问题
I need to to write a Java code that checks whether the user inputed number is in the Fibonacci sequence.
I have no issue writing the Fibonacci sequence to output, but (probably because its late at night) I'm struggling to think of the sequence of "whether" it is a Fibonacci number. I keep starting over and over again. Its really doing my head in.
What I currently have is the nth.
public static void main(String[] args)
{
ConsoleReader console = new ConsoleReader();
System.out.println("Enter the value for your n: ");
int num = (console.readInt());
System.out.println("\nThe largest nth fibonacci: "+fib(num));
System.out.println();
}
static int fib(int n){
int f = 0;
int g = 1;
int largeNum = -1;
for(int i = 0; i < n; i++)
{
if(i == (n-1))
largeNum = f;
System.out.print(f + " ");
f = f + g;
g = f - g;
}
return largeNum;
}
回答1:
Read the section titled "recognizing fibonacci numbers" on wikipedia.
Alternatively, a positive integer z is a Fibonacci number if and only if one of 5z^2 + 4 or 5z^2 − 4 is a perfect square.[17]
Alternatively, you can keep generating fibonacci numbers until one becomes equal to your number: if it does, then your number is a fibonacci number, if not, the numbers will eventually become bigger than your number, and you can stop. This is pretty inefficient however.
回答2:
If I understand correctly, what you need to do (instead of writing out the first n Fibonacci numbers) is to determine whether n is a Fibonacci number.
So you should modify your method to keep generating the Fibonacci sequence until you get a number >= n. If it equals, n is a Fibonacci number, otherwise not.
Update: bugged by @Moron's repeated claims about the formula based algorithm being superior in performance to the simple one above, I actually did a benchmark comparison - concretely between Jacopo's solution as generator algorithm and StevenH's last version as formula based algorithm. For reference, here is the exact code:
public static void main(String[] args) {
measureExecutionTimeForGeneratorAlgorithm(1);
measureExecutionTimeForFormulaAlgorithm(1);
measureExecutionTimeForGeneratorAlgorithm(10);
measureExecutionTimeForFormulaAlgorithm(10);
measureExecutionTimeForGeneratorAlgorithm(100);
measureExecutionTimeForFormulaAlgorithm(100);
measureExecutionTimeForGeneratorAlgorithm(1000);
measureExecutionTimeForFormulaAlgorithm(1000);
measureExecutionTimeForGeneratorAlgorithm(10000);
measureExecutionTimeForFormulaAlgorithm(10000);
measureExecutionTimeForGeneratorAlgorithm(100000);
measureExecutionTimeForFormulaAlgorithm(100000);
measureExecutionTimeForGeneratorAlgorithm(1000000);
measureExecutionTimeForFormulaAlgorithm(1000000);
measureExecutionTimeForGeneratorAlgorithm(10000000);
measureExecutionTimeForFormulaAlgorithm(10000000);
measureExecutionTimeForGeneratorAlgorithm(100000000);
measureExecutionTimeForFormulaAlgorithm(100000000);
measureExecutionTimeForGeneratorAlgorithm(1000000000);
measureExecutionTimeForFormulaAlgorithm(1000000000);
measureExecutionTimeForGeneratorAlgorithm(2000000000);
measureExecutionTimeForFormulaAlgorithm(2000000000);
}
static void measureExecutionTimeForGeneratorAlgorithm(int x) {
final int count = 1000000;
final long start = System.nanoTime();
for (int i = 0; i < count; i++) {
isFibByGeneration(x);
}
final double elapsedTimeInSec = (System.nanoTime() - start) * 1.0e-9;
System.out.println("Running generator algorithm " + count + " times for " + x + " took " +elapsedTimeInSec + " seconds");
}
static void measureExecutionTimeForFormulaAlgorithm(int x) {
final int count = 1000000;
final long start = System.nanoTime();
for (int i = 0; i < count; i++) {
isFibByFormula(x);
}
final double elapsedTimeInSec = (System.nanoTime() - start) * 1.0e-9;
System.out.println("Running formula algorithm " + count + " times for " + x + " took " +elapsedTimeInSec + " seconds");
}
static boolean isFibByGeneration(int x) {
int a=0;
int b=1;
int f=1;
while (b < x){
f = a + b;
a = b;
b = f;
}
return x == f;
}
private static boolean isFibByFormula(int num) {
double first = 5 * Math.pow((num), 2) + 4;
double second = 5 * Math.pow((num), 2) - 4;
return isWholeNumber(Math.sqrt(first)) || isWholeNumber(Math.sqrt(second));
}
private static boolean isWholeNumber(double num) {
return num - Math.round(num) == 0;
}
The results surprised even me:
Running generator algorithm 1000000 times for 1 took 0.007173537000000001 seconds
Running formula algorithm 1000000 times for 1 took 0.223365539 seconds
Running generator algorithm 1000000 times for 10 took 0.017330694 seconds
Running formula algorithm 1000000 times for 10 took 0.279445852 seconds
Running generator algorithm 1000000 times for 100 took 0.030283179 seconds
Running formula algorithm 1000000 times for 100 took 0.27773557800000004 seconds
Running generator algorithm 1000000 times for 1000 took 0.041044322 seconds
Running formula algorithm 1000000 times for 1000 took 0.277931134 seconds
Running generator algorithm 1000000 times for 10000 took 0.051103143000000004 seconds
Running formula algorithm 1000000 times for 10000 took 0.276980175 seconds
Running generator algorithm 1000000 times for 100000 took 0.062019335 seconds
Running formula algorithm 1000000 times for 100000 took 0.276227007 seconds
Running generator algorithm 1000000 times for 1000000 took 0.07422898800000001 seconds
Running formula algorithm 1000000 times for 1000000 took 0.275485013 seconds
Running generator algorithm 1000000 times for 10000000 took 0.085803922 seconds
Running formula algorithm 1000000 times for 10000000 took 0.27701090500000003 seconds
Running generator algorithm 1000000 times for 100000000 took 0.09543419600000001 seconds
Running formula algorithm 1000000 times for 100000000 took 0.274908403 seconds
Running generator algorithm 1000000 times for 1000000000 took 0.10683704200000001 seconds
Running formula algorithm 1000000 times for 1000000000 took 0.27524084800000004 seconds
Running generator algorithm 1000000 times for 2000000000 took 0.13019867100000002 seconds
Running formula algorithm 1000000 times for 2000000000 took 0.274846384 seconds
In short, the generator algorithm way outperforms the formula based solution on all positive int values - even close to the maximum int value it is more than twice as fast! So much for belief based performance optimization ;-)
For the record, modifying the above code to use long variables instead of int, the generator algorithm becomes slower (as expected, since it has to add up long values now), and cutover point where the formula starts to be faster is around 1000000000000L, i.e. 1012.
Update2: As IVlad and Moron noted, I am not quite an expert in floating point calculations :-) based on their suggestions I improved the formula to this:
private static boolean isFibByFormula(long num)
{
double power = (double)num * (double)num;
double first = 5 * power + 4;
double second = 5 * power - 4;
return isWholeNumber(Math.sqrt(first)) || isWholeNumber(Math.sqrt(second));
}
This brought down the cutover point to approx. 108 (for the long version - the generator with int is still faster for all int values). No doubt that replacing the sqrt calls with something like suggested by @Moron would push down the cutover point further.
My (and IVlad's) point was simply that there will always be a cutover point, below which the generator algorithm is faster. So claims about which one performs better have no meaning in general, only in a context.
回答3:
Instead of passing the index, n, write a function that takes a limit, and get it to generate the Fibonacci numbers up to and including this limit. Get it to return a Boolean depending on whether it hits or skips over the limit, and you can use this to check whether that value is in the sequence.
Since it's homework, a nudge like this is probably all we should be giving you...
回答4:
Ok. Since people claimed I am just talking thin air ('facts' vs 'guesses') without any data to back it up, I wrote a benchmark of my own.
Not java, but C# code below.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace SO
{
class Program
{
static void Main(string[] args)
{
AssertIsFibSqrt(100000000);
MeasureSequential(1);
MeasureSqrt(1);
MeasureSequential(10);
MeasureSqrt(10);
MeasureSequential(50);
MeasureSqrt(50);
MeasureSequential(100);
MeasureSqrt(100);
MeasureSequential(100000);
MeasureSqrt(100000);
MeasureSequential(100000000);
MeasureSqrt(100000000);
}
static void MeasureSequential(long n)
{
int count = 1000000;
DateTime start = DateTime.Now;
for (int i = 0; i < count; i++)
{
IsFibSequential(n);
}
DateTime end = DateTime.Now;
TimeSpan duration = end - start;
Console.WriteLine("Sequential for input = " + n +
" : " + duration.Ticks);
}
static void MeasureSqrt(long n)
{
int count = 1000000;
DateTime start = DateTime.Now;
for (int i = 0; i < count; i++)
{
IsFibSqrt(n);
}
DateTime end = DateTime.Now;
TimeSpan duration = end - start;
Console.WriteLine("Sqrt for input = " + n +
" : " + duration.Ticks);
}
static void AssertIsFibSqrt(long x)
{
Dictionary<long, bool> fibs = new Dictionary<long, bool>();
long a = 0;
long b = 1;
long f = 1;
while (b < x)
{
f = a + b;
a = b;
b = f;
fibs[a] = true;
fibs[b] = true;
}
for (long i = 1; i <= x; i++)
{
bool isFib = fibs.ContainsKey(i);
if (isFib && IsFibSqrt(i))
{
continue;
}
if (!isFib && !IsFibSqrt(i))
{
continue;
}
Console.WriteLine("Sqrt Fib test failed for: " + i);
}
}
static bool IsFibSequential(long x)
{
long a = 0;
long b = 1;
long f = 1;
while (b < x)
{
f = a + b;
a = b;
b = f;
}
return x == f;
}
static bool IsFibSqrt(long x)
{
long y = 5 * x * x + 4;
double doubleS = Math.Sqrt(y);
long s = (long)doubleS;
long sqr = s*s;
return (sqr == y || sqr == (y-8));
}
}
}
And here is the output
Sequential for input = 1 : 110011
Sqrt for input = 1 : 670067
Sequential for input = 10 : 560056
Sqrt for input = 10 : 540054
Sequential for input = 50 : 610061
Sqrt for input = 50 : 540054
Sequential for input = 100 : 730073
Sqrt for input = 100 : 540054
Sequential for input = 100000 : 1490149
Sqrt for input = 100000 : 540054
Sequential for input = 100000000 : 2180218
Sqrt for input = 100000000 : 540054
The sqrt method beats the naive method when n=50 itself, perhaps due to the presence of hardware support on my machine. Even if it was 10^8 (like in Peter's test), there are at most 40 fibonacci numbers under that cutoff, which could easily be put in a lookup table and still beat the naive version for the smaller values.
Also, Peter has a bad implementation of the SqrtVersion. He doesn't really need to compute two square roots or compute powers using Math.Pow. He could have atleast tried to make that better before publishing his benchmark results.
Anyway, I will let these facts speak for themselves, instead of the so called 'guesses'.
回答5:
A positive integer x is a Fibonacci number if and only if one of 5x^2 + 4 and 5x^2 - 4 is a perfect square
回答6:
There are a number of methods that can be employed to determine if a given number is in the fibonacci sequence, a selection of which can be seen on wikipedia.
Given what you've done already, however, I'd probably use a more brute-force approach, such as the following:
- Generate a fibonacci number
- If it's less than the target number, generate the next fibonacci and repeat
- If it is the target number, then success
- If it's bigger than the target number, then failure.
I'd probably use a recursive method, passing in a current n-value (ie. so it calculates the nth fibonacci number) and the target number.
回答7:
//Program begins
public class isANumberFibonacci {
public static int fibonacci(int seriesLength) {
if (seriesLength == 1 || seriesLength == 2) {
return 1;
} else {
return fibonacci(seriesLength - 1) + fibonacci(seriesLength - 2);
}
}
public static void main(String args[]) {
int number = 4101;
int i = 1;
while (i > 0) {
int fibnumber = fibonacci(i);
if (fibnumber != number) {
if (fibnumber > number) {
System.out.println("Not fib");
break;
} else {
i++;
}
} else {
System.out.println("The number is fibonacci");
break;
}
}
}
}
//Program ends
回答8:
If my Java is not too rusty...
static bool isFib(int x) {
int a=0;
int b=1;
int f=1;
while (b < x){
f = a + b;
a = b;
b = f;
}
return x == f;
}
回答9:
Trying to leverage the code you have already written I would propose the following first, as it is the simplest solution (but not the most efficient):
private static void main(string[] args)
{
//This will determnine which numbers between 1 & 100 are in the fibonacci series
//you can swop in code to read from console rather than 'i' being used from the for loop
for (int i = 0; i < 100; i++)
{
bool result = isFib(1);
if (result)
System.out.println(i + " is in the Fib series.");
System.out.println(result);
}
}
private static bool isFib(int num)
{
int counter = 0;
while (true)
{
if (fib(counter) < num)
{
counter++;
continue;
}
if (fib(counter) == num)
{
return true;
}
if (fib(counter) > num)
{
return false;
}
}
}
I would propose a more elegant solution in the generation of fibonacci numbers which leverages recursion like so:
public static long fib(int n)
{
if (n <= 1)
return n;
else
return fib(n-1) + fib(n-2);
}
For the extra credit read: http://en.wikipedia.org/wiki/Fibonacci_number#Recognizing_Fibonacci_numbers
You will see the that there are a few more efficient ways to test if a number is in the Fibonacci series namely: (5z^2 + 4 or 5z^2 − 4) = a perfect square.
//(5z^2 + 4 or 5z^2 − 4) = a perfect square
//perfect square = an integer that is the square of an integer
private static bool isFib(int num)
{
double first = 5 * Math.pow((num), 2) + 4;
double second = 5 * Math.pow((num), 2) - 4;
return isWholeNumber(Math.sqrt(first)) || isWholeNumber(Math.sqrt(second));
}
private static bool isWholeNumber(double num)
{
return num - Math.round(num) == 0;
}
回答10:
I don't know if there is an actual formula that you can apply to the user input however, you can generate the fibonacci sequence and check it against the user input until it has become smaller than the last number generated.
int userInput = n;
int a = 1, b = 1;
while (a < n) {
if (a == n)
return true;
int next = a + b;
b = a;
a = next;
}
return false;
回答11:
You can do this in two ways , the recursive and mathematical. the recursive way start generating fibonacci sequence until you hit the number or pass it the mathematical way nicely described here ... http://www.physicsforums.com/showthread.php?t=252798
good luck.
回答12:
Consider the sequence of Fibonacci numbers 1,1,2,3,5,8,13,21, etc. It is desired to build 3 stacks each of capacity 10 containing numbers from the above sequences as follows:
Stack 1: First 10 numbers from the sequence. Stack 2: First 10 prime numbers from the sequence. Stack 3: First 10 non-prime numbers from the sequence.
(i) Give an algorithm of the flowchart (ii) Write a program (in BASIC, C++ or Java) to implement this.
Output:As stack operations take place you should display in any convenient form the 3 stacks together with the values held in them.
回答13:
Finding out whether a number is Fibonacci based on formula:
public static boolean isNumberFromFibonacciSequence(int num){
if (num == 0 || num == 1){
return true;
}
else {
//5n^2 - 4 OR 5n^2 + 4 should be perfect squares
return isPerfectSquare( 5*num*num - 4) || isPerfectSquare(5*num*num - 4);
}
}
private static boolean isPerfectSquare(int num){
double sqrt = Math.sqrt(num);
return sqrt * sqrt == num;
}
回答14:
Thought it was simple until i had to rack my head on it a few minutes. Its quite different from generating a fibonacci sequence. This function returns 1 if is Fibonnaci or 0 if not
public static int isFibonacci (int n){
int isFib = 0;
int a = 0, b = 0, c = a + b; // set up the initial values
do
{
a = b;
b = c;
c = a + b;
if (c == n)
isFib = 1;
} while (c<=n && isFin == 0)
return isFib;
}
public static void main(String [] args){
System.out.println(isFibonacci(89));
}
来源：https://stackoverflow.com/questions/3139645/determining-whether-a-number-is-a-fibonacci-number