In the four queens problem, we ended up finding this solution:
. . Q .
Q . . .
. . . Q
. Q . .
In lecture, we produced this highly detailed version of the trace:
explore col 1
row 1
place
explore col 2
row 1--dead end (not safe)
row 2--dead end (not safe)
row 3
place
explore col 3
row 1--dead end (not safe)
row 2--dead end (not safe)
row 3--dead end (not safe)
row 4--dead end (not safe)
remove
row 4
place
explore col 3
row 1--dead end (not safe)
row 2
place
explore col 4
row 1--dead end (not safe)
row 2--dead end (not safe)
row 3--dead end (not safe)
row 4--dead end (not safe)
remove
row 3--dead end (not safe)
row 4--dead end (not safe)
remove
remove
row 2
place
explore col 2
row 1--dead end (not safe)
row 2--dead end (not safe)
row 3--dead end (not safe)
row 4
place
explore col 3
row 1
place
explore col 4
row 1--dead end (not safe)
row 2--dead end (not safe)
row 3
place
explore col 5
print (this is a solution)
. . Q .
Q . . .
. . . Q
. Q . .
remove
row 4--dead end (not safe)
remove
row 2--dead end (not safe)
row 3--dead end (not safe)
row 4--dead end (not safe)
remove
remove
row 3
place
explore col 2
row 1
place
explore col 3
row 1--dead end (not safe)
row 2--dead end (not safe)
row 3--dead end (not safe)
row 4
place
explore col 4
row 1--dead end (not safe)
row 2
place
explore col 5
print (this is a solution)
. Q . .
. . . Q
Q . . .
. . Q .
remove
row 3--dead end (not safe)
row 4--dead end (not safe)
remove
remove
row 2--dead end (not safe)
row 3--dead end (not safe)
row 4--dead end (not safe)
remove
row 4
place
explore col 2
row 1
place
explore col 3
row 1--dead end (not safe)
row 2--dead end (not safe)
row 3
place
explore col 4
row 1--dead end (not safe)
row 2--dead end (not safe)
row 3--dead end (not safe)
row 4--dead end (not safe)
remove
row 4--dead end (not safe)
remove
row 2
place
explore col 3
row 1--dead end (not safe)
row 2--dead end (not safe)
row 3--dead end (not safe)
row 4--dead end (not safe)
remove
row 3--dead end (not safe)
row 4--dead end (not safe)
remove
I mentioned that this trace along with a decision tree are included as
handout #18.Then I discussed the next programming assignment. The program searches for all ways to form anagrams of a particular phrase. For example, if the phrase is "george bush", then some combinations of words that constitute anagrams are [bee, go, shrug] or [bugs, go, here] or [go, he, bus, erg]. An anagram is a rearrangement of the letters. Each of these word combinations has the same set of letters as "george bush". The program uses a dictionary of words to find all combinations that are anagrams of the given phrase.
In solving this problem, one of the questions your program will have to consider is whether a particular word is relevant to the problem you're trying to solve. For example, if you were trying to find anagrams of "george bush" and you were wondering whether to consider the word "abash", the answer would be no. That word is not relevant to the problem of finding anagrams of "george bush" because there are two a's in "abash" but no a's in "george bush". So that's not a word that matters to us. But the word "bee" is relevant All of the letters of "bee" appear in "george bush". So that means that we need to explore this possibility. We do that by taking the letters from "bee" away from the letters for "george bush". That leads to a new set of letters that we could use to continue the process.
So how does this relate to backtracking? The potential solution space is the set of all combinations of words from the dictionary. We can think of this as a decision tree by thinking in terms of picking a first word, then picking a second word, then picking a third word, and so on.
To understand this better, I went through a small example in detail. We made a very short dictionary file that had just four words:
bee
go
gush
shrug
When we ran the program and asked for anagrams of "George Bush", the program
reported these answers:
[bee, go, shrug]
[bee, shrug, go]
[go, bee, shrug]
[go, shrug, bee]
[shrug, bee, go]
[shrug, go, bee]
So at the top of our decision tree, the choice we are making is what word
should come first and the various possibilities are the dictionary of words
we've been given:
first word?
|
+-------+---+---+-------+
| | | |
"bee" "go" "gush" "shrug"
We will only follow paths that make sense, but with this short dictionary,
we're going to try each of these words as a possible first word. So we first
explore choosing "bee". Once we've chosen "bee" as the first word, at the next
level of the tree, we consider all possible choices for a second word:
first word?
|
+-------+---+---+-------+
| | | |
"bee" "go" "gush" "shrug"
|
second word?
|
+-------+---+---+-------+
| | | |
"bee" "go" "gush" "shrug"
At each level, we go through all the words. Remember that backtracking
involves an exhaustive search, so you just try all possibilities that make
sense. Of course, at this second level, we wouldn't be dealing with the same
set of letters as before. At the first level we were looking for words that
could be used to form the phrase "george bush". At the second level, we've
already accounted for the letters in the first word ("bee"), so now we are
searching for something that matches what's left ("gorg ush").I mentioned that the low-level details for this problem involve keeping track of how many of each letter we have, subtracting the letters for a word from the letters for the phrase, figuring out whether that subtraction is going to work, and so on. All of these details are handled by the LetterInventory class that we wrote for assignment #1. The inventory objects keep track of how many of each letter you have and the "subtract" method sees whether you can subtract one set of letters from another set of letters.
So let's revisit some of this in terms of LetterInventory objects. Imagine making a LetterInventory object for "george bush". If we printed it, we'd get as output [beegghorsu]. When we subtract the inventory for the word "bee", we get [gghorsu]. As we explore various words we could subtract from this, we find that we can successfully subtract the inventory for "go" to get an inventory that prints as [ghrsu]. And from that we can subtract the inventory for "shrug" to get an inventory that prints as [] (in other words, an empty inventory).
For the anagram problem, finding an empty LetterInventory is like getting to column 9 in the 8 queens problem. It means we've found a solution. It means that in our various explorations, we've found a sequence of words that we can successfully subtract from the original to get down to an empty inventory. That means we've accounted for every letter of the original with the current combination of words that we're exploring, which means this is a solution that we'd want to report.
But we also encounter dead ends along the way. For example, when we start with [beegghorsu] and subtract "bee" to get [gghorsu], we end up with something that can't have a second occurrence of "bee". We'd find that when we try to subtract the inventory for "bee" a second time that it fails (subtract returns a null). That would represent a dead end, just as in the 8 queens problem when we would find that we couldn't place a queen on at a particular spot. Dead ends aren't a problem in backtrack. We just skip them and move on.
That's how the anagram problem can be solved with backtracking. Each level of the decision tree (which means each invocation of the recursive method) involves choosing one word. Which choices do we pursue? Only those where we can successfully subtract the word's inventory from the current inventory. And as we proceed to lower levels in the tree, we use the new inventories that the subtract method gives us as the problem to solve at that level. In other words, the overall problem involves a phrase like "george bush", but as we make specific choices, that leads to new smaller problems that involve a smaller inventory of letters because as we choose words, we account for more and more of the letters of the original phrase. And whenever we get to an empty inventory in our recursive backtracking, we know we've found a solution that should be printed.
I mentioned that in some sense this can be the easiest assignment of the quarter. The solution is short (mine was 45 lines long) and we're being very specific about how to approach it. Recursive backtracking is a very specific technique and we're telling you exactly how to apply it to this problem. So you could, theoretically, be done in an hour. Most likely it will take a lot longer because this is your first exposure to backtracking and it will take you a while to figure out how all of this works.
The short example we explored in lecture using a 4-word dictionary is helpful to study both to understand what the program is supposed to do and to test your own solution. It is small enough that you can debug your program step by step to make sure that it is making the right choices at each level of the recursion. But even this short example leads to a large solution space to consider. For those who want to explore this, I have posted a detailed trace of execution that shows step by step what it explores. In this trace, instead of using strings like "george bush" and "gorg ush", I have instead listed what the corresponding LetterInventory objects would return when you call their toString method ([beegghorsu] and [gghorsu]). So if you include println statements in your program, you can compare what your program is doing against this detailed trace to make sure it is working properly.
I have also posted a detailed diagram of the solution space that is explored. Solutions are listed in red inside a blue oval. The diagram is so large that when it is initialy displayed, you won't be able to read it. Click on it and it will expand to a size that is readable. The trace is more likely to be useful than the diagram, but the diagram might help you to understand what the backtracking is doing.
The writeup points out two optimizations that I am requiring you to include. First, it's clear that we will often be needing the LetterInventory for a particular word. There is no reason to do that more than once for any given word, so I'm asking you to "preprocess" the dictionary to compute a LetterInventory for each word. How would we store these results? For each word we end up with a LetterInventory and we want to be able to quickly access the LetterInventory for any particular word. This is a perfect application of a Map. Remember that a Map is composed of key/value pairs, associating each key with a specific value. Here the keys are the words and the values are the LetterInventory objects.
Another optimization I've asked you to perform is to prune the dictionary before you begin the recursive backtracking. In the examples above I was working with the entire dictionary for the backtracking, but you can cut this down considerably by going through the dictionary before you begin the recursion and picking out just the words that are relevant. For example, if we're working with the phrase "george bush", we'd throw out words like "abash" and "aura" right away because they could never be part of a solution.
Then I started a new topic called binary trees. I mentioned that they turn out to be very helpful in solving many different problems. There is an elegance and power that you get with binary trees that makes them extremely useful.
I started with some basic terminology. I drew a crude picture of a tree and pointed out that we refer to the base of the tree as its root and we talk about branches of the tree with leaves on the branches. Computer scientists view the world upside down, so we draw a tree with the root a the top and the leaves at the bottom. For example, we might draw this tree:
12
/ \
/ \
18 7
/ \
/ \
4 13
Just as with linked lists, we refer to each different value as a node. At the
top of this tree we have the root node storing the value 12. We would refer to
the nodes with 12 and 7 as branch nodes of the tree because they have values
stored under them. The nodes with values 18, 4 and 13 are leaf nodes of the
tree because each one has nothing under it.Another set of terms we use is parent and child. The root node is the ultimate ancestor of every other node. It is the parent of the nodes 18 and 7. Similarly, we'd say that the parents of 18 and 7 are the root node 12. We also sometimes refer to 18 and 7 as siblings.
I then gave a recursive definition of a tree. I said that a tree is either:
+-----------+
| root node |
+-----------+
/ \
/ \
/ \
/\ /\
/ \ / \
/ \ / \
/ left \ / right\
/ subtree\ / subtree\
+----------+ +----------+
That will be a useful way to think about trees as we write binary tree code.
Using our recursive definition, we discussed how you could form various kinds
of trees. The simplest kind of tree is an empty tree, which can't really be
drawn because it's empty.Once you have an empty tree, you can use the second part of the definition to make a new kind of tree that is composed of a root node with left and right subtrees that are both empty. In other words, that would be a single leaf node, which we could represent with a star:
*
Now that we have this as an option, we can use our recursive rule to say that a
tree could be a root node with a left tree that is empty and a right tree that
is a leaf:
*
\
*
Or we can have an empty right and a leaf to the left:
*
/
*
Or we could have leaf on either side:
*
/ \
* *
These now become possibilities to use for our recursive definition, allowing us
to construct even more kinds of trees, as in:
* * * *
/ \ / \ / \ / \
* * * * * * * *
/ \ / \ / \ \ / \ \
* * * * * * * * * *
Then I showed people the node class we'll be using for a simple binary tree of
ints:
public class IntTreeNode {
public int data;
public IntTreeNode left;
public IntTreeNode right;
public IntTreeNode(int data) {
this(data, null, null);
}
public IntTreeNode(int data, IntTreeNode left, IntTreeNode right) {
this.data = data;
this.left = left;
this.right = right;
}
}
As with our linked list node, this node is very simple in that it has just some
public data fields and a few constructors. The node has a data field of type
int and two links of type IntTreeNode for the left and right subtrees. The
first constructor constructs a leaf node (using null for left and right). The
second constructor would be appropriate for a branch node where you want to
specify the left and right subtrees.This node class is "messy" in the sense that it is not well encapsulated. We did something similar with the linked list nodes and I made the analogy that they are like the cans of paint that a contractor might use in painting your house. We want a cleaner interface for dealing with a client, so I mentioned that we'll have a second object for storing a tree. Any external client will deal with the IntTree object and won't ever see these tree node objects.
We need only one data field in the tree class: a reference to the root of the tree:
public class IntTree {
private IntTreeNode overallRoot;
...
}
I pointed out that I'm purposely using the name "overallRoot" to distinguish
this root from all of the other roots. There is only one overall root. But
each subtree is itself the root of a tree and in our recursive methods we'll
often use the parameter name "root" as a way to indicate that it can be any of
the roots.I then spent time discussing the idea of tree traversals. The idea is to "traverse" the tree in such a way that you visit each node exactly once. There are many different ways to do this. We generally prefer recursive approaches, so we want to traverse the entire left subtree without dealing with anything from the right and in a separate operation, traverse the entire right subtree without dealing with anything from the left. That leads to the classic binary tree traversals. We have a Western bias that we traverse the left subtree before the right subtree. The question becomes, where do you deal with the root of the tree?
There are three possible answers you might give. You can process the root before you traverse either subtree, after you traverse both subtrees or in between traversing the two subtrees. These three approaches are known as preorder, inorder and postorder traversals.
+------------------+
+-----------<-------------+ process the root +------------->-----------+
| +--------+---------+ |
| | |
| | |
V +-----------------------+ V +------------------------+ V
pre | traverse left subtree | in | traverse right subtree | post
+-----------------------+ +------------------------+
For example, given the following tree:
+---+
| 2 |
+---+
/ \
/ \
+---+ +---+
| 0 | | 3 |
+---+ +---+
/ / \
/ / \
+---+ +---+ +---+
| 7 | | 1 | | 9 |
+---+ +---+ +---+
/ \ / \
/ \ / \
+---+ +---+ +---+ +---+
| 6 | | 5 | | 8 | | 4 |
+---+ +---+ +---+ +---+
The traversals would be as follows: