Penrose dodecahedrons: (1) in preference permutations.

Links to sections:

• Classical light signals vs quantum correlations.
• Quantum rules of the 'magic dodecahedra'.
• Classically assumed independence from quantum correlations.
• Disproof of independence from quantum correlations.
• Dodecahedron's preference permutations of five choices.

Classical light signals vs quantum correlations.

Roger Penrose's popular accounts are manly attempts to give the curious, like myself, a less vague idea of his profession's work. But in the midst of all the technical glosses, he introduces us to the magic dodecahedra as a recreational puzzle. Any one can enjoy this brain teaser 'without knowing a proton from a crouton', to use another popular science author's expression.
This page proceeds in this spirit, with a minimum of back-ground explanation.

Penrose beguiled readers of popular science, in his 1994 book, Shadows Of The Mind, with the 'magic dodecahedra', here called the Penrose dodecahedrons. They are Penrose's improved way of showing a contradiction between assumptions of classical physics and quantum physics.

A crucial difference between the two view-points is in the role of the physical observer. The classical physicist is a passive observer of an objective reality. But the quantum physicist differently affects the reality he observes by how he observes it. So, the quantum physicist's choice of observation affects what he actually observes.

The Penrose dodecahedrons are about how the newer rules of quantum physics cannot be explained away by the assumptions of classical physics. Take two separate observers who make observations or measurements on two parts of a conserved physical system of basic particles that have moved far apart. The smallest constituents of matter are governed by the quantum rules. The effect, of measuring one of the parts, should have a compensating effect on the other part, no matter how far off, to conserve the system as a whole.

The classical view-point, first expressed in the Einstein-Podolsky-Rosen paradox, asked how could this be? How could one part of the conservative quantum system influence the other, if they were too far apart for a signal to reach from one to the other, in time? ( This is bearing in mind that special relativity tells us that no communication could take place faster than the speed of light. )
But experiments, as by Alain Aspect, have supported the quantum rules, involving 'non-locality', so-called because their influence goes beyond the limiting principle of 'local causes' assumed in classical physics, and by Einstein et al.

However, quantum theory does not violate special relativity, because there is no way that measuring one part of a system, to influence its separated other part, could produce a faster-than-light signal or communication, akin to morse code. In The Emperor's New Mind, Penrose explains why the relationship is not one of communication, while still admitting it is deeply puzzling.

Quantum rules of the 'magic dodecahedra'.

Leaving aside the physical problems that gave rise to the Penrose dodecahedrons, we may merely regard them as a recreational puzzle. Two dodecahedrons represent the fields of choice of measurement open to two observers. Penrose says one is here and the other is far away ( perhaps galaxies away ) implying that light speed influences are not at work.

The twenty vertices, of both of the two dodecahedrons, each have a button. The operating ( quantum ) rules are as follows.
The observer here and the observer far away may select any of the twenty vertices, independently of each other. But they do not press the button on that first vertex of choice. Instead, each observer presses the buttons, of the three adjacent vertices, in any order he or she chooses. The observers' respective orders of choice do not affect whether the bell rings. In any case, the bell may not ring at all. If a bell rings, that ends the sequence of button-pressings.

The ( quantum ) correlations or 'entanglements' that may arise between the two observers' measurements, that is the button presses, are:

( 1 ) With reference to figure one, of the two separated but aligned dodecahedrons, suppose the observer here and the observer far away select diametrically opposite vertices of their respective dodecahedrons, from the choice of 20 vertices each of them could possibly make. Each observer may press, in any order - depending on whether a bell stops further pressings - up to all three adjacent vertex buttons to their first selected vertex ( whose button is not pressed ). A bell may only ring if the button vertex pressed, on one dodecahedron, is diametrically opposite to a pressed button vertex on the other observer's dodecahedron.

Figure 1: diametrically opposite vertices ring together, if at all.

In this example, the observer, here, selects vertex G and the observer, far away, happens to select its diametric opposite, G*. From their respective adjacent vertex preferences, the figure shows bells ringing at diametrically opposite vertices I* and I:

( 2 ) With reference to figure two, the two observers happen to select corresponding vertices. Whatever orders they happen to choose to press their respective three adjacent vertex buttons, a bell will ring on at least one of the six vertices the two observers can choose between them.

Figure 2: At least one of the six adjacent-to-select buttons rings.

Classically assumed independence from quantum correlations.

Classical physics assumes that the only possible kind of influence between the two dodecahedrons is of the nature of a signal. This could not out-pace light, which has not had time to communicate the observed influences. Therefore, the classical reasoning is that the dodecahedrons are independent. For an observed dodecahedron, it would follow that (1) it has already been determined which buttons ring or not.
The classical absence-of-influence assumption implies a certain vertex is a pre-arranged bell ringer, in context of figure one: if one of the here observer's three button presses ( in whatever order ) rings a bell, then the far observer's first of three vertex button presses must also ring, if it happens to be the diametric opposite of the here observer's ringing vertex.

(2) No next-to-adjacent buttons can both be ringers. Remember, once a bell rings to one of the selected vertex's three adjacent vertex buttons, that stops the operation of pressing the three, in one's chosen order. With regard to figure 1, if two next-to-adjacent buttons were both bell ringers, then the order, one observer pressed them, would make a difference to which bell rings, if the other observer's select button was the diametrically opposite one on his dodecahedron.

(3) On, say, the here dodecahedron, for a selected vertex, at least one, either, of its three adjacent vertices' buttons, or, of its diametrically opposite vertex's three adjacent buttons, must ring.

The third point follows from combining the two conditions illustrated in the two figures. Re figure 2, if the far observer selects the corresponding vertex to the here observer's choice, then either one from the near or far set of three adjacent buttons must ring. If none of the near set rings, then one of the far set must ring. But then, re figure 1, that ring would be liable to set off a matching ring from a vertex that corresponds to its diametric opposite vertex on the near dodecahedron.

Disproof of independence from quantum correlations.

Penrose goes on to show that the three classical conditions dont hold together. Condition (1) of pre-assigning all the dodecahedron's vertices, as ringers or not, cannot be made to work under conditions (2) and (3).

Penrose says someone else may think of a 'snappier proof'. It's definitely not one 'straight from the book', as Paul Erdös would say.

Figure 3: dodecahedron with inscribed cube.

Re Penrose's 'non-locality' proof: The vertices are given Penrose's letters: A to E are the vertices of a pentagon facet. F to J are their adjacent vertices. A* to J* are their respective antipodal - or diametrically opposite - vertices. Re the preference permutations of 5 choices: Some vertices show equivalent labels to Penrose's, for the next section, on how a dodecahedron represents the 120 permutations of 5 choices. Apologies for the Escher-like appearance of the inscribed red cube, which has five possible orientations in the dodecahedron - just one cube representing the 24 permutations of 4 choices. And 5 times 24 derives the 120 permutations of 5 choices. )

Penrose's demonstration is as follows ( see figure 3, re Penrose's 'non-locality' proof ):

From condition (3) at least one vertex must ring. Assume this is A. Suppose neighboring vertex B also rings. If so, condition (2) forbids all A and B vertex's ten surrounding vertices ( C, D, E, J, H*, F, I*, G, J*, H ) to ring, because they are all next-to-adjacent to A or B.

Condition (3) requires of the anti-podal pairs, H and H*, that at least one, of their two sets of three adjacent pairs, rings. Of these six, only F* or C* ( or both ) are not already ruled out as ringers. Likewise, taking the anti-podal pairs, J and J*, only their adjacent vertices G* and E* ( or both ) are not already ruled out as bell ringers. But G* and E* are both next-to-adjacent to both F* and C*. So, condition (2) also rules them out as bell ringers.

Thus a pair of ringing vertices will not meet the conditions: B, as well as A, cannot ring. A must have non-ringing adjacent and next-to-adjacent vertices ( B, C, D, E, J, H*, F, I*, G ). From condition (3), one vertex must ring of 6 adjacents to the anti-podal pair, A and A*: B*, E*, F* are remaining possibilities.

If F* is the bell ringer, adjacent H and next-to-adjacent E* and G* do not ring. But anti-podal vertices J and J* now have no bell ringers among their six adjacent vertices. The same reasoning applies to E* as the bell ringer with respect to anti-podal pair H and H*; and B* as the ringer with respect to pair I and I*.
This completes Penrose's proof that one observer's vertices pre-arranged to ring, because ( light-signal ) independent of the other observer's dodecahedron, does not work under the rules of quantum correlation between parts of a conservative system.

Dodecahedron's preference permutations of five choices.

Classic Greek geometry proves there are only five regular solids. Three of these have triangular facets ( 4, 8 and 20 in number ). The other two are the cube, with 6 square facets, and the dodecahedron, with 12 pentagon facets.
The dodecahedron is familiar in the black and white chequered facets of the soccer ball.

Murray Gell-Mann's The Quark and the Jaguar ( also reviewed ) features systems analysis. The most interesting systems are complex enough to carry useful information but regular enough to make it of general application. The dodecahedron is rather like that.

An electoral interpretation of the Penrose dodecahedrons depends on the geometry of regular solids serving as illustrations of permutatons of choice. The basic idea can be given by a triangle, whose corners or vertices are labeled A, B, C. These letters might represent three candidates to choose from. You could put this triangle on a pivot and spin it from a fixed position or origin.

Say, corner A is at that origin and anti-clockwise round the triangle you read off the vertices, A, B, C. This represents one of the six possible orders of choice or preference permutations.
Now rotate the triangle anti-clockwise from A at the origin to B at the origin. This yields a second order: B, C, A. Another anti-clockwise turn to C at the origin yields: C, A, B.
Repeat this process in the clock-wise direction for the other three possible preferences: ACB; BAC; CBA.

The 24 possible permutations of choice for four candidates, A, B, C, D, can be represented by a cube. The five possible orientations of a cube inscribed in a dodecahedron can represent the 120 possible orders of choice for five candidates, A, B, C, D, E.
Permutations of choice are given by the factorial of the number of choices ( or candidates ). Three choices means factorial three ( 3! ) or six permutations: 3! = 3.2.1 = 6. Factorial four, 4! = 4.3.2.1 = 24. Factorial five, 5! = 5.4.3.2.1. = 120.

All the permutations of choice are equally valid possibilities. Likewise, the six operations, on the triangle, used a standard procedure of position and direction to represent each of the six permutations. This is still true of representations using three-dimensional regular polygons.

With reference to figure 3, see how the red cube represents 24 permutations of four choices, F, G, C, E. Using the equivalent labels shown in the figure 3: F = a, G = b, H = c, I = d, J = e, F* = a*, G* = b, H* = c*, I* = d*, J* = e*.
FGCE = abCE are also the corners of the cube's near face ( marked in thick red lines ). Its opposite face ( in thin red lines ) is marked C*, E*, a*, b*. The asterisked corners still represent the same choices, say of candidates, C, E, a, b. The reason why we have this duplication of choices is that it takes three cube faces, in three dimensions, rotated, clock-wise and anti-clockwise, to represent all 24 permutations of the four choices.

Just as a pivot was put thru the triangle to turn out permutations of its three corners A, B, C, so a pivot or axis can be put thru each of three mutually perpendicular faces of the cube.

Take the red cube's front face, E, C, a*, b*. Four anti-clockwise turns yield the four permutations: ECa*b*, b*ECa*, a*b*EC, Ca*b*E. Four clockwise turns yield their reverse permutations: Eb*a*C, CEb*a*, a*CEb*, b*a*CE.
Now take the red cube's top face EabC, again working from the E corner or vertex, anti-clockwise to yield four more permutations: EabC, CEab, bCEa, abCE. Four clockwise turns again yields their reverse permutations: ECba, aECb, baEC, CbaE.
The red cube's third dimension, sideways does not show up properly on figure 3. But yet again starting at E anti-clockwise, it is Eb*C*a. Its four anti-clockwise permutations are Eb*C*a, aEb*C*, C*aEb*, b*C*aE. Its four clockwise perms are: EaC*b*, b*EaC*, C*b*Ea.

This completes the 24 possible permutations of ECa*b*, which, notice, encloses the D vertex. D belongs to the dodecahedron facet ABCDE. The red cube can be oriented to enclose each of these five corners, A, B, C, and E, as well as D.

BEc*d* encloses A. As above, we can cycle these four labels to get four permutations and then get four more permutations by reversing them.
The other two cube faces which meet at vertex B can be derived by a rule implicit in the relation between the three faces meeting at vertex E:
For the second face, ECa*b* transformed to EabC, so transform BEc*d* into BcdE. For the third face, ECa*b* transformed into Eb*C*a, so transform BEc*d* into Bd*E*c.

Thus the orientation of the cube, with respect to the top dodecahedron facet at A, yields three square faces meeting at vertex B. These three faces each have four corners which yield four permutations and their four reverse permutations, totalling 24 possible permutations.

The cube similarly oriented to B, yields three more square faces meeting at vertex C; cube orientation to C yields three faces meeting at D; cube orientation to D yields three faces meeting at E; cube orientation to E yields three faces meeting at A.
Thus the five possible cube orientations to dodecahedron facet ABCDE yield 5 times 24 equals 120 possible permutations of those five choices. See table 1.

Richard Lung