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segment. Expressions are given for two side lengths. A regular polygon is always convex. Categories of polygons based on the number of sides, and whether they are concave or convex. Show Answer Example 3. Each page is designed with the latest design principles. So a rotation of 72, 144, 216, 288, 360,. Higher Tier. next solving the apothem of octagon. Sum of the interior angles of a polygon = (N - 2) x 180° The number of diagonals in a polygon = 1/2 N(N-3) The number of triangles (when you draw all the diagonals from one vertex) in a polygon = (N - 2) Polygon Parts. Khan Academy is a 501(c)(3) nonprofit organization. Draw the lines of symmetry for each regular polygon, fill in the table including an expression for the number of lines of symmetry in a n-sided polygon. Learn vocabulary, terms, and more with flashcards, games, and other study tools. A Lesson for Kindergarten, First, and Second Grade. In this worksheet, decide whether a polygon is regular or irregular. com, a math practice program for schools and individual families. Polygons are 2-dimensional shapes. All regular star polygons are non-convex polygons. If a regular polygon has x angles each measuring q degrees, then what is the value of q ? A. The Corbettmaths Practice Questions on Frequency polygons. For example, a triangle has 3 sides and 3 angles. Examples of regular polygon: In the adjoining figure of an equilateral triangle ABC there are three sides i. Like 0 Dislike 0 Reply Quote Follow. General Questions: 1. Common Core Standards: Grade 3 Geometry, Grade 4 Geometry. Polygon features One Click Demo Install, so demo content can be installed in minutes. Do you really know everything about these 2-D polygons and. MathScore EduFighter is one of the best math games on the Internet today. Videos, worksheets, 5-a-day and much more. First, take a look at the question itself: The hexagon above has interior angles whose measures are all equal. PRACTICE: Polygons – Assignment Worksheet Monday, 1/28/13 6-1: | {
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angles whose measures are all equal. PRACTICE: Polygons – Assignment Worksheet Monday, 1/28/13 6-1: Properties and Attributes of Polygons. A triangle with all sides and angles equal is known as an equilateral triangle. The following diagram gives the formula to find the area of a regular polygon using the perimeter and the apothem. We also compute the number of regions formed by. I passed my intro programming class with a B because my prof gave us 20 min. 3 A polygon has 2 or more sides and angles. Examples: Regular: Not regular: More Geometry Subjects Circle Polygons Quadrilaterals Triangles Pythagorean Theorem Perimeter Slope Surface Area Volume of a Box or Cube Volume and Surface Area of a Sphere Volume and Surface Area of a. Use MathJax to format equations. Use your knowledge of the sums of the interior and exterior angles of a polygon to answer the following questions. 360/x -10x Kudos for the right answer and explanation. So if we know the sum of the interior angles, we just need to divide that sum by the number of sides to find out the measure of any angle of the polygon: 180(n-2)/n Exterior Angles. A polygon in which all the sides are the same length and all the measures have the same measure is regular polygon. Grade 3 geometry worksheet: Properties of polygons Author: K5 Learning Subject: Grade 3 Geometry Worksheet Keywords: Grade 3 geometry worksheet 2D shapes, lines, angles, parallel, area, perimiter Created Date: 4/28/2017 12:38:52 PM. Convex polygon B. Do you really know everything about these 2-D polygons and. Properties of Decagon. 7) The area of a parallelogram with base of 10 ft and height of 2 ft is 20 square ft. Theorem: The area of a regular polygon. The vertices of an N-vertex polygon are located at the angles (2*Math. This quiz is incomplete! To play this quiz, please finish editing it. All sides are equal length placed around a common center so that all angles between sides are also equal. A polygon is called a REGULAR polygon when all of its | {
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that all angles between sides are also equal. A polygon is called a REGULAR polygon when all of its sides are of the same length and all of its angles are of the same measure. Access the answers to hundreds of Polygons questions that are explained in a way that's easy for you to understand. In addition to the examples given by Karen: octagons (eight sides), nonagons (nine sides), decagons (ten sides), hendecagons (eleven sides), dodecagons (twelve sides), tridecagons (thirteen sides), tetradecagons (fourteen sides), pendecagons (fift. Many of these shapes, or polygons, can be described as flat closed figures with 3 or more sides. Hand touch global map form lines and polygonal. Here are your FREE materials for this lesson. See our pages on circles and. It is possible to do better than a hexagon, if an irregular polygon is acceptable. A regular polygon is one that has equal sides. A rectangle is a polygon. To know more about different data collection methods, and statistics download BYJU'S -The Learning App. Examples of regular polygon: In the adjoining figure of an equilateral triangle ABC there are three sides i. Polygons come in many shapes and sizes and you will have to know your way around them with confidence in order to ace those SAT questions on test day. Polygons come in many shapes and sizes and you will have to know them inside and out in order to take on the many different types of polygon questions the ACT has to offer. If the measures of a polygon's angles or side lengths differ, then the polygon is called an irregular polygon. Identifying and Describing Polygons All Classroom Lessons. Start by filling in the missing lengths of the sides. The sum of its angles will be 180° × 4 = 720° The sum of interior angles in a hexagon is 720°. Identifying shape attributes, Understanding polygons. Videos, worksheets, 5-a-day and much more. A polygon is a regular polygon if and only if all of its interior angles are equal and all of its sides are of equal length. In | {
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if and only if all of its interior angles are equal and all of its sides are of equal length. In this polygon game you will classify different polygons based on their properties. Polygons are 2-dimensional shapes. Combine multiple words with dashes(-), and seperate tags with spaces. In this video I will take you through everything you need to know in order to answer basic questions about the angles of polygons. Irregular polygon C. The number of sides and the radius can be entered via a file, STDIN, or just a plain old variable. 5 (Geometry: n-sided regular polygon) An n-sided regular polygon's sides all have (i. This calculator is designed to give the angles of any regular polygon. Please leave workings so I understand how you get the answer. Let's say you have pixel bitmaps that look something like this: From this I can easily extract a contour, which will be a concave polygon defined by a set of 2D points. Make flat shapes using toothpicks and clay. Before we get too caught up on the excitement of quadrilaterals, let's take time to learn the names and basic properties of different polygons. Updated: Aug 5, 2019. - [Voiceover] So ,a few things…that you should know about polygons,…let's start with the most basic polygon…which is our friend the triangle. Let’s first find the measure of each one of those angles. all of these. In Year 6 children use this knowledge and the following formula to calculate the size of the angles. The sum of the angles of a polygon with n sides, where n is 3 or more, is 180° × (n - 2) degrees. As the concept of irregular polygons is extremely general, knowledge about this concept can be very valuable, both for problem solving and for simple problems in the real world. The simplest regular polygon is the equilateral triangle, which consists of three edges of equal length and three angles between each pair of edges to be 60 degrees. Consider, for instance, the ir regular pentagon below. How many sides does an octagon have? eight 2. Creating | {
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for instance, the ir regular pentagon below. How many sides does an octagon have? eight 2. Creating Polygons - investigating the Concept of Triangle and the Properties of Polygons ( from NCTM ) ; Figures and Polygons - definitions and examples of a large number of figures fiting the definition of polygon ; Identify Polygons - names of polygons by number of. For the regular polygons inscribed in the unit circle, what is the range of values that the apothem can have?. Higher Tier. Please leave workings so I understand how you get the answer. Examples of regular polygon: In the adjoining figure of an equilateral triangle ABC there are three sides i. shapes}; see section 51. A polygon is regular, if all its sides have the same length and all its angles have the same measure. , AB, BC and CA are equal and there are three angles i. Abstract polygon. If a polygon is regular, there is a point which we can define as the center. Then ask guiding questions to see if they recognize the pattern (every time you add a side, the interior angle sum increases by 180°). Our online polygon trivia quizzes can be adapted to suit your requirements for taking some of the top polygon quizzes. A regular three-sided polygon is called an equilateral triangle. An Apothem is a line segment from center point of side of a regular polygon to the midpoint of the regular polygon. Categories & Grades. 1) heptagon 2) decagon 3) nonagon 4) hexagon 5) pentagon 6) nonagon 7) hexagon 8) nonagon State if each polygon is concave or convex. † Can any quadrilateral tile the plane? † Can any irregular polygon tile the plane?. A square is a rhombus. " Knowing the names of the polygons is only the start. Feel free to always add Asymptote solution to any questions related to plotting, drawing, diagramming, etc because it is allowed by law in this site. Indeed, it is clear, that under the given conditions, a regular triangle is the only possible regular polygon with the smaller interior angle. Each student will | {
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triangle is the only possible regular polygon with the smaller interior angle. Each student will receive an Area of Irregular Polygons Mini Lesson Examples. Choose from the following regular polygons: Triangle,. Test comprehension with the questions that follow. A regular polygon has sides which are all the same length as well as angles which are all the same. Regular Polygon. Browse other questions tagged tikz-pgf or ask your own question. A regular polygon is one that has equal sides. Polygons can be regular or. Khan Academy is a 501(c)(3) nonprofit organization. Abstract background with polygonal shapes. 25 Mocks + 30 Sectionals for Rs. 7) The area of a parallelogram with base of 10 ft and height of 2 ft is 20 square ft. You can tell, just by looking at the picture, that $$\angle A and \angle B$$ are not congruent. The shape on the left has equal sides and angles, so it is a regular polygon. 👍If you like this resource, then please rate it and/or leave a comment💬. Questions for Students. A regular polygon is a closed, 2-dimensional figure consisting exclusively of line segments of equal length. Polygon abstract background. Exterior angle = 180-178 = 2° because the interior and exterior angle together make a straight line. If a regular polygon has x sides, then the degree measure of each exterior angle is 360 divided by x. The sum of the exterior angles of any polygon is 360°. If none of the sides, when extended, intersects the polygon, it is a convex polygon; otherwise it is concave. What is the latest version of SketchUp that Keyframe Animation will work with?. In this polygon game you will classify different polygons based on their properties. The applets below illustrate what it means for any polygon to be classified as a regular polygon. 9 degrees are in the measure of an interior angle of a regular seven sided polygon. A Polygon is a 2D shape which is made up of straight line segments. In our example, it's equal to 5 in. Start studying Quadrilateral & Polygon | {
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straight line segments. In our example, it's equal to 5 in. Start studying Quadrilateral & Polygon Unit Review Questions. Fortunately, as you'll see in the following practice questions, there's a handy formula that you can use to find a missing interior angle in a […]. We can make "pencilogons" by aligning multiple, identical pencils end-of-tip to start-of-tip together without leaving any gaps, as shown above, so that the enclosed area forms a regular polygon (the example above left is an 8-pencilogon). Convex polygon B. The resulting polygon is guaranteed to be simple (it does not intersect or touch itself). It is the point which is equidistant from each vertex. Upgrade your subscription to get access to this quiz, more lessons, and more practice questions. polygon: a 2D or flat, closed shape with straight sides. A regular polygon is a polygon whose sides are equal length and whose angles are all the same value. Polygons are 2-dimensional shapes with straight sides. Area= (2)(apothem)(perimeter) is the formula for finding the area of a regular polygon. Work out the identity of the missing regular. Tutors Answer Your Questions about Polygons (FREE) Get help from our free tutors ===> Algebra. Name the quadrilaterals shown: rhombus, parallelogram, square, rectangle, or trapezoid. Upgrade your subscription to get access to this quiz, more lessons, and more practice questions. Here we look at Regular Polygons only. I can classify a polygon as concave or convex and regular or irregular. The sum of the angles of a polygon with n sides, where n is 3 or more, is 180° × (n - 2) degrees. To do so, I would like to construct an irregular polygon, and from each of it's sides, construct a *regular* n-gon, on the outside of the irregular-gon. Sample Problem 4: Draw a figure that fits the description. Selecting Objects Inside Polygon Is there a way within a source drawing to select all objects within a predetermined boundary/polygon? The regular select routine is limited to | {
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select all objects within a predetermined boundary/polygon? The regular select routine is limited to defining a polygon but not selecting an existing one. This quiz will focus on different shapes classified as polygons. t = 360 o / 12 = 30 o; We now use the formula for the area when the side of the regular polygon is known Area = (1 / 4) n x 2 cot (180 o / n) Set n = 12 and x = 6 mm area = (1 / 4) (12) (6 mm) 2 cot (180 o / 12). One interior angle of a regular polygon - (n - 2). Diagonals of a Regular Octagon. Can be “regular” – all sides and all angles are equal to each other Two sides of equal length Three acute angles Sum of angles = 180° All sides equal length Three acute angles Sum of angles = 180° Is a regular polygon No sides are equal No angles are equal May have obtuse angle Sum of angles = 180 ° Opposite sides are parallel Opposite. A polygon is a shape that has any number of straight sides, such as a triangle, square or hexagon. Polygon or Not? This Is the Question : In this game, students will quickly decide whether a geometrical figure is a polygon or not a polygon. Round your answer to the nearest tenth if necessary. For the Independent Practice students will use the area formulas to help them complete the Finding the Finding the Area of Regular Polygons Worksheet. This is of course made far more obnoxious by the fact that IsPolygon is a valid rule token, so your rule will seem to be correct, and even pass the rule-checker, but silently fail when you try to actually repour the. Mathematics (Linear) - 1MA0 ANGLES: POLYGONS Answer the questions in the spaces provided - there may be more space than you need. A clock is constructed using a regular polygon with 60 sides the polygon rotates every minute how has the polygon rotated after 7 minutes. Using the area of regular polygon calculator: an example. Videos, worksheets, 5-a-day and much more. The angles have been calculated manually. Polygons are multi-sided shapes with different properties. Videos, | {
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have been calculated manually. Polygons are multi-sided shapes with different properties. Videos, worksheets, 5-a-day and much more. Solve these reasoning questions. 2019 23:53, kumarivishakha118. Always show your workings. A three-sided polygon is called a triangle. A regular polygon is both equilateral and equiangular. SplashLearn is an award winning math learning program used by more than 30 Million kids for fun math practice. The formula is (n - 2) times 180 divided by n , where n is the number of sides. Find the area of each of the following regular polygons. The outline itself is matched with the InPolygon rule (which is what you want). Area of Polygons. parallel lines: two lines that are always the same distance apart and never touch. They should not be confused with regular polygons. Is this a inductive reasoning or deductive re. A "Regular Polygon" has: all sides equal and. The only constructible regular polygons with an odd number of sides are those for which this number is a product of distinct Fermat primes (so for instance 15 = 3 times 5, 51 = 3 times 17), and the only ones with an even number of sides are those obtained by repeatedly doubling these numbers (including 1), thus:- (1,2), 4, 8, 16, 32, 64,. true false. The sum of the angles of a polygon with n sides, where n is 3 or more, is 180° × (n - 2) degrees. (Must be 18 years old to sign up. 1) 2) 3) regular 19-gon 4) regular 14-gon Find the measure of one exterior angle in each regular polygon. A polygon is called a REGULAR polygon when all of its sides are of the same length and all of its angles are of the same measure. The area of the regular polygon is 297 m 2. 2D SHAPES > REVISION > GCSE QUESTIONS. If someone can help me out that would be awesome. Videos, worksheets, 5-a-day and much more. Unit 7 Test Polygons And Quadrilaterals Answer Key. Unanswered Questions. A brief description of the worksheets is on each of the worksheet widgets. A "Polygon" object contains the polygon outline. A | {
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worksheets is on each of the worksheet widgets. A "Polygon" object contains the polygon outline. A diagonal of a polygon is any line segment that connects non-consecutive vertices of the polygon. Expressions are given for two side lengths. Award Information. Menu Skip to content. [1] 6) One interior angle of a regular polygon is 135°. The measure of an exterior angle of a convex regular polygon is 45. A polygon is regular, if all its sides have the same length and all its angles have the same measure. This figure shows the most common polygons. Improve your math knowledge with free questions in "Interior angles of polygons" and thousands of other math skills. Geometry Polygons questions and answers for CAT. 652 Responses. A polygon is irregular, if it is not regular. Hint: 360/No. A polygon in which all the sides are the same length and all the measures have the same measure is regular polygon. I will be focusing on convex regular. Download the set (5 Worksheets). TeX - LaTeX Stack Exchange is a question and answer site for users of TeX, LaTeX, ConTeXt, and related typesetting systems. You will have to read all the given answers and click over the correct answer. An n-sided regular polygon is rotationally symmetric around its centre. In Year 6 children use this knowledge and the following formula to calculate the size of the angles. Tags: Question 40. A regular polygon is a polygon all of whose sides are of the same length (equilateral) and all of whose internal angles are the same (equiangular). A polygon that is not a regular polygon is called an irregular polygon. Classify the polygon by the number of sides. Chapter 12. all angles equal. As a child, there's a point in life where shapes and what we now have come to refer as "geometry" are the most fascinating things in the world. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. ) Sign Up More Info Close. disproportional perimeters. As an example: A | {
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find and share information. ) Sign Up More Info Close. disproportional perimeters. As an example: A pentagon has five sides, 360/5 =72. This figure shows the most common polygons. If you are not sure about the answer then you can check the answer using Show Answer button. Type all answers in the boxes provided. Polygons Exercises ; Topics Which of the following look like regular polygons? Gimme a Hint. The size of each of the interior angle of a regular polygon = (n-2) x 180º ÷ n Where n is the number of sides. You Try It! How many sides does a regular polygon have if each interior angle measures 160 ? Example 5: Solving Algebraic Problems Find the value of x. I can classify a polygon as concave or convex and regular or irregular. (Construct it from scratch, or use a script. You will have to read all the given answers and click over the correct answer. We can see from the above examples that the number of triangles in a polygon is always two less than the number of sides of the polygon. A brief description of the worksheets is on each of the worksheet widgets. Polygons Author: act Last modified by: act Created Date: 7/10/2008 4:49:14 PM Document presentation format: On-screen Show (4:3) Other titles: Corbel Arial Wingdings 2 Wingdings Wingdings 3 Calibri Module 1_Module 2_Module 3_Module 4_Module 5_Module 6_Module Polygons What is a polygon?. An octagon is any eight-sided polygon, and the sum of its angles is 1080°, as we saw above. You can tell, just by looking at the picture, that $$\angle A and \angle B$$ are not congruent. PI*K)/N where K goes from 0 to N-1, inclusive. The sum of the interior angles, in degrees, of a regular polygon is given by the formula 180(n – 2), where n is the number of sides. Main Ideas/Questions POLYGON Notes/Examples A polygon isa figure formed by three or more line Date: Class: hexagon? EXAMPLES 9. 180(x-3)/x B. Formula for the sum of interior angles. A convex polygon has no angles pointing inwards. s = (160cm)/8 = 20 cm. Two congruent | {
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angles. A convex polygon has no angles pointing inwards. s = (160cm)/8 = 20 cm. Two congruent figures have equal areas. Print full size. Examples of regular polygons include the equilateral triangle and square. 652 Responses. What is the latest version of SketchUp that Keyframe Animation will work with?. To find the perimeter of a polygon, add the lengths of its sides. If none of the sides, when extended, intersects the polygon, it is a convex polygon; otherwise it is concave. This is part of our collection of Short Problems. Formula to find the sum of interior angles of a n-sided polygon is = (n - 2) ⋅ 180 ° By using the formula, sum of the interior angles of the above polygon is = (9 - 2) ⋅ 180 ° = 7 ⋅ 180 ° = 126 0 ° Formula to find the measure of each interior angle of a n-sided regular polygon is. There are multiple ways to found the Apothem of a Regular Polygon. I can find the measure of an interior angle of any regular polygon. Background polygon. You can now earn points by answering the unanswered questions listed. Background polygon. The applets below illustrate what it means for any polygon to be classified as a regular polygon. A polygon is irregular, if it is not regular. Along with it, the property of equal-length sides, this implies that every regular polygon also has an inscribed circle or incircle that is tangent to every side at the midpoint. (Construct it from scratch, or use a script. org are unblocked. Diagonals of a Regular Octagon. The vertices of an N-vertex polygon are located at the angles (2*Math. Identifying and Describing Polygons All Classroom Lessons. The sum of the exterior angles of any polygon is 360 degrees. What is the minimum interior angle possible for a regular polygon or why. A worksheet on angles in polygons. A Ogee curve is a _____ a) semi ellipse b) continuous double curve with convex and concave c) freehand curve which connects two parallel lines d) semi hyperbola View Answer. A polygon is a plane shape (two-dimensional) | {
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two parallel lines d) semi hyperbola View Answer. A polygon is a plane shape (two-dimensional) with straight sides. Tell whether the polygon is equilateral, equiangular, or regular. 2nd and 3rd Grades. Interact with these applets for a minute or two, then answer the questions that follow. of sides = Interior angle Attempt drawing a pentagon on your own first. A convex polygon has no angles pointing inwards. Concave polygon. A regular polygon is a polygon that is both equilateral and equiangular (having both. Polygon abstract background. All photos are included but can be easily changed to match your branding’s style and personality. Assume that the regular polygon has n sides (or angles). A regular polygon is a polygon all of whose sides are of the same length (equilateral) and all of whose internal angles are the same (equiangular). Solution to Problem 2: Let t be the size of angle AOB, hence t = 360 o / 5 = 72 o; The polygon is regular and OA = OB. Add ScratchPad. For example if a polygon has 41 sides, it would be called a 41-gon. The Corbettmaths Practice Questions on Frequency polygons. …And how about the pentagon…which kinda looks like a home plate. Each question will change subtly every time you take. The number of sides is also the number of angles and, in a regular polygon, all the angles are the same. Expressions are given for two side lengths. (Remade and reuploaded 9th February 2020) Calculating interior and exterior angles of polygons. Making statements based on opinion; back them up with references or personal experience. Grade 3 geometry worksheet: Properties of polygons Author: K5 Learning Subject: Grade 3 Geometry Worksheet Keywords: Grade 3 geometry worksheet 2D shapes, lines, angles, parallel, area, perimiter Created Date: 4/28/2017 12:38:52 PM. Find the area of the regular polygon. Sample Problem 5: Each figure is a regular polygon. of sides = 360°/20° = 18. So, if we know all the interior angles other than x, then we can find x. polygons | {
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= 360°/20° = 18. So, if we know all the interior angles other than x, then we can find x. polygons Questions #1 Relates the number of sides, angles, and vertices of a polygon Question #2 Describes attributes of polygons Question #3 Wow! Thoroughly describes. Shape And Formula Of A Polygon Quiz 10 Questions | By Swords31535 | Last updated: Oct 11, 2017 | Total Attempts: 9314 All questions 5 questions 6 questions 7 questions 8 questions 9 questions 10 questions. You may also be interested in our longer problems on Angles, Polygons and Geometrical Proof Age 11-14 and Age 14-16. Regular Polygons Printable regular polygon sheet which includes picture and name of each of the 10 polygons from triangle, through to dodecagon. A polygon is a flat shape with straight sides that are joined to form a closed shape. You will have to read all the given answers and click over the correct answer. 1) 2) 3) regular 19-gon 4) regular 14-gon Find the measure of one exterior angle in each regular polygon. Although students should work independently, they may want to discuss or ask questions of their group. An n-sided regular polygon is rotationally symmetric around its centre. A regular polygon is always convex. I can solve reasoning questions about comparing, classifying and finding unknown angles in polygons. no space at all. Design a class named RegularPolygon that contains: Aprivate int data field named n that defines the number of sides in the polygon with default value 3. where n - number of sides of octagon. Irregular polygons are polygons that have unequal angles and unequal sides, as opposed to regular polygons which are polygons that have equal sides and equal angles. How many angles does a triangle have? three 3. While we are given two sides - the base and the height - we do not know the hypotenuse. com Regular Polygons Grade 3 Geometry Worksheet Answers: Color the regular quadrilateral. Questions and Facts † What are the only edge-to-edge regular tilings? That is, what regular | {
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Questions and Facts † What are the only edge-to-edge regular tilings? That is, what regular polygons allow for edge-to-edge tiling of the plane? † Can we use more than one regular polygon, with different number of size, to tile the plane edge-to-edge? † Any triangle can tile the plane. Here we look at Regular Polygons only. Improve your math knowledge with free questions in "Regular and irregular polygons" and thousands of other math skills. Note that the polygon we are given is not a regular polygon, since the side lengths are not all equal. In a regular octagon, each angle = 1080°/8 = 135° That angle is the supplement of a 45° angle. To find the perimeter, multiply the length of one side by the total number of sides. In this video, we'll talk about a few things you need to know about polygons for the GMAT, some of their basic names of regular polygons, and the sum of the interior angles of a polygon. Draw all of the diagonals in each regular polygon. Two congruent figures have equal areas. angles of polygons section 8-1 jim smith jchs names of polygons sides triangle 3 quadrilateral 4 pentagon 5 hexagon 6 heptagon 7 octagon 8 nonagon 9 decagon 10 dodecagon 12 n - gon n interior angle sum each triangle has 180° if n is the number of sides then: int angle sum = (n - 2 ) 180° angles of polygons section 8-1 names of polygons sides triangle 3 quadrilateral 4 pentagon 5 hexagon. Restate for the class a clear definition of polygon: a 2D or flat, closed shape with straight sides. Formula for the sum of interior angles. The moral of this story- While you can use our formula to find the sum of. Polygon features One Click Demo Install, so demo content can be installed in minutes. I usually print these questions as an A5 booklet and issue them in class or give them out as a homework. There can't be a better way to introduce polygons, than what kids love the most - coloring! Direct kids to follow the color key provided to identify polygons like triangles, quadrilaterals, | {
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Direct kids to follow the color key provided to identify polygons like triangles, quadrilaterals, pentagons and more. A regular polygon has an exterior angle of 20 degrees. com For a complete lesson on regular polygons, go to https://www. The simplest regular polygon is the equilateral triangle, which consists of three edges of equal length and three angles between each pair of edges to be 60 degrees. Find the area of each regular polygon. The shape on the left has equal sides and angles, so it is a regular polygon. ) The area formula for a regular polygon is 1/2ap. $$\therefore$$ the area of one such triangle $$= \frac{1}{2} \times$$ sides of polygon $$\times$$ length of perpendicular from the center to any side of. 2D SHAPES > REVISION > GCSE QUESTIONS. The vertical coordinate can be calculated as a sine of the angle times the radius of the circumcircle; the horizontal coordinate is calculated the same way, except you need to multiply the radius by the cosine of the angle. In this regular polygon lesson plan, 10th graders find the area of a regular polygon in terms of the apothem and the perimeter. Regular Polygon. The regular polygons were known to the ancient Greeks, and the pentagram, a non-convex regular polygon (star polygon), appears on the vase of Aristophonus, Caere, dated to the 7th century B. Using the area of regular polygon calculator: an example. t = 360 o / 12 = 30 o; We now use the formula for the area when the side of the regular polygon is known Area = (1 / 4) n x 2 cot (180 o / n) Set n = 12 and x = 6 mm area = (1 / 4) (12) (6 mm) 2 cot (180 o / 12). The sum of the exterior angles of any polygon is 360 degrees. A Polygon is a 2D shape which is made up of straight line segments. Browse other questions tagged qgis polygon intersection or ask your own question. Polygons maths worksheet 3 accurately completes the polygons. 6 9 52) 14 14. A regular polygon has sides of equal length, and all its interior angles are of equal size. A regular polygon have | {
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has sides of equal length, and all its interior angles are of equal size. A regular polygon have the number of lines of symmetry equal to the number of sides of the regular polygon. The dodecagon would have sides of alternating lengths, with ratio $4:\sqrt{12}$; as a specific size for the 30-60-90 triangle is not given. Label the sides and angles on each polygon. Sum of interior angles = 180 (n – 2) where n = the number of sides in the polygon. A polygon is a 2D shape with at least three sides. A regular polygon has sides which are all the same length as well as angles which are all the same. a positive number representing the interior space of a polygon. all of these. Categories & Grades. Kahoot! is a free game-based learning platform that makes it fun to learn - any subject, in any language, on any device, for all ages! Unit 7 test polygons and quadrilaterals answer key. 25 Mocks + 30 Sectionals for Rs. If a regular polygon has x angles each measuring q degrees, then what is the value of q ? A. In geometry, polygons cover a lot of ground, so you can bet that some questions on the ACT Math exam will involve polygons—specifically, finding the interior angles of a polygon. This calculator is designed to give the angles of any regular polygon. Always show your workings. White neural texture abstract vector. Khan Academy is a 501(c)(3) nonprofit organization. Let's say you have pixel bitmaps that look something like this: From this I can easily extract a contour, which will be a concave polygon defined by a set of 2D points. of sides = 360°/20° = 18. The sum of the exterior angles of any polygon is 360°. An irregular polygons have sides of differing length and angles of differing measure. Find the value of. The following figures are polygons. Find the length of one side. Support your method with examples. This quiz is incomplete! To play this quiz, please finish editing it. Updated: Aug 5, 2019. (n – 2)*180 = sum of all interior angles (6 – 2)*180 = 720. There are | {
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it. Updated: Aug 5, 2019. (n – 2)*180 = sum of all interior angles (6 – 2)*180 = 720. There are various ways to solve this question, but each takes a bit of effort. For a polygon to be 'regular' it must have. Each angle in a regular decagon is equal to 144°. How to find the interior angle of a regular polygon. Polygons are two-dimensional objects, not solids. tells you the sum of the interior angles of a polygon, where n represents the number of sides. Polygons are 2-dimensional shapes. A polygon is called a REGULAR polygon when all of its sides are of the same length and all of its angles are of the same measure. x1, x2 ,y1 ,y2. com for more resources, like predicted GCSE Maths Papers, Topic Buster and challenge yourself against our Demon questions. The formula. This quiz will focus on different shapes classified as polygons. A triangle is a 3 sided polygon with interior angles that add to 180 degrees. In these applets, a pentagon and a triangle will be used to illustrate this definition. Question 1. Identifying and Describing Polygons All Classroom Lessons. Regular polygons. There can't be a better way to introduce polygons, than what kids love the most - coloring! Direct kids to follow the color key provided to identify polygons like triangles, quadrilaterals, pentagons and more. Area of Regular Polygons If radii are drawn from the center of a regular polygon to the vertices, congruent isosceles triangles are formed. In the limit, a sequence of regular polygons with an increasing number of sides approximates a circle, if the perimeter or area is fixed, or a regular apeirogon (effectively a straight line. Regular Polygon Formulas. Illustrated with examples of polygons in everyday life. Newest Active Followers. 120° ° 85° 53° 115° x° 117° 105° 120. In geometry, polygons cover a lot of ground, so you can bet that some questions on the ACT Math exam will involve polygons—specifically, finding the interior angles of a polygon. A hexagon (six-sided polygon) can be | {
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finding the interior angles of a polygon. A hexagon (six-sided polygon) can be divided into four triangles. The interactive lessons include: Polygons and Non-Polygons in which students will drag a series of shapes to either the bucket that says polygon or the bucket that says non-polygon. A rectangle is a special quadrilateral where opposite sides are congruent-- that is, the same length -- and each angle is a right angle. Expected time to solve, similar questions of 11 Plus exam practice papers. Find the length of one side. This is of course made far more obnoxious by the fact that IsPolygon is a valid rule token, so your rule will seem to be correct, and even pass the rule-checker, but silently fail when you try to actually repour the. Area is the negative space inside a polygon. A comprehensive database of more than 34 polygon quizzes online, test your knowledge with polygon quiz questions. +34,000 Free Graphic Resources. Figure 4 – The Interior angles of polygons that can tessellate the plane add up to 360 degrees. The sum of the exterior angles of any polygon is 360 degrees. I also make them available for a student who wants to do focused independent study on a topic. The polygon is a 18-sided regular polygon; a regular octadecagon. A worksheet on angles in polygons. I want to find the direction of s. Published on Jan 20, 2017. In Year 6 children use this knowledge and the following formula to calculate the size of the angles. See our pages on circles and. So if we know the sum of the interior angles, we just need to divide that sum by the number of sides to find out the measure of any angle of the polygon: 180(n-2)/n Exterior Angles. A regular net is defined as one in which the vertex figure is required by symmetry to be a regular polygon or polyhedron. Polygons maths worksheet 4 and polygon maths worksheet 5 work with tessellation. (n – 2)*180 = sum of all interior angles (6 – 2)*180 = 720. | {
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What's $1^3+2^3+3^3+\cdots+99^3$ modulo $3$?
What's the remainder when the sum
$$1^3+2^3+3^3+\cdots+99^3$$ is divided by $$3$$?
Background:
I saw this question on MSE but it was closed and I wanted to learn how to approach it. The help given to the asker had failed to bring the question up to an acceptable standard.
I can solve it but I doubt my methods are efficient.
One way is to start by removing every third term. Then the first two terms of the remaining sequence can be removed, and so on... a pattern may emerge.
Another way is to look at the expansion of $$(x+1)^3$$ and see what it does to the residues for each $$x$$, then sum over that by induction.
But I'm sure my inventions aren't very efficient.
• Use $n^3\equiv n\pmod3$. – Angina Seng Dec 27 '18 at 5:42
• math.stackexchange.com/questions/1328798 – Andrei Dec 27 '18 at 5:43
• In general, note that you can handle the terms in groups of $3$ as $n + 3 \equiv n \pmod 3$, and also use that $n^3 \equiv n \pmod 3$ as Lord Shark the Unknown stated above. – John Omielan Dec 27 '18 at 5:46
• @LordSharktheUnknown ah thanks. I guess the $(x+1)^3$ method would've revealed that. – samerivertwice Dec 27 '18 at 5:46
By Fermat's little theorem $$a^3\equiv a\pmod3$$. Then you have $$\sum_{k=1}^{99} k=\frac{(99)(100)}2=50\cdot 99\equiv 0\pmod3$$.
\begin{align} \sum_{i=1}^{99}i^3 &= \sum_{i=0}^{32} (3i+1)^3 + \sum_{i=0}^{32} (3i+2)^3 + \sum_{i=0}^{32} (3i+3)^3 \\ &\equiv \sum_{i=0}^{32} 1^3 + \sum_{i=0}^{32} 2^3 + \sum_{i=0}^{32} 0^3 \pmod{3}\\ &\equiv \sum_{i=0}^{32} 1^3 + \sum_{i=0}^{32} (-1)^3 + \sum_{i=0}^{32} 0^3 \pmod{3}\\ &\equiv 0 \pmod{3} \end{align}
Split the numbers from $$1,\ldots, 99$$ in
• $$33$$ with remainder $$1$$
• $$33$$ with remainder $$2$$
• $$33$$ with remainder $$0$$ $$1^3 + 2^3 + \cdots + 99^3 \equiv 33\cdot (1^3+2^3+0^3) \equiv 0 \mod 3$$
consider these observations:
$$1 = 1 \pmod 3$$ $$2 = -1 \pmod 3$$ $$3 = 0 \pmod 3$$
This becomes: | {
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consider these observations:
$$1 = 1 \pmod 3$$ $$2 = -1 \pmod 3$$ $$3 = 0 \pmod 3$$
This becomes:
$$33(1^3+(-1)^3+0) \equiv 33(1+(-1)+0) \equiv 33(0) \equiv 0 \pmod 3$$
Hint:
$$n^3\equiv n\pmod3$$ as $$n^3-n=n(n-1)(n+1)$$ is the product of three consecutive integers
We have $$\sum_{r=1}^mr^3\equiv\sum_{r=1}^mr\pmod3$$
$$\equiv \dfrac{m(m+1)}2$$
Alternatively $$\sum_{r=1}^m r^3=\dfrac{m^2(m+1)^2}4$$
From $$a^{3} + b^{3} = (a+b)(a^{2}-ab+b^{2})$$, you have $$a+b|a^{3} +b^{3}$$ and \begin{align} &1^{3} + 2^{3} + 3^{3} + \cdots + 97^{3} + 98^{3} + 99^{3} \\ &= (1^{3} + 2^{3}) + (4^{3} + 5^{3}) + \cdots + (97^{3}+98^{3}) + (3^{3}+6^{3} + \cdots + 99^{3}) \end{align} is a multiple of 3.
Another method is to use the identity
$$\sum_{k=1}^nk^3=\left(\sum_{k=1}^nk\right)^2$$ So, $$\sum_{k=1}^{99}k^3\equiv\left(\sum_{k=1}^{99}k\right)^2\equiv\frac{(99)^2(100)^2}{4}\equiv0\ (\mod3)$$
Yet another method is to use that $$(x-1)^3=x^3-3x^2+3x-1$$ and $$(x+1)^3=x^3+3x^2+3x+1$$, so $$(x-1)^3+x^3+(x+1)^3 = 3x^3+6x$$, which is divisible by three. This shows that any three consecutive numbers, when cubed, add up to zero modulus three.
You can also calculate that sum using
$$\sum_{i=0}^{i=n}i^3=\dfrac{n^2(n+1)^2}{4}$$
Here: $$\sum_{i=0}^{i=99}i^3=\dfrac{99^2(99+1)^2}{4}$$ which is clearly divisible by 3. | {
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Contest Duration: - (local time) (100 minutes) Back to Home
Official
## B - log2(N) Editorial by en_translator
There are various approaches to solve this problem. Here, we will explain some of them. Even if you got accepted during the contest, we recommend considering other solutions.
Note that the sample codes in this editorials are provided in C++.
1. Basic solution
2. Solution using logarithmic function, and numerical errors
3. Solution using binary representation of an integer
#### 1. Basic solution
First, as $$k$$ increases, obviously $$2^k$$ gets larger too. Therefore, we can see that $$2^k \le N$$ for all integers $$k$$ up to some boundary, or $$2^k>N$$ for $$k$$ greater than that value.
Thus, we can find the answer with a loop where $$k$$ is incremented one by one .
Note that one can $$2^k$$ efficiently compute $$2^k$$ during the loop of incrementing $$k$$ for a good computational complexity.
#include<bits/stdc++.h>
using namespace std;
int main(){
long long n;
cin >> n;
long long val=1;
for(long long i=0;i<=60;i++){
if(val>n){
cout << i-1 << '\n';
break;
}
val*=2;
}
return 0;
}
#### 2. Solution using logarithmic function, and numerical errors
By applying logarithmic function to $$2^k \le N$$ we obtain $$k \le \log_2(N)$$. Therefore, one should be able to output the $$\log_2(N)$$, rounded down to an integer, to get accepted.
However, this code results in WA (Wrong Answer).
#include<bits/stdc++.h>
using namespace std;
int main(){
long long n;
cin >> n;
cout << floor(log2(n)) << '\n';
return 0;
}
The cause of WA is lack of precision. Specifically, a precision occurs for the following case ($$N=2^{59}-1$$). (WA: 59, AC: 58)
576460752303423487
In order to avoid the error, one can use a type having more precisions than 64-bit floating point decimal. If you modify the last code as follows, you can get accepted for this problem.
#include<bits/stdc++.h>
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#include<bits/stdc++.h>
using namespace std;
int main(){
long long n;
cin >> n;
cout << floor(log2((long double)n)) << '\n';
return 0;
}
No that high-precision type is not always omnipotent. If the constraints for this problem were $$N \le 2 \times 10^{18}$$, the modified code still outputs a wrong answer for the following test case. (WA: 60, AC: 59)
1152921504606846975
In competitive programming, discarding the integral property in an integral problem and using only decimals is dangerous. We recommend you to process integral problems in integer-typed values as much as possible. When you inevitably use decimals, be very careful about the errors.
#### 3. Solution using binary representation of an integer
Consider the binary representation of the integer.
For example, we have the conversions $$6=110_{(2)}$$ and $$15=1111_{(2)}$$.
Here, the inequality $$2^{k} \le N < 2^{k+1}$$ can be transformed into binary notations as follows.
$$1 \underbrace{ 00 \dots 0 }_{ k }\ _{(2)} \le N < 1 \underbrace{ 00 \dots 0 }_{ k+1 }\ _{(2)}$$
Using this property, we can derive the following solution.
• If the binary notation of $$N$$ (without leading $$0$$’s) has $$k$$ digits (that is, if the most significant bit with value $$1$$ is the $$k$$-th least significant bit), then the answer is $$k-1$$.
There are various ways to implement this, while using bit operations simplifies it. (Note that the least significant digit is treated as the $$0$$-th digit.)
#include<bits/stdc++.h>
using namespace std;
int main(){
long long n;
cin >> n;
for(int i=60;i>=0;i--){
if(n&(1ll<<i)){cout << i << '\n';break;}
}
return 0;
}
posted:
last update: | {
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# CLRS Chapter 4. Divide and Conquer
source code directory
## 4.1. The maximum-subarray problem
### Implementation
A divide-and-conquer implementation of the maximal subarray finder is as follows:
from math import floor
infty = float("inf") # -infty is used as the universal lower bound
def maximal_subarray(A, lo, hi):
if lo == hi:
return (A[lo], lo, hi)
else:
mid = lo + floor((hi - lo)/2)
# the divide-and-conquer step
(lsum, llo, lhi) = maximal_subarray(A, lo, mid)
(rsum, rlo, rhi) = maximal_subarray(A, mid+1, hi)
(csum, clo, chi) = maximal_crossing_subarray(A, lo, mid, hi)
# pick the maximal result
if (lsum >= csum) and (lsum >= rsum):
return (lsum, llo, lhi)
elif (rsum >= lsum) and (rsum >= csum):
return (rsum, rlo, rhi)
else:
return (csum, clo, chi)
def maximal_crossing_subarray(A, lo, mid, hi):
"""search left of mid and right of mid separately, then add"""
(lsum, llo) = maximal_subarray_fixed_starting_point(A, mid, lo, -1)
(rsum, rhi) = maximal_subarray_fixed_starting_point(A, mid+1, hi, 1)
return (lsum + rsum, llo, rhi)
def maximal_subarray_fixed_starting_point(A, start, end, direction):
total, subtotal, max_index = -infty, 0, start
for i in range(start, end + direction, direction):
subtotal = subtotal + A[i]
if subtotal > total:
total, max_index = subtotal, i
return (total, max_index)
>>> A = [13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7]
>>> maximal_subarray(A, 0, len(A)-1)
(43, 7, 10)
### Exercises
4.1-1. What does find-maximaum-subarray return when all elements of $A$ are negative?
find-maximum-subarray in the text is equivalent to maximal_subarray above, so let us take a look at the latter. We shall show that maximal_subarray(A, 0, len(A)-1) returns the maximum element of $A$ whenever $A$ consists entirely of negative numbers. | {
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We first observe that maximal_subarray_fixed_starting_point(A, start, end, direction) returns (A[start], start) for all legal inputs of start and end. From this, we see that maximal_crossing_subarray(A, lo, mid, hi) returns (A[mid] + A[mid+1], mid, mid+1). Since $A$ consists entirely of negative numbers,
whence it follows that the return value of maximal_subarray cannot come from maximal_crossing_subarray.
The only return values of maximal_subarray that does not come from maximal_crossing_subarray are the terms of $A$. Since the if-elif-else block at the end of maximal_subarray returns the maximum, we conclude that maximal_subarray(A, 0, len(A)-1) returns the maximum element of $A$, as was to be shown. $\square$
4.1-2. Write pseudocode for the brute-force method of solving the maximum-subarray problem. Your procedure should run in $\Theta(n^2)$ time.
infty = float("inf") # -infty is used as the universal lower bound
def maximal_subarray_brute_force(A):
n = len(A)
max_sum = -infty
for i in range(n):
iterative_sum = 0
for j in range(i, n):
iterative_sum += A[j]
if iterative_sum > max_sum:
max_sum = iterative_sum
max_lo, max_hi = i, j
return (max_sum, max_lo, max_hi)
>>> A = [13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7]
>>> maximal_subarray_brute_force(A)
(43, 7, 10)
>>> A = [13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7]
>>> maximal_subarray(A, 0, len(A)-1)
(43, 7, 10)
It is not hard to see that maximal_subarray_brute_force is correct, as it compares every possible sub-sum.
Assuming that the if clause is never satisfied, we can compute the lower bound of the running time as follows:
The if clause only adds constant time to the computation, and so the upper bound of the running time is $O(n^2)$. It follows that maximal_subarray_brute_force runs in $\Theta(n^2)$ time. $\square$ | {
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4.1-3. Implement both the brute-force and recursive algorithms for the maximum-subarray problem on your own computer. What problem size $n_0$ gives the crossover point at which the recursive algorithm beats the brute-force algorithm? Then, change the base case of the recursive algorithm to use the brute-force algorithm whenever the problem size is less than $n_0$. Does that change the crossover point?
>>> import random
50 elements:
>>> L1 = random.sample(range(-1000, 1000), 50)
>>> L2 = random.sample(range(-1000, 1000), 50)
>>> L3 = random.sample(range(-1000, 1000), 50)
>>> %%timeit
... maximal_subarray(L1, 0, len(L1)-1)
... maximal_subarray(L2, 0, len(L2)-1)
... maximal_subarray(L3, 0, len(L3)-1)
231 µs ± 1.94 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
>>> %%timeit
... maximal_subarray_brute_force(L1)
... maximal_subarray_brute_force(L2)
... maximal_subarray_brute_force(L3)
227 µs ± 978 ns per loop (mean ± std. dev. of 7 runs, 1000 loops each)
60 elements:
>>> L1 = random.sample(range(-1000, 1000),60)
>>> L2 = random.sample(range(-1000, 1000),60)
>>> L3 = random.sample(range(-1000, 1000),60)
>>> %%timeit
... maximal_subarray(L1, 0, len(L1)-1)
... maximal_subarray(L2, 0, len(L2)-1)
... maximal_subarray(L3, 0, len(L3)-1)
285 µs ± 6.63 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
>>> %%timeit
... maximal_subarray_brute_force(L1)
... maximal_subarray_brute_force(L2)
... maximal_subarray_brute_force(L3)
312 µs ± 1.21 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
It follows that $% $ for this particular example. Let us now modify the recursive algorithm to include the brute-force algorithm for all lists of size at most, say, 55: | {
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def maximal_subarray_mixed(A, lo, hi):
if hi - lo <= 55:
(brute_sum, brute_lo, brute_hi) = maximal_subarray_brute_force(
A[lo:hi+1])
return (brute_sum, brute_lo + lo, brute_hi + hi)
else:
mid = lo + floor((hi-lo)/2)
(lsum, llo, lhi) = maximal_subarray(A, lo, mid)
(rsum, rlo, rhi) = maximal_subarray(A, mid+1, hi)
(csum, clo, chi) = maximal_crossing_subarray(A, lo, mid, hi)
if (lsum >= csum) and (lsum >= rsum):
return (lsum, llo, lhi)
elif (rsum >= lsum) and (rsum >= csum):
return (rsum, rlo, rhi)
else:
return (csum, clo, chi)
Let us test the new algorithm.
60 elements:
>>> L1 = random.sample(range(-1000, 1000),60)
>>> L2 = random.sample(range(-1000, 1000),60)
>>> L3 = random.sample(range(-1000, 1000),60)
>>> %%timeit
... maximal_subarray(L1, 0, len(L1)-1)
... maximal_subarray(L2, 0, len(L2)-1)
... maximal_subarray(L3, 0, len(L3)-1)
279 µs ± 3.09 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
>>> %%timeit
... maximal_subarray_mixed(L1, 0, len(L1)-1)
... maximal_subarray_mixed(L2, 0, len(L2)-1)
... maximal_subarray_mixed(L3, 0, len(L3)-1)
278 µs ± 989 ns per loop (mean ± std. dev. of 7 runs, 1000 loops each)
70 elements:
>>> L1 = random.sample(range(-1000, 1000),70)
>>> L2 = random.sample(range(-1000, 1000),70)
>>> L3 = random.sample(range(-1000, 1000),70)
>>> %%timeit
... maximal_subarray(L1, 0, len(L1)-1)
... maximal_subarray(L2, 0, len(L2)-1)
... maximal_subarray(L3, 0, len(L3)-1)
330 µs ± 1.61 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
>>> %%timeit
... maximal_subarray_mixed(L1, 0, len(L1)-1)
... maximal_subarray_mixed(L2, 0, len(L2)-1)
... maximal_subarray_mixed(L3, 0, len(L3)-1)
330 µs ± 2.47 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
We see that the crossover point increases by 10 or so, but this is hardly an improvement. $\square$ | {
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We see that the crossover point increases by 10 or so, but this is hardly an improvement. $\square$
4.1-4. Suppose we change the definition of the maximum-subarray problem to allow the result to be an empty subarray, where the sum of the values of an empty subarray is 0. How would you change any of the algorithms that do not allow empty subarrays to permit an empty subarray to be the result?
Only the main routine needs to be changed:
from math import floor
def maximal_subarray_allow_empty_subarray(A, lo, hi):
if lo == hi:
if A[lo] < 0:
return (0, -1, -1) # allow for empty array, labeled by -1
else:
return (A[lo], lo, hi)
else:
mid = lo + floor((hi - lo)/2)
(lsum, llo, lhi) = maximal_subarray(A, lo, mid)
(rsum, rlo, rhi) = maximal_subarray(A, mid+1, hi)
(csum, clo, chi) = maximal_crossing_subarray(A, lo, mid, hi)
if (lsum < 0) and (rsum < 0) and (csum < 0):
return (0, -1, -1) # allow for empty array, labeled by -1
elif (lsum >= csum) and (lsum >= rsum):
return (lsum, llo, lhi)
elif (rsum >= lsum) and (rsum >= csum):
return (rsum, rlo, rhi)
else:
return (csum, clo, chi)
>>> A = [-1, -2, -3, -4, -5, -6, -7]
>>> maximal_subarray_allow_empty_subarray(A, 0, len(A)-1)
(0, -1, -1)
$\square$
4.1-5. Use the following ideas to develop a nonrecursive, linear-time algorithm for the maximum-subarray problem. Start at the left end of the array, and progress toward the right, keeping track of the maximum subarray seen so far. Knowing a maximum subarray of A[1..j], extend the answer to find a maximum subarray ending at index $j+1$ by using the following observation: a maximum subarray of A[1..j+1] is either a maximum subarray of A[1..j] or a subarray A[i..j+1] for some $1 \leq i \leq j+1$. Determine a maximum subarray of the form A[i..j+1] in constant time based on knowing a maximum subarray ending at index $j$.
We fix an index $j \geq 1$ and suppose that $A[p:q]$ is a maximum subarray of $A[:j]$. We pick another index $i$ such that | {
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By the maximality of $A[p:q]$ in $A[:j]$, every maximum subarray of $A[:j+1]$ must either equal to $A[p:q]$ or contain $A[j]$. Now,
whence we conclude that
Using the above observation, we devise a linear-time algorithm for the maximum-subarray problem.
Given an array $A$, we observe that $[A[0]]$ is the (unique) maximum subarray of $A[:0+1]$. We now apply the above observation inductively to produce a maximum subarray of $AA$, a process known as Kadane’s algorithm.
def kadane(A):
lo, hi, total = 0, 0, 0
tail_lo, tail_hi, tail_total = 0, 0, 0
for j in range(1, len(A)-1):
if tail_total + A[j] <= A[j]:
tail_lo, tail_hi, tail_total = j, j, A[j]
else:
tail_lo, tail_hi, tail_total = tail_lo, j, tail_total+A[j]
if tail_total >= total:
lo, hi, total = tail_lo, tail_hi, tail_total
return (total, lo, hi)
>>> A = [13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7]
>>> maximal_subarray(A, 0, len(A)-1) # see the implementation subsection
(43, 7, 10)
>>> kadane(A)
(43, 7,1 0)
$\square$
## 4.2 Strassen’s algorithm for matrix multiplication
See my blog post for details.
### Exercises
4.2-1. Use Strassen’s algorithm to compute the matrix product
Show your work.
Observe first that
The product is
$% $ $\square$
4.2-2. Write pseudocode for Strassen’s algorithm
The implementation below makes use of NumPy to reduce the implementation overhead on basic matrix operations.
import numpy as np
def strassen(A, B):
n = len(A)
if n == 1:
return A * B
# block matrices; // is integer division
a = np.array([[A[:n//2, :n//2], A[:n//2, n//2:]],
A[n//2:, :n//2], A[n//2:, n//2:]]])
b = np.array([[B[:n//2, :n//2], B[:n//2, n//2:]],
B[n//2:, :n//2], B[n//2:, n//2:]])
s1 = b[0,1] - b[1,1]
s2 = a[0,0] + a[0,1]
s3 = a[1,0] + a[1,1]
s4 = b[1,0] - b[0,0]
s5 = a[0,0] + a[1,1]
s6 = b[0,0] + b[1,1]
s7 = a[0,1] - a[1,1]
s8 = b[1,0] + b[1,1]
s9 = a[0,0] - a[1,0]
s10 = b[0,0] + b[0,1] | {
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} |
p1 = strassen(a[0,0], s1)
p2 = strassen(s2, b[1,1])
p3 = strassen(s3, b[0,0])
p4 = strassen(a[1,1], s4)
p5 = strassen(s5, s6)
p6 = strassen(s7, s8)
p7 = strassen(s9, s10)
c = np.zeros((n, n)) # initialize an n-by-n matrix
c[:n//2, :n//2] = p5 + p4 - p2 + p6
c[:n//2, n//2:] = p1 + p2
c[n//2:, :n//2] = p3 + p4
c[n//2:, n//2:] = p5 + p1 - p3 - p7
return c
>>> A = np.array([[1, 3], [7, 5]])
>>> B = np.array([[6, 8], [4, 2]])
>>> np.dot(A, B) # matrix multiplication
[[18 14]
[62 66]]
>>> strassen(A, B)
[[18. 14.]
[62. 66.]]
>>> A = np.array([[5, 2, 6, 1],
... [0, 6, 2, 0],
... [3, 8, 1, 4],
... [1, 8, 5, 6]])
>>> B = np.array([[7, 5, 8, 0],
... [1, 8, 2, 6],
... [9, 4, 3, 8],
... [5, 3, 7, 9]])
>>> np.dot(A, B) # matrix multiplication
[[ 96 68 69 69]
[ 24 56 18 52]
[ 58 95 71 92]
[ 90 107 81 142]]
>> strassen(A, B)
[[ 96. 68. 69. 69.]
[ 24. 56. 18. 52.]
[ 58. 95. 71. 92.]
[ 90. 107. 81. 142.]]
$\square$
4.2-3. How would you modify Strassen’s algorithm to multiply $n \times n$ matrices in which $n$ is not an exact power of 2? Show that the resulting algorithm runs in time $\Theta(n^{\log 7})$.
For each $n \times n$ matrix $A$, we let $m$ be the unique integer such that $% $ and augment $A$ with an $(2^m - n) \times (2^m - n)$ identity matrix $I$ as follows:
We then have
from which we can extract $AB$ in constant time. It thus suffices to compute the time complexity of the matrix multiplication $\tilde{A} \tilde{B}$.
Observe that
by the monotonicity of $t \mapsto t^{\log 7}$. Similarly,
Now,
and so
It follows that $n^{\log 7} = \Theta((2^m)^{\log 7})$, whence applying Strassen’s algorithm to the augmented matrices does not incur any additional asymptotic time complexity. $\square$ | {
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4.2-4. What is the largest $k$ such that if you multiply $3 \times 3$ matrices using $k$ multiplications (not assuming commutativity of multiplication), then you can multiply $n \times n$ matrices in time $o(n^{\log 7})$? What would be the running time of this algorithm be?
The recurrence relation to solve is
Observe that $n^2 = \Theta(n^{\log_3 9})$. By the master theorem ($\S4.5$), we have
where $(\ast)$ is $k > 9$ and $k(n/3)^2 \leq cn^2$ for some $% $ and all sufficiently large $n$.
Since $\log 7 > \log_3 k$ implies that $% $, we see that $T(n) = o(n^{\log 7})$ whenever $k \leq 9$. On the other hand, if $k > 9$, then then inequality
never holds for any $n \geq 1$ reagrdless of which $% $ we choose. It follows that we must have $k \leq 9$ in order to multiply $n$-by-$n$ matrices in time $o(n^{\log 7})$, and the running time is
4.2-5. V. Pan has discovered a way of multiplying $68 \times 68$ matrices using 132,464 multiplications, a way of multiplying $70 \times 70$ matrices using 143,630 multiplications, and a way of multiplying $72 \times 72$ matrices using 155,424 multiplications. Which method yields the best asymptotic running time when used in a divide-and-conquer matrix-multiplication algorithm? How does it compare to Strassen’s algorithm?
The recurrence relations to solve are
Observing that $n^2 = \Theta(n^{\log_{68} 4624}){}_{}$, we invoke the master theorem ($\S4.5$) to conclude that
Since $n^{\log 7} \approx 2.807$, we conclude that $T_{68}{}_{}$ is better than the running time of Strassen.
Observe now that $n^2\Theta(n^{\log_{70} 4900}){}_{}$. The master theorem ($\S4.5$) implies that
Similarly as above, we conclude that $T_{72}{}_{}$ is better than the running time of Strassen.
Finally, we observe that $n^2 = \Theta(n^{\log_{72} 1944}){}_{}$, which implies that
by the master theorem ($\S4.5$). Similarly as above, we conclude that $T_{72}{}_{}$ is better than the running time of Strassen.
We conclude that | {
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We conclude that
where $T_{S}{}_{}$ refers to the runing time of Strassen. $\square$
4.2-6. How quickly can you multiply a $kn \times n$ matrix by an $n \times kn$ matrix, using Strassen’s algorithm as a subroutine? Answer the same question with the order of the input matrices reversed.
Using block matrix multiplication
we can multiply a $kn \times n$ matrix by an $n \times kn$ matrix through $k^2$ iterations of Strassen. It follows that the computation takes $\Theta(k^2n^{\log 7})$.
Similarly, we use block matrix multiplication
we can multiply an $n \times kn$ matrix by a $kn \times n$ matrix through $k$ iterations of STrassen. It follows that the computation takes $\Theta(k n^{\log 7})$. $\square$
4.2-7. Show how to multiply the complex numbers $a + bi$ and $c + di$ using only three multiplcations of real numbers. The algorithm should take $a$, $b$, $c$, and $d$ as input and produce the real component $ac - bd$ and the imaginary component $ad + bc$ separately.
We set $P_1 = ac$, $P_2 = bd$, and $P_3 = (a+b)(c+d)$. Observe that
and that
It follows that
which only requires three real multiplications.
This algorithm is commonly known as the Karatsuba algorithm. $\square$
## The substitution method for solving recurrences
### Exercises
4.3-1. Show that the solution of $T(n) = T(n-1) + n$ is $O(n^2)$.
We let $C > 0$ be an arbitrary constant, to be chosen later. We fix a postive integer $n$ and assume that
for all $% $. We shall show that
Observe that
By choosing $C = 1$, we see that $-2(C-1)n = 0$ for all $n \geq 0$, from which it follows that
as was to be shown.
To turn the above argument into a complete induction proof, we observe that
Since $C = 1$, it suffices to find $m_0$ such that
Choosing any $m_0 \geq \sqrt{2T(1)}$, we see that
as was to be shown. The desired result now follows from mathematical induction with base case $n = m_0$. $\square$
4.3-2. Show that the solution of $T(n) = T(\lceil n/2 \rceil) + 1$ is $O(\log n)$. | {
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4.3-2. Show that the solution of $T(n) = T(\lceil n/2 \rceil) + 1$ is $O(\log n)$.
We first examine the recurrence for $n \geq 2$.
We let $C > 0$ be an arbitrary constant, to be chosen later. We fix a positive integer $n$ and assume that
for all $% $. We shall show that
Observe that
We can bound the last expression by $C \log n$ if and only if
The above inequality is equivalent to
Recalling that $n \geq 2$, we note that
Choosing $C = 1/(1 + \log(2/3))$, we see that
It now follows that
as was to be shown.
To turn the above argument into a complete induction proof, we observe that
for all $k \geq 1.$ Since $C = 1/(1 + \log(2/3))$, it suffices to find $k_0$ such that
The above inequality is equivalent to
which, in turn, is equivalent to
We pick any $k_0$ that satisfies the last inequality and set $m_0 = 2^{k_0}$, so that
The desired result now follows from mathematical induction with base case $n = m_0$. $\square$
4.3-3. We saw that the solution of $T(n) = 2T(\lfloor n/2 \rfloor) + n$ is $O(n \log n)$. Show that the solution of this recurrence is also $\Omega(n \log n)$. Conclude that the solution is $\Theta(n \log n)$.
We first examine the recurrence for $n \geq 2$.
We let $C > 0$ be an arbitrary constant, to be chosen later. We fix a positive integer $n$ and assume that
for all $% $. We shall show that
Observe that
The inequality
is equivalent to
which, in turn, is equivalent to
Since $n \geq 2$, we see that
We claim that the function
is monotonically decreasing on $[2,\infty)$. Indeed, we have that
from which we conclude that $f(k) > f(k+1)$. Monotonicity now follows.
The monotonicity of $f(k)$ implies that
Setting $C = \frac{2}{3}$, we see that $(\ast \ast)$ holds, whence $(\ast)$ holds. We conclude that
with this choice of $C$, as was to be shown.
To turn the above argument into a complete induction proof, we observe that $m_0 = 2$ yields
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as $T$ is typically assumed to be nonnegative.
The desired result now follows from mathematical induction with base case $n =m_0$.
4.3-4. Show that by making a different inductive hypothesis, we can overcome the difficulty with the boundary condition $T(1) = 1$ for recurrence (4.19) without adjusting the boundary conditions for the inductive proof.
Recall (4.19):
We let $C > 0$ be an arbitrary constant, to be chosen later. We fix a positive integer $n$ and assume that
for all $% $. We shall show that
Observe that
$C \geq T(1) - 1$, we have $1-C \leq -T(1)$, so that
To turn the above argument into a complete induction proof, we observe that
The desired result now follows from mathematical induction with base case $n=1$. $\square$
4.3-5. Show that $\Theta(n \log n)$ is the solution to the “exact” recurrence (4.3) for merge sort.
Recall (4.3):
We let $c > 0$ and $C > 0$, to be chosen later. We fix a positive integer $n$ and assume that
for all $% $. We shall show that
Since $T(n) = T(\lceil n/2 \rceil) + T(\lfloor n/2 \rfloor) + \Theta(n)$, we have the double-ended estimate
where the constants $d > 0$ and $D > 0$ are to be chosen later.
We first establish the upper bound. Since $\lfloor n/2 \rfloor \leq \lceil n/2 \rceil$,
$\lceil n/2 \rceil \geq n/2$, and so $n/\lceil n/2 \rceil \leq n/(n/2) = 2$. It then follows that
whence any choice of $C$ and $D$ with $C \leq D$ yields
Let us now establish the lower bound. Since $\lceil n/2 \rceil \geq \lfloor n/2 \rfloor$,
$\lfloor n/2 \rfloor \leq n/2$, and so $n/\lfloor n/2 \rfloor \geq n/(n/2) = 2$. It then follows that
whence any choice of $c$ and $d$ with $c \leq d$ yields
We have thus shown that
under the assumption that
holds for all $% $.
To turn the above argument into a complete induction proof, we set $c = d = T(2)/4$ and $C = D = 2T(2)$ and observe that, for $n = 2$,
and that
It follows from mathematical induction that $T(n) = \Theta(n \log n)$ with base case $n = 2$. $\square$ | {
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4.3-6. Show that the solution to $T(n) = 2T(\lfloor n /2 \rfloor + 17) + n$ is $O(n \log n)$.
Assume that $n \geq 68$, so that $17 \leq n/4$.
We let $C > 0$, to be chosen later. We fix a positive integer $n$ and assume that
for all $% $. We shall show that
Observe that
Now, $n \geq 68$, and so
and
It follows that
Since $\log(4/3) > 0$, we have $% $, and so
It thus suffices to pick $C > 0$ such that
keeping in mind that $n \geq 68$. Since $n$ and $\log n$ are both positive under this condition, we need only to pick $C$ large enough that
Letting $C = 1/\log(4/3)$, we obtain
under the assumption that
for all $% $.
To turn the above argument into a complete induction proof, we must show that
But, of course,
It now follows from mathematical induction with base case $n = 68$ that
as was to be shown. $\square$
4.3-7. Using the master method in Section 4.5, you can show that the solution to the recurrence $T(n) = 4T(n/3) + n$ is $T(n) = \Theta(n^{\log_3 4})$. Show that a substitution proof with the assumption $T(n) \leq cn^{\log_3 4}$ fails. Then show how to subtract off a lower-order term to make a substitution proof work.
Since
a naïve substitution argument would yield
Since
for all $n \geq 1$, the substitution argument fails.
If we assume instead that
for all $% $, then
and the usual substitution proof goes through. $\square$
4.3-8. Using the master method in Section 4.5, you can show that the solution to the recurrence $T(n) = 4T(n/2) + n$ is $T(n) = \Theta(n^2)$. Show that a substitution proof with the assumption $T(n) \leq cn^2$ fails. Then show how to subtract off a lower-order term to make a substitution proof work.
Since $c(n/2)^2 = (c/4)n^2$, a naïve substitution argument would yield
which is larger than $cn^2$ for all $n \geq 1$.
If we assume instead that
for all $% $, then
and the usual substitution proof goes through. $\square$ | {
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for all $% $, then
and the usual substitution proof goes through. $\square$
4.3-9. Solve the recurrence $T(n) = 3T(\sqrt{n}) + \log n$ by makign a change of variables. Your solution should be asymptotically tight. Do not worry about whether values are integral.
Since
we set $S(m) = T(2^m)$ and solve instead
The master theorem ($\S4.5$) shows that
We shall show this by substitution.
We let $c > 0$ and $C > 0$, to be chosen later. We fix a positive integer $n$ and assume that
for all $% $. We shall show that
We first establish the upper bound:
Now, the lower bound:
To turn the above argument into a complete induction proof, we must show that
Setting $c = C = S(1) + 2$, we see that
and that
Mathematical induction with base case $m = 1$ now shows that
for all $m \geq 1$. Since the lower bound agreed with the upper bound, we in fact have the identity
Since $T(n) = S(\log n)$, we conclude that
## 4.4. The recursion-tree method for solving recurrences
### Exercises
4.4-1. Use a recursion tree to determine a good asymptotic upper bound on the recurrence $T(n) = 3T(\lfloor n/2 \rfloor) + n$. Use the substitution method to verify your answer.
Observe that
Let us verify our guess. To this end, we let $C > 0$, to be chosen later. We fix a positive integer $n$ and assume that
for all $% $. We shall show that
Indeed,
To complete the induction proof, we must show that
We set $C = T(1) + 2$ and observe that
as was to be shown. It now follows from mathematical induction with base case $n = 1$ that $T(n) = O(n^{\log(3/2)})$. $\square$
4.4-2. Use a recursion tree to determine a good asymptotic upper bound on the recurrence $T(n) = T(n/2) + n^2$. Use the substitution method to verify your answer.
Observe that
Let us verify our guess. To this end, we let $C > 0$, to be chosen later. We fix a positive integer $n$ and assume that
for all $% $. We shall show that
Observe that
So long as $C \geq 4$, we have $C/4 + 1 \leq C$, and so | {
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Observe that
So long as $C \geq 4$, we have $C/4 + 1 \leq C$, and so
To complete the induction proof, we must show that
To this end, we simply select $C = \max(4, T(1))$. The desired result now follows from induction. $\square$
4.4-3. Use a recursion tree to determine a good asymptotic upper bound on the recurrence $T(n) = 4T(n/2 + 2) + n$. Use the substitution method to verify your answer.
We first observe the the recurrence relation as it is written above is ill-defined. Indeed,
when $T$ is a function defined on $\mathbb{N}$. To remedy this issue, we shall consider instead
Observe that
Setting $n = 2^k$, we obtain
We therefore conjecture that
Since the problem only asks us to establish the upper bound, we shall prove that
We let $C > 0$ and $D > 0$, to be chosen later. We fix a positive integer $n$ and assume that
for all $% $. We shall show that
Observe that
If we choose $C$ and $D$ so that $4C -D + 1 \leq 0$ and $8C - 6D \leq 0$, then we have the bound
How do we choose such $C$ and $D$? The first inequality can be written as
It turns out that this is sufficient to achieve the second inequality:
We also note that the constants must be chosen to satisfy the base case
for an appropriate choice of $n_0$. Since the smallest $D$ that satisfies $(\ast)$ is $4C + 1$, the base case should be chosen so that $Cn_0^2 - (4C+1)n_0 > 0$. This only works if $n_0 \geq 5$. indeed, if $n_0 = 4$, then
On the other hand, if $n_0 = 5$, then
which is positive if $C > 4/5$.
Summarizing the above argument, we choose $C = T(5) + 1$ and $D = 4T(5) + 5$. For the base case $n = 5$, we have
as $T$ is generallly assumed to be nonnegative.
We now assume inductively that
holds for all $% $ such that $m \geq 5$. Now, the inductive hypothesis implies that
as was to be shown. We conclude that $T(n) = O(n^2)$. $\square$ | {
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as was to be shown. We conclude that $T(n) = O(n^2)$. $\square$
4.4-4. Use a recursion tree to determine a good asymptotic upper bound on the recurrence $T(n) = 2T(n-1) + 1$. Use the substitution method to verify your answer.
Observe that
Let us verify our guess. To this end, we let $C > 0$, to be chosen later. We fix a positive integer $n$ and assume that
for all $% $. We then have
By choosing $C = T(1)/2$, we have
for $n = 1$. It now follows from mathematical induction that $T(n) = O(2^n)$. $\square$
4.4-5. Use a recursion tree to determine a good asymptotic upper bound on the recurrence $T(n) = T(n-1) + T(n/2) + n$. Use the substitution method to verify your answer.
To make sense of the $n/2$ term, we consider instead
Since $T(n) - T(n-1) = T(\lfloor n/2 \rfloor) + n$, we see that
Therefore,
Applying $(\ast)$ to the sum indexed by $i_1$, we obtain
Applying $(\ast)$ recursively, we obtain
Let us verify our guess. To this end, we let $C > 0$ and $D > 0$, to be chosen later. We fix a positive integer $n$ and assume that
for all $% $. We shall show that
Observe that
Since
we see that
Since we must have
we see that
is required for all $n \geq 1$. This holds when $C \geq 2 + 2D$:
Finally, we must pick $C$ and $D$ so that
holds when $n = 1$. This requires
Setting $D = T(1)$ and $C = 2T(1) + 2$, we see that
The choice of $C$ and $D$ also shows that
whenever
for all $% $. It now follows from mathematical induction with base case $n = 1$ that $T(n) = O(n^{\log n})$. $\square$
4.4-6. Argue that the solution to the recurrence $T(n) = T(n/3) + T(2n/3) + cn$, where $c$ is a constant, is $\Omega(n \log n)$ by appealing to a recursion tree.
Observe that
where $\begin{pmatrix} p \\q \end{pmatrix} = \frac{p!}{q!(p-q)!}$ is the binomial coefficient with respect to $(p,q)$. It follows that $T(n) = \Omega(n \log n)$. $\square$ | {
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4.4-7. Draw the recursion tree for $T(n) = 4T(\lfloor n/2 \rfloor) + cn$, where $c$ is a constant, and provide a tight asymptotic bound on its solution. Verify your bound by the substitution method.
Observe that
We thus conjecture that $T(n) = \Theta(n^2)$.
We first establish the upper bound. To this end, we let $C > 0$ and $D > 0$, to be chosen later. We fix a positive integer $n$ and assume that
for all $% $. We shall show that
Observe that
Therefore, $D$ must be chosen so that $(c-D)n +2 \leq 0$ for all $n \geq 1$. This holds if $D \geq c + 2$.
We now set
and
so that
Moreover, if $T(m) \leq Cm^2 - Dm$ for all $% $, then
It now follows from induction with base case $n = 1$ that $T(n) = O(n^2)$.
It remains to show that $T(n) = \Omega(n^2)$.
We let $C > 0$ and $D > 0$, to be chosen later. We fix a positive integer $n$ and assume that $T(m) \geq Cm^2 + Dm$ for all $% $. Observe that
To ensure that the resulting expression is bounded below by $Cn^2 + Dn$, we must at least have $D-2C+c > 0$. This holds if $D > 2C - c$.
We assume for now that $c > 0$ and set $C = \min(c/2, T(1)/2)$ and $D = \min(c/4, T(1)/4)$, so that
Moreover,
and so it follows from mathematical induction with base case $n=1$ that $T(n) = \Omega(n^2)$.
We have thus shown that $T(n) = \Theta(n^2)$. $\square$
4.4-8. Use a recursion tree to give an asymptotically tight solution to the recurrence $T(n) = T(n-a) + T(a) + cn$, where $a \geq 1$ and $c > 0$ are constants.
Observe that
We thus conjecture that $T(n) = \Theta(n^2)$.
Since $% $ for $% $, the recurrence relation cannot hold for $% $. We thus assume that the values of $T$ for $n = 1,\ldots,\lfloor a \rfloor$ are given.
We first show the upper bound. To this end, we let $C > 0$, to be chosen later. We shall show that
for all $n \geq 2 \lfloor a \rfloor$. We first assume that $C \geq T(\lfloor a \rfloor) / (\lfloor a \rfloor)^2$, so that $(\ast)$ holds for $n = \lfloor a \rfloor$.
Observe that | {
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Observe that
If $C \geq c/(2 \lfloor a \rfloor)$, then
and so
We thus assume that
We now fix $n_0 > 2\lfloor a \rfloor$ and assume that $(\ast)$ holds for $% $. We shall show that $(\ast)$ holds for $n = n_0$.
Observe that
If $C \geq c$, then
Since $a \geq 1$, we have the estimate
Now, $n_0 > 2 \lfloor a \rfloor$, and so
It follows that $(\ast)$ holds for $n = n_0$ whenever $C \geq c$.
Setting
we see that $(\ast)$ holds for all $n \geq 2 \lfloor a \rfloor$. It follows that $T(n) = O(n^2)$.
We now establish the lower bound. To this end, we let $C > 0$, to be chosen later. We shall show that
for all $n \geq 2 \lfloor a \rfloor$. We first assume that $C \leq T(\lfloor a \rfloor) / (\lfloor a \rfloor)^2$, so that $(\ast \ast)$ holds for $n = \lfloor a \rfloor$.
Observe that
If $C \leq c/(2 \lfloor a \rfloor)$, then
and so
We thus assume that
We now fix $n_0 > 2\lfloor a \rfloor$ and assume that $(\ast \ast)$ holds for $% $. We shall show that $(\ast\ast)$ holds for $n = n_0$.
Observe that
If $C \leq c/2a$, then
Since $a \geq 0$, we have $2ac^2 \geq 0$, and so
It follows that $(\ast \ast)$ holds for $n = n_0$ whenever $C \leq c/2a$.
Setting
we have $(\ast \ast)$ for all $n \geq 2 \lfloor a \rfloor$. It follows that $T(n) = \Omega(n^2)$.
We now conclude that $T(n) = \Theta(n^2)$, as was to be shown. $\square$
4.4-9. Use a recursion tree to give an asymptotically tight solution to the recurrence $T(n) = T(\alpha n) + T((1-\alpha) n) + cn$, where $\alpha$ is a constant in the range $% $ and $c > 0$ is also a constant.
Observe that
We let $a = \max(\alpha, 1-\alpha)$ and set $p = \lceil \log_{1/\alpha} n \rceil$, so that
for all $0 \leq i \leq p$. The recursion-tree computation yields
We therefore conjecture that $T(n) = \Theta(n \log n).$
We first establish the upper bound. To this end, we let $C > 0$, to be chosen later. We assume that
for all $% $.
Observe that
Since $% $ and $% $, we have that $% $ and $% $. Therefore, any
would yield | {
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Observe that
Since $% $ and $% $, we have that $% $ and $% $. Therefore, any
would yield
We now set
so that $T(2) \leq C n \log n$, and that
whenever $T(m) \leq Cm \log m$ for all $% $. It now follows from induction that $T(n) = O(n \log n)$.
We now establish the lower bound. To this end, we let $C > 0$, to be chosen later. We assume that
for all $% $.
Observe that
Since $% $ and $% $, we have that $% $ and $% $. Therefore, any
would yield
We now set
so that $T(2) \geq C n \log n$, and that
whenever $T(m) \geq Cm \log m$ for all $% $. It now follows from induction that $T(n) = \Omega(n \log n)$.
We have thus shown that $T(n) = \Theta(n \log n)$. $\square$
## 4.5. The master method for solving recurrences
### The master theorem, as stated in CLRS, Theorem 4.1
Let $a \geq 1$ and $b > 1$ be constants, let $f(n)$ be a function, and let $T(n)$ be defined on the nonnegative integers by the recurrence
where we interpret $n/b$ to mean either $\lfloor n /b \rfloor$ or $\lceil n/b \rceil$. Then $T(n)$ has the following asymptotic bounds:
1. If $f(n) = O(n^{\log_b a - \epsilon})$ for some constant $\epsilon > 0$, then $T(n) = \Theta(n^{\log_b a})$.
2. If $f(n) = \Theta(n^{\log_b a})$, then $T(n) = \Theta(n^{\log_b a} \log n)$.
3. If $f(n) = \Omega(n^{\log_b a + \epsilon})$ for some constant $\epsilon > 0$, and if $a f(n/b) \leq cf(n)$ for some constant $% $ and all sufficiently large $n$, then $T(n) = \Theta(f(n))$.
### Exercises
4.5-1. Use the master method to give tight asymptotic bounds for the following recurrences.
a. $T(n) = 2T(n/4) + 1$.
b. $T(n) = 2T(n/4) + \sqrt{n}$.
c. $T(n) = 2T(n/4) + n$.
d. $T(n) = 2T(n/4) + n^2$.
a belongs to case 1, and so
b belongs to case 2, and so
c belongs to case 3, as
for all $n$. Therefore,
d belongs to case 3, as
for all $n$. Therefore, | {
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c belongs to case 3, as
for all $n$. Therefore,
d belongs to case 3, as
for all $n$. Therefore,
4.5-2. Professor Caesar wishes to develop a matrix-multiplication algorithm that is asymptotically faster than Strassen’s algorithm. His algorithm will use the divide-and-conquer method, dividing each matrix into pieces of size $n/4 \times n/4$, and the divide and combine steps together will take $\Theta(n^2)$ time. He needs to determine how many subproblems his algorithm has to create in order to beat Strassen’s algorithm. If his algorithm creates $a$ subproblems, then the recurrence for the running time $T(n)$ becomes $T(n) = aT(n/4) + \Theta(n^2)$. What is the largest integer value of $a$ for which Professor Caesar’s algorithm would be asymptotically faster than Strassen’s algorithm?
We recall that Strassen’s time complexity is $\Theta(n^{\log_2 7})$. In light of this, it suffices to find the maximum integral value of $a$ such that
subject to the restriction that $f(n) = n^2$ is asymptotically smaller than $n^{\log_b a}$.
Since $b = 4$, the restriction yields the lower bound
Observe that $\log_4 x = \log_2 7$ is equivalent to
which, in turn, is equivalent to
Since $\log_4 2 = 1/2$, we conclude that
It follows that we must have $% $, whence the optimal integral value of $a$ is $48$. $\square$
4.5-3. Use the master method to show that the solution to the binary-search recurrence $T(n) = T(n/2) + \Theta(1)$ is $\Theta(\log n)$. (See Exercise 2.3-5 for a description of binary search.)
Since $a = 1$ and $b = 2$, we have $\log_b a = 0$. Therefore, $f(n) = \Theta(1)$ is asymptotically equivalent to $n^{\log_b a}$, whence the master method implies that
as was to be shown. $\square$
4.5-4. Can the master method be applied to the recurrence $T(n) = 4T(n/2) + n^2 \log n$? Why or why not? Give an asymptotic upper bound for this recurrence. | {
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The $\log n$ term in $f(n) = n^2\log n$ makes it impossible to apply the master method here. $\log_b a = 2$, so $f(n) = \Omega(n^{\log_b a})$. Nevertheless, $\log n = o(n^p)$ for all $p > 0$, and so $f(n) = O(n^{\log b (a + \epsilon)})$ for all $\epsilon > 0$. It follows that none of the three cases of the master method is applicable.
We shall compute the recursion tree of $T(n)$. Observe:
Since
we conjecture that
To show this, we let $C > 0$, to be chosen later. We assume that
for all $% $. (We pick 16, so that $\log n > 4$.) We shall show that
Observe that
Since $\log n > 4$, we have
whenever $C \geq 1$.
We now set
so that, for $n = 16$,
Moreover, $C \geq 1$, and so
for all $% $
implies
It now follows from induction that $T(n) = O(n^2(\log n)^2)$. $\square$
4.5-5. Consider the regularity condition $af(n/b) \leq cf(n)$ for some constant $% $, which is part of case 3 of the master theorem. Give an example of constants $a \geq 1$ and $b > 1$ and a function $f(n)$ that satisfies all the conditions in case 3 of the master theorem except the regularity condition.
Let’s first rule out a general class of examples. Let $a \geq 1$, $b > 1$, $C > 0$, and $\epsilon > 0$ be arbitrary, and let
Since
we see that setting $c = a/(a+\epsilon)$ yields
for all $n \in \mathbb{N}$. $\epsilon > 0$, and so $% $.
This suggests that we need an oscillating term. Let us pick $a = 1$, $b = 2$, and $f(n) = 2n - n\cos(2 \pi n)$. Observe that
Whenever $n$ is odd, $\cos(\pi n) = -1$ and $\cos(2 \pi n) = 1$, and so
It follows that, whenever $% $, the inequality
cannot hold for all $n$. $\square$
## 4.6. Proof of the master theorem
### Exercises
4.6-1. Give a simple and exact expression for $n_j$ in equation (4.27) for the case in which $b$ is a positive integer instead of an arbitrary real number.
We shall prove a more general result: whenever $m,n \in \mathbb{N}$ and $x \in \mathbb{R}$, we have the identity
Observe that
This implies, in particular, that | {
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Observe that
This implies, in particular, that
The upper bound of $(\ast)$ implies that
We suppose for a contradiction that
We can find an integer $p$ such that
This implies that
so that
The lower bound of $(\ast\ast)$ implies that
It now follows from $(\ast\ast)$ that
which is evidently absurd. We conclude that
Let us now recall (4.27):
We shall show that
whenever $b$ is an integer. We proceed by mathematical induction. The $j = 0$ case is trivially true. If the $j = j_0 - 1$ case holds, then $(\ast\ast\ast)$ implies that
as was to be shown. $\square$
4.6-2. Show that if $f(n) = \Theta(n^{\log_b a} \log^k n)$, where $k \geq 0$, then the master recurrence has solution $T(n) = \Theta(n^{\log_b a} \log^{k+1} n)$. For simplicity, confine your analysis to exact powers of $b$.
We begin with a remark that
not
Fix $k \geq 0$. Recall from (4.21) that
Since
we see that
It thus suffices to show that
Since $b > 1$,
and so
$(\ast)$ follows from $(\ast\ast)$ if we can prove, in addition, that
To this end, we recall that $n$ is restricted to be of the form $b^p$ for some exponent $p$. We shall show that
for all $p \geq 1$, where $C$ is a constant independent of $p$.
We shall prove $(\ast\ast\ast)$ by induction on $p$.
We tackle the inductive step first, so as to determine an appropriate value of $C$. Let us fix $p_0 > 1$ and assume that
for all $% $, where $C > 0$ is a constant to be chosen later. We shall show that
To this end, we observe that
By way of the inductive hypothesis $(\star)$, we bound the above quantity below by
To show $(\star)$, it now suffices to establish the estimate
We write
for each $p$ and observe that
$b > 1$ implies that $\log^k(b) > 0$, and so we need only to show that
The above inequality is equivalent to
as $A_p > 1$ for all sufficiently large $p$. In fact, if $C \geq \frac{1}{\log b}$, then $A_p > 1$ for all $p > 1$.
It thus suffices to establish
If $C = \frac{1}{\log b}$, then
Since $p_0 > 1$, we see that | {
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It thus suffices to establish
If $C = \frac{1}{\log b}$, then
Since $p_0 > 1$, we see that
for all $k \geq 1$, establishing $(\star\star)$ for $C = \frac{1}{\log b}$.
It follows that $(\star)$ holds for $C = \frac{1}{\log b}$, which establishes the inductive case of $(\ast \ast \ast)$.
To complete the proof of $(\ast\ast\ast)$ by induction, we must establish the base case $p = 1$, i.e.,
Since we have chosen $C = \frac{1}{\log b}$, the base case holds with equality. The induction proof is now complete.
Finally, $(\ast)$ follows from $(\ast\ast)$ and $(\ast\ast\ast)$, and we have the desired estimate
Our work here is done. $\square$
4.6-3. Show that case 3 of the master theorem is overstated, in the sense that the regularity condition $af(n/b) \leq cf(n)$ for some constant $% $ implies that there exists a constant $\epsilon > 0$ such that $f(n) = \Omega(n^{\log_b a + \epsilon})$.
We recall from Theorem 4.1 that $a \geq 1$ and $b > 1$. We let $C = \frac{a}{c}$ and observe that the regularity condition can be rewritten as
Applying $(\ast)$ $\lceil \log_b n \rceil$ times, we obtain
Since $\log_b n \leq \lceil \log_b n \rceil$, we have the estimate
Now, $C^{\log_b n} = n^{\log_b C}$, and so
$C = \frac{a}{c}$ implies $\log_b C = \log_b a + \log_b \frac{1}{c}$, and so
where $\epsilon = \log_b \frac{1}{c}$. It follows that
Finally, we remark that $% $ implies $\epsilon > 0$. $\square$
## Problems
### 4-1. Recurrence examples
Give asymptotic upper and lower bounds for $T(n)$ in each of the following recurrences. Assume that $T(n)$ is constant for $n \leq 2$. Make your bounds as tight as possible, and justify your answers.
a. $T(n) = 2T(n/2) + n^4$
b. $T(n) = T(7n/10) + n$
c. $T(n) = 16T(n/4) + n^2$
d. $T(n) = 7T(n/3) + n^2$
e. $T(n) = 7T(n/2) + n^2$
f. $T(n) = 2T(n/4) + \sqrt{n}$
g. $T(n) = T(n-2) + n^2$
a-f can be solved directly by the master theorem (Theorem 4.1), which we restate below for ease of reference: | {
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Theorem 4.1 (Master theorem). Let $a \geq 1$ and $b > 1$ be constants, let $f(n)$ be a function, and let $T(n)$ be defined on the nonnegative integers by the recurrence
where we interpret $n/b$ to mean either $\lfloor n/b \rfloor$ or $\lceil n/b \rceil$. Then $T(n)$ has the following asymptotic bounds:
1. If $f(n) = O(n^{\log_b a - \epsilon})$ for some constant $\epsilon > 0$, then $T(n) = \Theta(n^{\log_b a})$.
2. If $f(n) = \Theta(n^{\log_b a})$, then $T(n) = \Theta(n^{\log_b a} \log n)$.
3. If $f(n) = \Omega(n^{\log_b a + \epsilon})$ for some constant $\epsilon > 0$, and if $af(n/b) \leq cf(n)$ for some constant $% $ and all sufficiently large $n$, then $T(n) = \Theta(f(n))$.
#### Problem 4-1a
$a = b = 2$, so that $\log_b a = 1$. Since
$f(n) = \Omega(n^{\log_b a + 3})$ and
we see that
#### Problem 4-1b
$a = 1$ and $b = \frac{10}{7}$, so that $\log_b a = 0$. Since
and
we see that
#### Problem 4-1c
$a = 16$ and $b = 4$, so that $\log_b a = 2$. Since $f(n) = \Theta(n^{\log_b a})$, we see that
#### Problem 4-1d
$a = 7$ and $b = 3$, so that $% $. Since
and
we see that
#### Problem 4-1e
$a = 7$ and $b = 2$, so that $% $. Since
we see that
#### Problem 4-1f
$a = 2$ and $b = 4$, so that $\log_b a = \frac{1}{2}$. Since $f(n) = \Theta(n^{\log_b a})$, we see that
#### Problem 4-1g
We cannot use the master theorem, as
is not of the form
Observe, however, that
Recalling that
we obtain
where $n' = \lfloor n/2 \rfloor$.Since $n' = \Theta(n)$, we have
### 4-2. Parameter-passing costs
Throughout this book, we assume that parameter passing during procedure calls takes constant time, even if an $N$-element array is being passed. This assumption is valid in most systems because a pointer to the array is passed, not the array itself. This problem examines the implications of three parameter-passing strategies:
1. An array is passed by pointer. Time = $\Theta(1)$. | {
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1. An array is passed by pointer. Time = $\Theta(1)$.
2. An array is passed by copying. Time = $\Theta(N)$, where $N$ is the size of the array.
3. An array is passed by copying only the subrange that might be accessed by the called procedure. Time = $\Theta(q-p+1)$ if the subarray $A[p..q]$ is passed.
a. Consider the recursive binary search algorithm for finding a number in a sorted array (see Exercise 2.3-5). Given recurrences for the worst-case running times of binary search when arrays are passed using each of the three methods above, and give good upper bounds on the solutions of the recurrences. Let $N$ be the size of the original problem and $n$ be the size of a subproblem.
b. Redo part (a) for the Merge-Sort algorithm from Section 2.3.1.
#### Problem 4-2a
Recall binary_search from Exercise 2.3-5, which we reproduce below:
from math import floor
def binary_search(A, p, q, v):
if p > q :
return -1
if p == q:
return p if A[p] == v else -1
mid = p + floor((q-p)/2)
if A[mid] == v:
return mid
else:
if A[mid] > v:
return binary_search(A, p, mid-1, v)
else: # A[mid] < v
return binary_search(A, mid+1, q, v)
Let $T(N)$ be the worst-case running time of binary_search on a list of length $N$. Since it takes constant time to execute all but the recursive step of the algorithm, we have
where $f(N,n)$ is the time it takes to pass a length-$n$ subarray of an array of length $N$.
If arrays are passed by pointer, then $f(N,n) = \Theta(1)$, and so
By the master theorem (see Section 4.5), we have
If arrays are passed by copying, then $f(N,n) = \Theta(N)$, and so
By the master theorem, we have
Finally, if arrays are passed by copying the appropriate subranges, then $f(N,n) = N/n = N/\lfloor N/2 \rfloor$. We thus have $f(N,n) = \Theta(1)$, and so
#### Problem 4-2b
We recall the non-sentinel version of merge sort from Exercise 2.3-2, which we reproduce below:
import math
def merge(A, p, q, r):
L, R = A[p,q], A[q:r]
n,m = q-p, r-q
i, j, k = 0, 0, p | {
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import math
def merge(A, p, q, r):
L, R = A[p,q], A[q:r]
n,m = q-p, r-q
i, j, k = 0, 0, p
while i < n and j < m:
if L[i] < R[j]:
A[k] = L[i]
i += 1
else:
j += 1
k += 1
while i < n or j < m:
if i < n:
A[k] = L[i]
else:
A[k] = R[j]
k += 1
def merge_sort(A, p, r):
if p+1 < r:
q = p + math.floor((r-p)/2)
merge_sort(A, p, q)
merge_sort(A, q, r)
merge(A, p, q, r)
Note that the time complexity of merge is $\Theta(r-p)$. Since we divide the input array in half, the time complexity of merge_sort is
where $f(a,b)$ is the time it takes to pass a length-$b$ subarray of an array of length $a$.
Since $f(n, n/2) = \Omega(n)$ regardless of how we pass in an array, we conclude that
It follows from the master theorem (see Section 4.5) that
### 4-3. More recurrence examples
Give asymptotic upper and lower bounds for $T(n)$ in each of the following recurrences. Assume that $T(n)$ is constant for sufficiently small $n$. Make your bounds as tight as possible, and justify your answers.
a. $T(n) = 4T(n/3) + n \log n$.
b. $T(n) = 3T(n/3) + n / \log n$.
c. $T(n) = 4T(n/2) + n^2 \sqrt{n}$.
d. $T(n) = 3T(n/3 - 2) + n/2$.
e. $T(n) = 2T(n/2) + n/\log n$.
f. $T(n) = T(n/2) + T(n/4) + T(n/8) + n$.
g. $T(n) = T(n-1) + 1/n$.
h. $T(n) = T(n-1) + \log n$.
i. $T(n) = T(n-2) + \log n$.
j. $T(n) = \sqrt{n} T(\sqrt{n}) + n$.
See Section 4.5 for the statement of the master theorem.
#### Problem 4-3a
Recall from Problem 3-2 that
for every $\epsilon > 0$. It follows that
for all $% $, whence the master theorem implies that
#### Problem 4-3b
We fist show that the master theorem cannot be used here. To this end, we shall show that
for any choice of $\epsilon > 0$,
and that
for any choice of $\epsilon > 0$.
Since $\log n \geq 1$ for all $n \geq 2$, we have that
for all $n \geq 2$. This implies $(\ast\ast)$. $(\ast)$ follows from $(\ast)$.
For $(\ast\ast\ast)$, we observe that each constant $C > 0$ furnishes an integer $M_C$ such that
for all $n \geq M_C$. It follows that | {
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for all $n \geq M_C$. It follows that
for all $n \geq M_C$, and $(\ast\ast\ast)$ follows.
Finally, we recall from Problem 3-2 that
for every $\rho > 0$. This implies that each choice of $\rho > 0$ and $C > 0$ furnishes an integer $N_{\rho, C}$ such that
for all $n \geq N_{\rho, C}$. Therefore, we see that
for all $n \geq N_{\epsilon, C}{}_{}$, and $(\ast\ast\ast\ast)$ follows.
Let us now proceed without the master theorem.
We let $k = \log_3 n$ and observe that
Since $% $, we see that
Moreover, $% $, and so
We thus have the estimates
In order to estimate the sum, we define $f(x) = \frac{1}{k-x}$ and observe that
whenever $i-1 \leq x \leq i$. Therefore,
for any choice of index $i$, whence it follows that
Since $k - \lfloor k \rfloor \geq 0$, we see that
We thus end up with the upper bound
for constants $C$, $D$, and $E$. We conclude that
To establish a lower bound, we set $f(x) = \frac{1}{k-x}$ once againd and observe that
whenever $i \leq x \leq i+1$. Therefore,
for any choice of index $i$, whence it follows that
Since $% $, we see that
We thus end up with the lower bound
for constants $C$ and $D$. We conclude that
We have now completed the proof of the estimate
#### Problem 4-3c
The recurrence relation in question can be written as
Observe that
Moreover,
for all $n$, and $% $. We can therefore use the master theorem to conclude that
#### Problem 4-3d
We cannot use the master theorem, as the recurrence relation is not of the form $T(n) = aT(n/b) + f(n)$. If, however, we drop the $-2$ term, we can use the master theorem on
to obtain
Since the $-2$ term shouldn’t affect the asymptotic behavior of $T$, we conjecture that
To prove $(\ast)$, we first fix $n > 6$, so that $\frac{n}{3} - 2 > 1$. Assume inductively that
for all $% $, where $C > 0$ is a constant to be chosen later. Observe that
If we choose $C > \frac{1}{2 \log 3}$, then we have
for all $n \geq 6$, and we have the desired bound
To establish the base case $n=6$, we choose | {
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for all $n \geq 6$, and we have the desired bound
To establish the base case $n=6$, we choose
so that the inductive step continues to holds, while
for $n = 6$. We now conclude by induction that
To prove $(\ast)$, it now remains to show that
To this end, we fix $n > 24$, so that $\frac{n}{3} - 2 > 2$ and $\frac{n}{3} - 2 > \frac{n}{4}$. Assume inductively that
for all $% $, where $C > 0$ is a constant to be chosen later. Observe that
Since $\frac{n}{3} - 2 > \frac{n}{4}$, we have
Choosing $C \leq \frac{1}{8}$ yields
so that
as was to be shown.
To establish the base case $n = 24$, we choose
so that the inductive step continues to hold, while
for $n = 24$. We now conclude by induction that
as was to be shown. This completes the proof that
#### Problem 4-3e
Generalizing the computation carried out in Problem 4-3b, we show that a recurrence relation of the form
satisfies the asymptotic estimate
We let $k = \log_a n$ and observe that
We have shown in Problem 4-3b that
and so we conclude that
#### Problem 4-3f
The recurrence relation is not in a form that the master theorem can be applied to, so we must produce an asymptotic estimate directly.
Observe that
From this, let us guess that the accumulation of cost is
for some $k>0$. Since
we conjecture that
To verify this, we fix $n > 8$ and assume inductively that
for all $% $, where $C > 0$ and $D > 0$ are constants to be chosen. Observe that
Analogously,
So long as $C \geq 7$ and $D \leq 7$, we have
$D \leq \frac{7}{8}(1+D) \leq \frac{7}{8}(1+C) \leq C,$\$
which yields
as was to be shown.
To establish the base case $n = 8$, we choose
so that the inductive case continues to hold, while
and
as was to be shown. This completes the proof of
#### Problem 4-3g
The recurrence relation is not in a form that the master theorem can be applied to, so we must produce an asymptotic estimate directly.
Observe that
Page 1154 of CLRS shows that
and so we conclude that
#### Problem 4-3h | {
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Observe that
Page 1154 of CLRS shows that
and so we conclude that
#### Problem 4-3h
The recurrence relation is not in a form that the master theorem can be applied to, so we must produce an asymptotic estimate directly.
Observe that
We have proved in Exercise 3.2-3 that $\log n! = \Theta(n \log n)$, and so we conclude that
#### Problem 4-3i
The recurrence relation is not in a form that the master theorem can be applied to, so we must produce an asymptotic estimate directly.
Observe that
where $k!!$ is the double factorial function
We let $m = \lfloor n/2 \rfloor$. If $m$ is even, then
In this case,
which is $\Theta(m \log m) = \Theta(n \log n)$.
If $m$ is odd, then
In this case,
We have proved in Exercise 3.2-3 that $\log k! = \Theta(k \log k)$, and so $\log m!!$ has the asymptotic estimate of $\Theta(m \log m) = \Theta(n \log n)$.
We conclude that
#### Problem 4-3j
The recurrence relation is not in a form that the master theorem can be applied to, so we must produce an asymptotic estimate directly.
Let us begin by making the substitution $n = 2^m$:
Setting $S(m) = T(2^m)$, we see that
Observe that
Since
we see that $S(m) = \Theta\left( 2^m \log m\right)$.
The identity $m = \log n$ now implies that
### 4-4. Fibonacci numbers
The problem develops properties of the Fibonacci numbers, which are defined by recurrence (3.22). We shall use the technique of generating functions to solve the Fibonacci recurrence. Define the generating function (or formal power series) $\mathscr{F}$ as
where $F_i$ is the $i$th Fibonacci number.
a. Show that $\mathscr{F}(z) = z + z \mathscr{F}(z) + z^2 \mathscr{F}(z)$.
b. Show that
where
and
c. Show that
d. Use part (c) to prove that $F_i = \phi^i / \sqrt{5}$ for $i > 0$, rounded to the nearest integer. (Hint: Observe that $% $.)
#### Problem 4-4a
Observe that
We then have
Since $F_0 = F_1 = 1$ and $F_i = F_{i-1} + F_{i-2}$ for all $i \geq 2$, we conclude that
#### Problem 4-4b | {
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#### Problem 4-4b
It follows at once from 4-4a that
By the quadratic formula,
and the second equality follows.
Finally, we observe that
The third equality now follows.
(To deduce the third equality directly from the second equality, we must carry out the partial fraction decomposition of the second expression.) $\square$
#### Problem 4-4c
Page 1147 of CLRS tells us that
Substituting $x = \phi z$, we get
Similarly,
The desired identity now follows from the third equality in 4-4b. $\square$
#### Problem 4-4d
We have shown in 4-4c that
for all indices $i$. Since $% $, we have
whence it follows that
In other words, $F_i$ is $\phi^i/\sqrt{5}$ rounded to the nearest integer. $\square$
### 4-5. Chip testing
Professor Diogenes has $n$ supposedly identical integrated-circuit chips that in principle are capable of testing each other. The professor’s test jig accommodates two chips at a time. When the jig is loaded, each chip tests the other and reports whether it is good or bad. A good chip always reports accurately whether the other chip is good or bad, but the professor cannot trust the answer of a bad chip. Thus, the four possible outcomes of a test are as follows:
Chip $A$ says Chip $B$ says Conclusion
$B$ is good $A$ is good both are good, or both are bad
$B$ is good $A$ is bad at least one is bad
$B$ is bad $A$ is good at least one is bad
$B$ is bad $A$ is bad at least one is bad
a. Show that if at least $n/2$ chips are bad, the professor cannot necessarily determine which chips are good using any strategy based on this kind of pairwise test. Assume that the bad chips can conspire to fool the professor.
b. Consider the problem of finding a single good chip from among $n$ chips, assuming that more than $n/2$ of the chips are good. Show that $\lfloor n/2 \rfloor$ pairwise tests are sufficient to reduce the problem to one of nearly half the size. | {
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c. Show that the good chips can be identified with $\Theta(n)$ pairwise tests, assuming that more than $n/2$ of the chips are good. Give and solve the recurrence that describes the number of tests.
#### Notational Remark
Given two chips $x$ and $y$, we write $T_x(y)$ to denote what $x$ has to say about $y$. The possible outcomes are, of course, good, and bad.
A test consists of inputting two chips $x$ and $y$ and receiving the output $(T_x(y), T_y(x)).$
#### Problem 4-5a
Assume that at least $n/2$ chips are bad. Let $\mathcal{G}$ be the set of all good chips, and let $\mathcal{B}$ be a set of bad chips of size $\vert \mathcal{G} \vert$. We let $\mathcal{R}$ be the remaining chips, if there are any.
Observe that, for each $x \in \mathcal{G}$,
Let us consider the following adversarial strategy: each $x \in \mathcal{B}$ returns
and each $x \in \mathcal{R}$ returns
We now partition the collection of chips by declaring two chips $x$ and $y$ to be in the same category if and only if
The partition process produces three categories: namely, $\mathcal{G}$, $\mathcal{B}$, and $\mathcal{R}$.
A test does not show which category a pair of chips belongs to. It merely shows whether two chips belong to the same category. Since $\mathcal{G}$ and $\mathcal{B}$ are of the same size, this information is insufficient to distinguish the two categories from one another. It follows that the chosen adversarial strategy prevents distinguishing good chips from bad chips. $\square$
#### Problem 4-5b
Let us assume that more than $n/2$ of the chips are good.
We assume that $n$ is even, and set $m = n/2$. If there is an odd number of chips, we discard one at random to make the number of chips even.
We divide the chips into two groups of equal sizes, $\mathcal{X} = \{x_1,\ldots,x_m\}$ and $\mathcal{Y} = \{y_1,\ldots,y_m\}$, assigning indices $1,\ldots,m$ arbitrarily.
For each index $i$, we declare the pair $(x_i, y_i)$ to be in the keep pile $\mathcal{K}$ if and only if | {
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In other words, either $x_i$ and $y_i$ are both good chips or they are both bad chips.
We also define the discard pile $\mathcal{D}$ to be the collection of all pairs $(x_i,y_i)$ that do not belong to $\mathcal{K}$. We denote by $\vert \mathcal{K} \vert$ and $\vert \mathcal{D} \vert$ the number of pairs in $\mathcal{K}$ and $\mathcal{D}$, respectively. We note here that
Moreover, each pair in $\mathcal{D}$ must contain at least one bad chip, and so
We now observe that $\mathcal{K}$ contains more than $\vert \mathcal{K} \vert /2$ good pairs. Indeed, if $\mathcal{K}$ contained at most $\vert \mathcal{K} \vert / 2$ good pairs, then
which is absurd.
In light of this observation, we define
Since $\mathcal{R}$ contains precisely $\vert \mathcal{K} \vert$ many chips, we see from $(\ast)$ that $\mathcal{R}$ cannot contain more than $n/2 - \vert \mathcal{D} \vert$ many chips. Moreover, more than half of the pairs in $\mathcal{K}$ are good pairs, and so more than half of the chips in $\mathcal{R}$ are good chips. We now recurse on $\mathcal{R}$. $\square$
The procedure described above can be implemented as follows:
"""We identify chips by unique nonnegative integers and assume that
chip_table, an immutable two-dimensional list of numbers, is given.
To see what a testing chip ("testing") has to say about another chip
("tested"), we call chip_table[testing][tested], which returns
True for 'good', and False for 'bad'.
We record the chips to be tested in a list of numbers called chip_list.
chip_list is changed at each recursive step to track the progress.
"""
def get_a_good_chip(chip_table, chip_list):
n = len(chip_list)
if n <= 2:
return chip_list[0]
if n % 2 != 0: # if length mod 2 is not zero, i.e., odd
chip_list = chip_list[:-1] # cut out the last element
n = n-1
m = n//2
list_x, list_y = chip_list[:m], chip_list[m:]
keep = []
for i in range(m):
x, y = list_x[i], list_y[i]
if chip_table[x][y] and chip_table[y][x]: # if both return True
keep.append((x, y)) | {
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new_chip_list = [pair[0] for pair in keep]
return get_a_good_chip(chip_table, new_chip_list)
>>> # chips 1, 2, 3 are good
>>> chip_table = [[True, False, True, False, True],
... [False, True, True, True, False],
... [False, True, True, True, False],
... [False, True, True, True, False],
... [True, False, False, False, True]]
>>> get_a_good_chip(chip_table, [0, 1, 2, 3, 4])
1
>>> # chips 2, 3, 4, 5 are good
>>> chip_table = [[True, True, False, False, False, False],
... [True, True, False, False, False, False],
... [False, False, True, True, True, True],
... [False, False, True, True, True, True],
... [False, False, True, True, True, True],
... [False, False, True, True, True, True]]
>>> get_a_good_chip(chip_table, [0, 1, 2, 3, 4, 5])
2
#### Problem 4-5c
Once one good chip is found, it suffices to test all chips against the good chip to identify all good chips. This takes $\Theta(n)$ pairwise tests and can be implemented as follows:
def get_all_good_chips(chip_table):
n = len(chip_table)
testing = get_a_good_chip(chip_table, range(n))
good_chips = [tested for tested in range(n)
if chip_table[testing][tested]]
return good_chips
>>> # chips 1, 2, 3 are good
>>> chip_table = [[True, False, True, False, True],
... [False, True, True, True, False],
... [False, True, True, True, False],
... [False, True, True, True, False],
... [True, False, False, False, True]]
>>> get_all_good_chips(chip_table)
[1, 2, 3] | {
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>>> # chips 2, 3, 4, 5 are good
>>> chip_table = [[True, True, False, False, False, False],
... [True, True, False, False, False, False],
... [False, False, True, True, True, True],
... [False, False, True, True, True, True],
... [False, False, True, True, True, True],
... [False, False, True, True, True, True]]
>>> get_all_good_chips(chip_table)
[2, 3, 4, 5]
The time complexity $S$ of get_all_good_chips is
where the time complexity $T$ of get_a_good_chip is
The master theorem now implies that
whence it follows that
### 4-6. Monge arrays
An $m \times n$ array $A$ of real numbers is a Monge array if for all $i$, $j$, $k$, and $l$ such that $% $ and $% $, we have
In other words, whenever we pick two rows and two columns of a Monge array and consider the four elements at the intersection of the rows and the columns, the sum of the upper-left and lower-right elements is less than or equal to the sum of the lower-left and upper-right elements. For example, the following array is Monge.
a. Prove that an array is Monge if and only if for all $i=1,2,\ldots, m-1$ and $j=1,2,\ldots,n-1$, we have
(Hint: For the “if” part, use induction separately on rows and columns.)
b. The following array is not Monge. Change one element in order to make it Monge. (Hint: Use part(a))
c. Let $f(i)$ be the index of the column containing the leftmost minimum element of row $i$. Prove that $f(1) \leq f(2) \leq \cdots \leq f(m)$ for any $m \times n$ Monge array.
d. Here is a description of a divide-and-conquer algorithm that computes the leftmost minimum element in each row of an $m \times n$ Monge array $A$:
Construct a submatrix $A'$ of $A$ consisting of the even-numbered rows of $A$. Recursively determine the leftmost minimum for each row of $A'$. Then compute the leftmost minimum in the odd-numbered rows of $A$. | {
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Explain how to compute the leftmost minimum in the odd-numbered rows of $A$ (given that the leftmost minimum of the even-numbered rows is known) in $O(m+n)$ time.
e. Write the recurrence describing the running time of the algorithm described in part (d). Show that its solution is $O(m + n \log m)$.
#### Notational remark
As stated in the preface, we shall use the 0-first indexing convention for this problem.
#### Problem 4-6a
$(\Rightarrow)$ If $A$ is Monge, it suffices to pick $k = i+1$ and $l = j+1$. $\square$
$(\Leftarrow)$ We suppose that
for all $i=1,2,\ldots,m-1$ and $j=1,2,\ldots,n-1$.
We fix $i$ and $j$ and show that
for all $p \geq 1$ and $q \geq 1$.
Observe that $(\ast)$ is the $p = q = 1$ case. We let $q = 1$, fix $p > 1$, and assume inductively that
for all $% $. By the inductive hypothesis,
$(\ast)$ implies that
whence it follows that
as was to be shown. This establishes $(\ast\ast)$ for all $p \geq 1$ and $q = 1$.
We now fix $p > 1$ and $q > 1$ and assume inductively that
for all $% $. By the inductive hypothesis,
The $q =1$ case of $(\ast\ast)$ implies that
whence it follows that
as was to be shown. This establishes $(\ast\ast)$ in full generality.
$(\ast)$ follows at once from $(\ast\ast)$. $\square$
#### Problem 4-6b
The following code tests whether a two-dimensional NumPy array is Monge, using the criterion developed in 4-6a:
"""A is assumed to be a two-dimensional NumPy array.
- The syntax for referencing an entry of A is A[i, j].
- A.shape returns (m, n), the size of the matrix.
"""
def test_monge(A):
"""Check if A is a Monge array using the equivalent definition
derived in 4-6a"""
num_rows, num_cols = A.shape
failed = []
for i in range(num_rows-1):
for j in range(num_cols-1):
if A[i, j]+A[i+1, j+1] > A[i, j+1]+A[i+1, j]:
failed.append((i, j))
if not failed:
print("Monge")
is_monge = True
else:
print("Not Monge")
print("Failed coordinates: ")
for pair in failed:
print(pair)
is_monge = False
return is_monge | {
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>>> test_monge(np.array([[10, 17, 13, 28, 23],
... [17, 22, 16, 29, 23],
... [24, 28, 22, 34, 24],
... [11, 13, 6, 17, 7],
... [45, 44, 32, 37, 23],
... [36, 33, 19, 21, 6],
... [75, 66, 51, 53, 34]])
Monge
>>> test_monge(np.array([[37, 23, 22, 32],
... [21, 6, 7, 10],
... [53, 34, 30, 31],
... [32, 13, 9, 6],
... [43, 21, 15, 8]])
Not Monge
Failed coordinates:
(0, 1)
As the program suggests, the inequality
fails to hold, as
Since we have
we can add up to 7 to $A[0,2]$ without breaking the inequality
We therefore assign
at which point we obtain
while maintaining
Since the modification affects no other inequality, the resulting array is Monge.
>>> test_monge(np.array([[37, 23, 29, 32],
... [21, 6, 7, 10],
... [53, 34, 30, 31],
... [32, 13, 9, 6],
... [43, 21, 15, 8]])
Monge
#### Problem 4-6c
Let $A$ be an $m \times n$ Monge array. We fix $% $ and assume for a contradiction that
on $A$. Since
we see that the inequality
can never hold unless it is also the case that
Since $f(i)$ is the index of the column containing the leftmost minimum element of row $i$, $(\ast)$ implies that we must always have
rendering the identity
false. We thus conclude that
Since the choice of $i$ was arbitrary, the desired result now follows. $\square$
#### Problem 4-6d
We assume that the values of $f(0),f(2),\ldots, f(2 \lfloor (n-1)/2 \rfloor)$ are known. We have seen in 4-6c that
for all $k$. Since we must examine at least one integer for each row, the number of integers we must examine to compute $f(2k+1)$ is bounded above by
It then follows that the number of integers we must examine for all the odd rows is bounded above by
We now observe that | {
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We now observe that
whence the desired result follows. $\square$
#### Problem 4-6e
We begin by observing that a submatrix $A'$ of a Monge array $A$ consisting of the rows of $A$ is a Monge array. This, in particular, implies that the ordering relation proved in 4-6c continues to hold on such submatrices.
The algorithm for computing the leftmost minimum of each row of a Monge array can be implemented as follows:
"""A is assumed to be a two-dimensional NumPy array.
- The syntax for referencing an entry of A is A[i, j].
- A.shape returns (m, n), the size of the matrix.
- A[::2,:] is the submatrix consisting of the even rows of A.
0, k, 2k, 3k, and so on.
"""
def find_left_min(A):
if not test_monge(A):
return None
num_rows, num_cols = A.shape
mins = [0 for _ in range(num_rows)] # initialize array of length A.shape[0]
return lm_recursion(A, mins)
def min_index(row, start_index, end_index):
"""Scan a row from start_index to end_index-1 to find the
index of the leftmost minimum element in the specified range.
"""
min_ind = start_index
for i in range(start_index, end_index):
if row[i] < row[min_ind]:
min_ind = i
return min_ind
def lm_recursion(A, mins):
num_rows, num_cols = A.shape
if num_rows == 1: # if A has only one row, use the min_index scan
mins[0] = min_index(A[0], 0, num_cols)
else:
## take the even rows and recurse
evens = A[::2, :]
even_mins = lm_recursion(evens, mins[::2])
## even rows have already been computed;
## use the min_index scan in the range specified by even_mins
## to figure out the min index for odd rows.
start_index = 0
for i in range(num_rows):
if i % 2 == 0:
mins[i] = even_mins[i//2]
else:
left = mins[i-1]
right = even_mins[(i+1)//2]+1 if i+1 < num_rows else num_cols
mins[i] = min_index(A[i], left, right)
return mins | {
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>>> find_left_min(np.array([[10, 17, 13, 28, 23],
... [17, 22, 16, 29, 23],
... [24, 28, 22, 34, 24],
... [11, 13, 6, 17, 7 ],
... [45, 44, 32, 37, 23],
... [36, 33, 19, 21, 6 ],
... [75, 66, 51, 53, 34]))
[0, 2, 2, 2, 4, 4, 4]
>>> find_left_min(np.array([[37, 23, 29, 32],
... [21, 6, 7, 10],
... [53, 34, 30, 31],
... [32, 13, 9, 6],
... [43, 21, 15, 8]]))
[1, 1, 2, 3, 3]
What is the running time $T(m)$ of find_left_min, where $m$ is the number of rows? For the purpose of this discussion, let us focus on analyzing lm_recursion.
Let $A$ be an $m \times n$ array, so that num_rows == m and num_cols == n. If $m=1$, then the running time of lm_recursion is precisely that of min_index. Since min_index scans through each column once, we see that the running time is $\Theta(n)$. We thus declare
As for the $m > 1$ case, we observe that lm_recursion consists of three steps:
1. Construct evens, the subarray of even rows, from $A$.
2. Recurse on evens.
3. Go through the $m$ rows and compute mins.
Since $A$ is never modified throughout lm_recursion, there is no need to copy the data of $A$ to construct evens, which becomes $A$ in the recursive step. The construction of even thus reduces to collecting pointers to the even rows, which can be done in $\Theta(m/2)$ time.
The recursion step is invoked on evens, an array of $m/2$ rows. Since the recursion step is invoked precisely once without any additional procedure, it takes $T(m/2)$ time to run.
Finally, we have shown in 4-6d that the computation of mins takes $\Theta(m+n)$ time. It follows that the running time of lm_recursion is
Since we do not know what $n$ is in relation to $m$, we cannot apply the master theorem. Observe, however, that
Since | {
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Since
for all choices of $I$, we see that
We have shown above that $T(1) = \Theta(m)$, and so
Let us now prove the above estimate by induction. To this end, we fix $n$ and choose constants $k,K > 0$ such that
for all choices of $m$.
Let us first establish the upper bound
Fix $m \geq 2$ and assume inductively that
for all $% $, where $C,D > 0$ are constants to be chosen.
Observe that
So long as
we have
The above condition is satisfied precisely when
To establish the base case $m = 1$, we only need the hypothesis
Indeed, this leads to
We thus choose
so that $(\ast\ast)$ is trivially established, and that
thereby establishing $(\ast)$. It now follows from induction that
It remains to establish the lower bound
We fix $m \geq 2$ and assume inductively that
for all $% $, where $C,D > 0$ are constants to be chosen. Observe that
In order to obtain
we must have
and
Observe that the second inequality, along with the added hypothesis that
implies
which is precisely the first inequality. The inductive step thus holds true so long as
To establish the base case $m = 1$, we only need the hypothesis
Indeed, this leads to
We therefore choose
so that $(\star)$ and $(\star\star)$ follow. We now conclude by induction that
which, along with the upper bound established above, implies
as was to be shown. $\square$
## Additional remarks and further results
### Generalizations of the master method
The master method does not apply to divide-and-conquer problems with multiple subproblems of substantially different sizes. For example,
is not covered by the master method.
Mohamad Akra and Louay Bazzi, in their 1996 paper “On the Soluton of Linear Recurrence Equations,” present a generalization of the master method that is applicable to recurrence relations of the form
with minor restrictions on $f$. The solution is given by the formula
where $p$ is the unique real number that satisfies the identity
For the above example, we have $p = 1$, and so | {
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For the above example, we have $p = 1$, and so
Using the Akra—Bazzi method, we can also solve more complicated recurrence relations like
In this case, we have
and so the solution to the recurrence relation is
There are, of course, several generalizations beside the Akra—Bazzi method. We refer the reader to Chee Yap’s 2011 paper, “A real elementary approach to the master recurrence and generalizations.” $\square$
### Generating functions
We have discussed briefly in Problem 4-4 the generating-function technique of solving recurrence relations. We pursue the subject further here, following the treatment in Chapter 3 of Sedgewick/Flajolet.
Given a sequence $a_0,a_1,\ldots,a_k,\ldots$, we construct its ordinary generating function
and its exponential generating function
The generating-function method of solving recurrences typically involves multiplying both sides of the recurrence by $z^n$ ($z^n/n!$ for exponential generating functions), summing on $n$, manipulating the resulting identity to derive an explicit formula for the generating function, and computing the power-series expansion of the generating function.
Take, for example,
with the initial condition $a_0 = 0$, $a_1 = 1$, $a_2 = 1$. We multiply both sides by $z^n$, sum over $n \geq 3$, and reindex the sums to obtain the identity
For notational simplicity, we let
the ordinary generating function of $(a_{n})_{n}$. Observe that
We can use the above computation to simplify $(\ast)$ as follows:
Using the initial condition $a_0 = 0$, $a_1 = 1$, $a_2 = 1$, we conclude that
Taking the partial fraction decomposition, we obtain
We now recall that
from which we conclude that
It follows that
for all $n \geq 0$.
We shall revisit the method of generating functions in later chapters, as more sophisticated recurrence relations arise. $\square$ | {
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# Why do integrals “start” at 0? [duplicate]
This is a dumb question and I don't really know how to word it. When you take an antiderivative and plug in number you are given the area under the curve starting at 0 (assuming C is 0). I can easily see how the derivative of an integral is given by the function value, but why does the integral start at 0 and not any other number? When I try to imagine the area of some curve starting at negative 1 for example the area under the curve would intuitively to me still be given by the antiderivative. 0 makes sense as a starting point but for some reason I can't visualize it. I'm not sure if that made any sense but if anyone could help me wrap my head around it I'd appreciate it.
## marked as duplicate by BallBoy, астон вілла олоф мэллбэрг, Claude Leibovici integration StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 10 '18 at 6:38 | {
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• It doesn't matter. It's just convenient. As for your example, $F(x)=\int_{-1}^x f(t)dt$ is another anti-derivative of $f$. – Yanko May 9 '18 at 19:09
• "given by the antiderivative" There is no "the antiderivative". There is only "an antiderivative". – Arthur May 9 '18 at 19:13
• $\int dx/x$ certainly doesn't start from $0$! – Chappers May 9 '18 at 19:15
• Starting from somewhere else amounts to changing the constant of integration. – amd May 9 '18 at 20:46
• @LSpice, I may have been misinterpreting what op meant. I was essentially thinking the confusion was about why an integral over a degenerate interval was zero. – jdods May 10 '18 at 3:11
When you take an antiderivative and plug in number you are given the area under the curve starting at 0 (assuming C is 0).
This is not true. In certain situations it may be the case, but not generally. I think the reason this confusion arises is that a common problem given to calculus students is to find the antiderivative of a polynomial, e.g.,
$$\int x^3 +2x \, dx = \frac{1}{4}x^4 + x^2 + C$$
and in this case, if we set $C = 0$ we get
$$\frac{1}{4}x^4 + x^2$$
which is the same as $$\int_0^x u^3 +2u \, du = \frac{1}{4}u^4 + u^2 \big|_{u=0}^{u=x}.$$
This will work whenever the form that the antiderivative $F$ of $f$ you get takes satisfies $F(0) = 0$. But in general, setting $C = 0$ will not get you the integral $\int_0^x f(t) \, dt$. For example, if you take $f(x) = e^x$ then $$\int e^x \, dx = e^x + C$$ but setting $C = 0$ gives you $e^x$, which is not the same as $$\int_0^x e^t \, dt = e^x - 1.$$
Note that "setting "C = 0" in the expression for the antiderivative" is not actually a well-defined operation. Different methods of antidifferentiation can give you different expressions when you set $C = 0$. It is important to remember that there is no single antiderivative, and no canonical way of writing it. $\int 2x \, dx = x^2 + 3 + C$ is just as valid as $\int 2x \, dx = x^2 + C$. | {
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• Ahh you're e^x example shows that my assumption was wrong. Thanks for clearing it up. – conyare May 9 '18 at 19:39
• My favourite example is arcsin(x)+C vs -arccos(x)+C, these are equal families, but which function is supposed to be C=0? – Serge Seredenko May 9 '18 at 20:27
• @SergeSeredenko, or $\sin(x)^2 + C$ versus $-\cos(x)^2 + C$. – LSpice May 10 '18 at 1:03
• Also see this recent question where the difference between two solutions doesn't even look like a constant, but it is: math.stackexchange.com/q/2764985/78887 – Euro Micelli May 10 '18 at 3:18
Integrals don't always start at $0$. Let's start from definite integrals and move to the indefinite integrals you asked about. We know $\int_a^b f(x)dx$ gives the area under the curve between $a$ and $b$. If $f(x)$ has an antiderivative $F(x)$, the Fundamental Theorem of Calculus tells us that $\int_a^b f(x) = F(b) - F(a)$. Thus, if we want the area under the curve from $a$ to $b$, we compute $F(b) - F(a)$. If we want the area under the curve from $0$ to $b$, we compute $F(b) - F(0)$. $F(b) - F(0)$, as a function of $b$, gives us the area from $0$ to $b$.
Now there are a lot of "natural" functions where $F(0) = 0$ (e.g., functions like $x^2$ or $\sin x$), so to $F(b)$ gives the area from $0$ to $b$. But that won't be the case if $F(0) \neq 0$.
The above should make it clear that there's no reason $0$ is special -- if you want the area from $-1$ to $b$ as a function of $b$, just use $F(b) - F(-1)$. | {
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• I assumed that the fundamental theorem worked because F(a) gets the area to A from 0 and F(b) gets the area to B from 0 so subtracting will get the desired area in-between them. – conyare May 9 '18 at 19:18
• @conyare No, it actually goes the other way around! It is true that if all you knew were the 0-to-b version of the fundamental theorem, you could recover the a-to-b version. But the subtraction of another value is "fundamental" (pun somewhat intended) to the fundamental theorem. Jeffery Opoku-Mensah hints at a reason why this is true: since $F$ is defined only up to a constant, $F(b)$ doesn't give you a definite answer, only up to a constant. Only once you subtract $F(b)-F(a)$ does the constant cancel and you get a definite area. – BallBoy May 9 '18 at 19:21
Well, integrals (by which I think you mean anti derivatives) don't always start at $0$. Indeed, if a function $f=f(x)$ is continuous in some interval $I=[a,b]$, then it has an anti derivative given by $$\int_c^x{f(t)dt},$$ where $c\in I$. The $0$ is usually chosen for $c$ only as a matter of convenience. A well-known function defined by an antiderivative that "starts" from (note that the notion of starting from for antiderivatives should not be taken too literally since the function is defined even for $x<c\in I$) $1$, not $0$ as usual, is the logarithm function $$\log x=\int_1^x{\frac{1}{t}dt}$$ defined for all $x>0$. | {
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The antiderivative is generally given as the "simplest" form, and often the simplest form has a y value of zero when x is zero. For instance, if the function is constant, then for the antiderivative, you need a line whose slop is equal to that constant value. The simplest way of writing that is y = mx+C. You could write y = mx+5-C, and that would be a valid antiderivative, but that would be needlessly complicated. Since you're looking for simple functions, zero will pop up a lot. But there are cases where the simplest function doesn't go through the origin. For instance, if you're taking the antiderivative of sin(x), the simplest form is cos(x)+C. If you want to "start" at zero, you'd have to do cos(x)-1+C. Similarly, the antiderivative of e^x is generally given as e^x+C.
There are some cases where the same function can have different antiderivatives that look very different, but are actually the same thing. For instance, $\frac{1}{\sqrt{1-x^2}}$ can be integrated as either arcsin(x) or -arccos(x). But these just differ by the constant $\pi /2$.
It depends on what antiderivative you take. For example, an antiderivative of $2x$ can be $x^2$, $x^2+1$, $x^2+100$, or whatever. By the Fundamental Theorem, it turns out $x^2$ gives the area under the curve starting at zero. But $x^2+1$ would also give the area of the curve starting at $-1$, for example.
• I'm afraid you're confusing derivatives with antiderivatives... – zipirovich May 9 '18 at 19:22
• Yea, I wrote this in a confused rush, but the idea is still there. Just switch the x^2 and 2x. – Jeffery Opoku-Mensah May 9 '18 at 21:59 | {
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Filling Space with Pursuit Polygons
I want to make a program which can fill a 2D space with "pursuit polygons".
You can also look up "pursuit curves" or mice problem or watch this gif
What I've tried so far:
First I tried to produce these polygons by rotation of polygons
the square for example
data = {{0, 0}, {1, 0}, {1, 1}, {0, 1}, {0, 0}};
Graphics[{Table[{Scale[Rotate[Line[data], 3*x Degree], {x, x}]}, {x,
0, 20}]}]
then I decided to use spirals
I realised that instead of using n spirals for every n-polygon,
I can use only one(!) and produce the effect I want with "fine tuning"
here are the "pursuit polygons" that I made
Triangle:
a = 41.9;
b = 100;
W = Table[{t^2*Cos[a*t], t^2*Sin[a*t]}, {t, 0, b}];
AppendTo[W, W[[-4]]];
ListPlot[W]
Graphics[Line[W]]
Square:
S = Table[{t^2*Cos[42.4*t], t^2*Sin[42.4*t]}, {t, 0, 160}];
AppendTo[S, S[[-4]]];
Graphics[Line[S]]
Pentagon
P = Table[{t^2*Cos[45.251*t], t^2*Sin[45.251*t]}, {t, 0, 220}];
AppendTo[P, P[[-5]]];
Graphics[Line[P]]
Then I tried to put all these together by rotating and scaling...
(but I'm not satisfied with the result)
notice that for every polygon I use also the "anti-clockwise" version of the polygon, which produces interesting results | {
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T = Table[{t^2*Cos[41.9*t], t^2*Sin[41.9*t]}, {t, 0, 100}];
AppendTo[T, T[[-4]]]; (*Triangle*)
S = Table[{t^2*Cos[42.4*t], t^2*Sin[42.4*t]}, {t, 0, 160}];
AppendTo[S, S[[-4]]]; (*square*)
S1 = Table[{t^2*Cos[-42.4*t], t^2*Sin[-42.4*t]}, {t, 0, 160}];
AppendTo[S1, S1[[-4]]]; (*Anti-clockwise Square*)
P = Table[{t^2*Cos[45.251*t], t^2*Sin[45.251*t]}, {t, 0, 220}];
AppendTo[P, P[[-5]]]; (*Pentagon*)
P1 = Table[{t^2*Cos[-45.251*t], t^2*Sin[-45.251*t]}, {t, 0, 220}];
AppendTo[P1, P1[[-5]]]; (*Anti-clockwise Pentagon*)
Graphics[{Translate[
Rotate[Scale[Line[(T)], {2.09, 2.09}], -67 Degree], {29500, 3800}],
Rotate[Scale[Line[S + 0], {1, 1}], -30 Degree],
Translate[
Rotate[Scale[Line[(S1)], {.98, .98}], -87.5 Degree], {41000,
24700}],
Translate[
Rotate[Scale[Line[P1], {.605, .665}], -20.5 Degree], {76500,
6500}], Scale[
Translate[
Rotate[Scale[
Line[(T)], {1.2, 1.7}], -108 Degree], {50900, -12500}], {1.91,
1.31}],
Translate[
Rotate[Scale[Line[(P)], {.525, .665}], -64 Degree], {3000, 36000}],
Translate[
Rotate[Scale[Line[(P)], {.61, .61}], -87 Degree], {59000, 58000}],
Translate[
Rotate[Scale[Line[(T)], {2.1, 1.8}], 12 Degree], {82500, 39500}],
Translate[
Rotate[Scale[Line[(T)], {2.3, 1.9}], -133 Degree], {32000,
67000}]}]
Can you find a way to divide and fill any given space with pursuit polygons?
The result would look better if this could work with ANY convex polygon and not only the regular polygons that I used... | {
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• Are there any other names for that? Because I think this was asked before. – Kuba May 2 '17 at 10:05
• I provided all the names I could think of. There are not many images outthere.Even the first picture is something that I made a while ago. – J42161217 May 2 '17 at 10:14
• @Kuba Lucas, E. (1877). Problème des trois chiens. Nouvelle Correspondance Mathématique, 3, 175–176. – Michael E2 May 2 '17 at 11:54
• A related question. – J. M. is away May 11 '17 at 13:15
• And a very related blog-post of mine. And if I'm not completely off here, the word "spacefilling" is incorrect in this context. – halirutan Jun 11 '17 at 15:01
The reason that these are called "pursuit polygons" is because they are formed from a dynamical system in which different agents pursue each other. Example:
In this image, one agent starts in each corner of the triangle. The agent starting in the lower right corner pursues the agent starting in the top corner, the agent in the top corner pursues the agent in the lower left corner, and the agent in the lower left corner pursues the agent in the lower right corner.
Drawing lines in between the agents yields a series of triangles which shrink as the agents get closer to each other, and also rotate:
The corresponding differential equations are: \begin{align*} \dot{\mathbf{x}}_i &= \mathbf{x}_{i+1} - \mathbf{x}_i,\ i\in\{1,\dots,N-1\}\\ \dot{\mathbf{x}}_i &= \mathbf{x}_1 - \mathbf{x}_N,\ i=N, \end{align*} where $N$ is the number of agents. The agents start in corners for the visualizations we are making, but in general they don't have to.
This code solves the differential equations using Euler integration and plots the lines between the agents:
integrate[pts_, h_] := NestList[Nest[step[#, h] &, #, 10] &, pts, 100]
step[pts_, h_] := pts - h (pts - RotateRight[pts])
wrapAround[pts_] := Append[pts, First[pts]]
PursuitPolygon[pts_, h_: 0.01] := Graphics[{
Line[wrapAround[pts]],
Line[wrapAround /@ integrate[pts, h]]
}] | {
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For regular polygons, this results in logarithmic spirals, as you have pointed out:
Grid@Partition[Table[PursuitPolygon[N@CirclePoints[n]], {n, 2, 10}], 2]
But this approach is more general and works for any convex polygons, as you required. The following is an example of how we can fill in a triangulated region with spirals using this system:
reg = DiscretizeRegion[Rectangle[], MaxCellMeasure -> 0.1]
Show[PursuitPolygon @@@ MeshPrimitives[reg, 2]]
Wikipedia's entry on the mice problem suggests that the mice move at unit speed, i.e. like this: \begin{align*} \dot{\mathbf{x}}_i &= \frac{\mathbf{x}_{i+1} - \mathbf{x}_i}{\|\mathbf{x}_{i+1} - \mathbf{x}_i\|},\ i\in\{1,\ldots,N-1\}\\ \dot{\mathbf{x}}_i &= \frac{\mathbf{x}_1 - \mathbf{x}_N}{\|\mathbf{x}_1 - \mathbf{x}_N\|},\ i=N, \end{align*} This turns out to be difficult to implement because the denominator grows large as the points approach each other. I spoke to halirutan and MichaelE2 about this. Here is MichaelE2's solution, which I implemented for triangles but it could be implemented for any polygon.
PursuitPolygon[{pt1_, pt2_, pt3_}] := Module[{},
{sol1x, sol1y, sol2x, sol2y, sol3x, sol3y} = NDSolveValue[{
{x1'[t], y1'[t]} == Normalize[{x2[t], y2[t]} - {x1[t], y1[t]}],
{x2'[t], y2'[t]} == Normalize[{x3[t], y3[t]} - {x2[t], y2[t]}],
{x3'[t], y3'[t]} == Normalize[{x1[t], y1[t]} - {x3[t], y3[t]}],
{x1[0], y1[0]} == pt1,
{x2[0], y2[0]} == pt2,
{x3[0], y3[0]} == pt3,
WhenEvent[Norm[{x2[t], y2[t]} - {x1[t], y1[t]}] < 1*^-8, "StopIntegration"]},
{x1, y1, x2, y2, x3, y3},
{t, 0, 10}
];
{tmin, tmax} = MinMax@sol1x["Grid"];
Graphics@Table[Line[{{sol1x[t], sol1y[t]}, {sol2x[t], sol2y[t]}, {sol3x[t], sol3y[t]}, {sol1x[t], sol1y[t]}}], {t, tmin, tmax, (tmax - tmin)/15}]
]
Show[PursuitPolygon @@@ MeshPrimitives[reg, 2]] | {
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Show[PursuitPolygon @@@ MeshPrimitives[reg, 2]]
The solution is qualitatively different when the agents move at unit speed because they no longer meet at the centroid. Note that the difference in the picture is also that the density of the lines is different, this density can be adjusted in both solutions but changing some of the fixed numbers I put in. 15 in (tmax - tmin)/15 controls the number of lines in the last solution.
Here's an extension of @C.E.'s code to non-triangular polygons - polygons taken from a Voronoi mesh of roughly equidistributed (https://mathematica.stackexchange.com/a/141215/3056) 31 points over a square. Code is awful, I didn't really have time to think.
ClearAll@PursuitPolygon;
PursuitPolygon[pts_] := Module[{vars, sols},
vars = Table[Unique[], Length@pts, 2];
sols = NDSolveValue[{Sequence @@ (((#'[t] & /@ #1) ==
Normalize[Subtract @@ Map[#[t] &, {#2, #1}, {2}]]) & @@@
Partition[Join[vars, {First@vars}], 2, 1]),
Sequence @@ MapThread[(#[0] & /@ #1) == #2 &, {vars, pts}],
WhenEvent[Norm[x] < 1*^-8, "StopIntegration"] /.
x -> Subtract @@ Map[#[t] &, Take[vars, 2], {2}]},
Flatten@vars, {t, 0, 10}];
{tmin, tmax} = MinMax@First[sols]["Grid"];
Graphics@
Table[Line[
Map[#[t] &, Join[Partition[sols, 2], {Take[sols, 2]}], {2}]], {t,
tmin, tmax, (tmax - tmin)/15}]]
ClearAll@reg;
reg =
With[{reg = Rectangle[]},
With[{points = 31, samples = 4000, iterations = 100},
Nest[With[{randoms = Join[#, RandomPoint[reg, samples]]},
RegionNearest[reg][
Mean@randoms[[#]] & /@
Values@PositionIndex@Nearest[#, randoms]]] &,
RandomPoint[reg, points], iterations]] //
VoronoiMesh[#, {{0, 1}, {0, 1}}] &];
Show[PursuitPolygon /@ MeshPrimitives[reg, 2][[All, 1]]] | {
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# Integration by Partial Fractions
1. May 20, 2014
1. The problem statement, all variables and given/known data
Find the indefinite integral of the below, using partial fractions.
$$\frac{4x^2+6x-1}{(x+3)(2x^2-1)}$$
2. Relevant equations
?
3. The attempt at a solution
First I want to say there is probably a much easier and quicker way to get around certain things I have done but I have just done it the only way I could see (I always seem to go the long way around). So if I am correct then I would really appreciate some tips on how to do it quicker (for in exams) and also if I am incorrect to point out my mistakes. Thanks :)
first I set up the partial fractions.
$\frac{4x^2+6x-1}{(x+3)(2x^2-1)}=\frac{A}{x+3}+\frac{Bx+C}{2x^2-1} \\ \rightarrow \,\,\, 4x^2+6x-1=A(2x^2-1)+(Bx+C)(x+3) \\ 4x^2+6x-1=2Ax^2-A+Bx^2+3Bx+Cx+3C$
And then equated the coefficients:
For X^2: $4=2A+B$
For X^1: $6=3B+C$
For X^0: $-1=3C-A$
Then what I did what multiply the third equation above by 2 to get $-2=6C-2A$ and then added it to the first equation to get $2=6C+B$ and solved for B and substituted it into the second equation which resulted in:
$6=3(2-6C)+C\\ 6=6-11C \\ 0=-11C \\ ∴ C=0$
And then knowing C=0 I found A=1 and then that B=2 .
$\frac{4x^2+6x-1}{(x+3)(2x^2-1)}=\frac{1}{x+3}+\frac{2x}{2x^2-1} \\ \frac{4x^2+6x-1}{(x+3)(2x^2-1)}=\frac{1}{x+3}+\frac{2x}{2x^2-1}\\ \int \frac{4x^2+6x-1}{(x+3)(2x^2-1)} \,\, dx = \ln{|x+3|} + \frac{1}{2} \ln{|2x^2-1|}$
I think I may have done something wrong. I had to use an integral calculator online to find the integral of $\frac{2x}{2x^2-1}$ as I had no idea how to do it. Which kind of defeats the point.
2. May 20, 2014
### Saitama
To evaluate the last integral, you can use the substitution $2x^2-1=t$.
3. May 20, 2014
### Panphobia
Its all right, except you need to add a + C. But by the way the integral of 2x/(2x^2-1) can be easily solved using u substitution.
If u = 2x^2 -1 then du = 4x dx so du/2 = 2x dx
(1/2)*(1/u) | {
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If u = 2x^2 -1 then du = 4x dx so du/2 = 2x dx
(1/2)*(1/u)
and the integral of this is
ln|u|/2
which is
ln|2x^2-1|/2
Last edited: May 20, 2014
4. May 20, 2014
### Curious3141
You're missing the constant of integration, which is very important in an indefinite integral.
The partial fraction decomposition is tedious. If the denominator was easily factored into linear factors, Heaviside cover up rule would help more. As it stands, the cover up rule helps to find A quite quickly, but you get bogged down with B and C, and you have to manipulate radicals. Pointless. I think your method is fine.
5. May 20, 2014
Ah yes, sorry I forgot as it was just rough workings copied up into latex.
I am quite poor when it comes to integration by substitution for some reason. I find integration by parts easier as its a set formula but I do realise that integration by substitution is sometimes a lot easier and quicker so I need to get more familiar with it. Any tips on u substitution?
I can usually find what to substitute, and then get to a du=... but its the following bits that confuse me a little.
6. May 20, 2014
### scurty
I try to use u-substitution if I see a power that is one higher in the denominator than in the numerator. A more general approach would to be to look for derivatives in the numerator from what is found in the denominator. I always look for u-substitution first because they tend to be the easiest when dealing with fractions. Partial fractions and trig substitution, while fun to compute, are more time consuming.
It's hard for me to explain because I've done so many integrals that knowing what method to use is more intuitive to me than anything else. It comes easier the more you practice integrals. A general rule to always follow for u substitution is to notice candidates for u and du. | {
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# Find an arbitrary power of a lower triangular matrix of size $3\times 3$
Let $F$ be a field and let $A=\begin{bmatrix}a&0&0\\1&a&0\\0&1&a\end{bmatrix}\in\mathscr{M}_{3\times 3}(F)$. Show that $$A^k=\begin{bmatrix}a^k&0&0\\ka^{k-1}&a^k&0\\\dfrac{1}{2}k(k-1)a^{k-2}&ka^{k-1}&a^k\end{bmatrix}$$ for all $k > 0$. (Exercise 797 from Golan, The Linear Algebra a Beginning Graduate Student Ought to Know.)
I know it can be proved by induction, but since the topic is about Krylov Spaces, eigenvalues and Jordan canonical form, I wonder if there is another way to solve this problem.
• Inducción, José! – Pedro Tamaroff May 24 '15 at 20:59
• it is spelled induction, and I'm asking for another way, thanks. – José May 24 '15 at 21:02
• This is a trivial matter using induction, really. I doubt there is any gain going in any other route... – Mariano Suárez-Álvarez May 24 '15 at 21:04
• Did you try Krylov spaces ? – Dietrich Burde May 24 '15 at 21:42
• @Dietrich Burde: what is Krylov space? Any good online refs? I've never heard of this approach for such problems . . . – Robert Lewis May 24 '15 at 22:26
There is extended reading giving formula for calculating functions such as polynomials on matrixes: http://en.wikipedia.org/wiki/Matrix_function#Jordan_decomposition
For a route different to induction, express the matrix as $$A=aI_3+G$$ where $I_3$ is the identity matrix and $G$ is given by $$G=\begin{bmatrix}0&0&0\\1&0&0\\0&1&0\end{bmatrix}$$ A binomial expansion will yield \begin{align}A^k&=(aI_3+G)^k\\&={k \choose 0}a^kI_3+{k \choose 1}a^{k-1}I_3G+{k \choose 2}a^{k-2}I_3G^2+\sum_{m=3}^k{k \choose m}a^{k-m}I_3G^m\\&=a^kI_3+ka^{k-1}I_3G+\frac{k(k-1)}{2}a^{k-2}I_3G^2+\sum_{m=3}^k{k \choose m}a^{k-m}I_3G^m\end{align} Now $I_3G$ is given by $$G=\begin{bmatrix}0&0&0\\1&0&0\\0&1&0\end{bmatrix}$$ while $I_3G^2$ is $$G=\begin{bmatrix}0&0&0\\0&0&0\\1&0&0\end{bmatrix}$$ and $I_3G^m$ for $m\geq3$ is the zero matrix.
Using these results, we arrive at the given expression for $A^k$. | {
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Using these results, we arrive at the given expression for $A^k$.
• For the binomial expansion, would not explicitly (using matrix multiplication) working out $I_3G$, $I_3G^2$, and $I_3G^3$ suffice. Once we show that $I_3G^3$ is the zero matrix, then $I_3G^m$ for $m>0$ will also be zero matrices as well. – Alijah Ahmed May 24 '15 at 21:30
• Obviously I like your methodology! Endorsed!!! – Robert Lewis May 24 '15 at 22:29
Set
$N = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}; \tag{1}$
then a simple calculation reveals that
$N^2 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix} \tag{2}$
and
$N^3 = 0. \tag{3}$
Also,
$A = aI + N. \tag{4}$
Since $N$ and $I$ commute, we may apply the ordinary binomial theorem to (4), and find
$A^k = (aI + N)^k =$ $(aI)^k + k(aI)^{k - 1}N + \dfrac{k(k - 1)}{2}(aI)^{k - 2}N^2$ $= a^kI^k + ka^{k - 1}I^{k - 1}N + \dfrac{k(k - 1)}{2}a^{k - 2}I^{k - 2}N^2$ $= a^kI + ka^{k - 1}N + \dfrac{k(k - 1)}{2} a^{k - 2}N^2, \tag{5}$
since the expansion is trunated after the third term by virtue of $N^3 = 0$. When (5) is written out explicitly, we see that
$A^k = \begin{bmatrix} a^k & 0 & 0 \\ ka^{k - 1} & a^k & 0 \\ \dfrac{k(k - 1)}{2} a^{ k - 2} & ka^{k - 1} & a^k \end{bmatrix}, \tag{6}$
as per request.
Nota Bene: a few words on induction. It has been rightly noted that in deploying the binomial theorem we do not in fact escape induction. The only other option I can see is to prove this result by direct matrix mutiplication, again using induction. Here we hide the induction by invoking the binomial theorem, doing so since the problem statement asks for "another way"; so I assumed direct induction should be avoided. But in truth, this is a statement concerning the integers, so induction will almost certainly enter in at some point, being as it is an essential defining property of the integers. End of Note | {
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• @Dietrich Burde: Yes, yes, yes . . . take 'er easy, Pal, still editing. And I'll say a few words about induction before I'm done! Cheers! – Robert Lewis May 24 '15 at 21:33
• Sorry, everything is good. I just think, why not prove the (easy) matrix formula - by induction. – Dietrich Burde May 24 '15 at 21:35
• @Dietrich Burde: me too. Cheers! – Robert Lewis May 24 '15 at 21:38 | {
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# Question about convexity: how do we prove that $\displaystyle \sum_{i=1}^{k}p_{i}b_{i}\geq\prod_{i=1}^{k}b^{p_{i}}_{i}$?
Let $$b_{1},b_{2},\ldots,b_{k}$$ be nonnegative numbers and $$p_{1} + p_{2} + \ldots + p_{k} = 1$$ where each $$p_{i}$$ is positive. Then
\begin{align*} \sum_{i=1}^{k}p_{i}b_{i}\geq\prod_{i=1}^{k}b^{p_{i}}_{i} \end{align*}
MY ATTEMPT
Since the logarithm function is strictly increasing, the proposed inequality is equivalent to \begin{align*} \ln\left(p_{1}b_{1} + p_{2}b_{2} + \ldots + p_{k}b_{k}\right) \geq p_{1}\ln(b_{1}) + p_{2}\ln(b_{2}) + \ldots + p_{k}\ln(b_{k}) \end{align*}
Once $$f''(x) < 0$$, where $$f(x) = \ln(x)$$, we conclude that $$f$$ is concave and the proposed inequality holds.
My question is: am I proving this result correctly? If this is the case, is there another way to prove it?
Any contribution is appreciated.
• You have done it absolutely correctly. Another method is by using weighted AM-GM as shown in my answer that I'm typing. Aug 6, 2020 at 22:17
Using $$b_1,b_2,\cdots,b_k\ge 0$$ with $$p_1,p_2,\cdots,p_k$$ as the respective weights, weighted AM-GM inequality gives $$\frac{\sum_{i=1}^k p_ib_i}{\sum_{i=1}^k p_i} \ge \left(\prod_{i=1}^k b_i^{p_i}\right)^{\dfrac{1}{\sum_{i=1}^k p_i}}$$ which gives the required inequality.
Note: This is not much different from your approach because the special case of Jensen's inequality for the function $$f(x)=\ln(x)$$ can be proved by weighted AM-GM or vice-versa, the weighted AM-GM inequality can be proved by the Jensen's inequality for $$f(x)=\ln(x)$$ as you have done, since the inequality you want to prove is directly the weighted AM-GM inequality in disguise. Also, these two can be proven independently without using each other. | {
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# Products of compact spaces with countable tightness
In the previous two posts, we discuss the definitions of the notion of tightness and its relation with free sequences. This post and the next post are to discuss the behavior of countable tightness under the product operation. In this post, we show that countable tightness behaves well in products of compact space. In particular we show that countable tightness is preserved in finite products and countable products of compact spaces. In the next post we show that countable tightness is easily destroyed in products of sequential fans and that the tightness of such a product can be dependent on extra set theory assumptions. All spaces are Hausdorff and regular.
The following theorems are the main results in this post.
Theorem 1
Let $X$ and $Y$ be countably tight spaces. If one of $X$ and $Y$ is compact, then $X \times Y$ is countably tight.
Theorem 2
The product of finitely many compact countably tight spaces is countably tight.
Theorem 3
Suppose that $X_1, X_2, X_3, \cdots$ are countably many compact spaces such that each $X_i$ has at least two points. If each $X_i$ is a countably tight space, then the product space $\prod_{i=1}^\infty X_i$ is countably tight.
____________________________________________________________________
Finite products
Before proving Theorem 1 and Theorem 2, we prove the following results.
Theorem 4
Let $f:Y_1 \rightarrow Y_2$ be a continuous and closed map from the space $Y_1$ onto the space $Y_2$. Suppose that the space $Y_2$ is countably tight and that each fiber of the map $f$ is countably tight. Then the space $Y_1$ is countably tight.
Proof of Theorem 4
Let $x \in Y_1$ and $x \in \overline{A}$ where $A \subset Y_1$. We proceed to find a countable $W \subset Y_1$ such that $x \in \overline{W}$. Choose $y \in Y_2$ such that $y=f(x)$. | {
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Let $M$ be the fiber of the map $f$ at the point $y$, i.e. $M=f^{-1}(y)$. By assumption, $M$ is countably tight. Call a point $w \in M$ countably reached by $A$ if there is some countable $C \subset A$ such that $w \in \overline{C}$. Let $G$ be the set of all points in $M$ that are countably reached by $A$. We claim that $x \in \overline{G}$.
Let $U \subset Y_1$ be open such that $x \in U$. Because the space $Y_1$ is regular, choose open $V \subset U$ such that $x \in V$ and $\overline{V} \subset U$. Then $V \cap A \ne \varnothing$. Furthermore, $x \in \overline{V \cap A}$. Let $C=f(V \cap A)$. By the continuity of $f$, we have $y \in \overline{C}$. Since $Y_2$ is countably tight, there exists some countable $D \subset C$ such that $y \in \overline{D}$. Choose a countable $E \subset V \cap A$ such that $f(E)=D$. It follows that $y \in \overline{f(E)}$.
We show that that $\overline{E} \cap M \ne \varnothing$. Since $E \subset \overline{E}$, we have $f(E) \subset f(\overline{E})$. Note that $f(\overline{E})$ is a closed set since $f$ is a closed map. Thus $\overline{f(E)} \subset f(\overline{E})$. As a result, $y \in f(\overline{E})$. Then $y=f(t)$ for some $t \in \overline{E}$. We have $t \in \overline{E} \cap M$.
By the definition of the set $G$, we have $\overline{E} \cap M \subset G$. Furthermore, $\overline{E} \cap M \subset \overline{V} \subset U$. Note that the arbitrary open neighborhood $U$ of $x$ contains points of $G$. This establishes the claim that $x \in \overline{G}$.
Since $M$ is a fiber of $f$, $M$ is countably tight by assumption. Choose some countable $T \subset G$ such that $x \in \overline{T}$. For each $t \in T$, choose a countable $W_t \subset A$ with $t \in \overline{W_t}$. Let $W=\bigcup_{t \in T} W_t$. Note that $W \subset A$ and $W$ is countable with $x \in \overline{W}$. This establishes the space $Y_1$ is countably tight at $x \in Y_1$. $\blacksquare$ | {
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Lemma 5
Let $f:X \times Y \rightarrow Y$ be the projection map. If $X$ is a compact space, then $f$ is a closed map.
Proof of Lemma 5
Let $A$ be a closed subset of $X \times Y$. Suppose that $f(A)$ is not closed. Let $y \in \overline{f(A)}-f(A)$. It follows that no point of $X \times \left\{y \right\}$ belongs to $A$. For each $x \in X$, choose open subset $O_x$ of $X \times Y$ such that $(x,y) \in O_x$ and $O_x \cap A=\varnothing$. The set of all $O_x$ is an open cover of the compact space $X \times \left\{y \right\}$. Then there exist finitely many $O_x$ that cover $X \times \left\{y \right\}$, say $O_{x_i}$ for $i=1,2,\cdots,n$.
Let $W=\bigcup_{i=1}^n O_{x_i}$. We have $X \times \left\{y \right\} \subset W$. Since $X$ is compact, we can then use the Tube Lemma which implies that there exists open $G \subset Y$ such that $X \times \left\{y \right\} \subset X \times G \subset W$. It follows that $G \cap f(A) \ne \varnothing$. Choose $t \in G \cap f(A)$. Then for some $x \in X$, $(x,t) \in A$. Since $t \in G$, $(x,t) \in W$, implying that $W \cap A \ne \varnothing$, a contradiction. Thus $f(A)$ must be a closed set in $Y$. This completes the proof of the lemma. $\blacksquare$
Proof of Theorem 1
Let $X$ be the factor that is compact. let $f: X \times Y \rightarrow Y$ be the projection map. The projection map is always continuous. Furthermore it is a closed map by Lemma 5. The range space $Y$ is countably tight by assumption. Each fiber of the projection map $f$ is of the form $X \times \left\{y \right\}$ where $y \in Y$, which is countably tight. Then use Theorem 4 to establish that $X \times Y$ is countably tight. $\blacksquare$
Proof of Theorem 2
This is a corollary of Theorem 1. According to Theorem 1, the product of two compact countably tight spaces is countably tight. By induction, the product of any finite number of compact countably tight spaces is countably tight. $\blacksquare$
____________________________________________________________________ | {
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____________________________________________________________________
Countable products
Our proof to establish that the product space $\prod_{i=1}^\infty X_i$ is countably tight is an indirect one and makes use of two non-trivial results. We first show that $\omega_1 \times \prod_{i=1}^\infty X_i$ is a closed subspace of a $\Sigma$-product that is normal. It follows from another result that the second factor $\prod_{i=1}^\infty X_i$ is countably tight. We now present all the necessary definitions and theorems.
Consider a product space $Y=\prod_{\alpha<\kappa} Y_\alpha$ where $\kappa$ is an infinite cardinal number. Fix a point $p \in Y$. The $\Sigma$-product of the spaces $Y_\alpha$ with $p$ as the base point is the following subspace of the product space $Y=\prod_{\alpha<\kappa} Y_\alpha$:
$\displaystyle \Sigma_{\alpha<\kappa} Y_\alpha=\left\{y \in \prod_{\alpha<\kappa} Y_\alpha: y_\alpha \ne p_\alpha \text{ for at most countably many } \alpha < \kappa \right\}$
The definition of the space $\Sigma_{\alpha<\kappa} Y_\alpha$ depends on the base point $p$. The discussion here is on properties of $\Sigma_{\alpha<\kappa} Y_\alpha$ that hold regardless of the choice of base point. If the factor spaces are indexed by a set $A$, the notation is $\Sigma_{\alpha \in A} Y_\alpha$.
If all factors $Y_\alpha$ are identical, say $Y_\alpha=Z$ for all $\alpha$, then we use the notation $\Sigma_{\alpha<\kappa} Z$ to denote the $\Sigma$-product. Once useful fact is that if there are $\omega_1$ many factors and each factor has at least 2 points, then the space $\omega_1$ can be embedded as a closed subspace of the $\Sigma$-product.
Theorem 6
For each $\alpha<\omega_1$, let $Y_\alpha$ be a space with at least two points. Then $\Sigma_{\alpha<\omega_1} Y_\alpha$ contains $\omega_1$ as a closed subspace. See Exercise 3 in this previous post. | {
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Now we discuss normality of $\Sigma$-products. This previous post shows that if each factor is a separable metric space, then the $\Sigma$-product is normal. It is also well known that if each factor is a metric space, the $\Sigma$-product is normal. The following theorem handles the case where each factor is a compact space.
Theorem 7
For each $\alpha<\kappa$, let $Y_\alpha$ be a compact space. Then the $\Sigma$-product $\Sigma_{\alpha<\kappa} Y_\alpha$ is normal if and only if each factor $Y_\alpha$ is countably tight.
Theorem 7 is Theorem 7.5 in page 821 of [1]. Theorem 7.5 in [1] is stated in a more general setting where each factor of the $\Sigma$-product is a paracompact p-space. We will not go into a discussion of p-space. It suffices to know that any compact Hausdorff space is a paracompact p-space. We also need the following theorem, which is proved in this previous post.
Theorem 8
Let $Y$ be a compact space. Then the product space $\omega_1 \times Y$ is normal if and only if $Y$ is countably tight.
We now prove Theorem 3.
Proof of Theorem 3
Let $\omega_1=\cup \left\{A_n: n \in \omega \right\}$, where for each $n$, $\lvert A_n \lvert=\omega_1$ and that $A_n \cap A_m=\varnothing$ if $n \ne m$. For each $n=1,2,3,\cdots$, let $S_n=\Sigma_{\alpha \in A_n} X_n$. By Theorem 7, each $S_n$ is normal. Let $S_0=\Sigma_{\alpha \in A_0} X_1$, which is also normal. By Theorem 6, the space $\omega_1$ of countable ordinals is a closed subspace of $S_0$. Let $T=\omega_1 \times X_1 \times X_2 \times X_3 \times \cdots$. We have the following derivation.
\displaystyle \begin{aligned} T&=\omega_1 \times X_1 \times X_2 \times X_3 \times \cdots \\&\subset S_0 \times S_1 \times S_2 \times S_3 \times \cdots \\&\cong W=\Sigma_{\alpha<\omega_1} W_\alpha \end{aligned} | {
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Recall that $\omega_1=\cup \left\{A_n: n \in \omega \right\}$. The space $W=\Sigma_{\alpha<\omega_1} W_\alpha$ is defined such that for each $n \ge 1$ and for each $\alpha \in A_n$, $W_\alpha=X_n$. Furthermore, for $n=0$, for each $\alpha \in A_0$, let $W_\alpha=X_1$. Thus $W$ is a $\Sigma$-product of compact countably tight spaces and is thus normal by Theorem 7. The space $T=\omega_1 \times \prod_{n=1}^\infty X_n$ is a closed subspace of the normal space $W$. By Theorem 8, the product space $\prod_{n=1}^\infty X_n$ must be countably tight. $\blacksquare$
____________________________________________________________________
Remarks
Theorem 2, as indicated above, is a corollary of Theorem 1. We also note that Theorem 2 is also a corollary of Theorem 3 since any finite product is a subspace of a countable product. To see this, let $X=X_1 \times X_2 \times \cdots \times X_n$.
\displaystyle \begin{aligned} X&=X_1 \times X_2 \times \cdots \times X_n \\&\cong X_1 \times X_2 \times \cdots \times X_n \times \left\{t_{n+1} \right\} \times \left\{t_{n+2} \right\} \times \cdots \\&\subset X_1 \times X_2 \times \cdots \times X_n \times X_{n+1} \times X_{n+2} \times \cdots \end{aligned}
In the above derivation, $t_m$ is a point of $X_m$ for all $m >n$. When the countable product space is countably tight, the finite product, being a subspace of a countably tight space, is also countably tight.
____________________________________________________________________
Exercise
Exercise 1
Let $f:X \times Y \rightarrow Y$ be the projection map. If $X$ is a countably compact space and $Y$ is a Frechet space, then $f$ is a closed map.
Exercise 2
Let $X$ and $Y$ be countably tight spaces. If one of $X$ and $Y$ is a countably compact space and the other space is a Frechet space, then $X \times Y$ is countably tight.
Exercise 2 is a variation of Theorem 1. One factor is weakened to “countably compact”. However, the other factor is strengthened to “Frechet”. | {
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____________________________________________________________________
Reference
1. Przymusinski, T. C., Products of Normal Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 781-826, 1984.
____________________________________________________________________
$\copyright \ 2015 \text{ by Dan Ma}$ | {
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# Rocket Ship Question
1. Nov 6, 2007
### Submission1
We are having a debate here at work and we need an answer from some smart folks (that's you).
If you have a rocket ship in space that takes 1 gallon of gas to reach 1000 mph then, negating friction, would it take less, more or the same amount to go from 1000 mph to 2000 mph.
Thanks,
Sub.
2. Nov 6, 2007
### mgb_phys
Kinetic energy, the energy needed to move something depends on the speed squared.
So to travel at 2000 mph 4x as much energy as 1000mph. Since you started at 1000mph you will need to add 3x as much extra energy to get to 2000mph than you did to get to 1000mph.
You also have to take into account that the mass of the rocket is changing as you use up the fuel so it takes less energy to accelarate the rocket since it weighs less - if you doing this for real.
Last edited: Nov 6, 2007
3. Nov 6, 2007
### Submission1
Cool. Thanks.
4. Nov 6, 2007
### chronon
But there's nothing special about the starting frame of reference. So it shouldn't take any more fuel to add an extra 1000mph starting in the 1000mph frame as to get to the 1000mph frame starting from the zero frame.
5. Nov 6, 2007
### mgb_phys
Does that apply if it's accelerating ?
6. Nov 6, 2007
### D H
Staff Emeritus
This topic ws recently discussed in this thread.
The governing equation is the Tsiolokovsky rocket equation. Applying this equation to a rocket that burns all of its fuel,
$$\Delta v = v_e\ln\left(\frac {m_\text{rocket}+m_\text{fuel}}{m_\text{rocket}}\right)$$
Suppose you find you have to load some rocket with a quantity of fuel $m_{\text{fuel}|1000\text{mph}}$ to make the rocket attain a velocity of 1000 mph after burnout. Atfer churning the crank on the rocket equation, the amount of fuel it would take to bring the rocket to 2000 mph is
$$m_{\text{fuel}|2000\text{mph}} = m_{\text{fuel}|1000\text{mph}}\left(2+ \frac{m_{\text{fuel}|1000\text{mph}}}{m_\text{rocket}}\right)$$ | {
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In other words, one must more than double the quantity of fuel to double a rocket's final velocity.
7. Nov 6, 2007
### mgb_phys
Thanks - I'm away for a week and look what I miss.
8. Nov 6, 2007
### chronon
Indeed. But more of that fuel will be burnt in getting it from 0 to 1000mph than from 1000mph to 2000mph.
9. Nov 6, 2007
### D H
Staff Emeritus
A little more detail regarding previous post:
The reason it takes more than double the fuel to go from 0 to 1000 mph than from 0 to 2000 mph (or whatever) is because the rocket carries its fuel.
Assume we have some rocket and want to make two test flights with it. The first test flight involves flying the rocket from rest to a final speed of 1000 mph, and the second, rest to 2000 mph. Both tests end with the rocket devoid of all fuel. The amount during the second test of fuel left in the rocket at the point the rocket reaches 1000 mph is exactly the same as the amount of fuel initially placed in the rocket for the first test. In this sense, it takes the same amount of fuel to go from 1000 to 2000 mph as it does from 0 to 1000 mph.
However, the rocket has to first achieve that 1000 mph during the second test. It is the first 1000 mph that costs more. At the start of the second test, the rocket comprises the dry mass of the rocket, the quantity fuel (call this f1) needed to bring the dry mass of the rocket from 1000 to 2000 mph plus the quantity of fuel needed to bring the dry mass of the rocket plus f1 from 0 to 1000 mph.
Last edited: Nov 6, 2007
10. Nov 6, 2007
### D H
Staff Emeritus
Indeed. We just cross-posted.
11. Nov 6, 2007
### rcgldr | {
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### D H
Staff Emeritus
Indeed. We just cross-posted.
11. Nov 6, 2007
### rcgldr
Mentioned in the other thread is that there is work also peformed on the spent fuel as it is ejected behind the rockets engine. If the kinetic energy of both spent fuel, and the rocket (plus it's remaining unspent fuel) are summed, and if the rate of fuel consumption is constant, then thrust will be constant, and the sum of the kinetic energies of spent fuel, rocket (and remaining fuel) will increase linearly with time. This means that the power involved is constant (with a constant thrust).
Also from the other thread, it's easier to grasp this if you consider the rocket to be held in place so that all of the work done by the engine is to acclerate the spent fuel. The terminal velocity of the fuel is constant, and the rate of fuel consumption (mass ejected) is constant, so the kinetic energy of the fuel increases linearly with time.
12. Nov 6, 2007
### Staff: Mentor
From this thread and the other I have a kind of vague impression that rockets are better understood in terms of conservation of momentum than conservation of energy. I think that is because conservation of momentum requires you to think of the exhaust whereas conservation of energy does not.
13. Nov 6, 2007
### TVP45
You always have to ask, 1000 mph relative to what? A rocket ship just traveling along at constant velocity will never know it. You always need a reference against which you measure your velocity. And, it is in that reference frame where it takes 3 times as much energy to go from 1000 mph to 2000 mph as it does from 0 to 1000 mph.
14. Nov 6, 2007
### D H | {
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14. Nov 6, 2007
### D H
Staff Emeritus
That is how I prefer to look at things. A conservation of energy viewpoint is tough to do properly. The rocket fuel remaining in the rocket gains kinetic energy as the rocket accelerates. Moreover, some of the released chemical potential energy is wasted in the form of hot exhaust. Failing to account for either leads to erroneous results.
Conservation of linear momentum is a much simpler proposition. Adding conservation of angular momentum makes things a bit hairier. Adding the fact that in a real rocket the center of mass and inertia tensor change as the rocket burns fuel makes things a lot hairer. I wouldn't dream of attacking a non-point mass rocket with fuel located away from the center of mass with a conservation of energy perspective -- unless I was doing so at a micro level and using a CFD model.
The driving reason for investigating a problem from the point of view of any of the conservation principles is that doing so makes the answer fall out. I don't have to use a full-blown CFD model to get a very, very good model of what firing a roket engine does to the state of a vehicle. A full-blown CFD model that takes advantage of all of the conservation laws is needed to characterize behavior. However, CFD models cannot "see the forest for the trees". CFD models examine the spots on a beetle that sits on a tree in a huge forest. Once the behavior has been properly characterized, a conservation of momentum model does wonders.
15. Nov 8, 2007
### alvaros
So the answer is ... ??
A 1000 kg rocket push a 1 Kg fuel at 1000 * 1000 mph -> the rocket increases its velocity in 1000 mph ( respect to the Frame of Reference it was before starting the engine )
Again
A 999 kg rocket ( the same rocket ) push a 1 Kg fuel at 1000 * 1000 mph -> the rocket increases its velocity in 1000 mph.
( I use very high energy fuel, so we dont have to discuss about carrying the fuel or using Tsiolokovsky formula )
16. Nov 8, 2007
### TVP45 | {
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16. Nov 8, 2007
### TVP45
A kind of new thought on this thread....
This kind of question pops up a lot and there always seems to be (in the mind of the questioner) some kind of disconnect between common-sense examples and the equations.
Is the work-energy theorem the best way to teach people? I didn't get it for many years after that freshman class.
Is there another way?
17. Nov 9, 2007
### alvaros
Im very sorry, but I dont understand your english.
18. Nov 9, 2007
### TVP45
Alvaros,
19. Nov 10, 2007
### alvaros
I think my example is clear enough:
It seems that the second kg of fuel gives more kinetic energy to the rocket than the first Kg ( ?? )
I made a thread "what is a IFR" but everybody seems to have it very clear. I think the main mistakes are representing forces with an arrow -> and talking about IFR as some abstract.
In this problem the forces are between the rocket and the fuel ejected:
fuel <-> rocket
And the IFR is the center of mass of the system ( rocket + fuel ). The IFR always refers to something material ( with mass ).
20. Nov 10, 2007
### Staff: Mentor
Yes. Don't forget the KE of the exhaust. It is better to use conservation of momentum principles because they force you to consider the exhaust.
I am trying out some new (for me) applications of old ideas here here, so feel free to point out any mistakes in my logic:
Since the exhaust velocity is constant (assuming the mass of the rocket is large relative to the mass of the fuel) each kg fuel spent gives the same Δp and therefore the same Δv to the rocket.
Each successive kg with its Δv results in a larger ΔKE for the rocket since KE is proportional to v^2.
"But conservation of energy you protest!" The PE of the fuel does not only go into KE of the rocket, but also into the rather large KE of the exhaust. | {
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Because the exhaust is going in the opposite direction of the rocket each successive kg of exhaust gains less KE, exactly compensating the more KE gained by the rocket so that the increased KE of the rocket-exhaust system is always equal to the PE in the spent fuel (neglecting heat).
Bottom line: for rockets always use conservation of momentum so that you cannot neglect the exhaust.
Last edited: Nov 10, 2007 | {
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# Rocket Ship Question
1. Nov 6, 2007
### Submission1
We are having a debate here at work and we need an answer from some smart folks (that's you).
If you have a rocket ship in space that takes 1 gallon of gas to reach 1000 mph then, negating friction, would it take less, more or the same amount to go from 1000 mph to 2000 mph.
Thanks,
Sub.
2. Nov 6, 2007
### mgb_phys
Kinetic energy, the energy needed to move something depends on the speed squared.
So to travel at 2000 mph 4x as much energy as 1000mph. Since you started at 1000mph you will need to add 3x as much extra energy to get to 2000mph than you did to get to 1000mph.
You also have to take into account that the mass of the rocket is changing as you use up the fuel so it takes less energy to accelarate the rocket since it weighs less - if you doing this for real.
Last edited: Nov 6, 2007
3. Nov 6, 2007
### Submission1
Cool. Thanks.
4. Nov 6, 2007
### chronon
But there's nothing special about the starting frame of reference. So it shouldn't take any more fuel to add an extra 1000mph starting in the 1000mph frame as to get to the 1000mph frame starting from the zero frame.
5. Nov 6, 2007
### mgb_phys
Does that apply if it's accelerating ?
6. Nov 6, 2007
### D H
Staff Emeritus
This topic ws recently discussed in this thread.
The governing equation is the Tsiolokovsky rocket equation. Applying this equation to a rocket that burns all of its fuel,
$$\Delta v = v_e\ln\left(\frac {m_\text{rocket}+m_\text{fuel}}{m_\text{rocket}}\right)$$
Suppose you find you have to load some rocket with a quantity of fuel $m_{\text{fuel}|1000\text{mph}}$ to make the rocket attain a velocity of 1000 mph after burnout. Atfer churning the crank on the rocket equation, the amount of fuel it would take to bring the rocket to 2000 mph is
$$m_{\text{fuel}|2000\text{mph}} = m_{\text{fuel}|1000\text{mph}}\left(2+ \frac{m_{\text{fuel}|1000\text{mph}}}{m_\text{rocket}}\right)$$ | {
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In other words, one must more than double the quantity of fuel to double a rocket's final velocity.
7. Nov 6, 2007
### mgb_phys
Thanks - I'm away for a week and look what I miss.
8. Nov 6, 2007
### chronon
Indeed. But more of that fuel will be burnt in getting it from 0 to 1000mph than from 1000mph to 2000mph.
9. Nov 6, 2007
### D H
Staff Emeritus
A little more detail regarding previous post:
The reason it takes more than double the fuel to go from 0 to 1000 mph than from 0 to 2000 mph (or whatever) is because the rocket carries its fuel.
Assume we have some rocket and want to make two test flights with it. The first test flight involves flying the rocket from rest to a final speed of 1000 mph, and the second, rest to 2000 mph. Both tests end with the rocket devoid of all fuel. The amount during the second test of fuel left in the rocket at the point the rocket reaches 1000 mph is exactly the same as the amount of fuel initially placed in the rocket for the first test. In this sense, it takes the same amount of fuel to go from 1000 to 2000 mph as it does from 0 to 1000 mph.
However, the rocket has to first achieve that 1000 mph during the second test. It is the first 1000 mph that costs more. At the start of the second test, the rocket comprises the dry mass of the rocket, the quantity fuel (call this f1) needed to bring the dry mass of the rocket from 1000 to 2000 mph plus the quantity of fuel needed to bring the dry mass of the rocket plus f1 from 0 to 1000 mph.
Last edited: Nov 6, 2007
10. Nov 6, 2007
### D H
Staff Emeritus
Indeed. We just cross-posted.
11. Nov 6, 2007
### rcgldr | {
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### D H
Staff Emeritus
Indeed. We just cross-posted.
11. Nov 6, 2007
### rcgldr
Mentioned in the other thread is that there is work also peformed on the spent fuel as it is ejected behind the rockets engine. If the kinetic energy of both spent fuel, and the rocket (plus it's remaining unspent fuel) are summed, and if the rate of fuel consumption is constant, then thrust will be constant, and the sum of the kinetic energies of spent fuel, rocket (and remaining fuel) will increase linearly with time. This means that the power involved is constant (with a constant thrust).
Also from the other thread, it's easier to grasp this if you consider the rocket to be held in place so that all of the work done by the engine is to acclerate the spent fuel. The terminal velocity of the fuel is constant, and the rate of fuel consumption (mass ejected) is constant, so the kinetic energy of the fuel increases linearly with time.
12. Nov 6, 2007
### Staff: Mentor
From this thread and the other I have a kind of vague impression that rockets are better understood in terms of conservation of momentum than conservation of energy. I think that is because conservation of momentum requires you to think of the exhaust whereas conservation of energy does not.
13. Nov 6, 2007
### TVP45
You always have to ask, 1000 mph relative to what? A rocket ship just traveling along at constant velocity will never know it. You always need a reference against which you measure your velocity. And, it is in that reference frame where it takes 3 times as much energy to go from 1000 mph to 2000 mph as it does from 0 to 1000 mph.
14. Nov 6, 2007
### D H | {
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14. Nov 6, 2007
### D H
Staff Emeritus
That is how I prefer to look at things. A conservation of energy viewpoint is tough to do properly. The rocket fuel remaining in the rocket gains kinetic energy as the rocket accelerates. Moreover, some of the released chemical potential energy is wasted in the form of hot exhaust. Failing to account for either leads to erroneous results.
Conservation of linear momentum is a much simpler proposition. Adding conservation of angular momentum makes things a bit hairier. Adding the fact that in a real rocket the center of mass and inertia tensor change as the rocket burns fuel makes things a lot hairer. I wouldn't dream of attacking a non-point mass rocket with fuel located away from the center of mass with a conservation of energy perspective -- unless I was doing so at a micro level and using a CFD model.
The driving reason for investigating a problem from the point of view of any of the conservation principles is that doing so makes the answer fall out. I don't have to use a full-blown CFD model to get a very, very good model of what firing a roket engine does to the state of a vehicle. A full-blown CFD model that takes advantage of all of the conservation laws is needed to characterize behavior. However, CFD models cannot "see the forest for the trees". CFD models examine the spots on a beetle that sits on a tree in a huge forest. Once the behavior has been properly characterized, a conservation of momentum model does wonders.
15. Nov 8, 2007
### alvaros
So the answer is ... ??
A 1000 kg rocket push a 1 Kg fuel at 1000 * 1000 mph -> the rocket increases its velocity in 1000 mph ( respect to the Frame of Reference it was before starting the engine )
Again
A 999 kg rocket ( the same rocket ) push a 1 Kg fuel at 1000 * 1000 mph -> the rocket increases its velocity in 1000 mph.
( I use very high energy fuel, so we dont have to discuss about carrying the fuel or using Tsiolokovsky formula )
16. Nov 8, 2007
### TVP45 | {
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16. Nov 8, 2007
### TVP45
A kind of new thought on this thread....
This kind of question pops up a lot and there always seems to be (in the mind of the questioner) some kind of disconnect between common-sense examples and the equations.
Is the work-energy theorem the best way to teach people? I didn't get it for many years after that freshman class.
Is there another way?
17. Nov 9, 2007
### alvaros
Im very sorry, but I dont understand your english.
18. Nov 9, 2007
### TVP45
Alvaros,
19. Nov 10, 2007
### alvaros
I think my example is clear enough:
It seems that the second kg of fuel gives more kinetic energy to the rocket than the first Kg ( ?? )
I made a thread "what is a IFR" but everybody seems to have it very clear. I think the main mistakes are representing forces with an arrow -> and talking about IFR as some abstract.
In this problem the forces are between the rocket and the fuel ejected:
fuel <-> rocket
And the IFR is the center of mass of the system ( rocket + fuel ). The IFR always refers to something material ( with mass ).
20. Nov 10, 2007
### Staff: Mentor
Yes. Don't forget the KE of the exhaust. It is better to use conservation of momentum principles because they force you to consider the exhaust.
I am trying out some new (for me) applications of old ideas here here, so feel free to point out any mistakes in my logic:
Since the exhaust velocity is constant (assuming the mass of the rocket is large relative to the mass of the fuel) each kg fuel spent gives the same Δp and therefore the same Δv to the rocket.
Each successive kg with its Δv results in a larger ΔKE for the rocket since KE is proportional to v^2.
"But conservation of energy you protest!" The PE of the fuel does not only go into KE of the rocket, but also into the rather large KE of the exhaust. | {
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Because the exhaust is going in the opposite direction of the rocket each successive kg of exhaust gains less KE, exactly compensating the more KE gained by the rocket so that the increased KE of the rocket-exhaust system is always equal to the PE in the spent fuel (neglecting heat).
Bottom line: for rockets always use conservation of momentum so that you cannot neglect the exhaust.
Last edited: Nov 10, 2007
21. Nov 10, 2007
### TVP45
DaleSpam,
This is the sort of example I was asking about. If I were a high school teacher (thank G-d I'm not; that's a hard job), how would I present this example so that students can grasp it? I'm not being in any way critical, any more than I was of Alvaros; I am simply curious about how this can be developed conceptually.
22. Nov 10, 2007
### Staff: Mentor
If I were a teacher in a high-school level physics course I would not even mention conservation of energy in this problem. I would stick entirely with conservation of momentum. You can solve for everything of interest that way, and the conservation of momentum principle does what a good conservation principle should: it simplifies things.
As a teacher I would not introduce conservation of energy into the problem since it does nothing to simplify the problem. If a student asked I would tell them that energy is conserved, but it doesn't make the problem any easier. I think students would understand that.
23. Nov 11, 2007
### TVP45
At first blush, I agree. Momentum is much easier to deal with than energy. Two questions occur to me:
(1) Can a student grasp the concept of momentum without being sidetracked by preconceptions about energy?
(2) Is the calculation of change of momentum rather than absolute momentum sensible?
24. Nov 11, 2007
### alvaros
DaleSpam:
Wrong, the change in kinetic energy on the fuel is 1000 times the change in kinetic energy on the rocket. | {
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Momentum is an abstract concept, its much easier the 3rd Newtons law: the burning fuel pushes the rocket and the rocket pushes the fuel fuel <-> rocket
( you see the double arrow <-> , all forces ( ? ) are double arrow )
( It seems, as I read here, that the conservation of momentum is more "universal" than 3rd Newtons law, but, in this case, both laws are the same and true )
But, at the end, there isnt any clear answer to the problem.
25. Nov 11, 2007
### D H
Staff Emeritus
Invoking Newton's third law on this problem is in a sense more ad-hoc and more abstract than using conservation of momentum with regard to this problem. The concept of force (Newton's second law) is very abstract; anything change to an object's momentum involves some force. Applying conservation of momentum is no less abstract than applying Newton's third law and yields a deeper answer.
Put the spaceship in deep space, far from any massive object, so that there are no measurable external forces acting on the vehicle. Now, what exactly is the force that is making the spaceship accelerate? By assumption, there are no external forces. You can say posit some ad-hoc force $F$ that results from burning the fuel and get an answer via Newton's second law. Note well: Since the rocket's mass is changing, the simpler form $F=ma$ is not valid here. We have to use $F=dp/dt$ instead.
$$F = \frac{dp_r} {dt} = \frac{d}{dt} ( m_r \, v_r ) = \dot m_r v_r} + m_r \, a_r$$
Solving for the rocket's acceleration,
$$a_r = \frac {F} {m_r} - \frac {\dot m_r} {m_r} v_r}$$
This is not very satisying. What exactly is this force? It's purely ad-hoc for one thing. Moreover, it will turn out that the force is frame-dependent. | {
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Here is how things turn out from conservation of momentum point of view. For this derivation, I will use some math that mathematicians don't particular like but physicists use willy-nilly -- things $\Delta v$. Things can be done more formally using continuum physics (classical treatment of gases), but that is a lot messier.
Nomenclature:
$$\vec v_r(t)$$ Rocket velocity at time $t$, inertial observer
$$\vec v_e(t)$$ Rocket exhaust velocity at time $t$, relative to vehicle velocity
$$m_r(t)$$ Rocket mass at time $t$
$$\dot m_f(t)$$ Rate at which rocket consumes fuel at$t$
Over a small time interval $\Delta t$, the rocket will eject a small mass of fuel $\dot m_f(t)\,\Delta t$. At the start of the interval, the rocket has mass $m_r(t)$, velocity $\vec v_r(t)$, and momentum $m_r(t) \, \vec v_r(t)$. At the end of the interval, the rocket has mass, velocity, and linear momentum
$$m_r(t+\Delta t) = m_r(t)-\dot m_f(t)\,\Delta t$$
$$\vec v_r(t+\Delta t) = \vec v_r(t)+\Delta \vec v_r(t)$$
$$\vec p_r(t+\Delta t) = (m_r(t) -\dot m_f(t)\,\Delta t)\, (\vec v_r(t)+\Delta \vec v_r(t))$$
The bit of ejected fuel carries some momentum from the vehicle. The mass, inertial observer velocity, and momentum of the exhaust are
$$\Delta m_e(t) = \dot m_f(t)\,\Delta t$$
$$\vec v_{e_{\text{inertial}}}) = \vec v_r(t)+\vec v_e(t)$$
$\Delta \vec p_e(t+\Delta t) = \dot m_f(t)\,\Delta t\, (\vec v_r(t)+\vec v_e(t))[/tex] The momentum of the rocket+exhaust at the end of the time interval is thus $$\vec p_{r+e}(t+\Delta t) = (m_r(t) -\dot m_f(t)\,\Delta t)\, (\vec v_r(t)+\Delta \vec v_r(t)) + \dot m_f(t)\,\Delta t\, (\vec v_r(t)+\vec v_e(t))$$ Dropping the second-order term [itex]\Delta t \Delta \vec v_r(t)$ and simplifying,
$$\vec p_{r+e}(t+\Delta t) = m_r(t)\,\vec v_r(t)+ m_r(t) \, \Delta \vec v_r(t) + \dot m_f(t)\,\Delta t\, \vec v_e(t)$$ | {
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Assuming no external forces act on the rocket and the ejected fuel during this time interval, the rocket and the ejected fuel form a closed system. Momentum is conserved in a closed system, so $\vec p_{r+e}(t+\Delta t)=\vec p_r(t)[/tex]: $$m_r(t)\,\vec v_r(t)+ m_r(t) \, \Delta \vec v_r(t) + \dot m_f(t)\,\Delta t\, \vec v_e(t) = m_r(t)\,\vec v_r(t)$$ or $$m_r(t) \, \Delta \vec v_r(t) + \dot m_f(t)\,\Delta t\, \vec v_e(t) = 0$$ Dividing by [itex]\Delta t$ and taking the limit $\Delta t \to 0$,
$$\frac {d\vec v_r(t)}{dt} = - \, \frac {\dot m_f(t)} {m_r(t)} \, \vec v_e(t)$$
This is the equation for the acceleration of the rocket at time $t$.
By conservation of mass, the time derivative of the rocket's mass is just the additive inverse of the fuel consumption rate: $\dot m_r(t) = -\dot m_f(t)$. If the relative exhaust velocity is a constant vector, both the left and right hand sides of the above acceleration equation are integrable. Integrating from some initial time $t_0$ to some final time $t_1$ yields
$$\vec v_r(t_1) - \vec v_r(t_0) = \ln\left(\frac{m_r(t_1)}{m_r(t_0)}\right) \vec v_e$$
This is the Tsiolokovsky rocket equation. Since the final mass is smaller than the initial mass, the logarithm will be negative. The change in velocity is directed against the exhaust direction.
Last edited: Nov 11, 2007 | {
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