USACO Training "hamming": Hamming Codes

-

USACOT "hamming": Hamming Code

Problem Statement

Given N, B, and D: Find a set of N codewords (1 <= N <= 64), each of length B bits (1 <= B <= 8), such that each of the codewords is at least Hamming distance of D (1 <= D <= 7) away from each of the other codewords. The Hamming distance between a pair of codewords is the number of binary bits that differ in their binary notation. Consider the two codewords 0x554 and 0x234 and their differences (0x554 means the hexadecimal number with hex digits 5, 5, and 4):

        0x554 = 0101 0101 0100
        0x234 = 0010 0011 0100
Bit differences: xxx  xx
Since five bits were different, the Hamming distance is 5.

PROGRAM NAME: hamming

INPUT FORMAT

N, B, D on a single line

SAMPLE INPUT (file hamming.in)

16 7 3
OUTPUT FORMAT

N codewords, sorted, in decimal, ten per line. In the case of multiple solutions, your program should output the solution which, if interpreted as a base 2^B integer, would have the least value.

SAMPLE OUTPUT (file hamming.out)

0 7 25 30 42 45 51 52 75 76
82 85 97 102 120 127

The Interpretation and Method

This problem asks us to find a sequence that has at least Hamming distance d away from all other elements. These elements are restricted to bit size B (or at most 2^B). We can find this by simply pruning from all numbers 1~2^B, checking if each number can be added to the list, and if so doing so.
This inspires a simple for loop idea: we first append 0 to the answer list, and then going through 0~2^B, checking the validity of each number by seeing if the Hamming distance between any number in the current list and the testing number is greater than or equal to D. If so, then add this number to the list; once the list has N elements, terminate the loop and print the values out.

The Code Itself

/*
ID: qlstudi3
PROG: hamming
LANG: C++14
*/

 #include <cstdio>
#include <bitset>
#include <vector>
using namespace std;
int dist(int x, int y)
{
    bitset<8> b(x^y);
    return b.count();
}

 int main(int argc, const char * argv[])
{
    freopen("hamming.in","r",stdin);
    freopen("hamming.out","w",stdout);
    int n,b,Li;
    vector<int> ans;
    ans.push_back(0);
    scanf("%d %d %d",&n,&b,&Li);
    for(int i=0;i!=1<<b;i++)
    {
        bool y=true;
        for(int x:ans)
            y=y&&(dist(i,x)>=Li);
        if(y) ans.push_back(i);
        if(ans.size()==n) break;
    }
    for(int i=0;i<ans.size();i++)
    {
        if(i&&(i%10==0)) printf("\n");
        else if(i) printf(" ");
        printf("%d",ans[i]);
    }
    printf("\n");
    return 0;
}

Comments

Post a Comment

Popular posts from this blog

USACO Training "subset": Subset Sums

-
USACO Training "subset": Subset Sums Problem Statement Subset Sums JRM For many sets of consecutive integers from 1 through N (1 <= N <= 39), one can partition the set into two sets whose sums are identical. For example, if N=3, one can partition the set {1, 2, 3} in one way so that the sums of both subsets are identical: {3} and {1,2} This counts as a single partitioning (i.e., reversing the order counts as the same partitioning and thus does not increase the count of partitions). If N=7, there are four ways to partition the set {1, 2, 3, ... 7} so that each partition has the same sum: {1,6,7} and {2,3,4,5} {2,5,7} and {1,3,4,6} {3,4,7} and {1,2,5,6} {1,2,4,7} and {3,5,6} Given N, your program should print the number of ways a set containing the integers from 1 through N can be partitioned into two sets whose sums are identical. Print 0 if there are no such ways. Your program must calculate the answer, not look it up from a table. PROGRAM NAME: s
Read more

USACO 2018 Open: Talent Show

-
Image
USACO 2018 Open Talent Show In Short Given N objects with value Vi and weight Wi, a collection of elements is said to have total value-to-weight ratio as (the sum of their value) divided by (the sum of their weight). Find the best possible value-to-weight ratio of these X objects, given the restraint that the collection must have total weight at least W. Return that ratio multiplied by 1000 and floored. Analysis It would be very difficult to calculate this optimal ratio - we'd have to either try all possible collections - O(2^n), where n is at most 250, about 10 to the power of 75 calculations, OR use a greedy heuristic - get the objects with the best ratio until it satisfies the weight limit. Neither of those are too to our liking, so how about searching for it? The search space is linear and not infinite - it asks for the ratio to the closest thousandths. The search space is sorted as well - if ratio X is achievable, then ratio Y < X is also achievable, if
Read more

OpenJudge 1768: Kadane's Algorithm on 2D Arrays

-
OpenJudge 1768: Kadane's Algorithm on 2-D Arrays Simple Version Given a 2-D matrix of integers, find the sub-matrix of this whose sum is maximized, and find that sum. The size of the matrix is no larger than 100 by 100. Analysis O(N^4) - Works, but naive     That's true, an naive O(N^4) version would work. We simply need to calculate (two dimensional!) prefix sums first. A quick review to what 2-D prefix sums are:     The 2-D prefix sum of two coordinates, X and Y, is the sum of all elements that have X-coordinate less than or equal to X, and Y coordinate less than or equal to Y.     It can be calculated through tabulation by removing overlapping areas.     Let P be the prefix sum array and V be the array of values, so the formula is as follows:     P[x, y] = V[x, y] - P[x-1, y] - P[x, y-1] + P[x-1, y-1].     We need to add the P[x-1, y-1] at the end since subtracting P[x-1, y] and P[x, y-1] we had overlapped the P[x-1, y-1]. Then, all we need to
Read more