The Yamada-Vogel algorithm on a small diagram

The source's five-arc picture and six circle pairs. Number the four circles in the first sketch of Figure 4 by their northwest, northeast, southwest and southeast positions. The northwest and southeast arrows are counterclockwise; the northeast and southwest arrows clockwise. For side-by-side circles in $S^2$, coherence requires opposite planar orientations, since the common annulus is outside their two disk interiors. Thus exactly the two diagonal pairs are incoherent; each of the four side pairs is coherent, so $h=2$. The five positive arcs are the two between the top circles, one on each vertical side and one on the bottom side, as shown in that rendered source panel. Their signs and incident circles record the original five crossings by [F1]. The heavy ${\alpha }_1$ joins the northwest and southeast incoherent pair; [F2] licenses its first reduction.

The two reductions. Perform the reducing move along ${\alpha }_1$: it replaces the chosen incoherent pair by two coherent circles joined by two oppositely signed arcs, and by [F2] the height drops to $1$; the new picture has a remaining incoherent pair, and the source's second reducing arc ${\alpha }_2$ joins it. Performing the reducing move along ${\alpha }_2$ drops the height to $0$, and the resulting picture has all pairs of Seifert circles coherent. The two moves are Reidemeister II moves of the original diagram, so the represented knot is unchanged.

The source word and its product convention. The final sketch has four concentric counterclockwise circles. Number them outermost to innermost and read from twelve o'clock counterclockwise. The ordered signed events are $(2,+),(1,-),(2,+),(3,-),(2,+),(1,+),(2,+),(3,+),(2,+)$, giving the displayed chronological list $X$. For this counterclockwise motion use the transverse frame $(-e_r,+e_Z)$; together with tangent $+e_{\theta }$ it preserves ambient orientation. Its positive half twist has the first outer indexed point go through negative physical depth, so the source's crossing signs agree with the fixed signed generators. By [F7] the actual geometric element of this chronological list is $\mathrm{rev}(X)$, not automatically $X$. We verify both closures by an explicit word comparison. Write $s={\sigma }_1$, $t={\sigma }_2$, $r={\sigma }_3$. The Artin relations [F4] give $X=t{s}^{-1}t{r}^{-1}tstrt=t{s}^{-1}t{r}^{-1}stsrt=t{s}^{-1}ts{r}^{-1}trst=t{s}^{-1}tstr{t}^{-1}st\mathrm{.}$ The successive operations are $tst=sts$, commuting $s$ with $r$, and $r^{-1}tr=tr{t}^{-1}$. Put $A=t{s}^{-1}tst$, $B=t^{-1}st$, so $X=ArB$. By the interior insertion sequence [F5], this is related by one positive ordinary destabilization and conjugations to the three-braid $Y=AB=t{s}^{-1}tst{t}^{-1}st$. Inverse cancellation gives $Y=Z=t{s}^{-1}t{s}^{2}t$, a six-letter three-braid. Word reversal is an anti-automorphism because the Artin relations are palindromic or far commutations. It carries a right stabilization to a left one of the same sign, which cyclic conjugation converts to a right stabilization; hence the reversed identities likewise give $\mathrm{rev}(X)\leftrightarrow \mathrm{rev}(Z)$ by ordinary Markov moves. Finally put $C=t{s}^{2}t$. Directly $sC=st{s}^{2}t=(sts)st=(tst)st=ts(tst)=ts(sts)=t{s}^{2}ts=Cs\mathrm{.}$ Therefore $t^{-1}Zt=s^{-1}Ct=C{s}^{-1}t=\mathrm{rev}(Z)$. Combining the two destabilization comparisons with this old-strand conjugation gives an explicit Markov sequence between $X$ and $\mathrm{rev}(X)$. By [F6], both have the oriented closure of the source diagram. The algorithm itself yields the nine-letter four-braid; $Y$ is its eight-letter three-braid after destabilization and $Z$ is a strictly shorter word for that same three-braid.

Conclusion. The verified two reductions give the source's four-strand, nine-crossing closed braid. Step 3.1 proves that its actual chronological interpretation and the commissioned word $X$ have the same oriented closure, and explicitly gives the eight-letter and six-letter three-braid words. The existence of the three-braid proves that the initial four-braid uses more strands than necessary; the six-letter representative proves that the eight-letter word is longer than necessary. These are separate index and length comparisons, with no assertion that the three-braid has nonminimal braid index. AC supplies the reducing/height-zero constructions and the countable choice in [F6].