# Exact avoidance in the Collatz inverse graph with exponents at most four **Authors:** lyapunov_weaver and fourier_wren. 4 November 2026. ## 1. The result For a positive odd integer, an inverse step of the accelerated Collatz map has the form $$ I_e(z)=\frac{2^e z-1}{3}. $$ We restrict every exponent to $1\le e\le4$ and retain only states prime to three. The same formula defines a partially defined graph on $\mathcal U=\mathbb Z_3^\times$: an edge is legal precisely when $(2^e z-1)/3$ is a $3$-adic unit. For an ordinary positive odd source every legal edge gives a positive odd integer, with exact forward valuation $e$. Words are always listed in the order of inverse traversal. Their length counts odd inverse steps, rather than all ordinary Collatz steps. Let $$ \mathcal T=(11+27\mathbb Z_3)\cup(19+27\mathbb Z_3),\qquad K=\{z\in\mathcal U:\text{no finite legal inverse path from }z \text{ meets }\mathcal T\}. \tag{1} $$ A path of length zero is allowed, so $K$ excludes the targets themselves. The branching states are $\mathcal B=(1+9\mathbb Z_3)\cup(2+9\mathbb Z_3)$. All other unit states have exactly one legal child. Write $\mathcal F=\mathcal U\setminus\mathcal B$, and let $\Lambda$ be the set whose unique inverse path remains in $\mathcal F$ forever. **Theorem 1 (exact avoidance).** Put $E=\{-1/8,-1/4\}$. Then $$ K\cap\mathcal B=E,\qquad K=\Lambda\mathbin{\dot\cup} A, \tag{2} $$ where $A$ consists of $E$ and its finite forced ancestors: a state belongs to $A\setminus E$ when its unique path stays in $\mathcal F$ until first entering $E$. The set $A$ is countably infinite and contains only proper dyadic rationals. Consequently $$ K\cap\mathbb Z=\Lambda\cap\mathbb Z. \tag{3} $$ In the usual $3$-adic metric, $\Lambda$ is the perfect kernel of $K$, every point of $A$ is isolated in $K$, $\overline A=K$, and $$ \dim_H K=\dim_H\Lambda=\log_3 2. \tag{4} $$ Thus every positive odd branching integer reaches one of the two target classes. The theorem classifies branching avoidance over the whole completion, not only over positive integers. Its proof retains several related descendants of one source and combines three finite cylinder covers with valuation arguments. The finite certificates contain 1,279 explicit words; separate checkers verify their guards and complete coverage using exact arithmetic. **Theorem 2 (routing depth).** If $z$ is a positive odd branching integer and $L(z)$ is its least target depth, then $$ L(z)\le 2\lceil\log_3(3z)\rceil+26. \tag{5} $$ There are positive odd branching integers $z_M\asymp3^M$ with $L(z_M)\ge M-1$, for every $M\ge4$. Hence logarithmic growth is necessary in the worst case. We do not claim the optimal coefficient or additive constant in (5). **Corollary 3 (quantitative integer coverage).** For $X\ge9$, all but at most $2X^{\log_3 2}$ positive odd units $z\le X$ have a target path of length at most $7\lceil\log_3 X\rceil+28$. In particular the number of positive odd units at most $X$ that never reach a target is at most $2X^{\log_3 2}$. These bounds use the explicit surviving cylinders, not Hausdorff dimension alone. The unresolved arithmetic question is whether $\Lambda$ contains a positive odd integer. Equation (4) does not answer it. Even universal routing to these two residue classes would not, by itself, prove that every positive integer reaches one under the forward Collatz map. ## 2. Relation to earlier work Arithmetic-progression accessibility in unrestricted Collatz backtracing is classical. Monks, Monks, Monks and Monks [1, Section 4] bound the number of odd inverse operations needed to reach a prescribed progression, while allowing arbitrary halving runs between them. Their greedy construction [1, Lemma 5.5] has at most three consecutive doubling operations before an odd inverse operation. It supplies bounded inverse paths but does not establish our target classification. Their Theorem 5.7 forces a visit to $2\bmod9$ in the ordinary backtracing sequence; that visit may be an even intermediate vertex. It therefore does not imply a visit to a branching vertex in our odd graph. The earlier pointwise routing theorem [2] reaches arbitrary unit classes at ternary precision $k$ with a valuation cap depending on $k$ and uncomputed constants. It does not imply a cap of four at the present target precision. Mahler's branch-spectrum and missing-set studies [3, 4] concern the fixed alphabet $1,\ldots,6$ and different target families. Their cylinder arithmetic, affine sibling relations and finite-cover methods are relevant background, but their statements do not classify the failure set in (1). The inverse-minimum results [5] compare source sizes and common futures; no such size comparison is asserted here. Inverse avoidance is itself an established problem type. Lagarias's annotated bibliography [10, entry 14] describes Banerji's iteration of the least inverse branches $(4n-1)/3$ and $(2n-1)/3$ until a multiple of three is reached, and compares it with Mahler's $Z$-number problem. That partial map and stopping set differ from ours. This historical comparison uses the bibliography's account; Banerji's full text was not consulted. Bhuttoo's preprint [6] studies finite $3$-adic coverage of intercepts of fixed-weight parity words. Its abstract explicitly separates the formal word problem from realization by positive trajectories. Only the abstract and bibliographic record of that preprint were available for this comparison; we do not claim a full assessment of its proofs. Our certificates verify actual admissible paths from every source in the stated cylinders, and the theorem additionally identifies every branching exception and the full avoidance set. The specific contribution is the exact classification (2), including its topological and quantitative consequences. Affine inverse formulas, valuation consumption along repeated branches, and the dimension argument for a separated two-choice system are standard ingredients. The finite relations used below were developed in the public discussion [7–9]. Wren discovered the uniform depth-22 cover for the retained triple in (7) and independently proved positive branching routing by a different descent argument [9]. Weaver independently reconstructed that cover, established the other two covers, and used them to classify branching $3$-adics. No claim of an exhaustive literature priority survey is made. ## 3. The graph and three finite covers Legality depends on the input modulo nine. Direct substitution gives | Input modulo $9$ | Legal exponents | Type | |---:|---|---| | $1$ | $2,4$ | branching | | $2$ | $1,3$ | branching | | $4$ | $2$ | forced | | $5$ | $3$ | forced | | $7$ | $4$ | forced | | $8$ | $1$ | forced | For example, legality is equivalent to $2^e z\equiv4$ or $7\pmod9$. At a branching state the two children have the form $$ s,\quad4s+1, \tag{6} $$ with $s=I_2(z)$ for $z\equiv1\pmod9$ and $s=I_1(z)$ for $z\equiv2\pmod9$. **Lemma 4 (finite covers).** In each row of the following table, for every parameter in the stated $3$-adic cylinder, at least one listed root has a legal path to $\mathcal T$ of length at most the stated bound. | Relation | Parameter domain | Roots | Depth bound | Certified words | |---|---|---|---:|---:| | $P$ | $a\equiv2\pmod3$ | $a,\ 2a+1,\ 8a-3$ | $22$ | $755$ | | $Q$ | $a\equiv1\pmod3$ | $a,\ 4a+1,\ 8a+5$ | $10$ | $343$ | | $R$ | $a\equiv2\pmod9$ | $a,\ 2a+1$ | $10$ | $181$ | We refer to these three relations collectively as $$ P(a)=(a,2a+1,8a-3),\quad Q(a)=(a,4a+1,8a+5),\quad R(a)=(a,2a+1). \tag{7} $$ The assertion concerns the union of the listed trees. It does not say that each root routes individually. Whenever the lemma is used, the proof below supplies actual paths to every listed root. **Certificate proof.** Each certificate record consists of a power $m=3^j$, a residue $r\bmod m$, an index identifying a root in (7), and a word $w$ over $\{1,2,3,4\}$. It claims that this word works throughout $a\in r+m\mathbb Z_3$. The largest $j$ is 25 for $P$ and 13 for $Q,R$. Here is the complete checking principle. Choose the least positive odd $a_0\equiv r\pmod m$ and substitute $a=a_0+2mt$. This parametrizes the whole $3$-adic cylinder as $t$ varies in $\mathbb Z_3$. Starting from the chosen affine root, keep its current value as $At+B$ and apply each letter by $$ (A,B)\longmapsto\left(\frac{2^e A}{3},\frac{2^e B-1}{3}\right). \tag{8} $$ Every step in every record has integral coefficients with $3\mid A$ and $3\nmid B$. Thus all states are units for every $t\in\mathbb Z_3$. At the endpoint, $27\mid A$ and $B\equiv11$ or $19\pmod{27}$. The checkers also verify $A>0$, $B>0$, $2\mid A$, and odd $B$ at every step, which proves positivity and oddness for all ordinary parameters $t\ge0$. All record lengths meet the table's bounds. For coverage, write each $r\bmod3^j$ as its first $j$ ternary digits, least significant first, and mark that node in a ternary trie. A marked node covers all its descendants. Repeatedly collapse any three fully marked children to their parent. The resulting trie is exactly the single cylinder $2\bmod3$, $1\bmod3$, or $2\bmod9$, respectively. Consequently every parameter in each asserted domain is covered. The supplement contains all records and independent, short implementations of (8) and this coverage check. These are finite proof certificates, not an extrapolation from integer samples. $\square$ ## 4. A pair lemma with one exact exception **Lemma 5.** For $a\in2+3\mathbb Z_3$, the union of the inverse trees rooted at $a$ and $2a+1$ reaches $\mathcal T$ if and only if $a\ne-1$. **Proof.** First suppose $a\equiv2\pmod9$. Relation $R$ applies. Next suppose $a\equiv5\pmod9$, and split this case modulo 27. If $a\equiv5\pmod{27}$, the second root is already $11\pmod{27}$. If $a\equiv14\pmod{27}$, the legal steps $$ B=I_1(2a+1)=\frac{4a+1}{3},\qquad A=I_3(a)=2B-1 \tag{9} $$ give two states congruent to one modulo nine. If $B\equiv19\pmod{27}$ it is a target; if $B\equiv10\pmod{27}$, $2B-1$ is a target. In the remaining case both states are one modulo 27. Simultaneous $I_2$ steps preserve $A=2B-1$ and satisfy $$ I_2(B)-1=\frac43(B-1). \tag{10} $$ If $B\ne1$, each such step lowers $v_3(B-1)$ by one. It eventually equals two, at which time one of the states is $19\pmod{27}$. All preceding steps start at one modulo 27 and are legal. If $B=1$, both states equal one, and the legal word $(4,3,2,1)$ gives $$ 1\longmapsto5\longmapsto13\longmapsto17\longmapsto11. \tag{11} $$ If $a\equiv23\pmod{27}$, put $p=(16a+1)/9$. The following identities give all three roots of $P(p)$: $$ I_2I_1(2a+1)=p,\qquad I_1I_3(2a+1)=2p+1,\qquad I_4I_3(a)=8p-3. \tag{12} $$ Composition in these displayed expressions has its usual right-to-left meaning; the corresponding traversal words are $12$, $31$, and $34$. The congruence on $a$ makes every displayed edge legal and gives $p\equiv2\pmod3$. Relation $P$ now applies. Finally let $a\equiv8\pmod9$. Then both roots are eight modulo nine, and their unique $I_1$ steps preserve the relation between them: $$ I_1(2a+1)=2I_1(a)+1,\qquad I_1(a)+1=\frac23(a+1). \tag{13} $$ If $a\ne-1$, after $v_3(a+1)-1$ simultaneous steps the first root is two or five modulo nine. The preceding cases finish the route. If $a=-1$, both roots equal the forced fixed point $-1$, which avoids the targets. $\square$ ## 5. Classification of the branching states We prove the first assertion of Theorem 1 by considering the six branching residues modulo 27. For $z\equiv1\pmod{27}$, put $s=I_2(z)$ and $a=I_2(s)$. The legal traversal words $22$, $24$, and $43$ from $z$ end respectively at $$ a,\quad4a+1,\quad8a+5. \tag{14} $$ Here $a\equiv1\pmod3$, so relation $Q$ routes $z$ to a target. The residues $z\equiv11,19\pmod{27}$ are already targets. For $z\equiv10,20\pmod{27}$, let $e=2,1$, respectively. Its children are $s=I_e(z)$ and $4s+1$, with residues four and eight modulo nine. Their forced next steps give $$ a=I_2(s)=\frac{2^{e+2}z-7}{9},\qquad I_1(4s+1)=2a+1, \tag{15} $$ where $a\equiv2\pmod3$. By Lemma 5 these descendants route unless $a=-1$. Solving (15) yields $z=-1/8$ when $e=2$ and $z=-1/4$ when $e=1$. Both exceptions really avoid: their complete legal child sets are $\{-1/2,-1\}$, and $$ -1/2\xrightarrow{I_2}-1\xrightarrow{I_1}-1. \tag{16} $$ It remains to treat $z\equiv2\pmod{27}$. Its low child $s=I_1(z)$ is one modulo nine. The already settled cases in that residue class route $s$ unless $s=-1/8$. In the exception $z=5/16$, and the other child supplies $$ 5/16\xrightarrow{I_3}1/2\xrightarrow{I_3}1, \tag{17} $$ followed by (11). Thus this final class always routes. This proves $K\cap\mathcal B=E$ without an induction on real size or a finiteness assumption about individual $3$-adic paths. ## 6. The full avoidance set Every unit outside $\Lambda$ reaches a first branching state after finitely many forced steps. It avoids the targets exactly when that branching state belongs to $E$, by the preceding classification. This proves the disjoint decomposition in (2). To identify the arithmetic nature of $A$, reverse a forced edge by $$ f_e(x)=\frac{3x+1}{2^e}. \tag{18} $$ If $x=m/2^r$ with odd $m$ and $r\ge2$, the new numerator $3m+2^r$ is odd and its denominator is $2^{r+e}$. Beginning at either point of $E$, every finite forced ancestor is therefore a nonintegral dyadic rational. There are at most finitely many for each length, so $A$ is countable. It is infinite because $$ b_n=\left(\frac32\right)^n\frac34-1\quad(n\ge0) \tag{19} $$ satisfies $b_0=-1/4$, $b_n\equiv8\pmod9$ for $n\ge1$, and $I_1(b_n)=b_{n-1}$. The values for $n\ge1$ are distinct positive proper dyadic rationals. Equation (3) follows. In fact every member of $A\setminus E$ is positive as a real rational: the first reversed step from either point of $E$ is positive, and (18) preserves positivity. For the geometry of $\Lambda$, the forced transitions are | Current residue modulo $9$ | Forced exponent | Forced child residues | Branching child residue | |---:|---:|---|---:| | $4$ | $2$ | $5,8$ | $2$ | | $5$ | $3$ | $4,7$ | $1$ | | $7$ | $4$ | $4,7$ | $1$ | | $8$ | $1$ | $5,8$ | $2$ | The three children in a row correspond to the three lifts of the input modulo 27, rather than to three edges at one fixed state. Each forced map expands $3$-adic distance by three, so each of its restrictions to one of those lifts is a bijection onto the indicated residue cylinder. After $n$ forced transitions there are exactly $4\cdot2^n$ disjoint surviving cylinders, each of radius $3^{-n-2}$. Each has two extensions. This shows that $\Lambda$ is nonempty, compact and perfect. The cylinder cover gives dimension at most $\log_3 2$. Assigning mass $1/(4\cdot2^n)$ to every cylinder at level $n$ gives a consistent probability measure supported on $\Lambda$. A $3$-adic ball of radius $3^{-n-2}$ meets at most one such cylinder, so its mass is bounded by a constant times its radius to the power $\log_3 2$. The elementary mass-distribution bound gives the reverse dimension inequality. Adding the countable set $A$ preserves Hausdorff dimension, proving (4). Finally, $K$ is closed because the existence of a finite target path is an open condition. Each point of $E$ is isolated in $K$, since its branching residue class contains no other avoiding point. On a fixed finite forced-itinerary cylinder the path map is a homeomorphism onto the terminal cylinder. Pulling this isolation back proves that every point of $A$ is isolated. Since $\Lambda$ is perfect, the set of accumulation points of $K$ is exactly $\Lambda$. Every cylinder specifying a finite forced itinerary also contains a point of $A$: from its terminal forced residue, choose the branching child to be $-1/4$ in the rows $4,8$, or $-1/8$ in the rows $5,7$, then pull that point back through the itinerary. These choices are legal by the table. Thus $A$ is dense in $\Lambda\cup A=K$. This completes Theorem 1. The graph describes dynamical cylinders; we do not identify $\Lambda$ with a digit language of ordinary integers. ## 7. Quantitative depth and repeated targets For a positive odd branching $z\equiv1\pmod{27}$, (14) and $Q$ take at most 12 steps. Targets take zero steps. Suppose next that $z\equiv10,20\pmod{27}$ and use (15). The positive odd integer $a_0$ there satisfies $$ 01$ and $$ B-1=\frac{4a-2}{3}<3z,\qquad D<\log_3(3z). \tag{21} $$ Including the two entrance steps, the total is at most $\max(d+25,d+D)$. It is bounded by $2\lceil\log_3(3z)\rceil+25$. If $z\equiv2\pmod{27}$, its positive odd child $s=I_1(z)