# GLWE Homomorphic Addition with a Constant

Often you don’t want to add *another encrypted value* you just want to add a **known public constant** (like `+1`, or a known polynomial offset) to an encrypted message.

With GLWE, this is extremely simple and (importantly) **does not increase noise** in the same way ciphertext - ciphertext addition does.

***

## 1) What we have: a GLWE ciphertext encrypting `M`

A GLWE ciphertext encrypting a plaintext polynomial $$M \in \mathcal{R}\_p$$ has the form:

$$C = (A\_0,\ldots,A\_{k-1},B) \in \mathcal{R}\_q^{k+1}$$

with body:

$$B = \sum\_{i=0}^{k-1} A\_i S\_i + \Delta M + E \pmod q$$

* $$\vec{S}=(S\_0,\ldots,S\_{k-1})$$ is the secret key
* $$E$$ is the noise polynomial
* $$\Delta = q/p$$ is the encoding scale factor

***

## 2) The goal: add a known constant polynomial

Let $$\Sigma \in \mathcal{R}\_p$$ be a **public plaintext polynomial** you want to add:

$$M ;\mapsto; M + \Sigma$$

Examples of $$\Sigma$$:

* “add 1 to the constant term” → $$\Sigma(X)=1$$
* “add 3 to every coefficient” → $$\Sigma(X)=3 + 3X + 3X^2 + \cdots$$ (depending on your plaintext convention)
* “add a known offset vector packed into a polynomial”

***

## 3) Trick: represent the constant as a *trivial GLWE ciphertext*

A **trivial GLWE ciphertext** is a GLWE-shaped object with:

* all mask polynomials set to zero
* body set to the encoded plaintext (no secret key mixing)

For the constant $$\Sigma$$, the trivial ciphertext is:

$$C\_{\text{triv}}(\Sigma) = (0,\ldots,0,\Delta\Sigma)$$

This is “ciphertext-shaped,” so we can add it to other GLWE ciphertexts cleanly.

***

## 4) Homomorphic addition with a constant (the actual operation)

Just add component-wise:

$$C' = C + C\_{\text{triv}}(\Sigma)$$

That means:

* masks stay unchanged (because you add zeros): $$A\_i' = A\_i$$
* body shifts by $\Delta\Sigma$: $$B' = B + \Delta\Sigma \pmod q$$

So the resulting ciphertext is:

$$C' = (A\_0,\ldots,A\_{k-1},; B+\Delta\Sigma)$$

***

## 5) Why it works (correctness)

Decrypt $$C'$$:

$$
\begin{aligned}
T' &= B' - \sum\_i A\_i' S\_i \pmod q \\
&= (B + \Delta\Sigma) - \sum\_i A\_i S\_i \pmod q \\
&= (\Delta M + E) + \Delta\Sigma \pmod q \\
&= \Delta(M+\Sigma) + E \pmod q
\end{aligned}
$$

After rounding/decoding by $$\Delta$$, you recover:

$$M' = M + \Sigma \quad \text{in } \mathcal{R}\_p$$

You get the plaintext addition you wanted.

***

## 6) Noise impact (why this is “cheap”)

Notice what changed:

* The noise term stayed **exactly** $$E$$.
* You did **not** add a second independent noise term (as in ciphertext + ciphertext).

So “add constant” is typically treated as a nearly-free operation in noise budgeting. (You’re just shifting the encoded plaintext.)

This exact “add a constant by adding a trivial ciphertext” trick is described in the deep dive.

***

## 7) Tiny toy example

Let:

* $$q=64$$, $$p=4$$ ⇒ $$\Delta = q/p = 16$$
* You have an encryption of plaintext polynomial $$M$$, so the decrypted pre-rounded value is: $$T = \Delta M + E$$
* You want to add constant polynomial $$\Sigma = 1$$ (i.e., add 1 to the plaintext)

Make trivial ciphertext: $$C\_{\text{triv}}(\Sigma) = (0,\ldots,0,\Delta\Sigma) = (0,\ldots,0,16)$$

Add it: $$B' = B + 16$$

After decryption: $$T' = T + 16 = \Delta M + E + \Delta\Sigma = \Delta(M+1)+E$$

So decoding yields $$M+1$$.

***

## 8) Practical notes 

* $$\Sigma$$ **must be in the same plaintext space** ($$\mathcal{R}\_p$$) as $$M$$.
* You must use the **same** $$\Delta$$ (same `p` and `q` parameters) when forming $$\Delta\Sigma$$.
* If your scheme uses rounding during encoding (non power-of-two settings), then $$\Delta\Sigma$$ follows the same encoding conventions.


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