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<h1 id="carryless-arithmetic"><a class="header" href="#carryless-arithmetic">Carryless arithmetic</a></h1>
<p>When learning arithmetic in grade school, you learned the base 10 system where digits range from 0-9. Whenever adding digits exceeds 9, you have to carry the 1. However, this is not the only way to do arithmetic; we can instead omit the carry and allow digits to exceed 9!</p>
<p>Why would we want to do this? FHE operations actually add and multiply <a href="/intro/why.html">polynomials</a> under the hood. If we treat each coefficient of the polynomial as a digit and use carryless arithmetic, we can trick them into behaving like numbers. This results in more efficient data types.</p>
<h2 id="addition"><a class="header" href="#addition">Addition</a></h2>
<p>Let's start with a simple carryless addition example. Adding 99 + 43 without propagating carries gives us:</p>
<pre><code class="language-ignore"> 9 9
+ 4 3
------
13 12
</code></pre>
<p>That is <code>13</code> in the 10s place and <code>12</code> in the 1s place which is <code>13 * 10 + 12 * 1 = 142</code>. This is the same value as if we had propagated carries (<code>1 * 100 + 4 * 10 + 2 * 1 = 142</code>). Under carryless arithmetic, <em>both</em> are valid ways to represent the value 142; representations are no longer unique!</p>
<p>Furthermore, we can represent negative values by negating each digit. For example, <code>-123</code> is <code>-1 * 100 + -2 * 10 + -3 * 1</code>. Mechanically, we simply treat each polynomial coefficient as <code>p</code>'s complement<sup class="footnote-reference"><a href="#1">1</a></sup> value to allow negative numbers.</p>
<p>We can easily extend this reasoning to base 2 (binary). Under carryless arithmetic, base 2 simply means that each digit is a multiple of a power of 2. For example, we can compute <code>3 + 2 + (-4)</code> as follows:</p>
<pre><code class="language-ignore">3 + 2 =
1 1 = 3
+ 1 0 = 2
-----
2 1 = 2*2^1 + 1*2^0 = 5
5 + (-4) =
0 2 1 = 5
+ -1 0 0 = -4
--------
-1 2 1 = -1*2^2 + 2*2^1 + 1*2^0 = 1
</code></pre>
<p>Again <code>-1 2 1</code> equals <code>1</code>, as does <code>0 0 1</code>; the representation is not unique.</p>
<div class="footnote-definition" id="1"><sup class="footnote-definition-label">1</sup>
<p><code>p</code> is the <a href="/advanced/plain_modulus/plain_modulus.html">plain modulus</a></p>
</div>
<h3 id="adding-polynomials"><a class="header" href="#adding-polynomials">Adding polynomials</a></h3>
<p>To see why carryless arithmetic is useful, let's take the values of our previous example (<code>3</code>, <code>2</code>, <code>-4</code>) and map their digits onto polynomials like so: </p>
<p>\[3 = 0\times 2^2 + 1\times 2^1 + 1\times 2^0 \rightarrow 0x^2 + 1x + 1\]</p>
<p>\[2 = 0\times 2^2 + 1\times 2^1 + 0\times 2^0 \rightarrow 0x^2 + 1x + 0\]</p>
<p>\[-4 = -1\times 2^2 + 0\times 2^1 + 0\times 2^0 \rightarrow -1x^2 + 0x + 0\]</p>
<p>To add polynomials, recall that you simply collect the terms:</p>
<p>\[3 + 2 \rightarrow (0x^2 + 1x + 1) + (0x^2 + 1x + 0) = 0x^2 + 2x + 1\]</p>
<p>Now, let's subtract 4:</p>
<p>\[5 + (-4) \rightarrow (0x^2 + 2x + 1) + (-1x^2 + 0x + 0) = -1x^2 + 2x + 1 \]</p>
<p>Note that the coefficients are the same as the digits in our carryless arithmetic result. We can convert the polynomial back into an integer by evaluating it at <code>x = 2</code>, yielding <code>1</code>!</p>
<h2 id="multiplication"><a class="header" href="#multiplication">Multiplication</a></h2>
<p>Now, lets go through a multiplication example with binary carryless arithmetic. Here, we multiply <code>7 * 13 = 91</code>:</p>
<pre><code class="language-ignore"> 0 1 1 1 = 7
* 1 1 0 1 = 13
---------------
0 1 1 1
+ 0 0 0 0
+ 0 1 1 1
+ 0 1 1 1
---------------
0 1 2 2 2 1 1 = 1*2^5 + 2*2^4 + 2*2^3 + 2*2^2 + 1*2^1 + 1*2^0 = 91
</code></pre>
<p>Notice that when we collected each place, we didn't propagate carries when values exceeded <code>1</code>.</p>
<p>As another example, when we square <code>1 0 0 0 0 = 16</code>, we get <code>1 0 0 0 0 0 0 0 0 = 256</code>. Compared to the previous example, the operands are larger (<code>16 * 16</code> vs. <code>7 * 13</code>), but the result's largest digit is smaller — <code>1</code> in <code>1 0 0 0 0 0 0 0 0</code> vs. <code>2</code> in <code>0 1 2 2 2 1 1</code>. This will come up again when we talk about overflow.</p>
<h3 id="multiplying-polynomials"><a class="header" href="#multiplying-polynomials">Multiplying polynomials</a></h3>
<p>As with addition, we'll now show how carryless multiplication directly maps to polynomial multiplication. Let's encode <code>7</code> and <code>13</code> as polynomials:</p>
<p>\[7 = 0\times 2^3 + 1\times 2^2 + 1\times 2^1 + 1\times 2^0 \rightarrow 0x^3 + 1x^2 + 1x^1 + 1\]</p>
<p>\[13 = 1\times 2^3 + 1\times 2^2 + 0\times 2^1 + 1\times 2^0 \rightarrow 1x^3 + 1x^2 + 0x^1 + 1\]</p>
<p>Let's multiply these polynomials:</p>
<p>\[7 \times 13 \rightarrow (0x^3 + 1x^2 + 1x^1 + 1)(1x^3 + 1x^2 + 0x^1 + 1) \]
\[= 1x^3(0x^3 + 1x^2 + 1x^1 + 1) + 1x^2(0x^3 + 1x^2 + 1x^1 + 1)\]
\[ + 0x^1(0x^3 + 1x^2 + 1x^1 + 1) + 1(0x^3 + 1x^2 + 1x^1 + 1) \]</p>
<p>\[=(0x^6 + 1x^5 + 1x^4 + 1x^3) + (0x^5 + 1x^4 + 1x^3 + 1x^2)\]
\[+(0x^4 + 0x^3 + 0x^2 + 0x^1) + (0x^3 + 1x^2 + 1x^1 + 1) \]</p>
<p>\[=0x^6 + 1x^5 + 2x^4 + 2x^3 + 2x^2 + 1x^1 + 1\]</p>
<p>Again, the coefficients of this polynomial exactly match our carryless addition result, so we can trivially map our polynomial back into a number to recover our answer!</p>
<p>You might be wondering: &quot;can polynomial coefficients grow indefinitely?&quot; </p>
<p>Unfortunately, they can't.</p>
<h2 id="overflow"><a class="header" href="#overflow">Overflow</a></h2>
<p>As with normal computer arithmetic, values in FHE can <em>overflow</em>. In Sunscreen, plaintext polynomials have <a href="https://en.wikipedia.org/wiki/Method_of_complements"><code>p</code>'s complement</a> coefficients, where <code>p</code> is the <a href="./plain_modulus/plain_modulus.html">plaintext modulus</a>. Assuming <code>p</code> is odd, this means each coefficient must be in the range \([\frac{-p}{2}, \frac{p}{2}] \) <sup class="footnote-reference"><a href="#2">2</a></sup>. The result of any operation that falls outside this range will wrap around — it overflows.</p>
<p>Suppose <code>p = 7</code> and we try to add <code>3 + 1</code>. Values should be within the range \([-3,3]\), but <code>3 + 1 = 4</code> is not and thus wraps around to <code>-3</code>! This example looks at a single value, but polynomials feature many coefficients, each of which has the same range restriction.</p>
<div class="footnote-definition" id="2"><sup class="footnote-definition-label">2</sup>
<p>When <code>p</code> is odd, \(\frac{p}{2}\) rounds down. If <code>p</code> is even, the interval is \([\frac{-p}{2}, \frac{p}{2})\), i.e. the upper bound opens.</p>
</div>
<h3 id="addition-1"><a class="header" href="#addition-1">Addition</a></h3>
<p>Let's look at an addition example, treating the coefficients as carryless arithmetic binary digits. Take <code>p = 9</code> (meaning digits are in the interval \([-4,4]\)) and add <code>15 = 4 0 -1</code><sup class="footnote-reference"><a href="#3">3</a></sup> with <code>8 = 2 2 -4</code>.</p>
<pre><code class="language-ignore"> 4 0 -1 = 15
+ 2 2 -4 = 8
--------
-3 2 4 = -3*2^2 + 2*2^1 + 4*2^0 = -4
</code></pre>
<p>From this example, we have two observations:</p>
<ul>
<li>We expected <code>6 2 -5 = 23</code>, but both the 4s and 1s places overflowed, giving us a very strange <code>-4</code>!</li>
<li>Operands' digits only contribute to their respective place in the result (i.e. the 4s place contributes only to the 4s place, the 2s place to only the 2s place, etc).</li>
</ul>
<p>If we increase <code>p</code> to <code>13</code> and repeat the example, we do get the correct answer because <code>6</code> and <code>-5</code> are within \([-6,6]\).</p>
<div class="footnote-definition" id="3"><sup class="footnote-definition-label">3</sup>
<p>Recall that values have many representations under carryless arithmetic — this example is typical after performing a few operations!</p>
</div>
<h3 id="multiplication-1"><a class="header" href="#multiplication-1">Multiplication</a></h3>
<p>Next, let's consider canonical representations of <code>31 = 1 1 1 1 1</code> and <code>15 = 0 1 1 1 1</code> when <code>p = 7</code> (i.e. digits in interval \([-3, 3]\)). Adding these numbers doesn't lead to overflow, but what about multiplication? Let's find out:</p>
<pre><code class="language-ignore"> 1 1 1 1 1 = 31
* 0 1 1 1 1 = 15
--------------------------
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
0 0 0 0 0
--------------------------
0 1 2 3 -1 -1 3 2 1
= 0*256 + 1*128 + 2*64 + 3*32 + -1*16 + -1*8 + 3*4 + 2*2 + 1
= 345
</code></pre>
<p>We draw 2 observations from this example:</p>
<ul>
<li>We expected <code>465 = 0 1 2 3 4 4 3 2 1</code>, but got <code>345</code> because the 16s and 8s places overflowed.</li>
<li>The number of non-zero digits in each operand contributes to the result digits' magnitudes.</li>
</ul>
<p>Increasing <code>p</code> to <code>9</code> eliminates the overflow since <code>4</code> is in the interval \([-4, 4]\).</p>
<p>If we multiply canonical representations of <code>32 = 1 0 0 0 0 0</code> and <code>16 = 0 1 0 0 0 0</code> when <code>p = 7</code>, surely this will overflow as well, right? After all, they're bigger numbers than in our previous example! As it turns out, this gives exactly the right answer: <code>512 = 1 0 0 0 0 0 0 0 0 0</code>. The operands feature far fewer non-zero digits, and thus are less impacted by our second observation.</p>
<p>We chose operand digits with value 1 in this example to reveal the second observation, but larger values further compound digit growth! You can see this by redoing the example with 4's: <code>124 = 4 4 4 4 4</code> times <code>60 = 0 4 4 4 4</code> equals <code>7440 = 0 16 32 48 64 64 48 32 16</code>.</p>
<h3 id="preventing-overflow"><a class="header" href="#preventing-overflow">Preventing overflow</a></h3>
<p>Overflow is a bit counterintuitive under carryless arithmetic, but you can prevent it — simply increase the <a href="./plain_modulus/plain_modulus.html">plaintext modulus</a>. Understanding your computation and knowing the range of your inputs can help you choose the appropriate value.</p>
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