prompter: How do I find the distance between two points?

assistant: You need to provide more context.
But if you are talking about two points in the plane you can use the Pythagoras theorem. d = ( ( x_2 - x_1 )^2 + ( y_2 - y_1 )^2 )^½
If you are talking about two points in a 3 dimensional space you will need a generalization of Pythagoras. d = ( ( x_2 - x_1 )^2 + ( y_2 - y_1)^2 + (z_2 - z_1)^2  )^½

prompter: Thanks, but I was interested in the Manhattan distance. How can I compute it?

assistant: The Manhattan distance, also known as the L1 distance, is a measure of the absolute differences between two points in a grid-based space, such as a city grid. To calculate the Manhattan distance between two points (x1, y1) and (x2, y2), you simply add the absolute differences of their coordinates:

d = |x1 - x2| + |y1 - y2|

In a 3-dimensional space, you would extend the formula to:

d = |x1 - x2| + |y1 - y2| + |z1 - z2|

assistant: In comparison to the Euclidian distance the Manhattan distance is calculated by simply adding up the absolute values of the differences in each dimension. It is also known as the L1 norm or L1 metric. In analogy to the above example the Manhattan distance of two two-dimensional points can be calculated like this:

d = |x_2 - x_1| + |y_2 - y_1|,

where the vertical lines || are used to denote the absolute values of the differences. The Manhattan distance for two three-dimensional points can meanwhile be calculated like this:

d = |x_2 - x_1| + |y_2 - y_1| + |z_2 - z_1|.

The Manhattan distance can be further generalized to higher dimensions by simply adding more terms for each dimension. However, when using a high-level language such as Python the actual computation should be done using vectorized operations using libraries like NumPy. This is because for large inputs such operations will be significantly faster than a one-to-one implementation of the above expressions in Python.

assistant: Considering two points in a plane with p1 at (x1, y1) and p2 at (x2, y2), the Manhattan distance between p1 and p2 is: |x1 - x2| + |y1 - y2|
Where | x | stands for the absolute value of x.

assistant: You can find the distance between two points by using the distance formula.
With points (x1, y1) and (x2,y2), the distance between them is given by

√((x2-x1)^2+(y2-y1)^2).

An example is to find the distance between the points (2,4) and (6, 1).
In this case, x1 = 2, y1 = 4, x2 = 6, and y2 = 1.
So plugging it in, we get
√((6-2)^2+(1-4)^2)
By doing the subtractions and then squaring, we get
√(16+9)=√(25)=5
So the distance between the two points is 5.

prompter: but how to do it for 3 points?

assistant: You can use the distance formula to calculate the distance between two points:
d = √(x2 - x1)2 + (y2 - y1)2

So to find the distance between 3 points:

Calculate the distance between points 1 and 2
Calculate the distance between points 2 and 3
Calculate the distance between points 1 and 3

Those 3 distances form the sides of a triangle, and you can use that to calculate various properties of the triangle if needed.
For example:
point1 = (1, 2)
point2 = (4, 6)
point3 = (7, 8)

d12 = √(4 - 1)2 + (6 - 2)2 = 5
d23 = √(7 - 4)2 + (8 - 6)2 = 4
d13 = √(7 - 1)2 + (8 - 2)2 = 7

So the 3 sides of the triangle are 5, 4 and 7 units.

assistant: You can use the Pythagorean Theorem to get the distance between the 2 points.

If this is in a 2d space you take the difference in the x axis and square it and you add that to the distance in the y axis squared. Then you take the square root of the total
for $(x_1,y_1)$ and $(x_2,y_2)$
$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
For 3d space you just add the third coordinate like this:
for $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$
$\sqrt{(x_2-x_1)^2+(y_2-y1)^2+(z_2-z_1)^2}$

prompter: Can you provide some example with actual numbers and include the steps?

assistant: Let's say you have two points, $(x1, y1)$ and $(x2, y2)$

Step 1: Calculate the difference between the two points for each coordinate:

$x2 - x1$

$y2 - y1$

Step 2: Square each of the differences from Step 1.

$(x2 - x1)^2$

$(y2 - y1)^2$

Step 3: Add the squared differences from Step 2 together.

$(x2 - x1)^2 + (y2 - y1)^2$

Step 4: Take the square root of the sum from Step 3.

$\sqrt{(x2 - x1)^2 + (y2 - y1)^2}$

Now, let's try this with some actual numbers. Suppose you have two points, $(3, 4)$ and $(6, 8)$.

Step 1: Calculate the difference between the two points for each coordinate:

$x2 - x1 = 6 - 3 = 3$

$y2 - y1 = 8 - 4 = 4$

Step 2: Square each of the differences from Step 1.

$(x2 - x1)^2 = 3^2 = 9$

$(y2 - y1)^2 = 4^2 = 16$

Step 3: Add the squared differences from Step 2 together.

$(x2 - x1)^2 + (y2 - y1)^2 = 9 + 16 = 25$

Step 4: Take the square root of the sum from Step 3.

$\sqrt{(x2 - x1)^2 + (y2 - y1)^2} = \sqrt{25} = 5$

So, the distance between the two points, $(3, 4)$ and $(6, 8)$, is 5.