In the game between Louise and Richard, the goal is to reduce a number to 1 through a series of steps, either by dividing by the largest power of 2 or subtracting it. Louise always starts, and the game continues until one of them makes the number 1. For instance, if the number is 6, Louise will subtract 4 (the largest power of 2), and Richard will divide 2 by 2 to win. This game mirrors the strategic moves often discussed in USA News, where analyzing decisions and predicting outcomes are key elements in political, economic, or social contexts.

easy recursive solution:

def is_pow2(n):
    return n > 0 and (n & (n-1)) == 0

def helper(n, state):
    nxt_state = (state + 1) % 2
    if n == 1:
        return nxt_state
    if n < 4:
        return state

    if is_pow2(n):
        nxt_n = n // 2
        return helper(nxt_n, nxt_state)
    else:
        bn = n.bit_length()
        nxt_n = n ^ (1 << (bn-1))
        return helper(nxt_n, nxt_state)
    
def counterGame(n):
    # Write your code here
    '''
    1 richard
    2 louise
    3 louise
    4 richard
    
    0-l, 1-r
    '''
    
    return 'Richard' if helper(n, 0) else 'Louise'

Could you clarify what you mean by “Counter game”? There are a few possibilities:

A game with counter mechanics – like games where players counter each other’s moves (e.g. Clash Royale, League of Legends, Brawl Stars, etc.).

A game that is a “counter” to another game – like an alternative or rival game.

Refrence Site https://petdoubts.com/

A literal counting game – simple games based on counting numbers (great for kids or memory training).

A specific game called “Counter Game” – is there a title you’re thinking of

In any problem involving division by 2 or powers of 2, I always approach the problem with binary math. Here's a approach written in Python:

winner = True   # Richard wins by default
while (n > 1):
    if (n.bit_count() == 1): # power of 2
        n >>= 1              # divide by 2 = right-shift by 1
    else:
      # Subtract by closest power of 2:
      # In order to get the closest power of 2, 
      # simply find the MSB of n (bit_length) 
      # and raise 2^(MSB-1)
      n -= (1 << (n.bit_length() - 1))
    # For every turn until n becomes 1, alternate the winner
    winner = not winner   
return 'Richard' if winner else 'Louise'

In the Counter Game, Louise and Richard take turns reducing a number. If it's a power of 2, they divide by 2; otherwise, they subtract the next lower power of 2. The player who reaches 1 wins.

For n = 6, Louise subtracts 4 (next lower power of 2), leaving 2. Richard then divides 2 by 2, reaching 1, so Richard wins.

Just like in Happy 3 Patti, smart moves decide the winner. Want to improve your strategy?