AngelCG09 Chapter 3 Better Interactive Programs

皇家骑士团命运之轮(psp)新手攻略基础信息职业技能随着角色等级增长，技能习得种类会逐渐增加，技能有共通技能和职业技能，共通技能如防御力UP,转职后可以继承，职业技能如法师冥想，转职成其他职业后变灰，即无法使用。

.技能学习每个战役结束都会分配EXP和SP，EXP按等级差分配，等级越低的角色分配的越多，SP 平均分配，用来学习技能。

.人物界面L,C,N分别表示人物性格LAW秩序,CHAOS混沌,NEUTRAL中立作用暂时不明.必杀技首先技能栏先学会某样武器技能，比如双手剑，不停地砍兵，技能等级升至2会习得该武器种类的必杀技以上装备道具1.武器的获得可通过商店购买、敌人掉落和偷盗获得。

2.武器的熟练度修正非常重要，这种提升比较缓慢。

必杀同样会获得熟练度，但是砍非生命体不会。

3.武器的适格可以翻页查看，有熟练度加成的和属性补正的武器是首选，其中，熟练度+2的各种饰品合成书在四巫女剧情神殿中掉落。

4.装备不仅修正你的能力，也修正你的WT。

风使完全可以不带副手武器。

5.武器有贯通/切断/打击属性，不同属性针对不同敌人。

6.有的武器会存在隐藏加成，比如非常给力的雷神配雷神弓，以及只对女性角色加成的独角兽枪，这种加成是可怕的7.弓箭可以攻击到射程以外，越高射的越远。

8.商店的更新取决于整体等级的不断更新，最高可至30级上下，更高级的武器则需要通过合成来获取。

9.合成需要的素材多需要偷盗，也可通过掉落，合成书中，一级二级为商店购买，三级为四巫女剧情神殿必定获得，四级为大森林、海贼墓场等掉落，究极的在死者宫殿请注意。

掉落的规则如下：掉落古文书的场所一定，即鞭古文书只会有第四章大森林的風致の丘掉落。

掉落的敌人是一定的，需要查看其所带武器，即一定要佩戴该种类武器。

但是弓古文书的弓兵可能是带弓也可能带弩必杀技能1.TP200的数值上限不随等级上涨，但有职业技能修正2.受到/造成攻击越大，给的TP越多3.特殊技能使用后，造成的伤害仍然计算入TP，比如消耗40TP附加龙系大伤害后，打出600点伤害，仍将获得大量TP，但必杀是完全清0，不会再次获得4.必杀技的习得要至少武器熟练度达到2(满8级)，每生两级会有一个新的必杀技，除了奥兹玛R4多一个鞭子必杀，这些必杀技的消耗是一样的，但是打击/贯通/切断属性和光暗等不尽相同，下表是按2468等级学习的，不再重复5.必杀技要消耗TP，TP满100后方可发动6.必杀技发动后TP值清0而不是扣除100点7.多余的TP可能用于命中补正或者威力补正8.必杀技可以被躲闪，但不会被反击或者招架9.死者の宮殿1.在四章バンハムーバの神殿剧情后(打谢利)，查看华伦报告出现，沿途会存在女龙使的剧情2.死者宫殿中2F可以收到新人物，需要其存活3.死者宫殿中3F要站在(5.2)处开启下一层，无需先出去后查看华伦报告在进入，3F后有分支4.死者宫殿中有1.2.3.5.22.41.74.1005.5F的剧情需要在通关前完成，C线需要通过女尸术士的剧情，其他线不需要6.死者宫殿共有100层，可能出现高过自己10级的怪物，可S/L掉7.第二次携带四风神器到100层可出秘宝ファイアクレスト和十弐神将の音叉(消耗品)，其中8.死者宫殿观光书同样是消耗品，某些素材需要在商店卖掉九头龙9死者宫殿分支地图如下:剧情:Chapter1僕にその手を汚せというのか1.港町ゴリアテ敌人配置：-胜利条件：-要点提示：剧情关，移动主角一次后过关选项：1.どうか僕らををお許しください。

天使Shininglore研究集合天使纪念 2009-05-01 09:41 阅读246 评论5字号：大中小绪言自较早在国内运营的全3D游戏《精灵》之后，在2002年7月，ShiningloreOnline（天使在线）作为更成熟的3D MMORPG（多人在线角色扮演游戏）网络游戏来到了中国。

游戏虽然在画面上较为细致，但其程序设计却问题重重，以致2003年10月正式结束运营，其在韩国的后续版本也停止了开发。

点此查看：天使历程三部曲一直以来，忠实的游戏爱好者一直为天使的复活而努力。

国内相对宽松的环境，孕育了各类个人服务器的诞生。

经过各种是是非非，转眼已经到2009年了。

7年的时光，天使仍然在苟延残喘。

我不想去打破这样一种平衡，不过随着时间流逝，各种技术也如过眼云烟逐渐淡出记忆。

在此，我仅将过去所整理的资料进行重组，有兴趣的爱好者可以延续。

唯一的希望就是不要再去破坏这个已经倒闭了7年的游戏，这种捣乱炫耀的成就感不值一钱，游戏的本意就是大家一起快乐。

游戏文件游戏运行所需要的程序：天使客户端，天使服务端，数据库程序网络游戏的特点就是需要在客户端和服务器间传送网络数据包，仅一个无网络的客户端是无法启动的。

为了防止数据篡改，一般服务端会进行数据传送的加密，也就是封装网络数据包（简称封包）。

现在就先不讨论这个敏感的话题，以下是对游戏内部文件的直接修改。

天使客户端由Slonline.exe作为启动程序，资源文件包含在data名称的文件夹内。

游戏的图形、声效、任务数据等经过压缩，分别封装为以下7个文件（分别以天使8英雄中的7个名字来命名）：Mene.sop，Eto.sop，Siena.sop包含游戏特效图形文件（PNX动作模型信息，DDS特效贴图，TGA贴图）Rune.sop：包含游戏地形构成文件（MAP格式的地形及NES格式的任务信息）Serine.sop：包含游戏音乐文件（MP3格式的BGM）Sandra.sop：包含游戏声效文件（WAV格式的游戏声效）Bio.sop：包含sox格式的天使游戏数据文件另外，Soda.dat为游戏所有文件的索引信息天使研究（1）－文件提取和替换：Soda.exe的使用“Soda：一种用BZip2方法压缩的文件列表。

密技在一般状态下同时按住[Ctrl] + [Insert]再输入密技秘籍作用ODDM战斗中按[F5] 键补 HP，MP，DP(每按一次，就可以使用必杀技且可以使死去的人复活)ODFULLSKILL能学会所有的必杀技，魔法，技能ODNOBAT不会遇到随机率战斗ODFREESAVE随时存档ODGETGOLD #增加金钱,#为金钱数值(要空一格)ODGETDNA #1 #2获得敌人DNA元素,#1是敌人代码,#2是数量(最多99)ODGETITEM #1 #2获得物品,#1是物品代码,#2是数量(最多99)无限攻击在战斗时的选单，AP行动值还没储满时，“攻击”指令的右边有个小小的箭头，把鼠标移到箭头上按空白键就可以无限的攻击敌人，敌人没有反击的机会(注意：一定要学到必杀技才会出现小箭头)。

最强武器所在地各角色的最强武器(以下的要在救了蒂娜后找妖精王时才可取得)：布雷德-天罪(回艾克拉亚镇到教堂中间点一下那个钉在十字架的人，然后跟罪天使对决，战胜后得到)西尔法-天之尾羽张(找水精灵王，要先打倒守护兽(深海游魔)，便可获得)盖落普-碎魂(找木精灵王，它会给你，得到后会跟三个制裁天使战斗)杰克 -罪与罚(去找火精灵王，它会给你)米雪儿-龙咬鞭(去峡之地原来遇到魔龙的地方与怨灵战斗三次，完成后得到)地精灵王有天使的首饰风精灵王有妖精羽衣道具合成天蛆+阵风兽 =治愈II邪命树+姆咪族 =防御II鬼面蝎+刃风狼 =魔力II刃风狼+黄蜂萝莉 =生化攻击II阵风兽+姆咪兽 =念力攻击II鬼面蝎+蠕虫 =念力攻击III特殊DNA嗜血皇后+歪天使=强袭(强力推荐给盖落普使用)嗜血皇后+嗜血皇后 =治愈III猎诱虫+赤膜 =念力攻击IIIDNA元素换取物品歪天使x1 新生体成体x2 诱猎虫x3 =拉蒙之魂地精兽x1 风动兽x2 结晶魔神x2 =裂精灵拿法1、在mumu村有个mumu会跟你要花束，选第二个选项大概要五六次，他就会送你一只木精灵。

天使心跳经典语录[标签:栏目] ，天使心跳经典语录1、要因这种扭曲的记忆而消失！我们活过的人生是真实存在的，是没有半点虚假的人生！大家都努力活过！就是这样记录下来的记忆，是拼命活过的记忆，无论那是什么样的记忆，都是我们所活过的人生！2、天使也是人，真的天使哪会在失落时候吃喜欢的麻婆豆腐来安慰自己的。

只是因为学生会长的义务，所以才不能对扰乱风纪的我们置之不理。

因为我们在Guild生产兵器，她才创造出GuardSkill与我们对抗，这就是事情的前因后果吗，真滑稽。

竟然还没找到任何关于神的头绪。

3、陷阱只不过能拖延一会时间罢了，我们追，前进。

4、成为她的同伴就是“过着愉快的校园生活，然后从这个世界消失”啊……这样啊……感觉真可笑！她太过可怜，太过凄惨，这个世界到底是什么鬼系统啊！5、活在现实中的人是错的，而那些哭泣的人才是正确的，孤独的我们才有人类的样子。

6、笨蛋们，都醒了没，Guild废弃，和天使一起爆破掉了通告全体人员，立刻到OldGuild来，武器的补充在那里迅速进行，在天使复活前全到OldGuild 来，再重复一遍，快点，笨蛋们。

7、本来我们应该把记忆注入这些只有形式的东西里去而获得新生的。

8、既然你这么说的话，那就这样吧。

9、在这遇到的你，不是假的由依，是真正的由依，无论在哪里遇到你，我都会喜欢你。

如果在六十亿分之一的概率下再次相遇的话，即使你那时候也是身体无法动弹，我也会和你结婚。

10、我父母总是一天到晚吵个不停，没有自己的房间，我只能在那刺耳的骂声中，缩到墙角捂住耳朵，我躲在自己的壳中，找不到安宁的地方。

11、我不是说了不管你有什么残疾的吗！即使无法走路，无法站立，甚至无法生育，即使如此我也会和你结婚，一生都陪伴着你，在这里遇到的你，不是假的，是真正的无论在哪里遇到你，我都会喜欢上你，如果在六十亿份之一的概率下再次相遇的话，即使那时侯你的身体依旧无法动弹，我也会和你结婚。

12、海角天涯我亦义无反顾，在这里我受益匪浅，名为幸福的梦想，终有美梦成真之时。

天使计划全攻略----------------------------------勇者的分格线----------------------大结局「勇者的轨迹」数值:温柔:600智力:800毅力:300勇气:600道德:700气质:500集中力:300决断力:300运动能力:800魔法能力:700属性值-99以上(人类-天使不能接近恶魔)必要条件:看过神.魔王.艾蜜莉亚的结局(必须继承存档)/取得圣剑十字.魔剑牺牲.勇者遗落织物(不用同时玩出)---------------------------------------------天界的分格线----------------------------------------------神数值:温柔:420智力:480毅力:500勇气:500道德:500气质:480集中力:400决断力:380运动能力:500魔法能力:400属性400必要条件:天界旅行20次以上没有达到「人界之王」的条件管家询问要让儿子「当人界的国王吗?」时,选「到天界去」大天使数值:温柔:400智力:450毅力:380勇气:200道德:400气质:380集中力:390决断力:280运动能力:400魔法能力:400属性300必要条件:天界旅行20次以上没有达到「神」的条件管家询问要让儿子「当人界的国王吗?」时,选「到天界去」天使数值:温柔:350智力:430毅力:260勇气:100道德:300气质:280集中力:380决断力:180运动能力:300魔法能力:200属性200必要条件:天界旅行20次以上没有达成「大天使」的条件管家询问要让儿子「当人界的国王吗?」时,选「到天界去」---------------------------魔界的分格线---------------------------------魔王数值:智力:480毅力:500勇气:500气质:100集中力:400决断力:380运动能力:500魔法能力:600罪孽:500属性-400必要条件:魔界旅行20次以上没有达到「人界之王」的条件管家询问要让儿子「当人界的国王吗」,选「到魔界去」恶魔数值:智力:450毅力:380勇气:200气质:50集中力:390决断力:140运动能力:400魔法能力:500罪孽:400属性-300必要条件:魔界旅行20次以上没有达到「魔王」的条件管家询问要让儿子「当人界的国王吗」,选「到魔界去」魔兽数值:智力:420毅力:260勇气:100集中力:380决断力:70运动能力:300魔法能力:400罪孽:300属性-200必要条件:魔界旅行20次以上没有达到「恶魔」的条件管家询问要让儿子「当人界的国王吗」,选「到魔界去」---------------------人界的分格线-----------------------------------人界之王数值:智力:450毅力:500勇气:450道德:500气质:700决断力:430偶像魅力:500属性-99以上(人类-天使不能接近恶魔)必要条件:某界旅行20次以上(人界.天界.魔界都可以只要有20次以上就对了)没有达到「勇者的轨迹」的条件管家询问要让儿子「当人界的国王吗」,选「成为国王」仆人或家庭教师达到名手----------------闍街职业系--------------------魔法师：智力500勇气300勇气150魔法能力650魔法高级以上没有满足宫廷魔法师和教祖的条件智力390毅力200勇气150道德525气质150偶像魅力290教会达到名手没有满足大司教和宫廷魔法师的条件沙龙经营者：毅力400道德比毅力低决断力100酒保达到名手没有满足闍街老大的条件闍街老大：智力300勇气300道德99以下运动能力450亲子的羁绊49以下赌场达到名手罪孽200盗贼：(1) 狩猎成功次数最多的情况：毅力250道德比毅力低偶像魅力299以下亲子的羁绊75以下罪孽260没有满足士兵、佣兵和保镳条件(2) 武术成功次数最多的情况：智力400毅力250道德比毅力低偶像魅力299以下亲子的羁绊75以下罪孽260没有满足将军、骑士、地方领主、士兵和佣兵的条件(3) 赌场成功次数最的情况：智力400毅力250道德比毅力低偶像魅力299以下亲子的羁绊75以下罪孽260没有满足闍街老大和沙龙经营者的条件(4) 打铁店成功次数最的情况：智力400毅力250道德比毅力低偶像魅力299以下亲子的羁绊75以下罪孽260没有满足矿山经营者和士兵的条件(5) 魔法成功次数最的情况：智力400毅力250道德比毅力低偶像魅力299以下亲子的羁绊75以下罪孽260没有满足宫廷魔法师、教祖、魔法师和占卜师的条件保镳：(1) 狩猎成功次数最多的情况：智力50毅力50勇气50运动能力400罪孽50武术中级以上没有士兵和佣兵的条件(2) 武术成功次数最多的情况：智力50毅力50勇气50运动能力400罪孽50武术中级以上没有满足将军、骑士、地方领主、士兵、佣兵、盗贼的条件佣兵：(1) 武术成功次数最多的情况：智力120毅力100勇气100运动能力500武术高级以上没有满足将军、骑士、地方领主和士兵的条件(2) 狩猎成功次数最多的情况：勇气120集中力100决断力100运动能力500魔法能力50狩猎达到名手没有满足士兵的条件------------------名流结局系---------------------学校教师：智力400道德100集中力100通识高级以上没有满足贵族的女婿、宰相、法官、执政官、哲学家和博士的条件哲学家：智力500道德200集中力200通识中坚以上没有满足贵族的女婿、宰相、法官、执政官和博士的条件博士：智力600道德300集中力300通识中坚以上没有满足贵族的女婿、宰相、法官和执政官的条件法官：智力520毅力比勇气低勇气500道德490集中力比勇气低偶像魅力300通识和礼仪高级以上没有满足宰相和执政官的条件执政官：智力560毅力300勇气450道德420集中力250偶像魅力370通识和礼仪毕业没有满足宰相的条件宰相：智力600毅力350勇气400道德350集中力300偶像魅力450通识和礼仪中坚以上地方领主：智力300毅力300勇气400道德600决断力200通识和礼仪中级各执行20次以上没有满足贵族的女婿、宰相、富豪的女婿、将军和骑士的条件富豪的女婿：智力300毅力600决断力400偶像魅力300家庭教师和和仆人中坚以上没有满足贵族的女婿、宰相和沙龙经营者的条件贵族的女婿：智力420毅力500道德430气质680决断力400偶像魅力400家庭教师和和仆人资深以上没有满足宰相的条件神父：智力300毅力200道德400偶像魅力100教会打工资深以上没有满足大司教和教祖的条件大司教：智力480毅力200勇气300道德750气质300偶像魅力480教会打工达到名手将军：温柔600智力500勇气420道德520气质350偶像魅力480武术和魔法成功次数各350次以上骑士：温柔600智力500勇气400毅力300道德400运动能力700魔法能力250罪孽99以下武术和魔法成功次数各300次以上宫廷魔法师：智力620勇气500气质500魔法能力700魔法帮忙以上魔法成功次数最多注：假如发生「魔女之恋」事件，不可以选择「帮忙」，不然所有名流结局都不能达成，但有关事件不一定需要引发，仍可达成名流结局，一旦引发则不可帮忙。

Angel: Interactive Computer Graphics, Fifth Edition Chapter 1 Solutions1.1 The main advantage of the pipeline is that each primitive can be processed independently. Not only does this architecture lead to fast performance, it reduces memory requirements because we need not keep all objects available. The main disadvantage is that we cannot handle most global effects such as shadows, reflections, and blending in a physically correct manner.1.3 We derive this algorithm later in Chapter 6. First, we can form the tetrahedron by finding four equally spaced points on a unit sphere centered at the origin. One approach is to start with one point on the z axis(0, 0, 1). We then can place the other three points in a plane of constant z. One of these three points can be placed on the y axis. To satisfy the requirement that the points be equidistant, the point must be at(0, 2p2/3,−1/3). The other two can be found by symmetry to be at(−p6/3,−p2/3,−1/3) and (p6/3,−p2/3,−1/3).We can subdivide each face of the tetrahedron into four equilateral triangles by bisecting the sides and connecting the bisectors. However, the bisectors of the sides are not on the unit circle so we must push thesepoints out to the unit circle by scaling the values. We can continue this process recursively on each of the triangles created by the bisection process.1.5 In Exercise 1.4, we saw that we could intersect the line of which theline segment is part independently against each of the sides of the window. We could do this process iteratively, each time shortening the line segment if it intersects one side of the window.1.7 In a one–point perspective, two faces of the cube is parallel to the projection plane, while in a two–point perspective only the edges of the cube in one direction are parallel to the projection. In the general case of a three–point perspective there are three vanishing points and none of the edges of the cube are parallel to the projection plane.1.9 Each frame for a 480 x 640 pixel video display contains only about300k pixels whereas the 2000 x 3000 pixel movie frame has 6M pixels, or about 18 times as many as the video display. Thus, it can take 18 times asmuch time to render each frame if there is a lot of pixel-level calculations.1.11 There are single beam CRTs. One scheme is to arrange the phosphors in vertical stripes (red, green, blue, red, green, ....). The major difficulty is that the beam must change very rapidly, approximately three times as fast a each beam in a three beam system. The electronics in such a system the electronic components must also be much faster (and more expensive). Chapter 2 Solutions2.9 We can solve this problem separately in the x and y directions. The transformation is linear, that is xs = ax + b, ys = cy + d. We must maintain proportions, so that xs in the same relative position in the viewport as x is in the window, hencex − xminxmax − xmin=xs − uw,xs = u + wx − xminxmax − xmin.Likewiseys = v + hx − xminymax − ymin.2.11 Most practical tests work on a line by line basis. Usually we use scanlines, each of which corresponds to a row of pixels in the frame buffer. If we compute the intersections of the edges of the polygon with a line passing through it, these intersections can be ordered. The first intersection begins a set of points inside the polygon. The second intersection leaves the polygon, the third reenters and so on.2.13 There are two fundamental approaches: vertex lists and edge lists. With vertex lists we store the vertex locations in an array. The mesh is represented as a list of interior polygons (those polygons with no otherpolygons inside them). Each interior polygon is represented as an array of pointers into the vertex array. To draw the mesh, we traverse the list of interior polygons, drawing each polygon.One disadvantage of the vertex list is that if we wish to draw the edges inthe mesh, by rendering each polygon shared edges are drawn twice. Wecan avoid this problem by forming an edge list or edge array, each elementis a pair of pointers to vertices in the vertex array. Thus, we can draw each edge once by simply traversing the edge list. However, the simple edge list has no information on polygons and thus if we want to render the mesh in some other way such as by filling interior polygons we must add somethingto this data structure that gives information as to which edges form each polygon.A flexible mesh representation would consist of an edge list, a vertex listand a polygon list with pointers so we could know which edges belong to which polygons and which polygons share a given vertex.2.15 The Maxwell triangle corresponds to the triangle that connects thered, green, and blue vertices in the color cube.2.19 Consider the lines defined by the sides of the polygon. We can assigna direction for each of these lines by traversing the vertices in acounter-clockwise order. One very simple test is obtained by noting thatany point inside the object is on the left of each of these lines. Thus, if we substitute the point into the equation for each of the lines (ax+by+c), we should always get the same sign.2.23 There are eight vertices and thus 256 = 28 possible black/white colorings. If we remove symmetries (black/white and rotational) there are14 unique cases. See Angel, Interactive Computer Graphics (Third Edition) or the paper by Lorensen and Kline in the references.Chapter 3 Solutions3.1 The general problem is how to describe a set of characters that might have thickness, curvature, and holes (such as in the letters a and q). Suppose that we consider a simple example where each character can be approximated by a sequence of line segments. One possibility is to use a move/line system where 0 is a move and 1 a line. Then a character can be described by a sequence of the form (x0, y0, b0), (x1, y1, b1), (x2, y2, b2), .....where bi is a 0 or 1. This approach is used in the example in the OpenGL Programming Guide. A more elaborate font can be developed by using polygons instead of line segments.3.11 There are a couple of potential problems. One is that the application program can map different points in object coordinates to the same point in screen coordinates. Second, a given position on the screen when transformed back into object coordinates may lie outside the user’s window.3.19 Each scan is allocated 1/60 second. For a given scan we have to take 10% of the time for the vertical retrace which means that we start to draw scan line n at .9n/(60*1024) seconds from the beginning of the refresh. But allocating 10% of this time for the horizontal retrace we are at pixel m on this line at time .81nm/(60*1024).3.25 When the display is changing, primitives that move or are removed from the display will leave a trace or motion blur on the display as the phosphors persist. Long persistence phosphors have been used in text only displays where motion blur is less of a problem and the long persistence gives a very stable flicker-free image.Chapter 4 Solutions4.1 If the scaling matrix is uniform thenRS = RS(α, α, α) = αR = SRConsider R x(θ), if we multiply and use the standard trigonometric identities for the sine and cosine of the sum of two angles, we findR x(θ)R x(φ) = R x(θ + φ)By simply multiplying the matrices we findT(x1, y1, z1)T(x2, y2, z2) = T(x1 + x2, y1 + y2, z1 + z2)4.5 There are 12 degrees of freedom in the three–dimensional affine transformation. Consider a point p = [x, y, z, 1]T that is transformed top_ = [x_y_, z_, 1]T by the matrix M. Hence we have the relationshipp_ = Mp where M has 12 unknown coefficients but p and p_ are known. Thus we have 3 equations in 12 unknowns (the fourth equation is simplythe identity 1=1). If we have 4 such pairs of points we will have 12equations in 12 unknowns which could be solved for the elements of M.Thus if we know how a quadrilateral is transformed we can determine theaffine transformation.In two dimensions, there are 6 degrees of freedom in M but p and p_ haveonly x and y components. Hence if we know 3 points both before and after transformation, we will have 6 equations in 6 unknowns and thus in two dimensions if we know how a triangle is transformed we can determine theaffine transformation.4.7 It is easy to show by simply multiplying the matrices that theconcatenation of two rotations yields a rotation and that the concatenationof two translations yields a translation. If we look at the product of arotation and a translation, we find that the left three columns of RT arethe left three columns of R and the right column of RT is the rightcolumn of the translation matrix. If we now consider RTR_ where R_ is arotation matrix, the left three columns are exactly the same as the leftthree columns of RR_ and the and right column still has 1 as its bottomelement. Thus, the form is the same as RT with an altered rotation (whichis the concatenation of the two rotations) and an altered translation.Inductively, we can see that any further concatenations with rotations and translations do not alter this form.4.9 If we do a translation by -h we convert the problem to reflection abouta line passing through the origin. From m we can find an angle by whichwe can rotate so the line is aligned with either the x or y axis. Now reflectabout the x or y axis. Finally we undo the rotation and translation so the sequence is of the form T−1R−1SRT.4.11 The most sensible place to put the shear is second so that the instance transformation becomes I = TRHS. We can see that this order makessense if we consider a cube centered at the origin whose sides are alignedwith the axes. The scale gives us the desired size and proportions. Theshear then converts the right parallelepiped to a general parallelepiped.Finally we can orient this parallelepiped with a rotation and place it wheredesired with a translation. Note that the order I = TRSH will work too.4.13R = R z(θz)R y(θy)R x(θx) =⎡⎢⎢⎢⎣cos θy cos θz cos θz sin θx sin θy −cos θx sin θz cos θx cos θz sin θy + sin θx sin θz 0cos θy sin θz cos θx cos θz + sin θx sin θy sin θz −cos θz sin θx + cos θx sin θy sin θz 0 −sin θy cos θy sin θx cos θx cos θy 00 0 0 1⎤⎥⎥⎥⎦4.17 One test is to use the first three vertices to find the equation of theplane ax + by + cz + d = 0. Although there are four coefficients in theequation only three are independent so we can select one arbitrarily ornormalize so that a2 + b2 + c2 = 1. Then we can successively evaluateax + bc + cz + d for the other vertices. A vertex will be on the plane if weevaluate to zero. An equivalent test is to form the matrix⎡⎢⎢⎢⎣1 1 1 1x1 x2 x3 x4y1 y2 y3 y4z1 z2 z3 z4⎤⎥⎥⎥⎦for each i = 4, ... If the determinant of this matrix is zero the ith vertex isin the plane determined by the first three.4.19 Although we will have the same number of degrees of freedom in theobjects we produce, the class of objects will be very different. For exampleif we rotate a square before we apply a nonuniform scale, we will shear the square, something we cannot do if we scale then rotate.4.21 The vector a = u ×v is orthogonal to u and v. The vector b = u ×a is orthogonal to u and a. Hence, u, a and b form an orthogonal coordinatesystem.4.23 Using r = cos θ2+ sin θ2v, with θ = 90 and v = (1, 0, 0), we find forrotation about the x-axisr =√22(1, 1, 0, 0).Likewise, for rotation about the y axisr =√22(1, 0, 1, 0).4.27 Possible reasons include (1) object-oriented systems are slower, (2)users are often comfortable working in world coordinates with higher-level objects and do not need the flexibility offered by a coordinate-free approach, (3) even a system that provides scalars, vectors, and points would have to have an underlying frame to use for the implementation. Chapter 5 Solutions5.1 Eclipses (both solar and lunar) are good examples of the projection of an object (the moon or the earth) onto a nonplanar surface. Any time a shadow is created on curved surface, there is a nonplanar projection. All the maps in an atlas are examples of the use of curved projectors. If the projectors were not curved we could not project the entire surface of a spherical object (the Earth) onto a rectangle.5.3 Suppose that we want the view of the Earth rotating about the sun. Before we draw the earth, we must rotate the Earth which is a rotation about the y axis. Next we translate the Earth away from the origin. Finally we do another rotation about the y axis to position the Earth in its desired location along its orbit. There are a number of interesting variants of this problem such as the view from the Earth of the rest of the solar system.5.5 Yes. Any sequence of rotations is equivalent to a single rotation abouta suitably chosen axis. One way to compute this rotation matrix is to form the matrix by sequence of simple rotations, such asR = RxRyRz.The desired axis is an eigenvector of this matrix.5.7 The result follows from the transformation being affine. We can also take a direct approach. Consider the line determined by the points(x1, y1, z1) and (x2, y2, z2). Any point along can be written parametrically as (_x1 + (1 − _)x2, _y1 + (1 − _)y2, _z1 + (1 − _)z2). Consider the simple projection of this point 1d(_z1+(1−_)z2) (_x1 + (1 − _)x2, _y1 + (1 − _)y2)which is of the form f(_)(_x1 + (1 − _)x2, _y1 + (1 − _)y2). This form describes a line because the slope is constant. Note that the function f(_) implies that we trace out the line at a nonlinear rate as _ increases from 0 to 1.5.9 The specification used in many graphics text is of the angles the projector makes with x,z and y, z planes, i.e the angles defined by the projection of a projector by a top view and a side view.Another approach is to specify the foreshortening of one or two sides of a cube aligned with the axes.5.11 The CORE system used this approach. Retained objects were kept in distorted form. Any transformation to any object that was defined with other than an orthographic view transformed the distorted object and the orthographic projection of the transformed distorted object was incorrect.5.15 If we use _ = _ = 45, we obtain the projection matrixP =266641 0 −1 00 1 −1 00 0 0 00 0 0 1377755.17 All the points on the projection of the point (x.y, z) in the direction dx, dy, dz) are of the form (x + _dx, y + _dy, z + _dz). Thus the shadow of the point (x, y, z) is found by determining the _ for which the line intersects the plane, that isaxs + bys + czs = dSubstituting and solving, we find_ =d − ax − by − czadx + bdy + cdz.However, what we want is a projection matrix, Using this value of _ we findxs = z + _dx =x(bdy + cdx) − dx(d − by − cz)adx + bdy + cdzwith similar equations for ys and zs. These results can be computed by multiplying the homogeneous coordinate point (x, y, z, 1) by the projection matrixM =26664bdy + cdz −bdx −cdx −ddx−ady adx + cdz −cdy −ddy−adz −bdz adx + bdy −ddz0 0 0 adx + bdy + cdz37775.5.21 Suppose that the average of the two eye positions is at (x, y, z) and the viewer is looking at the origin. We could form the images using the LookAt function twice, that isgluLookAt(x-dx/2, y, z, 0, 0, 0, 0, 1, 0);/* draw scene here *//* swap buffers and clear */gluLookAt(x+dx/2, y, z, 0, 0, 0, 0, 1, 0);/* draw scene again *//* swap buffers and clear */Chapter 6 Solutions6.1 Point sources produce a very harsh lighting. Such images are characterized by abrupt transitions between light and dark. The ambient light in a real scene is dependent on both the lights on the scene and the reflectivity properties of the objects in the scene, something that cannot be computed correctly with OpenGL. The Phong reflection term is not physically correct; the reflection term in the modified Phong model is even further from being physically correct.6.3 If we were to take into account a light source being obscured by an object, we would have to have all polygons available so as to test for this condition. Such a global calculation is incompatible with the pipeline model that assumes we can shade each polygon independently of all other polygons as it flows through the pipeline.6.5 Materials absorb light from sources. Thus, a surface that appears red under white light appears so because the surface absorbs all wavelengths of light except in the red range—a subtractive process. To be compatible with such a model, we should use surface absorbtion constants that define the materials for cyan, magenta and yellow, rather than red, green and blue. 6.7 Let ψ be the angle between the normal and the halfway vector, φ be the angle between the viewer and the reflection angle, and θ be the anglebetween the normal and the light source. If all the vectors lie in the same plane, the angle between the light source and the viewer can be computer either as φ + 2θ or as 2(θ + ψ). Setting the two equal, we find φ = 2ψ. Ifthe vectors are not coplanar then φ < 2ψ.6.13 Without loss of generality, we can consider the problem in two dimensions. Suppose that the first material has a velocity of light of v1 andthe second material has a light velocity of v2. Furthermore, assume thatthe axis y = 0 separates the two materials.Place a point light source at (0, h) where h > 0 and a viewer at (x, y)where y < 0. Light will travel in a straight line from the source to a point(t, 0) where it will leave the first material and enter the second. It willthen travel from this point in a straight line to (x, y). We must find the tthat minimizes the time travelled.Using some simple trigonometry, we find the line from the source to (t, 0)has length l1 = √h2 + t2 and the line from there to the viewer has length1l2 = _y2 + (x − t)2. The total time light travels is thus l1v1 + l2v2 .Minimizing over t gives desired result when we note the two desired sinesare sin θ1 = h√h2+t2 and sin θ2 = −y √(y2+(x−t)2 .6.19 Shading requires that when we transform normals and points, we maintain the angle between them or equivalently have the dot productp ·v = p_ ·v_ when p_ = Mp and n_ = Mp. If M T M is an identity matrix angles are preserved. Such a matrix (M−1 = M T ) is called orthogonal. Rotations and translations are orthogonal but scaling and shear are not.6.21 Probably the easiest approach to this problem is to rotate the givenplane to plane z = 0 and rotate the light source and objects in the sameway. Now we have the same problem we have solved and can rotate everything back at the end.6.23 A global rendering approach would generate all shadows correctly. Ina global renderer, as each point is shaded, a calculation is done to seewhich light sources shine on it. The projection approach assumes that wecan project each polygon onto all other polygons. If the shadow of a given polygon projects onto multiple polygons, we could not compute these shadow polygons very easily. In addition, we have not accounted for thedifferent shades we might see if there were intersecting shadows from multiple light sources.Chapter 7 Solutions7.1 First, consider the problem in two dimensions. We are looking for an _ and _ such that both parametric equations yield the same point, that isx(_) = (1 − _)x1 + _x2 = (1 − _)x3 + _x4,y(_) = (1 − _)y1 + _y2 = (1 − _)y3 + _y4.These are two equations in the two unknowns _ and _ and, as long as the line segments are not parallel (a condition that will lead to a division by zero), we can solve for _ _. If both these values are between 0 and 1, the segments intersect.If the equations are in 3D, we can solve two of them for the _ and _ where x and y meet. If when we use these values of the parameters in the two equations for z, the segments intersect if we get the same z from both equations.7.3 If we clip a convex region against a convex region, we produce the intersection of the two regions, that is the set of all points in both regions, which is a convex set and describes a convex region. To see this, consider any two points in the intersection. The line segment connecting them must be in both sets and therefore the intersection is convex.7.5 See Problem 6.22. Nonuniform scaling will not preserve the angle between the normal and other vectors.7.7 Note that we could use OpenGL to, produce a hidden line removed image by using the z buffer and drawing polygons with edges and interiors the same color as the background. But of course, this method was not used in pre–raster systems.Hidden–line removal algorithms work in object space, usually with either polygons or polyhedra. Back–facing polygons can be eliminated. In general, edges are intersected with polygons to determine any visible parts. Good algorithms (see Foley or Rogers) use various coherence strategies to minimize the number of intersections.7.9 The O(k) was based upon computing the intersection of rays with the planes containing the k polygons. We did not consider the cost of filling the polygons, which can be a large part of the rendering time. If we consider a scene which is viewed from a given point there will be some percentage of 1the area of the screen that is filled with polygons. As we move the viewer closer to the objects, fewer polygons will appear on the screen but eachwill occupy a larger area on the screen, thus leaving the area of the screen that is filled approximately the same. Thus the rendering time will be about the same even though there are fewer polygons displayed.7.11 There are a number of ways we can attempt to get O(k log k) performance. One is to use a better sorting algorithm for the depth sort. Other strategies are based on divide and conquer such a binary spatial partitioning.7.13 If we consider a ray tracer that only casts rays to the first intersection and does not compute shadow rays, reflected or transmitted rays, then the image produced using a Phong model at the point of intersection will be the same image as produced by our pipeline renderer. This approach is sometimes called ray casting and is used in volume rendering and CSG. However, the data are processed in a different order from the pipeline renderer. The ray tracer works ray by ray while the pipeline renderer works object by object.7.15 Consider a circle centered at the origin: x2 + y2 = r2. If we know thata point (x, y) is on the curve than, we also know (−x, y), (x,−y),(−x,−y), (y, x), (−y, x), (y,−x), and (−y,−x) are also on the curve. This observation is known as the eight–fold symmetry of the circle. Consequently, we need only generate 1/8 of the circle, a 45 degree wedge, and can obtain the rest by copying this part using the symmetries. If we consider the 45 degree wedge starting at the bottom, the slope of this curve starts at 0 and goes to 1, precisely the conditions used for Bresenham’s line algorithm. The tests are a bit more complex and we have to account for the possibility the slope will be one but the approach is the same as for line generation.7.17 Flood fill should work with arbitrary closed areas. In practice, we can get into trouble at corners if the edges are not clearly defined. Such can be the case with scanned images.7.19 Note that if we fill by scan lines vertical edges are not a problem. Probably the best way to handle the problem is to avoid it completely by never allowing vertices to be on scan lines. OpenGL does this by havingvertices placed halfway between scan lines. Other systems jitter the y value of any vertex where it is an integer.7.21 Although each pixel uses five rays, the total number of rays has only doubled, i.e. consider a second grid that is offset one half pixel in both the x and y directions.7.23 A mathematical answer can be investigated using the notion of reconstruction of a function from its samples (see Chapter 8). However, a very easy to see by simply drawing bitmap characters that small pixels lead to very unreadable characters. A readable character should have some overlap of the pixels.7.25 We want k levels between Imin and Imax that are distributed exponentially. Then I0 = Imin, I1 = Iminr,I2 = Iminr2, ..., Ik−1 = Imax = Iminrk−1. We can solve the last equation for the desired r = ( ImaxImin)1k−17.27 If there are very few levels, we cannot display a gradual change in brightness. Instead the viewer will see steps of intensity. A simple rule of thumb is that we need enough gray levels so that a change of one step is not visible. We can mitigate the problem by adding one bit of random noise to the least significant bit of a pixel. Thus if we have 3 bits (8 levels), the third bit will be noise. The effect of the noise will be to break up regions of almost constant intensity so the user will not be able to see a step because it will be masked by the noise. In a statistical sense the jittered image is a noisy (degraded) version of the original but in a visual sense it appears better.。

galgame汉化列表本汉化列表主要针对文字占有较高比重的男性向GALgame，且以汉化补丁或中文版出售为准（汉化质量太过拙劣的机翻也不会加入列表）开头红体字是公司名，后面是游戏和汉化信息。

善用搜索，可以快速找到自己所需。

注意字体颜色的意思，黑色和橙色的大致就可以玩了。

注：颜色意义：完全汉化大部分汉化完成，基本能打通游戏部分汉化07th-Expansion《海猫鸣泣之时》EP1~6BY 六轩岛调查队…………………………100％《寒蝉鸣泣之时》解谜篇四篇完整版汉化补丁V2 BY 台湾同好＆寒冰汉化社……100％《寒蝉鸣泣之时礼》繁简汉化版BY 寒冰汉化社…………………………100％13CM《猫猫病院》白色情人节汉化补丁BY KDAYS ……………………31.25%130CM《她们的流仪（彼女たちの流仪）》双线补丁BY 小真?爱舞………………50％Active Soft《DISCIPLINE》BY神猫在线&蓝猫组共同汉化.............................................100%Alco Entertainment《情归故里》BY新天地互动多媒体.......................................................100%ALCOT《Triptych》中文补丁Ver 0.1测试版Akabeesoft2《车轮の国、向日葵の少女》汉化版v1.1...........................................100%《G线上の魔王》汉化版 v1.0 BY G弦上的魔王汉化委员..........100% 《车轮之国悠久之少年少女》v0.5汉化补丁BY 固有结界汉化组.......50%AUGUST《东月西阳》保奈美单线简体中文测试版V1.02 BY 时痕汉化组(工程停止）…………20％《更胜黎明前的琉璃色》简体中文完整版Ver1.1.1 BY 月桂琉璃汉化组…………100%《更胜黎明前的琉璃色Moonlight Cradle》1.0公测汉化补丁正式版BY …100%AXL《恋爱少女与守护之盾》简体中文体验版 Ver0.1 BY 言叶党汉化组&指尖奶茶应援会BAROQUE《魔法少年》繁体中文版 BY 台湾信必优…………100％BaseSon《恋姬无双》简中汉化5.1版 BY风之回忆工作室...........................？？《真恋姫无双》汉化补丁0.1怨念版BY WAR3ANDC.S…………………………？？BROCCOLI《银河天使队月夜情人》繁体中文版BY 台湾光谱资讯…………100％《银河天使队永恒情人》繁体中文版BY 台湾光谱资讯…………100％《魔法少女未知的天空艾塔提亚》繁体中文版BY 台湾光谱资讯…………100％《DI GI Charat Fantasy～仲夏夜之梦～》繁体中文版BY 台湾TGL&仲夏夜之梦…100％BROCCOLI&CIRCUS&GAMECRAB《真实之泪》简体中文版 BY 北京YLT…………100％繁体中文版 BY 台湾光谱资讯…………100％C's ware《夜行侦探～零》BY 新天地互动多媒体......................................................100%《夜行侦探～迷失者》 BY 新天地互动多媒体..............................................100%《美人鱼的季节》BY 新天地互动多媒体......................................................100%CARAMEL-BOX《少女爱上姐姐》简体中文汉化版V2.0 BY 瑞穗姐姐推广协会WLGO…………100％CIRCUS《D.C.P.C》Ver14.1简体中文完全版BY 月樱华想汉化组＆妹乃萌汉化组…………100％《D.C.II ~featuring Yun!~》简繁双语中文版V1.0 BY 小熊工作室&PIOVA汉化组………100％《舞-HiME 命运的系统树修罗》简体中文测试版V0.9 BY 3DM 日文游戏工作组&51HiME修罗汉化组………99％《D.C.Ⅱ地方巡演先行版「春风无敌大作战」》简体中文完整版V1.10 BY 3DM日文游戏工作组& 绯月雪桜汉化组…………100％《D.C.II～ダ?カーポII～》 V9.9汉化补丁最终公测版 BY蘑菇汉化组…………100%《C.D.C.D.2》14（单线）汉化版BY 夏之空汉化协会………………………………30%《D.C.II Spring Celebration》测试补丁0.5 BY 蘑菇汉化组...........................60%《D.C.II To You》中文化简繁完全补丁BY 神樱中文化翻译社………………100％CIRCUS&PrincessSoft《SAKURA ～雪月华～》双线汉化版BY CK-GAL中文化协会…………70%CROSSENT《AYAKASHI -魂兽-》汉化版BY CK-GAL（废坑确定）…………30％CUBE《夏ノ雨》汉化测试补丁v0.01………………………………………………1％CUBETYPE《凉宫春日的逆转1》 BY L.S.汉化组………………100％Cyc-soft《花历》 BY 天人互动............................................................100%D.O.《加奈…おかえり》v0.01b .......................................1%elf《臭作》序章汉化版（工程停止）……………………………………………………10％《下级生》中文化Ver 1.0 正式版…………………………………………100％《同级生2》中音汉化版BY 欢乐盒………………………………………………100%《同级生2 Win版》……………………………………………………100％《媚肉之香》汉化测试补丁……………………40％EMU《为你点亮的明灯》汉化测试版.............................5%etude《そして明日の世界より》完整汉化补丁BY pro2汉化组...........100％Family Soft《初恋之白色情人节》 BY 智冠...........................................100%FAVORITE《星空のメモリア -Wish upon a shooting star-》 1.0版测试补丁 BY CK-GAL中文化小组....100%《星空のメモリア Eternal Heart》汉化测试补丁 BY CK-GAL中文化小组....1%Feng《染成茜色的坂道》v0.5汉化补丁BY KDays............................................50%FlyingShine《Cross +Channal》体验版 BY 桃华月惮众.........................................100%GAINAX《美少女梦工厂》全系列1～5 简&繁中文版BY 台湾精讯资讯&北京娱乐通&台湾光谱资讯…………100％《新世纪福音战士：绫波育成计划》BY 新天地互动多媒体..................100%GAMANIA《心动情缘》简体中文版BY 天人互动………………………………100％Gene X《梦幻奇缘》 BY大宇...............................................................100%《梦幻奇缘2》 BY光谱.............................................................100%《恋爱物语》 BY Takara ...........................................................100%《恋爱物语～魔法学院》...........................................................100%《恋爱物语 2》...........................................................................100%GIGA《女仆咖啡帕露菲》简体中文完整版V1.10 BY CK-GAL…………100％《青空下的约定》简体中文测试版V1.10 BY CK-GAL…………100％《フォセット - 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人物追求攻略：------------------------------------------------------------------ 马特乌斯（11月）毅力200决断力400亲密度500以上通识成功：400次以上=最好结局300次以上=普通结局未满300次=最差结局达成结局的必要条件：看过马特乌斯事件「朋友的羁绊」【邂逅】1.通识初级以上 2.通识成功的次数10次以上【Child、Low Teen时期事件】《朋友的契机》1.执行休息 2.看过「邂逅」事件《在街上1》1.执行休息 2.看过「朋友的契机」事件3.马特乌斯的亲密度100以上备注：正确选项「一起去」智力比运动能力高or智力100以上，亲密度+50不符合以上情形，亲密度+10《在街上2》1.执行休息 2.看过「在街上1」事件 3.马特乌斯的亲密度200以上备注：正确选项「来比吧！」农场达中坚以上，亲密度+50不符合以上情形，亲密度+10《在打工处相遇》 1.执行市场或旅馆打工 2.看过「朋友的契机」事件3.市场或旅馆打工成功次数50次以上《在打工处1》 1.执行市场或旅馆打工 2.看过「在打工处相遇」事件3.市场或旅馆打工成功次数50次以上《在打工处2》 1.执行市场或旅馆打工 2.看过「在打工处1」事件3.市场或旅馆打工成功次数75次以上《在打工处3》1.执行市场或旅馆打工 2.看过「在打工处2」事件3.市场或旅馆打工成功次数100次以上《在学习处1》1.执行通识 2.看过「邂逅」事件 3.通识成功次数25次以上《在学习处2》1.执行通识 2.看过「在学习处1」事件 3.通识成功次数50次以上《在学习处3》1.执行通识 2.看过「在学习处2」事件 3.通识成功次数75次以上【Teen时期事件】《在街上1》1.执行休息 2.看过「在街上2(Child、Low Teen)」事件3.马特乌斯的亲密度300以上备注：正确选项「一起去晃晃」偶像魅力+集中力比气质的2倍还高，亲密度+50不符合以上情形，亲密度+10《在街上2》1.执行休息 2.看过「在街上1(Teen)」事件3.马特乌斯的亲密度400以上备注：正确选项「一决胜负吧」运动能力比智力高，亲密度+50不符合以上情形，亲密度+10《在打工处1》1.执行市场或旅馆打工2.看过「在打工处3(Child、Low Teen)」事件3.市场或旅馆打工成功次数125次以上《在打工处2》1.执行市场或旅馆打工 2.看过「在打工处1(Teen)」事件3.市场或旅馆打工成功次数150次以上《在打工处3》1.执行市场或旅馆打工 2.看过「在打工处2(Teen)」事件3.市场或旅馆打工成功次数175次以上《在学习处1》1.执行通识2.看过「在学习处3(Child、Low Teen)」事件3.通识成功次数100次以上《在学习处2》1.执行通识 2.看过「在学习处1(Teen)」事件3.通识成功次数125次以上《在学习处3》1.执行通识 2.看过「在学习处2(Teen)」事件3.通识成功次数150次以上《朋友的羁绊》1.马特乌斯亲密度比其他朋友平均亲密度高2.看过10个以上马特乌斯事件【无时间限制事件】《上完课之后……》1.执行通识 2.亲密度为100的倍数------------------------------------------------------------------------------------- 巴希流士（10月）毅力200决断力400亲密度500以上【勇者玩偶、特制铅笔、托球玩具、五金工具组、玩具船、望远镜】绘画成功：400次以上=最好结局300次以上=普通结局未满300次=最差结局绘画1个半月——【邂逅】休息1次——【朋友的契机】《在学习处1》——【蓝绿混合】1.绘画1次，狩猎1次 2.看过「邂逅」3.绘画成功次数25次《在学习处2》——【帮忙】 1.绘画1次，狩猎1次 2.看过「在学习处1」3.绘画成功次数50次《在学习处3》——【巴希流士】1.绘画1次，狩猎1次 2.看过「在学习处2」3.绘画成功次数75次《在打工处相遇》 1.绘画1次，狩猎1次 2.看过「朋友的契机」事件3.伐木或狩猎打工成功次数50次以上《在打工处1》——【小心一点】1.绘画1次，狩猎1次 2.看过「在打工处相遇」3.伐木或狩猎打工成功次数50次以上《在打工处2》——【把它画下来】1. 绘画1次，狩猎1次 2.看过「在打工处1」3.伐木或狩猎打工成功次数75次以上《在打工处3》——【麻烦你了】1. 绘画1次，狩猎1次 2.看过「在打工处2」3.伐木或狩猎打工成功次数100次以上音乐——偶像魅力，气质；《在街上1》1.执行休息 2.看过「朋友的契机」事件 3.巴希流士的亲密度100以上备注：正确选项「帮他选」偶像魅力+气质比运动能力的10倍高，亲密度+50不符合以上情形，亲密度+10《在街上2》1.执行休息 2.看过「在街上1」事件3.巴希流士的亲密度200以上备注：正确选项「一决胜负吧」绘画高级以上，亲密度+50不符合以上情形，亲密度+10（1245年）《在打工处1》——【就是说啊】 1. 绘画1次，狩猎1次2.看过「在打工处3(Child、Low Teen)」3.伐木或狩猎打工成功次数125次以上《在打工处2》——【拿一半过来吧】 1. 绘画1次，狩猎1次2.看过「在打工处1(Teen)」事件3.伐木或狩猎打工成功次数150次以上《在打工处3》——【环视四周】 1. 绘画1次，狩猎1次2.看过「在打工处2(Teen)」3.伐木或狩猎打工成功次数175次以上《在学习处1》——【好啊】 1. 绘画1次，狩猎1次2.看过「在学习处3(Child、Low Teen)」3.绘画成功次数100次以上《在学习处2》——【学会喜欢人】 1. 绘画1次，狩猎1次2.看过「在学习处1(Teen)」3.绘画成功次数125次以上《在学习处3》——【警告他们】 1. 绘画1次，狩猎1次2.看过「在学习处2(Teen)」3.绘画成功次数150次以上《在街上1》1.执行休息 2.看过「在街上2(Child、Low Teen)」事件3.巴希流士的亲密度300以上备注：正确选项「帮忙」武术高级以上，亲密度+50不符合以上情形，亲密度+10《在街上2》1.执行休息 2.看过「在街上1(Teen)」事件3.巴希流士的亲密度400以上*备注：正确选项「一决胜负吧」狩猎达到名手，亲密度+50不符合以上情形，亲密度+10《朋友的羁绊》1.巴希流士亲密度比其他朋友平均亲密度高2.看过10个以上巴希流士事件-------------------------------------------------------------------------------------- 汤玛士（6月）毅力200 决断力400 亲密度500以上武术成功：400次以上=最好结局300次以上=普通结局未满300次=最差结局达成结局的必要条件：看过汤玛士事件「朋友的羁绊」【邂逅】1.武术初级以上 2.武术成功的次数10次以上【Child、Low Teen时期事件】《朋友的契机》1.执行休息 2.看过「邂逅」事件《在街上1》1.执行休息 2.看过「朋友的契机」事件 3.汤玛士的亲密度100以上*备注：正确选项「帮他选」偶像魅力+气质比运动能力的10倍高，亲密度+50不符合以上情形，亲密度+10《在街上2》1.执行休息 2.看过「在街上1」事件 3.汤玛士的亲密度200以上*备注：正确选项「来比吧」伐木或狩猎资深以上，亲密度+50不符合以上情形，亲密度+10《在打工处相遇》1.执行打铁店打工 2.看过「朋友的契机」事件3.打铁店打工成功次数50次以上《在打工处1》1.执行打铁店打工 2.看过「在打工处相遇」事件3.打铁店打工成功次数50次以上《在打工处2》1.执行打铁店打工 2.看过「在打工处1」事件3.打铁店打工成功次数75次以上《在打工处3》1.执行打铁店打工 2.看过「在打工处2」事件3.打铁店打工成功次数100次以上《在学习处1》1.执行武术 2.看过「邂逅」事件 3.武术成功次数25次以上《在学习处2》1.执行武术 2.看过「在学习处1」事件 3.武术成功次数50次以上《在学习处3》1.执行武术 2.看过「在学习处2」事件 3.武术成功次数75次以上【Teen时期事件】《在街上1》1.执行休息 2.看过「在街上2(Child、Low Teen)」事件3.汤玛士的亲密度300以上*备注：正确选项「换成图画书」通识高级以下，亲密度+50不符合以上情形，亲密度+10《在街上2》1.执行休息2.看过「在街上1(Teen)」事件3.汤玛士的亲密度400以上*备注：正确选项「一决胜负吧」运动能力150以上，亲密度+50不符合以上情形，亲密度+10《在打工处1》1.执行打铁店打工 2.看过「在打工处3(Child、Low Teen)」事件3.打铁店打工成功次数125次以上《在打工处2》1.执行打铁店打工 2.看过「在打工处1(Teen)」事件3.打铁店打工成功次数150次以上《在打工处3》1.执行打铁店打工 2.看过「在打工处2(Teen)」事件3.打铁店打工成功次数175次以上《在学习处1》1.执行武术 2.看过「在学习处3(Child、Low Teen)」事件3.武术成功次数100次以上《在学习处2》1.执行武术 2.看过「在学习处1(Teen)」事件3.武术成功次数125次以上《在学习处3》1.执行武术 2.看过「在学习处2(Teen)」事件3.武术成功次数150次以上《朋友的羁绊》1.汤玛士亲密度比其他朋友平均亲密度高2.看过10个以上汤玛士事件【无时间限制事件】《上完课之后……》1.执行武术 2.亲密度为100的倍数-------------------------------------------------------------------------------------- 卡萝莉娜（1月）毅力200 智力200 决断力400 亲密度500以上【鼓、玩偶、音乐盒、化妆品组、荷叶边洋伞、香水、名曲的乐谱】餐厅&烹饪成功：400次以上=最好结局300次以上=普通结局未满300次=最差烹饪2次——【邂逅】；休息1次——【朋友的契机】；农场——温柔、运动能力；市场——智力、运动能力；生日送【荷叶边洋伞】，情人节送【天界棉花糖、天界曲奇饼】；烹饪——6个月；餐厅——7个月；农场——温柔、运动能力；市场——智力、运动能力；《在街上1》1.执行休息 2.看过「朋友的契机」事件3.卡萝莉娜的亲密度100以上*备注：正确选项「杀价」偶像魅力+智力比气质的10倍高，亲密度+50不符合以上情形，亲密度+10《在街上2》1.执行休息 2.看过「在街上1」事件 3.卡萝莉娜的亲密度200以上*备注：正确选项「帮她选」智力+温柔比气质的10倍还高，亲密度+50不符合以上情形，亲密度+10《在打工处相遇》 1.烹饪1次，餐厅1次 2.看过「朋友的契机」3.餐厅打工成功次数50次以上（2个月）《在打工处1》——【不行哦】 1.烹饪1次，餐厅1次 2.看过「在打工处相遇」3.餐厅打工成功次数50次以上（2个月）《在打工处2》——【交给我吧】 1.烹饪1次，餐厅1次 2.看过「在打工处1」3.餐厅打工成功次数75次以上（3个月）《在打工处3》——【很好吃的样】1. 烹饪1次，餐厅1次 2.看过「在打工处2」3.餐厅打工成功次数100次以上（4个月）《在学习处1》——【帮忙】 1. 烹饪1次，餐厅1次 2.看过「邂逅」事件3.烹饪成功次数25次以上（1个月）《在学习处2》——【不能打开】 1.烹饪1次，餐厅1次 2.看过「在学习处1」3.烹饪成功次数50次以上（2个月）《在学习处3》——【我不在意】 1. 烹饪1次，餐厅1次 2.看过「在学习处2」3.烹饪成功次数75次以上（3个月）《带回家里》——【一起玩吧】 1.执行休息 2.看过3个以上卡萝莉娜事件3.卡萝莉娜亲密度比其他朋友平均亲密度高（1245年）《在打工处4》——【保护她】 1. 烹饪1次，餐厅1次 2.看过「在打工处3」3.餐厅打工成功次数125次以上（5个月）《在打工处5》——【去帮她】 1. 烹饪1次，餐厅1次 2.看过「在打工处4)」3.餐厅打工成功次数150次以上（6个月）《在打工处6》——【似乎很好吃】 1. 烹饪1次，餐厅1次 2.看过「在打工处5」3.餐厅打工成功次数175次以上（7个月）《在学习处4》——【帮她修补】 1. 烹饪1次，餐厅1次 2.看过「在学习处3」3.烹饪成功次数100次以上（4个月）《在学习处5》——【帮忙她】 1. 烹饪1次，餐厅1次 2.看过「在学习处4」3.烹饪成功次数125次以上（5个月）《在学习处6》——【好啊】 1. 烹饪1次，餐厅1次 2.看过「在学习处5」3.烹饪成功次数150次以上（6个月）《在街上1》1.执行休息 2.看过「在街上2(Child、Low Teen)」事件3.卡萝莉娜的亲密度300以上*备注：正确选项「我和你一起去吧」礼仪中级以上，亲密度+50不符合以上情形，亲密度+10《在街上2》1.执行休息 2.看过「在街上1(Teen)」事件3.卡萝莉娜的亲密度400以上*备注：正确选项「去带她」智力200以上，亲密度+50不符合以上情形，亲密度+10《朋友的羁绊》——【去吃吧】 1.卡萝莉娜亲密度比其他朋友平均亲密度高2.看过10个以上卡萝莉娜事件-------------------------------------------------------------------------------------- 泰瑞莎（3月）毅力200 决断力400 亲密度500以上【玩偶、榨汁机、裁缝工具、玻璃弹珠组、天使美声糖、万能隔热手套】音乐成功：400次以上=最好结局300次以上=普通结局未满300次=最差结局音乐1个半月——【邂逅】；休息1次——【朋友的契机】；绘画1次，烹饪1次，武术6次；音乐13个月；生日送【天使美声糖、万能隔热手套】，情人节送【天使美声糖】；通识——【家庭教师】开启；家庭教师14个月；《圣诞节事件》1.看过「邂逅」事件 2.泰瑞莎的亲密度在20以上3.泰瑞莎的亲密度比卡萝莉娜及尤莉亚都高《在学习处1》——【教她】 1.执行音乐 2.看过「邂逅」3.音乐成功次数25次以上《在学习处2》——【安慰她】 1.执行音乐 2.看过「在学习处1」3.音乐成功次数50次以上《在学习处3》——【不一样呢】 1.执行音乐 2.看过「在学习处2」3.音乐成功次数75次以上《在打工处相遇》1.执行家庭教师打工 2.看过「朋友的契机」3.家庭教师打工成功次数50次以上《在打工处1》——【可以啊】 1.音乐1次，教师1次 2.看过「在打工处相遇」3.家庭教师打工成功次数50次以上《在打工处2》——【好啊】 1.音乐1次，教师1次 2.看过「在打工处1」3.家庭教师打工成功次数75次以上《在打工处3》——【吃蛋糕】 1.音乐1次，教师1次 2.看过「在打工处2」3.家庭教师打工成功次数100次以上《在街上1》1.执行休息 2.看过「朋友的契机」事件 3.泰瑞莎的亲密度100以上*备注：正确选项「可以啊」偶像魅力+气质比运动能力的10倍高，亲密度+50不符合以上情形，亲密度+10《在街上2》1.执行休息 2.看过「在街上1」事件 3.泰瑞莎的亲密度200以上*备注：正确选项「一起唱」音乐高级以上，亲密度+50不符合以上情形，亲密度+10《带回家里》——【一起来玩吧】 1.执行休息 2.看过3个以上泰瑞莎娜事件3.泰瑞莎亲密度比其他朋友平均亲密度高打铁——毅力200；礼仪——高级；烹饪/狩猎——决断力400；（1245年）《在打工处4》——【那就拜托你了】 1.音乐1次，教师1次2.看过「在打工处3」3.家庭教师打工成功次数125次以上《在打工处5》——【全都擅长】 1.音乐1次，教师1次2.看过「在打工处4」3.家庭教师打工成功次数150次以上《在打工处6》——【我们一起去道歉】1.音乐1次，教师1次2.看过「在打工处5」3.家庭教师打工成功次数175次（14个月）《在学习处4》——【就用自己的唱法】 1.执行音乐 2.看过「在学习处3」3.音乐成功次数100次以上《在学习处5》——【借她】 1.执行音乐 2.看过「在学习处4」3.音乐成功次数125次以上《在学习处6》——【让我听听看】 1.执行音乐 2.看过「在学习处5」3.音乐成功次数150次以上（13个月）《在街上1》1.执行休息 2.看过「在街上2(Child、Low Teen)」事件3.泰瑞莎的亲密度300以上*备注：正确选项「不管哪一个都很适合你」偶像魅力+气质比运动能力的10倍还高，亲密度+50不符合以上情形，亲密度+10《在街上2》1.执行休息 2.看过「在街上1(Teen)」事件3.泰瑞莎的亲密度400以上*备注：正确选项「带路」不符合以上情形，亲密度+10《朋友的羁绊》——【留下传单】1.泰瑞莎亲密度比其他朋友平均亲密度高2.看过10个以上泰瑞莎事件-------------------------------------------------------------------------------------- 尤莉亚（12月）毅力200决断力400亲密度500以上【项鍊、戒指、面具、万花筒、妖精之笛】绘画成功：400次以上=最好结局300次以上=普通结局未满300次=最差结局达成结局的必要条件：看过尤莉亚事件「朋友的羁绊」，而且没有选「丢掉箱子」魔法1个月——【邂逅】休息1次——《朋友的契机》魔法——中级酒馆——【占卜】开启生日送【面具、万花筒】《在学习处1》——【陪她一起生气】1.执行魔法 2.看过「邂逅」事件 3.魔法成功次数25次以上《在学习处2》——【好啊】1.执行魔法 2.看过「在学习处1」事件 3.魔法成功次数50次以上《在学习处3》——【抓住青蛙】1.执行魔法 2.看过「在学习处2」事件 3.魔法成功次数75次以上《在打工处相遇》1.执行占卜打工 2.看过「朋友的契机」事件3.占卜打工成功次数25次以上*占卜需透过魔女海瑞娜才可开启《在打工处1》1.执行酒馆打工 2.看过「在打工处相遇」事件3.酒馆打工成功次数50次以上《在打工处2》1.执行酒馆 2.看过「在打工处1」事件3.酒馆打工成功次数75次以上《在打工处3》1.执行酒馆打工 2.看过「在打工处2」事件3.酒馆打工成功次数100次以上《在街上1》1.执行休息 2.看过「朋友的契机」事件 3.尤莉亚的亲密度100以上*备注：正确选项「让她摸摸看」农场达到名手，亲密度+50不符合以上情形，亲密度+10《在街上2》1.执行休息 2.看过「在街上1」事件 3.尤莉亚的亲密度200以上*备注：正确选项「送她回家」尤莉亚亲密度比其他朋友平均还高，亲密度+50【Teen时期事件】《在街上1》1.执行休息 2.看过「在街上2(Child、Low Teen)」事件3.尤莉亚的亲密度300以上*备注：正确选项「将猫引过来」温柔比智力高，亲密度+50不符合以上情形，亲密度+10《在街上2》1.执行休息 2.看过「在街上1(Teen)」事件3.尤莉亚的亲密度400以上*备注：正确选项「帮她」魔法+智力比运动能力的10倍还高，亲密度+50不符合以上情形，亲密度+10《在打工处1》1.执行酒馆打工 2.看过「在打工处3(Child、Low Teen)」事件3.酒馆打工成功次数125次以上《在打工处2》1.执行酒馆打工 2.看过「在打工处1(Teen)」事件3.酒馆打工成功次数150次以上《在打工处3》1.执行酒馆打工 2.看过「在打工处2(Teen)」事件3.酒馆打工成功次数175次以上《在学习处1》1.执行魔法 2.看过「在学习处3(Child、Low Teen)」事件3.魔法成功次数100次以上《在学习处2》1.执行魔法 2.看过「在学习处1(Teen)」事件3.魔法成功次数125次以上《在学习处3》1.执行魔法 2.看过「在学习处2(Teen)」事件3.魔法成功次数150次以上《朋友的羁绊》1.尤莉亚亲密度比其他朋友平均亲密度高2.看过10个以上尤莉亚事件【无时间限制事件】《上完课之后……》1.执行魔法 2.亲密度为100的倍数《带回家里》1.执行休息 2.看过3个以上尤莉亚事件3.尤莉亚亲密度比其他朋友平均亲密度高---------------------------------------------------------------------------------------------------------------------------------------------------------------------------- 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—运动能力62、偶像魅力、罪孽；+餐厅：（要求）温柔400、气质258、决断力—智力、毅力+旅馆：（要求）温柔280、气质143、运动能力—智力、偶像魅力+仆人：（要求）智力180、气质、道德—毅力50、决断力+市场：智力150、运动能力—偶像魅力、气质+家事：温柔150、道德、运动能力—偶像魅力、罪孽、勇气43+教会：温柔130、道德—运动能力、罪孽+农场：温柔、毅力、运动能力—智力、气质、偶像魅力+伐木：毅力、集中力、运动能力—智力、气质、偶像魅力+打铁：毅力、集中力、运动能力—气质、智力-狩猎：勇气、决断力、集中力、运动能力—智力、温柔-酒馆：毅力、集中力、偶像魅力—智力、气质、道德——>赌场、占卜、酒保-赌场：智力、决断力、偶像魅力—道德-占卜：智力、勇气、魔法能力、罪孽—温柔、道德、偶像魅力、运动能力<<<（魔法中级，酒馆成功100次后开启）-酒保：毅力、集中力、魔法能力、罪孽—智力、道德、气质、偶像魅力-寻宝：勇气、毅力、运动能力—智力、气质<<<休息+教师：智力、道德、集中力—运动能力、勇气--------------------------------------勇者的分格线-------------------------------大结局「勇者的轨迹」温柔:600 智力:800 毅力:300 勇气:600 道德:700 气质:500 决断力:300 魔法能力:700 集中力:300 运动能力:800 属性-99以上属性值-99以上(人类-天使不能接近恶魔)●看过神.魔王.公主艾蜜莉亚的结局(必须继承存档)●取得圣剑十字.魔剑牺牲.勇者遗落织物(不用同时玩出)●生日选在年初，爸爸，开始用撒娇抗不良，道德到350就不用搞亲子关系烹饪1次、音乐1次、绘画1次、武术3次——【比赛】；-魔法：魔法能力700、智力、罪孽、集中力—温柔、气质、毅力；+武术：毅力400、勇气、集中力、运动能力—温柔、气质、智力；-狩猎：勇气500、决断力、集中力、运动能力—智力、温柔+通识：智力800、道德—运动能力、偶像魅力、勇气；+礼仪：道德700、气质、温柔—运动能力62、偶像魅力、罪孽；+餐厅：温柔600、气质500、决断力300—智力、毅力【圣剑十字】武术新手、打铁中坚；1244年12月：武术，认识【维托斯】，【当骑士】；1245年1月：武术1次、打铁1次，遇见【维托斯】，【家里被监视】；1246年12月：选择【调查清楚】；1247年1月：【维托斯捐款】，10月：【武斗大会】胜利；【魔剑牺牲】酒馆——拉斯特出现, 一直买东西积累点数，金钱一直维持66666以上；到达一定程度，选【买魔剑】；【勇者遗落之物】打铁——毅力300，【米迦勒会提昨晚有龙卷风经过森林】；休息——遇到【妖精菲妮】；酒馆——金额到6666元，遇见【拉斯特】，买【妖精之笛】；礼仪——【妖精之笛事件】【想要特殊道具】；【魔女之恋】武术新手，酒馆名手（1245）酒馆1次，【酒保】开启；武术1次，认识【维托斯】；酒保——【魔女契约】【当然】【恶魔戒指】；休息——【海瑞娜来访】【当然】；休息——【魔女之恋】【帮忙她】；-------------------------------------天界的分格线--------------------------------神温柔:420 智力:480毅力:500勇气:500道德:500气质:480集中力:400决断力:380运动能力:500魔法能力:400属性4005月10月比赛：扣温柔、智力，加毅力、勇气、偶像魅力黄克蓝，红克黄，蓝克红●没有达到「人界之王」的条件●生日选最后一天，爸爸，点菜减疲劳●选择游玩，让孩子维持在撒娇状态●天界旅行20次以上●9月米迦勒生日，不安排度假●圣诞节送礼物，同意米迦勒的话●1247年11月不排度假（1240）绘画1次，礼仪5次，点餐；礼仪5次，狩猎1次，点餐；——5月【绘画比赛1次】；天界旅行开启；看见赛姬（荷叶边洋伞、沙漏、戒指、天界曲奇饼、天界棉花糖）；狩猎，每3个月点餐1次；（1241）狩猎——勇气500、决断力380、集中力400、运动能力500；武术6次；（1242）打铁——毅力580；（1243）天界旅行20次；（1244）魔法<服装>——魔法能力400、智力、罪孽、集中力—毅力500；（1245）礼仪<服装/眼镜>——道德500；（1246）通识——智力540；勇气500？（1247）餐厅——温柔390；智力480？毅力500？礼仪——温柔420、气质480；（1247年11月）管家询问，「当人界的国王吗?」,选「到天界去」大天使温柔:400智力:450毅力:380道德:400气质:380勇气:200集中力:390决断力:280 运动能力:400魔法能力:400属性300●生日选年初，爸爸，点菜减疲劳●9月米迦勒生日，不安排度假●圣诞节送礼物，同意米迦勒的话●1247年11月不排度假（1240）绘画1次，礼仪5次，点餐；礼仪5次，狩猎1次，点餐；——5月【绘画比赛1次】；天界旅行开启；看见赛姬（荷叶边洋伞、沙漏、戒指、天界曲奇饼、天界棉花糖）；狩猎——勇气200、决断力280；（1241）农场——毅力380、运动能力400；魔法<服装>——魔法能力400、集中力390；毅力380？（1242）天界旅行20次；魔法<服装>——魔法能力400、集中力390；毅力380？·【武门比赛】（1243）礼仪<服装/眼镜>——道德400；（1244）通识——智力450；勇气200？（1245）礼仪——温柔400、气质380；天界旅行/休息/礼仪；（1246年12月）管家询问，「当人界的国王吗?」,选「到天界去」天使温柔:350智力:430毅力:260勇气:100道德:300气质:280集中力:380决断力:180 运动能力:300魔法能力:200属性200●生日选年初，爸爸，点菜减疲劳●9月米迦勒生日，不安排度假●圣诞节送礼物，同意米迦勒的话●1247年11月不排度假（1240）绘画1次，礼仪5次，点餐；礼仪5次，狩猎1次，点餐；——5月【绘画比赛1次】；天界旅行开启；看见赛姬（荷叶边洋伞、沙漏、戒指、天界曲奇饼、天界棉花糖）；狩猎——决断力180、勇气100；（1241）农场——毅力260、运动能力300；魔法<服装>——魔法能力200；毅力260？——10月【武门比赛】（1242）绘画——集中力380；（1243）天界旅行20次；（1244）通识——智力430；勇气100？（1245）礼仪——温柔350、气质280、道德300；运动能力300？天界旅行/休息/礼仪；（1246年12月）管家询问，「当人界的国王吗?」,选「到天界去」--------------------------------------魔界的分格线--------------------------------- 魔王气质:100决断力:380运动能力:500集中力:400毅力:500智力:480勇气:500魔法能力:600罪孽:500属性-400 魔界旅行20次5月10月比赛：扣温柔、智力，加毅力、勇气、偶像魅力黄克蓝，红克黄，蓝克红●生日选在最后一天，爸爸，点菜减疲劳●不用搭理天界和米迦勒的事，魔界旅行20次，出【圣剑和魔王】●没有达到「人界之王」的条件●2月梅菲斯特生日，不安排度假●1247年11月不排度假，管家询问（1240）1月：武术10次；6月：狩猎/魔界度假——决断力380；9月【魔界度假】开启；（1241）看见【拉斯特】（望远镜、万花筒、面具）；【赌场】【暗街】开启；武术/魔界度假——高级；9月【魔界武门比赛】；（1242）魔术<服装>/魔界度假；（1243）10月：酒馆——毅力500，【占卜】开启；（1244）10月：占卜/魔界度假；（1245）1月：打铁1次，武术1次，认识【维托斯】；2月：占卜/魔界度假——罪孽500，智力480，勇气500，魔法能力600；（1246）1月【暗街道具店】（琢磨明镜）；2月梅菲斯特生日送（琢磨明镜）；7月音乐<服装>——气质100；（1247）魔界旅行20次（圣剑）+占卜/赌场；（1247年11月）管家询问，「当人界的国王吗?」,选「到魔界去」恶魔智力:450毅力:380勇气:200气质:50集中力:390决断力:140运动能力:400魔法能力:500罪孽:400属性-300必要条件:魔界旅行20次以上没有达到「魔王」的条件管家询问要让儿子「当人界的国王吗」,选「到魔界去」。

哔哩哔哩b站会员激活自选题答案1 下列动画作品中，那一部是由京都动画公司制作的。

o A.幸运星√o B。

灼眼的夏娜o C.AngelBeats！o D。

魔法少女小圆2 C语言中，如果ar是个数组，那么表达式ar［i］与什么等价？o A.&ar+io B。

＊ar+io C。

＆（ar+i）o D.*（ar+i）√3 霹雳布袋戏中的中原战神是o A。

素还真o B。

一页书o C.羽人非獍o D.燕归人√4 交响情人梦第二季片尾曲开头是什么古典音乐？o A.菠菜罗舞曲√o .B。

菠菜舞曲o C.波列罗舞曲o D。

.菠菠舞曲5 虚渊玄常被称为什么？o A.邪恶战士o B.爱的战士√o C。

美少女战士o D。

便当店长6 3dmax 中旋转视图按哪个键o A。

space+右键o B。

alt+鼠标中键√o C.鼠标左键o D.ctrl+鼠标左键7 奥特曼来自哪个星云？o A。

m10o B.m78√o C。

m97o D.m798 由Nintendo公司发售的8位游戏机的名称缩写是什么？o A.PS3 o B. PSP o C.FC√o D.XBOX 9 下列哪个不是PSP的按键？o A.Oo B.Xo C。

△10 AB是什么动画的简称?o A。

Angel Betaso B.Angel Bookso C.Angel Beats√o D.Angry Beats11 《angel beat》中天使立华奏最喜爱的食物是？o A.哈根达斯o B.煎鸡蛋o C。

麻婆豆腐√o D。

猪骨拉面12 《APH》的旁白是谁？o A。

甲斐田阳o B。

甲斐田野o C.甲斐田雪√o D.甲斐田雨13 钢之炼金术师初版是零几年出的?o A。

10o C.09o D。

03√14 初音岛上的万年萝莉是谁o A。

森园立夏o B.芳乃樱√o C.白河小鸟o D。

朝仓音梦15 《魔法少女小圆》的导演是谁?o A。

虚渊玄o B.新房昭之√o C.新海诚o D。

论文题目: 试论埃塞尔?威尔逊小说《沼泽天使》中动物意象的意义The Significance of Animal Imagery in Ethel Wilson’s Swamp Angel论文文摘(中文):埃塞尔·威尔逊（1888-1980）是20世纪加拿大为数不多的杰出女作家之一，她一生共发表了五部小说，若干短篇小说及很多其它文学作品。

1954年发表的《沼泽天使》被誉为其最杰出之作。

《沼泽天使》以其独具特色的写作风格受到文学评论家及读者的广泛喜爱，并从多个角度对这部小说加以了解读和分析，特别是其对于自然环境及动物细致入微，生动传神的刻画。

动物意象的象征意义是文学修辞的重要手段之一。

动物意象的设计往往能给整部文学作品带来活力，能够更为生动、更为形象地刻画人物，更能自然地、有机地推动情节的发展，从而成功地传达作者试图阐释的主题及所传递的信息。

到目前为止，笔者尚未发现有评论家对《沼泽天使》中动物意象的设计和意义进行深入而详尽的分析和研究。

本文从这一角度入手，对小说中频繁出现的动物意象进行细致的分析和讨论，试图探索动物意象对人物的心理、品性、行为及形象塑造所起的作用，从而达到更准确地理解女主人公玛吉成长的心路历程，更深刻地理解小说的主题，即人类与自然的亲密一体，人与人之间有着复杂但又不可分割的关系。

论文共分五章：第一章是论文的引论及文献综述，主要说明本文的研究重点，介绍小说作者的背景，并总结已存在的以动物意象为主题的文学研究的状况，阐述动物意象作为文学修辞方法的重要意义。

第二章至第四章是论文的讨论部分，女主人公从故事的开始到结束经历了一个心灵变化的过程，我将从三个方面讨论动物意象在此过程中起到的作用。

第二章论述动物意象如何帮助体现玛吉向往在大自然中获得自由的心态；第三章论述动物意象如何有助于表现玛吉的理想生活与现实生活的矛盾；第四章论述动物意象如何展示玛吉的成长与进步，如何体现女主人公玛吉的品性特征。

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人物简介CG事件送物品事件Dream_EventFree_Event 约会正篇攻略注意事項:1.6月末會發生暑假EVENT,在之前一定要把體重降至85kg以下,否則不能成功邀約2.爲了得到全部的獎勵，要鍛煉「文系」和「腕」3.暑假時好感度盡量升為粉紅色。

所以請讓每次假日時的約會全都成功(約之前請先save一下,用SaveLoad大法)4.楓的自由事件(實行次數發生的事件)發生在「一般教養」「文系」和「部活」「Marble」因爲生日前有其他的事件，而且「Marble」的事件是次數觸發的，所以之前要購買所有的生日禮物。

5.生日,七夕,聖誕禮物都可以予先購入免得以後忘記「まりも」(送同一樣的禮物不會有所影響)6.要體育祭順利舉行,運動一定要達到20以上,否則會因為雨天而中止。

所以盡可能在夏天時把運動升到20以上。

7.友情END和戀愛END的分歧在聖誕節。

友情END恋爱END引起未發生的自由事件・如果繼續社團活動，第13次時會發生『楓Event「修行の一環」』事件・如果經常去Marble、Marble第11回次的時候會發生『楓Event「好きな音楽」』事件KEYWORD:（イタリア）（友達）（コントロール）-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------少女的恋爱革命DS 中文攻略/cat_qiqi/blog/item/9319d6136125f3876538db0f.html各人物攻略一ノ瀬蓮事件攻略：/cat_qiqi/blog/item/2dfb92cad84f4941f21fe7f0.html華原雅紀事件攻略：/cat_qiqi/blog/item/8083c2d650f9dd2807088bf2.html深水颯大事件攻略：/cat_qiqi/blog/item/23003ef48f5921def2d385fc.html橘剣之助事件攻略：/cat_qiqi/blog/item/672b4cf70ed9db25720eecff.html神城綾人事件攻略：/cat_qiqi/blog/item/d72c2e60a6bfc443ebf8f8f8.html木野村透事件攻略：/cat_qiqi/blog/item/eae733099dbbefa82eddd4f9.html若月龍太郎事件攻略：/cat_qiqi/blog/item/f04be84b9f9d30f782025cf9.html-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------時田楓分支剧情時田楓，二年级177CM。

angel在电力行业中的意思摘要：1. Angels in the Electric Power Industry2.Definition and Origin of the Term "Angel"3.Role of Angels in Electric Power Industry4.Application of Angels in Power Plants and Equipment5.Benefits and Challenges of Using Angels6.Future of Angels in the Electric Power Industry正文：Angels in the Electric Power IndustryThe term "angel" has a diverse range of meanings in different contexts.In the electric power industry, angels refer to specialized protective devices that safeguard equipment and personnel from potential hazards.This article will explore the role of angels in the electric power industry, their application in power plants and equipment, and the benefits and challenges associated with their use.Definition and Origin of the Term "Angel"The term "angel" originated from the Greek word "angelos," which means "messenger" or "envoy." In the context of the electric power industry, the term refers to a protective device or mechanism that serves as a messenger or envoy between machinery and humans,safeguarding them from harm.Role of Angels in Electric Power IndustryAngels in the electric power industry play a crucial role in maintaining safety and preventing accidents.They are typically installed on power plants, transmission lines, and distribution systems to monitor and control electrical parameters such as voltage, current, and power factor.By ensuring that these parameters remain within acceptable limits, angels help prevent overvoltage, short circuits, and other electrical faults that could cause damage to equipment or harm to personnel.Application of Angels in Power Plants and EquipmentAngels are widely used in power plants and electrical equipment to ensure optimal performance and safety.Some common applications include:1.Circuit Breakers: Circuit breakers are equipped with angels to protect electrical circuits from overload and short circuits.These angels detect异常电流conditions and automatically trip the circuit breaker to prevent damage to the system.2.Relays: Relays use angels to monitor and control electrical circuits.When a fault is detected, the angel triggers the relay to disconnect the power supply, thus safeguarding the equipment and personnel.3.Power Factor Correction Devices: Power factor correction devices improve the efficiency of electrical systems by compensating for poor power factors.Angels in these devices monitor the power factor and adjust the output accordingly to prevent damage to equipment and reduce energy losses.4.Overvoltage Protection: Angels are used in overvoltage protection devices to monitor voltage levels.When the voltage exceeds safe levels, the angel triggers the protection device to limit the voltage, thus preventing damage to equipment and protecting personnel.Benefits and Challenges of Using AngelsThe use of angels in the electric power industry offers several benefits, such as:1.Enhanced Safety: Angels help prevent accidents and damage to equipment by monitoring and controlling electrical parameters.2.Improved System Reliability: By protecting against faults and abnormal conditions, angels contribute to a more reliable electrical system.3.Reduced Maintenance Costs: Angels detect and prevent problems early on, thus reducing the need for costly repairs and replacements.However, there are also challenges associated with using angels, such as:1.High Initial Costs: Investing in angel devices can be expensive,especially for older equipment or systems with limited budgets.plexity: Angels can be complex to install, maintain, and repair, requiring specialized knowledge and training.3.False Alarms: In some cases, angels may trigger false alarms, leading to unnecessary shutdowns or disruptions in service.Future of Angels in the Electric Power IndustryAs the electric power industry continues to evolve, angels are likely to play an increasingly important role in ensuring safety and reliability.With advancements in technology, angels are becoming more sophisticated, providing better detection and protection capabilities.Furthermore, the integration of smart grid technologies and Internet of Things (IoT) devices will enable angels to communicate and coordinate with each other, improving the overall efficiency and safety of electrical systems.In conclusion, angels in the electric power industry are essential protective devices that safeguard equipment and personnel from potential hazards.Their application is wide-ranging, from circuit breakers to power factor correction devices.While they offer numerous benefits, such as enhanced safety and system reliability, they also come with challenges such as high initial costs and complexity.As the industry continues to evolve, angels are set to play an even greater role in ensuring the safe and reliable operation of electrical systems.。

3女儿成就篇*小技巧每个工作都做24天,最后一天作什么成什么~ 公主系~<王国的公主>条件一:1. 气质魅力9992. 养育费1000以上3. 体力智力斗志自尊道德体贴感受500以上4. 武勇< 体贴条件二:1. 见到王子4次以上2. 魅力800以上3. 道德体贴300以上<方法>要当公主的话就要让她上礼仪,舞蹈和教堂. 教堂少上点, 舞蹈多上点. 然后再死命的去罗丝夫人家打工(就是女佣啦), 之后罗丝夫人会介绍你的女儿去王宫打工or在街上买东西时被介绍去王宫打工(因常到王宫打工比较可以遇到王子), 若你去王宫是由王子带去只要遇到三次,如果是被罗丝夫人介绍去就要四次......到最后魅力800以上道德体贴300以上,便可成为公主.or从商人开始, 一开始就去商店买一堆洋娃娃,改变女儿(不再那么骄傲, 之后就好养多了)... 然后..每年发薪水前(10月15日)先存档,如果那一个月的薪水不满意, 就一直读档....每年就可以拿到1000多的薪水... 最后专门去学校还有舞蹈教室,打工去家庭教师.这样就很容易了..反正就是从气质还有魅力下手!!!<南国公主>1. 魅力体力气质感受9XX以上(尤其是感受更要超高)2. 十六岁的夏天穿著"南国之服"外出旅游,遇见南国王子, 并发生求婚事件. 若兔国王子同时求婚时, 南国王子则优先於兔国王子.<方法>要成为南国公主, 女儿必须在16岁夏天穿上南国之衣连续出外旅游二次以上，南国王子就会向你的女儿求婚. 而失败的结局则是南国王子会送女儿一瓶妖精之蜜.<兔国公主>1. 气质魅力感受9xx以上2. 十六岁的春天穿著"兔女郎装"外出旅游,遇见兔国王子, 并发生求婚事件.若南国王子同时求婚时, 南国王子则优先於兔国王子.<方法>女儿要在春天出游二次相遇兔兔王子,才有可能获得求婚.<地底国公主>1. 十六岁时, 气质500, 魅力700以上2. 到矿山打工10次以上3. 发生落磐事件4. 遇到鼹鼠王子两次, 共发生求婚事件<方法>到矿山工作遇到落磐, 就会有一支大鼹鼠(鼹鼠王子)跑出来对她说些...之后再作一次矿工就会向你求婚了.<神秘结局一~ 猫国公主>猫公主要魅力、气质、感受都要高! 尤其是感受!!三月时让女儿穿上猫咪装自由行动, 有可能捡到小猫来养, 在猫猫生病时照顾它, 它就会来找你, 二次后就会跟你求婚.不过体贴值好像要高点(猫咪装偶尔会有一个商人在郊游向你兜售)~ 黑暗系~<魔界公主>1. 道德02. 穿魔界的衣服3. 魅力9994. 信赖4x5. 冬天or秋天穿黑之服去旅游时, 遇到小恶魔.<方法>用流浪者养魔界公主。

必要信息：4，5月时练减肥情报，到60~70左右，在4月内去5次游戏室和5次问屋（卖点心的那家）会得到2个5W的东西，卖后就不要再打工了，省下的时间去减肥~~~对了，本人最好的减肥方法是买一个划船机和一个骑车一样的东西（名字忘了，好象18000吧？）再买下治愈乐曲集，用这个就不要吃点心了，又省下不少钱，还不加体重~~~~之后就可以用那两个东西来减肥，效果超好！美容就用柠檬吧~大约30个内，美容机就不需要了。

本人用这招，每个GG（包括以后变GG的透君）都败在本人的石榴裙下~~~~~~~~~每次都可以在1月前值全达到100（学习还是留到暑假和寒假，别忘了笔和纸？是问屋有买的那两个，忘了叫什么了。

）体重也打到45kg或者趁周末去日本一间屋买豆制曲奇点心（回复率高，卡路里少，虽然贵些但很值，试过了这个最好）回去安排行程，先吃，吃到右上角之有一点点蓝色，再去美容院减肥。

压力小了自然减的多，一次性能打下去3公斤，我都是让她这么减的。

当然贵了些，不过一开始先去游戏厅和日本一间屋各三次，店老板会给东西，那个入手就有效果，之后就完全无用了，拿去卖掉，一个有10万，前期的减肥资金就都靠它们了。

暑假的时候容易没钱，就去打工赚些，熬到9月份打工次数足够的话爸爸会送运动衣来，把那个用到只有一次的时候拿去卖掉也有10万。

不想这么减的话就攒些钱周末去伊丽莎白买个自行车，也是在压力小的时候安排运动，一次掉个1-2公斤吧？也是很有效的。

2.美容-去美容店。

就是外出后右下角的白色房子文科理科体育-可以去那位拿着鱼的男人的那家店里买什么防疲劳的手帕和笔（貌似）一次买多一点。

这样学习的时候得到的数值就会多一倍。

减肥情报-在“减肥”选项中，右下角的选项就是增加减肥情报的。

可以去看电视或用电脑。

（其实在美容院里减肥的时候也可以获得减肥情报）金手指代码金钱最大0226A3D4 0098967F美貌MAX 1226A3DA 000003E8技术MAX 1226A3DC 000003E8脚力MAX 1226A3DE 000003E8力量MAX 1226A3E0 000003E8根性MAX 1226A3E2 000003E8减肥信息MAX 1226A3E4 000003E8普通教育MAX 1226A3E6 000003E8文系MAX 1226A3E8 000003E8理数系MAX 1226A3EA 000003E8运动系MAX 1226A3EC 000003E8深水飒大生日：4月17日礼物：大型的松鼠布偶|香水/凯利熊|相框木野村透生日：5月10日礼物：画集+7（周末买）皮质钱包+5羊绒袜子3双套+5時田楓生日：6月13日礼物：球藻+7 手帕+7香水/分离 +7橘剑之助生日：6月30日礼物：柔软的枕头+7香水/武士之石+7小刀+5鷹士生日：8月9日礼物：香水/鬼魂+7柔软的枕头+7皮质的钱包+7小刀+5怀表+5恶魔词典+5游戏软件/SLG +5若月龙太郎生日：11月22日礼物：银手镯(+7)香水/黑暗岁月（+7）相框(+5)闹钟(+5)游戏软件/ACT(+5)华原雅纪生日：1月17日礼物：游戏软件/RPG +7 香水/蓝天+5 相框+5一之濑莲生日：2月4日礼物：怀表（你要是没钱的话，送“柔软枕头”也凑合）香水/巴西帝紅游戏软件/PUZ神城綾人生日：2月24日礼物：小刀|香水/库丘林|游戏碟/ADV（1）曾经的美少女（自动发生）（2）鲜明的对比（自动发生）（3）你这个无可救药的家伙！（自动发生）【档案】生日：5月10日血型：A型星座：金牛身高：似乎是169cm左右……学年：2年级学生房间号：203俱乐部：漫画研究会，还很喜欢动画，漫画，游戏爱好：写剧本或者是画素描之类的吧擅长学科：很擅长社会学！擅长运动：乒乓球的话……最重要的：啊，那，那个……喜欢的音乐：安静的气氛吧喜欢的食物：文字烧！御好烧也非常喜欢讨厌的季节：夏天……我怕热，没空调的话，或许会溶化……【礼物】*给男生的礼物最好提前买，一般休息日出门可以买到。

Angel: Interactive Computer Graphics, Fifth Edition Chapter 1 SolutionsThe main advantage of the pipeline is that each primitive can be processed independently. Not only does this architecture lead to fast performance, it reduces memory requirements because we need not keep all objects available. The main disadvantage is that we cannot handle most global effects such as shadows, reflections, and blending in a physically correct manner.We derive this algorithm later in Chapter 6. First, we can form the tetrahedron by finding four equally spaced points on a unit sphere centeredat the origin. One approach is to start with one point on the z axis (0, 0, 1). We then can place the other three points in a plane of constant z.One of these three points can be placed on the y axis. To satisfy the requirement that the points be equidistant, the point must be at(0, 2p2/3,−1/3). The other two can be found by symmetry to be at(−p6/3,−p2/3,−1/3) and (p6/3,−p2/3,−1/3).We can subdivide each face of the tetrahedron into four equilateral triangles by bisecting the sides and connecting the bisectors. However, thebisectors of the sides are not on the unit circle so we must push these points out to the unit circle by scaling the values. We can continue this process recursively on each of the triangles created by the bisection process.In Exercise , we saw that we could intersect the line of which the line segment is part independently against each of the sides of the window. We could do this process iteratively, each time shortening the line segmentif it intersects one side of the window.In a one–point perspective, two faces of the cube is parallel to the projection plane, while in a two–point perspective only the edges of the cube in one direction are parallel to the projection. In the general case of athree–point perspective there are three vanishing points and none of the edges of the cube are parallel to the projection plane.Each frame for a 480 x 640 pixel video display contains only about 300k pixels whereas the 2000 x 3000 pixel movie frame has 6M pixels, or about 18 times as many as the video display. Thus, it can take 18 times asmuch time to render each frame if there is a lot of pixel-level calculations.There are single beam CRTs. One scheme is to arrange the phosphorsin vertical stripes (red, green, blue, red, green, ....). The major difficulty isthat the beam must change very rapidly, approximately three times as fast a each beam in a three beam system. The electronics in such a system the electronic components must also be much faster (and more expensive).Chapter 2 SolutionsWe can solve this problem separately in the x and y directions. The transformation is linear, that is xs = ax + b, ys = cy + d. We must maintain proportions, so that xs in the same relative position in the viewport as x is in the window, hencex − xminxmax − xmin=xs − uw,xs = u + wx − xminxmax − xmin.Likewiseys = v + hx − xminymax − ymin.Most practical tests work on a line by line basis. Usually we usescanlines, each of which corresponds to a row of pixels in the frame buffer.If we compute the intersections of the edges of the polygon with a line passing through it, these intersections can be ordered. The first intersection begins a set of points inside the polygon. The second intersection leaves the polygon, the third reenters and so on.There are two fundamental approaches: vertex lists and edge lists. With vertex lists we store the vertex locations in an array. The mesh is represented as a list of interior polygons (those polygons with no other polygons inside them). Each interior polygon is represented as an array ofpointers into the vertex array. To draw the mesh, we traverse the list ofinterior polygons, drawing each polygon.One disadvantage of the vertex list is that if we wish to draw the edges inthe mesh, by rendering each polygon shared edges are drawn twice. We can avoid this problem by forming an edge list or edge array, each element is a pair of pointers to vertices in the vertex array. Thus, we can draw eachedge once by simply traversing the edge list. However, the simple edge listhas no information on polygons and thus if we want to render the mesh in some other way such as by filling interior polygons we must add something to this data structure that gives information as to which edges form each polygon.A flexible mesh representation would consist of an edge list, a vertex listand a polygon list with pointers so we could know which edges belong to which polygons and which polygons share a given vertex.The Maxwell triangle corresponds to the triangle that connects the red, green, and blue vertices in the color cube.Consider the lines defined by the sides of the polygon. We can assign a direction for each of these lines by traversing the vertices in a counter-clockwise order. One very simple test is obtained by noting that any point inside the object is on the left of each of these lines. Thus, if wesubstitute the point into the equation for each of the lines (ax+by+c), weshould always get the same sign.There are eight vertices and thus 256 = 28 possible black/white colorings. If we remove symmetries (black/white and rotational) there are 14 unique cases. See Angel, Interactive Computer Graphics (Third Edition) or the paper by Lorensen and Kline in the references.Chapter 3 SolutionsThe general problem is how to describe a set of characters that might have thickness, curvature, and holes (such as in the letters a and q). Suppose that we consider a simple example where each character can be approximated by a sequence of line segments. One possibility is to use amove/line system where 0 is a move and 1 a line. Then a character can be described by a sequence of the form (x0, y0, b0), (x1, y1, b1), (x2, y2, b2), .....where bi is a 0 or 1. This approach is used in the example in the OpenGL Programming Guide. A more elaborate font can be developed by using polygons instead of line segments.There are a couple of potential problems. One is that the application program can map different points in object coordinates to the same point in screen coordinates. Second, a given position on the screen when transformed back into object coordinates may lie outside the user’s window.Each scan is allocated 1/60 second. For a given scan we have to take 10% of the time for the vertical retrace which means that we start to draw scan line n at .9n/(60*1024) seconds from the beginning of the refresh. But allocating 10% of this time for the horizontal retrace we are at pixel mon this line at time .81nm/(60*1024).When the display is changing, primitives that move or are removed from the display will leave a trace or motion blur on the display as the phosphors persist. Long persistence phosphors have been used in text only displays where motion blur is less of a problem and the long persistence gives a very stable flicker-free image.Chapter 4 SolutionsIf the scaling matrix is uniform thenRS = RS(α, α, α) = αR = SRConsider R x(θ), if we multiply and use the standard trigonometric identities for the sine and cosine of the sum of two angles, we findR x(θ)R x(φ) = R x(θ + φ)By simply multiplying the matrices we findT(x1, y1, z1)T(x2, y2, z2) = T(x1 + x2, y1 + y2, z1 + z2)There are 12 degrees of freedom in the three–dimensional affine transformation. Consider a point p = [x, y, z, 1]T that is transformed top_ = [x_y_, z_, 1]T by the matrix M. Hence we have the relationshipp_ = Mp where M has 12 unknown coefficients but p and p_ are known. Thus we have 3 equations in 12 unknowns (the fourth equation is simply the identity 1=1). If we have 4 such pairs of points we will have 12 equations in 12 unknowns which could be solved for the elements of M. Thus if we know how a quadrilateral is transformed we can determine the affine transformation.In two dimensions, there are 6 degrees of freedom in M but p and p_ have only x and y components. Hence if we know 3 points both before and after transformation, we will have 6 equations in 6 unknowns and thus in two dimensions if we know how a triangle is transformed we can determine the affine transformation.It is easy to show by simply multiplying the matrices that the concatenation of two rotations yields a rotation and that the concatenationof two translations yields a translation. If we look at the product of arotation and a translation, we find that the left three columns of RT are the left three columns of R and the right column of RT is the right column of the translation matrix. If we now consider RTR_ where R_ is a rotation matrix, the left three columns are exactly the same as the left three columns of RR_ and the and right column still has 1 as its bottom element. Thus, the form is the same as RT with an altered rotation (which is the concatenation of the two rotations) and an altered translation. Inductively, we can see that any further concatenations with rotations andtranslations do not alter this form.If we do a translation by -h we convert the problem to reflection about a line passing through the origin. From m we can find an angle by which we can rotate so the line is aligned with either the x or y axis. Now reflect about the x or y axis. Finally we undo the rotation and translation so thesequence is of the form T−1R−1SRT.The most sensible place to put the shear is second so that the instance transformation becomes I = TRHS. We can see that this order makes sense if we consider a cube centered at the origin whose sides are aligned with the axes. The scale gives us the desired size and proportions. The shear then converts the right parallelepiped to a general parallelepiped. Finally we can orient this parallelepiped with a rotation and place it wheredesired with a translation. Note that the order I = TRSH will work too. R = R z(θz)R y(θy)R x(θx) =⎡⎡⎡⎡⎡cos θy cos θz cos θz sin θx sin θy −cos θx sin θz cos θx cos θz sin θy + sin θx sin θz 0cos θy sin θz cos θx cos θz + sin θx sin θy sin θz −cos θz sin θx + cos θx sin θy sin θz 0−sin θy cos θy sin θx cos θx cos θy 00 0 0 1⎡⎡⎡⎡⎡One test is to use the first three vertices to find the equation of the plane ax + by + cz + d = 0. Although there are four coefficients in the equation only three are independent so we can select one arbitrarily or normalize so that a2 + b2 + c2 = 1. Then we can successively evaluate ax + bc + cz + d for the other vertices. A vertex will be on the plane if weevaluate to zero. An equivalent test is to form the matrix⎡⎡⎡⎡⎡1 1 1 1x1 x2 x3 x4y1 y2 y3 y4z1 z2 z3 z4⎡⎡⎡⎡⎡for each i = 4, ... If the determinant of this matrix is zero the ith vertex isin the plane determined by the first three.Although we will have the same number of degrees of freedom in the objects we produce, the class of objects will be very different. For exampleif we rotate a square before we apply a nonuniform scale, we will shear thesquare, something we cannot do if we scale then rotate.The vector a = u × v is orthogonal to u and v. The vector b = u × a isorthogonal to u and a. Hence, u, a and b form an orthogonal coordinate system.Using r = cos θ2+ sin θ2v, with θ = 90 and v = (1, 0, 0), we find forrotation about the x-axisr =√22(1, 1, 0, 0).Likewise, for rotation about the y axisr =√22(1, 0, 1, 0).Possible reasons include (1) object-oriented systems are slower, (2) users are often comfortable working in world coordinates withhigher-levelobjects and do not need the flexibility offered by a coordinate-free approach, (3) even a system that provides scalars, vectors, and points would have to have an underlying frame to use for the implementation.Chapter 5 SolutionsEclipses (both solar and lunar) are good examples of the projection ofan object (the moon or the earth) onto a nonplanar surface. Any time a shadow is created on curved surface, there is a nonplanar projection. All the maps in an atlas are examples of the use of curved projectors. If the projectors were not curved we could not project the entire surface of a spherical object (the Earth) onto a rectangle.Suppose that we want the view of the Earth rotating about the sun. Before we draw the earth, we must rotate the Earth which is a rotation about the y axis. Next we translate the Earth away from the origin. Finally we do another rotation about the y axis to position the Earth in itsdesired location along its orbit. There are a number of interesting variantsof this problem such as the view from the Earth of the rest of the solar system.Yes. Any sequence of rotations is equivalent to a single rotation about a suitably chosen axis. One way to compute this rotation matrix is to form the matrix by sequence of simple rotations, such asR = RxRyRz.The desired axis is an eigenvector of this matrix.The result follows from the transformation being affine. We can also take a direct approach. Consider the line determined by the points(x1, y1, z1) and (x2, y2, z2). Any point along can be written parametricallyas (_x1 + (1 −_)x2, _y1 + (1 −_)y2, _z1 + (1 −_)z2). Consider the simple projection of this point 1d(_z1+(1−_)z2) (_x1 + (1 − _)x2, _y1 + (1 − _)y2)which is of the form f(_)(_x1 + (1 − _)x2, _y1 + (1 − _)y2). This form describes a line because the slope is constant. Note that the function f(_)implies that we trace out the line at a nonlinear rate as _ increases from 0to 1.The specification used in many graphics text is of the angles the projector makes with x,z and y, z planes, the angles defined by the projection of a projector by a top view and a side view.Another approach is to specify the foreshortening of one or two sides of acube aligned with the axes.The CORE system used this approach. Retained objects were kept indistorted form. Any transformation to any object that was defined with other than an orthographic view transformed the distorted object and the orthographic projection of the transformed distorted object was incorrect.If we use _ = _ = 45, we obtain the projection matrixP =266641 0 −1 00 1 −1 00 0 0 00 0 0 137775All the points on the projection of the point , z) in the direction dx, dy, dz) are of the form (x + _dx, y + _dy, z + _dz). Thus the shadow ofthe point (x, y, z) is found by determining the _ for which the line intersects the plane, that isaxs + bys + czs = dSubstituting and solving, we find_ =d − ax − by − czadx + bdy + cdz.However, what we want is a projection matrix, Using this value of _ we findxs = z + _dx =x(bdy + cdx) − dx(d − by − cz)adx + bdy + cdzwith similar equations for ys and zs. These results can be computed by multiplying the homogeneous coordinate point (x, y, z, 1) by the projectionmatrixM =26664bdy + cdz −bdx −cdx −ddx−ady adx + cdz −cdy −ddy−adz −bdz adx + bdy −ddz0 0 0 adx + bdy + cdz37775.Suppose that the average of the two eye positions is at (x, y, z) and the viewer is looking at the origin. We could form the images using the LookAt function twice, that isgluLookAt(x-dx/2, y, z, 0, 0, 0, 0, 1, 0);/* draw scene here *//* swap buffers and clear */gluLookAt(x+dx/2, y, z, 0, 0, 0, 0, 1, 0);/* draw scene again *//* swap buffers and clear */Chapter 6 SolutionsPoint sources produce a very harsh lighting. Such images are characterized by abrupt transitions between light and dark. The ambient light in a real scene is dependent on both the lights on the scene and thereflectivity properties of the objects in the scene, something that cannot becomputed correctly with OpenGL. The Phong reflection term is not physically correct; the reflection term in the modified Phong model is evenfurther from being physically correct.If we were to take into account a light source being obscured by an object, we would have to have all polygons available so as to test for thiscondition. Such a global calculation is incompatible with the pipeline model that assumes we can shade each polygon independently of all other polygons as it flows through the pipeline.Materials absorb light from sources. Thus, a surface that appears red under white light appears so because the surface absorbs all wavelengths oflight except in the red range—a subtractive process. To be compatible withsuch a model, we should use surface absorbtion constants that define the materials for cyan, magenta and yellow, rather than red, green and blue.Let ψ be the angle between the normal and the halfway vector, φ be the angle between the viewer and the reflection angle, and θ be the anglebetween the normal and the light source. If all the vectors lie in the sameplane, the angle between the light source and the viewer can be computer either as φ + 2θ or as 2(θ + ψ). Setting the two equal, we find φ = 2ψ. Ifthe vectors are not coplanar then φ < 2ψ.Without loss of generality, we can consider the problem in two dimensions. Suppose that the first material has a velocity of light of v1 andthe second material has a light velocity of v2. Furthermore, assume that the axis y = 0 separates the two materials.Place a point light source at (0, h) where h > 0 and a viewer at (x, y) where y < 0. Light will travel in a straight line from the source to a point(t, 0) where it will leave the first material and enter the second. It willthen travel from this point in a straight line to (x, y). We must find the tthat minimizes the time travelled.Using some simple trigonometry, we find the line from the source to (t, 0)has length l1 = √h2 + t2 and the line from there to the viewer has length 1l2 = _y2 + (x − t)2. The total time light travels is thus l1v1 + l2v2 .Minimizing over t gives desired result when we note the two desired sines are sin θ1 = h√h2+t2 and sin θ2 = −y √(y2+(x−t)2 .Shading requires that when we transform normals and points, we maintain the angle between them or equivalently have the dot productp ·v = p_ ·v_ when p_ = Mp and n_ = Mp. If M T M is an identity matrix angles are preserved. Such a matrix (M−1 = M T ) is called orthogonal. Rotations and translations are orthogonal but scaling and shear are not.Probably the easiest approach to this problem is to rotate the given plane to plane z = 0 and rotate the light source and objects in the same way. Now we have the same problem we have solved and can rotate everything back at the end.A global rendering approach would generate all shadows correctly. Ina global renderer, as each point is shaded, a calculation is done to see which light sources shine on it. The projection approach assumes that we can project each polygon onto all other polygons. If the shadow of a given polygon projects onto multiple polygons, we could not compute these shadow polygons very easily. In addition, we have not accounted for the different shades we might see if there were intersecting shadows from multiple light sources.Chapter 7 SolutionsFirst, consider the problem in two dimensions. We are looking for an _ and _ such that both parametric equations yield the same point, that is x(_) = (1 − _)x1 + _x2 = (1 − _)x3 + _x4,y(_) = (1 − _)y1 + _y2 = (1 − _)y3 + _y4.These are two equations in the two unknowns _ and _ and, as long as the line segments are not parallel (a condition that will lead to a division byzero), we can solve for _ _. If both these values are between 0 and 1, thesegments intersect.If the equations are in 3D, we can solve two of them for the _ and _ where x and y meet. If when we use these values of the parameters in the two equations for z, the segments intersect if we get the same z from both equations.If we clip a convex region against a convex region, we produce the intersection of the two regions, that is the set of all points in both regions,which is a convex set and describes a convex region. To see this, consider any two points in the intersection. The line segment connecting them must be in both sets and therefore the intersection is convex.See Problem . Nonuniform scaling will not preserve the anglebetween the normal and other vectors.Note that we could use OpenGL to, produce a hidden line removed image by using the z buffer and drawing polygons with edges and interiors the same color as the background. But of course, this method was not used in pre–raster systems.Hidden–line removal algorithms work in object space, usually with either polygons or polyhedra. Back–facing polygons can be eliminated. In general, edges are intersected with polygons to determine any visible parts.Good algorithms (see Foley or Rogers) use various coherence strategies tominimize the number of intersections.The O(k) was based upon computing the intersection of rays with the planes containing the k polygons. We did not consider the cost of filling thepolygons, which can be a large part of the rendering time. If we consider ascene which is viewed from a given point there will be some percentage of1the area of the screen that is filled with polygons. As we move the viewer closer to the objects, fewer polygons will appear on the screen but each will occupy a larger area on the screen, thus leaving the area of the screen that is filled approximately the same. Thus the rendering time will be about the same even though there are fewer polygons displayed.There are a number of ways we can attempt to get O(k log k) performance. One is to use a better sorting algorithm for the depth sort. Other strategies are based on divide and conquer such a binary spatial partitioning.If we consider a ray tracer that only casts rays to the first intersection and does not compute shadow rays, reflected or transmitted rays, then the image produced using a Phong model at the point of intersection will be the same image as produced by our pipeline renderer. This approach is sometimes called ray casting and is used in volume rendering and CSG. However, the data are processed in a different order from the pipeline renderer. The ray tracer works ray by ray while the pipeline renderer works object by object.Consider a circle centered at the origin: x2 + y2 = r2. If we know that a point (x, y) is on the curve than, we also know (−x, y), (x,−y), (−x,−y), (y, x), (−y, x), (y,−x), and (−y,−x) are also on the curve. This observation is known as the eight–fold symmetry of the circle. Consequently, we need only generate 1/8 of the circle, a 45 degree wedge, and can obtain the rest by copying this part using the symmetries. If we consider the 45 degree wedge starting at the bottom, the slope of thiscurve starts at 0 and goes to 1, precisely the conditions used for Bresenham’s line algorithm. The tests are a bit more complex and we have to account for the possibility the slope will be one but the approach is thesame as for line generation.Flood fill should work with arbitrary closed areas. In practice, we can get into trouble at corners if the edges are not clearly defined. Such can bethe case with scanned images.Note that if we fill by scan lines vertical edges are not a problem. Probably the best way to handle the problem is to avoid it completely by never allowing vertices to be on scan lines. OpenGL does this by having vertices placed halfway between scan lines. Other systems jitter the y value of any vertex where it is an integer.Although each pixel uses five rays, the total number of rays has only doubled, . consider a second grid that is offset one half pixel in both thex and y directions.A mathematical answer can be investigated using the notion of reconstruction of a function from its samples (see Chapter 8). However, avery easy to see by simply drawing bitmap characters that small pixels lead to very unreadable characters. A readable character should have some overlap of the pixels.We want k levels between Imin and Imax that are distributed exponentially. Then I0 = Imin, I1 = Iminr,I2 = Iminr2, ..., Ik−1 = Imax = Iminrk−1. We can solve the last equation forthe desired r = ( ImaxImin)1k−1If there are very few levels, we cannot display a gradual change in brightness. Instead the viewer will see steps of intensity. A simple rule ofthumb is that we need enough gray levels so that a change of one step is not visible. We can mitigate the problem by adding one bit of random noise to the least significant bit of a pixel. Thus if we have 3 bits (8 levels),the third bit will be noise. The effect of the noise will be to break up regions of almost constant intensity so the user will not be able to see astep because it will be masked by the noise. In a statistical sense the jittered image is a noisy (degraded) version of the original but in a visualsense it appears better.。