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 Post subject: Repeat Diagonals D\ 1
PostPosted: Sat Jul 23, 2016 5:24 pm 
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Grand Master
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Repeat Diagonals D\ 1

It is latin square; not sudoku.

The black diagonal is standard non-repeat.
The blue diagonals contain exactly three numbers.
The red diagonals contain exactly two numbers.
The green diagonals contain exactly two numbers.
The yellow diagonals contain exactly two numbers.

If a single diagonal contains exactly two numbers they alternate along the diagonal.

Views on difficulty please.


Image

Solution:
128967435
291783654
816249573
972315846
684152397
739524168
465831729
357496281
543678912


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PostPosted: Sat Jul 30, 2016 1:06 am 
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Another ingenious puzzle! I made this much more difficult for myself at first than it was because I didn't realize that each of the same coloured diagonals contain the SAME repeated numbers. I got very stuck very quickly! Once I realized what was going on - it was quite straightforward. Thanks HATMAN!

Oops - Corrections courtesy of deficiencies pointed out by HATMAN.
Additions shown in red.
Deletions shown in italic blue.

Repeat Diagonals D\ 1 WT:
I abbreviate the colours as (R)ed, (G)reen, (B)lue, (Y)ellow.

1. r4 shows that the R, B, G numbers are all different from each other and one of the Y numbers.
Y\ at r1c6..r4c9 may be alternate numbers or all the same number.
If the latter then Y\ at r6c1..r9c4 is all the other Y number or alternate numbers.
-> At least one of r1c6, r6c1, r9c4 is a different number than the Y number at r4c9.
-> At least one of c6, c4, or r6 shows that the R, B, G numbers are all different from each other and the other of the Y numbers.

c6 shows the R, B, G numbers are all different from each other and the other of the Y numbers.
(Since r1c6 and r4c9 in Y are different).

-> All colours have different numbers and between them they have all the numbers 1 -> 9.

2. 4(2) = {13}
R\ at r1c2..r8c9 cannot be all the same number since that would leave no place for that number on main D\.
-> regardless of whether 4(2) is [13] or [31], r5c6 cannot be 1
-> Since r1c2 = r5c6 -> 3(2) = [12]. r1c2 = 2 = r3c4 = r5c6 = r7c8.
-> HS 2 in D\ -> r9c9 = 2
Other R number in r2c3, r4c5 etc. in D\ can only be in r1c1 = 1
-> R diags are both (2,1, etc.)
->4(2) = [31]
-> two of the three B numbers are (34).

3. 24(3) shows the two G numbers are from (789)
-> 22(3) shows the two G numbers must be (89) or (79)
-> The other B number in r3c7 from (56)
Since Y numbers cannot be the same as any of the numbers in 22(3), or from (1234)
-> Y numbers from (58) or (67)

4. In r5, r5c9 must be a Y number
-> In r6, r6c8 must be the other Y number
-> In 21(3) r5c9 + r6c8 = +13
-> r6c9 = 8
-> Y numbers are (67)
Also -> 22(3) = [958] (I.e., B numbers are (345)).
-> 24(3) = [978]
-> r1c5 (a Y number) = 6
Since r5c9 and r6c8 are the two different Y numbers -> r3c8 and r4c9 are also the two different Y numbers.
-> Y\ at r1c6..r4c9 are alternating numbers.

-> r1c6 = r3c8 = 7 and r2c7 = r4c9 = 6
-> 21(3) = [768]

5. Easy from here.
e.g., for Blue
r1c7 = 4
r5c5 = 5
-> r5c3 = 4
Also r4c4 = 3 (D\)
-> r6c4 = 5
-> r8c4 = 4
etc.


Last edited by wellbeback on Sat Aug 06, 2016 8:15 am, edited 3 times in total.

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PostPosted: Tue Aug 02, 2016 7:29 pm 
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wellbeback

proof 2: second line, as my maths masters always said to me "correct but lines missing from the proof - no points" consider D\@r1c2 could be 11111111, it needs the normal black diagonal to disprove it..

I liked the proof of a unique full number set. I did not see this proof or use it - so your solution was simpler.


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PostPosted: Fri Aug 05, 2016 10:16 pm 
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Oops - yes! :oops: Thanks HATMAN.

The deficiency you pointed out also applies to Step 1! I have updated the WT.


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PostPosted: Tue Nov 29, 2016 2:57 am 
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Thanks HATMAN for another interesting and ingenious puzzle.

This is clearly a puzzle where one has to read the rules exactly as stated, not reading any more into them which was a great risk in this case, and using all the information which they provide. I've made more comments in my walkthrough.

I was impressed by wellbeback's steps 1 and 4. :applause: They make his walkthrough significantly different from my longer and somewhat more routine one.

Here is my walkthrough:
This is a latin square, no nonets.
The black diagonal is standard non-repeat.
The blue diagonals contain exactly three numbers.
The red diagonals contain exactly two numbers.
The green diagonals contain exactly two numbers.
The yellow diagonals contain exactly two numbers.

If any diagonal contains exactly two numbers they alternate along the diagonal.

I think I might initially have been confused by the rules but hopefully would quickly have realised that since red, green and yellow diagonals contain exactly two numbers, it means that each colour contains exactly the same two numbers for the diagonals of that colour. I had initially assumed that this must mean that each diagonal of a colour must contain both numbers but, as HATMAN pointed out in a later post commenting on wellbeback’s walkthrough, this doesn’t necessarily follow. Therefore steps 2 and 4 have been clarified or slightly amended. Similarly I’d initially assumed that each of the blue diagonals contained the same three numbers but, as I found later, that wasn’t a valid assumption.

Prelims

a) R1C12 = {12}
b) R56C6 = {15/24}
c) R56C7 = {13}
d) 22(3) cage at R3C6 = {589/679}
e) 21(3) cage at R5C9 = {489/579/678}, no 1,2,3
f) 24(3) cage at R8C5 = {789}

Steps resulting from Prelims
1a. Naked pair {12} in R1C12, locked for R1
1b. Naked pair {13} in R56C7, locked for C7

2. R1C2 = {12}, R6C7 = {13} -> red diagonal at R1C2 must either contain eight 1s or two of 1,2,3, alternating but cannot contain eight 1s because one of the cells on D\ must contain 1 -> red diagonal at R2C1 must also contain two of 1,2,3 alternating
2a. More specifically R1C2 = {12} -> R3C4, R5C6 and R7C8 = {12}, clean-up: no 1,2 in R6C6
2b. and R6C7 = {13} -> R2C3, R4C5 and R8C9 = {13}
2c. Red diagonal at R2C1 must contain two of 1,2,3, no 1,3 in R8C7 -> R8C7 = 2 -> R2C1, R4C3 and R6C5 = 2
2d. R1C1 = 1 -> no other 1 on black diagonal at R1C1
2e. R1C2 = 2 -> R3C4, R5C6 and R7C8 = 2
2f. R3C2 = {13} -> R5C4, R7C6 and R9C8 = {13}
2g. Naked pair {13} in R5C47, locked for R5
2h. R9C9 = 2 (hidden single on black diagonal at R1C1)
2i. R5C6 = 2 -> R6C6 = 4 -> no other 4 on black diagonal at R1C1
[R6C6 = 4 means that 4 is one of the three numbers on blue diagonals, but at this stage I won’t try to interpret that as I might get it wrong. My current assumption is that each of the blue diagonals contains 4, but I might be misinterpreting HATMAN’s rules. Also I’ll have to remember the final rule, if any diagonal contains two numbers they must alternate; this may possibly apply to one or more of the blue diagonals.]

3. 24(3) cage at R8C5 = {789} -> green diagonal at R4C1 must contain two of 7,8,9 alternating -> green diagonal at R1C4 must also contain two of 7,8,9 alternating
3a. 22(3) cage at R3C6 = {589/679} -> R3C7 = {56}
3b. Naked triple {789} in R289C5, locked for C5
3c. 9 in 22(3) cage is on green diagonal at R1C4 which therefore contains 9 and one of 7,8 alternating -> green diagonal at R4C1 must contain 9 and one of 7,8 alternating
3d. Both green diagonals are in C456 -> 9 in one of R17C4, one of R28C5 and one of R39C6, no other 9 in C456
3e. Both green diagonals are in R456 -> 9 in one of R4C17, one of R5C28 and one of R6C39, no other 9 in R456
3f. 9 in one of R6C3 + R7C4, CPE no 9 in R7C3
3g. R3C7 = {56}, R5C7 = {13} and R6C6 = 4 are all on blue diagonals and provide three numbers for these diagonals -> no 7,8,9 on blue diagonals

4. Either yellow diagonal at R1C6 contains four identical numbers or two alternating numbers, similarly for yellow diagonal at R6C1
4a. No 1,4,9 in R1C6 -> no 1,4,9 in R3C8
4b. No 1,3 in R2C7 -> no 1,3 in R4C9
4c. No 9 in R4C9 -> no 9 in R2C7
4d. No 1,4,9 in R6C1 -> no 1,4,9 in R8C3
4e. No 9 in R9C4 -> no 9 in R7C2

[I ought to have spotted this earlier; it would have slightly simplified step 4.]
5. Red diagonal at R1C2 must be alternating 1,2 (cannot be alternating 2,3 because black diagonal at R1C1 must contain 3) -> R2C3, R4C5, R6C7 and R7C8 = 1
5a. Red diagonal at R2C1 must also be alternating 1,2 -> R3C2, R5C4, R7C6 and R9C8 = 1
5b. R5C7 = 3

6. R5C7 = 3, R6C6 = 4 means either that the blue diagonal at R3C9 contains alternating 3,4 or contains 3,4 and one of 5,6
6a. Blue diagonals can only contain one of 5,6 -> whichever of 5,6 is in R3C7 must be in R5C5, CPE no 5,6 in R3C5
6b. Blue diagonals don’t contain both of 5,6, R5C5 = {56} -> R5C3 cannot be {56} -> R5C3 = 4
6c. 21(3) cage at R5C9 = {579/678} -> no 7 in R5C8 because green diagonal must have two alternating digits -> no 7 in R1C4 + R3C6
6d. R5C5 = {56}, blue diagonals don’t contain both of 5,6 -> R4C4 = 3, placed for black diagonal
6e. R6C4 = {56}, blue diagonals don’t contain both of 5,6 -> R6C2 = 3
6f. R5C5 = {56}, blue diagonals don’t contain both of 5,6 -> no 5,6 in R7C5
6g. Naked pair {34} in R37C5, locked for C5
6h. R6C4 = {56}, blue diagonals don’t contain both of 5,6 -> no 5,6 in R8C4 -> R8C4 = 4
6i. R3C7 = {56}, blue diagonals don’t contain both of 5,6 -> no 5,6 in R3C9
6j. Naked pair {34} in R3C59, locked for R3
6k. R4C6 = {56}, blue diagonals don’t contain both of 5,6 -> no 5,6 in R4C8 -> R4C8 = 4
6l. R4C6 = {56}, blue diagonals don’t contain both of 5,6 -> no 5,6 in R2C6 -> R2C6 = 3
6m. R3C7 = {56}, blue diagonals don’t contain both of 5,6 -> no 5,6 in R1C7 -> R1C7 = 4
6n. R2C9 = 4 (hidden single in R2) -> R3C9 = 3, R3C5 = 4, R7C5 = 3
6o. Eight of the digits on blue diagonal at R1C9 must be the same one of {56}, there cannot be two different digits on this diagonal because they would have to alternate -> R9C1 = {56}
6p. R9C23 = [43] (hidden pair in R9), R78C1 = [43] (hidden pair in C1), R1C8 = 3 (hidden single in R1)
[I’d seen step 6 for some time, but it took me a long time to spot the continuation with 6b …]

7. Consider combinations for 21(3) cage at R5C9 = {579/678}
21(3) cage = {579} => R6C9 = 9 => R5C8 = 8 => R3C6 + R4C7 = [89] because of alternative digits on green diagonals => R3C7 = 5 (cage sum)
or 21(3) cage = {678} => 6 in R5C9 + R6C8 => digit on blue diagonal at R1C9 must be 5
-> all digits on blue diagonal must be 5
7a. R3C7 = 5 -> R3C6 + R4C7 = {89}, locked for green diagonals
7b. R1C5 = 6, R9C5 = 7 (hidden single in C5)
7c. 21(3) cage = {678} -> R6C9 = 8, R4C7 = 8, R2C5 = 8, R1C4, R3C6 and R5C8 = 9, R9C6 = 8

8. And finally the yellow diagonals
8a. R9C4 = 6 -> R2C4 = 7 -> R2C7 = 6
8b. R1C6 = 7, R2C7 = 6 -> red diagonal at R1C6 must contain alternative 7,6 -> R3C8 = 7, R4C9 = 6

and the rest is naked singles, without using the diagonals.


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