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南开大学算法导论第六章课件

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带权的区间调度

预处理：算法实现的过程中先对开始时间和结束时间排序，便于后面的处理一些编程语言，如Lisp，就可以自动实现备忘录的功能(built-in support for memorization)，因而有好的执行效率；但是在其他的一些语言中，比如Java，就没有实现这一功能。
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带权的区间调度
不同区间的影响是不一样的
a b c d e f g h
0 1 2 3 4 5 6 7 8 9 10 11
Time
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带权的区间调度

贪心算法情形下，区间的权重可以看成都是1 对于区间权重不同的情形，贪心算法就失效了
weight = 999
b
weight = 1
a Time
解决之道：更加灵活的调度策略

对于带权区间调度问题，我们可以不使用备忘录式的递归方法，而通过迭代算法直接计算M[]中的项。
,
Input: n, s1,…,sn
f1,…,fn
,
v1,…,vn
Sort jobs by finish times so that f1 ≤ f2 ≤ ... ≤ fn. Compute p(1), p(2), …, p(n) Iterative-Compute-Opt { M[0] = 0 for j = 1 to n M[j] = max(vj + M[p(j)], M[j-1]) } 自底向上
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带权的区间调度

递归的备忘录形式为了避免上面的重复计算，我们把中间计算的结果存储起来，需要的时候先查找是否计算过下面将用到一个数组M[0…n]保存中间计算结果
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北师大版选修六 Pompeii 说课课件

Difficulties

To comprehend the content of the text clearly and quickly
Methods

Task-based teaching method --- assign tasks according the Ss’ level and learn step by step Student-centered learning (pair work /group work)

2. Find out “time capsule” in the text. And find out the reason.
Time capsule preserves a frozen moment in history.
(On the 1st-2nd line in paragraph 4)

If still have time

Look at the pictures and feel the last moment.
People gathered together for protection in their last hours of life.
One person, sitting alone, looks like he was praying.
Teaching difficulties & keys

Keys

How to predict the content of the text according to the title and the pictures How to grasp the main idea of the article and each paragraph

Algebra-Chapter 1 Tools of Algebra

Unit 1 – Tools of AlgebraSection 1 – Using Variables(Refer to Pages 49 - 50, problem numbers 1 - 72 in your textbook for additional practice)Vocabulary and Examples:Variable – a letter used to hold the place of a number in a math problem Ex: 3x + 4 = 2Algebraic Expression – a math problem that only includes the front of the problem (numbers, variables and operations – NO “=, <, >,” etc)Ex: 3 more than g g + 3 (notice the back of the problem - the partbehind the equal sign - is not there)Algebraic Equation – a math problem that includes the back of the problem as well as the front of the problem (a complete math sentence containing “=, <, >, etc.)Ex: 3 more than g is 45 g + 3 = 45 (notice the front of the problem aswell as the back of the problem are there):Sum” or “Increased By” or “More Than” – all three mean additionex: 6 more than a number is 12 a + 6 = 12**when the problem asks for “THE SUM OF, THE PRODUCT OF, THE QUOTIENT OF, OR THE DIFFERENCE OF” put parenthesis around that part of the problem Ex: the sum of 4 and d minus 40 (4 + d) - 40Product – means multiplication – Remember that in algebra, a raised dot is used for the multiplication symbol (so as not to confuse the multiplication symbol of “x” with the letter (variable) “x”) ex: the product of 6 and a number is 12 6 y a = 12Quotient – means divisionex: the quotient of 6 and a number is 12 6a= 12Difference – means subtractionex: the difference between 6 and a number is 12 6 - a = 12Less than – means subtraction - REVERSE THE ORDER OF THE NUMBER AND VARIABLE WHEN WRITING A MATH PROBLEM THAT INCLUDES “LESS THAN”ex: 6 less than a number is 12 a – 6 = 12Twice – 2 times (2 y )ex: the sum of 6 and twice a number is 12 6 + 2 y a = 12variable since the number is unknown, assign ita variable (letter)“is” means “equals”quotient is typically expressed as a fraction 2 timesTo write a phrase for a given math problem - use the vocabulary word for the given operation Ex: 4x – 8 the product of 4 and some number minus 8 OR8 less than the product of 4 and some numberTo define a variables and write an equation to model the relationship in a table –a)assign a variable (letter) for each column in the tableb) determine what was done to the number in the first column to get the number in thesecond column for each row. Ex: let n be the “number of sales” and eTo go from 5 to $2.00, multiply by $0.4To go from 10 to $4.00, multiply by $0.4To go from 15 to $6.00, multiply by $0.4To go from 20 to $8.00, multiply by $0.4THEREFORE: the VARIABLES are: “n” and “e”let n be the “number of sales” and e be “total earnings in dollars”the EQUATION becomes: e = 0.4nPractice Problems – Unit 1 – Section 1For problems 1 - 6, write an algebraic expression for each phrase.1) 45 increased by a 2) n less than 93) the product of 4 and r 4) the sum of 8 and h minus 375) 8 minus the product of 12 and y6) the sum of the quotient r and 12 and the quotient of w and 8For problems 7 - 10, write a phrase for each expression.7) 5 – g 8) 8a + 5 9) k410) 6abFor problems 11 and 12, define variables and write an equation to model the relationship in each table.12)Unit 1 – Tools of AlgebraSection 2 – Exponents and Orders of Operations(Refer to Page 60, problem numbers 1 - 37 in your textbook for additional practice)Vocabulary and Examples:Exponent – tells how many times a number, the base, is used as a factor.Ex: 64 To solve: 6 • 6 • 6 • 6Power – consists of the base and the exponentex: 64 is read as “6 to the fourth power” Order of Operation – the order that every math problem is to be completed in -First: complete all P arenthesesSecond: complete any E xponentsThird: complete any M ultiplication and D ivision (in the order it appears in the problem from left to right)Fourth: complete any A ddition and S ubtraction (in the order it appears in the problem from left to right)(PEMDAS – Using “Please Excuse My Dear Aunt Sally “helps toremember order of operations)– 10 • 6 ÷ 2base exponentThe base multiplied 4 times 1296 Within the parentheses, there is an exponent, and addition. Follow the order of operation here too. Solve the exponent first, then add.Practice Problems – Unit 1 – Section 2For problems 1 -5, simplify each expression.1) 8 y 4 + 922) 21 + 49 ÷ 7 + 13) (102 – 4 y 8) ÷ (8 + 9) 4) 111 + 3 (2211+ 8) ÷ 55) 27[52 ÷ (42 + 32) + 2]For problems 6 - 9, evaluate each expression for a = 5, b = 12, and c = 2. 6) abc + ab7) 2b ÷ c + 3a 8) b2 – 4a 9) b[(ac + a) ÷ a] – b For problems 10 – 13, evaluate each expression for m = 3, p = 7, and q = 4.10) (mp)2 – q 11) m( p – q)2 12) qp2 + pq2 13) m(p2 – q)14) The formula for the volume of a sphere with radius r is V = 433πr. Find the volume of a ballthat has a radius 8.7 in. Round your answer to the nearest hundredths place. (HINT: Plug in 8.7 for r in the formula and use 3.14 for π. Remember to follow order of operation rules!!)15) In 1883, Jan Matzeliger invented the shoe-lasting machine to attach the upper part of a sole to its sole. Before that, each shoe was assembled and sewn by hand. You can estimate thewages at that time using the formula wpt=034., where w is the hourly wage in dollars, p is theprice of a pair of shoes in dollars, and t is the time in minutes to assemble a pair of shoes.In 1891, a worker who used the shoe-lasting machine could assemble on pair of shoes in two minutes. A pair of shoes cost about $.94. Estimate the worker’s hourly wage to the nearest cent. (HINT: Replace p with the cost of a pair of shoes in 1891, and t with the amount of time it took to assemble a pair of shoes in 1891. Remember to follow the order of operation rules!)Unit 1 – Tools of AlgebraSection 3 – Exploring Real Numbers(Refer to Page 6, problem numbers 9 - 22 in your textbook for additional practice) Vocabulary and Examples:Natural Number – are the “counting numbers” beginning with 1 to infinity, but NOT including decimals and any fractions that DO NOT reduce to a natural numberExample: 1, 2, 3, 4, ……. Fractions like : 255(because it reduces to 5, whichis a natural number)Whole Number – are all of the natural numbers and 0Example: 0, 1, 2, 3, 4, …….Integers – are all of the whole numbers and their opposites (negative numbers), but NOT including decimals or any fractions that DO NOT reduce to integersExample: ……….-3, -2, -1,. 0, 1, 2, 3, 4, ……. Fractions like : −255(because itreduces to -5, which is a natural number) Rational Number – any number that can be written as a fraction. This includes all of the natural numbers, whole numbers and integers. The only numbers that CANNOT be rational are those numbers that are decimals that have no pattern and never end (irrational numbers)Example: (a) 16 = 161or644(b) .145 = 145 1000(c) .3 = 1 3(d) 227= 18.85714286….. This is not rational because thedecimal does not end and there is no patternIrrationalNumber – any number that is not rationalExample:10 = 16227766….. This number is irrational because thedecimal does not end and there is no patternOther examples are: 2, 0.102003000400005...Real Number – every number in our number system. This includes all rational numbers, whole numbers, integers, rational numbers, and irrational numbersCounterexample – is an example that proves a statement to be falseExample: no fractions are whole numbers – this statement is false. A counterexample that proves that it is false could be305= 6 because 6 is a whole number and is also a fraction.Inequality – the math symbols: < which means “is less than”, > which means “is greater than”, ≤ which means “is less than or equal to”, and ≥ which means “is greater than or equal to”Opposites - are the numbers that are the same distance from zero on a number line but lie in opposite directions.Example: 3 and -3, −715 and 715Absolute Value of a Number – is the distance from 0 that a number lies on a number line.Example: |3| = 3 because 3 is 3 units away from zero on a number line Example: |-3| = 3 because -3 is 3 units away from zero on a number lineExample: |-6 + 5| - first solve what is inside the absolute value symbols because the symbols also act as grouping symbols (parenthesis) and order of operation rules require that the parenthesis be done first.|-6 + 5| = |-1| Then take the absolute value of -1 which is 1 -the final answer is 1THEREFORE: it doesn’t matter whether the number inside the absolute value symbols is positive or negative, the answer is ALWAYS positive UNLESS there is a negative sign OUTSIDE the absolute value symbols, then the answer becomes negative. Example: -|24 – 3 y -2| = -|24 – 3 y -2|-|24 – (-6)| -|30| = -30Practice Problems – Unit 1 – Section 3For problems 1 - 5, name the set(s) of numbers to which each number belongs. The response could be natural number, whole number, integer, rational number, irrational number, and real number. 1) -12)233) 7 4) 05) 5For problems 6 -7, tell whether each statement is true or false. If the statement is false, give a counterexample.6) All integers are rational numbers.7) All negative numbers are integers.8) Order the numbers from least to greatest:711, 0.63, 0.636For problems 9 -11, find the absolute value.9) |4| 10)−91411) −625.For problems 12 and 13, use <, =, or > to compare.12) −3121. 312.13)−810 −1620For problems 14 and 15, simplify each expression. (HINT: absolute value symbols are grouping symbols, so do what is inside of them first)14) |-6 + 4| + |3| 15) 4 + |3 y 4 – 2|Unit 1 – Tools of AlgebraSection 4 – Adding Real Numbers(Refer to Page 120, problem numbers 1 - 37 in your textbook for additional practice) Vocabulary and Examples:Identity Property of Addition – The sum of zero and any number is always the non-zero number.Example: 0 + 6 = 6 -6 + 0 = -6Inverse Property of Addition – The sum of any number and its opposite will always result in zeroExample: -16 + 16 = 0 23+ ()−23= 0Matrix – a rectangular arrangement of numbers in rows and columns. The size of amatrix is indicated by the number of rows and the number of columns. To add or subtract matrices (plural of matrix), add the two numbers that correspond in each matrix andrecord the answer in the same position in a new matrix.Example:This matrix is a ‘2 by 3’ because there are 2 rows and 3 columns. Only matrices with the same dimensions can be added or subtracted together.Practice Problems – Unit 1 – Section 4For problems 1 -5, simplify.1) -8.7 + (-10.3)2)45 + 2153) -7 + (-4) 4) −438+5) −+23466) A diver dives 47 ft. below the surface of the water and then rises 12 ft. Use addition to find the diver’s depth.7) The temperature at 6 A.M. is -6°F. The temperature rises 13 degrees Fahrenheit by noon. Use addition to find the temperature at noon.8) On two football plays a team gains 8 yds. and then loses 5 yds. Use addition to find the result of the two plays.For problems 9 and 14, simplify.9) 7 -8 -8 9.4 10) 1.3 26 0.5 -4-12 6.2 -9 1718 -2 −58911) -13.2 + 7 + (-6.8) 12) -8.02 + |-5.9| + 0.413) |-2| +│2/3│ + 314) -4.3 + 1.2 + (-5.7)Use the table for problems 15 - 1715) How many people aged 18 to 34 participate in photography?16) How many people aged 45 to 64 draw?17) Which of the activities is most popular?+ -Unit 1 – Tools of AlgebraSection 5 – Subtracting Real Numbers(Refer to Page 139, problem numbers 1 - 64 in your textbook for additional practice) Vocabulary and Examples:Subtracting Numbers – when subtracting integers (positive and negative numbers), you can change the subtraction problem to addition by adding the opposite.Example: 3 – 5 means the same as 3 + (-5) = -26 – (-6) means the same as 6 + 6 = 12**REMEMBER: when working an absolute value problem, completely solve the problem within the absolute value symbols, and then make the FINAL answer positive (unlessthere is a negative sign OUTSIDE of the absolute value symbols, which would result inthe FINAL answer being negative)Example 1: |5 – 11| =|-6| = 6Example 2: -|12 – (-6)| OR -|12 + 6| =-|18| = -18Practice Problems – Unit 1 – Section 5For problems 1 -5, simplify.1) 3 - 7 2) 2 – (-9) 3) 5.3 – (-8.4)4) −25-7105) |-3 – (-5)|For problems 6 – 8, evaluate each expression for x = 3, y = -4, and z = 6.6) -x – y 7) 2x – z 8) -z + y – xFor problems 9 – 14, evaluate each expression for a = -2, b = 3.5, and c = -4.9) -c – b + a 10) |a| + |b| 11) -|3 + a|12) |a| + |3b| 13) |a – c| - |c| 14) -4b - |a|15) Archaeologists found a 1500-year-old ship at the bottom of the Black Sea. The ship is well preserved because oxygen could not make the ship decay. The ship is at a depth of 1000 ft. below the surface. This is about 350 ft. below the boundary between surface water, which has oxygen, and water below, which does not have oxygen. At what depth is the boundary?Unit 1 – Tools of AlgebraSection 6 – Multiplying and Dividing Real Numbers(Refer to Page 135, problem numbers 1 – 36 and Page 144, problem numbers 1 – 51 in your textbook for additional practice)Vocabulary and Examples:Identity Property of Multiplication – multiplying any number by 1 will always results in the non-one number Example: 6 y 1 = 6 1 y -23 = -23Multiplication Property of Zero – For every real number n , n y 0 = 0 (multiplying any number by zero always results in zero) Example: 1 y 0 = 0 0 y -23 = 0Multiplication Property of -1 – for every real number n , -1 y n = -n(multiplying any POSITIVE number by -1 will always result in a negative answer) Example: -1 y 6 = -6 23 y -1 = -23Multiplying and Dividing Integers (positive and negative numbers) – any time two numbers that have the SAME sign are multiplied or divided, the answer is always a positive number Example: 6 y 2 = 12 -6 y -2 = 12623= −−=623Any time two numbers that have DIFFERENT signs are multiplied or divided, the answer is always a negative number Example: -6 y 2 = -12 6 y -2 = -10−=−623623−=−−=−623Inverse Property of Multiplication – For every nonzero number, a , there is a multiplicativeinverse (reciprocal), 1a, such that a ·1/a = 1 (any number times the multiplicative inverse(reciprocal) – flip that number upside down - will always equal 1) Example: 6· 1/6 = 1 1/15 ·15 = 1Practice Problems – Unit 1 – Section 6For problems 1 -11, simplify.1) 13(-6) 2) -20(-4)3) -(-2)3 (HINT: a negative sign with no number attached to it is -1 – the 1 is invisible but really there!)4) -52(-3)35) 6 3−6) 3142−−7)-56÷(4+3)8)2.25÷39) |-6(-9)| ÷ (-2) - HINT: remember to work everything inside the absolute value symbols first, then take the absolute value of the answer!10)(-2)(5)(-3)11)2/3−4/5For problems 12 - 14, evaluate each expression for x = -2, y = 3, and z = 3.512) (y + 3x) ÷ y 13) (3x + 2y) ÷ (2x + 3y)14) 8 + 6x ÷ 4y - 3z y15) A toll bridge in Maine in the early 1900’s charged 2¢ per person and 614¢ for a dozen sheep.How much would the toll for 3 people and 4 dozen sheep have been?Unit 1 – Tools of AlgebraSection 7 – The Distributive Property(Refer to Page 74, problem numbers 1 - 50 in your textbook for additional practice)Vocabulary and Examples:The Distributive Property – allows the number that is OUTSIDE of the parenthesis to be MULTIPLIED by every term INSIDE the parenthesis (usually this property is used when what is inside the parenthesis CANNOT be completed first, even though order of operation rules indicate that parenthesis must be done first – typically, if there is avariable inside the parenthesis, the distributive property MUST be used because letters and numbers CANNOT be added or subtracted since they are not like terms)Example (a): 6(a + 4) This problem can be written as: 6 y a and 6 y 4 (remember: there is an understood multiplicationsymbol between the 6 and the parenthesis)6 y 6 y 4 = 24 THEREFORE, the final answer becomes6a + 24Example (b): 4(a - 2) This problem can be written as: 4 y and 4 y -24 y a = 4a4 y 2 = -8 THEREFORE, the final answer becomes4a - 8y a and -1 y -8-1 y a = -1a-1 y -8 = 8 THEREFORE, the final answer becomes-1a + 8Combining Like Terms - adding or subtracting terms that have the same variableEx: 6a 2 – 5ab + 3ab – 12a 2Both 6a 2 and (-12a 2) are like terms since they both have a 2 attached to the coefficient (the number in front of the variable) - Combine 6a 2 – 12a 2 and the answer is: -6a 2Combine (-5ab) and 3ab since they are like terms because they both have the same variables (ab) attached to the coefficient - Combine (-5ab) and 3ab and the answer is: -2abThe final answer is: -6a 2 – 2ab – once there are no more like terms to combine, the problem is finishedREMEMBER: what makes terms alike is that the variables are 100% identical or there are no variables attached, only the coefficient exists Example: -6d – 4a + 3d – 6 – 12a + 20Example: 8m 2 – 5mz + 4mz – m 2+ 4To write and expression for a phrase – any time the word “quantity” appears in an expression, use parenthesis around the phraseExample: write an expression for “3 times the quantity x minus 5” 3(x – 5)Like termsLike terms Like termsLike termsPractice Problems – Unit 1 – Section 7 For problems 1 -12, simplify.1) 7(r – 4) 2) -2(n – 6) 3) (5b – 4)1 54) -(x + 3) 5) (4 – a)(-1) 6) -(2 – 7x) 7) 4g – 7g 8) w + 23w 9) 13q – 30q 10) -(8.4 + 300g – 512h) 11) 9 – 4f + 6y – 3f + 1012) 1.4b – 3b2 + 4c – 2b2 + cFor problems 13 – 15, write an expression for each phrase.13) 3 times the quantity m minus 714) -4 times the quantity 4 plus w15) the product of -11 and the quantity m minus 816) Suppose you buy 4 cans of tomatoes at $1.02 each, 3 cans of tuna for $.99 each, and 3 boxes of pasta at $.52 each. Find the total cost.Unit 1 – Tools of AlgebraSection 8 – Properties of Real Numbers(Refer to Page 68 - 69, problem numbers 1 - 24 in your textbook for additional practice) Vocabulary and Examples:Commutative Property of Addition – reversing the order of any numbers in an additionproblem does not affect the answerExample: 8 + 4 = 4 + 8Commutative Property of Multiplication – reversing the order of any numbers in amultiplication problem does not affect the answerExample: 8 y 4 = 4 y 8Associative Property of Addition – moving the parenthesis from around any numbers inan addition problem and putting them around different numbers does not affect theanswerExample: (8 + 4) + 6 = 8 + (4 + 6)Associative Property of Multiplication – moving the parenthesis from around any numbers in a multiplication problem and putting them around different numbers does not affect the answerExample: (8 y 4) y 6 = 8 y (4 y 6)Identity Property of Addition – adding 0 to any number will always results in the nonzero numberExample: 6 + 0 = 6Identity Property of Multiplication – multiplying any number by 1 will always results in the non-one numberExample: 6 y 1 = 6 1 y -23 = -23Inverse Property of Addition – Adding the positive and negative of the same number will always result in 0Example: 6 + (-6) = 0Inverse Property of Multiplication – For every nonzero number, a, there is a multiplicativeinverse (reciprocal), 1a, such that a ·1/2 = 1 (any number times the multiplicative inverse(reciprocal) – flip that number upside down - will always equal 1Example: 6·1/6 = 1 1/15 ·15 = 1The Distributive Property – allows the term OUTSIDE of the parenthesis to be MULTIPLIED by every term INSIDE of the parenthesis (usually this property is used when what is inside the parenthesis CANNOT be done first, even though order of operation rules require parenthesis to be done first. The distributive property solves that problem)Multiplication Property of Zero – For every real number n, n y 0 = 0(multiplying any number by zero always results in zero for the answer)Example: 1 y 0 = 0 0 y -23 = 0Multiplication Property of -1 – for every real number n, -1 y n = -n(multiplying any POSITIVE number by -1 will always result in a negative answer) Example: -1 y 6 = -6 23 y -1 = -23Practice Problems– Unit 1 – Section 8For problems 1 -9, name the property that each equation illustrates.1) −67+ 0 = −672) 8 + 43 = 43 + 83) 1 y 2123=21234) (-7 + 4) + 1 = -7 + (4 + 1)5) -0.3 + 0.3 = 0 6) 9(7.3) = 7.3(9)7) 5(12 – 4) = 5(12) – 5(4) 8) 8(9 y 11) = (8 y 9) y 119) −12y (-2) = 1For problems 10 – 15, tell whether the expression in each pair is equivalent. 10) 6m + 1 and 1 y 6 + m 11) 9y and 9 + y12) -(5 – 9) and 9 – 5 13) 6a – 4 and 2[(2 + 1)a -2] 14) vwx y yz and v y w y zxy 15) 3(5 + a) and 15 + aUnit 1 – Tools of AlgebraSection 9 – Graphing Data on the Coordinate Plan(Refer to Page 620, problem numbers 1 - 18 in your textbook for additional practice)REMEMBER: when graphing an ordered pair, first move left or right along the x-axis, then move up or down on the y-axis, and then put a dot to represent the point (ordered pair)Scatter Plot – a graph that has points that are scattered through the graph. Points on a scatter plot CANNOT be connected to create a straight line.Positive correlation – In general, both sets of data (along the x- and y-axis, increase together – notice all of the points are clustered in an upward direction Negative correlation – In general, one set of data decreases (the y-axis data) as the other set of data increases (the x-axis data) – notice all of the points are clustered in a downward direction No correlation – None of the data is related – notice the points are not clustered at all, but instead are scattered all over the coordinate plane origin coordinate plane x - coordinate y - coordinate the 4 quadrants, but instead, isconsidered to be on the axis itselfPractice Problems – Unit 1 – Section 9For problems 1 -6, name the coordinates of each point on the graph below.1) A 2) B 3) C 4) D 5) E 6) FFor problems 7 – 10, graph the points on the same coordinate plane.7) (3, 0) 8) (-1, 4) 9) (-2, -3) 10) (3, -3)For problems 11 – 14, in which quadrant or on which axis would the point lie?11) (-10, 4) 12) (-11, 0) 13) (8, -13) 14) (0, 25)For problems 15 – 17, describe the correlation represented in each scatter plot below.Algebra 1a Adult Education 2006 - 21 - Unit 1 – Section 9Algebra 1a Adult Education 2006 - 55 - Unit 3 – Section 1Ex 5: Graph 4≥m notice that the variable is not first, so flipping the entireproblem so that the variable is first will make it so thatthe rule of “as long as the variable is first, the inequalitywill point in the direction that the arrow should go on thegraph” is true.m ≤4 notice that the inequality also flips directions.In this example, it is a closed circle and the arrow should point to the leftEx 6: write an inequality for each graphFirst, pick any variable. Then, notice the direction of the arrow andwhether the circle is closed (colored in) open (not colored in). Last,name the numbera)variable - “a”arrow is pointing left and closed circle – means “less than or equal to”number - “-4”THEREFORE: a ≤ -4b)variable - “c”arrow is pointing right and open circle – means “greater than”number - “2”THEREFORE: c > 2-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10。

讲义_6 & lecture_6

Comments
. A planar response surface may not always be appropriate, but even when not it is often a good approximate descriptor of the regression function in “local” regions of the input space . The meaning of the parameters can be determined by taking partials 2
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Multiple Regression
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coeﬃcients. They represents the partial eﬀect of one predictor variable when the other predictor variable is included in the model and is held constant.
Multiple Regression
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Example
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Yang Feng (Columbia University)

高等数学课件-D32洛必达法则

例题二：判断函数性质问题
题目
判断函数 f(x) = (e^x - e^(-x)) / (e^x + e^(-x)) 的奇偶性。
解题思路
本题考察的是利用洛必达法则判断函数的性质。首先，我们需要判断函数在x=0处的值，然后利用洛必达法则求解函数在x→0时的极限值，最后根据奇偶性的定义进行判断。
例题二：判断函数性质问题
THANKS FOR WATCHING
感谢您的观看
总结回顾本次课程内容
洛必达法则的基本概念
洛必达法则是用于求解不定式极限的一种有效方法，通过分子分母分别求导的方式，简化极限的求解过程。
洛必达法则的适用条件
在使用洛必达法则时，需要满足一定的条件，如分子分母在某点的去心邻域内可导，且分母导数不为零等。
洛必达法则的求解步骤
首先验证是否满足适用条件，然后分别对分子分母求导，得到新的分子分母，再次判断是否满足适用条件，如此循环直至求出极限或判断极限不存在。
泰勒公式可以将函数展开为多项式形式，而洛必达法则可以解决多项式函数的极限问题。因此，可以将泰勒公式与洛必达法则结合使用，解决复杂函数的极限问题。
要点二
复杂函数极限的求解
对于复杂函数，可以先使用泰勒公式将其展开为多项式形式，然后应用洛必达法则进行求解。这种方法可以简化复杂函数的极限求解过程。
在复变函数中应用
证明过程
由于$varphi(x)$在点$a$附近单调且有界，因此存在极限 $lim_{x to a} varphi(x) = l$。又因为$frac{F'(x)}{G'(x)} to l$，所以$frac{F(x)}{G(x)} to l$。
03 洛必达法则在高等数学中应用

近年高考英语一轮复习第一部分教材重点全程攻略Unit4Globalwarming讲义新人教版选修6

2019版高考英语一轮复习第一部分教材重点全程攻略Unit 4 Global warming讲义新人教版选修6编辑整理：尊敬的读者朋友们：这里是精品文档编辑中心，本文档内容是由我和我的同事精心编辑整理后发布的，发布之前我们对文中内容进行仔细校对，但是难免会有疏漏的地方，但是任然希望（2019版高考英语一轮复习第一部分教材重点全程攻略Unit 4 Global warming讲义新人教版选修6）的内容能够给您的工作和学习带来便利。

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Unit 4 Global warming话题素材——全球变暖［话题词汇］1。

atmosphere n。

大气;气氛2。

recycle v. 回收利用3。

unbearable adj。

难以忍受的4.decline vi. 减少;降低；下降5。

undertake vt. 承担；从事；负责6。

raise people's awareness of environmental protection 提高人们的环保意识7.bring 。

.. under control 控制……8.die out 灭绝9.white pollution 白色污染10。

from bad to worse 每况愈下11.be faced with 面临12.do what we can 做我们能做的一切13。

reduce pollution 减少污染14.throw away 丢弃15.in harmony with nature 与自然和谐相处［经典佳句］1.There_is_no_doubt_that the earth is becoming warmer and thatit_is_human_activities_that have caused this global warming。

lecture 6 网易课教程字幕有机化学可汗学院

Lecture 6I think we’re ready now to tackle some more or even more complicated examples. So let’s draw something crazy here. So let’s see, let me draw a chain. Let me draw it like that. And so like we’ve done in all of the examples, you want to find the longest chain. We could count from here. One, two, three, four, five, six, seven, eight, or maybe it’s one, two, three, four, five, six, seven. No. or maybe it’s one, two, three, four, five, six, seven, eight, nine, ten. That is our longest chain. Let me make that in green. So our longest chain here is in green. So this decane—sorry, this backbone has ten carbons in it. The prefix for ten is dec-. It’s an alkane since it has all single bonds, so we can write decane for the backbone. And then it has a group right here and this group consists of one, two carbons attached to the backbone. The prefix for two carbons is eth-, so this is an ethyl group. The –yl is because it’s a group attached to the main alkane chain, so we call this ethyldecane. But we have to specify where the ethyl group is attached. And we want to give it as low of a number as possible, so we start counting on the side closest to it. So it’s one, two, three, four, five, so this is 5-ethyldecane.Now let’s complicate this a little bit more. So let me just to copy and paste this. Copy. And then paste. So I have pasted it there. And let me complicate this molecule a little bit more. Let me add another ethyl group to it. So let’s say we have another ethyl group over there. Now, what is this going to be? Well, the longest chain is still going to be that thing in green, so it’s still going to be a decane. But now we have two ethyl groups, one on the five carbon, one, two, three, four, five, and then one on the six carbon. So we write here—you might be tempted to write 5-ethyl-6-ethyldecane, which really wouldn’t be wrong, but it would just be maybe more letters than you want to write. Instead you write 5, 6-diethyldecane. The 5, 6- tells us the two carbons one the main backbone that the ethyl groups are attached to, and the di- says that we have two ethanes. Or two ethyl groups, I should say, not ethane groups. Two ethyl groups, one over here and one right over here. Now, let’s me make it even more complex.Let’s take our thing and let’s make it even more complex. So let’s copy it and let us paste it, edit, paste, just like that. Let’s paste it. Let me make it a little bit lower over here. So the exact same thing we just dealt with, but let’s and one more group to it. Let’s say it looks something like this. Let’s say on that five carbon—let’s say on that six carbon right there, I have a group that looks something like this. What is this is going to be called? And here we probably want to break it down a little bit. We still have the backbone, so it’s still going to be a decane. But let’s just do it step by step. So we know that this right here, this is an ethyl group. It’s on the five carbon, so that is a 5-ethyl. This is another ethyl group on the six carbon, so that is a 6-ethyl. Now, we have this group over here. So how many carbons are on this group? Well, we have one, two, three carbons on it, so we might want to just call it a propyl group. But notice something, we are attached to the carbon, and then it later branches off into two. It actually immediately branches off into two. So we could actually call this either sec-propyl, so this right here, we can either call it sec-propyl, or because it is connected away from the—I guess it’s connected directly to the branch the way a propyl group works. You could connect it and it immediately branches, because you only have three carbons. so just to clarify things, it’s sec- because we’re attached to two carbons right there. But sec-propyl isn’t what people normally call it. They normally call it isopropyl. And the iso- root comes from anything of the form. So if I have a group that looks like this, this would be isobutyl, because we have one, two, three, four carbons, and we are attached right here, so this iswhere we would attach to the main backbone. If I have a group that looks like this, this would be isopentyl. I have one, two, three, four, five carbons. I’m attached away from the branch. But if we go one step closer, so we’re attached just like this, we’re attached to the original group, this is isopropyl. One, two, three carbons. let me write—isopropyl, isobutyl, isopentyl. Notice what the isos—the isos are like Y’s, I guess is the best way of thinking of it, and you’re attached to the base of the Y. in this case, you’re almost attached to the base of a V, but, hopefully, you get the idea. so the common name for this group would be isopropyl, or if we wanted to use the systematic naming, we could just start at one as being—let me do this in a good color—as the carbon we are attached to, and then we only have two carbons right over there. So if we have two carbons, we’re dealing with an ethyl group. And then you could say that you have a methyl group attached to the first carbon, so it would be 1-methylethyl. So this could either be called an isopropyl group, or a 1-methylethyl group. And in either case—let me put that in parentheses. In either case, it’s attached to the six carbon. So the prefixes are going to be on the six carbon on our main—we have a 1-methylethyl, where the methyl is attached to the first carbon on the ethyl where the methyl is attached to the first carbon on the ethyl chain, or we could call this a 6-isopropyl group attached to the six carbon. Now, when we decide how to write it, how do we know—we’re going to have to write essentially the ethyl and isopropyl. We’re going to have to write all of that in front of the decane. How do we decide whether to write the eth- or the propyl- first? Well, here you just do it in alphabetical order. And actually these little prefixes in front don’t matter. So you compare the P to the E. eth-, E comes first in alphabetical order, so this would be 5,6-diethyl. So even though D—well, D comes even before E, but you don’t even count that prefix. The E is what matters. Diethyl, and we could say, 6-isopropyldecane. This is the common name: 6-isopropyl decane. This would be the common name for it. If we want to use the systematic name, we would replace the 6-isopropyl with 6-(1-methylethyl). So let me write that down. So let me just copy this part. So copy and paste right there. And then I can copy and paste this right here. Copy and paste. So it becomes 5,6-diethyl-6-(1-methylethyl). I know it’s very confusing when you see it, but when you break it down, you really could—you would be able to draw this from the structure. And then finally you finish with a decane. So hopefully, these examples don’t confuse you too much. In the next few videos, we’ll do more and more examples. Because I think with the nomenclature on the organic chemistry, the more examples you see, the better. This is really core, so in the future you don’t get confused when people throw out something like, 5,6-diethyl-6-(1-methylethyl)decane.。

OPTAVIA 业务培训手册说明书

3-WAY CONVERSATION TIPS•3-way conversations can happen over the phone, via web conferencing (such as Zoom), or face-to-face. Work with your Business Coach to determine the best method.•These conversations should take about 20-30 minutes with the Candidate doing most of the talking.• Take notes using the Health Assessment or Wellbeing Evaluation.• Debrief with your Business Coach after the conversation.»Go over the call to review what transpired.»Review your notes and ask your Business Coach questions. Decide on next steps with the Candidate.• Remember, the goal isn’t necessarily to start the Candidate (unless they are ready). It is to watch your BusinessCoach in action so that you can learn how to invite Candidates to learn more about OPTA VIA.• On the day you are scheduled to talk, confirm the time with your Candidate and Business Coach. This can be donevia group text, email to both, or private group message on Facebook. This way, after the conversation, the three ofyou will be connected and your Business Coach can help answer any questions and provide additional support.• Just before the conversation, provide some background information to your Business Coach about yourCandidate: how you know them, what they might be interested in/looking for, and anything else you think might make the conversation more meaningful.•At the start of the conversation, welcome and introduce both parties.• Introduce your Business Coach to your Candidate, edify who they are and explain how they have helped you andothers. By doing so, you increase his/her value in your Candidate’s eyes.»Example: Jim’s been a great resource and support for me on my journey to Optimal Health. He’s helped dozens of people get healthy and is very knowledgeable.•Introduce your Candidate to your Business Coach, edify who they are and their relationship to you.»Example: Susie is a bank manager and also my oldest childhood friend.• Highlight the reason you set up the call; reiterate the reason for bringing everyone together.»Example: Susie mentioned that she was looking to get healthier, sleep better, and reduce some of the stress in her life.•Your Business Coach will open up the dialogue with your Candidate. »Example: So, Susie, tell me a little bit more about yourself…•Let him/her guide most of the conversation. He/she may ask you to share your story and contribute.• Listen to the flow and content of the discussion. Observe how your Business Coach asks questions andprovides information.»Take notes on what you hear for your debrief afterward.• Have a Health Assessment in front of you to take notes about your Candidate. You will learn many new thingsabout your Candidate as you observe the conversation. Listening and writing down what you hear will help youget to know them better and build a stronger relationship, helping them live their best life! We encourage you to work with your Business Coach/Mentor support team throughout this process. Our Community is here to support you every step of the way. uestions? Check out OPTA VIA Answers , call 1.888.OTPA VIA, or fill out the Contact Us form on your OTPA VIA personalized website! OPTAVIA_LRN_3-Way-Conversation-Tips_090518©2018 OPTA VIA LLC. ALL RIGHTS RESERVED.1。

CGA讲义

transaction data that can be expressed in money terms.
Economic Entity – requires that activities of the entity be
kept separate and distinct from the activities of its owner and all other economic entities.
Preview of Chapter 1
Financial Accounting IFRS Second Edition Weygandt Kimmel Kieso
1-3
What is Accounting?
Accounting consists of three basic activities - it
companies record assets at their cost.
Fair Value Principle – states that assets and liabilities
should be reported at fair value (the price received to sell an asset or settle a liability).
1-9
LO 3 Understand why ethics is a fundamental business concept.
Ethics in Financial Reporting Question
Ethics are the standards of conduct by which one's actions are judged as: a. right or wrong. b. honest or dishonest. c. fair or not fair.

步步高配套课件选修6Unit4G

3.“We can’t predict the climate well enough to know what to expect,...我们预报气候不够准确以致于不能知道要发生的事，…… 考点提炼 not...enough to do sth.意为： “不够……以致于不能…… ”；enough表示它前面的形容词、副词的程度，不定式表示由此产生的结果或行动，enough后面不能 that从句。
Nadine glanced round to see if there was anyone that she knew. 纳丁快速环视了一下，看看是否有她认识的人。
He gave her a quick glance as she walked into the room.
One glance at her face told me that she was tired. 一看她的脸色，我就知道她累了。
Quantities of food and tents were sent to Wenchuan from Shandong. 大量的食品和帐篷被从山东运往汶川。
归纳总结
quantity n.量；数量 in quantity
quantities
of+
可数名词复数 +
许多的，大量的语动词
quantity
There will be an increase in the range of 1 to 3 percent. 将会有1到3 The show ranges over many settings,from 18th century sailing ships to concert halls. 这个节目涉及许多场景：从18世纪的帆船到音乐厅。

Derivation and Empirical Validation of a Refined Traffic Flow Model

2
120
100
80
V (r, t) (km/h)
60
40
20
0 6:30
7:00
7:30
8:00
8:30
9:00
9:30
10:00
t (h)
Figure 2: Temporal evolution of the mean velocity V (r, t) at subsequent cross-sections of the Dutch highway A9 from Haarlem to Amsterdam at October 14, 1994 (ﬁve minute averages of single vehicle data). The prescribed speed limit is 120 km/h. We observe a breakdown of velocity during the rush hours between 7:30 am and 9:30 am due to the overloading of the highway at r = r0 := 41.8 km (· · · ). At the subsequent cross-sections the traﬃc situation recovers (- - -: r = r0 +1 km; – –: r = r0 +2.2 km; —: r = r0 +4.2 km).
Abstract The gas-kinetic foundation of ﬂuid-dynamic traﬃc equations suggested in previous papers [Physica A 219, 375 and 391] is further reﬁned by applying the theory of dense gases and granular materials to the Boltzmann-like traﬃc model by Paveri-Fontana. It is shown that, despite the phenomenologically similar behavior of ordinary and granular ﬂuids, the relations for these cannot directly be transferred to vehicular traﬃc. The dissipative and anisotropic interactions of vehicles as well as their velocity-dependent space requirements lead to a considerably different structure of the macroscopic traﬃc equations, also in comparison with the previously suggested traﬃc ﬂow models. As a consequence, the instability mechanisms of emergent density waves are diﬀerent. Crucial assumptions are validated by empirical traﬃc data and essential results are illustrated by ﬁgures. PACS numbers: 47.50.+d,51.10.+y,47.55.-t,89.40.+k Key Words: Kinetic gas theory, macroscopic traﬃc models, traﬃc instability, dense nonuniform gases, granular ﬂow

CODES, SYSTEMS AND GRAPHICAL MODELS

I NSTITUTE FOR M ATHEMATICS AND ITS A PPLICATIONSUniversity of Minnesota400Lind Hall207Church Street S.E.Minneapolis,Minnesota55455FAX(612)626-7370telephone(612)624-6066e-mail:ima-staff@IMA Schedules viaﬁnger:finger seminar@Newsletters,Updates and preprints are available viaanonymous ftp:,www:/The IMA was founded by and receives major support from the National Science Foundation.IMA NEWSLETTER#276August1–31,19991999Summer ProgramCODES,SYSTEMS AND GRAPHICAL MODELSSee /csg/for a full description of the1999Summer Program on Codes,Systems,and Graphical Models.IMA schedules are subject to revision,particularly during workshops.See /˜seminar/sched and /newsltrs/for the latest scheduling information.IMA Summer Program:Codes,Systems,and Graphical ModelsAugust2–13,1999Organizers:G.David Forney,Jr.(Massachusetts Insitute of Technology)Brian Marcus(IBM Almaden Research Center)Joachim Rosenthal(University of Notre Dame)Alexander Vardy(University of California–San Diego)Partially supported by the National Security AgencySee also /csg/News and NotesThe IMA is gradually implementing internet-based programs to improve our service to you.As a re-sult,we have changed the way we send out the Newsletter.We would like to e-mail you a notice when the next IMA Newsletter is available on the IMA Web page,instead of mailing you a hard copy.The updated version of our Newsletter is available on our IMA Web Page:/newsltrs/PARTICIPATING INSTITUTIONS:Centre National de la Recherche Scientiﬁque,Consiglio Nazionale delle Ricerche,Georgia Institute of Technology,Indiana University, Iowa State University,Kent State University,Michigan State University,Mississippi State University,Northern Illinois University,Ohio State University,Pennsylvania State University,Purdue University,Seoul National University(RIM-GARC),Texas A&M University,University of Chicago,University of Cincinnati,University of Houston, University of Illinois(Urbana),University of Iowa,University of Kentucky,University of Maryland,University of Michigan,University of Minnesota,University of Notre Dame,University of Pittsburgh,University of Wisconsin,Wayne State University.PARTICIPATING CORPORATIONS:Eastman Kodak,EPRI,Ford,General Motors,Honeywell,IBM,Lockheed Martin,Lucent Technologies,Medtronic,Motorola,Siemens, Telcordia Technologies,3M.Version of August16,1999TheSchedule for August1–31,1999IMA Summer Program:Codes,Systems,and Graphical ModelsAugust2–13,1999Organizers:G.David Forney,Jr.(Massachusetts Insitute of Technology)Brian Marcus(IBM Almaden Research Center)Joachim Rosenthal(University of Notre Dame)Alexander Vardy(University of California–San DiegoSee also /csg/The invention of turbo codes and other capacity-approaching codes has led to an exciting cross-fertilization of ideas be-tween researchers from different backgrounds.The aim of the workshop is to bring together mathematicians,computer scientists,and electrical engineers in the area of coding theory,systems theory and symbolic dynamics so that the techniques from one area can be applied to problems in the other area.The two weeks of the workshop will be subdivided into two main focus areas:Week1:Codes on Graphs and Iterative DecodingWeek2:Connections Among Coding Theory,System Theory and Symbolic DynamicsThe organizers plan a number of invited tutorial lectures speciﬁcally for inter-specialty communication.Leading workers in eachﬁeld will also be invited to present surveys of current research,with less emphasis on solved problems than on open ones.Finally,there will be both invited and contributed papers presenting recent research results.We expect the attendees to represent electrical engineering,mathematics and computer science departments in both academia and industry.As coding theory is the glue that holds the two weeks together,we expect that it will mostly be a subset of the coding theory participants who will attend both weeks.Week1(August2–6,1999)Codes on Graphs and Iterative DecodingBelief propagation in Bayesian networks has been extensively studied in artiﬁcial intelligence since the work of Pearl a decade ago,and turbo codes have recently become a subject of much research in coding theory.In the past year or two it has been recognized that the iterative decoding algorithm used for turbo codes and other capacity-approaching schemes are instances of belief propagation.This has led to an explosion of work devoted to understanding and exploiting this connection.A related problem is that of representing a given code by a graph,such as a Bayesian network.A centralMonday,August2All talks are in Lecture Hall EE/CS3-180unless otherwise noted.History and Tutorials DayG.David Forney,Jr.(chair)8:30am Registration and Coffee Reception Room EE/CS3-176 9:10am Willard Miller,Fred Dullesand G.David ForneyIntroduction and Welcome9:30–10:30am R.Michael TannerU of California-Santa Cruz Error-Correcting Codes and Graph-based Algorithms: Origins,Successes,the Current QuestsAbstract:Graphs have been used in many forms to describe the structure of codes for achieving in practical application the channel capacity predicted by Shannon.While perhaps the most versatile application has been as the language for ﬁnite-state transition-output systems for the encoding mapping of information streams to coded streams and decoding4:00pm IMA Tea IMA East,400Lind HallA variety of appetizers and beverages will be served.Tuesday,August3All talks are in Lecture Hall EE/CS3-180unless otherwise noted.Low Density Parity Check Codes DayR.Michael Tanner(chair)9:15am Coffee Reception Room EE/CS3-1769:30am David J.C.MacKayCambridge UniversitySparse Graph CodesAbstract:Sparse graph codes are codes whose constraints are deﬁned by sparse random graphs.The best known decoding method for these codes is the sum-product algorithm.Sparse graph codes include Gallager’s low-density parity-check codes,turbo codes and repeat–accumulate codes.I will review old theoretical results for Gallager codes,and new empirical work.Gallager codes are record-breaking codes for low signal-to-noise applications with large block-length and low rate(e.g., and).Are they also competitive for high rate,small block-length problems?We have studied the empirical performance of high rate binary and non-binary Gallager codes on three channels:the binary input Gaussian channel,the binary symmetric channel,and the16-ary symmetric channel.Weﬁnd that Gallager codes with rate and block length bits outperform comparable BCH and Reed-Solomon codes(decoded by a hard input decoder)by more than a decibel on the Gaussian channel.Even on the16-ary symmetric channel,Gallager codes have outstanding performance.10:30am Coffee Break Reception Room EE/CS3-17611:00am–12:00pm Robert J.McElieceCalifornia Institute of TechnologySome Simple Codes that Are Good in Both Theory andPracticeAbstract:In an unsuccessful attempt to prove coding theorems for the ensemble of classical turbo codes,we were forced to invent a much simpler ensemble,which we call“repeat-accumulate”(RA)codes.RA codes are an almost degenerate special case of serially concatenated“turbo”codes,but they have just enough structure to allow us to prove coding theorems for them.For example,we can show that maximum-likelihood decoding of RA codes“achieves capacity”on the AWGN channel.The really surprising thing about RA codes,however,is that their performance with a simple practical iterative decoding algorithm(belief propagation on an appropriate Tanner graph)is almost as good as maximum likelihood.Thus in applications where decoder complexity is as important as nearness to the Shannon liMassachusetts Inst.of Techology, RA codes may give turbo codes and LDPC codes a run for their money.[Research in collaboration with Sam Dolinar,Dariush Divsalar,and Hui Jin].2:00–3:00pm Thomas J.Richardson andRuediger UrbankeLucent Bell LabsAnalysis and Design of Iterative Decoding SystemsAbstract:In this talk we will try to give an overview of the current state of knowledge(and lack thereof)of iterative decod-ing systems.We will speak primarily about LDPCCs and Turbo codes.We will start by showing that virtually all known iterative decoding systems exhibit a threshold phenomenon which characterizes their asymptotic performance.Thresholds may be determined via a process we call density evolution which describes the distributions of extrinsic information pro-duced by each iteration of decoding.In the case of belief propagation,density evolution exhibits several special properties. In particular,densities which arise this way satisfy a certain symmetry condition which we call the consistency condition.FurtherContributed TalksCapacity Achieving Low-Density Erasure Codes3:30pm Amin ShokrollahiBell LabsAbstract:We give an overview of the work of Luby,Mitzenmacher,Shokrollahi,Spielman,and Stemann(1997)dealing with the construction of capacity achieving sequences of low-density erasure codes that have linear time encoders and decoders.We remark that the erasure channel is the only known channel for which we know explicit sequences of irregular degree distributions whose codes achieve capacity.The explicit sequences of Luby et al.and a new class of sequences discovered recently by the author are the only known sequences that provably approach capacity on an erasure channel.Iterative Decoding of Cycle Codes of Graphs4:00pm Gilles ZemorENST,ParisAbstract:We show that for cycle codes of graphs,the performance of iterative decoding and that of maximum likelihood decoding coincide.Analysis of Turbo Codes Using Bifurcation Theory4:30pm Dakshi AgrawalUniversity of IllinoisAbstract:The focus of this talk is on analyzing the phase trajectories of the turbo decoding algorithm as a function of signal-to-noise ratio(SNR).By exploiting the large length of turbo codes,the turbo decoding algorithm is treated as a single-parameter dynamical system,parameterized(approximately)by the SNR.In conjunction with extensive simulations, this parameterization is used to show that the entire SNR range can be subdivided into three regions with the waterfall region in the middle.These three regions have distinctive phase trajectories,and in most cases,the transient behavior of a phase trajectory can be used to accurately predict its asymptotic behavior.The existence and the properties ofﬁxed points in these three SNR regions will also be discussed.It is shown that the turbo decoding algorithm has two main types ofﬁxed points.In a wide range of SNRs(corresponding to bit-error rates less than1E-1),the decoding algorithm has‘unequivocal’ﬁxed points which correspond to mostly correct decisions on the information bits.Within this range,towards the lower values of SNR,there is anotherﬁxed point which corresponds to many erroneous decision on the information bits.Fixed points of this type are referred to as‘indecisive’ﬁxed points.It is demonstrated that the indecisiveﬁxed points bifurcate and disappear for SNRs in the waterfall region. We associate the qualitative transition of phase trajectories in the waterfall region to the bifurcation of indecisiveﬁxed points.The bifurcation of theseﬁxed points explains the quasi-periodic and periodic phase trajectory of turbo decoding as observed in simulations.Wednesday,August4All talks are in Lecture Hall EE/CS3-180unless otherwise noted.Inference DayBrendan J.Frey(chair)9:15am Coffee Reception Room EE/CS3-1769:30am Tommi JaakkolaMassachusetts Institute of TechnologyVariational Methods for InferenceAbstract:Variational methods have a long history as principled approximations in physics,statistics,and otherﬁelds. Techniques such as meanﬁeld approximation andﬁnite element methods are naturally viewed as variational methods.The basic idea underlying these methods is a transformation from the problem of interest such as computation of marginal prob-abilities in factor graphs to a manageable optimization problem.The objective function used in the resulting optimization problem relates(monotonically)to the estimation accuracy of the desired quantities(marginal probabilities)yielding e.g. upper and lower bounds.The purpose of this tutorial talk is to introduce a class of variational methods and demonstrate their use in probabilistic inference calculations in factor graphs.We show in particular how these methods can be read-ily combined with exact inference algorithms to maximally exploit any feasible substructures in the graphs.Numerical examples come from a large scale inference problem in medical diagnosis.10:30am Coffee Break Reception Room EE/CS3-17611:00am–12:00pm Radford M.NealUniversity of TorontoSparse Matrix Methods and Probabilistic Inference Algo-rithmsAbstract:Algorithms for solving sparse systems of linear equations seem similar to algorithms for probability propagation in networks with sparse connections.The aim of this talk is to examine the similarities and differences between these problems,and to see whether methods from oneﬁeld might help in the other.2:00pm Brendan J.FreyUniversity of WaterlooandYair WeissUniversity of California at Berkeley The Sum-Product Algorithm in Gaussian Networks with CyclesAbstract:In an attempt to better understand the behavior of the sum-product algorithm in richly-connected graphical models,we study its application in Gaussian networks with cycles.Experimentally,weﬁnd that the algorithm occasionally diverges and sometimes reaches a dynamic equilibrium.However,in most cases the algorithm converges and it turns out that theﬁxed point of the means coincides with the exact solution.Each of us will present a different approach to analyzing this problem and reveal theorems onﬁxed points,the exactness of inference and convergence to theﬁxed point.3:00pm Coffee Break Reception Room EE/CS3-176Contributed Talks3:30pm John AndersonUniversity of LundProperties of the Tailbiting BCJR DecoderAbstrThursday,August5All talks are in Lecture Hall EE/CS3-180unless otherwise noted.Robert J.McEliece(chair)9:15am Coffee Reception Room EE/CS3-176Binary Decision Diagrams9:30am Randall E.BryantCarnegie Mellon UniversityAbstract:Ordered Binary Decision Diagrams(OBDDs)provide an effective means of representing and manipulating Boolean functions symbolically.Functions are represented as graphs,and operations such as functional application,com-position,and equation solution are implemented using graph algorithms.OBDDs are used as the underlying representation for a wide variety of tasks in digital system synthesis and veriﬁcation,ﬁnite-state system analysis,andﬁnite logic.In many cases,they have enabled the solution of much larger and more complex problems than was previously considered possible. This talk provides an introduction to BDDs,plus a survey of some of their applications.10:30am Coffee Break Reception Room EE/CS3-17611:00am–12:00pm John LaffertyCarnegie Mellon UniversityTrellises,Decision Diagrams,and Factor GraphsAbstract:Ordered binary decision diagrams are graph-based data structures for representing Boolean functions.They have found widespread use in computer-aided design and in formal veriﬁcation of hardware and software systems.This talk will survey the striking connections between binary decision diagrams and code trellises,highlighting the difference in emphasis and the complementary methods that have been developed in the computer engineering and coding theory communities.The techniques developed for coding and veriﬁcation have been most successful for classes of Boolean functions that have special structure.To address the exponential blowup in the sizes of decision diagrams and trellises for general functions and codes,it will be necessary to explore new graphical representations and algorithms.We will conclude this talk by introducing some recent work on”projection decoding”that attempts to make a step in this direction,using randomized constructions of factor graphs to represent codes that may not have an explicit sparse representation.[This talk is the result of joint work with Alexander Vardy and Dan Rockmore.]2:00–3:00pm James L.MasseyETH Zurich and Lund University Linear Systems over Fields and Rings,Linear Complexity, and Fourier TransformsAbstract:This mostly tutorial presentation will review the deﬁnition of the linear complexity of a sequence over aﬁeld or a commutative ring,give an elementary proof that the linear complexity is the smallest dimension of a single-output linear system that can produce the sequence as its zero-input response,present a simple necessary and sufﬁcient condition for the existence of a generalized discrete Fourier Transform of a prescribed length N,and show the connection between the linear complexity of a periodic sequence and the generalized Discrete Fourier Transform of itsﬁrst period.6:00pm Workshop DinnerBona RestaurantBona is a Chinese/Vietnamese restaurant located near the IMA and the Day’s Inn at802Washington Avenue,the south side of Washington very near the intersection of Washington and Oak St.Friday,August6All talks are in Lecture Hall EE/CS3-180unless otherwise noted.Coding Theory DayAlexander Vardy(chair)8:45am Coffee Reception Room EE/CS3-1769:00-10:00am G.David Forney,Jr.Massachusetts Institute of Technology Codes and Systems on Graphs:Generalized State Real-izationsAbstract:In coding theory or behavioral system theory,a code or system is simply a set of possible output sequences.In a state realization of a code/system(e.g.,a trellis),a set of state variables is deﬁned as well.Conventionally,state variables are deﬁned on a sequential time axis corresponding to a subset of the integers.In a generalized state realization of a code/system,state variables may be connected according to an arbitrary graph topol-ogy;i.e.,the time axis is represented by a general graph.If the graph has cycles,then a substantial reduction in state complexity may be obtained.For example,the state complexity of a single-cycle graph realization(tail-biting trellis)can be as little as the square root of the state complexity of a conventional state realization(trellis).For another example,Reed-Muller codes have very simple generalized state realizations,in general with cycles.In general,the question of minimal generalized state realizations is wide open.TheMonday,August9All talks are in Lecture Hall EE/CS3-180unless otherwise noted.8:30am Registration and Coffee Reception Room EE/CS3-1769:10am ler,F.Dulles,B.Marcus and J.Introduction and WelcomeRosenthalAutomata and SystemsSession Chair:Jorn JustesenDynamical Systems and their Associated Automata9:30am Roger BrockettHarvard UniversityAbstract:Digital circuit designers have developed highly reliable techniques for associating automata with certain classes of electrical circuits described by smooth differential equations.However,the usual explanations of their methods do not give a general context for the process nor do they shed much light on alternative possibilities for constructing such associations.Because noise is ever-present and because reliability is of paramount importance,the association must becontinuousTuesday,August10All talks are in Lecture Hall EE/CS3-180unless otherwise noted. 8:45am Coffee Reception Room EE/CS3-176Convolutional CodesSession Chair:Karl Petersen9:00am Rolf JohannessonUniversity of Lund Woven Convolutional Codes:Encoder Properties and Er-ror ExponentsAbstract:Encoders for convolutional codes with large free distances can be constructed by combining several but less powerful convolutional encoders.This paper is devoted to constructions in which the constituent convolutional codes are woven together in a manner that resembles the structure of a fabric.The general construction is called twill and it is described together with two special cases,viz.,woven convolutional encoders with outer and inner warp.4:00pm Dharmendra ModhaIBM Almaden Research Center Art of Constructing Low-complexity Encoders/Decoders for Constrained Block CodesAbstract:A rate p:q block encoder is a dataword-to-codeword assignment from p-bit datawords to q-bit codewords,and the corresponding block decoder is the inverse of the encoder.When designing block encoders/decoders for constrained systems,often,more than codewords are available.In this paper,as our main contribution,we propose efﬁcient heuris-tic computer algorithms to(i)eliminate the excess codewords;and(ii)to construct low hardware complexity block en-coders/decoders.For(0,4/4)and(0,3/6)PRML constraints,block encoders/decoders generated using the proposed algorithms are comparable in complexity to human-generated encoders/decoders,but are signiﬁcantly simpler than lexico-graphical encoders/decoders.(Joint work with Brian Marcus).4:30–5:00pm Natasha JonoskaUniversity of South FloridaOn Encoding in DNA wordsAbstract:We discuss the problem of designing DNA codes,sets of words over alphabet A,G,C,T that satisfy certain constraints.The motivation for this problem comes from the need to reliably store and retrieve information in synthetic DNA strands.This is in particular needed in the models of DNA based computations and the design of molecular bar codes in chemical libraries.Wednesday,August11All talks are in Lecture Hall EE/CS3-180unless otherwise noted.8:45am Coffee Reception Room EE/CS3-176Multidimensional SystemsSession Chair:Jonathan Hall9:00am Klaus SchmidtUniversity of ViennaMulti-dimensional Symbolic Dynamical SystemsAbstract:This lecture discusses multidimensional shifts ofﬁnite type and gives examples of such systems from statistical mechanics,cellular automata and other areas.Roughly speaking,a-dimensional shift ofﬁnite type(SFT)is a closed, shift-invariant subset,where is aﬁnite set,and where a point belongs to if and only if it satisﬁes certain purely local restrictions at each coordinate.The purpose of the talk is to point out some of the new phenomena which arise in the transition from classical shifts of ﬁnite type(where)to the case where.The most notorious of these are certain undecidability problems which don’t give much trouble in practice,but which effectively prevent a satisfactory general theory of multi-dimensional SFT’s.The main emphasis of the talk will,however,be on the rigidity properties of certain multi-dimensional SFT’s which range from scarcity of isomorphism and shift-invariant measures to the appearance of unexpected intrinsic algebraic structures for certain classical SFT’s.References[1]R.Berger,The undecidability of the Domino Problem,Mem.Amer.Math.Soc.66(1966).[2]C.Cohn,N.Elkies and J.Propp,Local statistics for random domino tilings of the Aztec diamond,Duke Math.J.85(1996),117–166.[3]W.Geller and J.Propp,The projective fundamental group of a-shift,Ergod.Th.&Dynam.Sys.15(1995),1091–1118.Thursday,August12All talks are in Lecture Hall EE/CS3-180unless otherwise noted. 8:45am Coffee Reception Room EE/CS3-176Symbolic Dynamics and ApplicationsSession Chair:Uwe Helmke9:00am M.Michael BoyleUniversity of Maryland Applications of Symbolic Dynamics to the Structure The-ory of Nonnegative MatricesAbstract:I’ll discuss the two most dramatic results of the last several years in which symbolic dynamics is contributing to a deeper understanding of the“asymptotic algebra”of nonnegative matrices.Below S represents a unital subring S of the reals R(e.g.Q,R,Z)and denotes the nonnegative elements of S.Deﬁnition1The nonzero spectrum of a square matrix A is the unordered k-tuple of nonzero complex numbers such that the characteristic polynomial of A has the form.Deﬁnition2Let E be a subset of R containing0and1.Say over E if there exist matrices U,V with entries in E such that A=UV, B=VU.Then strong shift equivalence over E is the equivalence relation on square matrices over E which is generated by the relation.(We are interested in the case E=.)Question10:00am Natasha JonaskaUniversity of South FloridaMultiplicities of Representations of Soﬁc ShiftsAbstract:Given aﬁnite directed labeled graph,there is a natural mapping f:S T from the system S of all bi-inﬁnite paths on the graph to the system T of all bi-inﬁnite label sequences.T is called a soﬁc shift and(S,f)is called an cover of T.We deﬁne the multiplicity of the cover(S,f)to be the largest number of f-preimages of a point.The intrinsic multiplicity of of a soﬁc shift T is the minimum of the multiplicities over all covers of T,denoted by m(T).Is m(T)computable?We do not answer this question.However the attempt to solve this problem led us toﬁnd sharp estimates for the intrinsic multiplicity, sharpen a result of S.Williams,and solve a problem of P.Trow.Joint work with Doris Fiebig and Ulf Fiebig.References:1.M.Boyle,B.Kitchens,B.Marcus,A Note on Minimal Covers for Soﬁc systems,Proceedings of the AMS,95No.3,(Nov.1985),403-411.2.N.Jonoska:Soﬁc Systems with Synchronizing Representations,Theoretical Computer Science,1581-2(1996)81-115.3.B.Kitchens:Symbolic Dynamics,Springer19984.D.Lind,B.Marcus:An Introduction to Symbolic Dynamics,Cambridge University Press,New York(1995).5.P.Trow:Lifting covers of soﬁc shifts,preprint.6.S.Williams:A soﬁc system with inﬁnitely many minimal covers,Proc.Amer.Math.Soc.98,No.3(1986)503-505.7.S.Williams:Covers of non-almost-ﬁnite-type systems,Proc.Amer.Math.Soc.104(1988),245-252.11:00am Coffee Break Reception Room EE/CS3-17611:30am–12:30pm Selim TuncelUniversity of WashingtonCodings of Markov Chains and Weighted GraphsAbstract:When the edge weights of a weighted graph lie in a free Abelian group,for instance when they are real numbers, they may be naturally represented by monomials in one or more variables.I will pass to this representation to discuss(i) some of the objects associated with weighted graphs,such as the weight-per-symbol polytope and subgraphs arising from its faces,(ii)some related positivity issues,and(iii)the signiﬁcance of these objects and issues for coding problems.Contributed TalksSession Chair:Brian Marcus2:00pm Marie-Pierre B´e alUniversit´e de Marne-la-Vall´e eA Finite State Version of the Kraft-McMillan TheoremAbstract:We introduce the notion of super-state automaton constructed from another automaton.This construction is used to solve an open question about enumerative sequences of leaves of rational trees.We prove that any-rational sequence of nonnegative integers satisfying the Kraft inequality is the enumerative sequence of leaves by height of a-ary rational tree.This result is aﬁnite state version of the Kraft-McMillan theorem.We then use it to completely characterize the series that are the enumerative sequences of nodes in a-ary rational tree.Joint work with Fr´e d´e rique Bassino and Dominique Perrin2:30pm Olivier CartonAsynchronous Sliding Block MapsUniversit´e de Marne-la-Vall´e eAbstract:We deﬁne a notion of asynchronous sliding block map that can be realized by transducers labeled in.We show that,under some conditions,it is possible to synchronize this transducer by state splitting,in order to get a transducer which deﬁnes the same sliding block map and which is labeled in,where is a constant integer.In the case of a transducer with a strongly connected graph,the synchronization process can be considered as an implementation of an algorithm of C.Frougny and J.Sakarovitch of synchronization of rational relations of bounded delay.The algorithm can be applied in the case where the transducer has a constant integer transmission rate on cycles and has a strongly connected graph.It keeps the locality of the input automaton of the transducer.We show that the size of the sliding window of the synchronous local map grows linearly during the process,but that the size of the transducer is intrinsically exponential. In the case of non strongly connected graphs,the algorithm of C.Frougny and J.Sakarovitch does not keep the locality of the input automaton of the transducer.We give another algorithm to solve this case without losing the good dynamic properties that guarantees the state splitting process.(joint work with M.-P.B´e al).3:00pm Coffee Break Reception Room EE/CS3-176Deterministic Synchronization of Finite State Encoders 3:30pm Christiane FrougnyLIAFA,ParisAbstract:We present an algorithm which,given an input deterministicﬁnite state encoder with bounded rate,resynchronize it into an on-lineﬁnite state encoder,with rate1:1after an initial period of time where it reads the input and output nothing, and still input deterministic.This construction has applications to number representation and computer arithmetic. (joint work with Jacques Sakarovitch,CNRS,ENST,Paris).The Key Equation for One-point Codes4:00pm Michael O’SullivanUniversity College,CorkAbstract:The decoding problem for BCH codes is often translated to the problem ofﬁnding a solution for a“key equation.”There is a natural generalization of this key equation to one-point codes which also expresses the decoding problem.A solution to the key equation is a pair,where is a function on the curve used to construct the code and is a differential, and both have poles only at the one-point.K¨o tter’s generalization of the Berlekamp-Massey algorithm may be used to iteratively compute solutions to the key equation.The key ingredient in deﬁning the algorithm is the existence of bases for ,the functionﬁeld on the curve,and,the module of differentials,which are dual relative to the operator which takes the residue of a differential at.Ambiguity in Codes4:30–5:00pm Fernando GuzmanBinghamton UniversityAbstract:In this talk we present an algebraic method for measuring/controlling the ambiguity of an ambiguous code.A relation is established between codes and monoids,in such a way that to each class of codes corresponds a variety of monoids.It is this variety what measures the ambiguity of the given class of codes.This concept includes several forms of decipherability of codes,like unique decipherability,multiset decipherability and set decipherability,as introduced by Lempel in1986.We will give some examples of codes satisfying these types of decipherability,and illustrate one of the main techniques, namely the construction of the Domino Graph of a code.6:00pm Workshop DinnerCampus clubThe Campus club is located on the4thﬂoor of Coffman Student Union and serves a wide-ranging buffet.Coffman Union is located on the opposite side of Washington Avenue from the IMA and slightly to the west.。

Amos Fiat

ntroduction
1.1 The Problem
This paper is concerned with nding e cient strategy for replicating and migrating data in presence of a dynamic pattern of reads and writes (see gure 1, details in Section 2). This is a well-studied problem in distributed systems DF82].
write
F F F F read
Figure 1: The le allocation problem. A subset of the network processors are holding copies of a le F . The le may be read or written by any of the processors. In order to achieve maximum e ciency, we need to maximally exploit the locality of reference in the underlying distributed program, and incorporate it with the spatial locality of the underlying network architecture. More speci cally, if many read requests to a speci c le are issued in some vicinity, it is advisable to copy the relevant le to, or near, that vicinity. However, this should be balanced by the relatively greater cost of moving an entire le versus the smaller cost of transferring only the data being read. On the other hand, if the le is frequently written, then it seems advisable to remove all replica but one, and locate this replica close to the vicinity of the writes. In this paper, we will avoid making any statistical assumptions about the location or frequency of read and write requests. The above con icting heuristics must somehow be balanced in a dynamic fashion, so that the resulting algorithm automatically adapts to changing access patterns. The goal here is nding an \almost optimal" dynamic policy, rather than \learning" the best 1

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• 假币问题 • 俄式乘法 • 约瑟夫斯问题
• 欧几里德算法 • 插值查找 • 二叉查找树
变治_实例化简
• 预排序
– 检验数组中元素的惟一性： n(n-1)/2, nlogn+n – 模式计算： n(n-1)/2+n-1, nlogn+Θ(n)
• 高斯消去法
– Partial pivoting – LU 分解 – 矩阵的逆
二进制幂
• 设n = bI…bi…b0是在二进制系统中，表示一个正整数n 的比特串，则可以通过以下多项式的值来计算n： p(x) = bI xI + … + bi xi + … + b0 其中x = 2。
• 应用霍纳法则计算p(2) p←1 // n>=1, 第一个数字总是1 for i ← I-1 downto 0 do p ←2 p + bi
• Solve P2 with Algo A • Solution of P2 transformed to P1
问题化简_lcm
• Lcm(24,60)=120 24=2*2*2*3 质数因子 60=2*2*3*5 Lcm(24,60)=(2*2*3) *2*5
• 缺乏效率，并且需要一个连续质数的列表。 • 问题化简：
霍纳法则
• 用一个两行的表来帮助计算：
– 第一行包含了该多项式的系数 – 第用二来行存中储，中除间了结第果一个单元用来存储an，其他单元都
– 用第二行的最后一个单元乘以x的值再加上第一行的下一个系数，来算出表格下一个单元的值
– 以这种方式算出的最后一个单元的值，就是该多项式的值。
*例：计算 p(x) = 2x4 - x3 + 3x2 + x – 5 在x=3时的值
系数 2 -1
3
1
-5
X=3 2 3*2+(-1)=5 3*5+3=18 3*18+1=55 3*55-5=160
霍纳法则
Horner(P[0..n],x) //用霍纳法则求一个多项式在一个给定点的值 //输入：一个n次多项式的系数数组P[0..n](从低到高存储)，
以及一个数字x //输出：多项式在x点的值
精品
AlgoDALectureNotesW6
Divide and Conquer
• 大整数乘法： n log 3 ≈ n1.59 • 矩阵乘法与STRASSEN 算法：nlog7 ≈ n2.81 • 最近点对：nlgn • 凸包：nlgn，n2
减治
• 插入排序: n2, n, n2/4 • 快速排序+插入排序 • 拓扑排序: 减一 • 生成排列+ Johnson-Trotter • 生成子集+比特串方法
• 凸包：点的相对位置。
• 解析几何：几何-代数
计算图中的路径数量
• 从图（无向图或有向图）中第i个顶点到第j 个顶点之间，长度为k>0的不同路径的数量等于Ak 的第(i,j)个原素，其中，A是该图的邻接矩阵。
• 例：
优化问题的化简
• 必须求某个函数f(x) 的最小值，并且我们知道一个求函数最大值的算法。 min f(x) = -max[-f(x)]
二进制幂
• a n = a p(2) : a p← a 1 for i ← I-1 downto 0 do a p ← a 2 p + bi
•另
a = = = 如果b = 0 2pbi
a2p abi
(ap)2 abi (a p ) 2
i
= (a p )2 a 如果bi = 1
二进制幂
LeftRightBinaryExponentiation(a,b(n)) //用从左至右二进制幂算法计算an //输入：一个数字a和二进制位bI,.., b0 的列表b(n),
p(x) =(…(anx + an-1)x + ..)x + a0
霍纳法则
对于多项式 p(x) = 2x4 - x3 +3x2 + x - 5, 有：
p(x) = 2x4 - x3 + 3x2 + x - 5 = x（2x3 - x2 + 3x + 1 ）- 5 = x（x（2 x2 - x +3）+1）-5 = x（x（x（2 x-1）+3）+1）-5
• max f(x) = -min[-f(x)]
• AVL树: 1.39logn, 1.0表现—改变表现 (Representation Change)
• 2-3 树 • 堆和堆排序
霍纳法则
• 问题：
针对一个给定的x 的多项式
求值
p(x) = anxn + an-1xn-1 + … + a1x + a0
• 霍纳法则是一个很好的改变表现技术的例子。它不断地把x作为公因子从降次以后的剩余多项式中提取出来：
1. p←P[n] 2. for i←n-1 downto 0 do 3. p←x*p+P[i] 4. return p
5. 乘法和加法次数均为 n
二进制幂
• 霍纳法则计算an时，它退化成了一种对a 自乘的蛮力算法，以及一些无用的加法。
• 两种基于改变表现思想的计算an 的算法：
– 从左至右处理二进制串（n的二进制表示） – 从右至左处理
二进制幂
• 因为该算法在每次重复它惟一循环的时候都要做一到两次的乘法，所以它在计算an时，总的乘法次数M(n)是
b-1 ≤ M(n) ≤2(b-1) b 是代表指数n 的比特串的长度
b-1 = log2 n
• Vs. 减半
问题化简（Reduction)
• P1 reduced to P2 which can be solved by Algo A
lcm(m,n)和gcd(m,n)的积把m 和n 的每一个因子都恰好包含了一次，因此就简单地等于m 和n 的积。
lcm(m,n) = m n gcd(m, n)
问题化简
• 综合除法（2x4 - x3 + 3x2 + x – 5）/（x-3）的商2 x3 + 5x2 + 18x +55；余数 160。
这些位来自于一个正整数n的二进制展开式 //输出：an 的值
1. product ←a 2. for i←I-1 downto 0 do 3. product ←product*product 4. if bi = 1 then product ←product*a 5. return product