Hello Everyone,

I would like to propose a few ideas related to the structure of the poem. This is preliminary, but I believe that some of this proposal has not been voiced before, and it seems non-coincidental to me.

To begin, a theory regarding our oddball letters:
We have “twelpve” and “tweleve” instead of “twelve”, “thru” as a possible deliberate substitution for “through”, “on” instead of “one”, and “faeries”, which is one of several spellings used for that word. Now, think back to “Repositorys”, willfully misspelled to make an anagram. If I recall correctly, MS later even mentioned the misspelling as a potential tip-off for an anagram. All of these oddball letters are transparent in a way – meaning that their presence or absence does not interfere with our understanding of the poem. I’m thinking at this point that we should see them as tip-offs for one of the following three situations:
1. We are supposed to count letters to derive something, and he needed to add or subtract in a transparent way in order to make the count right.
2. We are supposed to assign values to letters (e=5…), and he needed to add or subtract in a transparent way to make the final counts right.
3. We are supposed to anagram something. Each line, each group of 5 words (see below), or something.

Next, consider that the consensus poem has 130 words. This strikes me as interesting in two ways:

First, 130 is nicely divisible by 13. We probably have 13 separate puzzles to solve, and perhaps if the poem were divided into sets of 10 words, each set would be then used for something token-specific. For example, the first ten words are “HIDDEN WELL NO EYE CAN SEE BELOW SKY AND ABOVE”, and perhaps a function or derivative of this string should mean something to us in our attack on a particular puzzle. Assigning the letter counts to each word gives a string of 6423335335. Mean anything? I don’t know yet. One could also assign letter values to each letter and see what we get. I haven’t started that yet, and I also have not tried to anagram it yet. I’m not trying to put up an answer at this point, but rather an idea for how it might work. Any thoughts?

Second, 130 is nicely divisible by 26. We have done lots of thinking about inserting the alphabet (sans Q, usually) in a 5x5 grid, and then doing something with it. What if we broke the poem into 26 strings of 5 words, excluded the 17th (equating to Q), derived numbers from the strings in some fashion, and then plugged them into a 5x5 grid? In contrast to the previous paragraph, I did try this one. Using the first quintet, I assigned the count values of 64233 and added them up to get 18. Do the “reduce by nines” conversion to a single digit (8+1) gives a value of 9. The second quintet is 3+5+3+3+5=19, 1+9=10, and 1+0=1. Do this for the whole thing, and you get what’s below…

a. HIDDEN WELL NO EYE CAN…….............6 4 2 3 3....18....9
b. SEE BELOW SKY AND ABOVE……...........3 5 3 3 5….19….1
c. EARTH BE TREASURES TWELPVE FOR……5 2 9 7 3….26….8
d. YOU TO KEEP AROUND THE…….............3 2 4 6 3….18….9
e. STATES TWELEVE PLACES TO SEEK……6 7 6 2 4….25….7
f. NOT ON PRIVATE PROPERTY NOR……....3 2 7 8 3….23….5
g. IN ANYONE'S HANDS NOTHING TO…….2 7 5 7 2….23….5
h. BE DISTURBED THRU THE LANDS……...2 9 4 3 5….23….5
i. EACH IS OUTSIDE FOR YOU……............4 2 7 3 3….19….1
j. TO FIND NO DIGGING OR……...............2 4 2 7 2….17….8
k. PRYING JUST REACH INSIDE AN…….....6 4 5 6 2….23….5
l. EVEN CODE WITH ON PIECE……............4 4 4 2 5….19….1
m. NARY WILL SPELL OUT THE……............4 4 5 3 3….19….1
n. SANCTUARY WITHIN THE TEXT YOU……9 6 3 4 3….25….7
o. HAVE THE KEY FOR THE…….................4 3 3 3 3….16….7
p. ONE THAT IS MISSING YOU……...........3 4 2 7 3….19….1
q. DID NOT SEE (G?) A CODE……............3 3 3 1 4….14….5
r. OF NUMBERS FIVE TO A…….................2 7 4 2 1….16….7
s. SIDE REVEALS THE NAME WHERE……...4 7 3 4 5….23….5
t. THE TREASURES ABIDE WITH NAME……3 9 5 4 4….25….7
u. IN HAND THE FAERIES KNOW…….........2 4 3 7 4….20….2
v. AND THEY WILL SHOW YOU……............3 4 4 4 3….18….9
w. WHERE TO GO SEIZE YOUR……............5 2 2 5 4….18….9
x. TOKEN AND REACH YOUR GOAL……......5 3 5 4 4….21….3
y. THE SECRET IS NOT HALF……...............3 6 2 3 4….18….9
z. BUT IT IS NOT WHOLE……....................3 2 2 3 5….15….6

If we drop the one for Q, we conveniently – and perhaps not coincidentally - get to eliminate the confusing mark (Q, G, 9, whatever?) that we inherited from the acorn on p52. Put the remainders straight into a 5x5 grid, starting at the upper left, and you get:
9 1 8 9 7
5 5 5 1 8
5 1 1 7 7
1 7 5 7 2
9 9 3 9 6

As I said in an earlier thread over on the 5x5 Images page called something like “And another way to fill a grid…”, I had been sorta hoping that one could get to a numerical grid like this, find that the 12 strings of 5 (5 rows, 5 columns, 2 diagonals) matched 12 zip codes nicely scattered about the lower 48. Tried that with this grid also, and only 4 are, so it’s at least not a correct zip code grid, but maybe it will give one of you an idea.

I did notice that the distribution of numbers in the grid looked funny. Doing this reduction method keeps you from getting any zeroes, so I really expected that if it were a normal, random text, then each number would show up about 2 or 3 times (25/9=2.78). Instead, it came out very different. To wit:
1-5 2-1 3-1 4-0 5-5 6-1 7-5 8-2 9-5

At first, I thought this meant a lot, insofar as it looked like MS had perhaps constructed the “magic square of 12 zip codes” and used lots of 1s, 5s, 7s, and 9s to make it work out. Skeptically, however, I tried the whole process on the first 125 words of “Danny Boy” (resume reading when you stop crying) and that distribution looked even weirder:
1-5 2-0 3-0 4-3 5-4 6-1 7-3 8-2 9-7

I attribute at least some of this to the frequency of the word “I” in the song. Danny Boy had 3 or 4 zip codes in his grid, too, by the way.

In closing, that’s all I’ve got right now. I hope some of you pick this up and find something useful in it. That 130 word count has to mean something.

Pine