I have got a code that generates all possible correct strings of balanced brackets. So if the input is n = 4 there should be 4 brackets in the string and thus the answers the code will give are: {}{} and
{{}}.
Now, what I would like to do is print the number of possible strings. For example, for n = 4 the outcome would be 2.
Given my code, is this possible and how would I make that happen?
Just introduce a counter.
// Change prototype to return the counter
int findBalanced(int p,int n,int o,int c)
{
static char str[100];
// The counter
static int count = 0;
if (c == n) {
// Increment it on every printout
count ++;
printf("%s\n", str);
// Just return zero. This is not used anyway and will give
// Correct result for n=0
return 0;
} else {
if (o > c) {
str[p] = ')';
findBalanced(p + 1, n, o, c + 1);
}
if (o < n) {
str[p] = '(';
findBalanced(p + 1, n, o + 1, c);
}
}
// Return it
return count;
}
What you're looking for is the n-th Catalan number. You'll need to implement binomial coefficient to calculate it, but that's pretty much it.
This question has been bugging me for quite a while: is it possible to write a factorial function (in any programming language) without any if statement (or similar) which returns 1 when called with 0 as argument too?
Many factorial functions are something like this (Python):
def factorial(n):
for x in range(1, n):
n *= x
return n if n > 0 else 1
But I don't know if it can be done without distinction between varied values of n... What do you think? It is not a matter of speed and optimizing, just my curiosity.
0! is defined as 1.
Here are the results from my code.
0 factorial = 1
1 factorial = 1
2 factorial = 2
3 factorial = 6
10 factorial = 3628800
And here's Java code with no if statement,
package com.ggl.testing;
public class Factorial {
public static void main(String[] args) {
int n = 0;
System.out.println(n + " factorial = " + factorial(n));
n = 1;
System.out.println(n + " factorial = " + factorial(n));
n = 2;
System.out.println(n + " factorial = " + factorial(n));
n = 3;
System.out.println(n + " factorial = " + factorial(n));
n = 10;
System.out.println(n + " factorial = " + factorial(n));
}
public static long factorial(int n) {
long product = 1L;
for (int i = 2; i <= n; i++) {
product *= (long) i;
}
return product;
}
}
Here's a Haskell version:
factorial n = product [1 .. n]
Not sure if that's considered cheating, though.
There are many ways to calculate the factorial of a number with code.
The original question shows how to do it by recursion, one answer shows how to use a for loop. The same could be achieved by a while loop.
Another answer shows the product function of an array in Haskell (similar to prod(1:n) in Matlab).
Another way to calculate factorial of n (including n=0) in Javascript whithout using a loop is:
function factorial(n){
return (new Array(n)).join().split(",").map((x,i) => i+1).reduce((x,y) => x*y)
}
def factor(n):
fact=1
for i in range (1,n+1):
fact= fact*1
return fact
try:
x=int(input('Input a number to find the factiorial: '))
print(factor(x))
except:
print('Number should be an integer')
x=int(input('Input a number to find the factiorial: '))
print(factor(x))
Hope this helps you!
I have encountered the following problem:
N is positive non-zero integer and I have to calculate the product of : N*(N-1)^2*(N-2)^3*..*1^N.
My solution so far is as follows:
N*myFact(N-1)*fact(N-1)
The thing is I'm not allowed to use any helping functions, such as 'fact()'.
EDIT: Mathematically it can be represented as follows: N!*(N-1)! (N-2)!..*1!
This function is called the superfactorial. A recursive implementation is
long superFact(n) {
if (n < 2) return 1;
long last = superFact(n-1);
long prev = superFact(n-2);
return last * last / prev * n;
}
but this is very inefficient -- it takes about 3*F(n) recursive calls to find superFact(n), where F(n) is the n-th Fibonacci number. (The work grows exponentially.)
Try:
int myFact(int n) {
return n == 1 ? 1 : myFact(n-1)*n;
}
I assume this needs to be accomplished with 1 function i.e. you're not allowed to create a fact helper function yourself.
You can use the fact that myFact(n-1) / myFact(n-2) == (n-1)!
int myFact(int n)
{
if (n == 0 || n == 1) {
return 1
} else {
// (n - 1)!
int previousFact = myFact(n - 1) / myFact(n - 2);
return myFact(n - 1) * previousFact * n;
}
}
I know that the fibonacci algorithm can be programmed without recursion like this:
int fibo(int n){
if(n <= 1){
return n;
}
int fibo = 1;
int fiboPrev = 1;
for(int i = 2; i < n; ++i){
int temp = fibo;
fibo += fiboPrev;
fiboPrev = temp;
}
return fibo;
}
and also that the recursive fibonacci has a complexity of O(2^k) approximately, but for what I see the non-recursive algorithm is O(n); so it seems is way more efficient, is it ok my calculus or is there any hidden complexity on the non-recursive solution?
Evaluate the complexity of the implementation on its own. In this case, the complexity related to the input n is defined by the for loop, which is directly proportional to the size of n. Therefore, the complexity is O(n) - linear.
Project Euler problem 14:
The following iterative sequence is
defined for the set of positive
integers:
n → n/2 (n is even) n → 3n + 1 (n is
odd)
Using the rule above and starting with
13, we generate the following
sequence: 13 → 40 → 20 → 10 → 5 → 16 →
8 → 4 → 2 → 1
It can be seen that this sequence
(starting at 13 and finishing at 1)
contains 10 terms. Although it has not
been proved yet (Collatz Problem), it
is thought that all starting numbers
finish at 1.
Which starting number, under one
million, produces the longest chain?
My first instinct is to create a function to calculate the chains, and run it with every number between 1 and 1 million. Obviously, that takes a long time. Way longer than solving this should take, according to Project Euler's "About" page. I've found several problems on Project Euler that involve large groups of numbers that a program running for hours didn't finish. Clearly, I'm doing something wrong.
How can I handle large groups of numbers quickly?
What am I missing here?
Have a read about memoization. The key insight is that if you've got a sequence starting A that has length 1001, and then you get a sequence B that produces an A, you don't to repeat all that work again.
This is the code in Mathematica, using memoization and recursion. Just four lines :)
f[x_] := f[x] = If[x == 1, 1, 1 + f[If[EvenQ[x], x/2, (3 x + 1)]]];
Block[{$RecursionLimit = 1000, a = 0, j},
Do[If[a < f[i], a = f[i]; j = i], {i, Reverse#Range#10^6}];
Print#a; Print[j];
]
Output .... chain length´525´ and the number is ... ohhhh ... font too small ! :)
BTW, here you can see a plot of the frequency for each chain length
Starting with 1,000,000, generate the chain. Keep track of each number that was generated in the chain, as you know for sure that their chain is smaller than the chain for the starting number. Once you reach 1, store the starting number along with its chain length. Take the next biggest number that has not being generated before, and repeat the process.
This will give you the list of numbers and chain length. Take the greatest chain length, and that's your answer.
I'll make some code to clarify.
public static long nextInChain(long n) {
if (n==1) return 1;
if (n%2==0) {
return n/2;
} else {
return (3 * n) + 1;
}
}
public static void main(String[] args) {
long iniTime=System.currentTimeMillis();
HashSet<Long> numbers=new HashSet<Long>();
HashMap<Long,Long> lenghts=new HashMap<Long, Long>();
long currentTry=1000000l;
int i=0;
do {
doTry(currentTry,numbers, lenghts);
currentTry=findNext(currentTry,numbers);
i++;
} while (currentTry!=0);
Set<Long> longs = lenghts.keySet();
long max=0;
long key=0;
for (Long aLong : longs) {
if (max < lenghts.get(aLong)) {
key = aLong;
max = lenghts.get(aLong);
}
}
System.out.println("number = " + key);
System.out.println("chain lenght = " + max);
System.out.println("Elapsed = " + ((System.currentTimeMillis()-iniTime)/1000));
}
private static long findNext(long currentTry, HashSet<Long> numbers) {
for(currentTry=currentTry-1;currentTry>=0;currentTry--) {
if (!numbers.contains(currentTry)) return currentTry;
}
return 0;
}
private static void doTry(Long tryNumber,HashSet<Long> numbers, HashMap<Long, Long> lenghts) {
long i=1;
long n=tryNumber;
do {
numbers.add(n);
n=nextInChain(n);
i++;
} while (n!=1);
lenghts.put(tryNumber,i);
}
Suppose you have a function CalcDistance(i) that calculates the "distance" to 1. For instance, CalcDistance(1) == 0 and CalcDistance(13) == 9. Here is a naive recursive implementation of this function (in C#):
public static int CalcDistance(long i)
{
if (i == 1)
return 0;
return (i % 2 == 0) ? CalcDistance(i / 2) + 1 : CalcDistance(3 * i + 1) + 1;
}
The problem is that this function has to calculate the distance of many numbers over and over again. You can make it a little bit smarter (and a lot faster) by giving it a memory. For instance, lets create a static array that can store the distance for the first million numbers:
static int[] list = new int[1000000];
We prefill each value in the list with -1 to indicate that the value for that position is not yet calculated. After this, we can optimize the CalcDistance() function:
public static int CalcDistance(long i)
{
if (i == 1)
return 0;
if (i >= 1000000)
return (i % 2 == 0) ? CalcDistance(i / 2) + 1 : CalcDistance(3 * i + 1) + 1;
if (list[i] == -1)
list[i] = (i % 2 == 0) ? CalcDistance(i / 2) + 1: CalcDistance(3 * i + 1) + 1;
return list[i];
}
If i >= 1000000, then we cannot use our list, so we must always calculate it. If i < 1000000, then we check if the value is in the list. If not, we calculate it first and store it in the list. Otherwise we just return the value from the list. With this code, it took about ~120ms to process all million numbers.
This is a very simple example of memoization. I use a simple list to store intermediate values in this example. You can use more advanced data structures like hashtables, vectors or graphs when appropriate.
Minimize how many levels deep your loops are, and use an efficient data structure such as IList or IDictionary, that can auto-resize itself when it needs to expand. If you use plain arrays they need to be copied to larger arrays as they expand - not nearly as efficient.
This variant doesn't use an HashMap but tries only to not repeat the first 1000000 numbers. I don't use an hashmap because the biggest number found is around 56 billions, and an hash map could crash.
I have already done some premature optimization. Instead of / I use >>, instead of % I use &. Instead of * I use some +.
void Main()
{
var elements = new bool[1000000];
int longestStart = -1;
int longestRun = -1;
long biggest = 0;
for (int i = elements.Length - 1; i >= 1; i--) {
if (elements[i]) {
continue;
}
elements[i] = true;
int currentStart = i;
int currentRun = 1;
long current = i;
while (current != 1) {
if (current > biggest) {
biggest = current;
}
if ((current & 1) == 0) {
current = current >> 1;
} else {
current = current + current + current + 1;
}
currentRun++;
if (current < elements.Length) {
elements[current] = true;
}
}
if (currentRun > longestRun) {
longestStart = i;
longestRun = currentRun;
}
}
Console.WriteLine("Longest Start: {0}, Run {1}", longestStart, longestRun);
Console.WriteLine("Biggest number: {0}", biggest);
}