@atul your understanding of my recurrences are fine but of the question are
not. You cannot have 3 adjacent houses with same colour. GGY is a fine case
for this problem.
On 28 Dec 2013 20:44, "atul anand" <[email protected]> wrote:

> @saurabh : i did not get your algo for modified ques i.e "No 3 adjacent
> houses should get same color"
>
> according to your recurrence
>
> answer[i][j][1] = min(answer[i-1][t][0],answer[i-1][t][1]) + colorvalue[j] for
> all t from 1 to k except j.
>
>
> now suppose in some case *answer[i-1][t][0] *is found minimum in
> above recurrence .
> *answer[i-1][t][0] *means that i - 2 and i - 1 are of same color say for
> eg color green.
>
> answer[i][j][1] is not colored same as i-1 house (green in this case)
> ...now say it is colored with color yellow.
>
> hence,
> answer[i][j][1] is sum of house with color green+green+yellow.
>
> i am not able to understand , how it is taking care that 3 adjacent are
> colored
> different.
>
> could you please clarify my doubt.
>
>
>
> On Thu, Dec 26, 2013 at 5:25 PM, Saurabh Paliwal <
> [email protected]> wrote:
>
>> @atul anand :- no, we need not use all the colors.
>>
>> @kumar raja :- sorry dude for replying this late. Continuing with the
>> previous idea, I extend the solution for the modified problem as
>>
>> 1. let answer[i][j][0] represent minimum cost of coloring i houses when
>> ith house is colored with jth color and so is the (i-1)th house.
>>
>> 2. let answer[i][j][1] represent minimum cost of coloring i houses when
>> ith house is colored with jth color and (i-1)th is not.
>>
>> The recurrence will be
>>
>> 3. *answer[i][j][0] = answer[i-1][j][1] + colorvalue[j]*
>>
>> 4. *answer[i][j][1] = min(answer[i-1][t][0],answer[i-1][t][1]) +
>> colorvalue[j]* for all t from 1 to k except j.
>>
>> // In the first problem, I mistakenly wrote colorvalue[t] here. It will
>> be colorvalue[j] there. ( sorry for the confusion )
>>
>> 5. finally solution will be min(answer[n][j][0], answer[n][j][1]) for all
>> j from 1 to k.
>>
>> 6. initial settings -: answer[1][j][0] = answer[1][j][1] = colorvalue[j].
>>
>>
>> Also now that I read the problem, I guess colorvalue is not fixed for
>> every color, so that will be a 2-D matrix as well.
>>
>> *replace every colorvalue[j] with colorvalue[i][j] in the above
>> recurrences*. ( i is the house number, j is the color number )
>>
>>
>>  On Wed, Dec 18, 2013 at 10:16 PM, atul anand <[email protected]>wrote:
>>
>>> @kumar :  i have one doubt regarding this question.Do we have to use all
>>> K colors[K] to color all building.
>>>
>>> for example :-
>>>
>>> color[3]={1,2,10};
>>>
>>>  now if we have to color 6 building then i can use only 1st 2 color to
>>> color all building and never 10 ...is this allowed ???
>>> building[N]={1,2,1,2,1,2}
>>>
>>>
>>>  On Sun, Dec 15, 2013 at 12:44 AM, kumar raja 
>>> <[email protected]>wrote:
>>>
>>>> You have 'n' houses and 'k' colors. The cost of coloring a house with
>>>> one of the 'k' colors is different from coloring another house with same
>>>> color.  The cost of coloring the house with different colors are different.
>>>> So now how to colour the 'n' houses such that no two adjcent houses will
>>>> get the same color and the total coloring cost for 'n' houses is minimized.
>>>>
>>>>
>>>>
>>>> Regards,
>>>> Kumar Raja.
>>>>
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>>
>>
>>
>> --
>>  -    Saurabh Paliwal
>>
>>        B-Tech. Comp. Science and Engg.
>>
>>        IIT ROORKEE
>>
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