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  1. Prepare
  2. Algorithms
  3. Dynamic Programming
  4. Prime XOR
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Prime XOR

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  • staryounger012
     + 0 comments

    This problem involves finding the number of unique multisets formed from an array such that the bitwise XOR of all elements in the multiset is a prime number. The challenge lies in handling duplicates and efficiently counting multisets rather than sequences or subsets.

    Key Points:

    Multisets vs Subsets: Unlike regular subsets, multisets allow repeated elements. For elements that appear multiple times, the count of possible selections ranges from zero to their frequency, i.e., f + 1 choices per element.

    XOR Behavior with Multiplicity: Adding an element an even number of times to the XOR results in no change, while an odd number of times flips the XOR by that element's value. Thus, the parity of the count chosen for each element affects the XOR outcome.

    Dynamic Programming Approach: We use a DP array indexed by XOR values, where dp[x] counts the number of multisets with XOR x. For each unique element, we split the choices into two groups: those with even counts and those with odd counts, and update the DP accordingly.

    Prime Check: After processing all elements, we sum the counts of multisets whose XOR is prime. Efficient prime checking with a sieve up to the maximum XOR value (8192 here) optimizes the solution.

    Modulo Arithmetic: Since the number of multisets can be very large, we take all counts modulo 10^9 + 7.

    This approach ensures that the counting respects multiset properties and efficiently computes the required total under given constraints.

  • Tusenka
     + 1 comment

    For multiset there would be infinit sets, am I right? Lets b - prime. Then Example: a^a=0 0^b=b {b} a^a^b=b {a, a, b} The example is wrong

  • vladalx
     + 0 comments

    Since what is required is a MULTISET ( can contain duplicates ) that can be formed from the elements of the initial array ( not specified that should only be used once, meaning an element can be selected multiple times for a multiset ), then the result is : an infinite number of multisets. If the elements should only be used once, then what is the reason for the MULTISET specification ?

  • zoyashah41306
     + 1 comment

    Prime XOR is a fascinating concept in computer science and cryptography, involving the exclusive OR (XOR) operation applied to prime numbers. This technique leverages the unique properties of prime numbers to enhance data security and encryption methods. As developers and researchers explore Prime XOR, they uncover new potentials for creating robust encryption algorithms that safeguard information against unauthorized access. Printed tote bags featuring the Prime XOR theme become popular among enthusiasts, symbolizing their passion for cutting-edge technology and cryptographic advancements. These bags not only serve a practical purpose but also inspire conversations about the future of digital security and innovation.

  • dehaka
     + 0 comments
    long primeXor(vector<int> arr) {
    vector<bool> prime(8192, true);
    prime[0] = false, prime[1] = false;
    for (int p = 2; p*p <= 8191; p++) {
        if (prime[p] == true) {
            for (int i = p*p; i <= 8191; i += p)
                prime[i] = false;
        }
    }
    long total = 0, mod = 1000000007;
    vector<long> counter(8192);
    vector<int> dupe(4501), unique;
    for (int i=0; i < arr.size(); i++) {
        if (dupe[arr[i]] == 0) unique.push_back(arr[i]);
        dupe[arr[i]]++;
    }
    counter[0] = 1;
    for (int i=0; i < unique.size(); i++) {
        vector<long> temporary = counter;
        int F = dupe[unique[i]]/2;
        for (int j=0; j < 8192; j++) {
            int K = unique[i]^j;
            counter[j] = (counter[j] + temporary[j]*F) % mod;
            if (dupe[unique[i]] % 2 == 0) {
                counter[K] = (counter[K] + temporary[j]*F) % mod;
            } else {
                counter[K] = (counter[K] + temporary[j]*(F+1)) % mod;
            }
        }
    }
    for (int i=0; i < counter.size(); i++) {
        if (prime[i]) total = (total + counter[i]) % mod;
    }
    return total;
}