Of the trio of *"Fig.9"* forms that I presented elsewhere and recently referred to (where two are symmetric, and the 3rd is the form that rock-climbers & cavers have dubbed "Fig.9"),

Yet only

**one** ( the third / "red" ) of the three Fig.9 forms, shown in the first post, is

*symmetric* - when the knots are presented

*this* way, i.e., as "stoppers" ( as "knotted" open strings in between two aligned and pointing to opposite directions tensioned ends ). The other two ( the "white" and the "yellow") can not even be considered as

*symmetric-to-each-other*...

A naive "explanation" is that the parent closed knot ( the 6_3 ) happens to be "

*almost*" symmetric, and that the"cut" is sufficient to restore a complete symmetry, which was somehow

*hidden / broken* within the closed form - but which now is free to manifest itself. On the contrary, when the parent closed knots are completely symmetric right from the start ( as it happens in the 6_1 and 6_2 ), a "cut" can not but destroy the existing perfect symmetry. So, when "cut" and "opened up ", the perfectly symmetric closed knots become asymmetric open "stoppers"- while the "almost" symmetric one is helped to reveal its symmetric nature, which was hidden / broken by the initial closure.

We can test this "explanation" on the only non-symmetric of the 7 knots with 7 crossings - namely, the 7_6 ( see the attached picture ). When "cut" and presented as a "stopper", does it become symmetric, too - just like the 6_3 "stopper" ( the "red" one) ?

At first I though that it does, but now I am not satisfied by the degree of the manifested symmetry any more- so I have to suppose that the previous naive "explanation", even if it seemed reasonable in the case of the knots with 6 crossings, it does not work at the next level, the knots with 7 crossings...