ganesh_6's blog

What is the reason behind the algorithm in solving Nim Sum 2
By ganesh_6, history, 2 years ago, In English

I know Nim Game, Nim Sum (xor sum) and the theorem, lemmas behind that. Also, I solved the https://cses.fi/problemset/task/1730 After solving that, I moved to https://cses.fi/problemset/task/1098 I could not figure out the solution. So, I found one solution online and could not figure out the nuances behind the working of the algorithm. I can see that he has used a[i]%4 which is used for one pile game to find out the winner. However, in this case he combined Nim Game 1 algorithm and 1 pile game algorithm to derive the solution. Please explain the reason behind working of this?

#include<bits/stdc++.h>
using namespace std;

string solve(int n, vector<int> heaps){
    for(int i=0;i<n;i++)
    {
        heaps[i]%=4;
    }
    int xr=0;
    for(int i=0;i<n;i++)
    {
        xr^=heaps[i];
    }
    if(xr)
    {
        return "first";
    }
    else{
        return "second";
    }
}
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2 years ago, # |
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tl;dr: The terms for you to google are Grundy numbers and the Sprague-Grundy theorem.

I wrote a blog explaining this a few years ago: https://codeforces.com/blog/entry/66040

Jump to "Extension on Nim and Grundy Numbers"

I do intend to rewrite this blog sometime this year since my college-level writing is not the best. But if you find it unsatisfactory, you can just google for other articles using the above keywords

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10 months ago, # |
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When you do %4 it isn't changing anything the outcome is still the same. The only difference is that now it becomes just like nim game 1 because everything is below 4 and you can remove 1, 2, or 3. This is essentially nim game 1 where you can remove anything from any pile. Now you can do nim sum just like in nim game 1.

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