Boggle Game:

Backtracking Algorithm Implementation

CS 1501

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Recursive Implementation

• Things to implementation for a recursive algorithm:– Find the common behavior in the algorithm, and implement that as a

function.– Action: what to do for the current situation– Recursive call: how to trigger next step– Stop criteria: when to stop further recursion

function Test (data) { if data satisfies stop criteria ------- (1) return;

do sometime for data ------- (2)

for all new_data from data Test (new_data); ------- (3)}

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Recursive Boggle

• Each move on the board can be a recursive call– They accomplish similar tasks based on the string we have so far.

function AdvanceStep (pos, string) { new_string = string + letter[pos]; if new_string is not prefix, not word -------- (1) return;}

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Recursive Boggle

• Each move on the board can be a recursive call– They accomplish similar tasks based on the string we have so far.

function AdvanceStep (pos, string) { new_string = string + letter[pos]; if new_string is not prefix, not word -------- (1) return;

if new_string is a word -------- (2) output string}

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Recursive Boggle

• Each move on the board can be a recursive call– They accomplish similar tasks based on the string we have so far.

function AdvanceStep (pos, string) { new_string = string + letter[pos]; if new_string is not prefix, not word -------- (1) return;

if new_string is a word -------- (2) output string

if new_string is a prefix -------- (3) for all possible next step new_pos AdvanceStep (new_pos, new_string); return}

de la Briandias Tree:

Dealing with string prefix

CS 1501

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Note Structure

• DLB tree is used to organize key set (dictionary).• A path from root to leaf represents a word.

ChildPointer Char Sibling

Pointerb

e

c

a

d

f

abeacadf

$

$

$

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Node Structure

• Determine prefix?– The node has at least one child that is not a “end-of-string” sign.

• Determine word?– Node has a child with “end-of-string” sign.

ChildPointer Char Sibling

Pointerb

e

c

a

d

f

abeacadf

$

$

$

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An Example

• Construct a DLB tree for the following words:abc, abe, abet, abx, ace, acidhives, iodin, inval, zoo, zool, zurich

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Insert Operation

INSERTION (tree pointer TREE, word WORD) { out = 0; i = 0; letter = word [at position i]; current = TREE;

DO WHILE (( i <= length of WORD) and (out=0)) { IF (letter equals (current->key)) { //if we have a match i = i +1; letter = word[ at position i]; current = current->child_pointer; //follow the child } ELSE { //we have to check siblings IF (current->sibling_pointer does not equal Null) current = current->sibling; ELSE out=1; //no sibling, we add new node } } (end of DO WHILE)

…… // insertion

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Insert Operation

INSERTION (tree pointer TREE, word WORD) { …….. // search

// we're at the point of insertion of new node, // unless the word is already there:

IF (out = 0) EXIT //if the word was already there, exit

// otherwise add a new node with the current letter current->sibling_pointer = CREATE new NODE (letter); i = i + 1; // and move on to append the rest of the letters

FOR (m=i ; m<= length of WORD; m++) { current.child_pointer = CREATE new NODE ( WORD[at position m]); current = current->child_pointer; } // now we just add the $ marker for end of word

current.child_pointer = CREATE new TERMINAL $ MARKER NODE;

}

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Delete Operation

DELETE (tree pointer TREE, word WORD) { current = TREE; // we start with the first node again out=0; i=0;

letter = WORD [at position i];

DO WHILE (( i <= length of WORD) and (out=0)) { IF (letter is equal to current->key) current = current->child_pointer; //we move down one level i = i + 1; letter = WORD [at position i]; // we move onto next letter ELSE IF (there is a sibling node) { last_sibling = current; //store the last sibling current = current->sibling // move to the sibling node } IF (there is no sibling node) out =1; // EXIT with ERROR } DO WHILE

…… // deleteion

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Delete Operation

DELETE (tree pointer TREE, word WORD) { ……. // search

PUSH last_sibling->sibling_pointer onto STACK

PUSH all the children of last_sibling->sibling_pointer onto STACK one by one, until the $ marker

Set all of the pointers on STACK to Null. as you're POPPING them off the STACK

}

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Corner cases

• The tree is empty when insert a word• The word is the last one in the tree• …