Pippenger's Algorithm

Reduction of multi-scalar multiplication (MSM) into double and add operations

AKA "Bucket" method

Motivation

Improve upon Double-and-add by reducing the total number of double and add operations needed for MSM.

Algorithm Process

Introduction

Multi-Scalar Multiplication (MSM) refers to the efficient computation of expressions in the following form:

$$S = \sum_{i=1}^{n} k_i P_i$$

• $k_i$: the $i$-th $\lambda$-bit scalar

• $P_i$: the $i$-th point on an elliptic curve

Pippenger's algorithm speeds up MSM by dividing each scalar $k_i$ into multiple $s$-bit chunks, and then grouping and processing scalars by their corresponding bit positions (windows).

1. Scalar Decomposition

First, we decompose the scalars $k_i$:

$$\begin{align*} k_i &= k_{i,1} + 2^{s}k_{i,2} + ... + 2^{(\lceil\frac{\lambda}{s}\rceil -1)s}k_{i,\lceil\frac{\lambda}{s}\rceil } \\ k_i &= \sum_{j=1}^{\lceil\frac{\lambda}{s}\rceil }2^{(j - 1)s} k_{i,j} \end{align*}$$

• $s$ : the number of bits per window

• $\lceil\frac{\lambda}{s}\rceil$ : the total number of windows

Through this definition, we can rewrite the original MSM summation as follows:

$$\begin{align*} S &= \sum_{i=1}^{n} k_i P_i \\ &= \sum_{i=1}^{n} \sum_{j=1}^{\lceil\frac{\lambda}{s}\rceil } 2^{(j-1)s}k_{i,j} P_i \\ &= \sum_{j=1}^{\lceil\frac{\lambda}{s}\rceil } 2^{(j-1)s} \left( \sum_{i=1}^{n} k_{i,j} P_i \right) \\ &= \sum_{j=1}^{\lceil\frac{\lambda}{s}\rceil } 2^{(j-1)s} W_j \end{align*}$$

Note that $W_j$ is the sum for the $j$-th window.

2. Bucket Accumulation

Given our $W_j$, we denote buckets $B$ as the following:

$$B_{\ell, j} = \sum_{i: k_{ij} = \ell} P_i$$

Or in other words, $B_{\ell, j}$ refers to the $\ell$ bucket in window $W_j$ consisting of the summation of points $P_i$ with the same $k_{i,j}$.

Note that $k_{i,j}$ must be within the range $[0, 2^w - 1]$; however, if $k_{i,j}$ is 0, as $k_{i,j}$ is the scalar multiplier, the result of the bucket will be 0. Therefore, the practical range of $k_{i,j}$ is $[1, 2^w - 1]$.

3. Bucket Reduction

Thus, in total, a window $W_j$ will be re-structured as follows:

$$W_j = \sum_{i=1}^{n} k_{ij}P_i = \sum_{\ell=1}^{2^s - 1} \ell B_{\ell, j}$$

Note that the computation of a singular $\ell B_{\ell,j}$ is computed through a running sum of recursive summations and not through further scalar multiplications.

Complexity Cost:

• Original: $(2^s - 1) \times \mathsf{ScalarMul} + (2^s - 2) \times \mathsf{PointAdd}$

• Bucketed: $(2^{s+1} - 2) \times \mathsf{PointAdd}$

C++ Implementation

Bucket ReduceBucketPoints(const std::vector<Bucket>& buckets) {
  Bucket res = Bucket::Zero();
  Bucket sum = Bucket::Zero();
  for (size_t i = buckets.size() - 1; i != SIZE_MAX; --i) {
    sum += buckets[i];
    res += sum;
  }
  return res;
}

Example

$$W_1 = B_{1,1} + 2B_{2,1} + 3B_{3,1} + 4B_{4,1}$$

// Round 1
sum = B_(4,1);
res = B_(4,1);
// Round 2
sum = B_(3,1) + B_(4,1);
res = B_(4,1) + [B_(3,1) + B_(4,1)];
// Round 3
sum = B_(2,1) + B_(3,1) + B_(4,1);
res = B_(4,1) + [B_(3,1) + B_(4,1)] + [B_(2,1) + B_(3,1) + B_(4,1)];
// Round 3
sum = B_(1,1) + B_(2,1) + B_(3,1) + B_(4,1);
res = B_(4,1) + [B_(3,1) + B_(4,1)] + [B_(2,1) + B_(3,1) + B_(4,1)] + [B_(1,1) + B_(2,1) + B_(3,1) + B_(4,1)];

// Total:
res = B_(1,1) + 2*B_(2,1) + 3*B_(3,1) + 4*B_(4,1);

4. Window Reduction - Final MSM Result

With the finished individual window computations, we return to the total MSM summation and add all windows together:

$$S= \sum_{j=1}^{\lceil\frac{\lambda}{s}\rceil } 2^{(j-1)s} W_j$$

In practice, the computation is calculated recursively from $j = \left\lceil \frac{\lambda}{s} \right\rceil$ to $j = 1$ like so:

$$T_j = \begin{cases} 0 & j = \left\lceil \frac{\lambda}{s} \right\rceil \\ W_j + 2 ^sT_{j+1} & j \ge 1 \\ \end{cases}$$

Complexity Cost

$$\left\lceil \frac{\lambda}{s} \right\rceil(s \times \mathsf{PointDouble} + \mathsf{PointAdd})$$

C++ Implementation

Bucket ReduceWindows(uint32_t lambda, uint32_t s, const std::vector<Bucket>& buckets) {
  Bucket res = Bucket::Zero();
  uint32_t window_num = static_cast<uint32_t>(std::ceil(lambda / s));
  for (uint32_t j = window_num - 1; j != UINT32_MAX; --i) {
    for (uint32_t i = 0; i < s; ++i) {
      res = res * 2;
    }
    res += buckets[j];
  }
  return res;
}

Complexity Analysis

• To compute $W_j$, $n \times \mathsf{PointAdd}$ is required: $\left\lceil \frac{\lambda}{s} \right\rceil(n \times \mathsf{PointAdd})$

• Each $W_j$ needs $(2^{s+1} - 2) \times \mathsf{PointAdd}$: $\left\lceil \frac{\lambda}{s} \right\rceil((2^{s+1} - 2) \times \mathsf{PointAdd})$

• Final $Q$ needs: $\left\lceil \frac{\lambda}{s} \right\rceil(s \times \mathsf{PointDouble} + \mathsf{PointAdd})$

Total complexity

$$\left\lceil \frac{\lambda}{s} \right\rceil(s \times \mathsf{PointDouble} + (n + 2^{s+1} - 1) \times \mathsf{PointAdd}) \approx \\\lambda \times \mathsf{PointDouble} + \left\lceil \frac{\lambda}{s} \right\rceil(n + 2^{s+1} - 1)\times \mathsf{PointAdd}$$

Optimization Table (When , $n = 2^{20}$)

s	PointAdd Count
2	134,218,624
4	67,110,848
8	33,570,784
16	18,874,352
32	68,727,865,336
64	147,573,952,589,680,607,228
128	1,361,129,467,683,753,853,853,498,429,727,074,942,974

🔍 This clearly shows that selecting an optimal $s$ value is critical for performance.

Example

Let's try to run Pippenger's on the following simple example:

$$\begin{align*} S &= k_1P_1 + k_2P_2 + k_3P_3\\ &=12P_1 + 9P_2 + 13P_3 \end{align*}$$

1. Scalar Decomposition

Let's define our bits per window $s$ to be 2. In this case, since the maximum amount of bits needed to constrain the values 12, 9, and 13 is 4, we will have $\lceil\frac{\lambda}{s}\rceil =\mathsf{max\_bits} / s = 4/2 = 2$ windows.

Note that in binary we have:

$$k_1 =12 = \underbrace{0}_{2^0}\underbrace{0}_{2^1}\underbrace{1}_{2^2}\underbrace{1}_{2^3}\\ k_2 = 9 = \underbrace{1}_{2^0}\underbrace{0}_{2^1}\underbrace{0}_{2^2}\underbrace{1}_{2^3} \\ k_3 =13 = \underbrace{1}_{2^0}\underbrace{0}_{2^1}\underbrace{1}_{2^2}\underbrace{1}_{2^3}$$

Decomposing our $k_0$, $k_1$, and $k_2$ results in the following:

$$\begin{align*} k_i &= k_{i,1} + 2^{s}k_{i,2} + ... + 2^{(\lceil\frac{\lambda}{s}\rceil -1)s}k_{i,\lceil\frac{\lambda}{s}\rceil } \\ \end{align*}$$

$$\begin{align*} \begin{array}{rlrlrl} k_1 = 12 &= \overbrace{\underbrace{0}_{2^0}\underbrace{0}_{2^1}}^{\text{window 1}}\overbrace{\underbrace{1}_{2^2}\underbrace{1}_{2^3}}^{\text{window 2}} & \quad k_2 = 9 &= \overbrace{\underbrace{1}_{2^0}\underbrace{0}_{2^1}}^{\text{window 1}}\overbrace{\underbrace{0}_{2^2}\underbrace{1}_{2^3}}^{\text{window 2}} & \quad k_3 = 13 &= \overbrace{\underbrace{1}_{2^0}\underbrace{0}_{2^1}}^{\text{window 1}}\overbrace{\underbrace{1}_{2^2}\underbrace{1}_{2^3}}^{\text{window 2}} \\[1ex] &= k_{1,1} + 2^sk_{1,2} & \quad &= k_{2,1} + 2^sk_{2,2} & \quad &= k_{3,1} + 2^sk_{3,2} \\[1ex] &= 0 + 3(2^2) & \quad &= 1 + 2(2^2) & \quad &= 1 + 3(2^2) \\[1ex] \end{array} \end{align*}$$

2+3. Bucket Accumulation + Reduction

$$W_j = \sum_{i=1}^{n} k_{ij}P_i = \sum_{\ell=1}^{2^s - 1} \ell B_{\ell, j}$$

Given our $s$ of 2, we now have buckets of $[B_{1,j}, B_{2,j}, B_{3,j}]$ for a given window $W_j$. Our windows are now re-structured to buckets like so:

$$\begin{align*} W_1 &= \overbrace{1}^{k_{2,1}}P_2 + \overbrace{1}^{k_{3,1}}P_3 = 1(P_2+P_3) = B_{1,1}\\ W_2 &= \overbrace{3}^{k_{1,2}}P_1 + \overbrace{2}^{k_{2,2}}P_2 + \overbrace{3}^{k_{3,2}}P_3 = 2(P_2) + 3(P_1 + P_3) = 2B_{2,2} + 3B_{3,2} \end{align*}$$

From bottom up, in our bucket accumulation stage we compute the buckets:

Bucket	Computation
$B_{1,1}$	$P_2 + P_3$
$B_{2,2}$	$P_2$
$B_{3,2}$	$P_1+P_3$

...and in our bucket reduction stage we compute the windows:

$$\begin{align*} W_1 &= B_{1,1}\\ W_2 &= (B_{2,2} + B_{2,2}) + (B_{3,2} + B_{3,2} + B_{3,2}) \end{align*}$$

4. Window Reduction - Final MSM Result

Putting all the windows together for the MSM sum, we get the following:

$$\begin{align*} S &=\sum_{j=1}^{\lceil\frac{\lambda}{s}\rceil } 2^{(j-1)s} W_j \\ &= W_1 + 2^2W_2 \\ \end{align*}$$

References

• https://hackmd.io/lNOsNGikQgO0hLjYvH6HJA

• https://eprint.iacr.org/2022/1321.pdf

Written by ryan Kim &Ashley Jeong of A41