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The similarity system

  • #11,181
  • Oct 29, 2014 6:46am Oct 29, 2014 6:46am
Quoting vlady1974
Thanks Captain! (Exhales) Bloody hell this is INTENSE!! LOL There is so much power in this I can't even begin to start.... Glad I was sipping coffee whilst I read that. It is absolutely beyond me at the moment.... except for the example. K, I will definitely need your help!!
Please refute this

"
And why the high of the next bar can not be between 1.2222 and 1.2211?

Imagine candle with the high at 1.2222 closes at 1.2215. Why the next candle (opening at 1.2215) can not have a high at 1.2220?
"
  • #11,182
  • Oct 29, 2014 6:47am Oct 29, 2014 6:47am
"linearly"

Quoting Eurusdd
This theorem is not mine. There are about 22 different proofs of the theorem. If it is true - I cannot say till I prove it myself - then it is the most powerful tool apart from TZ that I have presented here. This theorem means MONEY, if true! The truth of the statement depends on what they mean by subsequence. If they mean a sequence defined linearly on n then I will say there are cases where it is not true. If they allow any possible sequence on n then, yes it is true. So, if the latter is the case then there is one more...
My Best Trade is Next Trade!
  • #11,183
  • Oct 29, 2014 6:47am Oct 29, 2014 6:47am
My version applicable to trading:::
Attached Image (click to enlarge)
Click to Enlarge Name: Capture.PNG Size: 38 KB


NOTE: The theorem is most powerful if subsequences - in the definition, is defined linearly on n. But in this case I have a problem with a theorem for which there are 22 different proofs.
  • #11,184
  • Oct 29, 2014 6:48am Oct 29, 2014 6:48am
Quoting vlady1974
K, I will definitely need your help!!
If I don't get a heart attack today, I am there for you!

k
  • #11,185
  • Oct 29, 2014 6:48am Oct 29, 2014 6:48am
puh thats deep..iīm sure most people (myself included) havent really figured out transient/recurrent to trade profitable.

Looks like i have to rent a mathematician to trade for me...
TWEE, revoke my membership.
  • #11,186
  • Oct 29, 2014 6:49am Oct 29, 2014 6:49am
Quoting DavidRP
{quote} Please refute this " And why the high of the next bar can not be between 1.2222 and 1.2211? Imagine candle with the high at 1.2222 closes at 1.2215. Why the next candle (opening at 1.2215) can not have a high at 1.2220? "
because he choose n=2 and the sub sequnce needs to be n+1 so 3 higher highs or contradiction
follow the flow
  • #11,187
  • Oct 29, 2014 6:52am Oct 29, 2014 6:52am
Quoting Atihun
{quote} because he choose n=2 and the sub sequnce needs to be n+1 so 3 higher highs or contradiction
I meant if that is impossible in market. Is it impossible for price to make a high at 1.2220?
  • #11,188
  • Oct 29, 2014 6:53am Oct 29, 2014 6:53am
OK, Now I'm awake!
  • #11,189
  • Oct 29, 2014 6:53am Oct 29, 2014 6:53am
Quoting DavidRP
{quote} I meant if that is impossible in market. Is it impossible for price to close at 1.2220?
i think we are looking at highs only.in the example atleast. but you can do it for lows/close/open too. but mixing isnt allowed.
follow the flow
  • #11,190
  • Oct 29, 2014 6:54am Oct 29, 2014 6:54am
Quoting Atihun
{quote} i think we are looking at highs only.in the example atleast. but you can do it for lows/close/open too. but mixing isnt allowed.
Sorry I meant high

He is misundestanding the theorem. What the theorem assumes, is that you can always find a n+1 subsequence and it doesn't need to be sorted

http://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Szekeres_theorem
  • #11,191
  • Oct 29, 2014 6:57am Oct 29, 2014 6:57am
Quoting DavidRP
{quote} Sorry I meant high
if we would have a close below maybe n will be extended means if n was 2 it should be 3 so nē+1 10 and n+1 4 so two new highs within 9 candles or two lows(or lower highs)??
just thinking loud.
follow the flow
  • #11,192
  • Oct 29, 2014 7:00am Oct 29, 2014 7:00am
Quoting Eurusdd
My version applicable to trading::: {image}
Okay. Then the subset B isn't a series of consecutive prices. But then when I look at the example, price could be everywhere to either form a n+1 de- or increasing sequence.

PS: Ah. David. This would make sense. So the prices within a sequence are not from consecutive bars. But then I can't see the power of the theorem. Puuuhhhh.
  • #11,193
  • Oct 29, 2014 7:06am Oct 29, 2014 7:06am
Quoting DavidRP
{quote} Sorry I meant high He is misundestanding the theorem. What the theorem assumes, is that you can always find a n+1 subsequence and it doesn't need to be sorted http://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Szekeres_theorem
Well I never said something had to be sorted. In fact, I think I posted a revision, using a set for the subsequence so that the problem of sorting out is eliminated.

Even in this case, one can make use of it. So, please read what I said carefully.
  • #11,194
  • Oct 29, 2014 7:06am Oct 29, 2014 7:06am
Quoting Atihun
{quote} if we would have a close below maybe n will be extended means if n was 2 it should be 3 so nē+1 10 and n+1 4 so two new highs withing 9 candles or two lows(or lower highs)?? just thinking loud.
when you have 0 1 2 3 4 5 , and n=2 you always can find a subsequence of 3 either increasing or decreasing but it doesnt have to be 0-1-2 or 2-3-4 or 3-4-5 it can be 0-2-5 or 1-3-4 or 0-4-5 so you can not predict that high must finish below or above something because a different unsorted sequence can be found without the need of the last point going below or above something
  • #11,195
  • Oct 29, 2014 7:06am Oct 29, 2014 7:06am
Fractal ?
n =1 Anē+1 = 3
n =2 Anē+1 = 5
n =3 Anē+1 = 7
etc....
MTH
  • #11,196
  • Oct 29, 2014 7:07am Oct 29, 2014 7:07am
Quoting Eurusdd
{quote} Well I never said something had to be sorted. In fact, I think I posted a revision, using a set for the subsequence so that the problem of sorting out is eliminated. Even in this case, one can make use of it. So, please read what I said carefully.
ok

But then, your example with 1.2211 is wrong

I read very carefully this and it's WRONG

"For example, suppose you chose your reals to be high of bars.
The on the daily chart if you set n=2 then you need to look in the DNA for 5 bars.
Let us suppose the highs of the last 4 bars are 1.2344, 1.2566, 1.2211, 1.2222, ?????
What will be ?????

Since there must be a decreasing or increasing 3-subsequence, the high of the next bar must be above 1.2222 or below 1.2211.

Just two options."
  • #11,197
  • Oct 29, 2014 7:11am Oct 29, 2014 7:11am
Quoting DavidRP
{quote} when you have 0 1 2 3 4 5 , and n=2 you always can find a subsequence of 3 either increasing or decreasing but it doesnt have to be 0-1-2 or 2-3-4 or 3-4-5 it can be 0-2-5 or 1-3-4 or 0-4-5 so you can not predict that high must finish below or above something because a different unsorted sequence can be found without the need of the last point going below or above something
The power of the theorem lies in this observation.
For a given n, look in the DNA for a set of n^2 distinct real numbers forming some sequence that ends on the current number you are interested in. That is, if we include the current number, we should have n^2+1 distinct elements. Now, if you set does not have an n+1-increasing or n+1-decreasing subsequence; find the lowest and the highest value in your set. The number you are interested in should be above the highest or below the lowest value!

The problem reduces to finding that n^2 set of distinct numbers with no n+1-increasing or n+1-decreasing subsequence. The if your current number is different from the n^2, it should create the sequence you are after.
  • #11,198
  • Oct 29, 2014 7:14am Oct 29, 2014 7:14am
Quoting DavidRP
{quote} ok But then, your example with 1.2211 is wrong I read very carefully this and it's WRONG "For example, suppose you chose your reals to be high of bars. The on the daily chart if you set n=2 then you need to look in the DNA for 5 bars. Let us suppose the highs of the last 4 bars are 1.2344, 1.2566, 1.2211, 1.2222, ????? What will be ????? Since there must be a decreasing or increasing 3-subsequence, the high of the next bar must be above 1.2222 or below 1.2211. Just two options."
I really do not get your problem. This example is for a the case where the sequence is listed and subsequence defined linearly on n. Lol. What is your stress. the application of the theorem or the special case i represented?
That is why I stated the theorem in the form below so that sorting out is eliminated.
Attached Image (click to enlarge)
Click to Enlarge Name: Capture.PNG Size: 38 KB
  • #11,199
  • Oct 29, 2014 7:17am Oct 29, 2014 7:17am
Quoting Eurusdd
{quote} The power of the theorem lies in this observation. For a given n, look in the DNA for a set of n^2 distinct real numbers forming some sequence that ends on the current number you are interested in. That is, if we include the current number, we should have n^2+1 elements. Now, if you set does not have an n+1-increasing or n+1-decreasing subsequence; find the lowest and the highest value in your set. The number you are interested in should be above the highest or below the lowest value!
Wrong again

with your example above 1.2344, 1.2566, 1.2211, 1.2222, xxxxx

the last number could be 1.2220 and then you have a decreasing subsequence: 1.2566-1.2222-1.2220. That is was the theorem says. Nothing says that the last number has to be above 1.22111 and 1.2222
  • #11,200
  • Oct 29, 2014 7:18am Oct 29, 2014 7:18am
Quoting Eurusdd
{quote} I really do not get your problem. This example is for a the case where the sequence is listed and subsequence defined linearly on n. Lol. What is your stress. the application of the theorem or the special case i represented?
should n be randomly choosen? what if we combine with h. a tz is formed after 15 candles we have n=4. we have 4 increasing highs.
what should happen to the tz?
follow the flow
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