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小学上册第十三次英语第三单元期中试卷

小学上册第十三次英语第三单元期中试卷

小学上册英语第三单元期中试卷考试时间:90分钟(总分:120)A卷一、综合题(共计100题共100分)1. 填空题:We can see __________ (山脉) from our house.2. 听力题:The capital of Solomon Islands is __________.3. 填空题:My cat loves to watch ______ (鸟) outside the window.4. 选择题:What is the name of the famous scientist known for his work on the principles of mechanics?A. Isaac NewtonB. Albert EinsteinC. Galileo GalileiD. Johannes Kepler答案: A5. compass) helps us find directions. 填空题:The ____6. 选择题:What is the smallest country in the world?A. MonacoB. Vatican CityC. San MarinoD. Liechtenstein答案:B7. 填空题:The butterfly is a symbol of ______ (美丽).What is the name of the famous American author who wrote "The Great Gatsby"?A. F. Scott FitzgeraldB. Ernest HemingwayC. Mark TwainD. John Steinbeck答案:A9. 选择题:What is the name of the main character in "The Little Mermaid"?A. ArielB. BelleC. CinderellaD. Elsa答案:A. Ariel10. 填空题:The goldfinch is a lovely ______ (鸟) that sings beautifully.11. 填空题:I can ______ (提高) my performance through practice.12. 填空题:I imagine taking my pet for long ______ in the park. We could play ______ together and enjoy the fresh air. I would love to teach it new ______ and tricks. I think it would be so much fun!13. 选择题:What do we call a series of events that happen one after another?A. SequenceB. ListC. IndexD. Collection答案:A14. 选择题:What is the main purpose of a camera?A. To take picturesB. To play musicC. To make callsD. To send messages15. 听力题:The _______ grows from a small seed.What is the term for a person who studies fossils?A. PaleontologistB. ArchaeologistC. GeologistD. Biologist答案:A17. 听力题:The boy is a good ________.18. 填空题:The ________ was a famous philosopher from ancient Greece.19. 填空题:I enjoy watching ______ on TV.20. 填空题:I have a big _______ (梦想) to travel around the world.21. 听力题:The chemical formula for ammonium hydroxide is _______.22. 选择题:What is the name of the famous detective created by Arthur Conan Doyle?A. Hercule PoirotB. Sherlock HolmesC. Miss MarpleD. Sam Spade23. 听力题:My friend plays ____ (baseball) every weekend.24. 选择题:What fruit is typically associated with keeping the doctor away?A. BananaB. OrangeC. AppleD. Grape答案:C25. 听力题:The tree is ______ in the wind. (swaying)26. 填空题:The __________ is beautiful with all the stars at night. (天空)What is the main ingredient in bread?A. SugarB. FlourC. MilkD. Water答案:B28. 填空题:The ferret has a slender ________________ (身体).29. 选择题:Which fruit is known for having seeds on the outside?A. StrawberryB. RaspberryC. BlackberryD. Blueberry答案:A30. 填空题:My dad is really into _______ (运动). 他希望能 _______ (动词).31. 听力题:The garden is ___. (growing)32. 听力题:The ____ hops around and has big, floppy ears.33. 填空题:My favorite stuffed animal is a ________.34. 填空题:The first Olympic Games were held in ancient ______ (希腊).35. 听力题:The chemical formula for ytterbium chloride is _____.36. 填空题:My ________ (玩具名称) is a superhero in disguise.37. 填空题:在古代的________ (societies) 中,农业是主要的生计方式。

小学上册第十四次英语第2单元真题(含答案)

小学上册第十四次英语第2单元真题(含答案)

小学上册英语第2单元真题(含答案)英语试题一、综合题(本题有50小题,每小题1分,共100分.每小题不选、错误,均不给分)1 Each plant has its own preferred ______ (环境).2 Which vegetable is known for making you cry?A. CarrotB. OnionC. PotatoD. Pepper答案:B3 The chemical symbol for rhodium is ______.4 The stars are _____ bright tonight. (very)5 The fruit salad is ________ (新鲜).6 I enjoy going to the ______ with my family.7 How many days are there in a week?A. FiveB. SevenC. TenD. Twelve答案:B8 The chemical formula for calcium nitrate is _____.9 The dog is __________ in the yard.10 The _______ of sound can create harmonics in musical instruments.11 The chemical formula for strontium carbonate is _____.12 I sing _____ (歌曲) in the shower.13 What is the capital of Japan?A. SeoulB. BeijingC. TokyoD. Bangkok答案:C14 They are _____ (riding) their bikes.15 We have ______ (很多) plants in our house.16 The cake has _______ (水果装饰).17 Many cultures have myths associated with ______.18 I often imagine my ________ (玩具名) coming to life and going on adventures with me!19 What do we call a young ant?A. AntlingB. LarvaC. PupD. Egg答案:B. Larva20 The fall of the Roman Empire was a turning point in ________ (历史).21 The ____ lives in trees and loves to eat fruits.22 I have a dream to learn how to _______ (技能). It seems very _______ (形容词) and useful.23 What is the capital of Saint Kitts and Nevis?A. BasseterreB. CharlestownC. KingstownD. Roseau答案: A. Basseterre24 The ________ was a key event in the development of democracy.25 Oxidation involves the loss of ______.26 The __________ was a significant event in the fight for civil rights in America. (黑人民权运动)27 What do you call a baby kangaroo?A. JoeyB. CubC. PupD. Kit28 What do we call the study of plants?A. BiologyB. BotanyC. ZoologyD. Ecology答案:B. Botany29 A thermometer measures ______ changes.30 I want to make a __________ (名词) with my __________ (玩具名).31 The ________ is strong and sturdy.32 Did you ever watch a _______ (小蛇) slither?33 __________ reactions can produce heat or light.34 I like to play ______ (games) on my tablet.35 The ________ was an era of great cultural change in Europe.36 Mountains are in ________ (北美). The Rock37 The Earth's surface changes due to natural processes like ______.38 The ________ has soft fur and gentle eyes.39 The _____ (温度) can affect how quickly seeds germinate.40 My sister enjoys __________ (学习) about animals.41 I see a _____ butterfly in the garden. (beautiful)42 My ________ (玩具名称) helps me with my social skills.43 My mom likes to go shopping for ____.44 A ____ lives in a den and is clever.45 The _____ (景观) can include many different plants.46 How many legs does an octopus have?A. SixB. EightC. FourD. Ten47 The _______ (The Civil Rights Movement) sought to end racial discrimination.48 The butterfly is ___ (flapping) its wings.49 Which month comes after January?A. MarchB. FebruaryC. April50 The boy likes to play ________.51 The ________ was a famous leader during the American Revolution.52 What do we call the time of year when it is very cold and often snows?A. SummerB. WinterC. SpringD. Fall答案:B53 What do we call a person who repairs cars?A. MechanicB. ElectricianC. PlumberD. Carpenter答案: A54 Which fruit is yellow and curved?A. BananaB. AppleC. GrapeD. Orange答案: A55 What is the opposite of 'wet'?A. DryB. CleanC. Soft56 A _______ (小斑点) can be found in many parts of the world.57 The squirrel gathers ______ for winter.58 My favorite dessert is ______ (brownies).59 I like to ride ______ with my friends. (horses)60 How many legs does a spider have?A. SixB. EightC. TenD. Four61 My aunt enjoys going to ____ (theater) shows.62 A reaction that requires high temperatures and pressures is called a ______ reaction.63 As the sun began to set, we packed our things and headed home. I felt tired but very ______ (8). Spending time in the park with my family is one of my favorite activities. I can't wait to go back again next week and make more ______ (9).64 The __________ was a famous ship that sank in 1912. (泰坦尼克号)65 I want to _______ (学会) how to surf.66 What is the name of the ocean located on the east coast of the United States?A. Atlantic OceanB. Pacific OceanC. Indian OceanD. Arctic Ocean答案: A67 The ______ (花粉) is vital for plant reproduction.68 The __________ is a major city known for its cultural significance. (伦敦)69 How many planets are in our solar system?A. 7B. 8C. 9D. 1070 The ____ enjoys basking in the sun on warm days.71 How many inches are in a foot?A. 10B. 12C. 14D. 1672 The crocodile lurks in the ________________ (水中).73 What shape has three sides?A. SquareB. TriangleC. CircleD. Rectangle答案:B74 My aunt is a ______. She loves to paint.75 What is the color of a grape?A. YellowB. PurpleC. BlueD. Green76 The chemical symbol for thallium is _______.77 A _______ (海马) is a fascinating creature to observe.78 The _____ (sun/moon) rises in the east.79 The ______ has sharp claws for hunting.80 What do we call the large landmass of ice floating on the ocean?A. GlacierB. IcebergC. Ice CapD. Snowfield答案:B81 What is the name of the largest desert in the world?A. SaharaB. GobiC. KalahariD. Arctic答案:A82 The chemical symbol for platinum is ____.83 The ________ (canyon) is carved by water.84 The ________ likes to sit on my shoulder.85 The fish is _____ around the tank. (swimming)86 The chemical formula for -hexanol is ______.87 The ____ hops quickly and loves to chase after things.88 My brother has a toy _____ that he plays with.89 I love to ______ (进行) volunteer work.90 The library is _______ (非常安静).91 A _____ (植被) cover protects the soil from erosion.92 My mom loves to collect __________ (纪念品).93 The Hubble Space Telescope helps us explore ______.94 The __________ (历史的影响) guides our actions.95 I can build anything with my __________ (玩具名).96 My _____ (表弟) is very funny.97 I like to write ______ (故事) and share them with my friends. It’s fun to come up with new characters and adventures.98 The main component of starch is ______.99 Which of these is a warm-blooded animal?A. FishB. LizardC. DogD. Frog答案:C100 The stork brings ______ (婴儿) in stories.。

小学上册Y卷英语第三单元全练全测(含答案)

小学上册Y卷英语第三单元全练全测(含答案)

小学上册英语第三单元全练全测(含答案)英语试题一、综合题(本题有100小题,每小题1分,共100分.每小题不选、错误,均不给分)1. A rabbit's ears are very ______ (长).2.She loves to ______ (explore) new places.3.I like to ______ my grandparents during holidays. (visit)4.The Earth's surface is subject to various natural ______.5.The _____ (elephant) has big ears.6. A butterfly starts as a ______.7.Chemical bonds hold atoms ______ together.8. not have ______ (腿). Snakes s9.Chemical bonds are formed due to electrostatic forces between charged _____ (particles).10.My dad loves to watch ____ (documentaries).11.An ion is an atom that has lost or gained ______.12.What do we call the study of the universe beyond Earth?A. AstrologyB. AstronomyC. GeologyD. Meteorology答案:B13.What do we call the longest river in the world?A. AmazonB. NileC. MississippiD. Yangtze14.The capital of Jamaica is ________ (金斯顿).15.I take care of my _______ (植物).16.The signing of the Magna Carta happened in _____.17. A garden has many different types of _____ (植物).18.The substance that is dissolved in a solution is called a ______.19.My grandma bakes cookies every ______ (星期六). They smell very ______ (好).20.She is _____ (happy) to see you.21.What do we call a person who studies the anatomy of the human body?A. AnatomistB. BiologistC. PhysiologistD. Pathologist答案:A22.Which of these is a vegetable?A. OrangeB. BroccoliC. GrapeD. Mango答案:B23.The anaconda is a giant ______ (蛇).24.My teacher is very __________ (专注).25.What do we call the study of fungi?A. MycologyB. BotanyC. ZoologyD. Bacteriology答案:A26.What is the color of a typical goldfish?A. SilverB. BlueC. OrangeD. Yellow答案:C27.The ____ has a colorful shell and is often seen in the garden.28.Which device do you use to call someone?A. ComputerB. PhoneC. TelevisionD. Radio答案:B29.What do we call a baby goat?A. CalfB. KidC. LambD. Pup30.What is the term for a creature that can produce its own food?A. AutotrophB. HeterotrophC. ConsumerD. Decomposer答案:A31.The chemical structure of a compound can be represented by a ______ formula.32.I saw a _______ (小狐狸) peeking out from behind a bush.33.What is the name of the famous ancient city in Iraq?A. BabylonB. UrukC. NinevehD. All of the above34.What do we call the study of the Earth's physical features?A. GeographyB. GeologyC. CartographyD. Meteorology答案:A Geography35.The director, ______ (导演), makes movies.36. A solvent is the substance that dissolves a ______.37.What do you use to tell time?A. ClockB. CalendarC. PhoneD. Watch答案:A38.What do you call the longest river in the world?A. AmazonB. NileC. MississippiD. Yangtze39.What is the largest mammal in the ocean?A. SharkB. WhaleC. DolphinD. Seal答案:B40.The ice is ______ in the glass. (cold)41.hill) is a raised area of land, smaller than a mountain. The ____42.The dog digs a hole in the ______ (沙子).43.In a battery, chemical energy is converted into ________ energy.44.My friend has a new ________ (拼图). It has 100 pieces and is very ________ (挑战).45.What is the name of the structure that protects your foot?A. ShoeB. SockC. SandalD. Boot答案:A46.Water is made up of _____ and oxygen.47.What type of animal is a shark?A. MammalB. ReptileC. FishD. Amphibian答案:C48.The kitten loves to chase a ______.49.The chemical formula for phosphoric acid is _____.50.What do we call the study of the Earth's physical features?A. GeographyB. GeologyC. CartographyD. Meteorology51.The first successful composite transplant was performed in ________.52.I have a ________ (日记) to write in.53. A frog can leap from one place to another very ______ (远).54.What do we call a group of sheep?A. FlockB. HerdC. PackD. School55.I like to _______ puzzles on rainy days.56.My brother collects ____ (coins) from around the world.57.I can ______ (为) my family meals.58.What do you call a baby cat?A. PuppyB. KittenC. CubD. Foal59.The study of how landscapes evolve is crucial for ______ planning.60.What is the name of the main character in "The Little Mermaid"?A. ArielB. BelleC. CinderellaD. Elsa答案:A Ariel61.My mom is ______ (cooking) dinner.62.What is the process of water vapor turning into liquid called?A. EvaporationB. PrecipitationC. CondensationD. Sublimation答案:C63.The ________ (生态影响评估工具) assists in decision-making.64.What do you call a story made up in your imagination?A. RealityB. FictionC. FactD. History65.My sister, ______ (我妹妹), loves to play with dolls.66.What is the capital of Finland?A. HelsinkiB. OsloC. StockholmD. Reykjavik答案:A67. A ____(survey study) gathers data on community needs.68.How many wheels does a bicycle have?A. 1B. 2C. 3D. 4答案:B 269.The ______ is an organ that helps us see.70.What is the name of the famous artist known for his colorful paintings?A. Claude MonetB. Vincent van GoghC. Pablo PicassoD. All of the above71.What is the name of our planet?A. MarsB. VenusC. EarthD. Jupiter答案:C72.The rain makes the ground _______.73.What is the color of grass?A. BlueB. YellowC. GreenD. Red74.This is my favorite _______ (玩具).75.What do you call the sound a dog makes?A. MeowB. RoarC. BarkD. Tweet76.Which animal is known for its strong sense of smell?A. CatB. DogC. HamsterD. Goldfish答案:B77.What do you call a baby horse?A. CalfB. PuppyC. FoalD. Kitten答案:C78. A chemical change involves the formation of new _____.79.The chemical formula for ammonia is ________.80. A ________ (青蛙) jumps from lily pad to lily pad in the pond.81. A __________ is a geological feature that can affect local ecosystems and human activities.82.The capital of the Maldives is ________ (马累).83.What do you wear on your feet?A. HatB. ShoesC. GlovesD. Belt答案:B84.What is the term for a human's outer covering?A. SkinB. MuscleC. BoneD. Fat答案:A85.The dog is _______ (wagging) its tail.86.Many insects live on _____ (花).87.The ________ was a famous explorer who sailed around the world.88.The __________ (历史的推动力) shapes progress.89. A _____ (园艺工具) makes planting easier.90.We go to the ______ (电影院) on weekends.91. A lever helps us lift _______ easily.92.The kitten is ___. (cute)93.What is the name of the famous landmark in Paris?A. Big BenB. ColosseumC. Eiffel TowerD. Statue of Liberty94.The ____ has large eyes and can see well at night.95.The ____ is often seen in parks searching for food.96.The starfish lives in the _________. (海洋)97.The flamingo's color comes from its _________ (饮食).98.My grandma loves to knit ____ (blankets).99.What do we wear on our heads?A. ShoesB. GlovesC. HatD. Scarf答案:C100.The _____ (music/dance) is fun.。

小学上册第十三次英语第2单元真题试卷(有答案)

小学上册第十三次英语第2单元真题试卷(有答案)

小学上册英语第2单元真题试卷(有答案)英语试题一、综合题(本题有100小题,每小题1分,共100分.每小题不选、错误,均不给分)1.The __________ (人类历史) is full of incredible stories.2.I have _____ (one/two) pet cats.3.My cousin is very good at ____ (science).4.What do you call a person who studies ancient civilizations?A. HistorianB. ArchaeologistC. GeographerD. Anthropologist答案:B5.The _______ of a magnet can be demonstrated with iron filings.6.Did you see the _____ (兔子) hopping around?7.I want to ___ my room. (clean)8.Metals are typically ______ conductors of heat.9.We are ___ to school. (walking)10.__________ are important for the ecosystem because they pollinate flowers.11.Which bird is known for its colorful feathers and ability to mimic sounds?A. SparrowB. ParrotC. EagleD. Owl答案: B12.The ________ comes in many shapes and sizes.13. A __________ is a geological feature that can attract tourists.14.The dog is ________ in the park.15.We visit the ______ (艺术家工作室) to see creations.16. A cat's whiskers help it sense ______ (环境).17. A _______ is a tool that can help to measure the pressure of gases.18.The chemical formula for sulfuric acid is ______.19.Matter can exist in different states, including solid, liquid, and ________.20.Newton's laws describe the relationship between _______ and motion.21.What is the capital of Comoros?A. MoroniB. MoutsamoudouC. DomoniD. Mitsamiouli答案: A22.What do we call the liquid that falls from the sky?A. SnowB. RainC. HailD. Sleet答案:B.Rain23.What is the name of the famous painting of a woman with a mysterious smile?A. Starry NightB. The ScreamC. Mona LisaD. The Last Supper答案:C.Mona Lisa24.The ________ (人工栽培) supports agriculture.25.They are going to ________ a movie.26.My favorite subject in school is ______.27.The chemical formula for iron (II) sulfate is _______.28.I have a ________ (拼图游戏) of famous buildings.29.flora) of a region consists of its plant life. The ____30.What do we call the force that pulls objects toward each other?A. FrictionB. GravityC. MagnetismD. Inertia答案:B31.What is the name of the fairy that helps Peter Pan?A. Tinker BellB. CinderellaC. BelleD. Ariel答案: A32.Which planet do we live on?A. MarsB. EarthC. JupiterD. Saturn答案:B33.I like to eat ________ for breakfast.34.The __________ is known for its significant wildlife population. (非洲)35.I am excited to join the school ________ (合唱团) and sing with my friends.36.The pig loves to play in the ______.37.My friend loves __________ (户外活动).38.My teacher gives us ______ (作业) to help us learn new things. I try to finish it as soon as possible.39. A __________ is a natural feature formed by erosion.40.My uncle is a fantastic ____ (storyteller).41.__________ are substances that can donate protons in a reaction.42.What do you call the main part of a plant that supports it?A. RootB. StemC. LeafD. Flower答案:B43.The __________ is known for its vibrant nightlife.44.We should plant more ______ (树木).45. A shooting star is actually a ______ entering the Earth's atmosphere.46.Chemical reactions can be affected by _____, concentration, and surface area.47.It’s fun to build a ______ (雪堡) in the snow.48.What is the time when the sun sets called?A. DawnB. NoonC. DuskD. Midnight答案: C49.What is 60 ÷ 3?A. 15B. 20C. 25D. 30答案:b50.What is the name of the famous ancient city in Syria?A. PalmyraB. AleppoC. DamascusD. Homs答案:A.Palmyra51. A ______ (湿润的气候) supports lush vegetation.52.The _______ (小飞鱼) jumps out of the water to escape predators.53.We have a ______ (丰富的) menu for lunch.54.The ancient Romans used _____ to build their structures.55.I find joy in small ______ (事物) like a sunny day or a nice cup of hot chocolate. Life is full of little pleasures.56.The city of Helsinki is the capital of _______.57.I can ______ (提升) my creativity through art.58. A rainbow appears when light passes through _______.59.In spring, the _____ blooms beautifully.60.The wolf howls at the ________________ (月亮).61.What do we call the outer layer of the Earth?A. CoreB. MantleC. CrustD. Inner core答案:C62.The first person to receive a Nobel Prize was ______ (居里夫人).63.I like to watch _____ on TV. (cartoons)64.The ability of a solution to resist changes in pH is known as _____.65. A _______ can be used to measure the intensity of light.66.My sister enjoys practicing ____ (martial arts).67.What is the capital of Afghanistan?A. KabulB. KandaharC. HeratD. Mazar-i-Sharif答案: A68.The __________ is a large area of deep water.69. A rabbit's ears help it detect ______ (危险).70.The process of changing from liquid to gas is called ______.71.What do we call a person who repairs cars?A. MechanicB. DriverC. PilotD. Engineer答案:A72.Creating a ______ (家庭花园) can yield fresh produce.73. A _______ helps to measure the amount of pressure exerted on a surface.74.The ancient Greeks developed ________ to teach philosophy.75.What is the name of the famous scientist who proposed the theory of evolution?A. Isaac NewtonB. Albert EinsteinC. Charles DarwinD. Louis Pasteur答案:C76.sustainability report) evaluates progress. The ____77.I love to cook ______ (美味的食物) for my family. It’s a way to show my love and care.78.The Milky Way galaxy is a barred spiral _______.79.They are ___ a movie. (watching)80.What do we call a story that is made up?A. NonfictionB. BiographyC. FictionD. History答案: C81.The ______ (植物的演变) is an important topic in biology.82. A ______ (花园) can be a great hobby.83.What is the name of the famous Italian city known for its canals?A. VeniceB. RomeC. FlorenceD. Milan答案:A84.Which animal is known for its speed?A. TortoiseB. CheetahC. ElephantD. Sloth答案:B.Cheetah85.The _______ (The Scientific Method) transformed science and experimentation.86. A diatomic molecule consists of ______ atoms.87. A __________ is a landform created by the deposition of sediment.88.What is the capital of Barbados?A. BridgetownB. NassauC. KingstonD. Port au Prince答案: A89.The process of a liquid turning into gas is called _______.90.The first modern Olympics were held in ________ (1896).91.The __________ is a famous landmark in the United States. (自由女神像)92. A bison is a large, grazing ________________ (动物).93.My friend is always the first to __________ (帮助) others.94.My ________ (表哥) is studying hard for his exams.95.The chemical formula for potassium nitrate is ______.96.I saw a _______ (蝴蝶) land on a flower.97.The ____ has a shiny shell and is often found in gardens.98.The _______ of a battery is measured in volts.99.My sister is a good __________. (学生)100.What do we call a baby llama?A. CriaB. KidC. CalfD. Foal答案: A。

小学上册第十四次英语第六单元自测题(含答案)

小学上册第十四次英语第六单元自测题(含答案)

小学上册英语第六单元自测题(含答案)英语试题一、综合题(本题有100小题,每小题1分,共100分.每小题不选、错误,均不给分)1.The chemical name for water is __________.2.I like to collect ________.3.The restaurant serves _____ (breakfast/lunch).4. A solute is the substance that gets ______ in a solution.5.My friend is a ______. He enjoys creating music.6.The ancient Greeks practiced democracy in _____.7.My brother is a big __________ of golf. (粉丝)8. A mole is a unit that measures the amount of ______.9.What is the opposite of "full"?A. EmptyB. PackedC. LargeD. Heavy答案: A10.My favorite activity at school is _______ (科学实验).11.He is good at ________ soccer.12.My mom loves to teach me __________ (生活技能).13.The __________ (历史的情感联系) draw us together.14.Which gas do plants take in during photosynthesis?A. OxygenB. NitrogenC. Carbon DioxideD. Hydrogen答案:C15.What is the capital of Thailand?A. BangkokB. PhuketC. Chiang MaiD. Pattaya答案: A16.What do we use to see underwater?A. BinocularsB. TelescopeC. GogglesD. Glasses答案: C17.The _______ helps to keep the soil healthy.18.The ________ (生态系统服务) supports life.19.Many _______ require specific care.20.I am excited to learn _______ (新事物) this year. School is always full of _______ (惊喜).21.My favorite outdoor activity i s __________ because it’s adventurous.22.__________ are used in batteries to store electrical energy.23.What do you call a vehicle that travels on tracks?A. CarB. TrainC. PlaneD. Boat答案: B24.I have a ______ robot that talks.25.阅读短文,选择正确答案。

Cayley定理的证明方法

Cayley定理的证明方法

Cayley 定理的证明方法摘要:本文对Cayley 定理:n K 的生成树共有2n n -棵,即2()n n K n τ-=。

的几种证明方法简单归纳。

关键词:Cayley 公式 标号树枝 生成树第一种证明方法通过确定标号树枝的个数来求生成树的个数,设生成树的数目为x 个,因为每个生成树的每一个点都能作为一个根,所以标号树枝的个数为nx 个,现在就是确定标号树枝的个数1n n -,这样一来就能确定2n x n -=。

下面我们就来证明标号树枝的个数为1n n -。

通过一步一步建立标号树枝,先拿出n 个点的无边土,此时这个图有n 个树枝森林,,现在往上加边,加第一条边后,树枝森林数减少一个,,当树枝数目为k 时,加下一条边新边(,)u v 的选择为(1)n k -,任意一个点都能当作u ,而v 必须连接不含u 的树枝的根,用这种方法构造标号树枝的数目应该为111()(1)!n n i n n i nn --=-=-∏,因为每个标号树枝含有1n -条边,有(1)!n -种顺序,也就是说每个标号树枝被构造了(1)!n -次,所以标号树枝的个数为1n n -。

证毕。

第二种证明方法设2n ≥,12,,,nd d d 是正整数,并且1222n d d d n +++=-,则在顶点集{1,2,,}n 上具有顶点度序列为12,,,n d d d 的树的个数是多项式展开如下:1121111,,,11(1)(1)2211n n n d d n d d d n d d n n a a d d --≥-++-=--⎛⎫=⎪--⎝⎭∑特别地,令每一个1i a =,得到为了计算顶点集{1,2,,}n 上的树的数目,必须将12,,,n d d d 是正整数并且其和等于2n -的具有顶点度序列12,,,n d d d 的所有树的数目全部加在一起. 从前面的事实有顶点集{1,2,,}n 上的树的数目121,,,122n n d d d d d n ≥++=-=∑(顶点集{1,2,,}n 上的具有给定的度序列12,,,n d d d 树的数目)于是Cayley 公式得到证明. 第三种证明方法 构建概率模型在公理化体系,一般用三元总体(,,)F p Ω表示一概率空间,为了证明Cayley 定理,令Ω={n K 的生成树全体},k S ={顶点0v 的度为k 的生成树全体},k=1,2,…,n-1.显然i S ∩j S =Φ(i j ≠),且11n k k S -==Ω⋃;记()k k P P S =,令kk S P =Ω,k=1,2,… ,n-1.其中.表示集合中的元素个数.易知,上述定义下的(,,)F p Ω成为一概率空间.若Cayley 定理及Clarke 定理成立, 则在上面定义的概率空间(,,)F p Ω中, 分别成立下 列等式:2n n-Ω=, ()1211n kk n S n k ---⎛⎫=- ⎪-⎝⎭, k=1,2, … n-1. 为求Ω,k S ,建立如下概率模型.假设n 阶完全图n K 的生成树按下述要求生成:向n 阶完全图n K 的n 个顶点里随机地投(n-1)个点,每次投且仅投一点,共投(n-1)次.点落在每个顶点里的可能性是相等的.每投一次(即一次试验),点就落在某一顶点中,从这里就会且只会长出一条新边,且与原来的边不重复;如果点落入其他顶点中,则长出的新边总是向没有点落进的顶点方向生长,且不成圈.按上述方法,投(n-1)个点(即做n-1次试验)后,一定可以得到一个n 阶完全图n K 的生成树.定理证明下面,将在以上建立的概率模型中,用古典模型的方法求Ω及k S 。

cayley

Cayley graphs and coset diagrams1IntroductionLet G be afinite group,and X a subset of G.The Cayley graph of G with respect to X,written Cay(G,X)has two different definitions in the literature.The vertex set of this graph is the group G.In one definition,there is an arc from g to xg for all g∈G and x∈X;in the other definition,for the same pairs(g,x),there is an arc from g to gx.Cayley graphs are generalised by coset graphs.For these we take a subgroup H of G as part of the data.Now the vertices of the graph may be either the left cosets or the right cosets of H in G;and an arc may join the coset containing g to the coset containing xg,or to the coset containing gx.Thus,there are four possible types of coset graphs.The two types of Cayley graph are in a certain sense equivalent,while the four types of coset graph fall into two distinct types:one type encompasses all vertex-transitive graphs,while the other is seldom vertex-transitive but has more specialised uses in the theory of regular maps.The purpose of this essay is to explain where the definitions come from and what purposes they serve.2Group actionsAn action of a group G on a setΩis a homomorphism from G to the symmetric group onΩ.This simply means that to each group element g is associated a per-mutationπG ofΩ,and composition of group elements corresponds to compositionof permutations:πg1g2=πg1◦πg2,where◦denotes composition.However,there are two different conventions for composing permutations. These arise from different ways of representing a permutationπ:the image of a point x under the permutationπmay be denoted byπ(x)or by xπ(or xπ).In the first case,we say the permutation acts on the left,in the second case on the right.Ifπ1andπ2are permutations,thenπ1◦π2may mean“applyfirstπ1,thenπ2”, or it may mean“applyfirstπ2,thenπ1”.Now the definition we choose of◦is closely connected with the way we choose to write permutations.If permutations act on the left,then we would like to have(π1◦π2)(x)=π1(π2(x)),The Encyclopaedia of Design Theory Cayley graphs and coset diagrams/1that is,“firstπ2,thenπ1”;while if they act on the right,we would preferx(π1◦π2)=(xπ1)π2or xπ1◦π2=(xπ1)π2.When there is a choice,I will use the right action,since the definition of com-position under the right action is so much more natural.I will simplify the notation by writing simply xg for the image of x under the permutationπg.The rule for an action becomes simply x(gh)=(xg)h.Now suppose that G acts(on the right)on a setΩ.The action is transitive if we can move from any point ofΩto any other by some permutation induced by G;that is,for anyα,β∈Ωthere is an element g∈G such thatαg=β.Two actions of G,on setsΩ1andΩ2,are said to be isomorphic if there is a bijectionφ:Ω1→Ω2(which I shall also write on the right)such that(αφ)g= (αg)φfor allα∈Ω1and g∈G.In other words,up to renaming the points of the set,the actions are identical.Now let H be a subgroup of G.The right cosets of H are the sets Hx={hx: h∈H},and the left cosets are the sets xH={xh:h∈H},as x runs through G. Two cosets of the same type are either equal or disjoint;two elements x1,x2lie in the same right coset if and only if x1x−12∈H,and lie in the same left coset if and only if x−12x1∈H.So the cosets of each type form a partition of G.The subgroup H is normal in G if and only if the partitions into left and right cosets coincide. It is worth noting that there is a natural bijection between the right cosets and the left cosets:to the right coset Hx corresponds the left cosetx−1H={g−1:g∈Hx}.The sets of left and right cosets of H are denoted by G/H and H\G respectively. (The position of G relative to H in the notation tells where the coset representative should be put.)Now there is an actionρof G on the set of right cosets of H by the rule(Hx)ρg=H(xg).This means that the permutationρg corresponding to g maps the coset Hx to the coset H(xg).(One must show that the image is independent of the choice of coset representative x,so that the map is well-defined;that the map is a permutation; and that the condition for an action holds.All of this is straightforward.)We can write this in our simpler notation as(Hx)g=H(xg)without too much confusion. Similarly,there is an actionλon the set of left cosets of H given by(xH)λg=(g−1x)H.The Encyclopaedia of Design Theory Cayley graphs and coset diagrams/2(The inverse is required to make the action a homomorphism.)These two actions are isomorphic:the bijection in the preceding paragraph satisfies the conditions for an isomorphism of actions.We will concentrate on the action on right cosets. Clearly this is transitive.Conversely,every transitive action is isomorphic to an action on the right cosets of a subgroup.If G acts onΩ,andα∈Ω,the stabiliser ofαis the subsetGα={g∈G:αg=α}.It is a subgroup of G.If G acts transitively onΩ,then for eachβ∈Ω,the setX(α,β)={g∈G:αg=β}is a right coset of Gα,and the mapβ→X(α,β)is an isomorphism from the given action to the action on right cosets of Gα.3Graphs and permutationsWe will consider directed graphs,whose arcs are ordered pairs of vertices.An undirected graph will be a directed graph with the property that if(v,w)is an arc then so is(w,v);if this holds we speak of the edge{v,w}.A function f on a setΩcan be regarded as the set of all ordered pairs(α,f(α)) forα∈Ω.We can regard these pairs as the arcs of a digraph,the functional digraphΦ(f).A digraph is a functional digraph if and only if every vertex has exactly one arc leaving it.The function f is a permutation if and only if every vertex ofΦ(f)has exactly one arc entering it.If so,thenΦ(f−1)is obtained simply by reversing all the arcs.Proposition1If f andπare permutations ofΩ,thenπis an automorphism of the functional digraphΦ(f)if and only ifπand f commute.Proof(απ,f(α)π)∈Φ(f)⇔f(α)π=f(απ).Note the asymmetry:we have written f on the left andπon the right.This makes the commuting condition appear more natural!See[1]for more information on graph automorphisms.The Encyclopaedia of Design Theory Cayley graphs and coset diagrams/34Cayley graphsAn action of G onΩis said to be regular if it is transitive and the stabiliser of a point consists of the identity element of G only.Now the cosets(left or right) of the identity subgroup are the singleton subsets of G,which can be naturally identified with the elements of G.So any regular action of G is isomorphic to the “action of G on itself by right multiplication”,given by xπg=xg for x,g∈G.The same result would be true if we used left multiplication.The Cayley graph of G with connection set X⊆G is defined to be the directed graph whose arc set is the union of the arc sets of the functional digraphsΦ(λx), for x∈X.In other words,(g,xg)is an arc for all g∈G and x∈X.We denote this graph by Cay(G,X).Proposition2A digraphΓwith vertex set G admits G(acting by right multipli-cation)as a group of automorphisms if and only ifΓis a Cayley graph Cay(G,X) for some X⊆G.Proof In one direction we use the observation that the associative law for a group G is a“commutative law”for the left and right multiplications:(xλg)ρh=(gx)ρh=gxh=(xh)λg=(xρh)λg.So ifΓis a Cayley graph,then right multiplication preserves the arc sets of all the functional digraphsΦ(λx)for x∈X,and hence the arc set ofΓ.Conversely,suppose thatΓadmits the right action of G.LetX={x∈G:(1,x)is an arc ofΓ}.Then applying g on the right we see that(g,xg)is an arc,for all x∈X and g∈G. Conversely,if(g,h)is an arc,then(1,hg−1)is an arc,so hg−1∈X.ThusΓ= Cay(G,X).Proposition3The Cayley digraph Cay(G,X)is loopless if and only if1/∈X;it is undirected if and only if X=X−1;and it is connected if and only if X generates G. The group G acts vertex-transitively on Cay(G,X).If“left”and“right”are reversed throughout this section(so that a Cayley graph is a union of functional digraphs for the right action,and automorphisms act on the left),then an equivalent theory is obtained.The literature is divided on this point!The Encyclopaedia of Design Theory Cayley graphs and coset diagrams/45Vertex-transitive graphsNow there are two ways of generalising Cayley graphs:•We may take the definition as a union of functional digraphs of permuta-tions,and use arbitrary permutations;•We may regard vertex-transitivity as the important property and impose that. In this section I will consider the second approach.First a small digression.If H is a subgroup of G,then an H-H double coset is a subset of G of the formHxH={h1xh2:h1,h2∈H}.As with right and left cosets,it holds that G is a disjoint union of double cosets. However,the double cosets do not all have the same size:we have|HxH|=|H|2|H∩x−1Hx|.For it is easy to show that the denominator is the number of ways of writing an element in the form h1xh2for h1,h2∈H.The set of double cosets is written H\G/H.Double cosets of different subgroups H and K can also be defined but we do not require this.Now suppose thatΓis a graph with vertex setΩ,and G a group of automor-phisms ofΓacting vertex-transitively onΓ.As we saw earlier,the action of G on Ωis isomorphic to its action on the set of right cosets of a subgroup H of G(the stabiliser of a pointαofΩ;so we can replaceΩby the set H\G of right cosets. How do we describe the arcs ofΓ?Following the proof in the preceding section,letX={x∈G:(α,αx)is an arc ofΓ}.Then the following facts are easily seen:•Hg1is joined to Hg2if and only if g2=xg1for some x∈X;•X is a union of H-H double cosets;•Γis loopless if and only if H⊆X;The Encyclopaedia of Design Theory Cayley graphs and coset diagrams/5•Γis undirected if and only if X=X−1;•Γis connected if and only if X generates G;•G acts arc-transitively onΓif and only if X consists of just one double coset.Conversely,if X is a union of H-H double cosets,and we define a digraph on H\G by the rule that Hg is joined to Hxg for all x∈X and g∈G,then the resulting digraph,which we denote byΓ(G,H,X),is vertex-transitive.All vertex-transitive digraphs(up to isomorphism)are thus produced by this construction: Proposition4Any vertex-transitive graph is isomorphic to a graphΓ(G,H,X), where G is a group,H a subgroup of G,and X a union of H-H double cosets.6Homomorphisms and Sabidussi’s TheoremA homomorphism from a digraphΓ1to a digraphΓ2is a map f from the vertex set ofΓ1to that ofΓ2with the property that,if(v,w)is an arc ofΓ1,then(f(v),f(w)) is an arc ofΓ2.(There is no requirement about the case when(v,w)is not an arc.) Theorem5(Sabidussi[3])Every vertex-transitive graph is a homomorphic im-age of a Cayley graph.Proof We can represent our vertex-transitive digraphΓasΓ(G,H,X).Now let Γ =Cay(G,X),and define f(g)=Hg.Any arc(g,xg)ofΓ is mapped to an arc (Hg,Hxg)ofΓ.For example,the Petersen graph is vertex-transitive but is not a Cayley graph, since its automorphism group has no transitive subgroup of order10.However, the dodecahedron is a Cayley graph for the Frobenius group of order20,and the map which identifies antipodal vertices induces a homomorphism from the dodecahedron to the Petersen graph.7Coset diagramsNow we turn to the other possible generalisation of Cayley graphs.Given a set X of permutations ofΩ,defineΦ(X)to be the digraph whose arc sets are the unions of the arc sets of the functional digraphsΦ(f),for f∈X.We can regard the The Encyclopaedia of Design Theory Cayley graphs and coset diagrams/6arcs ofΦ(X)to be coloured by the elements of X,so that the arc(α,f(α))has colour f.We call this graph a coset diagram.There is not much to be said about this construction in general.It is useful in the theory of regular maps.Suppose that M is a map embedded in a surface.A dart orflag of M is an ordered triple consisting of a mutually incident vertex,edge and face of M.IfΩis the set of darts of the map,then one can define three permutations ofΩ:•a:(v,e,f)→(v ,e,f),where v is the other end of the edge e;•b:(v,e,f)→(v,e ,f),where e is the other edge incident with v and f;•c:(v,e,f)→(v,e,f ),where f is the face on the other side of the edge e.Clearly a,b,c are involutions(that is,a2=b2=c2=1).Moreover,ab maps a dart to the dart obtained by a rotation of the face f by one step;bc maps a dart to the dart obtained by a rotation of the edges at the vertex v by one step;and ac maps a dart to the dart obtained by reflecting in the midpoint of the edge e.Thus, we havea2=b2=c2=(ab)m=(bc)n=(ac)2=1,where m and n are the least common multiples of the face sizes and vertex degrees of the map.The involutions a,b,c arefixed-point-free(acting on the set of darts), and the group they generate is transitive.Conversely,given threefixed-point-free involutions a,b,c satisfying these re-lations and generating a transitive group G,there is a map on some surface en-coded by the three permutations.The automorphism group of the map consists of permutations of the darts commuting with the permutations defining the map.Now the permutations a,b,c are represented by coset diagrams for H in G, where H is the stabiliser of a dart.The advantage of a well-drawn coset diagram is that it makes it easy to check the relations satisfied by the three permutations. For example,if the diagram is symmetric about the vertical axis and ac is the reflection,then clearly(ac)2=1.For afine example of a coset diagram,see the portrait of Graham Higman,by Norman Blamey,in the Mathematical Institute,Oxford.Higman was a pioneer of their use in this context[2].A recent survey[4]is also recommended.A coset diagram is not usually a vertex-transitive graph.In the special case where G acts regularly,a coset diagram for any set X of its elements is a Cayley graph Cay(G,X);and we have seen that it admits the vertex-transitive group G.A well-known result from permutation group theory asserts that if the centraliser The Encyclopaedia of Design Theory Cayley graphs and coset diagrams/7of a transitive group G is also transitive,then G is regular;so the only coset dia-grams which admit vertex-transitive groups of colour-preserving automorphisms are Cayley graphs.Note that the rule for a coset diagram is that Hg is joined to Hgx for all g∈G, x∈X;compare the definition of the coset graphΓ(G,H,X),where Hg is joined to Hxg.8Normal Cayley graphsAs we defined it,a Cayley graph for G is a graph on the vertex set G which admits the action of G on the right.Proposition6The following are equivalent for the Cayley graphΓ=Cay(G,X):(a)Γadmits the action of G on the left;(b)the connection set X is a normal subset of G,that is,g−1Xg=X for allg∈G;(c)the connection set X is a union of conjugacy classes in G.Proof Clearly(b)and(c)are equivalent.If(a)holds,then left multiplication takes the arc(1,x)to(g,gx);so gxg−1∈X for any g∈G,x∈X,whence X is a normal subset.The converse is shown in the same way.Obviously,every Cayley graph for an abelian group is normal;but this is false for any non-abelian group.The analogue of normal Cayley graphs for coset graphs has not been inves-tigated.There is no“left action”of G on the set of right cosets of a non-normal subgroup H.However,conditions(b)and(c)make sense even in this more general context.9Other developmentsIfΓis a Cayley graph for a group G,then certainly G is contained in the automor-phism group ofΓ.A question which received a lot of attention was:When is G the full automorphism group.The graphΓis said to be a graphical/digraphical The Encyclopaedia of Design Theory Cayley graphs and coset diagrams/8regular representation of G(GRR or DRR)if the full group of automorphisms of the graph or digraphΓis just G.Abelian groups with exponent greater than2never have GRRs.For ifΓ= Cay(G,X)is undirected,then X=X−1,and the map g→g−1is an automorphism ofΓnot lying in G.Similar remarks hold for generalised dicyclic groups,those of the form G= A,t ,where t2∈A,t2=1,and t−1at=a−1for all a∈A.Hetzel and Godsil showed that apart from these andfinitely many others(all determined explicitly),every group has a GRR.It is now thought that,apart from these two infinite families of exceptions,a random undirected Cayley graph for the group G is a GRR for G with high probability.Other research areas include catalogues of vertex-transitive and Cayley graphs, quasi-Cayley graphs(“Cayley graphs for quasigroups”),and the very important area of infinite Cayley graphs offinite valency.Various generalisations of vertex-transitive graphs,such as graphs with constant neighbourhood,walk-regular graphs, and compact graphs,have also been studied.See[1]for some pointers to the lit-erature.References[1]P.J.Cameron,Automorphisms of graphs,in Topics in Algebraic Graph The-ory(ed.L.W.Beineke and R.J.Wilson),Cambridge Univ.Press,Cambridge, 2004,pp.137–155.[2]G.Higman and Q.Mushtaq,Coset diagrams and relations for PSL(2,Z).ArabGulf J.Sci.Res.1(1983),159–164.[3]G.Sabidussi,Vertex-transitive graphs,Monatsh.Math.68(1964),426–438.[4]J.ˇSir´aˇn,Regular maps on a given surface:a survey,in Topics in Dis-crete Mathematics(ed.M.Klazar et al.),Algorithms and Combinatorics26, Springer,Berlin,2006,pp.591–609.Peter J.CameronAugust2,2006 The Encyclopaedia of Design Theory Cayley graphs and coset diagrams/9。

计算旋转对称图的生成树

Journal of Combinatorial Theory,Series A118(2011)1270–1290Contents lists available at ScienceDirectJournal of Combinatorial Theory,Series A/locate/jctaEnumeration of spanning trees of graphs with rotational symmetryWeigen Yan a,1,Fuji Zhang b,2a School of Sciences,Jimei University,Xiamen361021,Chinab School of Mathematical Science,Xiamen University,Xiamen361005,Chinaa r t i c l e i n f o ab s t r ac tArticle history:Received29December2009 Available online19January2011Keywords:Spanning treeMatrix-Tree TheoremAdjacency matrixLaplacian matrix Methods of enumeration of spanning trees in afinite graph, a problem related to various areas of mathematics and physics, have been investigated by many mathematicians and physicists.A graph G is said to be n-rotational symmetric if the cyclic group of order n is a subgroup of the automorphism group of G. Some recent studies on the enumeration of spanning trees and the calculation of their asymptotic growth constants on regular lattices with toroidal boundary condition were carried out by physicists.A natural question is to consider the problem of enumeration of spanning trees of lattices with cylindrical boundary condition,which are the so-called rotational symmetric graphs. Suppose G is a graph of order N with n-rotational symmetry and all orbits have size n,which has n isomorphic induced subgraphs G0,G1,...,G n−1.In this paper,we prove that if there exists no edge between G i and G j for j=i−1,i+1(mod n), then the number of spanning trees of G can be expressed in terms of the product of the weighted enumerations of spanning trees of n graphs D i’s for i=0,1,...,n−1,where D i has N/n vertices if i=0and N/n+1vertices otherwise.As applications we obtain explicit expressions for the numbers of spanning trees and asymptotic tree number entropies forfive lattices with cylindrical boundary condition in the context of physics.©2010Published by Elsevier Inc.E-mail addresses:weigenyan@(W.Yan),fjzhang@(F.Zhang).1The author was supported in part by NSFC Grant(10771086)and by Program for New Century Excellent Talents in Fujian Province University.2The author was supported in part by NSFC Grant#10831001.0097-3165/$–see front matter©2010Published by Elsevier Inc.doi:10.1016/j.jcta.2010.12.007W.Yan,F.Zhang /Journal of Combinatorial Theory,Series A 118(2011)1270–129012711.IntroductionSuppose that G =(V (G ),E (G ))is an edge-weighted graph (say weighted graph)with no multiple edges and no loops and with vertex set V (G )={v 1,v 2,...,v n }and edge set E (G ).The weight w (T )of a spanning tree T in G is defined as the product of weights of edges in T ,i.e.,w (T )=e ∈E (T )w (e ).The generating function t (G )of spanning trees is given byt (G )=Tw (T ),where the sum is over all spanning trees of G .Define d G (v i )to be the sum of weights of edges in-cident with vertex v i .Then the diagonal matrix of vertex degrees of G is diag (d G (v 1),d G (v 2),...,d G (v n )).Denote the adjacency matrix and Laplacian matrix of G by A (G )=(a ij )n ×n and L (G )=diag (d G (v 1),d G (v 2),...,d G (v n ))−A (G ),respectively,where a ij equals the weight of edge (v i ,v j )if (v i ,v j )is an edge of G and a ij equals zero otherwise.A well-known formula for t (G )is the Matrix-Tree Theorem (see e.g.[2,3]):t (G )=det L (G )ij,where L (G )ij is the submatrix obtained from L (G )by deleting the i th row and the j th col-umn.Let D =(V ,A )be a weighted digraph with no multiple arcs and no loops,with vertex set V ={v 1,v 2,...,v n }.Denote the sum of weights of arcs with initial vertex v i by d D (v i ).Then the diagonal matrix of vertex out-degrees of D is diag (d D (v 1),d D (v 2),...,d D (v n )).Denote the adjacency matrix and Laplacian matrix of D by A (D )=(b ij )n ×n and L (D )=diag (d D (v 1),d D (v 2),...,d D (v n ))−A (D ),respectively,where b ij equals the weight of (v i ,v j )if (v i ,v j )is an arc of D and zero otherwise (if D is a weighted digraph with loops and multiple arcs,then b ij (i =j )equals the sum of weights of arcs with initial vertex v i and terminal vertex v j if there exist such arcs and b ij (i =j )equals zero otherwise,and b ii equals the sum of weights of loops with initial vertex v i if there exist such loops and zero otherwise).A directed spanning tree of D with root v i is a spanning subdigraph of D the underlying graph of which is a spanning tree of the underlying graph of D and there exists a unique directed path from v j (=v i )to v i for each j =1,2,...,n .The weight w (T )of a directed spanning tree T in D is defined as the product of weights of arcs in T .Denote by t (D ,v i )the sum of weights of directed spanning trees of D with root v i .A well-known result for t (D ,v i )is as follows.Theorem 1.1(Tutte,1948).Let D be a weighted digraph with vertex set {v 1,v 2,...,v n }and L (D )the Lapla-cian matrix of D.Thent (D ,v i )=det L (D )i,where L (D )i is the submatrix of L (D )obtained by deleting the ith row and the ith column from L (D )for 1 i n.The following lemma is well known (see for example [2]).Lemma 1.1.Let 0<μ1 μ2 ··· μn −1be the Laplacian eigenvalues of a connected weighted graph G with n vertices.Thent (G )=μ1μ2...μn −1n.1272W.Yan,F.Zhang /Journal of Combinatorial Theory,Series A 118(2011)1270–1290The enumeration of spanning trees of a graph G was first considered by Kirchhoff in the analysis of electric circuits [16].It is a problem of fundamental interest in mathematics [2,4,8,15,18,24,30]and in physics [5,6,22,26].The number of spanning trees is closely related to the partition function of the q -state Potts model in statistical mechanics [12,28].Some recent studies on the enumeration of spanning trees and the calculation of their asymptotic growth constants on regular lattices were carried out in [6,7,21,25].Temperley [23]found a bijection between spanning trees of an m ×n grid and perfect matchings of a (2m +1)×(2n +1)grid with a corner removed.Propp and,independently,Burton and Pemantle [4]generalized this bijection to map spanning trees of general (undirected unweighted)plane graphs to perfect matchings of a related graph.Kenyon,Propp,and Wilson [15]extend this bijection to the directed weighted case.For the asymptotic property of spanning trees of graphs,Lyons [18]proved the following result.Theorem 1.2.(See Lyons [18].)Let {G n }be a tight sequence of finite connected graphs with bounded aver-age degree such that lim n →∞log t (G n)|V (G n)|=h.If {G n }is a sequence of connected subgraph of {G n }such thatlim n →∞|{v ∈V (G n );d G n (v )=d G n(v )}||V (G n )|=1,then lim n →∞log t (G n)|V (G n )|=h.By two methods (one algebraic,one combinatorial),Ciucu and the current authors [8]obtained afactorization theorem for the number of spanning trees of the plane graphs with reflective symmetry (all orbits have two vertices).Recently,the current authors [30]obtained a factorization theorem for the number of spanning trees of the more general graphs with reflective symmetry (i.e.,so-called the graphs with an involution,all orbits have one or two vertices).Shrock and Wu [21]gave a method of enumerating spanning trees of lattices with toroidal boundary condition and obtained the exact formulae of the numbers of the spanning trees of many lattices with toroidal boundary condition.It is of interest to enumerate spanning trees of lattices with cylindrical boundary condition,which are the so-called n -rotational symmetric graphs (see the definition in Section 2).In this paper,we prove that if G is a graph of order N with n -rotational symmetry and all orbits have size n ,which has n isomorphic induced subgraphs G 0,G 1,...,G n −1such that there exists no edge between G i and G j for j =i −1,i +1(mod n ),then the number of spanning trees of G can be expressed in terms of the product of the numbers of spanning trees of n graphs D 0,D 1,...,D n −1,where D i has N /n vertices if i =0and N /n +1vertices otherwise.As applications,we obtain closed formulae of the numbers of spanning trees and asymptotic tree number entropies of the hexago-nal,8.8.4,3.12.12,33.42and triangular lattices with cylindrical boundary condition in the context of physics.Remark 1.1.Jockusch [14]proved by mostly combinatorial arguments that the number of perfect matchings of a large class of graphs with 4-rotational symmetry is squarish.Based on the above bijection of spanning trees of a plane graph and perfect matchings of a related graph in [4,15],our approach to enumerate spanning trees may be used for enumeration of perfect matchings of some plane graphs with rotational symmetry.2.A factorization theoremThe automorphism group of a graph G characterizes its symmetries,and is therefore very useful for simplifying the computation of some of its invariants.One of the most successful results in this direction is to use the character of the automorphism group of a graph G to compute its characteristic polynomial (see for example Cvetkovi´c et al.[9]and Feng et al.[10]).A graph G is said to be n -rotational symmetric if the cyclic group C n of order n is a subgroup of the automorphism group of G .We call C n the rotational subgroup of an n -rotational symmetric graph G .If G is a weighted graph,the weight of each edge e of G is denoted by w (e ).If the underlying graph of G is n -rotational symmetric (whose rotational subgroup is the cyclic group C n = g )and for every e ∈E (G )we have w (e )=w (g i (e ))for i =0,1,2,...,n −1,that is,the weighted function onW.Yan,F.Zhang /Journal of Combinatorial Theory,Series A 118(2011)1270–12901273Fig.1.(a)A connected graph G with 6-rotational symmetry;(b)one component G 0of G .the edges is constant on the orbits of g ,then we say that G is a weighted graph with n -rotational symmetry (or g preserves weights).Let G be a weighted graph of order N with n -rotational symmetry,which has symmetric axes 0, 1,..., n −1passing through the rotation center O ,where O is not a vertex of G .Fig.1(a)shows an example of a 6-rotational symmetric graph.If G has some vertices lying on its symmetric axes i ’s,then we can slightly rotate the symmetric axes such that there is no vertex lying on its new symmet-ric axes.Hence we can assume that there exists no vertex of G lying on 0, 1,..., n −1.If we delete all the edges intersected by 0, 1,..., n −1,then n isomorphic components G i =(V (G i ),E (G i ))for 0 i n −1are obtained,where V (G i )={v (i )1,v (i )2,...,v (i )N /n }is the vertex set of G i .Obviously,φ:v (i )k →v (j )k(k =1,2,...,m )is an isomorphism between G i and G j ,i ,j =0,1,...,n −1.Weassume that there exists no edge between G i and G j for j =i −1,i +1(mod n )(see Fig.1(a)).Let E (G i −G i +1)denote the set of edges between G i and G i +1in G .Let D 0be the graph withN /n vertices obtained from G 0by add one edge (v (0)i ,v (0)j )with weight ω(e )for each edge of theform e =(v (0)i ,v (1)j )∈E (G 0−G 1)and i =j .For the graph G in Fig.1(a),there exist two edges in E (G 0−G 1)one of which has the form e =(v (0)i ,v (1)j )∈E (G 0−G 1)and i =j .Hence D 0can be shown in Fig.2(a).We also need to define n −1weighted digraphs D 1,D 2,...,D n −1with N /n +1vertices from G 0,which may contain multiple arcs,as follows.Note that the vertex set of G 0is V (G 0)={v (0)1,v (0)2,...,v (0)N /n }.Let ¯G0be the weighted digraph obtained from G 0by replacing each edge e =(v (0)i ,v (0)j )of G 0with two arcs (v (0)i ,v (0)j )and (v (0)j ,v (0)i )with weight w (e ),respectively.Let ωbe a primitive n th root of unity.For each k =1,2,...,n −1,define D k to be the weighted digraphobtained from ¯G0by the following procedure.(1)Add a new vertex u to ¯G0.(2)For each edge of the form e =(v (0)i ,v (1)j )∈E (G 0−G 1)and i =j ,add four arcs (v (0)i ,u ),(v (0)j ,u ),(v (0)i ,v (0)j ),and (v (0)j ,v (0)i)in D k with weights w (e )(1−ωk ),w (e )(1−ω−k ),w (e )ωk ,and w (e )ω−k ,where w (e )is the weight of e in G .1274W.Yan,F.Zhang /Journal of Combinatorial Theory,Series A 118(2011)1270–1290Fig.2.(a)The graph D 0;(b)the digraph D k (k =1,2,...,n −1);(c)the digraph D ∗k,where each edge without orientation in (b)and (c)represents two oppositely oriented arcs,and ωis a primitive n th root of unity.(3)For each edge of the form e =(v (0)i ,v (1)i )∈E (G 0−G 1),add an arc (v (0)i ,u )with weightw (e )(2−ωk −ω−k ),where w (e )is the weight of e in G .For the graph G shown in Fig.1(a),the corresponding weighted digraph D k (k =1,2,...,n −1)is shown in Fig.2(b),where each edge without orientation in the figure represents two oppositely oriented arcs,and ωis a primitive n th root of unity.Now we can state one of our main results as follows.Theorem 2.1(Tree Factorization Theorem).Let G be a connected weighted graph of order N with n-rotational symmetry and all orbits have the same size n.Let the G i ’s be the n graphs defined above.Suppose there exist no edges between G i and G j for i −j =1(mod n )and there exist no vertices of G lying on the axes of rotation l 0,l 1,...,l n −1passing through O.Let D k ’s be defined as above.Thent (G )=t (D 0)nn −1k =1t (D k ,u ),where t (D 0)is the weighted enumeration of spanning trees of D 0and t (D k ,u )is the weighted enumeration of directed spanning trees with root u of D k for k =1,2,...,n −1.Proof.Let A (G 0)be the adjacency matrix of G 0and let R denote the adjacency relation between V (G 0)and V (G 1).Note that G is an n -rotational symmetric graph.By a suitable labelling of vertices of G ,the Laplacian matrix L (G )has the following form:L (G )=⎛⎜⎜⎜⎜⎜⎜⎝F 0−A (G 0)−R 0···0−R T−R T F 0−A (G 0)−R...000−R T F 0−A (G 0) 0..................000···F 0−A (G 0)−R−R···−R TF 0−A (G 0)⎞⎟⎟⎟⎟⎟⎟⎠n ×n=I n ⊗F 0−A (G 0)−B ⊗R −B T ⊗R Twhere F 0=diag (d G (v (0)1),d G (v (0)2),...,d G (v (0)m ))is a diagonal matrix of order m =:N /n ,W.Yan,F.Zhang /Journal of Combinatorial Theory,Series A 118(2011)1270–12901275B =⎛⎜⎜⎜⎜⎝010 (000)1···00..................000 (01)100 0⎞⎟⎟⎟⎟⎠n ×n,and B ⊗R denotes the tensor product of two matrices B and R .Let g be a cyclic group of order n .Obviously,ρ:g i →B i is a faithful representation of this group.Note that the cyclic group of order n has exactly n (linear)characters χi ’s such that χi (g )=ωi ,where ωis a primitive n th root of unity.Hence there exists an invertible matrix T such that T BT −1=diag (1,ω,...,ωn −1).Since B T =B −1,we have(T ⊗I m )L (G )(T ⊗I m )−1=(T ⊗I m ) I n ⊗ F 0−A (G 0)−B ⊗R −B T ⊗R T (T ⊗I m )−1=I n ⊗ F 0−A (G 0)−diag 1,ω,...,ωn −1 ⊗R −diag 1,ω−1,...,ω−(n −1) ⊗R T=diagF 0−A (G 0)−R −R T ,F 0−A (G 0)−ωR −ω−1R T ,...,F 0−A (G 0)−ωn −1R −ω−(n −1)R T.Hence the Laplacian characteristic polynomial of G can be expressed byφ(G ,x )=:det xI N −L (G )=φ0(G ,x )φ1(G ,x )...φn −1(G ,x ),whereφt (G ,x )=det xI m −F 0+A (G 0)+ωt R +ω−t R T.Note thatd d xφ(G ,x )=φ0(G ,x )n −1 t =1φt (G ,x )+φ0(G ,x )n −1 j =1n −1t =1φt (G ,x )φj (G ,x )φj (G ,x ).Hence,by Lemma 1.1,Nt (G )=μ1μ2...μN −1=(−1)N −1d d xφ(G ,x )x =0=(−1)N −1φ0(G ,0)n −1 j =1φj (G ,0)+(−1)N −1φ0(G ,0) n −1 j =1n −1t =1φt (G ,x )φj (G ,x )φ j (G ,x )x =0,(2.1)where μ1,μ2,...,and μN −1are the nonzero Laplacian eigenvalues of G .Note that N =mn .From (2.1)it suffices to prove the following Claims 1–3:Claim 1.φ0(G ,0)=0.Claim 2.φj (G ,0)=(−1)m t (D j ,u )for j =1,2,...,n −1.Claim 3.φ 0(G ,0)=(−1)m −1mt (D 0).We need to prove the following claims:1276W.Yan,F.Zhang /Journal of Combinatorial Theory,Series A 118(2011)1270–1290Claim 4.The Laplacian matrix of the graph D 0equals F 0−A (G 0)−R −R T .Claim 5.F 0−A (G 0)−ωj R −ω−j R T is the submatrix of the Laplacian matrix L (D j )of D j obtained by deleting the row and column corresponding to vertex u.First we prove Claim 4.Let X =(x st )1 s ,t m =F 0−A (G 0)−R −R T .Set δst =1if s =t andδst =0otherwise.Note that x st =δst d G (v (0)s )−a st −r st −r ts if 1 s ,t m ,where d G (v (0)s )isthe degree of v (0)s in G ,A (G 0)=(a st )1 s ,t m is the adjacency matrix of G 0,and R =(r st )1 s ,t m(resp.R T )denotes the incident relation between G 0and G 1(resp.G 0and G n −1)in G .Hencex ss =d G (v (0)s )−2r ss .If there exists an edge e s =(v (0)s ,v (1)s )with weight r ss =:w (e s )between G 0andG 1in G ,then,by the definition of the rotational symmetry,there exists an edge (v (0)s ,v (n −1)s )withweight r ss =w (e s )between G 0and G n −1in G .Suppose the set of edges between v (0)s and G 1in G is{e t =:(v (0)s ,v (1)t )|t =s ,i 1,i 2,...,i p }and the set of edges between v (0)sand G n −1in G is {e t =:(v (0)s ,v (n −1)t)|t =s ,j 1,j 2,...,j q },where 0 p ,q m −1.By the definition of the rota-tional symmetry,then {e t =:(v (0)t ,v (1)s )|t =s ,j 1,j 2,...,j q }is a subset of edges between G 0and G 1and w (e t )=w (e t ).By the definition of D 0,(v (0)s ,v (0)i 1),(v (0)s ,v (0)i 2),...,(v (0)s ,v (0)i p)and (v (0)s ,v (0)j 1),(v (0)s ,v (0)j 2),...,(v (0)s ,v (0)j q )are “adding”edges with weights w (e 1),w (e 2),...,w (e p )andw (e 1),w (e 2),...,w (e q )in D 0.Hence d D 0(v (0)s )=d G (v (0)s )−2r ss − p t =1w (e t )− q t =1w (e t )+ p t =1w (e t )+ q t =1w (e t )=d G (v (0)s )−2r ss =x ss .If there exists no edge between v (0)sand v (1)s in G (hence,by the definition of the rotational symmetry,there exists no edge between v (0)sand v (n −1)s ),then r ss =0and hence x ss =d G (v (0)s ).Suppose the set of edges between v (0)s andG 1in G is {e t =:(v (0)s ,v (1)t )|t =i 1,i 2,...,i p }and the set of edges between v (0)s and G n −1in G is {e t =:(v (0)s ,v (n −1)t)|t =j 1,j 2,...,j q },where 0 p ,q m −1.Similarly,we can showthat d D 0(v (0)s )=d G (v (0)s )− p t =1w (e t )− q t =1w (e t )+ p t =1w (e t )+ q t =1w (e t )=d G (v (0)s )=x ss .Hence diag (x 11,x 22,...,x mm )=F 0−diag (2r 11,2r 22,...,2r mm )is the diagonal matrix of vertex de-grees of D 0.If s =t ,then −x st =a st +r st +r ts .Similarly,by the definition of D 0,we can show that A (G 0)+R +R T −diag (2r 11,2r 22,...,2r mm )equals the adjacency matrix of D 0.So we have proved that the Laplacian matrix of the graph D 0equals F 0−A (G 0)−R −R T and Claim 4fol-lows.Now we prove Claim 5.Let Y =(y st )1 s ,t m =F 0−A (G 0)−ωj R −ω−j R T .Note that y st =δst d G (v (0)s )−a st −ωj r st −ω−j r ts if 1 s ,t m .Hence y ss =d G (v (0)s )−(ωj +ω−j )r ss .If (v (0)s ,v (1)s )is an edge between G 0and G 1in G ,then,by the definition of the rotational symmetry,(v (0)s ,v (n −1)s )is an edge between G 0and G n −1in G .Suppose the set of edges between v (0)s and G 1inG is {e t =:(v (0)s ,v (1)t )|t =s ,i 1,i 2,...,i p }and the set of edges between v (0)s and G n −1inG is {e t =:(v (0)s ,v (n −1)t)|t =s ,j 1,j 2,...,j q },where 0 p ,q m −1.By the definition ofthe rotational symmetry,then {e t =:(v (0)t,v (1)s )|t =s ,j 1,j 2,...,j q }is a subset of edges be-tween G 0and G 1.For the edge e s =(v (0)s ,v (1)s )in G ,by the definition of D j ,we must addan arcs (v (0)s ,u )with weight (2−ωj −ω−j )w (e s )when we construct D j from G 0.Similarly,for each edge e t =(v (0)s ,v (1)t )for t =i 1,i 2,...,i p (resp.e t =(v (0)t,v (1)s )for t =j 1,j 2,...,j q ),we must add four arcs (v (0)s ,u ),(v (0)t ,u ),(v (0)s ,v (0)t )and (v (0)t ,v (0)s )with weights w (e t )(1−ωj ),w (e t )(1−ω−j ),w (e t )ωj and w (e t )ω−j (resp.(v (0)t ,u ),(v (0)s ,u ),(v (0)t ,v (0)s )and (v (0)s ,v (0)t )withweights w (e t )(1−ωj ),w (e t )(1−ω−j ),w (e t )ωj and w (e t)ω−j )when we construct D j from G 0.Hence the weight of arcs with initial vertex v (0)s in D j equals d G (v (0)s )−2w (e s )− p t =1w (e t )− q t =1w (e t )+(2−ωj −ω−j )w (e s )+ p t =1w (e t )(1−ωj )+ p t =1w (e t )ωj + qt =1w (e t )(1−ω−j )+ω−j q t =1w (e t )=d G (v (0)s )−(ωj +ω−j )w (e s )=d G (v (0)s )−(ωj +ω−j)r ss since w (e t )=w (e t )(1 t q )by the definition of rotational symmetry.Hence diag (y 11,y 22,...,y mm )is the submatrixof the diagonal matrix of vertex (out-)degrees of D j by deleting the row and column correspondingW.Yan,F.Zhang/Journal of Combinatorial Theory,Series A118(2011)1270–12901277 Fig.3.(a)The hexagonal lattice H t(n,m)with toroidal boundary condition;(b)the hexagonal lattice G c(n,m)with cylindricalboundary condition.vertex u.Similarly,we can show that−Y+diag(y11,y22,...,y mm)is the submatrix of adjacency matrix of D j by deleting the row and column corresponding to vertex u.Thus we have proved that Y is the submatrix of Laplacian matrix of D j by deleting the row and column corresponding to vertex u, which implies that Claim5holds.Claim1is immediate from Claim4,Claim2holds because of Theorem1.1and Claim5and Claim3 follows from Lemma1.1and Claim4.The theorem thus holds.2Keeping the notation as in Theorem2.1,let D∗jbe a weighted digraph obtained from D j by delet-ing the vertex u and adding a loop with weight−d Dj(v(0)s)to vertex v(0)s in D j.For the graph Gshown in Fig.1(a),the weighted digraph D∗jis shown in Fig.2(c).From the proof of Theorem2.1,we have:Corollary2.1.Keeping the notation as in Theorem2.1,let D∗jbe the digraph defined as above and let A(D∗j)denote the adjacency matrix of D∗j.Thent(G)=t(D0)nn−1j=1detAD∗j.The condition that there exists no edge between G i and G j for i−j=1(mod n)in Theorem2.1 and Corollary2.1can be relaxed.In fact,we can take L(G)to be any block circulant matrix and we can also express the number of spanning trees of an n-rotational symmetric graph in terms of the product of numbers of spanning trees of n graphs.Since it is cumbersome to define these graphs,we will not pursue this further.3.Applications3.1.The hexagonal latticeThe hexagonal lattices with toroidal and cylindrical boundary conditions,denoted by H t(n,m) and H c(n,m),are shown in Fig.3(a)and(b).For H t(n,m)in Fig.3(a),all a i’s,b i’s,c i’s,andc∗i’s are some vertices on the left,right,upper and lower boundaries,respectively,and the left and right(resp.the upper and lower)boundaries of the picture are identified such that (a1,b1),(a2,b2),...,(a m+1,b m+1),(a1,c∗1),(c1,c∗2),(c2,c∗3),...,(c n−1,c∗n),(c n,b m+1)are edges in H t(n,m).For H c(n,m)in Fig.3(b),all a i’s and b i’s are some vertices on the left and right bound-aries,and the left and right boundaries of the picture are identified such that(a1,b1),(a2,b2),..., (a m+1,b m+1)are edges in H c(n,m).Shrock and Wu[21]showed that the number of spanning trees and the asymptotic tree number entropy of H t(n,m)can be expressed as1278W.Yan,F.Zhang /Journal of Combinatorial Theory,Series A 118(2011)1270–1290Fig.4.(a)The graph D 0for H c (n ,m );(b)the digraph D ∗kfor H c (n ,m );(c)the digraph D ∗1k for H c (n ,m );(d)the digraph D ∗2kfor H c(n ,m ),where each edge without orientation in (b)–(d)represents two oppositely oriented arcs,and ωis a primitive (n +1)th root of unity.tH t(n ,m )=3(m +1)(n +1)n i =0m j =0(i ,j )=(0,0)6−2cos θ1−2cos θ2−2cos (θ1+θ2),andlimn ,m →∞1(2n +2)(m +1)log t H t (n ,m )=18π22π2π 0log6−2cos x −2cos y −2cos (x +y )d x d y ≈0.8077,(3.1)where θ1=2i πm +1,θ2=2j πn +1.Theorem 3.1.For the hexagonal lattice H c (n ,m )with cylindrical boundary condition,the number of spanning trees of H c (n ,m )can be expressed astH c(n ,m ) =2m +1n +1nk =1a b(c +√ab )m +1−(c −√ab )m +1,where a =1−cos (2k πn+1),b =7−cos (2k πn +1),c =3−cos (2k πn +1).Proof.For the hexagonal lattice H c (n ,m )with cylindrical boundary condition,let G 0be the subgraph of H c (n ,m )marked by the bold line in Fig.3(b),which is a path with 2m +2vertices.Obviously,H c (n ,m )can be regarded as an (n +1)-rotational symmetric graph.By the definitions of D 0and D ∗k(1 k n )in Corollary 2.1,for H c (n ,m )the corresponding D 0and D ∗kare shown in Fig.4(a)and (b),where each edge without orientation in D ∗k is replaced with two oppositely oriented arcs,and ωis aprimitive (n +1)th root of unity.Obviously,we havet (D 0)=2m +1.(3.2)Let D ∗1k(resp.D ∗2k )be the digraph obtained from D ∗kby deleting one vertex with weight −2(resp.by replacing weight −2of one vertex with weight −3),which is shown in Fig.4(c)(resp.Fig.4(d)).Denote the adjacency matrices of D ∗k,D ∗1k and D ∗2k by A (D ∗k),A (D ∗1k )and A (D ∗2k ).Set x m +1(k )=det A D ∗k,y m +1(k )=det A D ∗1k,z m +1(k )=det A D ∗2k.It is then not difficult to see that x m (k ),y m (k )and z m (k )satisfyW.Yan,F.Zhang/Journal of Combinatorial Theory,Series A118(2011)1270–12901279⎧⎪⎪⎪⎨⎪⎪⎪⎩x m(k)=−2y m(k)−(2+2cosθ)z m−1(k)for m 2, y m(k)=−3z m−1(k)−y m−1(k),for m 2, z m(k)=−3y m(k)−(2+2cosθ)z m−1(k),for m 2, z0(k)=−y0(k)=1,z1(k)=4−2cosθ,y1(k)=−2,whereθ=2kπ.This system of recursions may be solve by standard means resulting in the following expression:x m(k)=ab(c+√ab)m−ab(c−√ab)m.(3.3)The theorem is immediate from Corollary2.1and(3.2)and(3.3).2Corollary3.1.The asymptotic tree number entropy of the hexagonal lattice H c(n,m)with cylindrical boundary condition islim n,m→∞1(n+1)(2m+2)logtH c(n,m)=1 4π2πlog3−cos x+7−8cos x+cos2xd x≈0.8077,which equals the asymptotic tree number entropy of the hexagonal lattice H t(n,m)with toroidal boundary condition.Proof.By Theorem3.1,it is not difficult to show thatlim n,m→∞1(n+1)(2m+2)logtH c(n,m)=limn,m→∞1(n+1)(2m+2)nj=1log3−cos2jπn+1+7−8cos2jπn+1+cos22jπn+1m+1=1 4π2πlog3−cos x+7−8cos x+cos2xd x.Using the following integration formula,which is thefirst formula in Section2of[13]and is also [27,Eq.(5)],1 2π2πlog(2A+2B cos x+2C sin x)d x=logA+A2−B2−C2for A2 B2+C2,it is easy to verify that1 2π2πlog6−2cos x−2cos y−2cos(x+y)d y=log3−cos x+7−8cos x+cos2x.HenceFig.5.(a)The 8.8.4lattice G t (n ,m )with toroidal boundary condition;(b)the 8.8.4lattice G c (n ,m )with cylindrical boundary condition.18π22π2π 0log 6−2cos x −2cos y −2cos (x +y )d x d y=14π2πlog3−cos x +7−8cos x +cos 2xd x .The above equation can also be obtained from [27,Eq.(31)].Hence,by (3.1),the corollary holds.23.2.The 8.8.4latticeThe 8.8.4lattice G t (n ,m )with toroidal boundary condition is shown in Fig.5(a).For G t (n ,m )inFig.5(a),all a i ’s,a ∗i ’s,b i ’s,and b ∗i’s are some vertices on the left,right,upper and lower boundaries,respectively,and the left and right (resp.the upper and lower)boundaries of the picture are iden-tified such that (a 1,a ∗1),(a 2,a ∗2),...,(a n ,a ∗n )and (b 1,b ∗1),(b 2,b ∗2),...,(b m ,b ∗m )are edges in G t(n ,m ).The lattice G t (n ,m )is a finite subgraph of the 8.8.4tiling in the Euclidean plane which has been used to describe phase transitions in the layered hydrogen-bonded SnCl 2·2H 2O crystal [20]in phys-ical systems [1,19,20].The 8.8.4lattice G t (n ,m )is composed of mn quadrangles.The 8.8.4lattice G c (n ,m )with cylindrical boundary condition is the one obtained from G t (n ,m )by deleting edges(b 1,b ∗1),(b 2,b ∗2),...,(b m ,b ∗m )(see Fig.5(b)).Shrock and Wu [21]showed that the number of span-ning trees and the asymptotic tree number entropy of H t (n ,m )can be expressed astG t(n ,m ) =16nmn−1 i =0m −1 j =0(i ,j )=(0,0)4(7−3cos θ1−3cos θ2−cos θ1cos θ2),andlimn ,m →∞14nmlog t G t (n ,m )=14log 2+14ππ0log7−3cos θ+4sin (θ/2)√5−cos θd θ≈0.7867,(3.4)where θ1=2i π,θ2=2j π.Chang and Shrock [6]have derived the formula (3.4).。

小学上册第十一次英语第1单元综合卷

小学上册英语第1单元综合卷英语试题一、综合题(本题有100小题,每小题1分,共100分.每小题不选、错误,均不给分)1.The ant is a tiny ______ (昆虫).2.I want to ___ (learn/know) more about science.3.I want to learn to ________ (制作甜点) for fun.4. A _____ is a large expanse of sand.5.What is the largest land carnivore?A. LionB. TigerC. Polar bearD. Grizzly bear6.What do we use to write on a blackboard?A. MarkerB. ChalkC. PenD. Crayon7.n Tea Party was a protest against the ________ (茶税). The Brit8.The nurse, ______ (护士), works in the emergency room.9.My brother is a __________ (游戏玩家).10.We have a ______ (快乐的) celebration for achievements.11.An ecosystem is a community of living organisms interacting with their ______ environment.12.I can ___ my favorite song. (sing)13. A non-metal usually gains ______ in a reaction.14.I like to write ______ (科技) articles to share knowledge with others. It’s a great way to inform people.15.I want to learn to ________ (剪纸) for art class.16.What is the name of the famous American singer known for "Bad Romance"?A. Lady GagaB. Katy PerryC. RihannaD. Ariana GrandeA17.What is the opposite of hot?A. WarmB. ColdC. CoolD. HeatB18.The study of Earth's geology is essential for understanding ______ resources.19.I like learning about different cultures. It’s fascinating to discover how people in __________ celebrate and live their lives. I hope to travel and experience it firsthand.20.The first person to win the Nobel Prize in Physiology was _______. (埃米尔·冯·贝尔)21.Which shape has three sides?A. CircleB. SquareC. TriangleD. RectangleC22.I tell my __________ about my day. (妈妈)23.What do we call the force that pulls objects toward the Earth?A. FrictionB. MagnetismC. GravityD. InertiaC24.She is _____ (playing) a game.25.The __________ is a major shipping route in the world. (巴拿马运河)26.I enjoy _______ (参加)文化活动。

Cartesian product of graphs

Cartesian product ofgraphsThe Cartesian product of graphs.In graph theory,the Cartesian product G□H of graphsG and H is a graph such that•the vertex set of G□H is the Cartesian productV(G)×V(H);and•any two vertices(u,u')and(v,v')are adjacent in G□H if and only if either•u=v and u'is adjacent with v'in H,or•u'=v'and u is adjacent with v in G.Cartesian product graphs can be recognized efficiently,intime O(m log n)for a graph with m edges and n vertices(Aurenhammer,Hagauer&Imrich1992).The operationis commutative as an operation on isomorphism classesof graphs,and more strongly the graphs G□H and H□G are naturally isomorphic,but it is not commutativeas an operation on labeled graphs.The operation is alsoassociative,as the graphs(F□G)□H and F□(G□H)are naturally isomorphic.The notation G×H is occasionally also used for Cartesianproducts of graphs,but is more commonly used for an-other construction known as the tensor product of graphs.The square symbol is the more common and unambigu-ous notation for the Cartesian product of graphs.It showsvisually the four edges resulting from the Cartesian prod-uct of two edges.[1]The Cartesian product is not a product in the categoryof graphs.(The tensor product is the categorical prod-uct.)However,it is a product in the category of reflexivegraphs.The category of graphs does form a monoidalcategory under the Cartesian product.1Examples•The Cartesian product of two edges is a cycle on fourvertices:K2□K2=C4.•The Cartesian product of K2and a path graph is aladder graph.•The Cartesian product of two path graphs is a gridgraph.•The Cartesian product of n edges is a hypercube:(K2)□n=Q n.Thus,the Cartesian product of two hypercubegraphs is another hypercube:Q □Q =Q ₊ .•The Cartesian product of two median graphs is an-other median graph.•The graph of vertices and edges of an n-prism is theCartesian product graph K2□C .•The rook’s graph is the Cartesian product of twocomplete graphs.2PropertiesIf a connected graph is a Cartesian product,it can be fac-torized uniquely as a product of prime factors,graphs thatcannot themselves be decomposed as products of graphs(Sabidussi1960;Vizing1963).However,Imrich andKlavžar(2000)describe a disconnected graph that canbe expressed in two different ways as a Cartesian productof prime graphs:(K1+K2+K22)□(K1+K23)=(K1+K22+K24)□(K1+K2),where the plus sign denotes disjoint union and the super-scripts denote exponentiation over Cartesian products.A Cartesian product is vertex transitive if and only if eachof its factors is(Imrich and Klavžar,Theorem4.19).A Cartesian product is bipartite if and only if each of itsfactors is.More generally,the chromatic number of theCartesian product satisfies the equation127EXTERNAL LINKS□(Sabidussi1957).The Hedetniemi conjecture states a re-lated equality for the tensor product of graphs.The inde-pendence number of a Cartesian product is not so easily calculated,but as Vizing(1963)showed it satisfies the inequalitiesα(G)α(H)+min{|V(G)|-α(G),|V(H)|-α(H)}≤α(G□H)≤min{α(G)|V(H)|,α(H)|V(G)|}.The Vizing conjecture states that the domination number of a Cartesian product satisfies the inequalityγ(G□H)≥γ(G)γ(H).3Algebraic graph theoryAlgebraic graph theory can be used to analyse the Carte-sian graph product.If the graph G1has n1vertices and the n1×n1adjacency matrix A1,and the graph G2has n2vertices and the n2×n2adjacency matrix A2,then theadjacency matrix of the Cartesian product of both graphs is given byA1□2=A1⊗I n2+I n1⊗A2where⊗denotes the Kronecker product of matrices and I n denotes the n×n identity matrix.[2]4HistoryAccording to Klavžar and Imrich,Cartesian products of graphs were defined in1912by Whitehead and Russell. They were repeatedly rediscovered later,notably by Sabidussi in1960.5Notes[1]Hahn,Geňa;Sabidussi,Gert(1997),Graph symmetry:al-gebraic methods and applications,NATO Advanced Sci-ence Institutes Series497,Springer,p.116,ISBN978-0-7923-4668-5.[2]Kaveh,A.;Rahami,H.(2005),“A unified method foreigendecomposition of graph products”,Communications in Numerical Methods in Engineering with Biomedical Ap-plications21(7):377–388,doi:10.1002/cnm.753,MR 2151527.6References•Aurenhammer,F.;Hagauer,J.;Imrich,W.(1992).“Cartesian graph factorization at logarithmic cost per edge”plexity2(4):331–349.doi:10.1007/BF01200428.MR1215316.|first4= missing|last4=in Authors list(help)•Imrich,Wilfried;Klavžar,Sandi(2000).Product Graphs:Structure and Recognition.Wiley.ISBN 0-471-37039-8.•Imrich,Wilfried;Klavžar,Sandi;Rall,Douglas F.(2008).Graphs and their Cartesian Products.A.K.Peters.ISBN1-56881-429-1.•Sabidussi,G.(1957).“Graphs with given group and given graph-theoretical properties”.Canad.J.Math.9:515–525.doi:10.4153/CJM-1957-060-7.MR0094810.•Sabidussi,G.(1960).“Graph multiplication”.Math.Z.72:446–457.doi:10.1007/BF01162967.MR0209177.•Vizing,V.G.(1963).“The Cartesian product of graphs”.Vycisl.Sistemy9:30–43.MR0209178. 7External links•Weisstein,Eric W.,“Graph Cartesian Product”, MathWorld.3 8Text and image sources,contributors,and licenses8.1Text•Cartesian product of graphs Source:/wiki/Cartesian%20product%20of%20graphs?oldid=600584603Contribu-tors:Tedernst,Tomo,Giftlite,Waltpohl,Smimram,Zaslav,Tompw,Jérôme,Joriki,Ivan Bilenjkij,Silly rabbit,Mhym,Aguetz,Grievous Angel,David Eppstein,Koko90,Justin W Smith,Alexbot,Addbot,DOI bot,AnomieBOT,Citation bot,Cs32en,Citation bot1,Helpful Pixie Bot and Anonymous:98.2Images•File:Graph-Cartesian-product.svg Source:/wikipedia/commons/7/77/Graph-Cartesian-product.svg Li-cense:Public domain Contributors:Originally from en.wikipedia;description page is/was here.Original artist:Original uploader was David Eppstein at en.wikipedia8.3Content license•Creative Commons Attribution-Share Alike3.0。

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ENUMERATIONS OF CAYLEY GRAPHS
arXiv:math/0609462v1 [math.CO] 16 Sep 2006
DONGSEOK KIM, JIN HWAN KIM, JAEUN LEE, AND DIANJUN WANG Abstract. We characterize the equivalenf Cayley graphs for a finite group A. Using these characterizations, we find enumeration formulae of the equivalence classes and weak equivalence classes of Cayley graphs. As an application, we find the number of weak equivalence classes of circulant graphs.
1. Introduction Let A be a finite group with identity e and let Ω be a set of generators for A with the properties that Ω = Ω−1 and e ∈ Ω, where Ω−1 = {x−1 | x ∈ Ω}. The Cayley graph C (A, Ω) is a simple graph whose vertex-set and edge-set are defined as follows: V (C (A, Ω)) = A and E (C (A, Ω)) = {{g, h} | g −1h ∈ Ω}. Because of its rich connections with broad range of areas, it has been in the center of the research in graph theory [1, 4, 14, 15]. Spectral estimations of Cayley graphs have been studied [2, 7]. It plays a key role in the study of the geometry of hyperbolic groups [9]. Recently, Li has found wonderful results on edge-transitive Cayley graphs [12, 13]. The Cayley graph C (A, Ω) admits a natural A-action, · : A× C (A, Ω) → C (A, Ω) defined by g · g ′ = gg ′ for all g, g ′ ∈ A. A graph G with an A-action is called an A-graph. So, every Cayley graph C (A, Ω) is an A-graph. A graph isomorphism f : G → H between two A-graphs is weak equivalence if there exists a group automorphism α : A → A such that f (g · u) = α(g ) · f (u) for all g ∈ A and u ∈ V (G). When α is the identity automorphism, we say that f is an equivalence. If there is a weak equivalence between A-graphs G and H , we say G and H are weak equivalent. Similarly, if there is an equivalence between A-graphs G and H , we say G and H are equivalent. For standard terms and notations, we refer to [8]. Enumerations of the equivalence classes and weak equivalence classes of some graphs have been studied [6, 11]. The purpose of this article is to enumerate the equivalence classes and weak equivalence classes of Cayley graphs for a finite group A. The outline of this paper is as follows. In section 2, we characterize the weak equivalence of Cayley graphs for a finite group A. Using these characterizations, we find enumeration formulae of the equivalence classes and weak equivalence classes of Cayley graphs in section 3. As an application, we find the number of weak equivalence classes of circulant graphs in section 4.
2000 Mathematics Subject Classification. Primary 05C30; Secondary 05C25. Key words and phrases. Cayley graphs, weak equivalences, equivalences, circulant graphs. The third author was supported by Com2 MaC-KOSEF(R11-1999-054).
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DONGSEOK KIM, JIN HWAN KIM, JAEUN LEE, AND DIANJUN WANG
2. A characterization of Cayley graphs Our definition of an weak equivalence between two Cayley graphs can be interpolated as a color-consistence and direction preserving graph isomorphism [8, Section 1.2.4]. Theorem 2.1. Let C (A, Ω) and C (A, Ω′ ) be two Cayley graphs. Then the followings are equivalent. (1) C (A, Ω) and C (A, Ω′) are weakly equivalent, (2) There exists an isomorphism α : A → A such that α(Ω) = Ω′ . Proof. (1) ⇒ (2): Let f : C (A, Ω) → C (A, Ω′ ) be a weak equivalence. Then there exists a group automorphism τ : A → A such that f (g ) = τ (g )f (e) for each g ∈ A. Let x ∈ Ω. Then {e, x} is an edge in C (A, Ω). Since {f (e), f (x)} is an edge in C (A, Ω′ ), f (e)−1 f (x) = f (e)−1 τ (x)f (e) is an element of Ω′ . Hence the map α : A → A defined by α(g ) = f (e)−1 τ (g )f (e) is a group isomorphism such that α(Ω) = Ω′ . (2) ⇒ (1): Let α : A → A be a group automorphism such that α(Ω) = Ω′ . We define f : C (A, Ω) → C (A, Ω′ ) by f (g ) = α(g ). If {g, h} is an edge in C (A, Ω), then g −1h ∈ Ω and f (g )−1 f (h) = α(g )−1α(h) = α(g −1h) ∈ α(Ω) = Ω′ . Hence f is a graph isomorphism such that f (gg ′) = α(gg ′) = α(g )α(g ′) = α(g )f (g ′), i.e., f is a weak equivalence. By using a similar method in Theorem 2.1, we can have the following theorem. Theorem 2.2. Let C (A, Ω) and C (A, Ω′ ) be two Cayley graphs. Then the followings are equivalent. (1) C (A, Ω) and C (A, Ω′) are equivalent, (2) Ω and Ω′ are conjugate in A, i.e., there exists an element g ∈ A such that g −1 Ωg = Ω′ . 3. Enumeration formulae For a finite group A, let Gm (A) = {Ω ⊂ A : Ω−1 = Ω, < Ω >= A, |Ω| = m, e ∈ Ω}. Notice that Gm (A) contains all equivalence classes of Cayley graphs C (A, Ω) of degree m. Let H be a group of group automorphisms of A. H admits a natural action on Gm (A) by α · Ω = α(Ω). By Theorem 2.1, E w (A, m), the number of weak equivalence classes of Cayley graphs C (A, Ω) of degree m, is equal to the number of orbits of the Aut(A) action on Gm (A), where Aut(A) is the group of all group isomorphisms of A. Similarly, one can see that the number E (A, m) of equivalence classes of Cayley graphs C (A, Ω) of degree m is equal to the number of orbits of the Inn(A) action on Gm (A) by Theorem 2.2, where Inn(A) is the group of all inner automorphisms of A. For any subset S of A, let O2 (S ) = {g ∈ S : g 2 = e, g = e}. We observe that
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