The Anagrams of Clue 3
Good evening on a hot, humid night in the deep south!
I've been working though this again over the last couple of nights, and I wanted to present the work in its entirety. Clue 3 appears to be composed of a series of anagram puzzles. The solution to each anagram contains the necessary hints to collect the next set of letters to anagram. I found mistakes in my original thoughts, and I believe I have worked them out.
If the anagram approach is incorrect, shouldn't it lose coherence after one or two steps? Is it possible to collect large blocks of letters within the 10 x 10 grid and create just about anything?
These are questions I've been asking myself. I tried anagramming other random blocks of letters, but all I got was gibberish.
The solution to clue 3 appears to lead to Ana's token.
Step 1: Use the 10 x 10 coordinates by counting the flower petals. The trick to this step is that the grid is numbered two different ways:
1. For the top row of flowers, the columns are numbered LEFT - TO - RIGHT as 0-1-2-3-4-5-6-7-8-9.
2. For the right, bottom, and left sides of flowers, the columns are numbered RIGHT - TO - LEFT as 1-2-3-4-5-6-7-8-9-0
3. For all four sides, the rows are numbered TOP - TO - BOTTOM as 1-2-3-4-5-6-7-8-9-0.
Starting with the top left flower pair and going clockwise around clue 3:
0,2 = D
4,6 = S
4,9 = T
0,7 = U
1,8 = Y
2,5 = X
2,7 = N
1,5 = T
1,1 = E
1,6 = I
9,5 = H
0,3 = V
5,2 = A
0,9 = E
1,3 = I
4,5 = T
0,5 = N
7,3 = V
1,5 = T
Each side creates its own little mini-anagram. The final anagram:
STUDY TEN - X. I HAVE IT: TN - V.
Study the 10 x 10 grid. The answer to 'TN - V' gives us the next set of letters to anagram. 'TN' is an abbreviation for TEN. Using the earlier part of the clue about TEN - X, we've already been introduced to the concept of using 'X' as TEN.
We draw a big 'X' through our 10 x 10 grid. We collect the letters contained inside the 'V'--the top quadrant of the 'X'.
Step 2: Draw a big 'X' through the 10 x 10 grid and collect the letters contained by the 'V':
'ML MESSAGE' REVEALS ANA TOKEN'S QUEST
The trick to this puzzle is to realize that 'ML' is Roman numeral 1050. Break it down into its component pieces, 10 - 50, and reconvert it back to Roman numerals: X and L.
If you draw a big '+' through the 10 x 10 grid, the grid is separated into four separate 5 x 5 grids, one in each of the corners.
This represents the four legs of the 'X'.
The one that we use is given to us by the clue 'L'. If we draw the '+' through the 10 x 10 grid, the 'L' is the top right quadrant.
Our selection is confirmed by seeing 'ML MESSAGE' hiding within the letters of the upper right 5 x 5 grid.
Step 3: Anagram the letters in the top right quadrant of the grid:
E G E E E
A E O T H
M M L L I
S S L Y E
S T F X T
FT HILL ST MEM
SEE - L - X - EYE
The first clue gives us the name of the park that should contain Ana's token: FT HILL ST MEM.
The second clue tells us where to go to collect our next set of letters: GO EAST from our current 5 x 5 grid. We wrap around and wind up in the top left quadrant of the 10 x 10 grid.
Studying the letters of the upper left 5 x 5 grid, you should realize there's a series of letters corresponding to our clue GO EAST: there's a series of directions along the diagonal: N N N N S.
Continuing that diagonal in a wraparound fashion, you also collect a similar series of directions from the lower right quadrant: C E S N I.
The entire set of directions: N N N N S C E S N I
There are two oddballs here, C and I. We use the third clue to solve these oddballs: SEE - L - X - EYE. The C and I are spelled out as SEE and EYE.
The entire set of directions incorporating the substitutions:
N N N N S SEE E S N EYE
We collect the appropriate letters as suggested by the directions in the 5 x 5 grid. We collect the letter north of the letter 'N', south of the letter 'S', etc.
N - E
N - O
N - L
N - A
S - U
SEE - EOO
E - I
S - I
N - W
EYE - TT
Anagram the entire set of letters:
I USE ONLY TWINS SEEN AT ONE ONE - E E
We collect the letters that are 'twins' seen at 1,1 to 5,5; namely 1-1, 2-2, 3-3, 4-4, and 5-5 from each of the 5 x 5 grids contained in clue 3.
Step 4: Anagram the 'twins' consisting of 1,1 to 5,5:
A Q N K R
E E L Y T
T G D E P
U R S B L
RED BLU-GRN PETALS
KEY Q = T
We go back to the flowers around the border of clue 3 and collect coordinates by counting only the red and blue-green petals. The numbers consist of 0 - 4 and are coordinates for our original 5 x 5 grid.
The trick to this step is that our orginal 5 x 5 grid is Caesar shifted with the KEY Q = T, namely a +3 Caesar shift.
This gives us the following grid of letters:
This grid is numbered LEFT - TO - RIGHT as 1-2-3-4-0. It is numbered TOP - TO - BOTTOM as 0-1-2-3-4.
Starting with the top left flower and proceeding clockwise around clue 3 counting only the total of red and blue-green petals of each flower:
0,1 = M
2,2 = O
1,3 = S
0,1 = M
1,1 = I
1,1 = I
1,3 = S
1,2 = N
1,1 = I
1,1 = I
3,1 = K
0,1 = M
1,0 = D
0,5 = C
1,0 = D
1,2 = N
0,2 = R
3,0 = F
1,1 = I
(I think the 0,5 may have been a typo from the author.)
I FIND MC, MI, OR SIM KINDS
The trick to the puzzle is to realize that MC and MI are Roman numerals 1100 and 1001. These are binary numbers for '12' and '9'. The 'similar kinds' would be the other two numbers of the set: 6 and 3. 12:00, 3:00, 6:00 and 9:00 positions.
We collect the letters that are in the 12:00, 3:00, 6:00 and 9:00 positions in each 5 x 5 grid.
Step 5: Collect the letters appearing in the 12:00, 3:00, 6:00, and 9:00 positions in each 5 x 5 grid:
S V A A
E M I F
T R E J
E B I E
I BEAR FIVE - E MATES
The 'five - E mates' are 5 - 5 pairs. We use the portion of clue 3 containing coordinate pairs for the 5 x 5 grid: the number of leaves of the plants.
The second part of the clue 'J' means '10' as it's the tenth letter of the alphabet. '10' gives us the necessary clue to help understand the puzzle.
Step 6: Collect the letters from our original 5 x 5 grid using the number of leaves of the plants.
Beginning with the top left plant and proceeding clockwise around clue 3:
5,3 = O
3,5 = X
5,3 = O
3,3 = M
3,3 = M
2,3 = L
4,2 = I
3,3 = M
4,1 = D
3,3 = M
3,4 = S
3,5 = X
5,2 = J
4,2 = I
5,2 = J
3,3 = M
3,3 = M
4,2 = I
The completed sequence of letters:
O X O O M M L I M D M S X J I J M M I
These are NOT the letters to anagram. They're a puzzle.
The first key to the puzzle: The string of letters are primarily Roman numerals.
Converting the appropriate Roman numeral letters into numbers:
0 10 0 0 1000 1000 50 1 1000 500 1000 S 10 J 1 J 1000 1000 1
We have three oddballs contained in this sequence: the letter 'J', the letter 'S' and the number '5'.
Just as we did for the oddballs in step 3, we use the second part of the clue to figure out what we do: 'J'.
The correct association is J = '10'. It's a hint that we're going to be looking for a binary code consisting of 1's and 0's as well as telling us what to do with the letters J and S as well as the number 5 in our sequence:
They're each a manifestation of '10'.
J = 10th letter of the alphabet.
S = 19--19th letter of the alphabet = 10 (Add the digits together 1 + 9)
5 = 10 in base 5
The three oddballs in the sequence are all equivalent to '10'.
We substitute '10' for the letter J, the letter S, and number 5 in our sequence to produce a final binary code:
0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 1 1 0 0 0 1 0 0 0 1 0 0 0 1 0 1 0 1 0 1 1 0 1 0 0 0 1 0 0 0 1
We use this completed binary code to select letters from our 10 x 10 grid.
Write the code in the squares of clue 3, one digit per letter.
The top row of Clue 3 consists of A T S A S E G E E E and the first ten numbers of our binary code are 0 1 0 0 0 1 0 0 0 1:
A - 0
T - 1
S - 0
A - 0
S - 0
E - 1
G - 0
E - 0
E - 0
E - 1
We select the letters corresponding to the '1'.
So, for the top row of letters, we would be selecting T, E and E for the anagram puzzle.
Step 7: Collect the letters from the 10 x 10 grid that match the 1's in the binary code:
T E E N E O O M I N K S S Y A T
NONE IS KEY TO MATES
The trick to the puzzle is to realize that the letters of NONE are a set of directions. The 'N' and 'E' are 'north' and 'east', and the 'O' is a blank in the middle.
We collect the letters from each 5 x 5 grid that are N O N E, namely north - middle - north - east.
Step 8: Collect the letters from each 5 x 5 grid in clue 3 that are N O N E: north - middle - north - east:
S N S A
E L E I
T D T E
E S E I
I SEE TEN AT L - E SIDES
The trick to the puzzle is to realize that 'L - E SIDES' means 'left' and 'east' sides.
We collect the ten letters along the left and east sides of the 10 x 10 grid.
Step 9: Collect the ten letters along the left and east sides of the 10 x 10 grid:
ADVENTURER + EHIETIOIDL =
I LIE HIDDEN AT OVERTURE
We collect the letters that are 'OVERTURE', namely 'over T, U, R, and E'.
Step 10: Collect the letters that are OVER T, U, R, and E (OVERTURE):
Over T: A E S E N A E
Over U: S S T F
Over R: N E S W U R E
Over E: C E Z L G V A I H F R R U R I
Z-CLUE HAS FIVE FINGERS
The 'Z-clue' refers to coordinate pair 5,5 from our original 5 x 5 grid.
'Z-clue has five fingers' is telling us where to look for our next and final set of anagram letters.
There's one place we've seen that is showing coordinate pair 5,5 on 'five fingers', and it's not on clue 3.
It's a reference to a specific illustration in the book.
The fingers of a particular fairy are displayed showing '5,5': p. 44.
The first part of the clue, 'Z-clue has five fingers' tells us where to look: p. 44. The second part of the clue tells us which letters to collect for our final anagram puzzle:
We use the letters from the original Dragonfly solution: RICKETTS GLEN SP FALLS TRAIL SIGN.
The anagram solution appears to lead to a specific location within Ft. Hill State Memorial.
I don't know if this is correct or not--if it is, it's one SUPERBLY crafted puzzle.
I would have thought it should lose coherence after one or two steps, but these look pretty good to me.
I can't tell if I've forced these solutions or not. Constructive criticism is always welcome.
Keep the faith,
PS: I'm not headed to search for this mythical token anytime soon, and I recommend that you don't either unless you feel this is rock solid.
PPS: An intrepid trover named 'Angels' is responsible for finding this first anagram and sending it to me a very long time ago. See what you caused, Angels?
wow--- I am going to have to print this out and digest this.
if anyone wants to scan the web-
here is a link to start...
see 3/29/04 (OHIO)
Good evening on a very hot night in the middle of the desert city of Las Vegas!
I've had a couple of days to digest what I posted about clue 3, and I wanted to pass along a couple of thoughts.
I think it's going to be correct, and I'll tell you why.
The final step takes us to p. 44 in the book--the Dragonfly puzzle page.
Look at what we're taking with us when we hit that page:
1. I bear 5 x 5 mates.
2. None is key to mates.
3. I use only twins seen at 1,1 - 5,5.
4. I find MC, MI, or similar kinds.
5. Key: Q = T.
6. I have it: 10 - 5.
7. I see ten at left and east sides.
Another way of stating 'I bear 5 x 5 mates' is 'I have 25 mates': 12 mates, 12 couples, and 1 grand prize.
'None is the key to mates'--it's a reference to the concept of the 'blank in the middle', the 'blank' in the directional code, and the 5 becoming a 0 that I've been recently posting about.
Do you understand what I'm implying?
The anagrams of clue 3 contain Mr. Stadther's 'announcement' for his other tokens as well as the hints to solve the puzzles specifically on p. 44--the dragonfly's mate and couple.
The set of three unused leaves on p. 44 in the lower right corner were the basis for assigning letters N, S, E, W, and the blank in the middle. They're four Roman numerals if you remember what I had posted.
North = 3,3 = M
East = 3,5 = X
South = 4,2 = I
The missing component is the C to the west.
Do you understand the link to the anagram? 'I find MC, MI, or similar kinds'.
We're going to find the same thing on p. 44 when we work on the mates.
Clue 3 was meant to be the transition from the original Polybius puzzles into the remainder of the book.
There was a reason Mr. Stadther really wanted us to solve clue 3 a lot sooner than now.
I've been looking at your Clue 3 thread awhile and have a few comments.
I like 'none is key to creatures' mates'. 'None' is the perfect expression, in a single word, of a binary aspect to the remaining puzzles. None is nothing, or zero, and yet contains 'one'. Zero and one. Rather clever. I have to think that you are correct, at least up to this point. It fits too well not to be right.
Your third step solution,
'3. I use only twins seen at 1,1 - 5,5.',
is interesting. I have an interpretation that strays from yours, involving clocks. I'll get to that later.
Your solution number 5: '5. Key: Q = T.',
is interesting, because I've found many instances of 'E' equalling some letter value, possibly implying a Caesar shift. In each case it is an E from an EYE, coupled with the letter K. I wonder about the similarity in presentation, 'Key', followed by two letters. I made a 'joke' a couple years ago---"Question: if a tree falls on a Q in the forest, does it make a noise? Ans.: It goes 'EEEEEEEEEEEEEEEEEEEEE'". Ha ha. It's long been suspected that somehow Q=E, or vice-versa.
My personal opinion is that we haven't (as a group) sufficiently solved Clue 2 to proceed with any significant wherewithal and confidence regarding Clue 3.
I agree with you that there are 25 tokens.
Here's the clock problem---in Roman numerals, the 12 O'clock position is denoted by 'XII'. That equals 12 in base 10, and 22 in base 5. Normally, our clocks run twice around in a day, so that 12 x 2 = 24 hours. However, in the world of Zac and Ana, if their base-5 clocks ran twice around in a day, it's 44 'hours'. In a single rotation, in 'our' normal 12 o'clock position, the Roman enumeration is 'XII' (22 in base-5). In base-5, taking half of that, it would be 'VI', which equals eleven. Suddenly, we get a 2 hour difference per half day (11) in base-5, and a whopping difference in a whole day (44). Nevertheless, we know that '44' in base-5 equals 24 in base-10. I want to point out that a 'twinning' effect can be set up, clock-style, so that there is 11 on the left, and 11 on the right, as in Clue 2's stand-alone block, 'NOT APRIL ONE' which consists of 11 letters. This is a mirrorring effect, actually.
In the end we ought to expect that some sort of binary code is required. Base-5 notation is probably a good suspect, and the Dream Song's 'oNONE', ONE, ON, EYE ON ONE, etc., is probably required, as well as a clock. Let's say that Zac has his 'eye on one', meaning he knows he has to finish his task of collecting the jewels by noon (before one).
I might be unnessessarily complicating the picture, but I don't think so. MS wanted the first puzzles to be solved (non-lost creatures) before the lost creatures, so he made those puzzles 'easier' (relatively) by having a single grid, base-10 notation, etc. I'm not as convinced as I was earlier that the original solution provides a basis for the mates' solution, at least in the instance of a Zac, Ana, or Yorah puzzle. That is, I don't think we have to have had the solution in hand of the original puzzle to solve the second. It sounds good, but on reflection it doesn't make perfect sense. MS couldn't have discounted the possibility that someone could have derived everything regarding the location of a Creature's mate, with the sole exception of that mate's 'mate solution'. To be fair, he'd have had to allow the solution to the location of a lost mate to proceed. Once that solution method had been circulated, then the other lost mates would have fallen in rapid succession (presumably) as did the creatures themselves. Then we'd be arguing about what the lines in the leaves meant on page 44, because they didn't fit into whatever notion we had diveloped about the interpretation of '5-to-a-side', etc. The solution method to the creatures might then be as hard to comprehend as the mates, given the fact that the solution method to the mates might be much different. We'd have a hard-drive bias.
But I can't dismiss the 'twins' thing. 'paTTon's run', rickeTT's glen, fort prickeTT's state park, foSS sp, lake darnaNNeLLe', etc. I wonder if there's a way to reverse the process, reflexively. What if the second tier solution method had been discovered first, beyond great odds against it?
I've been on vacation a few days, and I've still been thinking about this.
If anyone has managed to slug through these anagrams, I'd like someone to check behind me for something.
After you remove all the letters from the 10 x 10 grid used in clue 3 that were part of the various anagrams, I got ten letters that weren't used.
Did anyone else get the same set of letters?
That appears to me to be a final set of instructions to do something with the top row of letters by aligning the 'crowns'.
Interesting. As we discussed last year, I reverse-engineered a Clue 3 solve from the unused 10x10 letters in the first strings of anagrams. I assumed MS wouldn't have leftover letters in this puzzle - consistent with completeness of earlier new clues.
You've come full circle, although the solution set is different. I'm curious why you deviated from anagramming "RICKETTS GLEN SP FALLS TRAIL SIGN", maybe it led you in this direction too.
Since I already explored this approach, I'll check your results tonight. Unless, of course, another trover pre-empts me.
I did that late last night, and I just redid it, and I got a little different set:
R I E N A L A N G C W S
It would be mucho appreciated if you (and anyone else checked behind this).
Images of Clue 3 strings of anagrams
I reviewed the original strings of anagrams and updated my images. My solution set differs by one letter.
I created a series of images for each step and posted them online (URLs below). If I missed how that "E" was used, please let me know.
All Steps: Leftover letters
Step: Flower petals
Step 2: Inside the "V"
Step 3a: Top-right quadrant
Step 3b: Directionals
Step 4a: Twins
Step 4b: No image, uses 5x5 grid (rot3)
Step 5: Clock positions (12, 3, 6 and 9)
Step 6: No image, uses 5x5 grid (rot0)
Step 7: Binary (1's only)
Step 8: NONE (north, middle, north, east)
Step 9: Left and East sides
Step 10: Over "T", "U", "R" and "E"