There are probably many better explanations out there, but I'll try one:
It's important t note that if you pick any point p in the Sierpinsky triangle and you calculate f1(p), f2(p) and f3(p) you get tree points that are also in the Sierpinsky triangle. [f1, f2, f3 as defined in the top comment]
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You pick any point in the plane, let's call it q, and you draw it in the screen.
You pick any point in the Sierpinsky triangle, let's call it p, p is a secret point.
The important number is the distance between p and q. You could have pick a better point in the Sierpinsky triangle but anyone you have choose is fine.
Now you select one of the functions f1, f2, f3, let's call it f. And you calculate p'=f(p) and q'=f(q) (the same functions for both).
Now you have two new points p' and q'. p' is a again a point in the Sierpinsky triangle as I said at the beginning. q' is a point that you draw in the screen.
Again, the important number is the distance between p' and q'. What happens if you compare the distance between p' and q' with the initial distance between p an q? The new distance is one half of the initial distance, so q' is closer to p'. You could have pick a better initial point p in the Sierpinsky triangle or pick now a better point p*in the Sierpinsky triangle but don't worry, p' is fine.
And now you repeat the process with cut&paste...
You select one of the functions f1, f2, f3, let's call it f. It may be the same functions as before or another one. You calculate p''=f(p') and q''=f(q') (the same functions for both).
Now you have two new points p'' and q''. p'' is a again a point in the Sierpinsky triangle as I said at the beginning. q'' is a point that you draw in the screen.
Again, the important number is the distance between p'' and q''. What happens if you compare the distance between p'' and q'' with the previos distance between p' and q' and the initial distance between p an q? The new distance is one half of the previous distance and one quarter of the initial distance, so q'' is closer to p''. You could have pick a better initial point p in the Sierpinsky triangle or pick now a better point p**in the Sierpinsky triangle but don't worry, p'' is fine.
And now you repeat the process with cut&paste...
...
After enough halvings, the distance between p'''''...'''' and q'''''...'''' is smaller that any level of precision you select initially, like 1E-15. If you need more precision because you are using a high zoom level, just add more steps and more halvings.
Now you have a sequence of points q, q', q'', q''', q'''', ... that are drawn in the screen and p, p', p'', p''', p'''', ... that are only in your imagination.
Except a few of the initial points, all the other points q'''''...'''' are very close to the Sierpinsky triangle. One trick is to not draw the initial 1000 points, so you don't get some points in weird locations. You should prove that 1000 is enough, or use a bigger number or just cross your fingers and hope nobody notices them.
Note that this method actually doesn't draw the Sierpinsky triangle. It just draw a cloud of points that very close to the Sierpinsky triangle. (Unless you are very lucky, and the initial point q is also in the Sierpinsky triangle or something like that.)
