A host communications equipment to connect the external device, a 32 byte of data (hexadecimal), when the external device receives the data will be sent back a set of 32 byte of response data, if the host comparative analysis of the received data is correct, the host to establish communications with external devices,
Attachment 32 byte as before each set of data should be a set of random Numbers (guess for the random number), after the two values are for the 32 byte checksum, now want to know from their phones to "send" 32 byte, how to calculate the "receive" 32 byte.
If there are any player can break out the algorithm, contact me first, into a devious! Button: the 6-8-7-1-0-3-0-2
Here is our simulations from 00 to FF data should be sent to the host and external equipment response data (length only sent two groups, can contact me if you need more data),
Note: "send" is to host the data, "receive" the response data sent back is the external equipment,
Send: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 A4 01
Receive: C9 1 c 9 d DF 43 00 41 E6 2 e 66 71 33 62 C1 BD 2 d EB A2 57 06 6 c A1 FD 3 a 9 f the F4 D2 3 d 2 e CE 3 f E1 1 b 81
Send: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 01 C0 64
Reception: 9 e 4 f 63 42 BC FC 60 D0 9 b 85 4 c F3 26 88 5 c and 5 b 14 42 B8 7 b 38 C2 04, 81, 80 D5 9 a BD AA 65 D4 98 CE
CodePudding user response:
If send fixed data, get reply is also fixed?What you say, is likely to be a "external certification", the principle of it, is the Shared secret key,
A possible agreement:
Includes two parts, the first random number 1 host send Rm1 and Rm2,
2 key encryption Rm1 peripherals with root, get the ciphertext as "key process",
3 peripherals with all process encryption random number, return ciphertext,
(in fact, the more rigorous protocol is two-way authentication:
3 peripherals Rm2 is encrypted with the key process, cipher text as part of the return data,
4 peripherals to generate a random number 16 - byte Rd,
5 peripherals will be 16 - byte ciphertext and 16 - byte random number returned,
6 host process should use the same key to encrypt Rm2 cipher text, compare, if consistent, host recognition peripherals,
7 host USES the same key encryption Rd, the process of sending ciphertext to peripherals,
8 peripherals to verify this cipher, if consistent, host,
This process is achieved by three one-way communication, also known as 3 - pass certification,)
If equipment factory of the commercial is a national or international standards of commercial cipher algorithm, with your conditions and resources can basically break (find out key), for example, international TDES, AES algorithm and so on are offered millions of dollars to find attacks (crack) method, even cut deciphering the exhaustive rounds one or two orders of magnitude, so far no one can the receiving of the reward,
If you can break, but also your "devious", "is the bounty of the United States directly,
If necessary and at no cost, do you have a host and peripherals, and hand and you confirm is a one-way authentication, can be established through exhaustive testing a "dictionary," is what is called a dictionary attack,
CodePudding user response:
Thank you doing upstairs player so detailed replyWhat you asked "if send fixed data, get reply is also fixed?"
Yes, if send a fixed data, the resulting response is also the same fixed values,
Hard,,,,,,,,,,
CodePudding user response:
MD5 just add some salt, lifetime couldn't break you,CodePudding user response:
Take on the program, debugging,Even the md5 with a little salt, is also possible to find play
CodePudding user response:
The