Logic Masters Deutschland e.V.

Queen's Gambit

(Published on 14. January 2022, 15:58 by Will Power)

Normal Sudoku rules apply. For numbers 1, 2, 3 and 4, the same number cannot be a chess QUEEN's move apart (ie: ALL diagonals can have a maximum of one each of numbers 1, 2, 3 and 4). Numbers with an X between them sum to 10. Numbers with a V between them sum to 5. Not all possible X's and V's are shown.

F-Puzzles

Solution code: Row 1 and column 5, no spaces.


Solved by Gullie, Steven R, SKORP17, Greg, BlackApolloX, jalebc, fpac, valdal, gige, XhcnoirX, Realshaggy, samuella, chippers, wjdrumm, Calvinball, skywalker, meowzzz, weiken, Raistlen, mathpesto, ... pms_headache, bereolosp, josemadre, Malsted, Baconator, DylanRay, lmunira, Supertaster, RadchenkoAleksandr, AngelaQ, Saskia11, ikaikaw, kangaroo, Rollie, ChampionAsh5357, Jastucreudo, ArmagedDan
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Comments

on 26. November 2023, 19:39 by cascadeshiker
This was a fun puzzle. Took me a while to figure out how to attack it though.

on 14. February 2023, 18:33 by Hydalin
Easy one but very nice set-up. Really liked it, definitely a one star for me

on 12. February 2022, 17:43 by haligonian
Oh, this was fun! For me, about a 2 star. But approachable once I remembered to keep checking that diagonal constraint ;)

on 16. January 2022, 08:07 by Will Power
Bruno's original understanding was correct. There is a maximum of one each 1, 2, 3, and 4 on ALL diagonals. None of the numbers 1, 2, 3, or 4 can be a chess queen's move apart.

on 15. January 2022, 22:08 by Bruno
I didn't understand the rules well, I thought it was all the numbers 1,2,3,4 that couldn't be a chess queen's move apart. But it was only in the 2 diagonals ;)

on 14. January 2022, 20:32 by Shearing
Clever little puzzle. 2-star difficulty for me. Thank you.

Difficulty:1
Rating:85 %
Solved:232 times
Observed:13 times
ID:0008T7

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