diff --git a/backend/postprocessing/rankings.py b/backend/postprocessing/rankings.py
new file mode 100644
index 00000000..38686f67
--- /dev/null
+++ b/backend/postprocessing/rankings.py
@@ -0,0 +1,141 @@
+# -*- coding: utf-8 -*-
+from typing import List
+
+import numpy as np
+
+
+def head_to_head_votes(ranks: List[List[int]]):
+    tallies = np.zeros((len(ranks[0]), len(ranks[0])))
+    names = sorted(ranks[0])
+    ranks = np.array(ranks)
+    # we want the sorted indices
+    ranks = np.argsort(ranks, axis=1)
+    for i in range(ranks.shape[1]):
+        for j in range(i + 1, ranks.shape[1]):
+            # now count the cases someone voted for i over j
+            over_j = np.sum(ranks[:, i] < ranks[:, j])
+            over_i = np.sum(ranks[:, j] < ranks[:, i])
+            tallies[i, j] = over_j
+            # tallies[i,j] = over_i
+            tallies[j, i] = over_i
+            # tallies[j,i] = over_j
+    return tallies, names
+
+
+def cycle_detect(pairs):
+    """Recursively detect cylces by removing condorcet losers until either only one pair is left or condorcet loosers no longer exist
+    This method upholds the invariant that in a ranking for all a,b either a>b or b>a for all a,b.
+
+
+    Returns
+    -------
+    out : False if the pairs do not contain a cycle, True if the pairs contain a cycle
+
+
+    """
+    # get all condorcet losers (pairs that loose to all other pairs)
+    # idea: filter all losers that are never winners
+    # print("pairs", pairs)
+    if len(pairs) <= 1:
+        return False
+    losers = [c_lose for c_lose in np.unique(pairs[:, 1]) if c_lose not in pairs[:, 0]]
+    if len(losers) == 0:
+        # if we recursively removed pairs, and at some point we did not have
+        # a condorcet loser, that means everything is both a winner and loser,
+        # yielding at least one (winner,loser), (loser,winner) pair
+        return True
+
+    new = []
+    for p in pairs:
+        if p[1] not in losers:
+            new.append(p)
+    return cycle_detect(np.array(new))
+
+
+def get_winner(pairs):
+    """
+    This returns _one_ concordant winner.
+    It could be that there are multiple concordant winners, but in our case
+    since we are interested in a ranking, we have to choose one at random.
+    """
+    losers = np.unique(pairs[:, 1]).astype(int)
+    winners = np.unique(pairs[:, 0]).astype(int)
+    for w in winners:
+        if w not in losers:
+            return w
+
+
+def get_ranking(pairs):
+    """
+    Abuses concordance property to get a (not necessarily unqiue) ranking.
+    The lack of uniqueness is due to the potential existance of multiple
+    equally ranked winners. We have to pick one, which is where
+    the non-uniqueness comes from
+    """
+    if len(pairs) == 1:
+        return list(pairs[0])
+    w = get_winner(pairs)
+    # now remove the winner from the list of pairs
+    p_new = np.array([(a, b) for a, b in pairs if a != w])
+    return [w] + get_ranking(p_new)
+
+
+def ranked_pairs(ranks: List[List[int]]):
+    """
+    Expects a list of rankings for an item like:
+        [("w","x","z","y") for _ in range(3)]
+        + [("w","y","x","z") for _ in range(2)]
+        + [("x","y","z","w") for _ in range(4)]
+        + [("x","z","w","y") for _ in range(5)]
+        + [("y","w","x","z") for _ in range(1)]
+    This code is quite brain melting, but the idea is the following:
+    1. create a head-to-head matrix that tallies up all win-lose combinations of preferences
+    2. take all combinations that win more than they loose and sort those by how often they win
+    3. use that to create an (implicit) directed graph
+    4. recursively extract nodes from the graph that do not have incoming edges
+    5. said recursive list is the ranking
+    """
+    tallies, names = head_to_head_votes(ranks)
+    tallies = tallies - tallies.T
+    # print(tallies)
+    # note: the resulting tally matrix should be skew-symmetric
+    # order by strenght of victory (using tideman's original method, don't think it would make a difference for us)
+    sorted_majorities = []
+    for i in range(len(ranks[0])):
+        for j in range(len(ranks[i])):
+            if tallies[i, j] > 0:
+                sorted_majorities.append((i, j, tallies[i, j]))
+    # we don't explicitly deal with tied majorities here
+    sorted_majorities = np.array(sorted(sorted_majorities, key=lambda x: x[2], reverse=True))
+    # now do lock ins
+    lock_ins = []
+    for (x, y, _) in sorted_majorities:
+        # invariant: lock_ins has no cycles here
+        lock_ins.append((x, y))
+        # print("lock ins are now",np.array(lock_ins))
+        if cycle_detect(np.array(lock_ins)):
+            # print("backup: cycle detected")
+            # if there's a cycle, delete the new addition and continue
+            lock_ins = lock_ins[:-1]
+    # now simply return all winners in order, and attach the losers
+    # to the back. This is because the overall loser might not be unique
+    # and (by concordance property) may never exist in any winning set to begin with.
+    # (otherwise he would either not be the loser, or cycles exist!)
+    # Since there could be multiple overall losers, we just return them in any order
+    # as we are unable to find a closer ranking
+    numerical_ranks = np.array(get_ranking(np.array(lock_ins))).astype(int)
+    conversion = [names[n] for n in numerical_ranks]
+    return conversion
+
+
+if __name__ == "__main__":
+    ranks = (
+        [("w", "x", "z", "y") for _ in range(1)]
+        + [("w", "y", "x", "z") for _ in range(2)]
+        # + [("x","y","z","w") for _ in range(4)]
+        + [("x", "z", "w", "y") for _ in range(5)]
+        + [("y", "w", "x", "z") for _ in range(1)]
+        # [("y","z","w","x") for _ in range(1000)]
+    )
+    rp = ranked_pairs(ranks)
+    print(rp)
diff --git a/backend/requirements.txt b/backend/requirements.txt
index 92668609..b882d594 100644
--- a/backend/requirements.txt
+++ b/backend/requirements.txt
@@ -1,6 +1,7 @@
 alembic==1.8.1
 fastapi==0.88.0
 loguru==0.6.0
+numpy==1.22.4
 psycopg2-binary==2.9.5
 pydantic==1.9.1
 python-dotenv==0.21.0