bigmoves / jsn-sum14-2 Goto Github PK

Homework #2

1) Simulating Arrays

[Difficulty: Easy]

Create an ordinary object which is not an actual Array but behaves (somewhat) like one. You may use a global variable array to store your pseudo-array. When creating array initially, have it represent an empty Array.

a) Give array a property named length which is always a non-negative integer, and three more properties pop, push, and join which are functions (i.e. methods) behaving exactly like the corresponding method of real Arrays.

You may use the enclosed template file to get started.

b) Assuming you've implemented array correctly, predict the outcome of the following:

array.length=0;
array.push('c');
array.push('b');
array.push('a');
array.pop();
console.log(array.join('a'));

c) Similarly, predict the outcome here:

array.length=0;
console.log(array.join(array.push(array.push('a'))));

2) Object Comparison

[Difficulty: Moderate]

a) Write a function copy(obj), which duplicates an object (not just copying a reference to it). You only need to duplicate only one level: if obj contains another object inner, the duplicate may share inner rather than copying it too.

Write another function to compare two objects: equal(objA,objB) should return true only when objA and objB have exactly the same properties with the same values. Note that two empty objects should be considered equal.

Write a third function: similar(objA,objB) should return true only when objA and objB have exactly the same properties, regardless of their values.

b) We can interpret objects as sets of properties, and merge those sets in various ways. Let's define three such merges:

Union: The union of objects A,B is a new object which contains all the properties found in either A or B. If a property is found in both, the merged property gets the shared key and the value (A[key] || B[key]). For example: the union of {a:1,b:0} and {a:0,c:0} is {a:1,b:0,c:0}.

Intersection: The intersection of objects A,B is a new object which contains only those properties found in BOTH A and B. The value of each intersecting property is (A[key] && B[key]). For example, the intersection of {a:1,b:0} and {a:0,c:0} is {a:0}.

Subtraction: The subtraction of B from A, aka "A minus B", is an object which contains all the properties of A which are NOT in B. Note that this merge is usually not symmetric: A minus B doesn't equal B minus A (except in one case, which you should identify!) For example, {a:1,b:0} minus {a:0,c:0} is {b:0}, and the reverse subtraction is {c:0}.

Using those definitions, implement a function for each:

• union(objA,objB)

• intersect(objA,objB)

• subtract(objA,objB)

Your functions may return undefined if either of their arguments is not an object.

c) Write five sample assertions to validate each of your three merging functions (15 total). Remember that when comparing your results to expected, you'll need to see if objects are equal() but not identical.

d) Finally: even if your functions implement perfectly the definitions above, intersection and union are still not symmetric. That is, similar(union(A,B),union(B,A)) will always be true, but equal(union(A,B),union(B,A)) may not be. Likewise with intersection. Explain!

3) Cards, improved!

[Difficulty: Easy]

Revisit your playing card functions from homework #1. Repackage them in a Toolkit pattern, as methods of a single master object. You may assign that object to any variable you like, but that variable should not appear in the definitions of your methods. You'll need to change both your method definitions and their calls to other methods, but their logic will remain the same.

You may adopt the enclosed template file. Make sure your code still passes all the assertions there!

4) Social network!

Assume a world in which no two people have the same name. Create an object people whose purpose is to remember everyone ever mentioned and the relationships between them.

[Difficulty: Moderate]

a) Write three methods for people:

• people.meet(nameA,nameB) should accept two names, update people, and return the total number of times those two have met, including this new meeting. If either person isn't yet represented in people, add them. Then increment a count of the meetings between them. Assume that the order of arguments doesn't matter (i.e. meet(A,B) is the same as meet(B,A)), and that meeting oneself (A==B) has no effect.

• people.haveMet(nameA,nameB) should return a number greater than 0 if those people have met, and some falseish value if they haven't or don't exist.

• people.friendsOf(name) should return a string listing the names of all people whom name has met at least once (or undefined if name doesn't exist). List the names in alphabetical order, and make sure each name appears only once.

You may use the enclosed template file to get started.

Hint: the people object should contain an index of all people, linking each name to an individual object for that person. Each such person-object should have two properties:

• name is a string for that person's name. (This redundant copy of the name isn't necessary for the solution, but it may help you debug.)

• friends is another index object, unique to each person, with multiple keys (one for each friend that person has met), each with a numeric value. Because meetings are symmetric (each person meets the other), each number is duplicated in a corresponding property in the friend's index; make sure you update both copies of the number during a meeting.

Here is a diagram showing the data structure after people is fully initialized but before any method calls:

Here is the data structure just after the first method call people.meet('Dan','Ben'):

[Difficulty: Hard]

b) Write another method people.friendsOfFriendsOf(name) which returns a string listing, in alphabetical order, all the names of people within two degrees of separation from name: they've met either name or at least one of name's friends. Your list may include name itself but no duplicates: any person should be listed only once regardless of the number of connections with name.

(Hint: the union of sets includes no duplicates! Perhaps you could recycle code from somewhere?)