Hello all,

Over the past few weeks, I’ve tried to find some convincing solutions to the Z32 cipher but, while the results were not uninteresting, some weaknesses remained, and they were not totally satisfactory.

This time, I feel like I have something rather convincing but, of course, I’ll let you be the judge !

As this post is a little long (but there is a map at the end…), here is the quick conclusion:

The text contained in the Z32 cipher could be “ON FEAR RD, ONE IN RADIAN EIGHT, WE ROAD” (I learned that using ROAD instead RODE is a common misspelling in English).

The place indicated in the cipher is situated in Alamo, Contra Costa County, near two elementary schools (within a distance of 0.5 mi ) and along the Donald D. Doyle Highway.

This solution is not obtained through guessing what the answer could/should be, but is based on the application of the key obtained from the Z340 cipher and only very few new/natural hypotheses are introduced.

First step: apply the Z340 key twice in a row to the Z32 cipher

My basic idea is that the key to decode the Z32 cipher can’t be a new key. If this were the case, it would be impossible to decode because the message is too short and there are too few repeated signs (unlike the Z408 and Z340 ciphers). And I can’t think the Zodiak gave us an undecodable message!

The only two keys already available are those of the Z408 and Z340 ciphers. The Z340 key is the best candidate: the Z340 signs have more similarities with those of the Z32 signs (the Z32 only has two new signs compared to the Z340) and the Zodiac might have been encouraged to reuse it because the Z340 didn’t have been solved at the time.

The most natural way to make a new key from an existing key is to apply it twice in a row. This is the method uses below :

Several points should be noted:

Expressions like “ON FEAR” on the left and ROAD, WE on the right, are clearly visible, which is rather encouraging.

The 2 new signs compared to the Z340 cipher appear in blue and are left undecoded at this stage.

There are three additional undecoded signs (represented with ?) corresponding to the "I" obtained after the first application of the key, and which give an unknown result after the second application.

Second step: find the right direction to read the signs

This is ultimately the most difficult step and the Z340 solution shows that the Zodiac can be very inventive…

The idea here is to split the Z32 into 4 block and to read the characters in the following order:

Then I assume that the ?, corresponding to the I obtained after the first application of the key, remain I after the second application. Which give (the parts to be read backwards are in parentheses) :
ON FEAR RD (ONE) (IN) RIDI(AN.) EI.IDT WE ROAD

Only remain the two new signs not included in the Z340 key: here I’m assuming they are just silent characters. It’s a bit surprising but think about it: 1/ these signs don’t belong to the Z340 key, 2/ the 1st line, which contains these two characters, has two extra characters compared to the 2nde one although we know that the Zodiac is very attached to having symmetrical ciphers. Maybe these characters were introduced just to keep a 1st line of 17 characters, as for the Z408 or Z340 ciphers ? or maybe they are just here to make things more complicated (if so, it’s successful) ?

In short, we therefore finally obtain:
ON FEAR RD (ONE) (IN) RIDI(AN) EIIDT WE ROAD

Of course, the words RIDI(AN) and EEIDT are problematic. But I’m not too surprised because we know that the Zodiac makes, from time to time, intentional or unintentional coding errors : applying a coding key twice in a row, as assumed here, further increases the risk of errors.

Reading RADIAN instead of RIDIAN seems pretty natural. Concerning EEIDT, you’d expect to read a digit: if it’s 5 letters long, starts with EI and ends with T, EIGHT is a pretty obvious solution.

The final result is therefore:
ON FEAR RD (ONE) (IN) RADI(AN) EIGHT WE ROAD

Third step : but where is it ?

The length is quite obvious: 1in on the Philips 66 Map is equivalent to 6.4mi / 10.3km.

Radian 8 is to be read as on a clock: 240°, to which must be added the 17° for the deviation from magnetic north (average value in 1970 in the San Francisco area) : the final angle is 257° measured from the geographic north.
The location indicated is in Alamo, less than 0.5mi from two elementary schools: Doris-Eaton School to the south and Alamo Elementary School to the north.

These two schools were already built in 1970 when the Zodiac cipher was sent. And these schools are located next to important roads (Donald D. Doyle Highway, Livorna Road and Stone Valley Road), which could be a good play to set in the bomb to which the Zodiac refers (in particular, one side of the highway is bordered by sloping embankments, as in Zodiac’s sketches of his explosive device).

Why I think it’s a not so bad solution?

1/ The deciphering method is not arbitrary and is based on the Z340 cipher key. It does not involve odd numerology or guessing a new key to decode what one has already decided to find there. Of course, the text contains errors but ultimately quite limited in number imo.

2 / Regarding the distance of 1in from Mt Diablo, it is relatively short compared to the size of the Philips 66 map, which is consistent with the fact that the angle was measured on site (it is difficult to correctly point a place when it’s far away). The fact that the location is situated at the intersection of major roads may also have made it easier: I can imagine that a highway could be more easily seen from the Mt Diablo summit than school buildings.

3 / Regarding the angle at 8 o’clock, it’s not a true radian, but I don’t find it so surprising. First, this is consistent with the drawing of a watch made by the Zodiac on the map accompanying the cipher and suggesting that “radians” actually correspond to an angle measured with hours. But above all, the angle at 8 o’clock is consistent with another letter of the Zodiac sent one month later (Little list letter) which points strongly in this direction:

I obviously have no certainty that this is the right solution, and it would be very pretentious; but I have the feeling that here are some interesting arguments: tell me what you think about it!