LeetCode - Easy - 292. Nim Game

Topic

Brainteaser
Minimax
Bit Manipulation

Description

You are playing the following Nim Game with your friend:

Initially, there is a heap of stones on the table.
You and your friend will alternate taking turns, and you go first.
On each turn, the person whose turn it is will remove 1 to 3 stones from the heap.
The one who removes the last stone is the winner.

Given n, the number of stones in the heap, return true if you can win the game assuming both you and your friend play optimally, otherwise return false.

Example 1:

Input: n = 4
Output: false
Explanation: These are the possible outcomes:
1. You remove 1 stone. Your friend removes 3 stones, including the last stone. Your friend wins.
2. You remove 2 stones. Your friend removes 2 stones, including the last stone. Your friend wins.
3. You remove 3 stones. Your friend removes the last stone. Your friend wins.
In all outcomes, your friend wins.

Example 2:

Input: n = 1
Output: true

Example 3:

Input: n = 2
Output: true

Constraints:

1 <= n <= 2³¹ - 1

Analysis

Theorem: The first one who got the number that is multiple of 4 (i.e. n % 4 == 0) will lost, otherwise he/she will win.

Proof:

the base case: when n = 4, as suggested by the hint from the problem, no matter which number that that first player, the second player would always be able to pick the remaining number.
For 1* 4 < n < 2 * 4, (n = 5, 6, 7), the first player can reduce the initial number into 4 accordingly, which will leave the death number 4 to the second player. i.e. The numbers 5, 6, 7 are winning numbers for any player who got it first.
Now to the beginning of the next cycle, n = 8, no matter which number that the first player picks, it would always leave the winning numbers (5, 6, 7) to the second player. Therefore, 8 % 4 == 0, again is a death number.
Following the second case, for numbers between (24 = 8) and (34=12), which are 9, 10, 11, are winning numbers for the first player again, because the first player can always reduce the number into the death number 8.

位操作小技巧：

2的n次方的整数倍和2的n次方减1按位与后值为0。

$(a\cdot2^n) \& (2^n-1)=0(a\in N, n\in N )$

例如：

n = 2，2ⁿ = 4， 2ⁿ - 1 = 3

0 & 3 = 0 & 11 = 0
4 & 3 = 100 & 11 = 0
8 & 3 = 1000 & 11 = 0
12 & 3 = 1100 & 11 = 0
16 & 3 = 10000 & 11 = 0
20 & 3 = 10100 & 11 = 0
24 & 3 = 11000 & 11 = 0
28 & 3 = 11100 & 11 = 0
32 & 3 = 100000 & 11 = 0
36 & 3 = 100100 & 11 = 0
40 & 3 = 101000 & 11 = 0
44 & 3 = 101100 & 11 = 0
48 & 3 = 110000 & 11 = 0
52 & 3 = 110100 & 11 = 0
56 & 3 = 111000 & 11 = 0
60 & 3 = 111100 & 11 = 0
64 & 3 = 1000000 & 11 = 0
68 & 3 = 1000100 & 11 = 0
72 & 3 = 1001000 & 11 = 0
76 & 3 = 1001100 & 11 = 0
80 & 3 = 1010000 & 11 = 0
84 & 3 = 1010100 & 11 = 0
88 & 3 = 1011000 & 11 = 0
92 & 3 = 1011100 & 11 = 0
96 & 3 = 1100000 & 11 = 0
...

又例如：

n = 4，2ⁿ = 16， 2ⁿ - 1 = 15

0 & 15 = 0 & 1111 = 0
16 & 15 = 10000 & 1111 = 0
32 & 15 = 100000 & 1111 = 0
48 & 15 = 110000 & 1111 = 0
64 & 15 = 1000000 & 1111 = 0
80 & 15 = 1010000 & 1111 = 0
96 & 15 = 1100000 & 1111 = 0
112 & 15 = 1110000 & 1111 = 0
128 & 15 = 10000000 & 1111 = 0
144 & 15 = 10010000 & 1111 = 0
160 & 15 = 10100000 & 1111 = 0
176 & 15 = 10110000 & 1111 = 0
192 & 15 = 11000000 & 1111 = 0
208 & 15 = 11010000 & 1111 = 0
224 & 15 = 11100000 & 1111 = 0
240 & 15 = 11110000 & 1111 = 0
256 & 15 = 100000000 & 1111 = 0
272 & 15 = 100010000 & 1111 = 0
288 & 15 = 100100000 & 1111 = 0
304 & 15 = 100110000 & 1111 = 0
320 & 15 = 101000000 & 1111 = 0
336 & 15 = 101010000 & 1111 = 0
352 & 15 = 101100000 & 1111 = 0
368 & 15 = 101110000 & 1111 = 0
384 & 15 = 110000000 & 1111 = 0
400 & 15 = 110010000 & 1111 = 0
416 & 15 = 110100000 & 1111 = 0
432 & 15 = 110110000 & 1111 = 0
448 & 15 = 111000000 & 1111 = 0
464 & 15 = 111010000 & 1111 = 0
480 & 15 = 111100000 & 1111 = 0
496 & 15 = 111110000 & 1111 = 0
512 & 15 = 1000000000 & 1111 = 0
528 & 15 = 1000010000 & 1111 = 0
544 & 15 = 1000100000 & 1111 = 0
560 & 15 = 1000110000 & 1111 = 0
576 & 15 = 1001000000 & 1111 = 0
592 & 15 = 1001010000 & 1111 = 0
608 & 15 = 1001100000 & 1111 = 0
624 & 15 = 1001110000 & 1111 = 0
640 & 15 = 1010000000 & 1111 = 0
656 & 15 = 1010010000 & 1111 = 0
672 & 15 = 1010100000 & 1111 = 0
688 & 15 = 1010110000 & 1111 = 0
704 & 15 = 1011000000 & 1111 = 0
720 & 15 = 1011010000 & 1111 = 0
736 & 15 = 1011100000 & 1111 = 0
752 & 15 = 1011110000 & 1111 = 0
768 & 15 = 1100000000 & 1111 = 0
784 & 15 = 1100010000 & 1111 = 0
800 & 15 = 1100100000 & 1111 = 0
816 & 15 = 1100110000 & 1111 = 0
832 & 15 = 1101000000 & 1111 = 0
848 & 15 = 1101010000 & 1111 = 0
864 & 15 = 1101100000 & 1111 = 0
880 & 15 = 1101110000 & 1111 = 0
896 & 15 = 1110000000 & 1111 = 0
912 & 15 = 1110010000 & 1111 = 0
928 & 15 = 1110100000 & 1111 = 0
944 & 15 = 1110110000 & 1111 = 0
960 & 15 = 1111000000 & 1111 = 0
976 & 15 = 1111010000 & 1111 = 0
992 & 15 = 1111100000 & 1111 = 0

日后，判断一个数a能否被2的n次方整除，除了用求余，还可以用刚介绍位操作小技巧：2的n次方的整数倍和2的n次方减1按位与后值为0。

$(a\%2^n)=(a\&(2^n-1))(a\in N,n\in N)$

Submission

public class NimGame {
    
    
	public boolean canWinNim1(int n) {
    
    
		return n % 4 != 0;
	}

	public boolean canWinNim2(int n) {
    
    
		return (n & 3) != 0;
	}

}

Test

import static org.junit.Assert.*;
import org.junit.Test;

public class NimGameTest {
    
    

	@Test
	public void test() {
    
    
		NimGame obj = new NimGame();

		for (int i = 0; i < 1000; i++)
			assertEquals(obj.canWinNim1(i), obj.canWinNim2(i));
	}
}