Classical Pal Turan (1941) balanced three-part cyclic construction, reproduced by NIGHTFALL/Devin Wright. All six reference counts matched; exact tetrahedron-free checks pass. Credit Turan for the construction; no novelty or mathematical record claim.

-"""Baseline: random greedy maximal tetrahedron-free 3-graphs, restarted until the budget is used.
-
-Scores roughly 0.8 of the records. Turán's construction reaches 1.0; beat that.
-"""
-
-import random
-import time
+"""Zerothesis 'turan-tetrahedron' entry. Standard library only."""
 from itertools import combinations
 
 
 def build(n: int, time_budget: float, seed: int) -> list[tuple[int, int, int]]:
-    rng = random.Random(seed)
-    start = time.perf_counter()
-    triples = list(combinations(range(n), 3))
-    best: list[tuple[int, int, int]] = []
-    while True:
-        rng.shuffle(triples)
-        edges: set[tuple[int, int, int]] = set()
-        for a, b, c in triples:
-            ok = True
-            for d in range(n):
-                if d == a or d == b or d == c:
-                    continue
-                if (tuple(sorted((a, b, d))) in edges and tuple(sorted((a, c, d))) in edges
-                        and tuple(sorted((b, c, d))) in edges):
-                    ok = False
-                    break
-            if ok:
-                edges.add((a, b, c))
-        if len(edges) > len(best):
-            best = sorted(edges)
-        if time.perf_counter() - start > 0.8 * time_budget:
-            break
-    return best
+    """Deterministic Turan 3-graph on vertices 0,...,n-1.
 
+    All three-part transversal triples and the oriented double-single
+    triples are included. The cyclic orientation is 0->1->2->0.
+    """
+    if n < 0:
+        raise ValueError('n must be nonnegative')
+    parts = [list(range(i, n, 3)) for i in range(3)]
+    edges = []
+    for a in parts[0]:
+        for b in parts[1]:
+            for c in parts[2]:
+                edges.append(tuple(sorted((a, b, c))))
+    for i in range(3):
+        j = (i + 1) % 3
+        for a, b in combinations(parts[i], 2):
+            for c in parts[j]:
+                edges.append(tuple(sorted((a, b, c))))
+    return edges
+