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Date: Fri, 27 Jan 2017 18:06:25 +0200
To: Leigh <leight@gmail.com>
Cc: Yasuo Ohgaki <yohgaki@ohgaki.net>, Niklas Keller <me@kelunik.com>,
 Nikita Popov <nikita.ppv@gmail.com>, internals@lists.php.net
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Message-ID: <22fd75b88e253b0eb58cc789fd800760@gmail.com>
User-Agent: Roundcube Webmail/1.2.3
Subject: Re: [PHP-DEV] Re: Improving mt_rand() seed
From: lauri.kentta@gmail.com (=?UTF-8?Q?Lauri_Kentt=C3=A4?=)

On 2017-01-27 17:13, Leigh wrote:
> On 27 January 2017 at 14:30, Lauri Kenttä <lauri.kentta@gmail.com> 
> wrote:
>>> This needs to be thought of as 2^32 possible _streams_ with a period
>>> of (2^19937)−1. Offset within the stream is as important as the 
>>> stream
>>> variation itself.
>> 
>> This is not true. There is one stream of period (2^19937)−1, and
>> the initial state defines the current position in that stream.
> 
> I'm not sure about this, the LCG constant used in the initial
> generator seems completely unrelated to the rest of the algorithm, so
> I don't see how this offsets the stream position.
> 
> If it is truly the case, I stand corrected.

I'm sorry, I oversimplified a bit in a hurry.
MT has (2^19937)−1 possible internal states, and I understand that
it's supposed to go through all these, hence the (2^19937)−1 period.
In each state, mt_rand returns 624 values of 32 bits before
"twisting" to the next state. Initializing with a different state
offsets the stream position just as it would in a LCG, but because
of the huge state and with the current MT initialization method,
it's very difficult to say what that offset actually would be.
But still, starting with mt_srand(0) and looping trough mt_rand,
you should eventually end up in the same state as mt_srand(1).
It might take around 2^(19937-32) * 624 mt_rand calls, though.

-- 
Lauri Kenttä