Sierpinski Triangle: A Dice Game Draws It Dot by Dot

Roll a die, jump halfway

Mark the corners of a large triangle A, B and C, and start on corner A. Roll a die: on 1 or 2, move halfway toward A; on 3 or 4, halfway toward B; on 5 or 6, halfway toward C. Mark a dot there. Roll again from the new dot.

Illustration

A hundred dots look scattered; 10,000 trace the Sierpinski triangle, with a hole in the middle, the same hole in each corner piece, and again at several smaller sizes. Wolfram MathWorld calls this game the chaos game.

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The drawing uses one fixed run of rolls. After 1,000 dots, 554 of the 729 small triangles in Figure 1 hold a dot (shared-corner dots count once); after 10,000, all 729 do. The counts for other sizes are in sierpinski-triangle-dice-game.csv (CSV; where this file comes from: source, version and license, JSON) Source data: Computed for this page: one run of the dice game from a fixed seed.

Cut, remove, repeat

Join the midpoints of a triangle's three sides to split it into four half-size triangles. Remove the middle one. Do the same to each of the three that remain, and keep going.

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After n rounds there are 3ⁿ triangles, each with a side (1/2)ⁿ of the original, covering (3/4)ⁿ of the area. Level 5 keeps 243 triangles and about 23.7% of the area. Their total edge length grows instead, by 3/2 each round, so level 10 has about 58 times the starting perimeter.

The Sierpinski triangle is what remains after infinitely many rounds: its area is zero, yet every edge drawn along the way stays in it. Each drawing here is a finite stage, and its caption says which. The triangle is named after Wacław Sierpiński. Wolfram MathWorld cites his 1915 description.

The numbers for levels 0 to 10 are in sierpinski-triangle-levels.csv (CSV; where this file comes from: source, version and license, JSON) Source data: Computed for this page from the definition of the Sierpinski triangle and exact values of every number in rows 0 to 1,023 of Pascal's triangle.

Why every dot lands on the triangle

A jump halfway toward A shrinks the whole triangle to half size, pinned at A, so the dot lands in the half-size copy at A. Jumps toward B and C do the same at their corners: these are the three copies the cutting rule keeps. So the last roll picks the half-size triangle that holds the dot, the roll before picks the quarter-size triangle inside it, and so on.

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Starting on corner A, the dot never leaves the Sierpinski triangle, because each jump maps the shape onto one of its own copies. We reran the game with a short program: all 10,000 dots lie in the triangles left at level 12. With a fair die, every sequence of six corners turns up sooner or later (with probability 1), so each of the 729 level-6 triangles gets a dot.

Start at the center instead, and no dot ever lands on the shape: the tenth dot sits in the middle hole of a triangle whose side is 1/1,024 of the original, and each jump halves that hole. Wolfram MathWorld's version of the game throws away the first few dots.

Odd numbers in Pascal's triangle

In Pascal's triangle, each number is the sum of the two above it. Shade the odd numbers: the first 4 rows draw level 2, the first 8 rows level 3, and the first 32 rows level 5.

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Only odd and even matter: odd plus odd is even, and odd plus even is odd. Row 4 is odd only at its two ends, so each end starts its own copy of rows 0 to 3, with even numbers between: three copies in all, as in the cutting rule. In row 8 the two lower copies meet, odd plus odd makes even, and row 8 is again odd only at its ends, so the step repeats at rows 16, 32 and on.

As a check, a short program computed every number in the first 1,024 rows exactly (the largest has 307 digits). The odd ones match level 10 cell by cell. There are 59,049 of them, which is 3¹⁰. Wolfram MathWorld lists the same pattern.

Three half-size copies: dimension about 1.585

Shrink a segment by half and 2 copies rebuild it; a filled triangle takes 4, and 4 = 2². The power of 2 is the dimension: 1 for the segment, 2 for the filled triangle. The Sierpinski triangle takes 3 half-size copies, so its dimension D solves 2^D = 3: D = log 3/log 2, about 1.585.

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A rectangle with sides in the ratio √2 to 1, the shape of A-series paper, halves into 2 copies at 1/√2 of the length, and (√2)² = 2, so its dimension is 2. A coastline's dimension is measured, not counted: Coastline Paradox: Seven Rulers, Seven Lengths for Britain fits one to rulers walked around a map of Britain, and it holds only across the ruler lengths used.