A python library for the fully homomorphic encryption scheme ACES

This package proposes an implementation for the ACES cryptosystem where we take parameters $\omega = 1$ and $\mathsf{deg}(u) = n$. Details are available in the preprint "Constructing a Fully Homomorphic Encryption Scheme with the Yoneda Lemma".

The name "PyACES" is pronounced "pisces," echoing a symbol synonymous with malleability. In the Pisces symbol, the two fishes represent a harmonious interplay between opposite directions, reflecting the dual nature of encryption and decryption.

Install the package using pip:

pip install pyaces

Then use the following import in your Python code.

import pyaces

To upgrade to the latest version:

pip install --upgrade pyaces==version_number

ACES depends on the generation of an arithmetic channel, which is a quadruple $\mathsf{C} = (p,q,\omega,u)$ where $p$, $q$ and $\omega$ are three positive integers and $u$ is a polynomial in $\mathbb{Z}[X]$ for which the equation $u(\omega) = q$ holds in $\mathbb{Z}$. While the integer $\omega$ remains fixed at $1$, and the ACES library automatically generates the polynomial $u$, the user is still required to manually select the values for $p$ $q$, $n = \mathsf{deg}(u)$ and $N$.

A brief review of section 5.1 of the research paper already informs us that the condition $p^2 < q$ must hold for ACES to be valid. In conjunction with this prerequisite, the accompanying guide, accessible here in this repository, provides further insights into considerations for adjusting the values of $p$, $q$ $n$ and $N$. Specficially, users should take

• $p^2 < q$
• $p$ and $q$ coprime
• $q \gg (K_0Np)^{2^{K+1}}$ for some constant $K_0$ to process at least $K$ layers of operations
• $n = \mathsf{deg}(u) > 4$

For simplicity, we recommend that users use the formula below, where $s \geq 2$, $t \geq 2$, $K_0 \geq 1$, $K_1 \geq 1$ and $K \geq 1$:

$$q = (K_0Np)^{t^{K+1} + K_1 } + 1$$

To create an arithmetic channel $\mathsf{C} = (p,q,\omega,u)$, use the class ArithChannel. The following snippet shows how to generate an arithmetic channel with

• $p=2^5 = 32$,
• $q= p^{2^2+1}+1 = 33554433$,
• $\mathsf{deg}(u) = 10 = n$
• and $N=2$

>>> from pyaces import *
>>> ac = ArithChannel(32,32**5+1,10,2)

To generate the public key of a scheme with fully homomorphic properties, use the following instance:

>>> (f0,f1,vanmod,intmod,dim,N,u,tensor) = ac.publish(fhe = True)

We can quickly verify that the condition $u(1)=q$ holds as follows.

>>> u
[1]^10+[15561488]^9+[10729359]^8+[5895107]^7+[10512515]^6+[13114310]^5+[31946593]^4+[17963261]^3+[-72168201]^2 (33554433)
>>> sum(u.coefs)-intmod
0

Below, we will access the private key $x$ via the property ac.x. It is important to note that the integers $p$ and $q$ used for initializing the arithmetic channel ac must satisfy the relationships $p^2 < q$ and $\mathsf{gcd}(p, q) = 1$. In cases where these conditions are not met, the class ArithChannel will generate a new value for $q$ as $p^2+1$.

As elaborated in the research paper and the preceding section, the homomorphism property imposes an upper limit on the maximum achievable level. This limit is determined by the ratio $q/p$, and it can be calculated as shown below:

>>> q_p = intmod/vanmod
>>> q_p
1048576.03125

To encrypt messages $m \in \mathbb{Z}_p$, use the class ACES as shown below.

>>> bob = ACES(f0,f1,vanmod,intmod,dim,N,u)

The following example illustrates an encryption of the message $m=3$. The encryption algorithm outputs a ciphertext $(c,c')$ and a vector of integers that can be used to calculate the level of the ciphertext. For the sake of exposition, we also display the polynomials making the vector $c \in \mathbb{Z}_q[X]_u^{(n)}$.

>>> enc3, k3 = bob.encrypt(3)
>>> k3
[2, 14]
>>> for d in enc3.dec:
...   d
...
[21379560]^9+[25028370]^8+[8694508]^7+[28102446]^6+[5172890]^5+[24589603]^4+[5543292]^3+[11274872]^2+[13629468]^1+[19271064]^0 (33554433)
[9108842]^9+[5039426]^8+[21244355]^7+[3175448]^6+[22655005]^5+[15962061]^4+[20588907]^3+[26006670]^2+[27534017]^1+[16266756]^0 (33554433)
[1458536]^9+[24643021]^8+[15203777]^7+[23778795]^6+[24103882]^5+[23940451]^4+[3773861]^3+[9206299]^2+[10862051]^1+[7246808]^0 (33554433)
[22153519]^9+[7780220]^8+[16360279]^7+[28545698]^6+[22573294]^5+[163643]^4+[27158635]^3+[8665100]^2+[5091255]^1+[4834865]^0 (33554433)
[12109157]^9+[4473554]^8+[24217584]^7+[17701710]^6+[1472804]^5+[8994119]^4+[30242395]^3+[3669714]^2+[11512159]^1+[4809361]^0 (33554433)
[16571218]^9+[31797286]^8+[21517446]^7+[9449064]^6+[31547175]^5+[19754057]^4+[22483530]^3+[22371627]^2+[6812861]^1+[20406855]^0 (33554433)
[6535938]^9+[20943309]^8+[10663544]^7+[9754641]^6+[6342352]^5+[25438883]^4+[29421614]^3+[30503775]^2+[14529486]^1+[456676]^0 (33554433)
[17806160]^9+[19626456]^8+[29126106]^7+[24033050]^6+[23506780]^5+[28391799]^4+[4000439]^3+[10046721]^2+[19523779]^1+[17246181]^0 (33554433)
[15096919]^9+[15509580]^8+[14202625]^7+[28337590]^6+[9646278]^5+[31062288]^4+[26781822]^3+[8778106]^2+[20152751]^1+[5062739]^0 (33554433)
[2363338]^9+[16535237]^8+[30285612]^7+[4934198]^6+[21283726]^5+[5267871]^4+[26811682]^3+[5040335]^2+[7297640]^1+[24881255]^0 (33554433)
>>> enc3.uplvl
64

In the previous example, the vector k3 can be used to determine the level of the ciphertext $(c,c')$. Specifically, this is done by computing the scalar product of k3 with the vector ac.lvl_e. The integer enc3.uplvl is known to all parties and represents an upper bound on the level of $(c,c')$. It is crucial to emphasize that, in practical scenarios, the vector k3 (and the vector ac.lvl_e) should remain confidential, and only the theoretical upper bound enc3.uplvl is deemed safe to share. In fact, it is strongly recommended to retain only integers $0 \leq k'_i \leq$k3[i] and securely erase k3 from the computer memory.

To decrypt an encrypted message, use the ACESReader class. The following example demonstrates how to decrypt the ciphertext enc3. As expected, we retrieve the message $m=3$.

>>> alice = ACESReader(ac)
>>> alice.decrypt(enc3)
3

To do arithmetic on encrypted information, use the class ACESAlgebra. The following example shows how to recover the homomorphism properties:

$$\mathsf{Dec}(\mathsf{Enc}(3) + \mathsf{Enc}(5)) = 3+5\quad\quad\mathsf{Dec}(\mathsf{Enc}(3) \cdot \mathsf{Enc}(5)) = 3 \cdot 5$$

>>> alg = ACESAlgebra(vanmod,intmod,dim,u,tensor)
>>> enc5, k5 = bob.encrypt(5)
>>> alice.decrypt(alg.add(enc3,enc5))
8
>>> alice.decrypt(alg.mult(enc3,enc5))
15

Now that we have explored the features of the leveled FHE underlying ACES, let us illustrate the FHE protocol described in section 5.3 of the research paper. To make this tutorial relevant to illustrating the refreshing step for the FHE protocol, we will now take the parameters

• $p=2^5 = 32$,
• $q=10\cdot p^{2^2+1}+1 = 335544321$,
• $\mathsf{deg}(u) = 10 = n$
• and $N=5$

>>> from pyaces import *
>>> ac = ArithChannel(32,10*32**5+1,10,5)
>>> (f0,f1,vanmod,intmod,dim,N,u,tensor) = ac.publish(fhe = True)
>>> bob = ACES(f0,f1,vanmod,intmod,dim,N,u)
>>> alice = ACESReader(ac)
>>> alg = ACESAlgebra(vanmod,intmod,dim,u,tensor)
>>> q_p = intmod/vanmod

Now, we want to consider a list array of data and its encryption through ACES. We can illustrate this with the following lines of code.

>>> import random as rd
>>> array = [rd.randint(0,5) for _ in range(8)]
>>> enc_array = [bob.encrypt(a) for a in array]

As seen earlier, the algorithm bob.encrypt() returns the ciphertext and its associated level. Since we cannot share levels, we want to separate them as follows.

>>> send_array, keep_array = map(list,zip(*enc_array))

We now have the following data:

>>> array
[5, 4, 5, 0, 3, 3, 1, 1]
>>> send_array
[<pyaces.aces.ACESCipher object at 0x7f7c63e9ff10>, <pyaces.aces.ACESCipher object at 0x7f7c8cecb610>, <pyaces.aces.ACESCipher object at 0x7f7c63ebd700>, <pyaces.aces.ACESCipher object at 0x7f7c63ebd070>, <pyaces.aces.ACESCipher object at 0x7f7c63ebdee0>, <pyaces.aces.ACESCipher object at 0x7f7c63ebdcd0>, <pyaces.aces.ACESCipher object at 0x7f7c63ebdc40>, <pyaces.aces.ACESCipher object at 0x7f7c63ec5f10>]
>>> keep_array
[[21, 8, 17, 14, 7], [13, 27, 18, 3, 31], [0, 3, 23, 4, 23], [16, 30, 6, 9, 13], [29, 28, 18, 18, 26], [14, 19, 3, 13, 22], [18, 1, 6, 23, 18], [20, 10, 6, 30, 0]]

Since operations on levels need to be dissociated form the algebraic operations done on ciphertext, we want to initiate the refresher algebra as follows.

>>> rfr = ACESRefresher(ac)

To simulate an FHE protocol, let us suppose that $\mathsf{Alice}$ has sent the encrypted data send_array on a remote server and consider a scenario where an algoritm of additions and multiplications as shown below runs on the encrypted data.

$$F:(x_0,x_1,x_2,\dots,x_7) \mapsto (x_0 \cdot x_1 + x_2 \cdot x_3 + x_4 \cdot x_5) \cdot x_6 + x_7$$

We define this algorithm on the non-encrypted data array (as a sanity check), on the list of encrypted data send_array and on the list of levels keep_array as instructed in section 5.3 of the research paper.

>>> true_fun = Algebra().compile("(0*1+2*3+4*5)*6+7")
>>> send_fun = alg.compile("(0*1+2*3+4*5)*6+7")
>>> keep_fun = rfr.compile("(0*1+2*3+4*5)*6+7")
...

First, let us see what the algorithm is meant to output for the values in array:

>>> truth = true_fun(array)
>>> truth % 32
30

Now, let us shift our focus to the server side. To recap, we previously set $q = 10\cdot p^{2^2+1}+1$, implying that ACESAlgebra can likely accommodate only a single layer of a sum of multiplications. However, the send_fun function incorporates two such layers. As a result, this configuration is sufficient to surpass the levels beyond $q/p$, resulting in a failure of the leveled FHE procedure, as demonstrated below.

>>> online = send_fun(send_array)
>>> online.uplvl
12582912160
>>> q_p - online.uplvl
-12572426399.96875
>>> alice.decrypt(online)
25

As explained in Section 5.3 of the research paper, it is necessary to decompose the algorithm $F$ into smaller layers, namely $F_1$ and $F_2$. For instance, the definition of $F_1$ is presented below.

>>> true_fun1 = Algebra().compile("0*1+2*3+4*5")
>>> send_fun1 = alg.compile("0*1+2*3+4*5")
>>> keep_fun1 = rfr.compile("0*1+2*3+4*5")

When we apply $F_1$ to the corresponding ciphertexts, the leveled homomorphism proves to be successful, as evidenced below.

>>> truth1 = true_fun1(array)
>>> truth1 % 32
29
>>> online1 = send_fun1(send_array)
>>> online1.uplvl
2457600
>>> q_p - online1.uplvl
8028160.03125
>>> alice.decrypt(online1)
29

Subsequently, we can calculate the equivalent of $F_1$ on the corresponding levels using keep_fun1. In practice, the resulting output level k1 (or in fact any positive integer less than k1) is transmitted to the server, while maintaining the secrecy of the levels in keep_array. Then, the server uses the integer k1 to refresh the ciphertext online1.

>>> k1 = keep_fun1(rfr.process(keep_array))
>>> c1 = alg.refresh(online1,k1)
>>> c1.uplvl
2380640
>>> alice.decrypt(c1)
29

As can be seen above, the refresh operation does not impact the message associated with the ciphertext; rather, it diminishes the upper bound level c1.uplvl. While this reduction may appear trivial when compared to the earlier value of online1.uplvl, the refresh operation nevertheless remains efficient in revitalizing the ciphertext.

To confirm the success of this refresh operation, we can now proceed to compute the second layer of the algorithm $F$. First, let us define the function $F_2$ associated with this the second layer.

>>> true_fun2 = Algebra().compile("8*6+7")
>>> send_fun2 = alg.compile("8*6+7")
>>> keep_fun2 = rfr.compile("8*6+7")

If we now combine this second layer with the output of the refresh operation, we can validate that the computed ciphertext accurately recovers the true value of the computation.

>>> truth2 = true_fun2(array+[truth1])
>>> truth2 % 32
30
>>> online2 = send_fun2(send_array + [c1])
>>> online2.uplvl
12188876960
>>> q_p - online2.uplvl
-12178391199.96875
>>> alice.decrypt(online2)
30

A limitation in our presentation was that our algorithm struggled to handle numerous layers of additions and multiplications. Although this might restrict the volume of information processed simultaneously, we could enhance the level range by setting $p=2$ and $q = 2^{2^5+5}+1 = 137438953473$.