CODEWITHSUNDEEP
pythonalgorithmdynamic-programming
user1179317
user1179317

Reputation: 2913

Dungeon Game Solution explanation

The dungeon game is described as:

The demons had captured the princess (P) and imprisoned her in the bottom-right corner of a dungeon. T he dungeon consists of M x N rooms laid out in a 2D grid. Our valiant knight (K) was initially positioned in the top-left room and must fight his way through the dungeon to rescue the princess.

The knight has an initial health point represented by a positive integer. If at any point his health point drops to 0 or below, he dies immediately.

Some of the rooms are guarded by demons, so the knight loses health (negative integers) upon entering these rooms; other rooms are either empty (0's) or contain magic orbs that increase the knight's health (positive integers).

In order to reach the princess as quickly as possible, the knight decides to move only rightward or downward in each step.

Write a function to determine the knight's minimum initial health so that he is able to rescue the princess.

For example, given the dungeon below, the initial health of the knight must be at least 7 if he follows the optimal path RIGHT-> RIGHT -> DOWN -> DOWN.

Notes:

The knight's health has no upper bound. Any room can contain threats or power-ups, even the first room the knight enters and the bottom-right room where the princess is imprisoned.

Example:

dungeon = [[-2,  -3,  4],
           [-6, -15,  0],
           [10,  25, -6]]

Answer: 8

The code solution is:

def dungeonGame(dungeon):
    dp = [float("inf") for _ in dungeon[0]] 
    dp[-1] = 1 

    for i in reversed(range(len(dungeon))):
        dp[-1] = max(dp[-1] - dungeon[i][-1], 1)
        for j in reversed(range(len(dungeon[i]) - 1)):
            min_HP_on_exit = min(dp[j], dp[j + 1])
            dp[j] = max(min_HP_on_exit - dungeon[i][j], 1)

    return dp[0]

Can somebody explain how the solution above is working? Why is the dp only len 3 with the provided example? Is it because there are only 3 steps required, excluding start and finish rooms? Why is it getting the minimum on the adjacent dp's and then the maximum? Also how come it seems that the last column is not being taken into consideration since dungeon[i][j], where j only goes up to 1 (taking the given example matrix). I know the solution is written well, just trying to understand how its taking all the path into consideration.

Upvotes: 3

Views: 2900

Answers (1)

Junuxx
Junuxx

Reputation: 14261

This algorithm works its way back from the bottom right, going left and then up, finding the optimal score for each step along the way. I recommend you execute the algorithm with pen and paper, writing down the current values of i, j and dp along the way. That should really clear things up.

(Start): No i and no j yet, dp = [inf inf 1]

You'll need at least 1 HP after reaching the bottom right in order to win.

(After entering the first loop): i=2, dp = [inf inf 7].

You need 7 health to survive the -6 of the bottom right square itself.

(After entering the inner loop): i=2, j=1, dp = [inf 1 7]

If you're in the bottom center square, the bare minimum 1 health is enough to survive that square's +25, and reach the adjacent square that requires at least 7. And so on.

This is the crucial line that chooses between going right (stored in the next element of the intermediate results, dp[j + 1]) or down, dp[j].

min_HP_on_exit = min(dp[j], dp[j + 1])

There are only three elements to the intermediate results because with the movement rules (only move right and down) and a dungeon with a diagonal of 3, there are only at most 3 places where you could be after any number of moves.

Every time the solver moves up a line, the last column is taken care of as a special case here:

dp[-1] = max(dp[-1] - dungeon[i][-1], 1)

Why? Well, it's different from the other columns in that you can't move right, only down.

Upvotes: 2

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