84 lines
2.4 KiB
Python
84 lines
2.4 KiB
Python
"""
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Conformal Geometric Algebra geometry on Cl(4,1).
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Signature: (+,+,+,+,-), with Euclidean coordinates on e1,e2,e3.
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The two conformal null directions are built from e4 and e5:
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n_o = 0.5 * (e5 - e4) # origin, n_o^2 = 0
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n_inf = e4 + e5 # infinity, n_inf^2 = 0
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n_o · n_inf = -1
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A Euclidean point x embeds as:
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X = x + n_o + 0.5 * |x|^2 * n_inf
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Then X·X = 0 and X·Y = -0.5 * ||x-y||^2.
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This is the ONLY distance metric in CORE-AI.
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No cosine similarity. No L2 norm. No approximate indexing.
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"""
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import numpy as np
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from .cl41 import geometric_product, scalar_part, basis_vector, N_COMPONENTS
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# Basis-vector component indices for e4/e5 inside the grade-1 block.
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# component 1=e1, 2=e2, 3=e3, 4=e4, 5=e5.
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_E4_IDX = 4
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_E5_IDX = 5
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def cga_inner(X: np.ndarray, Y: np.ndarray) -> float:
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"""
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Symmetric inner product: 0.5 * scalar_part(X*Y + Y*X).
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For null vectors representing conformal points: equals -d^2 / 2.
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"""
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XY = geometric_product(X, Y)
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YX = geometric_product(Y, X)
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return 0.5 * scalar_part(XY + YX)
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def outer_product(X: np.ndarray, Y: np.ndarray) -> np.ndarray:
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"""
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Outer (wedge) product: X ^ Y.
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For a prompt versor X_p and response versor X_r,
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X_p ^ X_r is a grade-2 object encoding their geometric relationship.
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"""
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XY = geometric_product(X, Y)
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YX = geometric_product(Y, X)
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return 0.5 * (XY - YX)
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def is_null(X: np.ndarray, tol: float = 1e-6) -> bool:
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"""Check if X lies on the null cone: X·X = 0."""
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return abs(cga_inner(X, X)) < tol
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def null_project(X: np.ndarray) -> np.ndarray:
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"""
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Re-project X onto the null cone by extracting its Euclidean part and
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re-embedding it with the canonical CGA point map.
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"""
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euclidean = np.asarray(X, dtype=np.float32)[1:4].copy()
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return embed_point(euclidean)
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def embed_point(x: np.ndarray) -> np.ndarray:
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"""
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Embed a Euclidean point x in R^3 into the CGA null cone.
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X = x + n_o + 0.5|x|^2 n_inf,
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where n_o = 0.5(e5-e4), n_inf = e4+e5.
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"""
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x = np.asarray(x, dtype=np.float32)
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assert len(x) == 3, "embed_point expects a 3D vector"
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x_sq = float(np.dot(x, x))
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result = np.zeros(N_COMPONENTS, dtype=np.float32)
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result[1:4] = x
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# n_o + 0.5|x|^2 n_inf
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# e4 coefficient: -0.5 + 0.5|x|^2
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# e5 coefficient: 0.5 + 0.5|x|^2
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result[_E4_IDX] = 0.5 * (x_sq - 1.0)
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result[_E5_IDX] = 0.5 * (x_sq + 1.0)
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return result
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