Tuesday, March 25, 2008

A 'Simple' Traversal through a Number Grid -- Patterns, Functions, Algebra Investigation Part I

Here is an activity for Prealgebra and Algebra students. This introductory activity is not meant to be a conundrum for our crack problem-solvers out there, but the extensions below may prove more challenging.

Target Audience: Grades 6-9 (Prealgebra through Algebra 1)

Major Standards/Objectives:
(1) Representing numerical relationships and patterns algebraically
(2) Recognizing, interpreting and developing function notation
(3) Applying remainder concepts

A 2-column number matrix (grid) is shown above and assumed to continue indefinitely. We will be visiting (traversing) the numbers in the grid starting in the upper left corner with 1. Following the arrows we see that the tour proceeds right, then down, followed by left, then down and repeats.

First, some examples of the function notation we will be using to describe this traversal:
T(1) = 1 denotes that the 1st cell visited contains the number 1.
T(4) = 3 denotes that the 4th cell visited contains the number 3.
Similarly, T(6) = 6.


(a) Determine T(1), T(5), T(9), T(13), T(17).
(b) 1, 5, 9, 13, 17, ... all leave a remainder of ___ when divided by 4. (Fill in the blank)
Therefore, these numbers can be represented algebraically as 4n + 1, n = 0,1,2,3,...
(c) Based on (a) and (b), it appears that T(4n+1) = _______, where n = 0,1,2,3...
(d) Determine T(2), T(6), T(10), T(14)
(e) 2,6,10,14,... all leave a remainder of ___ when divided by 4. Therefore, these numbers can be represented algebraically as ______, n = _________ (Fill in blanks)
(f) Based on (d) and (e), it appears that T( _____ ) = _____, n = __________.

Note: The instructor may choose to start n from zero or one throughout this activity. I will vary it depending on our needs. It is important for students to see how restrictions (domain of a variable) is critical for an accurate description and that more than one set of restrictions is possible (provided they are equivalent).

Since T(3) = 4 and T(4) = 3, we cannot say that T(n) = n for all n. The numbers 3 and 4 leave remainders of 3 and 0 respectively when divided by 4. We will need a different rule for these kinds of numbers. Let's collect some more data:

(g) By extending the table, determine T(7) and T(8); T(11) and T(12); T(15) and T(16)
(h) Without extending the table, make a conjecture about the values of T(35) and T(36).
(i) Numbers such as 4,8,12,16,... can be represented algebraically as ____, n= 1,2,3,...
(j) Numbers such as 3,7,11,15,... can be represented algebraically as ____, n = 1,2,3,...

Note: Again, the instructor may not like varying the restrictions here. Adjust as needed.

(h) Ok, so you're an expert now. Well, prove it:
T(100) = ______; T(153) = _____; T(999) = ______
Show or explain your method.


Surely, a 3-column number grid or even a 5-column number grid can't be that much more difficult to solve using the same kind of traversal (move to the right until you come to the end, go down, move left until you come to the end, move down, lather, rinse, repeat...). ENJOY!

Ok, for our experts: Try an n x n grid!

DISCLAIMER: As with all of the investigations I publish, these are essentially original creations and therefore have not been proofread or edited by others. You are the 'others!'. You may not only find errors but alternate and perhaps superior ways to present these ideas.
Also, please adhere to the Guidelines for Attribution in the sidebar.


Florian said...

This is very interesting!

For a table with width a and n=0a+1,2a+1,4a+1, ...
we get the following mapping for 1,2,3,...to T(n):

(going from left to right in the table)
n |-> n
n+ a-1 |-> n+ a-1
(going from right to left in the table)
n+ a |-> n+2a-1
n+2a-1 |-> n+ a

This isn't really handy for calculating T(n). But
if we knew the (x,y)-Position of each value n
in the table we could write T(n) as a formula.

Look sharp and see that for each value n=1,2,...
the x and y positions are:

x = (n-1) mod a +1 (with 1 =< x =< a)
y = floor((n-1)/a) +1 (with 1 =< y)

And that n can now be expressed as:

n = y*a + x.

If y is odd then T(n) = n (this is a row where we go from left to right)
and if y is even T(n) = y*a + (1-x) (this a row where we go from right to left)


1) a=2, n=100:

Step 1: Get (x,y)-Position of n:

x = 99 mod 2 +1 = 2
y = floor(99/2)+1 = 50

(This means that n is in the 50th row and 2nd column of the table.)

Step 2: Determine how to calculate T(n)

y is even so T(n) = y*a + (1-x)

Setp 3: Calculate T(n):

T(100) = 50*2+(1-2) = 99

2) a=2, n=153:

Step 1: Get (x,y)-Position of n:

x = 152 mod 2 +1 = 1
y = floor(152/2) +1= 77

(This means that n is in the 77th row and 1st column of the table.)

Step 2: Determine how to calculate T(n)

y is odd so T(n) = n.

Setp 3: Calculate T(n):

T(153) = 153.

3) a=2, n=999:

Step 1: x = 998 mod 2 +1 = 1
y = floor(998/2) +1 = 500
Step 2: y is even so T(n) = y*a + (1-x)
Step 3: T(999) = 500*2 +(1-1) = 1000.

4) The formula works for any integer a > 0.
a = 7, n=8: (By drawing we expect T(8)=14)

Step 1: x = 7 mod 7 +1 = 1
y = floor(7/7)+1 = 2
Step 2: y is even so T(n) = y*a + (1-x)
Step 3: T(8) = 2*7 +(1-1) = 14.
Which is the expected result.

I think a proof is not necessary here, the formulas
are very straight forward once you see how they work
with the indexes in the rows that read from right to left.

Eric Jablow said...

You should introduce these students to the new word of the day: boustrophedon.

adjective & adverb
(of written words) from right to left and from left to right in alternate lines.

Dave Marain said...

Nice analysis, Florian! I appreciate the effort and thinking that went into it as well as the logic and clarity of your explanation.

Your formula looks strong but the meaning of 'n' may be getting ambiguous. For example, when a=7, n=8, you computed x=1 and y=2, which corresponds to the 1st column of the 2nd row. Yes, the number 8 does appear there but, in my notation, n = 8 refers to the 8th cell visited in the matrix which ends up being in the 7th column of the 2nd row. The value of that cell is of course 14. I think I'm getting confused by your interpretation of n. Clearly your method works.

Dave Marain said...

Boustrophedon has to be my new favorite word - even better than quomodocumque! I may actually use this left-right right-left text method in my next post. Now if we also write palindromically, we'll really confuse everyone!

Joshua Zucker said...

On the subject of Boustrophedon, check out

or read the wikipedia page on Boustrophedon transform.

(I'm a big fan of Neil Sloane and the OEIS)

Florian said...


T(n) is the value of the cell we
visit after the n-th step. And
n is simply a value that is in
the table and of which we can
compute the position.

The idea is to find the position of
n and if it lies in a even row T(n)
is different than n (because we go
from right to left in even rows).

I hope that explains it better.

mathmom said...

Thanks for the vocab word, Eric. If I get a chance to do this with my middle schoolers, I'll definitely get their English teacher to use this as their word of the day too. :)