The chip game leaves us with a consecutive block of edge contacts. Everywhere else, multiplying a polygon vertex by the eigenvalue lands exactly on another vertex.
Follow those exact steps. How long until a chain reaches an edge contact? Which contact does it reach? Those questions determine the powers that will eventually appear in the boundary equations.
Their answer is encoded by the Euclidean algorithm. Its remainders organize the spatial returns; a companion addition rule counts the steps.
We will see this explicitly in an eight-site example, then prove the arithmetic behind it. The diagrams describe index patterns; deciding which such patterns admit extremal convex polygons is the geometric task of the next stop.
Cut the index rotation at the contact block
We continue under the hypotheses of the contact post: a nonreal, least-order radial extremum, in the orientation with positive second-endpoint coefficients. Here assume $0<|\lambda|<1$.
After the chip-game reduction, the edge-contact sources form one block. Write $\varphi$ for their number. Rotate the cyclic labels so their image points are
where $\alpha_j,\beta_j>0$ and $\alpha_j+\beta_j=1$. We have relabeled the weights by their target edges. All indices are taken modulo $n$.
The chipped sources are $1-k,\ldots,\varphi-k$. The corresponding target labels are $1,\ldots,\varphi$. At every other source,
Thus the permutation $i\mapsto i+k\pmod n$ describes the deterministic steps. When its next target enters $\{1,\ldots,\varphi\}$, that step is instead a weighted contact $\xi_j$.
For $n=8$, $k=3$, and $\varphi=2$, the chipped sources are $6$ and $7$. The chains are
and
Figure 1. The final amber arrow lands at an edge point $\xi_j$, rather than at the vertex $x_j$. The return lengths are three and five: each counts the deterministic steps and the final weighted step.
Why records matter
Put $g=\gcd(n,k)$. The modular orbit
has period $n/g$, with representatives in $\{0,\ldots,n-1\}$. A record maximum exceeds every earlier entry; include $a_0=0$ as the initial record.
The game gave us a necessary condition on a reduced block:
For our example,
has record maxima $0,3,6,7$. Their complementary gaps are
These are the candidate block sizes. They are arithmetic candidates, not a geometric sufficiency theorem. Our choice $\varphi=2$ comes from the orbit’s record entry $a_2=6=8-2$.
Figure 2. Amber bars mark record maxima; the initial zero is marked on the baseline. Their gaps from $8$ give $8,5,2,1$. The positive record minima, which will control the other side of the return calculation, are $3$ and $1$.
Euclid finds the records and their times
There is a useful way to find both kinds of records together. Keep four positive integers:
Here $L$ is the current positive record minimum, reached at time $p$, and $U$ is the gap above the current record maximum, reached at time $q$. Start with
The next record occurs at time $p+q$. Add the two congruences:
If $L>U$, this gives a new positive minimum. If $U>L$, it gives a new maximum, with smaller gap $U-L$. So the update is
Subtract the spatial gaps; add their hitting times. The gaps follow the subtractive Euclidean algorithm. The times travel with them.
For $n=8$, $k=3$:
| $L$ | $U$ | $p$ | $q$ | New information |
|---|---|---|---|---|
| $3$ | $5$ | $1$ | $1$ | $a_1=3$, initially both records |
| $3$ | $2$ | $1$ | $2$ | New maximum: $a_2=6$ |
| $1$ | $2$ | $3$ | $2$ | New minimum: $a_3=1$ |
| $1$ | $1$ | $3$ | $5$ | New maximum: $a_5=7$ |
When $L=U=g$, the next sum of times is the full period $n/g$. The corresponding residue is zero, so there is no further positive minimum or strict maximum.
Why can’t another record arrive before $p+q$? The next proof supplies that missing part.
A complete proof of the record update
Throughout the process,
These identities hold initially and are preserved by either update. For any integer pair $(j,x)$ with $x\equiv kj\pmod n$, define
The congruences make $A,B$ integers, and the boxed identity gives
If $j>0$ and $0<x<L$, then $A,B$ must both be positive. Indeed, opposite signs either give $x\leq0$ or $x\geq L$; both nonpositive give $j\leq0$. Hence $j\geq p+q$.
The same argument, with $-U<x<0$, proves that a residue above $n-U$ cannot occur at a positive time less than $p+q$. We use its negative representative $x=a_j-n$.
At time $p+q$, the signed residue is $L-U$. It is strictly inside one of those two record gaps unless $L=U$. This proves that the stated update produces exactly the next record. Induction proves the algorithm.
The gcd is preserved under subtraction. Thus it ends at $L=U=g$, and $pU+qL=n$ then gives $p+q=n/g$.
Grouping consecutive subtractions recovers the usual divisions:
| Euclidean division | Its record contribution |
|---|---|
| $8=2\cdot3+2$ | Maximum gaps $8\to5\to2$ |
| $3=1\cdot2+1$ | Positive minima $3\to1$ |
| $2=2\cdot1$ | Maximum gaps $2\to1$; omit zero |
More generally, write the Euclidean remainders as $r_1=n,r_2=k,r_3,\ldots,r_s=g$. Begin the gap list with $n$. For each even-position remainder $r_{2j}$, subtract it repeatedly from the preceding odd-position remainder $r_{2j-1}$, retaining the positive results. These are precisely the complementary record gaps—and hence the candidate block sizes.
Complementary record gaps: candidate block sizes φ. The outlined candidate is selected in the return explorer below.
The full orbit and all records stay visible. The dark outlines mark the selected pair of hitting times p and q. Changing n or k also updates the return explorer.
The first-return rule
Now choose a proper candidate block with $g<\varphi<n$. In the record algorithm, find the stage with
It always exists: after $U$ first becomes $\varphi$, subtract it from $L$ as needed. Since the gcd is $g<\varphi$, those subtractions leave a positive remainder smaller than $\varphi$.
Write that remainder as $r$. Grouping the subtractions identifies it as the unique even-position Euclidean remainder satisfying
Its accompanying times are characterized by
For each starting label $i\in\{1,\ldots,\varphi\}$, stop the index orbit at its first positive return to that set. Then:
| Starting label | Return time | Contact reached |
|---|---|---|
| $1\leq i\leq\varphi-r$ | $p$ | $\xi_{i+r}$ |
| $\varphi-r<i\leq\varphi$ | $p+q$ | $\xi_{i+r-\varphi}$ |
All earlier steps are deterministic. The last source is $x_{j-k}$, whose image is $\xi_j$. Thus the index returns give the geometric relations
For $8,3,2$, the selected pair is $L=1,U=2,p=3,q=2$. Thus $r=1$: label $1$ takes three steps to contact $2$; label $2$ takes five steps to contact $1$.
Why these are the first returns
Use the integer representation from the record proof at the selected stage $L=r,U=\varphi$. It shows
It also gives the stronger lower bound
For this second bound, if $0<x<r+\varphi$ and $0<j<p$, the representation $j=Ap+Bq$, $x=Ar-B\varphi$ is impossible: $A\geq1,B\geq0$ forces $j\geq p$; $A\geq1,B\leq-1$ forces $x\geq r+\varphi$; the other signs force $x\leq0$ or $j\leq0$.
Consequently $a_j+i$ cannot wrap into $\{1,\ldots,\varphi\}$ before time $p$, for any $0\leq i\leq\varphi$. At time $p$, $a_p=r$. The starting labels $i\leq\varphi-r$ therefore land at $i+r$ and stop.
For the remaining labels, $i+r>\varphi$. The first pair of bounds shows that they stay outside the target set for every $j<p+q$: their sums lie above $\varphi$ and at most $n$. At time $p+q$, the residue is $n-\varphi+r$, which wraps them to $i+r-\varphi$. These are exactly the claimed first returns.
There is no zero residue in the time range used here: $\varphi>r\geq g$ and $p\varphi+qr=n$ imply $p+q<n/g$.
Target vertex label Chipped source · Dark ring: current source vertex
All chains and their vertex counts
| Start | Vertex indices in order | Steps | Final contact |
|---|
The circle records indices, not complex-plane locations. The final weighted step is displayed as ξ, rather than as a jump to a target vertex. These integer patterns require a separate geometric realizability argument.
The chain lengths account for every vertex
The index permutation has $g$ cycles. Since $\varphi\geq g$, the consecutive target block meets every residue class modulo $g$. Cutting the weighted steps therefore breaks all cycles into paths, one starting at each of $1,\ldots,\varphi$.
These paths have no repeated or shared vertices, and together they contain all $n$ vertices. A path of return time $p$ contains $p$ vertices: the starting vertex and $p-1$ further deterministic vertices. Its next image is the edge point.
There are $\varphi-r$ short paths and $r$ long paths. Hence
Figure 3. The paths contain $\{1,4,7\}$ and $\{2,5,0,3,6\}$. These disjoint sets exhaust all eight indices. The contact points $\xi_1,\xi_2$ are images, so they are not extra polygon vertices.
This is why return arithmetic will control polynomial degrees. Its step counts already account for the order of the realizing matrix.
The gcd case has equal-length chains
If $\varphi=g$, each residue class modulo $g$ has exactly one target label. The index orbit can return to the target block only by returning to that same label, after $n/g$ steps. Thus
For $n=8,k=2$, take $\varphi=g=2$. The paths have index lists $1,3,5,7$ and $2,4,6,0$. Both return times are four.
Figure 4. One target in each parity class cuts each rotation cycle once. The resulting relations are $\lambda^4x_1=\xi_1$ and $\lambda^4x_2=\xi_2$.
The full block $\varphi=n$ is another direct case: every source already has an edge contact, so every first return takes one step. The two-length formula was stated only for $g<\varphi<n$.
What the arithmetic has prepared
Return to the eight-site example. The long path contains $x_0=\lambda^2x_2$, so its two contacts give
The generating vertices are nonzero. Eliminate them:
The exponents are already fixed. The weights have not yet been forced to agree. If the next geometric step gives $\alpha_1=\alpha_2=t$, then
one of the forms explained in the boundary-equations post.
That is what Euclid contributes to this proof: it turns a contact block into an exact inventory of powers and return relations. The next stop, Moving Polygons Until the Weights Agree, asks which arithmetic patterns survive geometric variation and how their separate coefficients collapse to a common parameter.
Why separate edge-contact weights $\alpha_j$ must collapse to a single common parameter $t$: Moving Polygons Until the Weights Agree →. Or continue to Stop 6 (Where the Boundary Equations Come From) and the series roadmap.
Source and scope
F. I. Karpelevich, On the characteristic roots of matrices with nonnegative elements, Izv. Akad. Nauk SSSR Ser. Mat. 15:4 (1951), 361–383. Original paper. The record arithmetic is in §3, Lemma 2, pp. 365–366; Theorems III–IV and the ensuing vertex-chain description occupy pp. 369–371.
This post gives new illustrations and an elementary proof of the record update and the return rules needed here. It continues to use the extremal contact and chip-game reductions from the earlier posts. All explorer data are exact integer calculations; they describe contact-index blueprints and do not certify geometric realizability or boundary status. Equality of weights and boundary-arc selection remain later steps.