First Contact Report: DeepSeek R1 is Pretty Impressive, And You Should Try It

First, I need to make sure I understand cyclotomic rings. Cyclotomic rings are rings of integers extended by roots of unity. The nth cyclotomic ring is Z[ζn], where ζn is a primitive nth root of unity. These rings are used in algebraic number theory and have properties that might help in discretizing angles, since roots of unity correspond to rotations by angles that are multiples of 2π/n.

The user’s approach seems to map each outer angle of a polygon to an element in such a ring. For a regular hexagon, each outer angle is 60 degrees, which is 2π/6 radians. If they’re using the 12th cyclotomic ring, that’s related to 12th roots of unity, which correspond to 30-degree rotations. But in their example, the hexagon is represented by six 2s. If each 2 corresponds to a rotation step in the cyclotomic ring, maybe each step is 60 degrees. Since 60 degrees is 2π/6, but the cyclotomic ring might be chosen based on the least common multiple needed for the angles. Wait, 60 degrees is a 6th root of unity, so maybe the cyclotomic ring should be the 6th? But they mention ZZ12. Hmm.

Wait, the integers mod 12 might be different from the 12th cyclotomic ring. Maybe there’s a confusion here. The 12th cyclotomic field is Q(ζ12), which has degree φ(12)=4 over Q. But the ring of integers would be Z[ζ12], which includes elements like ζ12, ζ12^5, etc. But perhaps the user is using a modular arithmetic approach, where angles are represented modulo 12. For example, each step in the angle is 2π * k /12, so each unit in the sequence corresponds to 30 degrees. Then, a hexagon with 60-degree angles would need two units per angle (since 2*30=60). Hence, the sequence [2,2,2,2,2,2] over ZZ12. That makes sense. So each element in the sequence is the multiple of the base angle (30 degrees here) that makes up the outer angle of the polygon.

So, for a regular polygon with outer angle θ, the sequence would be [k, k, …, k] where k = θ / (2π/m), with m being the modulus of the cyclotomic ring. Wait, but in the example, the modulus is 12, which corresponds to 12th roots of unity, so base angle 2π/12 = 30 degrees. Then each 2 in the sequence would be 2*30 = 60 degrees, which is the outer angle of a regular hexagon (since the outer angle is 2π/6 = 60 degrees). So the modulus here is 12, which allows representing 30-degree steps. Thus, the cyclotomic ring must be chosen such that all angles of the polygon can be expressed as integer multiples of the base angle (2π/m), where m is the order of the cyclotomic ring.

Therefore, for a polygon with outer angles that are multiples of 2π/m, the cyclotomic ring of order m would be suitable. The minimal m would be the least common multiple of the denominators when the angles are expressed in fractions of π. For example, 60 degrees is π/3, which is 2π/6. But since we’re using steps of 2π/m, to represent π/3, we need m to be a multiple of 6. However, in the example, they used m=12. Maybe to allow for finer subdivisions or other angles in more complex polygons.

Now, the user is asking about Penrose kite and dart tiles. Penrose tiles are aperiodic tiles that can tile the plane non-periodically. The kite and dart are one such pair. Their angles are derived from the golden ratio and involve angles that are multiples of 36 degrees, since they are based on a pentagonal symmetry. Specifically, the kite has angles 72°, 72°, 72°, and 144°, and the dart has angles 36°, 72°, 36°, and 216°, or something like that. Wait, let me recall: In the Penrose kite and dart, the kite is a quadrilateral with two angles of 36°, one of 72°, and one of 216°, while the dart has angles 72°, 72°, 144°, and 72°? Wait, maybe I need to check.

Alternatively, perhaps better to think in terms of the original Penrose tiling. The kite and dart are each composed of two types of rhombuses: a thin rhombus (dart) and a thick rhombus (kite). Wait, no, maybe I’m mixing up. The standard Penrose tiling with kites and darts: the kite is a quadrilateral with four edges, two of which are short and two long, meeting at angles of 72 and 144 degrees. The dart is a concave quadrilateral. Wait, perhaps I need to be precise.

Alternatively, maybe the actual angles are as follows. The kite has four vertices with angles 2π/5, 3π/5, 2π/5, and 3π/5 (or similar fractions). Let me verify. Penrose tiles are based on the golden ratio and pentagonal symmetry, so angles are multiples of π/5. The kite and dart each have angles that are multiples of 36 degrees (π/5 radians). Specifically:

• The kite has four angles: two of 72 degrees (2π/5 radians) and two of 144 degrees (4π/5 radians).

• The dart has four angles: two of 36 degrees (π/5 radians), one of 144 degrees (4π/5

• radians), and one of 324 degrees (9π/5 radians). Wait, but 324 degrees is equivalent to

• -36 degrees modulo 360, which might complicate things. However, since we are dealing

• with outer angles in polygons, perhaps we need to consider the turning angles when

• traversing the perimeter.

Alternatively, when constructing the outer angles for the tiles, each time you traverse an edge and turn by the outer angle. For a convex polygon, the outer angle is π - inner angle. For a concave polygon, the outer angle would be π + |inner angle| or something. Wait, no. The outer angle is defined as the angle one turns when walking along the perimeter. For a convex polygon, it’s supplementary to the inner angle. For a concave polygon, the turn is more than 180 degrees.

In any case, Penrose tiles (kite and dart) have angles that are multiples of 36 degrees. Therefore, to represent them in the cyclotomic ring system, we need a cyclotomic ring that can represent angles in steps of 36 degrees. 36 degrees is π/5 radians, which is 2π/10, so a 10th cyclotomic ring would correspond to 36-degree steps. However, let’s check:

If the base angle is 2π/m, then to get 36 degrees (π/5), we need 2π/m = π/5 ⇒ m = 10. But 2π/m = π/5 ⇒ m=10. So the 10th cyclotomic ring would have angles in increments of 36 degrees. However, the angles in the Penrose tiles are 72, 144, 36, etc. Let’s see:

• 36 degrees = 1 * 36 = 1 step (if step is 36 degrees)

• 72 degrees = 2 steps

• 144 degrees = 4 steps

• 216 degrees = 6 steps (but 216 is 636, which is equivalent to 6 modulo 10, since 10 steps would be 360). Wait, 216 is 636=216, which is 216 mod 360. However, in the cyclotomic ring modulo m, perhaps m is 10, since 10 steps of 36 degrees make a full circle. But when representing outer angles, which can be more than 180 degrees (for concave parts), but in modular arithmetic, angles wrap around. However, when building a polygon, the sum of outer angles must be exactly 2π (or 360 degrees). Wait, the sum of outer angles of any polygon is 2π. So if we have a polygon with outer angles represented as multiples of 2π/m, then the sum of those multiples times 2π/m should equal 2π. Therefore, the sum of the sequence elements should be equal to m.

In the example, the hexagon has six 2s, summing to 12, which is m=12. Then 12*(2π/12)=2π. That checks out. So the key is that the sum of the sequence must equal m, ensuring the total outer angles sum to 2π.

So for Penrose tiles, which are quadrilaterals (kite and dart), their outer angles must sum to 2π, i.e., the sum of their sequence should be m. Let’s figure out the outer angles for kite and dart.

First, let’s consider the inner angles and then compute the outer angles.

For the kite: typically, it’s a convex quadrilateral with two angles of 72 degrees and two angles of 144 degrees. Wait, but let me confirm. Alternatively, the inner angles of the kite are 72°, 72°, 72°, and 144°? No, that doesn’t sum to 360. Wait, the sum of inner angles of a quadrilateral is 360 degrees. If it’s a kite (traditional kite shape, which is convex), usually two angles are equal and the other two are equal. For the Penrose kite, it’s designed with angles related to the golden ratio. Looking it up (since I might be misremembering), the standard Penrose kite has four angles: three of 72° and one of 144°. But 3*72 + 144 = 216 + 144 = 360, which works.

Wait, but that would mean the inner angles are 72°, 72°, 72°, and 144°. Then the outer angles would be 180° - 72° = 108°, three times, and 180° - 144° = 36°. So the outer angles would be [108°, 108°, 108°, 36°]. Converting these to the base angle of 36°, each 108° is 3 steps (336=108), and 36° is 1 step. So the sequence would be [3,3,3,1]. Then the sum is 3+3+3+1=10, so m=10. That works because 1036=360°, and the sum of the outer angles must be 360°, which matches.

Similarly, for the dart: the dart is a concave quadrilateral. Its inner angles are 36°, 72°, 36°, and 216°, perhaps. Let’s check: 36+72+36+216=360. Then the outer angles would be 180° - inner angle for convex angles, and 180° + |inner angle| for concave angles. Wait, the dart has one concave angle. Suppose the inner angles are 36°, 72°, 36°, and 216°. The first three are convex, so their outer angles are 180 - 36 = 144°, 180 - 72 = 108°, 180 - 36 = 144°, and the concave angle’s outer angle would be 180 + (216 - 180) = 216°, but actually, the outer angle is defined as the amount you turn when going around the polygon. For a convex angle, you turn left by (180 - inner angle), and for a concave angle, you turn right by (inner angle - 180), which would be equivalent to a negative left turn. But since the user’s representation uses positive numbers for counterclockwise (left) turns, a concave angle would require a negative value. However, the user mentioned they only consider simple polygons with unit length segments meeting at angles possible in the ring. Maybe they are using absolute values or have a way to represent the concave turns as positive by considering the supplement.

Wait, but in the problem statement, the user says “positive means counterclockwise rotation”. So if you have a concave angle, the outer angle would effectively be a rotation greater than 180 degrees, but since we are working modulo m, maybe it’s represented as a positive number. Alternatively, perhaps they adjust the angle to be within [0, 360) by adding multiples of 360. However, in the case of outer angles, the sum must be exactly 360 degrees. For a concave polygon, one of the outer angles will be greater than 180 degrees. For example, a concave quadrilateral has one outer angle that is 180 + (180 - inner angle). Wait, maybe I need to think differently.

Wait, when you traverse the polygon, at each vertex, the outer angle is the angle you turn. For convex vertices, it’s a left turn (positive in the user’s system), and for concave vertices, it’s a right turn, which would be negative. But the user says “positive means counterclockwise rotation”, so concave angles would require negative values. However, the user mentions they only consider simple polygons that can be built from unit length segments meeting at angles possible in the ring. So maybe they avoid concave angles by some other means, or perhaps they encode them as positive angles by using modulo m.

But Penrose dart is a concave polygon, so it would have a negative outer angle. However, if we represent angles in the cyclotomic ring, which is discrete, maybe they take absolute values or use a different convention. Alternatively, they might represent the turn angle as a positive number by considering the total rotation needed. For instance, if you have a concave angle where you turn right by 144 degrees, that could be represented as a left turn by 360 - 144 = 216 degrees. But then, in the cyclotomic ring modulo m, 216 degrees would be equivalent to 216 - 360 = -144, but since we’re using positive numbers, perhaps they take the absolute value or adjust it.

Wait, this is getting complicated. Let me recall that the sum of the outer angles of any polygon is exactly 360 degrees. For convex polygons, all outer angles are positive and less than 180. For concave polygons, some outer angles are considered negative (if turning clockwise), but their sum still must be 360. However, if we represent all turns as positive (counterclockwise), then concave angles would effectively be subtracting from the total, but that complicates the sum. Therefore, perhaps the user is only considering convex polygons? But the Penrose dart is concave. Hmm.

Alternatively, maybe in their system, they allow angles greater than 180 degrees, which would correspond to a “negative” turn when subtracted from 360. For example, a turn of 216 degrees counterclockwise is equivalent to a turn of -144 degrees (i.e., a clockwise turn of 144 degrees). However, since the user uses positive numbers for counterclockwise, a 216-degree turn would still be represented as 216, but in the cyclotomic ring modulo m, this would be equivalent to 216 - 360 = -144, but since they might be working in a ring where angles wrap around, but the actual angle is still 216 degrees. However, the sum of all outer angles must be exactly 360 degrees. So if you have a polygon with an outer angle of 216 degrees, then the other angles must compensate to still sum to 360.

But let’s get back to the Penrose tiles. Let’s assume that we can represent the concave angle as a positive number in the cyclotomic ring by considering the total rotation. For the dart, which has a concave angle, the outer angles would be:

Suppose the inner angles of the dart are 36°, 72°, 36°, and 216°. Then the outer angles would be:

• At the first convex vertex (36° inner): outer angle = 180 - 36 = 144°

• At the second convex vertex (72° inner): outer angle = 180 - 72 = 108°

• At the third convex vertex (36° inner): outer angle = 180 - 36 = 144°

• At the concave vertex (216° inner): outer angle. Wait, the inner angle here is 216°, which is greater than 180°, so the outer angle would be 180 - (360 - 216) = 180 - 144 = 36°. Wait, no, the outer angle when you have a concave vertex is calculated differently. Let me recall that the outer angle is the angle you turn when traversing the polygon. For a convex vertex, you turn left by (180 - inner angle). For a concave vertex, you turn right by (inner angle - 180), which is equivalent to turning left by (180 - (inner angle - 180)) = 360 - inner angle. So the outer angle for a concave vertex is 360 - inner angle. Therefore, for an inner angle of 216°, the outer angle is 360 - 216 = 144°. But that seems contradictory because a concave vertex should result in a turn that subtracts from the total. Wait, let’s think carefully.

If you’re walking along the polygon and encounter a concave angle, you have to turn the other way. So if the inner angle is 216°, then the turn you make is 180° - 216° = -36°, which is a right turn of 36°. In terms of outer angles, which are defined as the amount turned, a right turn is considered negative. However, since the user’s system uses positive numbers for counterclockwise (left) turns, a right turn would be a negative value. But the user mentioned they only consider simple polygons built from unit length segments meeting at angles possible in the ring. Maybe they don’t allow negative angles, so they take the absolute value or represent the turn as a positive angle by adding 360°.

Alternatively, perhaps they handle it by considering that the total sum must be 360°, so even if one angle is a right turn (negative), when represented as a positive angle modulo m, it still sums to 360. For example, a right turn of 36° is equivalent to a left turn of 324°, which is 324 = 360 - 36. If we use modulo 10 (since 36° is 1 step in m=10), then 324° is 324/36 = 9 steps. But 9 modulo 10 is 9, which would be equivalent to -1. However, this might complicate the sum.

Alternatively, maybe the user’s system allows for angles greater than 180°, treating them as the required turn, regardless of direction. But in reality, the outer angle for a concave vertex is negative, but if represented as a positive angle in the cyclotomic ring, it would mess up the total sum. Therefore, perhaps the user’s system only allows convex polygons, which is contradicted by the mention of Penrose tiles, which include concave darts.

Alternatively, maybe in their system, they can represent concave angles by allowing angles greater than 180°, but since the sum of all outer angles must be exactly 360°, the sequence must still sum to m. Let’s proceed with this assumption.

For the Penrose dart: suppose the inner angles are 36°, 72°, 36°, and 216°. Then the outer angles would be:

• For the first three convex vertices: 144°, 108°, 144°, as before.

• For the concave vertex with inner angle 216°, the outer angle is 360 - 216 = 144°, as per the formula for outer angles (since outer angle = 360° - inner angle for concave vertices). Wait, but that can’t be right, because then the sum would be 144 + 108 + 144 + 144 = 540°, which is way more than 360. Clearly, something’s wrong here.

Wait, no, I think I mixed up the definition. Let’s recall that the sum of outer angles of any polygon is always 360°, regardless of convexity. For convex polygons, all outer angles are positive and less than 180°. For concave polygons, some outer angles are considered negative, but when summed, they still total 360°. However, if we represent all outer angles as positive numbers (by taking their absolute value or adjusting modulo m), the sum would exceed 360°, which is impossible. Therefore, the user’s system must account for negative angles (i.e., clockwise turns) with negative numbers in the sequence. However, the user mentioned “positive means counterclockwise rotation,” implying that negative numbers would mean clockwise. But since they are working in a cyclotomic ring, which is a ring in algebraic number theory (not modular arithmetic), negative numbers are allowed. But the example given was [2,2,2,2,2,2] for a hexagon, which are all positive. The user also mentioned they only consider simple polygons built from unit length segments meeting at angles possible in the ring. Maybe the tiles they’re considering have angles that can be represented as positive multiples in the cyclotomic ring, even if some turns are effectively clockwise (negative), but represented as positive by choosing an appropriate m.

Alternatively, perhaps the Penrose dart can be represented with a sequence that includes negative numbers, but since the user’s example uses all positive, maybe there’s a different approach. Let me look up the exact angles for Penrose kite and dart.

Upon checking, the standard Penrose kite and dart have the following angles:

• The kite is a convex quadrilateral with two angles of 72° and two angles of 144°.

• The dart is a concave quadrilateral with two angles of 36°, one angle of 72°, and one reflex angle of 216°.

But when calculating outer angles:

For the kite:

• Each inner angle of 72° gives an outer angle of 180° - 72° = 108°.

• Each inner angle of 144° gives an outer angle of 180° - 144° = 36°.

So the outer angles would be [108°, 108°, 36°, 36°]. Wait, but that sums to 108+108+36+36 = 288°, which is less than 360. That’s incorrect. Therefore, my approach must be wrong.

Wait, no, wait. The sum of outer angles of any polygon is 360°, so if this is a quadrilateral, sum must be 360°. Therefore, the outer angles can’t be [108,108,36,36]. Therefore, perhaps I have the inner angles wrong.

Alternatively, maybe the kite has four angles: let’s refer to a source. According to standard references, the Penrose kite has four angles: 72°, 72°, 72°, and 144°. Sum of inner angles: 72+72+72+144=360. Then outer angles would be:

For each 72° inner angle: outer angle 180-72=108°, three times.

For the 144° inner angle: outer angle 180-144=36°.

Sum: 108+108+108+36 = 360. Yes, that works. So the kite’s outer angles are [108°, 108°, 108°, 36°].

Similarly, the dart has inner angles 36°, 36°, 72°, and 216°. Sum: 36+36+72+216=360. Outer angles:

For 36° inner: 180-36=144°, two times.

For 72° inner: 180-72=108°.

For 216° inner: since it’s a reflex angle, the outer angle is 180 - (360 - 216) = 180 - 144 = 36°, but wait, according to the outer angle definition for reflex angles, it’s 180° - (inner angle - 180°) = 360° - inner angle. So for 216°, outer angle = 360 - 216 = 144°. But then sum is 144+144+108+144=540°, which is too much. Clearly something is wrong here.

Wait, no, the outer angle for a reflex angle is actually the amount you turn when going around the polygon. If the inner angle is 216°, then as you walk along the edge, you turn by 180° - 216° = -36°, which is a right turn of 36°. In terms of outer angles, which must sum to 360°, this right turn is equivalent to a left turn of -36°, which would subtract from the total. Therefore, the outer angles are:

For the two 36° inner angles: 180 - 36 = 144°.

For the 72° inner angle: 180 - 72 = 108°.

For the 216° inner angle: 180 - 216 = -36°, which is equivalent to +324° in positive terms (since -36 + 360 = 324). However, representing this as 324°, which is 9 steps of 36°, the sequence would be [144, 144, 108, 324]. But 324° is 9*36°, which is 9 in the sequence. However, in modulo 10 cyclotomic ring, 9 is equivalent to -1. But the user’s example uses all positive numbers, so maybe they allow angles up to 360°.

But the sum of the sequence should be m. For the kite, [108, 108, 108, 36] in degrees. Each 108° is 3 steps of 36°, and 36° is 1 step. So the sequence is [3,3,3,1], summing to 10. Therefore, m=10. For the dart, the outer angles are [144, 144, 108, -36]. To represent the -36 as a positive angle, it’s 324°, which is 9 steps. So the sequence would be [4,4,3,9], since 144=436, 108=336, 324=9*36. Sum: 4+4+3+9=20. But m should be such that sum of the sequence equals m. Wait, but 20 steps of 36° would be 720°, which is double 360°. This inconsistency arises because representing the negative angle as a positive 324° adds extra.

However, the sum of outer angles must be exactly 360°. If we represent the -36° as 9 steps (324°), then the total would be 144 + 144 + 108 + 324 = 720°, which is double. Therefore, this approach doesn’t work. Hence, the user’s system must accommodate negative numbers to represent clockwise turns. But their example uses all positive numbers. Maybe they only consider convex polygons, but Penrose tiles include concave darts.

Alternatively, perhaps the minimal cyclotomic ring that can represent both kite and dart is the 10th cyclotomic ring, since the angles involved are multiples of 36°, which is 2π/10 radians. However, the issue with the dart’s outer angle sum suggests that maybe a different approach is needed. Alternatively, perhaps the dart’s outer angles can be represented with negative numbers in the sequence, which would still be elements of the cyclotomic ring (since rings allow negatives). So for the dart, the sequence would be [4,4,3,-1], since -1 step corresponds to -36°, which is the clockwise turn. Sum: 4+4+3-1=10, which equals m=10. Then the total angle is (4+4+3-1)36=1036=360°, which works.

Therefore, if negative numbers are allowed in the sequence (as elements of the cyclotomic ring), then the dart can be represented as [4,4,3,-1] in ZZ10 (or the 10th cyclotomic ring). However, the user’s example uses all positive numbers, but maybe negative numbers are permissible as long as they are in the ring. The key is that the ring must include all the necessary coefficients. The 10th cyclotomic ring (which is Z[ζ10]) includes integers with coefficients for ζ10, but if we’re representing angles as multiples of 36°, then the coefficients are integers, positive or negative. So the minimal cyclotomic ring here would be the 10th, since it captures 36° increments.

Therefore, the sequences would be:

• Kite: [3,3,3,1] in Z[ζ10], summing to 10.

• Dart: [4,4,3,-1] in Z[ζ10], also summing to 10.

But the user asked for the sequences of the kite and dart Penrose tiles and the minimal suitable cyclotomic ring. Since the angles involved are multiples of 36°, the minimal cyclotomic ring is the 10th, corresponding to 36° increments. Therefore, the answer should be that the minimal cyclotomic ring is Z[ζ10] (or the 10th cyclotomic ring), and the sequences are [3,3,3,1] for the kite and [4,4,3,-1] for the dart. Alternatively, if negative numbers aren’t allowed, there might be a different representation, but based on the user’s example allowing positive integers, perhaps they adjust the sequence to use modulo m. However, with m=10, representing -1 as 9 would give [4,4,3,9], summing to 20, which is double m=10. Therefore, that’s invalid. Hence, negative numbers must be allowed, and the minimal ring is indeed the 10th cyclotomic ring.

Another way: if we use m=5, since 72° is 2π/5 radians. But 36° is π/5, which is 2π/10. So 10th cyclotomic ring is needed to represent 36° steps. Hence, m=10.

Therefore, the final answer is that the minimal cyclotomic ring is the 10th, and the sequences are [3,3,3,1] for the kite and [4,4,3,-1] for the dart. But we need to confirm the outer angles again to ensure the sequences are correct.

For the kite:

• Outer angles: 108°, 108°, 108°, 36°.

In steps of 36°, that’s 3, 3, 3, 1. Sum 10. Correct.

For the dart:

• Outer angles: 144°, 144°, 108°, -36°.

In steps of 36°, that’s 4, 4, 3, -1. Sum 10. Correct.