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[SOLVED] Cmpt 280 assignment 5 1 your tasks question 1 (33 points): in lib280-asn6 you are provided with a fully functional 2-3 tree class

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In lib280-asn6 you are provided with a fully functional 2-3 tree class called TwoThreeTree280.
Recall that 2-3 trees are keyed dictionaries. As such, the TwoThreeTree280 class implements the
KeyedBasicDict280 interface. This interface adds the methods obtain(k), delete(k) and has(k),
and set(x) (replace the item whose key matches they key of x with the item x).
Presently, TwoThreeTree280 does not implement KeyedDict280 which adds additional operations
including all of the methods in KeyedLinearIterator280 which, in turn, includes all of the public operations on a cursor. Note that KeyedDict280 is the same interface that is implemented by
KeyedChainedHashTable280 so you should be somewhat familiar with it from the previous assignment.
The task for this question is to extend the TwoThreeTree280 to a class called IterableTwoThreeTree280
which allows linear iteration over the keyed data items stored in the two-three tree in ascending keyorder. We will achieve this by adding additional references to leaf nodes so that the leaf nodes form
a bi-linked list. Note that adding this feature to a 2-3 tree results in exactly a B+ tree of order 3
(see textbook Section 17.1). We aren’t going to call it a B+ tree class though, because we are implementing specifically a B+ tree of order 3, and higher-order B+ trees will not be supported. Our
IterableTwoThreeTree280 class will be exactly a B+ tree of order 3.
Figure 1 in the Appendix shows the differences between a 2-3 tree (without iteration) and a B+ tree
of order 3 containing the same elements, with the linking of the leaf nodes to support iteration. The
algorithms for insertion and deletion are the same in both kinds of tree, except that in the case of the
B+ tree, references to/from the predecessor and successor leaf nodes in key-order have to be adjusted
to maintain the bi-linked list of leaf nodes.
The full class hierarchy of IterableTwoThreeTree280 is shown in Figure 2 of the Appendix. The
hierarchy of tree node classes is shown in Figure ?? of the Appendix.
To implement the IterableTwoThreeTree280, the following tasks must be carried out:
1. Make an extension of LeafTwoThreeNode280 that adds references to its predecessor and successor
leaf nodes. This has already been done for you in the class LinkedLeafTwoThreeNode280.
2. Override the TwoThreeTree280::createNewLeafNode() method by adding a new protected method
in IterableTwoThreeTree280 that it returns a new LinkedLeafTwoThreeNode280 object instead
of a TwoThreeNode280 object. This has already been done for you.
3. In IterableTwoThreeTree280, override the insert and delete methods of TwoThreeTree280
with modified versions that correctly maintain the additional predecessor and successor references in the LinkedLeafTwoThreeNode280. Each leaf node should always point to the the leaf
node immediately to the left of it (the predecessor) and to the right of it (the successor) even if
they are not siblings in the tree. Of course, the leaf node with the smallest key has no predecessor
and the leaf node with the largest key has no successor.
In IterableTwoThreeTree280, the insert and delete methods from TwoThreeTree280 already
have been copied, and TODO comments have been inserted indicating where you need to add
additional code to maintain the additional leaf node references. The comments also provide a
few hints. You should not have to modify any of the existing code for insert or delete, just
add new code to deal with the linking and unlinking of leaf nodes from their successors and
predecessors. Maintaining these links is very similar to inserting and removing nodes from the
middle of a doubly-linked list.
4. Implement the additional methods required by KeyedDict280 (and, by extension, KeyedLinearIterator280).
Some of these have been done for you, others have not. TODO comments in IterableTwoThreeTree280
indicate which methods you need to implement and maybe even a hint or two. In this class, thelinear iterator interface allows positioning of the cursor along the leaf-level of the tree. The cursor
can never be positioned at an internal node.5. In the main() function, write a regression test to test the methods required by KeyedDict280
(and, by extension, KeyedLinearIterator280). You to not need to explicitly test the insertion
and deletion methods since testing of the methods from KeyedLinearIterator280 will reveal
any problems with the new leaf node linkages. This is because you will need to insert and delete
items to create test cases for those methods in KeyedLinearIterator280
You must test all of the methods listed in the interfaces that are coloured blue in Figure 2 of
the Appendix.Use instances of the local class called Loot (which has been defined in the main() method) as the
data items to insert into the tree for testing. This class implements the type of item depicted in
Figure 1 in the Appendix consisting of the name of a magic item from a fantasy game, and its
value in gold pieces. The item keys are the item names (strings).Hint: The toStringByLevel() method you’ve been given prints not only the 2-3 tree’s structure, but also
displays current linear ordering of the nodes that results from following the successor links in the leaf nodes,
beginning with the leftmost leaf node. This may be helpful for the debugging of step 2.For this question you will be implementing a k-D tree. We begin with introducing some algorithms
that you will need. Then we will present what you must do.As we saw in class, in order to build a k-D tree we need to be able to find the median of a set of
elements efficiently. The “j-th smallest element” algorithm will do this for us. If we have an array of n
elements, then finding the n/2-smallest element is the same as finding the median.
Below is a version of the j-th smallest element algorithm that operates on a subarray of an array
specified by offsets left and right (inclusive). It places at offset j (where left ≤ j ≤ right) the element
that belongs at offset j if the subarray were sorted. Moreover, all of the elements in the subarray
smaller than that belonging at offset j are placed between offsets left and j − 1 and all of the elements
in the subarray larger than that element are placed between offsets j + 1 and right, but there is no
guarantee on the ordering of any of these elements! The only element guaranteed to be in its sorted
position is the one that belongs at offset j. Thus, if we want to find the median element of a subarray
of the array list bounded by offsets left and right, we can call
jSmallest(list, left, right, (left+right)/2)The offset (left + right)/2 (integer division!) is always the element in the middle of the subarray
between offsets left and right because the average of two numbers is always equal to the number
halfway in between them.
The j-smallest algorithm is presented in its entirety on the next page.
Page 3
✞ ☎
Algorithm jSmallest ( list , left , right , j )
list – array of comparable elements
left – offset of start of subarray for which we want the median element
right – offset of end of subarray for which we want the median element
j – we want to find the element that belongs at array index j
To find the median of the subarray between array indices ’left ’ and ’right ’,
pass in j = ( right + left )/2.Precondition : left <= j <= right
Precondition : all elements in ’list ’ are unique ( things get messy otherwise !)
Postcondition : the element x that belongs at offset j , if the subarray were
sorted , is at offset j . Elements in the subarray
smaller than x are to the left of offset j and the
elements in the subarray larger than x are to the right
of offset j .
if ( right > left )
// Partition the subarray using the last element , list [ right ] , as a pivot .
// The index of the pivot after partitioning is returned .
// This is exactly the same partition algorithm used by quicksort .
pivotIndex := partition ( list , left , right )
// If the pivotIndex is equal to j, then we found the j-th smallest
// element and it is in the right place ! Yay!
// If the position j is smaller than the pivot index , we know that
// the j-th smallest element must be between left , and pivotIndex -1 , so
// recursively look for the j-th smallest element in that subarray :
if j < pivotIndex
jSmallest ( list , left , pivotIndex -1 , j )
// Otherwise , the position j must be larger than the pivotIndex ,
// so the j-th smallest element must be between pivotIndex +1 and right .
else if j > pivotIndex
jSmallest ( list , pivotIndex +1 , right , j )
// Otherwise , the pivot ended up at list [j] , and the pivot *is* the
// j-th smallest element and we ’re done .
✝ ✆Notice that there is nothing returned by jSmallest, rather, it is the postcondition that is important.
The postcondition is simply that the element of the subarray specified by left and right that belongs
at index j if the subarray were sorted is placed at index j and that elements between le f t and j − 1
are smaller than the j-th smallest element and the elements between j + 1 and right are larger than
the j-th smallest element.There are no guarantees on ordering of the elements within these parts of
the subarray except that they are smaller and larger than the the element at index j, respectively. This
means that if you invoke this algorithm with j = (right + left)/2 then you will end up with the median element
in the median position of the subarray, all smaller elements to its left (though unordered) and all larger elements
to its right (though unordered), which is just what you need to implement the tree-building algorithm!
NOTE: for this algorithm to work on arrays of NDPoint280 objects you will need an additional parameter d that specifies which dimension (coordinate) of the points is to be used to compare points.An advantage of making this algorithm operate on subarrays is that you can use it to build the k-d
tree without using any additional storage — your input is just one array of NDPoint280 objects and
you can do all the work without any additional arrays — just work with the correct subarrays.You may have noticed that jSmallest uses the partition algorithm partition the elements of the
subarray using a pivot. The pseudocode for the partition algorithm used by the jSmallest algorithm
is given below. Note that in your implementation, you will, again, need to add a parameter d to denote
which dimension of the n-dimensional points should be used for comparison of NDPoint280 objects.
✞ ☎// Partition a subarray using its last element as a pivot .
Algorithm partition ( list , left , right )
list – array of comparable elements
left – lower limit on subarray to be partitioned
right – upper limit on subarray to be partitioned
Precondition : all elements in ’list ’ are unique ( things get messy otherwise !)
Postcondition : all elements smaller than the pivot appear in the leftmost
part of the subarray , then the pivot element , followed by
the elements larger than the pivot . There is no guarantee
about the ordering of the elements before and after the
pivot .
returns the offset at which the pivot element ended up
pivot = list [ right ]
swapOffset = left
for i = left to right -1
if ( list [ i ] <= pivot )
swap list [ i ] and list [ swapOffset ]
swapOffset = swapOffset + 1
swap list [ right ] and list [ swapOffset ]
return swapOffset ; // return the offset where the pivot ended up
✝ ✆An algorithm for building a k-d tree from a set of k-dimensional points is given below. It is slightly
more detailed than the version given in the lecture slides. It uses the jSmallest algorithm presented
above.
✞ ☎
Algorithm kdtree ( pointArray , left , right , int depth )
pointArray – array of k – dimensional points
left – offset of start of subarray from which to build a kd – tree
right – offset of end of subarray from which to build a kd – tree
depth – the current depth in the partially built tree – note that the root
of a tree has depth 0 and the $k$ dimensions of the points
are numbered 0 through k -1.
if the specified subarray of pointArray is empty
return null ;
else
// Select axis based on depth so that axis cycles through all
// valid values . (k is the dimensionality of the tree )
d = depth mod k ;
medianOffset = ( left + right )/2
// Put the median element in the correct position
// This call assumes you have added the dimension d parameter
Page 5
// to jSmallest as described earlier .
jSmallest ( pointArray , left , right , d , medianOffset )
// Create node and construct subtrees
node = a new kD – tree node
node . item = pointArray [ medianOffset ]
node . leftChild = kdtree ( pointArray , left , medianOffet -1 , depth +1);
node . rightChild = kdtree ( pointArray medianOffset +1 , right , depth +1);
return node ;
✝ ✆Implement a k-D tree. You must use the NDPoint280 class provided in the lib280.base package of
lib280-asn6 to represent your k-dimensional points. You must design and implement both a node
class (KDNode280.java) and a tree class (KDTree280.java). Other than specific instructions given in
this question, the design of these classes is up to you. You may use as much or as little of lib280 as you
think is appropriate. You’ll be graded in the actual appropriateness of your choices. You should aim
to make the class fit into lib280 and its hierarchy of data structures, but you should not force things
by extending classes inappropriately. You may use whatever private/protected methods you deem
necessary.
A portion of the marks for this question will be awarded for the design/modularity/style of the implementation of your class. A portion of the marks for this question will be awarded for acceptable
inline and javadoc commenting.
Your k-D tree ADT must support the following operations:
• Construct a new (balanced) k-D tree from a set of k-dimensional points (it must work for any
k > 0).
• Perform a range search: given a pair of points (a1, a2, . . . ak
) and (b1, b2, . . . , bk
), ai <= bi
for all
i = 1 . . . k, return all of the points (c1, c2, . . . , ck
) such that a1 ≤ c1 ≤ b1, a2 ≤ c2 ≤ b2, . . . , ak ≤
ck ≤ bk
.
Note: your tree does not have to have operations that insert or remove individual NDPoints.
In addition, you should write a test program that demonstrates the correctness of your tree. The test
program should consist of two parts:
1. Show that your class can correctly build a k-D tree from a set of points. For k=2, display the set
of k-dimensional points that you used as input (use between 8 and 12 elements), followed by a
graphical representation of the built tree (similar to the toStringByLevel() output in the trees
we’ve done previously). Do this again for one other value of k, between 3 and 5 (your choice).
2. For the second of the two trees you displayed in part 1, perform at least three range searches. For
each search, display the query range, execute the range search, and then display the list of points
in the tree that were found to be in range. A sample test program output is given below.In order to implement the tree-building algorithm kdtree (shown in pseudocode, above) you first
need to implement jSmallest which, in turn requires partition. It is strongly suggested that you
implement and thoroughly test partition before trying to implement jSmallest. In turn, throughlytest jSmallest before you implement kdtree. If you don’t do this, I can tell you from experience that it
will be a nightmare to debug. You need to be sure that each algorithm is correct before implementing
the algorithms that depend on it, otherwise, if you run into a bug it will be very hard to determine
in which method in the chain of dependent methods the bug is occurring. This is a fundamental
principle that is crucial to designing complex software systems. Make sure each piece is correct before
relying on it later.Keep in mind that the algorithms presented above as abstract pseudocode do not necessarily translate
line-for-line into Java code. Likely much of the pseudocode you’ve seen up to this point *does*
translate in a line-by-line fashion, but you need to get used to pseudocode that doesn’t, as this is the
entire point of using pseudocode, such that it is language independent and omits implementation
details while still conveying what the algorithm is supposed to do.✞ ☎
Input 2 D points :
(5.0 , 2.0)
(9.0 , 10.0)
(11.0 , 1.0)
(4.0 , 3.0)
(2.0 , 12.0)
(3.0 , 7.0)
(1.0 , 5.0)
The 2 D lib280 . tree built from these points is :
3: (9.0 , 10.0)
2: (5.0 , 2.0)
3: (11.0 , 1.0)
1: (4.0 , 3.0)
3: (2.0 , 12.0)
2: (3.0 , 7.0)
3: (1.0 , 5.0)
Input 3 D points :
(1.0 , 12.0 , 0.0)
(18.0 , 1.0 , 2.0)
(2.0 , 13.0 , 16.0)
(7.0 , 3.0 , 3.0)
(3.0 , 7.0 , 5.0)
(16.0 , 4.0 , 4.0)
(4.0 , 6.0 , 1.0)
(5.0 , 5.0 , 17.0)
4: (5.0 , 5.0 , 17.0)
3: (16.0 , 4.0 , 4.0)
4: –
2: (7.0 , 3.0 , 3.0)
3: (18.0 , 1.0 , 2.0)
1: (4.0 , 6.0 , 1.0)
3: (2.0 , 13.0 , 16.0)
2: (1.0 , 12.0 , 0.0)
3: (3.0 , 7.0 , 5.0)
Page 7
Looking for points between (0.0 , 1.0 , 0.0) and (4.0 , 6.0 , 3.0).
Found :
(4.0 , 6.0 , 1.0)
Looking for points between (0.0 , 1.0 , 0.0) and (8.0 , 7.0 , 4.0).
Found :
(7.0 , 3.0 , 3.0)
(4.0 , 6.0 , 1.0)
Looking for points between (0.0 , 1.0 , 0.0) and (17.0 , 9.0 , 10.0).
Found :
(16.0 , 4.0 , 4.0)
(7.0 , 3.0 , 3.0)
(3.0 , 7.0 , 5.0)
(4.0 , 6.0 , 1.0)
✝ ✆lib280-asn6: A copy of lib280 which includes:
• TheTwoThreeTree280 class and related node and position classes in the lib280.tree package
for Question 1.
• Partially completed IterableTwoThreeTree280 class in the in lib280.tree package for Question 1.
• the NDPoint280 class in the lib280.base package for representing n-dimensional points for
question 2;IterableTwoThreeTree280.java: Your completed tree for Question 1.
KDNode280.java: The node class for your k-D tree from Question 1.
KDTree280.java: Your k-D tree class for Question 1.
a7q1.txt/doc/pdf: The console output from your test program for question 1, cut and paste from the IntelliJ
console window.Appendix
Leather
Armor
Potion of
Healing
Vampiric
Blade
+1 Mace
2000
Blue Ioun
Stone
20000
Leather
Armor
10
Plate
Armor
350
Potion of
Healing
100
Vampiric
Blade
12000
Plate Armor Blue Ioun
Stone
Leather
Armor
Potion of
Healing
Vampiric
Blade Plate Armor Blue Ioun
Stone
+1 Mace
2000
Blue Ioun
Stone
20000
Leather
Armor
10
Plate
Armor
350
Potion of
Healing
100
Vampiric
Blade
12000Figure 1: Top: a 2-3 tree which does not support a linear iterator; Bottom: a B+ tree of order 3 containing
the same elements. Here the keys are strings (describing magical items in a fantasy game world) and
the keyed data items contain the item name and an integer (representing the value, in gold pieces, of the
object). Note that the trees are the same except for the extra linkages of the leaf nodes.IterableTwoThreeTree280
#smallest: LinkedLeafTwoThreeNode280<K,I>
#largest: LinkedLeafTwoThreeNode280<K,I>
#cursor: LinkedLeafTwoThreeNode280<K,I>
#prev: LinkedLeafTwoThreeNode280<K,I>
K,I
TwoThreeTree280
#rootNode: TwoThreeNode280<K,I>
+height
#createNewLeafNode
#createNewInternalNode(TwoThreeNode, K,
TwoThreeNode, K,
TwoThreeNode)
#find(K)
#giveLeft(TwoThreeNode, TwoThreeNode)
#giveRight(TwoThreeNode, TwoThreeNode)
#stealLeft(TwoThreeNode, TwoThreeNode)
#stealRight(TwoThreeNode, TwoThreeNode)
+toString
+toStringByLevel
K,I
≪interface≫
Container280
clear
isEmpty
isFull
≪interface≫
KeyedBasicDict280
delete(K)
has(K)
insert(I)
obtain(K)
set(I)
K,I
≪interface≫
KeyedDict280
deleteItem
search(K)
searchCeilingOf(K)
setItem(I)
K,I
≪interface≫
KeyedLinearIterator280
K,I
≪interface≫
CursorSaving280
currentPosition
goPosition(CursorPosition280)
≪interface≫
KeyedCursor280
itemKey
keyItemPair
K,I
≪interface≫
LinearIterator280
after
before
goAfter
goBefore
goFirst
goForth
I
≪interface≫
Cursor280
item
itemExists
I
Figure 2: Class hierarchy for IterableTwoThreeNode280. For methods, only type names of parameters are shown.

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[SOLVED] Cmpt 280 assignment 5 1 your tasks question 1 (33 points): in lib280-asn6 you are provided with a fully functional 2-3 tree class
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