The Math & Science Behind Eidos

“The important thing is to never stop questioning.” – Albert Einstein

1. The Math Behind Eidos
2. Cognitive Science and Points of Strategy

The Math behind Eidos.

Henry Kandel

Now that we’ve had so much fun playing this game, let’s go deeper into it – let’s get inspired and generate some questions!   

The core idea behind the game is that every pair of cards must have exactly one and only one pair of matching symbols to every other card, so no matter what two cards you pull from the large deck there is a match.

You can examine the cards in this deck and verify for yourself that every pair of cards contains one and only one pair of matching symbols! Maybe this is surprising – would you have suspected that a deck with this many cards and symbols could have this property? It’s not obvious! When we created this deck, how much freedom did we have in terms of how many symbols and cards would be required? Does our 9-symbol deck have to have exactly 73 cards? Is there a simple mathematical relationship between number of symbols and number of cards? Could a deck with this special property be arbitrarily large? (...before it became impossible to keep them in stores, or ship them, of course.)

The answer will depend on how many symbols exist on each card. (Let’s assume each card contains the same number of symbols.) The cards in this particular deck have 9 symbols per card – To answer the question, How many total cards/symbols?, let’s analyze some simpler cases... 

What if there are only 2 symbols per card?  Consider the first card – Call the symbols A and B:

To produce the next card, we need to match one of the symbols (let’s say A), but the other one has to be different, so we need to invent a new symbol, C:

It’s easy to see that the deck can only contain one more card:

This deck is now complete.  (And not too many people would want to play this game at this point!)

How do we know the deck is complete?  Well, the 3 cards contain all possible combinations of the 3 symbols A, B, and C.  So if there were any more cards, there’d have to be at least one additional symbol (call it D). 

…But what symbol could possibly take the place of ? on our first card with the symbol D?

None! Choosing A or B or C to replace the ? would leave one other card with no matches at all. ...And if a brand-new symbol is chosen for the card with D, then the card won’t have a match with any of the first three cards!

So, our super-simple deck is complete: 2 symbols per card means a 3-card deck, and 3 symbols in total.  Let’s consider the next case, a deck with 3 symbols per card:

Let’s organize the deck, so that cards that share A with this first card are placed in a line, all the cards that share B on another line, and C on a third line:

What could the two other symbols on the first card in the A row be? Well, they couldn’t be A (since symbols don’t appear twice on the same card). But they couldn’t be B or C either, because then they'd have two matches with the first card. So they have to have two entirely new symbols, which we’ll call 1 and 2.

Now, how about the next card in the row? They couldn't be A, B or C, for the same reasons we just saw. And they couldn't be 1 or 2 either, because they already have a match with the first card in the row. So they too need two new symbols (4 and 5).

Consider the first column of this rectangular array.  Since the cards do not share their letter, they must share one of the other symbols (let’s say 1):

But the rules specify a pair of cards must share exactly one pair of symbols, not more, so we have to be very careful completing the cards in the array so far:

Done![1] Now, each pair of cards share exactly one pair of symbols. How do we know this deck is complete? Well, consider adding just one more card (in the A row, but of course it could be in any row) – to abide by the game rules, we’ll need to invent two more symbols:

But we can see immediately that this new card fails to produce a match with any of the cards not in the A row. So, this new card cannot exist, nor any new card; the deck is complete![2]

 The situation with 4 symbols per card is more complex – It takes longer to produce all the cards, but here they are:

The argument for why there can’t be any more cards is the same as in the simpler case – The deck is complete!  So far here’s how things look:

 

Symbols per Card	2	3	4
# of Cards	3	7	13
Total # of Symbols	3	7	13

 

Great mathematicians of the past (such as Johannes Kepler) have found general formulas just by looking at charts such as this for long periods of time — staring at the data until something clicked. But we have an easier job, because of how we laid out the cards in the above examples:

Let’s call the number of symbols per card N. (In this particular example, N=4.)

 Our array of cards contains N-1 columns, with N cards per column, for a total of (N-1) x N.  Add the “first card,” and we have:

TOTAL # of cards = (N-1) x N + 1     

Do a little algebra and you can produce the formula:

TOTAL # of cards = N² – N + 1

So if the deck you’re holding contains cards with 9 symbols per card (N=9), we’ll need 9² – 9 + 1, or 73 total cards — Count ’em and see!

To calculate the total number of symbols needed, consider the A row of the array above:

TOTAL # of symbols = N + (N-1) x (N-1) ...Which a little algebra will convince you is an identical formula to the one we just derived!

TOTAL # of symbols = N² – N + 1

The cool deck you’re holding, then, must have 73 cards and 73 unique symbols. It does – Great!

...But if you’re an ultra-rigorous mathematician type, you’ll demand an answer to this question:

“Ok, so you created the full decks for the N=2, 3, and 4 cases, and you gave a convincing argument for why the decks in each case could contain no more cards (than N² – N + 1), but how can you be absolutely sure that decks with N>4 can be made? Is there a way to prove that all (N² – N + 1) cards in these larger decks can be made?”

Here we must rely on geometry, a branch of mathematics – but not the regular old geometry most people think of, the geometry most of us encountered in school. That particular geometry is Euclidean Geometry. It proceeds from a set of axioms originated by Euclid, the Ancient Greek mathematician who lived in the 3rd-4th Centuries BCE—axioms that refer to points and lines (objects of which we think we have an intuitive grasp).

The deductive methods and conclusions of Euclid’s geometry were so prized in Ancient Greece that upon the gate of Plato’s famous academy was reportedly engraved, “Let no one ignorant of geometry enter!”

It’s easy to see how we can make a geometric model of our game. Each point will represent a card, and each line will represent a symbol. A perfect model of our game is made, once we specify that each point must have exactly N lines emanating from it, and that each pair of points shares only one line. Let’s start with the simple case where N = 2:

See how the model shows the cards, the pairs of symbols, and the fact that the deck is complete (for N=2, there are 3 cards altogether)?

...But try to get just a little more complex...try to create a similar model for N=3, and you’ll run into problems. (Try it!) In order to make it work, you’ll have to bend at least one of the lines – This particular example is called the “Fano Plane”:

Can you see how the rules of our game are built into this model?  Each point (card) has exactly three lines (symbols) emanating from it, and each pair of points shares exactly one line.

But we have (by necessity) bent one of the lines, into a circle in fact.  And a circle does not qualify as a line!  Another way of saying this is that Euclidean Geometry will no longer suffice, for N>2. If geometry will still be useful, it will have to be of a non-Euclidean kind.

Non-Euclidean geometries are relatively recent in the history of math.  They turn out to be essential for modeling the physical Universe. Non-Euclidean geometries produce results that will surprise you. For example, you may have learned in school that the sum of the angles of a triangle will always equal 180°. But this is only if the geometry you’re using is Euclidean!

Here you can see that the triangle’s angles, in a spherical geometry, generally add up to more than 180° while the triangle’s angles, in a hyperbolic geometry, add up to less than 180°.

(The connection between non-Euclidean geometry and the physical Universe isn’t obvious, since locally our Universe is very close to Euclidean. In fact, Euclid himself almost certainly believed his geometry was more than just math, and also served to describe the physical world. But it turns out, according to Einstein’s Theory of General Relativity, that the presence of mass/energy bends spacetime, so that, for instance, the path of a light beam is actually curved.)

What kind of exotic geometry will help us extend the model of our game? It is called projective geometry, and its axioms were laid down and its conclusions explored by many mathematicians over the centuries. Its axioms can perfectly model our game! When our game is modeled with projective geometry, it can be proven that a complete deck (with N² – N + 1 total cards and unique symbols) is possible if N can be expressed as a prime number, raised to a natural-number exponent.

For our particular deck, N = 9, which can be expressed as 3².  3 is a prime number, and it is raised to the exponent 2, which is a natural number.  So we’re good! 

As far as the most general case goes — whether it’s possible to make a deck where N is a number that can’t be expressed this way — well, that’s an unsolved problem in mathematics!  So at the heart of our game we get a glimpse at the forefront of mathematics.  There’s millions of dollars to be earned in solving related problems[3].  Get to work, kids!

Cognitive Science and Strategy:

Now let’s raise some questions about how the game is actually played! 

After all, when we play the game, we don’t just require cards and symbols to work in certain ways: We want to pick out the exactly one pair of symbols, as quickly as possible!

How can we improve our speed on such a task?

One answer is to reduce the amount of conscious processing our brains are doing.  The less thinking (conscious processing), the better!

...So the question becomes, “How can we reduce thinking, so that we pick out pairs of cards using processes that are not conscious?”  There are two auspicious possibilities:

1. Practice (repetition)
2. Meditation (relaxation)

A jazz musician reflects on his instrumental practice, noting how the vast number of progressions he commits to memory through repetition become accessible to him when he improvises a solo for a live show. He may spend a while designing (thinking consciously about) the individual progressions, but once they have been adequately practiced, they can be accessed in an instant: witness the flurry of rapid notes in the improvisation of a jazz master!

A left fielder in softball “shags” flies, one after the other, day in and day out, so that when a similar fly (to any of these particulars) is launched in a high-stakes game, she can run to the proper location along the quickest path, without thinking! 

So perhaps if you practice this cool card game, you will find you can snag pairs without having to go symbol-by-symbol, thinking, “Ok, does this one appear on another card...”

What about Meditation?

Meditation, a process that builds focus and enables relaxation, has been shown to improve reaction times for certain tasks, and improve sensory capacities.  For example, Zen meditation cultivates joriki, which translates roughly as “powerful concentration.” In such as state, the mind is essentially free of thinking. One Zen master describes a deer who has just heard a rustling in the nearby leaves; it freezes and all its senses are focused on the location of the sound. It is not thinking, “What if it’s a wolf? What direction would I run?” Any thinking would slow down its subsequent decisions, which would in turn threaten its survival.

Would a period of meditation before the game improve your performance?  There’s only one way to find out!

...And there may be further ways to improve your speed with the game, that are quite surprising...

What if you blur your vision slightly? Why would this possibly help? Well, it might cause you to rely on certain types of visual processing rather than others, which might improve your speed. Consider the famous Stroop Test. The goal is to say the color, not the word.  When does your performance slow down?  And can you prevent this slow-down by blurring your vision?

Is this surprising? If the words were in a totally unfamiliar language, you’d probably go way faster! (Because inadvertent thinking about their meanings would be impossible.)

What if you look slightly off to the side? An accomplished artist reveals how she performs better on tests involving distinguishing very similar colors, when she doesn’t look directly at the image! Could this be related to the well-known fact that you can see, in the periphery of your vision, stars in the sky that are too dim to see if you look at them directly? (Try it!)

What if you play the game in low lighting? This would drastically reduce the factor of color. Why might this improve your speed? Well, in WW1, the British Air Force found that colorblind pilots performed better on certain tasks at night, because of their superior sense of contrast. Could this apply to our game?

What a satisfying run of questions! We hope we’ve done Einstein proud, but maybe now it’s time to set questioning aside and play the game some more!  Let’s conclude this riff with a meta-question:

What other questions can you come up with that relate to this cool card game???

About Henry Kandel   

Henry Kandel teaches physics and many other things at Saint Ann’s School in Brooklyn.  He loves games, and uses them to explore a wide range of topics in the math & science classrooms:  Ecosystems, political primaries, sports, aesthetics, social dynamics, extraterrestrial civilizations and more!

Footnotes

1 For the first column (on the left), we just completed the cards with “2,” “3,” and “4.”  For the card at the bottom of the second column (on the right), we chose the only symbols possible: “3” and “4.”  For the remaining two cards, we employed the “brute force” method of trial and error; and because the case is so simple, it didn’t take long!

2 When we arrange it this way, you can see the number of rows must equal the number of symbols (in this case, 3).  You can also see the number of new symbols (beyond “A,” “B,” and “C”) cannot exceed 4.  Since each card with a single letter must additionally have 2 new symbols, you can see – by examining the bottom row – that only two columns are possible.

In the general case, with N symbols per card, there must be N rows.  Since our arrangement requires one new symbol (in the case shown, “1”) be shared by every card in the left-most column, we’ll require (N-1) new symbols for each card.  Since there are N rows, (N-1) new symbols per card (in the left-most) column, will require Nx(N-2) new symbols.  (In the case shown, with 3 original symbols, the total number of new symbols required – beyond “A,” “B,” “C,” and “1,” is 3x1=3.  These new symbols are “2,” “3,” and “4.”)  Don’t forget about the original symbols on the first card!  So the total number of symbols will be Nx(N-2) plus N!  A little algebra reduces that to N^2 – N.  If you divide by the N rows (since the number of items in a rectangular grid is equal to the number of columns times the number of rows), you have a total of (N-1) columns!

3 Our problem is, that while it’s possible to prove that if N is a “prime power” (prime number raised to an integer exponent, which in terms of physical card decks, must be positive), you CAN make the deck – it has not yet been proven that those are the ONLY decks that can be made!

There are numerous open questions in math that relate to prime numbers, and they’re some of the most famous open questions. Super-famous cases like the Goldbach Conjecture refer to sums of primes, rather than prime powers. Answering the Eidos question could earn you millions of dollars, see https://www.claymath.org/millennium-problems