Draw cube vertices with fewest number of steps - geometry

Question

What's the fewest number of steps needed to draw all of the cube's vertices, without picking up the pen from the paper?
So far I have reduced it to 16 steps:
0, 0, 0
0, 0, 1
0, 1, 1
1, 1, 1
1, 1, 0
0, 1, 0
0, 0, 0
1, 0, 0
1, 0, 1
0, 0, 1
0, 1, 1
0, 1, 0
1, 1, 0
1, 0, 0
1, 0, 1
1, 1, 1
I presume it can be reduced less than 16 steps as there are only 12 vertices to be drawn
You can view a working example in three.js javascript here:
http://jsfiddle.net/kmturley/5aeucehf/show/

Answer 1

Well I encoded a small brute force solver for this
the best solution is with 16 vertexes
took about 11.6 sec to compute
all is in C++ (visualization by OpenGL)
First the cube representation:
//---------------------------------------------------------------------------
#define a 0.5
double pnt[]=
{
-a,-a,-a, // point 0
-a,-a,+a,
-a,+a,-a,
-a,+a,+a,
+a,-a,-a,
+a,-a,+a,
+a,+a,-a,
+a,+a,+a, // point 7
1e101,1e101,1e101, // end tag
};
#undef a
int lin[]=
{
0,1,
0,2,
0,4,
1,3,
1,5,
2,3,
2,6,
3,7,
4,5,
4,6,
5,7,
6,7,
-1,-1, // end tag
};
// int solution[]={ 0, 1, 3, 1, 5, 4, 0, 2, 3, 7, 5, 4, 6, 2, 6, 7, -1 }; // found polyline solution
//---------------------------------------------------------------------------
void draw_lin(double *pnt,int *lin)
{
glBegin(GL_LINES);
for (int i=0;lin[i]>=0;)
{
glVertex3dv(pnt+(lin[i]*3)); i++;
glVertex3dv(pnt+(lin[i]*3)); i++;
}
glEnd();
}
//---------------------------------------------------------------------------
void draw_pol(double *pnt,int *pol)
{
glBegin(GL_LINE_STRIP);
for (int i=0;pol[i]>=0;i++) glVertex3dv(pnt+(pol[i]*3));
glEnd();
}
//---------------------------------------------------------------------------
Now the solver:
//---------------------------------------------------------------------------
struct _vtx // vertex
{
List<int> i; // connected to (vertexes...)
_vtx(){}; _vtx(_vtx& a){ *this=a; }; ~_vtx(){}; _vtx* operator = (const _vtx *a) { *this=*a; return this; }; /*_vtx* operator = (const _vtx &a) { ...copy... return this; };*/
};
const int _max=16; // know solution size (do not bother to find longer solutions)
int use[_max],uses=0; // temp line usage flag
int pol[_max],pols=0; // temp solution
int sol[_max+2],sols=0; // best found solution
List<_vtx> vtx; // model vertexes + connection info
//---------------------------------------------------------------------------
void _solve(int a)
{
_vtx *v; int i,j,k,l,a0,a1,b0,b1;
// add point to actual polyline
pol[pols]=a; pols++; v=&vtx[a];
// test for solution
for (l=0,i=0;i<uses;i++) use[i]=0;
for (a0=pol[0],a1=pol[1],i=1;i<pols;i++,a0=a1,a1=pol[i])
for (j=0,k=0;k<uses;k++)
{
b0=lin[j]; j++;
b1=lin[j]; j++;
if (!use[k]) if (((a0==b0)&&(a1==b1))||((a0==b1)&&(a1==b0))) { use[k]=1; l++; }
}
if (l==uses) // better solution found
if ((pols<sols)||(sol[0]==-1))
for (sols=0;sols<pols;sols++) sol[sols]=pol[sols];
// recursion only if pol not too big
if (pols+1<sols) for (i=0;i<v->i.num;i++) _solve(v->i.dat[i]);
// back to previous state
pols--; pol[pols]=-1;
}
//---------------------------------------------------------------------------
void solve(double *pnt,int *lin)
{
int i,j,a0,a1;
// init sizes
for (i=0;i<_max;i++) { use[i]=0; pol[i]=-1; sol[i]=-1; }
for(i=0,j=0;pnt[i]<1e100;i+=3,j++); vtx.allocate(j); vtx.num=j;
for(i=0;i<vtx.num;i++) vtx[i].i.num=0;
// init connections
for(uses=0,i=0;lin[i]>=0;uses++)
{
a0=lin[i]; i++;
a1=lin[i]; i++;
vtx[a0].i.add(a1);
vtx[a1].i.add(a0);
}
// start actual solution (does not matter which vertex on cube is first)
pols=0; sols=_max+1; _solve(0);
sol[sols]=-1; if (sol[0]<0) sols=0;
}
//---------------------------------------------------------------------------
Usage:
solve(pnt,lin); // call once to compute the solution
glColor3f(0.2,0.2,0.2); draw_lin(pnt,lin); // draw gray outline
glColor3f(1.0,1.0,1.0); draw_pol(pnt,sol); // overwrite by solution to visually check correctness (Z-buffer must pass also on equal values!!!)
List
is just mine template for dynamic array
List<int> x is equivalent to int x[]
x.add(5) ... adds 5 to the end of list
x.num is the used size of list in entries
x.allocate(100) preallocate list size to 100 entries (to avoid relocations slowdowns)
solve(pnt,lin) algorithm
first prepare vertex data
each vertex vtx[i] corresponds to point i-th point in pnt table
i[] list contains the index of each vertex connected to this vertex
start with vertex 0 (on cube is irrelevant the start point
otherwise there would be for loop through every vertex as start point
_solve(a)
it adds a vertex index to actual solution pol[pols]
then test how many lines is present in actual solution
and if all lines from lin[] are drawn and solution is smaller than already found one
copy it as new solution
after test if actual solution is not too long recursively add next vertex
as one of the vertex that is connected to last vertex used
to limit the number of combinations
at the end sol[sols] hold the solution vertex index list
sols is the number of vertexes used (lines-1)
[Notes]
the code is not very clean but it works (sorry for that)
hope I did not forget to copy something