If you have already tied ropes 1 and 2 then their final length would be 23 now 
you are left with two ropes o length 23 and 5 which suns out to be 28. This 
remains the same in each of your two examples. 

Why did you add 13 twice ? I mean you can tie them in 28 cost if you first tie 
any two and then tie with the left one. Why use 3 knots when it can be done in 
2 !!

Sent from my iPhone

On Mar 28, 2011, at 8:36 PM, Raunak Agrawal <[email protected]> wrote:

> I think the possible solution is :
> 
> Tie the highest two ropes at the end of the rope.
> 
> This is because of the following reason:
> 
> Eg: Rope 1: Size 10 mtr
>       Rope 2: Size 13 mtr
>       Rope 3: Size 5 mtr
> 
> Rope1--> Rope2--> Rope3       Cost: (10+13) + (13+5) = 41
> 
> Rope1-->Rope3--> Rope2      Cost: (10+5) + (5+13) = 33
> 
> 
> So the optimum cost is : 1st Highest length Rope + 2*(Length of all other 
> ropes other that two having the highest size) + 2nd Highest Rope.
> 
> 
> On Mon, Mar 28, 2011 at 7:49 PM, Gunjan Sharma <[email protected]> 
> wrote:
> The question seems to be correct. Think again....
> 
> 
> On Mon, Mar 28, 2011 at 5:24 PM, kunal srivastav <[email protected]> 
> wrote:
> if you tie all of them and the cost is sum of invidual lengths then in the 
> end the cost will be sum of all lengths irrespective of any order that we tie 
> them in..
> i think the ques would req you to say that the cost is the longer of the 
> two..plz check
> 
> 
> On Mon, Mar 28, 2011 at 12:11 PM, bittu <[email protected]> wrote:
> you are given n ropes,maybe of different length. the cost of tying two
> ropes is the sum of their lengths.Find a way to tie these ropes
> together so that the cost is minimum.
> 
> 
> 
> Thanks
> Shashank
> 
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