《行尸走肉米琼恩》试玩视频
好玩的手机单机游戏前十名排行榜
好玩的手机单机游戏前十名排行榜好玩的手机单机游戏前十名推荐。
随着手机屏幕不断的增大，越来越多的玩家喜欢在手机上玩游戏。
其中单机游戏更是手机党的最爱，那么大家知道哪些单机游戏不需要联网就可以玩。
下面我马上给大家分享排行榜前十名不用联网的单机手机游戏。
排行榜前十名手机单机游戏推荐手游排行榜NO1：《刺客信条：本色》游戏类型：角色扮演类这款游戏和这个系列的其他游戏一样，都是一款角色扮演类手游。
游戏整体的背景是意大利文艺复兴时期，玩家在游戏中需要扮演一名刺客。
玩家可以在游戏中的四个职业中选择一个，不同的职业拥有的技能都是不一样的。
小伙伴们可以根据自己的喜欢，然后去选择职业。
游戏中的四个职业都是根据刺客这个职业衍生而来的，玩家在游戏需要完成一个又一个的刺杀任务。
游戏采用的是故事推进形式，并且每一个故事都包含了若干个的任务。
游戏场景：游戏采用的是触摸移动和虚拟双摇杆这两种操作方式，玩家可以自行选择操作方式，在游戏主界面右下角有一个走动和跑动切换键。
为了方便玩家的操作，一般在玩家附近的敌人头上都会有骷髅标示。
玩家可以悄悄的移动到敌人的身边，然后在不知不觉中将敌人刺杀。
手游排行榜NO2：《伊苏编年史2》游戏类型：角色扮演在伊苏编年史1中，熟悉的勇士亚特鲁在经历了无与伦比的冒险旅途之后，身负重伤，有幸在2代游戏中被萌妹子莉莉娅所救。
然而，勇士的探险怎么会因为一点挫折而停下脚步，亚特鲁将会继续肩负着自己的使命，从不知名的黑暗和邪恶势力中解救伊苏城市中的人民于水火之中。
游戏封面截图：如果说伊苏编年史2 iOS和PC端游的最大不同之处，那想必就是两种完全不同的操作方法。
为了适配移动端的操作体验，这款移植佳作简化了人物角色的攻击技能，玩家只要通过滑动屏幕的方式指挥我们的主角向着敌人狂奔而去即可展开自主攻击，颇有点勇往直前的意味在其中。
画风质朴：在伊苏编年史2 iOS的游戏中，游戏开发商非常用心地推出了两种不同的游戏画质供玩家们选择，一种是较为久远的粗糙画面，另一种是打磨过的游戏画面，两种画面分别对应着该游戏在PC平台发行时的年代特征。
歌剧简介
1 啊，花已凋零——贝利尼歌剧《梦游女》中阿米娜的咏叹调（1楼）2 啊，上帝，给我力量——比才歌剧《卡门》中米卡埃拉的咏叹调（4楼）3 爱情乘着玫瑰翅膀——威尔第歌剧《游吟诗人》中丽昂诺拉的咏叹调（6楼）4 求爱神给我安慰——莫扎特歌剧《费加罗的婚礼》中伯爵夫人的咏叹调（7楼）5 别说我,尊敬的人——莫扎特歌剧《唐璜》中安娜的咏叹调（9楼）6 纯净的天使——威尔第歌剧《命运之力》中莱奥诺拉的咏叹调（10楼）7 多奇怪——威尔第歌剧《茶花女》中薇奥列塔的咏叹调（12楼）8 多谢,好友们——威尔第歌剧《西西里晚祷》中爱蕾娜的咏叹调（21楼）9 厄尔南尼,我们私奔吧——威尔第歌剧《厄尔南尼》中爱尔薇拉的咏叹调（22楼）10 愤怒,我的愤怒——佩格莱西歌剧《女仆主妇》中塞尔皮娜的咏叹调（23楼）11 复仇的痛苦——莫扎特歌剧《魔笛》中夜后的咏叹调（24楼）12 姑娘的秋波——多尼采蒂歌剧《唐·帕斯夸勒》中诺莉娅的咏叹调（25楼）13 挂满了柔软帐幕——普契尼歌剧《曼侬·列斯库》中曼侬的咏叹调（26楼）14 她抬头仰望——威尔第歌剧《假面舞会》中奥斯卡的咏叹调（27楼）15 可爱的名字——威尔第歌剧《弄臣》中吉尔达的咏叹调（31楼）16 离家去远方——卡塔拉尼歌剧《瓦蕾》中瓦蕾的咏叹调（32楼）17 妈妈被杀害——乔尔达诺歌剧《安德烈·谢尼埃》中玛达蕾娜的咏叹调（33楼）18 漫步街上——歌剧《绣花女》中穆塞塔的咏叹调（34楼）19 美妙时刻已来临——莫扎特歌剧《费加罗婚礼》种苏珊娜的咏叹调（7楼）20 那天深夜在大海上——博依托歌剧《梅菲斯特》中玛格丽塔的咏叹调（36楼）21 你鞭打我吧——莫扎特歌剧《唐璜》中彩琳娜的咏叹调（9楼）22 你那冰凉的心——普契尼歌剧《图兰朵》中柳儿的咏叹调（37楼）23 你无情的背叛了我——莫扎特歌剧《唐璜》中爱尔薇拉的咏叹调（9楼）24 我亲爱的爸爸——普契尼歌剧《加尼·斯基基》中劳莱塔的咏叹调（38楼）25 晴朗的一天——普契尼歌剧《蝴蝶夫人》中巧巧桑的咏叹调（44楼）26 人们叫我咪咪——普契尼歌剧《绣花女》中咪咪的咏叹调（34楼）27 上帝啊!给我安宁——威尔第歌剧《命运之力》中莱奥诺拉的咏叹调（10楼）28 胜利归来——威尔第歌剧《阿依达》中阿依达的咏叹调（45楼）29 圣洁女神——贝利尼歌剧《诺尔玛》中诺尔玛的咏叹调（46楼）30 圣母玛丽亚——威尔第歌剧《奥赛罗》中苔丝狄蒙娜的咏叹调（47楼）31 失去妈妈的孩子——普契尼歌剧《修女安吉里卡》中安吉里卡的咏叹调（48楼）32 是分手的时候了——多尼采蒂歌剧《军中女郎》中玛丽亚的咏叹调（49楼）33 死亡啊!——蓬皮耶利歌剧《乔贡达》中乔贡达的咏叹调（50楼）34 他那温柔的声音——贝利尼歌剧《清教徒》中爱尔薇拉的咏叹调（51楼）35 万籁无声——多尼采蒂歌剧《拉美摩尔的露契亚》中露契亚的咏叹调（57楼）36 为艺术,为爱情——普契尼歌剧《托斯卡》中托斯卡的咏叹调（58楼）37 我去死,先求你——威尔第歌剧《假面舞会》中阿美莉娅的咏叹调（27楼）38 我是作家忠实的仆人——契雷亚歌剧《阿德里亚娜·莱库弗》（59楼）39 夜深沉——威尔第歌剧《游吟诗人》中莱昂诺拉的咏叹调（6楼）40 一个女孩正十五——莫扎特歌剧《女人心》中黛斯碧娜的咏叹调（60楼）41 永别了,快乐的梦——威尔第歌剧《茶花女》中维奥列塔的咏叹调（12楼[/url]）42 在这座宫殿里——普契尼歌剧《图兰朵》中图兰朵的咏叹调（37楼）。
手机游戏奥秘之家之珍妮房间通关秘籍
手机游戏奥秘之家之珍妮房间通关秘籍1.在书架的音乐盒里拆卸几节电池2.床上中间靠背后找到夜视仪组件13.床左边找到塑料拖把，拆卸后得到长棍4.使用长棍从床下得到夜视仪组件2——镜头5.组合夜视仪6.给夜视仪装上电池7.夜视仪看床下，发现A<B的提示，后面使用8.床右面书桌上画像拼图，拼好后使用夜视仪观看发现提示9.对应书架右面墙上的四幅图，找到对应的数字：247910.使用上述密码打开书架中间密码箱，获得门卡、胶水、盘子碎片11.使用胶水和盘子碎片组合拼好盘子。
12.将盘子放在书架第二层的支架上，点击查看提示。
13.提示的意思是同时按下书架中间蓝色长方形、绿色圆形、黄色正方形、红色长方形亮灯的按钮，书架移动，发现密室门。
14.点击密室门，使用夜视仪发现蝴蝶图案。
15.（只有发现蝴蝶图案后才能返回旋转四幅画）返回到墙上移动四幅画形成蝴蝶图案。
16.在移动四幅画，四个角形成心形。
（由房间窗帘后的门上提示得知）17.计算XYAB的计算题，得到1523或5123两种，试验后确定，这里需要用到A<B的提示18.使用密码打开密码箱得到螺丝刀19.点击墙上类似电视的图案，发现电盒，用螺丝刀打开电盒，露出电线，点击电线，进入密室。
20.先点击墙上星形玄机图案，退回后才能点击椅子上的东西，发现一个类似茶壶的东西，装上电池，墙上投射时钟。
21.解时钟难题，将时钟旋转至15:55分，得到密码。
22.使用密码打开书桌上的密码盒，得到咒符、水晶球、明信片23.在墙上挂的地图上使用明信片，得到一个坐标卡。
24.在墙上星形玄机图案上使用咒符退出房间。
25.使用门卡打开房门得到一个精密仪器，使用坐标卡。
26.恭喜通关。
神秘博士 第7季 第9集 迷藏 剧情介绍 对白摘录 看美剧学英语
来吧 和我交谈
Come to me. Speak to me.
让我指引你回家
Let me show you the way home.
噢 啊
Oh, ah!
让我指引你回家吧
Let me show you the way home!
啊
Ah!
Emma
Emma?
她那么...
是助手
Assistant. 源自现在是1974年 你是他的助手和"非物装备"
It's 1974 - you're the assistant and "Non-objective equipment".
就是"灵媒"的意思
Meaning "Psychic".
我知道 不过还是谢谢你
Getting that. Bless you, though.
Erm, a stray FM broadcast, possibly?
不过我安装了一些铁素干扰抑制器和射频抗流圈 以防万一
But I've fitted some ferrite suppressors and some RF chokes. Just in case.
你确定你要这样做吗
什么博士
Doctor what?
(Doctor what与电影《持续尖叫(Carry On Screaming)》[1]中的Doctor Watt谐音 该剧中曾用调侃称Doctor Who为其叔叔)
你想那么叫也行 这位是Clara
If you like. And this is Clara.
waters质谱masslynx软件使用说明
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MassLynx NT Users Guide
Contents
MassLynx NT User’s Guide............................................................................
惠普 LaserJet Pro MFP M25-M27 用户指南说明书
DIN 7975-2016自攻螺钉连接 - 芯孔直径和使用指南
2
Blechschrauben der Form R nach DIN EN ISO 1478 mit Spitze (auch Suchspitze genannt) werden vorzugsweise verwendet.
Preisgruppe 11 www.din.de www.beuth.de
!%LrA"
2417930
DIN 7975:2016-04
Inhalt
Seite
Vorwort ...................................................................................................................................................................................... 3 1 2 3 3.1 3.2 3.3 3.4 4 5 6 6.1 6.2 6.3 6.4 6.5 Anwendungsbereich ............................................................................................................................................... 4 Normative Verweisungen ..................................................................................................................................... 4 Grenzen der Anwendung einfacher Schraubenverbindungen mit Blechschrauben ...................... 4 Allgemeines ............................................................................................................................................................... 4 Untere Grenze der Blechdicke ............................................................................................................................ 4 Obere Grenze der Blechdicke.............................................................................................................................. 5 Mindestwert der Gesamtdicke der zu verschraubenden Bleche ........................................................... 5 Kernlochdurchmesser ........................................................................................................................................... 7 Durchmesser des Durchgangsloches.............................................................................................................. 14 Sonderfälle ............................................................................................................................................................... 14 Allgemeines ............................................................................................................................................................. 14 Gestanzte Löcher.................................................................................................................................................... 14 Schrauben mit Überzügen .................................................................................................................................. 14 Verschrauben in austenitischem Chrom-Nickel-Stahl ............................................................................. 14 Verbindungen mit Blechschrauben aus nichtrostendem Stahl ............................................................ 14
DINEN EN 10268 2013_12_DE版本
Frühere Ausgaben DIN EN 10268: 1999-02, 2006-10 DIN EN 10268 Berichtigung 1: 2007-02
2
EUROPÄISCHE NORM EUROPEAN STANDARD NORME EUROPÉENNE
ICS 77.140.50
Cold rolled steel flat products with high yield strength for cold forming – Technical delivery conditions; German version EN 10268:2006+A1:2013 Produits plats laminés à froid à haute limite d’élasticité pour formage à froid – Conditions techniques de livraison; Version allemande EN 10268:2006+A1:2013
© 2013 CEN
Alle Rechte der Verwertung, gleich in welcher Form und in welchem Verfahren, sind weltweit den nationalen Mitgliedern von CEN vorbehalten.
Ref. Nr. EN 10268:2006+A1:2013 D
Preisgruppe 10 www.din.de www.beuth.de
!$Ö+u"
1970882
QUELLE: NOLIS (Norm vor Anwendung auf Aktualität prüfen!/Check standard for current issue prior to usage)
《终极刺客：契约》沉默刺客评价攻略
等待多时的《终极刺客：契约》终于出现，游戏将讲述刺客Agent47的回忆史，在世界各地执行艰难的任务。
游戏任务只有12关，只有前几关是新设关卡，后几关则几乎是从一、二代中挑选出来进行改动而成的，有些竟然是地图完全相同，任务不同，让人觉得有点赶工的嫌疑。
不过，游戏的自由度比以往两部都高，每关的过关方法有好多种，获得沉默刺客评价的难度也比以往降低了不少。
游戏有三种难度，建议大家用专家模式来玩，才能玩出游戏的精髓。
简明流程第一关：收容所（Asylum Aftermath）任务：逃离收容所（Escape Sanitarium）一开始出现在白色的房间里，从地上的尸体中搜出一把汽车钥匙，接着沿着路一直跑过去，路上全都是疯子，他们不会怀疑你。
坐电梯到2楼，在门口一堆尸体中换上疯子的衣服，并从电梯对面的大门穿过去，下楼梯能到1楼。
下面有4个SWAT守侯在那里，进入左边的那个房间，等一个SWAT上楼一段时间后，立即跑上去躲在门边，当他出来后，从后面用麻醉针将他麻醉，换上他的衣服，从房间外面的的楼梯爬下，上车逃离。
第二关：肉王的舞会（The Meat King's Party）任务：暗杀肉王（Assassinate Meat King Campbell Sturrock）暗杀律师（Assassinate Lawyer Andrel Puscus）救出委托人的女儿（Resuce The Client's Daughter）一开始出现在冷藏车上，先将身上的枪支全扔掉，换上屠夫的衣服，走出冷藏车后记得将门关上，然后往舞会的入口走去，经过门口时要搜身，铁钩不会被搜出来。
来到厨房，厨师让你去送鸡肉给肉王，你将铁钩偷偷藏在肉里面，然后从下面的楼梯上2楼，那里同样要搜身（如果把铁钩放到身边会被发现），经过搜身后再把铁钩藏在身上。
来到肉王的房间，把窗帘放下，趁肉王吃肉时，拿出铁钩将他了结。
回到1楼，从上面的楼梯来到2楼，经过几个房间后能找委托人的女儿，要注意躲开附近砍人的厨师，委托人的女儿死了，你只能带走她的一只手臂。
《行尸走肉》系列正版Steam购买教程
《⾏⼫⾛⾁》系列正版Steam购买教程
《⾏⼫⾛⾁：⽶琼恩》发布后，有许多玩家不清楚游戏该如何购买《⾏⼫⾛⾁》系列游戏。
今天为⼤家带来S t e a m购买正版《⾏⼫⾛⾁》的教程，不会购买游戏的玩家快来看看吧。
你需要⼀个s t e a m，请⾃⾏百度并且下载(注意，s t e a m危害极⼤，严重者可能失去双⼿!请您做好剁⼿截肢准备!)，下载完之后注册账号，登录。
然后你就这样登录了主界⾯，点击商店页⾯⾥⾯的搜索(看图)，然后输⼊游戏英⽂名。
就是这样，你会看见很多⼩框框(注意：如果⼏乎每⼀项旁边都有绿条条请及时下线，否则您可能会失去您的双⼿)，上⾯的两个⼩框框就是⾏⼫⾛⾁，进去后会让您选择年龄，⾄少要18岁，不要作死，否则进不去了。
当然你也可以点击搜索，这两个框框是⼀⼆季，400天是D L C存在在第⼀季⾥⾯。
拿第⼀季来说(第⼆季我已经买了)，进去之后是这样的，如果您和我⼀样看不见图⽚，那么您也是移动⽹络的受害者。
往下拉会看到这个，第⼀个是第⼀季原版，第⼆个是T社⼤包吧，第三个是D L C400D AY S，现在新国区很便宜吧，以前⽼国区好像第⼆季价格
24.99美元吧
拿原版来说，添加⾄购物车，为⾃⼰购买或作为礼物。
购买页⾯是这样的，⽀付宝15年12⽉初开始不能⽤，你只能使⽤银联了。
进去之后就可以购买了，但是
购买完之后——要等跳转s t e a m界⾯!——要等跳转s t e a m界⾯!——要等跳转s t e a m界⾯!
逗游⽹——中国2亿游戏⽤户⼀致选择的”⼀站式“游戏服务平台。
AZombiesLife0.8版全人物完美剧情攻略分享
AZombiesLife0.8版全人物完美剧情攻略分享今天小编为大家带来A Zombie’s Life 0.8版全人物完美剧情攻略分享：先说说这次更新的内容：1、新增打僵尸的模式2、新增莱斯利的剧情3、新的cg因为新出的cg关系很乱，现在先整理一下：妈妈和女警姐姐和阿姨姐姐和老师阿姨和女警1、打僵尸现在打僵尸可以换上近战武器与僵尸肉搏，←↑→防御僵尸的攻击，↓攻击僵尸。
近战武器我没找到在哪里获得，我直接回想屋里拿的。
我发现一个比较有趣的机制。
就是装备手枪的时候上满子弹，然后换近战武器，这样还是可以发射子弹远程攻击僵尸。
2、新增剧情这版本莱斯利有个剧情，在酒店下方新开了个地图，我把它叫做工厂地图，进去之后发现莱斯利在里面，对话后出现一个无法杀死的僵尸，这时候把他引到工厂右边的坑旁，把他推进去，就可以了。
人物剧情攻略妈妈：1、一开始交谈（好感：1）2、送酒，晚上在房里睡觉触发剧情（好感度：2）3、早上起来对话（好感度：3）4、叫妈妈出去搜集物资，自己也跟出去后找到妈妈，触发接吻剧情（好感度：4）5、回到家再和妈妈对话（好感度：5）6、断水之后和妈妈一起洗澡触发剧情（好感度：6）7、送酒给妈妈，晚上在自己房里睡觉触发第一次夜袭（好感度：7）8、早上和妈妈对话送酒，晚上睡觉触发第二次夜袭（好感度：8）9、和妈妈对话，找到戒指送给妈妈，戒指在办公楼的顶楼的电脑旁（好感度：9）10、送酒给妈妈，晚上睡觉，触发最后一个夜袭（好感度：10）11、晚上去妈妈房里睡觉触发剧情（好感度：11）12、继续去妈妈房里睡（好感度：12）13、再去，触发后入（好感度：13）14、送酒给妈妈，晚上去姐姐房里触发15、送酒给阿姨，晚上去妈妈房里触发16、送酒给女警，晚上去妈妈房里触发姐姐：1、对话（好感度：1）2、送酒，晚上在自己房里睡觉（好感度：2）3、对话（好感度：3）4、让姐姐去搜集物资，自己也跟出去后找到姐姐触发剧情（好感度：4）5、对话（好感度：5）6、晚上不睡觉，第二天中午和家人吃饭触发咬剧情。
Micro Focus z Server消息和诊断说明书
z/Server Messages and Diagnosticsz/Server Messages and DiagnosticsMicro FocusThe Lawn22-30 Old Bath RoadNewbury, Berkshire RG14 1QNUKCopyright © 2011-2014 Micro Focus. All rights reserved.MICRO FOCUS and the Micro Focus logo are trademarks or registered trademarks of Micro Focus or its subsidiaries or affiliatedcompanies in the United States, United Kingdom and other countries. All other marks are the property of their respective owners.z/Server Messages and Diagnostics Contents1 Introduction (27)1.1 Message format (27)1.2 Error message example (28)1.3 Warning message example (29)1.4 Information message example (29)1.5 Messages and GTF (29)2 Messages (30)2.1 API - Application Programming Interface (30)2.1.1 API0001I (30)2.1.2 API0002I (31)2.1.3 API0007E (31)2.1.4 API0008I (31)2.1.5 API0009E (31)2.1.6 API0010I (31)2.1.7 API0011I (32)2.1.8 API0015I (32)2.1.9 API0054S (32)2.1.10 API0065I (32)2.1.11 API0069E (32)2.1.12 API0072W (33)2.1.13 API0099I (33)2.1.14 API0106W (33)2.1.15 API0107E (33)2.1.16 API0108W (33)2.1.17 API0109I (34)2.1.18 API0110I (34)2.1.19 API0111I (34)2.2 BND - IEWBIND (Binder) Interface (35)2.2.1 BND0001E (35)2.2.2 BND0002I (35)2.2.3 BND0003I (35)2.2.4 BND0004E (35)z/Server Messages and Diagnostics2.2.5 BND0005E (35)2.2.6 BND0006I (36)2.3 CSI - Catalog Search Interface (37)2.3.1 CSI0001E (37)2.3.2 CSI0002I (37)2.3.3 CSI0003E (37)2.3.4 CSI0004I (37)2.3.5 CSI0005I (37)2.3.6 CSI0006I (38)2.3.7 CSI0007I (38)2.3.8 CSI0008E (38)2.3.9 CSI0009I (38)2.3.10 CSI0010E (39)2.3.11 CSI0011I (39)2.3.12 CSI0012E (39)2.3.13 CSI0013E (39)2.3.14 CSI0014W (39)2.3.15 CSI0015I (39)2.3.16 CSI0016E (40)2.3.17 CSI0017I (40)2.3.18 CSI0018I (40)2.3.19 CSI0019I (40)2.3.20 CSI0020E (40)2.3.21 CSI0021E (40)2.3.22 CSI0022E (41)2.3.23 CSI0023E (41)2.3.24 CSI0024E (41)2.3.25 CSI0025E (41)2.3.26 CSI0026I (41)2.3.27 CSI0027I (42)2.3.28 CSI0028I (42)2.3.29 CSI0029E (42)2.3.30 CSI0030W (42)2.3.31 CSI0031I (42)z/Server Messages and Diagnostics2.3.32 CSI0032I (43)2.3.33 CSI0033E (43)2.3.34 CSI0034I (43)2.3.35 CSI0035E (43)2.3.36 CSI0036E (43)2.3.37 CSI0037E (43)2.3.38 CSI0038E (44)2.3.39 CSI0039E (44)2.3.40 CSI0040I (44)2.3.41 CSI0041I (44)2.3.42 CSI0042E (44)2.3.43 CSI0043I (44)2.3.44 CSI0044E (45)2.3.45 CSI0045E (45)2.3.46 CSI0046E (45)2.3.47 CSI0047E (45)2.3.48 CSI0048E (45)2.3.49 CSI0049I (45)2.3.50 CSI0050E (46)2.3.51 CSI0051E (46)2.3.52 CSI0052E (46)2.3.53 CSI0053E (46)2.3.54 CSI0054I (46)2.3.55 CSI0055E (46)2.3.56 CSI0056E (47)2.3.57 CSI0057W (47)2.3.58 CSI0058E (47)2.3.59 CSI0059E (47)2.3.60 CSI0060E (47)2.3.61 CSI0061I (47)2.3.62 CSI0062I (48)2.3.63 CSI0063E (48)2.4 DSC - Holder Task Configuration (49)2.4.1 DSC0001E (49)z/Server Messages and Diagnostics2.4.2 DSC0002E (49)2.4.3 DSC0003E (49)2.4.4 DSC0004E (49)2.4.5 DSC0005E (49)2.4.6 DSC0008E (49)2.4.7 DSC0009E (50)2.4.8 DSC0020E (50)2.4.9 DSC0022E (50)2.4.10 DSC0023 (50)2.4.11 DSC0024 (50)2.4.12 DSC0025 (50)2.4.13 DSC0026W (50)2.4.14 DSC0027E (50)2.4.15 DSC0099I (51)2.4.16 DSC0100E (51)2.4.17 DSC0101I (51)2.4.18 DSC0102I (51)2.4.19 DSC0200E (51)2.5 EMC - Extended MCS Interface (52)2.5.1 EMC0001E (52)2.5.2 EMC0002I (52)2.5.3 EMC0003E (52)2.5.4 EMC0004E (53)2.5.5 EMC0005I (53)2.5.6 EMC0006E (53)2.5.7 EMC0007E (53)2.5.8 EMC0008I (53)2.5.9 EMC0009I (54)2.5.10 EMC0010W (54)2.5.11 EMC0011E (54)2.5.12 EMC0012E (54)2.5.13 EMC0013I (54)2.5.14 EMC0014E (54)2.5.15 EMC0015I (55)z/Server Messages and Diagnostics2.5.16 EMC0016I (55)2.5.17 EMC0017I (55)2.5.18 EMC0018I (55)2.5.19 EMC0019I (55)2.5.20 EMC0020I (55)2.5.21 EMC0021I (56)2.5.22 EMC0022E (56)2.5.23 EMC0023E (56)2.6 HLD - Holder Task Interface (57)2.6.1 HLD0001I (57)2.6.2 HLD0002I (57)2.6.3 HLD0003I (57)2.6.4 HLD0004I (57)2.6.5 HLD0005I (58)2.6.6 HLD0006I (58)2.6.7 HLD0007I (58)2.6.8 HLD0008E (58)2.6.9 HLD0009I (59)2.6.10 HLD0010E (59)2.6.11 HLD0011I (59)2.6.12 HLD0012I (59)2.6.13 HLD0013E (59)2.6.14 HLD0014E (59)2.6.15 HLD0015E (60)2.6.16 HLD0016W (60)2.6.17 HLD0017E (60)2.6.18 HLD0018I (60)2.6.19 HLD0019I (60)2.6.20 HLD0020I (61)2.6.21 HLD0021I (61)2.6.22 HLD0022I (61)2.6.23 HLD0024I (61)2.6.24 HLD0025I (61)2.6.25 HLD0026I (62)z/Server Messages and Diagnostics2.6.26 HLD0027I (62)2.6.27 HLD0028I (62)2.6.28 HLD0029I (62)2.6.29 HLD0030I (63)2.6.30 HLD0031I (64)2.6.31 HLD0032I (64)2.6.32 HLD0033I (64)2.6.33 HLD0034I (64)2.6.34 HLD0035I (64)2.6.35 HLD0036E (64)2.6.36 HLD0038E (65)2.6.37 HLD0039I (65)2.6.38 HLD0040E (65)2.6.39 HLD0041I (65)2.6.40 HLD0042I (65)2.6.41 HLD0043I (66)2.6.42 HLD0044I (66)2.6.43 HLD0045E (66)2.6.44 HLD0046E (66)2.6.45 HLD0047I (66)2.6.46 HLD0048I (66)2.6.47 HLD0049I (67)2.6.48 HLD0050I (67)2.6.49 HLD0051I (67)2.6.50 HLD0052I (67)2.6.51 HLD0053E (67)2.6.52 HLD0054I (67)2.6.53 HLD0055E (68)2.6.54 HLD0056E (68)2.6.55 HLD0057W (68)2.6.56 HLD0058E (68)2.6.57 HLD0059E (68)2.6.58 HLD0060I (69)2.6.59 HLD0061E (69)z/Server Messages and Diagnostics2.6.60 HLD0062I (69)2.6.61 HLD0063I (69)2.6.62 HLD0064I (69)2.6.63 HLD0065I (69)2.6.64 HLD0066I (70)2.6.65 HLD0067E (70)2.6.66 HLD0066I (70)2.6.67 HLD0066I (70)2.6.68 HLD2001W (70)2.7 IPC - Command Task (71)2.7.1 IPC0001W (71)2.7.2 IPC0002W (71)2.7.3 IPC0003W (71)2.7.4 IPC0004W (71)2.7.5 IPC0005I (71)2.7.6 IPC0006W (71)2.7.7 IPC0007W (72)2.7.8 IPC0040I (72)2.7.9 IPC0042I (72)2.7.10 IPC0043I (72)2.7.11 IPC0044I (72)2.7.12 IPC0052I (72)2.7.13 IPC0061I (73)2.7.14 IPC0062I (73)2.7.15 IPC0063I (73)2.7.16 IPC0064I (73)2.7.17 IPC0065E (73)2.7.18 IPC0066I (73)2.7.19 IPC0067I (74)2.7.20 IPC0068I (74)2.7.21 IPC0069E (74)2.7.22 IPC0070E (74)2.7.23 IPC0071I (74)2.7.24 IPC0072E (74)z/Server Messages and Diagnostics2.7.25 IPC0073I (75)2.7.26 IPC0074I (75)2.7.27 IPC0075I (75)2.7.28 IPC0076I (75)2.7.29 IPC0077E (75)2.7.30 IPC0078E (75)2.7.31 IPC0079I (76)2.7.32 IPC0080I (76)2.7.33 IPC0081E (76)2.7.34 IPC0082I (76)2.7.35 IPC0083E (76)2.7.36 IPC0084I (76)2.7.37 IPC0085E (77)2.7.38 IPC0086W (77)2.7.39 IPC0087E (77)2.7.40 IPC0088E (77)2.7.41 IPC0090I (78)2.7.42 IPC0091I (78)2.7.43 IPC0092E (78)2.7.44 IPC0093E (78)2.7.45 IPC0094I (78)2.7.46 IPC0095I (78)2.7.47 IPC0096I (79)2.7.48 IPC0097E (79)2.7.49 IPC0098I (79)2.7.50 IPC0099I (79)2.7.51 IPC0100E (79)2.7.52 IPC0101I (79)2.8 JES - Job Entry Subsystem Interface (80)2.8.1 JES0001I (80)2.8.2 JES0002I (80)2.8.3 JES0003E (80)2.8.4 JES0004E (81)2.8.5 JES0005E (81)2.8.7 JES0007E (81)2.8.8 JES0008I (82)2.8.9 JES0009I (82)2.8.10 JES0010I (82)2.9 LIC - License Interface (83)2.9.1 LIC0001E (83)2.9.2 LIC0002W (83)2.9.3 LIC0003I (83)2.9.4 LIC0004W (83)2.9.5 LIC0005E (83)2.9.6 LIC0006E (84)2.9.7 LIC0007W (84)2.9.8 LIC0008E (84)2.9.9 LIC0009I (84)2.9.10 LIC0010E (84)2.9.11 LIC0011I (84)2.10 MAL - Email Interface (86)2.10.1 MAL0001E (86)2.10.2 MAL0002E (86)2.10.3 MAL0003E (86)2.10.4 MAL0004E (86)2.10.5 MAL0006I (86)2.10.6 MAL0007I (86)2.10.7 MAL0008I (87)2.10.8 MAL0009I (87)2.10.9 MAL0010I (87)2.10.10 MAL0011E (87)2.10.11 MAL0012E (87)2.10.12 MAL0013E (88)2.10.13 MAL0014I (88)2.11 REC - Recovery Exception Handler (89)2.11.1 REC0001E (89)2.11.2 REC0002E (89)2.11.4 REC0004I (90)2.11.5 REC0005I (90)2.11.6 REC0006I (90)2.11.7 REC0007I (90)2.11.8 REC0008I (90)2.11.9 REC0009E (90)2.11.10 REC0010W (91)2.11.11 REC0011E (91)2.11.12 REC0012E (91)2.11.13 REC0013E (91)2.11.14 REC0014W (91)2.11.15 REC0015I (92)2.11.16 REC0016I (92)2.11.17 REC0017I (92)2.11.18 REC0018I (92)2.11.19 REC0019I (92)2.12 REX - REXX Interface (93)2.12.1 REX0001E (93)2.12.2 REX0002I (93)2.12.3 REX0003I (93)2.12.4 REX0004I (93)2.12.5 REX0005E (94)2.12.6 REX0006E (94)2.12.7 REX0007I (94)2.12.8 REX0008E (94)2.12.9 REX0009E (94)2.12.10 REX0010E (94)2.12.11 REX0011E (95)2.12.12 REX0012I (95)2.12.13 REX0013W (95)2.12.14 REX0014I (95)2.12.15 REX0015E (95)2.12.16 REX0016W (96)2.12.18 REX0018E (96)2.12.19 REX0019S (96)2.12.20 REX0020I (96)2.12.21 REX0021I (96)2.12.22 REX0022E (97)2.12.23 REX0023S (97)2.12.24 REX0024I (97)2.12.25 REX0025E (97)2.12.26 REX0026I (97)2.12.27 REX0027I (98)2.12.28 REX0028I (98)2.12.29 REX0029E (98)2.12.30 REX0030E (98)2.12.31 REX0031I (98)2.12.32 REX0032E (98)2.12.33 REX0033I (99)2.12.34 REX0034I (99)2.12.35 REX0035E (99)2.12.36 REX0036I (99)2.12.37 REX0037I (99)2.12.38 REX0038I (100)2.12.39 REX0039I (100)2.12.40 REX0040I (100)2.12.41 REX0041I (100)2.12.42 REX0042E (100)2.12.43 REX0043I (100)2.12.44 REX0044I (101)2.12.45 REX0045I (101)2.12.46 REX0046I (101)2.12.47 REX0047I (101)2.12.48 REX0048E (101)2.12.49 REX0049E (101)2.12.50 REX0050E (102)2.12.52 REX0052E (102)2.12.53 REX0053I (102)2.12.54 REX0054E (102)2.12.55 REX0055I (102)2.12.56 REX0056I (103)2.12.57 REX0057E (103)2.12.58 REX0058I (103)2.12.59 REX0059I (103)2.12.60 REX0060I (103)2.12.61 REX0062E (103)2.12.62 REX0063I (104)2.12.63 REX0064I (104)2.12.64 REX0065E (104)2.12.65 REX0066W (104)2.12.66 REX0067I (105)2.12.67 REX0068I (105)2.12.68 REX0069E (105)2.12.69 REX0070I (105)2.12.70 REX0071E (105)2.12.71 REX0072E (105)2.13 SJB - Started Job (107)2.13.1 SJB0001E (107)2.13.2 SJB0002E (107)2.13.3 SJB0003E (107)2.13.4 SJB0004I (107)2.13.5 SJB0005E (108)2.13.6 SJB0006I (108)2.14 SLR - TSO Scheduler Interface (109)2.14.1 SLR0001I (109)2.14.2 SLR0002I (109)2.14.3 SLR0003I (109)2.14.4 SLR0004I (109)2.14.5 SLR0005E (109)2.14.7 SLR0007I (110)2.14.8 SLR0008E (110)2.14.9 SLR0009E (110)2.14.10 SLR0010E (110)2.14.11 SLR0011E (110)2.14.12 SLR0012E (110)2.14.13 SLR0013E (110)2.14.14 SLR0014E (111)2.14.15 SLR0015E (111)2.14.16 SLR0016E (111)2.14.17 SLR0017E (111)2.14.18 SLR0018I (111)2.14.19 SLR0019I (111)2.14.20 SLR0020S (112)2.14.21 SLR0021S (112)2.14.22 SLR0021S (112)2.14.23 SLR0023I (112)2.14.24 SLR0024E (112)2.14.25 SLR0025E (112)2.14.26 SLR0026I (113)2.14.27 SLR0027W (113)2.14.28 SLR0028E (113)2.14.29 SLR0029W (113)2.14.30 SLR0030I (113)2.14.31 SLR0031S (113)2.14.32 SLR0032E (113)2.14.33 SLR0033E (114)2.14.34 SLR0034E (114)2.14.35 SLR0035I (114)2.14.36 SLR0036I (114)2.14.37 SLR0037I (114)2.14.38 SLR0038W (115)2.14.39 SLR0039I (115)2.14.41 SLR0041I (115)2.14.42 SLR0042I (116)2.14.43 SLR0043I (116)2.14.44 SLR0044I (116)2.14.45 SLR0045E (116)2.14.46 SLR0046I (116)2.14.47 SLR0047I (117)2.14.48 SLR0048I (117)2.14.49 SLR0049I (117)2.14.50 SLR0050E (117)2.14.51 SLR0051E (117)2.14.52 SLR0052E (118)2.14.53 SLR0053W (118)2.14.54 SLR0054I (118)2.14.55 SLR0055E (118)2.14.56 SLR0056E (118)2.14.57 SLR0057I (118)2.14.58 SLR0058I (119)2.14.59 SLR0059E (119)2.14.60 SLR0060E (119)2.14.61 SLR0061E (119)2.14.62 SLR0062E (119)2.14.63 SLR0063E (119)2.14.64 SLR0064I (120)2.14.65 SLR0065E (120)2.14.66 SLR0066E (120)2.14.67 SLR0067E (120)2.14.68 SLR0068I (120)2.14.69 SLR0069E (120)2.14.70 SLR0070I (120)2.14.71 SLR0071E (121)2.14.72 SLR0072E (121)2.14.73 SLR0073E (121)2.14.75 SLR0075I (121)2.14.76 SLR0076I (122)2.14.77 SLR0077I (122)2.14.78 SLR0078E (122)2.14.79 SLR0079E (122)2.14.80 SLR0080E (122)2.14.81 SLR0081I (123)2.14.82 SLR0082E (123)2.14.83 SLR0083I (123)2.14.84 SLR0084E (123)2.14.85 SLR0085E (123)2.14.86 SLR0086I (124)2.14.87 SLR0087E (124)2.14.88 SLR0088E (124)2.14.89 SLR0089E (124)2.14.90 SLR0090E (124)2.14.91 SLR0091E (125)2.14.92 SLR0092E (125)2.14.93 SLR0093E (125)2.14.94 SLR0094E (125)2.14.95 SLR0095E (125)2.14.96 SLR0096E (126)2.14.97 SLR0097E (126)2.14.98 SLR0098E (126)2.14.99 SLR0099E (126)2.14.100 SLR0100E (126)2.14.101 SLR0101I (127)2.14.102 SLR0102W (127)2.14.103 SLR0103E (127)2.14.104 SLR0104E (127)2.14.105 SLR0105E (127)2.14.106 SLR0106E (127)2.14.107 SLR0107I (128)2.14.109 SLR0109I (128)2.14.110 SLR0110I (128)2.14.111 SLR0111I (128)2.14.112 SLR0112E (129)2.14.113 SLR0113I (129)2.14.114 SLR0114E (129)2.14.115 SLR0115I (129)2.14.116 SLR0116I (129)2.14.117 SLR0117E (129)2.14.118 SLR0118E (130)2.14.119 SLR0119E (130)2.14.120 SLR0120E (130)2.14.121 SLR0121E (130)2.14.122 SLR0122E (130)2.14.123 SLR0123E (131)2.14.124 SLR0124E (131)2.14.125 SLR0125E (131)2.14.126 SLR0126I (131)2.14.127 SLR0127E (131)2.14.128 SLR0128E (131)2.14.129 SLR0129E (132)2.14.130 SLR0130E (132)2.14.131 SLR0131E (132)2.14.132 SLR0132E (132)2.14.133 SLR0133E (132)2.14.134 SLR0134E (133)2.14.135 SLR0135S (133)2.14.136 SLR0136I (133)2.14.137 SLR0137E (133)2.14.138 SLR0139E (133)2.14.139 SLR0140E (134)2.14.140 SLR0141E (134)2.14.141 SLR0142E (134)2.15 SPR - Standard Pool Routine (135)2.15.1 SPR0001I (135)2.15.2 SPR0002E (135)2.15.3 SPR0003E (135)2.15.4 SPR0004I (135)2.15.5 SPR0005I (136)2.15.6 SPR0006I (136)2.15.7 SPR0007E (136)2.15.8 SPR0008W (136)2.15.9 SPR0009E (136)2.15.10 SPR0010S (136)2.15.11 SPR0011I (137)2.15.12 SPR0012I (137)2.15.13 SPR0013I (137)2.15.14 SPR0014I (137)2.15.15 SPR0015I (137)2.15.16 SPR0016I (138)2.16 SRV - Server Interface (139)2.16.1 SRV0001I (139)2.16.2 SRV0002I (139)2.16.3 SRV0003E (139)2.16.4 SRV0004W (139)2.16.5 SRV0005I (139)2.16.6 SRV0006I (140)2.16.7 SRV0007I (140)2.16.8 SRV0008I (140)2.16.9 SRV0009I (140)2.17 SSI - Subsystem Interface Routine (141)2.17.1 SSI0001I (141)2.17.2 SSI0002I (141)2.17.3 SSI0003I (141)2.17.4 SSI0004I (141)2.17.5 SSI0005I (141)2.17.7 SSI0007I (142)2.17.8 SSI0008I (142)2.17.9 SSI0009I (142)2.18 STA - Started Task Interface (143)2.18.1 STA0001E (143)2.18.2 STA0002E (143)2.18.3 STA0003I (143)2.18.4 STA0004E (143)2.18.5 STA0005E (143)2.18.6 STA0006I (143)2.18.7 STA0007E (144)2.19 TAU – Eclipse Client Interface (145)2.19.1 TAU0001I (145)2.19.2 TAU0002I (145)2.19.3 TAU0003I (145)2.19.4 TAU0004I (145)2.19.5 TAU0005I (145)2.19.6 TAU0006I (146)2.19.7 TAU0007E (146)2.19.8 TAU0008I (146)2.19.9 TAU0009E (146)2.19.10 TAU0010I (146)2.19.11 TAU0011I (147)2.19.12 TAU0012E (147)2.19.13 TAU0013I (147)2.19.14 TAU0014I (147)2.19.15 TAU0015I (147)2.19.16 TAU0016I (147)2.19.17 TAU0017I (147)2.19.18 TAU0018I (148)2.19.19 TAU0019W (148)2.19.20 TAU0020E (148)2.19.21 TAU0021I (148)2.19.23 TAU0023I (148)2.19.24 TAU0024I (148)2.19.25 TAU0025I (149)2.19.26 TAU0026W (149)2.19.27 TAU0027E (149)2.19.28 TAU0028W (149)2.19.29 TAU0029I (149)2.19.30 TAU0030E (150)2.19.31 TAU0031E (150)2.19.32 TAU0032E (150)2.19.33 TAU0033E (150)2.19.34 TAU0034E (150)2.19.35 TAU0035E (151)2.19.36 TAU0036I (151)2.19.37 TAU0037I (151)2.19.38 TAU0038I (151)2.19.39 TAU0039I (151)2.19.40 TAU0040E (151)2.19.41 TAU0041I (152)2.19.42 TAU0042I (152)2.19.43 TAU0043E (152)2.19.44 TAU0044E (152)2.19.45 TAU0045I (152)2.19.46 TAU0046I (152)2.19.47 TAU0047S (153)2.19.48 TAU0048W (153)2.19.49 TAU0049W (153)2.19.50 TAU0050E (153)2.19.51 TAU0051W (153)2.19.52 TAU0052I (153)2.19.53 TAU0053I (154)2.19.54 TAU0054S (154)2.19.55 TAU0055I (154)2.19.57 TAU0057E (154)2.19.58 TAU0058E (154)2.19.59 TAU0059I (155)2.19.60 TAU0060I (155)2.19.61 TAU0061I (155)2.19.62 TAU0062E (155)2.19.63 TAU0063S (155)2.19.64 TAU0064W (156)2.19.65 TAU0065I (156)2.19.66 TAU0066I (156)2.19.67 TAU0067I (156)2.19.68 TAU0068I (156)2.19.69 TAU0069E (156)2.19.70 TAU0070I (157)2.19.71 TAU0071I (157)2.19.72 TAU0072W (157)2.19.73 TAU0073S (157)2.19.74 TAU0074I (157)2.19.75 TAU0075W (157)2.19.76 TAU0076I (158)2.19.77 TAU0077I (158)2.19.78 TAU0078E (158)2.19.79 TAU0079E (158)2.19.80 TAU0080E (159)2.19.81 TAU0081E (159)2.19.82 TAU0082E (159)2.19.83 TAU0083W (159)2.19.84 TAU0084E (159)2.19.85 TAU0085E (159)2.19.86 TAU0086I (160)2.19.87 TAU0087I (160)2.19.88 TAU0088S (160)2.19.89 TAU0089E (160)2.19.91 TAU0091E (161)2.19.92 TAU0092I (161)2.19.93 TAU0093I (161)2.19.94 TAU0099I (161)2.19.95 TAU0100E (161)2.19.96 TAU0101I (162)2.19.97 TAU0102I (162)2.19.98 TAU0103W (162)2.19.99 TAU0104I (162)2.19.100 TAU0105E (162)2.19.101 TAU0106W (163)2.19.102 TAU0107I (163)2.19.103 TAU0108E (163)2.19.104 TAU0109W (163)2.19.105 TAU0110I (163)2.19.106 TAU0111I (164)2.19.107 TAU0112I (164)2.19.108 TAU0113E (164)2.19.109 TAU0114I (164)2.19.110 TAU0115E (164)2.19.111 TAU0116I (165)2.19.112 TAU0117E (165)2.19.113 TAU0118I (165)2.19.114 TAU0119E (165)2.19.115 TAU0120E (165)2.19.116 TAU0121E (165)2.19.117 TAU0122W (166)2.19.118 TAU0123S (166)2.19.119 TAU0124E (166)2.19.120 TAU0125I (166)2.19.121 TAU0126I (166)2.19.122 TAU0127E (167)2.19.123 TAU0128E (167)2.19.125 TAU0130E (167)2.19.126 TAU0131I (167)2.19.127 TAU0132E (168)2.19.128 TAU0133E (168)2.19.129 TAU0134I (168)2.19.130 TAU0135S (168)2.19.131 TAU0136S (168)2.19.132 TAU0137I (169)2.19.133 TAU0138E (169)2.19.134 TAU0139E (169)2.19.135 TAU0140E (170)2.19.136 TAU0141I (170)2.19.137 TAU0142I (170)2.19.138 TAU0143S (170)2.19.139 TAU0144W (170)2.19.140 TAU0145I (171)2.19.141 TAU0146I (171)2.19.142 TAU0147I (171)2.19.143 TAU0148E (171)2.19.144 TAU0149E (171)2.19.145 TAU0150I (172)2.19.146 TAU0151I (172)2.19.147 TAU0152I (172)2.19.148 TAU0153I (172)2.19.149 TAU0154S (172)2.19.150 TAU0155W (172)2.19.151 TAU0156E (173)2.19.152 TAU0157I (173)2.20 UNI - Unicode code conversion services (174)2.20.1 UNI0001I (174)2.20.2 UNI0002I (174)2.20.3 UNI0003I (174)2.20.4 UNI0004W (174)2.20.6 UNI0006W (175)2.20.7 UNI0007E (175)2.20.8 UNI0008W (175)2.20.9 UNI0009W (176)2.20.10 UNI0010W (176)2.20.11 UNI0011W (176)2.20.12 UNI0012E (176)2.20.13 UNI0013I (176)2.21 XML - XML System Parser Interface (177)2.21.1 XML0001E (177)2.21.2 XML0002I (177)2.21.3 XML0003E (177)2.21.4 XML0004E (177)2.21.5 XML0005E (178)2.21.6 XML0006I (178)2.21.7 XML0007E (178)2.21.8 XML0008E (178)2.21.9 XML0009I (178)2.21.10 XML0010I (178)2.21.11 XML0011I (179)2.21.12 XML0012I (179)2.21.13 XML0013E (179)2.21.14 XML0014E (179)2.21.15 XML0015E (179)2.21.16 XML0016E (179)2.21.17 XML0017I (179)2.21.18 XML0018I (180)2.21.19 XML0019E (180)2.21.20 XML0020E (180)3 Abend Codes (181)3.1 ABEND Code U2222 (181)3.1.1 Reason Code 1 (181)3.2 Language Environment Abends (181)4 Troubleshooting (182)4.1 Holder (182)4.1.1 MAINTASK (182)4.1.2 SYSTSPRINT (182)4.1.3 DSPPRT (182)4.2 Scheduler (182)4.2.1 MAINTASK (183)4.2.2 CMDTASK (183)4.2.3 SRVTASK (183)4.2.4 T000000x (183)4.2.5 SYSTSPRT (184)4.2.6 ZCOTSPRT (184)4.3 User Server (184)4.3.1 MAINTASK (184)4.3.2 CMDTASK (184)4.3.3 SRVTASK (185)4.3.4 T0000001 (185)4.3.5 SYSTSPRT (185)4.3.6 ZCOTSPRT (185)4.3.7 SYSPRINT (185)4.3.8 ISPLOG (185)4.4 Trace Level (185)4.5 Required Information for Customer Support (186)1 IntroductionThis document provides a description of messages issued by z/Server including possible corrective actions. Every message has an 8-character identification and an associated message text.Troubleshooting often requires knowledge of configuration settings, such as the name of the holder task, the scheduler task, and the user server task(s). Please see the z/Server Installation Guide for details about these settings. Most messages only appear in the JESx job log in different DD names and not in the z/OS hardcopy log. True hardcopy log messages are marked.1.1 Message formatThe message identification has the form: pfxnnnns, where:pfx The z/Server component prefix (facility ID)The following facility IDs exist:nnnn The message number for the component, for example, 0051. The message numbers are unique within a component.s The message severity:I InformationW WarningE ErrorS Severe errorInformational messages are normally used for tracing or debugging purposes. Errors apply to the associated service and are typically caused by incorrect arguments. Severe errors may require z/Server to be restarted.The message description provides information to identify the cause of the problem. Many messages contain placeholders ([mmm], where mmm is a consecutive number; for example, [002] is the second placeholder). These placeholders are replaced with specific values in the actual message.The message description may contain messages, return and reason codes, etc. passed from invoked services (for example, the XML system parser or the security product). In these cases, the IBM manual to be consulted for details of the problem are given.With very few exceptions, the first placeholder ([001]) of each message is the timestamp when the message was issued (hh:mm:ss.ttt):hh hourmm minutess secondsttt millisecondsThe explanation for any message describes if the message is only issued at a certain IPTRACE level. This documented trace level always implies that the message is issued at this trace level and up (since a higher trace level includes all lower trace levels).1.2 Error message exampleError messages have severity E or S. There are three general classes of error:∙Errors caused by incorrect user input, such as incorrect password (SLR0071E) or filter did not find any data sets (CSI0051E). Such errors can be rectified by correcting the input and retrying the action. Some of these error messages can only occur due to incorrect input to the set of z/Server APIs, typically from an eclipse client.∙Errors that indicate problems in the configuration file, such as EMC0011E (port is not available). Such errors may be corrected by changing the configuration (a task for the system administrator), although they can also indicate some other error that needs to be investigated (also a task for the system administrator).Errors that indicate an internal z/Server error, such as API0007E (error during call to EZASOKET). If the message information provided does not help to identify the problem, please contact the support hotline.As an example, the CSI0051E message is issued by the CSI facility (CSI prefix) and has a severity of E (=error).The message is shown as follows in this book:CSI0051E [001] No data sets for filter [002] foundThis message has two placeholders: [001] and [002]. The placeholders are replaced with the timestamp and the associated filter. An actual CSI0051E error message would look like this:CSI0051E 07:48:53.935 No data sets for filter SYS2 foundThe two placeholders have been replaced with the time (07:48) and the filter (SYS2) that could not be resolved. In this case, using a filter that covers available data sets would prevent the error message.1.3 Warning message exampleWarning messages have severity W. Such messages are normally written to the associated JESx DD name, see below. Warnings normally indicate that some standard action has been taken. If this action or default is not required or wanted, change the appropriate setting.TAU0019W 09:14:05.443 Environment-Variable REATTACH was not specified, using defaults !TAU0019W 09:14:05.651 Environment-Variable CCSID was not specified, using defaults !TAU0019W 09:14:05.652 Environment-Variable TSOE_JOB_CLASS was not specified, using defaults !TAU0019W 09:14:05.652 Environment-Variable TSOE_SCHED_TAB was not specified, using defaults !LIC0007W 09:14:05.659 This server may only be used through a licensed client.1.4 Information message exampleInformation messages have severity I. Such messages are normally written to the associated JESx DD name, see below.SLR0111I 09:14:05.998 TSO/E-Scheduler Init/Term-Routine called with function STARTSLR0001I 09:14:06.170 TSO-Scheduler initializationHLD0030I 09:14:06.188 User administration ended with RC 00000000 (hex)1.5 Messages and GTFFor debugging purposes, it is recommended to start GTF and trace user records x'3E8'. In such a case, there will be significantly less messages written to the JESx spool. Most messages will be written to GTF only, regardless of the IPTRACE level being set. It may get difficult to even follow what happens withinz/Server using just the remaining messages to hardcopy log and JESx job log.2 Messages2.1 API - Application Programming InterfaceAll API messages are issued b y TAURIP03. It is used as an interface for all TCPIP communication. Standard return codes are described in the table below. This interface is called using z/Server API calls, both internally and externally.2.1.1 API0001IAPI0001I hh:mm:ss.ttt Before call IP03 starting call in [002] environmentFunction-Code : [003]Task-Area : [004] [005]Explanation:[002] is either server or client.[003] is described in SC31-8788 z/OS Communications Server IP Sockets Application Programming Interface Guide and Reference.[004] is the address of the task area.[005] is global (if [002] is server), otherwise local (if [002] is client)User response: None.Issued by: TAURIP032.1.2 API0002IAPI0002I hh:mm:ss.ttt After Call IP03 ending call. Rc is : [002] Explanation: This is an informational message due to IPTRACE level 4 being set.User response: None.Issued by: TAURIP032.1.3 API0007EAPI0007E hh:mm:ss.ttt Error during call to EZASOKETFunction : [002]Reason-Code : [003]Explanation: This message is issued when IPTRACE level 4 is set. See SC31-8788 z/OS Communications Server IP Sockets Application Programming Interface Guide and Reference for details about the function used and the returned error.System administrator response: Determine why the error occurred. If necessary, contact customer support.Issued by: TAURIP032.1.4 API0008IAPI0008I hh:mm:ss.ttt Timeout for SELECT is [002] seconds and [003] microseconds.Explanation: This is an informational message due to IPTRACE level 4 being set.User response: None.Issued by: TAURIP032.1.5 API0009EAPI0009E hh:mm:ss.ttt For function PeekMsg the sixth Parameter must be 'P'. Explanation: This message is issued in response to an API call.User response: Correct the calling program.Issued by: TAURIP032.1.6 API0010IAPI0010I hh:mm:ss.ttt Using static z/Server environment for client processing at address [002]Explanation: This is an informational message due to IPTRACE level 4 being set.User response: None.Issued by: TAURIP03。
A Practical Guide to Quantitative Finance Interview (Part)
Table of ContentsChapter 1 General Principles (1)1. Build a broad knowledge base (1)2. Practice your interview skills (1)3. Listen carefully (2)4. Speak your mind (2)5. Make reasonable assumptions (2)Chapter 2 Brain Teasers (3)2.1 Problem Simplification (3)Screwy pirates (3)Tiger and sheep (4)2.2 Logic Reasoning (5)River crossing (5)Birthday problem (5)Card game (6)Burning ropes (7)Defective ball (7)Trailing zeros (9)Horse race (9)Infinite sequence (10)2.3 Thinking Out of the Box (10)Box packing (10)Calendar cubes (11)Door to offer (12)Message delivery (13)Last ball (13)Light switches (14)Quant salary (15)2.4 Application of Symmetry (15)Coin piles (15)Mislabeled bags (16)Wise men (17)2.5 Series Summation (17)Clock pieces (18)Missing integers (18)Counterfeit coins I (19)2.6 The Pigeon Hole Principle (20)Matching socks (21)Handshakes (21)Have we met before? (21)Ants on a square (22)Counterfeit coins II (22)Contentsii 2.7 Modular Arithmetic (23)Prisoner problem (24)Division by 9 (25)Chameleon colors (26)2.8 Math Induction (27)Coin split problem (27)Chocolate bar problem (28)Race track (29)2.9 Proof by Contradiction (31)Irrational number (31)Rainbow hats (31)Chapter 3 Calculus and Linear Algebra (33)3.1 Limits and Derivatives (33)Basics of derivatives (33)Maximum and minimum (34)L’Hospital’s rule (35)3.2 Integration (36)Basics of integration (36)Applications of integration (38)Expected value using integration (40)3.3 Partial Derivatives and Multiple Integrals (40)3.4 Important Calculus Methods (41)Taylor’s series (41)Newton’s method (44)Lagrange multipliers (45)3.5 Ordinary Differential Equations (46)Separable differential equations (47)First-order linear differential equations (47)Homogeneous linear equations (48)Nonhomogeneous linear equations (49)3.6 Linear Algebra (50)Vectors (50)QR decomposition (52)Determinant, eigenvalue and eigenvector (53)Positive semidefinite/definite matrix (56)LU decomposition and Cholesky decomposition (57)Chapter 4 Probability Theory (59)4.1 Basic Probability Definitions and Set Operations (59)Coin toss game (61)Card game (61)Drunk passenger (62)A Practical Guide To Quantitative Finance InterviewsN points on a circle (63)4.2 Combinatorial Analysis (64)Poker hands (65)Hopping rabbit (66)Screwy pirates 2 (67)Chess tournament (68)Application letters (69)Birthday problem (71)100th digit (71)Cubic of integer (72)4.3 Conditional Probability and Bayes’ formula (72)Boys and girls (73)All-girl world? (74)Unfair coin (74)Fair probability from an unfair coin (75)Dart game (75)Birthday line (76)Dice order (78)Monty Hall problem (78)Amoeba population (79)Candies in a jar (79)Coin toss game (80)Russian roulette series (81)Aces (82)Gambler’s ruin problem (83)Basketball scores (84)Cars on road (85)4.4 Discrete and Continuous Distributions (86)Meeting probability (88)Probability of triangle (89)Property of Poisson process (90)Moments of normal distribution (91)4.5 Expected Value, Variance & Covariance (92)Connecting noodles (93)Optimal hedge ratio (94)Dice game (94)Card game (95)Sum of random variables (95)Coupon collection (97)Joint default probability (98)4.6 Order Statistics (99)Expected value of max and min (99)Correlation of max and min (100)Random ants (102)Chapter 5 Stochastic Process and Stochastic Calculus (105)iiiContentsiv 5.1 Markov Chain (105)Gambler’s ruin problem (107)Dice question (108)Coin triplets (109)Color balls (113)5.2 Martingale and Random walk (115)Drunk man (116)Dice game (117)Ticket line (117)Coin sequence (119)5.3 Dynamic Programming (121)Dynamic programming (DP) algorithm (122)Dice game (123)World series (123)Dynamic dice game (126)Dynamic card game (127)5.4 Brownian Motion and Stochastic Calculus (129)Brownian motion (129)Stopping time/ first passage time (131)Ito’s lemma (135)Chapter 6 Finance (137)6.1. Option Pricing (137)Price direction of options (137)Put-call parity (138)American v.s. European options (139)Black-Scholes-Merton differential equation (142)Black-Scholes formula (143)6.2. The Greeks (149)Delta (149)Gamma (152)Theta (154)Vega (156)6.3. Option Portfolios and Exotic Options (158)Bull spread (159)Straddle (159)Binary options (160)Exchange options (161)6.4. Other Finance Questions (163)Portfolio optimization (163)Value at risk (164)Duration and convexity (165)Forward and futures (167)Interest rate models (168)A Practical Guide To Quantitative Finance Interviews Chapter 7 Algorithms and Numerical Methods (171)7.1. Algorithms (171)Number swap (172)Unique elements (173)Horner's algorithm (174)Moving average (174)Sorting algorithm (174)Random permutation (176)Search algorithm (177)Fibonacci numbers (179)Maximum contiguous subarray (180)7.2. The Power of Two (182)Power of 2? (182)Multiplication by 7 (182)Probability simulation (182)Poisonous wine (183)7.3 Numerical Methods (184)Monte Carlo simulation (184)Finite difference method (189)vChapter 2 Brain TeasersIn this chapter, we cover problems that only require common sense, logic, reasoning, and basic—no more than high school level—math knowledge to solve. In a sense, they are real brain teasers as opposed to mathematical problems in disguise. Although these brain teasers do not require specific math knowledge, they are no less difficult than other quantitative interview problems. Some of these problems test your analytical and general problem-solving skills; some require you to think out of the box; while others ask you to solve the problems using fundamental math techniques in a creative way. In this chapter, we review some interview problems to explain the general themes of brain teasers that you are likely to encounter in quantitative interviews.2.1 Problem SimplificationIf the original problem is so complex that you cannot come up with an immediate solution, try to identify a simplified version of the problem and start with it. Usually you can start with the simplest sub-problem and gradually increase the complexity. You do not need to have a defined plan at the beginning. Just try to solve the simplest cases and analyze your reasoning. More often than not, you will find a pattern that will guide you through the whole problem.Screwy piratesFive pirates looted a chest full of 100 gold coins. Being a bunch of democratic pirates, they agree on the following method to divide the loot:The most senior pirate will propose a distribution of the coins. All pirates, including the most senior pirate, will then vote. If at least 50% of the pirates (3 pirates in this case) accept the proposal, the gold is divided as proposed. If not, the most senior pirate will be fed to shark and the process starts over with the next most senior pirate… The process is repeated until a plan is approved. You can assume that all pirates are perfectly rational: they want to stay alive first and to get as much gold as possible second. Finally, being blood-thirsty pirates, they want to have fewer pirates on the boat if given a choice between otherwise equal outcomes.How will the gold coins be divided in the end?Solution: If you have not studied game theory or dynamic programming, this strategy problem may appear to be daunting. If the problem with 5 pirates seems complex, we can always start with a simplified version of the problem by reducing the number of pirates. Since the solution to 1-pirate case is trivial, let’s start with 2 pirates. The seniorBrain Teasers4pirate (labeled as 2) can claim all the gold since he will always get 50% of the votes from himself and pirate 1 is left with nothing.Let’s add a more senior pirate, 3. He knows that if his plan is voted down, pirate 1 will get nothing. But if he offers private 1 nothing, pirate 1 will be happy to kill him. So pirate 3 will offer private 1 one coin and keep the remaining 99 coins, in which strategy the plan will have 2 votes from pirate 1 and 3.If pirate 4 is added, he knows that if his plan is voted down, pirate 2 will get nothing. So pirate 2 will settle for one coin if pirate 4 offers one. So pirate 4 should offer pirate 2 one coin and keep the remaining 99 coins and his plan will be approved with 50% of the votes from pirate 2 and 4.Now we finally come to the 5-pirate case. He knows that if his plan is voted down, both pirate 3 and pirate 1 will get nothing. So he only needs to offer pirate 1 and pirate 3 one coin each to get their votes and keep the remaining 98 coins. If he divides the coins this way, he will have three out of the five votes: from pirates 1 and 3 as well as himself.Once we start with a simplified version and add complexity to it, the answer becomes obvious. Actually after the case 5,n a clear pattern has emerged and we do not need to stop at 5 pirates. For any 21n pirate case (n should be less than 99 though), the most senior pirate will offer pirates 1,3,, and 21n each one coin and keep the rest for himself.Tiger and sheepOne hundred tigers and one sheep are put on a magic island that only has grass. Tigers can eat grass, but they would rather eat sheep. Assume: A . Each time only one tiger can eat one sheep, and that tiger itself will become a sheep after it eats the sheep. B . All tigers are smart and perfectly rational and they want to survive. So will the sheep be eaten?Solution: 100 is a large number, so again let’s start with a simplified version of the problem . If there is only 1 tiger (1n ), surely it will eat the sheep since it does not need to worry about being eaten. How about 2 tigers? Since both tigers are perfectly rational, either tiger probably would do some thinking as to what will happen if it eats the sheep. Either tiger is probably thinking: if I eat the sheep, I will become a sheep; and then I will be eaten by the other tiger. So to guarantee the highest likelihood of survival, neither tiger will eat the sheep.If there are 3 tigers, the sheep will be eaten since each tiger will realize that once it changes to a sheep, there will be 2 tigers left and it will not be eaten. So the first tiger that thinks this through will eat the sheep. If there are 4 tigers, each tiger will understandA Practical Guide To Quantitative Finance Interviews5that if it eats the sheep, it will turn to a sheep. Since there are 3 other tigers, it will be eaten. So to guarantee the highest likelihood of survival, no tiger will eat the sheep.Following the same logic, we can naturally show that if the number of tigers is even, the sheep will not be eaten. If the number is odd, the sheep will be eaten. For the case 100,n the sheep will not be eaten.2.2 Logic ReasoningRiver crossingFour people, A ,B ,C and D need to get across a river. The only way to cross the river is by an old bridge, which holds at most 2 people at a time. Being dark, they can't cross the bridge without a torch, of which they only have one. So each pair can only walk at the speed of the slower person. They need to get all of them across to the other side as quickly as possible. A is the slowest and takes 10 minutes to cross; B takes 5 minutes; C takes 2 minutes; and D takes 1 minute.What is the minimum time to get all of them across to the other side?1Solution: The key point is to realize that the 10-minute person should go with the 5-minute person and this should not happen in the first crossing, otherwise one of them have to go back. So C and D should go across first (2 min); then send D back (1min); A and B go across (10 min); send C back (2min); C and D go across again (2 min). It takes 17 minutes in total. Alternatively, we can send C back first and then D back in the second round, which takes 17 minutes as well.Birthday problemYou and your colleagues know that your boss A ’s birthday is one of the following 10 dates:Mar 4, Mar 5, Mar 8Jun 4, Jun 7Sep 1, Sep 5Dec 1, Dec 2, Dec 8A told you only the month of his birthday, and told your colleague C only the day. After that, you first said: “I don’t know A ’s birthday; C doesn’t know it either.” After hearing 1 Hint: The key is to realize that A andB should get across the bridge together.Brain Teasers6what you said, C replied: “I didn’t know A ’s birthday, but now I know it.” You smiled and said: “Now I know it, too.” After looking at the 10 dates and hearing your comments, your administrative assistant wrote down A ’s birthday without asking any questions. So what did the assistant write?Solution: Don’t let the “he said, she said” part confuses you. Just interpret the logic behind each individual’s comments and try your best to derive useful information from these comments.Let D be the day of the month of A ’s birthday, we have {1,2,4,5,7,8}.D If the birthday is on a unique day, C will know the A ’s birthday immediately. Among possible D s, 2 and 7 are unique days. Considering that you are sure that C does not know A ’s birthday, you must infer that the day the C was told of is not 2 or 7. C onclusion: the month is not June or December. (If the month had been June, the day C was told of may have been 2; if the month had been December, the day C was told of may have been 7.) Now C knows that the month must be either March or September. He immediately figures out A ’s birthday, which means the day must be unique in the March and September list. It means A ’s birthday cannot be Mar 5, or Sep 5. C onclusion: the birthday must be Mar 4, Mar 8 or Sep 1.Among these three possibilities left, Mar 4 and Mar 8 have the same month. So if the month you have is March, you still cannot figure out A ’s birthday. Since you can figure out A ’s birthday, A ’s birthday must be Sep 1. Hence, the assistant must have written Sep 1.Card gameA casino offers a card game using a normal deck of 52 cards. The rule is that you turn over two cards each time. For each pair, if both are black, they go to the dealer’s pile; if both are red, they go to your pile; if one black and one red, they are discarded. The process is repeated until you two go through all 52 cards. If you have more cards in your pile, you win $100; otherwise (including ties) you get nothing. The casino allows you to negotiate the price you want to pay for the game. How much would you be willing to pay to play this game?2Solution: This surely is an insidious casino. No matter how the cards are arranged, you and the dealer will always have the same number of cards in your piles. Why? Because each pair of discarded cards have one black card and one red card, so equal number of2Hint: Try to approach the problem using symmetry. Each discarded pair has one black and one red card. What does that tell you as to the number of black and red cards in the rest two piles?A Practical Guide To Quantitative Finance Interviews7red and black cards are discarded. As a result, the number of red cards left for you and the number of black cards left for the dealer are always the same. The dealer always wins! So we should not pay anything to play the game.Burning ropesYou have two ropes, each of which takes 1 hour to burn. But either rope has different densities at different points, so there's no guarantee of consistency in the time it takes different sections within the rope to burn. How do you use these two ropes to measure 45 minutes?Solution: This is a classic brain teaser question. For a rope that takes x minutes to burn, if you light both ends of the rope simultaneously, it takes /2x minutes to burn. So we should light both ends of the first rope and light one end of the second rope. 30 minutes later, the first rope will get completely burned, while that second rope now becomes a 30-min rope. At that moment, we can light the second rope at the other end (with the first end still burning), and when it is burned out, the total time is exactly 45 minutes. Defective ballYou have 12 identical balls. One of the balls is heavier OR lighter than the rest (you don't know which). Using just a balance that can only show you which side of the tray is heavier, how can you determine which ball is the defective one with 3 measurements?3Solution: This weighing problem is another classic brain teaser and is still being asked by many interviewers. The total number of balls often ranges from 8 to more than 100.Here we use 12nto show the fundamental approach. The key is to separate the original group (as well as any intermediate subgroups) into three sets instead of two. The reason is that the comparison of the first two groups always gives information about the third group.Considering that the solution is wordy to explain, I draw a tree diagram in Figure 2.1 to show the approach in detail. Label the balls 1 through 12 and separate them to three groups with 4 balls each. Weigh balls 1, 2, 3, 4 against balls 5, 6, 7, 8. Then we go on to explore two possible scenarios: two groups balance, as expressed using an “=” sign, or 1, 3Hint: First do it for 9 identical balls and use only 2 measurements, knowing that one is heavier than the rest.Brain Teasers82, 3, 4 are lighter than 5, 6, 7, 8, as expressed using an “<” sign. There is no need to explain the scenario that 1, 2, 3, 4 are heavier than 5, 6, 7, 8. (Why?4)If the two groups balance, this immediately tells us that the defective ball is in 9, 10, 11 and 12, and it is either lighter (L ) or heavier (H ) than other balls. Then we take 9, 10 and 11 from group 3 and compare balls 9, 10 with 8, 11. Here we have already figured out that 8 is a normal ball. If 9, 10 are lighter, it must mean either 9 or 10 is L or 11 is H . In which case, we just compare 9 with 10. If 9 is lighter, 9 is the defective one and it is L ; if 9 and 10 balance, then 11 must be defective and H ; If 9 is heavier, 10 is the defective one and it is L . If 9, 10 and 8, 11 balance, 12 is the defective one. If 9, 10 is heavier, than either 9 or 10 is H, or 11 is L.You can easily follow the tree in Figure 2.1 for further analysis and it is clear from the tree that all possible scenarios can be resolved in 3 measurements. In general if you have the information as to whether the defective ball is heavier or 4 Here is where the symmetry idea comes in. Nothing makes the 1, 2, 3, 4 or 5, 6, 7, 8 labels special. If 1, 2, 3, 4 are heavier than 5, 6, 7, 8, let’s just exchange the labels of these two groups. Again we have the case of 1, 2, 3, 4 being lighter than 5, 6, 7, 8.1/2L or 6H5H or 3L4L or 7/8H 12L or 12H 9/10H or 11L 9/10L or 11H 5,6,7,8+?1,2,53,6,9_1/2/3/4 L or 5/6/7/8 H1L 6H 2L 8H 4L 7H 3L 5H 2_18_79_39,108,11_9/10/11/12 L or H 9L 11H 10L 12H 11L 12L 10H 9H_912_8_9Figure 2.1 Tree diagram to identify the defective ball in 12 ballsA Practical Guide To Quantitative Finance Interviews9lighter, you can identify the defective ball among up to 3n balls using no more than n measurements since each weighing reduces the problem size by 2/3. If you have no information as to whether the defective ball is heavier or lighter, you can identify the defective ball among up to (33)/2n balls using no more than n measurements. Trailing zerosHow many trailing zeros are there in 100! (factorial of 100)?Solution: This is an easy problem. We know that each pair of 2 and 5 will give a trailing zero. If we perform prime number decomposition on all the numbers in 100!, it is obvious that the frequency of 2 will far outnumber of the frequency of 5. So the frequency of 5 determines the number of trailing zeros. Among numbers 1,2,,99, and 100, 20 numbers are divisible by 5 (5,10,,100 ). Among these 20 numbers, 4 are divisible by 52 (25,50,75,100). So the total frequency of 5 is 24 and there are 24 trailing zeros.Horse raceThere are 25 horses, each of which runs at a constant speed that is different from the other horses’. Since the track only has 5 lanes, each race can have at most 5 horses. If you need to find the 3 fastest horses, what is the minimum number of races needed to identify them?Solution: This problem tests your basic analytical skills. To find the 3 fastest horses, surely all horses need to be tested. So a natural first step is to divide the horses to 5 groups (with horses 1-5, 6-10, 11-15, 16-20, 21-25 in each group). After 5 races, we will have the order within each group, let’s assume the order follows the order of numbers (e.g., 6 is the fastest and 10 is the slowest in the 6-10 group)5. That means 1, 6, 11, 16 and 21 are the fastest within each group.Surely the last two horses within each group are eliminated. What else can we infer? We know that within each group, if the fastest horse ranks 5th or 4th among 25 horses, then all horses in that group cannot be in top 3; if it ranks the 3rd, no other horse in that group can be in the top 3; if it ranks the 2nd, then one other horse in that group may be in top 3; if it ranks the first, then two other horses in that group may be in top 3. 5 Such an assumption does not affect the generality of the solution. If the order is not as described, just change the labels of the horses.Chapter 4 Probability TheoryC hances are that you will face at least a couple of probability problems in most quantitative interviews. Probability theory is the foundation of every aspect of quantitative finance. As a result, it has become a popular topic in quantitative interviews. Although good intuition and logic can help you solve many of the probability problems, having a thorough understanding of basic probability theory will provide you with clear and concise solutions to most of the problems you are likely to encounter. Furthermore, probability theory is extremely valuable in explaining some of the seemingly-counterintuitive results. Armed with a little knowledge, you will find that many of the interview problems are no more than disguised textbook problems.So we dedicate this chapter to reviewing basic probability theory that is not only broadly tested in interviews but also likely to be helpful for your future career. 1 The knowledge is applied to real interview problems to demonstrate the power of probability theory. Nevertheless, the necessity of knowledge in no way downplays the role of intuition and logic. Quite the contrary, common sense and sound judgment are always crucial for analyzing and solving either interview or real-life problems. As you will see in the following sections, all the techniques we discussed in Chapter 2 still play a vital role in solving many of the probability problems.Let’s have some fun playing the odds.4.1 Basic Probability Definitions and Set OperationsFirst let’s begin with some basic definitions and notations used in probability. These definitions and notations may seem dry without examples—which we will present momentarily—yet they are crucial to our understanding of probability theory. In addition, it will lay a solid ground for us to systematically approach probability problems.Outcome (Ȧ):the outcome of an experiment or trial.Sample space/Probability space ( ):the set of all possible outcomes of an experiment.1 As I have emphasized in Chapter 3, this book does not teach probability or any other math topics due to the space limit—it is not my goal to do so, either. The book gives a summary of the frequently-tested knowledge and shows how it can be applied to a wide range of real interview problems. The knowledge used in this chapter is covered by most introductory probability books. It is always helpful to pick up one or two classic probability books in case you want to refresh your memory on some of the topics. My personal favorites are First Course in Probability by Sheldon Ross and Introduction to Probability by Dimitri P. Bertsekas and John N. Tsitsiklis.Probability Theory60()P Z : Probability of an outcome (()0,,()1P P Z Z Z Z :t : ¦).Event:A set of outcomes and a subset of the sample space.()P A : Probability of an event A , ()()AP A P Z Z ¦.A B : Union A B is the set of outcomes in event A or in event B (or both).A B or AB : Intersection A B (or AB ) is the set of outcomes in both A and B .c A : The complement of A , which is the event “not A ”.Mutually Exclusive :A B ) where ) is an empty set.For any mutually exclusive events 12,,N E E E ,11().N N i i i i P E P E §· ¨¸©¹¦ Random variable: A function that maps each outcome (Ȧ) in the sample space ( ) into the set of real numbers.Let’s use the rolling of a six-sided dice to explain these definitions and notations. A roll of a dice has 6 possible outcomes (mapped to a random variable): 1, 2, 3, 4, 5, or 6. Sothe sample space :is {1,2,3,4,5,6} and the probability of each outcome is 1/6 (assuming a fair dice). We can define an event A representing the event that the outcomeis an odd number {1,3,5},Athen the complement of A is {2,4,6}.c A Clearly ()P A (1)(3)(5)1/2.P P P Let B be the event that the outcome is larger than 3: {4,5,6}.B Then the union is {1,3,4,5,6}A B and the intersection is {5}.A B One popular random variable called indicator variable (a binary dummy variable) for event A is defined as the following:1,{1,3,5}.0,{1,3,5}A if x I if x ® ¯ Basically 1A I when A occurs and 0A I if c A occurs. The expected value of A I is []()A E I P A .Now, time for some examples.A Practical Guide To Quantitative Finance Interviews61Coin toss gameTwo gamblers are playing a coin toss game. Gambler A has (1)n fair coins; B has n fair coins. What is the probability that A will have more heads than B if both flip all their coins?2Solution: We have yet to cover all the powerful tools probability theory offers. What do we have now? Outcomes, events, event probabilities, and surely our reasoning capabilities! The one extra coin makes A different from B . If we remove a coin from A ,A and B will become symmetric. Not surprisingly, the symmetry will give us a lot of nice properties. So let’s remove the last coin of A and compare the number of heads in A ’s first n coins with B ’s n coins. There are three possible outcomes:1E :A ’s n coins have more heads than B ’s n coins;2E :A ’s n coins have equal number of heads as B ’s n coins;3E :A ’s n coins have fewer heads than B ’s n coins.By symmetry, the probability that A has more heads is equal to the probability that B has more heads. So we have 13()().P E P E Let’s denote 13()()P E P E x and 2().P E y Since ()1,P Z Z :¦ we have 2 1.x y For event 1,E A will always have more headsthan B no matter what A ’s (1)n th coin’s side is; for event 3E ,A will have no moreheads than B no matter what A ’s (1)n th coin’s side is. For event 2E ,A ’s (1)n thcoin does make a difference. If it’s a head, which happens with probability 0.5, it will make A have more heads than B . So the (1)n th coin increases the probability that A has more heads than B by 0.5y and the total probability that A has more heads is 0.50.5(12)0.5x y x x when A has (1)n coins.Card gameA casino offers a simple card game. There are 52 cards in a deck with 4 cards for each value jack queen king ace 2,3,4,5,6,7,8,9,10,,,,J Q K A . Each time the cards are thoroughly shuffled (so each card has equal probability of being selected). You pick up a card from the deck and the dealer picks another one without replacement. If you have a larger number, you win; if the numbers are equal or yours is smaller, the house wins—as in all other casinos, the house always has better odds of winning. What is your probability of winning? 2 Hint: What are the possible results (events) if we compare the number of heads in A ’s first n coins withB ’s n coins? By making the number of coins equal, we can take advantage of symmetry. For each event, what will happen if A ’s last coin is a head? Or a tail?。
动物世界中的三口之家
Easy Adventures 行走《优美的长颈鹿家庭》长颈鹿通常生一对角，终生不会脱掉，皮肤上的花斑网纹则为一种天然的保护色。
长颈鹿喜欢群居，一般十多头生活在一起，有时多到几十头一大群。
长颈鹿是谨慎胆小的动物，每当遇到天敌时，立即逃跑。
它能以每小时72公里的极快速度奔跑。
当跑不掉时，它那铁锤似的巨蹄就是很有力的武器。
成年长颈鹿的蹄子足可以将狮子肋骨踢断。
长颈鹿繁殖期不固定，孕期14-15个月，每胎产1仔，生下来的幼仔身高1.85米，出生后20分钟即能站立，几天后便能奔驰如飞，哺乳期约9个月，3.5-4.5岁性成熟，寿命约30年。
《安博塞利的灰冠鹤一家》灰冠鹤一般栖息在沼泽地上或低平的树顶上。
非洲人民对灰冠鹤十分喜爱，尤其喜欢听它那轻柔的鸣声，由于鸣叫时间都在黎明、中午和子夜，农民把它看成是准确的生物钟。
灰冠鹤有金光闪烁的“皇冠”，喉部有红色的肉垂。
《时尚的斑马家庭》在繁殖季节，斑马表现得十分紧张而活跃，雄兽之间互相会毫不客气地进行激烈得争斗，打斗的方式为互相碰击颈部、用嘴咬、用前蹄来踢等。
败者狼狈逃窜，获胜者则与雌兽一起生活一段时间，通过亲昵、嬉戏等行为，然后交配。
每只获胜的雄兽每年要交配数只雌兽。
雌兽的妊娠期约为11-13个月，每隔3年生产1次。
每胎产1仔。
幼仔出生后不久，即可站立和走路。
哺乳期约为6个月，3.5-4岁时性成熟。
寿命约为20-30年。
《安波塞利大象一家》大象栖息于多种环境，尤喜丛林、草原和河谷地带。
群居，雄性偶有独栖。
以植物为食，食量极大，每日食量225千克以上。
大象的求爱方式比较复杂，每当繁殖期到来，雌象便开始寻找安静僻静之处，用鼻子挖坑，建筑新房，然后摆上礼品。
雄象四处漫步，用长鼻子在雌象身上来回抚摸，接着用鼻子互相纠缠，有时把鼻尖塞到对方的嘴里。
大象繁殖期不固定，孕期约22个月，每产1仔，13-14岁性成熟，寿命可达七八十年。
Easy Adventures 行走。
EndersGame《安德的游戏》人物分析
威列德（Vlad）： 来自白俄罗斯的战斗学校学员，曾任安德属下 飞龙战队 B 组组长。在艾洛斯决战时，是第二个 陷入崩溃的支队长。
达普尔（Ducheval ‘Shovel’）： 来自拉丁美洲的战斗学校学员，曾任安德属下飞龙
战队 E 组组长。达普尔非常崇拜安德，后成为艾洛斯上 安德团队成员。他是第三个因承受不了巨大压力陷入崩 溃的支队长。
师，传说中虫族战争英雄。第二次虫族战争中，曾 力挽危局，指挥舰队独立消灭入侵的虫族。是在安 德之前，最了解虫族的人。为了能将自己掌握的知 识传递给后继者，此后一直在宇宙中以光速往返飞 行，直到联合舰队挖掘出安德。在培养并辅助安德 彻底消灭虫族后，雷汉作为舰长，加入了殖民舰队， 奔赴远方。
安德森上校（M ajor Gwen Anderson）： 格拉夫上校在战斗学校的副手。主要负责在战斗室
“豆子”朱里安·德尔菲克(Julian 'Bean' Delphiki)： 生于荷兰鹿特丹、具有希腊和尼日利亚血统的
战斗学校学员，安德在飞龙战队时最信任的骨干队 员之一。豆子智力超群，五岁就被送进战斗学校， 因成绩优异，被认为是仅次于安德的天才。安德在 战斗学校指挥的最后一场战斗中，豆子领导他的小 队，起到了决定胜负的关键作用。在艾洛斯，豆子 被国际舰队暗定为安德的替补者，在安德指挥舰队 摧毁虫族母星之战中，发挥了重要作用。
战斗学校拒绝。安德未去战斗学校前，华伦蒂在暴虐的 彼得和善良的安德中间，充当调解人角色。华伦蒂深爱 着安德，始终是安德的情感寄托，安德去战斗学校后， 曾两次帮助安德克服遇到的心理问题。后成为底层民众 的舆论领袖“德摩斯梯 尼”，帮助彼得成 为世界联盟的 霸主。虫族战争后，和安德一起去虫族行星殖民，成为 一个历史学家。
丁·米克(Dink Meeker)： 来自荷兰的战斗学校老兵，安德的好友之一。米克