diff --git a/src/tuplesumfilter/messaging.py b/src/tuplesumfilter/messaging.py
new file mode 100644
index 0000000..909336e
--- /dev/null
+++ b/src/tuplesumfilter/messaging.py
@@ -0,0 +1,76 @@
+"""before we design a messaging system, 
+we need to search for a given word
+or set of words in all our messages
+"""
+
+from typing import List, Set
+
+Id = int
+
+'''
+id  | message                                             | other fields...  
+----------------------------------------------------------|-----------------
+1   | Hi Fred, how are you?                               |
+2   | How do I add a new method to the API                |
+3   | You need to follow the process defined on the wiki  |
+.     ...
+.     ...
+1B  | ...                                                 |
+
+      a
+you: [1, 3, 6, 10]
+the: [2, 3, 5, 6, 11]
+      b
+'''
+
+
+
+def search(word: str, messages: dict[Id,str]) -> List[Id]:
+    """We'd really do a post-write job to populate a 2nd DB 
+    table that 1:many mapped word->[message_id] so we 
+    could do fast lookup
+    """
+    return [idx for idx, mess in messages if word in message]
+
+
+def intersect_original(A: List[Id], B: List[Id]) -> List[Id]
+    """O(n*m) where n and m are lengths of the lists of ids"""
+    C = []
+    for a in A:
+        for b in B:
+            if a == b:
+                C.append(a)
+    return C
+
+
+def intersect_better(A: List[Id], B: List[Id]) -> Set[Id]:
+    """O(n) + O(m) + O(m) where n and m are lengths of the lists of ids
+    requires extra storage for sets"""
+    return set(A).intersection(set(B))
+
+    #creating a set O(n) + O(m) + fo))
+
+
+def intersect_even_better(A: List[Id], B: List[Id]) -> List[Id]:
+    """Assuming ids are already sorted, we can do O(m*n)
+    without the extra storage of a set
+    We track 2 list indicies and shuffle 
+    when there we've exhasted the available less than nums
+    """
+    # assume that the lists are already sorted
+    # by someone doing
+    # sort(A)
+    # sort(B)
+    
+    C = []
+    
+    while ia < len(A) and ib < len(B):
+        if A[ia] == B[ib]:
+            C.append(A[ia])
+            ia += 1
+            ib += 1
+        elif A[ia] > B[ib]:
+            ib += 1
+        else:
+            ia += 1
+    return C