四项一起对得上吗

运行下面这段程序：

def perm(a):
    res = []
    path = []
    used = [False] * len(a)
    def dfs():
        if len(path) == len(a):
            res.append(list(path))
            return
        for i in range(len(a)):
            if used[i]:
                continue
            used[i] = True
            path.append(a[i])
            dfs()
            used[i] = False
            path.pop()
    dfs()
    return res

def subs(a):
    res = []
    path = []
    def dfs(s):
        res.append(list(path))
        for i in range(s, len(a)):
            path.append(a[i])
            dfs(i + 1)
            path.pop()
    dfs(0)
    return res

def pruned(n):
    st = {"c": 0, "v": 0}
    path = []
    used = [False] * (n + 1)
    def dfs():
        st["v"] += 1
        if len(path) == n:
            st["c"] += 1
            return
        for x in range(1, n + 1):
            if used[x]:
                continue
            if path and abs(path[-1] - x) == 1:
                continue
            used[x] = True
            path.append(x)
            dfs()
            path.pop()
            used[x] = False
    dfs()
    return st["c"], st["v"]

def ok(col, c):
    r = len(col)
    for i in range(r):
        if col[i] == c or abs(col[i] - c) == r - i:
            return False
    return True

def queens(n):
    st = {"c": 0, "v": 0}
    col = []
    def dfs():
        st["v"] += 1
        if len(col) == n:
            st["c"] += 1
            return
        for c in range(n):
            if not ok(col, c):
                continue
            col.append(c)
            dfs()
            col.pop()
    dfs()
    return st["c"], st["v"]

ok4 = (len(perm([1, 2, 3])) == 6
       and len(subs([1, 2, 3])) == 8
       and pruned(5) == (14, 70)
       and queens(4)[0] == 2)
print(ok4)