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April 7, 2013 07:35
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scala coursera course week 2
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package funsets | |
import common._ | |
/** | |
* 2. Purely Functional Sets. | |
*/ | |
object FunSets { | |
/** | |
* We represent a set by its characteristic function, i.e. | |
* its `contains` predicate. | |
*/ | |
type Set = Int => Boolean | |
/** | |
* Indicates whether a set contains a given element. | |
*/ | |
def contains(s: Set, elem: Int): Boolean = s(elem) | |
/** | |
* Returns the set of the one given element. | |
*/ | |
def singletonSet(elem: Int): Set = | |
return (x: Int) => | |
x == elem | |
/** | |
* Returns the union of the two given sets, | |
* the sets of all elements that are in either `s` or `t`. | |
*/ | |
def union(s: Set, t: Set): Set = | |
return (x: Int) => | |
(contains(s, x) || contains(t, x)) | |
/** | |
* Returns the intersection of the two given sets, | |
* the set of all elements that are both in `s` and `t`. | |
*/ | |
def intersect(s: Set, t: Set): Set = | |
return (x: Int) => | |
(contains(s, x) && contains(t, x)) | |
/** | |
* Returns the difference of the two given sets, | |
* the set of all elements of `s` that are not in `t`. | |
*/ | |
def diff(s: Set, t: Set): Set = | |
return (x: Int) => | |
(contains(s, x) && !(contains(t, x))) | |
/** | |
* Returns the subset of `s` for which `p` holds. | |
*/ | |
def filter(s: Set, p: Int => Boolean): Set = | |
return (x: Int) => p(x) | |
/** | |
* The bounds for `forall` and `exists` are +/- 1000. | |
*/ | |
val bound = 1000 | |
/** | |
* Returns whether all bounded integers within `s` satisfy `p`. | |
*/ | |
def forall(s: Set, p: Int => Boolean): Boolean = { | |
def iter(a: Int): Boolean = { | |
if (a > bound) true | |
else if (contains(s, a) && !p(a)) false | |
else iter(a+1) | |
} | |
iter(-bound) | |
} | |
/** | |
* Returns whether there exists a bounded integer within `s` | |
* that satisfies `p`. | |
*/ | |
def exists(s: Set, p: Int => Boolean): Boolean = !forall(s, x => !p(x)) | |
/** | |
* Returns a set transformed by applying `f` to each element of `s`. | |
*/ | |
def map(s: Set, f: Int => Int): Set = { | |
def iterate(s:Set, accumulateSet: Set, a:Int): Set = { | |
if (a > bound) accumulateSet | |
else if (contains(s, a)) iterate(s, union(accumulateSet, x=>f(a) == x), a+1) | |
else iterate(s, accumulateSet, a+1) | |
} | |
iterate(s, x=>false, -3) | |
} | |
/** | |
* Displays the contents of a set | |
*/ | |
def toString(s: Set): String = { | |
val xs = for (i <- -bound to bound if contains(s, i)) yield i | |
xs.mkString("{", ",", "}") | |
} | |
/** | |
* Prints the contents of a set on the console. | |
*/ | |
def printSet(s: Set) { | |
println(toString(s)) | |
} | |
} |
def map(s: FunSet, f: Int => Int): FunSet = y => exists(s, x => y == f(x))
translates to when you get a y, check if there is a value x in my set, for which f(x)== y
there is a bug in this code
filter should be defined as :
def filter(s: FunSet, p: Int => Boolean): FunSet = (a: Int) => if (contains(s,a)) p(a) else false
otherwise it returns all elements within the bounds. the contains check is necessary, we cannot just blindly apply p(a)
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This failed my tests ...
is that an infinite loop? 🤔