N things to select for N people, you were given a NxN matrix and cost at each element, you needed to find the one combination with max total weight, such that each person gets exactly one thing.
I found difficulty in making its dp state.
please help me and if possible then also write code for it
C++ style code:
double max_rec(int n, int r, int* c, double** m, bool* f)
{
if (r < n)
{
double max_v = 0.0;
int max_i = -1;
for (int i = 0; i < n; i++)
{
if (f[i] == false)
{
f[i] = true;
double value = m[r][i] + max_rec(n, r + 1, c, m, f);
if (value > max_v)
{
max_v = value;
max_i = i;
}
f[i] = false;
}
}
c[i] = max_i;
return max_v;
}
return 0.0;
}
int* max_comb(int n, double** m)
{
bool* f = new bool[n];
int* c = new int[n];
max_rec(n, 0, c, m, f);
delete [] f;
return c;
}
Call max_comb with N and your NxN matrix (2d array). Returns the column indices of the maximum combination.
Time complexity: O(N!)
I know this is bad but the problem does not have a greedy structure.
And as #mszalbach said, try to attempt the problem yourself before asking.
EDIT: can reduce to polynomial time by memoizing.
Related
ı want to calculate determinant of matrix with thread but i have a error "term does not eveluate to a function taking 0 arguments" ı want to solve big matrix with thread and parsing matrix,what can ı do
int determinant(int f[1000][1000], int x)
{
int pr, c[1000], d = 0, b[1000][1000], j, p, q, t;
if (x == 2)
{
d = 0;
d = (f[1][1] * f[2][2]) - (f[1][2] * f[2][1]);
return(d);
}
else
{
for (j = 1; j <= x; j++)
{
int r = 1, s = 1;
for (p = 1; p <= x; p++)
{
for (q = 1; q <= x; q++)
{
if (p != 1 && q != j)
{
b[r][s] = f[p][q];
s++;
if (s > x - 1)
{
r++;
s = 1;
}
}
}
}
for (t = 1, pr = 1; t <= (1 + j); t++)
pr = (-1)*pr;
c[j] = pr*determinant(b, x - 1);
}
for (j = 1, d = 0; j <= x; j++)
{
d = d + (f[1][j] * c[j]);
}
return(d);
}
}
int main()
{
srand(time_t(NULL));
int i, j;
printf("\n\nEnter order of matrix : ");
scanf_s("%d", &m);
printf("\nEnter the elements of matrix\n");
for (i = 1; i <= m; i++)
{
for (j = 1; j <= m; j++)
{
a[i][j] = rand() % 10;
}
}
thread t(determinant(a, m));
t.join();
printf("\n Determinant of Matrix A is %d .", determinant(a, m));
}
The immediate problem is that here: thread t(determinant(a, m)); you pass the result of calling determinant(a, m) as the function to execute, and zero arguments to call that function with - but an int is not a function or other callable object, which is what the error you got complains about.
std::thread's constructor takes the function to run and the arguments to supply separately, so you would need to call std::thread(determinant, a, m).
Now we have another problem, std::thread doesn't provide a way to retrieve the return value, and so you calculate it again here: printf("\n Determinant of Matrix A is %d .", determinant(a, m));.
To fix this, we can use std::async from the <future> header, which will manage the thread handling for us, and lets us retrieve the result later:
auto result = std::async(std::launch::async, determinant, a, m);
int det = result.get()
This will run determinant(a,m) on a new thread, and return a std::future<int> into which the return value may eventually be placed.
We can then try to retrieve that value with std::future::get(), which will block until the value can be retrieved (or until an exception occurs in the thread).
In this example, we still execute determinant in a pretty serial fashion, since we delegate the work to a thread, then wait for that thread to finish its work before continuing.
However we are now free to store the future, and defer calling std::future::get() until we actually need the value, potentially much later in your program.
There are a few other problems in the rest of your code:
all your array indexing is off by one (array indices run from 0 to N-1 in C and C++)
a few of the variables you're using don't exist (like a and m)
C-arrays are passed by pointer, so if you ever change the code not to block on the thread right there, the array will go out of scope and your thread may read garbage from the dangling pointer. If you use a proper container like std::array or std::vector, you can pass it by value so your thread will own the data to operate on for its entire lifetime.
Given a value N, if we want to make change for N cents, and we have infinite supply of each of S = { S1, S2, .. , Sm} valued coins, how many ways can we make the change? The order of coins doesn’t matter.There is additional restriction though: you can only give change with exactly K coins.
For example, for N = 4, k = 2 and S = {1,2,3}, there are two solutions: {2,2},{1,3}. So output should be 2.
Solution:
int getways(int coins, int target, int total_coins, int *denomination, int size, int idx)
{
int sum = 0, i;
if (coins > target || total_coins < 0)
return 0;
if (target == coins && total_coins == 0)
return 1;
if (target == coins && total_coins < 0)
return 0;
for (i=idx;i<size;i++) {
sum += getways(coins+denomination[i], target, total_coins-1, denomination, size, i);
}
return sum;
}
int main()
{
int target = 49;
int total_coins = 15;
int denomination[] = {1, 2, 3, 4, 5};
int size = sizeof(denomination)/sizeof(denomination[0]);
printf("%d\n", getways(0, target, total_coins, denomination, size, 0));
}
Above is recursive solution. However i need help with my dynamic programming solution:
Let dp[i][j][k] represent sum up to i with j elements and k coins.
So,
dp[i][j][k] = dp[i][j-1][k] + dp[i-a[j]][j][k-1]
Is my recurrence relation right?
I don't really understand your recurrence relation:
Let dp[i][j][k] represent sum up to i with j elements and k coins.
I think you're on the right track, but I suggest simply dropping the middle dimension [j], and use dp[sum][coinsLeft] as follows:
dp[0][0] = 1 // coins: 0, desired sum: 0 => 1 solution
dp[i][0] = 0 // coins: 0, desired sum: i => 0 solutions
dp[sum][coinsLeft] = dp[sum - S1][coinsLeft-1]
+ dp[sum - S2][coinsLeft-1]
+ ...
+ dp[sum - SM][coinsLeft-1]
The answer is then to be found at dp[N][K] (= number of ways to add K coins to get N cents)
Here's some sample code (I advice you to not look until you've tried to solve it yourself. It's a good exercise):
public static int combinations(int numCoinsToUse, int targetSum, int[] denom) {
// dp[numCoins][sum] == ways to get sum using numCoins
int[][] dp = new int[numCoinsToUse+1][targetSum];
// Any sum (except 0) is impossible with 0 coins
for (int sum = 0; sum < targetSum; sum++) {
dp[0][sum] = sum == 0 ? 1 : 0;
}
// Gradually increase number of coins
for (int c = 1; c <= numCoinsToUse; c++)
for (int sum = 0; sum < targetSum; sum++)
for (int d : denom)
if (sum >= d)
dp[c][sum] += dp[c-1][sum - d];
return dp[numCoinsToUse][targetSum-1];
}
Using your example input:
combinations(2, 4, new int[] {1, 2, 3} ) // gives 2
I want to know efficient approach for the New Lottery Game problem.
The Lottery is changing! The Lottery used to have a machine to generate a random winning number. But due to cheating problems, the Lottery has decided to add another machine. The new winning number will be the result of the bitwise-AND operation between the two random numbers generated by the two machines.
To find the bitwise-AND of X and Y, write them both in binary; then a bit in the result in binary has a 1 if the corresponding bits of X and Y were both 1, and a 0 otherwise. In most programming languages, the bitwise-AND of X and Y is written X&Y.
For example:
The old machine generates the number 7 = 0111.
The new machine generates the number 11 = 1011.
The winning number will be (7 AND 11) = (0111 AND 1011) = 0011 = 3.
With this measure, the Lottery expects to reduce the cases of fraudulent claims, but unfortunately an employee from the Lottery company has leaked the following information: the old machine will always generate a non-negative integer less than A and the new one will always generate a non-negative integer less than B.
Catalina wants to win this lottery and to give it a try she decided to buy all non-negative integers less than K.
Given A, B and K, Catalina would like to know in how many different ways the machines can generate a pair of numbers that will make her a winner.
For small input we can check all possible pairs but how to do it with large inputs. I guess we represent the binary number into string first and then check permutations which would give answer less than K. But I can't seem to figure out how to calculate possible permutations of 2 binary strings.
I used a general DP technique that I described in a lot of detail in another answer.
We want to count the pairs (a, b) such that a < A, b < B and a & b < K.
The first step is to convert the numbers to binary and to pad them to the same size by adding leading zeroes. I just padded them to a fixed size of 40. The idea is to build up the valid a and b bit by bit.
Let f(i, loA, loB, loK) be the number of valid suffix pairs of a and b of size 40 - i. If loA is true, it means that the prefix up to i is already strictly smaller than the corresponding prefix of A. In that case there is no restriction on the next possible bit for a. If loA ist false, A[i] is an upper bound on the next bit we can place at the end of the current prefix. loB and loK have an analogous meaning.
Now we have the following transition:
long long f(int i, bool loA, bool loB, bool loK) {
// TODO add memoization
if (i == 40)
return loA && loB && loK;
int hiA = loA ? 1: A[i]-'0'; // upper bound on the next bit in a
int hiB = loB ? 1: B[i]-'0'; // upper bound on the next bit in b
int hiK = loK ? 1: K[i]-'0'; // upper bound on the next bit in a & b
long long res = 0;
for (int a = 0; a <= hiA; ++a)
for (int b = 0; b <= hiB; ++b) {
int k = a & b;
if (k > hiK) continue;
res += f(i+1, loA || a < A[i]-'0',
loB || b < B[i]-'0',
loK || k < K[i]-'0');
}
return res;
}
The result is f(0, false, false, false).
The runtime is O(max(log A, log B)) if memoization is added to ensure that every subproblem is only solved once.
What I did was just to identify when the answer is A * B.
Otherwise, just brute force the rest, this code passed the large input.
// for each test cases
long count = 0;
if ((K > A) || (K > B)) {
count = A * B;
continue; // print count and go to the next test case
}
count = A * B - (A-K) * (B-K);
for (int i = K; i < A; i++) {
for (int j = K; j < B; j++) {
if ((i&j) < K) count++;
}
}
I hope this helps!
just as Niklas B. said.
the whole answer is.
#include <algorithm>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <map>
#include <sstream>
#include <string>
#include <vector>
using namespace std;
#define MAX_SIZE 32
int A, B, K;
int arr_a[MAX_SIZE];
int arr_b[MAX_SIZE];
int arr_k[MAX_SIZE];
bool flag [MAX_SIZE][2][2][2];
long long matrix[MAX_SIZE][2][2][2];
long long
get_result();
int main(int argc, char *argv[])
{
int case_amount = 0;
cin >> case_amount;
for (int i = 0; i < case_amount; ++i)
{
const long long result = get_result();
cout << "Case #" << 1 + i << ": " << result << endl;
}
return 0;
}
long long
dp(const int h,
const bool can_A_choose_1,
const bool can_B_choose_1,
const bool can_K_choose_1)
{
if (MAX_SIZE == h)
return can_A_choose_1 && can_B_choose_1 && can_K_choose_1;
if (flag[h][can_A_choose_1][can_B_choose_1][can_K_choose_1])
return matrix[h][can_A_choose_1][can_B_choose_1][can_K_choose_1];
int cnt_A_max = arr_a[h];
int cnt_B_max = arr_b[h];
int cnt_K_max = arr_k[h];
if (can_A_choose_1)
cnt_A_max = 1;
if (can_B_choose_1)
cnt_B_max = 1;
if (can_K_choose_1)
cnt_K_max = 1;
long long res = 0;
for (int i = 0; i <= cnt_A_max; ++i)
{
for (int j = 0; j <= cnt_B_max; ++j)
{
int k = i & j;
if (k > cnt_K_max)
continue;
res += dp(h + 1,
can_A_choose_1 || (i < cnt_A_max),
can_B_choose_1 || (j < cnt_B_max),
can_K_choose_1 || (k < cnt_K_max));
}
}
flag[h][can_A_choose_1][can_B_choose_1][can_K_choose_1] = true;
matrix[h][can_A_choose_1][can_B_choose_1][can_K_choose_1] = res;
return res;
}
long long
get_result()
{
cin >> A >> B >> K;
memset(arr_a, 0, sizeof(arr_a));
memset(arr_b, 0, sizeof(arr_b));
memset(arr_k, 0, sizeof(arr_k));
memset(flag, 0, sizeof(flag));
memset(matrix, 0, sizeof(matrix));
int i = 31;
while (i >= 1)
{
arr_a[i] = A % 2;
A /= 2;
arr_b[i] = B % 2;
B /= 2;
arr_k[i] = K % 2;
K /= 2;
i--;
}
return dp(1, 0, 0, 0);
}
I can't figure out how to generate all compositions (http://en.wikipedia.org/wiki/Composition_%28number_theory%29) of an integer N into K parts, but only doing it one at a time. That is, I need a function that given the previous composition generated, returns the next one in the sequence. The reason is that memory is limited for my application. This would be much easier if I could use Python and its generator functionality, but I'm stuck with C++.
This is similar to Next Composition of n into k parts - does anyone have a working algorithm?
Any assistance would be greatly appreciated.
Preliminary remarks
First start from the observation that [1,1,...,1,n-k+1] is the first composition (in lexicographic order) of n over k parts, and [n-k+1,1,1,...,1] is the last one.
Now consider an exemple: the composition [2,4,3,1,1], here n = 11 and k=5. Which is the next one in lexicographic order? Obviously the rightmost part to be incremented is 4, because [3,1,1] is the last composition of 5 over 3 parts.
4 is at the left of 3, the rightmost part different from 1.
So turn 4 into 5, and replace [3,1,1] by [1,1,2], the first composition of the remainder (3+1+1)-1 , giving [2,5,1,1,2]
Generation program (in C)
The following C program shows how to compute such compositions on demand in lexicographic order
#include <stdio.h>
#include <stdbool.h>
bool get_first_composition(int n, int k, int composition[k])
{
if (n < k) {
return false;
}
for (int i = 0; i < k - 1; i++) {
composition[i] = 1;
}
composition[k - 1] = n - k + 1;
return true;
}
bool get_next_composition(int n, int k, int composition[k])
{
if (composition[0] == n - k + 1) {
return false;
}
// there'a an i with composition[i] > 1, and it is not 0.
// find the last one
int last = k - 1;
while (composition[last] == 1) {
last--;
}
// turn a b ... y z 1 1 ... 1
// ^ last
// into a b ... (y+1) 1 1 1 ... (z-1)
// be careful, there may be no 1's at the end
int z = composition[last];
composition[last - 1] += 1;
composition[last] = 1;
composition[k - 1] = z - 1;
return true;
}
void display_composition(int k, int composition[k])
{
char *separator = "[";
for (int i = 0; i < k; i++) {
printf("%s%d", separator, composition[i]);
separator = ",";
}
printf("]\n");
}
void display_all_compositions(int n, int k)
{
int composition[k]; // VLA. Please don't use silly values for k
for (bool exists = get_first_composition(n, k, composition);
exists;
exists = get_next_composition(n, k, composition)) {
display_composition(k, composition);
}
}
int main()
{
display_all_compositions(5, 3);
}
Results
[1,1,3]
[1,2,2]
[1,3,1]
[2,1,2]
[2,2,1]
[3,1,1]
Weak compositions
A similar algorithm works for weak compositions (where 0 is allowed).
bool get_first_weak_composition(int n, int k, int composition[k])
{
if (n < k) {
return false;
}
for (int i = 0; i < k - 1; i++) {
composition[i] = 0;
}
composition[k - 1] = n;
return true;
}
bool get_next_weak_composition(int n, int k, int composition[k])
{
if (composition[0] == n) {
return false;
}
// there'a an i with composition[i] > 0, and it is not 0.
// find the last one
int last = k - 1;
while (composition[last] == 0) {
last--;
}
// turn a b ... y z 0 0 ... 0
// ^ last
// into a b ... (y+1) 0 0 0 ... (z-1)
// be careful, there may be no 0's at the end
int z = composition[last];
composition[last - 1] += 1;
composition[last] = 0;
composition[k - 1] = z - 1;
return true;
}
Results for n=5 k=3
[0,0,5]
[0,1,4]
[0,2,3]
[0,3,2]
[0,4,1]
[0,5,0]
[1,0,4]
[1,1,3]
[1,2,2]
[1,3,1]
[1,4,0]
[2,0,3]
[2,1,2]
[2,2,1]
[2,3,0]
[3,0,2]
[3,1,1]
[3,2,0]
[4,0,1]
[4,1,0]
[5,0,0]
Similar algorithms can be written for compositions of n into k parts greater than some fixed value.
You could try something like this:
start with the array [1,1,...,1,N-k+1] of (K-1) ones and 1 entry with the remainder. The next composition can be created by incrementing the (K-1)th element and decreasing the last element. Do this trick as long as the last element is bigger than the second to last.
When the last element becomes smaller, increment the (K-2)th element, set the (K-1)th element to the same value and set the last element to the remainder again. Repeat the process and apply the same principle for the other elements when necessary.
You end up with a constantly sorted array that avoids duplicate compositions
I want to write a recursive merge sort program in C++. The problem is, I don't know how to get the base case idea working recursively. Can anybody please tell me what would be the base case for Merg Function(), Split Function() and MergSort() function. I would be thankful to you.
void Merg(int A[], int s1, int e1, int s2, int e2)
{
int B[8];
int i=0;
while (A[s1] < A[s2])
B[i] = B[s1];
i++;
s1++;
if (s1 == e1)
{
B[i] = A[s2];
i++;
s2++;
}
while (A[s2] < A[s1])
B[i] = B[s2];
i++;
s2++;
if (s2 == e2)
{
B[i] = A[s1];
i++;
s1++;
}
}
void Split(int A[], int s, int e)
{
int mid = (s+e)/2;
if (s < e && mid != 0)
{
Split(A, s, mid);
Split(A, mid+1, e);
}
Merg(A, s, mid, mid+1, e);
}
int main()
{
int A[8] = {10,4,8,12,11,2,7,5};
Split(A, 0, 7);
return 0;
}
The base case is an array that is guaranteed to be sorted, so either an empty array or an array of length 1.
Your merge function is not correct, but at least contains most of the right ideas. All you need there is a further wrapping loop and a few conditions to prevent your merge running past the end of the arrays. The split function is totally off, splitting is not recursive, further splits happen inside the recursive mergeSort calls.
if length(A) < 2 return // already sorted
split A in lower half L and upper half H
merge-sort L
merge-sort H
merge the sorted L and H
done