Tag Archives: recreational mathematics

Tri Again (Again (Again))

by in Fractals, Geometry, Mathematics, Rep-Tiles and tagged , , , , , , , , , , | Leave a comment

Like the moon, mathematics is a harsh mistress. In mathematics, as on the moon, the slightest misstep can lead to disaster — as I’ve discovered again and again. My latest discovery came when I was looking at a shape called the L-tromino, created from three squares set in an L-shape. It’s a rep-tile, because it can be tiled with four smaller copies of itself, like this:

Rep-4 L-tromino

And if it can be tiled with four copies of itself, it can also be tiled with sixteen copies of itself, like this:

Rep-16 L-tromino

My misstep came when I was trying to do to a rep-16 L-tromino what I’d already done to a rep-4 L-tromino. And what had I already done? I’d created a beautiful shape called the hourglass fractal by dividing-and-discarding sub-copies of a rep-4 L-tromino. That is, I divided the L-tromino into four sub-copies, discarded one of the sub-copies, then repeated the process with the sub-sub-copies of the sub-copies, then the sub-sub-sub-copies of the sub-sub-copies, and so on:

Creating an hourglass fractal #1

Creating an hourglass fractal #2

Creating an hourglass fractal #3

Creating an hourglass fractal #4

Creating an hourglass fractal #5

Creating an hourglass fractal #6

Creating an hourglass fractal #7

Creating an hourglass fractal #8

Creating an hourglass fractal #9

Creating an hourglass fractal #10

Creating an hourglass fractal (animated)

The hourglass fractal

Next I wanted to create an hourglass fractal from a rep-16 L-tromino, so I reasoned like this:

• If one sub-copy of four is discarded from a rep-4 L-tromino to create the hourglass fractal, that means you need 3/4 of the rep-4 L-tromino. Therefore you’ll need 3/4 * 16 = 12/16 of a rep-16 L-tromino to create an hourglass fractal.

So I set up the rep-16 L-tromino with twelve sub-copies in the right pattern and began dividing-and-discarding:

A failed attempt at an hourglass fractal #1

A failed attempt at an hourglass fractal #2

A failed attempt at an hourglass fractal #3

A failed attempt at an hourglass fractal #4

A failed attempt at an hourglass fractal #5

A failed attempt at an hourglass fractal (animated)

Whoops! What I’d failed to take into account is that the rep-16 L-tromino is actually the second stage of the rep-4 triomino, i.e. that 4 * 4 = 16. It follows, therefore, that 3/4 of the rep-4 L-tromino will actually be 9/16 = 3/4 * 3/4 of the rep-16 L-tromino. So I tried again, setting up a rep-16 L-tromino with nine sub-copies, then dividing-and-discarding:

A third attempt at an hourglass fractal #1

A third attempt at an hourglass fractal #2

A third attempt at an hourglass fractal #3

A third attempt at an hourglass fractal #4

A third attempt at an hourglass fractal #5

A third attempt at an hourglass fractal #6

A third attempt at an hourglass fractal (animated)

Previously (and passionately) pre-posted:

• Tri Again
• Tri Again (Again)

Count Amounts

by in Base Behavior, Integer Sequences, Mathematics and tagged , , , , , , , , , , , , | Leave a comment

One of my favourite integer sequences is what I call the digit-line. You create it by taking this very familiar integer sequence:

• 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20…

And turning it into this one:

• 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 0, 1, 1, 1, 2, 1, 3, 1, 4, 1, 5, 1, 6, 1, 7, 1, 8, 1, 9, 2, 0… (A033307 in the Online Encyclopedia of Integer Sequences)

You simply chop all numbers into single digits. What could be simpler? Well, creating the digit-line couldn’t be simpler, but it is in fact a very complex object. There are hidden depths in its patterns, as even a brief look will uncover. For example, you can try counting the digits as they appear one-by-one in the line and seeing whether the digit-counts compare. Do the 1s of the digit-line always outnumber the 0s, as you might expect? Yes, they do (unless you start the digit-line 0, 1, 2, 3…). But do the 2s always outnumber the 0s? No: at position 2, there’s a 2, and at position 11 there’s a 0. So that’s one 2 and one 0. Does it happen again? Yes, it happens again at the 222nd digit of the digit-line, as below:

1, 2count=1, 3, 4, 5, 6, 7, 8, 9, 1, 0count=1, 1, 1, 1, 22, 1, 3, 1, 4, 1, 5, 1, 6, 1, 7, 1, 8, 1, 9, 23, 02, 24, 1, 25, 26, 27, 3, 28, 4, 29, 5, 210, 6, 211, 7, 212, 8, 213, 9, 3, 03, 3, 1, 3, 214, 3, 3, 3, 4, 3, 5, 3, 6, 3, 7, 3, 8, 3, 9, 4, 04, 4, 1, 4, 215, 4, 3, 4, 4, 4, 5, 4, 6, 4, 7, 4, 8, 4, 9, 5, 05, 5, 1, 5, 216, 5, 3, 5,4, 5, 5, 5, 6, 5, 7, 5, 8, 5, 9, 6, 06, 6, 1, 6, 217, 6, 3, 6, 4, 6, 5, 6, 6, 6, 7, 6, 8, 6, 9, 7, 07, 7, 1, 7, 218, 7, 3, 7, 4, 7, 5, 7, 6, 7, 7, 7, 8, 7, 9, 8, 08, 8, 1, 8, 219, 8, 3, 8, 4, 8, 5, 8, 6, 8, 7, 8, 8, 8, 9, 9, 09, 9, 1, 9, 220, 9, 3, 9, 4, 9, 5, 9, 6, 9, 7, 9, 8, 9, 9, 1, 010, 011, 1, 012, 1, 1, 013, 221, 1, 014, 3, 1, 015, 4, 1, 016, 5, 1, 017, 6, 1, 018, 7, 1, 019, 8, 1, 020, 9, 1, 1, 021…

So count(2) = count(0) = 1 at digit 11 of the digit-line in the 0 of what was originally 10. And count(2) = count(0) = 21 @ digit 222 in the 0 of what was originally 110. Is a pattern starting to emerge? Yes, it is. Here are the first few points at which the count(2) = count(0) in the digit-line of base 10:

1 @ 11 in 10
21 @ 222 in 110
321 @ 3333 in 1110
4321 @ 44444 in 11110
54321 @ 555555 in 111110
654321 @ 6666666 in 1111110
7654321 @ 77777777 in 11111110
87654321 @ 888888888 in 111111110
987654321 @ 9999999999 in 1111111110
10987654321 @ 111111111110 in 11111111110
120987654321 @ 1222222222221 in 111111111110
[...]

The count(2) = count(0) = 321 at position 3333 in the digit-line, and 4321 at position 44444, and 54321 at position 555555, and so on. I don’t understand why these patterns occur, but you can predict the count-and-position of 2s and 0s easily until position 9999999999, after which things become more complicated. Related patterns for 2 and 0 occur in all other bases except binary (which doesn’t have a 2 digit). Here’s base 6:

1 @ 11 in 10 (1 @ 7 in 6)
21 @ 222 in 110 (13 @ 86 in 42)
321 @ 3333 in 1110 (121 @ 777 in 258)
4321 @ 44444 in 11110 (985 @ 6220 in 1554)
54321 @ 555555 in 111110 (7465 @ 46655 in 9330)
1054321 @ 11111110 in 1111110 (54121 @ 335922 in 55986)
12054321 @ 122222221 in 11111110 (380713 @ 2351461 in 335922)
132054321 @ 1333333332 in 111111110 (2620201 @ 16124312 in 2015538)
1432054321 @ 14444444443 in 1111111110 (17736745 @ 108839115 in 12093234)
15432054321 @ 155555555554 in 11111111110 (118513705 @ 725594110 in 72559410)
205432054321 @ 2111111111105 in 111111111110 (783641641 @ 4788921137 in 435356466)
2205432054321 @ 22222222222220 in 1111111111110 (5137206313 @ 31345665636 in 2612138802)

And what about comparing other pairs of digits? In fact, the count of all digits except 0 matches infinitely often. To write the numbers 1..9 takes one of each digit (except 0). To write the numbers 1 to 99 takes twenty of each digit (except 0). Here’s the proof:

11, 21, 31, 41, 51, 61, 71, 81, 91, 12, 01, 13, 14, 15, 22, 16, 32, 17, 42, 18, 52, 19, 62, 110, 72, 111, 82, 112, 92, 23, 02, 24, 113, 25, 26, 27, 33, 28, 43, 29, 53, 210, 63, 211, 73, 212, 83, 213, 93, 34, 03, 35, 114, 36, 214, 37, 38, 39, 44, 310, 54, 311, 64, 312, 74, 313, 84, 314, 94, 45, 04, 46, 115, 47, 215, 48, 315, 49, 410, 411, 55, 412, 65, 413, 75, 414, 85, 415, 95, 56, 05, 57, 116, 58, 216, 59, 316, 510, 416, 511, 512, 513, 66, 514, 76, 515, 86, 516, 96, 67, 06, 68, 117, 69, 217, 610, 317, 6
11, 417, 612, 517, 613, 614, 615, 77, 616, 87, 617, 97, 78, 07, 79, 118, 710, 218, 711, 318, 712, 418, 713, 518, 714, 618, 715, 716, 717, 88, 718, 98, 89, 08, 810, 119, 811, 219, 812, 319, 813, 419, 814, 519, 815, 619, 816, 719, 817, 818, 819, 99, 910, 09, 911, 120, 912, 220, 913, 320, 914, 420, 915, 520, 916, 620, 917, 720, 918, 820, 919, 920…

And what about writing 1..999, 1..9999, and so on? If you think about it, for every pair of non-zero digits, d1 and d2, all numbers containing one digit can be matched with a number containing the other. 100 → 200, 111 → 222, 314 → 324, 561189571 → 562289572, and so on. So in counting 1..999, 1..9999, 1..99999, you use the same number of non-zero digits. And once again a pattern emerges:

count(0) = 0; count(1) = 1; count(2) = 1; count(3) = 1; count(4) = 1; count(5) = 1; count(6) = 1; count(7) = 1; count(8) = 1; count(9) = 1 (writing 1..9)
count(0) = 9; count(1) = 20; count(2) = 20; count(3) = 20; count(4) = 20; count(5) = 20; count(6) = 20; count(7) = 20; count(8) = 20; count(9) = 20 (writing 1..99)
0: 189; 1: 300; 2: 300; 3: 300; 4: 300; 5: 300; 6: 300; 7: 300; 8: 300; 9: 300 (writing 1..999)
0: 2889; 1: 4000; 2: 4000; 3: 4000; 4: 4000; 5: 4000; 6: 4000; 7: 4000; 8: 4000; 9: 4000 (writing 1..9999)
0: 38889; 1: 50000; 2: 50000; 3: 50000; 4: 50000; 5: 50000; 6: 50000; 7: 50000; 8: 50000; 9: 50000 (writing 1..99999)
0: 488889; 1: 600000; 2: 600000; 3: 600000; 4: 600000; 5: 600000; 6: 600000; 7: 600000; 8: 600000; 9: 600000 (writing 1..999999)
0: 5888889; 1: 7000000; 2: 7000000; 3: 7000000; 4: 7000000; 5: 7000000; 6: 7000000; 7: 7000000; 8: 7000000; 9: 7000000 (writing 1..9999999)
[...]

And here’s base 6 again:

0: 0; 1: 1; 2: 1; 3: 1; 4: 1; 5: 1 (writing 1..5)
0: 5; 1: 20; 2: 20; 3: 20; 4: 20; 5: 20 (writing 1..55 in base 6)
0: 145; 1: 300; 2: 300; 3: 300; 4: 300; 5: 300 (writing 1..555)
0: 2445; 1: 4000; 2: 4000; 3: 4000; 4: 4000; 5: 4000 (writing 1..5555)
0: 34445; 1: 50000; 2: 50000; 3: 50000; 4: 50000; 5: 50000 (writing 1..55555)
0: 444445; 1: 1000000; 2: 1000000; 3: 1000000; 4: 1000000; 5: 1000000 (writing 1..555555)
0: 5444445; 1: 11000000; 2: 11000000; 3: 11000000; 4: 11000000; 5: 11000000 (writing 1..5555555)
0: 104444445; 1: 120000000; 2: 120000000; 3: 120000000; 4: 120000000; 5: 120000000 (writing 1..55555555)
0: 1144444445; 1: 1300000000; 2: 1300000000; 3: 1300000000; 4: 1300000000; 5: 1300000000 (writing 1..555555555)

At the Mountings of Mathness

by in Art, Fractals, H.P. Lovecraft, Mathematics and tagged , , , , , , , , , , , , , | Leave a comment

Mounting n. a backing or setting on which a photograph, work of art, gem, etc. is set for display. — Oxford English Dictionary

Viewer’s advisory: If you are sensitive to flashing or flickering images, you should be careful when you look at the final fourth and fifth of the animated gifs below.

H.P. Lovecraft in some Mountings of Mathness

Paradoxical Puzzle Pair

by in Mathematics, Puzzles and tagged , , , , , , , , , , , , , , | 2 Comments

Two interesting puzzles, one of which looks hard and is in fact easy, while the other looks easy and is in fact hard.

1. Three Cards

The values attached to a deck of bridge cards start with the Two of Clubs as lowest, with Diamonds, Hearts and Ace of Spades as highest.

If you draw three cards at random from the deck, what is the probability that they will be drawn in order of increasing value? (Answer 1)

2. The Hungry Hunter

A hunter, having run out of food, met two shepherds. One of the shepherd had three loaves of bread and the other had five loaves. When the hunter asked for food, the shepherds agreed to divide the eight identical loaves equally between the three of them. The hunter thanked them and gave them $8. How should the shepherds divide the money? (Answer 2)

• Puzzles and answers from Erwin Brecher’s How Do You Survive a Duel? And Other Mathematical Diversions, Puzzles and Brainteasers (Carlton Books 2018)

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Answer #1: The puzzle sounds far more complicated than it is. The deck of cards is a red herring. The question reduces to this: Take three cards, say 2, 3 and 4 of clubs, facedown. What is the probability of turning them over in the order 2, 3, 4? There are six possible ways of arranging three cards. Therefore the probability is one-sixth.

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Answer #2: It would be wrong to split the money into $3 and $5. Each of the three ended up with 2⅔ loaves. In other words, the first shepherd parted with ⅓ of a loaf and the other shepherd with 2⅓ or 7/3 loaves. The first shepherd should therefore get $1 and the second shepherd $7.

Agnathous Analysis

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In Mandibular Metamorphosis, I looked at two distinct fractals and how you could turn one into the other in one smooth sweep. The Sierpiński triangle was one of the fractals:

Sierpiński triangle

The T-square fractal was the other:

T-square fractal (or part thereof)

And here they are turning into each other:

Sierpiński ↔ T-square (anim)
(Open in new window if distorted)

But what exactly is going on? To answer that, you need to see how the two fractals are created. Here are the stages for one way of constructing the Sierpiński triangle:

Sierpiński triangle #1

Sierpiński triangle #2

Sierpiński triangle #3

Sierpiński triangle #4

Sierpiński triangle #5

Sierpiński triangle #6

Sierpiński triangle #7

Sierpiński triangle #8

Sierpiński triangle #9

When you take away all the construction lines, you’re left with a simple Sierpiński triangle:

Constructing a Sierpiński triangle (anim)

Now here’s the construction of a T-square fractal:

T-square fractal #1

T-square fractal #2

T-square fractal #3

T-square fractal #4

T-square fractal #5

T-square fractal #6

T-square fractal #7

T-square fractal #8

T-square fractal #9

Take away the construction lines and you’re left with a simple T-square fractal:

T-square fractal

Constructing a T-square fractal (anim)

And now it’s easy to see how one turns into the other:

Sierpiński → T-square #1

Sierpiński → T-square #2

Sierpiński → T-square #3

Sierpiński → T-square #4

Sierpiński → T-square #5

Sierpiński → T-square #6

Sierpiński → T-square #7

Sierpiński → T-square #8

Sierpiński → T-square #9

Sierpiński → T-square #10

Sierpiński → T-square #11

Sierpiński → T-square #12

Sierpiński → T-square #13

Sierpiński ↔ T-square (anim)
(Open in new window if distorted)

Post-Performative Post-Scriptum

Mandibular Metamorphosis also looked at a third fractal, the mandibles or jaws fractal. Because I haven’t included the jaws fractal in this analysis, the analysis is therefore agnathous, from Ancient Greek ἀ-, a-, “without”, + γνάθ-, gnath-, “jaw”.

Mandibular Metamorphosis

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Here’s the famous Sierpiński triangle:

Sierpiński triangle

And here’s the less famous T-square fractal:

T-square fractal (or part of it, at least)

How do you get from one to the other? Very easily, as it happens:

From Sierpiński triangle to T-square (and back again) (animated)
(Open in new window if distorted)

Now, here are the Sierpiński triangle, the T-square fractal and what I call the mandibles or jaws fractal:

Sierpiński triangle

T-square fractal

Mandibles / Jaws fractal

How do you cycle between them? Again, very easily:

From Sierpiński triangle to T-square to Mandibles (and back again) (animated)
(Open in new window if distorted)

Elsewhere other-accessible…

• Agnathous Analysis — a closer look at these shapes

FractAlphic Frolix

by in A.E. Housman, Artificial alphabets, Artificial scripts, Linguistics, Poetry and tagged , , , , , , , , , , , , , , , , , , , , , , , , , | Leave a comment

A fractal is a shape that contains smaller (and smaller) versions of itself, like this:

The hourglass fractal

Fractals also occur in nature. For example, part of a tree looks like the tree as whole. Part of a cloud or a lung looks like the cloud or lung as a whole. So trees, clouds and lungs are fractals. The letters of an alphabet don’t usually look like that, but I decided to create a fractal alphabet — or fractalphabet — that does.

The fractalphabet starts with this minimal standard Roman alphabet in upper case, where each letter is created by filling selected squares in a 3×3 grid:

The above is stage 1 of the fractalphabet, when it isn’t actually a fractal alphabet at all. But if each filled square of the letter “A”, say, is replaced by the letter itself, the “A” turns into a fractal, like this:

Fractal A (animated)

Here’s the whole alphabet being turned into fractals:

Full fractalphabet (black-and-white)

Full fractalphabet (color)

Full fractalphabet (b&w animated)

Full fractalphabet (color animated)

Now take a full word like “THE”:

You can turn each letter into a fractal using smaller copies of itself:

Fractal THE (b&w animated)

Fractal THE (color animated)

But you can also create a fractal from “THE” by compressing the “H” into the “T”, then the “E” into the “H”, like this:

Compressed THE (animated)

The compressed “THE” has a unique appearance and is both a letter and a word. Now try a complete sentence, “THE CAT BIT THE RAT”. This is the sentence in stage 1 of the fractalphabet:

And stage 2:

And further stages:

Fractal CAT (b&w animated)

Fractal CAT (color animated)

But, as we saw with “THE” above, that’s not the only fractal you can create from “THE CAT BIT THE RAT”. Here’s what I call a 2-compression of the sentence, where every second letter has been compressed into the letter that precedes it:

THE CAT BIT THE RAT (2-comp color)

THE CAT BIT THE RAT (2-comp b&w)

And here’s a 3-compression of the sentence, where every third letter has been compressed into every second letter, and every second-and-third letter has been compressed into the preceding letter:

THE CAT BIT THE RAT (3-comp color)

THE CAT BIT THE RAT (3-comp b&w)

As you can see above, each word of the original sentence is now a unique single letter of the fractalphabet. Theoretically, there’s no limit to the compression: you could fit every word of a book in the standard Roman alphabet into a single letter of the fractalphabet. Or you could fit an entire book into a single letter of the fractalphabet (with additional symbols for punctuation, which I haven’t bothered with here).

To see what the fractalphabeting of a longer text in the standard Roman alphabet might look like, take the first verse of a poem by A.E. Housman:

On Wenlock Edge the wood’s in trouble;
His forest fleece the Wrekin heaves;
The gale it plies the saplings double,
And thick on Severn snow the leaves. (“Poem XXXI” of A Shropshire Lad, 1896)

The first line looks like this in stage 1 of the fractalphabet:

Here’s stage 2 of the standard fractalphabet, where each letter is divided into smaller copies of itself:

And here’s stage 3 of the standard fractalphabet:

Now examine a colour version of the first line in stage 1 of the fractalphabet:

As with “THE” above, let’s try compressing each second letter into the letter that precedes it:

And here’s a 3-comp of the first line:

Finally, here’s the full first verse of Housman’s poem in 2-comp and 3-comp forms:

On Wenlock Edge the wood’s in trouble;
His forest fleece the Wrekin heaves;
The gale it plies the saplings double,
And thick on Severn snow the leaves. (“Poem XXXI of A Shropshire Lad, 1896)

“On Wenlock Edge” (2-comp)

“On Wenlock Edge” (3-comp)

Appendix

This is a possible lower-case version of the fractalphabet:

Controlled Chaos

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The chaos game is a simple mathematical technique for creating fractals. Suppose a point jumps over and over again 1/2 of the distance towards a randomly chosen vertex of a triangle. This shape appears, the so-called Sierpiński triangle:

Sierpiński triangle created by the chaos game

But the jumps don’t have to be random: you can use an array to find every possible combination of jumps and so create a more even image. I call this controlled chaos. However, if you try the chaos game (controlled or otherwise) with a square, no fractal appears unless you restrict the vertex chosen in some way. For example, if the point can’t jump towards the same vertex twice or more in a row, this fractal appears:

Ban on jumping towards previously chosen vertex, i.e. v + 0

And if the point can’t jump towards the vertex one place clockwise of the previously chosen vertex, this fractal appears:

Ban on v + 1

If the point can’t jump towards the vertex two places clockwise of the previously chosen vertex, this fractal appears:

Ban on v + 2

If the point can’t jump towards the vertex three places clockwise, or one place anticlockwise, of the previously chosen vertex, this fractal appears (compare v + 1 above):

Ban on v + 3

You can also ban vertices based on how close the point is to them at any given moment. Suppose that the point can’t jump towards the nearest vertex, which means that it must choose to jump towards either the 2nd-nearest, 3rd-nearest or 4th-nearest vertex. A fractal we’ve already seen appears:

Must jump towards vertex at distance 2, 3 or 4

In effect, not jumping towards the nearest vertex means not jumping towards a vertex twice or more in a row. Another familiar fractal appears if the point can’t jump towards the most distant vertex:

d = 1,2,3

But new fractals also appear when the jumps are determined by distance:

d = 1,2,4

d = 1,3,4

And you can add more targets for the jumping point midway between the vertices of the square:

d = 1,2,8

d = 1,4,6

d = 1,6,8

d = 1,7,8

d = 2,3,6

d = 2,3,8

d = 2,4,8

d = 2,5,6

And what if you choose the next vertex by incrementing the previously chosen vertex? Suppose the initial vertex is 1 and the possible increments are 1, 2 and 2. This new fractal appears:

increment = 1,2,2 (for example, 1 + 1 = 2, 2 + 2 = 4, 4 + 2 = 6, and 6 is adjusted thus: 6 – 4 = 2)

And with this set of increments, it’s déjà vu all over again:

i = 2,2,3

And again:

i = 2,3,2

With more possible increments, familiar fractals appear in unfamiliar ways:

i = 1,3,2,3

i = 1,3,3,2

i = 1,4,3,3

i = 2,1,2,2

i = 2,1,3,4

i = 2,2,3,4

i = 3,1,1,2

Now try increments with midpoints on the sides:

v = 4 + midpoints, i = 1,2,4

As we saw above, this incremental fractal can also be created from a square with four vertices and no midpoints:

i = 1,3,3; initial vertex = 1

But the fractal changes when the initial vertex is set to 2, i.e. to one of the midpoints:

i = 1,3,3; initial vertex = 2

And here are more inc-fractals with midpoints:

i = 1,4,2 (cf. inc-fractal 1,2,4 above)

i = 1,4,8

i = 2,6,3

i = 3,2,6

i = 4,7,8

i = 1,2,3,5

i = 1,4,5,4

i = 6,2,4,1

i = 7,6,2,2

i = 7,8,2,4

i = 7,8,4,2

Allus Pour, Horic

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*As a rotating animated gif (optimized at ezGIF).

Performativizing Paronomasticity

The title of this incendiary intervention is a paronomasia on Shakespeare’s “Alas, poor Yorick!” (Hamlet, Act 5, scene 1). “Allus” is a northern form of “always”, “pour” has its standard meaning, and “Horic” is from the Greek ὡρῐκός, hōrikos, which strictly speaking means “in one’s prime, blooming”. However, it could also be interpreted as meaning “hourly”. So the paronomasia means “Always pour, O Hourly One!” (i.e. hourglass).

Squooh You

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Here’s an interesting little puzzle:

Winnie-the-Pooh and Piglet set out to visit one another. They leave their houses at the same time and walk along the same road. But Piglet is absorbed in counting the birds overhead, and Winnie-the-Pooh is composing a new “hum,” so they pass one another without noticing. One minute after the meeting, Winnie-the-Pooh is at Piglet’s house, and 4 minutes after the meeting Piglet is at Winnie-the-Pooh’s. How long has each of them walked? — “A puzzle by S. Sefibekov” viâ Futility Closet

If you’re good at maths, you should see the answer in an intuitive instant. I’m not good at maths, so it took me much longer, because I didn’t understand what was going on. But I can explain the answer like this. Pooh is obviously walking faster than Piglet. Therefore, Pooh and Piglet can’t have met after one minute, because that would mean Pooh takes one minute to walk the distance walked by Piglet in one minute.

So let’s suppose Pooh and Piglet met after two minutes. If Pooh takes one minute to walk the distance walked by Piglet in two minutes, then Pooh is walking twice as fast as Piglet. Does that work? Yes, because Piglet walks Pooh’s distance in four minutes, which is twice as long as Pooh took. Therefore Piglet is walking twice as slowly as Pooh. It’s symmetrical and we can conclude that they did indeed meet after two minutes. Pooh then walks another minute, for three minutes in total, and Piglet walks another four minutes, for six minutes in total.

But guessing is not a good way to find the answer to the puzzle. Let’s try to reason it through properly. Suppose that Pooh and Piglet meet after one unit of time, during which Piglet has walked one unit of distance and Pooh has walked x units of distance, where x > 1. In other words, Pooh is walking x times faster than Piglet. The distances they walk before meeting are therefore in the ratio:

1 : x

Next, note that Pooh will cover the distance Piglet has already walked in 1 unit / x = 1 minute, while Piglet covers the distance Pooh has already walked in x / 1 = 4 minutes. The times they take are therefore in the ratio:

1 / x : x → 1 : x^2 → 1 : 4

And if 1 : x^2 is 1 : 4 (the ratio of the minutes they walk after meeting), then 1 : x (the ratio of the distances they walk before meeting) = 1 : √(x^2) = 1 : √4 = 1 : 2. Pooh is therefore walking 2x faster than Piglet and Piglet is walking 2x slower than Pooh. If Pooh covers Piglet’s distance in 1 minute, Piglet must have taken 2 minutes to walk that distance. And if Piglet covers Pooh’s distance in 4 minutes, Pooh must have taken 2 minutes to walk that distance.

Therefore, when they meet, each of them has been walking for 2 minutes. Pooh therefore walks 2 + 1 = 3 minutes in total and Piglet walks 2 + 4 = 6 minutes in total.

The result can be generalized for all relative speeds. Suppose that Pooh and Piglet meet after m1 minutes and that Pooh then takes m2 minutes to walk the distance Piglet walked in m1 minutes, while Piglet takes m3 minutes to walk the distance Pooh walked in m1 minutes. The time they walk before they meet, m1 minutes, is therefore supplied by this simple equation:

m1 = √(m3 / m2)

And you can call √(m3 / m2), the square root of m3 / m2, the squooh function:

m1 = √(m3 / m2) = squooh(m2,m3)

Now suppose the distance between Pooh’s and Piglet’s houses houses is 12 units of distance and that Piglet always walks at 1 unit a minute. If Pooh walks at the same speed as Piglet, i.e. 1 unit a minute, then:

After they meet, Pooh walks 6 more min = m2, Piglet walks 6 more min = m3.

How long do they walk before they meet?

m1 = m3 / m2 = 1, √1 * 6 = 6

They meet after 6 min.

Now suppose that after they meet, Pooh walks 2 more min, Piglet walks 8 more min.

Therefore, m3 / m2 = 4, √4 * 2 = 2 * 2 = 4 = m1

Therefore they walk for 4 min before they meet and Pooh walks 2x faster than Piglet.

After they meet, Pooh walks 1 more min, Piglet walks 9 more min (m3 / m2 = 9, √9 * 1 = 3)

Therefore they walk for 3 min before they meet and Pooh walks 3x faster than Piglet.

After they meet, Pooh walks 0·6 more min, Piglet walks 9·6 more min (m3 / m2 = 16, √16 * 0·6 = 4 * 0·6 = 2·4)

Therefore they walk for 2·4 min before they meet and Pooh walks 4x faster than Piglet:

After they meet, Pooh walks 0·4 more min, Piglet walks 10 more min (m3 / m2 = 25, √25 * 0·4 = 5 * 0·4 = 2)

Therefore they walk for 2 min before they meet and Pooh walks 5x faster than Piglet.

And so on.