The New Sudoku Players' Forum

[Richard S]

I have made two new cross sums sudoku's, this time with the letter H (with a bit of fantasy...).

Two different toroidal patterns, two different sum-possibilities.

1.
Fill in the digits 0 and 1 to 8 in every row, column and outlined coloured toroidal region. Outside the grid you find the sums of the digits you come across first from that side. The first number (in brackets) is the number of digits (cells) you have to add up, the second number is the sum of those digits/cells.

Rules of kakuro and toroidal sudoku's appear both.






2.
Different toroidal pattern, different digits to fill in: the zero appears twice, so fill in 0, 0, 1 to 7 in every row, column and outlined region.




Enjoy it.
Richard

[udosuk]

Here are pics with considerably smaller sizes:





Richard, using Excel and Paint and saving the files as .png format, you can create light-sized images with very good quality...

Pics reuploaded

[Richard S]

Thanks Udosuk,

The pics look far better than my scanned images. I will use MSPaint to post my next puzzles soon.

How do you like them?

Regards, Richard

[udosuk]

How do you like them?

Looks tough... Got some projects in hand (and tons of puzzles waiting) so will probably have to put them on hold until the next holiday...

Some more tips: from Excel you could directly copy & paste an array of cells to Paint, and then you could use the "image attributes" to fix the width and height of the pic. Alternatively you could press "Shift-PrtSc" (the button should be on the top right corner of your keyboard) to copy the image of your whole screen (1280x1024 or whatever) and then paste it on Paint. Then you can select whatever area you want and copy and paste it to a new image...

[Richard S]

Also for this two puzzles I've made a walkthrough. Both in tiny text. Copy and paste in a text editor to view.

First for the 007-puzzle:


(4)4 in R3 = 0013
R3C4 = 1
R3C123 = 003

R3C89 = 26
R3C567 = 457

(2)12 in R1 = 57
R3C1 = 3 because of (3)14
R3C23 = 00

R1C1 = 5
R1C2 = 7
R2C1 = 6

(2)9 in C3 = 27
R1C3 = 2
R2C3 = 7

R6C9 = 3
R4C2 = 3
R4C3 = 4
R4C1 = 1
R9C1 = 0
R5C2 = 4

R4567C8 = 0156
R3C8 = 2
R3C9 = 6

(5)9 in R5 = 00135
R5C3 = 6

4 in nonet in R189C4
4 in nonet in R6789C5
4 in nonet in R67C6
4 in nonet:> R3C7

R3C6 cannot be 5 :>R3C6 = 7

R3C5 = 5
R1C6 = 3
R2C6 = 5
R2C9 = 0

(5)9 in row 5 = 00135
R5C9 = 5
R4C9 = 7
R6C2 =2



And for the 018 puzzle:


Sum of R4567C6 = 21
Sum of R3456C4 = 8
Sum of R456C4 = 4
R3C4 = 4

R5C789 = 468
R4C9 = 7
R5C123 = 015
R5C4 = 3
R5C6 = 2

R46C4 = 01
R467C6 = 568

0 in nonet in R6789C5
R89C6 = 34
R123C6 = 017
R3C689 = 123
R3C6 = 1

Cage (4)14 contains a 4
R5C9 = 6

0 in nonet in R6-9C5
0 in nonet in R3-6C7
R67C9 = 12
R6C9 = 2
R7C9 = 1
R3C9 = 3
R3C8 = 2
R12C9 = 04

R46C7 = 03
(3)7 in R2 = 124
R2C3 = 4
R2C12 = 12
R1C9 = 4
R2C9 = 0
R1C6 = 0
R2C6 = 7

2 in nonet in R189C1
R2C1 = 1
R2C2 = 2

4 in nonet in R6-9C5
R8C6 = 3
R9C6 = 4
R8C5 = 4
R9C5 = 4
R9C5 = 0

R9C2 = 3
R8C2 = 1
R5C3 = 1

R67C2 = 04
R6C2 = 4
R7C2 = 0
R5C2 = 5

R9C9 = 5
R8C9 = 8

Don't hesitate to ask if you still got stuck...

Realise both puzzles are toroidal, meaning they have repeating numbers in the left and right leg of the H.
In this case, in both the puzzles two numbers are repeated in both legs.