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Let $G = (V, E)$ be a simple undirected graph, and $s$ be a particular vertex in it called the source. For $x \in V$, let $d(x)$ denote the shortest distance in $G$ from $s$ to $x$. A breadth first search (BFS) is performed starting at $s$. Let $T$ be the resultant BFS tree. If $(u, v)$ is an edge of $G$ that is not in $T$, then which one of the following CANNOT be the value of $d(u) - d(v)$?

1. $-1$
2. $0$
3. $1$
4. $2$
edited | 1.9k views
Q. - > Why edge weight is considered as 1 ?

Ans. -> I graph edge weight is 1 then BFS could be used for finding single source shortest path. If edge weight is not assumed to be then all options are possible. Please correct me if i am wrong.

$2$ is the answer.

$d(u) - d(v) = 0$ is possible when both $u$ and $v$ have an edge from $t$ and $t$ is in the shortest path from $s$ to $u$ or $v$.

$d(u) - d(v) = 1$ is possible when $v$ and $t$ are in the shortest path from $s$ to $u$ and both $t$ and $v$ are siblings- same distance from $s$ to both $t$ and $v$ causing $t-u$ edge to be in BFS tree and not $v-u$.

$d(u) - d(v) = -1$ is possible as explained above by interchanging $u$ and $v$.

$d(u) - d(v) = 2$ is not possible. This is because on BFS traversal we either visit $u$ first or $v$. Let's take $u$ first. Now, we put all neighbors of $u$ on queue. Since $v$ is a neighbour and $v$ is not visited before as assumed, $d(v)$ will become $d(u) + 1$. Similarly, for $v$ being visited first.

answered by Veteran (319k points) 581 1452 2970
edited by
edge weight is not defined. There is never a place where they say that edge weight is unity.

Isn't this an issue with this question?
I suppose that is a fair assumption given what all others we assume :)
how will we think without knowing that edge weight is unity? Is it default weight of any graph?
IIT key was ?

(B)Take distance from S

d(U) =1 from S

d(V)=1 from S

So, d(U) - d(V) =0 , d(X) is one shortest distance of the graph

and BFS traversal will not take edge UV

(C) Now for assuming d(U) and d(V) has a distance 1, we similarly calculate the distance from S(in next figure)

(A)In previous figure change the position of U and V, so get

d(U)-d(V) = -1

(D) but 2 not possible as there is a direct edge from U to V always, So, no graph possible with min distance 2. The max distance could be 1

So, ans is (D)

answered by Veteran (65.2k points) 35 223 631
edited by

By Option Elimination:-

Here (2,3) or (3,2)and (3,4) or (4,3) pairs eligible for (u,v) pair as per question bcz these edges are not part of BFS Tree.

lets take (2,3) here d(2)-d(3)=1-1=0 So Option B is feasible

now (3,4) here d(3)-d(4)=1-2=-1 , So Option A is feasible

now (4,3) here d(4)-d(3)=2-1=1 So Option C is feasible

Hence Option D is Ans.

answered by Veteran (20.3k points) 13 78 207

By having known this property, answer is straight away 2 i.e. option D.

answered by Loyal (2.8k points) 3 25 58
answered by Active (1.5k points) 2 8 21
option d... thanku arjun sir
answered by Loyal (4.9k points) 16 36 50
edited
both u and v can be child nodes of the same parent rt?