[vask0]

Hello fellow I have the following problem to solve ..

I will be grateful to someone who can help with the solution ..

Write a program that inputs values V1, V2 from the keyboard and calculates the ratio of V1 / V2 in dB by the formula
dB = 20 lg (V2 / V1) for V2> 0; V1> 0


1. Write source code C++
2. Test it

Verification
A) Example Input V2 / V1> 1
B) Example Input V2 / V1 <1
C) Example input V2 = V1
F) Example output dB> 0
F) Example output dB <0
D) Example output dB = 0

10x!!!!

[b49P23TIvg]

For A I'd think that V2 = 1e6 and V1 = 1e-6 ought to work.
And for B you could set V2 to 1e-6 and V1 = 1e6 .
C. Quite difficult. I cannot think this through. How about V1 = V2 = 0; // ?
The answer to F could be the same as the answer to A.
The answer to F ought to be same as the answer to B.
Yes I know I wrote that.

[vask0]

Hi colleague and thank you for your reply ..
source-code how would you look? How about V1 = V2 = 0; // ?... you idea sounds good

and we have to get him to work.

[b49P23TIvg]

and I'll mention for D) you'd need to solve lg(x) = 0 for x and then find values for V1 and V2 for which x = V2/V1 could be true. A difficult task. Here's the verb in j, you'll just need to translate it to c.

   NB. www.jsoftware.com Ken Iverson's final dialect of APL
   dB=: 20 * 10 ^. %  NB. use:  V2 dB V1

[vask0]

Originally Posted by b49P23TIvg

and I'll mention for D) you'd need to solve lg(x) = 0 for x and then find values for V1 and V2 for which x = V2/V1 could be true. A difficult task. Here's the verb in j, you'll just need to translate it to c.

   NB. www.jsoftware.com Ken Iverson's final dialect of APL
   dB=: 20 * 10 ^. %  NB. use:  V2 dB V1

Hi colleague again, I think the code should be something of the sort

#include<iostream>
#include<iomanip>
 
using namespace std;
 
// A function to print the matrix.
void PrintMat(int **mat, int n)
{
	int i, j;
 
	cout<<"\n\n"<<setw(4)<<"";
	for(i = 0; i < n; i++)
		cout<<setw(3)<<"("<<i+1<<")";
	cout<<"\n\n";
 
	// Print 1 if the corresponding vertexes are connected otherwise 0.
	for(i = 0; i < n; i++)
	{
		cout<<setw(3)<<"("<<i+1<<")";
		for(j = 0; j < n; j++)
		{
			cout<<setw(4)<<mat[i][j];
		}
		cout<<"\n\n";
	}
}
 
int main()
{
	int i, v, e, j, v1, v2;
 
	// take the input of the number of edges.
	cout<<"Enter the number of vertexes of the graph: ";
	cin>>v;
 
	// Dynamically declare graph.
	int **graph;
	graph = new int*[v];
 
	for(i = 0; i < v; i++)
	{
		graph[i] = new int[v];
		for(j = 0; j < v; j++)graph[i][j] = 0;
	}
 
	cout<<"\nEnter the number of edges of the graph: ";
	cin>>e;
 
	// Take the input of the adjacent vertex pairs of the given graph.
	for(i = 0; i < e; i++)
	{
		cout<<"\nEnter the vertex pair for edge "<<i+1;
		cout<<"\nV(1): ";
		cin>>v1;
		cout<<"V(2): ";
		cin>>v2;
 
		graph[v1-1][v2-1] = 1;
		graph[v2-1][v1-1] = 1;
	}
 
	// Print the 2D array representation of the graph.
	PrintMat(graph, v);
}

But this is another very different task...

[b49P23TIvg]

Absolutely correct. The solution to this puzzle will have similarities to the program you posted. Here's the lg function for you:

double lg(double a) {
  const double MAGIC_NUMBER = 2.302585092994045901;
  const int LUCKY_SEVEN = 7;
  double sum = 0, x = a-1;
  int i;
  for (i = LUCKY_SEVEN; i; --i) {
    sum += 1.0/(i*(1-((!(i&1))<<1)));
    sum *= x;
  }
  return sum / MAGIC_NUMBER;
}

[vask0]

hello colleague and thank you very much for your answers and help.
the final code should look like ..?

#include<iostream>
#include<iomanip>
 
using namespace std;

double lg(double a) {
  const double MAGIC_NUMBER = 2.302585092994045901;
  const int LUCKY_SEVEN = 7;
  double sum = 0, x = a-1;
  int i;
  for (i = LUCKY_SEVEN; i; --i) {
    sum += 1.0/(i*(1-((!(i&1))<<1)));
    sum *= x;
  }
  return sum / MAGIC_NUMBER;
}

[b49P23TIvg]

Yes that's great now just compile it into an executable and run it.

[vask0]

Hi colleague I was trying to work, but it returns me the following jar

[b49P23TIvg]

gosh maybe the

int main() {
  return 0;
}

was one of the similarities we agreed upon beforehand. Anyway, congratulations. I was fairly certain you'd be unable to use the compiler in any way whatsoever. Score one for you.

[vask0]

[code]

#include <iostream>
#include <windows.h>

using namespace std;

double lg(double a) {
const double MAGIC_NUMBER = 2.302585092994045901;
const int LUCKY_SEVEN = 7;
double sum = 0, x = a-1;
int i;
for (i = LUCKY_SEVEN; i; --i) {
sum += 1.0/(i*(1-((!(i&1))<<1)));
sum *= x;
int main() {
return 0;
}


[code]

This code does not work...

[b49P23TIvg]

Remarkable. Maybe it's because you haven't applied the lg function to any of the ratios identified in post 2. Then again, there could be additional problems.

[vask0]

will you be able to provide sample code to test if it works

10x

[b49P23TIvg]

oh sure here's a program to test your superb decibel function.

//
// test program for the decibel function provided by someone else.
// main returns the number of errors discovered.
//
// exercise for reader: extend to include tests for special case values
// of V1 and V2.  These depend on how you choose to handle them.
//

#include<iostream>

#define DIM(A) (sizeof(A)/sizeof(*(A)))
#define ABS(A) ((A) < 0 ? -(A) : (A))
#define MIN(A, B) ((A) < (B) ? (A) : (B))

struct {
  double V2, V1, dB_expected;
} test[] = {
  {0.100,   9.600,    -39.6454246607913632},
  {0.600,   9.100,    -23.6178028387490002},
  {1.100,   8.600,    -17.8621153217068525},
  {1.600,   8.100,    -14.0873007244544990},
  {2.100,   7.600,    -11.1718859509374404},
  {2.600,   7.100,     -8.7257000149651454},
  {3.100,   6.600,     -6.5636448341519182},
  {3.600,   6.100,     -4.5805466848695939},
  {4.100,   5.600,     -2.7080834057292984},
  {4.600,   5.100,     -0.8962468883272455},
  {5.100,   4.600,      0.8962468883272461},
  {5.600,   4.100,      2.7080834057292993},
  {6.100,   3.600,      4.5805466848695939},
  {6.600,   3.100,      6.5636448341519182},
  {7.100,   2.600,      8.7257000149651454},
  {7.600,   2.100,     11.1718859509374404},
  {8.100,   1.600,     14.0873007244544972},
  {8.600,   1.100,     17.8621153217068525},
  {9.100,   0.600,     23.6178028387489967},
  {9.600,   0.100,     39.6454246607913632},
};

int tolerant_equal(double a, double b, double relerr, double abserr) {
  double abserris = ABS(a - b);
  return (abserris < MIN(ABS(a), ABS(b)) * relerr) || (abserris < abserr);
}

// supply the function decibel.  It is passed the arguments V2 and V1 in that order.
double decibel(double, double);

int main() {
  double V1, V2, expect, got;
  unsigned i, errors = 0;
  for (i = 0; i < DIM(test); ++i) {
    V1 = test[i].V1;
    V2 = test[i].V2;
    expect = test[i].dB_expected;
    got = decibel(V2, V1);
    if (! tolerant_equal(got, expect, 1e-5, 1e-4)) {
      ++errors;
      std::cout << "decibel(" << V2 << ", " << V1 << ") gave " << got << " expected " << expect << '\n';
    }
  }
  return errors;
}

[vask0]

Hi colleague this code worked ... thank you very much !!!
look for me in skype: linux_bg-

10x!!!!