Logic Masters Deutschland e.V.

Hooks (Anchor Sudoku)

(Published on 26. June 2021, 07:58 by FryTheGuy)

Hooks

This is the sixth puzzle in a series of seven anchor sudoku puzzles.

This is by far the hardest of the seven puzzles. Its probably similar in difficulty to Poison Arrow which a few people rated as a 6 for difficulty.

Originally I planned for this to be the final puzzle, but thought it would be a little depressing if the final puzzle did not get any solves, so I am moving it become second to last.

Like the previous puzzle, I am included some minor hints to push you in the right direction.

Anchors are like arrows, except you subtract the smallest digit from the sum of the other digits. So for example an anchor with a 5 in the square, and 4,2,3 in the chain would be valid because 5=4+3-2. Repeat digits are allowed so 5,5,1,1 would also be valid because 5=5+1-1.

Previous Puzzles
Rules

Standard sudoku rules apply.

Subtracting the smallest number along an anchor from the sum of the other numbers equals the number in the square cell.

The numbers in the cage must sum to the given total in the top left.

Solve the puzzle at https://f-puzzles.com/?id=yebqnwms

Hints

Hint 1 for break-in: The break-in uses set theory

Hint 2 for break-in: Try to put a bound on the maximum value for the digits which will be subtracted from the four anchors that go into box 5

Solution code: Rows 1 and 2


Solved by Dentones, bigger, Piatato, Vebby, Bobbobert
Full list

Comments

on 7. May 2022, 21:32 by Vebby
Absolutely bonkers, what a puzzle! :O

on 1. July 2021, 18:41 by FryTheGuy
@Piatato

I am happy you got through this one! I was nervous that it would be more frustrating than fun.

on 29. June 2021, 16:12 by Piatato
Great puzzle! I'm looking forward to one more anchor puzzle! :-D

on 27. June 2021, 04:08 by FryTheGuy
@bigger

The next puzzle is easier than this one, and the next puzzle is the final puzzle. Sorry for my poor choice of words.

on 26. June 2021, 19:40 by bigger
By far, as in the next one would be harder? Or there would be more than 7?

Difficulty:5
Rating:N/A
Solved:5 times
Observed:8 times
ID:0006S7

Enter solution

Solution code:

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