Hodge dual: Difference between revisions

@@ Line 94: / Line 94: @@
 # Let '''C''' be the ''k''-combinations of the numbers 1 through ''n'' in lexicographic order. '''C''' will have the same length as '''V''' and '''M'''.
 # For each combination <math>C_i</math>, compute <math>S_i = \sum C_i - \frac{k (k+1)}{2}</math>.
-# Multiply the ''i''-th element of '''V''' by <math>(-1)^{S_i}</math>
+# Multiply the ''i''-th element of '''V''' by <math>(-1)^{S_i}</math>.
 # Reverse the elements of '''V'''.
 To find an unknown '''V''' from a known '''M''', first reverse '''M''' and then adjust the signs.
+Python implementation of the above algorithm:
+{{Databox| Code |
+<syntaxhighlight lang="python">
+import itertools
+import math
+def hodge_dual(n, k, v):
+    N = math.comb(n, k)
+    if len(v) != N:
+        raise ValueError(f"Length of v must be {N}")
+    # Generate lex-ordered k-indices (tuples)
+    indices_list = list(itertools.combinations(range(1, n + 1), k))
+    # k-th triangular number
+    T0 = k * (k + 1) // 2
+    w = [0] * N
+    for i, I in enumerate(indices_list):
+        total = sum(I)
+        exponent = total - T0
+        s = 1 if exponent % 2 == 0 else -1
+        j = N - 1 - i  # Position in dual vector (reverse lex order)
+        w[j] = s * v[i]
+    return w
+if __name__ == "__main__":
+    n = 3
+    k = 2
+    # Output: [4, -4, 1]
+    print(hodge_dual(n, k, [1, 4, 4]))
+</syntaxhighlight>
+}}
 == Applications ==