See picture here
Hello,
I need to convert the formula above to code (Java or C#)
I did a lot of tries without success.
Can anyone help?
Here's your answer in Java:
public static int f(int n) {
if(n == 0)
return 1;
int sum = 0;
for(int i = 0; i < n; i++)
sum += f(i) * f(n - 1 - i);
return sum;
}
Enjoy.
Below code is in php, in java or c# would have same logic
assuming f(x) return x+1, for n=3, sum would be 10
<?php
function fun($x)
{
echo $x+1;
return $x+1;
}
$n=3;
$sum=0;
if($n>0)
{
$i = $n-1;
while($i>=0)
{
$sum = $sum +(fun($i) * fun($n-1-$i));
$i= $i-1;
}
}
else if($n=0)
{
$sum = 1;
}
echo $sum;
?>
Related
I am trying to calculate the euclidean distance for KNN but in parallel using dpc++. the training dataset contains 5 features and 1600 rows, while I want to calculate the distance between the current test point and each training point on the grid in parallel, but I keep getting an error regarding sycl kernal.
code for the function:
code
std::vector<double> distance_calculation_FPGA(queue& q,const std::vector<std::vector<double>>& dataset,const std::vector<double>& curr_test) {
range<1> num_items{ dataset.size()};
std::vector<double>res;
res.resize(dataset.size());
buffer dataset_buf(dataset);
buffer curr_test_buf(curr_test);
buffer res_buf(res.data(), num_items);
q.submit([&](handler& h) {
accessor a(dataset_buf, h, read_only);
accessor b(curr_test_buf, h, read_only);
accessor dif(res_buf, h, write_only, no_init);
h.parallel_for(num_items, [=](auto i) {
for (int j = 0; j <(const int) a[i].size(); ++j) {
dif[i] += (a[i][j] - b[j]) * (a[i][j] - b[j]) ;
}
});
});
for (int i = 0; i < res.size(); ++i) {
std::cout << res[i] << std::endl;
}
//old distance calculation (serial)
//for (int i = 0; i < dataset.size(); ++i) {
// double dis = 0;
// for (int j = 0; j < dataset[i].size(); ++j) {
// dis += (curr_test[j] - dataset[i][j]) * (curr_test[j] - dataset[i][j]);
//}
//res.push_back(dis);
//}
return res;
}
the error I am receiving:
SYCL kernel cannot call a variadic function
SYCL kernel cannot call an undefined function without SYCL_EXTERNAL attribute
Would be extremely grateful for any help!
Thanks
We tried running your code by creating dummy 'dataset' and 'curr_test' variables. We were able to run the program successfully. Please refer this thread
Please refer to the complete code attached below.
#include <CL/sycl.hpp>
#include <iostream>
using namespace sycl;
std::vector<double> distance_calculation_FPGA(queue& q,const std::vector<std::vector<double>>& dataset,const std::vector<double>& curr_test)
{
range<1> num_items{ dataset.size()};
std::vector<double>res;
res.resize(dataset.size());
buffer dataset_buf(dataset);
buffer curr_test_buf(curr_test);
buffer res_buf(res.data(), num_items);
q.submit([&](handler& h) {
accessor a(dataset_buf, h, read_only);
accessor b(curr_test_buf, h, read_only);
accessor dif(res_buf, h, write_only, no_init);
h.parallel_for(num_items, [=](auto i) {
for (int j = 0; j <(const int) a[i].size(); ++j) {
// dif[i] += (a[i][j] - b[j]) * (a[i][j] - b[j]) ;
dif[i]+=a[i][j];
}
});
});
q.wait(); //We have added this line of code for synchronization.
for (int i : res) {
std::cout <<i<< std::endl;
}
return res;
}
int main(){
std::vector<std::vector<double>> dataset;
for(int i=0;i<5;i++)
{
std::vector<double> d;
for(int j=0;j<1600;j++)
{
d.push_back((double)j);
}
dataset.push_back(d);
}
std::vector<double> curr_test;
for(int i=0;i<1600;i++)
{
curr_test.push_back((double)i);
}
queue q;
std::cout << "Running on "<<
q.get_device().get_info<sycl::info::device::name>()<< std::endl;
//print the device name as a test to check the parallelisation
distance_calculation_FPGA(q,dataset,curr_test);
return 0;
}
Apologies for the basic question, I'm new to java and have been stuck on this for days.
I need firstly to convert letters to numbers and then using recursion to get the sum of those numbers. I think I am close but I'm also aware it very messy
public static void main(String[] arg) {
String str= "11";
//////////////////////////
String s = "helloworld";
String t = "";
for (int i = 0; i < s.length(); ++i) {
char ch = s.charAt(i);
if (!t.isEmpty()) {
t += " ";
}
int n = (int)ch - (int)'a' + 1;
t += String.valueOf(n);
}
System.out.println(t);
//////////////////////////////
int sum=0;
int x=Integer.parseInt(t);
int y=recursion(x);
System.out.println("The Sum of the digits is: "+ y);
}
public static int recursion(int y) {
if(y/10>=1) {
int tempvar =y%10;
int remain=y/10;
return tempvar + recursion(remain);
}
else {
return y;
}
}}
Ok so first of all, this line: int x=Integer.parseInt(t); will crash the program in runtime because the string t has spaces in it. So you need to remove this:
if (!t.isEmpty()) {
t += " ";
}
Second, parsing the string t to int is a problem, because the number in string t can get like very, very large. Parsing this very..very large number to an int will also produce an exception during runtime. So a better way to do this is leaving it as a string, loop over it and just add the digits in it.
I have two solutions here:
I loop on t and add the digits.
I assume you have some constraint on the size of t so as it can be parsed to int (or long), and then use recursion as you want.
public class Main {
public static void main(String[] args) {
String s = "hew";
String t = "";
for (int i = 0; i < s.length(); ++i) {
char ch = s.charAt(i);
int n = (int)ch - (int)'a' + 1;
t += String.valueOf(n);
}
System.out.println("t: "+t);
System.out.println("sum using string: " + getSumUsingString(t));
System.out.println("sum using int: " + getSumUsingLong(Long.parseLong(t), 0));
}
// the string function
private static long getSumUsingString(String t) {
long sum = 0;
for (int i = 0; i < t.length(); i++) {
sum += t.charAt(i)-48;
}
return sum;
}
// recursive function
private static long getSumUsingLong(long num, long sum) {
if (num==0) return sum;
sum += num % 10;
return getSumUsingLong(num / 10, sum);
}
}
Note:
You can use for example, BigInteger class in Java to deal with very large numbers if you really need to parse this string t to a number.
I got "Error Reading characters of string error" in runtime. And it is not handling with try catch,
void process(const BYTE* pBodyIndexBuffer){
m_pCoordinateMapper->MapCameraPointToDepthSpace(m_pJoints[JointType_ShoulderLeft].Position, &p);
dx = static_cast<int>(p.X + 0.5);
dy = static_cast<int>(p.Y + 0.5);
try
{
if (p.X < 500 && p.X >= 0 && p.Y <= 410 && p.Y >= 0)
{
pPoint = pBodyIndexBuffer[dx + (dy*cDepthWidth)];
while (1) {
if (pPoint == 0xff) break;
pPoint = pBodyIndexBuffer[dx + (dy * cDepthWidth)];
dx -= 1;
dy -= 1;
p.X -= 1;
p.Y -= 1;
OutputDebugString(L"Moved \n");
}
}
m_pBodyEdgeswidth[ShoulderLeft] = getDistance(m_pJoints[JointType_ShoulderLeft].Position, p, distance);
}
catch (const std::exception&)
{
OutputDebugString(L"Error Occured");
}
}
I provided the part of code.
Here is the issue I am facing,
Please explain how to resolve this issue.
But sometimes it starts running without error.
how I am passing is,
BYTE *bodyIndex = NULL;
cm->getBodyIndexStream(&bodyIndex);
if(bodyIndex) process(bodyIndex);
Thanks in advance .
You will need to pass a valid non-null buffer to getBodyIndexStream. You are passing null buffer, which satisfies compiler and will definitely fail at runtime. It is like:
int* ptr = NULL;
// Set value
*ptr = 120; // crash
The following is a recursive heapify function for an array based priority queue/binary heap.
Can anybody tell me why I'm going into an infinite recursion?
private static void heapify(int i){
if(i < b_heap.size()/2){
int left = i * 2;
int right = left++;
if(b_heap.get(i).getP() > b_heap.get(left).getP()){
swap(i,left);
}
if(b_heap.get(i).getP() > b_heap.get(right).getP()){
swap(i,right);
}
heapify(i++);
heapify(i+2);
}
else{
return;
}
}
ok so I fixed the infinite loop, but the function still doesn't heapify correctly.
here is the new code,
private static void heapify(int i){
DecimalFormat df = new DecimalFormat("0.00");
if(i < b_heap.size()/2){
int left = i * 2;
int right = i * 2 +1;
if(b_heap.get(i).getP() > b_heap.get(left).getP()){
swap(i,left);
}
if(b_heap.get(i).getP() > b_heap.get(right).getP()){
swap(i,right);
}
i++;
heapify(i);
}
else{
return;
}
}
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I am preparing for my interview tomorrow -- I need the answer to this question:
How can you print 1 to 10 & 10 to 1 by using recursion with single variable
void print_recursive(int n) {
printf("%d\n", n);
if (n < 10)
print_recursive(n+1);
printf("%d\n", n);
}
With one function and one variable only:
void recurs(int num) {
printf("%d\n", num);
if (num < 10) {
recurs(num + 1);
}
printf("%d\n", num);
}
int main() {
recurs(1);
return 0;
}
I'm going to get downvoted I just know it but here is (a) solution. Not the best one but you should be able to make it better yourself.
class Program
{
static void Main(string[] args)
{
printNumDown(10);
}
static void printNumDown(int num)
{
Console.WriteLine(num.ToString());
if (num > 1)
printNumDown(num - 1);
else
printNumUp(num + 1);
}
static void printNumUp(int num)
{
Console.WriteLine(num.ToString());
if (num < 10)
printNumUp(num + 1);
}
}
You should be able to figure this out yourself.
Hint: Make a method that takes a 10 as a parameter, then prints the parameter and calls itself with 9.
Here's a sneaky way:
#include <stdio.h>
static void recur_up (int n) {
if (n > 1)
recur_up (n - 1);
printf ("%d\n", n);
}
static void recur_down (int n) {
printf ("%d\n", n);
if (n > 1)
recur_down (n - 1);
}
int main (void) {
recur_up (10);
recur_down (10);
return 0;
}
which generates:
1
2
3
4
5
6
7
8
9
10
10
9
8
7
6
5
4
3
2
1
It would have been a lot more elegant going down then up since you could do that with a single function:
static void recur_both (int n) {
printf ("%d\n", n);
if (n > 1)
recur_down (n - 1);
printf ("%d\n", n);
}
Javascript version:
printNumber(1);
function printNumber(num){
document.write(num);
if (num < 10)
printNumber(num + 1);
document.write(num);
}
Why are you guys all being so difficult? In Pseudocode:
function recurfunc(n) {
if (n < 10) {
echo (-1 * (floor(abs(n)) - 10));
recurfunc(n+1);
}
}
Then call recurfunc with -9.5 as its start.
Seems kind of obvious to me that the answer is using absolute value.
Here's one in Ruby:
puts (r = ->n=1 { if n<=10 then [n] + r.(n+1) + [n] else [] end }).()
Note: before you conclude that Ruby is an unreadable mess even worse than Perl, let me assure you: this is not idiomatic Ruby. Idiomatic Ruby would be more like
def recursive_count_up_and_down(n=1)
return [] unless n<=10
[n] +
recursive_count_up_and_down(n + 1) +
[n]
end
puts recursive_count_up_and_down
Of course, Ruby is an imperative language, so doing it really idiomatically would not use recursion:
1.upto(10) do |i| puts i end
10.downto(1) do |i| puts i end
Here's another neat one that unfortunately doesn't use recursion, either:
puts Array.new(20) {|i| if i < 10 then i+1 else 20-i end }
BTW: all the solutions so far, including mine, are actually cheating, because technically they use two variables, because
function foo {}
is equivalent to
var foo = λ{}
So, in my example above, there are two variables: recursive_count_up_and_down and n. We could eliminate both of those by writing in a tacit point-free style (for example in the SK-calculus), but I'll leave that as an exercise to the reader. (Meaning I can't figure it out :-) )
#include<stdio.h>
void c(int n)
{
static int i;
if(i<n)
{
i++;
printf("%d",i);
c(n);
}
}
int main()
{
c(5);
}
import java.util.Scanner;
public class PrintNumberFrom1To10{
public static void printNumber(int num){
if(num<=10){
System.out.println(num);
printNumber(num+1);
}
}
public static void main(String[] args) {
System.out.println("Enter Your Number: ");
Scanner sc = new Scanner(System.in);
int num = sc.nextInt();
printNumber(num);
}
}