[Java] ^연산자(XOR)란? (chatGPT)

저는 부끄럽게도 개발경력이 1년이 넘어가는데도 아직도 bit 개념이 낯섭니다. 이번에 codewars 사이트를 처음으로 사용해 보았는데 거기에서 Find the odd int라는 문제를 풀게 되었습니다. 제가 푼 방법과 다른 방식으로 풀으신 분들의 문제가 이해가 가지 않더라구요. 구글링을 해도 원하는 지식을 습득할 수 없어서 답답하여 chatGPT를 활용해 보았습니다. chatGPT에게 물어가면서 습득한 방법에 대해 작성해 보도록 하겠습니다.

faced with a problem

codewars에서 풀게된 문제인 Find the odd int 입니다.

codewars

Given an array of integers, find the one that appears an odd number of times.

There will always be only one integer that appears an odd number of times.

Examples

[7] should return 7, because it occurs 1 time (which is odd).

[0] should return 0, because it occurs 1 time (which is odd).

[1,1,2] should return 2, because it occurs 1 time (which is odd).

[0,1,0,1,0] should return 0, because it occurs 3 times (which is odd).

[1,2,2,3,3,3,4,3,3,3,2,2,1] should return 4, because it appears 1 time (which is odd).

my code

제가 푼 코드입니다.

public static int findIt(int[] a) {
        Map<Integer, Integer> map = new HashMap<>();
        for (int j : a) {
            map.put(j, map.getOrDefault(j, 0) + 1);
        }

        for (Integer key : map.keySet()) {
            if (map.get(key) % 2 != 0) {
                return key;
            }
        }

        return -1;
    }

their code

많은 사람들이 푼 방법입니다.

public static int findIt(int[] A) {
    int xor = 0;
    for (int i = 0; i < A.length; i++) {
        xor ^= A[i];
    }
    return xor;
}

저는 처음에 저 코드를 마주하고 전혀 이해가 가지 않더군요. 저게 무엇이길래 딸랑 저렇게만 해도 정답이 나온느건지 도저히 이해가 가지 않았습니다. 구글링을 열심히 해 보았지만 해당 코드에서 사용된 ^ 연산자의 뜻이 이해가 가지 않더라구요.(저의 구글링 실력에 한계가...) 그래서 chatGPT에게 물어보았습니다.

ChatGPT

what is XOR and ^

한심한 영어실력이라...

Q: i am studying java operation. i don't understant ^ operation. could you explain what is ^ operation in java?

A: In Java, the ^ operator is the bitwise exclusive OR operator. It performs a binary XOR operation on the bits of two integer values.

Here's how the ^ operator works:

• For each bit position, the ^ operator returns 1 if exactly one of the bits is 1, and 0 otherwise.
• The resulting value is an integer whose bits represent the bitwise XOR of the input values.

Here's an example:

javaCopy codeint a = 5;    // 0101 in binary
int b = 3;    // 0011 in binary
int c = a ^ b; // 0110 in binary (6 in decimal)

In this example, the ^ operator is used to compute the bitwise XOR of the integer values a and b. The resulting value c is 6 in decimal, which is 0110 in binary.

You can also use the ^ operator with the assignment operator (^=) to perform an in-place bitwise XOR operation. For example:

javaCopy codeint a = 5;   // 0101 in binary
int b = 3;   // 0011 in binary
a ^= b;      // a is now 0110 in binary (6 in decimal)

In this example, the ^= operator is used to perform an in-place bitwise XOR operation on a and b. The resulting value of a is 6 in decimal, which is 0110 in binary.

java의 ^ 연산자에 대해서 질문해 보았습니다.

2진수(비트)로 a, b를 비교하고 2진수에서 같은 숫자면 0을, 한개라도 다르면 1을 return하는 연산자입니다. 위에 예시가 있으니 좀 더 이해하기 쉽습니다.

그런데 이 때 XOR이라는 단어가 등장했습니다. 예전 leetCode에서 easy 단계로 XOR 관련 문제를 한번 접한적이 있는데 그때도 구글링에 실패해서 결국 못풀었던 기억이 나더군요... 그래서 이참에 XOR도 이해하고 넘어가기 위해 chatGPT에게 XOR에 관해 질문해 보았습니다.

Q: i don't understand xor operation. what is that mean?

A: The XOR operation, also known as exclusive OR, is a logical operation that is performed on two binary values (bits). The result of an XOR operation is 1 only when the two bits being compared are different. Otherwise, the result is 0.

Here are the possible outcomes of an XOR operation:

• 0 XOR 0 = 0
• 0 XOR 1 = 1
• 1 XOR 0 = 1
• 1 XOR 1 = 0

In other words, if the bits being compared are the same, the result of the XOR operation is 0. If the bits being compared are different, the result is 1.

In the context of programming, the XOR operation is often used in bitwise operations to compare and manipulate individual bits in integers. It can be used to toggle a specific bit (i.e., flip the value of the bit) or to check whether two bits are different.

For example, suppose we have two integers: a = 5 (binary 0101) and b = 3 (binary 0011). If we perform the XOR operation on these two integers (a XOR b), the result will be 6 (binary 0110). This is because the binary representation of a XOR b is 0110, which corresponds to the bits that are different in the binary representation of a and b.

XOR은 exclusive OR이라고 표현하기도 하고 위에 언급했듯 동일한 숫자면 0을 하나라도 다르면 1을 return하는 연산자를 뜻합니다. 자바의 ^가 XOR 였습니다... XOR을 이해하지 못하니 ^ 도 같이 이해를 못했던 거였습니다 ㅠ

처음 접했을 때는 그래도 잘 이해가 되지 않아 조금 더 많은 예제들을 보여달라고 요청했습니다.

Q: show me more example

A: Sure, here are some examples of the XOR operation in action:

1. 0b0101 ^ 0b0011 (decimal 5 XOR decimal 3) = 0b0110 (decimal 6)

This example XORs the binary values 0b0101 (decimal 5) and 0b0011 (decimal 3) to produce the binary value 0b0110 (decimal 6).

Here's how the binary values are XORed:

markdownCopy code  0101
X 0011
------
  0110

As you can see, the bits that are different between the two values are set to 1 in the result, while the bits that are the same are set to 0.

1. 0b1010 ^ 0b0101 (decimal 10 XOR decimal 5) = 0b1111 (decimal 15)

This example XORs the binary values 0b1010 (decimal 10) and 0b0101 (decimal 5) to produce the binary value 0b1111 (decimal 15).

Here's how the binary values are XORed:

markdownCopy code  1010
X 0101
------
  1111

Again, the bits that are different between the two values are set to 1 in the result, while the bits that are the same are set to 0.

1. 0b1100 ^ 0b1010 (decimal 12 XOR decimal 10) = 0b0110 (decimal 6)

This example XORs the binary values 0b1100 (decimal 12) and 0b1010 (decimal 10) to produce the binary value 0b0110 (decimal 6).

Here's how the binary values are XORed:

markdownCopy code  1100
X 1010
------
  0110

As before, the bits that are different between the two values are set to 1 in the result, while the bits that are the same are set to 0.

이제 어느정도 XOR에 대해 습득했습니다. 이제 본론으로 넘어가 문제의 코드를 이해하기 위해 질문해 보았습니다.

How the code works

A: ok. i understand. there is a problem.

Given an array of integers, find the one that appears an odd number of times.

There will always be only one integer that appears an odd number of times.

Examples

[7] should return 7, because it occurs 1 time (which is odd).

[0] should return 0, because it occurs 1 time (which is odd).

[1,1,2] should return 2, because it occurs 1 time (which is odd).

[0,1,0,1,0] should return 0, because it occurs 3 times (which is odd).

[1,2,2,3,3,3,4,3,3,3,2,2,1] should return 4, because it appears 1 time (which is odd).

A: to solve this problem, you can use the XOR operator. Since each number occurs an even number of times except for the one that occurs an odd number of times, you can XOR all the numbers in the array together. The result will be the number that occurs an odd number of times.

Here's some sample Java code that implements this solution:

javaCopy codepublic static int findOdd(int[] nums) {
    int result = 0;
    for (int num : nums) {
        result ^= num;
    }
    return result;
}

This code iterates over each number in the array and XORs it with the current result. Since XOR is commutative and associative, the order of the numbers doesn't matter, and the result will be the same regardless of the order in which they are processed.

At the end of the loop, the result will be the number that occurs an odd number of times, since all the other numbers will have canceled each other out. This number is then returned as the solution to the problem.

제가 XOR 연산자를 사용하는 방법에 대해서 묻지 않았는데 위의 대화를 통해 알아서 유추한 후 대답해 준 듯 하다는 사실에 놀랐습니다.ㅎㅎ

위의 코드가 어떻게 맞게 돌아가는지 조금 더 자세히 물어 보았습니다.

Q: i don't understand. could you tell me more specific how it works?

A: Sure, I can explain the solution in more detail.

Let's take the example array [1, 2, 2, 3, 3, 3, 4, 3, 3, 3, 2, 2, 1]. To find the number that occurs an odd number of times, we'll use the XOR operator (^).

We start with an initial value of 0:

Copy code
0

Then, we XOR it with the first number in the array, which is 1:

Copy code
0 ^ 1 = 1

Next, we XOR it with the second number, which is 2:

Copy code
1 ^ 2 = 3

We continue in this way, XORing the current result with each number in the array:

Copy code3 ^ 2 = 1
1 ^ 3 = 2
2 ^ 3 = 1
1 ^ 4 = 5
5 ^ 3 = 6
6 ^ 3 = 5
5 ^ 3 = 6
6 ^ 3 = 5
5 ^ 2 = 7
7 ^ 2 = 5
5 ^ 1 = 4
4 ^ 1 = 5

At the end, we're left with the number 5. This is the number that occurs an odd number of times in the array, since all the other numbers appear an even number of times. Specifically, 1, 2, and 3 each appear 4 times, while 4 appears 1 time.

So the answer to the problem is 5.

I hope this helps clarify the solution. Let me know if you have any more questions!

음... 저렇게만 보아서는 이해가 안가더군요. 그래서 다시 한번 더 물어 보았습니다.

Q: well still i don't get it. how is it work????

A: I apologize for the confusion. Let me try to explain the XOR solution in more detail with an example.

Suppose we have an array [2, 4, 2, 4, 5, 6, 5]. The goal is to find the number that appears an odd number of times.

First, we initialize a variable result to 0.

makefileCopy code
result = 0

Next, we XOR each number in the array with result.

javascriptCopy coderesult = result ^ 2 // result is now 2
result = result ^ 4 // result is now 6
result = result ^ 2 // result is now 4
result = result ^ 4 // result is now 0
result = result ^ 5 // result is now 5
result = result ^ 6 // result is now 3
result = result ^ 5 // result is now 6

When we XOR two bits together, the result is 1 if the bits are different and 0 if they are the same. In other words, the XOR operator returns 1 if and only if exactly one of the two bits is 1.

Now, let's examine what happens when we XOR each number in the array with result.

The first number is 2. The binary representation of 2 is 10. When we XOR this with result, which is currently 0, we get:

sqlCopy coderesult = 0 ^ 2
       = 10 (in binary)

Next, we XOR 4 with result. The binary representation of 4 is 100. When we XOR this with result, we get:

sqlCopy coderesult = 10 ^ 100
       = 110 (in binary)

The next number is 2 again. When we XOR this with result, we get:

sqlCopy coderesult = 110 ^ 10
       = 100 (in binary)

Next, we XOR 4 with result again. This brings us back to 0:

sqlCopy coderesult = 100 ^ 100
       = 0 (in binary)

The next number is 5. The binary representation of 5 is 101. When we XOR this with result, we get:

sqlCopy coderesult = 0 ^ 101
       = 101 (in binary)

The next number is 6. The binary representation of 6 is 110. When we XOR this with result, we get:

sqlCopy coderesult = 101 ^ 110
       = 11 (in binary)

Finally, we XOR 5 with result. This brings us back to 6:

sqlCopy coderesult = 11 ^ 101
       = 110 (in binary)

At the end of the loop, result is equal to 6. This means that the number that appears an odd number of times in the array is 6. In this case, all of the other numbers appear an even number of times, but if there were multiple numbers that appeared an odd number of times, this method would not work.

I hope this explanation helps you understand the XOR solution better. Let me know if you have any more questions!

이때 겨우 이해가 되더군요(머리가 나쁨). 비트로 연산했을 시 같은 숫자가 더해지면 결국 0이 된다는 것을요!

위에 bit가 같을 경우 0 아닐 경우 1을 return한다. 는 설명이 결국 같은 수가 짝수번으로 더해지면 결과는 0이 나오게 됩니다.

chatGPT에게 무엇인가를 배운 적은 많지만 이렇게 처음부터 끝까지 배운 적은 처음인 듯 합니다. google과는 확실히 느낌이 많이 다르군요.

많은 분들이 chatGPT에게 잘못된 정보를 전해줄 때가 많다고 하니 적절한 분별력을 갖추는 것도 좋은 방법일 듯 합니다.

chatGPT 덕분에 ^ 연산자와 XOR이 무엇을 뜻하는지 배울 수 있었습니다.

긴 글 읽어 주셔서 감사합니다.

즐거운 코딩 되세요!