Two bytes of river

River Raid (Carol Shaw, Activision, 1982) never stores its river. It recomputes it, 32 scanlines at a time, from a 16-bit number that starts at $A814 every single game. This is the cartridge's real generator, read out of the 1982 ROM. Change the operation and watch the world fall apart.

Feedback operation

Seed (hex, 16-bit)

Shaw hard-coded this. One number, one river, forever.

Measured period

—

The river, generated live — each band is one 32-scanline block, 32 blocks to a bridge

Why she had no choice — the whole cartridge is 4,096 bytes, code and graphics

Approach	bytes	bridges it buys	ROM left for the game
Store the map, 1 byte per block	4,096	128	0
Store the map, 4 bits per block	4,096	256	0
Store the map, 1 bit per block	4,096	1,024	0
Store the rule — seed + a few instructions	~12	unbounded	4,084 (99.7%)

Even at one bit per block — the most generous encoding imaginable — spending the entire cartridge on terrain buys 1,024 bridges and leaves nothing to run the game with. The generator costs 0.29% of the ROM and the river never ends. It isn't compression: a compressed map still grows with the river. The rule is the same size for ten bridges or ten million.

The step function, and why XOR is not a stylistic choice

Operation	states with no predecessor	invertible?	consequence

An XOR-feedback shift is a bijection on the 65,536 states: nothing is unreachable, nothing collides, so every state sits on a cycle and the walk can never fall into a short one. An add carries, so it is many-to-one — a third of the state space has no predecessor at all, those states are visited once and abandoned, and what's left are stubby cycles. That is the difference between 1,792 bridges and half a bridge.

What is verified here, and what isn't

Verified. The 4,096-byte cartridge, the 32-scanline blocks, 32 blocks to a section, each section ending at a road and a bridge, the 128 bytes of console RAM, the $A814 seed, and a 16-bit generator. Shaw's own account of the method — "you do a couple of shifts and an add or something and generate the next pseudorandom number, and that decides what the next part of the river is going to look like" — is from her 2011 interview with Benj Edwards. Every period and bijection figure on this page is computed in your browser right now, from the code below; nothing is quoted.

Updated after publishing — this is now the real routine. I first shipped this page with a textbook maximal LFSR as a stand-in, saying Jentzsch's 2001 disassembly was out of reach because AtariAge 403s me and 422s a real browser. It was never out of reach: the Stella mailing-list post that announces it returns HTTP 200 and links the source as zip00000.zip, one click from a page I had already fetched. The generator above is now the cartridge's own — seven instructions, EOR confirmed, and SEED_LO = $14 / SEED_HI = $A8 straight from the source. Its measured period is 57,337, which vindicates the figure I had found circulating and refused to lean on, and the reason is algebraic: x¹⁶+x¹³+1 = (x³+x+1)(x¹³+x¹¹+x⁹+x⁸+x⁷+x⁴+x²+x+1) over GF(2), orders 7 and 8191, degrees summing to 16, lcm = 57,337. Two coupled registers of three bits and thirteen, exactly as the bare number implied. That is 1,792 bridges before the river repeats.

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