) head;
index = 0;
}
// returns true if there is a next node
// this is always should return true if the list has something in it
public boolean hasNext() {
// TODO Auto-generated method stub
return size != 0;
}
// advances the iterator to the next item
// handles wrapping around back to the head automatically for you
public E next() {
// TODO Auto-generated method stub
prev = nextItem;
nextItem = nextItem.next;
index = (index + 1) % size;
return prev.item;
}
// removed the last node was visted by the .next() call
// for example if we had just created a iterator
// the following calls would remove the item at index 1 (the second person in the ring)
// next() next() remove()
public void remove() {
int target;
if(nextItem == head) {
target = size - 1;
} else{
target = index - 1;
index--;
}
CircularLinkedList.this.remove(target); //calls the above class
}
this is SolitaireEncryption.java
public class SolitaireEncryption {
public static char encryptChar(char letter, int key) {
int value = letter - 'a';
int encryptedValue = (value + key) % 26;
char encryptedChar = (char) (encryptedValue+'a');
return encryptedChar;
}
public static char decryptChar(char letter, int key) {
int value = letter - 'a';
int decryptedValue = (value + (26-key)) % 26;
char decryptedChar = (char) (decryptedValue+'a');
return decryptedChar;
}
public int getKey(CircularLinkedList deck){ // calls the steps methods
return -1;
}
private static void step1(CircularLinkedList deck){
}
private static void step2(CircularLinkedList deck){
}
private static void step3(CircularLinkedList deck){
}
private static void step4(CircularLinkedList deck){
}
private static int step5(CircularLinkedList deck){
return -1;
}
public static void main(String[] args) {
CircularLinkedList deck = new CircularLinkedList<>();
}
}
}
// It's easiest if you keep it a singly linked list
// SO DON'T CHANGE IT UNLESS YOU WANT TO MAKE IT HARDER
private static class Node{
E item;
Node next;
public Node(E item) {
this.item = item;
}
}
public static void main(String[] args){
}
}
Extracted text: 5 Assignment Use CircularLinkedList.java as a starting point to make a new program that reads in a 'deck' of 28 numbers (from a file, from a command line, or an array, your choice), asks the user for one or more messages to decrypt, and decrypts them using the modified Solitaire algorithm described above. Note that if your program is decrypting multiple messages, all but the first should be decrypted using the deck as it exists after the decryption of the previous message. (The first uses the deck provided, of course.) 5.1 Why a Circular Linked List? We use Circular Linked List here for a few reasons • Good practice solving and thinking for the upcoming exam. • The wrap around nature of the Circular Linked List allows us to not worry about Steps 1 and 2. • Linked Lists are ideal for these kinds of manipulations, as we can move entire sections of the list around much, much quicker than an ArrayList. For example, in step 3, we can split our deck into multiple smaller decks and recombine them very quickly. 5.2 Output Your output will be just the decrypted messages lists of characters without spaces or punctuation.
Extracted text: 3 Example As usual, an example will really help make sense of the algorithm. Let's say that this is the original ordering of our half-deck of cards: 1 4 7 10 13 16 19 22 25 28 3 6 9 12 15 18 21 24 27 2 5 8 11 14 17 20 23 26 Step 1 Swap 27 with the value following it. So, we swap 27 and 2: 1 4 7 10 13 16 19 22 25 28 3 6 9 12 15 18 21 24 2 27 5 8 11 14 17 20 23 26 Step 2 Move 28 two places down the list. It ends up between 6 and 9: 1 4 7 10 13 16 19 22 25 3 6 28 9 12 15 18 21 24 2 27 5 8 11 14 17 20 23 26 Step 3 Do the triple cut. Everything above the first joker (28 in this case) goes to the bottom of the deck, and everything below the second (27) goes to the top: 5 8 11 14 17 20 23 26 28 9 12 15 18 21 24 2 27 1 4 7 10 13 16 19 22 25 3 6 Step 4 The bottom card is 6. The first 6 cards of the deck are 5, 8, 11, 14, 17, and 20. They go just ahead of 6 at the bottom end of the deck: 23 26 28 9 12 15 18 21 24 2 27 1 4 7 10 13 16 19 22 25 3 5 8 11 14 17 20 6 Step 5 The top card is 23. Thus, our generated keystream value is the 24 th card, which is 11. Repeat Self Test: What is the next keystream value? The answer is provided at the end of this document.