Stock market puzzles :) - page 7
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OK, let's do this :)
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What variations of the relationship between the numbers are possible? ><= a total of three possibilities.
We have four digits how many relations between them - three,
It takes two bits to encode three states, but since four variants fit into two bits, let's simply label 11 as a forbidden state, for example.
That is, the type of bar is defined by three zones of two bits each - XX XX XX .
**
But the task is somewhat more than just coding a bar, we now have to consider a double bar. When the close price of the first one is EVEN the open price of the second one. If close price is not equal to open price then such bars are invalid and they MUST be converted (either first or second) to the standard variant. That is when the open price equals the close price. QUESTION - how many variants of such double bars will there be now, since we have only OHLOHLC instead of OHLC-OHLC ?
your picture is not quite correct, it would be better to have this
and then we get a relationship like this
1. H>O
2. H=O
3. L<0
4. L=0
5. H>C
6. H=C
7. L<C
8. L=C
9. C>O
10. C<O
11. C=O
then calculate all options....
theme ++;
(laughs)
makes sense.
Do you see a chipmunk? It's not there! :)
Your picture isn't quite right, it's better like this.
and then we get a relationship like this.
1. H>O
2. H=O
3. L<0
4. L=0
5. H>C
6. H=C
7. L<C
8. L=C
9. C>O
10. C<O
11. C=O
then calculate all the options....
Decide on prices O and C:
1. C > O
2. C < O
3. C = O
we need 2 bits to encode the result.
Then compare H with the maximum of O and C, and L with the minimum of O and C:
4. H >
5. H =
we will need 1 bit to encode the result;
6. L <
7. L =
we will need 1 bit to encode the result.
Total - 4 bits are required.
---
PS. Only it's not clear why it's needed.
Decide on the O and C prices:
1. C > O
2. C < O
3. C = O
we need 2 bits to encode the result.
Then compare H with the maximum of O and C, and L with the minimum of O and C:
4. H >
5. H =
we will need 1 bit to encode the result;
6. L <
7. L =
we will need 1 bit to encode the result.
Total - 4 bits are required.
---
PS. Only it's not clear why it's needed.
Once you introduce a comparison, then the whole coding theory makes no sense. Say for a T1 bar there will be one maximum, for a T10 barthis maximum will be the minimum, but the actual shadow or body size will be the same ... but the candlestick's con is different.
If you do the coding, it can be relative, but the parameter relative to which the coding is done should be a constant.
As soon as you introduce a comparison, the whole coding theory makes no sense. Say for a T1 bar there will be one maximum, for a T10 bar, that maximum will be a minimum, but the actual shadow or body size will be the same ... but the candlestick's con is different.
If you make coding, it can be relative but the parameter relative to which the coding is made must be a constant.
Firstly, it was not me who suggested the comparison.
Secondly, this shows loss-of-information compression. There is nothing unusual about loss of information, it is filtering. This, personally, begs the question "why do we need it?
Firstly, it was not me who suggested the comparison.
Secondly, this shows loss-of-information compression. There is nothing unusual about loss of information, it is filtering. This, personally, begs the question "why do we need it?