# Arithmetic & Geometry

Standard

After a month of the unexpected break from mathematics, I will resume the regular weekly blog posts. It’s a kind of relaunch of this blog, and I  will begin with the discussion of an arithmetic problem with a geometric solution.  This is problem 103 from  The USSR Olympiad Problem Book:

Prove that
$\displaystyle{t_1+t_2+\ldots + t_n = \left\lfloor\frac{n}{1}\right\rfloor + \left\lfloor\frac{n}{2}\right\rfloor + \left\lfloor\frac{n}{3}\right\rfloor + \ldots + \left\lfloor\frac{n}{n}\right\rfloor}$
where $t_k$ is the number of divisors of the natural number $k$.

One can solve this problem by using the principle of mathematical induction or using the fact that the number of integers of the sequence $1,2,3,\ldots , n$ which are divisible by any chosen integer $k$ is equal to $\left\lfloor \frac{n}{k}\right\rfloor$. The second approach of counting suggests an elegant geometric solution.

Consider the equilateral hyperbola $\displaystyle{y = \frac{k}{x}}$ , (of which we shall take only the branches in the first quadrant):

We note all the points in the first quadrant which have integer coordinates as the intersection point of the dotted lines. Now, if an integer $x$ is a divisor of the integer $k$, then the point $(x,y)$ is a point on the graph of the hyperbola $xy=k$. Conversely, if the hyperbola $xy=k$ contains an integer point, then the x-coordinate is a divisor of $k$. Hence the number of integers $t_k$ is  equal to the number of the integer points lying on the hyperbola $xy=k$. The number $t_k$ is thus equal to the number of absciassas of integer points lying on the hyperbola $xy=k$. Now, we make use of the fact that the hyperbola $xy=n$ lies “farther out” in the quadrant than do $xy=1, xy=2, xy=3, \ldots, xy=n-1$. The following implication hold:

The sum $t_1+t_2\ldots + t_n$ is equal to the number of integer points lying under or on the hyperbola $xy=n$. Each such point will lie on a hyperbola $xy=k$, where $k\leq n$. The number of integer points with abscissa $k$ located under the hyperbola is equal to the integer part of the length of the segment $\overline{AB}$ [in figure above $k=3$]. That is $\left\lfloor\frac{n}{k}\right\rfloor$, since $\displaystyle{|\overline{AB}|=\frac{n}{k}}$, i.e. ordinate of point $A$ on hyperbola $xy=n$ for abscissa $k$. Thus, we obtain

$\displaystyle{t_1+t_2+\ldots + t_n = \left\lfloor\frac{n}{1}\right\rfloor + \left\lfloor\frac{n}{2}\right\rfloor + \left\lfloor\frac{n}{3}\right\rfloor + \ldots + \left\lfloor\frac{n}{n}\right\rfloor}$

Caution: Excess of anything is harmful, even mathematics.

# Counting Cycles

Standard

During one of my reading projects in 2015, I read about the Enigma cipher machine. While reading about it, I came to know that the number of possible keys of this machine is 7, 156, 755, 732, 750, 624, 000. One can see the counting procedure at pp. 22 of this document. But the counting procedure was not found to be satisfactory by most members of the audience (during my presentation). My failure to convince the audience that the counting procedure was correct, lead to my distrust in the counting arguments in general. Many times, I still, find the counting procedures controversial.

So, in an attempt to regain the trust, I will present two counting procedures for counting the number of cycles of length $r$ when $n$ objects (colours/beads/numbers) are given.

Procedure A: Using multiplication principle
Step 1: Choose $r$ objects from the $n$ choices.
Step 2: Arrange the selected $r$ objects in a cyclic order.

1. Since the Step 1 and Step 2 are independent of each other but should be performed together, we will multiply the results (i.e. use the multiplication principle). From Step 1 we will get $\binom{n}{r}$ and from Step 2, we will get $(r-1)!$ as per the circular permutation formula. Hence we get:

$\displaystyle{\# r-\text{cycles from } n \text{ objects} =\binom{n}{r}\times(r-1)! = \frac{n!}{r (n-r)!}}$

Procedure B: Using division principle
Step 1: Permute $r$ of the $n$ objects.
Step 2: Realise the mistake that you counted the permutations $r$ extra times because these circular permutations of objects are equivalent since the circle can be rotated.

Since in Step 2 we want to correct the overcounting mistake of Step 1 performed for different objects simultaneously, we will divide the result of Step 1 by the result of Step 2. From Step 1 we will get $^n P_r$ and from Step 2 we will get $r$. Hence we get:

$\displaystyle{\# r-\text{cycles from } n \text{ objects} =\frac{^n P_r}{r} = \frac{n!}{r (n-r)!}}$

I am still not happy with the Procedure B, so if you have a better way of stating it please let me know.

# Real vs Complex numbers

Standard

I want to talk about the algebraic and analytic differences between real and complex numbers. Firstly, let’s have a look at following beautiful explanation by Richard Feynman (from his QED lectures) about similarities between real and complex numbers:

From Chapter 2 of the book “QED – The Strange Theory of Light and Matter” © Richard P. Feynman, 1985.

Before reading this explanation, I used to believe that the need to establish “Fundamental theorem Algebra” (read this beautiful paper by Daniel J. Velleman to learn about proof of this theorem) was only way to motivate study of complex numbers.

The fundamental difference between real and complex numbers is

Real numbers form an ordered field, but complex numbers can’t form an ordered field. [Proof]

Where we define ordered field as follows:

Let $\mathbf{F}$ be a field. Suppose that there is a set $\mathcal{P} \subset \mathbf{F}$ which satisfies the following properties:

• For each $x \in \mathbf{F}$, exactly one of the following statements holds: $x \in \mathcal{P}$, $-x \in \mathcal{P}$, $x =0$.
• For $x,y \in \mathcal{P}$, $xy \in \mathcal{P}$ and $x+y \in \mathcal{P}$.

If such a $\mathcal{P}$ exists, then $\mathbf{F}$ is an ordered field. Moreover, we define $x \le y \Leftrightarrow y -x \in \mathcal{P} \vee x = y$.

Note that, without retaining the vector space structure of complex numbers we CAN establish the order for complex numbers [Proof], but that is useless. I find this consequence pretty interesting, because though $\mathbb{R}$ and $\mathbb{C}$ are isomorphic as additive groups (and as vector spaces over $\mathbb{Q}$) but not isomorphic as rings (and hence not isomorphic as fields).

Now let’s have a look at the consequence of the difference between the two number systems due to the order structure.

Though both real and complex numbers form a complete field (a property of topological spaces), but only real numbers have least upper bound property.

Where we define least upper bound property as follows:

Let $\mathcal{S}$ be a non-empty set of real numbers.

• A real number $x$ is called an upper bound for $\mathcal{S}$ if $x \geq s$ for all $s\in \mathcal{S}$.
• A real number $x$ is the least upper bound (or supremum) for $\mathcal{S}$ if $x$ is an upper bound for $\mathcal{S}$ and $x \leq y$ for every upper bound $y$ of $\mathcal{S}$ .

The least-upper-bound property states that any non-empty set of real numbers that has an upper bound must have a least upper bound in real numbers.
This least upper bound property is referred to as Dedekind completeness. Therefore, though both $\mathbb{R}$ and $\mathbb{C}$ are complete as a metric space [proof] but only $\mathbb{R}$ is Dedekind complete.

In an arbitrary ordered field one has the notion of Dedekind completeness — every nonempty bounded above subset has a least upper bound — and also the notion of sequential completeness — every Cauchy sequence converges. The main theorem relating these two notions of completeness is as follows [source]:

For an ordered field $\mathbf{F}$, the following are equivalent:
(i) $\mathbf{F}$ is Dedekind complete.
(ii) $\mathbf{F}$ is sequentially complete and Archimedean.

Where we defined an Archimedean field as an ordered field such that for each element there exists a finite expression $1+1+\ldots+1$ whose value is greater than that element, that is, there are no infinite elements.

As remarked earlier, $\mathbb{C}$ is not an ordered field and hence can’t be Archimedean. Therefore, $\mathbb{C}$  can’t have least-upper-bound property, though it’s complete in topological sense. So, the consequence of all this is:

We can’t use complex numbers for counting.

But still, complex numbers are very important part of modern arithmetic (number-theory), because they enable us to view properties of numbers from a geometric point of view [source].

# Cross Diagonal Cover – VI

Standard

The problem has finally been solved by Matthew Scroggs.

${\text{\# filled squares} = (\mathrm{lcm}(m-1,n-1)+1 )- \frac12\left(\frac{\mathrm{lcm}(m-1,n-1)}{m-1}-1\right)\left(\frac{\mathrm{lcm}(m-1,n-1)}{n-1}-1\right)}$

He and I, independently,  found a counterexample  for the conjecture by replacing lowest common multiple by greatest common factor using the relation: $ab=\mathrm{lcm}(a,b)\gcd(a,b)$.

In 15×5, there are 26 filled squares and gcd(15,5)=5, so 15×5 is a counterexample to the conjecture!

As it turns out, SAGE is giving float(26/5) = 5.0 when I run it inside the program, but return 5.2 when I run it independently. Leading to wrong conjecture.

The solution is pretty beautiful and  this is the outline:

There are 3 levels of simplifications of the problem:

Original Grid –> Dual Grid –> Mirror Grid –> Inner-point Grid

Counting the “Corners visited twice” (the subtracted term) was something I wasn’t able to do. Corners visited was essentially what I called “number of steps needed for algorithm to stop“. So, his mirror argument provides a proof without words of that result.

# Cross Diagonal Cover – V

Standard

This has been an exciting week! Prof. Sukanta Pati proved an interesting theorem that enables us to get decomposition of $2m-1\times n$ grids into simpler grids, hence simplifying counting to large extent (note that $m=n$ is also allowed). It enables us to surpass the difficulty posed by “more than two crosses in one square”, thus supporting the idea of colouring (i.e. not giving importance to two crosses in a square).

Given $\dpi{300}\inline m\times n$ grid which has consisting of $\dpi{300}\inline m$ rows and $\dpi{300}\inline n$ columns such that the terminating corner square lie in $\dpi{300}\inline m^{th}$ row. Let the number of covered squares in $\dpi{300}\inline m\times n$ grid be $\dpi{300}\inline C(m,n)$ . Then for  $\dpi{300}\inline 2m-1\times n$ grid we have $\dpi{300}\inline C(2m-1, n)= 2C(m,n)-\beta(m,n)$  where $\dpi{300}\inline \beta(m,n)$ is the number of covered squares in $\dpi{300}\inline m^{th}$ row of $\dpi{300}\inline m\times n$ grid.

Instead of giving its proof, I will give an  example when $m=5, n=10$.

I omitted the repetitions of crosses (x) since the number of x in each square doesn’t matter. Note that if we reflect 5×10 about the middle of 5th row we will get 9×10.

Here, $C(5,10) = 25$ and $C(9,10) = 2\times 25 - 5 = 45$. Using this example you can easily prove the generalized statement by exploiting the bilateral symmetry of grid and x about ythe dotted line.

Though  I suspected that least common multiple of $m-1, n-1$ determine the number of steps my algorithm evaluates, Pritam Laskar  found the exact formula.

Given a $m\times n$ grid, the Cross Diagonal Cover algorithm terminates after  $lcm(m-1, n-1) + 1$  steps.

For proof, follow the arguments of grant93jr in his/her comment (he/she mistakenly called their lcm to be their gcd).

Using the SageMath Program, about which I wrote in previous post, I am pretty sure that $\gcd(m,n)$ always divides the number of filled squares. So, now the question is

Why the greatest common divisor of m and n always divides the number of filled squares?

# Cross Diagonal Cover – III

Standard

I found many counterexamples to my conjecture, like

• for Case 2 in 7 by 5 grid we have 12, 11 by 5 grid we have 19 and 15 by 5 grid we have 26 filled squares
• for Case 3 in 4 by 10 grid we have 10 filled squares

In a comment by grant93jr   following part of of my initial question, to determine the number of squares in $m\times n$ grid that can be covered by crosses (x) by following Cross Diagonal Cover Algorithm, was proved:

Given m>n, whenever $n-1$ divides $m-1$ we have $m$ filled squares.

Examples of such grids are: 3 by 5, 4 by 7, 4 by 10, ….

Also in that comment it has been suggested that the algorithm terminates after $k(m-1) = l(n-1)$ steps where $k, l$ are natural numbers. It is certainly true, but I haven’t been able to use this idea for counting number of filled squares when $n-1$ does not divide $m-1$ since in that case some squares are filled twice.

There is another unexplained observation:

Why no filled square has more than 2 crosses?

Frustrated by so many unanswered questions I started colouring squares, so that number of times I visited a square is not visible. Since I really don’t care about how many times I visited  given square while counting the number of filled squares this may help in understanding the underlying symmetry.

Replacing cross (x) by any colour and applying Cross Diagonal Cover Algorithm

After observing above diagrams I suspect that my algorithm leads to a non-deterministic cellular automata.

So, the question which still remains to be answered is:

How many filled squares are there when  $m>n$ and $n-1$ does not divide $m-1$ ?

Examples of such grids are: 7 by 5, 11 by 5, 15 by 5 …