Ricerca per interpolazione
Dato un array ordinato di n valori distribuiti uniformemente arr[] scrive una funzione per cercare un particolare elemento x nell'array.
La ricerca lineare trova l'elemento nel tempo O(n). Salta la ricerca impiega tempo O(n) e Ricerca binaria impiega tempo O(log n).
La ricerca per interpolazione rappresenta un miglioramento Ricerca binaria per i casi in cui i valori in una matrice ordinata sono distribuiti uniformemente. L'interpolazione costruisce nuovi punti dati all'interno dell'intervallo di un insieme discreto di punti dati noti. La ricerca binaria va sempre all'elemento centrale per verificare. D'altra parte la ricerca per interpolazione può andare in posizioni diverse a seconda del valore della chiave da cercare. Ad esempio, se il valore della chiave è più vicino all'ultimo elemento, è probabile che la ricerca con interpolazione inizi la ricerca verso il lato finale.
Per trovare la posizione da ricercare utilizza la seguente formula.
// L'idea della formula è restituire un valore più alto di pos
// quando l'elemento da cercare è più vicino ad arr[hi]. E
// valore più piccolo quando è più vicino ad arr[lo]arr[] ==> Array in cui devono essere cercati gli elementi
x ==> Elemento da cercare
lo ==> Indice iniziale in arr[]
ciao ==> Indice finale in arr[]
dopo = il +
Esistono molti metodi di interpolazione diversi e uno di questi è noto come interpolazione lineare. L'interpolazione lineare prende due punti dati che assumiamo come (x1y1) e (x2y2) e la formula è: al punto(xy).
Questo algoritmo funziona nel modo in cui cerchiamo una parola in un dizionario. L'algoritmo di ricerca per interpolazione migliora l'algoritmo di ricerca binaria. La formula per trovare un valore è: K = > K è una costante che viene utilizzata per restringere lo spazio di ricerca. Nel caso della ricerca binaria il valore di questa costante è: K=(basso+alto)/2.
La formula per pos può essere derivata come segue.
Let's assume that the elements of the array are linearly distributed.
General equation of line : y = m*x + c.
y is the value in the array and x is its index.
Now putting value of lohi and x in the equation
arr[hi] = m*hi+c ----(1)
arr[lo] = m*lo+c ----(2)
x = m*pos + c ----(3)
m = (arr[hi] - arr[lo] )/ (hi - lo)
subtracting eqxn (2) from (3)
x - arr[lo] = m * (pos - lo)
lo + (x - arr[lo])/m = pos
pos = lo + (x - arr[lo]) *(hi - lo)/(arr[hi] - arr[lo])Algoritmo
Il resto dell'algoritmo di interpolazione è lo stesso ad eccezione della logica di partizione di cui sopra.
- Passaggio 1: In un ciclo calcolare il valore di "pos" utilizzando la formula della posizione della sonda.
- Passaggio 2: Se è una corrispondenza, restituisce l'indice dell'elemento ed esce.
- Passaggio 3: Se l'elemento è inferiore a arr[pos] calcola la posizione della sonda del sottoarray sinistro. Altrimenti calcola lo stesso nel sottoarray destro.
- Passaggio 4: Ripetere finché non viene trovata una corrispondenza o il sottoarray si riduce a zero.
Di seguito è riportata l'implementazione dell'algoritmo.
// C++ program to implement interpolation // search with recursion #include using namespace std ; // If x is present in arr[0..n-1] then returns // index of it else returns -1. int interpolationSearch ( int arr [] int lo int hi int x ) { int pos ; // Since array is sorted an element present // in array must be in range defined by corner if ( lo <= hi && x >= arr [ lo ] && x <= arr [ hi ]) { // Probing the position with keeping // uniform distribution in mind. pos = lo + ((( double )( hi - lo ) / ( arr [ hi ] - arr [ lo ])) * ( x - arr [ lo ])); // Condition of target found if ( arr [ pos ] == x ) return pos ; // If x is larger x is in right sub array if ( arr [ pos ] < x ) return interpolationSearch ( arr pos + 1 hi x ); // If x is smaller x is in left sub array if ( arr [ pos ] > x ) return interpolationSearch ( arr lo pos - 1 x ); } return -1 ; } // Driver Code int main () { // Array of items on which search will // be conducted. int arr [] = { 10 12 13 16 18 19 20 21 22 23 24 33 35 42 47 }; int n = sizeof ( arr ) / sizeof ( arr [ 0 ]); // Element to be searched int x = 18 ; int index = interpolationSearch ( arr 0 n - 1 x ); // If element was found if ( index != -1 ) cout < < 'Element found at index ' < < index ; else cout < < 'Element not found.' ; return 0 ; } // This code is contributed by equbalzeeshan
C // C program to implement interpolation search // with recursion #include // If x is present in arr[0..n-1] then returns // index of it else returns -1. int interpolationSearch ( int arr [] int lo int hi int x ) { int pos ; // Since array is sorted an element present // in array must be in range defined by corner if ( lo <= hi && x >= arr [ lo ] && x <= arr [ hi ]) { // Probing the position with keeping // uniform distribution in mind. pos = lo + ((( double )( hi - lo ) / ( arr [ hi ] - arr [ lo ])) * ( x - arr [ lo ])); // Condition of target found if ( arr [ pos ] == x ) return pos ; // If x is larger x is in right sub array if ( arr [ pos ] < x ) return interpolationSearch ( arr pos + 1 hi x ); // If x is smaller x is in left sub array if ( arr [ pos ] > x ) return interpolationSearch ( arr lo pos - 1 x ); } return -1 ; } // Driver Code int main () { // Array of items on which search will // be conducted. int arr [] = { 10 12 13 16 18 19 20 21 22 23 24 33 35 42 47 }; int n = sizeof ( arr ) / sizeof ( arr [ 0 ]); int x = 18 ; // Element to be searched int index = interpolationSearch ( arr 0 n - 1 x ); // If element was found if ( index != -1 ) printf ( 'Element found at index %d' index ); else printf ( 'Element not found.' ); return 0 ; }
Java // Java program to implement interpolation // search with recursion import java.util.* ; class GFG { // If x is present in arr[0..n-1] then returns // index of it else returns -1. public static int interpolationSearch ( int arr [] int lo int hi int x ) { int pos ; // Since array is sorted an element // present in array must be in range // defined by corner if ( lo <= hi && x >= arr [ lo ] && x <= arr [ hi ] ) { // Probing the position with keeping // uniform distribution in mind. pos = lo + ((( hi - lo ) / ( arr [ hi ] - arr [ lo ] )) * ( x - arr [ lo ] )); // Condition of target found if ( arr [ pos ] == x ) return pos ; // If x is larger x is in right sub array if ( arr [ pos ] < x ) return interpolationSearch ( arr pos + 1 hi x ); // If x is smaller x is in left sub array if ( arr [ pos ] > x ) return interpolationSearch ( arr lo pos - 1 x ); } return - 1 ; } // Driver Code public static void main ( String [] args ) { // Array of items on which search will // be conducted. int arr [] = { 10 12 13 16 18 19 20 21 22 23 24 33 35 42 47 }; int n = arr . length ; // Element to be searched int x = 18 ; int index = interpolationSearch ( arr 0 n - 1 x ); // If element was found if ( index != - 1 ) System . out . println ( 'Element found at index ' + index ); else System . out . println ( 'Element not found.' ); } } // This code is contributed by equbalzeeshan
Python # Python3 program to implement # interpolation search # with recursion # If x is present in arr[0..n-1] then # returns index of it else returns -1. def interpolationSearch ( arr lo hi x ): # Since array is sorted an element present # in array must be in range defined by corner if ( lo <= hi and x >= arr [ lo ] and x <= arr [ hi ]): # Probing the position with keeping # uniform distribution in mind. pos = lo + (( hi - lo ) // ( arr [ hi ] - arr [ lo ]) * ( x - arr [ lo ])) # Condition of target found if arr [ pos ] == x : return pos # If x is larger x is in right subarray if arr [ pos ] < x : return interpolationSearch ( arr pos + 1 hi x ) # If x is smaller x is in left subarray if arr [ pos ] > x : return interpolationSearch ( arr lo pos - 1 x ) return - 1 # Driver code # Array of items in which # search will be conducted arr = [ 10 12 13 16 18 19 20 21 22 23 24 33 35 42 47 ] n = len ( arr ) # Element to be searched x = 18 index = interpolationSearch ( arr 0 n - 1 x ) if index != - 1 : print ( 'Element found at index' index ) else : print ( 'Element not found' ) # This code is contributed by Hardik Jain
C# // C# program to implement // interpolation search using System ; class GFG { // If x is present in // arr[0..n-1] then // returns index of it // else returns -1. static int interpolationSearch ( int [] arr int lo int hi int x ) { int pos ; // Since array is sorted an element // present in array must be in range // defined by corner if ( lo <= hi && x >= arr [ lo ] && x <= arr [ hi ]) { // Probing the position // with keeping uniform // distribution in mind. pos = lo + ((( hi - lo ) / ( arr [ hi ] - arr [ lo ])) * ( x - arr [ lo ])); // Condition of // target found if ( arr [ pos ] == x ) return pos ; // If x is larger x is in right sub array if ( arr [ pos ] < x ) return interpolationSearch ( arr pos + 1 hi x ); // If x is smaller x is in left sub array if ( arr [ pos ] > x ) return interpolationSearch ( arr lo pos - 1 x ); } return - 1 ; } // Driver Code public static void Main () { // Array of items on which search will // be conducted. int [] arr = new int []{ 10 12 13 16 18 19 20 21 22 23 24 33 35 42 47 }; // Element to be searched int x = 18 ; int n = arr . Length ; int index = interpolationSearch ( arr 0 n - 1 x ); // If element was found if ( index != - 1 ) Console . WriteLine ( 'Element found at index ' + index ); else Console . WriteLine ( 'Element not found.' ); } } // This code is contributed by equbalzeeshan
JavaScript < script > // Javascript program to implement Interpolation Search // If x is present in arr[0..n-1] then returns // index of it else returns -1. function interpolationSearch ( arr lo hi x ){ let pos ; // Since array is sorted an element present // in array must be in range defined by corner if ( lo <= hi && x >= arr [ lo ] && x <= arr [ hi ]) { // Probing the position with keeping // uniform distribution in mind. pos = lo + Math . floor ((( hi - lo ) / ( arr [ hi ] - arr [ lo ])) * ( x - arr [ lo ]));; // Condition of target found if ( arr [ pos ] == x ){ return pos ; } // If x is larger x is in right sub array if ( arr [ pos ] < x ){ return interpolationSearch ( arr pos + 1 hi x ); } // If x is smaller x is in left sub array if ( arr [ pos ] > x ){ return interpolationSearch ( arr lo pos - 1 x ); } } return - 1 ; } // Driver Code let arr = [ 10 12 13 16 18 19 20 21 22 23 24 33 35 42 47 ]; let n = arr . length ; // Element to be searched let x = 18 let index = interpolationSearch ( arr 0 n - 1 x ); // If element was found if ( index != - 1 ){ document . write ( `Element found at index ${ index } ` ) } else { document . write ( 'Element not found' ); } // This code is contributed by _saurabh_jaiswal < /script>
PHP // PHP program to implement $erpolation search // with recursion // If x is present in arr[0..n-1] then returns // index of it else returns -1. function interpolationSearch ( $arr $lo $hi $x ) { // Since array is sorted an element present // in array must be in range defined by corner if ( $lo <= $hi && $x >= $arr [ $lo ] && $x <= $arr [ $hi ]) { // Probing the position with keeping // uniform distribution in mind. $pos = ( int )( $lo + ((( double )( $hi - $lo ) / ( $arr [ $hi ] - $arr [ $lo ])) * ( $x - $arr [ $lo ]))); // Condition of target found if ( $arr [ $pos ] == $x ) return $pos ; // If x is larger x is in right sub array if ( $arr [ $pos ] < $x ) return interpolationSearch ( $arr $pos + 1 $hi $x ); // If x is smaller x is in left sub array if ( $arr [ $pos ] > $x ) return interpolationSearch ( $arr $lo $pos - 1 $x ); } return - 1 ; } // Driver Code // Array of items on which search will // be conducted. $arr = array ( 10 12 13 16 18 19 20 21 22 23 24 33 35 42 47 ); $n = sizeof ( $arr ); $x = 47 ; // Element to be searched $index = interpolationSearch ( $arr 0 $n - 1 $x ); // If element was found if ( $index != - 1 ) echo 'Element found at index ' . $index ; else echo 'Element not found.' ; return 0 ; #This code is contributed by Susobhan Akhuli ?>
Produzione
Element found at index 4
Complessità temporale: O(log 2 (tronco d'albero 2 n)) per il caso medio e O(n) per il caso peggiore
Complessità dello spazio ausiliario: O(1)
Un altro approccio: -
Questo è l'approccio iterativo per la ricerca di interpolazione.
- Passaggio 1: In un ciclo calcolare il valore di "pos" utilizzando la formula della posizione della sonda.
- Passaggio 2: Se è una corrispondenza, restituisce l'indice dell'elemento ed esce.
- Passaggio 3: Se l'elemento è inferiore a arr[pos] calcola la posizione della sonda del sottoarray sinistro. Altrimenti calcola lo stesso nel sottoarray destro.
- Passaggio 4: Ripetere finché non viene trovata una corrispondenza o il sottoarray si riduce a zero.
Di seguito è riportata l'implementazione dell'algoritmo.
C++ // C++ program to implement interpolation search by using iteration approach #include using namespace std ; int interpolationSearch ( int arr [] int n int x ) { // Find indexes of two corners int low = 0 high = ( n - 1 ); // Since array is sorted an element present // in array must be in range defined by corner while ( low <= high && x >= arr [ low ] && x <= arr [ high ]) { if ( low == high ) { if ( arr [ low ] == x ) return low ; return -1 ; } // Probing the position with keeping // uniform distribution in mind. int pos = low + ((( double )( high - low ) / ( arr [ high ] - arr [ low ])) * ( x - arr [ low ])); // Condition of target found if ( arr [ pos ] == x ) return pos ; // If x is larger x is in upper part if ( arr [ pos ] < x ) low = pos + 1 ; // If x is smaller x is in the lower part else high = pos - 1 ; } return -1 ; } // Main function int main () { // Array of items on whighch search will // be conducted. int arr [] = { 10 12 13 16 18 19 20 21 22 23 24 33 35 42 47 }; int n = sizeof ( arr ) / sizeof ( arr [ 0 ]); int x = 18 ; // Element to be searched int index = interpolationSearch ( arr n x ); // If element was found if ( index != -1 ) cout < < 'Element found at index ' < < index ; else cout < < 'Element not found.' ; return 0 ; } //this code contributed by Ajay Singh
Java // Java program to implement interpolation // search with recursion import java.util.* ; class GFG { // If x is present in arr[0..n-1] then returns // index of it else returns -1. public static int interpolationSearch ( int arr [] int lo int hi int x ) { int pos ; if ( lo <= hi && x >= arr [ lo ] && x <= arr [ hi ] ) { // Probing the position with keeping // uniform distribution in mind. pos = lo + ((( hi - lo ) / ( arr [ hi ] - arr [ lo ] )) * ( x - arr [ lo ] )); // Condition of target found if ( arr [ pos ] == x ) return pos ; // If x is larger x is in right sub array if ( arr [ pos ] < x ) return interpolationSearch ( arr pos + 1 hi x ); // If x is smaller x is in left sub array if ( arr [ pos ] > x ) return interpolationSearch ( arr lo pos - 1 x ); } return - 1 ; } // Driver Code public static void main ( String [] args ) { // Array of items on which search will // be conducted. int arr [] = { 10 12 13 16 18 19 20 21 22 23 24 33 35 42 47 }; int n = arr . length ; // Element to be searched int x = 18 ; int index = interpolationSearch ( arr 0 n - 1 x ); // If element was found if ( index != - 1 ) System . out . println ( 'Element found at index ' + index ); else System . out . println ( 'Element not found.' ); } }
Python # Python equivalent of above C++ code # Python program to implement interpolation search by using iteration approach def interpolationSearch ( arr n x ): # Find indexes of two corners low = 0 high = ( n - 1 ) # Since array is sorted an element present # in array must be in range defined by corner while low <= high and x >= arr [ low ] and x <= arr [ high ]: if low == high : if arr [ low ] == x : return low ; return - 1 ; # Probing the position with keeping # uniform distribution in mind. pos = int ( low + ((( float ( high - low ) / ( arr [ high ] - arr [ low ])) * ( x - arr [ low ])))) # Condition of target found if arr [ pos ] == x : return pos # If x is larger x is in upper part if arr [ pos ] < x : low = pos + 1 ; # If x is smaller x is in lower part else : high = pos - 1 ; return - 1 # Main function if __name__ == '__main__' : # Array of items on whighch search will # be conducted. arr = [ 10 12 13 16 18 19 20 21 22 23 24 33 35 42 47 ] n = len ( arr ) x = 18 # Element to be searched index = interpolationSearch ( arr n x ) # If element was found if index != - 1 : print ( 'Element found at index' index ) else : print ( 'Element not found' )
C# // C# program to implement interpolation search by using // iteration approach using System ; class Program { // Interpolation Search function static int InterpolationSearch ( int [] arr int n int x ) { int low = 0 ; int high = n - 1 ; while ( low <= high && x >= arr [ low ] && x <= arr [ high ]) { if ( low == high ) { if ( arr [ low ] == x ) return low ; return - 1 ; } int pos = low + ( int )((( float )( high - low ) / ( arr [ high ] - arr [ low ])) * ( x - arr [ low ])); if ( arr [ pos ] == x ) return pos ; if ( arr [ pos ] < x ) low = pos + 1 ; else high = pos - 1 ; } return - 1 ; } // Main function static void Main ( string [] args ) { int [] arr = { 10 12 13 16 18 19 20 21 22 23 24 33 35 42 47 }; int n = arr . Length ; int x = 18 ; int index = InterpolationSearch ( arr n x ); if ( index != - 1 ) Console . WriteLine ( 'Element found at index ' + index ); else Console . WriteLine ( 'Element not found' ); } } // This code is contributed by Susobhan Akhuli
JavaScript // JavaScript program to implement interpolation search by using iteration approach function interpolationSearch ( arr n x ) { // Find indexes of two corners let low = 0 ; let high = n - 1 ; // Since array is sorted an element present // in array must be in range defined by corner while ( low <= high && x >= arr [ low ] && x <= arr [ high ]) { if ( low == high ) { if ( arr [ low ] == x ) { return low ; } return - 1 ; } // Probing the position with keeping // uniform distribution in mind. let pos = Math . floor ( low + ((( high - low ) / ( arr [ high ] - arr [ low ])) * ( x - arr [ low ]))); // Condition of target found if ( arr [ pos ] == x ) { return pos ; } // If x is larger x is in upper part if ( arr [ pos ] < x ) { low = pos + 1 ; } // If x is smaller x is in lower part else { high = pos - 1 ; } } return - 1 ; } // Main function let arr = [ 10 12 13 16 18 19 20 21 22 23 24 33 35 42 47 ]; let n = arr . length ; let x = 18 ; // Element to be searched let index = interpolationSearch ( arr n x ); // If element was found if ( index != - 1 ) { console . log ( 'Element found at index' index ); } else { console . log ( 'Element not found' ); }
Produzione
Element found at index 4
Complessità temporale: O(log2(log2 n)) per il caso medio e O(n) per il caso peggiore
Complessità dello spazio ausiliario: O(1)
