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'''(B) $T' = T$ with total weight $t' < t^2$'''
 
'''(B) $T' = T$ with total weight $t' < t^2$'''
 
   
 
   
(C) $T' \neq T$ but total weight $t' = t2$
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(C) $T' \neq T$ but total weight $t' = t^2$
 
   
 
   
 
(D) None of the above
 
(D) None of the above
  
  
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==={{Template:Author|Arjun Suresh|{{arjunweb}} }}===
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When the edge weights are squared the minimum spanning tree won't change.
  
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$t'$ < $t^2$, because sum of squares is always less than the square of the sums except for a single element case.
  
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Hence, B is the general answer and A is also true for a single edge graph. Hence, in GATE 2012, marks were given to all.
  
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[[Category:Algorithms & Data Structures Questions from GATE]]
 
[[Category:GATE2012]]
 
[[Category:GATE2012]]
[[Category: Some of the previous year GATE questions]]
 

Latest revision as of 11:29, 15 July 2014

Let $G$ be a weighted graph with edge weights greater than one and $G'$ be the graph constructed by squaring the weights of edges in $G$. Let $T$ and $T'$ be the minimum spanning trees of $G$ and $G'$, respectively, with total weights $t$ and $t'$. Which of the following statements is TRUE?

(A) $T' = T$ with total weight $t' = t^2 $

(B) $T' = T$ with total weight $t' < t^2$

(C) $T' \neq T$ but total weight $t' = t^2$

(D) None of the above


Solution by Arjun Suresh

When the edge weights are squared the minimum spanning tree won't change.

$t'$ < $t^2$, because sum of squares is always less than the square of the sums except for a single element case.

Hence, B is the general answer and A is also true for a single edge graph. Hence, in GATE 2012, marks were given to all.





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Let $G$ be a weighted graph with edge weights greater than one and $G'$ be the graph constructed by squaring the weights of edges in $G$. Let $T$ and $T'$ be the minimum spanning trees of $G$ and $G'$, respectively, with total weights $t$ and $t'$. Which of the following statements is TRUE?

(A) $T' = T$ with total weight $t' = t^2 $

(B) $T' = T$ with total weight $t' < t^2$

(C) $T' \neq T$ but total weight $t' = t2$

(D) None of the above




blog comments powered by Disqus