You are hired by a game design company and one of their most popular games is The Journey. The game has a ton of quests, and for a player to win, the player must finish all the quests. There are a total of N quests in the game. Here is how the game works: the player can arbitrarily pick one of the N quests to start from. Once the player completes a quest, they unlock some other quests. The player can then choose one of the unlocked quests and complete it, and so on.
For instance, let's say that this game had only 4 quests: A, B, C, and D. Let's say that after you complete
• quest A, you unlock quests [B, D). .
• quest B, you unlock quests [C, D).
• quest C, you unlock nothing [ ].
• quest D, you unlock quest [C].
Is this game winnable?
Yes, because of the following scenario:
The player picks quest A to start with. At the end of the quest A, the unlocked list contains [B, D). Say that player chooses to do quest B, then the unlocked list will contain [C, D). Say that player chooses to complete quest C, then the unlocked list will contain quest [D]. Finally, player finishes quest D.
Note that if the player had started with quest C instead of quest A, they would have lost this game, because they wouldn't have unlocked any quest and would be stuck. But the game is still winnable, because there is at least one starting quest which makes the player win. The Journey has N quests, enumerated as {nj, N2, . ng}. We construct a directed graph G for the game, where each node in G represents a quest. If completing quest n; will unlock quest n; then there is a directed edge from n; to n; in the graph. Suppose the total number of edges in the graph is M. (1) We call a quest as a lucky quest if starting from this quest will allow the player to win the game. In the example above, quest A is a lucky quest. Show that (a) All lucky quests are in the same strongly connected component of G, and that (b) Every quest in that component is a lucky quest. [We are expecting a short but rigorous proof for both claims] (2) Suppose there is at least one lucky quest. Give an algorithm that runs in time O(N+M) and finds a lucky quest. You may use any algorithm we have seen in class as subroutine. [We are expecting pseudocode or a description of your algorithm, and a short justification of the runtime]

A semi in any graph 'G' is defined by a way to align element if this can be crossed from the beginning of any link in that pixel or if this can be stated that there is a path across each organized node pair within this subgram.

Consecrated pair means (ni, nj) and (nj, ni) that 2 different pairs would be regarded.

In point a:

We're going to be using contradictions for this segment. They start with the assumption that two lucky journeys are not even in the same strongly interconnected component.

For instance, qi and qj are providing special quests that were not in the very strongly linked element and which implies that, if we start to cross from qi or qi, then qj cannot be approached if we begin to move through qj and as we established if qi or qj is fortunate contests, we may reach any other hunts. Which implies qj from qi or conversely should be reachable. Or we might claim that qi is part of the strongly linked portion of qj or vice versa with this situation.

Consequently, we would not be capable of forming part of a different and strongly linked element for two successful scientists; they must have the same strongly related to the element.

In point b:

Its definition of its strong, line segment indicates that all other searches within such a strongly coordinate system can be made possible from any quest. Because if all quests are accessible from any quest then a lucky search is named, and all other quests can be accessed from any quest in a very coordinated system. So, all contests are fortunate contests in a strongly connected element.

Algorithm:

Build 'n' size range named 'visited' wherein 'n' is the number of graphic nodes that keep records of nodes already frequented. Running DFS out of every unknown vertex or mark all edges as seen. The lucky search will be the last unexplored peak.

Psuedo-Code:

Requires this functionality to also be called Solution, it requires 2 reasons as an input, V is a set of objects in graph 'G' and 'G' is a chart itself.

Solution(V, G):

visited[V] = {false}//assign value

for v: 0 to 'V'-1: //using for loop

if not visited[v]: //use if block

DFS(v, visited, V, G) //use DFS method

ans = v //holding value

return ans //return value