Posted on 04/06/2025 5:53:46 AM PDT by nikos1121
ZRG WXGQTLBXGY JBUG BM XKZ OKSZR JBIBXA. —MKNSTZGM
The way it works is a letter stands for another letter. For example: AXYDLBAAXR is LONGFELLOW (does not apply to today's cryptogram).
Beware, the game is very addictive. If this is your first time, don't be intimidated, you’ll be solving them all within a few days. If you’re stumped, take a break and return to it.
PLEASE DO NOT post the answer in general comments, but DO post your time and how you made out.
You can certainly send your solution to my private reply, or if you need a hint for today’s Cryptogram ASK THE GROUP FOR HELP!
I suggest printing these out and work them on paper. If you need a little help you can copy and paste it to Hal’s Helper below.
You can then work on the puzzle without using pen and paper, but I recommend that you do NOT look at the letter counter.
HAL'S CRYPTOGRAM HELPER
One last request. Feel free to post a fun or clever clue, the more tangential to the quotation the better, but please don’t put the actual words of the quote in the clue.
Short and sweet today. Can you break two minutes?
MTF IFXM DHFDUHUMWJL EJH MJSJHHJQ WX ZJWLB NJCH IFXM MJZUN. ---T. YUORXJL IHJQL YHSolution to previous Puzzle: (select the yellow text with your cursor to read):
THE BEST PREPARATION FOR TOMORROW IS DOING YOUR BEST TODAY. ---H. JACKSON BROWN JR
HAL'S CRYPTOGRAM HELPER
Can you break two minutes?
I think can, I think I can...
I got this one pretty quick - but I disagree with “short and sweet” - the short ones are generally the hardest.
This one was easy because the short words are common words and everything fills in easily. But yes, with HAL, long ones are easy because a few lucky guesses and the whole thing just lights up.
Thanks.
Plus, there are only several quotable people with only one name.
The two-letter short word gives the author away, granted. I did that, too.
“Don’t look back. Something might be gaining on you.” — Satchel Paige.
Another thing about the long ones - the odds of encountering a word that is a dead giveaway - like a single letter word - increase exponentially as the quote gets longer.
One of my favorites that I always check for is a four letter word beginning and ending with the same letter - usually “that”, which then reveals the ubiquitous ‘the’ and all the other ‘th’ words.
The probability of encountering any given word, including short ones, increases linearly, not exponentially, with cipher length. I was going to avoid the topic but you might find this interesting: https://en.wikipedia.org/wiki/Unicity_distance
“The probability of encountering any given word, including short ones, increases linearly, not exponentially,”
That sounds right now that I think about it. But it’s not just the odds of encountering any given word that I’m talking about - it’s the odds of encountering any given clue - and a clue is often not merely a guessable word, but rather a relationship between words and punctuation, shared features between words, or word juxtaposition and possible sentence structure, etc..
Since the number of possible relationships does increase exponentially relative to the number of elements, my gut tells me the odds of encountering a useful clue also increases exponentially as a cryptogram gets longer, but I will not attempt to support my gut feelings mathematically.
If you read the link, it shows that the probability of having more than one valid solution decreases as NL, where L is the message length, and N is the number symbols (letters). The brute force method tries all possible keys. In a cryptogram there are 26!. (26 factorial, = 26 x 25 x 24 x 23 ..x 2 x 1) possible keys. Therefore the unicity index is approximately log(26!)/log(26) If all letters in English were equally likely, the average unicity distance for a monoalphabetic substitution, i.e., a cryptogram, would be about 19 characters. In practice it is actually smaller. Besides cryptograms divulge word lengths and punctuation.
In attacking the Enigma, Bletchley Park used a modified brute force approach, where they made a stab at the contents, using it as a crib. For instance reports from weather stations usually contained the phrase WETTERBERICHT. Certain stations always began their reports with the phrase "HEIL HITLER". Willkommen, danke sehr, said Bletchley Park. The enigma, needless to say had a LOT more than 26! keys.
To bore you further, WETTERBERICHT was a particularly juicy speculative crib. One cryptological weakness of the Enigma was that no letter could substitute for itself. If a 13 character run from a weather station never had the same letters in positions 3,4,13 (T) or 1,2 (E) or 6,9, it as a candidate. Selecting those runs and letting the Bombe run while testing all such 13 character runs as candidates for WETTERBERICHT could give away that day's Enigma setting for the entire German Army.
That makes sense for the brute force method, where a computer tries all 25! possible solutions, but that’s not the way humans solve a cryptogram.
The question its intended to answer is HOW LONG does a cryptogram have to be before you can solve it, unambiguously, on average. About 19 characters. The familiar pattern ABCA, which is “always” that has 79 valid solution. The theorem does not say that all 19 character phrases are unambiguously crackable, only that on average that out of phrases composed of valid English words, ones at least 19 characters long should be crackable. Solutions for “that”:
THAT
EASE
HIGH
ESTE
TANT
TENT
PROP
ERIE
TEST
SEAS
SONS
AREA
ARIA
EDGE
DIED
SHES
NOUN
SAYS
DEAD
PREP
ELSE
TEXT
ASIA
GREG
ONTO
TACT
REAR
BOMB
THET
AURA
HATH
SETS
ALTA
EINE
GANG
SUNS
SITS
ALVA
KICK
OHIO
SINS
ALBA
ROAR
TOUT
HUGH
EYRE
ALMA
SANS
TROT
KIRK
PUMP
TINT
POMP
AQUA
HUSH
ELBE
RODR
OSLO
PULP
HASH
SARS
GONG
SAMS
TILT
SUMS
BULB
TAIT
URDU
TUFT
SOUS
XBOX
KRAK
SOBS
MALM
SUIS
SIMS
NEON
DYED
TAFT
SIRS
I didn’t say ABCA was always THAT - only that it happens often enough to make it well worth checking for other words that might be TH words, especially THE.
I would be interested to know how many times each of the 79 ABCA words on your list appeared in Today’s Cryptogram over the last year - and when they did - how useful they were at finding the solution.
My guess is that about 40 appeared 0 times, 38 appeared 1-10 times, and THAT appeared about 60 times.
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