8 replies to this topic

[rainwater_stillicide]

Write a function f which takes an int as an argument and where f(f(x)) = -x

eg.
my_function( my_function ( 5 ) ) = -5

You can use any language you like. Feel free to post any partial solutions you come up with also.

[AlexZ]

Write a function f which takes an int as an argument and where f(f(x)) = -x

eg.
my_function( my_function ( 5 ) ) = -5

You can use any language you like. Feel free to post any partial solutions you come up with also.

Here my C++ solution.
Excuse me, but I've solved my problem and I've posted the correct solution in the same post to take clear the topic!!

#include <iostream>
using namespace std;
int opposite ( int num ) ;

int main()
{
    int n;
    cout << "Insert a number " << endl;
    cin >> n;

    cout << "Opposite number  = " << opposite ( n ) << endl;

    return 0;

}

int opposite ( int num )
{
    num =~ num;

    return num+1;
}

Edited by AlexZ, 07 March 2010 - 05:46 PM.

[rainwater_stillicide]

Write a function f which takes an int as an argument and where f(f(x)) = -x

eg.
my_function( my_function ( 5 ) ) = -5

You can use any language you like. Feel free to post any partial solutions you come up with also.

Here my C++ solution.
Excuse me, but I've solved my problem and I've posted the correct solution in the same post to take clear the topic!!

#include <iostream>
using namespace std;
int opposite ( int num ) ;

int main()
{
    int n;
    cout << "Insert a number " << endl;
    cin >> n;

    cout << "Opposite number  = " << opposite ( n ) << endl;

    return 0;

}

int opposite ( int num )
{
    num =~ num;

    return num+1;
}

I'm afraid that is not correct. You are only applying opposite once, the challenge is to write a function which you apply twice to get the negative of the original argument:
opposite ( opposite ( x ) ) = -x

[AlexZ]

Write a function f which takes an int as an argument and where f(f(x)) = -x

eg.
my_function( my_function ( 5 ) ) = -5

You can use any language you like. Feel free to post any partial solutions you come up with also.

Here my C++ solution.
Excuse me, but I've solved my problem and I've posted the correct solution in the same post to take clear the topic!!

#include <iostream>
using namespace std;
int opposite ( int num ) ;

int main()
{
    int n;
    cout << "Insert a number " << endl;
    cin >> n;

    cout << "Opposite number  = " << opposite ( n ) << endl;

    return 0;

}

int opposite ( int num )
{
    num =~ num;

    return num+1;
}

I'm afraid that is not correct. You are only applying opposite once, the challenge is to write a function which you apply twice to get the negative of the original argument:
opposite ( opposite ( x ) ) = -x

Oh yes you're right!!Sorry...I'll try to do another function, but can you explain me better your exercise?
EDIT:
But this code works fine just like you've suggested to me

#include <iostream>
using namespace std;
int opposite ( int num ) ;

int main()
{
    int n;
    cout << "Insert a number " << endl;
    cin >> n;

    cout << "Opposite number  = " << opposite ( opposite(n) ) << endl;

    return 0;

}

int opposite ( int num )
{
    if ( num > 0 )
    {
        num =~ num;
        return num+1;
    }
    else { return num; }
}

Edited by AlexZ, 08 March 2010 - 08:31 AM.

[JBu92]

well if python worked with imaginary numbers:

Spoiler

import math
i = math.sqrt(-1)
def doubinv(num):
    num = num*i
x = input()
print doubinv(doubinv(x))

but it doesn't... :-(
edit: here's a cheaty, non-math way to do it... also python

Spoiler

import math
i = 0
def doubinv(num):
    i += 1
    if (i = 2):
        num = num*-1
x = input()
print doubinv(doubinv(x))

Edited by JBu92, 17 September 2010 - 08:46 AM.

[systmkor]

Here is my solution in C.

Spoiler

#include <stdio.h>
#include <math.h>

int inverso( int num ) {
  return - fabs(num);
}

int main(void) {
  int num_in, num_out;

  num_in = 0;
  num_out = 0;
  printf("Input Number: ");
  scanf("%d", &num_in);
  num_out = inverso(inverso(5));
  printf("Opposite: %d\n", num_out);

  return(0);
}

[JBu92]

Here is my solution in C.

Spoiler

#include <stdio.h>
#include <math.h>

int inverso( int num ) {
  return - fabs(num);
}

int main(void) {
  int num_in, num_out;

  num_in = 0;
  num_out = 0;
  printf("Input Number: ");
  scanf("%d", &num_in);
  num_out = inverso(inverso(5));
  printf("Opposite: %d\n", num_out);

  return(0);
}

counterexample: -1

[codered22]

Here's mine.

Spoiler

// File: main.cpp
// Title: BinRev BASIC002
// Description: Write a function f which takes an int as an argument and where
//          	f(f(x)) = -x
// Author: codered22
// Date: 2010-10-23

#include<iostream>

using namespace std;

int getNegative(int _num) {
	if (_num <= 0)
		return _num;
	else
		return _num * -1;
}

int main() {
	int x;

	cin >> x;
	
	cout << getNegative(getNegative(x));
	
	return 0;
}

[heisenbug]

Spoiler

$inverse = inverse($number);

sub inverse() {

	shift;

	$return_number = $_ * -1;

	return $retun_number;

	}