into the online RFC archives by Guillaume Lahaye and  ]
[ John Hewes 6/97 ]
```