A good solution for this can be seen at this
link<http://prismoskills.appspot.com/lessons/Dynamic_Programming/Chapter_04_-_Minimum_no_of_coins_to_make_a_sum.jsp>


On Wed, Apr 23, 2014 at 12:42 AM, bujji jajala <[email protected]>wrote:

> Hi Dave/ Ravi Rajan,
>
>                               I think If conditions that Dave has
> mentioned are satisfied, greedy approach can be proved to work
> with help of induction.  I tried to prove with coin denominations 1 and 3.
>
> f(n) = min coins required to get change for n rupees with one rupee and 3
> rupee coins(:-) only.
> f(0) =0
> Necessary condition for greedy approach to give optimal solution.
> f(n-3) >= f(n-1)  whenever n > =3
>
> f(m) <= f(m+2)    for all m> 1
>
> Assumption:
> The f(m) <= f(m+2)  condition holds for all m <=N
> ----------------------(1)
>
> F(N+1) = min { 1+ f(N), 1+ f(N-2) } = 1+f(N-2)  ( This follows with
> induction assumption (1) )  -------------(2)
>
> f(N+3)  = min{ 1+ f(N+2), 1+ f(N) } = 1+ f(N)  ( This too follows with
> induction assumption (1) )  -------------(3)
>
> (2) and (3) imply
>
> f(N+1) <= f(N+3)
>
> Verifying base conditions is simple.
>
> Please let me n=know if there is any flaw in proof.
>
> @Dave  Can you give some intuition behind your statement?
>
> -Thanks
>  Bujji
>
>
>
>
> On Wed, May 29, 2013 at 8:54 PM, Ravi Ranjan <[email protected]>wrote:
>
>> @DAve
>>
>> any proper reason or link to proof that at least twice can be solved
>> using greedy but others are not??
>>
>>
>> On Tue, May 28, 2013 at 12:41 PM, Shashwat Kumar <
>> [email protected]> wrote:
>>
>>> This seems to be Coin Change problem. Just google that.
>>>
>>>
>>> On Tue, May 28, 2013 at 12:42 AM, Adolfo Ccanto <[email protected]>wrote:
>>>
>>>> Can you write the constraints for this problem?
>>>> thanks.
>>>> Adolfo
>>>>
>>>>
>>>> 2013/5/18 rahul sharma <[email protected]>
>>>>
>>>>> Minimum of coins to sum s
>>>>>
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>>>
>>>
>>> --
>>> Shashwat Kumar
>>> Third year Undergraduate student
>>> Department of Computer Science and Engineering
>>> IIT Kharagpur
>>>
>>>
>>>
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