# Support Vector Machine basics

## Homework Statement

If I have a 3*3 grid, or 3*3 matrix, which records clicked points.
I.e.

Pattern T =

[1,1] = 1
[1,2] = 1
[1,3] = 1
[2,1] = -1
[2,2] = 1
[2,3] = -1
[3,1] = -1
[3,2] = 1
[3,3] = -1

or

1 1 1
-1 1 -1
-1 1 -1

What is the best way to statistically match x = 1 vs x = not 1

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Mark44
Mentor

## Homework Statement

If I have a 3*3 grid, or 3*3 matrix, which records clicked points.
I.e.

Pattern T =

[1,1] = 1
[1,2] = 1
[1,3] = 1
[2,1] = -1
[2,2] = 1
[2,3] = -1
[3,1] = -1
[3,2] = 1
[3,3] = -1

or

1 1 1
-1 1 -1
-1 1 -1

What is the best way to statistically match x = 1 vs x = not 1

What do you mean "statistically match"?

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I.e.

A new sequence is entered and we wish to identify group membership to T or C based on sample data for T and C. SInce T and C won't necessarily consist of 2 example database( 1 for each) but rather 1000 for T and 1000 C examples, how does one accurately calculate a match?

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Mark44
Mentor
I still don't have a clue what you're trying to do.

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My question is related to optical character recognition. So if I draw a T, how do I match it to T that accurately reflects the entire training data?

I.e. if I have pattern T1, pattern T2, .., are the sequences for n T's.
C1, C2,..,CN is C training points for n C's.

If I feed a single C sequence in now, how do I weigh my decision most accurately?

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Mark44
Mentor
OK, now I understand. At the risk of oversimplification, let me define T this way:
Pattern T =

1 1 1
0 1 0
0 1 0

To match this pattern, a candidate pattern A should have a distance of 0 from this pattern, with distance calculated as
$$\sqrt{(a_0 - t_0)^2 + (a_1 - t_1)^2 + (a_2 - t_2)^2 + ... + (a_8 - t_8)^2}$$

In this formula I have flattened out your matrix to a one-dimensional array. The ti values are from pattern T, and the ai values are from the candidate pattern A.

If you get a "distance" of 0, the two patterns match exactly. If you have several candidate patterns with nonzero distances, pick the one with the smaller "distance."

That's how I would approach it.