Given Data

• Set $A = {1, 2, 3, 4}$
• Number of elements: $n = 4$

STEP 1: Definition of Power Set

The power set of a set $A$ is the set of all subsets of $A$, including the empty set $\emptyset$ and the set $A$ itself.

It is denoted by $P(A)$ or $2^A$.

Key property: If $|A| = n$, then the power set contains $2^n$ subsets, i.e. $$|P(A)| = 2^n$$

STEP 2: Power Set of $A = {1, 2, 3, 4}$

Since $n = 4$, the number of subsets is: $$|P(A)| = 2^4 = 16$$

Listing subsets by size:

Count check: $1 + 4 + 6 + 4 + 1 = 16$ ✓ (matches $\binom{4}{0}+\binom{4}{1}+\binom{4}{2}+\binom{4}{3}+\binom{4}{4}$)

Final Answer:

$$ P(A) = {\ \emptyset,\ {1},\ {2},\ {3},\ {4},\ {1,2},\ {1,3},\ {1,4}, $$ $$ {2,3},\ {2,4},\ {3,4},\ {1,2,3},\ {1,2,4},\ {1,3,4},\ {2,3,4},\ {1,2,3,4}\ } $$

Total number of elements in $P(A) = 16$.

STEP 1 - EXTRACT

Given data:

• Modulus: $m = 11$
• Set: $\mathbb{Z}_{11} = {0, 1, 2, \ldots, 10}$
• Compute $7 +_{11} 9$
• Compute $7 *_{11} 9$

All required data present.

STEP 2 - SOLVE

Arithmetic Modulo m

Arithmetic modulo m is arithmetic carried out on the finite set

$$\mathbb{Z}_m = {0, 1, 2, \ldots, m-1}$$

where every result of addition or multiplication is replaced by its remainder upon division by $m$. This keeps all results within the set $\mathbb{Z}_m$ (closure).

Addition in $\mathbb{Z}_m$: $$a +_m b = (a + b) \bmod m$$

Multiplication in $\mathbb{Z}_m$: $$a *_m b = (a \cdot b) \bmod m$$

Compute $7 +_{11} 9$

$$7 +_{11} 9 = (7 + 9) \bmod 11 = 16 \bmod 11$$

Since $16 = 1 \times 11 + 5$:

$$7 +_{11} 9 = 5$$

Compute $7 *_{11} 9$

$$7 *_{11} 9 = (7 \times 9) \bmod 11 = 63 \bmod 11$$

Since $63 = 5 \times 11 + 8$:

$$7 *_{11} 9 = 8$$

Summary

Generalized Pigeonhole Principle

Given Data

• Number of students in class: $24$
• Grading system: standard grades assumed as ${A, B, C, D, F}$, so number of grade categories $k = 5$
• Required: number of gradings needed to guarantee at least two students share the same grade

Note: The number of distinct grades is not stated in the question. The standard university grading scheme with 5 grades (A, B, C, D, F) is assumed. If a different number of grades is intended, the answer changes accordingly.

Definition

Generalized Pigeonhole Principle:

If $N$ objects are placed into $k$ boxes, then at least one box contains at least $\left\lceil \dfrac{N}{k} \right\rceil$ objects.

Basic Pigeonhole Principle (special case): If $N > k$, then at least one box contains at least $2$ objects.

Solution

Here the grades are the pigeonholes and the students are the objects.

We want at least two students in the same grade category, i.e.

$$\left\lceil \frac{N}{k} \right\rceil \geq 2$$

This is guaranteed when

$$N \geq k + 1$$

Worst-case reasoning: Assign grades one by one. The first $5$ students could each receive a different grade (one per category). The very next student ($6^{th}$) must repeat a grade.

With $k = 5$:

$$N = k + 1 = 5 + 1 = 6$$

$$\boxed{\text{At most } 6 \text{ gradings guarantee two students with the same grade}}$$

Interpretation of "24 students"

The class has $24$ students, which far exceeds the $6$ required. So the class of $24$ students certainly contains repeated grades. Indeed, distributing $24$ students among $5$ grades:

$$\left\lceil \frac{24}{5} \right\rceil = \lceil 4.8 \rceil = 5$$

So with all $24$ students, at least one grade is shared by at least $5$ students. But to merely guarantee two students with the same grade, grading only $6$ students suffices.

Answer: The maximum number of gradings needed is $6$ (using $k = 5$ grades). The number $24$ simply confirms a repetition is unavoidable.

Dijkstra's Algorithm for Shortest Path

Definition

Dijkstra's algorithm is a greedy algorithm that finds the shortest path from a source node to all other nodes (or a specific destination node) in a weighted graph with non-negative edge weights.

Algorithm Steps

DIJKSTRA(Graph G, source s):

1. Initialize:
   - dist[s] = 0          (distance to source is 0)
   - dist[v] = ∞          (distance to all other vertices is infinity)
   - prev[v] = NULL        (previous node in optimal path)
   - Q = set of all vertices (unvisited set / priority queue)

2. While Q is not empty:
   a. u = vertex in Q with minimum dist[u]   (extract-min)
   b. Remove u from Q
   c. For each neighbor v of u (still in Q):
      - alt = dist[u] + weight(u, v)
      - If alt < dist[v]:
          dist[v] = alt        (relaxation step)
          prev[v] = u

3. Return dist[], prev[]

Pseudocode (Formal)

function Dijkstra(G, s):
    for each vertex v in G:
        dist[v] ← ∞
        prev[v] ← UNDEFINED
    dist[s] ← 0
    Q ← all vertices of G

    while Q ≠ ∅:
        u ← vertex in Q with min dist[u]
        remove u from Q

        for each neighbor v of u:
            if v ∈ Q:
                alt ← dist[u] + w(u, v)
                if alt < dist[v]:
                    dist[v] ← alt
                    prev[v] ← u

    return dist[], prev[]

Worked Example

Consider the graph:

    A --4-- B
    |       |
    2       1
    |       |
    C --3-- D
     \     /
      5   2
       \ /
        E

Find shortest path from A to E.

Shortest path: A → C → D → E with cost = 7

Key Properties

Important Notes

• The relaxation step (if alt < dist[v]) is the core operation
• Dijkstra's algorithm fails with negative weight edges (use Bellman-Ford instead)
• It produces a shortest path tree rooted at the source node