# Why the VC dimension of triangles in 2D space is not greater than 7?

I understand that there are sets of 7 points on a circle that can be fully shattered using triangles.But, it is not clear to me why it cannot shatter 8 points.

Is there any intuitive way of arriving at the conclusion that it can't shatter 8 points? Is there a simple explanation without using much advanced geometry tools?

• The tags seem unmotivated by anything in the question. statistics? classification? machine-learning? – Gerry Myerson Oct 4 '18 at 23:54
• Sorry for my ignorance... The answer below gives me a guess of what "shattering" may mean, but then I don't see how a triangle can "shatter" 7 pts on a circle... Can you please define shattering, or give a 7 pt example? Thx. – Yaakov Baruch Oct 5 '18 at 10:08
• 7 point example: if you have 7 points along the perimeter of a circle and say they are randomly labelled + and -. Using a triangle I will be able to separate them into positive and negative classes. This is because, with 7 points, you can have max of 3 contiguous blocks of -. (Note that they are arranged in circular way). Now, you can separate the three blocks using 3 corners of your triangle. Hence 7 points can be shattered. – Jyotish Robin Oct 5 '18 at 16:03
• @JyotishRobin. Thank you. That also clarifies what shattering means. – Yaakov Baruch Oct 6 '18 at 19:37

Let $$x_1, \dots, x_8$$ be $$8$$ points on a circle in this order. Then there is no triangle containing exactly $$x_2, x_4, x_6, x_8$$ in its interior and the other four points in its exterior. Intuitively, it would have to intersect the circle in at least $$8$$ points, but any triangle intersects a circle in at most $$6$$ points.