[seqfan] Re: Replacing chunks by their average (explanation)
Eric Angelini
Eric.Angelini at kntv.be
Mon Mar 30 12:41:15 CEST 2015
Hmmmm, let me check in a couple of hours (I'm far from my PC now)
Best,
É.
Catapulté de mon aPhone
> Le 30 mars 2015 à 10:32, "Lars Blomberg" <lars.blomberg at visit.se> a écrit :
>
> Eric,
> You state that "Occams principle" should be used so that for a large "k" we should first try "2, 2k-2".
> It seems to me that this is in conflict with the lexico-first requirement since "2,2,2k-4" is smaller than "2,2k-2".
> /Lars
>
> -----Ursprungligt meddelande-----
> Från: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] För Eric Angelini
> Skickat: den 29 mars 2015 23:11
> Till: Sequence Discussion list
> Ämne: [seqfan] Re: Replacing chunks by their average (explanation)
>
>
> Hello Jim, Lars, Aai, Maximilian and
> the others;
> Here is a word of explanation.
>
> I didn't want a trivial seq for S
> but a seq S showing the "collapse"
> of a chunk of same-parity integers
> into a single integer.
>
> So let's start S with a(1)=1 and a(2)=2.
>
> S=1,2,...
>
> If we want to illustrate a collapse
> we'd better show it as quickly as
> possible; so a(3)=2 (and not 1 or 3,
> meaning no immediate collapse);
>
> S=1,2,2,...
>
> We must keep in mind that S will
> collapse into itself (by definition);
> this means that what we see here is
> also S' -- the result of a former
> collapse; thus the first "1" is replacing a chunk of integers -- but which one?
>
> S =1,2,2,..
> S'=1, 2, 2,..
>
> The collapsing process wants a
> chunk of same-parity integers
> to be replaced by a single value
> (integer) which is the average of
> the said chunk (for example, the
> chunk of even intergers 2,4,4,4,6
> will be replaced by a single "4"
> (i.e. will "collapse" into a single "4"), this "4" being the result of 2+4+4+4+6 divided by 5).
>
> We thus write a list of "ancestors"
> ("chunk-ancestors") of "1" and see
> what happens:
>
> The chunk "1,1,1,1...1" will collapse
> into "1";
> The chunk "0,2" also;
> The chunk "2,0" also;
> The chunk "1,0,0,3" also;
> etc.
> All those are "ancestor-chunks" of "1".
> But if we accept zeros we will have
> problems later; as those zeros must,
> at some point, be themselves the
> average of another "ancestor-chunk";
> and a zero-average means that the
> said ancestor is made by a lot of
> other zeros, etc. -- producing triviality again, which we precisely want to avoid [this is the Zero Law: no zeros can be part of S].
>
> We bump thus into the First Law:
> - a "1"comes from another single "1",
> period.
>
> (This comes from a well-known implicit
> Occam Principle: keep always the ancestor as simple as possible in order not to get stuck.)
>
> Let's list now the ancestors of "2";
> they are of the type "1,3" and "3,1"
> and of the type "2,2,2,2,2,2...2".
> [We cannot mix even and odd terms
> in the "ancestor-chunk" -- as the said
> chunk would be parity-inhomogeneous
> (like "1,2,3"), which is forbidden].
> The "1,3" and "3,1" types could also
> look like "1,3,3,1" or "3,1,3,1,3,1" of
> course.
>
> We are almost equipped now to
> build simultaneously S and S' -- the
> only missing thing being the Second
> Law:
> - Two "1"´s _must_ always be separated
> by an odd quantity of integers.
> [It took me almost an hour to find and
> understand this (now) obvious 2d Law...]
>
> As we want S to be the lexico-first
> "collapsing to itself" seq, we will
> always try to produce ancestors-chunks
> of "k" of the type "1, 2k-1", else
> "2, 2k-2" (this is due again to the
> Occam Principle: simple things first).
>
> Let's try this start for S:
>
> S =1,2,2,1,3,...
> S'=1 2 2
>
> This looks promising but... the First Law has not being respected and problems arise immediately; if S starts like 1,2,2,1,3,...
> this means that S' will look the same
> and will "dictate" the "ancestors" of S
> (the S' line is the line with the averages):
>
> S =1,2,2,1,3,...
> S'=1 2 2 1 3
>
> The second "1" of S' dictates a "1" in S:
>
> S =1,2,2,1,3, 1,...
> S'=1 2 2 1 3
>
> But we see here that this "1" is of the
> same parity as the chunk "1,3" -- thus
> ruining the process, forcing us to
> write S and S' like this:
>
> S =1,2,2,1,3, 1,...
> S'=1 2 3
>
> ... they are not the same anymore --
> contradiction.
>
> This comes from the break of the First Law: there were only two integers between the two "1"'s of S.
> Can we fix that like this (inverting
> "1,3" into "3,1"):
>
> S =1,2,2,3,1,...
> S'=1 2 2 3 1
>
> Well, no; another contradiction will
> appear because of the last "1" of S'
> which will dictate this behaviour of
> S:
>
> S =1,2,2,3,1, ... ,1
> S'=1 2 2 3 1
>
> We see here that the "3" of S' cannot
> produce an "ancestory" in S that will
> fit _between_ the "1"'s of S; this is
> because of the Third Law:
>
> - An odd "average" cannot produce a
> chunk of even integers that are in odd quantity in the said chunk.
> That is, "3" cannot be the average
> of a chunk like [a,b,c,...n] where all
> terms are even AND where those
> terms a,b,c,...n are in a odd quantity
> (those two requirements are dictated
> by the First and the Second Law,
> the First saying that à "1" in S' gives
> a "1" in S and the Second saying
> that between two "1"'s of S there
> _must_ be an odd quantity of terms.
>
> We thus try this start for S (and S'):
>
> S =1,2,2,2,1,3,...
> S'=1 2 2 2 1 3
>
> This will work and produce the S I've
> submitted two days ago.
>
> Hope this is clear now -- and fun!
> Best,
> É.
> ---- ---- ---- ---- ---- ----
>> S=
> 1,2,2,2,1,3,2,2,1,2,4,3,1,2,1,2,1,7,2,4,1,2,1,2,1,2,12,3,1,4,1,2,1,2,1,2,1,23,
> 2,4,1,4,1,2,1,2,1,2,1,2,44,3,14,14,1,2,1,2,1,2,1,2,1,87,2,4,1,4,1,4,1,2,1,2,1,
> 2,1,2,1,2,172,3,1,4,1,4,1,4,1,2,1,2,1,2,1,2,1,2,1,343,...
>
>> Hello SeqFans,
> Replace in S all chunks of same-parity integers by the average of the said chunk and you will get S back.
>
> Example:
> - the first chunk is made by 1, -- replace it with 1;
> - the second chunk is 2,2,2, -- replace it with 2; -the third chunk is 1,3, -- replace it with 2; -the fourth chunk is 2,2, -- replace it with 2; -the fifth chunk is made by 1, -- replace it with 1; -the sixth chunk is made by 2,4, -- replace it by 3; etc.
> The replacements so far have been
> 1,2,2,2,3,1... which rebuilds S.
>
> One quickly finds this array where the rightmost diagonal is A023105.
>
> ---
> 1,
> 2,
> 2,
> 2,1,3,
> 2,2,1,2,4,
> 3,1,2,1,2,1,7,
> 2,4,1,2,1,2,1,2,12,
> 3,1,4,1,2,1,2,1,2,1,23,
> 2,4,1,4,1,2,1,2,1,2,1,2,44,
> 3,1,4,1,4,1,2,1,2,1,2,1,2,1,87,
> 2,4,1,4,1,4,1,2,1,2,1,2,1,2,1,2,172,
> 3,1,4,1,4,1,4,1,2,1,2,1,2,1,2,1,2,1,343,
> ...
>
> Best,
> É.
> Catapulté de mon aPhone
>
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