squid_game_core.py

# squid_game.py

from functools import lru_cache
import math

def parse_tier_map(tier_str: str):
    """
    Example tier_str:
      \"1-4:1\n5-6:3\"
    means:
      if a player has 1..4 squids => each worth x1 => total = k*1
      if a player has 5..6 squids => each worth x3 => total = k*3
    """
    lines = tier_str.strip().splitlines()
    tier_map = []
    for line in lines:
        range_part, mult_part = line.split(":")
        per_squid_mult = float(mult_part.strip())

        if "-" in range_part:
            low_str, high_str = range_part.split("-")
            low, high = int(low_str), int(high_str)
        else:
            low = high = int(range_part.strip())

        tier_map.append((low, high, per_squid_mult))
    tier_map.sort(key=lambda x: x[0])
    return tier_map

def tierValue(k: int, tier_map) -> float:
    """
    For k squids, find bracket => return k * bracket_multiplier.
    If k <= 0 => 0.
    """
    if k <= 0:
        return 0.0
    for (low, high, mult) in tier_map:
        if low <= k <= high:
            return k * mult
    if tier_map and k > tier_map[-1][1]:
        return k * tier_map[-1][2]
    return 0.0  # fallback if no bracket matches

def is_terminal(distribution, remaining) -> bool:
    """
    Game ends if exactly one zero-squid player or no squids remain.
    """
    zero_count = sum(1 for x in distribution if x == 0)
    if zero_count == 1:
        return True
    if remaining == 0:
        return True
    return False

def compute_final_payout(distribution, tier_map):
    """
    distribution: e.g. (0,0,4)
    
    RULES:
      - If exactly 1 zero-squid => that one pays sum_of_winners_tier
      - If multiple zeros => each zero-squid pays sum_of_winners_tier individually
        => each winner gets (number_of_zero_squids * winner_tier_value).
    """
    n = len(distribution)
    zero_indices = [i for i, x in enumerate(distribution) if x == 0]
    m = len(zero_indices)  # number of zero-squid players
    winner_indices = [i for i, x in enumerate(distribution) if x > 0]

    # sum of each winner's bracketed total
    winner_values = [tierValue(distribution[w], tier_map) for w in winner_indices]
    sum_winner_values = sum(winner_values)

    payoffs = [0.0]*n
    if m == 0:
        # No zeros => no payment => everyone gets 0
        return payoffs
    else:
        # Each zero pays sum_winner_values
        for z in zero_indices:
            payoffs[z] = -sum_winner_values
        # Each winner gets their tierValue (if single loser)
        # or m * their tierValue (if multiple losers)
        for i, w in enumerate(winner_indices):
            payoffs[w] = winner_values[i] if m == 1 else m * winner_values[i]
        return payoffs

def format_state(distribution, remaining):
    """Format a game state for display"""
    return f"({','.join(map(str, distribution))}, {remaining})"

@lru_cache(None)
def get_expected_value(distribution, remaining, tier_map_tuple):
    """
    Memoized DP: returns an N-tuple of payoffs from state=(distribution, remaining).
    distribution: tuple of ints
    remaining: int (squids left)
    tier_map_tuple: bracket info (like ( (1,4,1.0), (5,6,3.0) ))
    """
    if is_terminal(distribution, remaining):
        final_pay = compute_final_payout(distribution, tier_map_tuple)
        return tuple(final_pay)
    
    n = len(distribution)
    accumulated = [0.0]*n
    for winner in range(n):
        new_dist = list(distribution)
        new_dist[winner] += 1
        sub_ev = get_expected_value(tuple(new_dist), remaining-1, tier_map_tuple)
        for i in range(n):
            accumulated[i] += sub_ev[i]
    # average
    for i in range(n):
        accumulated[i] /= n
    return tuple(accumulated)

def get_expected_value_forced_win(
    i, 
    distribution, 
    leftover, 
    tier_map_tuple
):
    """
    假设下一只乌贼 100% 给玩家 i。
    则先把 distribution[i] += 1, leftover -=1,
    然后对 (distribution', leftover') 做完全随机的 get_expected_value(...)。
    返回：一个长度 N 的 tuple，表示每个玩家在这种强制赢前提下的期望最终收益。
    """
    if leftover <= 0:
        # 没乌贼剩了，也可能是某些奇怪边界，直接算终局：
        return get_expected_value(distribution, leftover, tier_map_tuple)

    dist_forced = list(distribution)
    dist_forced[i] += 1
    new_dist = tuple(dist_forced)
    return get_expected_value(new_dist, leftover - 1, tier_map_tuple)


def get_expected_value_forced_lose(
    i, 
    distribution, 
    leftover, 
    tier_map_tuple
):
    """
    假设下一只乌贼 100% 不会给玩家 i，
    即本轮发乌贼只在其余 (n-1) 人中随机选 winner，
    然后后续 (leftover-1) 轮恢复正常 n 人随机。

    做法：遍历所有 winner != i (prob=1/(n-1))，发给那个 winner，
    然后 leftover-1 的状态再用 get_expected_value 完全随机。

    返回：一个长度 N 的 tuple (每个玩家最终EV)
    """
    n = len(distribution)
    if leftover <= 0:
        return get_expected_value(distribution, leftover, tier_map_tuple)

    # 如果 n=1，那就无可比了……(此处不太可能)
    # 一般 n>=2, leftover>=1
    
    # 假设我们这里显式地做一次 "下一只的发放" 的平均
    # winner只能在 [0..n-1] - {i} 之中。
    # Probability = 1/(n-1)

    accumulated = [0.0]*n
    valid_winners = [w for w in range(n) if w != i]

    for w in valid_winners:
        dist_next = list(distribution)
        dist_next[w] += 1
        sub_ev = get_expected_value(tuple(dist_next), leftover-1, tier_map_tuple)
        for p in range(n):
            accumulated[p] += sub_ev[p]

    # 做平均
    for p in range(n):
        accumulated[p] /= len(valid_winners)  # == (n-1)

    return tuple(accumulated)

def compute_ev_win_lose_two_extremes(distribution, leftover, tier_map_tuple):
    """
    返回一个数据结构,记录每个玩家 i 在:
      - forced_win    时的期望收益
      - forced_lose   时的期望收益
      - difference    = forced_win - forced_lose
    """
    n = len(distribution)
    results = []

    for i in range(n):
        forced_win_vec = get_expected_value_forced_win(i, distribution, leftover, tier_map_tuple)
        forced_lose_vec = get_expected_value_forced_lose(i, distribution, leftover, tier_map_tuple)

        # 我们可能只关心玩家 i 本人的比较, 也可以把全部人都算,
        # 这里演示只关心 i
        forced_win_i = forced_win_vec[i]
        forced_lose_i = forced_lose_vec[i]
        diff_i = forced_win_i - forced_lose_i

        results.append({
            'player': i,
            'forcedWinEV': forced_win_i,
            'forcedLoseEV': forced_lose_i,
            'difference': diff_i
        })
    return results

# Infinite Squid Game Implementation
def infinite_squid_game_expected_counts_partial(distribution):
    """
    For the "infinite squid + cumulative score, stop when only 1 player has 0" game,
    given the current distribution (list/tuple, representing each player's current count),
    returns a list representing the expected number of squids each player will have at the end of the game.
    
    Rule summary:
      - If the current number of players with 0 squids = z:
        * z=1 => Game stops immediately, no more squids distributed => final value = current holdings
        * z=0 => "Exactly 1 player with 0" can never occur, infinite loop => return inf
        * z>=2 => Expected to distribute n*(H_z -1) more times (H_z=1+1/2+...+1/z),
                  Expected number of times each player is selected = (H_z-1),
                  Therefore final = distribution[i] + (H_z-1).
    """
    n = len(distribution)
    zero_count = sum(1 for x in distribution if x == 0)
    
    # Already satisfied or exceeded "end/cannot end" condition
    if zero_count == 1:
        # Game ends immediately => maintain current state
        return list(distribution)
    if zero_count == 0:
        # "Exactly 1 player with 0" can never occur => game never ends => infinite score
        return [math.inf]*n
    
    # zero_count >= 2 => continue distributing squids until z=1
    H_z = sum(1.0 / k for k in range(1, zero_count+1))  # Harmonic number H_z
    increment = (H_z - 1.0)
    return [x + increment for x in distribution]

# Finite Squid Game Implementation
def get_bitmask(n):
    """
    Returns a list [0,1,2,...,(1<<n)-1], purely for iteration reference
    """
    return list(range(1 << n))

def count_ones(x):
    """Count how many 1s are in the binary representation of x"""
    return bin(x).count("1")

@lru_cache(None)
def dp_ev(n, bitmask_u, leftover):
    """
    Returns a length-n tuple (prob0, prob1, ..., prob_{n-1}),
    representing the "final probability" of each player getting a squid in state (U=bitmask_u, leftover).
    
    - n: total number of players
    - bitmask_u: which players haven't received a squid yet (binary mask)
    - leftover: remaining squids that can be distributed
    """
    # 1) Terminal condition
    num_u = count_ones(bitmask_u)  # Number of players who haven't received a squid yet
    if num_u <= 1 or leftover == 0:
        # Game ends immediately
        # For players who already have a squid, their probability=1
        # For those who don't, probability=0
        ret = []
        for i in range(n):
            # If i is in bitmask_u => probability=0, otherwise =1
            if (bitmask_u & (1 << i)) != 0:
                ret.append(0.0)
            else:
                ret.append(1.0)
        return tuple(ret)
    
    # 2) Not terminated => randomly select 1 person from U to give a squid
    ret_accum = [0.0]*n
    # Randomly select w from U with equal probability
    for w in range(n):
        bit_w = (1 << w)
        if (bitmask_u & bit_w) != 0:
            # w is in U
            # Next state: after w gets a squid, remove w from bitmask
            next_u = bitmask_u ^ bit_w  # Set to 0
            sub_ev = dp_ev(n, next_u, leftover - 1)
            
            for i in range(n):
                ret_accum[i] += sub_ev[i] / num_u
    
    # 3) Return accumulated result
    return tuple(ret_accum)

def finite_squid_game_probabilities(n, r):
    """
    Returns a length-n probability vector: prob[i] = probability of player i getting a squid
    Based on dp_ev().
    
    - n: number of players
    - r: number of squids to distribute
    """
    # Initial U = (1<<n)-1, meaning no one has a squid yet
    bitmask_u = (1 << n) - 1
    return dp_ev(n, bitmask_u, r)