diff --git a/2015/09/9.md b/2015/09/9.md
new file mode 100644
index 0000000..cad67be
--- /dev/null
+++ b/2015/09/9.md
@@ -0,0 +1,58 @@
+## \-\-- Day 9: All in a Single Night \-\--
+
+Every year, Santa manages to deliver all of his presents in a single
+night.
+
+This year, however, he has some [new
+locations]{title="Bonus points if you recognize all of the locations."}
+to visit; his elves have provided him the distances between every pair
+of locations. He can start and end at any two (different) locations he
+wants, but he must visit each location exactly once. What is the
+*shortest distance* he can travel to achieve this?
+
+For example, given the following distances:
+
+    London to Dublin = 464
+    London to Belfast = 518
+    Dublin to Belfast = 141
+
+The possible routes are therefore:
+
+    Dublin -> London -> Belfast = 982
+    London -> Dublin -> Belfast = 605
+    London -> Belfast -> Dublin = 659
+    Dublin -> Belfast -> London = 659
+    Belfast -> Dublin -> London = 605
+    Belfast -> London -> Dublin = 982
+
+The shortest of these is `London -> Dublin -> Belfast = 605`, and so the
+answer is `605` in this example.
+
+What is the distance of the shortest route?
+
+Your puzzle answer was `117`.
+
+## \-\-- Part Two \-\-- {#part2}
+
+The next year, just to show off, Santa decides to take the route with
+the *longest distance* instead.
+
+He can still start and end at any two (different) locations he wants,
+and he still must visit each location exactly once.
+
+For example, given the distances above, the longest route would be `982`
+via (for example) `Dublin -> London -> Belfast`.
+
+What is the distance of the longest route?
+
+Your puzzle answer was `909`.
+
+Both parts of this puzzle are complete! They provide two gold stars:
+\*\*
+
+At this point, you should [return to your Advent calendar](/2015) and
+try another puzzle.
+
+If you still want to see it, you can [get your puzzle
+input](9/input).
+
diff --git a/2015/09/solution.py b/2015/09/solution.py
new file mode 100644
index 0000000..cacbe7f
--- /dev/null
+++ b/2015/09/solution.py
@@ -0,0 +1,74 @@
+#!/bin/python3
+import sys,time,re,json
+from pprint import pprint
+
+sys.path.insert(0, '../../')
+from fred import list2int,get_re,nprint,lprint,loadFile,dprint,TSP
+start_time = time.time()
+
+input_f = 'input'
+
+part = 2
+#########################################
+#                                       #
+#              Part 1                   #
+#                                       #
+#########################################
+
+def constructGraph(input_f):
+    """
+        The graph of cities should look something like this
+
+        graph = {
+            "node": [("dist1", distance1), ("dist", distance2)],
+        }
+    """
+    inst = []
+    graph = {}
+    with open(input_f) as file:
+        for line in file:
+            inst = line.rstrip().split(' = ')
+            inst = inst[0].split(' to ') + [int(inst[1])]
+            if inst[0] not in graph:
+                graph[inst[0]] = []
+            if inst[1] not in graph:
+               graph[inst[1]] = []
+            graph[inst[0]].append((inst[1],inst[2]))
+            graph[inst[1]].append((inst[0],inst[2]))
+            
+    return graph
+    
+if part == 1:
+    graph = constructGraph(input_f)
+    shortest = float('inf')
+    routes = []
+    cities = set(graph.keys()) 
+    for neighbors in graph.values():
+        for neighbor, _ in neighbors:
+            cities.add(neighbor) 
+    for c in cities:
+        print(TSP(graph, c))
+        shortest = min(shortest,TSP(graph, c)[1])
+        
+    print(shortest)
+
+#########################################
+#                                       #
+#              Part 2                   #
+#                                       #
+#########################################
+if part == 2:
+    graph = constructGraph(input_f)
+    #print(graph)
+    longest = 0
+    routes = []
+    cities = set(graph.keys()) 
+    for neighbors in graph.values():
+        for neighbor, _ in neighbors:
+            cities.add(neighbor) 
+    for c in cities:
+        longest = max(longest,TSP(graph, c,'longest')[1])
+        
+    print(longest)
+
+print("--- %s seconds ---" % (time.time() - start_time))
\ No newline at end of file
diff --git a/2024/06/solution.py b/2024/06/solution.py
index 2351a11..7c69b06 100644
--- a/2024/06/solution.py
+++ b/2024/06/solution.py
@@ -1,8 +1,9 @@
 #!/bin/python3
-import sys,re
+import sys,time,re
 from pprint import pprint
 sys.path.insert(0, '../../')
 from fred import list2int,get_re,nprint,lprint,loadFile,toGrid,get_value_in_direction,grid_valid
+start_time = time.time()
 
 input_f = 'input'
 
@@ -130,6 +131,9 @@ if part == 2:
     result = 0
     for idx,i in enumerate(steps):
         grid[i[0]][i[1]] = '#'
+        print(idx)
         result += isLoop(grid,start,'up')
         grid[i[0]][i[1]] = '.'
-    print(result)
\ No newline at end of file
+    print(result)
+
+print("--- %s seconds ---" % (time.time() - start_time))
\ No newline at end of file
diff --git a/__pycache__/fred.cpython-311.pyc b/__pycache__/fred.cpython-311.pyc
index 0783001..1c60b63 100644
Binary files a/__pycache__/fred.cpython-311.pyc and b/__pycache__/fred.cpython-311.pyc differ
diff --git a/fred.py b/fred.py
index e544805..1a99389 100644
--- a/fred.py
+++ b/fred.py
@@ -1,5 +1,5 @@
-import re
-import sys
+import sys,re,heapq
+from itertools import permutations
 
 def loadFile(input_f):
     """
@@ -139,6 +139,28 @@ def findDupes(input: list):
         raise TypeError("Input must be a list.")
     return [item for item in set(input) if input.count(item) > 1]
 
+def dprint(graph: dict):
+    """
+    Prints a dictionary in a formatted way, with each key-value pair on a new line.
+
+    Args:
+        graph (dict): The dictionary to be printed.
+
+    Returns:
+        None
+
+    Raises:
+        TypeError: If the input is not a dictionary.
+    """
+    if not isinstance(graph, dict):
+        raise TypeError("Input must be a dictionary.")
+    
+    try:
+        print("\n".join("{}\t\t{}".format(k, v) for k, v in graph.items()))
+    except Exception as e:
+        raise RuntimeError(f"An error occurred while printing the dictionary: {e}")
+
+
 def lprint(x: str, log: bool):
     """
     Prints a string if logging is enabled.
@@ -390,3 +412,115 @@ def get_value_in_direction(grid, position, direction=None, length=1, type: str =
         return ''.join(values)
     else:
         return values
+
+def dijkstra(graph, start, end):
+    """
+    Dijkstra's Algorithm to find the shortest path between two nodes in a graph.
+
+    Parameters:
+    - graph: A dictionary where each key is a node, and the value is a list of tuples. 
+             Each tuple represents (neighbor, distance).
+    - start: The starting node.
+    - end: The destination node.
+
+    Returns:
+    - path: A list of nodes that represent the shortest path from start to end.
+    - total_distance: The total distance of the shortest path.
+    """
+
+    priority_queue = [(0, start)]  # Initially, we start with the starting node, with distance 0.
+
+    distances = {node: float('inf') for node in graph} 
+    distances[start] = 0  # The distance to the start node is 0 (we're already there).
+
+    previous_nodes = {node: None for node in graph}
+
+    while priority_queue:
+    
+        current_distance, current_node = heapq.heappop(priority_queue) # heapq.heappop() removes and returns the node with the smallest distance.
+
+        # If we're at the destination node, we can stop. We've found the shortest path.
+        if current_node == end:
+            break
+
+        for neighbor, weight in graph[current_node]:
+            distance = current_distance + weight
+
+            if distance < distances[neighbor]:
+                distances[neighbor] = distance 
+                previous_nodes[neighbor] = current_node  
+
+                heapq.heappush(priority_queue, (distance, neighbor))
+
+    path = []  # This will store the cities in the shortest path.
+    while end is not None:  # Start from the destination and trace backward.
+        path.append(end)  # Add the current node to the path.
+        end = previous_nodes[end]  # Move to the previous node on the path.
+
+    path.reverse()
+
+    return path, distances[path[-1]]
+
+def TSP(graph, start, path_type='shortest'):
+    """
+    Solves TSP (Traveling Salesperson Problem) using brute force by generating all possible paths.
+    Handles missing edges by skipping invalid paths.
+
+    Parameters:
+    - graph: A dictionary where each key is a node, and the value is a list of tuples (neighbor, distance).
+    - start: The starting node.
+    - path_type: Either 'shortest' or 'longest' to find the shortest or longest path.
+    
+    Returns:
+    - shortest_path: The shortest path visiting all nodes exactly once and returning to the start.
+    - shortest_distance: The total distance of this path.
+    """
+    # Validate path_type
+    if path_type not in ['shortest', 'longest']:
+        raise ValueError("path_type must be either 'shortest' or 'longest'.")
+
+    # Extract all unique cities (keys + neighbors)
+    cities = set(graph.keys())
+    for neighbors in graph.values():
+        for neighbor, _ in neighbors:
+            cities.add(neighbor)
+
+    # Ensure the start node is included
+    if start not in cities:
+        raise ValueError(f"Start location '{start}' not found in the graph.")
+
+    # Remove the start city from the set for permutation
+    cities.remove(start)
+
+    # Set the initial best_distance to the appropriate extreme (inf for shortest, -inf for longest)
+    best_distance = float('inf') if path_type == 'shortest' else -float('inf')
+    best_path = None
+
+    # Generate all permutations of the remaining cities
+    for perm in permutations(cities):
+        # Create a complete path starting and ending at the start node
+        path = [start] + list(perm)
+
+        # Calculate the total distance of this path
+        total_distance = 0
+        valid_path = True
+        for i in range(len(path) - 1):
+            # Check if the edge exists in the graph
+            neighbors = dict(graph.get(path[i], []))
+            if path[i + 1] in neighbors:
+                total_distance += neighbors[path[i + 1]]
+            else:
+                valid_path = False
+                break  # Stop checking this path if it's invalid
+
+        # If the path is valid, update the best_path based on the required path_type
+        if valid_path:
+            if (path_type == 'shortest' and total_distance < best_distance) or \
+               (path_type == 'longest' and total_distance > best_distance):
+                best_distance = total_distance
+                best_path = path
+
+    if not best_path:
+        print(f"No valid path found starting at '{start}'. Ensure all cities are connected.")
+    
+    return best_path, best_distance if best_path else None
\ No newline at end of file