It is necessary to remake from the incidence matrix in the adjacency matrix of the adjustment graph. For non-versatile algorithms in general, it is necessary: ​​it is necessary in the first matrix in each row to find all the elements with the value 1 and put 1 in the second matrix at the intersection of these elements. Posted by the program, but the aochy does not bring it quite true, please help

{1, 1, 0, 0}, {1,0,1,0}, {0.0 0.1}, {0,1,1,0}}
Redo from this matrix of incident in the arrangement matrix.
   for (int j = 1; j & lt; = v; j ++)
      {
        for (int i = 1; i & lt; = v; i ++)
        {
          IF (i [i] [j] == 1) {
            COUNT + = 1;
          }
        }
      }
      INT COUNT2 = COUNT / 2;
      COUT & LT; & LT; COUNT2;
      COUT & LT; & LT; "\ nmatritsa SMEJNOSTI \ N";
      INT SM [100] [100];
      INT P1, P2;
      For (int i = 1; i & lt; = count2; i ++) {
        For (int j = 1; j & lt; = count2; j ++) {
          SM [i] [j] = 0;
        }
      }
      COUT & LT; & LT; "\ NCALC \ N";
      For (int i = 1; i & lt; = count2; i ++) {
        For (int j = 1; j & lt; = count2; j ++) {
          IF (i [i] [j] == 1) {
            For (int k = j + 1; k & lt; = count2; k ++) {
              if (i [i] [k] == 1) {
                SM [k] [j] = 1;
            SM [j] [k] = 1;
              }
            }
          }
        }
      }
      // _________________________ Output
      COUT & LT; & LT; "\ NSM Matrix: \ N" & lt; & lt; Endl;
      For (int j = 1; j & lt; = count2; j ++)
      {
        for (int i = 1; i & lt; = count2; i ++)
        {
          COUT & LT; & LT; "" & lt; & lt; SM [i] [J];
        }
        COUT & LT; & LT; Endl;
      }
      COUT & LT; & LT; Endl;
      // End.
      _Getch ();
    }

Answer 1, Authority 100%

in size of the original matrix determine V and E – the number of vertices and edge (in the program for some reason implies their equality).

Next, go through all single pairs of rows (i.e., find a pair of ribs, incident to this top)

for (v = 1; v & lt; = v; v ++)
  For (E1 = 1; E1 & lt; = E-1; E1 ++)
    IF (I [V] [E1])
       For (E2 = E1 + 1; E2 & LT; = E; E2 ++)
         IF (I [V] [E2]) {
           Ar [E1] [E2] = 1;
           Ar [E2] [E1] = 1;
         }

Answer 2, AUTHORITY 60%

a little ahead of MBO 🙂 – While I painted all comments. Well, not to disappear code …

If you have C++, it is better to use vectors.

# include & lt; vector & gt;
#Include & lt; iostream & gt;
Using Namespace STD;
Vector & lt; vector & lt; int & gt; & gt; Rebuild (Const Vector & LT; vector & lt; int & gt; & gt; & amp; g)
{
  Size_t VXS = G [0] .size (); // Number of columns
  // here you need to make sure that
  // in each row the same number of columns
  Vector & lt; vector & lt; int & gt; & gt; R (VXS, vector & lt; int & gt; (VXS, 0)); // Result
  // Passage by lines
  For (Const Auto & amp; S: G)
  {
    Vector & lt; Size_t & gt; IX; // Collect nonzero values
    For (Size_t i = 0; i & lt; s.size (); ++ i)
      if (s [i]) ix.push_back (i);
    // if their at least 2 ...
    if (ix.size () & gt; 1)
    {
      // each with each - arrange units
      For (Size_t i = 0; i & lt; ix.size () - 1; ++ i)
        For (Size_t j = i + 1; j & lt; ix.size (); ++ j)
          R [IX [i]] [IX [J]] = R [IX [J]] [IX [I]] = 1;
    }
  }
  Return R;
}
INT MAIN ([[Maybe_Unused]] int argc,
     [[Maybe_Unused]] Const Char * ArgV [])
{ 
Vector & lt; vector & lt; int & gt; & gt; G = {{1,1,0,0}, {1,0,1,1}, {0,0,1,1}, {0,1,1,0}};
   Vector & lt; vector & lt; int & gt; & gt; R = Rebuild (G);
   For (Size_t i = 0; i & lt; r.size (); ++ i)
   {
     for (Size_t j = 0; j & lt; r [i] .size (); ++ j)
       COUT & LT; & LT; R [i] [j] & lt; & lt; "";
     COUT & LT; & LT; Endl;
   }
}