Post the puzzle or solving technique that's causing you trouble and someone will help

Hello Friends,
I'm learning to solve expert level sudoku during this lockdown. I've been stuck with the attached puzzle for almost 50mins. I've applied all the techniques I know of. Like Naked singles , hidden singles , naked pairs , pointing pairs , claiming pairs , naked triples , X wing , hidden pair , naked quad. I'm so eager to know how to get one more number. Please help. Thanks in advance.
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abhilashnoudu

Posts: 1
Joined: 21 April 2020

Hi abhilashnoudu,

You've missed one hidden single (5r2c3), but adding that won't probably help you much. What you need after that is at least one chain, unless I'm missing something. Here's one that will solve the puzzle:

Code: Select all
`.------------------.-----------.----------------.| 479   14     147 | 8   3   5 |  49     2    6 || 49    3      5   | 2   19  6 |  149    7    8 || 2     8      6   | 7   19  4 |  139    139  5 |:------------------+-----------+----------------:| 47    6      478 | 9   2   3 |  58    d45   1 || 5     9      3   | 4   8   1 |  7      6    2 || 1     2     b48  | 5   6   7 |  89    c49   3 |:------------------+-----------+----------------:| 6     145  a[1]4 | 13  7   2 |  35-1   8    9 || 8     7      2   | 13  5   9 |  6      13   4 || 3     5-1    9   | 6   4   8 |  2    e(1)5  7 |'------------------'-----------'----------------'`

(1=4)r7c3 - r6c3 = r6c8 - (4=5)r4c8 - (5=1)r9c8 => -1 r7c7,r9c2; stte

compressed: Show
(14)r76c3 = (451)r649c8 => -1 r7c7,r9c2; stte

In other words, the chain proves that there must be a 1 in either r7c3 or r9c8 or both. Thus it can't be in either r7c7 or r9c2. After those eliminations it's just singles. There are many other chains and eliminations available that would help too, but that's the one I found first (and it's enough).

I don't think there are much simpler options available, unless you find XY-Chains easier. The same eliminations with a bit longer XY-Chain:

Code: Select all
`.------------------.-----------.----------------.| 479   14     147 | 8   3   5 |  49     2    6 || 49    3      5   | 2   19  6 |  149    7    8 || 2     8      6   | 7   19  4 |  139    139  5 |:------------------+-----------+----------------:| 47    6      478 | 9   2   3 |  58    e45   1 || 5     9      3   | 4   8   1 |  7      6    2 || 1     2     b48  | 5   6   7 | c89    d49   3 |:------------------+-----------+----------------:| 6     145  a[1]4 | 13  7   2 |  35-1   8    9 || 8     7      2   | 13  5   9 |  6      13   4 || 3     5-1    9   | 6   4   8 |  2    f(1)5  7 |'------------------'-----------'----------------'`

(1=4)r7c3 - (4=8)r6c3 - (8=9)r6c7 - (9=4)r6c8 - (4=5)r4c8 - (5=1)r9c8 => -1 r7c7,r9c2; stte
-SpAce-: Show
Code: Select all
`   *             |    |               |    |    *        *        |=()=|    /  _  \    |=()=|               *            *    |    |   |-=( )=-|   |    |      *     *                     \  ¯  /                   *    `

"If one is to understand the great mystery, one must study all its aspects, not just the dogmatic narrow view of the Jedi."

SpAce

Posts: 2581
Joined: 22 May 2017

abhilashnoudu wrote: I've been stuck with the attached puzzle for almost 50mins. I've applied all the techniques I know of. Like Naked singles , hidden singles , naked pairs , pointing pairs , claiming pairs , naked triples , X wing , hidden pair , naked quad. I'm so eager to know how to get one more number. Please help. Thanks in advance.

After all the singles, you can solve with two short bivalue-chains
Code: Select all
`biv-chain[3]: r1c7{n4 n9} - b6n9{r6c7 r6c8} - r6n4{c8 c3} ==> r1c3 ≠ 4biv-chain[4]: r1c7{n4 n9} - b6n9{r6c7 r6c8} - r6n4{c8 c3} - b7n4{r7c3 r7c2} ==> r1c2 ≠ 4stte`

or two short xy-chains, if you prefer to start learning chains without bilocation:
Code: Select all
`biv-chain-rc[3]: r1c7{n4 n9} - r6c7{n9 n8} - r6c3{n8 n4} ==> r1c3 ≠ 4biv-chain-rc[5]: r1c2{n1 n4} - r1c7{n4 n9} - r6c7{n9 n8} - r6c3{n8 n4} - r7c3{n4 n1} ==> r1c3 ≠ 1, r7c2 ≠ 1, r9c2 ≠ 1stte`
denis_berthier
2010 Supporter

Posts: 1691
Joined: 19 June 2007
Location: Paris

Hi Denis,

denis_berthier wrote:
Hidden Text: Show
After all the singles, you can solve with two short bivalue-chains
Code: Select all
`biv-chain[3]: r1c7{n4 n9} - b6n9{r6c7 r6c8} - r6n4{c8 c3} ==> r1c3 ≠ 4biv-chain[4]: r1c7{n4 n9} - b6n9{r6c7 r6c8} - r6n4{c8 c3} - b7n4{r7c3 r7c2} ==> r1c2 ≠ 4stte`

or two short xy-chains, if you prefer to start learning chains without bilocation:
Code: Select all
`biv-chain-rc[3]: r1c7{n4 n9} - r6c7{n9 n8} - r6c3{n8 n4} ==> r1c3 ≠ 4biv-chain-rc[5]: r1c2{n1 n4} - r1c7{n4 n9} - r6c7{n9 n8} - r6c3{n8 n4} - r7c3{n4 n1} ==> r1c3 ≠ 1, r7c2 ≠ 1, r9c2 ≠ 1stte`

I don't think you need two chains either way. The second chain is both available and effective alone in both cases, as far as I see:

biv-chain[4]: r1c7{n4 n9} - b6n9{r6c7 r6c8} - r6n4{c8 c3} - b7n4{r7c3 r7c2} ==> r1c2 ≠ 4; stte

Eureka translation: Show
(4=9)r1c7 - r6c7 = (9-4)r6c8 = r6c3 - r7c3 = (4)r7c2 => -4 r1c2; stte

or:

biv-chain-rc[5]: r1c2{n1 n4} - r1c7{n4 n9} - r6c7{n9 n8} - r6c3{n8 n4} - r7c3{n4 n1} ==> r1c3 ≠ 1, r7c2 ≠ 1, r9c2 ≠ 1; stte

Eureka translation: Show
(1=4)r1c2 - (4=9)r1c7 - (9=8)r6c7 - (8=4)r6c3 - (4=1)r7c3 => -1 r1c3,r79c2; stte

SpAce

Posts: 2581
Joined: 22 May 2017

SpAce wrote:I don't think you need two chains either way. The second chain is both available and effective alone in both cases, as far as I see:
biv-chain[4]: r1c7{n4 n9} - b6n9{r6c7 r6c8} - r6n4{c8 c3} - b7n4{r7c3 r7c2} ==> r1c2 ≠ 4; stte
or:
biv-chain-rc[5]: r1c2{n1 n4} - r1c7{n4 n9} - r6c7{n9 n8} - r6c3{n8 n4} - r7c3{n4 n1} ==> r1c3 ≠ 1, r7c2 ≠ 1, r9c2 ≠ 1; stte

Hi SpAce

You're right.
I can't get the alternative paths automatically in CSP-Rules, as the simplest-first strategy is fixed; but I can check afterwards:
- in the 3D case:
Code: Select all
`(bind ?*simulated-eliminations* (create\$ 412))(solve "...835.26.3.2.6.782867.4..5.6.923..159348176212.567..36...72.89872.596.43.96482.7")stte`

- in the 2D case:
Code: Select all
`(bind ?*simulated-eliminations* (create\$ 113 172 192))(solve "...835.26.3.2.6.782867.4..5.6.923..159348176212.567..36...72.89872.596.43.96482.7")stte`

At any stage, SudoRules always chooses the simplest rule applicable; that's why I call it the simplest-first strategy.
A real player has generally no fixed strategy. He takes whatever comes. If he finds a long chain, he will use it. I don't think, once he has found a solution, he will try to eliminate useless steps. But I understand that, for a few people in this forum, it's a game of itself to do this (and indeed, sometimes, it leads to paths with fewer steps).

Here is a third solution, again with typed-chains, each in a single 2D space. Maximum length is 4.
Code: Select all
`biv-chain-rc[3]: r1c7{n4 n9} - r6c7{n9 n8} - r6c3{n8 n4} ==> r1c3 ≠ 4biv-chain-rn[4]: r4n5{c8 c7} - r7n5{c7 c2} - r7n4{c2 c3} - r6n4{c3 c8} ==> r4c8 ≠ 4`

Again, the first chain can be discarded.

As a result, the puzzle can be solved with biv-chains of maximum length 4 in the rn-space, though it requires length 5 in the rc-space.
denis_berthier
2010 Supporter

Posts: 1691
Joined: 19 June 2007
Location: Paris