Hirzebruch, in the paper 'Arrangements of Lines and Algebraic Surfaces'
constructs a special $K3$ surface out of a 'complete quadrilateral' in 
 $CP^2$. A complete quadritlateral consists of 
 4 points in general position and the $6$ lines joining them.  

Over each line Hirzebruch forms the local 2:1 fold cover to get
a new surface which comes as a branched
covering of $CP^2$, branched over the $6$ lines.
Away from the lines the covering has degree $2^{6-1}$.
This surface has singularities of conical type at the original $4$ points
(At these points 3 lines are coincident. )
Blow up the singularities coming from these 4 points.. The resulting smooth surface is Hirzebruch's $K3$.


Viewed from a different perspective, I believe that I can get this same
$K3$ has an elliptic surface over $CP^1$ with $6$ singular fibers.
I also believe that each of the singular fibers are of $A\_1$ type, meaning two $CP^1$'s intersecting transversally
(as in $xy = 0$), but am less sure of this. 
The corresponding singular points on $CP^1$ can be taken to be the vertices of the octahedron.  

And I believe that the manifest symmetry group of order $4! = 24$ seen in Hirzebruch's construction (permute the 
original 4 points) agrees with the symmetry group of the octahedron. 


Questions. Do you know this second K3? If so, could you give me a reference for it?
Have you seen a place which shows that the second $K3$ is the same as Hirzebruch's?


More generally, what are the first few 'simplest' elliptic $K3$'s? 
By 'elliptic' I mean expressed as elliptic surface $f: X \to CP^1$, over
$CP^1$. By 'simplest' I mean a small number of singular fibers whose singularities
are as 'simple' as possible. For example, if all singular fibers are of $A\_1$-type,
what is the fewest number of fibers? Must this number be $6$? 
(I have looked in Barth-Hulek-Peters-van de Ven's 'Compact Complex Surfaces',
 esp. ch. V, sec. 2 and suppose this information is buried there somehow
or other, but is rather beyond me to untangle it from there. Neither did I
find this 2nd $K3$ in Gompf and Stipsicz's book )



A K3 over $P^1$  with six singular $A_1$- fibers?
[**EDITED** to give likely identification of the OP's surface at the end]


An elliptic K3 surface has discriminant $\Delta$ of degree $24$, and $\Delta$ has valuation 2 or 3 at an $A\_1$ fiber depending on whether the Kodaira type is I$\_2$ or III, so if that's the only kind of singular fiber there must be at least $24/3=8$ such fibers (and that is possible, e.g. $y^2 = x^3 + ax$ over the $t$-line, with $a=a(t)$ a polynomial of degree $8$ with distinct roots).


Your description of an $A\_1$ fiber is not quite right: it's two lines meeting at *two* points, counted with multiplicity, so either transversely at two different points (type I$\_2$) or at the same point with multiplicity 2 (type III). The intersection graph is the *extended* Dynkin diagram $\tilde A\_1$. Likewise for the other A-D-E fiber types.


For the K3 surface you're studying: in general a K3 surface can have many (albeit finitely many) non-isomorphic elliptic fibrations, so without further information I can't compute the configuration of singular fibers and other invariants of your fibration. How did you obtain it? Do you have an explicit formula?


[**added the next day**] On further thought, there's a natural elliptic fibration on this surface that's probably what the OP had in mind, with octahedral symmetry respecting the $S\_4$ symmetry of Hirzebruch's construction, and it is indeed a well-known fibration: the universal elliptic curve over the modular curve ${\rm X}(4)$, i.e. the universal curve with full level-4 structure. This fibration in fact respects all $2^5 4!$ symmetries of the Hirzebruch construction ($4!$ permuting the four vertices, and $2^5$ deck transformations): the octahedral symmetries give ${\rm PSL}\_2({\bf Z}/4{\bf Z}) \cong S\_4$, and translation by the $4^2$ torsion points together with inversion in the group law gives the factor of $2^5$. The six reducible fibers, coming from the six cusps of ${\rm X}(4)$, are of type I$\_4$, so contribute $A\_3$ each to the Néron-Severi lattice. I don't know whether this identification is in the literature already; if anybody would know whether it's been published already, Dolgachev would.


To get this fibration, start with the pencil of conics through the four special points $P\_1,\ldots,P\_4$. A generic such conic is a rational curve on which Hirzebruch's $2^5:1$ cover restricts to a $2^3:1$ cover branched at four points, giving an elliptic curve isomorphic with the double cover branched at the same points. That's not quite a fibration of our K3, because the missing factor of $2^2$ comes from extracting square roots of functions that are constant but not necessarily square on the generic conic; we need a $2^2:1$ cover of the base curve of the pencil. This base change is branched at three points, corresponding to the three reducible conics in the pencil (unions of lines {$P\_1 P\_2$ and $P\_3 P\_4$}, {$P\_1 P\_3$ and $P\_4 P\_2$}, or {$P\_1 P\_4$ and $P\_2 P\_3$}). Thus the actual base curve has six cusps, with $S\_4$ symmetry coming from the $S\_3$ permutations of the reducible conics together with the $2^2:1$ cover.


To do this explicitly it is convenient to put the $P\_m$ at the four points with projective coordinates $(\pm 1 : \pm 1 : \pm 1)$. The action of $S\_4$ is then by signed permutation matrices.
The six lines $P\_m P\_n$ are $x = \pm y$, $y = \pm z$, and $z = \pm x$. The pencil of conics is $Ax^2 + By^2 + Cz^2 = 0$ with $(A:B:C)$ on the line $A+B+C=0$. The $2^2:1$ base change is $(A:B:C) = (a^2:b^2:c^2)$.


