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What is the minimum number of lines to separate the sets?

enter image description here

a) 2 b) 3 c) 4 d) 1 e) 5

Observe the graph below, it is possible to separate linearly with a line at least two of the classes?

enter image description here

a) Yes, just pass a straight line on the vertical axis.
b) Yes, just pass a straight line on the horizontal axis.
c) Yes, with all points separated.
d) No, because the classes mix.
e) No, because there are four quadrants.

How many circles are needed to group all points into three groups with 2 points and a group with 4 points?

enter image description here

a) 3 circles
b) 5 circles
c) 4 circles
d) Points would remain
e) There is no answer

I believe the answers are

  1. 3
  2. No, because the classes mix
  3. Points would remain

These challenges were proposed to me by a discreet math teacher and, as they are "challenges", I don't imagine that the answers are so simple.

Source

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    $\begingroup$ You said that similar questions were from a particular book; are these also from that book? If so, that source must be cited here. (Additionally, you need to be more precise on what "separating" means; and this also might be skirting the line on what exactly counts as a 'puzzle'.) $\endgroup$
    – Deusovi
    Jan 1 at 1:59
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    $\begingroup$ In the last question 3x2 + 4 = 10 which is less than 12 so it's easy to see you have points leftover. $\endgroup$
    – hexomino
    Jan 1 at 3:44
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Answer 1

3, as follows
enter image description here
It's not too difficult to show that 2 is impossible. The four blue points near the y-axis cannot possibly be grouped together which necessitates at least two lines and the blue point on the left can be grouped with at most one other blue point which leads to the necessity for the third line.

Answer 2

If squares represent points with their centres then I think it is possible to linearly separate the black and blue classes as follows
enter image description here
but that does not seem to quite match up with the multiple choice.

Answer 3

3x2 + 4 = 10 so there are, at most, 10 points in our groupings but there are 12 points displayed. Hence, I think the appropriate answer is "Points would remain", although that is not much fun.

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