补全：真正能发多少

👁️ 2 人浏览 💬 0 人评论 ❤️ 添加收藏

本节的拥塞窗口小工具：

cwnd_trace(nrounds, ssthresh, events={})   cwnd 从 1 段起：小于 ssthresh 每轮翻倍（慢启动），否则每轮加 1（拥塞避免）；
                           events[轮] = "timeout"：ssthresh = cwnd // 2、cwnd 归 1；"dup"（三个重复 ACK）：ssthresh = cwnd // 2、cwnd = ssthresh；交回每轮开始时的 cwnd
effective(cwnd, rwnd)      真正能发的窗口 = 两者里小的

补全 real_windows(cwnds, rwnd)：每轮真正能发的窗口 = 拥塞窗口和接收窗口里小的；交回列表。拥塞窗口走 1、2、4、8、9、10，接收窗口一直是 6：

提交你的答案

请登录后提交答案。

去登录

← 超时和三个重复 ACK 的区别补全：发完要几轮 →

更多题目

代码编辑器

语言： Ctrl + Enter 运行

TABLE = {
    ("CLOSED", "listen"):        ("LISTEN", None),
    ("CLOSED", "connect"):       ("SYN-SENT", "SYN"),
    ("LISTEN", "SYN"):           ("SYN-RECV", "SYN-ACK"),
    ("SYN-SENT", "SYN-ACK"):     ("ESTAB", "ACK"),
    ("SYN-RECV", "ACK"):         ("ESTAB", None),
    ("ESTAB", "close"):          ("FIN-WAIT-1", "FIN"),
    ("ESTAB", "FIN"):            ("CLOSE-WAIT", "ACK"),
    ("FIN-WAIT-1", "ACK"):       ("FIN-WAIT-2", None),
    ("FIN-WAIT-1", "FIN"):       ("CLOSING", "ACK"),
    ("FIN-WAIT-2", "FIN"):       ("TIME-WAIT", "ACK"),
    ("CLOSING", "ACK"):          ("TIME-WAIT", None),
    ("CLOSE-WAIT", "close"):     ("LAST-ACK", "FIN"),
    ("LAST-ACK", "ACK"):         ("CLOSED", None),
    ("LAST-ACK", "retrans"):     ("LAST-ACK", "FIN"),
    ("TIME-WAIT", "FIN"):        ("TIME-WAIT", "ACK"),
    ("TIME-WAIT", "timeout"):    ("CLOSED", None),
    ("CLOSED", "FIN"):           ("CLOSED", "RST"),
    ("LAST-ACK", "RST"):         ("CLOSED", None),
}

class Host:
    def __init__(self, name):
        self.name = name
        self.state = "CLOSED"

def step(self, event):
        """交回要发出去的段（没有就 None）；没这一行的 (状态, 事件) 一律不动。"""
        key = (self.state, event)
        if key not in TABLE:
            return None
        self.state, out = TABLE[key]
        return out

def talk(script, drop=()):
    """script: ["B:listen", "A:connect", "A:close", "B:close", "A:timeout", ...]。
    一台主机发出的段立刻送到对面（除非它的序号在 drop 里），对面的回信再送回来。
    交回 (段的轨迹 ["A>B SYN", ...], {"A": 状态, "B": 状态})。"""
    hosts = {"A": Host("A"), "B": Host("B")}
    trace = []

def send(frm, seg):
        to = "B" if frm == "A" else "A"
        trace.append(frm + ">" + to + " " + seg)
        if len(trace) in drop:
            trace[-1] += " x"
            return
        reply = hosts[to].step(seg)
        if reply:
            send(to, reply)

for item in script:
        who, event = item.split(":")
        seg = hosts[who].step(event)
        if seg:
            send(who, seg)
    return trace, {k: h.state for k, h in hosts.items()}

def segments(nbytes, mss, start=1):
    """把 nbytes 字节切成段，交回 [(序号, 长度)]。序号是这段第一个字节的编号（SYN 占掉了 0 号，数据从 1 起）。"""
    out = []
    seq = start
    while nbytes > 0:
        size = min(mss, nbytes)
        out.append((seq, size))
        seq += size
        nbytes -= size
    return out

def acks(segs, arrived):
    """接收方的累计确认：每收到一段就回一个 ack = 下一个想要的字节号。乱序到的先存着，序号补齐了 ack 才往前跳。
    arrived: 和 segs 等长的 True/False。交回每次到达后回的 ack 列表（丢掉的段没有 ack）。"""
    expect = segs[0][0]
    have = set()
    out = []
    for (seq, size), ok in zip(segs, arrived):
        if not ok:
            continue
        have.add((seq, size))
        while any(s == expect for s, _ in have):
            expect += next(n for s, n in have if s == expect)
        out.append(expect)
    return out

def rto_trace(samples):
    """RFC 6298 的整数版：第一次 SRTT = R，RTTVAR = R // 2；之后 RTTVAR = (3 * RTTVAR + |SRTT - R|) // 4，SRTT = (7 * SRTT + R) // 8；
    RTO = SRTT + 4 * RTTVAR。交回每个样本之后的 (SRTT, RTTVAR, RTO)。"""
    out = []
    srtt = rttvar = None
    for r in samples:
        if srtt is None:
            srtt, rttvar = r, r // 2
        else:
            rttvar = (3 * rttvar + abs(srtt - r)) // 4
            srtt = (7 * srtt + r) // 8
        out.append((srtt, rttvar, srtt + 4 * rttvar))
    return out

def backoff(rto, n):
    """连续超时 n 次之后的 RTO：每次翻倍。"""
    return rto * (2 ** n)

def resend(segs, plan, rtt, rto):
    """停等 + 超时重传：一段一段发，收到 ack 才发下一段；plan 作用在每一次发送上（. 到  x 丢），丢了等 rto 再发（rto 每连续丢一次翻倍）。
    交回 (一共发了几次, 一共花了多少时间)。时间单位随 rtt / rto。"""
    sends = 0
    clock = 0
    k = 0
    for _ in segs:
        cur = rto
        while True:
            act = plan[k] if k < len(plan) else "."
            k += 1
            sends += 1
            if act == "x":
                clock += cur
                cur *= 2
                continue
            clock += rtt
            break
    return sends, clock

def fast_retransmit(ack_list):
    """看接收方回的 ack 序列：同一个 ack 连着出现第 4 次（原始 + 3 个重复）时触发快速重传，交回触发时的位置（从 1 数）；没触发交回 0。"""
    for i in range(3, len(ack_list)):
        if ack_list[i] == ack_list[i - 1] == ack_list[i - 2] == ack_list[i - 3]:
            return i + 1
    return 0

def rounds(nsegs, wnd):
    """窗口一次能放 wnd 段：nsegs 段要几轮（一轮 = 一个 RTT）。停等就是 wnd = 1。"""
    return (nsegs + wnd - 1) // wnd

def flow(nsegs, buf, reads):
    """流控：接收缓冲装得下 buf 段；reads 是应用每轮从缓冲里读走几段的清单（用完就照最后一个值）。
    每轮：发送方按上一轮通告的窗口发（不超过剩余），到了先进缓冲，应用再读，接收方通告 rwnd = buf - 缓冲里的。
    交回 (每轮通告的 rwnd 列表, 用了几轮)。通告是 0 的那一轮之后，发送方只能干等。最多算 50 轮。"""
    inbuf = 0
    left = nsegs
    rwnd = buf
    adv = []
    n = 0
    while left > 0 or inbuf > 0:
        send = min(rwnd, left)
        left -= send
        inbuf += send
        inbuf -= min(reads[n] if n < len(reads) else reads[-1], inbuf)
        rwnd = buf - inbuf
        adv.append(rwnd)
        n += 1
        if n >= 50:
            break
    return adv, n

def cwnd_trace(nrounds, ssthresh, events=None):
    """cwnd 从 1 段起：cwnd < ssthresh 时每轮翻倍（慢启动），否则每轮加 1（拥塞避免）。
    events: {轮: "timeout" | "dup"}：超时 → ssthresh = cwnd // 2，cwnd 归 1；三个重复 ACK → ssthresh = cwnd // 2，cwnd = ssthresh（Reno 的快速恢复）。
    交回每一轮开始时 cwnd 的列表（长度 nrounds）。"""
    events = events or {}
    cwnd = 1
    out = []
    for r in range(1, nrounds + 1):
        out.append(cwnd)
        ev = events.get(r)
        if ev == "timeout":
            ssthresh = max(cwnd // 2, 1)
            cwnd = 1
        elif ev == "dup":
            ssthresh = max(cwnd // 2, 1)
            cwnd = ssthresh
        elif cwnd < ssthresh:
            cwnd = min(cwnd * 2, ssthresh)
        else:
            cwnd += 1
    return out

def effective(cwnd, rwnd):
    """真正能发的窗口：拥塞窗口和接收窗口里小的那个。"""
    return min(cwnd, rwnd)

SS_SAMPLE = """State      Recv-Q Send-Q Local Address:Port  Peer Address:Port Process
LISTEN     1      5          127.0.0.1:25001      0.0.0.0:*     users:(("python3",pid=46,fd=3))
LISTEN     0      5          127.0.0.1:25000      0.0.0.0:*     users:(("python3",pid=27,fd=3))
TIME-WAIT  0      0          127.0.0.1:31010    127.0.0.1:25002
FIN-WAIT-2 5000   0          127.0.0.1:31001    127.0.0.1:25000 users:(("python3",pid=43,fd=3))
CLOSE-WAIT 1      0          127.0.0.1:25000    127.0.0.1:31001 users:(("python3",pid=27,fd=5))
ESTAB      5      0          127.0.0.1:25001    127.0.0.1:31003
TIME-WAIT  0      0          127.0.0.1:25002    127.0.0.1:31011
ESTAB      0      0          127.0.0.1:31003    127.0.0.1:25001 users:(("python3",pid=49,fd=3))
ESTAB      3976   0          127.0.0.1:31000    127.0.0.1:25000 users:(("python3",pid=30,fd=3))
ESTAB      0      0          127.0.0.1:25000    127.0.0.1:31000 users:(("python3",pid=27,fd=4))
"""

def parse_ss(text):
    """把 ss -tanp 的输出拆成一行一个字典：state / recvq / sendq / local / lport / peer / pport / pid（没有进程就是 None）。"""
    rows = []
    for line in text.splitlines()[1:]:
        parts = line.split()
        if len(parts) < 5:
            continue
        lh, lp = parts[3].rsplit(":", 1)
        ph, pp = parts[4].rsplit(":", 1)
        pid = None
        if len(parts) > 5 and "pid=" in parts[5]:
            pid = int(parts[5].split("pid=")[1].split(",")[0])
        rows.append(dict(state=parts[0], recvq=int(parts[1]), sendq=int(parts[2]),
                         local=lh, lport=int(lp), peer=ph, pport=pp if pp == "*" else int(pp), pid=pid))
    return rows

def by_state(rows):
    out = {}
    for r in rows:
        out[r["state"]] = out.get(r["state"], 0) + 1
    return out

def listeners(rows):
    return sorted(r["lport"] for r in rows if r["state"] == "LISTEN")

def peers_of(rows, port):
    """服务端口 port 上（Local 是它）的非 LISTEN 连接，交回 [(状态, 对端端口)]。"""
    return [(r["state"], r["pport"]) for r in rows if r["lport"] == port and r["state"] != "LISTEN"]

def real_windows(cwnds, rwnd):
    # TODO: 用 effective
    return []

w = real_windows(cwnd_trace(6, 8), 6)
print(",".join(str(x) for x in w) + "/" + str(sum(w)))

本次输入：

输出：

👩‍🏫

AI

💬 题目评论

补全：真正能发多少

更多题目

全部评论