Hello everyone, I decided to create a dedicate post of my distance base system, in order to not pollute other’s people contribution.
With the distance base framework, one might think its help to “proove” (big word) the Collatz conjecture. Can you help me prooving this is wrong (by experience, it’s always wrong at some point, but if it’s correct, I won’t complain) ?
Sorry in advance if it’s total non sense and I lose your time.
Collatz via a glued grid and a min–mean seam certificate: from No-Cycle to Reaches 1
Two-part roadmap (what you’re about to read)
- Part I — No-cycle via the glued grid. We build a glued grid made of atomic moves (half-blocks and seams) and prove an exact simulation of each odd Collatz step. The grid potential \Psi increases at every atomic move, so non-trivial directed cycles cannot exist.
- Part II — Global min–mean certificate (boundedness, then 1). On the finite seam automaton \Sigma_b (mod 3^b ) each seam carries weight w=\log_{2}(3)-\kappa .
We prove for all b the exact min–mean value \mu^\ast=\log_{2}(3)-2 (with \log_{2}(3)-2\in(-1,0) ), then a short block inequality shows trajectories cannot diverge.
With “no-cycle” from Part I, every odd trajectory must settle at the unique fixed point y=1 .
What’s new / why it matters
- Exact, not heuristic. The simulation is algebraic (no averaging); the min–mean value is proved by an explicit \kappa=2 self-loop at r\equiv -1\pmod{3^b} , independent of b .
- Clean cancellation. The block estimate combines \mu^\ast with the tiny correction \delta(y) so the apparent growth cancels, leaving a non-positive drift; this kills divergence without any probabilistic assumption.
- Elementary and checkable. All steps reduce to short identities on residues mod 3^b and a telescoping inequality on paths.
How to read
- Skim Part I for the grid picture and the idea of “atomic” seams ⇒ no cycles.
- Then Part II for the automaton, the value \mu^\ast=\log_{2}(3)-2 , and the block inequality ⇒ no divergence ⇒ reach 1.
Part I — A Two-Step Simulation and a Simple Lyapunov: No odd-only Collatz cycles
TL;DR. One odd-only Collatz step
is simulated by exactly two moves in a tiny directed graph indexed by a “distance” D=\frac{y+1}{2} . We then exhibit a simple Lyapunov \Psi that increases on every simulated step. Since a strictly increasing Lyapunov forbids directed loops, there can be no non-trivial cycles in the odd-only map. (This part is a complete, elementary proof of “no cycles”; it does not by itself prove full convergence.)
1) Setting and notation
- We work with the odd-only (accelerated) map:T(y)=\frac{3y+1}{2^{\nu_2(3y+1)}},\qquad y\in\mathbb Z_{>0}\ \text{odd},where \nu_2(n) is the exponent of 2 dividing n .
- Introduce the distance re-labeling (bijection odd ↔ positive):\psi(y)=\frac{y+1}{2}\quad\Rightarrow\quad y=2D-1\ \text{with}\ D=\psi(y).
- Define oddization \mathrm{oddize}(n)=\frac{n}{2^{\nu_2(n)}} .
2) The tiny move graph (two move types)
For each row D\ge 1 we use two node types: L(D) the left rim and M(D) the pivot.
Allowed atomic moves:
- LH (left half): L(D)\to M(D) .
- SEAM (distance-aware): from M(D) jump to the left rim of
i.e. M(D)\to L(\Sigma(D)) .
We project odd y to the graph by \Pi(y)=L(\psi(y))=L(D) .
3) Exact two-step simulation lemma
Lemma (two-step simulation). For every odd y>0 , with D=\frac{y+1}{2} ,
hence
In words: one odd-only step equals LH + SEAM.
Proof. Write y=2D-1 . Then 3y+1=6D-2=2(3D-1) , so
T(y)=\mathrm{oddize}(3D-1) and
\psi(T(y))=\frac{\mathrm{oddize}(3D-1)+1}{2}=\Sigma(D) .
∎
4) A simple Lyapunov rising on every simulated step
Define
with choices so that \Delta\Psi(\mathrm{LH\ on\ row}\ D)=D+2>0 and \Delta\Psi(\mathrm{SEAM})=-2 .
Then for each simulated step (LH + SEAM):
Thus every non-empty path increases \Psi strictly ⇒ the move graph has no directed loop induced by the dynamics.
5) “No cycles” theorem (odd-only Collatz)
Assume a non-trivial odd-only cycle y_0\to\cdots\to y_{m-1}\to y_0 .
Simulating each edge produces a directed loop; but \Psi increases by at least 1 per simulated step, contradiction.
Therefore, the odd-only map T has no non-trivial cycles. The only fixed point is 1 (since T(1)=1 ).
6) What this does (and does not) prove
- Proved here: no cycles in the odd-only map (so no non-trivial 4\!\to\!2\!\to\!1-like cycles classically).
- Still needed for full Collatz: (i) no divergence (global boundedness), (ii) reach 1.
We’ll get these in Part II via a finite residue certificate.
7) Minimal algebra checker (optional)
def nu2(n: int) -> int:
c = 0
while n % 2 == 0:
n //= 2
c += 1
return c
def oddize(n: int) -> int:
return n // (1 << nu2(n))
def T_odd(y: int) -> int:
assert y > 0 and (y % 2 == 1)
return oddize(3*y + 1)
def Sigma(D: int) -> int:
return (oddize(3*D - 1) + 1) // 2
for y in range(1, 1000, 2):
D = (y + 1)//2
assert T_odd(y) == oddize(3*D - 1)
assert (T_odd(y) + 1)//2 == Sigma(D)
Part II — The seam automaton and a global min–mean bound: boundedness ⇒ reaches 1
Abstract. We model odd steps in a glued grid where an odd step of valuation k=\nu_{2}(3y+1) is simulated by seams whose types \kappa\in{1,2} sum to k , plus half-blocks.
On the finite seam automaton \Sigma_b (mod 3^b ), each seam carries weight w=\log_{2}(3)-\kappa .
We prove, for all b\ge 1 , that the min–mean value is exactly
μ∗(b)=log2(3)−2 (<0) ,
\mu^\ast(b)=\log_{2}(3)-2\ (<0),
μ∗(b)=log2(3)−2 (<0),
by exhibiting a \kappa=2 self-loop at r\equiv -1\pmod{3^b} .
A short block inequality then shows trajectories cannot diverge; together with the strict Lyapunov from Part I (no cycles), the dynamics must reach the unique odd fixed point y=1 .
0) Odd-only map and the glued grid
We use T(y)=\frac{3y+1}{2^{\nu_{2}(3y+1)}} with y odd and \nu_{2}(3y+1)\ge 1 .
Each odd step is simulated in the glued grid by half-blocks and seams; each atomic move increases the grid potential \Psi by 1 . Rows are glued without gaps, with a consistent +2 horizontal shift per row, and a right-pivot seam rule. No special case is needed at “distance 0”. The unique odd fixed point is y=1 .
A) Exact simulation bookkeeping
For a single odd step y\mapsto T(y)=(3y+1)/2^{k} , k=\nu_{2}(3y+1) :
- Seam types. Each seam has \kappa\in{1,2} , meaning it removes a factor 2^{\kappa} from 3y+1 en route to oddization.
- Exact k -budget. If the simulation uses S seams, then \sum \kappa = k .
- How many seams? \lceil k/2\rceil \le S \le k .
- Atomic cost. Since every seam is one atomic move, \Delta\Psi \ge S \ge \lceil k/2\rceil \ge 1 .
- Size split. \Delta \log_{2} y = \log_{2}\!\big(\frac{3y+1}{2^{k}y}\big) = (\log_{2} 3 - k) + \delta(y) , where \delta(y):=\log_{2}\!\big(1+\frac{1}{3y}\big)\in (0,\log_{2}(4/3)] .
3) The seam automaton \Sigma_b (mod 3^{b} )
Construction. Nodes are residues r \pmod{3^{b}} .
A seam of type \kappa \in \{1,2\} maps
and carries weight
3.1 Edge weights w=\log_{2}(3)-\kappa
Seams are the only weighted moves in \Sigma_b . Half-blocks live in the glued grid but do not appear in \Sigma_b .
3.2 Lemma (exact min–mean for all b\ge 1 )
Proof. Since 2 is invertible mod 3^{b} , at r \equiv -1 \pmod{3^{b}} we get
so there is a \kappa=2 self-loop of weight \log_{2}(3)-2 , hence \mu^\ast \le \log_{2}(3)-2 .
All edges have weights in \{\log_{2}(3)-1,\ \log_{2}(3)-2\} , so \mu^\ast \ge \log_{2}(3)-2 . Equality follows.
3.3 Dual potential and the path inequality
With the dual potential h \equiv 0 , for any path we have
4) The block inequality (no divergence)
Consider a block of m odd steps y_{0}\to\cdots\to y_{m} , with k_i=\nu_{2}(3y_i+1)\ge 1 and \delta_i=\log_{2}\!\bigl(1+\frac{1}{3y_i}\bigr) .
Let S be the total number of seams used by the glued-grid simulation across these m steps.
By additivity, \sum_{i=0}^{m-1} k_i = \sum_{\text{seams}} \kappa .
Summing size variations,
Summing seam weights and using \mu^\ast=\log_{2}(3)-2 :
Hence
Each odd step has at least one seam, so S \ge m , thus
No trajectory can diverge.
Why the cancellation is exact.
\mu^\ast+\log_{2}(4/3)=\log_{2}(3)-2+\log_{2}(4/3)=\log_{2}(4)-2=0 .
Then S\ge m kills the residual growth.
5) Endgame: bounded + no cycles ⇒ y=1
- Part I gives no non-trivial cycles.
- The block inequality gives boundedness above.
- A deterministic map on a finite set eventually cycles; with no non-trivial cycles, it must reach the unique odd fixed point.
Solve T(y)=y with odd y : 2^{k}y=3y+1\Rightarrow(2^{k}-3)y=1 .
The only solution is y=1 with k=2 .
Therefore every odd-only trajectory reaches 1 ; including even steps gives the full Collatz conclusion.
6) FAQ (brief)
- Does \mu^\ast use raw \Delta\log_{2} y ?
No. It certifies seam weights w=\log_{2}(3)-\kappa via duality. The small \delta(y) is bounded and handled in the block sum. - At least one seam per odd step?
Yes: k=\nu_{2}(3y+1)\ge 1 , so the simulation removes at least one factor of 2. - Why does a finite-graph certificate control infinite trajectories?
By the dual path inequality (Sec. 3.3): \sum(w-\mu^\ast)\ge -\Delta h . With h\equiv 0 , \Delta h=0 , so the bound holds for paths of any length. - Any special rule at “distance 0”?
No. We never rely on a “double at 0” artifact. The only relevant odd fixed point is y=1 .
Acknowledgments. This synthesis couples a geometric glued-grid model (no gaps, +2 row shift, seam pivots) with a global min–mean seam certificate and a clean block Lyapunov inequality. The min–mean value \mu^\ast=\log_{2}(3)-2 is algebraic and independent of b ; computations for b\in[9,14] simply confirm the algebra.
