sort ropes according to lengths say 1 .2...3.......n
now tie like N-----------1-----------(N-1)======N+1+1+N-1=2N+1;
(N-1)----------2----------(N-2) ===N-1+2+2+(N-2)=2N+1;
N-2-----------3-----------N-3====2n+1;
hope i am correct
On Mon, Mar 28, 2011 at 10:47 PM, Gunjan Sharma <gunjan.khan...@gmail.com>wrote:
> The second cost will be 5+23 irrespective of the end. Correct me u think
> otherwise.
>
>
> On Mon, Mar 28, 2011 at 10:44 PM, Raunak Agrawal
> <raunak.ra...@gmail.com>wrote:
>
>> @Kunal:
>>
>>
>> Eg: Rope 1: Size 10 mtr
>> Rope 2: Size 13 mtr
>> Rope 3: Size 5 mtr
>>
>>
>> Cost of tying rope 1 and rope 2 = 10 +13 = 23
>>
>> Now we have tow end...one for rope 1 and another for rope 2 which can be
>> tied with rope 3.
>>
>> So tie Rope 3 end with rope 2 end : (Length of rope 2 + 3) = (5 + 13) = 18
>>
>> So total cost = 23 +18 = 41.
>>
>>
>> So for different combination we can have different cost....please let me
>> know in case there is any doubt.
>>
>> On Mon, Mar 28, 2011 at 10:35 PM, Kunal <kunal.shrivas...@gmail.com>wrote:
>>
>>> 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 <raunak.ra...@gmail.com>
>>> 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 <<gunjan.khan...@gmail.com>
>>> gunjan.khan...@gmail.com> wrote:
>>>
>>>> The question seems to be correct. Think again....
>>>>
>>>>
>>>> On Mon, Mar 28, 2011 at 5:24 PM, kunal srivastav
>>>> <<kunal.shrivas...@gmail.com>
>>>> kunal.shrivas...@gmail.com> 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 < <shashank7andr...@gmail.com>
>>>>> shashank7andr...@gmail.com> 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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>>>> Chairman IEEE Students Chapter IIT Roorkee
>>>> B.Tech IV year CSE
>>>>
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>>>>
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>
>
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> Regards
> Gunjan Sharma
> Chairman IEEE Students Chapter IIT Roorkee
> B.Tech IV year CSE
>
> Contact No- +91 9997767077
>
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