Chalkboard XOR Game
Identifying Problem Isomorphism
“Chalkboard XOR Game” can be mapped approximately to “Nim Game”.
In “Chalkboard XOR Game”, an array of numbers is given, and two players take turns to remove a number from the array. The game ends when the XOR of all the numbers of the array is zero.
Similarly, in “Nim Game”, there are piles of stones, and two players take turns removing any number of stones from a single pile, with the game ending when no stones are left.
Both problems share the concept of two players taking turns to remove elements (be it numbers or stones) until a certain condition is met (XOR becomes zero or no stones are left). The player who cannot make a move loses.
“Chalkboard XOR Game” has a more complex structure than “Nim Game” because it involves bitwise XOR operation and has an additional array data structure to deal with. “Nim Game” has a simpler structure as it deals directly with counts of stones without needing bitwise operations. “Nim Game” is simpler.
The “Chalkboard XOR Game” problem (LeetCode 810) involves both bitwise XOR operation and game theory. Here are some problems that could be solved before it to understand the basic concepts involved:
Single Number (LeetCode 136): This is a simple problem that introduces the concept of XOR operation.
Single Number II (LeetCode 137): A slightly more advanced problem that also uses XOR operation.
Single Number III (LeetCode 260): Another problem in the Single Number series. It’s good practice for using XOR in more complex ways.
Missing Number (LeetCode 268): This problem can be solved with XOR operation and it involves manipulating indices and values.
Find the Difference (LeetCode 389): This problem requires you to find the difference between two strings and can be solved using XOR operation.
Bitwise AND of Numbers Range (LeetCode 201): This problem will help you understand bitwise operations more thoroughly.
Counting Bits (LeetCode 338): In this problem, you need to count the number of 1 bits for each number in a range. This problem will help you get comfortable with bitwise manipulation.
Total Hamming Distance (LeetCode 477): This problem requires a deep understanding of bitwise operations, and will help solidify your understanding.
Nim Game (LeetCode 292): This problem introduces you to game theory.
Flip Game II (LeetCode 294): This problem builds on the Nim Game problem and introduces more complex game theory.
These problems will provide you with a solid foundation in bitwise operations and game theory, which are both required for solving the “Chalkboard XOR Game” problem.


Problem Classification
The problem falls into the domain of Game Theory and Bit Manipulation. It presents a twoplayer game involving Alice and Bob who take turns erasing a number from a given chalkboard (array of integers). The winner of the game is determined based on the state of the bitwise XOR of all the elements on the chalkboard.
Here are the ‘What’ components of the problem:
Array of Integers: The problem involves an array of integers,
nums
, which represents numbers written on a chalkboard.Two Players: The problem involves two players, Alice and Bob, who take turns to play.
Erasing a Number: In each turn, a player erases exactly one number from the chalkboard.
Game Rules: The rules of the game are as follows:
 If erasing a number makes the bitwise XOR of all elements become 0, the player who did the erasing loses.
 If the bitwise XOR of all elements is 0 at the start of a player’s turn, that player wins.
Winning Condition: The task is to determine if Alice can win the game, assuming both players play optimally.
Output: The output of the problem is a boolean value indicating whether Alice can win (True) or not (False).
Classifying the problem:
Bit Manipulation: The problem involves computation of bitwise XOR, which is a common operation in the field of bit manipulation.
Game Theory: The problem involves two players who alternately make moves. The outcome of the game depends on the choices of both players, and we are asked to determine the outcome assuming optimal play. This fits into the realm of game theory.
Dynamic Programming / Greedy: Since the optimal strategy depends on the current state of the board and the actions of the other player, this problem could potentially involve elements of dynamic programming or greedy algorithms, where we make the optimal choice at each step with the hope that these local optimal choices will lead to a global optimum. However, this is more of a speculation based on the “optimal play” phrase in the problem description, and we would need to explore the problem further to confirm if this classification is correct.
Clarification Questions
What are the clarification questions we can ask about this problem?
Identifying Problem Isomorphism
Can you help me with finding the isomorphism for this problem?
Which problem does this problem map to the corresponding isomorphic problem on Leetcode ?
Q&A
Why [1,2,3] returns true?
How can Alice win? No matter what she erase, Bob can erase any number and Alice will loss.
In the game described in the problem, the XOR of all elements in the array represents the current “value” of the game. If this value is 0 at the start of a player’s turn, they win. If a player’s move results in this value being 0, they lose. Let’s analyze the [1,2,3] scenario:
The initial XOR of [1,2,3] is 0 (since 1^2^3 equals 0). Remember, Alice always starts first.
Here are the possible scenarios:
If Alice removes 1, the remaining array is [2,3], which gives a XOR of 1 (since 2^3 equals 1). Then no matter what number Bob removes, Alice can remove the last number and win.
If Alice removes 2, the remaining array is [1,3], which gives a XOR of 2 (since 1^3 equals 2). Then no matter what number Bob removes, Alice can remove the last number and win.
If Alice removes 3, the remaining array is [1,2], which gives a XOR of 3 (since 1^2 equals 3). Then no matter what number Bob removes, Alice can remove the last number and win.
So in all cases, Alice can ensure that she wins, hence the function should return true for the input [1,2,3]. This is under the assumption that both players are playing optimally.
Less Algebraic Way
Suppose at current step we have xor to be w != 0. So, we lose immediately after one move only if we pick a number equal to w. So, we don’t lose if there exist a number distinct than w. And lose if all numbers are equal to w right now. So, if currently there are even number of numbers, if they are currently all equal then total xor would be 0 and we actually win already. As told before if exist two non equal numbers we can perform the move.
That’s a bit less algebraic way of deriving what was told in official solution.
Exactly, you’ve captured the essence of the strategy that should be used to solve this problem.
For any given state of the game, the player whose turn it is loses if and only if the XOR of all elements is 0 or all elements are equal (and nonzero).
The only chance for a player to win is to leave an array of numbers whose XOR is nonzero and at least two of the numbers are different after their turn. This forces the next player to leave a winning position (XOR of 0 or all equal nonzero numbers) after their turn.
If the XOR of all numbers initially is 0, Alice wins outright since it’s her turn first. If not, Alice should aim to remove a number such that the XOR of the remaining numbers is not zero and not all remaining numbers are equal. Since Alice moves first, she can always ensure this condition unless all numbers are initially equal. Hence, if the initial XOR is nonzero, Alice wins if and only if not all numbers are equal.
So the overall strategy can be condensed into a simple rule:
 Alice wins if and only if the XOR of all numbers is 0 or not all numbers are equal.
To implement this strategy in code, you’d just need to compute the XOR of all numbers and check whether they are all equal. If the XOR is 0 or not all numbers are equal, return True, otherwise return False. This solution is very efficient since it requires just one pass over the array and uses O(1) additional space.
Array Length Parity
nums = [x_1, x_2, …, x_n], S = x_1 ^ x_2 ^ … ^ x_n If S == 0 at first, then Alice wins with no doubt. If S != 0 at first. if n % 2 == 0: then there exists at least one x_i that makes next S != 0, (S=S ^ x_i). Why? because otherwise, if for all x_i, next S == 0, then (S^x_1)^…^(S^x_n) = (S ^ … ^S)^(x_1^…^x_n) = 0^S != 0, which contradicts with all next S == 0, (next S = S ^ x_i). So, Alice can make the number of nums odd safely. At last, Alice wins. if n % 2 != 0: Alice just can make it even, then Bob wins as case n % 2 == 0.
Yes, your understanding is correct. This game, at its core, is about the XOR operation properties and array length parity.
When the bitwise XOR (S) of all the elements is 0, Alice wins since it’s her turn, and Bob would be the one to force the XOR to become 0 again after Alice’s move. Hence, Alice would win before Bob has a chance to play.
When S is not equal to 0 and the number of elements (n) is even, Alice can always select a number that does not make S equal to 0, forcing Bob to play with an oddsized array. Alice can keep forcing Bob to play with an oddsized array until the last turn, where Bob is forced to take the last number and lose.
However, when S is not equal to 0 and n is odd, Alice can only make it evensized for Bob. As Bob has the same optimal strategy as Alice, he will keep giving an oddsized array back to Alice, forcing her to lose the game.
This insight allows us to solve the problem with just one pass over the array. If S (the XOR of all numbers) is 0 or the length of the array is even, Alice wins, otherwise, she loses. So, the Python code would look something like this:


Why even never loses?
Let me give you a simple example to let you know: why even never lose?
if some number pass to you (you are alice, the pretty girl), and the xor of them are not 0 (as if they are 0, then Bob lose, Alice win) So, what would the numbers looks like?
as this is a xor issue, I will write binary so you could see more clearer num1: 2  0 0 0 1 0 num2: 1  0 0 0 0 1 num3: 7  0 0 1 1 1 XOR : 4  0 0 1 0 0
So, the XOR would be 0 if all digit int the xor number are 0. Even if one digit is not 0, make the xor not 0. Then, we let one digit not be 0, be 1. How does xor of a specific digit be 1? odd number of items that has bit 1 in that bit.(this is the hardest setence in this explanation, if you understand, you pass) Ok! I dicide to give a expmple! Just the above graph, only the 3rd digit xor 1, as there are only 1 item num3 has 1 in digit 3. And if there are 1,3,5,7,9… items with that digit 1, xor is always 1(maybe too clear and seems stupid)
So, the number pass to you must looks like above case! Then what? You must not lose, means you must delete a item, the left xor is still not 0, And if above case is still guaranteed, then it is solved!
So!, we delete other items than num3, as we want to leave the only item that has 1 in the 3rd digit. Then it is simple, just delete num1 or num2.
Now, it seems still there isn’t anything related with “even/odd” condition appears as in the solution code…
Think! what if there are no item1 and item2… Then, when you delete item3 which you must as it is the only thing, you lost! Think more!!! I said, it could be 1,3,5,7,9… items with that digit 1. Like: num1: 4  0 0 1 0 0 num2: 4  0 0 1 0 0 num3: 4  0 0 1 0 0 Yeah, same number is a good example, but there are more, I will show you another one, (I am not laike those show one case that satisfied themselves and hurry for conclusion) num1: 6  0 0 1 1 0 num2: 5  0 0 1 0 1 num3: 4  0 0 1 0 0
The same number example, you see that, if you move any number, you lose! The below example save you from dying, but you didn’t win, you need to play longer to see…
So……….What is the condition to make you not lose? Give me at least one number besides the odd items with specified digit 1!!! if the odd number is 3, then, give me another not related num4.
At least one beside the odd items which isssssssssss even items even items comes to you, you must not die, as its xor is not 0, so it must have a digit that only odd item has 1, so you always can pick the one other than those as there are even items.
A game, 2 players, even means not dead, then odd always give to the opponent, if he is lucky, he can suvive longer, the longest is we pass him the last item, he delete it, he lose, we win.
Thanks, guys. My first time to write my ideas. Maybe too long, but the logic I think is correct enough. But as the “Solution” mentioned, there are more cases, that makes us must win even we have odd items. Which means we make a dicision machine, and here is a poke give to us, we think that is not good accroding to the logic of our machine, but the truth is that poke is awsome. Out solution didn’t include this case, make us not wrong, but partly true.
But that part could be hardto deduce, and I choose to stop here.
Your detailed explanation is quite insightful and largely correct. You’re essentially using properties of the XOR operation and game theory to argue why Alice, given an even number of elements, can always win if the XOR of all elements is not 0.
In your example, you’ve shown that an array with an odd number of elements where XOR is not 0 leads to Alice’s loss. If Alice has an array with an even number of elements where XOR is not 0, she can always remove an element such that the remaining elements’ XOR is still not 0. This is because there must be at least one element that does not contribute to the ‘1’ in the XOR result. By removing such an element, the XOR of the remaining elements is still not 0, and now Bob is forced to deal with an odd number of elements.
You also correctly pointed out that in some cases, Alice can still win even with an odd number of elements, but deriving these cases can be complex.
In a nutshell, your understanding is in line with the problem’s solution: if the XOR of all numbers is 0 or the number of elements is even, Alice wins. Otherwise, she loses. This conclusion nicely combines the properties of XOR operation, parity of the element count, and the turnbased game nature.
Can firsthand always flight to last move?
The solution relies on the fact that the first player always has a valid move when there are even number of entries. I am wondering whether it is also true for odd number of entries with at least two distinct elements. let A=s[0]^…s[n1] WLOG, we can assume s[0]!=s[1]. let B=A^s[0] if B!=0, then s[0] is a safe move. otherwise, s[1] must be a safe move since A^s[1]=B^(s[0]^s[1])=0^(s[0]^s[1])!=0 when s[0]!=s[1].
The problem did not guarantee the numbers are distinct, but we can always reduce it to such without changing the outcome. => strategy: If opponent take one of the two equal numbers, then take the other one, otherwise, follow the same way the player would play in the reduced board. so xorGame(s[0]^…s[n1])==xorGame(s[3]^…s[n1]) if s[0]==s[1]
If my assumption above is right, does it mean the first player can always fight to the last move provided that the game start with at least two distinct numbers?
You’re absolutely correct in your analysis. In essence, when it’s Alice’s turn, as long as there are at least two distinct numbers in the array, Alice can always make a move such that the bitwise XOR of the remaining numbers is not zero. Hence, Alice can continue the game until the very last move.
Your proof relies on two essential points:
If the XOR of all elements is not zero, there exists a number that Alice can remove such that the XOR of the remaining numbers is still not zero.
If the XOR of all elements is zero, Alice has already won. She doesn’t need to continue the game. But if she does continue, she can always ensure that the XOR of the remaining numbers is not zero.
With these two points, Alice can control the game until the very last move as long as the array has at least two distinct numbers.
However, remember that having control until the last move doesn’t always mean winning. Alice wins if and only if the XOR of all numbers is zero or the number of elements is even. This is because if Alice ends up with an array with an odd number of elements where the XOR is not zero, Bob will win instead.
Constraints
The problem statement is a turnbased game scenario where Alice and Bob take turns to remove numbers from an array and the winner is determined based on the XOR of the remaining numbers in the array. Here are the key characteristics or conditions that we can exploit for an efficient solution:
Bit Manipulation: The game is based on the XOR operation, which is a binary operation. Bit manipulation is a powerful tool that can be exploited for an efficient solution. Understanding how XOR operation works is the key to solving this problem.
Constraints: The constraints of the problem (0 <= nums[i] < 2^16, 1 <= nums.length <= 1000) suggest that the numbers are bounded and of moderate size. This makes bit manipulation feasible and efficient for this problem.
Parity of the array length: The problem statement suggests that the parity (evenness or oddness) of the number of elements in the array plays a role in determining the winner. If the total XOR of the array is zero, then Alice wins since the XOR of an even number of same elements is zero. If the total XOR of the array is nonzero and the number of elements is even, Alice can still make a move that ensures the XOR of the remaining numbers is nonzero, effectively ensuring her victory.
Optimal Strategy: Given that both players play optimally, we can assume that they will make the best possible move at each step. This means that the solution should account for the best strategy that each player can employ based on the current state of the game.
By recognizing and leveraging these characteristics, we can devise an efficient solution strategy for this problem.
Solution Approach and Analysis
Let’s use the game metaphor to explain the solution.
Imagine that Alice and Bob are playing a game of “Digital Chess”. Each move, they remove a chess piece (number) from the board (array). The overall board state is represented by a binary value (XOR of all numbers), and if a player’s move results in a board state of 0, they lose.
The key to this game is understanding how the XOR operation affects the board state. XOR is a binary operation that returns 1 if the number of 1’s is odd, and 0 if it’s even.
First, we calculate the overall board state by taking the XOR of all the numbers. This represents the initial board state.
Alice goes first. If the initial board state is 0 (i.e., the XOR of all numbers is 0), then Alice wins because the board state is already 0 and it’s Bob’s turn.
If the initial board state is not 0 and the number of pieces (numbers) is even, Alice can always make a move to ensure the board state remains nonzero. This is because the board state is nonzero only when there’s at least one binary digit where the number of 1’s is odd. Since the total number of numbers is even, Alice can always avoid removing a number with a 1 in that digit, leaving an odd number of such numbers for Bob and ensuring the board state remains nonzero.
Now, let’s explain why the parity of numbers matter using an example:
 Suppose we have numbers [3, 3, 2], their XOR is 2. It’s Alice’s turn and no matter which number she removes, the XOR of the remaining numbers is always 0 (Bob wins).
 However, if we have an even number of numbers, say [3, 3, 2, 2], the XOR is 0 (Alice wins directly) or if the XOR is not 0, Alice can always make a move to keep the XOR nonzero by removing a number which doesn’t affect the digit where the number of 1’s is odd in the XOR.
Finally, how would specific operations or changes in the problem’s parameters affect the solution?
 If we increase the range of the numbers or the size of the array, it wouldn’t affect the solution’s correctness, but it could impact the efficiency if the solution is not optimized for large inputs.
 If the game rule changes such that the player who makes the XOR of all numbers to 0 wins, then the solution would need to be adjusted accordingly.
This problem is a great example of how understanding bitwise operations and parity can be used to solve complex problems.
Thought Process
Given the problem statement, the cues and insights that we can extract are:
 We are dealing with bitwise XOR operation.
 XOR operation has certain properties that we could leverage: a. XOR of a number with itself is 0. b. XOR of a number with 0 is the number itself. c. XOR results in a 1 bit only if the corresponding bit in the input numbers was 1 an odd number of times.
 Alice and Bob are playing optimally, which means they will make the best possible move at each step.
 We need to determine if Alice can win the game, i.e., if there’s a strategy that Alice can follow to win.
 Alice moves first and they take turns erasing exactly one number.
The problem involves bitwise operations and game theory. A key observation is that if the XOR of all numbers is 0, the current player will win. This is because XOR of a number with itself is 0, and hence, the remaining numbers (after the current player makes a move) cannot be XORed to 0.
Given these insights, here is a stepbystep approach:
Calculate the XOR of all the numbers. If it’s 0, Alice wins because she is the first to move.
If the XOR is not 0 and the number of elements is odd, Bob will win because no matter what Alice chooses to erase first, Bob can always mimic Alice’s moves on the remaining turns and leave Alice with a XOR of 0.
If the XOR is not 0 and the number of elements is even, Alice can always make a move such that the XOR of the remaining numbers is not 0. This is because if the XOR is not 0, there must be at least one bit where the number of 1’s is odd, and Alice can avoid removing a number with a 1 in that bit. This leaves Bob with an odd number of numbers that have a 1 in that bit, ensuring that the XOR remains nonzero.
Here is the Python code implementing the above approach:


This function first checks if the length of the input list nums
is even. If it is, the function immediately returns True
, as Alice will always win. If the length of nums
is odd, the function computes the XOR of all numbers in nums
. If the XOR is 0, Alice wins and the function returns True
; otherwise, it returns False
.
This solution takes advantage of the properties of XOR operation and the parity of the number of elements in nums
to solve the problem efficiently. The time complexity is O(n), where n is the length of nums
, and the space complexity is O(1), as we only need a constant amount of space to store the XOR and the length of nums
.
Language Agnostic Coding Drills
Dissecting the Code:
Here are the distinct coding concepts contained in the code snippet:
 Variable Initialization: The concept of creating and initializing a variable to a specific value. In this code, this is represented by initializing
xor
to 0.  List Length Calculation: The concept of determining the number of elements in a list. This is accomplished using
len(nums)
.  Conditional Statements: The concept of executing different blocks of code depending on whether a certain condition is met. This is achieved by using the
ifelse
statement.  Modulo Operation: The concept of using the modulo operator
%
to find the remainder of division. This operator is used to check if the length ofnums
is even.  Looping Through Lists: The concept of iterating over each element in a list. This is done using a
for
loop to iterate overnums
.  Bitwise XOR Operation: The concept of using the bitwise XOR operator
^=
to modify a value based on each element in the list.
 Variable Initialization: The concept of creating and initializing a variable to a specific value. In this code, this is represented by initializing
Ordering Coding Concepts by Difficulty:
Here are the identified concepts, listed in increasing order of difficulty:
 Variable Initialization: This is one of the simplest concepts in programming and is typically one of the first things beginners learn.
 List Length Calculation: While this concept requires knowledge of lists, it’s a straightforward concept and widely used.
 Conditional Statements: This concept involves logical thinking and control flow understanding, making it slightly more complex than variable initialization and list length calculation.
 Looping Through Lists: Understanding loops requires a grasp of control flow and is a key component of most programming tasks.
 Modulo Operation: Understanding the modulo operation requires some mathematical knowledge, making it a slightly more advanced concept than simple arithmetic operations.
 Bitwise XOR Operation: This is the most advanced concept identified in this code. It involves understanding binary numbers and bitwise operations, which are relatively advanced topics in computer science.
Describing the ProblemSolving Approach:
This problem involves determining whether Alice, who plays first in a game involving bitwise XOR operations, wins or not. Here’s how the identified coding concepts contribute to the overall solution:
 First, the List Length Calculation concept is used to determine the number of elements in
nums
.  The Modulo Operation concept is then used to check if the length of
nums
is even. If it is, Alice will always win, so Conditional Statements are used to returnTrue
in this case.  If the length of
nums
is odd, we use the Variable Initialization and Looping Through Lists concepts to calculate the XOR of all the elements in the list.  Finally, Conditional Statements are used again to check if the XOR of all the elements is zero. If it is, Alice loses (so return
False
), otherwise, Alice wins (so returnTrue
).
Note: This is a more complex problem that involves some nuanced understanding of bitwise operations and game theory. The problemsolving approach relies on the properties of the XOR operation and the strategy of Alice and Bob. Alice can always win if the number of elements is even, and if the number of elements is odd, Alice can only win if the XOR of all the elements is not zero. This strategy is based on mathematical proofs and may not be intuitively obvious.
 First, the List Length Calculation concept is used to determine the number of elements in
Targeted Drills in Python
Implementing Python Coding Drills for General Concepts:
 Variable Initialization
1 2
xor = 0 print(xor)
 List Length Calculation
1 2 3
nums = [1, 2, 3, 4, 5] length = len(nums) print(length)
 Conditional Statements
1 2 3 4 5
num = 5 if num % 2 == 0: print("Even") else: print("Odd")
 Looping Through Lists
1 2 3
nums = [1, 2, 3, 4, 5] for num in nums: print(num)
 Modulo Operation
1 2 3
num = 5 remainder = num % 2 print(remainder)
 Bitwise XOR Operation
1 2 3 4 5
xor = 0 nums = [1, 2, 3, 4, 5] for num in nums: xor ^= num print(xor)
 Variable Initialization
Implementing Python Coding Drills for ProblemSpecific Concepts:
The problemspecific concept in this problem is the use of bitwise XOR operation in the context of a game. It is essential for our problem because the XOR operation is the foundation of the game. Here is a drill for this concept:
 XOR Game Logic
1 2 3 4 5 6 7 8
nums = [1, 2, 3, 4, 5] xor = 0 for num in nums: xor ^= num if len(nums) % 2 == 0 or xor == 0: print("Alice wins") else: print("Bob wins")
 XOR Game Logic
Assembling the Drills:
To solve the initial problem, these pieces can be integrated together in the following order:
 Variable Initialization: Initialize the variable
xor
to 0.  List Length Calculation: Calculate the length of the list
nums
.  Conditional Statements and Modulo Operation: Check if the length of
nums
is even. If it is, returnTrue
(Alice wins).  Looping Through Lists and Bitwise XOR Operation: If the length of
nums
is not even, calculate the XOR of all the elements innums
.  Conditional Statements: Check if the XOR of all the elements is zero. If it is, return
False
(Bob wins), otherwise returnTrue
(Alice wins).
The final solution would look something like this:
1 2 3 4 5 6 7 8 9 10 11
from typing import List class Solution: def xorGame(self, nums: List[int]) > bool: xor = 0 for num in nums: xor ^= num if len(nums) % 2 == 0 or xor == 0: return True else: return False
Each drill contributes to building up to this final solution by teaching a specific coding concept that is necessary for solving the problem.
 Variable Initialization: Initialize the variable