How balanced do you think the puzzle types of this test were?	Fairly balanced
What was your opinion of the distribution of easy/hard puzzles?	Too many medium difficult puzzles
What did you think about the puzzle quality of the test?	Fairly Nice
How did you feel about the length / time limit for this test?	A bit too many puzzles
Of the puzzles you solved/attempted, how well did the point values reflect the difficulty?	Most puzzles were worth the right amount
What was your opinion of the booklet formatting and printing?	Just right

How balanced do you think the puzzle types of this test were?	Perfectly balanced
What was your opinion of the distribution of easy/hard puzzles?	Just right
What did you think about the puzzle quality of the test?	Very nice
How did you feel about the length / time limit for this test?	A bit too many puzzles
Of the puzzles you solved/attempted, how well did the point values reflect the difficulty?	Most puzzles were worth the right amount
What was your opinion of the booklet formatting and printing?	Just right

Thanks Bob! I wish you a lot of holidays with lots of puzzle fun!

FYI: Breinbrekers doesn't exist anymore. It is taken off the market by the end of 2013. After that the Dutch Puzzle Association was erected. (Website: www.wcpn.nl) where you can solve a Breinbreker-like puzzle daily. If you are interested in my puzzles, check out the forum post about my Sudoku Variants Series or my author's profile about my friday-noon-puzzles, both on the website of Logic Masters Germany, where I am active for about 5/6 years now.



I have choosen exactly the same cell-size as in my previous tests; there were no complaints back then. Maybe it is an optical effect because there are now four puzzles on each page? At least writing space should be the same.



Thanks Tawan!
I think you have made an interesting point here. Writing puzzles starts with a lot of thinking for me. (Especially for tests or championships.) On the way back in my car from work to home I tend to do a lot of thinking about puzzles. In the weeks before I actually start writing the puzzles, numerous ideas have crossed my mind about what 'tricks' I can apply. I like to explore the possibilities a variant offers.
Especially for this 7x7-test I can add that it wasn't always easy to use my ideas. The fact that you have only seven digits and not much room makes it harder (than I thought), so quite some ideas didn't lead to a puzzle. I also used much more T&E to get there, so to say. Constructing 9x9-puzzles is easier.


Thanks Swaroop. Your notes have made me think, and what I have written as response to Tawan's lines comes into sight here. (My) thinking about puzzles mostly concentrates on
a) the opening or
b) a by-pass in the solving route that can make a huge difference in solving times between top players and the rest of the field.
I think you can call these solving steps indeed 'tricky', and I like to distinguish this from 'hard'.

By putting these 'tricks' in my puzzles, I also hope that people learn something while solving my puzzles, so in the end I have contributed a small piece in the development of the puzzle sport. The notes to the puzzles that I will write in the next forum thread will be helpful if players didn't recognize the tricks during solving.

As promised yesterday, here are some notes to a selection of the puzzles in the set. I didn't re-read to check typos, so if you find one: just let me know!
I hope these hints are helpful and I like to read feedback to confirm the intended route through the puzzle, or to report a shortcut that I missed...

Law of leftovers
For most of the players this will be a very trivial rule, but for those with not so much experience in solving irregular grids I will write a few lines about this.
Law of leftovers is a rule that works in all irregular sudokus and it means that an area of ‘N’ rows or columns needs ‘N’ irregular areas. Looking to the first puzzle in the test: Rows 1, 2 and 3 contain two full irregular areas (top right and top middle) and one area that is ‘sticking out’ of those three rows; The F-shape top left has two cells sticking out of rows 1,2,3. At the same time, rows 1,2,3 have two cells ‘sticking in’; these are the cells in R3C5 and R3C7. The cells sticking into rows 1,2,3 must be the same as the cells sticking out of the area.
The same rule works also for scattered grids. Looking to puzzle 3 in the booklet, there are three grey cells in rows 1 and 2. There are also three cells ‘sticking out’ of the first three rows, namely R4C123. Those cells sticking out contain the same digits as the three grey cells in rows 1 and 2. The fun part with scattered sudokus is that this works horizontally and vertically; The digits in R1C7 + R2C6 stick out of rows 1,2 and 3 as well as out of columns 5,6 and 7.

Scattered
I noticed some players had trouble with this puzzle from the basics page. There is a nice shortcut in the puzzle that some of you might have missed. 2 in R3 goes in R3C6; R7C6 = 2. R6C4 = 7; R4C6 = 5; R1C7 = 7; all 7’s; R2C4 = 5; R12C5 = {34}. R1C6 + R2C7 = {16}. This means that the grey cell in R2C6 can only be 3. Now the puzzle runs smooth.

Quad Sums
R7C1=4; R1C7 = 4
QS in R12C12 = 1+2+3=6; R34C1 = {57}
2 in R2 in C12; 3 in C1 in R12
QS in R67C67 = 1+2+3=6; R45C7 = {57}
2 in C7 in R67; 3 in R6 in C67
3 cannot go in R6C6 because of QS in R56C56. R6C7 = 3; R7C7 = 2
6 cannot go in R6C6. R6C6 = 1; R7C6 =6
R56C1 = {26}
R2C2 = 2; R1C2 = 6
R23C7 = {16}; R1C1 = 1; R2C1 = 3
2 in shape in R56C4. Since 2 in C1 is in R56, QS R56C56 cannot contain 2. R5C5 = 1, R6C6 = 6.
R5C1 = 6; R6C1 = 2; R5C4 = 2; R4C6 = 2; R3C5 = 2
Only place for 6 in R4 = R4C3; R4C4 = 1; R5C3 = 3

Toroidal Kropki
One quick opening can be found in row 5. There are 3 white dots ‘covering’ 6 different cells, between C1 and C2; C4 and C5; C6 and C7. This means that the digit in R5C3 must be odd. Since there are a black and a white dot on both sides of this cell, that digit can only be 3.

Consecutive
Pair {24} in R2C4 + R4C2.
R2 contains 2 dots which must have pairs {12} and {45}
R2C13 = {67}; R13C2 = {15}
The dot between R67C2 must have odd digit 7, which goes in R7C2; R6C2 = 6, R6C3 = 5.
5 in R4C6.
Consecutive pair in R2C67 = {12}; R2C4 = 4; R2C5 = 5

OddMax
7 locked in R7C56; R5C3 = 7; R6C2 = 7
7 in C1 locked in R234; 7 in C7 locked in R123; 7 in R1 locked in C567. :=> R1C7 = 7
R2C5 = 7, R7C6 = 7, R3C4 = 7; R4C1 = 7
2 locked in R1C234; R3C6 = 2
5 in C6 in R6; 5 in C5 in R7; all 5’s
3 in shape in R5C6; 3 in shape in R67C3; R3C2 = 3; R1C5 = 3

Point to next
R2C7 = 7; R2C6 = 6; R1C4 = 6; R2C5 = 1; R2C4 = 5
R6C3 = 6; R6C5 = 7

Palindrome
A shortcut in the puzzle could be that the digit in R3C3 must be the same as in R5C5 and thus in R4C4 too.

Equal
A shortcut in this puzzle could be that the shape in the middle contains 5 digits; The number of odd digits must be even, so the number of even digits is odd, in a grid of 7x7 all even digits are included. Only possible combination is 2,4,6 + 5,7.

Consecutive Circles
Every circle must contain a 4!
Law of leftovers: R13C3 = R7C45; R57C5 = R1C34
R1C4 = 3; R3C3 = 3; R5C5 = 2; R7C4 = 2
Center circle does not contain 2 and 3 so must be 4-5-6-7. R4C4 = 1.
Bottom circle has {23} in R6C67; 4 in R7C67; 4 in R56C2; 4 in R12C1.
2 in U-shape left locked in R4C12; 2 in U-shape right locked in R23C7
R6C6 = 2; R6C7 = 3; R7C7 = 4
R4C6 = 3; R5C1 = 3; R2C2 = 3; R2C1 = 4; R1C2 = 6; R1C1 = 5
R23C6 = {56}; R7C6 = 1; R1C6 = 4; R5C6 = 7
R2C7 = 2; R56C2 = {14}
4 in R3 locked in C45 :=> R4C5 <>4 :=> 4 in C5 in R3C5 and 4 in R4C3.
R4C5 = 6; R5C4 = 5; R3C4 = 7

Killer
Counting of cages in rows 1/2; 6/7; columns 1/2; 6/7 leads to information on the sum of the three cells in those rows/columns that are not included in the cages.
R1C1 + R2C12 have a sum of 6; R3C2 = 4
R6C67 + R7C7 have a sum of 11; R5C6 = 1
R67C7 have a sum of 11, combination of {47} or {56}
R6C6 + R7C67 have a sum of 14. R7C6 is three higher than R6C7. R6C7 = 4; R7C6 = 7; R5C7 = 7
R6C6 + R7C7 = {25}; R67C5 = {36}
4 locked in R7C34; R4C1 = 4
7 locked in R1C45; R2C3 = 7; R1C23 = {56}
4 in C3 in R7C3
4 locked in R2C45; R1C6 = 4; R1C7 = 3; R2C67 = {56}
Cage in R34C67 = {1236}; R3C5 = 5
1 locked in R6C34; cage [19] is {1567}
R5C1 + R7C4 = {23}
R2C1 + R5C1 form a pair on {23}; R1C1 = 1; R7C2 = 1
5 locked in R67C1
Cage [15] in R45C12 cannot contain 5. R1C2 = 5; R2C1 = 2

Low
R6C5 = 1; R3C5 = 2; R3C3 = 1; R3C2 = 3; R24C2 = {12}
R2C5 = 7; R2C1 = 6; R1C3 = 7; R4C4 = 7.
R5C7 = 1; R7C4 = 1; R1C1 = 1; R2C6 = 1; R4C2 = 1; R2C2 = 2.
R5C4 = 2; R1C6 = 2; R4C1 = 2; R7C3 = 2.
3 locked in R12C4; R6C3 = 3; R6C1 = 4; R7C1 = 3; all 3’s
4 locked in R57C2; R1C2 = 5; R2C3 = 4; R2C7 = 5

Sum 100
So far, I have seen only sum 100 sudokus in normal 9x9-grids; you can’t use all cells to be part of grey cell-combinations because of the total sum of 45. Therefore I deliberately wanted to create a puzzle where all cells of at least one row were included in sums, since the total sum of 28 can perfectly be split in 20 + 8(0); R1C146 have a sum of 8 and R1C2357 have a sum of 20. For R7 you can also determine the sums of 8 and 20 in the same way.
Row 3 and 5 contain two sum-combinations. The digits in the left cells of these combinations add up to 9, the right cells add up to 10.
R5C4 = 2. Now the combination of the digits in R5C1 + R5C5 can only by {45} since {36} would block both possibilities to build 10 in R5C2 + R5C6. So R5C1 = 4; R5C5 = 5. R5C26 = {37}; R5C37 = {16}
Law of leftovers: R15C3 = R7C45. Since the 1 in the bottom row must be in the sum of 8 and thus in R7C136, 1 is not in R5C3. R5C3 = 6, R5C7 = 1.
Law of leftovers: R45C7 = R3C15. R3C5 = 1.
1 must go in R1C4 (in both row and column it is the only place left).

How balanced do you think the puzzle types of this test were?	Perfectly balanced
What was your opinion of the distribution of easy/hard puzzles?	Just right
What did you think about the puzzle quality of the test?	Very nice
How did you feel about the length / time limit for this test?	Way too many puzzles (too little time)
Of the puzzles you solved/attempted, how well did the point values reflect the difficulty?	Most puzzles were worth the right amount
What was your opinion of the booklet formatting and printing?	Just right

Can the password please be removed? I didn't have a chance to compete in this contest, but still would like to solve the puzzles. Thanks!

Will the scores in this test be included in the ratings

It will be removed. For now: RS7x7=49

Have fun!

Yes, but it may take some time before it is done.

Thanks Richard! They were fun. I do like the 7x7 format!

Attempted the test . Just loved it ..... but now want to solve the puzzles at leisure - PW to the PUZZLE BOOKLET PLEASE !

Removed the password.

thanks...
Will try to go thru the Solving tips now

Am working slowly thru the 7*7 Sudoku tips.
Vide Quad sums . A. could not understand R7C1=4; R1C7 = 4 ( 4 in C1 R 6 and 7 , similarly 4 in C 7 R 1 and 2 ) . B. How was 2 ruled out in R67C6 ( you mentioned 2 in C7 in R67; 3 in R6 in C67 ) - 3 is understood in R6C67 .
Vide Consequtive . R2 contains 2 dots which must have pairs {12} and {45} . How were pairs 56 , 67 ruled out ? Then next how was [ R2C13 = {67}; R13C2 = {15} ] derived ?

I only saw vjain's questions today, so here is a belated response:



There is a 4 in R3C2. Because of that, the 4 in R1 must be somewhere in R1C3-7. Those cells are all in the same irregular area, so the 4 is ‘locked’ there. Hence, the 4 is not in R23C7. There is also a 4 in R5C6; because of being in that shape, no 4 in R4567C7. The only possibility for the 4 in C7 is now R1C7. This works exactly the same for R7C1.




The 2 in C7 must be in the irregular area bottom right, more specific, in R4567C7. The sum-dot in that irregular area must contain the series 1-2-3-6 since the 4 is already placed elsewhere in that area. Combining both previous conclusions, the 2 in that irregular area goes in R67C7.



There is a given 3 in row 2 without a dot next to it. This means that the 2 and the 4 in row 2 have to go in two of the four remaining cells in that row. There are two dots there, both connecting two different cells. The 2 and 4 cannot be placed next to the same dot, so this means that in those 4 cells you find the pairs {12} and {45}, without knowing which pair goes where at this stage.
The 6 and 7 in row 2 have to go in R2C13.
Since the 2 and 4 also cannot be placed above or below the given 3 (no dots there too), the 1 and 5 have to go there.


Hope this helps, and sorry for the delay. Must be my holiday that I didn't notice your questions before!