# Generating Maximum Length Sequence using Galois LFSR

Started by July 4, 2009
```Hello--

I am trying to generate a Maximum Length Sequence (MLS) using a Linear
Feedback Shift Register (LFSR).  Assuming that the taps have been loaded
into variable "taps," and that "lfsr" is the variable being used
in the
shift register computation, I am trying to generate the sequence using
code similar to that found on Wikipedia at the page
http://en.wikipedia.org/wiki/Linear_feedback_shift_register.  Here is a
code snippet:

// generate sequence
for(int i = 0; i < L; i++)
{
lfsr = (lfsr >> 1) ^ (-(lfsr & 1u) & taps);

// if I have an output array with L elements,
// what do I set it equal to in this loop?
// output[i] = ??
}

However, how do I determine the output of the LFSR?  For m = 12, it
follows that L = (2^m) - 1 = 4095.  How would I generate an MLS sequence
with a length of 4095 using the above code?  The loop should repeat 4095
times.  What is the "output stream"? Is it simply the lowest bit in the
sequence?  Is this the bit that is fed back into the sequence after
passing through the logic gates?

Is there a way to test if my output is an MLS?  Does the Galois LFSR
generate exactly the same output as a Fibonacci LFSR?

```
```
Nicholas Kinar wrote:

> Hello--
>
> I am trying to generate a Maximum Length Sequence (MLS) using a Linear
> Feedback Shift Register (LFSR).  Assuming that the taps have been loaded
> into variable "taps," and that "lfsr" is the variable being
used in the
>  shift register computation, I am trying to generate the sequence using
> code similar to that found on Wikipedia at the page
> http://en.wikipedia.org/wiki/Linear_feedback_shift_register.  Here is a
> code snippet:
>
>
> // generate sequence
> for(int i = 0; i < L; i++)
> {
>     lfsr = (lfsr >> 1) ^ (-(lfsr & 1u) & taps);
>
>     // if I have an output array with L elements,
>     // what do I set it equal to in this loop?
>     // output[i] = ??
> }
>
>
> However, how do I determine the output of the LFSR?  For m = 12, it
> follows that L = (2^m) - 1 = 4095.  How would I generate an MLS sequence
> with a length of 4095 using the above code?  The loop should repeat 4095
> times.  What is the "output stream"? Is it simply the lowest bit in
the
> sequence?  Is this the bit that is fed back into the sequence after
> passing through the logic gates?
>
> Is there a way to test if my output is an MLS?  Does the Galois LFSR
> generate exactly the same output as a Fibonacci LFSR?
>
>
>
```
```don't mind Vlad.  he's sorta mean.  at least with inferiors (i'm sure
glad i don't work for him).

On Jul 4, 12:55&#2013266080;am, Nicholas Kinar <n.ki...@usask.ca> wrote:
>
> I am trying to generate a Maximum Length Sequence (MLS) using a Linear
> Feedback Shift Register (LFSR). &#2013266080;Assuming that the taps have
> into variable "taps," and that "lfsr" is the variable being
used in the
> &#2013266080; shift register computation, I am trying to generate the
sequence using
> code similar to that found on Wikipedia at the
pagehttp://en.wikipedia.org/wiki/Linear_feedback_shift_register.
&#2013266080;Here is a
> code snippet:
>
> // generate sequence
> for(int i = 0; i < L; i++)
> {
> &#2013266080; &#2013266080; &#2013266080; &#2013266080; lfsr =
(lfsr >> 1) ^ (-(lfsr & 1u) & taps);

i dunno what the minus sign is for.  i don't think it belongs there.
normally, if the shift register state is zero (all zero bits), the
shift and conditional XOR operation should return you to the same zero
state.  if it's an MLS, that zero state is the *only* state not in the
MLS.  i.e. if it's an MLS *any* non-zero state is an acceptable
initial state and *all* other non-zero states will occur before you
get back to your original non-zero state.

> &#2013266080; &#2013266080; &#2013266080; &#2013266080; // if I
have an output array with L elements,
> &#2013266080; &#2013266080; &#2013266080; &#2013266080; // what
do I set it equal to in this loop?
> &#2013266080; &#2013266080; &#2013266080; &#2013266080; //
output[i] = ??
>
> }
>

your L "elements" are single bits.  the "S" of MLS is a sequence
of
*bits*, not of words.

> However, how do I determine the output of the LFSR?

pick any bit of the shift register (your "lfsr") and use it.  i
usually pick the same bit that falls off the edge

> &#2013266080;For m = 12, it
> follows that L = (2^m) - 1 = 4095. &#2013266080;How would I generate an MLS
sequence
> with a length of 4095 using the above code?

you need to select your taps word correctly.  first of all, for m=12,
then taps > 2048, but there is more to it.  somewhere on the web,
there is a resource of "primitive polynomials".  find that and use one
of them (ignore the least significant bit, which is always 1, the
other m bits of the primitive polynomial are the m bits of your taps
word).  for m=12, you could write a program that guesses at a 12-bit
word, try it out (see if it goes through all 2^12 - 1 non-zero
states), and if it fails, try another word for taps.

> &#2013266080;The loop should repeat [after] 4095 times.

yes.  that's how you confirm it's MLS. start with a state of 0x0001
and count how many iterations it takes to get back to 0x0001.  if it's
4095 and your "taps" word has zeros in the bits other than the least-
significant 12 bits, then it's an MLS

> &#2013266080;What is the "output stream"? Is it simply the lowest
bit in the
> sequence?

sure.  any bit will do.

> &#2013266080;Is this the bit that is fed back into the sequence after
> passing through the logic gates?

it's the bit that triggers the XOR operation in your code, so in that
way, it feeds back.

> Is there a way to test if my output is an MLS? &#2013266080;Does the Galois
LFSR
> generate exactly the same output as a Fibonacci LFSR?

fuck if i know.

r b-j
```
```robert bristow-johnson  <rbj@audioimagination.com> wrote:

>On Jul 4, 12:55&#2013266080;am, Nicholas Kinar <n.ki...@usask.ca>
wrote:

>> for(int i = 0; i < L; i++)

>>    lfsr = (lfsr >> 1) ^ (-(lfsr & 1u) & taps);

>i dunno what the minus sign is for.  i don't think it belongs there.

Looks right to me.  Read the code again.

Steve
```
```robert bristow-johnson  <rbj@audioimagination.com> wrote:

>you need to select your taps word correctly.  first of all, for m=12,
>then taps > 2048, but there is more to it.  somewhere on the web,
>there is a resource of "primitive polynomials".  find that and use
one
>of them (ignore the least significant bit, which is always 1, the
>other m bits of the primitive polynomial are the m bits of your taps
>word).  for m=12, you could write a program that guesses at a 12-bit
>word, try it out (see if it goes through all 2^12 - 1 non-zero
>states), and if it fails, try another word for taps.

Here is the simplest one for m=12:

0x05eb

It is probably bit reversed from the OP's set-up,
The leadign coefficient of one is deleted from
this constant.

(Unlike the OP, I always put the LS term of the polynomial
in the lsb of a binary word, but after looking at OP's
code it seems more compact to do it his way.)

Steve
```
```Hi Robert--

Thank you for your reply!  The code that I found on Wikipedia was
different enough to make me wonder exactly what was happening.  There's
additional reference code for MLS sequence generation that can be found
here:

http://libmls0.sourceforge.net/

> your L "elements" are single bits.  the "S" of MLS is a
sequence of
> *bits*, not of words.
>

>
> pick any bit of the shift register (your "lfsr") and use it.  i
> usually pick the same bit that falls off the edge
>

>
> you need to select your taps word correctly.  first of all, for m=12,
> then taps > 2048, but there is more to it.  somewhere on the web,
> there is a resource of "primitive polynomials".  find that and use
one
> of them (ignore the least significant bit, which is always 1, the
> other m bits of the primitive polynomial are the m bits of your taps
> word).  for m=12, you could write a program that guesses at a 12-bit
> word, try it out (see if it goes through all 2^12 - 1 non-zero
> states), and if it fails, try another word for taps.
>

There are a few resources on "primitive polynomials" currently on the
web, but perhaps you are thinking about the following paper?

Stahnke, W. 1973. "Primitive Binary Polynomials," Mathematics of
Computation 27(124): 977-980.

The paper gives a table of exponents of terms of primitive binary
polynomial terms up to m = 168, where the length of the sequence is L =
(2^m) - 1.  I've tested this table up to m = 30, and it seems to
generate an MLS.

>>  The loop should repeat [after] 4095 times.
>
> yes.  that's how you confirm it's MLS. start with a state of 0x0001
> and count how many iterations it takes to get back to 0x0001.  if it's
> 4095 and your "taps" word has zeros in the bits other than the
least-
> significant 12 bits, then it's an MLS
>

You know, that's an interesting way of checking whether something is an MLS.

>>  What is the "output stream"? Is it simply the lowest bit in the
>> sequence?
>
> sure.  any bit will do.
>

>>  Is this the bit that is fed back into the sequence after
>> passing through the logic gates?
>
> it's the bit that triggers the XOR operation in your code, so in that
> way, it feeds back.
>

>> Is there a way to test if my output is an MLS?  Does the Galois LFSR
>> generate exactly the same output as a Fibonacci LFSR?
>

Perhaps a quick test of whether the sequence is an MLS could be found here:

http://www.libinst.com/mlsmeas.htm

A shifted version of the sequence is simply lined up and multiplied.
There's probably also a theoretical way of proof regarding whether
something is an MLS, but that can be left to the mathematicians.  It
just makes me wonder how you would test your own code.

Thanks, Robert.
```
```Thanks, Steve!

> Here is the simplest one for m=12:
>
> 0x05eb
>
> It is probably bit reversed from the OP's set-up,
> The leadign coefficient of one is deleted from
> this constant.
>

Yes, it appears that the Galois shift register requires bit reversed
versions.

> (Unlike the OP, I always put the LS term of the polynomial
> in the lsb of a binary word, but after looking at OP's
> code it seems more compact to do it his way.)
>
> Steve

Yeah, it's kind of an interesting way of doing things.
```
```Nicholas Kinar  <n.kinar@usask.ca> wrote:

>Thanks, Steve!

No problemo.

>> Here is the simplest one for m=12:
>>
>> 0x05eb
>>
>> It is probably bit reversed from the OP's set-up,
>> The leading coefficient of one is deleted from
>> this constant.

>Yes, it appears that the Galois shift register requires bit reversed
>versions.

The way you're doing it, yes.

>> (Unlike the OP, I always put the LS term of the polynomial
>> in the lsb of a binary word, but after looking at OP's
>> code it seems more compact to do it his way.)

>Yeah, it's kind of an interesting way of doing things.

It's good to keep track of the algebra.  For a generating
polynomial G of degree m over GF(2), you are computing the remainder

x^n mod G

for a series of values n=0, n=1,....

This remainder is a polynomial over GF(2) of (at most) degree m-1.
How you represent it in a bit field is totally up to you,
and your method is as correct as any.

Steve
```
```On Jul 4, 8:02 pm, spop...@speedymail.org (Steve Pope) wrote:
> robert bristow-johnson  <r...@audioimagination.com> wrote:
>
> >On Jul 4, 12:55 am, Nicholas Kinar <n.ki...@usask.ca> wrote:
> >> for(int i = 0; i < L; i++)
> >>    lfsr = (lfsr >> 1) ^ (-(lfsr & 1u) & taps);
> >i dunno what the minus sign is for.  i don't think it belongs there.
>
> Looks right to me.  Read the code again.

yeah, you're right.  the code is correct.  i almost went and changed
it at Wikipedia, but thought better.

On Jul 5, 10:26 am, Nicholas Kinar <n.ki...@usask.ca> wrote:
>
>
> > your L "elements" are single bits.  the "S" of MLS is
a sequence of
> > *bits*, not of words.
>
> > pick any bit of the shift register (your "lfsr") and use it.  i
> > usually pick the same bit that falls off the edge

i forgot to mention that to get a virtually zero-mean MLS signal for
the purposes of measuring impulse response (or just for some noise
that is guaranteed to be "white" in some sense of the word), then you
take that bit, (we'll call it a[n] where n is discrete time) which is
0 or 1 and create this signed MLS signal:

x[n] = (-1)^a[n]

so, it's nearly zero mean.  there are L/2 ones and L/2-1 zeros in a[n]
(because the zero state in the shift register is the only state
missing and if it were included, there would be an equal L/2 ones and
zeros, but it's not).  then there is one more -1 than +1 in x[n] over
L samples.  the mean is -1/L

this way, when you autocorrelate, you get

Rx[k] = MEAN{ x[n] * x[n+k] }

( "*" means simple multiplication.)

which is

Rx[k] = MEAN{ (-1)^a[n] * (-1)^a[n+k] }

= MEAN{ (-1)^(a[n] + a[n+k]) }

which happens to be the same as

Rx[k] = MEAN{ (-1)^(a[n] XOR a[n+k]) }

now it turns out that if you take that periodic MLS bit sequence, copy
it, circularly rotate one of the copies by k samples, and XOR the two
MLSes together, it turns out that you can show that the same recursive
generating algorithm that produces the MLS also governs that XOR
product of two with a lag in between.  except, we know for MLS, the
zero state gets mapped back to the zero state.  for an MLS, those are
the two loops; zero goes straight to zero, but any non-zero state of
the shift register goes through the whole MLS, which are all the other
non-zero states, before it gets back to your original non-zero state.
so that MEAN above for k not a multiple of L (nor zero), is -1/L.  but
the mean for k=0 or any other integer multiple of L is the mean of a
sequence of ones.

> > you need to select your taps word correctly.  first of all, for m=12,
> > then taps > 2048, but there is more to it.  somewhere on the web,
> > there is a resource of "primitive polynomials".  find that and
use one
> > of them (ignore the least significant bit, which is always 1, the
> > other m bits of the primitive polynomial are the m bits of your taps
> > word).  for m=12, you could write a program that guesses at a 12-bit
> > word, try it out (see if it goes through all 2^12 - 1 non-zero
> > states), and if it fails, try another word for taps.
>
> There are a few resources on "primitive polynomials" currently on
the
> web, but perhaps you are thinking about the following paper?
>
> Stahnke, W. 1973. "Primitive Binary Polynomials," Mathematics of
> Computation 27(124): 977-980.
>
> The paper gives a table of exponents of terms of primitive binary
> polynomial terms up to m = 168, where the length of the sequence is L =
> (2^m) - 1.  I've tested this table up to m = 30, and it seems to
> generate an MLS.

right, and you will see m+1 "coefficients" (that are either 0 or 1) in
the primitive polynomial.  but the two end bits are always 1 and the
bit-reverse sequence can be shown to also be a primitive polynomial.
so, depending on which direction you're shifting, you leave off one of
the bits.  you were shifting right so the "1" bit in the left is kept,
and you use the m most-significant bits on your word "taps" (right
justified).

> >>  The loop should repeat [after] 4095 times.
>
> > yes.  that's how you confirm it's MLS. start with a state of 0x0001
> > and count how many iterations it takes to get back to 0x0001.  if it's
> > 4095 and your "taps" word has zeros in the bits other than the
least-
> > significant 12 bits, then it's an MLS
>
> You know, that's an interesting way of checking whether something is an MLS.

it's similar to a recent experience i had where i missed that the
simplest and obvious way to display the step response of a filter (IIR
or FIR), assuming you know the coefficients, is to simple bang a step
input to a simulated filter with the same coefficients.  somebody at
music-dsp had to point that out to me, where i was thinking of doing
Heaviside partial fraction expansion of the transfer function times 1/
(z-1), inverse Z transform, and getting it from there.

> Thanks, Robert.

FWIW

BTW, if you Google "Maximum Length Sequence" and look at the link
right below the one you referred to, there is a Little MLS Tutorial
that explains the math behind most of this.  it doesn't derive
primitive polynomials, but does prove how MLS can get your time-
aliased impulse response from where you can get your frequency
response.

--

r b-j                  rbj@audioimagination.com

"Imagination is more important than knowledge."

```
```robert bristow-johnson  <rbj@audioimagination.com> wrote:

>yeah, you're right.  the code is correct.  i almost went and changed
>it at Wikipedia, but thought better.

example, that one is calculating a remainder.  I do not
know anyone who uses the terminology Fibonacci LFSR,
and the polynomial in the "Fibonacci" section is reversed
from any normal usage.  There is no mathematical difference
between the sequences described in the "Fibonacci" and
to the "conventional LFSR" as though it is something different
than the "Galois" one, which it calls "alternate".

Maybe I will input a complaint on the talk page.

Steve
```
```On Jul 9, 9:31&#2013266080;am, robert bristow-johnson
<r...@audioimagination.com>
wrote:
> On Jul 8, 10:23&#2013266080;am, Clay <c...@claysturner.com> wrote:
>
>
>
>
>
> > On Jul 8, 8:36&#2013266080;am, robert bristow-johnson
<r...@audioimagination.com>
> > wrote:
>
> > > On Jul 7, 2:08&#2013266080;pm, Clay <c...@claysturner.com>
wrote:
>
> > > ...
>
> > > > When I think about generating a MAXIMUM length sequence, I
woudn't
> > > > even consider something so short as just 4095 states before
repeating.
> > > > I often use a Mersenne Twister, which has a period of 2^19937.
Yes
> > > > that's correct.
>
> > > how does it work? &#2013266080;i cannot imagine it having a
period of 2^m without
> > > it having a state that is at least m bits wide. &#2013266080;does
it stroll
> > > through 19937 bits repeatedly?
>
> > Of course it has 19937 bits in memory - that is only 624 words (32
> > bits each). This may be an issue in a memory limited application, but
> > on a general computer not a problem at all. It uses a basic shift
> > register with feedback, but unlike common schemes, the feed back is
> > "twisted."
>
> > The Wiki article may help describe it for you. The pseudo code is
> > pretty simple to follow
>
> >http://en.wikipedia.org/wiki/Mersenne_twister
>
> so a pseudo-statement like:
>
> "int y := 32nd bit of(MT[i]) + last 31 bits of(MT[(i+1) mod 624])"
>
> means that the 32nd bit of MT[i] remains in the 32nd place in y (the
> MSB) and the lower 31 bits of (MT[(i+1) mod 624] go into the lower 31
> bits of y? &#2013266080;and does the multiplication in
initializeGenerator()
> involve a (long long) or (unsigned long long) type?
>
> it also appears that extractNumber() is essentially a table lookup
> (with a pretty damn big table) in that it is only a function of MT
> [index] and it does not affect the generation of MT[]. &#2013266080;is that
bit
> scrambling really necessary? &#2013266080;then, if it *is* necessary (the
words
> would not look random enough without the bit scrambling), the question
> is, is it sufficient?
>
> just curious (and ignorant).
>
> r b-j- Hide quoted text -
>
> - Show quoted text -

Hello Robert,

1st we note that 19937 bits is 623 complete 32 bit words plus 1 extra
bit.

Basically the MT does bit scrambling on each of the 623 words ontil
which time you shift down (but not simple linear shifting) the whole
19937 bits.

I would say that both of these stages are necessary to ensure the
statistical properties.

One of the problems with standard linear congruence modulo generators,
was you could take pairs of numbers and then use one as an x value and
the other as a y value and plot a spot at (x,y). With an LCM
generator, the plotted spots would all appear on straight lines. So
there was a strong sample to sample correlation. And a lot has been
done to make scramblers to remove this correlation.

The MT picks consecutive numbers from different points in the 623
word chain. So this removes the sample to sample correlation. Once you
have made 623 picks, then  a new 623 word chain is formed.

Note the efficiency in this process, that the overall data chain
"shifting" occures once every 623 calls! Common "C"
implementations of
the MT actually run faster than the system supplied rand() function.
And yes unsigned 32 bit numbers are being used.

In the authors' seminal paper on the MT, the paper's title alludes to
the lack of sample to sample correlation for sets of up to 623 - 32
bit samples!

IHTH,

Clay

```
```On Jul 8, 10:23=A0am, Clay <c...@claysturner.com> wrote:
> On Jul 8, 8:36=A0am, robert bristow-johnson <r...@audioimagination.com>
> wrote:
>
>
>
> > On Jul 7, 2:08=A0pm, Clay <c...@claysturner.com> wrote:
>
> > ...
>
> > > When I think about generating a MAXIMUM length sequence, I woudn't
> > > even consider something so short as just 4095 states before
repeating=
.
> > > I often use a Mersenne Twister, which has a period of 2^19937. Yes
> > > that's correct.
>
> > how does it work? =A0i cannot imagine it having a period of 2^m without
> > it having a state that is at least m bits wide. =A0does it stroll
> > through 19937 bits repeatedly?
>
>
> Of course it has 19937 bits in memory - that is only 624 words (32
> bits each). This may be an issue in a memory limited application, but
> on a general computer not a problem at all. It uses a basic shift
> register with feedback, but unlike common schemes, the feed back is
> "twisted."
>
> The Wiki article may help describe it for you. The pseudo code is
> pretty simple to follow
>
> http://en.wikipedia.org/wiki/Mersenne_twister

so a pseudo-statement like:

"int y :=3D 32nd bit of(MT[i]) + last 31 bits of(MT[(i+1) mod 624])"

means that the 32nd bit of MT[i] remains in the 32nd place in y (the
MSB) and the lower 31 bits of (MT[(i+1) mod 624] go into the lower 31
bits of y?  and does the multiplication in initializeGenerator()
involve a (long long) or (unsigned long long) type?

it also appears that extractNumber() is essentially a table lookup
(with a pretty damn big table) in that it is only a function of MT
[index] and it does not affect the generation of MT[].  is that bit
scrambling really necessary?  then, if it *is* necessary (the words
would not look random enough without the bit scrambling), the question
is, is it sufficient?

just curious (and ignorant).

r b-j

```
```On Jul 8, 8:36&#2013266080;am, robert bristow-johnson
<r...@audioimagination.com>
wrote:
> On Jul 7, 2:08&#2013266080;pm, Clay <c...@claysturner.com> wrote:
>
>
>
> ...
>
> > When I think about generating a MAXIMUM length sequence, I woudn't
> > even consider something so short as just 4095 states before repeating.
> > I often use a Mersenne Twister, which has a period of 2^19937. Yes
> > that's correct.
>
> how does it work? &#2013266080;i cannot imagine it having a period of 2^m
without
> it having a state that is at least m bits wide. &#2013266080;does it
stroll
> through 19937 bits repeatedly?
>
> r b-j

Hello Robert,

Of course it has 19937 bits in memory - that is only 624 words (32
bits each). This may be an issue in a memory limited application, but
on a general computer not a problem at all. It uses a basic shift
register with feedback, but unlike common schemes, the feed back is
"twisted."

The Wiki article may help describe it for you. The pseudo code is
pretty simple to follow

http://en.wikipedia.org/wiki/Mersenne_twister

The paper "Mersenne twister: a 623-dimensionally equidistributed
uniform pseudo-random number generator" is the one where  Makoto
Matsumoto  and Takuji Nishimura first describe the method.

You may find the paper here:

http://www.math.sci.hiroshima-u.ac.jp/~m-mat/MT/ARTICLES/mt.pdf

Enjoy!

Clay

```
```On Jul 7, 2:08&#2013266080;pm, Clay <c...@claysturner.com> wrote:
>
...
>
> When I think about generating a MAXIMUM length sequence, I woudn't
> even consider something so short as just 4095 states before repeating.
> I often use a Mersenne Twister, which has a period of 2^19937. Yes
> that's correct.

how does it work?  i cannot imagine it having a period of 2^m without
it having a state that is at least m bits wide.  does it stroll
through 19937 bits repeatedly?

r b-j

```
```>
> When I think about generating a MAXIMUM length sequence, I woudn't
> even consider something so short as just 4095 states before repeating.
> I often use a Mersenne Twister, which has a period of 2^19937. Yes
> that's correct. And it is easy to implement and free code exist on the
> web for it. The inventor (Makoto Matsumoto )  put it (code and
> algorithm) into the public domain, so you are free to use it without
> worry about IP issues. And statistically, the Twister's output fairs
> better on statistics tests than stuff from a simple loopback of a
> shift register with some exclusive ors. You may want to look into
> this.
>
> Clay

Thanks Clay; that's a really neat algorithm!  It's certainly something
to think about when doing statistical tests.  The large period is just
amazing.

```
```On Jul 7, 3:42&#2013266080;pm, spop...@speedymail.org (Steve Pope) wrote:
> Clay &#2013266080;<c...@claysturner.com> wrote:
> >On Jul 7, 2:53&#2013266080;pm, spop...@speedymail.org (Steve Pope)
wrote:
> >> For example, with a maximal-length sequence formed from a
> >> a prescribed point in the middle of the sequence, or to
> >> examine the register contents and figure out where you
> >> are in the sequence. &#2013266080;Can a Mersenne twister do that?
> >Yes you can start the twister anywhere you want in the sequence.
> >Likewise a small number (compared to the length) of observations can
> >tell you the starting state. It is not for cryptographic use because
> >of these properties. I'm not sure how one would measure the distance a
> >current state is away from a defined starting state without clocking
> >it through all of the intervening states. Here its long period makes
> >it take a while!
>
> Thanks.
>
> I've never used the Twister acutally. &#2013266080;If I need something
> stronger or more random than an LFSR I usually go with DES or AES, or
> something derived from them. &#2013266080;
>
> Steve

I use it quite a bit. It works very well as a random number generator
for games. Also I use it for all sorts of monte carlo testing. It is
even good for monte carlo integration.

Clay
```
```Clay  <clay@claysturner.com> wrote:

>On Jul 7, 2:53&#2013266080;pm, spop...@speedymail.org (Steve Pope) wrote:

>> For example, with a maximal-length sequence formed from a
>> a prescribed point in the middle of the sequence, or to
>> examine the register contents and figure out where you
>> are in the sequence. &#2013266080;Can a Mersenne twister do that?

>Yes you can start the twister anywhere you want in the sequence.
>Likewise a small number (compared to the length) of observations can
>tell you the starting state. It is not for cryptographic use because
>of these properties. I'm not sure how one would measure the distance a
>current state is away from a defined starting state without clocking
>it through all of the intervening states. Here its long period makes
>it take a while!

Thanks.

I've never used the Twister acutally.  If I need something
stronger or more random than an LFSR I usually go with DES or AES, or
something derived from them.

Steve
```
```On Jul 7, 2:53&#2013266080;pm, spop...@speedymail.org (Steve Pope) wrote:
> Clay &#2013266080;<c...@claysturner.com> wrote:
> >On Jul 4, 12:55&#2013266080;am, Nicholas Kinar <n.ki...@usask.ca>
wrote:
> >> I am trying to generate a Maximum Length Sequence (MLS) using a
Linear
> >> Feedback Shift Register (LFSR).
> >When I think about generating a MAXIMUM length sequence, I woudn't
> >even consider something so short as just 4095 states before repeating.
> >I often use a Mersenne Twister, which has a period of 2^19937. Yes
> >that's correct. And it is easy to implement and free code exist on the
> >web for it.
>
> Turns out that engineering requirements come in all forms
> and sometimes you do not want, or cannot use, the more
> elaborate solution.
>
> For example, with a maximal-length sequence formed from a
> a prescribed point in the middle of the sequence, or to
> examine the register contents and figure out where you
> are in the sequence. &#2013266080;Can a Mersenne twister do that?
>
> Steve

Hello Steve,

Yes you can start the twister anywhere you want in the sequence.
Likewise a small number (compared to the length) of observations can
tell you the starting state. It is not for cryptographic use because
of these properties. I'm not sure how one would measure the distance a
current state is away from a defined starting state without clocking
it through all of the intervening states. Here its long period makes
it take a while!

Clay

```
```Clay  <clay@claysturner.com> wrote:

>On Jul 4, 12:55&#2013266080;am, Nicholas Kinar <n.ki...@usask.ca>
wrote:

>> I am trying to generate a Maximum Length Sequence (MLS) using a Linear
>> Feedback Shift Register (LFSR).

>When I think about generating a MAXIMUM length sequence, I woudn't
>even consider something so short as just 4095 states before repeating.
>I often use a Mersenne Twister, which has a period of 2^19937. Yes
>that's correct. And it is easy to implement and free code exist on the
>web for it.

Turns out that engineering requirements come in all forms
and sometimes you do not want, or cannot use, the more
elaborate solution.

For example, with a maximal-length sequence formed from a
a prescribed point in the middle of the sequence, or to
examine the register contents and figure out where you
are in the sequence.  Can a Mersenne twister do that?

Steve
```
```On Jul 4, 12:55&#2013266080;am, Nicholas Kinar <n.ki...@usask.ca> wrote:
> Hello--
>
> I am trying to generate a Maximum Length Sequence (MLS) using a Linear
> Feedback Shift Register (LFSR). &#2013266080;Assuming that the taps have
> into variable "taps," and that "lfsr" is the variable being
used in the
> &#2013266080; shift register computation, I am trying to generate the
sequence using
> code similar to that found on Wikipedia at the
pagehttp://en.wikipedia.org/wiki/Linear_feedback_shift_register.
&#2013266080;Here is a
> code snippet:
>
> // generate sequence
> for(int i = 0; i < L; i++)
> {
> &#2013266080; &#2013266080; &#2013266080; &#2013266080; lfsr =
(lfsr >> 1) ^ (-(lfsr & 1u) & taps);
>
> &#2013266080; &#2013266080; &#2013266080; &#2013266080; // if I
have an output array with L elements,
> &#2013266080; &#2013266080; &#2013266080; &#2013266080; // what
do I set it equal to in this loop?
> &#2013266080; &#2013266080; &#2013266080; &#2013266080; //
output[i] = ??
>
> }
>
> However, how do I determine the output of the LFSR? &#2013266080;For m =
12, it
> follows that L = (2^m) - 1 = 4095. &#2013266080;How would I generate an MLS
sequence
> with a length of 4095 using the above code? &#2013266080;The loop should
repeat 4095
> times. &#2013266080;What is the "output stream"? Is it simply the
lowest bit in the
> sequence? &#2013266080;Is this the bit that is fed back into the sequence
after
> passing through the logic gates?
>
> Is there a way to test if my output is an MLS? &#2013266080;Does the Galois
LFSR
> generate exactly the same output as a Fibonacci LFSR?

Hello Nicholas,

When I think about generating a MAXIMUM length sequence, I woudn't
even consider something so short as just 4095 states before repeating.
I often use a Mersenne Twister, which has a period of 2^19937. Yes
that's correct. And it is easy to implement and free code exist on the
web for it. The inventor (Makoto Matsumoto )  put it (code and
algorithm) into the public domain, so you are free to use it without
worry about IP issues. And statistically, the Twister's output fairs
better on statistics tests than stuff from a simple loopback of a
shift register with some exclusive ors. You may want to look into
this.

Clay
```