AngelCG09 Chapter 3 Better Interactive Programs
psp皇家骑士团命运之轮(新手攻略)
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SOX文件编辑-01
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。
堕落天使秘籍
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
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
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。
恋上危险校医
⌛️
许墨的承诺
• 对林峰承诺，会处理好这段感情，不会给他带来麻烦
• 让林峰相信自己，愿意为他承担一切
校医的神秘过去
林峰偶然得知许墨的过去
• 许墨曾经有过一段刻骨铭心的爱情
• 为了这段感情，许墨付出了巨大的代价
林峰的关心
• 询问许墨关于他过去的事情
• 关心许墨的伤痛和心情
许墨的倾诉
• 向林峰讲述了自己的过去
• 明白了爱情需要付出和坚持
林峰的觉醒
• 意识到爱情的力量可以战胜一切困难
• 相信自励
• 鼓励林峰勇敢面对挑战
• 告诉林峰，他会永远支持他
校医对爱情的执着与坚持
许墨对爱情的执着
• 知道自己对林峰的感情是真挚的
• 愿意为这段感情付出一切努力
许墨的坚持
• 面临困境时，依然不放弃对林峰的爱
04
命运的捉弄与现实的冲击
校医的意外受伤与主人公的担当
许墨在一次意外中受伤
• 林峰得知后，立刻赶往现场
• 为许墨提供了及时的救治
林峰的担当
• 承担起照顾许墨的责任
• 让许墨感受到了温暖和关爱
许墨的感激
• 对林峰的关心表示感谢
• 觉得自己找到了值得信赖的人
⌛️
主人公与校医的正面冲突
林峰与许墨因为过去而产生冲突
• 优秀的学生素质
主人公：林峰，一名普通大学生
• 对医学充满热情
• 性格内向，不善交际
• 对爱情抱有美好的憧憬
校医的神秘气质
许墨的医术高
超
许墨的性格独
特
许墨的背景神
秘 -很少有人
知道他的过去
01
02
交互式计算机图形学(第五版)1-7章课后题答案
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
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。
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• Change the viewing parameters so that only those primitives near the cursor are in the altered view volume改变视图参数，使得只有靠近鼠标指针的对象
• Retained mode graphics图形的记忆模式保留模式
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3.8 Picking选取
(1) • Identify a user-defined object on the display从显
示结果中识别用户定义的对象
• In principle, it should be simple because
读取缓冲区中的颜色
• The returned color gives the id of the object根据
返回的颜色确定是哪个对象
• Back buffer is not displayed后缓冲区不可见
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3.13 Logic Operations (Writing Modes像素的写入模式-1)
一的颜色(有时是在颜色索引模式中)
• We then render the scene to a color buffer other than the front buffer so the results of the rendering are not visible然后把场景输入到不是前缓
冲区的一个颜色缓冲区中，这样就不会看到显示出来 的结果
-Logic Operations逻辑运算 section 3.13 -Rubberbanding橡皮条Interactive drawing of lines and rectangles直线与矩形的交互绘制方法 -Display Lists显示列表section 3.4 Display Lists
-Cannot write temporary lines this way because we cannot recover what was “under” the line in a fast simple way用这种方法不能绘制一条临时 直线，因为我们不能用快速简单的方法恢复在临 时直线下面的内容
Better Interactive Programs 更好交互程序设计
原著Ed Angel Professor of Computer Science, Electrical and Computer Engineering, and Media Arts University of New Mexico 编辑 武汉大学计算机学院图形学课程组
位于新的视景体中 -Use gluPickMatrix (see text for details)
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第二种方法 Using Regions of the Screen应用屏幕上的区域
• Many applications use a simple rectangular arrangement of the screen在许多应用程序中，屏幕采用简单
OpenGL用下述三种模式显示内容 -GL_RENDER: normal rendering to the frame buffer (default)正常显示到帧缓冲区中(缺省模式)
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第一种方法 3.8.1 Picking and Selection Model (Rendering Modes显示输出模式-2)
application
bitwise logical operation逐位逻辑操作 ‘
frame buffer帧缓冲区
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3.13 Logic Operations (XOR write写入模式-1)
• Usual (default) mode: source replaces destination (d’ = s)缺省的写入模式是直接用 源像素取代目标像素即d’=s
Writing mode写入模式 • Copy复制模式（or replacement替换 ） • Exclusive OR异或模式 （or XOR）
• Source pixel源像素 • Destination pixel目标像素
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3.13 Logic Operations (Writing Modes像素的写入模式-2)
第一种方法
• Hit list击中(hit)列表
-Most general approach but most difficult to implement最一般性的方法，实现也最困难 -OpenGL supports this approach OpenGL支持
• 本ppt主要介绍的
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3.8 Picking选取 (Three Approaches三种方法-2)
• id is set by application program to identify objectsid是由应用程序设置，用以识别对象
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第一种方法 3.8.1 Picking and Selection Model (Using Selection Mode选择模式的应用) • Initialize name buffer初始化名称缓冲区 • Enter selection mode (using mouse)进入选择模式
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第一种方法 3.8.1 Picking and Selection Model (Selection Mode Functions选择模式中用到的函数)
• glSelectBuffer(): specifies name buffer指定名称 缓冲区 • glInitNames(): initializes name buffer初始化名称 缓冲区 • glPushName(id): push id on name buffer把id放到 名称缓冲区中 • glPopName(): pop top of name buffer把名称缓冲区 顶部的名称弹出 • glLoadName(id): replace top name on buffer取代 在名称缓冲区顶部的名称
位置，确定屏幕的区域就更简单了
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第三种方法 Using another buffer and colors for picking通过另 外的缓冲区和颜色进行选取-1
• For a small number of objects, we can assign a unique color (often in color index mode) to each object对于很少数目的对象，可以给每个对象赋以唯
(例如：应用鼠标)
• Render scene with user-defined identifiers显示输
出场景，场景中对象具有用户指定的标识 • Reenter normal render mode重新进入正常的显示 输出模式 -This operation returns number of hits该操作返回击中对 象个数 • Examine contents of name buffer (hit records)检 查名称缓冲区的内容(击中记录) -Hit records include id and depth information击中记录包 含id与深度信息
-The mouse gives the position and -We should be able to determine to which object(s) a position corresponds
• 原则上说要达到这个目标是很简单的，因为鼠标可以 提供位置信息，我们因当能够根据位置确定对应的是 哪个对象
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第三种方法 Using another buffer and colors for picking通过另 外的缓冲区和颜色进行选取-2
• We then get the mouse position and use
glReadPixels() to read the color in the buffer we just wrote at the position of the mouse然后获到鼠标位置,
-GL_FEEDBACK: provides list of primitives rendered but no output to the frame buffer提供要显示的几何列表， 但并不输出到帧缓冲区中 -GL_SELECTION: Each primitive in the view volume generates a hit record that is placed in a name stack which can be examined later在视景体中的每个几何体 创建一个击中记录，这个记录放到名称堆栈中，稍后要 被检测
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第一种方法 3.8.1 Picking and Selection Model (Selection Mode and Picking选择模式与选取)
• As we just described it, selection mode won’t work for picking because every primitive in the view volume will generate a hit在选择模式中不能进行选取
绘制对象时，利用后缓冲区或者其它缓冲区存贮对象 的标识
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第一种方法 3.8.1 Picking and Selection Model (Rendering Modes显示输出模式-1)
• OpenGL can render in one of three modes selected by glRenderMode(mode) mode)可以设置