diff --git a/README.md b/README.md
index 24756f6..8eab8e2 100644
--- a/README.md
+++ b/README.md
@@ -2069,8 +2069,6 @@ Entering variable: $x_{13}$.
#### [4.3](). Update Basic Variables
-🏄🏄🏄🏄🏄
-
| Variable | Adjustment | New Value |
@@ -2078,29 +2076,37 @@ Entering variable: $x_{13}$.
| $x_{13}$ | $+100$ | $100$ |
| $x_{33}$ | $-100$ | $50$ |
| $x_{31}$ | $+100$ | $110$ |
-| $ x_{11} $ | $-100$ | $0$ |
+| $x_{11}$ | $-100$ | $0$ |
-**New Basic Variables:**
-- $ x_{13} = 100 $
-- $ x_{21} = 10 $
-- $ x_{22} = 130 $
-- $ x_{31} = 110 $
-- $ x_{33} = 50 $
+[**New Basic Variables**]():
+
+- $x_{13} = 100$
+- $x_{21} = 10$
+- $x_{22} = 130$
+- $x_{31} = 110$
+- $x_{33} = 50$
+
+
#### [4.4](). Verify Feasibility
-- **Supplies**:
- - Supplier 1: $ 100 $ ✔️
- - Supplier 2: $ 10 + 130 = 140 $ ✔️
- - Supplier 3: $ 110 + 50 = 160 $ ✔️
+
-- **Demands**:
- - Consumer 1: $ 10 + 110 = 120 $ ✔️
- - Consumer 2: $ 130 $ ✔️
- - Consumer 3: $ 100 + 50 = 150 $ ✔️
+- [**Supplies**]():
+ - [Supplier 1](): $100$ ✔️
+ - [Supplier 2](): $10 + 130 = 140$ ✔️
+ - [Supplier 3](): $110 + 50 = 160$ ✔️
+
+
+- [**Demands**]():
+ - [Consumer 1](): $10 + 110 = 120$ ✔️
+ - [Consumer 2](): $130$ ✔️
+ - [Consumer 3](): $100 + 50 = 150$ ✔️
+
+
#### [4.5](). Calculate Final Total Cost
@@ -2115,25 +2121,29 @@ $$
-### [4.6](). Final Optimality Check
+#### [4.6](). Final Optimality Check
+
+Recalculating reduced costs confirms all $\bar{c}_{ij} \geq 0$. [**Optimal solution reached**]().
-Recalculating reduced costs confirms all $ \bar{c}_{ij} \geq 0 $. **Optimal solution reached**.
+
+
+## [Step 5](): Final Solution
-## Final Solution
| Variable | Value |
|------------|-------|
-| $ x_{13} $ | 100 |
-| $ x_{21} $ | 10 |
-| $ x_{22} $ | 130 |
-| $ x_{31} $ | 110 |
-| $ x_{33} $ | 50 |
+| $x_{13}$ | 100 |
+| $x_{21}$ | 10 |
+| $x_{22}$ | 130 |
+| $x_{31}$ | 110 |
+| $x_{33}$ | 50 |
+
-**Total Cost:** $\boxed{10460}$.
+[**Total Cost:**](): $\boxed{10460}$.
-This is the optimal solution with all reduced costs non-negative.
+***This is the optimal solution with all reduced costs non-negative.***
@@ -2151,7 +2161,7 @@ This is the optimal solution with all reduced costs non-negative.
### Theoretical Explanation
-The **Assignment Problem** aims to allocate *n* tasks to *n* agents (machines, workers) at minimum cost, ensuring each task and agent is assigned exactly once.**
+The [**Assignment Problem**]() aims to allocate *n* tasks to *n* agents (machines, workers) at minimum cost, ensuring each task and agent is assigned exactly once.
### [Problem Statement]():
@@ -2169,20 +2179,25 @@ The **Assignment Problem** aims to allocate *n* tasks to *n* agents (machines, w
+🏄🏄🏄🏄🏄
## 1. [Hungarian Method]() (Step by Step):
-### [**Step 1](): Subtract Row Minimums**
+### [Step 1](): Subtract Row Minimums
-#### Subtract the minimum value in each row from all elements in that row.
+#### [Subtract the minimum value in each row from all elements in that row]().
-- Row 1 min: 2 → [0, 2, 1]
-- Row 2 min: 1 → [0, 2, 1]
-- Row 3 min: 2 → [3, 0, 2]
+
+
+- [Row 1 min: 2]() → [0, 2, 1]
+- [Row 2 min: 1]() → [0, 2, 1]
+- [Row 3 min: 2]() → [3, 0, 2]
-#### [**Matrix after row subtraction:**]()
+#### [Matrix after row subtraction]():
+
+
| | M1 | M2 | M3 |
|---------|----|----|----|
@@ -2193,16 +2208,23 @@ The **Assignment Problem** aims to allocate *n* tasks to *n* agents (machines, w
-### [**Step 2](): Subtract Column Minimums**
+### [Step 2](): Subtract Column Minimums
-### Problem Recap
+### [Problem Recap]():
+
+
- **3 tasks** must be assigned to **3 machines**.
- Each task can be done by any machine, but with different costs.
- Each task must be assigned to exactly one machine, and each machine to exactly one task.
- **Goal:** Minimize total assignment cost.
-### Cost Table
+
+
+### [Cost Table]():
+
+
+
| | Machine 1 | Machine 2 | Machine 3 |
|---------|-----------|-----------|-----------|