حل آموزشی تمرین‌های فصل 13Chapter 13 exercise solutions, explained

11 تمرین ۱ — شناسهٔ سرتاسری در عملExercise 1 — End-to-end IDs in practice

بازگشت به صورت تمرینBack to the exercise · ارزیابی و سطحAssessment and level

پاسخ ۶۰ثانیه‌ایThe 60-second answer

صورت مسئله و فرض‌هاProblem and assumptions

چرا این روش و این عددها؟Why this method and these numbers?

مدل ذهنیMental model

حل گام‌به‌گامStep by step

طراحی مشخصConcrete design

مثال محاسبه‌شدهWorked example

دانلود نمودار مستقلDownload standalone diagram

دانلود SVG مستقلDownload standalone SVG

دانلود نمودار مستقلDownload standalone diagram

گزینه‌های جایگزینAlternatives

سناریوهای خرابیFailure scenarios

کد قابل اجراRunnable code

اجراشده و تأییدشده در همین محیطExecuted and verified in this environment

این کد آموزشی و ساده‌سازی‌شده است؛ فقط برای آشنایی با ایده. کد واقعی پروداکشن ملاحظات بسیار بیشتری دارد و هیچ‌کس این‌شکلی کد پروداکشن نمی‌زند.Simplified teaching code for building intuition only — not production code; nobody ships production code like this.

import java.util.*;

// Teaching demo: end-to-end request IDs (Example 13-2 pattern).
// One atomic commit binds UNIQUE(request_id) + debit/credit + input log.
// A lost commit reply + user retry must move money exactly once.
public class TransferDemo {
    static class Row { String payload; String status; String result; Row(String p){payload=p;status="PENDING";} }
    Map<String, Row> requests = new LinkedHashMap<>();
    Map<String, Long> balances = new LinkedHashMap<>(Map.of("A", 1000L, "B", 500L));
    List<String> inputLog = new ArrayList<>();

    // Returns the stored/fresh result string. Simulates one DB transaction.
    synchronized String submit(String id, String from, String to, long amount) {
        String payload = from + "->" + to + ":" + amount;
        Row old = requests.get(id);
        if (old != null) {
            if (!old.payload.equals(payload)) return "REJECTED: id reused with different payload";
            return "DEDUP: " + old.result; // same id + same payload -> stored result
        }
        Row row = new Row(payload);
        requests.put(id, row); // UNIQUE(request_id): a concurrent second insert fails here
        if (balances.get(from) < amount) { row.status = "FAILED"; row.result = "insufficient funds"; inputLog.add(id + "=FAIL"); return row.result; }
        balances.put(from, balances.get(from) - amount);
        balances.put(to, balances.get(to) + amount);
        row.status = "SUCCESS"; row.result = "moved " + amount + " " + from + "->" + to;
        inputLog.add(id + "=" + payload);
        return row.result;
    }

    // Independent consumer: rebuild balances ONLY from the input log.
    static Map<String, Long> replay(List<String> log) {
        Map<String, Long> b = new LinkedHashMap<>(Map.of("A", 1000L, "B", 500L));
        for (String e : log) {
            if (e.endsWith("=FAIL")) continue;
            String p = e.substring(e.indexOf('=') + 1); // from->to:amount
            String[] u = p.split("->|:");
            b.put(u[0], b.get(u[0]) - Long.parseLong(u[2]));
            b.put(u[1], b.get(u[1]) + Long.parseLong(u[2]));
        }
        return b;
    }

    public static void main(String[] a) {
        TransferDemo db = new TransferDemo();
        String r1 = db.submit("req-7", "A", "B", 200);
        System.out.println("first submit          = " + r1 + "   (reply LOST on the wire)");
        String r2 = db.submit("req-7", "A", "B", 200); // user retry, same id + same payload
        System.out.println("retry same id+payload = " + r2);
        System.out.println("reuse id, new payload = " + db.submit("req-7", "A", "B", 999));
        System.out.println("same payload, new id  = " + db.submit("req-8", "A", "B", 200) + " (independent transfer allowed)");
        System.out.println("balances              = " + db.balances + " (each id moved money exactly once)");
        Map<String, Long> rebuilt = replay(db.inputLog);
        System.out.println("replay from log       = " + rebuilt);
        System.out.println(rebuilt.equals(db.balances) ? "OK: replay matches live state" : "MISMATCH");
    }
}

آزمون صحت و معیار عملیاتیCorrectness checks and operational criteria

پاسخ سؤال‌های پیگیریFollow-up answers

اشتباهات رایجCommon mistakes

منابع:Sources: AWS · safe retries with idempotent APIs · PostgreSQL 18 · transaction isolation

22 تمرین ۲ — مهاجرتِ دوگانهExercise 2 — Dual migration

بازگشت به صورت تمرینBack to the exercise · ارزیابی و سطحAssessment and level

پاسخ ۶۰ثانیه‌ایThe 60-second answer

صورت مسئله و فرض‌هاProblem and assumptions

چرا این روش و این عددها؟Why this method and these numbers?

مدل ذهنیMental model

حل گام‌به‌گامStep by step

طراحی مشخصConcrete design

مثال محاسبه‌شدهWorked example

دانلود نمودار مستقلDownload standalone diagram

دانلود نمودار مستقلDownload standalone diagram

گزینه‌های جایگزینAlternatives

سناریوهای خرابیFailure scenarios

کد قابل اجراRunnable code

اجراشده و تأییدشده در همین محیطExecuted and verified in this environment

این کد آموزشی و ساده‌سازی‌شده است؛ فقط برای آشنایی با ایده. کد واقعی پروداکشن ملاحظات بسیار بیشتری دارد و هیچ‌کس این‌شکلی کد پروداکشن نمی‌زند.Simplified teaching code for building intuition only — not production code; nobody ships production code like this.

import java.util.*;

// Teaching demo: dual-schema migration from one shared input log.
// Old flat projection vs new (table + details) projection, % traffic,
// shadow compare, deliberate bug in the new projection, rollback.
public class MigrationDemo {
    record Upsert(String id, String name, long price, String detail) {}
    // old view: flat row per product. new view: row + details row.
    Map<String, String> oldView = new LinkedHashMap<>();
    Map<String, String> newView = new LinkedHashMap<>();
    Map<String, String> newDetails = new LinkedHashMap<>();
    List<Upsert> log = new ArrayList<>(List.of(
        new Upsert("p1", "Lamp", 30, "warm LED"),
        new Upsert("p2", "Chair", 120, "oak"),
        new Upsert("p3", "Desk", 250, "standing")));

    void projectOld(Upsert u) { oldView.put(u.id, u.name + "|" + u.price); }
    void projectNew(Upsert u, boolean buggy) {
        newView.put(u.id, u.name + "|" + u.price);
        // BUG (injected): details dropped when price > 100
        if (!(buggy && u.price > 100)) newDetails.put(u.id, u.detail);
    }

    public static void main(String[] a) {
        MigrationDemo m = new MigrationDemo();
        // backfill + catch-up: both projections consume the same log
        for (Upsert u : m.log) { m.projectOld(u); m.projectNew(u, true); }
        System.out.println("backfill done: old=" + m.oldView.size() + " new=" + m.newView.size());
        // shadow read: compare every read against the old view (0% live traffic yet)
        int mism = 0;
        for (String id : m.oldView.keySet())
            if (!m.newDetails.containsKey(id)) { mism++; System.out.println("shadow mismatch: " + id + " details missing in new view"); }
        System.out.println("shadow mismatches = " + mism + " -> canary BLOCKED, stays 0%");
        // fix projection, re-project from the same log (replay, not patch)
        m.newView.clear(); m.newDetails.clear();
        for (Upsert u : m.log) m.projectNew(u, false);
        mism = 0;
        for (String id : m.oldView.keySet()) if (!m.newDetails.containsKey(id)) mism++;
        System.out.println("after fix: mismatches = " + mism + " -> canary 5% -> 50% -> 100%");
        // rollback drill: flag back to 100% old; old view untouched throughout
        String flag = "new";
        flag = "old"; // one flag flip
        System.out.println("rollback: traffic flag = " + flag + " (old view served reads all along: " + m.oldView.size() + " rows)");
        System.out.println("OK: write path never moved; old view retired only after cutover criteria held");
    }
}

آزمون صحت و معیار عملیاتیCorrectness checks and operational criteria

پاسخ سؤال‌های پیگیریFollow-up answers

اشتباهات رایجCommon mistakes

منابع:Sources: Atlassian · engineering interview handbook · Amazon · SDE III interview preparation

33 تمرین ۳ — پرداختِ چندshardیِ مینیاتوریExercise 3 — Miniature multishard payment

بازگشت به صورت تمرینBack to the exercise · ارزیابی و سطحAssessment and level

پاسخ ۶۰ثانیه‌ایThe 60-second answer

صورت مسئله و فرض‌هاProblem and assumptions

چرا این روش و این عددها؟Why this method and these numbers?

مدل ذهنیMental model

حل گام‌به‌گامStep by step

طراحی مشخصConcrete design

مثال محاسبه‌شدهWorked example

دانلود نمودار مستقلDownload standalone diagram

دانلود SVG مستقلDownload standalone SVG

دانلود نمودار مستقلDownload standalone diagram

گزینه‌های جایگزینAlternatives

سناریوهای خرابیFailure scenarios

کد قابل اجراRunnable code

اجراشده و تأییدشده در همین محیطExecuted and verified in this environment

این کد آموزشی و ساده‌سازی‌شده است؛ فقط برای آشنایی با ایده. کد واقعی پروداکشن ملاحظات بسیار بیشتری دارد و هیچ‌کس این‌شکلی کد پروداکشن نمی‌زند.Simplified teaching code for building intuition only — not production code; nobody ships production code like this.

import java.io.*;
import java.nio.*;
import java.nio.channels.*;
import java.nio.charset.StandardCharsets;
import java.nio.file.*;
import java.util.*;
import java.util.zip.CRC32;

// Teaching crash model: fsynced files survive a processor restart.
// The harness can inspect all logs at one logical boundary; this is NOT an
// atomic remote balance API or proof against power loss/directory-entry loss.
public class ShardPayDemo {
    record Decision(String id, long amount, long fee, boolean approved) {}
    final Path dir;
    final Map<String, Decision> decisions = new LinkedHashMap<>();
    final Map<String, Long> balances = new LinkedHashMap<>();
    final Set<String> seen = new HashSet<>(), settled = new HashSet<>();
    ShardPayDemo(Path dir) throws IOException {
        this.dir=dir; Files.createDirectories(dir);
        balances.put("src",1000L); balances.put("dst",100L); balances.put("fee",0L);
        for (String row : read("src")) {
            String[] t=row.split(" ");
            if (t[0].equals("D")) {
                Decision d=new Decision(t[1],Long.parseLong(t[2]),Long.parseLong(t[3]),Boolean.parseBoolean(t[4]));
                if (decisions.putIfAbsent(d.id,d)==null && d.approved)
                    balances.merge("src",-(d.amount+d.fee),Long::sum);
            } else if (t[0].equals("S")) settled.add(t[1]);
        }
        for (String shard : List.of("dst","fee"))
            for (String row : read(shard)) {
                String[] t=row.split(" "); String key=shard+":"+t[1];
                if (seen.add(key)) balances.merge(shard,Long.parseLong(t[2]),Long::sum);
            }
    }
    void append(String shard,String payload) throws IOException {
        byte[] p=payload.getBytes(StandardCharsets.UTF_8); CRC32 crc=new CRC32(); crc.update(p);
        ByteBuffer b=ByteBuffer.allocate(p.length+12).putInt(p.length).put(p).putLong(crc.getValue()); b.flip();
        try (FileChannel ch=FileChannel.open(dir.resolve(shard+".log"),StandardOpenOption.CREATE,StandardOpenOption.WRITE,StandardOpenOption.APPEND)) {
            while(b.hasRemaining()) ch.write(b);
            ch.force(true);
        }
    }
    List<String> read(String shard) throws IOException {
        Path path=dir.resolve(shard+".log"); List<String> rows=new ArrayList<>();
        if (!Files.exists(path)) return rows;
        ByteBuffer b=ByteBuffer.wrap(Files.readAllBytes(path)); int validEnd=0;
        while(b.hasRemaining()) {
            if(b.remaining()<4) break; // incomplete final frame
            int length=b.getInt();
            if(length<0 || length>1_048_576) throw new IOException("invalid length: fail closed");
            if(b.remaining()<length+8) break;
            byte[] payload=new byte[length]; b.get(payload); long expected=b.getLong();
            CRC32 crc=new CRC32(); crc.update(payload);
            if(crc.getValue()!=expected) throw new IOException("checksum mismatch: stop recovery");
            rows.add(new String(payload,StandardCharsets.UTF_8)); validEnd=b.position();
        }
        // Repair an incomplete suffix before accepting any new append.
        if(validEnd!=b.limit()) try(FileChannel ch=FileChannel.open(path,StandardOpenOption.WRITE)) {
            ch.truncate(validEnd); ch.force(true);
        }
        return rows;
    }
    void credit(String shard,String id,long amount) throws IOException {
        String key=shard+":"+id;
        if(seen.contains(key)) return;
        // The log record is the authority for BOTH the effect and dedup on replay.
        append(shard,"C "+id+" "+amount);
        seen.add(key); balances.merge(shard,amount,Long::sum);
    }
    void process(String id,long amount,long fee,int crashAfterLegs) throws IOException {
        if(amount<0 || fee<0) throw new IllegalArgumentException();
        Decision d=decisions.get(id);
        if(d==null) {
            d=new Decision(id,amount,fee,balances.get("src")>=amount+fee);
            // ONE durable record describes decision, reservation AND pending outputs.
            append("src","D "+id+" "+amount+" "+fee+" "+d.approved);
            decisions.put(id,d);
            if(d.approved) balances.merge("src",-(amount+fee),Long::sum);
        }
        if(d.amount!=amount || d.fee!=fee) throw new IllegalArgumentException("ID reused with different payload");
        if(!d.approved || settled.contains(id)) return;
        if(crashAfterLegs==0) return;
        credit("dst",id,d.amount); if(crashAfterLegs==1) return;
        credit("fee",id,d.fee); if(crashAfterLegs==2) return;
        append("src","S "+id); settled.add(id);
    }
    long inTransitAtCommonBoundary() {
        long pending=0;
        for(Decision d:decisions.values()) if(d.approved) {
            if(!seen.contains("dst:"+d.id)) pending+=d.amount;
            if(!seen.contains("fee:"+d.id)) pending+=d.fee;
        }
        return pending;
    }
    void checkConservation() {
        long total=balances.values().stream().mapToLong(Long::longValue).sum()+inTransitAtCommonBoundary();
        if(total!=1100) throw new AssertionError("conservation: "+total);
    }
    public static void main(String[] args) throws IOException {
        for(int cut=0;cut<=3;cut++) {
            Path dir=Files.createTempDirectory("shardpay-");
            ShardPayDemo first=new ShardPayDemo(dir); first.process("pay-1",200,5,cut);
            first.checkConservation();
            // Fresh processor: NO copying maps or dedup from the previous instance.
            ShardPayDemo restarted=new ShardPayDemo(dir); restarted.checkConservation();
            restarted.process("pay-1",200,5,3); restarted.process("pay-1",200,5,3);
            restarted.checkConservation();
            if(!restarted.balances.equals(Map.of("src",795L,"dst",300L,"fee",5L)) || restarted.inTransitAtCommonBoundary()!=0)
                throw new AssertionError("replay changed effects");
            System.out.println("PASS cut="+cut+" recovered="+restarted.balances+" pending=0");
        }
        Path dir=Files.createTempDirectory("shardpay-tail-");
        ShardPayDemo p=new ShardPayDemo(dir); p.process("pay-1",200,5,1);
        Files.write(dir.resolve("src.log"),new byte[]{0,0},StandardOpenOption.APPEND);
        ShardPayDemo repaired=new ShardPayDemo(dir); repaired.process("pay-1",200,5,3); repaired.checkConservation();
        byte[] corrupt=Files.readAllBytes(dir.resolve("src.log")); corrupt[4]^=1;
        Files.write(dir.resolve("src.log"),corrupt);
        boolean rejected=false;
        try { new ShardPayDemo(dir); } catch(IOException e) { rejected=true; }
        if(!rejected) throw new AssertionError("corruption silently accepted");
        System.out.println("PASS truncated suffix repaired; checksum corruption rejected");
    }
}

آزمون صحت و معیار عملیاتیCorrectness checks and operational criteria

پاسخ سؤال‌های پیگیریFollow-up answers

اشتباهات رایجCommon mistakes

منابع:Sources: AWS · safe retries with idempotent APIs · Jepsen · consistency models

44 تمرین ۴ — حسابرسیِ سرتاسریExercise 4 — End-to-end auditing

بازگشت به صورت تمرینBack to the exercise · ارزیابی و سطحAssessment and level

پاسخ ۶۰ثانیه‌ایThe 60-second answer

صورت مسئله و فرض‌هاProblem and assumptions

چرا این روش و این عددها؟Why this method and these numbers?

مدل ذهنیMental model

حل گام‌به‌گامStep by step

طراحی مشخصConcrete design

مثال محاسبه‌شدهWorked example

دانلود نمودار مستقلDownload standalone diagram

دانلود SVG مستقلDownload standalone SVG

دانلود نمودار مستقلDownload standalone diagram

گزینه‌های جایگزینAlternatives

سناریوهای خرابیFailure scenarios

کد قابل اجراRunnable code

اجراشده و تأییدشده در همین محیطExecuted and verified in this environment

این کد آموزشی و ساده‌سازی‌شده است؛ فقط برای آشنایی با ایده. کد واقعی پروداکشن ملاحظات بسیار بیشتری دارد و هیچ‌کس این‌شکلی کد پروداکشن نمی‌زند.Simplified teaching code for building intuition only — not production code; nobody ships production code like this.

import java.util.*;

// Teaching demo: end-to-end audit checker over transfer histories.
// Independent recomputation to a STATED checkpoint; same-checkpoint compare;
// lag vs corruption distinguished; missing update injected; incomplete log admitted.
public class AuditDemo {
    record Xfer(String id, String from, String to, long amount) {}
    List<Xfer> log = new ArrayList<>(List.of(
        new Xfer("t1", "A", "B", 100), new Xfer("t2", "B", "C", 40),
        new Xfer("t3", "A", "C", 50), new Xfer("t4", "C", "A", 10)));

    // The REAL derivation under audit (has an injected silent bug: t3 skipped).
    Map<String, Long> derivedWithBug(int checkpoint) {
        Map<String, Long> b = new LinkedHashMap<>(Map.of("A", 500L, "B", 300L, "C", 200L));
        for (int i = 0; i < checkpoint; i++) {
            Xfer x = log.get(i);
            if (x.id.equals("t3")) continue; // SILENT DROPPED UPDATE (injected)
            b.put(x.from, b.get(x.from) - x.amount);
            b.put(x.to, b.get(x.to) + x.amount);
        }
        return b;
    }

    // The CHECKER: written independently, straight from the log. No shared code.
    static Map<String, Long> recompute(List<Xfer> log, int checkpoint) {
        Map<String, Long> b = new LinkedHashMap<>(Map.of("A", 500L, "B", 300L, "C", 200L));
        for (int i = 0; i < checkpoint; i++) {
            Xfer x = log.get(i);
            b.put(x.from, b.get(x.from) - x.amount);
            b.put(x.to, b.get(x.to) + x.amount);
        }
        return b;
    }

    public static void main(String[] a) {
        AuditDemo d = new AuditDemo();
        int cp = 4; // stated checkpoint: log offset 4
        Map<String, Long> actual = d.derivedWithBug(cp);
        Map<String, Long> expect = recompute(d.log, cp);
        System.out.println("checkpoint=" + cp + " actual=" + actual + " expected=" + expect);
        for (String k : expect.keySet())
            if (!expect.get(k).equals(actual.get(k)))
                System.out.println("MISMATCH at " + k + ": live=" + actual.get(k) + " recomputed=" + expect.get(k));
        // lag, not corruption: derived state is just older (checkpoint 2 vs 4)
        Map<String, Long> older = d.derivedWithBug(2);
        System.out.println("stale read cp=2 vs checkpoint 4 -> " + (older.equals(expect) ? "match" : "LAG (behind, not corrupt): " + older));
        // honest limit: if the input log itself is incomplete, the checker is blind
        List<Xfer> partial = d.log.subList(0, 3);
        System.out.println("log truncated to 3 entries: checker recomputes " + recompute(partial, 3));
        System.out.println("OK: caught silent t3 at same checkpoint; lag separated; incomplete log admitted");
    }
}

آزمون صحت و معیار عملیاتیCorrectness checks and operational criteria

پاسخ سؤال‌های پیگیریFollow-up answers

اشتباهات رایجCommon mistakes

منابع:Sources: Jepsen · consistency models · Atlassian · engineering interview handbook