A Proof of the Limits of the Four-Colour Theorem
Rey.BEng 2026 11th September 2026
Not a replacement for Inoue–Kawarabayashi–Miyashita–Mohar–Thomassen–Thorup (arXiv:2603.24880). A statement of why the theorem stops where it stops.
Theorem (limits).
On a planar contact surface, four colours suffice. Five is not a colour. Two is not a completed integer. The quadratic cost O(n2) is the tax of treating 2 as closed. The near-linear cost O(n log n) is the count of ticks.
0. Objects
Let (G) be a finite planar triangulation (every face a triangle, including the outer face). Vertices are contact patches. Edges are shared residual. Faces are the even lock of two patches.
Even residual: two standing orders (second and fourth).
Odd leftover: local cycle that has not flopped.
Tick: one counted reduction.
Completed 2: the pairing that writes every vertex against every other as if the patch had split.
0i2 (k.g.s2) = r2 m
State A — open residual
E = 2c / h
State B — locked residual
E = hbar / c
Energy with the tick visible:
E = ^m (c2)
(m) is counted. (c2) is the even package. The exponent is not a finished 2.
1. Why four, not five (colour limit)
A colour is a quality of contact on an occupied face. Four qualities are the even lock: two faces × two even orders.
A fifth colour would be a completed odd grade sitting in the interval. That grade does not persist as a standing axis (Minkowski third order is virtual; it lives on infinity, units s3/m2.
Hence: four colours suffice on the plane. Five is not admitted as a paint.
This is the colour limit of the theorem. It is not an accident of computer search. It is the statement that the odd does not occupy a seat.
2. Why five remains (obstruction limit)
Kawarabayashi et al. do not delete 5. They prove every planar triangulation contains linearly many pairwise non-touching reducible configurations or pairwise non-crossing obstructing cycles of length at most 5.
C5 is the shortest closed kink that can still block a simultaneous reduction. It is the local odd leftover — Z-kink, unoccupied parity, glueball-style force lock without a matter seat.
You cannot replace C5 by C4 inside the obstruction and keep the plane honest: is the non-planar witness. Five stays as a bound on cycle length, not as a fifth colour.
Obstruction limit: cap the odd at 5; reduce everywhere else at once.
3. Why (n) suffices and n2 does not (cost limit)
Old proofs guaranteed one reducible object. Each reduction was then a global rebuild. Cost O(n2): every vertex paired as if 2 had closed.
New structure: linearly many non-touching reductions in the flat parts (combinatorial curvature zero). Flat = the surface that is topologically flat each tick. Many locks fire together. Cost O(n log n): (n) ticks, logarithmic bookkeeping.
n2 is the completed-square tax.
(n) is the counted residual.
log n is the arena check — the third order not in the map.
This is the same limit as
2 = 1.999…
There is no finished 2. There is 1, stretched. Writing n2 pretends the stretch closed. Writing (n) counts the stretch.
4. Why the theorem cannot be simpler inside the can
The four-colour statement is a Rest-Mass theorem: one plane, one finite graph, one list of colours. Inside that can you will always need either a computer or a huge unavoidable set, because the odd leftover is not a coordinate you can cancel by hand.
The limit proof is Rest Time: even orders persist; odd is local and capped; cost is counted (m), not completed m^2.
A one-page human proof that erases 5 and erases the machine would require the odd to vanish from the plane. It does not vanish. It is bounded.
5. What would falsify this reading
- A planar graph that requires five colours. (Kills 4CT and this limit statement together.)
- A 4-colouring algorithm whose necessary cost is even when linearly many disjoint reductions exist. (Kills the (n) vs n2 limit.)
- An obstructing family whose shortest non-reducible cycles are forced to length while remaining planar and finite. (Kills the “5 is the kink bound” clause.)
- A standing fifth colour that is not a C5-type obstruction. (Kills the even-lock reading.)
Corollary
The four-colour theorem is true and limited.
Four is the even lock.
Five is the leftover cycle.
Two is not closed.
(n) is sufficient.
n^2 was the false completion.
That is the proof of the limits of the theorem.
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