| The multiple rollers of a coding machine are each embossed with the 26 letters and wrapped in order similar to an odometer.
|
|
With the rollers displaying the message:
"FOUR IS A SQUARE",
each roller position is advanced by the value of successive digits of the irrational √2 to yield:
"GSVV KT D XWWDYH".
If the rollers are locked as one cylinder and rotated through the display window one may view 25 other encryptions.
Using a related but not identical process, and just the values of
Π (= 3.141592), the following is the encrypted result:
"PAZYPMEPTUNYRXEWHFGNJRQMK".
Determine the original phrase.
Notes:
1. Spaces do not exist on the machine and are provided here (above example)for readability.
2. If it helps, each vowel occurs at least once.
3. [√2 = 1.41421356237]
(In reply to
re(2): Hint by brianjn)
The ‘√2’ example illustrates a direct link between the given phrase and encryption, but the reference to ’25 other encryptions’ suggests to me that the desired phrase for the longer problem using pi will only become evident by ’locking wheels and rotating the cylinder’ through all 26 resulting encryptions. Applying 3.141592… to the 25 character encryption in the exact same manner as the √2 example, however, whether rotating the wheels forward, backward, or even rotating the odd numbers in one direction and even numbers in the other, quickly got me nowhere.<o:p></o:p>
<o:p> </o:p>
The emphasis on ‘Pieces’ suggests to me two ways of proceeding. If this means using the 3 line pieces of the pi symbol itself in some manner, two such pieces could indicate either 11 or 7. I’ve played with multiplying the individual pi digits by these numbers (rotating the wheels in both directions), and subtracting the digits from both 11 and 7 to get new numbers. (Where I got a negative result with 7, I rotated the wheel in the opposite direction). Again, no luck, and since the instructions indicate using ‘just the values of pi’ = 3.141592, I don’t think creating a set of new numbers fits the bill here anyway.<o:p></o:p>
<o:p> </o:p>
So the other approach would be using ‘pieces’ of the value for pi in some combination. Some approaches I’ve tried (rotating the wheels through the ‘window’ in both directions) include:<o:p></o:p>
<o:p> </o:p>
Using two consecutive numbers to dictate the number of each roller advance: 31/14/41/15/59/92/26 …. as well as 31/41/59/26/53/58/97…. <o:p></o:p>
<o:p> </o:p>
And using three number combinations:<o:p></o:p>
314/141/415/159/592/926… and 314/159/265/358/979/323…. Since each letter returns to itself after 26 ‘clicks’, the first ‘314’ really reduces to 2 further clicks (since 12 full rotations x 26 = 312), thus ending up at either ‘R’ turning in one direction or ‘N’ the other way, for starters.<o:p></o:p>
<o:p> </o:p>
So far again, no luck determining the phrase. Haven’t tried any 4-number combinations yet. Sure, 3141 might reduce to 21 using the above approach (yielding either ‘K’ or ‘U’). But I’m doing this manually without a computer and it’s starting to give me a severe brain cramp.<o:p></o:p>
<o:p> </o:p>
brianjn… as Dej Mar alludes to, I can’t help but feel that the range of possible permutations to solve this (using such a long number as pi) is incredibly daunting even with the information given.
Hope someone has some ideas here and could respond to this!
re(3): Hint
