Can You Solve This Without Trial and Error?

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Re: Can You Solve This Without Trial and Error?

Postby saul » Wed Dec 11, 2013 3:31 pm

denis_berthier wrote:
We (in France) even had kings named Henri. (But that's no longer a good job here - too risky !)


Yes, I remember all the attempts on Henri IV's life. We used to have a king in the States too, but it didn't work out.

I just did H5701 without trial and error. It seems as though I really have improved, though I can't say in what way. I don't seem to be using a battery of techniques that I wasn't using originally. Perhaps I'm just not overlooking as many simple inferences as I once did. In this case, I was about to resort to trial and error when I realized I had overlooked a hidden single. I've also noticed that, now that I'm trying hard puzzles, when I go back to the mediums, they seem a good deal easier than before, though not trivial, by any means.
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Re: Can You Solve This Without Trial and Error?

Postby denis_berthier » Wed Dec 11, 2013 11:59 pm

saul wrote:I just did H5701 without trial and error. It seems as though I really have improved, though I can't say in what way. I don't seem to be using a battery of techniques that I wasn't using originally. Perhaps I'm just not overlooking as many simple inferences as I once did. In this case, I was about to resort to trial and error when I realized I had overlooked a hidden single. I've also noticed that, now that I'm trying hard puzzles, when I go back to the mediums, they seem a good deal easier than before, though not trivial, by any means.


Yes, I think you've really improved. This one is in W7, i.e. pretty hard.
Do you use hidden singles for magic digits in non-magic sectors?

saul wrote:I don't seem to be using a battery of techniques that I wasn't using originally.

There doesn't seem to be a great lot of pattern-based techniques for Kakuro. At least, I haven't found many on the web.
Are you using any kind of chain rules ?


It seems I've a large part of the hard atk collection in my database (a few hundred). I had H5711 (an iso of H5701 by vertical symmetry) I also had the previous 2 you proposed.
When I try to find a new one, I often fall on an iso of one I've already solved.
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Re: Can You Solve This Without Trial and Error?

Postby saul » Thu Dec 12, 2013 7:08 pm

Hi,

I need to think a bit before responding to your last post, Denis. Meanwhile, here's puzzle I spent a lot of time on. I got to this point on H82561 without trial and error:

Image

At this point, I couldn't see what to do, so I fell back on trial an error. It's apparent that the assumption r3c6 = 5 has lots of consequences, but I couldn't keep them all in my head. The assumption does lead to a contradiction, but not quickly. I got interested in how many cells I could fill in, without violating any rules, and here's what I came up with:

Image

Perhaps this is another way of gauging the difficulty of a puzzle: how large an incorrect solution exists? Unfortunately, this seems to be even harder than solving the puzzle. (Also, incorrect solution has to be defined properly, to avoid quibbles. I mean that no segment has a repeated digit, and all segments that are completely filled in have the correct sum. Also, it cannot be extended to the correct solution. )

Not every cell above is a consequence of the assumption that r3c6=5. I eventually got stuck and just guessed that r12c6=4, since I knew that that was the correct answer.

Even after this trial an error, the puzzle is no stroll in the park. Once we know r3c6 = 1, we have that r6c6 must be 2 or 5. Since there is no way to make 35 in 6 using a 1 together with a 2 or 5, we can eliminate the 1's from the first segment in r6. Since we can't make 12 in 4 without a 1, r5c3 = 1, and this gets us moving.

Here is the correct answer, for comparison with the incorrect solution above:

Image
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Re: Can You Solve This Without Trial and Error?

Postby denis_berthier » Fri Dec 13, 2013 12:45 am

saul wrote:here's puzzle I spent a lot of time on.

This is one more puzzle of which I had an iso in my collection (H82541 - iso of H82561 via horizontal symmetry).
It is in W4 (with only 1 whip[4]), i.e. moderately hard. But, as you know, it suffices to miss 1 elimination to make it look harder.
Here is my solution in case you want to check if you've missed an easy elimination (it has nothing noticeable, i.e. no rare pattern):
Hidden Text: Show
Code: Select all
naked-single ==> r13c12 = 4
naked-single ==> r12c4 = 2
naked-single ==> r13c4 = 1
naked-single ==> r11c3 = 5
naked-single ==> r9c11 = 4
naked-single ==> r9c4 = 4
naked-single ==> r8c11 = 3
naked-single ==> r4c10 = 5
naked-single ==> r5c10 = 6
naked-single ==> r3c11 = 5
naked-single ==> r2c4 = 5
naked-single ==> r2c1 = 6
naked-single ==> r4c1 = 4
naked-single ==> hr4c0 = 49
naked-single ==> r4c2 = 9
naked-single ==> hr3c10 = 59
naked-single ==> r3c12 = 9
naked-single ==> hr5c7 = 126
naked-single ==> hr9c3 = 134
naked-single ==> hr9c10 = 489
naked-single ==> r9c13 = 8
naked-single ==> r9c12 = 9
naked-single ==> vr8c13 = 12348
naked-single ==> hr11c1 = 15
naked-single ==> r11c2 = 1
naked-single ==> hr13c11 = 14
naked-single ==> r13c13 = 1
hidden-single-in-magic-verti-sector ==> r3c1 = 3
hidden-single-for-magic-digit-in-verti-sector ==> r3c7 = 9
naked-single ==> r3c3 = 7
naked-single ==> r2c3 = 9
hidden-single-for-magic-digit-in-verti-sector ==> r11c5 = 1
naked-single ==> r11c11 = 3
naked-single ==> r12c11 = 1
hidden-single-for-magic-digit-in-verti-sector ==> r9c6 = 1
naked-single ==> r9c5 = 3
naked-single ==> r5c5 = 1
naked-single ==> r5c6 = 3
hidden-single-for-magic-digit-in-verti-sector ==> r13c6 = 3
hidden-single-for-magic-digit-in-verti-sector ==> r11c6 = 2
naked-single ==> r11c13 = 4
hidden-single-for-magic-digit-in-verti-sector ==> r8c1 = 1
ctr-to-verti-sector  ==> r8c2 ≠ 2
ctr-to-verti-sector  ==> r13c2 ≠ 2
hidden-single-for-magic-digit-in-horiz-sector ==> r13c5 = 2
naked-single ==> vr10c5 = 125
naked-single ==> r12c5 = 5
hidden-single-for-magic-digit-in-horiz-sector ==> r13c2 = 4
naked-single ==> vr7c2 = 145789
cell-to-horiz-ctr  ==> hr12c8 ≠ 14568
cell-to-horiz-ctr  ==> hr10c11 ≠ 16
ctr-to-horiz-sector  ==> r10c12 ≠ 6
naked-single ==> r10c12 = 5
naked-single ==> hr10c11 = 25
naked-single ==> r10c13 = 2
naked-single ==> r12c13 = 3
ctr-to-horiz-sector  ==> r12c9 ≠ 2
ctr-to-horiz-sector  ==> r12c10 ≠ 2
cell-to-horiz-ctr  ==> hr6c8 ≠ 34689
cell-to-horiz-ctr  ==> hr6c8 ≠ 35679
ctr-to-horiz-sector  ==> r6c13 ≠ 3
ctr-to-horiz-sector  ==> r6c12 ≠ 3
ctr-to-horiz-sector  ==> r6c9 ≠ 3
cell-to-horiz-ctr  ==> hr6c8 ≠ 45678
cell-to-horiz-ctr  ==> hr1c7 ≠ 245689
cell-to-verti-ctr  ==> vr3c13 ≠ 458
cell-to-verti-ctr  ==> vr3c13 ≠ 467
ctr-to-verti-sector  ==> r6c13 ≠ 4
ctr-to-verti-sector  ==> r5c13 ≠ 4
cell-to-verti-ctr  ==> vr6c8 ≠ 24689
cell-to-verti-ctr  ==> vr6c8 ≠ 25679
verti-sector-to-ctr  ==> vr7c1 ≠ 125
naked-single ==> vr7c1 = 134
cell-to-horiz-ctr  ==> hr9c0 ≠ 57
ctr-to-horiz-sector  ==> r9c2 ≠ 5
ctr-to-horiz-sector  ==> r9c2 ≠ 7
cell-to-horiz-ctr  ==> hr10c0 ≠ 5679
ctr-to-horiz-sector  ==> r10c2 ≠ 5
verti-sector-to-ctr  ==> vr9c7 ≠ 279
verti-sector-to-ctr  ==> vr9c7 ≠ 378
verti-sector-to-ctr  ==> vr9c7 ≠ 369
verti-sector-to-ctr  ==> vr9c7 ≠ 189
verti-sector-to-ctr  ==> vr1c7 ≠ 589
naked-single ==> vr1c7 = 679
naked-pairs-in-horiz-sector: r2{c7 c10}{n6 n7} ==> r2c12 ≠ 7, r2c12 ≠ 6, r2c9 ≠ 7, r2c9 ≠ 6
cell-to-verti-ctr  ==> vr0c9 ≠ 678
ctr-to-verti-sector  ==> r1c9 ≠ 6
ctr-to-verti-sector  ==> r3c9 ≠ 6
naked-pairs-in-horiz-sector: r2{c7 c10}{n6 n7} ==> r2c8 ≠ 6
biv-chain[2]: vr0c10{n69 n78} - r2c10{n6 n7} ==> r1c10 ≠ 7, r1c10 ≠ 6
biv-chain[2]: hr13c0{n123467 n123458} - r13c3{n7 n8} ==> r13c1 ≠ 8
biv-chain[2]: r13c1{n6 n5} - vr11c1{n68 n59} ==> r12c1 ≠ 6
biv-chain[2]: r12n6{c7 c6} - r12n4{c6 c7} ==> r12c7 ≠ 9, r12c7 ≠ 8, r12c7 ≠ 7
biv-chain[2]: r12n4{c6 c7} - r12n6{c7 c6} ==> r12c6 ≠ 7
biv-chain[2]: r12c6{n6 n4} - vr8c6{n12356 n12347} ==> r10c6 ≠ 6
biv-chain[2]: hr4c3{n12347 n12356} - r4c7{n7 n6} ==> r4c8 ≠ 6, r4c5 ≠ 6
biv-chain[2]: hr4c3{n12356 n12347} - r4c7{n6 n7} ==> r4c5 ≠ 7
biv-chain[2]: hr1c0{n15 n24} - r1c1{n1 n2} ==> r1c2 ≠ 2
biv-chain[2]: hr1c0{n24 n15} - r1c1{n2 n1} ==> r1c2 ≠ 1
cell-to-verti-ctr  ==> vr0c2 ≠ 126789
biv-chain[2]: vr0c9{n489 n579} - r2c9{n4 n5} ==> r3c9 ≠ 5
cell-to-verti-ctr  ==> vr0c9 ≠ 579
naked-single ==> vr0c9 = 489
naked-single ==> r2c9 = 4
naked-single ==> r3c9 = 8
naked-single ==> r1c9 = 9
naked-single ==> r1c10 = 8
naked-single ==> vr0c10 = 78
naked-single ==> r2c10 = 7
naked-single ==> r2c7 = 6
naked-single ==> r4c7 = 7
naked-single ==> hr4c3 = 12347
hidden-single-in-magic-horiz-sector ==> r2c12 = 5
ctr-to-verti-sector  ==> r1c12 ≠ 1
ctr-to-verti-sector  ==> r4c12 ≠ 1
ctr-to-verti-sector  ==> r6c12 ≠ 1
horiz-sector-to-ctr  ==> hr1c7 ≠ 145789
ctr-to-horiz-sector  ==> r1c8 ≠ 4
ctr-to-horiz-sector  ==> r1c12 ≠ 4
horiz-sector-to-ctr  ==> hr1c7 ≠ 235789
naked-single ==> hr1c7 = 136789
naked-pairs-in-horiz-sector: r1{c11 c13}{n1 n3} ==> r1c12 ≠ 3, r1c8 ≠ 3, r1c8 ≠ 1
naked-single ==> r1c8 = 6
naked-single ==> r1c12 = 7
naked-single ==> vr0c12 = 245789
naked-single ==> r4c12 = 2
hidden-single-in-magic-verti-sector ==> r2c11 = 2
cell-to-horiz-ctr  ==> hr5c11 ≠ 39
ctr-to-horiz-sector  ==> r5c13 ≠ 3
ctr-to-horiz-sector  ==> r5c13 ≠ 9
cell-to-horiz-ctr  ==> hr5c11 ≠ 57
naked-single ==> hr5c11 = 48
naked-single ==> r5c13 = 8
naked-single ==> r5c12 = 4
naked-single ==> r6c12 = 8
ctr-to-verti-sector  ==> r4c13 ≠ 1
naked-single ==> r4c13 = 3
naked-single ==> r4c11 = 1
naked-single ==> r1c11 = 3
naked-single ==> r1c13 = 1
naked-single ==> r2c13 = 3
naked-single ==> r2c8 = 1
naked-single ==> r5c8 = 2
naked-single ==> r3c8 = 4
naked-single ==> r4c8 = 3
naked-single ==> r5c9 = 1
naked-single ==> r9c9 = 3
naked-single ==> r9c8 = 1
naked-single ==> vr3c13 = 368
naked-single ==> r6c13 = 6
naked-single ==> hr6c8 = 25689
naked-single ==> r6c11 = 2
naked-single ==> r7c11 = 1
naked-single ==> r6c10 = 9
naked-single ==> r6c9 = 5
hidden-single-in-magic-verti-sector ==> r1c4 = 3
naked-single ==> hr1c3 = 34
naked-single ==> r1c5 = 4
naked-single ==> r4c5 = 2
naked-single ==> r4c4 = 1
naked-single ==> r3c4 = 2
naked-single ==> r4c6 = 4
ctr-to-verti-sector  ==> r7c8 ≠ 2
ctr-to-verti-sector  ==> r7c8 ≠ 3
cell-to-verti-ctr  ==> vr0c2 ≠ 234789
whip[2]: hr6c1{n236789 n146789} - r6c6{n2 .} ==> r6c3 ≠ 1
hidden-single-for-magic-digit-in-verti-sector ==> r5c3 = 1
naked-single ==> r5c1 = 2
naked-single ==> r1c1 = 1
naked-single ==> r5c2 = 4
naked-single ==> r1c2 = 5
naked-single ==> hr1c0 = 15
ctr-to-verti-sector  ==> r6c2 ≠ 3
biv-chain[2]: vr0c2{n245679 n145689} - r2c2{n7 n8} ==> r6c2 ≠ 8
biv-chain[2]: r2c2{n7 n8} - vr0c2{n245679 n145689} ==> r6c2 ≠ 7
whip[2]: hr6c1{n236789 n146789} - r6c6{n2 .} ==> r6c2 ≠ 1
whip[2]: hr12c8{n13479 n13569} - r12c12{n7 .} ==> r12c10 ≠ 6, r12c9 ≠ 6
whip[2]: hr13c8{n18 n36} - r13c9{n2 .} ==> r13c10 ≠ 6
whip[2]: hr13c8{n18 n45} - r13c9{n2 .} ==> r13c10 ≠ 4
whip[2]: hr13c8{n27 n18} - r13c9{n2 .} ==> r13c10 ≠ 8
naked-triplets-in-verti-sector: c2{r9 r10 r12}{n9 n8 n7} ==> r8c2 ≠ 9
cell-to-horiz-ctr  ==> hr8c0 ≠ 12469
naked-triplets-in-verti-sector: c2{r9 r10 r12}{n9 n8 n7} ==> r8c2 ≠ 8
cell-to-horiz-ctr  ==> hr8c0 ≠ 13468
naked-triplets-in-verti-sector: c2{r9 r10 r12}{n9 n8 n7} ==> r8c2 ≠ 7
naked-single ==> r8c2 = 5
whip[2]: r8c4{n8 n9} - r8c5{n9 .} ==> hr8c0 ≠ 13459
ctr-to-horiz-sector  ==> r8c3 ≠ 4
ctr-to-horiz-sector  ==> r8c5 ≠ 9
ctr-to-horiz-sector  ==> r8c4 ≠ 9
biv-chain[3]: c6n1{r6 r3} - r3c2{n1 n6} - r6c2{n6 n2} ==> hr6c1 ≠ 345689, r6c6 ≠ 2
hidden-single-in-magic-verti-sector ==> r7c6 = 2
hidden-single-in-magic-verti-sector ==> r13c9 = 2
naked-single ==> hr13c8 = 27
naked-single ==> r13c10 = 7
cell-to-horiz-ctr  ==> hr6c1 ≠ 236789
ctr-to-horiz-sector  ==> r6c3 ≠ 3
ctr-to-horiz-sector  ==> r6c7 ≠ 3
verti-sector-to-ctr  ==> vr9c10 ≠ 1789
verti-sector-to-ctr  ==> vr9c10 ≠ 3679
biv-chain[2]: r6c6{n5 n1} - hr6c1{n245789 n146789} ==> r6c7 ≠ 5, r6c5 ≠ 5, r6c3 ≠ 5
biv-chain[2]: hr6c1{n245789 n146789} - r6c2{n2 n6} ==> r6c7 ≠ 6, r6c5 ≠ 6, r6c4 ≠ 6, r6c3 ≠ 6, r6c3 ≠ 2
naked-single ==> r6c3 = 4
naked-single ==> vr4c3 = 1245
naked-single ==> r7c3 = 5
naked-single ==> r8c3 = 2
naked-single ==> hr8c0 = 12568
hidden-single-in-magic-verti-sector ==> r3c5 = 5
naked-single ==> r3c6 = 1
naked-single ==> r3c2 = 6
naked-single ==> r6c2 = 2
naked-single ==> r6c6 = 5
naked-single ==> hr6c1 = 245789
naked-single ==> vr0c2 = 245679
naked-single ==> r2c2 = 7
naked-single ==> r2c5 = 8
naked-single ==> r8c5 = 6
naked-single ==> r8c4 = 8
hidden-single-in-magic-horiz-sector ==> r7c7 = 3
naked-single ==> vr5c7 = 389
hidden-single-in-magic-horiz-sector ==> r8c8 = 5
naked-single ==> vr6c8 = 15689
hidden-single-in-magic-horiz-sector ==> r7c9 = 4
hidden-single-for-magic-digit-in-horiz-sector ==> r6c7 = 8
naked-single ==> r8c7 = 9
whip[3]: hr12c8{n13479 n13578} - r12c12{n6 n7} - r12c9{n7 .} ==> r12c10 ≠ 8
whip[3]: vr9c10{n4678 n4579} - r11c10{n8 n5} - r10c10{n5 .} ==> r12c10 ≠ 9
whip[3]: vr9c10{n4678 n4579} - r11c10{n8 n9} - r10c10{n9 .} ==> r12c10 ≠ 5
naked-single ==> r12c10 = 4
naked-single ==> hr12c8 = 13479
naked-single ==> r12c12 = 7
naked-single ==> r12c9 = 9
whip[4]: r10c6{n7 n5} - vr8c6{n12347 n12356} - r12n4{c6 c7} - vr9c7{n567 .} ==> r10c7 ≠ 7
biv-chain[4]: r12n6{c7 c6} - vr8c6{n12347 n12356} - r10n7{c6 c9} - r11n7{c9 c7} ==> vr9c7 ≠ 459
ctr-to-verti-sector  ==> r11c7 ≠ 9
ctr-to-verti-sector  ==> r10c7 ≠ 9
x-wing-in-horiz-sectors: n9{r10 r11}{c8 c10} ==> r7c8 ≠ 9
x-wing-in-horiz-sectors: n9{r6 r7}{c4 c5} ==> r10c4 ≠ 9
biv-chain[2]: r10c4{n7 n6} - hr10c0{n3789 n4689} ==> r10c3 ≠ 7, r10c2 ≠ 7
naked-pairs-in-verti-sector: c2{r9 r10}{n8 n9} ==> r12c2 ≠ 9, r12c2 ≠ 8
naked-single ==> r12c2 = 7
hidden-single-in-magic-verti-sector ==> r13c3 = 7
naked-singles to the end


saul wrote:Perhaps this is another way of gauging the difficulty of a puzzle: how large an incorrect solution exists? Unfortunately, this seems to be even harder than solving the puzzle.

Proving that a puzzle (your "incorrect solution") has no solution is finding some contradiction in the (extended set of) givens. This is basically the same thing as when we try to eliminate a candidate by contradiction. It can therefore be as hard (wrt any rating of type "hardest step") as finding a solution. In the meantime (before the contradiction is found), there is an underlying incorrect partial solution, the size of which depends on how hard it is to reach the contradiction.
So, there's a clear relationship but it isn't really another way of rating the puzzle: the least number of additional values you must add to the resolution state before you reach a contradiction by direct constraints propagation is exactly the length of the smallest elimination allowed by some braid.
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Re: Can You Solve This Without Trial and Error?

Postby Mani » Fri Dec 13, 2013 4:30 am

Re. Consider the 35-in-6 row in the middle. If you choose 1 in the first or second row, you find it leads to an impossible situation, as 1+5 (from 5th cell) can't leave 29-in-4 behind which needs 5 again. So, both cells need to forego the 1. That makes the 12-in-4 vertical choose its 1 in the top cell. Likewise, the 5 can be eliminated from a couple of cells in the 35-in-6 row as 6542 adds to 17 and leaves 18 in 2 cells that is not feasible, and that should lead to solving the puzzle, hopefully. I would call this intuitive solving, not trial and error.

And, it is often easier to start off all over again and solve a sum than ask someone to continue where one left.
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Re: Can You Solve This Without Trial and Error?

Postby Mani » Fri Dec 13, 2013 2:18 pm

I just solved the sum now, took a while but was not too tough. Your bold letter answers in H82561 grid picture below are all correct.
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Re: Can You Solve This Without Trial and Error?

Postby saul » Fri Dec 13, 2013 4:49 pm

Mani wrote:Re. Consider the 35-in-6 row in the middle. If you choose 1 in the first or second column (Saul edited), you find it leads to an impossible situation, as 1+5 (from 5th cell) can't leave 29-in-4 behind which needs 5 again. So, both cells need to forego the 1. That makes the 12-in-4 vertical choose its 1 in the top cell. Likewise, the 5 can be eliminated from a couple of cells in the 35-in-6 row as 6542 adds to 17 and leaves 18 in 2 cells that is not feasible, and that should lead to solving the puzzle, hopefully. I would call this intuitive solving, not trial and error.


Excellent. I agree that this is certainly not trial and error. I have no doubt that this leads to a solution, though I haven't tried it.

It's good to hear from someone else interested in kakuro. Welcome to the forum.
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Re: Can You Solve This Without Trial and Error?

Postby Mani » Sun Dec 15, 2013 8:05 am

Thanks for the good words.

You have probably gotten over this particular puzzle that you should now turn your attention to another that kept me struggling the last couple of days. When I struggle so much, I write to ATK for a clue, and the clue is to tell me if there is a valid solution or not, not any bit of the solution itself. Often these days, before I get a reply, and at times when I get one saying there is indeed, I tug harder at it to reach the answer.

I just landed the solution to this one, you may like to try it - H72451.
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Re: Can You Solve This Without Trial and Error?

Postby denis_berthier » Sun Dec 15, 2013 11:34 am

Hi Mani,
Welcome to the forum.

Mani wrote:When I struggle so much, I write to ATK for a clue, and the clue is to tell me if there is a valid solution or not, not any bit of the solution itself. Often these days, before I get a reply, and at times when I get one saying there is indeed, I tug harder at it to reach the answer.

I wonder: have you ever got the answer that there is no solution ?

Mani wrote:I just landed the solution to this one, you may like to try it - H72451.

One more puzzle isomorphic to one I had solved (H72411) - a pretty hard one (in W7).
It may help to know that there are pairs and triplets (both naked and hidden) - but even so, it remains in W7.
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Re: Can You Solve This Without Trial and Error?

Postby Mani » Mon Dec 16, 2013 5:33 am

To be fair to ATK, 99% of the occasions, I have had a positive reply and I found it to be true.
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Re: Can You Solve This Without Trial and Error?

Postby denis_berthier » Mon Dec 16, 2013 5:56 am

Could any one of you try M44252 ?
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Re: Can You Solve This Without Trial and Error?

Postby Mani » Mon Dec 16, 2013 4:49 pm

Just solved M44252. Was tough in its own way!

How to copy an image in here?

I didn't understand some of your notations, perhaps you have a gist of those?
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Re: Can You Solve This Without Trial and Error?

Postby saul » Mon Dec 16, 2013 5:01 pm

denis_berthier wrote:Could any one of you try M44252 ?


I did it without much difficulty:
r2c2 - r4c4 + r5c5 - r7c7 = 8

Bill pointed out this cut to me in a previous post about a puzzle with this same configuration of cells, if I'm not mistaken. Anyway, it was a configuration very like this, with the the four cells in the running along the main diagonal.
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Re: Can You Solve This Without Trial and Error?

Postby saul » Mon Dec 16, 2013 6:08 pm

Mani wrote:How to copy an image in here?

First, you have to have posted the image somewhere on the Web, like dropbox. Then you click the "Img" button above, and insert the URL for your image between the "img" and "/img" tags. There will be square brackets around the tags. Leave those in. I had to delete them in this post or the bulletin board application messes up the formatting. If you preview the post, it will show the image, so you can tell if you've done it correctly.
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Re: Can You Solve This Without Trial and Error?

Postby denis_berthier » Mon Dec 16, 2013 10:15 pm

Mani wrote:I didn't understand some of your notations, perhaps you have a gist of those?

I'm using the notations and resolution rules defined in my last book "Pattern-Based Constraint Satisfaction ..." (http://www.carva.org/denis.berthier/PBCS/index.html), available in pdf. There's a chapter on Kakuro.

In a few words, there are the natural rc variables (those associated with the white cells, the values of which we want to find), plus variables representing the combinations allowed by the sums in the horizontal and vertical sectors (the hrc and vrc variables). I attach the latter to the black cell at the start of the sector, but this is only a naming convention.
As for the main pattern used (whip), it represents a continuous chain of inferences from a candidate (the target); most of the time, it mixes variables of the above different types.
In a whip, a block V{llc rlc} represents a variable V with two candidate-values: llc with negative valence and rlc with positive valence (when the target is granted positive valence). On the grid, there may be more candidates for this variable, but it doesn't matter as long as they are linked to the target (z-candidates) or to a previous rlc (t-candidates). Bivalue-chains are a particular case of whips, in which there are no such additional candidates.
In my approach, whips are the main chain pattern, they allow to avoid T&E.

Except for the special x-wing case, whips of length <= 2, remain in a single sector and deal with interactions between rc and hrc (or vrc) variables. In Kakuro, they can be considered as obvious steps.
Last edited by denis_berthier on Mon Dec 16, 2013 10:26 pm, edited 2 times in total.
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