MORC

Posted on Aug 12

Two Penrose Tilings That Coincide Exactly, from the Same Ten-Pentagon Ring

#geometry #ai #science

This is part 2. Part 1 introduced the unit — ten regular pentagons in a closed ring — and the two ways two such rings can be joined.

The claim, up front

From rings of ten pentagons, two Penrose tilings appear:

one made of pentagons and rhombi — the figure the rings themselves draw
another made of pentagons and rhombi — the figure obtained by joining the ring centres

The two coincide exactly, at a scale ratio of φ³.

And when one of them runs out of moves, the other supplies the next one.

Every decision is made in integers. Real numbers appear only when drawing to the screen.

1. Two rings can be joined in only two ways

Fig. 1 — Combinations of ten-pentagon rings.

Take ten regular pentagons of circumradius 1 around a centre. Neighbours share an edge, and a regular decagon is left as a hole in the middle. Call this a ring. Its radius is φ², and the distance between pentagon centres is φ.

There are only two ways to place two rings.

Name	Centre distance	Pentagons shared
adjacent	φ²·\|ζ⁴−1\| ≈ 4.9798	two neighbouring
skip	φ²·\|ζ²−1\| ≈ 3.0777	two with two between them

The ratio is exactly φ. These two, and nothing else, are used to grow the figure outward while keeping five-fold rotational symmetry. That is the whole construction.

On vocabulary. This article is written in the words of the construction itself — ring, address, adjacent, skip, pentagram. The correspondence with existing mathematics (Penrose tilings, inflation) is deferred to §10, where it is labelled as such.

2. Address and orientation

The orientation of a pentagon is fixed by its bearing from the ring centre and nothing else. Call the pentagon in the k-th direction address k; its orientation depends only on the parity of k. There is no further freedom.

A ring maps onto itself under a 36° rotation, so a ring carries no rotational freedom of its own. A ring is determined by its centre coordinate alone.

Addresses sit directly on the phase integers of b13phase:

BASE = 3120 = 60 × 52
36° = 312 = 6 × STEP_MIN
address k → phase k × 312

3. Two tilings

Two Penrose tilings can be drawn.

Fig. 2 — First layer. **A* (left): skip connections, a pentagon at the centre. B (right): adjacent connections, a pentagram at the centre.*

Inside 2-B a pentagram appears — the star-shaped gap enclosed by five pentagons. There is a strong rule here, so this is the side to follow first.

Around a pentagram, five rings appear, of exactly the same form as the first layer.

As an integer identity, with g the centre of the pentagram:

enclosing pentagons     g + φ  · ζ^(k+2i)     (address parity is (k+1) mod 2)
enclosing ring centres  g + φ³ · ζ^(k+2i)     (i = 0..4)

The first layer is this rule with g = 0.

Turning the rule takes the rings from 25 → 45 → 55, and the pentagrams from 6 → 16 → 21. The rule then stops.

The reason it stops is plain. At 55 rings, all 21 pentagrams are already fully enclosed by five rings each. The number of new rings the rule demands is zero. Nothing here is computationally hard; the rule simply has no work left.

Fig. 3 — Growth of the pentagram-centred figure.

4. Joining the ring centres

The rings were being laid down; but looking only at their centres, a different tiling appears.

Take the ring centres alone and join them with lines.

What appears is another tiling. The pentagons were placed. The rhombi were not — they are simply what the ring centres leave between them.

regular pentagons: circumradius φ³, every edge an adjacent connection
rhombi: edge φ³·|ζ²−1|, short diagonal a skip connection

And the centroid of each regular pentagon lands exactly on a pentagram of the original figure. Where a star was, a large pentagon arrives.

A skip connection is the short diagonal of one of these rhombi. If skip connections are drawn as edges when taking the faces of the centre figure, each rhombus splits into two triangles. Take the faces using adjacent connections only.

5. The rhombi face the centre

There is one further condition on the construction.

Within a layer, the skip connections do not overlap. Under this condition every rhombus points toward the centre coordinate.

The centroid of a rhombus is the midpoint of two rings, and its long axis is always perpendicular to the connection vector. So:

long axis points at the centre ⟺ (C₂−C₁) ⊥ (C₁+C₂) ⟺ |C₁| = |C₂|

Two rings joined by a skip connection must have equal radius. This is what defines a layer.

The condition is not cosmetic. Building the figure without it and comparing:

	Faces of the centre figure
condition held	closes with 21 regular pentagons + 10 rhombi
condition broken	15 rhombi / 41 pentagons / five 15-gons / five invalid pentagons

Breaking it wrecks the centre figure: 15-gons appear, which no Penrose tiling contains.

And the condition is stronger than mirror symmetry — the broken figure was still closed under reflection. Symmetry alone is not enough.

6. Getting past the dead end

![Sixty rings with all ten large rhombi completed]

Fig. 4 — 60 rings and the completed large rhombi.

At 55 rings the growth rule is exhausted. So look at the centre figure instead.

Only five of the large rhombi were complete. The five skip-connected pairs at radius 15.8262 were each missing one vertex.

That missing vertex is the next ring centre.

It is pinned down by an integer identity:

C_new = a + φ²(ζ⁴−1)·ζ^k     (choose the k that also makes an adjacent connection from b)
→ five points at radius 20.5623

The point is unique. Over Z[ζ₁₀] there are always exactly two points adjacent-connected to both a and b — the two vertices of the rhombus. For the five completed pairs both are already in the figure; for the five incomplete pairs one already is, so the missing one is determined. That point also satisfies the address-parity condition and the zero-overlap condition. There is no choice to make.

	55 rings	60 rings
rings	55	60
pentagons	385	420
large rhombi	5 (5 pairs incomplete)	all 10 complete
overlaps	0	0
address-parity mismatches	0	0
skip pairs of unequal radius	0	0

Two rules alternate. When the rule that places rings around pentagrams runs out, the rule that fills the gaps in the large rhombi produces the next step. Where one stops, the other moves.

7. The two figures coincide

By now the point of Fig. 2-A and Fig. 2-B may already be visible: they are the same figure at different scales.

Fig. 5 — The pentagon-centred tiling (Fig. 2-A) grown out: 55 rings, 385 pentagons.

Overlay the two at 60 rings.

Fig. 6 — The two tilings overlaid.

Centre figure	Original figure	Count	Match
pentagon centroids	pentagrams	21 ↔ 21	integer tuples identical
rhombus centres	thin rhombi	10 ↔ 10	integer tuples identical

This match is not read off the picture, and it is not a floating-point tolerance. Centroids are held as integer 4-tuples scaled by 5, rhombus centres as the sum of a skip pair, and they are compared with == without rounding.

Scale ratio:

pentagon circumradius   4.236068 / 1        = 4.236068
rhombus edge            4.979797 / 1.175571 = 4.236066
                                       φ³ = 4.236068

The small rhombus sits exactly inside the large one.

Measured from the same pentagram, the ladder of powers of φ is visible directly:

five pentagon centres around a pentagram …… radius φ
                                          …… radius φ²
                                          …… radius φ³
five ring centres around a pentagram      …… radius φ³   ← lands on the third rung

8. Writing it with integers only

With real coordinates, overlap detection needs a threshold, and thresholds stop working as the figure grows. So the whole thing is written in integers from the start.

Coordinates are integer 4-tuples over Z[ζ₁₀], with ζ = e^{iπ/5}:

Zeta = tuple[int, int, int, int]     # a + bζ + cζ² + dζ³
PHI  = (1, 0, 1, -1)                 # φ, with φ² − φ − 1 = 0 exactly

Reduce with ζ⁴ = −1 + ζ − ζ² + ζ³. Multiplication is an integer convolution and nothing more.

Distances are held as squared lengths, as integer pairs (p, q) = p + qφ over Z[φ]. Two distances are equal iff the tuples are equal. No approximation enters.

def norm2(a: Zeta) -> tuple[int, int]:
    z = zmul(a, zconj(a))
    assert z[1] == 0 and z[3] == -z[2]      # realness falls out structurally
    return (z[0] - z[2], z[2])

That assert is not a sanity check but a structural statement. Multiplying by the conjugate kills the imaginary part by construction, so the squared length necessarily lands in the real subfield Z[φ]. Its basis there is 1 and φ, which is exactly why z[1] == 0 and z[3] == -z[2] holds and the coefficients (p, q) can be read straight out.

Comparison also closes over the integers. Since 2(p+qφ) = (2p+q) + q√5, the sign is decided by comparing (2p+q)² against 5q².

def phi_sign(x):
    A, B = 2*x[0] + x[1], x[1]
    if A == 0 and B == 0: return 0
    if A >= 0 and B >= 0: return 1
    if A <= 0 and B <= 0: return -1
    return (1 if A*A > 5*B*B else -1) if A > 0 else (-1 if A*A > 5*B*B else 1)

Orientation uses the b13phase phase integers with BASE = 3120: address k → k × 312.

Real numbers are used only to order the faces cyclically, and to draw. Never to decide anything.

The constants.py of b13phase states that where a calculation needs closure under φ, Z[ζ₁₀] must be used rather than BASE-unit rounded integers. This implementation follows that: phase (orientation) from b13phase, coordinates from Z[ζ₁₀].

Verification output:

composition: {'rings': 60, 'pentagons': 420, 'pentagrams': 21}

  overlaps == 0:                        True
  parity mismatches == 0:               True
  skip pairs:                           10
  all equal radius:                     True
  faces of centre figure:               {'pentagon': 21, 'rhombus': 10}
  pentagon centroids == pentagrams:     True
  rhombus centres == thin rhombi:       True
  squared edge length of rhombus:       (7, 11)
  φ⁶ × small rhombus edge:              True

(7, 11) means 7 + 11φ. That equals φ⁶ times the edge of the small rhombus, as an integer identity.

9. Connecting to b13phase (BASE = 3120)

Every structural angle lands on a lattice of 156 BASE units.

Expressing all bearings in BASE units — the 10 connection directions, the 10 ring-to-pentagon directions, the 5 pentagram-to-ring directions — gives the 20 values {0, 156, 312, 468, …, 2964}. Their greatest common divisor is exactly 156, and 3120 ÷ 156 = 20, which divides.

The quantum of angle turned out to be a multiple of 13 throughout.

	BASE units
6° (STEP_MIN)	52	= 4 × 13
18° (connection-direction step)	156	= 12 × 13
36° (one address)	312	= 24 × 13
72° (five-fold)	624	= 48 × 13
360°	3120	= 240 × 13

The angular lattice on the Penrose side is a 20-fold division (18° steps), and it divides 3120 by 156, leaving a cofactor of 12·13.

And the newly derived layer landed exactly on it. Of the 60 ring centres, only 20 have a bearing that is a BASE integer. In squared radius these are (5,8)=φ⁶, (34,55)=φ¹⁰, (52,84), and (117,189) — that last one being the five points at radius 20.5623 derived by filling the gaps in the large rhombi. Their bearings are multiples of 312, that is, exact multiples of 36°. They fall on the side that sits on the b13phase lattice.

Fig. 7 — Overlay with b13phase.

10. What is obvious here, and what is not

Drawing the line honestly.

The obvious side. That two figures coincide at a scale ratio is the self-similarity (inflation) of Penrose tilings itself. The correspondence putting a higher-level pentagon where a star was is a known composition rule. Nothing mathematically surprising is happening.

The precise form of "necessary in order to continue". The condition that rhombi face the centre is not necessary for the tiling to continue. Even in the figure that broke the condition, pentagon overlap was zero. What the condition is necessary for is continuing while preserving self-similarity — keeping the property that the centre figure remains a tiling and keeps supplying the next layer. And in fact the layer after 60 rings came only from this condition.

The non-obvious side. The unit — a ring of ten pentagons — was not defined from Penrose theory. It was found by drawing. That the ring centres, a derived set of points, land exactly on the vertices of the higher tiling was not knowable until it was checked.

11. Attribution

Construction, observation, and the rules found — all from drawing, by the author. The ring as a unit; the two connection types, skip and adjacent; the condition that rhombi face the centre; the idea of joining the ring centres; and the plan of getting the next layer out of it.
Checking, counterexamples, and integerisation — the AI (Claude). Building the figure that breaks the condition and showing the centre figure wrecked; confirming the zero-error match; the implementation over Z[ζ₁₀].

This division of labour worked in one direction only. Every attempt to swap the roles — to have the AI aim and the author check — missed.

Code

Everything is in b13_two_tilings.py. Place b13phase v0.6.9 at the same level and run it.

b13_two_tilings.py
b13phase_v069/
    b13phase/

python3 b13_two_tilings.py

What comes next

At 60 rings the skip pairs stay at 10 and stop increasing, so the large rhombi all complete and the figure halts again. The next question is what mechanism produces new skip pairs.

This figure is intended as the base figure of a computing structure — the pentagon processor — with rings as computing units, pentagons as internal phase and memory cells, and the connections between rings as channels. That is for a separate article.

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