Logic Masters Deutschland e.V.

Double Dutch Sudoku Advent (2) - Search Nine vs. Edge Sums

(Published on 2. December 2014, 12:00 by Eisbär)

Two years ago, Richard and I have created a sudoku advent calender with sudoku variants that were (relatively) unknown in the puzzle portal back then. This year we have created another advent calender, which has a different theme: every sudoku contains a combination of two different variants. We have chosen 12 well known variants and 12 relatively unknown variants: every day we combine two of these in the schedule AB BC CD … KL LA. While we were creating and test-solving these combination puzzles we both learned a lot. We strongly believe that combining two variants leads to interesting and surprising new solving techniques which really adds to the solving fun. We hope players will enjoy and learn as much as we did.

Logically solvable
All puzzles can be solved completely logically although the logic is sometimes well hidden and is inherent to the combination of restrictions that the different types offer. For that reason we have written some solving hints for most of the puzzles, published in a very tiny font. If you want to read the hints, simply copy these in a text editor and enlarge the font size.

Search Nine
Place the digits from 1 tot 9 in every row, column and 3x3-block. Every arrow is pointing at number 9 in the respective row or column. The number in the arrow is the distance (with respect to the number of cells) from the arrow to the 9.

Edge Sums
The numbers outside of the grid are the sums of the first and last digit in the respective row or column.


Solving hints

R1C2 = 9; R2C2 = 1; R4C2 = 3
9 in R2C5; R2C8 = 3
9 in R3C8; R3C7 = 1; R3C5 = 3; R3C3 = 5
3 in R5C6
9 in R4C3; R4C4 = 1; R4C8 = 5; 5 in C4 in R1C4; All 5’s can be placed
9 in middle 3x3-block in R6C46; R2C4 or R2C6 = 4; edge sums in R2 = {28}
3 in R8C9; 7 in R8C1; 9 in R8 in R8C46
R2C46 = {46}; R2C3 = 7
R4C6 = 2/4
R4C19 = 2/4/6/8; R6C19 = 2/4/6/8/
R7C19 = {19}; R9C1 = 1/9R1C1 = 3; R3C2 = 8; R2C1 = 2; R2C9 = 8
Edge sum R3 = {46}; pair {27} in R3C46
R9C3 = 3; R5C3 = 2
R4C5 not 2 since it would block both possibilities for edge sum 10 in R4
2 in C5 in R9C5

Solution code: Row 5, followed by column 5

Last changed on on 5. December 2014, 12:19

Solved by NikolaZ, Richard, garganega, zorant, HWHW, Statistica, fridgrer, tuace, r45, Fred76, marcmees, zuzanina, sandmoppe, flaemmchen, rätselhaft, Zzzyxas, MrLiang, lutzreimer, kishy72, Goodie, Kwaka, ... Carolin, RobertBe, rcg, Saskia, dm_litv, skypper, Marian, yusaku, sojaboon, jirk, Julianl, Matt, bob, cdwg2000, tamz29, qw014052, amitsowani, keelyc27, EKBM, geronimo92, Krokant, rubbeng, misko
Full list

Comments

on 2. December 2014, 13:40 by Eisbär
Text-error fixed in English version

Difficulty:3
Rating:87 %
Solved:125 times
Observed:12 times
ID:000243

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