crypt8 definitions?

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dale_para_bajo

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Please do not read bellow this line. Thank for your curosity.

This thread is NOT about CSA RBT Chains. NO Creation, merging, modifying, HDD or space requirement stuff,etc.
So if that is what you are looking for:
Please do not read bellow this line. Thank for your curosity.

I am aware of Search Button use, but Please notice it does not work without java script. If You are comming witghout a Good Hart and willing to help:
Please do not read bellow this line. Thank for your curosity.

Now if you are still here and willing to help I wellcome you and beg you to clarify my doubght. Received my thanks ahead.

This thread is about clarifying the diferent types of payload size of crypt8.
But more important it is NOT about How you use them or where they apply. So NO CSA RBT stuff here please.
Instead of How they are define in 1 single 188 TS line. Please Notice how Important is that I am not speaking about Chains.
I am ONLY interested in 1 single 188 TS line. How it looks encrypted or decrtypted?

Now one more time if you feel you do not understand me or feels need to speak about CSA RBT:
Please do not read bellow this line. Thank for your curosity.


----------------------------------------------------------------------------------------------
I will post my question in simple words so that the one left and willing to help understand me.

A Full TS line is 188 bytes, or 4 byte of Header + 184 Payload. All bytes number are given in Decimal. So Bytes 000 - 003 are the header.

Now lets be more Specific in crypt8:

*B8hxFFh & B8hx00h definition are the Original crypt8 types.
*B8hx = 184 Payload.


The crypt8 B8hx00h, is found in Ecrypted TS bytes 004-011.
A Plain Line just looks like. ALL "00"
47 3F 41 12 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
Click to expand...



The crypt8 B8hxFFh, is found in Ecrypted TS bytes 004-011.
A Plain Line just looks like. ALL "FF"
47 3F 41 10 FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF
FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF
FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF
FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF
Click to expand...


Corrupted or fake crypt8 as example a Null packet got encrypted inside a Packed DTT.
Please note that this line reapeats a lot making CSA RBT belive is a valid crypt8 but it bytes 004-187 are NOY ALL "00" or "FF".
So "47 1F FF 12" makes it a "FAKE" not valid.
47 22 BE 1C 47 1F FF 12 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
Click to expand...

1-So can any one show me the other samples?[/QUOTE]
 
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dale_para_bajo

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Here is what I Imagine but that I am not totally sure about it. Please confirm or correct me.
Lets start from 08hx00h. The table that ooOO_SORGOS_OOoo loves to use.

I wil asume clear still looks like
47 3F 41 12 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
Click to expand...

But the crypt8 bytes are innstead located at the end.

But can a mixt of bytes be consider as on of this type?
47 3F 41 10 FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF
FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF
FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF
FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF 00 00 00 00 00 00 00 00
Click to expand...

Thanks ahead.

PD:
I will appreciate if some one post a log containing crypt8 types valid for 08hx00h or 08hxFFh. That will help me understand it.
 
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xosef1234

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Your example is not correct. You need to consider the adaptation field. For 08h c8 the total length of the AF should be in the range of 169 to 176.
 
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dale_para_bajo

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Yes xosef1234 I thank you for acknowledging that. The adaptation Filed.

I did not post an encrypted TS line as I thought it only look like a bunch of meaningless random hex bytes. But that bather some people. SO they where correct becouse In the header the adaptation filed show inportant info.

This is how Klim explained to me. I still need to run some test as to see what is encrypted and what is not or even where it start.

I hope klim do not mine I post her answer here so that all interested can learn too.

klim said:
Let's come back to biss.
I'm in this thing since few time, but just to complete the things told in my post, we can continue as follow...

We've seen the example of a c8 in pl size 184 as

47 02 00 98 1A CC A9 70 E1 04 A1 5B A8 E5 22 D5 C0 DB EA F0 D0 2F F3 95 70 AF 58 09 3A 92 95 48 03 64 27 4E 2B 10 69 42 A6 0E CE D3 FA 63 1A 80 62 29 04 7F 08 2B F2 DE 51 90 D4 10 F2 DC 57 76 3B 0E 34 51 88 38 5C C2 F0 6D F9 0C 2A 16 6A 8F AA 3A E2 99 9A CE AD C3 39 02 72 6C 6A BA F4 9A B9 E0 1A 55 F7 EF 56 25 C1 2B 3A 2F D9 C1 F1 C7 9A 56 D4 0E C3 F2 46 7A 28 D3 CC EE 02 73 1E 87 C2 EE E5 64 68 E1 C3 B1 A7 35 E5 3F 94 9C 26 BD 44 87 7E 10 C9 75 38 C9 BD F1 65 2D BD 69 88 95 C0 04 C1 4F F4 17 CA 6E 76 7A 0A CE 46 09 E2 C8 11 38 80 42 2E 7C D6 0B 0E FC 25 F3

where
47 02 00 98 = header (separetor byte, pid, ..)
1A CC A9 70 E1 04 A1 5B A8 E5 --> c8
... restant bytes

For other pls size things works as follow.
Pls size of the complete instruction is ever of 188 bytes (4 header + 184 command ), but missing part (because pls size is shorter are filled by ff bytes). While the indicator byte has the first bit in B

So look as this pls size 48 as example:

47 02 00 BA 80 00 FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF A9 C3 F2 CD D8 74 1E 28 7A 0F 3F CF CB 68 0E DE 3C 3A 2E FA D9 70 48 EF 28 50 3D 7B 7D 9E 13 CB 7C 9B C1 2B 82 B5 82 CD F8 B2 95 3F AC 1E CD 2B 55 E9 97 D9 84 7A B1

where you can see a similare header with

47 = separator byte
02 00 = video pid
BX = indicator byte
80 = lenght of null FF bytes
00 = separator byte
FF FF (80 hex = 128 dec. bytes )
A9 C3 F2 CD D8 74 1E 28 --> C8
...7A 0F 3F CF CB 68 0E DE 3C 3A 2E FA D9 70 48 EF 28 50 3D 7B 7D 9E 13 CB 7C 9B C1 2B 82 B5 82 CD F8 B2 95 3F AC 1E CD 2B 55 E9 97 D9 84 7A B1 (restant bytes corpus)


FF bytes lenght can be different, but in any case, total lenght is ever done by 4 header bytes + 184 bytes of command and our c8 is ever at the end of internal FF bytes...

In other pls as pls 8 i.e. it works in the same way...

47 02 00 B8 AC 00 FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF 3E 7C 0D 60 00 41 7B F9 A2 3F 3D

with an header
47 02 00 BX

then

AC = FF bytes lenght
00 = null bytes
3E 7C 0D 60 00 41 7B F9 --> C8
...A2 3F 3D (resting instruction corpus)

***
@klim
Click to expand...
 
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dale_para_bajo

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Now I need to know how to distinguish payload size.
lets take klim samples.

She call pls size 48 to
Code: 
47 02 00 BA 80 00 FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF 
FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF 
FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF A9 C3 F2 CD D8 74 1E 28 
7A 0F 3F CF CB 68 0E DE 3C 3A 2E FA D9 70 48 EF 28 50 3D 7B 7D 9E 13 CB 7C 9B C1 2B 82 B5 82 CD F8 B2 95 3F AC 1E CD 2B 55 E9 97 D9 84 7A B1

80 = lenght of null FF bytes
80 = 128 or I guess -> 1 0x00 + 127 0xFF
Bytes Left are 55. -> There are 184 - 1(80) - 128 bytes

Now what makes it "pls size 48"


The other sample:
Code: 
47 02 00 B8 AC 00 FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF 
FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF 
FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF 
FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF 3E 7C 0D 60 00 41 7B F9 A2 3F 3D

AC = lenght of null FF bytes
AC = 172 or I guess -> 1 0x00 + 171 0xFF
Bytes Left are 11. -> There are 184 - 1(AC) - 172 bytes

Now what makes it "pls size 8"

I will apreciate the explanation.

-------------------------
PD:
Upsss! xosef1234 said!!

For 08h c8 the total length of the AF should be in the range of 169 to 176.
Click to expand...
 
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xosef1234

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Yes, but adaptation field does not necessarily have to be these 00h and ffh bytes but can be any bytes.
 
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dale_para_bajo

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What I recall is that it is use to introduce timing reference on the PID so that the main App can maintain synchronous between video and audio. Something like that.

So this means that this types of TS are very limited! Synck will keep changing that mean no crypt8 repeating.

Now I need to understand how a Payload A0 can help in DTT. As the process of packing in DTT do in fact create it self many 1/2 False crypt8. So I guess 1/4 of those may meet A0 Payload size crypt8.

So this may explain the quote
Version 1.18
For what are the new A0h plain types useful
As written ...

With the A0h plain type I was able to find the CWs.

But for all other standard BISS provider it think the A0h plain type is useless.
Click to expand...

Interesting....
 
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