A board game, and two views of a unique dice are shown below. This unique dice has only three faces with numbers one, two and three. Consider all the ladders are advantages that allow the player to jump from initial to final point. What is the minimum number of dice throws that is required to move from Point A to Point B using all given advantages on the board?
Show Hint
Sometimes, taking the earliest available ladder (like the one at cell 3) is a trap because it bypasses better shortcuts later in the game. Always calculate alternative paths that skip early advantages.
Step 1: Understanding the Question:
We are given a board game numbering from 1 (Point A) to 50 (Point B).
The movement is determined by rolling a unique 3-sided dice that only outputs values $\{1, 2, 3\}$.
Ladders are present on the board to help players jump forward.
We need to find the minimum number of throws needed to go from 1 to 50 by utilizing the most optimal path with the ladders.
Step 2: Key Formula or Approach:
We map the board and locate the ladders:
1. Bottom Ladders: Connecting $\{3, 4, 5\}$ to $\{18, 17, 16\}$.
2. Middle Ladders: Connecting $\{14, 13, 12\}$ to $\{27, 28, 29\}$.
3. Top Ladders: Connecting $\{37, 36, 35\}$ to $\{44, 45, 46\}$.
We can use a dynamic programming approach or state-space search to find the shortest path from cell 1 to cell 50.
Step 3: Detailed Explanation:
Let's analyze the transitions:
- Start at Cell 1.
- Since we can choose any dice roll from $\{1, 2, 3\}$ to find the minimum path, let us evaluate the two main strategies:
Strategy 1: Using the first and third ladders:
- Throw 1: Roll 2 to land on Cell 3 $\rightarrow$ climbs to Cell 18.
- Throw 2: Roll 3 to land on Cell 21.
- Throw 3: Roll 3 to land on Cell 24.
- Throw 4: Roll 3 to land on Cell 27.
- Throw 5: Roll 3 to land on Cell 30.
- Throw 6: Roll 3 to land on Cell 33.
- Throw 7: Roll 3 to land on Cell 36 $\rightarrow$ climbs to Cell 45.
- Throw 8: Roll 3 to land on Cell 48.
- Throw 9: Roll 2 to land on Cell 50.
- Total = 9 throws.
Strategy 2: Bypassing the first ladder to use the second and third ladders:
- Throw 1: Roll 3 to land on Cell 4.
- Throw 2: Roll 3 to land on Cell 7.
- Throw 3: Roll 3 to land on Cell 10.
- Throw 4: Roll 2 to land on Cell 12 $\rightarrow$ climbs to Cell 29.
- Throw 5: Roll 3 to land on Cell 32 (Note: the path goes $29 \rightarrow 30 \rightarrow 31 \rightarrow 32$, which is 3 steps).
- Throw 6: Roll 3 to land on Cell 35 $\rightarrow$ climbs to Cell 46.
- Throw 7: Roll 3 to land on Cell 49.
- Throw 8: Roll 1 to land on Cell 50.
- Total = 8 throws.
Comparing the two strategies, Strategy 2 is the most optimal, requiring only 8 throws.
Step 4: Final Answer:
The minimum number of dice throws required is 8.
