Develop Reference

Overriding the Hashcode func for Sliding Puzzle problem

I am assigning each state of a sliding puzzle a hashcode. The code I currently have is returning the same hashcode for different states. Does anyone know an efficient way to uniquely hash states of a sliding puzzle?
public int hashCode() {
int total = 2;
if(board[0][0] == -1) {
total += 9;
}
else{
total += 9 * board[0][0];
}
if(board[0][1] == 1){
total += 1;
}
else{
total += 1 * board[0][1];
}
if(board[0][2] == 2){
total += 2;
}
else{
total += 2 * board[0][2];
}
if(board[1][0] == 3){
total += 3;
}
else{
total += 3 * board[1][0];
}
if(board[1][1] == 4){
total += 4;
}
else{
total += 4 * board[1][1];
}
..etc

Hash codes can be non-unique. Hash collisions are a fact of life that implementations should be able to deal with.
What you are looking for is more of a representation of the state. A state of this puzzle is a permutation of the numbers 0 (? meaning empty), 1-8. So you're looking for a unique numeric representation of a permutation.
This would be called a Lehmer code:
https://en.wikipedia.org/wiki/Permutation#Numbering_permutations
So much for theory but how do you implement that? Here is a suggested algorithm:
https://www.researchgate.net/figure/The-Lehmer-code-A-complete-translation-from-permutation-to-decimal-by-way-of-the_fig1_230831447

Calculating a hash code for an array is a common task the Java API provides a method for:
Arrays.deepHashCode(board)
(If you have additional knowledge about the content of the array you may be able to craft a better hashcode function, but the generic one is quite decent, and your time may be better spent optimizing elsewhere)
If you're curious how the Java API does this, here is the salient part of the implementation:
public static int hashCode(int a[]) {
if (a == null)
return 0;
int result = 1;
for (int element : a)
result = 31 * result + element;
return result;
}
As you can see, the implementation multiplies the previous result with a constant before adding the next element. If we imagine the result stored in a positional number system with base 31, this shifts the digits of the previous result one position to the left, before adding information about the new element in the lowest digit. Because the previous digits have been moved out of the way, the is no overlap between the old and new digits, and the information can be reconstructed perfectly as long as the result does not overflow. And even if the result does overflow, it will be reduced by power of 2, and since 31 is coprime to 2, some information about the old digits will remain.
If you know that your array contains small numbers (for instance, less than 15), multiplying by 15 rather than 31 would be preferable, as it can encode more elements before starting to overflow, resulting in slightly better spread hash values.

Word frequency hash table

Ok, I have a project that requires me to have a dynamic hash table that counts the frequency of words in a file. I must use java, however, we are not allowed to use any built in data types or built in classes at all except standard arrays. Also, I am not allowed to use any hash functions off the internet that are known to be fast. I have to make my own hash functions. Lastly, my instructor also wants my table to start as size "1" and double in size every time a new key is added.
My first idea was to sum the ASCII values of the letters composing a word and use that to make a hash function, but different words with the same letters will equal the same value.
How can I get started? Is the ASCII idea on the right track?

A hash table isn't expected to have in general a one-to-one mapping between a value and a hash. A hash table is expected to have collisions. That is, the domain of the hash-function is expected to be larger than the range (i.e., the hash value). However, the general idea is that you come up with a hash function where the probability of collision is drastically small. If your hash-function is uniform, i.e., if you have it designed such that each possible hash-value has the same probability of being generated, then you can minimize collisions this way.
Getting a collision isn't the end of the world. That just means that you have to search the list of values for that hash. If your hashing function is good, overall your performance for lookup should still be O(1).
Generating hashing functions is a subject of its own, and there is no one answer. But a good place for you to start could be to work with the bitwise representations of the characters in the string, and perform some sort of convolution operations on them (rotate, shift, XOR) in series. You could perform these in some way based on some initial seed-value, and then use the output of the first step of hashing as a seed for the next step. This way you can end up magnifying the effects of your convolution.
For example, let's say you get the character A, which is 41 in hex, or 0100 0001 in binary. You could designate each bit to mean some operation (maybe bit 0 is a ROR when it is 0, and a ROL when it is 1; bit 1 is an OR when it is 0, and a XOR when it is 1, etc.). You could even decide how much convolution you want to do based on the value itself. For example, you could say that the lower nibble specifies how much right-rotation you will do, and the upper nibble specifies how much left rotation you will do. Then once you have the final value, you will use that as the seed for the next character. These are just some ideas. Use your imagination as see what you get!

It does not matter how good your hash function is, you will always have collisions you need to resolve.
If you want to keep your approach by using the ASCII values of the you shouldn't just add the values this would lead to a lot collisions. You should work with the power of the values, for example for the word "Help" you just go like: 'H' * 256 + 'e' * 256 + 'l' * 256² + 'p' * 256³. Or in pseudocode:
int hash(String word, int hashSize)
int res = 0
int count = 0;
for char c in word
res += 'c' * 256^count
count++
count = count mod 5
return res mod hashSize
Now you just have to write your own Hashtable:
class WordCounterMap
Entry[] entrys = new Entry[1]
void add(String s)
int hash = hash(s, entrys.length)
if(entrys[hash] == null{
Entry[] temp = new Entry[entry.length * 2]
for(Entry e : entrys){
if(e != null)
int hash = hash(e.word, temp.length)
temp[hash] = e;
entrys = temp;
hash = hash(s, entrys.length)
while(true)
if(entrys[hash] != null)
if(entrys[hash].word.equals(s))
entrys[hash].count++
break
else
entrys[hash] = new Entry(s)
hash++
hash = hash mod entrys.length
int getCount(String s)
int hash = hash(s, length)
if(entrys[hash] == null)
return 0
while(true)
if(entrys[hash].word.equals(s))
entrys[hash].count++
break
hash++
hash = hash mod entrys.length
class Entry
int count
String word
Entry(String s)
this.word = s
count = 1

How can I get the position of bits

I have a decimal number which I need to convert to binary and then find the position of one's in that binary representation.
Input is 5 whose binary is 101 and Output should be
1
3
Below is my code which only provides output as 2 instead I want to provide the position of one's in binary representation. How can I also get position of set bits starting from 1?
public static void main(String args[]) throws Exception {
System.out.println(countBits(5));
}
private static int countBits(int number) {
boolean flag = false;
if (number < 0) {
flag = true;
number = ~number;
}
int result = 0;
while (number != 0) {
result += number & 1;
number = number >> 1;
}
return flag ? (32 - result) : result;
}

Your idea of having countBits return the result, instead of putting a System.out.println inside the method, is generally the best approach. If you want it to return a list of bit positions, the analogue would be to have your method return an array or some kind of List, like:
private static List<Integer> bitPositions(int number) {
As I mentioned in my comments, you will make life a lot easier for yourself if you use >>> and get rid of the special code to check for negatives. Doing this, and adapting the code you already have, gives you something like
private static List<Integer> bitPositions(int number) {
List<Integer> positions = new ArrayList<>();
int position = 1;
while (number != 0) {
if (number & 1 != 0) {
positions.add(position);
}
position++;
number = number >>> 1;
}
return positions;
}
Now the caller can do what it wants to print the positions out. If you use System.out.println on it, the output will be [1, 3]. If you want each output on a separate line:
for (Integer position : bitPositions(5)) {
System.out.println(position);
}
In any case, the decision about how to print the positions (or whatever else you want to do with them) is kept separate from the logic that computes the positions, because the method returns the whole list and doesn't have its own println.
(By the way, as Alex said, it's most common to think of the lower-order bit as "bit 0" instead of "bit 1", although I've seen hardware manuals that call the low-order bit "bit 31" and the high-order bit "bit 0". The advantage of calling it "bit 0" is that a 1 bit in position N represents the value 2N, making things simple. My code example calls it "bit 1" as you requested in your question; but if you want to change it to 0, just change the initial value of position.)

Binary representation: Your number, like anything on a modern day (non-quantum) computer, is already a binary representation in memory, as a sequence of bits of a given size.
Bit operations
You can use bit shifting, bit masking, 'AND', 'OR', 'NOT' and 'XOR' bitwise operations to manipulate them and get information about them on the level of individual bits.
Your example
For your example number of 5 (101) you mentioned that your expected output would be 1, 3. This is a bit odd, because generally speaking one would start counting at 0, e.g. for 5 as a byte (8 bit number):
76543210 <-- bit index
5 00000101
So I would expect the output to be 0 and 2 because the bits at those bit indexes are set (1).
Your sample implementation shows the code for the function
private static int countBits(int number)
Its name and signature imply the following behavior for any implementation:
It takes an integer value number and returns a single output value.
It is intended to count how many bits are set in the input number.
I.e. it does not match at all with what you described as your intended functionality.
A solution
You can solve your problem using a combination of a 'bit shift' (>>) and an AND (&) operation.
int index = 0; // start at bit index 0
while (inputNumber != 0) { // If the number is 0, no bits are set
// check if the bit at the current index 0 is set
if ((inputNumber & 1) == 1)
System.out.println(index); // it is, print its bit index.
// advance to the next bit position to check
inputNumber = inputNumber >> 1; // shift all bits one position to the right
index = index + 1; // so we are now looking at the next index.
}
If we were to run this for your example input number '5', we would see the following:
iteration input 76543210 index result
1 5 00000101 0 1 => bit set.
2 2 00000010 1 0 => bit not set.
3 1 00000001 2 1 => bit set.
4 0 00000000 3 Stop, because inputNumber is 0

You'll need to keep track of what position you're on, and when number & 1 results in 1, print out that position. It look something like:
...
int position = 1;
while (number != 0) {
if((number & 1)==1)
System.out.println(position);
result += number & 1;
position += 1;
number = number >> 1;
}
...

There is a way around working with bit-wise operations to solve your problem.
Integer.toBinaryString(int number) converts an integer to a String composed of zeros and ones. This is handy in your case because you could instead have:
public static void main(String args[]) throws Exception {
countBits(5);
}
public static void countBits(int x) {
String binaryStr = Integer.toBinaryString(x);
int length = binaryStr.length();
for(int i=0; i<length; i++) {
if(binaryStr.charAt(i)=='1')
System.out.println(length-1);
}
}
It bypasses what you might be trying to do (learn bitwise operations in Java), but makes the code look cleaner in my opinion.

The combination of Integer.lowestOneBit and Integer.numberOfTrailingZeros instantly gives the position of the lowest 1-Bit, and returns 32 iff the number is 0.
Therefore, the following code returns the positions of 1-Bits of the number number in ascending order:
public static List<Integer> BitOccurencesAscending(int number)
{
LinkedList<Integer> out = new LinkedList<>();
int x = number;
while(number>0)
{
x = Integer.lowestOneBit(number);
number -= x;
x = Integer.numberOfTrailingZeros(x);
out.add(x);
}
return out;
}

Algorithm to efficiently determine the [n][n] element in a matrix

This is a question regarding a piece of coursework so would rather you didn't fully answer the question but rather give tips to improve the run time complexity of my current algorithm.
I have been given the following information:
A function g(n) is given by g(n) = f(n,n) where f may be defined recursively by
I have implemented this algorithm recursively with the following code:
public static double f(int i, int j)
{
if (i == 0 && j == 0) {
return 0;
}
if (i ==0 || j == 0) {
return 1;
}
return ((f(i-1, j)) + (f(i-1, j-1)) + (f(i, j-1)))/3;
}
This algorithm gives the results I am looking for, but it is extremely inefficient and I am now tasked to improve the run time complexity.
I wrote an algorithm to create an n*n matrix and it then computes every element up to the [n][n] element in which it then returns the [n][n] element, for example f(1,1) would return 0.6 recurring. The [n][n] element is 0.6 recurring because it is the result of (1+0+1)/3.
I have also created a spreadsheet of the result from f(0,0) to f(7,7) which can be seen below:
Now although this is much faster than my recursive algorithm, it has a huge overhead of creating a n*n matrix.
Any suggestions to how I can improve this algorithm will be greatly appreciated!
I can now see that is it possible to make the algorithm O(n) complexity, but is it possible to work out the result without creating a [n][n] 2D array?
I have created a solution in Java that runs in O(n) time and O(n) space and will post the solution after I have handed in my coursework to stop any plagiarism.

This is another one of those questions where it's better to examine it, before diving in and writing code.
The first thing i'd say you should do is look at a grid of the numbers, and to not represent them as decimals, but fractions instead.
The first thing that should be obvious is that the total number of you have is just a measure of the distance from the origin, .
If you look at a grid in this way, you can get all of the denominators:
Note that the first row and column are not all 1s - they've been chosen to follow the pattern, and the general formula which works for all of the other squares.
The numerators are a little bit more tricky, but still doable. As with most problems like this, the answer is related to combinations, factorials, and then some more complicated things. Typical entries here include Catalan numbers, Stirling's numbers, Pascal's triangle, and you will nearly always see Hypergeometric functions used.
Unless you do a lot of maths, it's unlikely you're familiar with all of these, and there is a hell of a lot of literature. So I have an easier way to find out the relations you need, which nearly always works. It goes like this:
Write a naive, inefficient algorithm to get the sequence you want.
Copy a reasonably large amount of the numbers into google.
Hope that a result from the Online Encyclopedia of Integer Sequences pops up.
3.b. If one doesn't, then look at some differences in your sequence, or some other sequence related to your data.
Use the information you find to implement said sequence.
So, following this logic, here are the numerators:
Now, unfortunately, googling those yielded nothing. However, there are a few things you can notice about them, the main being that the first row/column are just powers of 3, and that the second row/column are one less than powers of three. This kind boundary is exactly the same as Pascal's triangle, and a lot of related sequences.
Here is the matrix of differences between the numerators and denominators:
Where we've decided that the f(0,0) element shall just follow the same pattern. These numbers already look much simpler. Also note though - rather interestingly, that these numbers follow the same rules as the initial numbers - except the that the first number is one (and they are offset by a column and a row). T(i,j) = T(i-1,j) + T(i,j-1) + 3*T(i-1,j-1):
1
1 1
1 5 1
1 9 9 1
1 13 33 13 1
1 17 73 73 17 1
1 21 129 245 192 21 1
1 25 201 593 593 201 25 1
This looks more like the sequences you see a lot in combinatorics.
If you google numbers from this matrix, you do get a hit.
And then if you cut off the link to the raw data, you get sequence A081578, which is described as a "Pascal-(1,3,1) array", which exactly makes sense - if you rotate the matrix, so that the 0,0 element is at the top, and the elements form a triangle, then you take 1* the left element, 3* the above element, and 1* the right element.
The question now is implementing the formulae used to generate the numbers.
Unfortunately, this is often easier said than done. For example, the formula given on the page:
T(n,k)=sum{j=0..n, C(k,j-k)*C(n+k-j,k)*3^(j-k)}
is wrong, and it takes a fair bit of reading the paper (linked on the page) to work out the correct formula. The sections you want are proposition 26, corollary 28. The sequence is mentioned in Table 2 after proposition 13. Note that r=4
The correct formula is given in proposition 26, but there is also a typo there :/. The k=0 in the sum should be a j=0:
Where T is the triangular matrix containing the coefficients.
The OEIS page does give a couple of implementations to calculate the numbers, but neither of them are in java, and neither of them can be easily transcribed to java:
There is a mathematica example:
Table[ Hypergeometric2F1[-k, k-n, 1, 4], {n, 0, 10}, {k, 0, n}] // Flatten
which, as always, is ridiculously succinct. And there is also a Haskell version, which is equally terse:
a081578 n k = a081578_tabl !! n !! k
a081578_row n = a081578_tabl !! n
a081578_tabl = map fst $ iterate
(\(us, vs) -> (vs, zipWith (+) (map (* 3) ([0] ++ us ++ [0])) $
zipWith (+) ([0] ++ vs) (vs ++ [0]))) ([1], [1, 1])
I know you're doing this in java, but i could not be bothered to transcribe my answer to java (sorry). Here's a python implementation:
from __future__ import division
import math
#
# Helper functions
#
def cache(function):
cachedResults = {}
def wrapper(*args):
if args in cachedResults:
return cachedResults[args]
else:
result = function(*args)
cachedResults[args] = result
return result
return wrapper
#cache
def fact(n):
return math.factorial(n)
#cache
def binomial(n,k):
if n < k: return 0
return fact(n) / ( fact(k) * fact(n-k) )
def numerator(i,j):
"""
Naive way to calculate numerator
"""
if i == j == 0:
return 0
elif i == 0 or j == 0:
return 3**(max(i,j)-1)
else:
return numerator(i-1,j) + numerator(i,j-1) + 3*numerator(i-1,j-1)
def denominator(i,j):
return 3**(i+j-1)
def A081578(n,k):
"""
http://oeis.org/A081578
"""
total = 0
for j in range(n-k+1):
total += binomial(k, j) * binomial(n-k, j) * 4**(j)
return int(total)
def diff(i,j):
"""
Difference between the numerator, and the denominator.
Answer will then be 1-diff/denom.
"""
if i == j == 0:
return 1/3
elif i==0 or j==0:
return 0
else:
return A081578(j+i-2,i-1)
def answer(i,j):
return 1 - diff(i,j) / denominator(i,j)
# And a little bit at the end to demonstrate it works.
N, M = 10,10
for i in range(N):
row = "%10.5f"*M % tuple([numerator(i,j)/denominator(i,j) for j in range(M)])
print row
print ""
for i in range(N):
row = "%10.5f"*M % tuple([answer(i,j) for j in range(M)])
print row
So, for a closed form:
Where the are just binomial coefficients.
Here's the result:
One final addition, if you are looking to do this for large numbers, then you're going to need to compute the binomial coefficients a different way, as you'll overflow the integers. Your answers are lal floating point though, and since you're apparently interested in large f(n) = T(n,n) then I guess you could use Stirling's approximation or something.

Well for starters here are some things to keep in mind:
This condition can only occur once, yet you test it every time through every loop.
if (x == 0 && y == 0) {
matrix[x][y] = 0;
}
You should instead: matrix[0][0] = 0; right before you enter your first loop and set x to 1. Since you know x will never be 0 you can remove the first part of your second condition x == 0 :
for(int x = 1; x <= i; x++)
{
for(int y = 0; y <= j; y++)
{
if (y == 0) {
matrix[x][y] = 1;
}
else
matrix[x][y] = (matrix[x-1][y] + matrix[x-1][y-1] + matrix[x][y-1])/3;
}
}
No point in declaring row and column since you only use it once. double[][] matrix = new double[i+1][j+1];

This algorithm has a minimum complexity of Ω(n) because you just need to multiply the values in the first column and row of the matrix with some factors and then add them up. The factors stem from unwinding the recursion n times.
However you therefore need to do the unwinding of the recursion. That itself has a complexity of O(n^2). But by balancing unwinding and evaluation of recursion, you should be able to reduce complexity to O(n^x) where 1 <= x <= 2. This is some kind of similiar to algorithms for matrix-matrix multiplication, where the naive case has a complexity of O(n^3) but Strassens's algorithm is for example O(n^2.807).
Another point is the fact that the original formula uses a factor of 1/3. Since this is not accurately representable by fixed point numbers or ieee 754 floating points, the error increases when evaluating the recursion successively. Therefore unwinding the recursion could give you higher accuracy as a nice side effect.
For example when you unwind the recursion sqr(n) times then you have complexity O((sqr(n))^2+(n/sqr(n))^2). The first part is for unwinding and the second part is for evaluating a new matrix of size n/sqr(n). That new complexity actually can be simplified to O(n).

To describe time complexity we usually use a big O notation. It is important to remember that it only describes the growth given the input. O(n) is linear time complexity, but it doesn't say how quickly (or slowly) the time grows when we increase input. For example:
n=3 -> 30 seconds
n=4 -> 40 seconds
n=5 -> 50 seconds
This is O(n), we can clearly see that every increase of n increases the time by 10 seconds.
n=3 -> 60 seconds
n=4 -> 80 seconds
n=5 -> 100 seconds
This is also O(n), even though for every n we need twice that much time, and the raise is 20 seconds for every increase of n, the time complexity grows linearly.
So if you have O(n*n) time complexity and you will half the number of operations you perform, you will get O(0.5*n*n) which is equal to O(n*n) - i.e. your time complexity won't change.
This is theory, in practice the number of operations sometimes makes a difference. Because you have a grid n by n, you need to fill n*n cells, so the best time complexity you can achieve is O(n*n), but there are a few optimizations you can do:
Cells on the edges of the grid could be filled in separate loops. Currently in majority of the cases you have two unnecessary conditions for i and j equal to 0.
You grid has a line of symmetry, you could utilize it to calculate only half of it and then copy the results onto the other half. For every i and j grid[i][j] = grid[j][i]
On final note, the clarity and readability of the code is much more important than performance - if you can read and understand the code, you can change it, but if the code is so ugly that you cannot understand it, you cannot optimize it. That's why I would do only first optimization (it also increases readability), but wouldn't do the second one - it would make the code much more difficult to understand.
As a rule of thumb, don't optimize the code, unless the performance is really causing problems. As William Wulf said:
More computing sins are committed in the name of efficiency (without necessarily achieving it) than for any other single reason - including blind stupidity.
EDIT:
I think it may be possible to implement this function with O(1) complexity. Although it gives no benefits when you need to fill entire grid, with O(1) time complexity you can instantly get any value without having a grid at all.
A few observations:
denominator is equal to 3 ^ (i + j - 1)
if i = 2 or j = 2, numerator is one less than denominator
EDIT 2:
The numerator can be expressed with the following function:
public static int n(int i, int j) {
if (i == 1 || j == 1) {
return 1;
} else {
return 3 * n(i - 1, j - 1) + n(i - 1, j) + n(i, j - 1);
}
}
Very similar to original problem, but no division and all numbers are integers.

If the question is about how to output all values of the function for 0<=i<N, 0<=j<N, here is a solution in time O(N²) and space O(N). The time behavior is optimal.
Use a temporary array T of N numbers and set it to all ones, except for the first element.
Then row by row,
use a temporary element TT and set it to 1,
then column by column, assign simultaneously T[I-1], TT = TT, (TT + T[I-1] + T[I])/3.

Thanks to will's (first) answer, I had this idea:
Consider that any positive solution comes only from the 1's along the x and y axes. Each of the recursive calls to f divides each component of the solution by 3, which means we can sum, combinatorially, how many ways each 1 features as a component of the solution, and consider it's "distance" (measured as how many calls of f it is from the target) as a negative power of 3.
JavaScript code:
function f(n){
var result = 0;
for (var d=n; d<2*n; d++){
var temp = 0;
for (var NE=0; NE<2*n-d; NE++){
temp += choose(n,NE);
}
result += choose(d - 1,d - n) * temp / Math.pow(3,d);
}
return 2 * result;
}
function choose(n,k){
if (k == 0 || n == k){
return 1;
}
var product = n;
for (var i=2; i<=k; i++){
product *= (n + 1 - i) / i
}
return product;
}
Output:
for (var i=1; i<8; i++){
console.log("F(" + i + "," + i + ") = " + f(i));
}
F(1,1) = 0.6666666666666666
F(2,2) = 0.8148148148148148
F(3,3) = 0.8641975308641975
F(4,4) = 0.8879743941472337
F(5,5) = 0.9024030889600163
F(6,6) = 0.9123609205913732
F(7,7) = 0.9197747256986194

How to find if two numbers are consecutive numbers in gray code sequence

I am trying to come up with a solution to the problem that given two numbers, find if they are the consecutive numbers in the gray code sequence i.e., if they are gray code neighbors assuming that the gray code sequence is not mentioned.
I searched on various forums but couldn't get the right answer. It would be great if you can provide a solution for this.
My attempt to the problem - Convert two integers to binary and add the digits in both the numbers separately and find the difference between the sum of the digits in two numbers. If the difference is one then they are gray code neighbors.
But I feel this wont work for all cases. Any help is highly appreciated. Thanks a lot in advance!!!

Actually, several of the other answers seem wrong: it's true that two binary reflected Gray code neighbours differ by only one bit (I assume that by « the » Gray code sequence, you mean the original binary reflected Gray code sequence as described by Frank Gray). However, that does not mean that two Gray codes differing by one bit are neighbours (a => b does not mean that b => a). For example, the Gray codes 1000 and 1010 differ by only one bit but are not neighbours (1000 and 1010 are respectively 15 and 12 in decimal).
If you want to know whether two Gray codes a and b are neighbours, you have to check whether previous(a) = b OR next(a) = b. For a given Gray code, you get one neighbour by flipping the rightmost bit and the other neighbour bit by flipping the bit at the left of the rightmost set bit. For the Gray code 1010, the neighbours are 1011 and 1110 (1000 is not one of them).
Whether you get the previous or the next neighbour by flipping one of these bits actually depends on the parity of the Gray code. However, since we want both neighbours, we don't have to check for parity. The following pseudo-code should tell you whether two Gray codes are neighbours (using C-like bitwise operations):
function are_gray_neighbours(a: gray, b: gray) -> boolean
return b = a ^ 1 OR
b = a ^ ((a & -a) << 1)
end
Bit trick above: a & -a isolates the rigthmost set bit in a number. We shift that bit by one position to the left to get the bit we need to flip.

Assumptions:
Inputs a and b are grey code sequences in binary reflected gray code.
i.e a's and b's bit encoding is binary gray code representations.
#convert from greycode bits into regular binary bits
def gTob(num): #num is binary graycode
mask = num >> 1
while mask!=0:
num = num^mask
mask >>= 1
return num; #num is converted
#check if a and b are consecutive gray code encodings
def areGrayNeighbors(a,b):
return abs(gTob(a) - gTob(b)) == 1
Few Test cases:
areGrayNeighbors(9,11) --> True (since (1001, 1011) differ in only one
bit and are consecutive numbers in decimal representation)
areGrayNeighbors(9,10) --> False
areGrayNeighbors(14,10) --> True
References:
method gTob() used above is from rodrigo in this post The neighbors in Gray code

public int checkConsecutive2(int b1, int b2){
int x = (b1 ^ b2);
if((x & (x - 1)) !=0){
return 0;
}else{
return 1;
}
}

If two numbers are in gray code sequence, they differ by one binary digit. i.e the exclusive OR on the two numbers returns a power of 2. So, find XOR and check if the result is a power of two.
This solution works well for the all the test cases written by CodeKaichu above. I would love to know if it fails in any cases.
public boolean grayCheck(int x, int y) {
int z = x^y;
return (z&z-1)==0;
}

An obvious answer, but it works.
Convert each gray code into its respective Binary form, subtract the two. If you answer is a binary equivalent of +1 or -1 then the two gray codes are adjacent.
This seems like an over kill, but when you're siting in an interview and don't know the correct method, this works. Also to optimize, one can check the single bit difference filter, so we don't waste time converting and subtracting numbers that we know for sure aren't adjacent.

If you just want to check if the input numbers differ by just one bit:
public boolean checkIfDifferByOneBit(int a, int b){
int diff = 0;
while(a > 0 && b > 0){
if(a & 1 != b & 1)
diff++;
a = a >> 1;
b = b >> 1;
}
if (a > 0 || b > 0) // a correction in the solution provided by David Jones
return diff == 0 // In the case when a or b become zero before the other
return diff == 1;
}

You can check if two numbers differ by one bit or not as follows. In this method, difference in the length of binary numbers are taken care of. Eg, the output for 11 (1011) and 3 (11) will be returned as true.
Also, this does not solve the second criteria for Gray code adjacency. But if you only want to check if the numbers differ by one bit, the code below will help.
class Graycode{
public static boolean graycheck(int one, int two){
int differences = 0;
while (one > 0 || two > 0){
// Checking if the rightmost bit is same
if ((one & 1) != (two & 1)){
differences++;
}
one >>= 1;
two >>= 1;
}
return differences == 1;
}
public static void main(String[] args){
int one = Integer.parseInt(args[0]);
int two = Integer.parseInt(args[1]);
System.out.println(graycheck(one,two));
}
}

If two numbers are in gray code sequence, they differ by one binary digit. i.e the exclusive OR on the two numbers returns a power of 2. So, find XOR and check if the result is a power of two.
python 3.8
a=int(input())
b=int(input())
x=a^b
if((x and (not(x & (x - 1))) )):
print("True")
else:
print("False")

I've had to solve this question in an interview as well. One of the conditions for the two values to be a gray code sequence is that their values only differ by 1 bit. Here is a solution to this problem:
def isGrayCode(num1, num2):
differences = 0
while (num1 > 0 or num2 > 0):
if ((num1 & 1) != (num2 & 1)):
differences++
num1 >>= 1
num2 >>= 1
return differences == 1