Prev: tetration, matrix^matrix, log(matrix)*matrix ormatrix*log(matrix)
Next: WIKIPEDIA Bum wrong again
From: I.N. Galidakis on 5 Sep 2009 03:17
Tim Little wrote:
> On 2009-09-04, I.N. Galidakis <morpheus(a)olympus.mons> wrote:
>> If I put your streams in my Maple generator to generate more digits
>> (so that more digits are encoded), the initial segments are
>> predicted correctly:
>
> They were not predicted at all. LC(L) was identical in each case, yet
> you chose different ways to process the data because you knew what the
> result ought to be. That's not an algorithm, and cannot be made to be
> one.
>
> Having pointed out the same property in two different ways without
> success, I wish you good luck in analyzing your system by yourself.

I wrote a second post which you probably did not see. I explain the problem
there and I also agree that unique prediction in this case is not possible.

> - Tim
--
Ioannis

From: Tim Little on 5 Sep 2009 03:49
On 2009-09-04, I.N. Galidakis <morpheus(a)olympus.mons> wrote:
> However, I think it is fairly obvious that each combination choice
> determines a unique sequence and all choices can be enumerated
> easily

It is fairly obvious that each combination choice determines a unique
sequence. However it is not at all obvious that the set of possible
choices is easy to enumerate, or even finite. Hint: there are many
more than just the 3 "original messages" yielding L when truncated
than the ones I presented. Can you find them all?


- Tim
From: Tim Little on 5 Sep 2009 04:10
On 2009-09-05, I.N. Galidakis <morpheus(a)olympus.mons> wrote:
> I wrote a second post which you probably did not see. I explain the
> problem there and I also agree that unique prediction in this case
> is not possible.

My apologies, I did not see that post until after I had posted.


- Tim
From: I.N. Galidakis on 5 Sep 2009 05:23
Tim Little wrote:

> On 2009-09-04, I.N. Galidakis <morpheus(a)olympus.mons> wrote:
>> However, I think it is fairly obvious that each combination choice
>> determines a unique sequence and all choices can be enumerated
>> easily
>
> It is fairly obvious that each combination choice determines a unique
> sequence. However it is not at all obvious that the set of possible
> choices is easy to enumerate, or even finite. Hint: there are many
> more than just the 3 "original messages" yielding L when truncated
> than the ones I presented. Can you find them all?

I don't know. I am not even sure exactly yet how the process works. I can
certainly give it a try and see what I get:

1) Picking the block [2,5,4,3,2,1] along with the bottleneck point 7, leads to
your first stream.
2) Picking the block [5,4,3,2,1] seems to lead to your second stream.
3) Picking the block [4,3,2,1] seems to lead to your third stream.
4) Picking the block [3,2,1]:

L:=[3,2,1,2,7,5,4,7,3,1,4,5,0,9,3,9,1,2,8,1,2,
1,7,8,7,7,3,3,9,5,0,5,4,2,4,0,5,6,7,7,7,6,5,3,2,3,4,2,
1,4,4,1,6,5,6,7,3,5];

LC(L) = [3, 2, 5, 4,|,3, 2, 1, 2, 7, 5, 4, 7, 3, 1, 4, 5] (predicted left of |)

which seems to lead to:

[5, 3, 2, 5, 4, 3, 2, 1, 2, 7, 5, 4, 7, 3, 1, 4, 5]

5) Picking the block [2,1]:

L:=[2,1,2,7,5,4,7,3,1,4,5,0,9,3,9,1,2,8,1,2,
> 1,7,8,7,7,3,3,9,5,0,5,4,2,4,0,5,6,7,7,7,6,5,3,2,3,4,2,
> 1,4,4,1,6,5,6,7,3,5];

LC(L);
[2, 2, 5, 4, 3,|,2, 1, 2, 7, 5, 4, 7, 3, 1, 4, 5] (predicted left of |)

which seems to lead to:
[6, 2, 2, 5, 4, 3, 2, 1, 2, 7, 5, 4, 7, 3, 1, 4, 5]

6) Picking the block [1]:

L:=[1,2,7,5,4,7,3,1,4,5,0,9,3,9,1,2,8,1,2,
> 1,7,8,7,7,3,3,9,5,0,5,4,2,4,0,5,6,7,7,7,6,5,3,2,3,4,2,
> 1,4,4,1,6,5,6,7,3,5];

LC(L);
[1, 2, 5, 4, 3, 2, 1,|, 2, 7, 5, 4, 7, 3, 1, 4, 5] (predicted left of |)

which seems to lead to:
[7, 1, 2, 5, 4, 3, 2, 1, 2, 7, 5, 4, 7, 3, 1, 4, 5]

7) Picking the block [2,5,4,3,2,1]:

L:=[2,5,4,3,2,1,2,7,5,4,7,3,1,4,5,0,9,3,9,1,2,8,1,2,
> 1,7,8,7,7,3,3,9,5,0,5,4,2,4,0,5,6,7,7,7,6,5,3,2,3,4,2,
> 1,4,4,1,6,5,6,7,3,5];

LC(L);
[2, 4,| 2, 5, 4, 3, 2, 1, 2, 7, 5, 4, 7, 3, 1, 4, 5] (predicted left of |)

which seems to lead to:
[3, 2, 4, 2, 5, 4, 3, 2, 1, 2, 7, 5, 4, 7, 3, 1, 4, 5]

Now, an obvious note of caution: As far as I can see, every time there are k
predicted digits, at any stage, I have the freedom to either try a bottleneck
digit equal to k+1 or add 1<=j<=k new digits to L and update and continue.
Apparently there seem to be many more sequences than those above, if I follow
the latter tactic, but it seems to me all combinations can be enumerated, based
on the size of j. For example, in the last example, I added the bottleneck 3 and
it was over with, with the predicted digits [2,4].

Let's see what happens if I add one digit at a time, so we add the 4:

L:=[4,2,5,4,3,2,1,2,7,5,4,7,3,1,4,5,0,9,3,9,1,2,8,1,2,
> 1,7,8,7,7,3,3,9,5,0,5,4,2,4,0,5,6,7,7,7,6,5,3,2,3,4,2,
> 1,4,4,1,6,5,6,7,3,5];

LC(L);
[4, 3, 2, 5, 4, 3, 2, 1, 2, 7, 5, 4, 7, 3, 1, 4, 5]

I think this deadlocks, because L does not show up in LC(L), so we can stop
here. Let's try adding [2,4] instead:

L:=[2,4,2,5,4,3,2,1,2,7,5,4,7,3,1,4,5,0,9,3,9,1,2,8,1,2,
> 1,7,8,7,7,3,3,9,5,0,5,4,2,4,0,5,6,7,7,7,6,5,3,2,3,4,2,
> 1,4,4,1,6,5,6,7,3,5];

LC(L);
[2,|,2, 4, 2, 5, 4, 3, 2, 1, 2, 7, 5, 4, 7, 3, 1, 4, 5] (predicted left of |)

If we add the digit 2 now, we get:

L:=[2,2,4,2,5,4,3,2,1,2,7,5,4,7,3,1,4,5,0,9,3,9,1,2,8,1,2,
> 1,7,8,7,7,3,3,9,5,0,5,4,2,4,0,5,6,7,7,7,6,5,3,2,3,4,2,
> 1,4,4,1,6,5,6,7,3,5];

LC(L);
[2, 4, 3, 2, 5, 4, 3, 2, 1, 2, 7, 5, 4, 7, 3, 1, 4, 5]

I think this deadlocks again, because L does not show in LC(L). So in this case
the prediction leads only to the last example 7). For the other cases there may
be many more sequences.

> - Tim
--
Ioannis

From: I.N. Galidakis on 5 Sep 2009 05:25
Tim Little wrote:
> On 2009-09-05, I.N. Galidakis <morpheus(a)olympus.mons> wrote:
>> I wrote a second post which you probably did not see. I explain the
>> problem there and I also agree that unique prediction in this case
>> is not possible.
>
> My apologies, I did not see that post until after I had posted.

No problem. Many thanks for helping me debug this :-)

> - Tim
--
Ioannis
First  |  Prev  |  Next  |  Last
Pages: 1 2 3
Prev: tetration, matrix^matrix, log(matrix)*matrix ormatrix*log(matrix)
Next: WIKIPEDIA Bum wrong again