#### 6.1. Use Case Description

Let us assume to use an MFA system as a combination with biometric factors, a password, and a physical key. At first, let this system has two biometric factors: fingerprint and facial recognition.

These factors will be represented as two points with the coordinates $({x}_{1},{y}_{1}),({x}_{2},{y}_{2})$. By using Lagrange and Newton interpolation formulas, we calculate a function (in this particular case, linear function) $f\left(x\right)$, which goes through two these points. Then, we choose password and key values as some points on this curve $({x}_{key},{y}_{key}),({x}_{pass},{y}_{pass})$. Finally, we have a curve $f\left(x\right),degf\left(x\right)=1$ with four factors that are represented as points $({x}_{1},{y}_{1}),({x}_{2},{y}_{2}),({x}_{key},{y}_{key}),$ and $({x}_{pass},{y}_{pass})$ on it.

It becomes straightforward to construct a threshold system $k=2$ from $n=4$ by using such a curve.

Assume, we wish to add to this MFA system one more biometric factor represented as the point with the coordinates $({x}_{3},{y}_{3})$. In this case, we have to calculate new function $\varphi \left(x\right)$, which goes through three points $({x}_{1},{y}_{1}),({x}_{2},{y}_{2}),$ and $({x}_{3},{y}_{3})$ associated with the corresponding three biometric factors. Therefore, we should replace the function $f\left(x\right)$ with function $\varphi \left(x\right)$. After that, in the same way as in the previous case, we choose password and key values as some points $({x}_{ke{y}^{\prime}},{y}_{ke{y}^{\prime}}),({x}_{pas{s}^{\prime}},$ and ${y}_{pas{s}^{\prime}})$ on this curve.

Finally, we have a curve $\varphi \left(x\right),deg\varphi \left(x\right)=2$ with five factors which are represented as points $({x}_{1},{y}_{1}),({x}_{2},{y}_{2}),({x}_{3},{y}_{3}),({x}_{ke{y}^{\prime}},{y}_{ke{y}^{\prime}}),$ and $({x}_{pas{s}^{\prime}},{y}_{pas{s}^{\prime}})$ in it.

Using this curve allows us to construct a threshold system $k=3$ from $n=5$.

Now, we will resolve this case in terms of Lagrange and Newton interpolation formulas.

To construct a new curve associated with the function $\varphi \left(x\right),deg\left(\varphi \right(x\left)\right)=2$, we should use two previous known points $({x}_{1},{y}_{1}),({x}_{2},{y}_{2})$ and one new point $({x}_{3},{y}_{3})$. Using Lagrange interpolation formula, it is necessary to change the polynomial, and thereby all basis polynomials must be recalculated. In Newton’s polynomial form, there is an advantage that only the last term must be found to obtain the new polynomial. Next, we show some simple examples of adding one new biometric factor using two variants of polynomials.

#### 6.2. Lagrange Interpolation Example

This section shows a simple example of how to add a new point using a Lagrange interpolation formula.

At first, let us assume that there are two known points

$({x}_{1},{y}_{1}),({x}_{2},{y}_{2})$ associated with two biometric factors, and it is required to find the function

$f\left(x\right)$, which goes through the corresponding two points.

x | ${x}_{1}$ | ${x}_{2}$ |

| −1 | 0 |

$f\left(x\right)$ | ${y}_{1}$ | ${y}_{2}$ |

| 4 | 2 |

To find the function

$f\left(x\right)$, we use the Lagrange interpolation formula

Then, we simply calculate

$f\left(x\right)$:

If we add a third biometric factor as a point

$({x}_{3},{y}_{3})$ on the second step, we obtain the following.

x | ${x}_{1}$ | ${x}_{2}$ | ${x}_{3}$ |

| −1 | 0 | 1 |

$f\left(x\right)$ | ${y}_{1}$ | ${y}_{2}$ | ${y}_{3}$ |

| 4 | 2 | −2 |

To find a new function

$\varphi \left(x\right)$ with degree 2, we should recalculate all coefficients in the following way.

Here, we demonstrated how to find a new interpolation function using the dataset extended by one biometric factor. In this case, all basis polynomials must be recalculated.

#### 6.3. Newton Interpolation Example

This subsection presents a simple example of how to add a point

$({x}_{3},{y}_{3})=(1,-2)$ using Newton interpolation formula. The same example as in the previous subsection is used.

x | ${x}_{1}$ | ${x}_{2}$ |

| −1 | 0 |

$f\left(x\right)$ | ${y}_{1}$ | ${y}_{2}$ |

| 4 | 2 |

To find the function

$f\left(x\right)$, we use Newton interpolation formula

Let us find divided differences:

Next, we calculate the function

$f\left(x\right)$ as

If we add a third biometric factor as a point

$({x}_{3},{y}_{3})$ at the second step, we obtain the following.

x | ${x}_{1}$ | ${x}_{2}$ | ${x}_{3}$ |

| −1 | 0 | 1 |

$f\left(x\right)$ | ${y}_{1}$ | ${y}_{2}$ | ${y}_{3}$ |

| 4 | 2 | −2 |

Then, we only have to find a new divided difference instead of full recalculation as

and next, we just need to calculate the polynomial

This simple example clearly shows the features of two interpolation algorithms. The presented calculations demonstrate the fundamental difference between the Newton formula and the Lagrange formula: when the number of interpolation points increases in the Newton formula, it is unnecessary to recalculate all the previous components of the interpolation formula.