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HW 4 EECS 110

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Problem #2 For this problem, you will implement a (text-based) menu of options for analyzing a list of stock prices (or any list of floating-point values). For this problem, you will implement a (text-based) menu of options for analyzing a list of stock prices (or any list of floating-point values). The top-level function to write is called tts() -- it takes no inputs. Instead, it offers the user a menu with these choices: The top-level function to write is called tts() -- it takes no inputs. Instead, it offers the user a menu with these choices: (0) Input a new list (1) Print the current list (2) Find the average price (3) Find the standard deviation (4) Find the min and its day (5) Find the max and its day (6) Your TT investment plan (9) Quit

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Problem #2 (0) Input a new list Hint: use the input() (instead of raw_input) so the user can input a list so the user can input a list (1) Print the current list Hint: print "%3d %5.2f" % (i, L[i])

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Problem #2 (2) Find the average price … (3) Find the standard deviation

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Problem #2 (4) Find the min and its day … (5) Find the max and its day … (6) Your TT investment plan The time-travel strategy: For menu option 6, you will want to find the best day on which to buy and sell the stock in question in order to maximize the profit earned. However, the sell day must be greater than or equal to the buy day. You may want to adapt the example function diff from class in order to find this maximum profit.

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Problem #2 Other menu options… print a warning message if the integer is not a valid menu option print a warning message if the integer is not a valid menu option quit if the user inputs quit if the user inputs #ask for input #get input While… #menu options #ask for input #get input

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Problem #3 Imagine a circle inscribed within a square that spans the area where -1 ≤ x ≤ 1 and -1 ≤ y ≤ 1. The area of the inscribed circle, whose radius is 1.0 would be pi If you were to throw darts at random locations in the square, only some of them would hit the circle inscribed within it. The ratio area of the circle / area of the square can be estimated by the ratio # of darts that hit the circle / total # of darts thrown As the number of darts increases, the second ratio, above, gets closer and closer to the first ratio. Since three of the four quantities involved are known, they can be used to approximate the area of the circle - this in turn can be used to approximate pi

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Problem #3 The forPi( n ) function takes in a positive integer n and should "throw" n darts at the square. Each time a dart is thrown, the function should print The forPi( n ) function takes in a positive integer n and should "throw" n darts at the square. Each time a dart is thrown, the function should print the number of darts thrown so far the number of darts thrown so far the number of darts thrown so far that have hit the circle the number of darts thrown so far that have hit the circle the resulting estimate of pi the resulting estimate of pi def forPi(n) #initiate variables for i in ….: #use random to find an x and a y between -1 & 1 #make equation to find if the (x,y) coord is in the circle #if its in the circle, you’ve got a hit! #print the num of throws, num. of hits, estimate of pi

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Problem #3 The whilePi( error ) function, on the other hand, will take in a positive floating-point value, error, and should then continue to throw darts at the dartboard (the square) until the resulting estimate of pi is less than error. As with the forPi function, this function should print The whilePi( error ) function, on the other hand, will take in a positive floating-point value, error, and should then continue to throw darts at the dartboard (the square) until the resulting estimate of pi is less than error. As with the forPi function, this function should print the number of darts thrown so far the number of darts thrown so far the number of darts thrown so far that have hit the circle the number of darts thrown so far that have hit the circle the resulting estimate of pi the resulting estimate of pi def whilePi(error): #initiate variables while (difference between your pi estimate and pi is greater than error): #find x and y between -1 and 1 #calculate if x & y are in circle # if in circle you’ve got a hit! #print the num of throws, num. of hits, estimate of pi

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Problem #3 Hints… Hints… The distance equation can be useful… The distance equation can be useful… Remember that random.uniform( -1.0, 1.0 ) Remember that random.uniform( -1.0, 1.0 ) returns a floating pt. number between -1 & 1

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Extra Credit: Problem #4 Part a This problem investigates the "Read-it-and- weep" sequence: The sequence starts like this: This problem investigates the "Read-it-and- weep" sequence: The sequence starts like this: 1, 11, 21, 1211, 111221, 312211, 13112221,... This sequence is called the Look and Say sequence, or sometimes, the "Read it and Weep" sequence. (Read it out loud and you will understand why.) Look and SayLook and Say

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Problem #4 Part a Write a helper function readOne( s ) that takes in a string of digits s and returns the string that represents the "reading" of s. For example: Write a helper function readOne( s ) that takes in a string of digits s and returns the string that represents the "reading" of s. For example: >>> readOne( '1' ) '11' >>> readOne( '11' ) '21' >>> readOne( '21' ) '1211' >>> readOne( '1211' ) '111221' A special case to consider for readOne is the case when the input, s, has length 1. A special case to consider for readOne is the case when the input, s, has length 1.

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Problem #4 Part a Write a function called Weep(n) that generates and prints the first n terms in the sequence. >>> Weep(10) 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 '11131221133112132113212221'

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Problem 4 Part b The harmonic series is defined as: The harmonic series is defined as: 1/1 + 1/2 + 1/3 + 1/4 +... 1/1 + 1/2 + 1/3 + 1/4 +... This series can be shown to diverge; that is, the series goes to infinity. However, this series diverges very slowly. Write a function called This series can be shown to diverge; that is, the series goes to infinity. However, this series diverges very slowly. Write a function called harmonicN( numToReach ) that returns the least number of terms that must be summed in order to reach or exceed numToReach. harmonicN( numToReach ) that returns the least number of terms that must be summed in order to reach or exceed numToReach.

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Problem 4 Part b While this series goes to infinity, python will have trouble as the terms get very small. To handle very small numbers, you need to use a special representation that is not built into Python but available in a package called decimal. from decimal import * getcontext().prec = 20 Decimal(1) / Decimal(3) Decimal("0.33333333333333333333")

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Problem 4 Part c While the harmonic series diverges, a very slight modification to this series turns it into one that converges! For any digit d between 0 and 9, if you remove those terms that contain the digit d anywhere in the denominator, then this series converges. Example, w/ digit as 2… if we remove the terms 1/2, 1/12, 1/20, 1/21, 1/22, 1/23,..., then this series will converge! While this may seem counter-intuitive, note that as the numbers in the denominator get large, the denominators without d become quite sparse, and the series eventually converges (although very slowly).

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Problem 4 Part c Write a function called withoutd( d, Numterms ) that evaluates this new series with respect to digit d by adding up the first Numterms terms of the series and returning this value. Note that Numterms does not include the terms that were "thrown out" - it is the actual number of terms that were added. Here are a few things that will be useful to you: A number of type Decimal can be cast into a string. For example: >>> x = 1/Decimal(3) >>> x Decimal("0.33333333333333333333") >>> str(x) '0.33333333333333333333' You can determine how many times a particular symbol occurs in a string using the built- in count method. It works like this: If s is some string then s.count("5") will return the number of times that the symbol 5 occurs in string s. Notice that the argument to count is a string itself! For example, "foo".count("o") will return 2.

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