사고 트리 탐색

연쇄에서 트리로

class ThoughtNode:
    def __init__(self, state, parent=None):
        self.state = state      # partial solution / reasoning so far
        self.parent = parent
        self.children = []
        self.value = None       # evaluator score

ToT의 네 가지 구성 요소

tot = {
    'decompose': step_definition,   # e.g. one equation, one move
    'generate':  propose_thoughts,  # sampling or proposal prompt
    'evaluate':  score_state,        # value/vote prompt
    'search':    bfs_with_beam,      # BFS | DFS | best-first
}

후보 사고 생성하기

def propose_thoughts(state, k=4):
    prompt = (
        'Given the partial solution below, propose ' + str(k) +
        ' DISTINCT possible next steps.\n' + state
    )
    return parse_list(llm(prompt, temperature=0.7))

상태 평가하기

def score_state(state):
    prompt = (
        'Rate how likely this partial solution leads to a correct '
        'final answer. Reply sure / likely / impossible.\n' + state
    )
    label = llm(prompt, temperature=0).strip().lower()
    return {'sure': 1.0, 'likely': 0.5, 'impossible': 0.0}.get(label, 0.3)

빔 탐색을 사용하는 BFS

def bfs_with_beam(root, depth, branch, beam):
    frontier = [root]
    for _ in range(depth):
        nxt = []
        for node in frontier:
            for t in propose_thoughts(node.state, branch):
                child = ThoughtNode(node.state + '\n' + t, node)
                child.value = score_state(child.state)
                nxt.append(child)
        frontier = sorted(nxt, key=lambda n: -n.value)[:beam]
    return max(frontier, key=lambda n: n.value)

백트래킹을 사용하는 DFS

def dfs(node, depth, branch, prune=0.2):
    if depth == 0 or is_solution(node.state):
        return node
    for t in propose_thoughts(node.state, branch):
        child = ThoughtNode(node.state + '\n' + t, node)
        child.value = score_state(child.state)
        if child.value < prune:
            continue                      # backtrack: prune dead end
        res = dfs(child, depth - 1, branch, prune)
        if res and is_solution(res.state):
            return res
    return None

ToT와 자기 일관성 비교

# Rule of thumb
# - reachable by diverse single passes  -> self-consistency
# - needs lookahead / pruning / backtrack -> tree-of-thought
# ToT cost ~ branch * depth * beam * (gen + eval) LLM calls

비용 폭증과 예산

import heapq

def best_first(root, max_calls):
    heap = [(-root.value, root)]
    best, calls = root, 0
    while heap and calls < max_calls:
        _, node = heapq.heappop(heap)
        for t in propose_thoughts(node.state, 3):
            calls += 1
            child = ThoughtNode(node.state + '\n' + t, node)
            child.value = score_state(child.state); calls += 1
            if child.value > best.value:
                best = child
            heapq.heappush(heap, (-child.value, child))
    return best

평가기의 신뢰성

def robust_eval(state, votes=3):
    scores = [score_state(state) for _ in range(votes)]
    return sum(scores) / votes        # average to reduce judge noise
# Even better: replace with a deterministic verifier when available

실제 적용 가능성

def choose_strategy(task):
    if task.requires_search and task.step_verifiable:
        return 'tree-of-thought'
    if task.discrete_answer:
        return 'self-consistency'
    return 'single chain-of-thought'

최소한의 ToT 해결기

def solve(problem, branch=4, depth=3, beam=5, budget=200):
    root = ThoughtNode(problem)
    root.value = score_state(root.state)
    node = bfs_with_beam(root, depth, branch, beam)
    return extract_solution(node.state)

빠른 확인

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