Množenje matrice | Rekurzivno
Zadane su dvije matrice A i B. Zadatak je rekurzivno pomnožiti matricu A i matricu B. Ako matrica A i matrica B nisu multiplikativno kompatibilne, generirajte izlaz 'Nije moguće'.
Primjeri:
Input: A = 12 56
45 78
B = 2 6
5 8
Output: 304 520
480 894
Input: A = 1 2 3
4 5 6
7 8 9
B = 1 2 3
4 5 6
7 8 9
Output: 30 36 42
66 81 96
102 126 150
Preporuča se prvo uputiti Iterativno množenje matrice .
Prvo provjerite je li množenje između matrica moguće ili ne. Za ovo provjerite je li broj stupaca prve matrice jednak broju redaka druge matrice ili ne. Ako su oba jednaka, nastavite dalje, inače generirajte izlaz 'Nije moguće'.
U rekurzivnom množenju matrica implementiramo tri petlje iteracije kroz rekurzivne pozive. Unutarnji najrekurzivniji poziv od pomnožiMatrix() je ponoviti k (col1 ili row2). Drugi rekurzivni poziv od pomnožiMatrix() je mijenjanje stupaca, a najudaljeniji rekurzivni poziv je mijenjanje redaka.
Ispod je kod za rekurzivno matrično množenje.
C++Java// Recursive code for Matrix Multiplication #includeconst int MAX = 100 ; void multiplyMatrixRec ( int row1 int col1 int A [][ MAX ] int row2 int col2 int B [][ MAX ] int C [][ MAX ]) { // Note that below variables are static // i and j are used to know current cell of // result matrix C[][]. k is used to know // current column number of A[][] and row // number of B[][] to be multiplied static int i = 0 j = 0 k = 0 ; // If all rows traversed. if ( i >= row1 ) return ; // If i < row1 if ( j < col2 ) { if ( k < col1 ) { C [ i ][ j ] += A [ i ][ k ] * B [ k ][ j ]; k ++ ; multiplyMatrixRec ( row1 col1 A row2 col2 B C ); } k = 0 ; j ++ ; multiplyMatrixRec ( row1 col1 A row2 col2 B C ); } j = 0 ; i ++ ; multiplyMatrixRec ( row1 col1 A row2 col2 B C ); } // Function to multiply two matrices A[][] and B[][] void multiplyMatrix ( int row1 int col1 int A [][ MAX ] int row2 int col2 int B [][ MAX ]) { if ( row2 != col1 ) { printf ( 'Not Possible n ' ); return ; } int C [ MAX ][ MAX ] = { 0 }; multiplyMatrixRec ( row1 col1 A row2 col2 B C ); // Print the result for ( int i = 0 ; i < row1 ; i ++ ) { for ( int j = 0 ; j < col2 ; j ++ ) printf ( '%d ' C [ i ][ j ]); printf ( ' n ' ); } } // Driven Program int main () { int A [][ MAX ] = { { 1 2 3 } { 4 5 6 } { 7 8 9 } }; int B [][ MAX ] = { { 1 2 3 } { 4 5 6 } { 7 8 9 } }; int row1 = 3 col1 = 3 row2 = 3 col2 = 3 ; multiplyMatrix ( row1 col1 A row2 col2 B ); return 0 ; } // This code is contributed by Aarti_Rathi Python3// Java recursive code for Matrix Multiplication class GFG { public static int MAX = 100 ; // Note that below variables are static // i and j are used to know current cell of // result matrix C[][]. k is used to know // current column number of A[][] and row // number of B[][] to be multiplied public static int i = 0 j = 0 k = 0 ; static void multiplyMatrixRec ( int row1 int col1 int A [][] int row2 int col2 int B [][] int C [][] ) { // If all rows traversed if ( i >= row1 ) return ; // If i < row1 if ( j < col2 ) { if ( k < col1 ) { C [ i ][ j ] += A [ i ][ k ] * B [ k ][ j ] ; k ++ ; multiplyMatrixRec ( row1 col1 A row2 col2 B C ); } k = 0 ; j ++ ; multiplyMatrixRec ( row1 col1 A row2 col2 B C ); } j = 0 ; i ++ ; multiplyMatrixRec ( row1 col1 A row2 col2 B C ); } // Function to multiply two matrices A[][] and B[][] static void multiplyMatrix ( int row1 int col1 int A [][] int row2 int col2 int B [][] ) { if ( row2 != col1 ) { System . out . println ( 'Not Possiblen' ); return ; } int [][] C = new int [ MAX ][ MAX ] ; multiplyMatrixRec ( row1 col1 A row2 col2 B C ); // Print the result for ( int i = 0 ; i < row1 ; i ++ ) { for ( int j = 0 ; j < col2 ; j ++ ) System . out . print ( C [ i ][ j ]+ ' ' ); System . out . println (); } } // driver program public static void main ( String [] args ) { int row1 = 3 col1 = 3 row2 = 3 col2 = 3 ; int A [][] = { { 1 2 3 } { 4 5 6 } { 7 8 9 }}; int B [][] = { { 1 2 3 } { 4 5 6 } { 7 8 9 } }; multiplyMatrix ( row1 col1 A row2 col2 B ); } } // Contributed by Pramod KumarC## Recursive code for Matrix Multiplication MAX = 100 i = 0 j = 0 k = 0 def multiplyMatrixRec ( row1 col1 A row2 col2 B C ): # Note that below variables are static # i and j are used to know current cell of # result matrix C[][]. k is used to know # current column number of A[][] and row # number of B[][] to be multiplied global i global j global k # If all rows traversed. if ( i >= row1 ): return # If i < row1 if ( j < col2 ): if ( k < col1 ): C [ i ][ j ] += A [ i ][ k ] * B [ k ][ j ] k += 1 multiplyMatrixRec ( row1 col1 A row2 col2 B C ) k = 0 j += 1 multiplyMatrixRec ( row1 col1 A row2 col2 B C ) j = 0 i += 1 multiplyMatrixRec ( row1 col1 A row2 col2 B C ) # Function to multiply two matrices # A[][] and B[][] def multiplyMatrix ( row1 col1 A row2 col2 B ): if ( row2 != col1 ): print ( 'Not Possible' ) return C = [[ 0 for i in range ( MAX )] for i in range ( MAX )] multiplyMatrixRec ( row1 col1 A row2 col2 B C ) # Print the result for i in range ( row1 ): for j in range ( col2 ): print ( C [ i ][ j ] end = ' ' ) print () # Driver Code A = [[ 1 2 3 ] [ 4 5 6 ] [ 7 8 9 ]] B = [[ 1 2 3 ] [ 4 5 6 ] [ 7 8 9 ]] row1 = 3 col1 = 3 row2 = 3 col2 = 3 multiplyMatrix ( row1 col1 A row2 col2 B ) # This code is contributed by sahilshelangiaJavaScript// C# recursive code for // Matrix Multiplication using System ; class GFG { public static int MAX = 100 ; // Note that below variables // are static i and j are used // to know current cell of result // matrix C[][]. k is used to // know current column number of // A[][] and row number of B[][] // to be multiplied public static int i = 0 j = 0 k = 0 ; static void multiplyMatrixRec ( int row1 int col1 int [] A int row2 int col2 int [] B int [] C ) { // If all rows traversed if ( i >= row1 ) return ; // If i < row1 if ( j < col2 ) { if ( k < col1 ) { C [ i j ] += A [ i k ] * B [ k j ]; k ++ ; multiplyMatrixRec ( row1 col1 A row2 col2 B C ); } k = 0 ; j ++ ; multiplyMatrixRec ( row1 col1 A row2 col2 B C ); } j = 0 ; i ++ ; multiplyMatrixRec ( row1 col1 A row2 col2 B C ); } // Function to multiply two // matrices A[][] and B[][] static void multiplyMatrix ( int row1 int col1 int [] A int row2 int col2 int [] B ) { if ( row2 != col1 ) { Console . WriteLine ( 'Not Possiblen' ); return ; } int [] C = new int [ MAX MAX ]; multiplyMatrixRec ( row1 col1 A row2 col2 B C ); // Print the result for ( int i = 0 ; i < row1 ; i ++ ) { for ( int j = 0 ; j < col2 ; j ++ ) Console . Write ( C [ i j ] + ' ' ); Console . WriteLine (); } } // Driver Code static public void Main () { int row1 = 3 col1 = 3 row2 = 3 col2 = 3 ; int [] A = {{ 1 2 3 } { 4 5 6 } { 7 8 9 }}; int [] B = {{ 1 2 3 } { 4 5 6 } { 7 8 9 }}; multiplyMatrix ( row1 col1 A row2 col2 B ); } } // This code is contributed by m_kit< script > // Javascript recursive code for Matrix Multiplication let MAX = 100 ; // Note that below variables are static // i and j are used to know current cell of // result matrix C[][]. k is used to know // current column number of A[][] and row // number of B[][] to be multiplied let i = 0 j = 0 k = 0 ; function multiplyMatrixRec ( row1 col1 A row2 col2 B C ) { // If all rows traversed if ( i >= row1 ) return ; // If i < row1 if ( j < col2 ) { if ( k < col1 ) { C [ i ][ j ] += A [ i ][ k ] * B [ k ][ j ]; k ++ ; multiplyMatrixRec ( row1 col1 A row2 col2 B C ); } k = 0 ; j ++ ; multiplyMatrixRec ( row1 col1 A row2 col2 B C ); } j = 0 ; i ++ ; multiplyMatrixRec ( row1 col1 A row2 col2 B C ); } // Function to multiply two matrices A[][] and B[][] function multiplyMatrix ( row1 col1 A row2 col2 B ) { if ( row2 != col1 ) { document . write ( 'Not Possible' + ' ' ); return ; } let C = new Array ( MAX ); for ( let i = 0 ; i < MAX ; i ++ ) { C [ i ] = new Array ( MAX ); for ( let j = 0 ; j < MAX ; j ++ ) { C [ i ][ j ] = 0 ; } } multiplyMatrixRec ( row1 col1 A row2 col2 B C ); // Print the result for ( let i = 0 ; i < row1 ; i ++ ) { for ( let j = 0 ; j < col2 ; j ++ ) document . write ( C [ i ][ j ] + ' ' ); document . write ( ' ' ); } } let row1 = 3 col1 = 3 row2 = 3 col2 = 3 ; let A = [ [ 1 2 3 ] [ 4 5 6 ] [ 7 8 9 ] ]; let B = [ [ 1 2 3 ] [ 4 5 6 ] [ 7 8 9 ] ]; multiplyMatrix ( row1 col1 A row2 col2 B ); < /script>
Izlaz30 36 42 66 81 96 102 126 150Vremenska složenost: O(redak1 * stupac2* stupac1)
Pomoćni prostor: O(log (max(row1col2)) Kao implicitni stog koristi se zbog rekurzijeNapravi kviz