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I use the following code to generate a key and then use it to encrypt a simple string message. The output of the function Encrypt is an encrypted object. But I would rather convert this into binary digits (a long sequence of zeros and ones). Unfortunately, I am not able to do so. I tried to use the function IntegerDigits[] but that did not work. Thanks in advance!

key = GenerateSymmetricKey[];
message = "Hello world";
Encrypt[key, message]
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  • $\begingroup$ Is this what you are after? Encrypt[key, message]["Data"] // Normal // IntegerDigits[#, 2, 8] & // Flatten $\endgroup$
    – Kuba
    Oct 21, 2017 at 10:44
  • $\begingroup$ Thanks for the prompt reply! Well, yes that is what I am after. But then after the conversion to binary digits, how may I take back those same digits and decrypt them? Apparently, the Decrypt[] function is also expecting an encrypted object, not a sequence of binary digits. $\endgroup$ Oct 21, 2017 at 11:57

1 Answer 1

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I will turn Kuba's comment into an answer. The EncryptedObject you create using Encrypt contains the encrypted data and the initialization vector of the encryption method as a ByteArray. These can be turned into an array of digits using Normal and IntegerDigits.

key = GenerateSymmetricKey[];
message = "Hello world";
eobj = Encrypt[key, message];
bin = eobj["Data"] // Normal // IntegerDigits[#, 2, 8] & // Flatten;
initvec = 
  eobj["InitializationVector"] // Normal // IntegerDigits[#, 2, 8] & // Flatten;

Now you send bin and initvec to the receiver (or do what ever else you wanted to do with the binary digits). To decrypt the message you can create a new EncryptedObject and decrypt it:

eobj2 = EncryptedObject[Association[
     "Data" -> ByteArray[(FromDigits[#1, 2] & ) /@ ArrayReshape[bin, {16, 8}]], 
     "InitializationVector" -> ByteArray[(FromDigits[#1, 2] & ) /@ 
           ArrayReshape[initvec, {16, 8}]], 
     "OriginalForm" -> String]];
Decrypt[key, eobj2]

Note that I used ByteArray, FromDigits, and ArrayReshape to reverse Kuba's Normal, IntegerDigits and Flatten, respectively.

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  • $\begingroup$ Thank you for your help! $\endgroup$ Oct 23, 2017 at 9:20

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