The Binary Engine

Forget the numbers. Watch the bits.

Every positive integer is a string of binary digits: 27 = 11011. The Collatz map does two very different things to these bits:

• Even step (÷2): Shift right. The rightmost bit falls off. One bit destroyed.
• Odd step (×3+1): Multiply by 3 and add 1. The carry chain ripples through the bits. Bit length grows by at most 1.

Step through it yourself:

Start:

Step 0start275 bits

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Starting value: 27 in binary. Click Step → to apply the Collatz rule.

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Watch the bit-length counter. Even steps always shrink it. Odd steps can grow it by 1 — but the next step after an odd number is always even (since 3n+1 is even when n is odd). So the worst case is: grow by 1, then shrink by 1. Net: zero.

But it's usually better than zero. Try 27 and watch: the bit-length starts at 5, climbs to ~14 at the peak, then steadily falls to 1. The bits are being consumed.

The dropping sets

Not all numbers drop at the same rate. Some take 3 steps (fast), others take 13 steps (slow), others even more. Numbers that take the same number of steps to drop form a dropping set.

Explore the patterns:

From: Show: 100 numbers

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Set 1 (50.0%) Set 3 (25.0%) Set 6 (7.0%) Set 8 (6.0%) Set 11 (3.0%) Set 13 (3.0%) Set 73 (1.0%) Set 83 (1.0%) Set 88 (2.0%) Set 91 (1.0%) Set 96 (1.0%)

Click any number to see its dropping set properties.

Notice something remarkable: numbers in the same dropping set form arithmetic progressions. All of Set₃ (green) are exactly the numbers ≡ 1 (mod 4). All of Set₆ (blue) are ≡ 3 (mod 16). This isn't a coincidence — it's the affine orbit structure: within each set, the destination is an exact linear function of the starting value.

The contraction ratio

Each dropping set has a contraction ratio $3^s / 2^{k-s}$ where $s$ is the number of odd steps and $k$ is the total steps:

Set	Steps (k)	Odd steps (s)	Ratio	Speed
Set₃	3	1	3/4 = 0.75	Fast
Set₆	6	2	9/16 = 0.56	Fast
Set₈	8	3	27/32 = 0.84	Moderate
Set₁₃	13	5	243/256 = 0.95	Slow

Set₁₃ barely shrinks — it contracts by only 5%. Could an orbit get stuck visiting slow sets like this?

The bit destruction identity

Every ratio is less than 1. Every single one. This is because $\log_2 3$ is irrational — there's no way for $3^s$ to exactly equal $2^{k-s}$. The bits destroyed per drop are:

$$\beta(s) = \lceil s \cdot \log_2 3 \rceil - s \cdot \log_2 3 > 0$$

This is always positive. No drop is ever lossless. But some are very close to zero (Set₁₃ has $\beta = 0.075$). The slow sets correspond to the best rational approximations of $\log_2 3$ — the convergents of its continued fraction.

By Roth's theorem from Diophantine approximation: $\beta(s) > c/s$ for some constant $c$. The approximations can't get too good, so the slow sets can't get too slow.

What we know so far

✅ Every drop destroys $\beta > 0$ bits — proved ✅ Average destruction: 0.45 bits per drop — computed ✅ No set is "too slow" — Roth's theorem

But two questions remain:

1. Can the orbit loop (visit the same values forever)?
2. Can the orbit systematically dodge the fast drops?

Next: we eliminate loops.

← The PuzzleNext: No Loops →