Algorithms

shaibee wrote:

have a double value describing rotation angle for a text box object. I need to convert it to quaternion. I have searched the web, and all I get is converters that conver Euler angle to Quaternion.

I would appreciate any formulae or algorithm.
And this double value is the only information I have for rotation.

73Zeppelin wrote:

rotation is given by a unit vector (ax, ay, az) and the angle theta is in radians

/ write_item(X) by T

write_item(X)
{
     while(true)
     {
          if( write_TS(X) > TS(T) || read_TS(X) > TS(T) )  // <--THIS LINE
          {
                abort(T); // break or return
          }
          else if( write_TS(X) < TS(T) )
          {
               wait until the transaction T', which TS(T) > TS(T'), has
               either committed or aborted; // write_TS(X) == TS(T')
                  continue;   
          }
          else
          {
                write_item(X);
                write_TS(X) = TS(T);
                break; // or return                
          }
     }
} 

// rmax(X): the max timestamp of a transaction that read that item.
// wmax(X): the max timestamp of a transaction that wrote the item.

// read X:

if t >= wmax(X)
    rmax(X) <- max(t, rmax(X))
    wait for a committed value of X
    perform read
else 
    abort

// write X:

if t >= rmax(X) and t >= wmax(X)
    wmax(X) <- t
    wait until X value is a committed value and
        pending reads are done
    perform write
else
    if t < rmax(X)
    abort
else
    do nothing

<br />
// there often are two nested loops<br />
// this "main loop" represent both loops<br />
main loop {<br />
    int i = ...;<br />
    int j = ...;<br />
    if (A[i] > A[j])<br />
        swap(A[i],A[j]);<br />
    // there might be more than one comparison-swap pair<br />
    // per iteration of the main loop, but you get the idea<br />
}<br />

<br />
main loop {<br />
    int i = ...;<br />
    int j = ...;<br />
    if (A[i] == A[j])<br />
        return A[i];    // repeated element found<br />
    else if (A[i] > A[j])<br />
    swap(A[i],A[j]);<br />
}<br />