The paper Fully Homomorphic Encryption over the Integers talks about a super simple symmetric key implementation on page 1 and 2.

It says that to generate a key, you pick a random odd number between $[2^{N-1} ,2^N)$ where N is the size of your key in bits.

To encrypt a bit (a 0 or a 1), you do this:
$c = pq +2r + m$

In other words:
$CipherBit = Key * RandomIntegerA + 2 * RandomIntegerB + PlainTextBit$

RandomIntegerA is there to hide the key better and RandomIntegerB is a small number to add noise to the encrypted value (to make it a "learning with error" problem).

To recover an encrypted bit you do this:
$PlainTextBit = (CipherBit \% Key) \% 2$

You can do an XOR by adding two cipher bits together, and you can do an AND by multiplying them together.

I've simplified my implementation a bit to make it easier to explain in a presentation, so i have RandomIntegerA hard coded as 1, and RandomIntegerB hard coded at 0.

This simplifies encryption (yes, making it insecure, but more easily explained) to this operation:
$CipherBit = Key + PlainTextBit$

The decryption, AND and XOR operations remain the same.

Now that the explanation is out of the way, here is my issue...

When Key is 9 (1001 in binary), let's say i have 3 cipherbits A,B and C that are valued at 10, 10 and 114100 respectively.

A quick check when decrypting them shows that they are all true bits:
$(10 \% 9) \%2 == 1$
$(114100 \% 9) \% 2 == 1$

I hit a problem though when I calculate A xor B xor C. To do this, I add them all up and get 114120, which decrypts to a zero bit:
$(114120 \% 9) \% 2 == 0$

However, 1 xor 1 xor 1 is 1!

Something I've noticed is that before doing the % 2 on A,B,C their remainders are 1, 1, 7 respectively.

When adding them together, it effectively adds (and mods) their remainders, so it makes sense that (1+1+7)%9 would be 0.

Can anyone see where I've gone wrong?


EDIT: Here is an even simpler counter example With a key of 9, say we have two encrypted bits: 1 and 8, which decrypt to 1 and 0 respective.

To do an XOR, we should be able to add them. When we add them, we get 9.

If we then decrypt 9, 9%9 is 0, so it says that 1 xor 0 is 0, which is wrong!

It seems like this scheme doesn't work if you hit these perfect "roll over" points.

Like if your key is 533 and you want to xor two encrypted bits: 532 (0) and 1 (1), you get 533, which decrypts to zero.

Or if your key is 5 and you want to xor 4 (0) and 1 (1), you add and get 5, which decrypts again to zero!

I've seen plenty of other papers extending the homomorphic encryption over integers... surely they can't all be wrong :P Not sure what I'm messing up though...

  • $\begingroup$ Could it be order of operations? (1 xor 1) xor 1 = 0 xor 1 = 1 whereas an xor(1,1,1) = 0 for an xor operator that takes more than 2 inputs. $\endgroup$
    – mikeazo
    Aug 27, 2015 at 19:08
  • $\begingroup$ Thanks for the response. Its a good idea but the order that xor is done (or addition) doesn't change the result. 1 xor 1 xor 1 comes up with a result of 1 no matter what order you do it in. Also 10+10+114100 comes up with the same result no matter what order you do it in. $\endgroup$
    – Alan Wolfe
    Aug 27, 2015 at 19:11
  • $\begingroup$ Maybe it is just a corner case that is so unlikely to happen with real parameters and proper use? Like in RSA if you try to encrypt a multiple of one of the prime factors, the encryption is the identity: crypto.stackexchange.com/questions/1004/… $\endgroup$
    – mikeazo
    Aug 27, 2015 at 19:25
  • $\begingroup$ It could be, but I've never seen it mentioned before in the papers I've read so far, and it seems like a pretty big hole. I'd go far as to say this is not really fully homomorphic encryption if it has a chance of giving the wrong answer! :P $\endgroup$
    – Alan Wolfe
    Aug 27, 2015 at 19:28
  • $\begingroup$ and in the case of RSA doing identity encryption, a person usually has no way to tell that the data isn't encrypted, so i'd say it's still working, even if not working as well, versus this being a malfunction. Room for argument there, but IMO! :P $\endgroup$
    – Alan Wolfe
    Aug 27, 2015 at 19:37

1 Answer 1


The problem is that you're getting an "overflow" of the errors relative to the secret key.

We have $10 \bmod 9 = 1$, and $114100 \bmod 9 = 7$, so the "errors" in your ciphertexts are $1$, $1$, and $7$, which are all odd, hence the plaintext bits are $1$ (as desired). When you add the ciphertexts, the errors add correspondingly, so you get a ciphertext with error $9$, which normally would decrypt to $1$. However, in this case your secret key is also $9$, so in fact decryption "sees" an error of $0$, i.e., the "intended" error overflows back to zero. This is why your final ciphertext decrypts to $0$ instead of $1$.

As the paper explains, the accumulated errors must stay smaller than the secret key in order for decryption to work. (For simplicity, throughout this answer I'm assuming only non-negative errors, and that the modular reduction returns a non-negative representative.)

  • $\begingroup$ oh I see... I was under the false impression that by not adding any error to the encryptions, that it would remain error free, and then it would only be limited by the size of the integers I used (These numbers with error are the results of other AND and XOR operations). Unfortunate but totally understandable. Thank you so much Chris, it makes a lot of sense. $\endgroup$
    – Alan Wolfe
    Aug 27, 2015 at 20:10
  • 3
    $\begingroup$ The entire $(2r+m)$ component, i.e., the residue modulo $p$, plays the role of the "error," so your example ciphertexts still have error in them. $\endgroup$ Aug 27, 2015 at 20:13
  • $\begingroup$ you know, it's really too bad the key has to be odd. I understand why (the parity of the cipher and plain text matches with an even key, which makes encryption useless!), but with an even key, when the residue gets larger than the key, the answers are still correct. If someone could figure out a way to let that happen, seems like we wouldn't need things like boot strapping and modulus switching to get FHE! $\endgroup$
    – Alan Wolfe
    Aug 28, 2015 at 16:44

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