trovare l'algoritmo di submatrix più grande

Question

Ho una matrice N * N (N = 2 a 10000) di numeri che possono variare da 0 a 1000. Come posso trovare la sottomatrice più grande (rettangolare) che consiste dello stesso numero?

Esempio:

 1 2 3 4 5 
    -- -- -- -- -- 
1 | 10 9 9 9 80 
2 | 5 9 9 9 10 
3 | 85 86 54 45 45 
4 | 15 21 5 1 0 
5 | 5 6 88 11 10 

L'output dovrebbe essere l'area della sottomatrice, seguito da coordinate 1-base del suo elemento superiore sinistro. Per l'esempio, sarebbe (6, 2, 1) perché ci sono sei 9 s situate a colonna 2, riga 1.

Answer 1

Questo è un work in progress

ho pensato a questo problema e credo di avere un Algoritmo O(w*h).

L'idea è questa:

• per qualsiasi (i,j) calcolare il maggior numero di cellule con lo stesso valore nella colonna j partire da (i,j). Memorizza questi valori come heights[i][j].
• creare un vettore vuoto di matrice sub (a LIFO)
• per tutti consecutive: i
  • per tutti colonna: j
    • pop tutte le matrici sub cui height > heights[i][j]. Poiché la sottomatrice con altezza>heights[i][j] non può rimanere in questa cella
    • spinta sottomatrice definita da (i,j,heights[i][j]) dove j è la più lontana coordinata dove possiamo inserire una sottomatrice di altezza: heights[i][j]
    • aggiornamento sub matrice corrente max

la parte difficile è nel ciclo interno. Io uso qualcosa di simile all'algoritmo della sottofinestra max per garantire che sia in media O(1) per ogni cella.

Proverò a formulare una prova ma nel frattempo ecco il codice.

#include <algorithm> 
#include <iterator> 
#include <iostream> 
#include <ostream> 
#include <vector> 

typedef std::vector<int> row_t; 
typedef std::vector<row_t> matrix_t; 

std::size_t height(matrix_t const& M) { return M.size(); } 
std::size_t width (matrix_t const& M) { return M.size() ? M[0].size() : 0u; } 

std::ostream& operator<<(std::ostream& out, matrix_t const& M) { 
    for(unsigned i=0; i<height(M); ++i) { 
    std::copy(begin(M[i]), end(M[i]), 
      std::ostream_iterator<int>(out, ", ")); 
    out << std::endl; 
    } 
    return out; 
} 

struct sub_matrix_t { 
    int i, j, h, w; 
    sub_matrix_t(): i(0),j(0),h(0),w(1) {} 
    sub_matrix_t(int i_,int j_,int h_,int w_):i(i_),j(j_),h(h_),w(w_) {} 
    bool operator<(sub_matrix_t const& rhs) const { return (w*h)<(rhs.w*rhs.h); } 
}; 


// Pop all sub_matrix from the vector keeping only those with an height 
// inferior to the passed height. 
// Compute the max sub matrix while removing sub matrix with height > h 
void pop_sub_m(std::vector<sub_matrix_t>& subs, 
      int i, int j, int h, sub_matrix_t& max_m) { 

    sub_matrix_t sub_m(i, j, h, 1); 

    while(subs.size() && subs.back().h >= h) { 
    sub_m = subs.back(); 
    subs.pop_back(); 
    sub_m.w = j-sub_m.j; 
    max_m = std::max(max_m, sub_m); 
    } 

    // Now sub_m.{i,j} is updated to the farest coordinates where there is a 
    // submatrix with heights >= h 

    // If we don't cut the current height (because we changed value) update 
    // the current max submatrix 
    if(h > 0) { 
    sub_m.h = h; 
    sub_m.w = j-sub_m.j+1; 
    max_m = std::max(max_m, sub_m); 
    subs.push_back(sub_m); 
    } 
} 

void push_sub_m(std::vector<sub_matrix_t>& subs, 
     int i, int j, int h, sub_matrix_t& max_m) { 
    if(subs.empty() || subs.back().h < h) 
    subs.emplace_back(i, j, h, 1); 
} 

void solve(matrix_t const& M, sub_matrix_t& max_m) { 
    // Initialize answer suitable for an empty matrix 
    max_m = sub_matrix_t(); 
    if(height(M) == 0 || width(M) == 0) return; 

    // 1) Compute the heights of columns of the same values 
    matrix_t heights(height(M), row_t(width(M), 1)); 
    for(unsigned i=height(M)-1; i>0; --i) 
    for(unsigned j=0; j<width(M); ++j) 
     if(M[i-1][j]==M[i][j]) 
    heights[i-1][j] = heights[i][j]+1; 

    // 2) Run through all columns heights to compute local sub matrices 
    std::vector<sub_matrix_t> subs; 
    for(int i=height(M)-1; i>=0; --i) { 
    push_sub_m(subs, i, 0, heights[i][0], max_m); 
    for(unsigned j=1; j<width(M); ++j) { 
     bool same_val = (M[i][j]==M[i][j-1]); 
     int pop_height = (same_val) ? heights[i][j] : 0; 
     int pop_j  = (same_val) ? j    : j-1; 
     pop_sub_m (subs, i, pop_j, pop_height, max_m); 
     push_sub_m(subs, i, j,  heights[i][j], max_m); 
    } 
    pop_sub_m(subs, i, width(M)-1, 0, max_m); 
    } 
} 

matrix_t M1{ 
    {10, 9, 9, 9, 80}, 
    { 5, 9, 9, 9, 10}, 
    {85, 86, 54, 45, 45}, 
    {15, 21, 5, 1, 0}, 
    { 5, 6, 88, 11, 10}, 
}; 

matrix_t M2{ 
    {10, 19, 9, 29, 80}, 
    { 5, 9, 9, 9, 10}, 
    { 9, 9, 54, 45, 45}, 
    { 9, 9, 5, 1, 0}, 
    { 5, 6, 88, 11, 10}, 
}; 


int main() { 
    sub_matrix_t answer; 

    std::cout << M1 << std::endl; 
    solve(M1, answer); 
    std::cout << '(' << (answer.w*answer.h) 
     << ',' << (answer.j+1) << ',' << (answer.i+1) << ')' 
     << std::endl; 

    answer = sub_matrix_t(); 
    std::cout << M2 << std::endl; 
    solve(M2, answer); 
    std::cout << '(' << (answer.w*answer.h) 
     << ',' << (answer.j+1) << ',' << (answer.i+1) << ')' 
     << std::endl; 
} 

Answer 2

Questo è un ordine righe * colonne Soluzione

Agisce

• partenza dal fondo della matrice, e determinare quanti elementi indicati ogni numero partita in una colonna. Questo è fatto in tempo O (MN) (molto banalmente)
• Quindi va da cima a fondo & da sinistra a destra e vede se un dato numero corrisponde al numero a sinistra. Se è così, tiene traccia di come le altezze si correlano l'una all'altra per tracciare le possibili forme rettangolari

Ecco una implementazione Python funzionante.Scuse dato che non sono sicuro di come ottenere l'evidenziazione della sintassi di lavoro

# this program finds the largest area in an array where all the elements have the same value 
# It solves in O(rows * columns) time using O(rows*columns) space using dynamic programming 




def max_area_subarray(array): 

    rows = len(array) 
    if (rows == 0): 
     return [[]] 
    columns = len(array[0]) 


    # initialize a blank new array 
    # this will hold max elements of the same value in a column 
    new_array = [] 
    for i in range(0,rows-1): 
     new_array.append([0] * columns) 

    # start with the bottom row, these all of 1 element of the same type 
    # below them, including themselves 
    new_array.append([1] * columns) 

    # go from the second to bottom row up, finding how many contiguous 
    # elements of the same type there are 
    for i in range(rows-2,-1,-1): 
     for j in range(columns-1,-1,-1): 
      if (array[i][j] == array[i+1][j]): 
       new_array[i][j] = new_array[i+1][j]+1 
      else: 
       new_array[i][j] = 1 


    # go left to right and match up the max areas 
    max_area = 0 
    top = 0 
    bottom = 0 
    left = 0 
    right = 0 
    for i in range(0,rows): 
     running_height =[[0,0,0]] 
     for j in range(0,columns): 

      matched = False 
      if (j > 0): # if this isn't the leftmost column 
       if (array[i][j] == array[i][j-1]): 
        # this matches the array to the left 
        # keep track of if this is a longer column, shorter column, or same as 
        # the one on the left 
        matched = True 

        while(new_array[i][j] < running_height[-1][0]): 
         # this is less than the one on the left, pop that running 
         # height from the list, and add it's columns to the smaller 
         # running height below it 
         if (running_height[-1][1] > max_area): 
          max_area = running_height[-1][1] 
          top = i 
          right = j-1 
          bottom = i + running_height[-1][0]-1 
          left = j - running_height[-1][2] 

         previous_column = running_height.pop() 
         num_columns = previous_column[2] 

         if (len(running_height) > 0): 
          running_height[-1][1] += running_height[-1][0] * (num_columns) 
          running_height[-1][2] += num_columns 

         else: 
          # for instance, if we have heights 2,2,1 
          # this will trigger on the 1 after we pop the 2 out, and save the current 
          # height of 1, the running area of 3, and running columsn of 3 
          running_height.append([new_array[i][j],new_array[i][j]*(num_columns),num_columns]) 


        if (new_array[i][j] > running_height[-1][0]): 
         # longer then the one on the left 
         # append this height and area 
         running_height.append([new_array[i][j],new_array[i][j],1]) 
        elif (new_array[i][j] == running_height[-1][0]): 
         # same as the one on the left, add this area to the one on the left 
         running_height[-1][1] += new_array[i][j] 
         running_height[-1][2] += 1 



      if (matched == False or j == columns -1): 
       while(running_height): 
        # unwind the maximums & see if this is the new max area 
        if (running_height[-1][1] > max_area): 
         max_area = running_height[-1][1] 
         top = i 
         right = j 
         bottom = i + running_height[-1][0]-1 
         left = j - running_height[-1][2]+1 

         # this wasn't a match, so move everything one bay to the left 
         if (matched== False): 
          right = right-1 
          left = left-1 


        previous_column = running_height.pop() 
        num_columns = previous_column[2] 
        if (len(running_height) > 0): 
         running_height[-1][1] += running_height[-1][0] * num_columns 
         running_height[-1][2] += num_columns 

      if (matched == False): 
       # this is either the left column, or we don't match to the column to the left, so reset 
       running_height = [[new_array[i][j],new_array[i][j],1]] 
       if (running_height[-1][1] > max_area): 
        max_area = running_height[-1][1] 
        top = i 
        right = j 
        bottom = i + running_height[-1][0]-1 
        left = j - running_height[-1][2]+1 


    max_array = [] 
    for i in range(top,bottom+1): 
     max_array.append(array[i][left:right+1]) 


    return max_array 



numbers = [[6,4,1,9],[5,2,2,7],[2,2,2,1],[2,3,1,5]] 

for row in numbers: 
    print row 

print 
print 

max_array = max_area_subarray(numbers)  


max_area = len(max_array) * len(max_array[0]) 
print 'max area is ',max_area 
print 
for row in max_array: 
    print row