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+5 votes

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. $2$
asked in Algorithms by Veteran (39.8k points) 258 1313 1945
edited by | 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.

6 Answers

+18 votes
Best answer

$2$ 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 :)
they forget to add 'Unweighted' word i think.i wasted lot of time thinking about this question how to solve
how will we think without knowing that edge weight is unity? Is it default weight of any graph?
IIT key was ?
+12 votes

(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
+4 votes

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
+3 votes

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

Source :

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

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