Don, if you are talking about cards ordering then it will be n! which looks 
pretty high to compute ( consider even for n=10). and then for every count 
, you will need to compute every single color and values , i.e. worst case 
will be 25 n! ~ O(n!) . Can we do better ?

On Wednesday, 2 July 2014 01:20:07 UTC+5:30, Don wrote:
>
> Initially there are 5! = 120 possible orders for the cards. You should ask 
> for the hint which will reduce that number by the largest amount. So in 
> your example above, asking which cards have value 1 would not reduce the 
> number at all. Asking which card is Green would reduce the possible orders 
> to 24. The algorithm is to make a list of the possible orders. Then for 
> each question, count the number of orders in the list would give each 
> possible answer. The largest count is the score for that question. Ask the 
> question with the smallest score. Then based on the answer, remove any 
> orders from the list which are not consistent with that answer. Repeat the 
> process until you have only one order remaining.
> Don
>
> On Tuesday, July 1, 2014 12:15:14 PM UTC-4, vikas wrote:
>>
>> One of my friend posted this question to me and I am unable to get 
>> anywhere. Could you guys please help 
>>
>> A game of cards has 5 colors and 5 values. Thus a total of 25 cards are 
>> possible. Player 1 will choose n cards and tell you exactly , what all 
>> cards he has. Then he will spread cards in front of you with face downwards 
>> in random fashion.
>> So you know the card values but not the ordering. Player 1 will provide 
>> you 2 types of hints , he can tell you what all cards have some color or he 
>> can tell you what all cards have some value. you need to tell the ordering 
>> of cards with the help of these hints.
>> challenge is to find out minimum number of hints to be provided by player 
>> 1 so that you are able to make it for all cards.
>>
>> e.g. say player chosen n=5 cards
>> and tells you the values
>>
>> B1 Y1 W1 G1 R1
>>
>> he can either tell you that what all cards are 1 (in this case all cards)
>> or can tell you different color like leftmost is color 'B" and next after 
>> 'Y' and so on. So minimun hints in this case is 4
>>
>>  
>> another example
>> if n=4
>> G4 R3 R4 B3
>>
>> Here he can tell you the card 1st and 3rd are of  values 4 and cards 2 
>> and 3 are of color 'R'. This will be sufficient to infer the card values. 
>> so answer is 2
>>
>>
>> I am not interested in code , please suggest plain simple algorithms, if 
>> possible with arguments that why it is correct ?
>>
>>

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