Pyramid Mysteries. Ehrhard Behrends

Transcription

1 Pyramid Mysteries Ehrhard Behrends In this article we describe the 3D version of a phenomenon that has been published under the title Triangle Mysteries [1]. There one starts with a row of n little squares that are coloured red, green and blue. By a simple rule they generate a row of n 1 coloured squares, and one continues until finally one square remains. The mystery: for certain n the colour of the final square can easily be predicted. In the present paper we begin with an n n-grid of little coloured cubes. By a certain rule this gives rise to an n 1 n 1-grid of coloured cubes that form the second layer of a pyramid. One continues with an n 2 n 2-grid, then an n 3 n 3-grid, etc., until finally a pyramid is constructed. We will show that for suitable n one can predict the colour of the top cube very simply from the first n n-grid. A variant is also discussed. There one starts with a triangle formed by coloured cubes that is considered as the base of a pyramid. The various layers are again constructed by a simple rule. And also here one can predict in certain cases the colour of the cube on the top. The proofs are elementary, they only use known properties of binomial coefficients n k when n is the power of a prime. Our results cover the case of arbitrarily many dimensions, but we will mainly concentrate on the 3D case. For the formal approach we simply need a finite set the set of colours and a map that associates with 3 resp. 4 colours another one, i.e., φ is a map from 3 resp. 4 to ; it will be used to determine the colour of the next cube that will be put on top of three resp. four coloured cubes of the preceding layer of the pyramid. Once φ is given one can start to build pyramids. Pyramids where the base is a square Let, φ 4 : 4 and an n 2 be given. We start with the base of our pyramid, the first layer. It consists of a square made of n 2 little cubes. The cubes are coloured with the elements of. Formally such an arrangement is a square matrix x i,j i,j=0,...,n 1 with x i,j. Here is an example with n = 10 and = {r, g, b} where these letters stand for the colours red, green, and blue. The first layer could look like this. The first layer, the colours are chosen at random. Next we are going to construct the second layer. The rule is the following: 1

2 Where four cubes with colours c 1, c 2, c 3, c 4 of the first layer meet put on top of them in the middle another cube the colour of which is φ 4 c 1, c 2, c 3, c 4. It will be convenient here to consider only such φ 4 where all permutations of c 1, c 2, c 3, c 4 are mapped to the same element of. Thus it is not necessary to specify which of the four cubes is associated with c 1 etc. This means that we generate a new square of cubes with colours y i,j i,j=0,...,n 2 by the formula y i,j := φ 4 x i,j, x i+1,j, x i,j+1, x i+1,j+1. For our example see the preceding picture we consider as φ 4 the following map: The elements of are identified with the numbers 0, 1, 2 of the group Z 3 by 0 = r, 1 = g, 2 = b, and φ 4 c 1, c 2, c 3, c 4 := c 1 + c 2 + c 3 + c 4 mod 3. Thus, e.g., φ 4 b, b, r, g = b. For this particular φ 4 the second layer of our pyramid looks like this: The second layer. It should be clear how to continue. We construct smaller and smaller square layers of cubes until finally we arrive at the single top cube. The question: Is it possible to predict in a simple way the colour of this top cube just by checking the colours of the first layer? In the next picture we see on the right the final pyramid from the top. The cube on the last layer is red: could it be possible to know this in advance? One, two and ten layers from left to right. Motivated by the investigations in [1] we will call an integer n φ 4 -simple if the top colour c is always just the φ 4 -value of the corner cubes of the ground layer, i.e., if c = φ 4 x 0,0, x n 1,0, x 0,n 1, x n 1,n 1 holds. Thus, trivially, n = 2 is always φ 4 -simple. It will be shown that n = 10 is φ 4 -simple for the φ 4 defined above so that since φ 4 b, g, r, r = r it is no surprise that red is the top color. 2

3 Pyramids where the base is a triangle This time we start with, a φ 3 : 3 and an n 2. First we choose any triangular base of our pyramid, the first layer. It consists of a triangle made of little cubes: 1 in the first row, then 2 etc. until finally there is a row of n cubes. The cubes are coloured with the elements of, i.e., we are given a triangular matrix x i,j 0 j i n 1 with x i,j. x i,j is the colour of the j th cube in the i th row. As before we work with = {r, g, b}, and in the example we are going to discuss we have n = 4. The first layer was produced with the help of a random generator, it can be seen on the left in the next picture. We continue similarly as in the case of pyramids with a square base: Where three cubes with colours c 1, c 2, c 3 of the first layer meet put on top of them another cube the colour of which is φ 3 c 1, c 2, c 3. This means that we generate a new triangle of cubes with colours y i,j 0 j i n 2 by the formula y i,j := φ 3 x i,j, x i+1,j, x i+1,j+1. In our example we identify with Z 3 and define φ 3 by φ 3 c 1, c 2, c 3 := c 1 + c 2 + c 3 mod 3 so that, e.g., φ 3 b, b, r = g. We continue to construct smaller and smaller layers until finally there is only one single little cube on the top. The four layers of the pyramid of our example can be seen here: The layers of the pyramid with triangular base in the case n = 4. As for pyramids with a square base we are interested in situations where the top colour can simply be determined from the corner colours of the base: the number n will be called φ 3 -simple if for arbitrary choices of x i,j 0 j i n 1 the top color c is given by the formula c = φ 3 x 0,0, x n 1,0, x n 1,n 1. It will follow from the results of the next sections that 4 is φ 3 -simple for our φ 3 so that one can predict immediately in our example that the top colour must be red. Our main results concerning φ-simple n can be found in the next sections, and in the last section we discuss some generalizations. Pyramids where the base is a square Let us now set the stage for our theoretical development. From now on we will assume that is a finite abelian group with respect to the operation +, and has at least two elements. 3

4 The mapping φ 4 : 4 is defined by φ + 4 : c 1, c 2, c 3, c 4 c 1 + c 2 + c 3 + c 4. We note that in the above example we worked with φ 4 = φ + 4, and was the group Z 3, +. To prepare the following investigations we introduce some notation. Here m, n 2 denote arbitrary integers. The mappings Φ + m Φ + m : m2 m 12 is the map that defines the next layer: we have Φ + m xi,j i,j=0,...,m 1 := yi,j i,j=0,...,m 2, where y i,j := φ + 4 x i,j, x i+1,j, x i,j+1, x i+1,j+1. With this definition the top colour c when starting with x i,j i,j=0,...,n 1 is c = Ψ + n xi,j i,j=0,...,n 1 := Φ + 2 Φ + n 1 Φ+ n xi,j i,j=0,...,n 1. The σ m,+ For x and k, l = 0,..., m 1 we denote by S k,l;m,x that element x i,j m2 where x k,l = x and the other x i,j are zero. σ m,+ is defined to be the top colour when we work with φ + 4, i.e. σm,+ = Ψ+ ms k,l;m,x. It will be clear soon why it is convenient to extend this definition: we put σ m,+ := 0 for k, l { 1, m} { 1,..., m} and k, l { 1,..., m} { 1, m} 1. The σ n,+ can be explicitly determined: Lemma 1: For k, l = 0,..., n 1 one has σ n,+ = n 1 k n 1 l Proof. The proof is by induction on n. In the case n = 2 we have σ 2,+ = φ + 4 x, 0, 0, 0 = x = 1 k 1 l x for k, l {0, 1}. Now suppose that the lemma is proved for some number n 1 with n 3. Let us analyze the first step in the calculation of σ n,+ : we pass from the first layer S k,l;n,x to Φ n S k,l;n,x. This is an n 1 n 1-matrix with the entry x at one, two or four positions the other entries are zero. For example, if 0 < k, l < n 1, then the x are at the four positions k, l, k 1, l 1, k 1, l, and k, l 1. Since Φ + n 1 is a homomorphism we arrive at the equation x. σ n,+ = σn 1,+ + σ n 1,+ k 1,l;x + σn 1,+ k,l 1;x + σn 1,+ k 1,l 1;x. 1 These indices extend the {0,..., m 1} 2 -pattern to a { 1,..., m} 2 -pattern, and the σ for the new indices are zero. 4

5 We note that we can use the same formula if we adopt the extended definition of the σ n,+ : e.g., σn,+ 0,0;x = σn 1,+ 0,0;x, and this also is covered by the formula since σ n,+ 1,0;x = σn,+ 0, 1;x = σn,+ 1, 1;x = 0. By the induction hypothesis we know that the σ on the right-hand side can be expressed by binomial coefficients so that σ n,+ = n 2 n 2 x. k =k 1,k; l =l 1,l n 1 k l But this sum coincides with n 1 x which can easily be deduced from the identity n 2 m 1 + n 2 m = n 1 m. We now are going to show that the φ + 4 -simple integers can be characterized: Proposition 2: Suppose that, + is isomorphic to Z p, + d for a prime p and an integer d. Then a number n is φ + 4 -simple iff there is an s N such that n = p s + 1. Proof. The key tool will be as in [1] Balak Ram s result [3] on binomial coefficients: Let p be a prime and m an integer. Then all m l for l = 1,..., m 1 are divisible by p iff there is an s such m = p s. Let m, r be integers such that m > r > 0. If r divides all m l for l = 1,..., m 1 then r is a prime and by the first part m is of the form r s. A proof can be found in [1] and [3], and for a far-reaching generalization we refer the reader to [2]. Now let n be given. We observe that Ψ + n : n2 is a group homomorphism when we consider n2 as a product group. This has the following consequence: xi,j i,j = Ψ + n S i,j;n,xi,j Ψ + n = i,j i,j k Ψ + n Si,j;n,xi,j l = i,j = i,j σ n,+ i,j,x i,j n 1 n 1 x i,j. i j By definition an n is φ + 4 -simple iff the preceding sum always coincides with φ + 4 x 0,0, x n 1,0, x 0,n 1, x n 1,n 1 = x 0,0 + x n 1,0 + x 0,n 1 + x n 1,n 1, i.e., if n 1 n 1 i,j / A i j xi,j = 0, where A denotes the set containing the four elements 0, 0, n 1, 0, 0, n 1, n 1, n 1. And this is obviously true iff x = 0 for all x and for all i, j with i, j / A. n 1 i n 1 j 5

6 Suppose that n is of the form p s + 1. Then, by the first part of Ram s result, p divides all n 1 l for l = 1,..., n 2, and consequently all n 1 n 1 i j with i, j / A are divisible by p. But px = 0 for all x since, + is isomorphic with Z p, + d. This shows that the p s + 1 are φ + 4 -simple. Suppose now that n is not of the form p s + 1. By Ram s result we find a k {1,..., n 2} such that n 1 k is not divisible by p so that α := n 1 k 0 in Z p. Let x be any nonzero element in Z p d. Then φ + 4 Hence n is not φ + 4 -simple. 0, 0, 0, 0 = 0 α 2 x = Ψ + 4 S k,k;x,+. Now we understand why our examples in the first section worked with n = 4 and n = 10 : the good n here are the n of the form 3 s + 1. Here is another example, we work with = Z 2, + 0 is red and 1 is green. Note that in this case the φ + 4 -simple n are the integers of the form 2 s + 1, and therefore it is no surprise that the top colour is red once one has seen the base: A pyramid where n = 3 with = Z 2 and φ + 4. In this case it is simple to translate the definition of φ + 4 without using the arithmetic in Z 2 : find c = φ + 4 c 1, c 2, c 4, c 4 such that the total number of green balls among c, c 1, c 2, c 4, c 4 is even. Suppose that n is not φ + 4 simple in the preceding case. Then it might happen that Ψ + n xi,j φ + 4 x 0,0, x n 1,0, x 0,n 1, x n 1,n 1. How often will this be the case? The answer can be found in Proposition 3: Let, + be again isomorphic to Z p, + d for a prime p. Suppose that n is not φ + 4 -simple. If N = pd denotes the cardinality of, then the following is true: the cardinality of the x i,j n2 where Ψ + n xi,j = φ + 4 x 0,0, x n 1,0, x 0,n 1, x n 1,n 1 holds divided by the cardinality of n2 is precisely 1/N. To state it otherwise: if one wants to predict the top colour by giving the guess φ + 4 x 0,0, x n 1,0, x 0,n 1, x n 1,n 1 one has a chance of 1/N to be correct. 6

7 Proof. Let n be an integer and x i,j i,j n2 be arbitrary. The prediction that the top colour is φ + 4 x 0,0, x n 1,0, x 0,n 1, x n 1,n 1 will be correct iff n 1 n 1 x i,j = 0 i j i,j,i,j / A the set A of indices is as in the proof of proposition 2.2. We observe that the map Ψ : x i,j n 1 n 1 i,j,i,j / A i j xi,j is a group homomorphism, and since n is not φ + 4 -simple it is not the trivial homomorphism. It follows that there must be a pair k, l / A and an x such that n 1 n 1 i j x 0. This implies that r := n 1 n 1 i j mod p 0, and since Zp is a field we may select r Z p with rr = 1 mod p. It is now easy to show that x i,j n 1 n 1 i,j,i,j / A i j xi,j from n2 to is onto: a given y has S k,l;n,r y as a preimage. From this we may conclude that the number of elements in the kernel of Ψ is N n2 4 /N, and since there are N 4 possible choices for the x i,j with i, j A the proposition is proved. Only the groups considered in the preceding propositions admit φ + 4 -simple integers: Proposition 4: Suppose that, + is not isomorphic to any of the groups of the preceding proposition. Then there are no φ + 4 -simple integers n > 2. Proof., + is a finite commutative group so that it has the form J α=1 Z p α sα with different primes p 1,..., p J. Note that J 2 by our assumption. Now suppose that a φ + 4 -simple integer n exists. For i, j / A we must have n 1 n 1 i j mod pα = 0 since otherwise it would be easy to find x i,j i,j such that Ψ x i,j 0 in contrast to the fact that n is φ + 4 -simple. We conclude from the second part of Ram s result that n can be written as p sα α + 1 for every α and suitable s α. But this is surely not possible for more than one p α. Pyramids where the base is a triangle All the results of the preceding paragraph have an analogue. The modifications are the following: For the finite abelian group, + with at least two elements we consider the mapping φ + 3 : 3 defined by c 1, c 2, c 3 c 1 + c 2 + c 3. A triangular pattern the layers of the pyramid of elements of needs m+m = mm+1/2 =: δm entries. Thus a typical pattern is given by x i,j 0 j i m 1, this describes a triangle where the last row has m elements. Consequently the passage to the next layer is described by a map Φ + m : δm δm 1 ; it is defined by Φ + m xi,j 0 j i m 1 = y i,j 0 j i m 2, where y i,j = φ + 3 x i,j, x i+1,j, x i+1,j+1. 7

8 Ψ + n, the map that assigns the top colour to the pattern of the ground layer is again defined as Φ + 2 Φ+ n : δn. The number n will be called φ + 3 -simple if always Ψ+ n xi,j 0 j i n 1 = φ + 3 x 0,0, x n 1,0, x n 1,n 1 holds. For 0 l k m 1 the pattern S,m δm is defined to have x at the position k, l and all other entries are zero. σ m,+ denotes the element Ψ + m S,m. It will then be crucial that σ m,+ = m 1 k k l x for all k, l. This is again proved by induction by using simple properties of binomial coefficients. With these preparations at hand one can prove the following results for pyramids with a triangular base similarly as in the previous section: Proposition 5: Suppose that, + is isomorphic to Z p, + d for a prime p and an integer d. Then a number n is φ + 3 -simple iff there is an s N such that n = p s + 1. In the next picture we illustrate this proposition by a pyramid built with balls 2 : note that 4 is φ + 3 -simple in this example. A pyramid: triangular base, n = 4, = Z 3 and φ = φ + 3. Proposition 6: Let, + be again isomorphic to Z p, + d for a prime p. Suppose that n is not φ + 3 -simple. If N = pd denotes the cardinality of, then the following is true: the cardinality of the x i,j δn where Ψ + n xi,j = φ + 3 x 0,0, x n 1,0, x n 1,n 1 holds divided by the cardinality of δn is precisely 1/N. Proposition 7: Suppose that, + is not isomorphic to any of the groups of the preceding proposition. Then there are no φ + 3 -simple n > 2. Magic in hyperspace and more general φ In the first part of this section we note that one can naturally generalize our theory to arbitrarily many dimensions D: magicians in hyperspace can present 2 The colours r, g, b correspond to 0, 1, 2 as above. 8

9 the same tricks! The preceding results of the present paper and the results of [1] correspond to D = 3 and D = 2, respectively. Hyperpyramids where the base is a hypersquare Let us fix an integer D, an n N, and a finite abelian group, + with at least two elements. We define, with D := 2 D 1, a mapping φ + D, φ D : D by c i i D i c i. We are going to construct a hyperpyramid consisting of little coloured hypercubes. The ground layer is made from n D hypercubes, their colours are given by x i1,i 2,...,i D 1 i1,...,i D 1 =0,...,n 1 nd 1. Wherever D hypercubes meet we put on top of them another hypercube with colour determined by φ + D. More formally: we map the pattern x i 1,i 2,...,i D 1 i1,...,i D 1 =0,...,n 1 to y i1,i 2,...,i D 1 i1,...,i D 1 =0,...,n 2, where y i1,i 2,...,i D 1 := φ + D xi1+j 1,i 2+j 2,...,i D 1 +j D 1 j1,...,j D 1 {0,1}. In this way we continue, after n steps we arrive at the top layer that consists of only one hypercube. If its colour is always φ + D xi1,i 2,...,x D 1 i1,...,i D 1 {0,n 1} then we will call n a φ + D -simple integer: the top colour can be predicted easily from the corner colours of the first layer. It can be shown with similar techniques as in the preceding sections that the φ + D -simple n are precisely the ps + 1 if, + is a power of Z p and that there are no such n > 2 for other. The key result here is the fact that the final colour is n 1 n 1 n 1 x i 1 i 2 i D 1 if the ground layer has colour x at the position i 1,..., i D 1 and colour zero at the other places. Hyperpyramids where the base is hypertriangular A hypertriangular array of size n is a family x i1,...,i D 1 0 id 1 i D 2 i 2 i 1 n 1 where the x are in. We map such an array to one of size n 1 by using a rule that generalizes the rule for 3-dimensional pyramids with a triangular base, and after n 1 steps we arrive at the top of a D-dimensional hyperpyramid with base colours x i1,...,i D 1. Again we are able to identify those n where one can predict the top colour in a simple way: as before precisely the p s + 1 have this property in the case = Z p, + d, the crucial lemma is the assertion that the top colour is n 1 i1 id 2 x i 1 i 2 i D 1 9

10 if the first layer has x at the position i 1,..., i D 1 and the other entries are zero. More general φ In the preceding sections we always have assumed that φ is of type φ + sum of the colours. This approach has the advantage that the definition of φ is commutative : all permutations of the entries lead to the same value. This makes it easier to translate the rules to an audience of non-mathematicians but it is in fact not neccessary. Let us explain what we have in mind at the simplest example where D = 2 and = Z p, +. In [1] we worked with the φ + 2 : 2 defined by c 1, c 2 c 1 +c 2 and c 1, c 2 c 1 c 2. Now fix any α, β Z p and define φ α,β : 2 by x, y αx+βy. Here we consider as a vector space over Z p, and we note that φ α,β is the most general group homomorphism from 2 to ; the preceding cases correspond to the choice α = β = 1 and α = β = 1, respectively. Fix an n and a family x i i=0,...,n 1 n. We will consider these x i as in [1] as the colours of the first row of coloured squares of a triangle, and with the help of φ α,β one builds a second row, then a third one and so on: finally a single square, the bottom square of the triangle, will be found. Surprisingly one can predict this colour as before for the n of the form p s + 1, the colour will be φ α,β x 0, x n 1. For the proof one has to combine the following facts: Consider a starting pattern x i i=0,...,n 1 where there is colour x at position k and all other x i are zero. Then the top colour is n 1 k α n 1 k β k x. This can be proved by induction. The little Fermat theorem : one has γ p = γ for all γ Z p. It is then possible to show the claim with the same techniques that we have used above. Similarly one can pass in the case of three dimensions, e.g., from our map φ + 4 of the first section to φ α,β,γ,δ c 1, c 2, c 3, c 4 := αc 1 + βc 2 + γc 3 + δc 4, where α, β, γ, δ Z p are fixed elements. The same results will hold: the n of the form p s + 1 are always good numbers, and if all α, β, γ, δ are different from zero there are no others. Here is an example: a pyramid with a square base where n = 4. We work with = Z 3 and the usual translation: 0, 1, 2 correspond to r, g, b, and we consider φ defined by φ = φ 4 : c 1, c 2, c 3, c 4 c 1 c 2 c 3 c 4. It is no surprise that the top colour must be blue.: A pyramid: square base, n = 4, = Z 3 and φ = φ 4. 10

11 The next step would be to treat similar generalizations for D dimensions, but we omit to outline the clumsy technical details here. Acknowledgement: The author wishes to express his gratitude to Steve Humble and the readers of the New York Times who discussed the results of [1] in May 2013 in a blog google triangle mysteries new york times or go to The present research was motivated by these contributions. References [1] E. Behrends, St. Humble. Triangle Mysteries. The Mathematical Intelligencer, pp [2] H. Joris, C. Oestreicher, and J. Steinig. The greatest common divisor of certain sets of binomial coefficients. J. Number Theory, pp n! [3] Balak Ram. Common Factors of m!n m!, m = 1,..., n 1. Journal of the Indian Mathematical Club, pp Ehrhard Behrends Mathematisches Institut, Freie Universität Berlin Arnimallee 6 D Berlin Germany behrends@math.fu-berlin.de 11