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  • Triangles and quadrilaterals should be built on all possible points: A , B , red points , and green points.

  • Expressions of "point on diameter" and "point on arc" are ambiguous, because example points A and B are common to both. Let me ask another way, because I am still not sure if we all talk about the same thing:

    Having

    • two points, A and B marked as black,
    • many (n >= 3) points on arc (other than A and B) marked as red,
    • however many points on a diameter (other than A and B) marked as green, zero or more,

    is a triangle or a quadrilateral built on red points only, without green, and without black, legal, or not?

    It could be easier if I just attempted the kata, but having this clear would also be nice, I think :)

  • @Blind4Basics I agree, points of the diameter

  • Yeap, you are right, not all the triangles and quadrilaterals should have a point or points which lay on the diameter

  • mmh... Maybe I haven't been clear enough about what part I was talking about. It's not about opinions here, it's about saying something that doesn't match the actual task ;)

    "All the triangles and quadrilaterals always have, at least, a point of the diameter".

    this is incorrect.

  • @Blind4Basics I said, in my opinion. I know your opinion. The instructions, don't show an example with n >= 4.

  • nope, again, this would be wrong. ;)

  • IMO it should say something similar to "All the triangles and quadrilaterals always have, at least, a point of the diameter". It isn´t enough, the condition that m > 0, because if n >= 3 or n >= 4, it's possible to generate them. The users above are requiring this explanation.

  • It seems that both shapes need, at least, one element of both kind of points.

    to enforce the answer of the author: no. But there is a "trick".

  • In the image example it is not shown, but yes, for example, if we would have 3 points on arc let's say A , B, and C , it is possible to built a triangle ABC, the same way with quadrilaterals.

  • In the instructions its written that m>0, (m number of points on the diamete).Also we do not have in test cases situation where m=0. In fact it is possible m to be 0, but do not forget that we at least have 2 endpoints on the diameter which belong to both diameter and arc. So, in the framework of this kata I decided to have at least 1 point on the diameter except end points.

  • It seems that both shapes need, at least, one element of both kind of points. Let´s see the author's answer, this should be explained in the instructions.

  • yes, or any other way

  • Just to make it clear: is it OK if a triangle or a quadrilateral is built on points located solely on arc?

  • I`ll try it today.

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