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I already have the solution for this crack the lock puzzle but am stuck on how to get there. The smaller puzzles I've solved use a different style of deductive reasoning.

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Rules:

  • For this specific question the answer contains 16 characters.
  • No character can be used twice in the solution.
  • The two numbers at the end of the code tell you two things:
    • The first number tells you how many characters from that row will appear in the answer
      • Ex: 9 characters out of the 16 that appear in the first row will be in the answer.
    • The second number (furthest right) tells you how many of the characters used are in the correct place
      • Ex: we know 9 of characters from the first row will be in the answer and of those 9, 7 of them are in their correct position for the answer.

The last line's clues are 9, 2. So 9 used - 2 in the right spot

I am interested in HOW people get to the answer and hope to learn how to solve future puzzles!

If you want more puzzles like this look for @ali_imashli on TikTok. He makes and post these puzzles regularly. This is the link directly to this puzzle. Beware the answer may be in the comments

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This is a fairly lengthy solution, so you may want to have a copy of the puzzle handy to follow along.

The top and bottom rows have 8 elements in common. Because the bottom row has only 2 Right, at most 2 of the common elements may be Right. Because the top row has only 2 Wrong (correct but in the wrong position), at most 2 of the common elements are Wrong. That means of the 8 not-common elements, the top row has 5-7 Right among [T!B=+6KW], and the bottom row has 5-7 Wrong among [7M?KWS@!].

Now compare the top row to the 4th row. Remember that among [T!B=+6KW] there were 5-7 Right. Five of this set, [!B6KW], are in the same position in the 4th row. Since the 4th row clue has 2 Right, at most 2 of the common elements may be Right. The only way for 5-7 of the larger set to be Right is for [T=+] to all be Right on the top row and 2 of the common elements [!B6KW] to be Right. On the 4th row, since the 2 Right have been found in the set [!B6KW], exactly 0 of the other letters are Right. In particular, the 8 in position 1 is not Right, meaning D is Right. Since S is not Right in position 2, and D and + have already been placed, that means 3 must be Right in position 2.

So far 5 symbols have been placed:

D3T.....=..+....

From before, 5-7 among [7M?KWS@!] in the bottom row are Wrong. On the 2nd row, 2 Rights (DT) and 2 Wrongs (+=) have been found, leaving 4 more Rights. Of the remaining letters, the set [@SM?!K] intersects with the bottom row's set [7M?KWS@!] which contains 5-7 Wrong letters. Since the first set is 2 smaller than the other, at least 3 letters must be Correct in the first set, and therefore at least 3 Rights (since there are only Rights left). Since S is located where = is already placed, it can be excluded, leaving at 3 Rights out of the set [@M?!K].

There is an important property of row 2, which is that only Rights remain to be found, and no Wrongs. This means that for all of row 2's letters, none of those letters may be Right in any other row except if it appears in the same position as row 2, because that would make the letter in row 2 be a Wrong, of which there are none left (+ and = have already been found). Using this fact, in row 1 where we narrowed 2 Rights to the set [!B6KW], that set can be further narrowed to [B6W]. Using the rule on the other symbols in row 1, the last Right in row 1 must be in the set [AR4]. Using the same rule on row 3 to find where the 4 Rights are, there must be 4 Rights in the set [C4W7A]. This set intersects row 1's set at 3 places: [CWA/AR4]. Because 4 of set [C4W7A] must be Right and one of [CWA] is not Right because row 1 is Right there, the two symbols 4 and 7 in row 3 must be Right. Also, since two of CWA and one of AR4 must be Right, the one of AR4 can't be A because that would force the two other symbols to be WA which duplicates A. Therefore, C must be Right, and we have the intersection [WA/R4], where one of each row must be Right.

We now have: D3T...C4=..+.7..
The 7 that we found occupies one of the positions from the previous row 2 set [@M?!K], leaving 3 Right out of the set [@M?K]. Row 1 has 2 Right out of the set [B6W] which intersects the row 2 set at [?K/6W]. Similar to the previous logic, since 3 of 4 of [@M?K] are Right and 2 of 3 of [B6W] are Right, each must occupy exactly one of the two common places in [?K/6W], leaving the not-common symbols Right. These are @M of row 2 and B of row 1.

We now have: [email protected]=YM+X7XY
Where XX are [?K/6W] (one each), and YY are [WA/R4] (one each).

There is only one unknown position, and that must come from row 2 where there is still one Right yet to be found. This must be 5.

We now have: D3T@5BC4=YM+X7XY
Where XX are [?K/6W], and YY are [WA/R4]

Looking at row 4 and removing symbols that either are known to be correct or symbols that are known not to be in the solution, we are left with 3 Correct out of [RA6KW] (2 of 5 incorrect). Looking at row 5 and doing the same, we are left with 4 Correct out [?AKRW] (1 of 5 incorrect). Since these two intersect at [RAKW], and at least one of those must be incorrect because row 4 has 2 of 5 incorrect, that means row 5's ? must be Correct. That makes the XX set be ? and W. Since W is now taken, YY must be R and A.

The final solution is: D3T@5BC4=RM+?7WA

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  • $\begingroup$ I really appreciate you taking the time to explain this me!! It was so helpful! $\endgroup$
    – Rcyr1010
    Aug 9, 2023 at 17:09

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