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TK Sin 19

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HATMAN

Post subject: TK Sin 19

Posted: Mon Mar 27, 2023 6:45 pm

Grand Master

Joined: Wed Apr 30, 2008 9:45 pm
Posts: 694
Location: Saudi Arabia

Triankle Sindoku 19

This is a Triankle Puzzle with Sindoku clues. They work well together. This is the easiest one I have created - a straightforward paper solvable once you are used to the format.

Triankle:
The numbers are from 0 to 9.
There are 8 triangular nonets covering twelve rows and 11 columns.
The two different numbers in the centre two grey cells repeat horizontally and vertically. I.e. in row six there are eleven numbers so nine different numbers with the grey number twice and row seven the same. In column 6 there are 12 numbers eight different numbers with the two grey numbers twice.
The other eight numbers do not repeat anywhere.
The repeating numbers are present in every nonet.
Each of the other eight numbers is absent in one and only one nonet.
So Semi-symmetric.
There are five pairs of numbers which are unknown.
If a cell contains a number the opposite cell must contain it or its partner.

Sindoku
The clues are based on singletons i.e. a number with no adjacent numbers.
The clues at the corners refer to the four adjacent cells
If one number is given it is a singleton and the other three numbers are consecutive.
If two numbers are given they are singletons and the other two numbers are consecutive.
If the corner is blue all four numbers are singletons.
If the corner is pink there are two pairs of consecutive numbers.
If the corner is red there are four consecutive numbers.
if the corner is green there is a repeat number or perhaps two.
If nothing is given at a corner - nothing is implied.

Note 0 is consecutive to 1 and 9 so 901 is a sequence.

(I will use the colours in later ones.)

I attach the Triankle base file in JSudoku.

Attachments:

TKSin Base.zip [3.05 KiB]
Downloaded 693 times

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Andrew

Post subject: Re: TK Sin 19

Posted: Thu Nov 09, 2023 3:16 am

Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada

Thanks HATMAN

Semi-symmetric isn't my favourite type of clue, often harder than symmetric.

After an easy start it gets slow later when using semi-symmetry and groups of three adjacent numbers but I think I've managed to solve it.

Take care with semi-symmetry and groups of three adjacent numbers, it's easy to go wrong with them, and remember that each of the numbers in R6C6 and R7C6 must be in each of the nonets/triangles; if I'd remembered that sooner my solving path (not posted) would have been a bit shorter.

Here is my solution:

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