Nim Game
Identifying Problem Isomorphism
“Nim Game” has a close isomorphic relationship with “Divisor Game”.
In “Nim Game”, there is a single pile of stones, and two players take turns removing 1 to 3 stones from the pile. The person who removes the last stone wins. The solution of the game depends on the initial number of stones. If the number is a multiple of 4, the first player will lose, otherwise, the first player can always win by keeping the number of stones remaining in the pile a multiple of 4.
Similarly, in the “Divisor Game”, two players take turns choosing a positive divisor (less than the number itself) of a number and subtracting it from that number. The player who is left with the number 1 wins. Again, the game’s outcome depends on the initial number. If it’s even, the first player wins. Otherwise, they lose.
Both problems involve two players, and the strategies for winning the games are based on keeping or bringing the game to a state that leaves the second player with a disadvantageous position (multiple of 4 in Nim Game, even number in Divisor Game). This structural similarity establishes their isomorphism.
“Divisor Game” is simpler due to the straightforward concept of even and odd numbers, while “Nim Game” needs a little bit of deeper understanding around multiples of a particular number (4 in this case).


In this game, there are some rocks and on each turn, a player can take 1, 2 or 3 rocks. The player who takes the last rock loses the game.
The idea here is to think about what happens after your turn, putting yourself in your friend’s shoes. Let’s look at an example:
If there are 4 rocks left, and it’s your turn, no matter how many rocks you take (1, 2, or 3), your friend will be in a winning position. This is because after your turn, your friend can start with 3, 2, or 1 rocks, respectively, and can always take enough rocks on their turn to win the game.
Therefore, if there are 4 rocks left at the start of your turn, you’re destined to lose the game, assuming your friend plays optimally.
The interesting observation here is that if there are 1, 2, or 3 rocks left at the start of your turn, you’re in a winning position. That’s because you can take all remaining rocks and win the game immediately.
This pattern repeats for multiples of 4. If the number of rocks left at the start of your turn is a multiple of 4, you’re destined to lose, because no matter how many rocks you take, your friend can start their turn with 1, 2, or 3 rocks less than a multiple of 4, which is a winning position.
Therefore, the best strategy for this game is to always leave a number of rocks that is a multiple of 4 after your turn. In terms of coding, this can be checked by returning n%4 in C language. If n%4 equals 0, you’re going to lose the game, if played optimally by both players. If n%4 is not 0, then you can win by taking the right number of rocks to leave a multiple of 4 for your friend’s turn.
Let’s discuss a game strategy based on the SpragueGrundy theorem, a concept in game theory. It can seem a little complex, so let’s break it down.
In many games, like the one in this explanation where you can remove 1, 2, or 3 stones from a pile, we can use a strategy based on “nimbers” and “subgames.” A subgame is just a part of the game that we can analyze separately.
A “nimber” is a value we calculate for each subgame, which tells us whether we’re likely to win or lose. The nimber is zero if we’ve already lost the subgame. Otherwise, it’s the smallest number that isn’t the nimber of a game state we can reach in our next move. Basically, it’s a number that represents the state of the game.
Here’s where things get interesting: we can predict whether the first player will win or lose by taking the nimbers of all the subgames and applying a mathematical operation called “XOR” (exclusive OR). If the result is zero, the first player will lose; if it’s not zero, they will win.
Now, back to the stonepicking game. In this case, the whole game is one subgame, so we only have one nimber to consider. We can find the nimber of a game state (a pile of stones) by examining what states we can move to.
If there are no stones (empty pile), we’ve already lost, so the nimber is 0. If there’s one stone, we can remove it and reach an empty pile (nimber 0), so the nimber for one stone is 1. For two stones, we can remove one or two and reach a state with nimber 0 or 1, so the nimber for two stones is 2. For three stones, we can reach states with nimber 0, 1 or 2, so the nimber for three stones is 3.
Here’s the interesting bit: for four stones, we can only reach states with nimber 1, 2 or 3. Since we can’t reach a state with nimber 0, that’s our nimber for four stones. So, the nimber of a pile of stones is the number of stones modulo 4 (the remainder when you divide the number of stones by 4).
This means the first player will win if and only if the number of stones isn’t a multiple of 4, because the nimber (and thus the result of the XOR operation) will be nonzero only in that case.
So the SpragueGrundy theorem gives us a smart strategy: always leave a number of stones that is a multiple of 4, and you’ll win the game (as long as your opponent also plays optimally).
Problem Classification
The problem can be classified into the following categories:
Game Theory: The problem involves two players taking turns to make moves, with the goal of winning the game. Both players are assumed to play optimally.
Mathematical Problem: The solution to the problem relies on discovering and applying a mathematical pattern or rule.
Decisionmaking Problem: The goal is to determine a boolean outcome  whether or not you can win the game.
Combinatorial Problem: The problem involves determining the number of ways a certain outcome can occur, in this case, the removal of 1 to 3 stones in each turn.
Language Agnostic Coding Drills
Understanding Boolean: The base concept here is understanding what a Boolean is. It is a data type that can be either
True
orFalse
.Basic Operations and Control Flow: Understand the basic operations such as comparison (
==
,!=
,<=
,>=
), and control flow tools (if
,for
loops).Array/List manipulation: Understand how to create and manipulate arrays/lists, such as creating an array of a certain size and accessing and updating its elements.
Mathematical Concepts: Understand the basic math concept of less than, greater than, and equal to.
Dynamic Programming: The solution uses a dynamic programming approach where an array (memo) is used to store the result for each number of stones (n) and this array is filled up incrementally.
Problemspecific Concept  Game rules: Understand the rules of the Nim game. If the number of stones n is less than or equal to 3, you can win by taking all the stones. For other cases, you need to figure out a strategy.
Approach to solving:
The problem can be approached in the following steps:
If n is less than or equal to 3, return True because you can win by taking all the stones.
Create a memoization array of size n + 1 and initialize the first four elements to True (since you can win if n is less than or equal to 3).
Iterate through the rest of the memoization array. For each index i, iterate through the last three indices (ij, where j ranges from 1 to 3). If any of those indices corresponds to a losing game (False), set the current index to True (since you can force the other player into a losing position). Otherwise, set the current index to False (since the other player can force you into a losing position).
The final element in the memoization array (index n) will represent whether you can win or not when there are n stones.
Targeted Drills in Python
Understanding Boolean:
1 2 3 4
a = True b = False print(a == b) # Should print False print(a != b) # Should print True
Basic Operations and Control Flow:
1 2
for i in range(5): # Should print numbers 0 to 4 print(i)
Array/List manipulation:
1 2 3
arr = [False] * 5 # Creates an array of size 5 with all elements as False arr[0] = True # Sets the first element to True print(arr) # Should print [True, False, False, False, False]
Mathematical Concepts:
1 2 3
print(4 > 3) # Should print True print(2 <= 1) # Should print False print(3 == 3) # Should print True
Dynamic Programming:
1 2 3 4 5 6 7 8 9
memo = [False] * 10 for i in range(10): if i < 4: memo[i] = True else: for j in range(1,4): if memo[ij] == False: memo[i] = True print(memo) # Should print first four True and then alternate True, False
You will run into memory limit exceeded with this solution. The issue here seems to be that you are using a bottomup dynamic programming approach to solve the problem, which requires an array of size n
. This can lead to memory exhaustion when n
is very large.
However, for the Nim Game problem, there is a much simpler mathematical solution. According to the game’s rule, you can always win a Nim game if the number of stones n
in the pile is not divisible by 4. That’s because no matter how many stones you remove (1, 2, or 3), your opponent can always remove enough stones to make the total number of stones removed in that round equal to 4. Therefore, as long as n
is not divisible by 4, you can always leave your opponent with a pile of size 4k, which guarantees your victory.
Here’s how you can modify your function to implement this solution:


This function now only uses constant memory, so it should no longer cause a Memory Limit Exceeded error.
 Problemspecific Concept  Game rules:
1 2 3 4 5
n = 10 # Number of stones if n <= 3: print(True) # You can win by taking all the stones else: print(False) # Requires further calculation
By understanding these drills, you should have the skills necessary to integrate them and solve the Nim game problem.