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by **surbier** » Tue Dec 23, 2008 8:21 pm

Thanks all,

aran wrote:
A non-forcing but longer alternative to Hobiwan's solution :

146r156c8=8r1c8-(8=1)r2c9-(1=3)r3c9-(3=5)r3c8-(5=14.#8r1c8)r7c8=>pair14r57c8 : => <14>r9c8

I cannot follow the first part of this AIC, could you please give a tiny hint ?

Typo '.#' should be a 5 ?

surbier

by **aran** » Wed Dec 24, 2008 3:58 am

surbier wrote:Thanks all,

aran wrote:
A non-forcing but longer alternative to Hobiwan's solution :

146r156c8=8r1c8-(8=1)r2c9-(1=3)r3c9-(3=5)r3c8-(5=14.#8r1c8)r7c8=>pair14r57c8 : => <14>r9c8

I cannot follow the first part of this AIC, could you please give a tiny hint ?

Typo '.#' should be a 5 ?

surbier

Surbier
It's a hiiden triple move.
The logic is : either r156c8 is the triple 146, or if not then r1c8 is 8 (this is the only possible alternative to the triple) etc through the chain to the conclusion that r57c8 is the pair 14. So since at least one endpoint of the chain is true, 14 is eliminated in r9c8.
As to the notation : the symbol # was intended : it's semi-standard (not doubt some will disagree with that) meaning "and taking into account the fact that in this chain we already know that r1c8 is 8).
Hope this is clearer.

by **surbier** » Mon Dec 29, 2008 9:45 am

Thanks aran, I see now the ALS in the AICs.

At this point cannot follow:

A :6r9c2=6r9c6-(6=8)r8c6-(8=3)r8c9-(3=1)r3c9 : <1>r3c2
B :6r9c2=(6-2)r8c1=2r1c2 : <2>r3c2
=><6>r1c2

In a discontinuous AIC, I would expect for <1>r3c2 in A: two weak links 1r3c2-1r3c9 and 1r3c2-6r9c2; the latter one does not exist.
I suppose it's something trivial I overlook.

by **aran** » Mon Dec 29, 2008 10:14 pm

surbier wrote:Thanks aran, I see now the ALS in the AICs.

At this point cannot follow:

A :6r9c2=6r9c6-(6=8)r8c6-(8=3)r8c9-(3=1)r3c9 : <1>r3c2
B :6r9c2=(6-2)r8c1=2r1c2 : <2>r3c2
=><6>r1c2

In a discontinuous AIC, I would expect for <1>r3c2 in A: two weak links 1r3c2-1r3c9 and 1r3c2-6r9c2; the latter one does not exist.
I suppose it's something trivial I overlook.

Surbier
I'll tell you how the move works as I wrote it :
A : if r9c2 is not 6, then r3c2 cannot be 1
B : if r9c2 is not 6, then r3c2 cannot be 2
Consequently if r9c2 is not 6, then r3c2 must be 6.
So the now-established strong link on 6 between r9c2 and r3c2 means : <6>r1c2.
As such it's not a discontinuous loop elimination.

by **surbier** » Tue Dec 30, 2008 10:53 pm

Thanks, I got it now.

On A: since r9c2 is already a bi-value cell of 1,6
then if r9c2 is not 6 , it must be 1, hence <1>r137c2 directly.

by **surbier** » Thu Jan 01, 2009 9:47 am

- and pattern overlay deletes the 1 in r1c7

I would like to add something on a different topic.
When I started to follow the POM elimination
in an earlier step of this sudoku I transposed the grid into POM space, which
became rather large (just copy and paste the code section into a html
file to view it in the browser. It fits onto letter/A4 paper when using
small fonts and layout=landscape).

Code: Select all: <html><head></head><body><h1>POM merge grid</h1><table border="1"> <tr> <td>3(abc) 6(ab) 8(abc) </td><td>1(abcdef) 3(def) 6(cd) 8(defg) </td><td>1(ghijkl) 3(gh) 8(hi) 9(ab) </td><td>1(mnopqrstuvwxyzA) 3(ijk) 4(abc) 7(a) </td><td><center>5</center></td><td>1(BCDEFGHIJ) 7(b) 9(cd) </td><td>1(KLM) 4(defghijklmn) 6(ef) 7(cd) </td><td>4(opqrstuv) 6(gh) 8(jk) </td><td> <center>2</center></td></td><tr><td>2(a) 3(ijk) 6(eg) 8(jk) </td><td><center>7</center></td><td>1(mnopqBCDK) 5(abc) </td><td>3(abcdefgh) 4(defop) </td><td> 4(ghijklmnqrstuv) 6(acfh) </td><td>1(abcghiLM) 2(b) </td><td>1(djrsEF) 4(abc) 5(d) 6(bd) </td><td> <center>9</center></td><td>1(efkltuvwxyzAGHIJ) 8(abcdefghi) </td></td><tr><td><center>4</center></td><td>1(rtuvwEGHL) 2(b) 6(fh) </td><td>1(sxyzAFIJM) 5(d) 9(cd) </td><td><center>8</center></td><td>6(bdeg) 7(c) </td><td>1(defjklK) 2(a) 7(d) 9(ab) </td><td> 1(agmB) 5(a) 6(ac) 7(ab) </td><td>3(abdegij) 5(bc) </td><td>1(bchinopqCD) 3(cfhk) </td></td><tr><td> <center>1</center></td><td>3(abcghijk) 4(abdeghij) </td><td> <center>7</center></td><td><center>6</center></td><td>3(def) 9(ac) </td><td><center>5</center></td><td><center>8</center> </td><td><center>2</center></td><td>4(cfklmnopqrstuv) 9(bd) </td></td><tr><td><center>9</center></td><td> 4(cfklmnopqrstuv) 8(abcjk) </td><td><center>6</center></td> <td><center>2</center></td><td>7(abd) 8(defhi) </td><td> 1(mnopqrstuvwxyzA) 7(c) 8(g) </td><td><center>3</center> </td><td>1(abcdefghijklBCDEFGHIJKLM) 4(abdeghij) </td><td> <center>5</center></td></td><tr><td>3(def) 8(defghi) </td><td><center>5</center></td><td><center>2</center></td><td> 1(abcdefghijklBCDEFGHIJKLM) 9(bd) </td><td>3(abcghijk) 8(abcjk) </td><td><center>4</center></td><td>1(notuxy) 6(gh) 9(ac) </td><td>1(mpqrsvwzA) 6(abcdef) </td><td> <center>7</center></td></td><tr><td><center>7</center></td><td> 1(ghjkmnpsxzBCFIKM) 2(a) 8(h) </td><td>1(abdertvEGL) 4(ghkqr) 8(adjk) </td><td>4(ijlst) 5(b) 9(ac) </td> <td>4(abcdefop) 8(g) 9(bd) </td><td><center>3</center> </td><td>1(cfilqwADHJ) 2(b) 4(uv) 5(c) </td><td> 1(ouy) 4(mn) 5(ad) 8(bcefi) </td><td><center>6</center> </td></td><tr><td>2(b) 3(gh) 6(cdfh) </td><td> <center>9</center></td><td>3(acdfik) 4(acdfilmosu) 8(beg) </td><td>4(gknqv) 5(acd) </td><td><center>1</center></td> <td>6(abeg) 8(acdfhij) </td><td>2(a) 4(prt) 5(b) </td><td><center>7</center></td><td>3(bej) 4(behj) 8(k) </td></td><tr><td><center>5</center></td><td>1(iloqyADJ) 6(abeg) 8(i) </td><td>1(cfuwH) 3(bej) 4(bejnptv) 8(cf) </td><td>4(hmru) 7(bcd) </td><td><center>2</center> </td><td>6(cdfh) 7(a) 8(bek) </td><td>1(behkpvzCGI) 4(oqs) 9(bd) </td><td>1(ntx) 3(cfhk) 4(cfkl) 8(adgh) </td><td>1(adgjmrsBEFKLM) 3(adgi) 4(adgi) 8(j) 9(ac) </td></td></table></body></html>

with the following candidate patterns:

Code: Select all: 1 = abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLM 2 = ab 3 = abcdefghijk 4 = abcdefghijklmnopqrstuv 5 = abcd 6 = abcdefgh 7 = abcd 8 = abcdefghijk 9 = abcd

Since I didn't spot all neccessary vulnerable pairs I wrote some code
which found the following vulnerable pairs:

vulnerable pair in r1c4 + r5c8 for 1: patterns 4a and 4b can be eliminated
vulnerable pair in r2c6 + r7c2 for 2: patterns 1g 1h 1M can be eliminated
vulnerable pair in r3c6 + r7c7 for 2: patterns 1f 1l can be eliminated
vulnerable pair in r2c4 + r4c2 for 3: patterns 4d 4e can be eliminated
vulnerable pair in r2c3 + r7c8 for 5: pattern 1o can be eliminated
vulnerable pair in r2c9 + r7c3 for 8: pattern 1e 1t 1v 1G can be eliminated
vulnerable pair in r4c9 + r7c4 for 9: pattern 4l 4s 4t can be eliminated
vulnerable pair in r6c4 + r9c9 for 9: patterns 1a 1d 1j 1B 1E 1F 1K 1L can be eliminated

resulting in the following merge grid:

Code: Select all: <html><head></head><body><h1>POM merge grid</h1><table border="1"><tr> <td>3(abc) 6(ab) 8(abc) </td><td>1(bc) 3(def) 6(cd) 8(defg) </td><td>1(ik) 3(gh) 8(hi) 9(ab) </td><td>1(mnpqrsuwxyzA) 3(ijk) 4(c) 7(a) </td> <td><center>5</center></td><td>1(CDHIJ) 7(b) 9(cd) </td><td>1() 4(fghijkmn) 6(ef) 7(cd) </td><td> 4(opqruv) 6(gh) 8(jk) </td><td><center>2</center></td> </td><tr><td>2(a) 3(ijk) 6(eg) 8(jk) </td><td> <center>7</center></td><td>1(mnpqCD) 5(abc) </td><td> 3(abcdefgh) 4(fop) </td><td>4(ghijkmnqruv) 6(acfh) </td><td>1(bci) 2(b) </td><td>1(rs) 4(c) 5(d) 6(bd) </td><td><center>9</center></td><td>1(kuwxyzAHIJ) 8(abcdefghi) </td></td><tr><td><center>4</center></td><td> 1(ruwH) 2(b) 6(fh) </td><td>1(sxyzAIJ) 5(d) 9(cd) </td><td><center>8</center></td><td>6(bdeg) 7(c) </td><td>1(k) 2(a) 7(d) 9(ab) </td><td>1(m) 5(a) 6(ac) 7(ab) </td><td>3(abdegij) 5(bc) </td> <td>1(bcinpqCD) 3(cfhk) </td></td><tr><td><center>1</center> </td><td>3(abcghijk) 4(ghij) </td><td><center>7</center> </td><td><center>6</center></td><td>3(def) 9(ac) </td> <td><center>5</center></td><td><center>8</center></td><td> <center>2</center></td><td>4(cfkmnopqruv) 9(bd) </td></td> <tr><td><center>9</center></td><td>4(cfkmnopqruv) 8(abcjk) </td><td><center>6</center></td><td><center>2</center></td> <td>7(abd) 8(defhi) </td><td>1(mnpqrsuwxyzA) 7(c) 8(g) </td><td><center>3</center></td><td>1(bcikCDHIJ) 4(ghij) </td><td><center>5</center></td></td><tr><td>3(def) 8(defghi) </td><td><center>5</center></td><td><center>2</center> </td><td>1(bcikCDHIJ) 9(bd) </td><td>3(abcghijk) 8(abcjk) </td><td><center>4</center></td><td>1(nuxy) 6(gh) 9(ac) </td><td>1(mpqrswzA) 6(abcdef) </td><td> <center>7</center></td></td><tr><td><center>7</center></td><td> 1(kmnpsxzCI) 2(a) 8(h) </td><td>1(br) 4(ghkqr) 8(adjk) </td><td>4(ij) 5(b) 9(ac) </td><td> 4(cfop) 8(g) 9(bd) </td><td><center>3</center></td><td> 1(ciqwADHJ) 2(b) 4(uv) 5(c) </td><td>1(uy) 4(mn) 5(ad) 8(bcefi) </td><td><center>6</center></td></td><tr><td> 2(b) 3(gh) 6(cdfh) </td><td><center>9</center></td><td> 3(acdfik) 4(cfimou) 8(beg) </td><td>4(gknqv) 5(acd) </td><td><center>1</center></td><td>6(abeg) 8(acdfhij) </td> <td>2(a) 4(pr) 5(b) </td><td><center>7</center></td><td> 3(bej) 4(hj) 8(k) </td></td><tr><td><center>5</center> </td><td>1(iqyADJ) 6(abeg) 8(i) </td><td>1(cuwH) 3(bej) 4(jnpv) 8(cf) </td><td>4(hmru) 7(bcd) </td> <td><center>2</center></td><td>6(cdfh) 7(a) 8(bek) </td> <td>1(bkpzCI) 4(oq) 9(bd) </td><td>1(nx) 3(cfhk) 4(cfk) 8(adgh) </td><td>1(mrs) 3(adgi) 4(gi) 8(j) 9(ac) </td></td></table></body></html>

with remaining candidate patterns:

Code: Select all: 1 = bcikmnpqrsuwxyzACDHIJ 2 = ab 3 = abcdefghijk 4 = cfghijkmnopqruv 5 = abcd 6 = abcdefgh 7 = abcd 8 = abcdefghijk 9 = abcd

and no pattern left for candidate 1 in r1c7.

I was looking for vulnerable triplets and quads, to continue eliminations in pattern space,
but couldn't find any further pattern eliminations for this particular sudoku.

surbier

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SE=8.4 with 23 clues