You may NOT ASSUME that "halt" will always translate to 0935 and "garble" will always translate to 5555. We will test your project 3 on a different "codebook.txt" than you have been using. In other words, if you HARDCODED these values, your project will fail our tests. The ONLY thing you can hardcode is if you encounter a word that you can not translate (it is not a plural or part of an expression) you may replace it in checkcodebook() in Encoder with the word "garble". Then when you look up "garble" in codebook.txt (your bst) it should always tranlate to the same 4-digit code (which is 5555 right now but will be changed for projects 3 and 4 tests.)

YES!

You should initialize "curr" or "child" to a COPY of the root and initialize "parent" to NULL since the root has no parent. In ModBst, you only pass in "curr" (a copy of the root) because each node already knows where its parent is located.

No! The staff discussed this and determined that since there were no metro shutdowns, no power outages, and no hurricanes, there is no extension.

Glad you asked. They are handled differently in each. Let's handle one at a time:

• Let's say this is the incoming message:
  
        I raise the falg each morning.
  
  
  
  • Each word except "falg" is found in codebook.
  • Once you are certain is is not part of an expression or a plural, you translate it to "garble"
  • This means that your original dll would look like
    
    i -> raise -> the -> falg -> each -> morning -> halt ->
  
  
  
  • After checkcodebook, your dll looks like
    
    i -> raise -> the -> garble -> each -> morning -> halt ->
  
  
  
  • When you encode this, each word translates to its 4-digit equivalent except "garble" which translates to "5555" (the current code for garble). Do not hardcode this as it may change in codebook.txt.
  
  
• Now let's say that you have a message to decode and it looks like this:
  
        5922 6201 9705 7117 5555 3290 0935
  
  
  • Most of this can be translated:
    
    go up above the garble ???? halt
  
  
  
  • So what do we do with this in Decoder? Task 4 says to translate to a garble, but I think it is better to throw an exception.
  • It turns out that this is also a garble but since garble is 5555, we can either translate to garble again or throw an exception. For project 3, either is fine. We will make it more specific for project 4.
  • If you tranlate to "garble" your printout would be:
    
    go up above the garble garble.
  
  
  
  • If you throw an exception your printout would look like:
    
    go up above the garble
    GarbledCode() exception thrown

No! You can call locate() (which is recursive) to find where the node and its parent are, then determine which case you have -deleting a leaf, deleting a node with a right subtree, deleting a node with a left subtree, or deleting a node with 2 subtrees. By the way, if you start out with the node to be deleted has 2 subtrees, with right successor or left predecessor, you then will end up deleting only a leaf or a node with one subtree.

As long as either the function itself is recursive or it calls a recursive function, you have satisfied the recursion.

The pointer that is passed in is the root of some tree or subtree. The node passed in will point to the smallest node of that subtree after calling findMin() or will point to the largest node in that subtree after calling findMax(). Warning: yes it will change where root is pointing if you pass root into it.

The same as in Codebook: each pair is on a separate line. plaintext word followed by a single blank followed by 4-digit code followed by endl.

The comment was changed (see faq immediately below this one) to say that the function that calls this sets the root to NULL.

• Comments in some header files have changed.
  • Decoder.h
  • Pair.h
  • Encoder.h
  • Bst.h
  • ModBst.h
  
  
• The official version of TNode has been posted. Use it in Project 3.

  Use this version in Project 3 so that everything will match. It is required that your version and our TNode.h match.
  
  

• Can you explain a bit more about ModBst and why we need it?

  You are writing a modified Node, ModNode, and a modified Bst, ModBst, so that we will be able to use it without having to do the extra work in the next project. It will all become clear there. So for now, please just trust that it is better to have this completed before project 4. I can clarify some things though. The manner in which we have to do this without inheritance is rather tedious. For example, Encoder must be changed just so it says #include ModBst.h" instead of "#include Bst.h". Also, the search and locate functions now take only 2 arguments instead of 3 (as in Bst.h). main.cpp must have some #includes modified. So when you put p3mod in your makefile, make copies of Encoder.* (called ModEncoder.*) and main.cpp - call it Modmain.cpp. Then compile these modified versions.
  
  

• Look in the project description under Task 6 for a specific example of the type of expression we are testing for (with respect to extra credit.)

  The example, hopefully, will clear up any questions about the extra credit in Task 6. However, this should be the least of your concerns right now. First get everything else running and test it well. To spend time on an extra credit worth 5 points and end up loosing 10 or 20 points on other parts that will get tested more thoroughly does not make good sense.
  
  

• Look in the project description under Task 4 for more information about "fixing halt".

  Anyone who did not read the comments in Encoder above encode() missed the explanation of "fixing halt".
  
  

• In what format should the Bst be printed?

  Print the Bst Pairs in the same format as you printed the Codebook: the word/expression, followed by a single blank, followed by the 4-digit code.
  
  

• I have tryed to run my project with proj3's primary and secondary inputs and I realized that there is a code in the secondary input (0928) that is not in the codebook.txt.

  Did you read the description in comments above encode() in Encoder? "halt" no longer remains static (always 0935), rather it changes by -7 each time it is encountered. If this were a real message, seeing 0935 every 8 to 12 words would be a "dead give-away" to anyone trying to "break" our code that 0935 is a separater or demarcation of some sort.
  
  

• How about copy() in ModBst?

  Yes "curr" needs a reference there too, I changed ModBst.h in the posting account to reflect this.
  
  

• How can we make a copy of other in Bst in the copy() function if curr is passed in as a pointer instead of a reference to a pointer?

  Thanks for noticing this. I left it out by mistake. I changed Bst.h in the posting account to reflect this.
  
  

• You didn't say what inorder(), preorder() and postorder() are supposed to do.

  Whoops! I should have said that preorder() prints the preorder traversal of the tree, inorder() prints the inorder traversal of the tree and postorder() prints the postorder traversal of the tree.
  
  

• Did you change some of the header file comments?

  Yes! I changed Bst.h and ModBst.h comments only. Please download the newest versions so they will diff with ours.
  
  

• What happens in search() in Bst if the string X is not found?

  Then "curr" would be NULL and "parent" would be the parent of the new node that will be inserted (that contains X). "curr" is the node whose string you are comparing with X. Since you can only insert a leaf node into a Bst, if X is not found then "curr" is NULL and "parent" is the node that will connect up to the new node containing X that you have created.
  
  

• How should we handle garbles?

  Any word you can't translate, after checking for expressions to the right or left, and plurals, just replace the word in the message with "garble". When garble is looked up in your Bst, it will translate to 5555.
  
  

• When encountering complicated expressions, if "Y" words aren't in the codebook but are part of an expression, could they be treated as garbles?

  Any word that someone can't handle should be treated as a garble. However, the "complicated expression" test needs to be translated correctly in order to get 5 extra points (Task 6). So anyone who handles the expression as one containing garbles will not receive the extra credit.
  
  

• When is Project #3 due?

  Project #3 is due by 11pm on Thursday, October 30th, 2003.
  Note: any project may be turned in late up to 2 days as specified on the class syllabus.
  For details regarding the late policy and any penalties associated with submitting late projects please see the class syllabus.

See the class syllabus for policies concerning email Last Modified: Tue Oct 14, 2003

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