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Assignment 1_ “Introduction To Haskell” and Recursion
Assignment 1_ “Introduction To Haskell” and Recursion
Problem 1: An “ellipsoid” is a three-dimensional shape that can roughly be thought of as a sphere that may have been “squashed” or elongated along one or more of three perpendicular axes. Here’s a diagram to illustrate:
This diagram was copied from http://www.web-formulas.com/Math_Formulas/Geometry_Volume_of_Ellipsoid.aspx and you may refer to this web site if you would like more information about ellipsoids.
If the lengths of all three of the axes are equal, the ellipsoid is called a sphere. For the purposes of this assignment, let’s define a “football” as an ellipsoid in which exactly two of the axis are equal.
Write a function called shape that takes three parameters, the lengths of the three axis of an ellipsoid, and returns a character describing the kind of ellipsoid it is: ‘S’ for a sphere, ‘F’ for a football, ‘E’ for an ellipsoid which is not a sphere or football, and ‘X’ if any of the parameters are equal to or less than zero. The parameters for this function should all be Doubles and the result should be a Char.
Example Run Using My Solution: *Assignment1 shape 3 3 3 'S' *Assignment1 shape 7.3 7.3 7.3 'S' *Assignment1 shape 2 5.1 5.1 'F' *Assignment1 shape 9 9 3 'F' *Assignment1 shape 3 4 3 'F' *Assignment1 shape 7 2 4 'E' *Assignment1 shape 3 5 9 'E' *Assignment1 shape 0 3 4 'X' *Assignment1 shape 5 (-2) 3 'X' *Assignment1 shape 7 6 (-5) 'X'
Note (referring to the last two test cases): In Haskell it is often necessary to put parenthesis around negative numbers. If you write shape 5 -2 3, Haskell will interpret this as shape (5-2) 3, which is not what you probably meant at all!
Problem 2: If the lengths of the three axes of an ellipsoid are a, b, and c (as in the diagram for Problem 1), the volume of the ellipsoid is 4/3 * * a * b * c. In the case of a sphere, where a and b and c are all equal to the same value r, this leads to the familiar formula 4/3 r3 for the volume of a sphere.
Write a Haskell function called volume which takes three parameters of type Double (the lengths of the three axes) and returns a result of type Double (the volume of an ellipsoid with those axes).
If any of the axes are negative, your function should abort with an error message instead of returning a result. If one or more of the axes are zero (but none are negative), your function should just return zero.
Note: There is a constant called pi built into Haskell. Please use it instead of typing a value of your own like 3.14 or 3.14159. The answers we test for will assume you’re using Haskell’s value for pi.
Example Run With My Solution:
*Assignment1 volume 1.1 2.2 3.3 33.45167857542412 *Assignment1 volume 8 7 7 1642.0057602762652 *Assignment1 volume 6 0 3 0.0 *Assignment1 volume (-1) 9 4 *** Exception: ellipsoid with negative side(s) CallStack (from HasCallStack): error, called at assn1Solution.hs:18:33 in main:Assignment1
Problem 3: Write a function called logSum that takes two Int parameters a and b. Its value must be the sums of the natural logarithms of all the numbers between a and b inclusive. In other words, the value must be log a + log (a+1) + log (a+2) + log (a+3) + … + log (b-1) + log b. If a == b, the value should be log a. If a b, the value should be zero (since there are no integers between a and b in this case). The result of logSum should be a Double.
To compute natural logarithms, use the Haskell log function.
In a programming language like Java or Python, you’d use a loop to add up all of these logs, but you can’t do that in Haskell. You’ll have to use recursion to get the same effect.
An additional challenge in Haskell is that it is very precise about types. This often is useful and prevents certain kinds of errors, but this is one of the times when it can be a bit of a pain. The Haskell log function expects its parameter to be a Double, and you’ll be getting Ints as parameters to your functions. If you pass an Int to the Haskell log function you’ll get a type error. The solution is to use the fromIntegral function to convert an Int to a Double. We’ll be talking about the details of the Haskell type system later; for now, just take my word for it that fromIntegral will help you out here.
Example Run With My Solution:
*Assignment1 logSum 1 4 3.1780538303479458 *Assignment1 logSum 5 20 39.15756263040554 *Assignment1 logSum 3 3 1.0986122886681098 *Assignment1 logSum 5 4 0.0
Problem 4: Astronauts have discovered a strange new kind of life form that grows in isolated spots on some other planets in the solar system. They’ve named these beings “ET”s and have brought some back to earth to study. Scientists have concluded that ETs have peculiar grown patterns that seem to obey the following rules in Earth gravity:
•
If the mass of an ET is less than 1 gram its mass will double in one day. For example, if an ET weighs 0.75 grams today it will weight 1.5 grams tomorrow.
•
If the mass of an ET is at least 1 gram but less than 20 grams its mass after one day will be 1.5 times the starting mass plus 2 grams. For example, an ET that weighs 2 grams today will weigh 5 grams tomorrow (1.5 *2 + 2)
•
If the mass of an ET is at least 20 grams but less than 100 grams its mass after one day will be 1.2 times its starting mass plus 1 gram. For example, an ET that weighs 50 grams today will weigh 61 grams tomorrow (50*1.2 + 1)
•
If the mass of an ET is 100 grams or more its mass will be 1.1 times its starting mass plus 0.5 grams. For example, an ET that weighs 200 grams today will weigh 220.5 grams tomorrow (200*1.1 + 0.5). The scientists aren’t very good at programming and have asked you to provide them with a Haskell function that will predict the growth of an ET in a given number of days. This function should be called growET and should take two parameters (in the order given): 1. The initial mass of the ET in grams (a Double) 2. The number of days the ET will be allowed to grow (an Int) It should return the predicted mass of the ET in grams (a Double) at the end of the growth period.
The function should abort with a descriptive error message instead of calculating a result if any of the following conditions are true:
•
The initial mass is less than zero
•
The initial mass is non-negative but the number of days is less than 0. (Zero days is not an error; the ET just doesn’t grow at all.)
•
The initial mass is non-negative but the number of days is more than 100. (This computation would not be useful because no ET has been observed to survive for more than 100 days.)
By “descriptive error message”, I mean an error message that tells the user what the problem is. For example, “initial mass less than zero” is descriptive. “ERROR” or “Error #3” are not descriptive.
Important Restriction: You must solve this problem using recursion. Don’t waste time looking for a way to do it with exponentiation or logarithm functions.
Problem 4 example run on the next page….
Example Run With My Solution: *Assignment1 growET 5 1 9.5 *Assignment1 growET 17 1 27.5 *Assignment1 growET 55 1 67.0 *Assignment1 growET 102 1 112.7 *Assignment1 growET 3 0 3.0 *Assignment1 growET 0.4 6 24.35 *Assignment1 growET 5 8 73.07104 *Assignment1 growET 82 4 146.5888 *Assignment1 growET 0.2 100 381629.6082808138
Problem 1: An “ellipsoid” is a three-dimensional shape that can roughly be thought of as a sphere that may have been “squashed” or elongated along one or more of three perpendicular axes. Here’s a diagram to illustrate:
This diagram was copied from http://www.web-formulas.com/Math_Formulas/Geometry_Volume_of_Ellipsoid.aspx and you may refer to this web site if you would like more information about ellipsoids.
If the lengths of all three of the axes are equal, the ellipsoid is called a sphere. For the purposes of this assignment, let’s define a “football” as an ellipsoid in which exactly two of the axis are equal.
Write a function called shape that takes three parameters, the lengths of the three axis of an ellipsoid, and returns a character describing the kind of ellipsoid it is: ‘S’ for a sphere, ‘F’ for a football, ‘E’ for an ellipsoid which is not a sphere or football, and ‘X’ if any of the parameters are equal to or less than zero. The parameters for this function should all be Doubles and the result should be a Char.
Example Run Using My Solution: *Assignment1 shape 3 3 3 'S' *Assignment1 shape 7.3 7.3 7.3 'S' *Assignment1 shape 2 5.1 5.1 'F' *Assignment1 shape 9 9 3 'F' *Assignment1 shape 3 4 3 'F' *Assignment1 shape 7 2 4 'E' *Assignment1 shape 3 5 9 'E' *Assignment1 shape 0 3 4 'X' *Assignment1 shape 5 (-2) 3 'X' *Assignment1 shape 7 6 (-5) 'X'
Note (referring to the last two test cases): In Haskell it is often necessary to put parenthesis around negative numbers. If you write shape 5 -2 3, Haskell will interpret this as shape (5-2) 3, which is not what you probably meant at all!
Problem 2: If the lengths of the three axes of an ellipsoid are a, b, and c (as in the diagram for Problem 1), the volume of the ellipsoid is 4/3 * * a * b * c. In the case of a sphere, where a and b and c are all equal to the same value r, this leads to the familiar formula 4/3 r3 for the volume of a sphere.
Write a Haskell function called volume which takes three parameters of type Double (the lengths of the three axes) and returns a result of type Double (the volume of an ellipsoid with those axes).
If any of the axes are negative, your function should abort with an error message instead of returning a result. If one or more of the axes are zero (but none are negative), your function should just return zero.
Note: There is a constant called pi built into Haskell. Please use it instead of typing a value of your own like 3.14 or 3.14159. The answers we test for will assume you’re using Haskell’s value for pi.
Example Run With My Solution:
*Assignment1 volume 1.1 2.2 3.3 33.45167857542412 *Assignment1 volume 8 7 7 1642.0057602762652 *Assignment1 volume 6 0 3 0.0 *Assignment1 volume (-1) 9 4 *** Exception: ellipsoid with negative side(s) CallStack (from HasCallStack): error, called at assn1Solution.hs:18:33 in main:Assignment1
Problem 3: Write a function called logSum that takes two Int parameters a and b. Its value must be the sums of the natural logarithms of all the numbers between a and b inclusive. In other words, the value must be log a + log (a+1) + log (a+2) + log (a+3) + … + log (b-1) + log b. If a == b, the value should be log a. If a b, the value should be zero (since there are no integers between a and b in this case). The result of logSum should be a Double.
To compute natural logarithms, use the Haskell log function.
In a programming language like Java or Python, you’d use a loop to add up all of these logs, but you can’t do that in Haskell. You’ll have to use recursion to get the same effect.
An additional challenge in Haskell is that it is very precise about types. This often is useful and prevents certain kinds of errors, but this is one of the times when it can be a bit of a pain. The Haskell log function expects its parameter to be a Double, and you’ll be getting Ints as parameters to your functions. If you pass an Int to the Haskell log function you’ll get a type error. The solution is to use the fromIntegral function to convert an Int to a Double. We’ll be talking about the details of the Haskell type system later; for now, just take my word for it that fromIntegral will help you out here.
Example Run With My Solution:
*Assignment1 logSum 1 4 3.1780538303479458 *Assignment1 logSum 5 20 39.15756263040554 *Assignment1 logSum 3 3 1.0986122886681098 *Assignment1 logSum 5 4 0.0
Problem 4: Astronauts have discovered a strange new kind of life form that grows in isolated spots on some other planets in the solar system. They’ve named these beings “ET”s and have brought some back to earth to study. Scientists have concluded that ETs have peculiar grown patterns that seem to obey the following rules in Earth gravity:
•
If the mass of an ET is less than 1 gram its mass will double in one day. For example, if an ET weighs 0.75 grams today it will weight 1.5 grams tomorrow.
•
If the mass of an ET is at least 1 gram but less than 20 grams its mass after one day will be 1.5 times the starting mass plus 2 grams. For example, an ET that weighs 2 grams today will weigh 5 grams tomorrow (1.5 *2 + 2)
•
If the mass of an ET is at least 20 grams but less than 100 grams its mass after one day will be 1.2 times its starting mass plus 1 gram. For example, an ET that weighs 50 grams today will weigh 61 grams tomorrow (50*1.2 + 1)
•
If the mass of an ET is 100 grams or more its mass will be 1.1 times its starting mass plus 0.5 grams. For example, an ET that weighs 200 grams today will weigh 220.5 grams tomorrow (200*1.1 + 0.5). The scientists aren’t very good at programming and have asked you to provide them with a Haskell function that will predict the growth of an ET in a given number of days. This function should be called growET and should take two parameters (in the order given): 1. The initial mass of the ET in grams (a Double) 2. The number of days the ET will be allowed to grow (an Int) It should return the predicted mass of the ET in grams (a Double) at the end of the growth period.
The function should abort with a descriptive error message instead of calculating a result if any of the following conditions are true:
•
The initial mass is less than zero
•
The initial mass is non-negative but the number of days is less than 0. (Zero days is not an error; the ET just doesn’t grow at all.)
•
The initial mass is non-negative but the number of days is more than 100. (This computation would not be useful because no ET has been observed to survive for more than 100 days.)
By “descriptive error message”, I mean an error message that tells the user what the problem is. For example, “initial mass less than zero” is descriptive. “ERROR” or “Error #3” are not descriptive.
Important Restriction: You must solve this problem using recursion. Don’t waste time looking for a way to do it with exponentiation or logarithm functions.
Problem 4 example run on the next page….
Example Run With My Solution: *Assignment1 growET 5 1 9.5 *Assignment1 growET 17 1 27.5 *Assignment1 growET 55 1 67.0 *Assignment1 growET 102 1 112.7 *Assignment1 growET 3 0 3.0 *Assignment1 growET 0.4 6 24.35 *Assignment1 growET 5 8 73.07104 *Assignment1 growET 82 4 146.5888 *Assignment1 growET 0.2 100 381629.6082808138
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