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1. Warmup

"I'm thinking of five letters," I smiled.

"Are they ... G, B, K, J, and N?" you asked.

"No, none of those."

"C, X, Q, A, and N?"

I gulped. "Four of those."

"K, I, Y, T, and M?"

"One of those."

"S, D, Z, I, and K?"

"One of those. Now that's enough guessing; you have enough information to figure it out!"


This is of course a simplified version of the word game Jotto, with the difference that here the answers and guesses are not limited to real words. This also cuts out the occasional anagram-solving component of late-game Jotto.

You might also call it a non-word, white-peg-only version of Word Mastermind. Or an alphabetical cows-only version of Bulls and Cows.

Here are three more, in approximately increasing difficulty. I made sure they were all solvable without a computer, but have at it however you like.

2. Easy

"I Y L H C?" --"None."

"O H K E F?" --"One."

"U R I Q E?" --"Two."

"Y N A W R?" --"Three."

"J N H L X?" --"None."

"J Q O V C?" --"Two. That's enough!"

3. Less easy

"W Q P Z N?" --"One."

"V Y H A Z?" --"Three."

"B R I Q C?" --"One."

"B A T K F?" --"Two."

"V U M R J?" --"None."

"V N H F I?" --"Three."

4. Least easy

"V X K H S?" --"Two."

"T C I O L?" --"One."

"G V U F M?" --"None."

"L Q P F S?" --"Two."

"V Z X Q S?" --"One."

"H O Q V R?" --"Three."

"K Z O A T?" --"Two."

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1

These puzzles can be solved by narrowing down the clues gradually. For example, let's remove all the letters in the "none" clue. This gives:

CXQA 4
IYTM 1
SDZI 1
Now C X Q A must all be chosen. There's only one letter left, so it must be the letter the last two rows have in common, which is I.

2

Removing the letters in a "none" gives the following:
OKEF 1
URQE 2
AWR 3
QOV 2
So A W R must be three of the letters. Removing those from other clues (and decrementing the counts appropriately) gives:

OKEF 1
UQE 1
QOV 2
And now both of the remaining letters have to be in QOV - so we can take out all letters that don't appear there.

O 1
Q 1
QOV 2
So O Q are the remaining letters.

3

Start by removing the letters in the "none" clue again. This gives:

WQPZN 1
VYHAZ 3
BIQC 1
BATKF 2
NHFI 3
Note the two 3 clues; the only letter they share is H. If the H is not chosen, there are at least six letters chosen total. So the H must be chosen, and the remaining letters must be in one of the two 3 clues.

Removing the unnecessary letters (and the H):

ZN 1
VYAZ 2
I 1
AF 2
NFI 2

So I must be another chosen letter. Using the earlier "can't pick too many letters" argument on rows 2 and 4, A must be chosen as well.

ZN 1
VYZ 1
F 1
NF 1
And now the remaining two letters must be F and Z.

4

XKHS 2
TCIOL 1
LQPS 2
ZXQS 1
HOQR 3
KZOAT 2

Say H is not chosen - that is, O, Q, and R are chosen. Then the first row implies that at least one of {X,S} is chosen; but means that we choose at least two from ZXQS. So H must be chosen.

XKS 1
TCIOL 1
LQPS 2
ZXQS 1
OQR 2
KZOAT 2
Now, LQPS and KZOAT are disjoint, and their clues are both 2. So all 4 of our remaining letters must be from those two sets, and any letter besides those can be discarded.

KS 1
TOL 1
LQPS 2
ZQS 1
OQ 2
KZOAT 2

Now O and Q must be chosen. So T/L/Z/S are not chosen, and the clues simplify to
K 1
P 1
KAT 1
So K and P are the remaining two letters.

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  • $\begingroup$ Thanks! So elegantly explained that it might help me code a solver program. $\endgroup$ Oct 25 at 23:17
  • $\begingroup$ Nice. In the second listing from your solution to puzzle 3, you have the line AF 2, so you can jump straight to including A and F from there. $\endgroup$
    – Ergwun
    Oct 26 at 3:08
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Four of ACNQX, one of IKMTY; note that those two sets are disjoint, so those are all five of our letters. One of DIKSZ; the only one of those mentioned above is I, so we have that plus four of ACNQX. We have none of BGJKN; in particular we don't have N so that's the absent one from ACNQX. So our five letters are ACIQX.

None of ICHJLNXY. Three of ANRWY but neither of NY, so all of ARW. Two of CJOQV, disjoint from ARW so these are all our letters. One of EFHKO which must therefore be O. Two of EIQRU; one is R, already known; the other must be Q. So our letters are AOQRW.

Three of AHVYZ and three of FHINV, which come to more than our total number of letters, so we have H or V. We don't have any of JMRUV so in fact we have H but not V. And with an overlap of just one between AHVYZ and FHINV, these must be all our letters: H, two of AYZ, two of FIN. We have one of BCIQR, which can only be I. We have two of ABFKT, which must be AF. (So now we have our two of FIN and one from AYZ.) And we have one of NPQWZ, which must be N or Z, so the remaining letter is Z. AFHIZ.

None of FGMUV so let's remove those from the other info we have: HKSX:2 CILOT:1 LPQS:2 QSXZ:1 HOQR:3 AKOTZ:2. If we don't have H then we have KSX:2, OQR:3 but then there's no possible way to have AKOTZ:2. So we must have H. That leaves KSX:1 CILOT:1 LPQS:2 QSXZ:1 OQR:2 AKOTZ:2, and four letters to find. If we don't have O then we have QR:2 and AKTZ:2; then CILOT:1 means we have T so at this point we have HQRT+1 and KSX:1 LPS:1 AKZ:1 and, alas, those don't have a common letter; so we do after all have O. So we have HO+3, and KSX:1 CILT:0 LPQS:2 QSXZ:1 QR:1 AKTZ:1. CILT:0 means we can reduce the others to KSX:1 PQS:2 QSXZ:1 QR:1 AKZ:1. If not Q then PRS are all immediate and then AKZ:1 is impossible; so we have HOQ+2, and KSX:1 PS:1 AKZ:1; but also SXZ:0 so these reduce to K:1 P:1 AK:1, which means our last two letters are KP. So the full set is HKOPQ.

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  • $\begingroup$ Apparently I'm supposed to avoid comments like "+1" and "thanks"! But +1, and thanks! I think you might have beat Deusovi to the punch by a few minutes, too. Both excellent answers. $\endgroup$ Oct 25 at 23:23
  • $\begingroup$ Yes, I was about five minutes quicker (or maybe started about five minutes earlier) than Deusovi. (For the avoidance of doubt, you are not required to give the checkmark to the earlier answer, and I am not suggesting that you should have..) $\endgroup$
    – Gareth McCaughan
    Oct 26 at 11:19

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