Point to be noted-
1. Always Aditi starts the game.
2. A player loses the game if he/she has to remove the last item.
3. A player always removes 1 or 2 items with a winning strategy.
Method 1: shortcut- time complexity- O(1)
n = total no of tokens.
If n mod 3 - 1 = 0, then Bharat has a winning strategy (1 is the last token being taken by the loser)
if n mod 3 - 1 = 1 or 2, then Aditi has a winning strategy.
The trick here is with 4 tokens whoever plays next is not having a winning strategy. So, the strategy must be to ensure that the opponent always takes his turn with $3n+1$ tokens. This is possible because if the opponent takes 1, I take 2 and vice verse.
Let's check each option-
[Here my strategy of solving this would be just contradicting the each option by considering all the possibilities].
A. For n =3. // Here I am trying to make Bharat win, Let's see if he can win or not.
Aditi - removes 2 item , n = 3-2 = 1
Now only 1 item remains, so Bharat is forced to remove the last item and loses the game. Hence for n=3, Aditi has a winning strategy.
The first part of A - False.
For n =4. // Here I am trying to make Aditi win, Let's see if she can win or not.
Case -1: case 2:
Aditi - removes 1 item , n = 4-1 = 3 Aditi - removes 2 item , n = 4-2 = 1
Bharat- removes 2,item , n = 3 -2 =1 Bharat- removes 1,item , n = 2 -1 =1
Now only 1 item remains in both the case, so Aditi is forced to remove the last item and loses the game. Hence for n=4, Bharat has a winning strategy.
The 2nd part of A - True.
Hence, A- False.
B. For n =7. // Here I am trying to make Aditi win, Let's see if she can win or not.
Case 1: Case 2:
Aditi - removes 1 item , n = 7- 1 = 6 Aditi - removes 2 item , n = 7- 2 = 5
Bharat- removes 2 item , n = 6 -2 = 4. Bharat- removes 1 item , n = 5 -1 = 4.
Now remaining tokens in both the case, n = 4.
case 1: case 2:
Aditi - removes 1 item ,n=4-1 = 3 Aditi - removes 2 item , n = 4-2= 2.
Bharat- removes 2,item ,n=3-2=1, Bharat- removes 1,item , n = 2-1 =1.
Now in both the case, only 1 item remains, so Aditi is forced to remove the last item and loses the game. Hence for n=7, Bharat has a winning strategy. Aditi could not win the game even taking all possibility of Aditi's turn.
For n = 8. // Here I am trying to make Aditi win taking all possibility of Bharat's turn, Let's see if she can win or not.
Aditi - removes 1 item , remaining item n = 8 - 1 = 7
Case 1: case 2:
Bharat- removes 1 item , n = 7 - 1 = 6. Bharat- removes 2 item ,n = 7 - 2 = 5.
Aditi - removes 2 item , n = 6 - 2 = 4 , Aditi - removes 1 item , n = 5-1 = 4.
Now In both cases, remaining items = 4.
case 1 case 2
Bharat- removes 1,item , n = 4-1 = 3 Bharat- removes 2,item , n = 4-2 = 2.
Aditi - removes 2 item , n = 3-2 = 1 , Aditi - removes 1 item , n = 2-1 = 1.
Now in both the case, only 1 item is remaining so Bharat is forced to remove the last item and loses the game. Hence for n=8, Aditi has a winning strategy.
B- True.
C. For both n=3 and n=4, Aditi has a winning strategy.
See explanation for A. where Bharat has a winning strategy for both n=3 and n=4.
Hence, False.
D. For both n=7 and n=8, Bharat has a winning strategy.
See explanation for B. where Aditi has a winning strategy for both n=8.
Hence, False.
E. Bharat never has a winning strategy.
See explanation for B where Aditi has a winning strategy for both n=8.
Hence, False.
Ans - B.