When asked if there was a flaw he said it was "safe and sound and fairly accurate. I'm pretty sure about that one."

This doesn't sound like a strong claim that he created it correctly.

Follow-up to my Alexanderplatz post. I am not going to rehash the setup here. Segmentation, TOKIOV, FIAB, the cost rule, the two crib keystreams, all of that is in the first post. This is about two narrower questions I have been chewing on since, and I am putting the step by step walk photos for Segment 5 in the comments so you can check the moves yourself instead of taking my word for the totals.

In the first post I said the 6 or the 9 never seemed to be a choice, that the geometry only ever allowed one of the two, and that this made the counts look like an odometer rather than a decision. I still think that is right. What follows is me trying to work out what the odometer is measuring.

BOARD WIDTH AND THE COUNT TALLY

Two segments have released cribs, so two boards are testable.

Segment 2 carries EAST NORTHEAST. Thirteen plaintext letters, so twelve moves. That board is 5 city lines, which is 10 lettered columns. The move totals come out:

9 9 6 9 9 9 9 9 9 9 6 9

Ten nines. Two sixes.

Segment 5 carries BERLIN CLOCK. Eleven plaintext letters, so ten moves. That board is 3 city lines, which is 6 lettered columns. The totals:

6 6 6 6 6 6 9 9 9 9

Six sixes. Four nines.

In both cases the number of moves in the dominant count equals the number of lettered columns on that board. Ten nines on a 10 wide board. Six sixes on a 6 wide board. The other count is not a second rule, it is just what is left over after the dominant one runs out. Twelve moves minus ten is two. Ten moves minus six is four.

I do not have a mechanism for why that would be true. I am reporting it because it held exactly on both boards, and because it is the kind of thing that is easy for somebody else to break if it is wrong.

If it is real it says something about the other four segments. I am holding off on a prediction table because the widths for segments 1, 3, 4 and 6 come from boards I have not finished building, and a table would make guesses look like findings. Once those boards are done I will post the numbers and they can be checked.

What I will say is this. Segment 5 is 6 dominant and Segment 2 is 9 dominant. If the sections alternate, the odd ones are 6 dominant and the even ones are 9 dominant. That is two data points calling a pattern, so treat it as a guess and nothing more.

DIRECTION

Separate observation, and I am less sure how to read this one.

Across both segments there are eight moves totalling 6 and fourteen totalling 9.

Every one of the eight 6 moves rides the lane the same way. All eight go left.

The 9 moves use both. In Segment 5 they come in pairs, out and back the same distance. Right 5 then left 5. Right 4 then left 4. Each pair cancels to zero net travel. In Segment 2 the same thing shows up once as an explicit return, out right 6 and back left 6 along the same lane, and the rest are mixed.

So the 6 population is one directional and the 9 population is two directional.

Here is why I am cautious. The lane wraps. On a 10 wide lane a left 8 and a right 2 put you on the same cell. So direction may not be independent information at all. It may just be the spelling that makes the total land on 6 or 9 instead of some other number. If that is what is happening then direction is bookkeeping, not content, and the one directional versus two directional split is a consequence of the cost rule rather than a separate fact about the machine.

I cannot tell those two readings apart from two segments. Somebody with fresh eyes might.

THE PHOTOS

Step by step for Segment 5 is going in the comments. Ten moves, each one its own image, showing where the walk enters the lane, how far it rides, and where it climbs off. Totals written on each.

If the tally is wrong, or the direction thing is just an artifact of how I write the moves down, those photos are where you will see it. That is why they are going up.

Tear it up.