Понастоящем изучавам структурите на данни в университета и се натъкнах на въпроса за сложността на рекурсията.
Като се има предвид този код:
Unsigned func (unsigned n)
{
if (n ==0) return 1;
if(n==1) return 2;
\do somthing(in O(1) )
return func(n-1) + func(n-1);
}
Знам какво прави кодът. Знам, че във формата, която сега е, сложността на времето е О (2 ^ n).
Въпросът ми обаче е дали ще се промени сложността на времето, ако вместо последното обаждане за връщане на кода ще напиша: return 2*func(n-1) ?
Знам, че що се отнася до сложността на паметта, става дума за значително намаляване на пространството, което се поема от рекурсията, но що се отнася до сложността на времето, ще има ли някаква промяна?
Направих математика с помощта на рекурсивна функция и стигнах до разбирането, че няма да има промяна в сложността на времето, нали?
Отговори:
4 за отговор № 1Този метод има само O(n), защото ако го пуснете с 5, той отива на рекурсия с 4, след това с 3 и т.н.
Unsigned func (unsigned n)
{
if (n ==0) return 1;
if(n==1) return 2;
\do somthing(in O(1) )
return 2*func(n-1);
}
Но какво да кажем за това:
Unsigned func (unsigned n)
{
if (n ==0) return 1;
if(n==1) return 2;
\do somthing(in O(1) )
return func(n-1) + func(n-1);
}
Например func (5) ще изпълни първо, както следва
5 -> 4 -> 3 -> 2 -> 1
След това се връща на 2, но там ще изпълни втората част, така че изглежда целият процес
5 -> 4 -> 3 -> 2-> 1; 2-> 1; 3->2->1; etc.
Следователно тя ДА променя сложността драстично от O(n) да се O(2^n)
Опитайте практическия пример за този код:
public class Complexity {
private static int counter;
public static void main(String[] args) {
for (int i = 0; i < 20; i++) {
counter = 0;
func(i);
System.out.println("For n="+i+" of 2*func the number of cycles is " + counter);
counter = 0;
func2(i);
System.out.println("For n="+i+" of func + func the number of cycles is " + counter);
}
}
public static int func(int n) {
counter++;
if (n == 0) {
return 1;
}
if (n == 1) {
return 2;
}
return 2 * func(n - 1);
}
public static int func2(int n) {
counter++;
if (n == 0) {
return 1;
}
if (n == 1) {
return 2;
}
return func2(n - 1) + func2(n - 1);
}
}
С този изход:
For n=0 of 2*func the number of cycles is 1
For n=0 of func + func the number of cycles is 1
For n=1 of 2*func the number of cycles is 1
For n=1 of func + func the number of cycles is 1
For n=2 of 2*func the number of cycles is 2
For n=2 of func + func the number of cycles is 3
For n=3 of 2*func the number of cycles is 3
For n=3 of func + func the number of cycles is 7
For n=4 of 2*func the number of cycles is 4
For n=4 of func + func the number of cycles is 15
For n=5 of 2*func the number of cycles is 5
For n=5 of func + func the number of cycles is 31
For n=6 of 2*func the number of cycles is 6
For n=6 of func + func the number of cycles is 63
For n=7 of 2*func the number of cycles is 7
For n=7 of func + func the number of cycles is 127
For n=8 of 2*func the number of cycles is 8
For n=8 of func + func the number of cycles is 255
For n=9 of 2*func the number of cycles is 9
For n=9 of func + func the number of cycles is 511
For n=10 of 2*func the number of cycles is 10
For n=10 of func + func the number of cycles is 1023
For n=11 of 2*func the number of cycles is 11
For n=11 of func + func the number of cycles is 2047
For n=12 of 2*func the number of cycles is 12
For n=12 of func + func the number of cycles is 4095
For n=13 of 2*func the number of cycles is 13
For n=13 of func + func the number of cycles is 8191
For n=14 of 2*func the number of cycles is 14
For n=14 of func + func the number of cycles is 16383
For n=15 of 2*func the number of cycles is 15
For n=15 of func + func the number of cycles is 32767
For n=16 of 2*func the number of cycles is 16
For n=16 of func + func the number of cycles is 65535
For n=17 of 2*func the number of cycles is 17
For n=17 of func + func the number of cycles is 131071
For n=18 of 2*func the number of cycles is 18
For n=18 of func + func the number of cycles is 262143
For n=19 of 2*func the number of cycles is 19
For n=19 of func + func the number of cycles is 524287
Но ако си спомните вече изчислените резултати, сложността е все още O(n) дори и при втория подход:
public class Complexity {
private static int counter;
private static int[] results;
public static void main(String[] args) {
for (int i = 0; i < 20; i++) {
counter = 0;
func(i);
System.out.println("For n="+i+" of 2*func the number of cycles is " + counter);
counter = 0;
func2(i);
System.out.println("For n="+i+" of func + func the number of cycles is " + counter);
counter = 0;
results = new int[i+1];
func3(i);
System.out.println("For n="+i+" of func + func with remembering the number of cycles is " + counter);
}
}
public static int func(int n) {
counter++;
if (n == 0) {
return 1;
}
if (n == 1) {
return 2;
}
return 2 * func(n - 1);
}
public static int func2(int n) {
counter++;
if (n == 0) {
return 1;
}
if (n == 1) {
return 2;
}
return func2(n - 1) + func2(n - 1);
}
public static int func3(int n) {
counter++;
if (n == 0) {
return 1;
}
if (n == 1) {
return 2;
}
if (results[n] == 0){
results[n] = func3(n - 1) + func3(n - 1);
}
return results[n];
}
}
С този изход:
For n=0 of 2*func the number of cycles is 1
For n=0 of func + func the number of cycles is 1
For n=0 of func + func with remembering the number of cycles is 1
For n=1 of 2*func the number of cycles is 1
For n=1 of func + func the number of cycles is 1
For n=1 of func + func with remembering the number of cycles is 1
For n=2 of 2*func the number of cycles is 2
For n=2 of func + func the number of cycles is 3
For n=2 of func + func with remembering the number of cycles is 3
For n=3 of 2*func the number of cycles is 3
For n=3 of func + func the number of cycles is 7
For n=3 of func + func with remembering the number of cycles is 5
For n=4 of 2*func the number of cycles is 4
For n=4 of func + func the number of cycles is 15
For n=4 of func + func with remembering the number of cycles is 7
For n=5 of 2*func the number of cycles is 5
For n=5 of func + func the number of cycles is 31
For n=5 of func + func with remembering the number of cycles is 9
For n=6 of 2*func the number of cycles is 6
For n=6 of func + func the number of cycles is 63
For n=6 of func + func with remembering the number of cycles is 11
For n=7 of 2*func the number of cycles is 7
For n=7 of func + func the number of cycles is 127
For n=7 of func + func with remembering the number of cycles is 13
For n=8 of 2*func the number of cycles is 8
For n=8 of func + func the number of cycles is 255
For n=8 of func + func with remembering the number of cycles is 15
For n=9 of 2*func the number of cycles is 9
For n=9 of func + func the number of cycles is 511
For n=9 of func + func with remembering the number of cycles is 17
For n=10 of 2*func the number of cycles is 10
For n=10 of func + func the number of cycles is 1023
For n=10 of func + func with remembering the number of cycles is 19
For n=11 of 2*func the number of cycles is 11
For n=11 of func + func the number of cycles is 2047
For n=11 of func + func with remembering the number of cycles is 21
For n=12 of 2*func the number of cycles is 12
For n=12 of func + func the number of cycles is 4095
For n=12 of func + func with remembering the number of cycles is 23
For n=13 of 2*func the number of cycles is 13
For n=13 of func + func the number of cycles is 8191
For n=13 of func + func with remembering the number of cycles is 25
For n=14 of 2*func the number of cycles is 14
For n=14 of func + func the number of cycles is 16383
For n=14 of func + func with remembering the number of cycles is 27
For n=15 of 2*func the number of cycles is 15
For n=15 of func + func the number of cycles is 32767
For n=15 of func + func with remembering the number of cycles is 29
For n=16 of 2*func the number of cycles is 16
For n=16 of func + func the number of cycles is 65535
For n=16 of func + func with remembering the number of cycles is 31
For n=17 of 2*func the number of cycles is 17
For n=17 of func + func the number of cycles is 131071
For n=17 of func + func with remembering the number of cycles is 33
For n=18 of 2*func the number of cycles is 18
For n=18 of func + func the number of cycles is 262143
For n=18 of func + func with remembering the number of cycles is 35
For n=19 of 2*func the number of cycles is 19
For n=19 of func + func the number of cycles is 524287
For n=19 of func + func with remembering the number of cycles is 37
2 за отговор № 2
Това зависи от семантиката на вашия език за програмиране / алгоритъм.
Ако чрез f(n) искаш да кажеш "да се обади на функцията, независимо дали е биланаречен със същия аргумент преди "(както е случаят с повечето програмни езици), тогава вашата промяна ще намали сложността драстично на O (n). Имате едно O (1) функционално повикване за намаляване на аргумента.
В противен случай (ако говорите за чисти функции и за запаметяване на известни резултати), двата алгоритъма вече имат сложност O (n).