Can You Solve This Without Trial and Error?

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Re: Can You Solve This Without Trial and Error?

Postby denis_berthier » Fri May 17, 2013 6:53 am

saul wrote:I've never seen a puzzle that you could separate into 5 parts at the very beginning, using "difference doublarities." As you've pointed out, there are lots of puzzles that break into 5 sub-puzzles at the start using 1-cuts, but this is the first I've ever seen, or at least recognized, like this.


The X-shape and the low density of white cells are very particular also.
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Re: Can You Solve This Without Trial and Error?

Postby Smythe Dakota » Sat May 18, 2013 3:33 am

saul wrote: .... I've never seen a puzzle that you could separate into 5 parts at the very beginning, using "difference doublarities." As you've pointed out, there are lots of puzzles that break into 5 sub-puzzles at the start using 1-cuts, but this is the first I've ever seen, or at least recognized, like this.

Hmm, you might be right. It's probably difficult, in a Mepham-sized puzzle, to have four difference doubularities dividing the puzzle into five parts.

Today's (May 17, 2013) Mepham puzzle has two sum doubularities, one each in two corners, that chop a few cells away from the main puzzle. It also has two difference doubularities, in the other two corners, which divide the remaining puzzle into three roughly equal parts.

Of course, the part of a puzzle chopped off by a doubularity, unlike the part chopped off by a singularity, doesn't necessarily allow solving that part of the puzzle by itself. In this case, though, they both did.

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Re: Can You Solve This Without Trial and Error?

Postby saul » Mon May 27, 2013 1:49 pm

Denis,

I wonder how ATK's M1672 would rate on your difficulty scale? It's just an 8-by-8, but I can't get anywhere at all with it.
Image

I can't believe that this is rated medium. Just when I start to think that I'm getting better at these puzzles, one like this comes along.
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g-whips example

Postby denis_berthier » Mon May 27, 2013 3:33 pm

saul wrote:I wonder how ATK's M1672 would rate on your difficulty scale?

Rating it M is inconsistent with the other M puzzles.
It can't be solved by whips alone. It requires g-whips and it has gW rating 3.
A g-whip is much like a whip, but in a g-whip, you can have several right-linking candidates, provided that all are linked to the next left-linking candidate.
I give you the beginning of my solution:

Hidden Text: Show
Code: Select all
*****  KakuRules 1.2 based on CSP-Rules 1.2, config: W-g2W   *****
naked-singles ==> r7c8 = 9, r7c1 = 7, r8c1 = 9, r8c8 = 8, r8c7 = 9, r1c8 = 6, vr0c8 = 68, r2c8 = 8
cell-to-horiz-ctr  ==> hr3c1 <> 567
cell-to-horiz-ctr  ==> hr4c0 <> 35
ctr-to-horiz-sector  ==> r4c2 <> 3
ctr-to-horiz-sector  ==> r4c2 <> 5
cell-to-verti-ctr  ==> vr0c6 <> 2457
verti-sector-to-ctr  ==> vr0c5 <> 18
ctr-to-verti-sector  ==> r1c5 <> 1
ctr-to-verti-sector  ==> r2c5 <> 1
biv-chain[2]: hr4c0{n17 n26} - r4c1{n1 n2} ==> r4c2 <> 2
biv-chain[2]: r4c1{n1 n2} - hr4c0{n17 n26} ==> r4c2 <> 1
whip[2]: hr1c0{n25 n16} - r1c1{n2 .} ==> r1c2 <> 1
whip[2]: r3c4{n7 n1} - r3c3{n1 .} ==> hr3c1 <> 189
ctr-to-horiz-sector  ==> r3c2 <> 1
ctr-to-horiz-sector  ==> r3c3 <> 1
ctr-to-horiz-sector  ==> r3c4 <> 1
whip[2]: hr3c1{n369 n279} - r3c3{n3 .} ==> r3c4 <> 2, r3c2 <> 2
whip[2]: hr5c6{n16 n25} - r5c7{n1 .} ==> r5c8 <> 5
whip[2]: vr0c1{n15 n24} - r2c1{n1 .} ==> r1c1 <> 2
whip[2]: hr1c0{n16 n25} - r1c1{n4 .} ==> r1c2 <> 5
whip[2]: vr6c4{n16 n25} - r8c4{n3 .} ==> r7c4 <> 5
whip[2]: vr6c4{n25 n16} - r8c4{n3 .} ==> r7c4 <> 6
whip[2]: vr0c5{n27 n36} - r1c5{n2 .} ==> r2c5 <> 3
whip[2]: vr0c5{n36 n27} - r1c5{n3 .} ==> r2c5 <> 2
whip[2]: vr5c6{n29 n47} - r7c6{n2 .} ==> r6c6 <> 4
whip[2]: vr5c6{n29 n38} - r7c6{n2 .} ==> r6c6 <> 3
whip[2]: vr5c6{n38 n29} - r7c6{n3 .} ==> r6c6 <> 2
horiz-sector-to-ctr  ==> hr6c4 <> 278
horiz-sector-to-ctr  ==> hr6c4 <> 269
whip[2]: vr3c8{n25 n16} - r4c8{n3 .} ==> r5c8 <> 6
whip[2]: hr5c6{n25 n16} - r5c8{n2 .} ==> r5c7 <> 1
entering level g2W2 with <Fact-38763>
g2-whip[2]: r3c3{n3 n2|n4} - hr3c1{n378 .} ==> r3c2 <> 3, r3c4 <> 3
g2-whip[2]: r3c3{n4 n2|n3} - hr3c1{n468 .} ==> r3c4 <> 4

After this, you can find more whips[2] and g-whips[2] in column 4 and row 3
After you'll need whips[3]
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Re: g-whips example

Postby saul » Mon May 27, 2013 4:05 pm

denis_berthier wrote:I give you the beginning of my solution:

Thank you!
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Re: Can You Solve This Without Trial and Error?

Postby Smythe Dakota » Thu Jun 06, 2013 8:01 am

saul wrote: .... I can't believe that this is rated medium. Just when I start to think that I'm getting better at these puzzles, one like this comes along.

By contrast, how about ATK's M32872? It, too, is 8x8, but it has four singularities, and is duck-soup simple.

I think ATK must be assigning difficulty levels by drawing random numbers out of a hat.

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Re: Can You Solve This Without Trial and Error?

Postby saul » Thu Jun 06, 2013 3:01 pm

Smythe Dakota wrote:By contrast, how about ATK's M32872? It, too, is 8x8, but it has four singularities, and is duck-soup simple.
I think ATK must be assigning difficulty levels by drawing random numbers out of a hat.

It certainly seems that they don't take surface sums into account in assigning difficulty levels.
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Re: Can You Solve This Without Trial and Error?

Postby denis_berthier » Fri Jun 07, 2013 6:18 am

Smythe Dakota wrote:By contrast, how about ATK's M32872? It, too, is 8x8, but it has four singularities, and is duck-soup simple.

Don't you count 6 "singularities", with two 2x2 included in larger ones?

saul wrote:It certainly seems that they don't take surface sums into account in assigning difficulty levels.

I haven't explored the Medium puzzles in as much detail as the Hard ones, but in the latter (which are generally larger and sometimes very hard), surface sums rarely appear. It may just be because their generation software isn't specifically biased to create puzzles with surface sums. I think these must be rare in large puzzles if black cells are put randomly on the grid (with some reasonable density).

If their rating system is based on chains, it isn't very consistent either: some H ones are very easy.
But it remains my preferred website for Kakuro.
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Re: Can You Solve This Without Trial and Error?

Postby Smythe Dakota » Fri Jun 07, 2013 11:20 am

denis_berthier wrote: .... Don't you count 6 "singularities", with two 2x2 included in larger ones? ....

Once you remove the 4 obvious singularities, one near each corner, all you're left with is doubularities -- zillions of them. But they're all trivial. To be precise, there are 6, one in each corner piece, and two in the remaining center piece. Solving each one requires no special techniques, other than tiny amounts of T&E, too small to even really be called T&E.

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Re: Can You Solve This Without Trial and Error?

Postby denis_berthier » Fri Jun 07, 2013 11:45 am

Smythe Dakota wrote:
denis_berthier wrote: .... Don't you count 6 "singularities", with two 2x2 included in larger ones? ....

Once you remove the 4 obvious singularities, one near each corner, all you're left with is doubularities -- zillions of them.

OK, I had forgotten we are not counting the same things. I was thinking of 6 six 1-cuts.
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Re: Can You Solve This Without Trial and Error?

Postby Smythe Dakota » Sat Jun 08, 2013 8:55 am

denis_berthier wrote: .... OK, I had forgotten we are not counting the same things. I was thinking of 6 six 1-cuts.

I assume my 4 singularities are 4 of your 1-cuts. They are at r2c3, r4c7, r5c2, r7c6. What are the other 2?

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Re: Can You Solve This Without Trial and Error?

Postby denis_berthier » Sat Jun 08, 2013 10:15 am

Smythe Dakota wrote:
denis_berthier wrote: .... OK, I had forgotten we are not counting the same things. I was thinking of 6 six 1-cuts.

I assume my 4 singularities are 4 of your 1-cuts. They are at r2c3, r4c7, r5c2, r7c6. What are the other 2?

Bill Smythe


Actually, my numbering is different because I take the 1-cut cell inside the smaller surface:
r2c2, r2c6, r3c7, r7c3, r6c2, r7c7
The 2 you don't count are the small 2x2 squares inside the bigger ones. But I agree they are not useful.
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Re: Can You Solve This Without Trial and Error?

Postby Smythe Dakota » Sat Jun 08, 2013 8:53 pm

denis_berthier wrote: .... Actually, my numbering is different because I take the 1-cut cell inside the smaller surface:
r2c2, r2c6, r3c7, r7c3, r6c2, r7c7
The 2 you don't count are the small 2x2 squares inside the bigger ones. But I agree they are not useful.

I don't see how you could call, for example, r7c7 a 1-cut in any helpful way. You could just as easily call r7c8 a 1-cut.

In fact, if you interchange the 17 and 14 sums above r7c7 and r7c8, you have an entirely equivalent puzzle.

Cutting r7c7 out of the puzzle would, indeed, divide the remaining puzzle topologically, or graph-theory-wise, into two separate pieces of paper. But these two pieces are not separate kakuro-wise! r7c6 and r7c8 remain connected when r7c7 is removed, just as r7c6 and r7c7 remain connected when r7c8 is removed, because they're in the same sum!

The graph-theory notion of 1-cut is, quite simply, irrelevant in kakuro. I really think you need to use my "singularities" instead -- but you can still call them 1-cuts, if you prefer.

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Re: Can You Solve This Without Trial and Error?

Postby denis_berthier » Sun Jun 09, 2013 4:14 am

Smythe Dakota wrote:I don't see how you could call, for example, r7c7 a 1-cut in any helpful way.

It immediately gives r7c6 = (19+17)-(17+14)=5
I thought you would have considered this as a "singularity".

Smythe Dakota wrote:You could just as easily call r7c8 a 1-cut.

No, it doesn't separate the graph.
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Re: Can You Solve This Without Trial and Error?

Postby saul » Sun Jun 09, 2013 1:33 pm

Smythe Dakota wrote:The graph-theory notion of 1-cut is, quite simply, irrelevant in kakuro.
Bill Smythe

Not so.
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