Added doctest and more explanation about Dijkstra execution.

.travis.yml

@@ -26,6 +26,7 @@ script:
       matrix
       networking_flow
       other
+      project_euler
       searches
       sorts
       strings

data_structures/hashing/double_hash.py

@@ -1,6 +1,6 @@
 #!/usr/bin/env python3
 
-from .hash_table import HashTable
+from hash_table import HashTable
 from number_theory.prime_numbers import next_prime, check_prime
 
 

data_structures/hashing/hash_table_with_linked_list.py

@@ -1,4 +1,4 @@
-from .hash_table import HashTable
+from hash_table import HashTable
 from collections import deque
 
 

data_structures/hashing/quadratic_probing.py

@@ -1,6 +1,6 @@
 #!/usr/bin/env python3
 
-from .hash_table import HashTable
+from hash_table import HashTable
 
 
 class QuadraticProbing(HashTable):

data_structures/stacks/infix_to_postfix_conversion.py

@@ -2,7 +2,7 @@
 from __future__ import absolute_import
 import string
 
-from .Stack import Stack
+from stack import Stack
 
 __author__ = 'Omkar Pathak'
 

graphs/dijkstra.py

@@ -1,24 +1,50 @@
 """pseudo-code"""
 
 """
-DIJKSTRA(graph G, start vertex s,destination vertex d):
-// all nodes initially unexplored
-let H = min heap data structure, initialized with 0 and s [here 0 indicates the distance from start vertex]
-while H is non-empty:
-    remove the first node and cost of H, call it U and cost
-    if U is not explored
-    mark U as explored
-    if U is d:
-        return cost // total cost from start to destination vertex
-    for each edge(U, V): c=cost of edge(u,V) // for V in graph[U]
-        if V unexplored:
-            next=cost+c
-            add next,V to H (at the end)
+DIJKSTRA(graph G, start vertex s, destination vertex d):
+
+//all nodes initially unexplored
+
+1 -  let H = min heap data structure, initialized with 0 and s [here 0 indicates
+     the distance from start vertex s]
+2 -  while H is non-empty:
+3 -    remove the first node and cost of H, call it U and cost
+4 -    if U has been previously explored:
+5 -      go to the while loop, line 2 //Once a node is explored there is no need
+         to make it again
+6 -    mark U as explored
+7 -    if U is d:
+8 -      return cost // total cost from start to destination vertex
+9 -    for each edge(U, V): c=cost of edge(U,V) // for V in graph[U]
+10 -     if V explored:
+11 -       go to next V in line 9
+12 -     total_cost = cost + c
+13 -     add (total_cost,V) to H
+
+You can think at cost as a distance where Dijkstra finds the shortest distance
+between vertexes s and v in a graph G. The use of a min heap as H guarantees
+that if a vertex has already been explored there will be no other path with
+shortest distance, that happens because heapq.heappop will always return the
+next vertex with the shortest distance, considering that the heap stores not
+only the distance between previous vertex and current vertex but the entire
+distance between each vertex that makes up the path from start vertex to target
+vertex.
 """
+
 import heapq
 
 
 def dijkstra(graph, start, end):
+    """Return the cost of the shortest path between vertexes start and end.
+
+    >>> dijkstra(G, "E", "C")
+    6
+    >>> dijkstra(G2, "E", "F")
+    3
+    >>> dijkstra(G3, "E", "F")
+    3
+    """
+
     heap = [(0, start)]  # cost from start node,end node
     visited = set()
     while heap:
@@ -28,20 +54,65 @@ def dijkstra(graph, start, end):
         visited.add(u)
         if u == end:
             return cost
-        for v, c in G[u]:
+        for v, c in graph[u]:
             if v in visited:
                 continue
             next = cost + c
             heapq.heappush(heap, (next, v))
-    return (-1, -1)
+    return -1
+
+
+G = {
+    "A": [["B", 2], ["C", 5]],
+    "B": [["A", 2], ["D", 3], ["E", 1], ["F", 1]],
+    "C": [["A", 5], ["F", 3]],
+    "D": [["B", 3]],
+    "E": [["B", 4], ["F", 3]],
+    "F": [["C", 3], ["E", 3]],
+}
+
+"""
+Layout of G2:
+
+E -- 1 --> B -- 1 --> C -- 1 --> D -- 1 --> F
+ \                                         /\
+  \                                        ||
+    ----------------- 3 --------------------
+"""
+G2 = {
+    "B": [["C", 1]],
+    "C": [["D", 1]],
+    "D": [["F", 1]],
+    "E": [["B", 1], ["F", 3]],
+    "F": [],
+}
+
+"""
+Layout of G3:
+
+E -- 1 --> B -- 1 --> C -- 1 --> D -- 1 --> F
+ \                                         /\
+  \                                        ||
+    -------- 2 ---------> G ------- 1 ------
+"""
+G3 = {
+    "B": [["C", 1]],
+    "C": [["D", 1]],
+    "D": [["F", 1]],
+    "E": [["B", 1], ["G", 2]],
+    "F": [],
+    "G": [["F", 1]],
+}
+
+shortDistance = dijkstra(G, "E", "C")
+print(shortDistance)  # E -- 3 --> F -- 3 --> C == 6
 
+shortDistance = dijkstra(G2, "E", "F")
+print(shortDistance)  # E -- 3 --> F == 3
 
-G = {'A': [['B', 2], ['C', 5]],
-     'B': [['A', 2], ['D', 3], ['E', 1]],
-     'C': [['A', 5], ['F', 3]],
-     'D': [['B', 3]],
-     'E': [['B', 1], ['F', 3]],
-     'F': [['C', 3], ['E', 3]]}
+shortDistance = dijkstra(G3, "E", "F")
+print(shortDistance)  # E -- 2 --> G -- 1 --> F == 3
 
-shortDistance = dijkstra(G, 'E', 'C')
-print(shortDistance)
+if __name__ == "__main__":
+    import doctest
+    doctest.testmod()

project_euler/problem_01/sol1.py

@@ -1,13 +1,34 @@
-'''
+"""
 Problem Statement:
 If we list all the natural numbers below 10 that are multiples of 3 or 5,
 we get 3,5,6 and 9. The sum of these multiples is 23.
 Find the sum of all the multiples of 3 or 5 below N.
-'''
+"""
 from __future__ import print_function
+
 try:
-    raw_input          # Python 2
+    raw_input  # Python 2
 except NameError:
     raw_input = input  # Python 3
-n = int(raw_input().strip())
-print(sum([e for e in range(3, n) if e % 3 == 0 or e % 5 == 0]))
+
+
+def solution(n):
+    """Returns the sum of all the multiples of 3 or 5 below n.
+
+    >>> solution(3)
+    0
+    >>> solution(4)
+    3
+    >>> solution(10)
+    23
+    >>> solution(600)
+    83700
+    >>> solution(-7)
+    0
+    """
+
+    return sum([e for e in range(3, n) if e % 3 == 0 or e % 5 == 0])
+
+
+if __name__ == "__main__":
+    print(solution(int(raw_input().strip())))

project_euler/problem_01/sol2.py

@@ -1,20 +1,39 @@
-'''
+"""
 Problem Statement:
 If we list all the natural numbers below 10 that are multiples of 3 or 5,
 we get 3,5,6 and 9. The sum of these multiples is 23.
 Find the sum of all the multiples of 3 or 5 below N.
-'''
+"""
 from __future__ import print_function
+
 try:
-    raw_input          # Python 2
+    raw_input  # Python 2
 except NameError:
     raw_input = input  # Python 3
-n = int(raw_input().strip())
-sum = 0
-terms = (n-1)//3
-sum+= ((terms)*(6+(terms-1)*3))//2 #sum of an A.P.
-terms = (n-1)//5
-sum+= ((terms)*(10+(terms-1)*5))//2
-terms = (n-1)//15
-sum-= ((terms)*(30+(terms-1)*15))//2
-print(sum)
+
+
+def solution(n):
+    """Returns the sum of all the multiples of 3 or 5 below n.
+
+    >>> solution(3)
+    0
+    >>> solution(4)
+    3
+    >>> solution(10)
+    23
+    >>> solution(600)
+    83700
+    """
+
+    sum = 0
+    terms = (n - 1) // 3
+    sum += ((terms) * (6 + (terms - 1) * 3)) // 2  # sum of an A.P.
+    terms = (n - 1) // 5
+    sum += ((terms) * (10 + (terms - 1) * 5)) // 2
+    terms = (n - 1) // 15
+    sum -= ((terms) * (30 + (terms - 1) * 15)) // 2
+    return sum
+
+
+if __name__ == "__main__":
+    print(solution(int(raw_input().strip())))

project_euler/problem_01/sol3.py

@@ -1,50 +1,66 @@
-from __future__ import print_function
-
-'''
+"""
 Problem Statement:
 If we list all the natural numbers below 10 that are multiples of 3 or 5,
 we get 3,5,6 and 9. The sum of these multiples is 23.
 Find the sum of all the multiples of 3 or 5 below N.
-'''
-'''
-This solution is based on the pattern that the successive numbers in the series follow: 0+3,+2,+1,+3,+1,+2,+3.
-'''
+"""
+from __future__ import print_function
 
 try:
-    raw_input          # Python 2
+    raw_input  # Python 2
 except NameError:
     raw_input = input  # Python 3
-n = int(raw_input().strip())
-sum=0
-num=0
-while(1):
-    num+=3
-    if(num>=n):
-        break
-    sum+=num
-    num+=2
-    if(num>=n):
-        break
-    sum+=num
-    num+=1
-    if(num>=n):
-        break
-    sum+=num
-    num+=3
-    if(num>=n):
-        break
-    sum+=num
-    num+=1
-    if(num>=n):
-        break
-    sum+=num
-    num+=2
-    if(num>=n):
-        break
-    sum+=num
-    num+=3
-    if(num>=n):
-        break
-    sum+=num
 
-print(sum);
+
+def solution(n):
+    """
+    This solution is based on the pattern that the successive numbers in the
+    series follow: 0+3,+2,+1,+3,+1,+2,+3.
+    Returns the sum of all the multiples of 3 or 5 below n.
+
+    >>> solution(3)
+    0
+    >>> solution(4)
+    3
+    >>> solution(10)
+    23
+    >>> solution(600)
+    83700
+    """
+
+    sum = 0
+    num = 0
+    while 1:
+        num += 3
+        if num >= n:
+            break
+        sum += num
+        num += 2
+        if num >= n:
+            break
+        sum += num
+        num += 1
+        if num >= n:
+            break
+        sum += num
+        num += 3
+        if num >= n:
+            break
+        sum += num
+        num += 1
+        if num >= n:
+            break
+        sum += num
+        num += 2
+        if num >= n:
+            break
+        sum += num
+        num += 3
+        if num >= n:
+            break
+        sum += num
+    return sum
+
+
+if __name__ == "__main__":
+    print(solution(int(raw_input().strip())))