Talk:Huzita–Hatori axioms

I'm doing a project on the relationship between math and origami and this page is very useful, except that it doesn't ever say what "s" is in the equations (the very first one is F(s) = p1 + s(p2 - p1) and they never tell you what the F or the s stand for. The s is bothering me more.) I'm so frustrated. Could anyone answer this as soon as possible please? Thanks!!

The s is the independent variable in the parametric equation. I found it pretty confusing at first, too, until I learned some more about parametric equations, and then it made more sense. The F is just an arbitrary name for the function, sort of like in normal algebraic functions like f(x) = x². Anyhow, the idea in those parametric equations is that s varies continuously from minus infinity to infinity; when s=5, the first equation says that a point on the line is p1+5(p2-p1). Read up on parametric equations if this still doesn't make sense :-) The MathWorld site is a really good resource for math-related stuff. Also, the last axiom (Axiom 6) currently does not have a parametric equation because I was unable to find a source that provided one. It's likely to be even more complex than the one for Axiom 5, since it finds a line that is tangent to two parabolas. -- Wapcaplet 01:33, 17 Jan 2004 (UTC)

See http://www.nytimes.com/2004/06/22/science/22orig.html about non straight-fold origami -- Anon

Interesting! I've seen some other stuff about curved folds in origami; very impressive stuff. Though, as far as I know it has nothing to do with Huzita's axioms, which only pertain to straight folds. I wonder if there are similar axioms or equations that could describe such curved folds; they'd certainly be far more complex! The equation of a line is fairly simple, but the equation for an arbitrary curve... definitely above my head, mathematically speaking. -- Wapcaplet 19:00, 22 Jun 2004 (UTC)

Edit: I see now that you posted this by way of explanation for adding the phrase "straight-fold" to the article. Good call. -- Wapcaplet 19:09, 22 Jun 2004 (UTC)

The seventh axiom?

This page talks about the Huzita-Hatori axioms, including a 7th axiomatic fold discovered by Koshiro Hatori in 2002 that is independent of Huzita's 6 established axioms.

I'm not in a position to judge the validity of this page, and haven't found any other online references. I hope someone in a better position to judge will update the article as appropriate. Hv 03:02, 2 September 2005 (UTC)[reply]

Here's the link to Hatori's own discussion. Hv 03:09, 2 September 2005 (UTC)[reply]

Parametric Equations and Inner Products

This page is seriously broken. The Axioms don't specify a model, so one should not assume that one is working with an inner product space or a space where parametric equations make sense. For example, if a piece of paper is not convex, one cannot apply the algorithm in the exposition of Axiom 2 to find folds mapping any two points to each other. Other algorithms can be broken easily as well.

I don't know enough about this field to fix the article myself, but it is either missing hypotheses used in the field or is taking mathematical liberties without justification. It is not obvious that every model for these axioms is convex. Indeed, that is false. I also take issue with the assumption that every model of the axioms is an inner product space. -- poopdeville

I've added to the article two of the references I remember using. My math isn't too great (in fact, I didn't understand most of what you just said) so I hope someone else can tidy it up. I thought it was pretty straightforward, since the axioms only deal with a 2D cartesian plane; is it more complicated than that? -- Wapcaplet 23:37, 20 September 2005 (UTC)[reply]

- The problem is that I can think of a lot of structures that satisfy the axioms but for which you can't use inner products. As a simple example of what I mean, consider the group axioms. Any set that satisfies them is a "group." Everything that can be proved about groups (in general -- that is, statements of the form "If G is a group, then G...") follows from the group axioms. Similarly, anything that can be proved about sets that satisfy Huzita's Axioms can be proved from Huzita's Axioms. Unfortunately, it doesn't follow from the axioms that every set satisfying the axioms is convex.

Intuitively, if you take a star-shaped sheet of paper, you'll see that you can (1) fold it so that any two points coincide, (2) fold it so that there is a single line going between them, (3) fold it so that any line coincides with any other, and (4) so on. But you can't use the algorithms presented in the article to find these folds because there is no guarantee that the parametric function F(s) exists everywhere.

I thought the intended model was supposed to be the whole plane. If a piece of paper is not convex, that only means it's not a model of these axioms, not that the axioms are "seriously broken". Michael Hardy 23:46, 15 October 2005 (UTC)[reply]

I don't get it

If Axiom 5 and 6 can have 0 solutions, how are they true?--Kuciwalker 16:30, 17 January 2006 (UTC)[reply]

Not axioms, but operations

Kuciwalker, I had a similar feeling of 'how can they be axioms if they're not always true?' However, on Robert Lang's origami page (http://www.langorigami.com/science/hha/hha.php4) it reads,

"Humiaki Huzita and Benedetto Scimemi presented a paper [...] in which they identified six distinctly different ways one could create a single crease by aligning one or more combinations of points and lines [...] on a sheet of paper. Those six operations became known as the Huzita axioms. The Huzita axioms provided the first formal description of what types of geometric constructions were possible with origami."

So I think the axioms describe the ways in which folds may be made, but do not necessarily imply that they can be made in all situations. I agree with you that the way the axioms are presented, as unequivocal statements, is confusing.

7th axiom

What does this mean?

Robert J. Lang has proven that this list of axioms completes the axioms of origami.

In what sense are the 7 axioms complete?

And in what sense is the seventh axiom independent? As far as I know, the first 6 axioms can construct all points that you get by (iteratively) taking square roots and cube roots. Does axiom 7 add anything to that?

194.24.138.3 (talk) 23:26, 18 March 2010 (UTC)[reply]

It looks to me like the origami-version of the parallel postulate; without it you might have a hyperbolic or spherical space instead of a Euclidean one. Of course, this wouldn't change the class of constructable angles, just the metric of the points. (I don't know if anyone has ever considered constructable points in the sphere/hyperbolic plane. Ben Standeven (talk) 04:32, 20 April 2012 (UTC)[reply]

These aren't a minimal set of axioms, in fact one could get away with just axiom 6. They are the complete set of single folds one can do in origami. Dmcq (talk) 14:26, 20 April 2012 (UTC)[reply]

360 degree protractor construction

I would like to see it mentioned that a 360-gon can be created through paper folding while it cannot be created by compass and straight edge. Since the 360 gon is used for the division of 1 degree it might be nice to show that paper folding provides a potential for constructing it. Also, if anyone has a link to someone constructing 1 degree with paper folding... I'd love to see it. — Preceding unsigned comment added by Peawormsworth (talk • contribs) 19:35, 4 September 2012 (UTC)[reply]

Things need to be in a reliable source first, Wikipedia is most definitely a follower not a place to put one's own ideas. Dmcq (talk) 10:37, 5 September 2012 (UTC)[reply]