I have opened an account for Ridit, one of 7-years-old students learning Java at SPOJ. The first task i gave to him was PALIN -The Next Palindrome. Here is the link to this problem- PALIN- The next Palindrome- SPOJAfter i explained it to him, he was able to solve it mostly except removing the leading zeros, which i did. Following is his solution of the problem -
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
// TODO Auto-generated method stub
try {
Scanner in = new Scanner(System.in);
int t = Integer.parseInt(in.nextLine());
String[] numbersInString = new String[t];
for (int i = 0; i <t; i++) {
String str = in.nextLine();
numbersInString[i] = removeLeadingZeros(str);
}
for (int i = 0 ; i<t; i++) {
int K = Integer.parseInt(numbersInString[i]);
int answer = findTheNextPalindrome(K);
System.out.println(answer);
}
}catch(Exception e) {
return;
}
}
static boolean isPalindrome(int x) {
String str = Integer.toString(x);
int length = str.length();
StringBuffer strBuff = new StringBuffer();
for(int i = length - 1;i>=0;i--) {
char ch = str.charAt(i);
strBuff.append(ch);
}
String str1 = strBuff.toString();
if(str.equals(str1)) {
return true;
}
return false;
}
static int findTheNextPalindrome(int K) {
for(int i = K+1;i<9999999; i++) {
if(isPalindrome(i) == true) {
return i;
}
}
return -1;
}
static String removeLeadingZeros(String str) {
String retString = str;
if(str.charAt(0) != '0') {
return retString;
}
return removeLeadingZeros(str.substring(1));
}
}
It is giving correct answer in Eclipse on his computer, but it is failing in SPOJ. If someone helps this little boy in his first submission, it will definitely make him very happy. I couldn't find any problem with this solution... Thank you in advance...
This might be helpful
import java.io.IOException;
import java.util.Scanner;
public class ThenNextPallindrom2 {
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
int t = 0;
Scanner sc = new Scanner(System.in);
if(sc.hasNextInt()) {
t = sc.nextInt();
}
sc.nextLine();
int[] arr, arr2;
while(t > 0) {
t--;
String s = sc.nextLine();
arr = getStringToNumArray(s);
if(all9(arr)) {
arr2 = new int[arr.length + 1];
arr2[0] = 1;
for(int i=0;i<arr.length;i++) {
arr2[i+1] = 0;
}
arr2[arr2.length -1] = 1;
arr = arr2;
} else{
int mid = arr.length/ 2;
int left = mid-1;
int right = arr.length % 2 == 1 ? mid + 1 : mid;
boolean left_small = false;
while(left >= 0 && arr[left] == arr[right]) {
left--;
right++;
}
if(left < 0 || arr[left] < arr[right]) left_small = true;
if(!left_small) {
while(left >= 0) {
arr[right++] = arr[left--];
}
} else {
mid = arr.length/ 2;
left = mid-1;
int carry = 1;
if(arr.length % 2 == 0) {
right = mid;
} else {
arr[mid] += carry;
carry = arr[mid]/10;
arr[mid] %= 10;
right = mid + 1;
}
while(left >= 0) {
arr[left] += carry;
carry = arr[left] / 10;
arr[left] %= 10;
arr[right++] = arr[left--];
}
}
}
printArray(arr);
}
}
public static boolean all9(int[] arr) {
for(int i=0;i<arr.length;i++) {
if(arr[i] != 9)return false;
}
return true;
}
public static void printArray(int[] arr) {
for(int i=0;i<arr.length;i++) {
System.out.print(arr[i]);
}
System.out.println();
}
public static int[] getStringToNumArray(String s) {
int[] arr = new int[s.length()];
for(int i=0; i<s.length();i++) {
arr[i] = Integer.parseInt(String.valueOf(s.charAt(i)));
}
return arr;
}
}
Prompt: Write a program that reads five cards from the user, then analyzes the cards and prints out the category of hand that they represent.
Poker hands are categorized according to the following labels: Straight flush, four of a kind, full house, flush, straight, three of a kind, two pairs, pair, high card.
I currently have my program set as follows, first prompting the user for 5 cards, 2-9, then sorting the cards in ascending order. I set up my program to prompt the user and then go through several if else statements calling methods. I am having issues though where its not identifying three or four of a kind.
Example, if I enter 1, 3, 2, 1, 1, it identifies it as TWO PAIRS instead of Three of a Kind.
Same for entering 1, 1,1, 1, 4, it identifies as three of kind instead of 4.
Any suggestions to my code?
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
final int HAND_SIZE = 5;
int[] hand = new int[HAND_SIZE];
getHand(hand); //Prompt user for hand
sortHand(hand);//Sort hand in ascending order
if(containsFullHouse(hand))
{
System.out.print("FULL HOUSE!");
}
else if(containsStraight(hand))
{
System.out.print("STRAIGHT!");
}
else if(containsFourOfAKind(hand))
{
System.out.print("FOUR OF A KIND!");
}
else if(containsThreeOfAKind(hand))
{
System.out.println("THREE OF A KIND!");
}
else if(containsTwoPair(hand))
{
System.out.println("TWO PAIRS!");
}
else if(containsPair(hand))
{
System.out.println("PAIR!");
}
else
System.out.println("High Card!");
}
public static void getHand(int[] hand)
{
Scanner input = new Scanner(System.in);
System.out.println("Enter five numeric cards, 2-9, no face cards please");
for(int index = 0; index < hand.length; index++)
{
System.out.print("Card " + (index + 1) + ": ");
hand[index] = input.nextInt();
}
}
public static void sortHand(int[] hand)
{
int startScan, index, minIndex, minValue;
for(startScan = 0; startScan < (hand.length-1); startScan++)
{
minIndex = startScan;
minValue = hand[startScan];
for(index = startScan + 1; index <hand.length; index++)
{
if(hand[index] < minValue)
{
minValue = hand[index];
minIndex = index;
}
}
hand[minIndex] = hand[startScan];
hand[startScan] = minValue;
}
}
public static boolean containsPair(int hand[])
{
boolean pairFound = false;
int pairCount = 0;
int startCheck = hand[0];
for(int index = 1; index < hand.length; index++)
{
if((hand[index] - startCheck) == 0)
{
pairCount++;
}
startCheck = hand[index];
}
if (pairCount == 1)
{
pairFound = true;
}
else if(pairCount !=1)
{
pairFound = false;
}
return pairFound;
}
public static boolean containsTwoPair(int hand[])
{
boolean twoPairFound = false;
int twoPairCount = 0;
int startCheck = hand[0];
for(int index = 1; index < hand.length; index++)
{
if((hand[index] - startCheck) == 0)
{
twoPairCount++;
}
startCheck = hand[index];
}
if (twoPairCount == 2)
{
twoPairFound = true;
}
else if(twoPairCount != 2)
{
twoPairFound = false;
}
return twoPairFound;
}
public static boolean containsThreeOfAKind(int hand[])
{
boolean threeFound = false;
int threeKind = 0;
int startCheck = hand[0];
for(int index = 1; index < hand.length; index++)
{
if((hand[index] - startCheck) == 0)
{
threeKind++;
}
startCheck = hand[index];
}
if(threeKind == 3)
{
threeFound = true;
}
else if(threeKind !=3)
{
threeFound = false;
}
return threeFound;
}
public static boolean containsStraight(int hand[])
{
boolean straightFound = false;
int straight = 0;
int startCheck = hand[0];
for(int index = 1; index < hand.length; index++)
{
if((hand[index] - startCheck) == 1)
{
straight++;
}
startCheck = hand[index];
}
if(straight == 4)
{
straightFound = true;
}
return straightFound;
}
public static boolean containsFullHouse(int hand[])
{
boolean fullHouseFound = false;
int pairCheck = 0;
int startPairCheck = hand[0];
for(int index = 1; index < hand.length; index++)
{
if((hand[index] - startPairCheck) == 0)
{
pairCheck++;
}
startPairCheck = hand[index];
}
int threeOfKindCheck = 0;
int startThreeKindCheck = hand[0];
for(int index = 1; index < hand.length; index++)
{
if((hand[index] - startThreeKindCheck) == 0)
{
threeOfKindCheck++;
}
startThreeKindCheck = hand[index];
}
if(pairCheck == 1 && startThreeKindCheck == 3)
{
fullHouseFound = true;
}
return fullHouseFound;
}
public static boolean containsFourOfAKind(int hand[])
{
boolean fourFound = false;
int fourKind = 0;
int startCheck = hand[0];
for(int index = 1; index < hand.length; index++)
{
if((hand[index] - startCheck) == 0)
{
fourKind++;
}
startCheck = hand[index];
}
if(fourKind == 1)
{
fourFound = true;
}
else if(fourKind !=4)
{
fourFound = false;
}
return fourFound;
}
}
Some hints.
Start with the highest hand. This eliminates lots of logic.
I.e if you check for pairs first, than you also have to check to make sure that your pair is the only pair, and not three of a kind.
But if you already ruled all of those out your code would be check card 1and2 23 34 and 45.
Hellow, For some off reason I cannot get my hash table to fill with the items and keys when I insert. It seems like it is being added when run through the driver, but nothing is stored, and there are absolutely no error messages.
I am suppose to create a hash table that utilizes quadratic probing. I am suppose to count the number of transversals which cannot be > 20.
Here is the code:
public class HashTable {
String[] table;
String[] keys;
int currentSize, maxSize, numOfEntries;
int traverse = 0;
double capacity;
public HashTable(int size, double load){
if(size <= 0){
System.out.println("Size must be greater then 0");
}
else if(load < 0 || load > 1){
System.out.println("Load must be between 0 and 1. EX: 0.75");
}else{
this.currentSize = 0;
table = new String[size];
keys = new String[size];
this.capacity = load * size;
this.maxSize = size;
}
}
public int hash(String num){
return (2 * Integer.parseInt(num) + 5) % table.length;
}
public int getLength(){
return table.length;
}
public int getMaxSize(){
return maxSize;
}
public int probe(String num){
int temp = hash(num);
int calc = 0;
numOfEntries = 0;
while(table[temp] != null){
traverse++;
temp = (int)((temp + (float)calc/2 + (float) (calc * calc)) % maxSize);
calc++;
numOfEntries++;
}
if(traverse >= 20){
System.out.println("Insert Failed : Reached 20 Traversal Limit!");
return 20;
}
return temp;
}
public void resize(){
String [] tempTable = table;
if(table.length >= capacity){
table = new String[tempTable.length * 2];
for(int i = 0; i < tempTable.length; i++){
if(tempTable[i] != null){
insert(tempTable[i]);
}
}
}
}
public void insert(String num){
int temp = probe(num);
table[temp] = num;
currentSize++;
if(currentSize >= capacity){
resize();
}
}
public String searchKey(String key){
String temp = "";
numOfEntries = 0;
for(int i = 0; i < keys.length; i++){
if(key == keys[i]){
temp = table[i];
numOfEntries++;
break;
}
numOfEntries++;
}
if(temp == ""){
System.out.println("No items that match that key!");
}
return temp;
}
public int numOfEntries(){
return numOfEntries;
}
public void print(){
System.out.println("Key\t:\tValue");
for(int i = 0; i < table.length; i++){
if(keys[i] != null){
System.out.println(keys[i] + "\t:\t" + table[i]);
}
}
System.out.println();
}
}
Here is the Driver:
import java.util.*;
public class HashTableDriver {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.println("Enter Table Size:\t");
int size = scan.nextInt();
System.out.println("Enter Load Factor (Between 0 and 1");
double load = scan.nextDouble();
int temp = 0;
HashTable table = new HashTable(size,load);
System.out.println(table.getLength());
System.out.println(table.getMaxSize());
while(temp != 4){
System.out.println();
System.out.println("i)(1) Insert an item to the Hash Table" +
"\nii)(2) Search A Specific Key" +
"\niii)(3) Display the Table" +
"\nvii)(4) Quit");
String input = scan.next();
switch(input){
case "1":
System.out.println();
System.out.println("Enter value to add to table");
String desKey1 = scan.next();
table.insert(desKey1);
continue;
case "2":
System.out.println("Please enter specific key desired:\t");
String desKey2 = scan.next();
if(table.searchKey(desKey2).equals(desKey2)){
System.out.println("Key Found");
System.out.println("Number of cells accessed:\t" + table.numOfEntries());
}
else{
System.out.println("Key Not Found");
}
continue;
case "3":
table.print();
continue;
case "4":
temp = 4;
break;
}
}
}
}
I have to make a program that converts Roman numbers to decimal. I am confused about how to write the conditions for the Roman numbers, such as IV (4), IX (9), XL (40) and CM(900). The code that I wrote works for all the other numbers.
public static void main(String[] args) {
System.out.print("Enter a roman numeral: ");
Scanner in = new Scanner(System.in);
String Roman = in.next();
int largo = Roman.length();
char Roman2[] = new char[largo];
int Roman3[] = new int[largo];
for (int i = 0; i < largo; i++) {
Roman2[i] = Roman.charAt(i);
}
for (int i = 0; i < largo; i++) {
if (Roman2[i] == 'I') {
Roman3[i] = 1;
} else if (Roman2[i] == 'V') {
Roman3[i] = 5;
} else if (Roman2[i] == 'X') {
Roman3[i] = 10;
} else if (Roman2[i] == 'L') {
Roman3[i] = 50;
} else if (Roman2[i] == 'C') {
Roman3[i] = 100;
} else if (Roman2[i] == 'M') {
Roman3[i] = 1000;
}
}
int total = 0;
for (int m = 0; m < Roman3.length; m++) {
total += Roman3[m];
}
System.out.println("The Roman is equal to " + total);
}
You can check the previous digit.
For example, I added the condition that detects IV :
if (Roman2[i]=='I'){
Roman3[i]=1;
} else if (Roman2[i]=='V'){
Roman3[i]=5;
if (i>0 && Roman2[i-1]=='I') { // check for IV
Roman3[i]=4;
Roman3[i-1]=0;
}
} else if (Roman2[i]=='X'){
Roman3[i]=10;
} else if (Roman2[i]=='L'){
Roman3[i]=50;
} else if (Roman2[i]=='C'){
Roman3[i]=100;
} else if (Roman2[i]=='M'){
Roman3[i]=1000;
}
Define enum like below:
public enum RomanSymbol {
I(1), V(5), X(10), L(50), C(100), D(500), M(1000);
private final int value;
private RomanSymbol(final int value) {
this.value = value;
}
public int getValue() {
return this.value;
}
public int calculateIntEquivalent(final int lastArabicNumber, final int totalArabicResult) {
if (lastArabicNumber > this.value) {
return totalArabicResult - this.value;
} else {
return totalArabicResult + this.value;
}
}
}
And use it like RomanSymbol.I.getValue() which will return 1 and similarly for other.
So if you accept character from user, you can get the values as:
char symbol = 'I';//lets assume this is what user has entered.
RomanSymbol rSymbol = RomanSymbol.valueOf(String.valueOf(symbol));
int invalue = rSymbol.getValue();
And if you have string like IV, then you could calculate on something like for example:
int lastValue = rSymbol.calculateIntEquivalent(intValue, 0);
lastValue = rSymbol.calculateIntEquivalent(intValue, lastValue); //and so on
I am struggling with a "find supersequence" algorithm.
The input is for set of strings
String A = "caagccacctacatca";
String B = "cgagccatccgtaaagttg";
String C = "agaacctgctaaatgctaga";
the result would be properly aligned set of strings (and next step should be merge)
String E = "ca ag cca cc ta cat c a";
String F = "c gag ccat ccgtaaa g tt g";
String G = " aga acc tgc taaatgc t a ga";
Thank you for any advice (I am sitting on this task for more than a day)
after merge the superstring would be
cagagaccatgccgtaaatgcattacga
The definition of supersequence in "this case" would be something like
The string R is contained in supersequence S if and only if all characters in a string R are present in supersequence S in the order in which they occur in the input sequence R.
The "solution" i tried (and again its the wrong way of doing it) is:
public class Solution4
{
static boolean[][] map = null;
static int size = 0;
public static void main(String[] args)
{
String A = "caagccacctacatca";
String B = "cgagccatccgtaaagttg";
String C = "agaacctgctaaatgctaga";
Stack data = new Stack();
data.push(A);
data.push(B);
data.push(C);
Stack clone1 = data.clone();
Stack clone2 = data.clone();
int length = 26;
size = max_size(data);
System.out.println(size+" "+length);
map = new boolean[26][size];
char[] result = new char[size];
HashSet<String> chunks = new HashSet<String>();
while(!clone1.isEmpty())
{
String a = clone1.pop();
char[] residue = make_residue(a);
System.out.println("---");
System.out.println("OLD : "+a);
System.out.println("RESIDUE : "+String.valueOf(residue));
String[] r = String.valueOf(residue).split(" ");
for(int i=0; i<r.length; i++)
{
if(r[i].equals(" ")) continue;
//chunks.add(spaces.substring(0,i)+r[i]);
chunks.add(r[i]);
}
}
for(String chunk : chunks)
{
System.out.println("CHUNK : "+chunk);
}
}
static char[] make_residue(String candidate)
{
char[] result = new char[size];
for(int i=0; i<candidate.length(); i++)
{
int pos = find_position_for(candidate.charAt(i),i);
for(int j=i; j<pos; j++) result[j]=' ';
if(pos==-1) result[candidate.length()-1] = candidate.charAt(i);
else result[pos] = candidate.charAt(i);
}
return result;
}
static int find_position_for(char character, int offset)
{
character-=((int)'a');
for(int i=offset; i<size; i++)
{
// System.out.println("checking "+String.valueOf((char)(character+((int)'a')))+" at "+i);
if(!map[character][i])
{
map[character][i]=true;
return i;
}
}
return -1;
}
static String move_right(String a, int from)
{
return a.substring(0, from)+" "+a.substring(from);
}
static boolean taken(int character, int position)
{ return map[character][position]; }
static void take(char character, int position)
{
//System.out.println("taking "+String.valueOf(character)+" at "+position+" (char_index-"+(character-((int)'a'))+")");
map[character-((int)'a')][position]=true;
}
static int max_size(Stack stack)
{
int max=0;
while(!stack.isEmpty())
{
String s = stack.pop();
if(s.length()>max) max=s.length();
}
return max;
}
}
Finding any common supersequence is not a difficult task:
In your example possible solution would be something like:
public class SuperSequenceTest {
public static void main(String[] args) {
String A = "caagccacctacatca";
String B = "cgagccatccgtaaagttg";
String C = "agaacctgctaaatgctaga";
int iA = 0;
int iB = 0;
int iC = 0;
char[] a = A.toCharArray();
char[] b = B.toCharArray();
char[] c = C.toCharArray();
StringBuilder sb = new StringBuilder();
while (iA < a.length || iB < b.length || iC < c.length) {
if (iA < a.length && iB < b.length && iC < c.length && (a[iA] == b[iB]) && (a[iA] == c[iC])) {
sb.append(a[iA]);
iA++;
iB++;
iC++;
}
else if (iA < a.length && iB < b.length && a[iA] == b[iB]) {
sb.append(a[iA]);
iA++;
iB++;
}
else if (iA < a.length && iC < c.length && a[iA] == c[iC]) {
sb.append(a[iA]);
iA++;
iC++;
}
else if (iB < b.length && iC < c.length && b[iB] == c[iC]) {
sb.append(b[iB]);
iB++;
iC++;
} else {
if (iC < c.length) {
sb.append(c[iC]);
iC++;
}
else if (iB < b.length) {
sb.append(b[iB]);
iB++;
} else if (iA < a.length) {
sb.append(a[iA]);
iA++;
}
}
}
System.out.println("SUPERSEQUENCE " + sb.toString());
}
}
However the real problem to solve is to find the solution for the known problem of Shortest Common Supersequence http://en.wikipedia.org/wiki/Shortest_common_supersequence,
which is not that easy.
There is a lot of researches which concern the topic.
See for instance:
http://www.csd.uwo.ca/~lila/pdfs/Towards%20a%20DNA%20solution%20to%20the%20Shortest%20Common%20Superstring%20Problem.pdf
http://www.ncbi.nlm.nih.gov/pubmed/14534185
You can try finding the shortest combination like this
static final char[] CHARS = "acgt".toCharArray();
public static void main(String[] ignored) {
String A = "caagccacctacatca";
String B = "cgagccatccgtaaagttg";
String C = "agaacctgctaaatgctaga";
String expected = "cagagaccatgccgtaaatgcattacga";
List<String> ABC = new Combination(A, B, C).findShortest();
System.out.println("expected: " + expected.length());
System.out.println("Merged: " + ABC.get(0).length() + " " + ABC);
}
static class Combination {
int shortest = Integer.MAX_VALUE;
List<String> shortestStr = new ArrayList<>();
char[][] chars;
int[] pos;
int count = 0;
Combination(String... strs) {
chars = new char[strs.length][];
pos = new int[strs.length];
for (int i = 0; i < strs.length; i++) {
chars[i] = strs[i].toCharArray();
}
}
public List<String> findShortest() {
findShortest0(new StringBuilder(), pos);
return shortestStr;
}
private void findShortest0(StringBuilder sb, int[] pos) {
if (allDone(pos)) {
if (sb.length() < shortest) {
shortestStr.clear();
shortest = sb.length();
}
if (sb.length() <= shortest)
shortestStr.add(sb.toString());
count++;
if (++count % 100 == 1)
System.out.println("Searched " + count + " shortest " + shortest);
return;
}
if (sb.length() + maxLeft(pos) > shortest)
return;
int[] pos2 = new int[pos.length];
int i = sb.length();
sb.append(' ');
for (char c : CHARS) {
if (!tryChar(pos, pos2, c)) continue;
sb.setCharAt(i, c);
findShortest0(sb, pos2);
}
sb.setLength(i);
}
private int maxLeft(int[] pos) {
int maxLeft = 0;
for (int i = 0; i < pos.length; i++) {
int left = chars[i].length - pos[i];
if (left > maxLeft)
maxLeft = left;
}
return maxLeft;
}
private boolean allDone(int[] pos) {
for (int i = 0; i < chars.length; i++)
if (pos[i] < chars[i].length)
return false;
return true;
}
private boolean tryChar(int[] pos, int[] pos2, char c) {
boolean matched = false;
for (int i = 0; i < chars.length; i++) {
pos2[i] = pos[i];
if (pos[i] >= chars[i].length) continue;
if (chars[i][pos[i]] == c) {
pos2[i]++;
matched = true;
}
}
return matched;
}
}
prints many solutions which are shorter than the one suggested.
expected: 28
Merged: 27 [acgaagccatccgctaaatgctatcga, acgaagccatccgctaaatgctatgca, acgaagccatccgctaacagtgctaga, acgaagccatccgctaacatgctatga, acgaagccatccgctaacatgcttaga, acgaagccatccgctaacatgtctaga, acgaagccatccgctacaagtgctaga, acgaagccatccgctacaatgctatga, acgaagccatccgctacaatgcttaga, acgaagccatccgctacaatgtctaga, acgaagccatcgcgtaaatgctatcga, acgaagccatcgcgtaaatgctatgca, acgaagccatcgcgtaacagtgctaga, acgaagccatcgcgtaacatgctatga, acgaagccatcgcgtaacatgcttaga, acgaagccatcgcgtaacatgtctaga, acgaagccatcgcgtacaagtgctaga, acgaagccatcgcgtacaatgctatga, acgaagccatcgcgtacaatgcttaga, acgaagccatcgcgtacaatgtctaga, acgaagccatgccgtaaatgctatcga, acgaagccatgccgtaaatgctatgca, acgaagccatgccgtaacagtgctaga, acgaagccatgccgtaacatgctatga, acgaagccatgccgtaacatgcttaga, acgaagccatgccgtaacatgtctaga, acgaagccatgccgtacaagtgctaga, acgaagccatgccgtacaatgctatga, acgaagccatgccgtacaatgcttaga, acgaagccatgccgtacaatgtctaga, cagaagccatccgctaaatgctatcga, cagaagccatccgctaaatgctatgca, cagaagccatccgctaacagtgctaga, cagaagccatccgctaacatgctatga, cagaagccatccgctaacatgcttaga, cagaagccatccgctaacatgtctaga, cagaagccatccgctacaagtgctaga, cagaagccatccgctacaatgctatga, cagaagccatccgctacaatgcttaga, cagaagccatccgctacaatgtctaga, cagaagccatcgcgtaaatgctatcga, cagaagccatcgcgtaaatgctatgca, cagaagccatcgcgtaacagtgctaga, cagaagccatcgcgtaacatgctatga, cagaagccatcgcgtaacatgcttaga, cagaagccatcgcgtaacatgtctaga, cagaagccatcgcgtacaagtgctaga, cagaagccatcgcgtacaatgctatga, cagaagccatcgcgtacaatgcttaga, cagaagccatcgcgtacaatgtctaga, cagaagccatgccgtaaatgctatcga, cagaagccatgccgtaaatgctatgca, cagaagccatgccgtaacagtgctaga, cagaagccatgccgtaacatgctatga, cagaagccatgccgtaacatgcttaga, cagaagccatgccgtaacatgtctaga, cagaagccatgccgtacaagtgctaga, cagaagccatgccgtacaatgctatga, cagaagccatgccgtacaatgcttaga, cagaagccatgccgtacaatgtctaga, cagagaccatccgctaaatgctatcga, cagagaccatccgctaaatgctatgca, cagagaccatccgctaacagtgctaga, cagagaccatccgctaacatgctatga, cagagaccatccgctaacatgcttaga, cagagaccatccgctaacatgtctaga, cagagaccatccgctacaagtgctaga, cagagaccatccgctacaatgctatga, cagagaccatccgctacaatgcttaga, cagagaccatccgctacaatgtctaga, cagagaccatcgcgtaaatgctatcga, cagagaccatcgcgtaaatgctatgca, cagagaccatcgcgtaacagtgctaga, cagagaccatcgcgtaacatgctatga, cagagaccatcgcgtaacatgcttaga, cagagaccatcgcgtaacatgtctaga, cagagaccatcgcgtacaagtgctaga, cagagaccatcgcgtacaatgctatga, cagagaccatcgcgtacaatgcttaga, cagagaccatcgcgtacaatgtctaga, cagagaccatgccgtaaatgctatcga, cagagaccatgccgtaaatgctatgca, cagagaccatgccgtaacagtgctaga, cagagaccatgccgtaacatgctatga, cagagaccatgccgtaacatgcttaga, cagagaccatgccgtaacatgtctaga, cagagaccatgccgtacaagtgctaga, cagagaccatgccgtacaatgctatga, cagagaccatgccgtacaatgcttaga, cagagaccatgccgtacaatgtctaga, cagagccatcctagctaaagtgctaga, cagagccatcctagctaaatgctatga, cagagccatcctagctaaatgcttaga, cagagccatcctagctaaatgtctaga, cagagccatcctgactaaagtgctaga, cagagccatcctgactaaatgctatga, cagagccatcctgactaaatgcttaga, cagagccatcctgactaaatgtctaga, cagagccatcctgctaaatgctatcga, cagagccatcctgctaaatgctatgca, cagagccatcctgctaacagtgctaga, cagagccatcctgctaacatgctatga, cagagccatcctgctaacatgcttaga, cagagccatcctgctaacatgtctaga, cagagccatcctgctacaagtgctaga, cagagccatcctgctacaatgctatga, cagagccatcctgctacaatgcttaga, cagagccatcctgctacaatgtctaga]