Answer for regression of the answer for the article on regression with the answer…

For the cross-stitch curve, how long is the final perimeter, and how large is the enclosed area?

If the cross-stitch is built to extend outward from the center of the square, the perimeter is infinite, but the area is twice that of the original square. If the stitch extends inward towards the center, the perimeter is still infinite, but the area goes to zero.



When I first started going through the Gardner Colossal book, I made notes of things that I wanted to follow up on, or include in the blog entries. One of which was David Cope’s EMI (Experiments in Musical Intelligence). I’m not sure I’ll be able to do much with this program in the short-term, so I wanted to put the link in the blog now, in case anyone wants to check it out.

Answer Rotations

Answers for rotation and reflection:
1) Oliver Lee, age 44, lives at 312 Main Street. He asked the city to give him the license plate 337-31770 for his car. Why?

Rotated 180 degrees, the plate reads “OLLIE LEE”.

2) A basket contains more than 6 eggs, which are either white or brown. If x is the number of white eggs and y is the number of brown eggs, then what values are needed such that the sum of x and y turned upside-down is the product of x and y? That is, how many eggs are there?

x = 9 and y = 9. x + y = 18. x * y = 81. Turn 18 upside-down and you get 81.

If the problem allowed any number of eggs, then x = 3 and y = 3 (6 and 9) would also work.

Colossal Gardner, ch. 13

Hypercubes. George Gamow and Martin Gardner both put a lot of effort into describing higher dimensions, since the idea of curved (or warped) space is tied to universe systems of 4 dimensions or more. While String Theory never shows up in Gardner’s book, there is one mention of M-Theory in the addendum for the Hyperspheres chapter, which speculates on 10 or 11 dimensions, most of them wrapped up in a really small space. If M-Theory ever pans out, it could relate gravity to the strong and weak forces, so there’s a reason to consider 4 dimensions and up kind of seriously. On the other hand, Gamow and Einstein argued that our universe is closed in such a way that a spaceship traveling in a straight line out away from our sun would eventually return from the opposite direction. Gamow’s discussions indicate that there might be a left-for-right mirror flip along the way, so that the pilot would be able to put a left-handed glove on his right hand (or that he might be upside-down compared to when he left).

Most of the discussions of hypersolids seem to come down to explaining what a tesseract is, and how it would look to 3D observers. If you’re not familiar with the idea, first take a point on a piece of paper. If you extend the point in one direction, you get a line segment – with 2 points, 1 line, 0 squares, 0 cubes and 0 tesseracts. If you take that line segment and extend it in a direction perpendicular to the segment, you get a square – with 4 points, 4 lines, 1 square, 0 cubes and 0 tesseracts. Extend the square in a direction normal to the plane and you get a cube – with 8 points, 12 lines, 6 squares, 1 cube and 0 tesseracts.

(Rotating tesseract from a 3D perspective, from the wiki article.)

The trick is to then extend the cube in a direction normal to the 3D space, which in effect makes 8 cubes that overlap without touching each other (there’s the original cube, 6 cubes extending from the 6 faces of the original cube, and the final cube you get when you stop pushing in the 4th direction) – with 16 points, 32 lines, 24 squares, 8 cubes and one tesseract.

You can calculate the number of faces, etc., for a hypercube of any dimension “n” by using the simple binomial (2x+1)^n. So, for a 4D cube, write out (2x + 1)(2x + 1)(2x + 1)(2x + 1):
(4x^2 + 2x + 2x + 1)*(4x^2 + 2x + 2x + 1)
(4x^2 + 4x + 1)*(4x^2 + 4x + 1)
16x^4 + 16x^3 + 4x^2 + 16x^3 + 16x^2 + 4x + 4x^2 + 4x + 1
16x^4 + 32x^3 + 24x^2 + 8x + 1
(where the x^4 coefficient gives the number of points, and the x^3 coefficient gives the number of squares, etc.)

To go to a 5D hypercube, just multiply the above equation by (2x + 1) again.

The rest of Gardner’s article is just an examination of what the hypercube would look like to us if we use 3D projections or slices. He goes on to give examples of hypercubes in art, including Dali’s Corpus Hypercubus, and Heinlein’s “-And He Built a Crooked House” short story.

(Dali’s Corpus Hypercubus, from the wiki entry.)

Gardner closes by saying that Heinlein’s idea of the house dropping out of 3D space could have a possible analogue in giant stars that undergo gravitational collapse, according to J. A. Wheeler. The density of a quasar could be great enough to curve space to the point where the mass drops out of space-time “releasing energy as it vanishes.” This might explain the enormous amount of energy emanating from quasi-stellar radio sources.

If the longest line that can fit into a unit square is the diagonal, with a length of sqrt(2), and the largest square that can fit into a unit cube has an area of 9/8 and a side of 3/4 * sqrt(2), then what’s the largest cube that can fit into a tesseract?

(All rights belong to their owners. Images used here for review purposes only.)

Colossal Gardner, ch. 7

The next article for Plane Geometry is on Penrose Tiling.

Unlike the earlier discussion of Rep-Tiles, which are periodic, Penrose Tilings are nonperiodic, and were first discovered by Roger Penrose. Periodic tilings can be made by shifting the pattern in specific directions without rotation or reflection. These are used extensively in the pictures by M. C. Escher. Nonperiodic tilings, at a minimum then, involve rotating or mirror reflecting the tiles. A very simple nonperiodic tile is the triangle. put two triangles back to back to form either a larger triangle, or a 4-sided rhomboid; the tiles themselves get rotated or flipped, but the patterns they make may themselves be periodic. Periodic tiles can be made nonperiodic by giving the edges different colors and requiring that only tiles with matching colors can be adjacent to each other.

Gardner laments that Escher died before learning about Penrose tiles, but I think M.C.’s butterflies print qualifies. Which is interesting to me, because Roger’s father, geneticist L.S. Penrose, invented the unending Penrose Staircase, which Escher depicted in his “Ascending and Descending” woodblock print. (Roger himself worked in general relativity and quantum mechanics.) In 1973, Roger found 6 tiles that are nonperiodic, with notches and tabs that forced the tiles to only fit one way. That was lowered to two triangular tiles, called “kites” and “darts”. Before making them public, he filed patents in the U.K., U.S. and Japan. You can now buy them as a game from Kadon Enterprises.

Penrose tiling, from the wiki entry. (All rights belong to their owners. Images used here for review purposes only.)

In 1993, John Conway came up with a 3D object that could be used to tile an enclosed volume. Called a “biprism”, these objects stack on each other, but in an irregular pattern. Currently referred to as a Schmitt-Conway-Danzer biprism, the cut-out pattern shown below was created for Gardner by Doris Schattschneider, co-creator of the M. C. Escher Kaleidocycles, and first female editor of Mathematics Magazine.

An example of tiling Penrose chickens (from the Martin Gardner book)

The Penrose Tiles are tied to the discovery that quasicrystals, crystals with five-fold symmetry, are possible.

Make your own Conway biprism

I tried to make 4 biprisms, and all four failed in exactly the same way. Granted, I was using very flimsy copy paper, but the final fold when I glued the arms together, always pointed at least 5 degrees off-center, making me think that maybe there’s something wrong with the original pattern. It might be better to use thicker paper stock, and then cut the sides into separate pieces and hold then together with cloth tape. Anyway, the instructions to make these are – use a dried-out pallpoint pen to score along all the lines. Then fold the arms marked “u” upward (valley folds) so that those arms pair up to make a prism on the top of the central rhomboid. And, fold the arms marked “d” downward (mountain folds) to have a matching biprism on the bottom side of the rhomboid.

You may have more fun playing with the Kadon Enterprises game sets, or with your own set of Penrose chickens.

Wednesday Answer

Solution: What is the smallest convex area in which a line segment of length 1 can be rotated 360 degrees? (A convex figure is one in which a straight line, joining any two of its points, lies entirely on the figure. Examples include circles and squares.)

Answer, an equilateral triangle of height 1. For the line segment to rotate, the sides have to be at least length 1. Of all convex figures with widths of 1, the equilateral triangle has the smallest area. Try taking a toothpick and a cardboard cutout of a triangle and check for yourself. To make it easier, glue a second toothpick in the middle of the first one to make an axle for rotating the first toothpick within the triangle cutout.

Colossal Gardner, ch. 4

We now move into Plane Geometry and the concept of Curves of Constant Width. Actually, the main content of Gardner’s article is captured in the wiki entry. His starting point is that you can use Reuleaux triangles in the place of wheels in a roller-based system, and a platform placed on top of the rollers will remain level and steady as the rollers rotate across a flat surface. So, it’s not necessary to use circular wheels if you don’t want to. (If you want pictures, go to the wiki article.)

The more interesting application of the Reuleaux triangle was in the Watts Brothers Tool Works patent for a drill and chuck system capable of drilling square holes. They also manufacture drill bits for pentagonal and hexagonal holes that have sharper corners. Although the outer edge of a curve of constant width is in contact with its bounding space at all times (as with a square for the Reuleaux triangle), the center of rotation moves around all over the place. That means that for the Watts square-hole bit, a special chuck is required to trace out the correct rotation to ensure that the hole actually comes out square. You can see the patents (filed in 1917) at Google patents: US1241175, US1241176 and US1241177.

The puzzle this time is based on the Kakeya Needle problem. What is the smallest convex area in which a line segment of length 1 can be rotated 360 degrees? (A convex figure is one in which a straight line, joining any two of its points, lies entirely on the figure. Examples include circles and squares.) Check the wiki article on Kakeya sets for an illustration of the Kakeya Needle.


Wednesday answer

Wednesday answer:
Find a copy of J. A. Lindon’s “Doppelganger“.


Colossal Gardner, ch. 3

The last chapter of the arithmetic and algebra section is on palindromes. I absolutely love the “near miss” palindrome at the front of the article, attributed to Ethel Merperson, in Son of the Giant Sea Tortoise, edited by Mary Ann Madden (Viking, 1975) –

A man, a plan, a canal – Suez!

Palindromes can take many forms, from words and sentences that can be read the same forward and backward, numbers that can be rotated, palindromic primes, and even photos of things (like a bird in flight, going from wing tip to wing tip).

Here’s a game. Start with any positive integer. Reverse it and add the two numbers together. There’s a conjecture that you’ll get a palindrome after a finite number of steps.


121  <— End



13431 <- End

There have been papers written on the existence of palindromic primes and powers. You can play with palindromic roots to get palindromic squares, such as 121^2 = 14641.

You can find many of the same language examples in the wiki article. Yreka City in California used to have the Yreka Bakery and Yrella Gallery. Then there was the former premier of Cambodia, Lon Nol. And, you can have sentences where the word order is palindromic: from J. A. Lindon – “You can cage a swallow, can’t you, but you can’t swallow a cage, can you?”

Note: Lindon was a pioneer in the recreational mathematics field of anti-magic squares. In magic squares, the rows and columns all add up to the same number. With anti-magic squares, the sums of the rows, columns and diagonals are all different. He died in 1979.

Challenge: Can you find a copy of J. A. Lindon’s “Doppelganger,” and can you write a longer palindromic poem yourself?

Comments: I love palindromes, and I had a book collection of them at one time. Back when I first had a 4-banger calculator, I’d tried to find palindromic numbers. I’ve never heard of anti-magic squares before, but they could be fun to play with.

Coconuts answer

There are actually two variants for the coconuts puzzle. In the older, easier one the monkey gets an extra coconut at the end. Versions of this one apparently go back to the middle ages. The newer version used by William, where the monkey only gets 5 coconuts, is more difficult to solve.

The Diophantine approach is to build up a polynomial as follows (for the older version of the puzzle):
N = 5A + 1
4A = 5B + 1
4B = 5C + 1
4C = 5D + 1
4D = 5E + 1
4E = 5F + 1

N is the original number of coconuts, and F is the number that each of the 5 sailors received at the end. The way to read this is, at the beginning, sailor A divided the pile “N” 5 ways, kept one pile of size “A”, and had 1 coconut left over that he gave to the monkey. Now, there is a pile of 4*A coconuts that is divided 5 ways by sailor B. He keeps one pile of size “B”, has one coconut left over for the monkey, and puts the other four B-sized piles back together again, etc.

If we turn this into one equation by normalizing A, B, C, D and E (i.e. – E = (5F+1)/4; D = (5E+1)/4 = (5F+1)*5/16; etc.) we get 1,024*N = 15,625*F + 11529.

We want the smallest positive value of N such that both N and F are integers. This is easy to do in VBScript as a simple program to run N from 1 to 10,000, but not so easy to derive by hand. (My solution is N = 15,621, F = 1,023.)

One of the interesting things about the Diophantine equations is that some of them have only one solution, others have infinite solutions, and others have none. In our case, the coconut problem has an infinite number of solutions, so we want the smallest one.

However, Gardner presents a twist approach using negative coconuts, which dates back to Norman Manning in 1912. Alternatively, you could take some of the coconuts and paint them blue to be easier to track. The reasoning goes – if you’re dividing up the coconuts into 5 equal piles 6 times, with no leftover coconut for the monkey at the end, then the smallest number that would work is 5^6 = 15625. Now, take four of those 15,625 coconuts, paint them blue and put them in the bushes. This leaves you with 15,621 coconuts, which you CAN divide into 5 piles, with one for the monkey for round 1. Put the four blue coconuts plus the other 4 piles together to make one pile of 5^5 coconuts. This can obviously be divided by 5. But, we pull the 4 blue ones and give the extra to the monkey. In Gardner’s words “This procedure – borrowing the blue coconuts only long enough to see that an even division into fifths can be made, then putting them aside again, is repeated at each division.” After the last round, the blue coconuts are left in the bushes, not belonging to anyone.

The use of blue, or negative, coconuts explains something I discovered when I ran my program. One of the solutions is for N = -4, and F = -1. Adding 5^6 (15,625) to this N gives 15,621, which is the next solution my program found, with f = 1,023. All subsequent solutions have the form N = k*5^6 – 4, and F = k*4^5 – 1 (where k runs from 1 to infinity).

The above description is for the variant of the puzzle where the monkey gets that 6th coconut in the final round of dividing the coconuts into 5 piles. For William’s version, where the last round doesn’t have the extra coconut, the Diophantine equation is:
1,024*N = 15,625*F + 8404

There’s a different general equation that can be used here, depending on whether “n”, the number of sailors, is even or odd:
# Coconuts = (1 + nk)*n^n – (n-1)  : for even n
# Coconuts = (n – 1 + nk)*n^n – (n-1) : for odd n

k is the multiplier used above to get infinite solutions as multiples of N, and for the lowest positive solution, k=0.
For n = 5 sailors,
N = # Coconuts = (1 + 0*5)*5^5 – (5-1)
N = 1*3125 – 4
N = 3,121
The last round of divvying will give 5 piles of 204 coconuts, and nothing extra for the monkey.

Finally, the last puzzle on Monday had three sailors finding the pile of coconuts. The first sailor takes 1/2 of the pile, plus half a coconut. The second sailor takes half of the remaining pile plus half a coconut. The third sailor does the same thing. Left over is exactly one coconut, which they give to the monkey. How many coconuts did they start with?
If you work backwards,
(1 + 1/2) * 2 = 3
(3 + 1/2) * 2 = 7
(7 + 1/2) * 2 = 15
Answer: 15 coconuts

Comments: Man, I never expected this chapter of the book to give me so many problems. I’d initially thought I’d spend an hour writing it up and that’d be the end of it. But, I kept making mistakes in forming the Diophantine equations for each puzzle, and both my math, and my VBScript program kept coming out wrong. I spent close to 2 days on this one.

Obviously, I’m not as good at recreational math puzzles as I’d liked to think I was.