The New Sudoku Players' Forum

[Mani]

H22361, I just cracked after going astray a couple of times. It requires careful examination of opportunities all around, unlike in some sums where a quadrant can be independently solved first.

[saul]

The second toughie I just cracked, with some external guidance, was H51271. Every Kakuro afficianado should give this a try.

The interesting part was the top-right corner. Other than that the puzzle was fairly straightforward, except for the bottom-left which needed T&E.

This was quite an enjoyable puzzle. I found the same thing as Authority. I had nothing but the SE quadrant, and it looked very tough, but then I noticed a hidden pair in the NE corner, and that gave me everything but the SW corner. The SW corner needed a lot of T&E for such a small area, I thought.

[denis_berthier]

The second toughie I just cracked, with some external guidance, was H51271. Every Kakuro afficianado should give this a try.

The interesting part was the top-right corner. Other than that the puzzle was fairly straightforward, except for the bottom-left which needed T&E.

This was quite an enjoyable puzzle. I found the same thing as Authority. I had nothing but the SE quadrant, and it looked very tough, but then I noticed a hidden pair in the NE corner, and that gave me everything but the SW corner. The SW corner needed a lot of T&E for such a small area, I thought.

Interesting puzzle. It is not solvable by whips (at least by whips of length < 20). But it is solvable by g-whips[2], triplets and whips[3], with no T&E.

There are unusually many g-whip[2] eliminations. These are interesting and easy patterns, because, as the whips[2] (except the x-wings), they remain inside a sector:
g-whip[2]: r1c12{n8 n679} - vr0c12{n1348 .} ==> r4c12 ≠ 8
g-whip[2]: r13c5{n4 n1237} - vr10c5{n468 .} ==> r12c5 ≠ 4
g-whip[2]: vr0c13{n259 n1236789} - r2c13{n5 .} ==> r3c13 ≠ 6, r1c13 ≠ 6
g-whip[2]: r1c13{n7 n1289} - vr0c13{n457 .} ==> r3c13 ≠ 7
g-whip[2]: r1c12{n6 n789} - vr0c12{n2356 .} ==> r2c12 ≠ 6, r3c12 ≠ 6
g-whip[2]: r1c12{n7 n689} - vr0c12{n2347 .} ==> r3c12 ≠ 7

Here are the triplets:
naked-triplets-in-horiz-sector: r3{c1 c4 c9}{n3 n2 n1}
naked-triplets-in-horiz-sector: r2{c9 c10 c11}{n3 n1 n2}

[saul]

[quote="denis_berthier"]
There are unusually many g-whip[2] eliminations. These are interesting and easy patterns, because, as the whips[2] (except the x-wings), they remain inside a sector:
g-whip[2]: r1c12{n8 n679} - vr0c12{n1348 .} ==> r4c12 ≠ 8
<snip>

Which numbering system are you using? Is the top row row 0 or row 1?

[denis_berthier]

There are unusually many g-whip[2] eliminations. These are interesting and easy patterns, because, as the whips[2] (except the x-wings), they remain inside a sector:
g-whip[2]: r1c12{n8 n679} - vr0c12{n1348 .} ==> r4c12 ≠ 8
<snip>

Which numbering system are you using? Is the top row row 0 or row 1?

top (black) row is r0

[Mani]

H6-1942
Like to share my recent success with this problem (ATK). Being an 11X11 puzzle (I don't usually try below hard 13x13), it was unusually tough, not giving in for quite a while, sticking at last with the right bottom quarter. Yet, am happy I solved it without T&E!

[saul]

Bill,

Here's a surface sum I think you'll enjoy. I had gotten to this point on H41862:

H41862.png (113.41 KiB) Viewed 1998 times

and I wanted to finish off the SE corner without using T&E.

Here's what I eventually came up with. Looking at columns, we have r12c7 + r13c7 + r12c8 + r13c8 = 19. Looking at the row, we have r13c7 + r13c8 + r13c9 = 18, so we infer r12c7 + r12c8 = r13c9 + 1. This means that r12c7 and r12c8 sum to 8 or 10. But if the sum is 10, the summands must be 4 and 6, a possibility precluded by the candidates for r12c13. So we get three cells at once: r12c7 = 3, r12c8 = 5, r13c9 = 7, and it's easy to finish the puzzle.

Of course, in hindsight, it's also possible to deduce a contradiction by assuming that r12c7 = 4, but either I didn't try that, or I made a mistake in tracing the implications mentally.

[Mani]

I too reached the same position. Thereafter, there was a simple surface calculation that was needed, (42+18) - 5 - 4 - 24(from 8+79+79) giving 17. Eliminate then 3 from the 356 at the bottom of 14 vertical as 7or9 + 3 + 4or6 cannot give 17. Now, 14 vertical gets resolved as 56,56 and 1,2. Whereafter, it is a breeze to solve the rest. I just completed it, don't want to post answer here though.

[kakuroatk]

Added more puzzles to the medium level in version 2.21... M9**** Series .... Enjoy
(may need to clear browser history to load v2.21)

[Smythe Dakota]

Is there any way to ask specifically for the M9**** puzzles? I tried entering M9**** and M9???? and a few other things, but that didn't work. I had to just choose randomly until a 9**** puzzle came up, which was only about 1 time in 10 (as you would expect).

Bill Smythe

[Authority]

Added more puzzles to the medium level in version 2.21... M9**** Series .... Enjoy
(may need to clear browser history to load v2.21)

Tried one of these out. I don't know if it's the case for all of them, but the one I did was chock-full of naked doubles and triples. Is that what separates them from the easy puzzles?

[kakuroatk]

Is there any way to ask specifically for the M9**** puzzles? I tried entering M9**** and M9???? and a few other things, but that didn't work. I had to just choose randomly until a 9**** puzzle came up, which was only about 1 time in 10 (as you would expect).

I plan on adding that to a future release where you can type M9 and it will give you a random puzzle in the M9****, same with H1 etc...
anytime I release a new version I add the description to the new feature to the bottom of the page...

[saul]

Try M73131. It seems terribly difficult for a 10-by-10 medium. Not only did I have to use trial and error, but I had trouble finding an assumption that would lead to a contradiction with only one level of T&E. When I did find one, the contradiction involved a rather large surface sum.

[kakuroatk]

Try M73131. It seems terribly difficult for a 10-by-10 medium. Not only did I have to use trial and error, but I had trouble finding an assumption that would lead to a contradiction with only one level of T&E. When I did find one, the contradiction involved a rather large surface sum.

Did it myself in about 30 minutes... yes it is a more difficult medium however I was able to solve it logically removing pencilmarks w/o T&E/guessing... I find it easier myself to use the keyboard arrow keys, number keys and caps lock to toggle pencilmark on/off. I also use "C" - copy cell and "V" - paste cell for easier use ... I also use keyboard keys "A" and "D" alot

[saul]

Did it myself in about 30 minutes... yes it is a more difficult medium however I was able to solve it logically removing pencilmarks w/o T&E/guessing... I find it easier myself to use the keyboard arrow keys, number keys and caps lock to toggle pencilmark on/off. I also use "C" - copy cell and "V" - paste cell for easier use ... I also use keyboard keys "A" and "D" alot

30 minutes? I'm impressed. I'm intrigued that you could do it without T&E. Maybe I'll try it again.

I didn't realize you could use caps lock to toggle the state. I use "P". In an app I wrote for KenKen, I didn't use a state. If you hold the shift key down when pressing a number key, it's an answer; otherwise, it's a candidate. I like that a lot.

I use "A" and "D" a lot, too. I hadn't realized until just now that they work if you're positioned in one of the white cells. I had thought you had to be positioned on the clues. is this a recent enhancement, or have I just never tried it out before?