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While this does open up the code (sans descriptive text) to external attacks, I rather feel that once you have a logic analyzer or compromised microcontroller o

While this does open up the code (sans descriptive text) to external attacks, I rather feel that once you have a logic analyzer or compromised microcontroller on the logic bus of your secure device you've got the attacker on the wrong side of the airtight hatchway. I'm personally much more interested in what it means for 'attackers' who wish to use it to open up their own hardware. Perhaps that might not align with the goals of Xilinx or the OEM, but it's great for their customers!

Nokinside 7y ago
Attacker needs physical access to just one device in the whole product line that uses the same encryption key. After that, all you need is to get the device to update using the bitstream you made. In some cases this could be remote attack.
- IshKebab 7y ago
  I'm not sure this is true. This attack allows reading the encrypted bitstream, but it doesn't say anything about allowing you to sign modified bitstreams.
  - nebulous1 7y ago
    > In this paper, we introduce novel low-cost attacks against the Xilinx 7-Series (and Virtex-6) bitstream encryption, resulting in the total loss of authenticity and confidentiality This seems to be saying that it's possible
  - teraflop 7y ago
    As the paper explains, there is no "signing" involved (in the sense of a public-key cryptosystem). Each encrypted bitstream includes an HMAC, but because the HMAC key is part of the encrypted bitstream itself, it basically only acts like a non-cryptographic checksum. An attacker who knows the encryption key can simply choose an arbitrary HMAC key and generate a valid HMAC for arbitrary data. EDIT: I should clarify that this attack doesn't appear to actually let someone extract the AES encryption key. But they can use an FPGA that has the key programmed as a decryption oracle. And a weakness of CBC mode is that a decryption oracle can be used for encryption as well.
    - IshKebab 7y ago
      Ah interesting, thanks!

Nokinside 7y ago

Attacker needs physical access to just one device in the whole product line that uses the same encryption key. After that, all you need is to get the device to update using the bitstream you made. In some cases this could be remote attack.

IshKebab 7y ago
I'm not sure this is true. This attack allows reading the encrypted bitstream, but it doesn't say anything about allowing you to sign modified bitstreams.
- nebulous1 7y ago
  > In this paper, we introduce novel low-cost attacks against the Xilinx 7-Series (and Virtex-6) bitstream encryption, resulting in the total loss of authenticity and confidentiality This seems to be saying that it's possible
- teraflop 7y ago
  As the paper explains, there is no "signing" involved (in the sense of a public-key cryptosystem). Each encrypted bitstream includes an HMAC, but because the HMAC key is part of the encrypted bitstream itself, it basically only acts like a non-cryptographic checksum. An attacker who knows the encryption key can simply choose an arbitrary HMAC key and generate a valid HMAC for arbitrary data. EDIT: I should clarify that this attack doesn't appear to actually let someone extract the AES encryption key. But they can use an FPGA that has the key programmed as a decryption oracle. And a weakness of CBC mode is that a decryption oracle can be used for encryption as well.
  - IshKebab 7y ago
    Ah interesting, thanks!