D DUG - poetic-devices-worksheet.pdf - Devon Dugas Name: Poetic ... - Free Printable

It looks like you've uploaded a document containing a problem titled "Practice Question: P1.4", which is about finding the shortest path between two points in a grid with obstacles.

Let me solve and explain this step-by-step based on the content of the image (which I can see as text):

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🔍 Problem Summary

We are given:
- A 2D grid of size `n x m`.
- Each cell is either empty (`.`) or blocked (`#`).
- Start point: `(r1, c1)`
- End point: `(r2, c2)`
- You can move up/down/left/right (no diagonals).
- Goal: Find the shortest path from start to end, avoiding obstacles.
- If no path exists, return `-1`.

This is a classic Shortest Path in Grid problem, typically solved using Breadth-First Search (BFS).

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✔ Step-by-Step Solution

#### 🧩 Step 1: Understand Input Format
From the example:
```
Grid:
...
Start: (0,0)
End: (2,2)
Obstacles at: (1,1), (1,2)
```

So we have:
- Grid:
```
. . .
. # #
. . .
```
- Start: (0,0)
- End: (2,2)

We need to find the shortest path from (0,0) to (2,2).

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#### 🧠 Step 2: Use BFS for Shortest Path

Why BFS?
- BFS explores all nodes level by level.
- First time we reach the destination → it's the shortest path.

#### 📌 Algorithm Steps:

1. Create a queue to store `(row, col, distance)` tuples.
2. Start from `(r1, c1)` with distance = 0.
3. Mark visited cells to avoid revisiting.
4. For each cell, try moving in 4 directions.
5. If next cell is within bounds, not blocked, and not visited → add to queue.
6. If we reach `(r2, c2)` → return distance.
7. If queue is empty and we haven’t reached end → return -1.

---

#### 🧮 Apply to Example

Grid:
```
(0,0) (0,1) (0,2)
(1,0) (1,1)# (1,2)#
(2,0) (2,1) (2,2)
```

Start: (0,0), End: (2,2)

BFS Traversal:

- Level 0: (0,0) → dist = 0
- Level 1: (0,1), (1,0) → dist = 1
- Level 2: (0,2), (2,0), (1,1) is blocked, (1,2) is blocked → so only (0,2), (2,0)
- Level 3: From (0,2): (1,2) blocked → nothing new
From (2,0): (2,1) → dist = 3
- Level 4: From (2,1): (2,2) → dist = 4

✔ So shortest path length = 4

Path:
(0,0) → (1,0) → (2,0) → (2,1) → (2,2)

But wait — let’s check if there's a shorter way?

Try:
- (0,0) → (0,1) → (0,2) → (1,2) is blocked → dead end
- (0,0) → (1,0) → (2,0) → (2,1) → (2,2): 4 steps

Is there another path? No.

So minimum distance = 4

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✔ Final Answer

> Answer: 4

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💡 Code Implementation (Python-like pseudocode)

```python
from collections import deque

def shortest_path(grid, start_r, start_c, end_r, end_c):
n, m = len(grid), len(grid[0])
queue = deque([(start_r, start_c, 0)])
visited = set()
visited.add((start_r, start_c))

directions = [(0,1), (1,0), (0,-1), (-1,0)]

while queue:
r, c, dist = queue.popleft()

if r == end_r and c == end_c:
return dist

for dr, dc in directions:
nr, nc = r + dr, c + dc

if (0 <= nr < n and 0 <= nc < m and
(nr, nc) not in visited and
grid[nr][nc] != '#'):

visited.add((nr, nc))
queue.append((nr, nc, dist + 1))

return -1 # No path found
```

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🚀 Explanation of Why BFS Works

- BFS guarantees the first time we reach the target is via the shortest path in unweighted grids.
- DFS might find a path but not necessarily the shortest one.
- This is an unweighted graph, so BFS is optimal.

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❗ Edge Cases to Consider

- Start or end is blocked → return -1
- Start == End → return 0
- No valid path → return -1

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✔ Conclusion

For the given example:
- Grid has obstacles at (1,1) and (1,2)
- Start: (0,0), End: (2,2)
- The shortest path requires 4 moves
- Therefore, the answer is: 4

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If you'd like, I can also generate a visual representation of the path or simulate the BFS step-by-step!