The Problem

Given a set of $N$ observations

(x_i,y_i)_{i=1}^N

(x_i,y_i)_{i=1}^N

0

0

6

6

coming from a true (solid blue) but unknown function

The Problem

Given a set of $N$ observations

(x_i,y_i)_{i=1}^N

(x_i,y_i)_{i=1}^N

0

0

6

6

coming from a true (solid blue) but unknown function

Approximate the function from the observations

Approaches

Linear models (using the basis of standard Polynomials)

Neural Networks (MLP)

\vdots

\vdots

Support Vector Machines (SVM)

taking a linear combination of the bases of (a standard) Polynomial function $(1,x,x^2,\cdots,x^n)$

Approximate the function from the observations

f(x)=w_0+w_1x+w_2x^2+\cdots+w_nx^n

f(x)=w_0+w_1x+w_2x^2+\cdots+w_nx^n

f(x)=\sum \limits_{p=1}^nw_p \color{blue}\phi_p(x)

f(x)=\sum \limits_{p=1}^nw_p \color{blue}\phi_p(x)

Given a set of $N$ observations

\{(x_i,y_i)\}_{i=1}^N

\{(x_i,y_i)\}_{i=1}^N

where

\color{red}\phi_p(x)=x^p

\color{red}\phi_p(x)=x^p

Polynomials

0

0

6

6

Basis : Global (fixed)

coming from a true (solid blue) but unknown function

\color{blue}\phi_p(x)=exp(-c(x-\mu_p)^2)

\color{blue}\phi_p(x)=exp(-c(x-\mu_p)^2)

Look for bases that are local or adaptive to the training data

and B-splines

Bases are global functions of input variable (a small change around the region of a data point changes the approximated function globally )

Suffers a lot from Curse of Dimensionality (COD)

Neural Networks

Sigmoidal

For example,

Image source: PRML, Bishop

Neural Networks ( Universal Approximation Theorem UAT)

f(x)=\sum \limits_{p=1}^nw_p \color{blue}\phi_p(x)

f(x)=\sum \limits_{p=1}^nw_p \color{blue}\phi_p(x)

Here, we let the neural network to learn the complex feature transformation $\phi(x)$ via adaptive basis functions

With $N=50$ data points

Learned Bases (Activation functions) of a neuron in a hidden layer with 3 units

Neuron-1

Neuron-2

Neuron-3

Image source: PRML, Bishop

Neural Networks ( Universal Approximation Theorem UAT)

f(x)=\sum \limits_{p=1}^nw_p \color{blue}\phi_p(x)

f(x)=\sum \limits_{p=1}^nw_p \color{blue}\phi_p(x)

Here, we let the neural network to learn the complex feature transformation $\phi(x)$ via adaptive basis functions

With $N=50$ data points

Learned Bases (Activation functions) of a neuron in a hidden layer with 3 units

Neuron-2

Neuron-3

Neuron-1

Ultimately, it is all about combining the activation functions to approximates the true function at observed points

Image source: PRML, Bishop

MLP-Neural Networks ( Universal Approximation Theorem UAT)

UAT: With One (hidden) layer NN

\hat{f}(\mathbf{x}) \approx \sum \limits_{i=1}^{N(\epsilon)}\alpha_i\sigma(W_i^T\mathbf{x}+b)

\hat{f}(\mathbf{x}) \approx \sum \limits_{i=1}^{N(\epsilon)}\alpha_i\sigma(W_i^T\mathbf{x}+b)

We can approximate any continuous function to desired precision $\epsilon$

z=w^Tx

z=w^Tx

For a single neuron

z

z

Here the activation (basis) function $\sigma$ is Fixed

The adaptability to the data is due to the learnable parameters $w$

f(x)=\sum \limits_{p=1}^nw_p \color{blue}\phi_p(x)

f(x)=\sum \limits_{p=1}^nw_p \color{blue}\phi_p(x)

Here, we let the neural network to learn the complex feature transformation $\phi(x)$ via adaptive basis functions

Image source: DL Lectures by Mitesh Khapra

So, what are the problems with MLPs?

Interpretability is difficult with the vast number of learnable weights $w$

Continual learning without re-training is difficult

Catastrophic forgetting due to fixed and non-local activation functions

Sparsification and pruning are difficult due to large number of parameters

KAN overcomes (at least in toy settings) all of these

Adaptive Basis functions are all we need

Consider a multivariate function $f(\mathbf{x})$ with $\mathbf{x} \in \mathbb{R}^n$ , $(x_1,x_2,\cdots,x_n)$

x_1 \in \mathbb{R}

x_1 \in \mathbb{R}

{z}_1 \in \mathbb{R}

{z}_1 \in \mathbb{R}

z_2 \in \mathbb{R}

z_2 \in \mathbb{R}

Bold cliam: Any continuous smooth multivariate function $f(\mathbf{x})$ can be represented using $n$ univariate functions $\phi_p(x_p), p=1,2,\cdots,n$

\phi_2(x_2)

\phi_2(x_2)

\phi_1(x_1)

\phi_1(x_1)

x_2 \in \mathbb{R}

x_2 \in \mathbb{R}

Is $\phi_p(x_p)$ fixed or learnable? Learnable

How do we combine the output from the $n$ univariate functions to get $f(\mathbf{x})$ ?

\cdots

\cdots

\phi_n(x_n)

\phi_n(x_n)

z_n \in \mathbb{R}

z_n \in \mathbb{R}

x_n \in \mathbb{R}

x_n \in \mathbb{R}

\cdots

\cdots

\cdots

\cdots

f(x_1,x_2,\cdots,x_n)=

f(x_1,x_2,\cdots,x_n)=

\sum \limits_{q=1}^{2n+1}\Phi_q

\sum \limits_{q=1}^{2n+1}\Phi_q

\sum \limits_{p=1}^{n}\color{blue}\phi_{p}(x_p)

\sum \limits_{p=1}^{n}\color{blue}\phi_{p}(x_p)

Kolmogorov Representation Theorem

x_1 \in \mathbb{R}

x_1 \in \mathbb{R}

\cdots

\cdots

{z}_1 \in \mathbb{R}

{z}_1 \in \mathbb{R}

z_2 \in \mathbb{R}

z_2 \in \mathbb{R}

\phi_n(x_n)

\phi_n(x_n)

\phi_2(x_2)

\phi_2(x_2)

\phi_1(x_1)

\phi_1(x_1)

z_n \in \mathbb{R}

z_n \in \mathbb{R}

x_2 \in \mathbb{R}

x_2 \in \mathbb{R}

x_n \in \mathbb{R}

x_n \in \mathbb{R}

\cdots

\cdots

\cdots

\cdots

Sum the outputs from all univariate functions

In reality, we need $(n(2n+1))$ $\phi$ s and ( $2n+1$ ) $\Phi_q$ s, totaly $2n^2+3n+1$ learnable univarate functions (not just $n$ as claimed in the previous slide)

\bigg(\sum \limits_{p=1}^{n}\color{blue}\phi_{qp}(x_p) \color{black}\bigg)

\bigg(\sum \limits_{p=1}^{n}\color{blue}\phi_{qp}(x_p) \color{black}\bigg)

\mathbb{R}

\mathbb{R}

[0,1]

[0,1]

\mathbb{R}

\mathbb{R}

\mathbb{R}

\mathbb{R}

Well, How do we make a learnable 1D function?

Hey, B-splines

B-Spline Functions

\phi(x)=\sum \limits_{i=0}^nB_{i}(x)c_i

\phi(x)=\sum \limits_{i=0}^nB_{i}(x)c_i

A function in the basis $B$

$c_i \in \mathbb{R}$ : spline coefficeints (control points)

$B_i(x)$ : The basis function of order $k$ (degree $d=k-1$ )

$t=(t_{-k},\cdots,t_{G+k})$ : Uniformly spaced knots (think of this as a uniformly spaced grid of points defined by $G$ )

$c_i$ 's are learnable parameters (initialized randomly based on the problem)

G=5

G=5

[t_{-k+i},t_{i+1}]

[t_{-k+i},t_{i+1}]

Basis support:

G+k-1

G+k-1

d=?

d=?

k=3

k=3

G=5

G=5

x \in [-1,1]

x \in [-1,1]

h= \frac{x[1]-x[0]}{G}=\frac{2}{5}=0.4

h= \frac{x[1]-x[0]}{G}=\frac{2}{5}=0.4

t=[-k:G+k]=[-3,-2,-1,\cdots,8]

t=[-k:G+k]=[-3,-2,-1,\cdots,8]

h=0.4

h=0.4

(k+1)*h=1.6

(k+1)*h=1.6

Spline width = $(k+1)*h=1.6$

Increasing the value of $G$ decreases $h$ and hence the width of the splines ( sort of controlling the resolution)

?

?

k=3

k=3

G=5

G=5

x \in [-1,1]

x \in [-1,1]

h= \frac{x[1]-x[0]}{G}=\frac{2}{5}=0.4

h= \frac{x[1]-x[0]}{G}=\frac{2}{5}=0.4

t=[-k:G+k]=[-3,-2,-1,\cdots,8]

t=[-k:G+k]=[-3,-2,-1,\cdots,8]

grid=t*h+ (-1)

grid=t*h+ (-1)

offset by starting value of x

[-2.2,-1.8,-1.4,-1,-0.6, \cdots 1, 1.4,1.8,2.2]

[-2.2,-1.8,-1.4,-1,-0.6, \cdots 1, 1.4,1.8,2.2]

[-2.2,-1.8,-1.4,-1,\cdots

[-2.2,-1.8,-1.4,-1,\cdots

The grid extends on both sides with the basis function

h=0.4

h=0.4

Building KAN with $n=2$ inputs

x_{01}

x_{01}

x_{02}

x_{02}

w_{ijk}

w_{ijk}

Each* learnable weight $w_{ijk}$ in MLP is replaced by an learnable activation $\phi_{ijk}$ as shown in the figure right

Each node sums the incoming values

*Not replacing bias (as it is not function of input)

\sigma

\sigma

\sigma

\sigma

\sigma

\sigma

\sigma

\sigma

\sigma

\sigma

\sum \limits_{q=1}^{5}\Phi_q

\sum \limits_{q=1}^{5}\Phi_q

\bigg(\sum \limits_{p=1}^{2}\color{blue}\phi_{qp}(x_p) \color{black}\bigg)

\bigg(\sum \limits_{p=1}^{2}\color{blue}\phi_{qp}(x_p) \color{black}\bigg)

\Phi_5\bigg(\phi_{51}(x_1)+\phi_{52}(x_2)\bigg)

\Phi_5\bigg(\phi_{51}(x_1)+\phi_{52}(x_2)\bigg)

A term in the outer summation

\sigma

\sigma

KAN-Layer-1

$n \to 2n+1$

KAN-Layer-2

$2n+1 \to 1$

n

n

2n+1

2n+1

1

1

x_{01}

x_{01}

x_{02}

x_{02}

w_{ijk}

w_{ijk}

Number of parameters:

*Not replacing bias (as it is not function of input)

\sigma

\sigma

\sigma

\sigma

\sigma

\sigma

\sigma

\sigma

\sigma

\sigma

\sigma

\sigma

15 (ignoring bias)

k=3,G=5

k=3,G=5

For each function

Therefore, 15*(5+3)=120

Number of coefficients: $c=G+k$

What more parameters than MLP?

Building KAN with $n=2$ inputs

\Phi_1=\begin{bmatrix} \phi_{11}() & \phi_{12}() \\ \phi_{21}() & \phi_{22}() \\ \vdots \\ \phi_{51}() & \phi_{52}()\end{bmatrix}

\Phi_1=\begin{bmatrix} \phi_{11}() & \phi_{12}() \\ \phi_{21}() & \phi_{22}() \\ \vdots \\ \phi_{51}() & \phi_{52}()\end{bmatrix}

\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}

\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}

x_{01}

x_{01}

x_{02}

x_{02}

w_{ijk}

w_{ijk}

Number of parameters:

*Not replacing bias (as it is not function of input)

\sigma

\sigma

\sigma

\sigma

\sigma

\sigma

\sigma

\sigma

\sigma

\sigma

\sigma

\sigma

15 (ignoring bias)

15*(8)=120

What more parameters than MLP?

However, often, it requires less number of KAN layers than MLP Layers

Building KAN with $n=2$ inputs

Generalized KAN

$2n^2+3n+1$ learnable functions are sufficient if we are able to find the correct $\phi(x)$ that represents $\mathbf{f(x)}$ .

But, $\mathbf{f(x)}$ is not known. Moreover, we choose smooth B-splines (among other non-smooth alternatives like fractals) to learn $\phi(x)$

Therefore, to get a better approximation, one can build KAN with arbitrary width and depth ( replacing weights by learnable activations)

The author mentions "To the best of our knowledge, there is not yet a “generalized” version of the theorem that corresponds to deeper KANs"

However, they do derive the approximation theory for KAT assuming $k-th$ order B-splines for all $\Phi$ s.

Deep MLP vs KAN

MLP(\mathbf{x})= (W_3 \circ \sigma_2 \circ W_2 \circ \sigma_1 \circ W_1)\mathbf{x}

MLP(\mathbf{x})= (W_3 \circ \sigma_2 \circ W_2 \circ \sigma_1 \circ W_1)\mathbf{x}

KAN(\mathbf{x})= (\Phi_3 \circ \Phi_2 \circ \Phi_1)\mathbf{x}

KAN(\mathbf{x})= (\Phi_3 \circ \Phi_2 \circ \Phi_1)\mathbf{x}

Optimization: Backpropagation

Optimization : Backpropagation, (LBFGS for small networks)

\phi(x)=w(b(x)+spline(x))

\phi(x)=w(b(x)+spline(x))

Training tricks: Residual connection

Training tricks:

b(x)=\frac{x}{1+e^{-x}}

b(x)=\frac{x}{1+e^{-x}}

(silu), $w$ learnable

Initialization: Typically, Xavier

Initialization: Xavier for $w$ , $N(0,0.1)$ for $c_is$

KAN and Scaling Law

Loss: $L \propto N^{-\alpha}$

The higher the value of $\alpha$ for the model parameters $N$ , the lower the loss. We can get quick improvement by scaling the model parameters

How do we predict $\alpha$ ?

Different theories: relating to intricate dimensionality of data $d$ (need to ponder over this), as a function of the class activations.

Function class: Order $k$ of piecewise polynomial like ReLU ( $k=1$ ), (Maxout?)

\alpha=\frac{k+1}{d}

\alpha=\frac{k+1}{d}

For KAN: B-Splines with order $k=3$ gives $\alpha = 4$ (independent of $d$ )

Increasing the number of parameters in KAN reduces the loss quickly than MLP (albeit with more computations)

Fine Graining by Grid Extension

(G+k-1)

(G+k-1)

t_{-k}

t_{-k}

t_{G+k}

t_{G+k}

Uniform Grid intervals (knots)

Splines can adjust its resolution (like wavelets)

Coarse Resolution

Fine Resolution

One can increase the grid size (therefore number of parameters) on the fly during training (without re-training from scratch)

As per scaling law for KAN, the loss $L \propto N^{-4}$ should reduces quickly

Unique to KAN

by varying the Grid size $G$

Validated in a toy setting

Fine Graining by Grid Extension

(G+k-1)

(G+k-1)

t_{-k}

t_{-k}

t_{G+k}

t_{G+k}

Uniform Grid intervals (knots)

Splines can adjust its resolution (like wavelets)

by varying the Grid size $G$

Target

An estimator

Use least squares to solve this

Fine Graining by Grid Extension

(G+k-1)

(G+k-1)

t_{-k}

t_{-k}

t_{G+k}

t_{G+k}

Uniform Grid intervals (knots)

Splines can adjust its resolution (like wavelets) with the Grid size $G$

One can increase the grid size (therefore number of parameters) on the fly during training (without re-training from scratch)

f(x,y)=exp(sin(\pi x)+y^2)

f(x,y)=exp(sin(\pi x)+y^2)

Unique to KAN

As per scaling law for KAN, the loss $L \propto N^{-4}$ should reduces quickly

Fine Graining by Grid Extension

(G+k-1)

(G+k-1)

t_{-k}

t_{-k}

t_{G+k}

t_{G+k}

Uniform Grid intervals (knots)

Splines can adjust its resolution (like wavelets) with the Grid size $G$

One can increase the grid size (therefore number of parameters) on the fly during training (without re-training from scratch)

As per scaling law for KAN, this should reduces the loss quickly

Unique to KAN

KAN trains the network continually without catastrophic forgetting (Caution: not trained and tested on NLP tasks)

Continual Learning

WE can play around with different basis

FastKAN

ChebyKAN

External Vs Internal Degree of Freedom (DoF)

Refer to page 10 and 11 of the paper for details

External DoF: Composition (like MLP) of layers

Internal DoF: Grid size (Unique to KAN)

Sparsification and Pruning

Given the samples from known function structure, the network learns the structure via sparsification and pruning.

Since we do not have linear weights. The $L_1$ norm is computed for each KAN layer $\Phi()$

Here the target function contains three (activation) functions (exp,sin,parabola)

However, the KAN layers have 15 activations

Can it only learn those three out of 15 activation functions?

Sparsification and Pruning

Given the samples from known function structure, the network learns the structure via sparsification and pruning.

Since we do not have linear weights. The $L_1$ norm is computed on $\Phi()$ matrices.

Sparsification and Pruning

Given the samples from known function structure, the network learns the structure via sparsification and pruning.

Since we do not have linear weights. The $L_1$ norm is computed on $\Phi()$ matrices.

Interpretability

Suppose the function structure is unkown

After training, we end up with this simple structure

Then we could infer that the target function contains only three univariate functions

Continual learning and Catastrophic forgetting

Humans are good at multi-tasking without fogetting any of the tasks when learning a new task

human brains have functionally distinct modules placed locally in space

Most artificial neural networks, including MLPs, do not have this notion of locality, which is probably the reason for catastrophic forgetting.

KANs have local plasticity and can avoid catastrophic forgetting by leveraging the locality of splines (a sample will only affect a few nearby spline coefficients, leaving far-away coefficients intact)

Let's validate this via a simple 1D regression task composed of 5 Gaussian peaks

Continual learning and Catastrophic forgetting

Data around each peak is presented sequentially (instead of all at once)

Continual learning and Catastrophic forgetting

Data around each peak is presented sequentially (instead of all at once)

Continual learning and Catastrophic forgetting

Data around each peak is presented sequentially (instead of all at once)

Slow to train.KANs are usually 10x slower than MLPs, given the same number of parameters

Limitations

Findings are validated using Toy experimental settings that are closely related to maths and physics.

They left it to engineers to optimize (efficient implementations have already started emerging)

When to use KAN?

Playground to get started with

https://kindxiaoming.github.io/pykan/examples.html

What is your take on KAN?

What are Splines?

Splines with control points

c_0

c_0

c_1

c_1

(1-t)c_0+t c_1

(1-t)c_0+t c_1

t \in [0,1]

t \in [0,1]

t=0

t=0

t=0.5

t=0.5

t=1

t=1

A line segment is parameterized by $t$ (convex combination of the control points ( $c_0,c_1$ ))

q(t|c_0,c_1,t_0,t_1)=\frac{t_1-t}{t_1-t_0}c_0+\frac{t-t_0}{t_1-t_0}c_1

q(t|c_0,c_1,t_0,t_1)=\frac{t_1-t}{t_1-t_0}c_0+\frac{t-t_0}{t_1-t_0}c_1

t \in [t_0,t_1]

t \in [t_0,t_1]

and

t_0< t_1]

t_0< t_1]

A line segment

q_{0,1}(t)=q(t|c_0,c_1,t_0,t_1)=\frac{t_1-t}{t_1-t_0}c_0+\frac{t-t_0}{t_1-t_0}c_1

q_{0,1}(t)=q(t|c_0,c_1,t_0,t_1)=\frac{t_1-t}{t_1-t_0}c_0+\frac{t-t_0}{t_1-t_0}c_1

A quadradic curve

Control points : $c_0,c_1,c_2$

t : $t_0,t_1,t_2$ , $t_0<t_1<t_2$

c_0

c_0

c_1

c_1

c_2

c_2

q_{1,2}(t)=q(t|c_1,c_2,t_1,t_2)=\frac{t_1-t}{t_1-t_0}c_0+\frac{t-t_0}{t_1-t_0}c_1

q_{1,2}(t)=q(t|c_1,c_2,t_1,t_2)=\frac{t_1-t}{t_1-t_0}c_0+\frac{t-t_0}{t_1-t_0}c_1

Construct two line segments

Then the curve is given by

q_{0,2}(t)=q(t|c_0,c_1,c_2,t_0,t_1,t_2)=\frac{t_2-t}{t_2-t_0} \color{blue} q_{0,1}(t) \color{black}+\frac{t-t_0}{t_2-t_0}\color{blue}q_{1,2}(t)

q_{0,2}(t)=q(t|c_0,c_1,c_2,t_0,t_1,t_2)=\frac{t_2-t}{t_2-t_0} \color{blue} q_{0,1}(t) \color{black}+\frac{t-t_0}{t_2-t_0}\color{blue}q_{1,2}(t)

q_{0,1}(t)

q_{0,1}(t)

q_{1,2}(t)

q_{1,2}(t)

Splines with control points

q_{0,1}(t)=q(t|c_0,c_1,t_0,t_1)=\frac{t_1-t}{t_1-t_0}c_0+\frac{t-t_0}{t_1-t_0}c_1

q_{0,1}(t)=q(t|c_0,c_1,t_0,t_1)=\frac{t_1-t}{t_1-t_0}c_0+\frac{t-t_0}{t_1-t_0}c_1

A quadradic curve

Control points : $c_0,c_1,c_2$

t : $t_0,t_1,t_2$ , $t_0<t_1<t_2$

c_0

c_0

c_1

c_1

c_2

c_2

q_{1,2}(t)=q(t|c_1,c_2,t_1,t_2)=\frac{t_1-t}{t_1-t_0}c_0+\frac{t-t_0}{t_1-t_0}c_1

q_{1,2}(t)=q(t|c_1,c_2,t_1,t_2)=\frac{t_1-t}{t_1-t_0}c_0+\frac{t-t_0}{t_1-t_0}c_1

Construct two line segments

Then the curve is given by

q_{0,2}(t)=q(t|c_0,c_1,c_2,t_0,t_1,t_2)=\frac{t_2-t}{t_2-t_0} \color{blue} q_{0,1}(t) \color{black}+\frac{t-t_0}{t_2-t_0}\color{blue}q_{1,2}(t)

q_{0,2}(t)=q(t|c_0,c_1,c_2,t_0,t_1,t_2)=\frac{t_2-t}{t_2-t_0} \color{blue} q_{0,1}(t) \color{black}+\frac{t-t_0}{t_2-t_0}\color{blue}q_{1,2}(t)

t=(0,1,2)

t=(0,1,2)

Splines with control points

q_{0,1}(t)=q(t|c_0,c_1,t_0,t_1)=\frac{t_1-t}{t_1-t_0}c_0+\frac{t-t_0}{t_1-t_0}c_1

q_{0,1}(t)=q(t|c_0,c_1,t_0,t_1)=\frac{t_1-t}{t_1-t_0}c_0+\frac{t-t_0}{t_1-t_0}c_1

A quadradic curve

Control points : $c_0,c_1,c_2$

t : $t_0,t_1,t_2$ , $t_0<t_1<t_2$

c_0

c_0

c_1

c_1

c_2

c_2

q_{1,2}(t)=q(t|c_1,c_2,t_1,t_2)=\frac{t_1-t}{t_1-t_0}c_0+\frac{t-t_0}{t_1-t_0}c_1

q_{1,2}(t)=q(t|c_1,c_2,t_1,t_2)=\frac{t_1-t}{t_1-t_0}c_0+\frac{t-t_0}{t_1-t_0}c_1

Construct two line segments

Then the curve is given by

q_{0,2}(t)=q(t|c_0,c_1,c_2,t_0,t_1,t_2)=\frac{t_2-t}{t_2-t_0} \color{blue} q_{0,1}(t) \color{black}+\frac{t-t_0}{t_2-t_0}\color{blue}q_{1,2}(t)

q_{0,2}(t)=q(t|c_0,c_1,c_2,t_0,t_1,t_2)=\frac{t_2-t}{t_2-t_0} \color{blue} q_{0,1}(t) \color{black}+\frac{t-t_0}{t_2-t_0}\color{blue}q_{1,2}(t)

t=(0,0.5,2)

t=(0,0.5,2)

unlike for line segment, choice of $t$ (parameters) affects the shape of curve

Splines with control points

Polynomial Splines may not always be smooth

We need to be carefult about the choice of intervals to get a smooth curves

Solution?

Splines - Bezier Curves

Control points $n=3$ : $c_i, i=0,1,2$

t : $t_0,t_1,t_2$

The interval is restricted to [0,1] (convex combination)

p_{0,1}(t)=(1-t)c_0+tc_1

p_{0,1}(t)=(1-t)c_0+tc_1

p_{1,2}(t)=(1-t)c_1+tc_2

p_{1,2}(t)=(1-t)c_1+tc_2

FInal curve is the convex combination of above

p_{0,2}(t)=(1-t)p_{0,1}(t)+tp_{1,2}(t)

p_{0,2}(t)=(1-t)p_{0,1}(t)+tp_{1,2}(t)

p_{0,2}(t)=(1-t)^2c_0+2t(1-t)c_1+t^2c_2

p_{0,2}(t)=(1-t)^2c_0+2t(1-t)c_1+t^2c_2

Substitute the expresssion for $p_{0,1},p_{1,2}$

c_0

c_0

c_1

c_1

c_2

c_2

p_{0,2}(t)=b_{02}(t)c_0+b_{12}(t)c_1+b_{22}(t)c_2

p_{0,2}(t)=b_{02}(t)c_0+b_{12}(t)c_1+b_{22}(t)c_2

\text{degree},d=2

\text{degree},d=2

p_{0,2}(t)=\sum \limits_{i=0}^2b_{i,d}(t)c_i

p_{0,2}(t)=\sum \limits_{i=0}^2b_{i,d}(t)c_i

It is smooth

p_{0,2}(t)=\sum \limits_{i=0}^2b_{i,d}(t)c_i

p_{0,2}(t)=\sum \limits_{i=0}^2b_{i,d}(t)c_i

We can generalize this to degree $d$ and $n$ control points

P(t)=\sum \limits_{i=0}^nB_{i,d}(t)c_i

P(t)=\sum \limits_{i=0}^nB_{i,d}(t)c_i

Where $B_{i,d}(t)$ is a basis function for $c_i$

d \choose i

d \choose i

B_{i,d}(t)=

B_{i,d}(t)=

t^i(1-t)^{d-i}

t^i(1-t)^{d-i}

For example, it is a Bernstein Polynomials for Bezier curves

Basis-Spline

p_{0,2}(t)=(1-t)^2c_0+2t(1-t)c_1+t^2c_2

p_{0,2}(t)=(1-t)^2c_0+2t(1-t)c_1+t^2c_2

p_{0,2}(t)=b_{02}(t)c_0+b_{12}(t)c_1+b_{22}(t)c_2

p_{0,2}(t)=b_{02}(t)c_0+b_{12}(t)c_1+b_{22}(t)c_2

Quick Check for $n=3, d=n-1$

p_{0,2}(t)=\sum \limits_{i=0}^2b_{i,d}(t)c_i

p_{0,2}(t)=\sum \limits_{i=0}^2b_{i,d}(t)c_i

We can generalize this to degree $d$ and $n+1$ control points

P(t)=\sum \limits_{i=0}^nB_{i,d}(t)c_i

P(t)=\sum \limits_{i=0}^nB_{i,d}(t)c_i

In the case of B-Splines, it is a called Blending functions described by order $k=d+1$ with a non-decreasing sequence of numbers called knots $t_i: i=0,1,2,\cdots n+k$

Let us try to understand this with illustrations (borrowed from Kenneth)

From curves to B-Spline Functions

Let us start with first order $k=1$ constant $d=0$ piecewise polynomial functions

B-Splines $B_{i,0}(t)$ , knots $t_i$ are uniformly spaced in the interval [0,1].

$B_{0,0}(t)$

$B_{1,0}(t)$

The first degree ( $d=1$ ) spline is obtained by combining the first order splines

The function for the $i-th$ control point (or knot) is simply a shifted version of $B_{0,d}(t)$ by $i$ units

Image source: Lectures by Kenneth I Joy

B-Splines $B_{i,1}(t)$ , knots $t_i$ are uniformly spaced in the interval [0,1].

Note the increase in support for the function as the order increases

The quadratic (third order) spline is obtained by combining the second order splines

$B_{0,1}(t)$

Image source: Lectures by Kenneth I Joy

B-Splines $B_{i,2}(t)$ , knots $t_i$ are uniformly spaced in the interval [0,1].

$B_{0,2}(t)$

$B_{1,2}(t)$

Image source: Lectures by Kenneth I Joy

Source

P(t)=\sum \limits_{i=0}^nB_{i,d}(t)c_i

P(t)=\sum \limits_{i=0}^nB_{i,d}(t)c_i

Now treat $c_i \in \mathbb{R}$ as a real number

Now, the curve $P(t)$ is learnable via $c_i$

The number of control points $c_i$ and the order $k=d+1$ are independent.

B-Spline Functions

Now, $c_i$ is called B-Spline coefficient

For rigours definition and proofs: here

References

Paper: KAN: Kolmogorov–Arnold Networks

Introduction to Splines

Kolmogorov Arnold Network

Be a Game Changer?

Arun Prakash

AI4BHĀRAT, IIT Madras