diff --git a/pytest/pure.py b/pytest/pure.py
new file mode 100644
index 0000000..1c7ad77
--- /dev/null
+++ b/pytest/pure.py
@@ -0,0 +1,75 @@
+import itertools
+from math import inf
+
+n = 16
+k = 6
+p = 2
+
+def in_set_a(x):
+    return x.bit_count() == k
+
+def in_set_b(y):
+    return y.bit_count() == k + 1
+
+def related(a, b):
+    return (x ^ y).bit_count() == 1
+
+
+m = inf
+m_prime = inf
+
+window = (1 << p) - 1
+l = 0
+l_prime = 0
+
+# Look at all the `x`s, find the `x` with least number of relations, and
+# what is the `x` with most relations (under the window condition)
+for x in range(2**n):
+    if not in_set_a(x):
+        continue
+
+    relations_count = 0
+
+    for y in range(2**n):
+        if not in_set_b(y):
+            continue
+
+        if not related(x, y):
+            continue
+
+        relations_count += 1
+
+        if (x & window) != (y & window):
+            l += 1
+
+    # This `x` has less relations than all `x` we've looked at so far
+    if relations_count < m:
+        m = relations_count
+
+for y in range(2**n):
+    if not in_set_b(y):
+        continue
+
+    relations_count = 0
+
+    for x in range(2**n):
+        if not in_set_a(x):
+            continue
+
+        if not related(x, y):
+            continue
+
+        relations_count += 1
+
+        if (x & window) != (y & window):
+            l_prime += 1
+
+    # This `y` has less relations than all `y` we've looked at so far
+    if relations_count < m_prime:
+        m_prime = relations_count
+
+# Finally, see if we can get a better `l_max` under this window than
+# what we've seen so far
+l_max = l * l_prime
+
+print(m, m_prime, l_max)