16_1057125

The "beast" in full beauty:

I used the 4-cable (knot) with (blackboard) framing 16 and pattern the braid [−2−3−1−2−1], i.e., the unframed cable with 4-braid pattern [(123)⁶⁴−2−3−1−2−1] (=[1(123)⁶²] up to conjugacy; "±i" stands for σ_i^±1).

The polynomial until m-degree 4 is given below, in the format where the first row records the m-degrees (0 to 4) and the first two integers in every next row the l-degrees for given (even) m-degree. (Clearly, powers of both l and m must be even in every monomial.)

 261 1057125     0     4
  220  252                        107      -22019      338106      902878     -718680    -3767346    -2410797     2607686     3813661      885472     -889806     -543819      -41263       37391        9639         739          16
  218  254            112     -224220    13177877   -77912165  -158848554   435850104  1093610477   174760000 -1319003747 -1179669367   -56397912   376191658   174320733     8332538   -12598957    -3396172     -337832      -12978         -94
  218  256        -347866    99428343  1674801372 -1409217546  1414376011 -2040756822  -523606069 -1649687580  -734358094 -1114907086 -2123161090   997162016  1942589839  -476514278  1750181528   452980766    47861085     2234154       35523          76   

Here, like elsewhere on my pages, the "mirrored" (with l and l⁻¹ exchanged) Lickorish-Millett convention is used. The coefficients were calculated as 4-byte machine integers (integers modulo 2³² taken between −2³¹ and 2³¹−1).

To rule out braid index 5, by MFW, we need a term in l-degree (at least) 256. The problem with this knot is that such a term occurs only for m-degree 4. For all its "siblings", degree at most 2 sufficed. At a diagram of 261 crossings, the difference between m-degrees 2 and 4 is one between 6 weeks and 13 years of CPU time.

You can see the hardware I used on this photo of my desk (Oct 2022):

I had a computation server (right, front) with 20 3.4-GHz CPUs and another desktop (left, back) with 8 4.2-GHz CPUs. This is a solid equipment of the state of the art of early 2010s, so nothing impressive. I requested no supercomputer timeslots or the like. I had more hardware available, but I very deliberately tried to limit myself to these 2 machines. (I did use my – also 10-year old – laptop for consistency test calculations, though.)

With the parallelized truncated cabled MFW, the calculation took quite exactly half a year, between 6 Nov 2022 and 5 May 2023. (My estimates show that on the computation server without parallelizing, i.e., on a single CPU, I would have needed about 13.4 years.)

The truncation algorithm yields lower m-degree terms essentially for free, which can be calculated independently; this gives a consistency test.

Another "checksum" test comes from the cabling property of the Alexander-Conway polynomial ∇. It can be calculated pretty (at least far more) easily that the z-degree 4 coefficient of ∇ is 2,284,396. This must (and does) appear (modulo 2³²) as the alternating sum of the coefficients in the last row. The first coefficient in the row (third number) must be taken with '−' sign, as the sum of l-degree (218) and m-degree (4) of its monomial is not divisible by 4. (This checksum test works for lower m-degrees as well, with degree 0 yielding alternating sum 1 and degree 2 yielding 2705.)

I cannot be certain that 3-cable HOMFLY fails, but the terms I'd need to compute promised less for their cost than the 4-cable. (2-cable is way too helpless.)

Also, I added 5 (negative) crossings to get a 4-cable knot basically because this was technically my laziest way to produce a cable knot diagram. (I used it for 4-cabling some of the other knots as well.) Of course, I could have added only 3 pattern crossings, or none at all and treated a cable link. But with fewer crossings, these diagrams have a larger writhe, which ostensibly impacts a skein method (I use). So it is not very clear which would have been faster. And trying to experiment with multiple cable patterns is hardly reasonable at this scale of complexity.

I hope this gives a bit of insight into how I work, and in particular draws away from the impression that I am a computer maniac. (This conclusion is corroborated by a very respectable person who even recently said that I live in the "stony age", as I went to the railway station to buy tickets and do not use an app.)

I would be interested if alternative means (Khovanov-Rozansky?) can simplify this great ordeal.

My experience with determining the braid index somewhat suggests the following very "wild" guess, which I use the occasion to forward here:

Conjecture.
(a) The n-cabled MFW for all n will yield the braid index for every knot.
(b) For every n, there is a knot whose braid index is not determined by n'-cabled MFW for n'≤n.