Just for Fun

7月 162018

My colleague Robert Allison recently blogged about using the diameter of Texas as a unit of measurement. The largest distance across Texas is about 801 miles, so Robert wanted to find the set of all points such that the distance from the point to Texas is less than or equal to 801 miles.

Robert's implementation was complicated by the fact that he was interested in points on the round earth that are within 801 miles from Texas as measured along a geodesic. However, the idea of "thickening" or "inflating" a polygonal shape is related to a concept in computational geometry called the offset polygon or the inflated polygon. A general algorithm to inflate a polygon is complicated, but this article demonstrates the basic ideas that are involved. This article discusses offset regions for convex and nonconvex polygons in the plane. The article concludes by drawing a planar region for a Texas-shaped polygon that has been inflated by the diameter of the polygon. And, of course, I supply the SAS programs for all computations and images.

Offset regions as a union of circles and rectangles

Assume that a simple polygon is defined by listing its vertices in counterclockwise order. (Recall that a simple polygon is a closed, nonintersecting, shape that has no holes.) You can define the offset region of radius r as the union of the following shapes:

  • The polygon itself
  • Circles of radius r centered at each vertex
  • Rectangles of width r positions outside the polygon along each edge

The following graphic shows the offset region (r = 0.5) for a convex "house-shaped" polygon. The left side of the image shows the polygon with an overlay of circles centered at each vertex and outward-pointing rectangles along each edge. The right side of the graphic shows the union of the offset regions (blue) and the original polygon (red):

The image on the right shows why the process is sometimes called an "inflating" a polygon. For a convex polygon, the edges are pushed out by a distance r and the vertices become rounded. For large values of r, the offset region becomes a nearly circular blob, although the boundary is always the union of line segments and arcs of circles.

You can draw a similar image for a nonconvex polygon. The inflated region near a convex (left turning) vertex looks the same as before. However, for the nonconvex (right turning) vertices, the circles do not contribute to the offset region. Computing the offset region for a nonconvex polygon is tricky because if the distance r is greater than the minimum distance between vertices, nonlocal effects can occur. For example, the following graphic shows a nonconvex polygon that has two "prongs." Let r0 be the distance between the prongs. When you inflate the polygon by an amount r > r0/2, the offset region can contain a hole, as shown. Furthermore, the boundary of the offset regions is not a simple polygon. For larger values of r, the hole can disappear. This demonstrates why it is difficult to construct the boundary of an offset region for nonconvex polygons.

Inflating a Texas-shaped polygon

The shape of the Texas mainland is nonconvex. I used PROC GREDUCE on the MAPS.US data set in SAS to approximate the shape of Texas by a 36-sided polygon. The polygon is in a standardized coordinate system and has a diameter (maximum distance between vertices) of r = 0.2036. I then constructed the inflated region by using the same technique as shown above. The polygon and its inflated region are shown below.

The image on the left, which shows 36 circles and 36 rectangles, is almost indecipherable. However, the image on the right is almost an exact replica of the region that appears in Robert Allison's post. Remember, though, that the distances in Robert's article are geodesic distances on a sphere whereas these distances are Euclidean distances in the plane. For the planar problem, you can classify a point as within the offset region by testing whether it is inside the polygon itself, inside any of the 36 rectangles, or within a distance r of a vertex. That computation is relatively fast because it is linear in the number of vertices in the polygon.

I don't want to dwell on the computation, but I do want to mention that it requires fewer than 20 SAS/IML statements! The key part of the computation uses vector operations to construct the outward-facing normal vector of length r to each edge of the polygon. If v is the vector that connects the i_th and (i+1)_th vertex of the polygon, then the outward-facing normal vector is given by the concise vector expression r * (v / ||v||) * M, where M is a rotation matrix that rotates by 90 degrees. You can download the SAS program that computes all the images in this article.

In conclusion, you can use a SAS program to construct the offset region for an arbitrary simple polygon. The offset region is the union of circles, rectangles, and the original polygon, which means that it is easy to test whether an arbitrary point in the plane is in the offset region. That is, you can test whether any point is within a distance r to an arbitrary polygon.

The post Offset regions: Find all points within a specified distance from a polygon appeared first on The DO Loop.

12月 142016
Animated image of falling Koch snowflakes
Out of the bosom of the Air,
      Out of the cloud-folds of her garments shaken,
Over the woodlands brown and bare,
      Over the harvest-fields forsaken,
            Silent, and soft, and slow
            Descends the snow.

"Snow-flakes" by Henry Wadsworth Longfellow

Happy holidays to all my readers! In my last post I showed how to create a well-known fractal called the Koch snowflake. The snowflake is aptly named because it has six-fold symmetry. But as Longfellow noted, a real snowflake is not stationary, but descends "silent, and soft, and slow."

As a gift to my readers, I decided to create an animated greeting card entirely in SAS. The animated GIF (click to enlarge) uses some of the SAS techniques that I have blogged about in 2016. The falling and rotating snowflakes were created by using matrix computations in the SAS/IML language. The animated GIF was created by using ODS graphics and PROC SGPLOT.

Create an animated greeting card with #SAS
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Techniques used to create the animation

If you want to learn how I created this animated GIF with SAS, read on. I've blogged about all the essential techniques in previous posts:

You can download the SAS program that creates the greeting card. Let me know if you adapt this program to create other animated images.

If you like SAS statistical programming or want to learn more about it, subscribe to this blog. In most articles I show how to use SAS for serious pursuits like statistical modeling, data analysis, optimization, and more. But programming with SAS can also be fun, and sometimes it takes a less-serious application to make people say, "Wow! That's cool! I didn't know SAS could do that!"

tags: Fractal, Just for Fun

The post Animate snowfall in SAS appeared first on The DO Loop.

12月 102016
Koch Snowflake in SAS

I have a fondness for fractals.

In previous articles, I've used SAS to create some of my favorite fractals, including a fractal Christmas tree and the "devil's staircase" (Cantor ) function. Because winter is almost here, I think it is time to construct the Koch snowflake fractal in SAS.

A Koch snowflake is shown to the right. The Koch snowflake has a simple construction. Begin with an equilateral triangle. For every line segment, replace the segment by four shorter segments, each of which is one-third the length of the original segment. The two middle segments are offset by 60 degrees from the direction of the original segment. You then iterate this process, with each step replacing every line segment with four others.

The Koch division process

One step in the construction of a Koch Snowflake

For simplicity, the graph at right shows the first step of the construction on a single line segment that has one endpoint at (0,0) and another endpoint at (1,0). The original horizontal line segment of length one is replaced by four line segments of length 1/3. The first and fourth segments are in the same direction as the original segment (horizontal). The second segment is rotated -60 degrees from the original direction. It, too, is 1/3 the original length. The third segment connects the second and fourth line segments.

The following program defines a function that uses the vector capabilities of SAS/IML software to carry out the fundamental step on a line segment. The function takes two row vectors as arguments, which are the (x,y) coordinates of the endpoints of the line segment. The function returns a 5 x 2 matrix, where the rows contain the endpoints of the four line segments that result from the Koch construction.

Two statements in the function might be unfamiliar. The direct product operator (@) quickly constructs points that are k/3 units along the original line segment for k=0, 1, 2, and 3. The ATAN2 function computes the direction angle for a two-dimensional vector.

proc iml;
/* divide a line segment (2 pts) into 4 segments (5 pts).  
   Create middle point by rotating through -60 degrees */ 
start KochDivide(A, E);                    /* (x,y) coords of endpoints */
   segs = j(5, 2, .);                      /* matrix to hold 4 shorter segs */
   v = (E-A) / 3;                          /* vector 1/3 as long as orig */
   segs[{1 2 4 5}, ] = A + v @ T(0:3);     /* endpoints of new segs */
   /* Now compute middle point. Use ATAN2 to find direction angle. */
   theta = -constant("pi")/3 + atan2(v[2], v[1]); /* change angle by pi/3 */
   w = cos(theta) || sin(theta);           /* vector to middle point */
   segs[3,] = segs[2,] + norm(v)*w;
   return segs;
/* test on line segment from (0,0) to (1,0) */
s = KochDivide({0 0}, {1 0});
title  "Fundamental Step in the Koch Snowflake Construction";
ods graphics / width=600  height=300; 
call series(s[,1], s[,2]) procopt="aspect=0.333";

Construct the Koch snowflake in SAS

Now that we have defined a function that can replace a segment by four shorter segments, let's define a function that applies that function repeatedly to each segment of a polygon. The following function takes two arguments: the polygon and the number of iterations. An N-sided closed polygon is represented as an (N+1) x 2 matrix of vertices.

/* create Koch Snowflake from and equilateral triangle */
start KochPoly(P0, iters=5);    
P = P0;
do j=1 to iters;
   N = nrow(P) - 1;             /* old number of segments */
   newP = j(4*N+1, 2);          /* new number of segments + 1 */
   do i=1 to N;                 /* for each segment... */
      idx = (4*i-3):(4*i+1);                    /* rows for 4 new segments */
      newP[idx, ] = KochDivide(P[i,], P[i+1,]); /* generate new segments */
   P = newP;                    /* update polygon and iterate */
return P;

To test the function, define an equilateral triangle. The call to the KochPoly function creates a Koch snowflake from the equilateral triangle.

/* create equilateral triangle as base for snowflake */
pi = constant("pi");
angles = -pi/6 // pi/2 // 7*pi/6;  /* angles for equilateral triangle */
P = cos(angles) || sin(angles);    /* vertices of equilateral triangle */
P = P // P[1,];                    /* append first vertex to close triangle */
K = KochPoly(P);                   /* create the Koch snowflake */

The Koch snowflake that results from this iteration is shown at the top of this article. For completeness, here are the statements that write the snowflake to a SAS data set and graph it by using PROC SGPLOT:

S = j(nrow(K), 3, 1);      /* add ID variable with constant value 1 */
S[ ,1:2] = K;
create Koch from S[colname={X Y ID}];  append from S;  close;
ods graphics / width=400px height=400px;
footnote justify=Center "Koch Snowflake";
proc sgplot data=Koch;
   styleattrs wallcolor=CXD6EAF8;               /* light blue */
   polygon x=x y=y ID=ID / outline fill fillattrs=(color=white);
   xaxis display=none;
   yaxis display=none;

Generalizations of the Koch snowflake

The KochPoly function does not check whether the polygon argument is an equilateral triangle. Consequently, use the function to "Koch-ify" any polygon! For example pass in the rectangle with vertices P = {0 0, 2 0, 2 1, 0 1, 0 0}; to create a "Koch box."

Enjoy playing with the program. Let me know if you generate an interesting shape!

tags: Fractal, Just for Fun

The post Create a Koch snowflake with SAS appeared first on The DO Loop.

7月 052016

Last week I blogged about how to draw the Cantor function in SAS. The Cantor function is used in mathematics as a pathological example of a function that is constant almost everywhere yet somehow manages to "climb upwards," thus earning the nickname "the devil's staircase."

The Cantor function has three properties: (1) it is continuous, (2) it is non-decreasing, and (3) F(0)=0 and F(1)=1. There is a theorem that says that any function with these properties is the cumulative distribution function (CDF) for some random variable. In theory, therefore, you can generate a large number of random variates (a large sample), then use PROC UNIVARIATE to plot the empirical CDF. The empirical CDF should be the devil's staircase function.

A random sample from the Cantor set

My last article showed that every point in the Cantor set can be written as the infinite sum x = Σi ai3-i where the ai equal 0 or 2. You can approximate the Cantor set by truncating the series at a certain number of terms. You can generate random points from the (approximate) Cantor set by choosing the coefficients randomly from the Bernoulli distribution with probability 0.5. The following DATA step generates 10,000 points from the Cantor set by using random values for the first 20 coefficients:

data Cantor;
call streaminit(123456);
k = 20;             /* maximum number of coefficients */
n = 10000;          /* sample size to generate */
do i = 1 to n;      /* generate n random points in Cantor set */
   x = 0;
   do j = 1 to k;   /* each point is sum of powers of 1/3 */
      b = 2 * rand("Bernoulli", 0.5);  /* random value 0 or 2 */
      x + b * 3**(-j);                 /* power of 1/3 */
keep x;
proc univariate data=Cantor;
   cdf x;

The call to PROC UNIVARIATE displays the empirical CDF for the random sample of points. And voilà! The Cantor distribution function appears! As a bonus, the output from PROC UNIVARIATE (not shown) indicates that the mean of the distribution is 1/2 (which is obvious by symmetry) and the variance is 1/8, which you might not have expected.

This short SAS program does two things: it shows how to simulate a random sample uniformly from the (approximate) Cantor set, and it indicates that the devil's staircase function is the distribution function for this a uniform random variable.

Did you know that the Cantor staircase function is a cumulative distribution? Simulate it in #SAS!
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A random sample in SAS/IML

Alternatively, you can generate the same random sample by using SAS/IML and matrix computations. My previous article drew the Cantor function by systematically using all combinations of 0s and 2s to construct the elements of the Cantor set. However, you can use the same matrix-based approach but generate random coefficients, therefore obtaining a random sample from the Cantor set:

proc iml;
k = 20;          /* maximum number of coefficients */
n = 10##4;       /* sample size to generate */
B = 2 * randfun(n||k, "Bernoulli", 0.5);  /* random n x k matrix of 0/2 */
v = 3##(-(1:k)); /* powers of 1/3 */
x = B * v`;      /* sum of random terms that are either 0*power or 2*power */

Recall that ECDF is a step function that plots the ith ordered datum at the height i/n. You can approximate the empirical CDF in SAS/IML by using the SERIES subroutine. Technically, the lines that appear in the line plot have a nonzero slope, but the approximation looks very similar to the PROC UNIVARIATE output:

call sort(x, 1);             /* order elements in the Cantor set */
x = 0 // x // 1;             /* append end points */
y = do(0, 1, 1/(nrow(x)-1)); /* empirical CDF */
title "Empirical Distribution of a Uniform Sample from the Cantor Set";
call series(x, y);

Interpretation in probability

It is interesting that you can actually define the Cantor set in terms of rolling a fair six-sided die. Suppose that you roll a die infinitely often and we adopt the following rules:

  • If the die shows a 1 or 2, you get zero points.
  • If the die shows a 3 or 4, you get one point.
  • If the die shows a 5 or 6, you get two points.

This game is closely related to the Cantor set. Recall that the Cantor set can be written as the set of all base-3 decimal values in [0,1] for which the decimal expansion does not contain a 1. In this game, each point in the Cantor set corresponds to a sequence of rolls that never contain a 3 or 4. (Equivalently, the score is always even.) Obviously this would never happen in real life, which is an intuitive way of saying that the Cantor set has "measure zero."


The first time I saw the Cantor function depicted as an empirical distribution function was when I saw a very compact MATLAB formula like this:

stairs([min(x) sort(x)],0:1/length(x):1) % Plot the c.d.f of x

The previous formula is equivalent to my SAS/IML program, but in my program I broke the formula apart so that the components could be understood more easily. This formula appears in a set of probability demos by Peter Doyle. I had the privilege to interact with Professor Doyle at The Geometry Center (U. MN) in the early 1990s, so perhaps he was responsible for showing me the Cantor distribution.

These UNC course notes from Jan Hannig also discuss the Cantor distribution, but the original source is not cited.

tags: Just for Fun, Math

The post Cantor sets, the devil's staircase, and probability appeared first on The DO Loop.

6月 292016

I was a freshman in college the first time I saw the Cantor middle-thirds set and the related Cantor "Devil's staircase" function. (Shown at left.) These constructions expanded my mind and led me to study fractals, real analysis, topology, and other mathematical areas.

The Cantor function and the Cantor middle-thirds set are often used as counter-examples to mathematical conjectures. The Cantor set is defined by a recursive process that requires infinitely many steps. However, you can approximate these pathological objects in a matrix-vector language such as SAS/IML with only a few lines of code!

Construction of the Cantor middle-thirds set

The Cantor middle-thirds set is defined by the following iterative algorithm. The algorithm starts with the closed interval [0,1], then does the following:

  1. In the first step, remove the open middle-thirds of the interval. The two remaining intervals are [0,1/3] and [2/3, 1], which each have length 1/3.
  2. In the ith step, remove the open middle-thirds of all intervals from the (i-1)th step. You now have twice as many intervals and each interval is 1/3 as long as the intervals in the previous step.
  3. Continue this process forever.

After two steps you have four intervals: [0,1/9], [2/9,1/3], [2/3, 7/9], and [8/9,1]. After three steps you have eight intervals of length 1/27. After k steps you have 2k intervals of length 3-k.

The Cantor set is the set of all points that never get removed during this infinite process. The Cantor set clearly contains infinitely many points because it contains the endpoints of the intervals that are removed: 0, 1, 1/3, 2/3, 1/9. 2/9, 7/9, 8/9, and so forth. Less intuitive is the fact that the cardinality of the Cantor set is uncountably infinite even though it is a set of measure zero.

Construction of the Cantor function

The Cantor function F: [0,1] → [0,1] can be defined iteratively in a way that reflects the construction of the Cantor middle-thirds set. The function is shown at the top of this article.

  • At step 0, define the function on the endpoints of the [0,1] interval by F(0)=0 and F(1)=1.
  • At step 1, define the function on the closed interval [1/3, 2/3] to be 1/2.
  • At step 2, define the function on [1/9, 2/9] to be 1/4. Define the function on [7/9, 8/9] to be 3/4.
  • Continue this construction. At each step, the function is defined on the closure of the middle-thirds intervals so that the function value is halfway between adjacent values. The function looks like a staircase with steps of difference heights and widths. The function is constant on every middle-thirds interval, but nevertheless is continuous and monotonically nondecreasing.

Visualize the Cantor staircase function in #SAS.
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Visualizing the Cantor function in SAS

This is a SAS-related blog, so I want to visualize the Cantor function in SAS. The middle-third intervals during the kth step of the construction have length 3-k, so you can stop the construction after a small number of iterations and get a decent approximation. I'll use k=8 steps.

Although the Cantor set and function were defined geometrically, they have an equivalent definition in terms of decimal expansion. The Cantor set is the set of decimal values that can be written in base 3 without using the '1' digit. In other words, elements of the Cantor set have the form x = 0.a1a2a3... (base 3), where ai equals 0 or 2.

An equivalent definition in terms of fractions is x = Σi ai3-i where ai equals 0 or 2. Although the sum is infinite, you can approximate the Cantor set by truncating the series after finitely many terms. A sum like this can be expressed as an inner product x = a*v` where a is a k-element row vector that contains 0s and 2s and v is a vector that contains the elements {1/3, 1/9, 1/27, ..., 1/3-k}.

You can define B to be a matrix with k columns and 2k rows that contains all combinations of 0s and 2s. Then the matrix product B*v is an approximation to the Cantor set after k steps of the construction. It contains the right-hand endpoints of the middle-third intervals.

In SAS/IML you can use the EXPANDGRID function to create a matrix whose rows contain all combinations of 0s and 2s. The ## operator raises an element to a power. Therefore the following statements construct and visualize the Cantor function. With a little more effort, you can write a few more statements that improve the approximation and add fractional tick marks to the axes, as shown in the graph at the top of this article.

proc iml;
/* rows of B contain all 8-digit combinations of 0s and 2s */ 
B = expandgrid({0 2}, {0 2}, {0 2}, {0 2}, {0 2}, {0 2}, {0 2}, {0 2});
B = B[2:nrow(B),];      /* remove first row of zeros */
k = ncol(B);            /* k = 8 */
v = 3##(-(1:k));       /* vector of powers 3^{-i} */
t = B * v`;            /* x values: right-hand endpts of middle-third intervals */
u = 2##(-(1:k));       /* vector of powers 2^{-i} */
f = B/2 * u`;          /* y values: Cantor function on Cantor set */
call series(t, f);     /* graph the Cantor function */

I think this is a very cool construction. Although the Cantor function is defined iteratively, there are no loops in this program. The loops are replaced by matrix multiplication and vectors. The power of a matrix language is that it enables you to compute complex quantities with only a few lines of programming.

Do you have a favorite example from math or statistics that changed the way that you look at the world? Leave a comment.


This short article cannot discuss all the mathematically interesting features of the Cantor set and Cantor function. The following references are provided for the readers who want additional information:

tags: Just for Fun, vectorization

The post Visualize the Cantor function in SAS appeared first on The DO Loop.

12月 162015
Lo how a rose e'er blooming
From tender stem hath sprung.

As I write this blog post, a radio station is playing Chrismas music. One of my favorite Christmas songs is the old German hymn that many of us know as "Lo, How a Rose E're Blooming." I was humming the song recently when someone asked me how to use SAS to draw curves in polar coordinates. I immediately thought about "polar roses," mathematical parametric curves of the form r = cos(k θ).

In the spirit of Christmas, I present "Lo, how a polar rose e'er blooming, from SGPLOT hath sprung."

Plotting polar coordinates in SAS

It is easy to graph a polar curve in SAS. Here's a quick overview, or a "polar express" tour, if you prefer.

Many familiar polar equations are parametric curves. The simplest are relations are of the form r = f(θ), where θ is the polar angle. The following SAS DATA step creates evenly spaced values of theta in the range [0, 2π]. For each value of theta, it computes r = cos(k θ), which is the equation for a rose with k petals. The polar coordinates (r, θ) are converted to Euclidean coordinates through the usual transformation x = r*cos(theta) and y = r*sin(theta).

After creating the points along the curve, you can visualize it by using PROC SGPLOT. The SERIES statement is used to display the parametric curve in Euclidean coordinates. The XAXIS and YAXIS statements use the MIN= and MAX= options to ensure that the image appears in a square region, and the ASPECT=1 option is used to ensure that the aspect ratio of the plot does not distort the geometry of the rose.

%let k = 5;
/* draw the curve r=cos(k * theta) in polar coords, which is a k-leaved rose */
data Rose;
do theta = 0 to 2*constant("pi") by 0.01;
   r = cos(&k * theta);       /* rose */
   x = r*cos(theta);          /* convert to Euclidean coords */
   y = r*sin(theta);
title "Polar Rose: r = cos(&k theta)";;
proc sgplot data=Rose aspect=1;
   refline 0 / axis=x;
   refline 0 / axis=y;
   series x=x y=y;
   xaxis min=-1 max=1;
   yaxis min=-1 max=1;

Graphing generalized roses

The Wikipedia page for the polar rose contains a brief discussion about what the curve r = cos(k θ) looks like when k is a rational number n/d. When k is not an integer, most of the curves do not look like a rose—or any other kind of flower! Only a myopic mathematician would call some of these images roses.

Regardless, let's see how to get some of these "roses" to bloom in SAS. The following DATA step generates all positive rational numbers of the form n/d, where n ≤ 7 and d ≤ 9 are integers. For each value of k, the program computes the generalize rose equation and converts the values from polar to Euclidean coordinates:

data Roses;
do n = 1 to 7;
   do d = 1 to 9;
      k = n / d;
      /* draw the rose r=cos(n/d * theta) */
      do theta = 0 to 2*lcm(n,d)*constant("pi") by 0.05;
         r = cos(k * theta);      /* generalized rose */
         x = r*cos(theta);        /* convert to Euclidean coords */
         y = r*sin(theta);

It is not any harder to visualize 63 roses than it is to visualize one. The SGPANEL procedures was designed to display multiple cells, each with the same type of graph. Therefore, to display an entire bouquet of roses, you merely have to specify the N and D variables on the PANELBY statement, and SAS handles the rest:

title "Polar Roses: r = cos(n/d theta)";
proc sgpanel data=Roses aspect=1;
   panelby n d / layout=lattice onepanel rowheaderpos=left;
   refline 0 / axis=x transparency=0.5;
   refline 0 / axis=y transparency=0.5;
   series x=x y=y;
   colaxis display=none;
   rowaxis display=none;

You can see certain numeric patterns in the lattice display. In particular, when n/d is a reducible fraction (such as 2/4, 3/6, and 4/8), the image is identical to the reduced fraction (such as 1/2). Do you see other patterns? Leave a comment.

These roses are beautiful, like the song that inspired me to create them. Since I get up early in the morning to write my blog, a few additional lyrics seem appropriate:

It came a blossom bright.
Amid the cold of winter
When half spent was the night.

Happy holidays to all my readers. I am taking a short vacation and will return in 2016. Until then, keep your eyes open: math, statistics, and beauty are all around you, ever blooming!

tags: Just for Fun, Math

The post Lo, how a polar rose e'er blooming appeared first on The DO Loop.

9月 042015

In my previous blog post, I showed how you can use SAS to program a "weaving" algorithm that takes an image, cuts it into strips, and weaves the strips together to create mathematical art. I used matrices and heat maps for the computations and visualization.

At the end of the post, I presented an image of woven strips (shown at left) and challenged the reader to "deduce the original image from the weave." This article shows that you can "unweave" the image, reassemble the strips, and use interpolation to obtain an approximation to the original image. You can download the SAS/IML program that computes and visualizes the preimage.

Unweaving a weave

The first step in reconstructing the original image is to "pull apart" the horizontal and vertical strips, rotate the horizontal ones, put them next to the vertical strips, and close all the gaps. This requires writing the inverse of the weave algorithm. When you do this, you obtain the following image:


Notice that there are gaps in our knowledge about the original image. Due to the weaving process, there are places where a vertical strip was covered by a horizontal strip, or the other way around. These missing values are shown in white. Mathematically, the weaving algorithm represents a many-to-one function, so there is no inverse function that can restore the preimage from the image.

Nevertheless, the unweaving reveals the main features of the original image. And we can do more. If we know that the original image was created from a continuous mathematical function, you can use interpolation to try to fill in the gaps. For this image you can use the information from nearby cells (north, south, east, and west) to estimate a value for the missing cells. The following uses a simple estimation scheme, which is to assign each missing cell the average value of its neighbors. You can proceed in five steps:

  1. Approximate the corners. For the lower left corner, use the average of the cell above and the cell to the right. For the lower right corner, use the average of the cell above and the cell to the left.
  2. Approximate the left edge. For each missing cell, average the cells above, below, and to the right.
  3. Approximate the right edge. For each missing cell, average the cells above, below, and to the left.
  4. Approximate the bottom edge. For each missing cell, average the cells above, to the left, and to the right.
  5. Approximate the interior cells. For each missing cell, average the cells above, below, to the left, and to the right.

After implementing this process, you obtain the following image, which approximates the original image:


How well does this image approximate the original image? Very well! The mean square error is very small, and the maximum error values are within 20% of the true values. For comparison, here is the original image. You can see that it is very similar to the reconstructed image.


The approximation could be improved by using a more sophisticated interpolation scheme, such as bivariate linear interpolation.

Feel free to download the SAS/IML program and play with other weaves. How well can you recover an original image from its weave?
tags: Just for Fun

The post Mathematical art (part 2): Unweaving matrices appeared first on The DO Loop.

9月 022015

An artist friend of mine recently created a beautiful abstract image and described the process on her blog. She says that "after painting my initial square, I cut it into strips and split them down the middle, then wove them together.... I had no idea when I started piecing these together again what they would look like."

I marveled at the beauty of the finished result, but I also started thinking about the process of constructing the artwork. You cut an initial picture into n vertical strips. (For simplicity, assume that n is even.) Then you take the first n/2 strips and space them out so that there are gaps between them. You take the last n/2 strips and turn them sideways, weaving them in and out of the vertical strips.

As my friend said in a private communication, the final image "depends entirely on the arrangement of the elements.... There are so many versions of the final outcome." By this she means that the process is deterministic, although the artist can make choices regarding the number of strips, the orientation of the strips, and the weaving pattern. Yes, it is difficult to predict the final image while staring at the initial image, but this is the sort of algorithm that you could ask a computer to carry out.

So, of course, I had to learn to make images like this with SAS!

Weaving matrices

To make sure that I understood the algorithm, I started with a 10 x 10 numeric matrix X that contains the values 1–100 in row-major order. That is, the first row contains 1–10, the second row contains 11–20, and so forth. I then created a 10 x 10 matrix W (for "weave") that contained all missing values. I copied the first five columns of X into the even columns of W. I then copied the last five columns of X into the even rows of W. I then went back and "wove" the columns and rows by re-copying certain elements from the first five columns of X. You can view this construction process for an example matrix.

The process can be encapsulated into a SAS/IML function, which is shown below:

proc iml;
start weave(x);
   n = nrow(x);             /* n must be even */
   y = j(n, n, .);
   j = 1:n/2;               /* vertical strips */
   y[, 2*j] = x[, j];       /* copy to every other column */
   i = (n/2+1):n;           /* horizontal strips */
   y[2*i-n,] = x[, i]`;     /* copy to every other row */
   k = do(1, n/2, 2);       /* weaving: raise certain rows */
   y[2*k, 2*k] = x[2*k, k]; /* every other for even rows */
   k = do(2, n/2, 2);
   y[2*k, 2*k] = x[2*k,k];  /* every other for odd rows */
   return( y );
x = shape(1:100, 10);       /* original 10x10 matrix */
w = weave(x);
print w;

Weaving heat maps

Now the fun can begin. Recall that a heat map is essentially a visualization of a matrix of numbers. A heat map can be used to visualize mathematical functions of two variables, and you can use the HEATMAPCONT function in SAS/IML 13.1 to easily create a heat map.

You can use the EXPANDGRID function to define a grid of evenly spaced points in the square [-1, 1] x [-1, 1]. You can evaluate a mathematical function on the grid and reshape the results into a matrix. You can then create a heat map of the original function and the "weave" of the function. For details, you can download the complete SAS/IML program that I used to construct the images in this article.

For example, in the following image, the function on the left is f(x,y) = x. The heat map that visualizes the function shows that the function is constant on vertical strips. You can see that each strip is a constant color, and the image on the right has taken those strips and woven them together.


The woven image is beautiful, but you can make more interesting images by choosing more interesting functions to use as the original image. The left portion of the following image shows a heat map for the quadratic function f(x,y) = x2 + y2. The image on the right is formed by weaving the vertical strips from the left image.


You get the idea: You can visualize any mathematical function and then form "mathematical art" by cutting the function along strips and weaving the strips back together.

If you are adventurous, you can play with different color palettes for the images. I chose a muted rainbow palette, but you can experiment with other color ramps. The program for these examples include other color ramps that change the aesthetics of the images.


I'll leave you with one more image. It is the weave for a common function from multivariable calculus. Can you deduce the original image from the weave? If not, look in the program that creates these examples..

These images show flexibility of the SAS/IML language for creating a custom algorithm. These simple images lack the depth and complexity of "real" art, but they show how matrices can be used to illustrate the main ideas behind a simple artistic process.

Try it yourself! What images can you make? If you create an interesting picture, leave a comment that includes the mathematical formula.

tags: Just for Fun

The post Mathematical art: Weaving matrices appeared first on The DO Loop.

4月 152015

Last week I was chatting with some mathematicians and I mentioned the blog post that I wrote last year on the distribution of Pythagorean triples.

In my previous article, I showed that there is an algorithm that uses matrix multiplication to generate every primitive Pythagorean triple by starting with the simple (3,4,5) right triangle. The algorithm does not systematically generate triangles with large legs from triangles with smaller legs. Instead, it might generate several triangles with large legs, followed by one or more triangles with small legs. Given a specific radius r, it would be interesting to prove how many iterations of the algorithm are required to guarantee that the algorithm has generated all primitive Pythagorean triples whose hypothesis is less than r.

From a statistical point of view, it is interesting to estimate the distribution of the smallest angle in primitive Pythagorean right triangles. Suppose that you consider only the primitive triangles whose hypotenuse is less than 106. The following histogram shows the distribution of the smallest angle (arctan(b/a) if a is the longer leg) for angles ranging from 0 to 45 degrees in increments of 0.5 degrees. The histogram is formed from 98,151 triangles, which is considerably less than the 159,139 primitive right triangles whose hypotenuses are less than one million.


The histogram has several interesting features:

  • Apparently there are relatively few long-and-skinny triangles. In this sample, few have angles less than 5 degrees. The smallest angle is 3.47 degrees, which corresponds to the (33, 544, 545) right triangle.
  • The distribution is not uniform. In fact, there are noticeable gaps in the density distribution. The biggest gaps are near 37, 23, and 28 degrees.
  • Some Pythagorean triangles that are nearly isosceles. In this sample, the (137903, 137904, 195025) triangle has an angle equal to 44.99979 degrees. This triangle is an example of a twin-leg Pythagorean triple in which the legs lengths differ by 1.

The gaps are very interesting. I conjecture that the gaps in the distribution are not random, but that they are related to low-order triangles. The following statements compute the smallest angle of some right triangles whose legs are small integers:

proc iml;
v = {  4  3  5, 
      12  5 13,
      15  8 17,
      24  7 25,
      21 20 29,
      35 12 37 };
a = atan2(v[,2], v[,1]) * 180/constant('pi');
print v[L="" c={"a" "b" "c"}] a[L="Angle"];

If you overlay lines at each of the angles above, they fall right into the gaps of the histogram. The (3,4,5) triangle is centered at the biggest gap (37 degrees). The (5, 12, 13) triangle is responsible for the next largest gap (23 degrees), and so forth. You can overlay these lines on the histogram, or overlay the lines on a fringe plot, as shown in the following image in which the red lines represent the angles of six low-order triangles:


If this histogram is representative of the true density distribution of the smallest angle in a Pythagorean triangle, the low-order triangles seem to be surrounded by a region of reduced density. If I were to study this problem further, I would conjecture that the gap induced by a low-order triangle is related to Farey sequences. Farey sequences arise in the darndest places, ranging from the dynamics of planetary objects to the study of Diophantine equations. In fact, the fringe plot reminds me of the gaps and divisions in Saturn's rings, which is another place where Farey sequences have an application. Of course, these gaps could also be an artifact of the algorithm that I am using to generate the primitive Pythagorean triangles, but that can be checked by using a different algorithm to generate the Pythagorean triples. Personally, I think the gaps are produced by a real number-theoretic phenomenon.

The density distribution of the smallest angle is interesting, don't you think? Mathematicians who work in number theory can probably explain why the gaps exist, but they were a surprise to me. Who would have guessed that Pythagorean triangles would lead to such an interesting distribution? It's a beautiful connection between probability, statistics, and high-school geometry.

tags: Just for Fun, Math

The post The distribution of Pythagorean triples by angle appeared first on The DO Loop.

3月 132015

Saturday, March 14, 2015, is Pi Day, and this year is a super-special Pi Day! This is your once-in-a-lifetime chance to celebrate the first 10 digits of pi (π) by doing something special on 3/14/15 at 9:26:53. Apologies to my European friends, but Pi Day requires that you represent dates with the month placed first in order to match the sequence 3.141592653....

Last year I celebrated Pi Day by using SAS to explore properties of the continued fraction expansion of pi. This year, I will examine statistical properties of the first 10 million digits of pi. In particular, I will show that the digits of pi exhibit statistical properties that are inherent in a random sequence of integers.

Reading 10 million digits of pi

I have no desire to type in 10 million digits, so I will use SAS to read a text file at a Princeton University URL. The following statements use the FILENAME statement to point to the URL:

/* read data over the internet from a URL */
filename rawurl url "http://www.cs.princeton.edu/introcs/data/pi-10million.txt"
                /* proxy='http://yourproxy.company.com:80' */ ;
data PiDigits;
   infile rawurl lrecl=10000000;
   input Digit 1. @@;
   Position = _n_;
   Diff = dif(digit);      /* compute difference between adjacent digits */
proc print data=PiDigits(obs=9);
   var Digit;

The PiDigits data set contains 10 million rows. The call to PROC PRINT displays the first few decimal digits, which are (skipping the 3) 141592653....

For other ways to use SAS to download data from the internet, search Chris Hemedinger's blog, The SAS Dummy for "PROC HTTP" and you will find several examples of how to download data from a URL.

The distribution of digits of pi

You can run many statistical tests on these numbers. It is conjectured that the digits of pi are randomly uniformly distributed in the sense that the digits 0 through 9 appear equally often, as do pairs of digits, trios of digits, and so forth.

You can call PROC FREQ to compute the frequency distribution of the first 10 million digits of pi and to test whether the digits appear to be uniformly distributed:

/* Are the digits 0-9 equally distributed? */
proc freq data = PiDigits;
tables Digit/ chisq out=DigitFreq;

The frequency analysis of the first 10 million digits shows that each digit appears about one million times. A chi-square test indicates that the digits appear to be uniformly distributed. If you turn on ODS graphics, PROC FREQ also produces a deviation plot that shows that the deviations from uniformity are tiny.

A "pi chart" of the distribution of the digits of pi

As an advocate of the #OneLessPie Chart Initiative, I am intellectually opposed to creating pie charts. However, I am going to grit my teeth and make an exception for this super-special Pi Day. You can use the Graph Template Language (GTL) to create a pie chart. Even simpler, Sanjay Matange has written a SAS macro that creates a pie chart with minimal effort. The following DATA step create a percentage variable and then calls Sanjay's macro:

data DigitFreq;
   set DigitFreq;
   Pct = Percent/100; 
   format Pct PERCENT8.2;
/* macro from http://blogs.sas.com/content/graphicallyspeaking/2012/08/26/how-about-some-pie/ */
%GTLPieChartMacro(data=DigitFreq, category=Digit, response=Pct,
         title=Distribution of First 10 Million Digits of Pi,

The pie chart appears at the top of this article. It shows that the digits 0 through 9 are equally distributed.

Any autocorrelation in the sequence?

In the DATA step that read the digits of pi, I calculated the difference between adjacent digits. You can use the SGPLOT procedure to create a histogram that shows the distribution of this quantity:

proc sgplot data=PiDigits(obs=1000000);  
   vbar Diff;

That's a pretty cool triangular distribution! I won't bore you with mathematical details, but this shape arises when you examine the difference between two independent discrete uniform random variables, which suggests that the even digits of pi are independent of the odd digits of pi.

In fact, more is true. You can run a formal test to check for autocorrelation in the sequence of numbers. The Durbin-Watson statistic, which is available in PROC REG and PROC AUTOREG, has a value near 2 if a series of values has no autocorrelation. The following call to PROC AUTOREG requests the Durbin-Watson statistic for first-order through fifth-order autocorrelation for the first one million digits of pi. The results show that there is no detectable autocorrelation through fifth order. To the Durban-Watson test, the digits of pi are indistinguishable from a random sequence:

proc autoreg data=PiDigits(obs=1000000);  
   model Digit = / dw=5 dwprob;

Are the digits of pi random?

Researchers have run dozens of statistical tests for randomness on the digits of pi. They all reach the same conclusion. Statistically speaking, the digits of pi seems to be the realization of a process that spits out digits uniformly at random.

Nevertheless, mathematicians have not yet been able to prove that the digits of pi are random. One of the leading researchers in the quest commented that if they are random then you can find in the sequence (appropriately converted into letters) the "entire works of Shakespeare" or any other message that you can imagine (Bailey and Borwein, 2013). For example, if I assign numeric values to the letters of "Pi Day" (P=16, I=9, D=4, A=1, Y=25), then the sequence "1694125" should appear somewhere in the decimal expansion of pi. I wrote a SAS program to search the decimal expansion of pi for the seven-digit "Pi Day" sequence. Here's what I found:

proc print noobs data=PiDigits(firstobs=4686485 obs=4686491);
   var Position Digit;

There it is! The numeric representation of "Pi Day" appears near the 4.7 millionth decimal place of pi. Other "messages" might not appear in the first 10 million digits, but this one did. Finding Shakespearian sonnets and plays will probably require computing more digits of pi than the current world record.

The digits of pi pass every test for randomness, yet pi is a precise mathematical value that describes the relationship between the circumference of a circle and its diameter. This dichotomy between "very random" and "very structured" is fascinating! Happy Pi Day to everyone!

tags: Data Analysis, Just for Fun