# Numbers and Logic

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I am a big fan of number theory. I find the answer to Hilbert’s Tenth Problem fascinating. I was introduced to this problem, a couple of years ago, via the documentary titled : “Julia Robinson and Hilbert’s Tenth Problem“, here is the trailer:

You can read more about it here. Also for the sake of completeness, let me state Hilbert’s Tenth Problem:

Does there exist an algorithm to determine whether a given Diophantine equation has a solution in rational integers?

In 1970, Yuri Matiyasevich completed the solution of this problem by using the concept of Turing Machine. This short video provides a nice overview about Turing Machines in general

The answer to Hilbert’s Tenth Problem problem is

No such algorithm exists.

This interplay of number theory and logic is really interesting, isn’t it? But I can’t discuss solution of Hilbert’s Tenth Problem here, since I have never read it. But there is nice overview at Wikipedia.

I will rather discuss a puzzle from Boris A. Kordemsky’s book which illustrates the idea of this interplay.

Ask a friend to pick a number from 1 through 1000. After asking him/her ten questions that can be answered yes or no, you tell him/her the number. What kind of question?

The key to the solution is that 2 to the tenth power is 1024 (that is, over 1000). With each question you knock out half the remaining numbers, and after ten questions only the thought number is left.

I welcome you to think of a number and write the corresponding yes/no questions as a comment below.

# Call for Submissions

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I will be hosting 135th Carnival of Mathematics on 15 June 2016.

The Carnival of Mathematics is a monthly blogging round up hosted by a different blog each month. The Aperiodical is responsible for organising a host each month.

The Carnival of Mathematics accepts any mathematics-related blog posts: explanations of serious mathematics, puzzles, writing about mathematics education, mathematical anecdotes, refutations of bad mathematics, applications, reviews, etc. Sufficiently mathematized portions of other disciplines are also acceptable.

Last date for submission is 10 June 2016

# Revision 1: Why I love Mathematics?

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Unlike Mathematics itself, which is forever, philosophy of mathematics keeps on changing. Philosophy of mathematics includes basic definitions and ideologies. About one and a half years ago I wrote about “Why I love Mathematics?” and in these one and a half years my ideologies changed.

Now, for me love means

a feeling of awesomeness for something/ someone.

I believe that this definition captures the general idea. Now let me update the answer to the question.

Mathematics is not a human being, so it can’t accept or reject me. I like logical, concrete and unambiguous statements which are provided by mathematics. Hence I study it.

Unfortunately, contrary to what I believed earlier, mathematics doesn’t have its own language.

The language of mathematics is the language being spoken by the citizens of the “world center(s) of mathematics” of that time.

Let me illustrate my point:

• The ancient mathematics was communicated in Greek, Arabic and Sanskrit (symbolic languages of Babylonians and Mayans are yet to be fully deciphered).
• The medieval mathematics (14th century to 18th Century) was communicated in Italic languages like Latin, Italian, French etc. A good supporter of this argument is the fact that Carl Friedrich Gauss being German wrote his most celebrated book, Disquisitiones Arithmeticae, in Latin.
• The before-my-birth mathematics (18th Century to 20th Century) was communicated in Germanic languages like English and German, since then Cambridge (UK) and Göttingen were the “world centers of mathematics”. A good supporter of this argument is the fact that Paul Erdős being Hungarian wrote his first paper in German (though, he and his friends also published in Hungarian). Interestingly, Russian was the language of many beautiful olympiad problem books until disintegration of USSR.
• The after-my-birth mathematics (21st Century onwards) is communicated in English (Germanic Language) and French (Italic Language) since today’s world centers of mathematics are USA and France. German lost its position as scientific language because of Adolf Hitler‘s dictatorship.

I am not claiming that today mathematics doesn’t exist in any languages other than English or French, but these are the languages in which we today consider publishing our work.  For example, there are more than 550 million people speaking Spanish so it is obvious that their mathematics textbooks are written in their languages (some spanish books). Even in  India, though we have more than 260 million people speaking Hindi and high school mathematics textbooks in Hindi (see: KhanAcademy in Hindi), but still at college level English is only official mode of instruction so that the students have access to the latest discoveries.

# Cross Diagonal Cover – V

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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 – IV

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While discussing this problem with Dr. Shailesh Shirali, he commented that there has to be a way to phrase the problem in terms of a ray of light being reflected off the walls of the rectangle, bouncing around, proceeding from one corner to some other corner.

The problem in re-stating my problem in “light reflection” form was that reflections must be from center of squares and this is not the way light “actually” reflects from surfaces. But, I clubbed that idea with “graph theory” (directed graphs), which actually answers the first question I asked in my previous post:

Why no filled square has more than 2 crosses?

As per my “Cross diagonal cover algorithm” we can’t retrace a path and each square has just two diagonals. Hence, if we replace all the squares by their centers then each center is of degree two (i.e. can allow intersection of at most two distinct paths). Hence proving that no filled square can have more than 2 crosses. For example:

Now moving to the bigger question of counting the total number of filled squares, there hasn’t been much progress yet (even on arriving at a concrete conjecture). But, Prof. Amritanshu Prasad wrote a SageMath  program for my algorithm which enables us to find number of the filled squares for individual cases without actually drawing them! For Example: for m=101 , n=102 grid, there will be 5151 = (101 * 102 )/ 2 filled squares. Pretty cool 🙂

SageMath is is an open source implementation of mathematics and scientific software based on Python 2.  Unfortunately, since the SageMath program is essentially a Python script I am not allowed to embed  it in my WordPress.com blog. But, motivated by my discussions with Ms. Marina Ibrishimova, I tweaked Prof. Prasad’s original source-code and added comments (initialized by #) to make it self explanatory.

Here is the SageMath (or Python) code to count the number of filled squares for $2 \leq m\leq n\leq 10$

#-----START OF PYTHON FUNCTION FOR CROSS DIAGONAL COVER ALGORITHM-----

def cover(m,n):
#a function to count the number of squares covered in a grid with m rows
#and n columns
xinc = 1
#we will use this variable to increment (or decrement) the value of
#x-coordinate (row position) by 1 unit after each step
yinc = 1
#we will use this variable to increment (or decrement) the value of
#y-coordinate (column position) by 1 unit after each step
pos = [1, 1]
#we are assuming our grid to have at least 2 rows and 2 columns
#and will initialize counting from (1,1) position.
visited = [[0, 0], [1, 1]]
#since we start from (1,1) position, we implicitly assume to have
#counted (0,0) position.
while pos != [m-1, 0] and pos != [0, n-1] and pos != [m - 1, n - 1]:
#whenever we reach any corner the algorithm terminates, so need to include
#position of other three corners apart from starting corner in condition.
if pos[0] in [0, m-1]:
#evaluates to true if x-coordinate of current position is a
#member of the collection [0,m-1]
xinc = - xinc
#if x-coordinate is either 0 or m-1 then we will switch diagonal
#by switching the sign of xinc
if pos[1] in [0, n-1]:
#evaluates to true if y-coordinate of current position is a member
#of the collection [0,n-1]
yinc = -yinc
#if y-coordinate is either 0 or n-1 then we will switch diagonal
#by switching the sign of yinc
pos[0] += xinc
#update the x-coordinate of new position as x(new) = x(old) + xinc,
# where xinc = 1 or -1.
pos[1] += yinc
#update the y-coordinate of new position as y(new) = y(old) + yinc,
# where yinc = 1 or -1.
if not (pos in visited):
#if this new state is not there in visited positions list
visited.append([pos[0], pos[1]])
#then add this new position to the list of visited positions.
return visited

#---------------END OF FUNCTION---------------

#we will write a small code to be able to use above function to find number of
#filled squares for all grids where m,n lie between 1 and 11 (both excluded).

for i in range (2, 11):
#The range() function provides an easy way to construct a list of integers.
#The range() function only does numbers from the first to the last,
#not including the last.
for j in range(i, 11):
print i, j, len(cover(i, j))
# len(indata) counts the bytes of data in indata here it
#counts the number of elements is visited list.


To run the above code, please copy-paste it in SageMathCell and click evaluate button. You will get the list displaying m, n and the number of filled squares in each of the grid (with 1<m,n<11).

In case you want to understand above Python function, here is an illustration when m=3 and n=4:

Note that though (1,1) position is encountered twice but it is added to “visited squares” list only once.

So far I haven’t been able to derive a useful interpretation even after getting access to lot of data. I hope to formulate a conjecture soon…