Assassin 49

Our weekly <a href="http://www.sudocue.net/weeklykiller.php">Killer Sudokus</a> should not be taken too lightly. Don't turn your back on them.
CathyW
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Assassin 49

Post by CathyW »

An accurate description of this week's puzzle. It's taken me a couple of hours but have just finished it. :D

Will post my 32 step walkthrough later.
Para
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Post by Para »

Hi

Ok finished this one. There's a part in the middle where i was just looking for something to break open this puzzle. Took a few steps but it finally cracked. But i think the step that helped me was there a while already.

Walk-through Assassin 49

1. R12C1 = {19/28/37/46}: no 5

2. R12C5 and R12C9 = {16/25/34}: no 7,8,9

3. R23C3 and R23C7 = {59/68}: no 1,2,3,4,7

4. 11(3) in R3C8 = {128/137/146/236/245}: no 9

5. 13(4) in R4C3 = {1237/1246/1345}: no 8,9; 1 locked in 13(4) cage: R5C12: no 1

6. 23(3) in R6C8 = {689}: {689} locked in 23(3) cage: R89C8: no 6,8 or 9
7. R78C3, R89C1 and R89C5 = {29/38/47/56}: no 1

8. R78C7 = {39/48/57}: no 1,2,6

9. R89C9 = {17/26/35}: no 4,8,9

10. 45 on R12: 2 innies: R2C37 = {58} -->> locked for R2
10a. Clean up: R3C37 = {69} -->> locked for R3
10b. Clean up: R1C1: no 2; R1C59: no 2

11. 45 on R89: 2 outies: R78C37 = 8 = {35} -->> locked for R7
11a. Clean up: R8C3 = {68}, R8C7 = {79}

12. R23C3 = [59]: [86] clashes with R8C3
12a. R23C7 = [86]; R78C3 = [38]; R78C7 = [57]

13. R89C9 = {26} -->> locked for C9 and N9
13a. Naked Pair {89} in R7C89 -->> locked for R7, N9 and 23(3) cage in R6C8
13b. R6C8 = 6
13c. R12C9 = {34} -->> locked for C9 and N3

14. Clean up: R12C1: no 1; R9C1: no 4; R9C5: no 3,4

15. {134} locked in R8C8+R9C78 in 22(5) in R8C6
15a. 22(5) = {13459/13468}
15b. R89C6 = {59}/[68]
15c. R89C5: no {56}: clashes with R89C6

16. 9 in N3 locked for 27(5) cage in R1C6 -->> R12C6: no 9

17. 9 in N2 locked for C4

18. 21(4) in R4C9 needs 1 of {89} and 2 of {157} in R456C9 -->> 21(4) = {1479/1578}
18a. R5C8 = {14578}
18b. 1 and 7 locked in 21(4) cage for N6 (nowhere else in N6)

19. 21(4) in R4C9 needs 3 of {2349} in R456C7 -->> 21(4) = {2379/2469/3459}
19a. R5C6 = {567}
19b. 9 locked in 21(4) cage in R456C7 -->> locked for C7 and N6
19c. Hidden Single: R7C9 = 9; R7C8 = 8

20. 21(4) in R4C9 = {1578} -->> locked for N6

21. 11(3) in R3C8 = [254]/{17}[3] -->> R3C8: no 5, R4C8: no 2
21a. 2 in N6 locked for C7
21b. R1C7 = 1
21c. Clean up: R2C5: no 6
21d. 1 in N9 locked for C8

22. 11(3) in R3C8 = [254]
22a. Hidden single: R5C8 = 5; R9C7 = 4
22b. 21(4) in R4C7 = {2379}; -->> R5C6 = 7

23. {159} locked in 27(5) in R1C6 -->> 27(5) = 1{2789/4679} -->> R12C6 = {46}/[82]
23a. R89C6 = {59}: [68] clashes with R12C6 -->> {59} locked for C6 and N8
23b. Clean up : R89C5 = [38/47]
23c. R12C5: no {34}: clashes with R8C5

24. 16(3) in R4C5 = {169/259}: 9 locked in N5 in 16(3) and {349} clashes with R8C5
24a. 45 on C5: 2 innies: R37C5 = 11 = {47} : locked for C5
24b. R89C5 = [38]; R89C8 = [13]

25. 15(3) in R3C1 needs 2 of {13478} in R3C12 -->> 15(3) = {168/348/357} -->> R4C2 = {3568}

26. 14(3) in R6C2 needs 2 of {12467} in R7C12 -->> 14(3) = {[9]14/167/[8]24/[5]27}: [3]{47} would clash with R7C5 -->> R6C2 = {15789}

27. 22(5) needs 3 of {12467} in R7C456 and one of {47} in R7C5: 22(5) = {12478/13468/13567/23467}
27a. 22(5) needs at least one of {67} in R7C456 -->> 14(3) in R6C2 can’t have both {67} in R7C12: 14(3) = {[9]14/[7]{16}/[8]24/[5]27}: R6C2: no 1

28. 13(4) in R4C3 = {1237/1246} -->> 2 locked in 13(4) cage: R5C12: no 2

29. 9 in N2 locked in 26(5) in R12C4 -->> 26(5) = {12689/13589/13679/14579/23489/23589/24569}
29a. 26(5): no combinations with {15} allowed: 26(5) uses 5: R1C4 = 5 -->> R12C5 = [61] -->> no room for 1 in 26(5)
29b. 26(5) = {12689/13679/23489/23579/24569}

30. R7C4: no 4: sees all 4’s in N5

31. Can’t have both {48} in R3C456: clashes with R12C6 -->> R3C12 at least needs one of {48}
31a. 15(3) = {18[6]/{348}: no 5,7; 8 locked in 15(3) -->> R1C2: no 8
31b. 7 in R3 locked for N2
31c. 7 in R3 locked in 23(5) in R3C4 -->> 23(5) = {12578/13478/14567/23567}
31d. 23(5) needs 3 of {13478} in R3C456 -->> 23(5): no {23567}

Finally found what I was looking for:
32. 2 in R1 locked in 26(5) in R1C2: R2C24: no 2 -->> 26(5) = {12689/23489/23579/24569}
32a. 26(5) needs one of {58}, 5 and 8 only in R1C4 -->> R1C4 = {58}
32b. 2 in R1 locked for N1
32c. Clean up: R1C1: no 8
32d. 8 in N1 locked for R3
32e. 8 in N1 locked for 15(3) in R3C1: R4C2: no 8

33. Killer Pair {26} in R12C5 + R12C6 locked for N2

34. R89C1: no [47] clashes with R12C1

35. 45 on N4: 2 innies – 1 outie: R46C2 – R5C4 = 9
35a. 13(4) in R4C3 = {127[3]/1246}-->> R5C5 = 3: 13(4) = {1237} otherwise 13(4) = {1246}
35b. R5C4 = 1 -->> R46C2 = [37]
35c. R5C4 = 2 -->> R46C2 = [38] (6 needed in 13(4))
35d. R5C4 = 3 -->> R46C2 = [39]
35d. R5C4 = 4 : no combination possible
35e. R5C4 = 6 -->> R46C2 = [68]
35f. Conclusion: R5C4: no 4; R6C2: no 5

36. 4 in N5 locked for R6
36a. 4 in N5 locked for 22(5) in R6C4: R7C56: no 4
36b. R37C5 = [47]; R12C6 = [82]; R12C5 = [61]; R3C46 = [73]; R12C4 = [59]; R12C8 = [97]
36c. Hidden singles: R8C4 = 4; R6C6 = 4

37. 15(3) = {18}[6] -->> R4C2 = 6
And the rest is singles and simple cage sums
greetings

Para
Last edited by Para on Thu May 31, 2007 1:24 pm, edited 1 time in total.
Jean-Christophe
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Post by Jean-Christophe »

You find it too easy :?:
Here is a V2. Harder, but JSudoku could solve it by logic

Image

3x3::k:2304:6913:6913:6913:2052:6917:6917:6917:1288:2304:6913:2827:6913:2052:6917:3087:6917:1288:4114:4114:2827:6165:6165:6165:3087:3609:3609:6171:4114:3613:6165:4383:6165:6689:3609:4387:6171:6171:3613:3613:4383:6689:6689:4387:4387:6171:2606:3613:6704:4383:6704:6689:4660:4387:2606:2606:3640:6704:6704:6704:1340:4660:4660:2623:6208:3640:6208:2627:7236:1340:7236:2375:2623:6208:6208:6208:2627:7236:7236:7236:2375:

Enjoy :twisted:
rcbroughton
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Post by rcbroughton »

I agree with Para, a bit more straightforward than that 48Heavie we've been struggling with.

Here's my walkthrough pretty much as I found it. only really sticky move was around step 31.

1. 45 on r12 r2c37=13={58} locked for r2
1a. 13c37={69} locked for r3
1b. cleanup - no 2 at r1c1, r1c5, r1c9

2. 45 on r89 r8c37=15=[69]/{78}
2a. cleanup - 11(2)n7=[56]/[47]/[38]
2b. cleanup - 12(2)n9=[57]/[48]/[39]

3. from 2a. 11(2)n7 can't be [56] as it blocks 14(2)n1
3a. cleanup from 2 - 12(2)n9 can't be [39]

4. outies of r89 - r7c37=8 can only be [35]
4a. 11(2)n7=[38]
4b. 12(2)n9=[57]
4c. 14(2)n1=[59]
4d. 14(2)n3=[86]

5. cleanup
5a no 4 r9c1,
5b no 3,4 r9c5
5c 8(2)n9={62} locked for c9 and n9
5d no 1 10(2)n1
5e. no 1,5 r9c1
5f. 7(2)n3={34} locked for n3 and c9

6. 23(3)n69 = 6{89} - {89} locked for r7 and n9

7. 45 on c5 - r37c5=11 - r3c5={47}/[56]

8. 45 on c9 - innies - outies = 9 - no 9 at r5c8

9. 45 on c1 - innies = outies+1 - no 1 at r5c2

10. 45 on n1 - outies total 20 - max r12c4=17 - so no 1,2 at r4c2

11. 45 on n9 - outies r89c6 total 14 - no 1,2,3,4,7 ={59}/[68]

12. 11(3)n36=[218]/[173]/[713]/[254] - no 5 at r3c8, no 1,2,5,7 at r4c8

13. 11(2)n8 ={29}/[38] - other combos blocked by 7(2)n2 and/or r37c5=11(2)
13a. 4 locked in 7(2)n2 and r37c5 for c5

14. 13(4)n45={1237}/{1246} - no 5,7 at r5c4
14a. must use {12} - no 1,2 at r5c1 and no 2 at r5c2

15. 21(4)n6={1389}/{1479}/{1578} - no 2, 9 at r5c8 and 1 locked for n6

16. 21(4)n56 ={2379}/{2469}/{3459} - r5c6=5/6/7 only
16a. must use 9 locked in r456c7 for c7 and n6

17. from 15 - 21(4)n6={1578} - locked for n6

18. 45 on n34 - outies = 31, but r89c6=14, r125c6 total 17
18a. -> r1c6={4/6/7/8/9}
18b. -> r2c6={1/2/3/4/6/7}

19. 45 on r89 - innies total 20
19a. only combo in r1289c8 is {79}{13} - locked for c8, {79} locked for n3 an 27(5), {13} locked for n9
cleanup
19b. r9c7=4
19c. 11(3)n36 = [254]
19d. r1c7=1
19e. r5c8=5
19f. 23(3)n69=[689]
19g. r5c6=7

20. 27(2)n23 - no 3 at r2c6
20a. r12c6={46}/[82]
20b. r89c6={59} - locked for n8 and c6

21. 11{2}n8=[38]
21a. r89c8=[13]
21b. 9 locked in 16(3)n5 for n5 and c5
21c. 7(2)n2=[52]/[61]

22. {12569} locked in 7(2) & 16(3)c5 - remove 6 from r7c5

23. 14(3)n47=9{14}/{167}/8{42}/5{27}/3{47} - no combo with 2,4 in r6c2

24. 15(3)n14={18}6/{34}8/{37}5 - no 3,7,9 at r4c2

25. 2,6 locked in 7(2) and r12c6 in n2 - nowhere else in n2

26. 45 on n1 - outies = 20 - r4c2=5/6/8 - r12c4=12,14,15
26a. -> no 4,7 r1c3
26b. -> no 1 r2c3

27. 2 locked in n2 c56 for r2
27a. cleanup no 8 r1c1

28. 45 on r7 - outies r6c246=16={178}/[952]/{358} - no 2 at r6c4

29. 45 on r3 - outies r4c246=15={168}/{258} - no 3
29a r4c246 - 8 locked for r4

30, 45 on n7 - outies total 14
30a. r89c4 can only equal 13,11,9,7,6,5 - no possible 6 at r9c4

31. 26(5)n12 combos:
{12689} - ok
{24569} - ok
{13679}/{14579} - blocked by {79} r12c8
{14678} - {46}[817] blocked by r1c6 {67}[814] blocked by 10(2)n1
{23489} - blocked by 10(2)n1 or 7(2)n3
{23678} - blocked by 10(2)n1
{24578} - [825]{74} - blocked by 7(2)n2
{34568} - blocked by 10(2)n1
{23579} - ok but 9 locked at r2c4
31a. no 7 at r2c3

32. 7 now locked in c456 for r3
32a. cleanup 15(3)n14 - no 5
32b. from step 26 r12c4=[59]/{39}/[84] - no 3 at r2c4

33. 23(5)n25 -
{348}{26} blocked by r3c12
{12578}/{14567} - must put 5 at r4c4
{13478} - ok
33a. no 2,6 at r4c4

34. 11(2)n9 - can't use [47] because of 10(2)n1

35. 15(3)n14 - must use 8 - removes 8 from r1c2
35a. 8 now locked in r3c12 for n1 - locked for r3 and for 15(3)={18}6
35b. {18} locked in r3c12 for r3 and n1
35c. from step 31 {12689}/{24569}/{23579} - 9 locked in r2c4

36. 23(5)n25 = {347}{18}
36a. - {347} locked for n2
36b. - {18} locked for r4, n5
cleanup
36c. r4c9=7
36d. r12c6=[82]
36e. 7(2)n2=[61]
36f. r1c4=5
36g. from step 35c. 26(5)=[27539]
36h. 10(2)n1=[46]

37. 13(4)n45=2{14}6
37a r9c3=6

and the rest is hidden singles and cage sums

I'll take a look at this V2 some time . . . I love a challenge.

Rgds
Richard
Last edited by rcbroughton on Sun Jul 01, 2007 3:53 pm, edited 1 time in total.
CathyW
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Post by CathyW »

As promised, here's my walkthrough - steps in the order I did them including a few (possibly) redundant outies. Not sure without doing it again if it would have made much difference if my step 26 had been done earlier:

Edit: Typo corrections and some clarifications, especially of last step before it falls out.
Note I have assumed obvious inclusions and exclusions from cage sums e.g. 23(3) must be {689}, no 5 in 10(2).

1. Innies r12 -> r2c37 = 13 –> only option is {58} due to 14(2) cages. 5, 8 not elsewhere in r2.

2. Outies r12 -> r3c37 = 15 = {69}. 6, 9 not elsewhere in r3.

3. Clean up odd combinations from steps 1 and 2: r1c159 <> 2.

4. Innies r89 -> r8c37 = 15 -> 69/78/87

5. Outies r89 -> r7c37 = 8 -> must be {35}, not elsewhere in r7 -> r8c3 = 6/8, r8c7 = 7/9.

6. If 11(2) in N7/c3 = {56}, no options left for 14(2) in N3/c3
-> r7c3 = 3, r8c3 = 8, r7c7 = 5, r8c7 = 7
-> r2c3 = 5, r3c3 = 9, r2c7 = 8, r3c7 = 6.

7. Clean up odd combinations from placements:
10(2) in N1 <> 1
7(2) in N3 <> 1
8(2) in N9 = {26} -> 2,6 not elsewhere in N9/c9 -> r6c8 = 6
-> 7(2) in N3 = {34} -> 3,4 not elsewhere in N3/c9
-> 89 naked pair in N9 -> not elsewhere in N9/r7.
r9c1 <> 4
r9c5 <>3,4

8. 22(5) in N8/9: within N9 = {134} -> r89c6 = 14 = 59/95/68

9. Innies c5 -> r37c5 = 11 = 47/74/56
If r37c5 = {47}, 7(2) in r12c5 = {16/25}
If r37c5 = 56, 7(2) in r12c5 = {34}
-> 4 not in 16(3) or 11(2) of c5 -> r9c5 <> 7.

10. Since 7(2) in r12c9 = {34}, 7(2) in r12c5 cannot also be {34} as this would lead to two solutions for the puzzle -> 7(2) in r12c5 = {16/25} -> split 11(2) in r37c5 = {47}, not elsewhere in c5, 11(2) in r89c5 cannot be {56}.

11. 11(2) in r89c5 = {29/38}, split 14(2) in r89c6 = {59/68} -> 8, 9 not elsewhere in N8.

12. Combination options for 11(3) in r3c89 + r4c8: {128/137/245} -> r3c8 <> 5, r4c8 cannot have 1257 -> r3c8 = 127, r4c8 = 348.

13. Outies N3 -> r12c6 + r4c8 = 14

14. Outies N1 -> r12c4 + r4c2 = 20 -> r12c4, r4c2 <> 1, r4c2 <> 2

15. Outies N7 -> r6c2 + r89c4 = 14

16. Innies c12 -> r1289c2 = 17

17. Innies c89 -> r1289c8 = 20
-> If r89c8 = {13}, r12c8 = {79}
-> r89c8 cannot be {14} since can’t make 15 from available candidates in r12c8.
-> r89c8 cannot be {34} since can’t make 13 from available candidates in r12c8.
-> r89c9 = {13} -> r9c7 = 4;
r12c8 = {79} -> 7, 9 not elsewhere in N3/c8/cage 27(5), r1c7 = 1/2, r3c8 = 1/2, -> r3c9 = 5
-> r3c8 = 2, r1c7 = 1, r4c8 = 4 -> r12c6 = 46/64/82
-> NT {239} in r456c7 -> r5c6 = 7
-> 21(4) in N6 = {1578} -> r5c8 = 5
-> r7c9 = 9, r7c8 = 8.

18. Clean up odd combination from placements: r2c5 <> 6.

19. 13(4) in r456c3 + r5c4 must have 1 and 2. Combinations available 12 + 37/46.

20. 15(3) in r3c12 + r4c2 = {186/348/357} -> r4c2 <> 9 -> r12c4 <> 2 (from complex cage 20(3) – step 14) -> 2 locked to r2c56 -> r2c12 <> 2 -> r1c1 <> 8

21. 9 locked to r12c4 -> not elsewhere in c4
-> both 26(5) in N1/2 and complex 20(3) must have 9.
Options for complex cage 20(3) in r12c4+r4c2 = {389/479/569}.
2 locked to r1c23 within 26(5) -> 26(5) must have 2 and 9 -> options 12689/23489/23579/24569.

22. 14(3) in r6c2 + r7c12: Combination options {149/167/248/257}.
{347} not possible due to r7c5 = 4/7 -> r6c2 <> 3, analysis of remaining options -> r6c2 <> 2, 4

23. 10(2) in r12c1 = {37/46} -> 11(2) in r89c1 can’t be {47}
-> within r8, 4 locked to 23(5) cage -> combination options {12479/13469/14567/23459}

24. Outies – Innies N4 -> r46c2 – r5c4 = 9 -> r6c2 <> 1.

25. Innies c1234 -> r3467c4 = 20

26. Innies c6789 -> r3467c6 = 14 -> r3467c6 <> 9
-> 16(3) in N5/c5 must have 9: {169/259} -> r8c5 = 3, r9c5 = 8
-> r8c8 = 1, r9c8 = 3, -> split 14(2) in r89c6 = {59} -> r46c6 <> 5, r89c4 <> 5.

27. 1 now locked to r9 within 23(5) cage -> combination options 14 + 279/369/567.

28. Outies r123 -> r4c246 = 15. Combination option analysis {168/258/267/357}-> r4c2 <> 3, r4c4 <> 3
-> 3 now locked to 23(4) in N4 -> 23(4) <> 1, 2 -> 1/2 locked to 13(4) within c3 -> r5c4 <> 1, 2, r1c3 <> 2, r9c3 <> 1, 2

29. Outies r789 -> r6c246 = 16. Combination option analysis {178/259/349/358/457} -> r6c4 <> 2

30. HS r1c2 = 2

31. 8 locked to r1c46/r3c12 -> r3c46, r4c2 <> 8.

32. Combination options for 26(5) in r1c234+r2c24: r1c2=2 / {1689/3489/3579/4569}
If 1689 -> r2c2 = 1, r1c4 = 8, r2c4 = 9, r1c3 = 6
If 3489 -> r2c2 = 3, r1c4 = 8, r2c4 = 9, r1c3 = 4
If 3579 -> r2c2 = 3, r1c4 = 5, r2c4 = 9, r1c3 = 7
If 4569 -> r2c2 = 4/6, r1c4 = 5, r2c4 = 9, r1c3 = 4/6
-> r2c4 = 9
-> r2c8 = 7, r1c8 = 9
-> remaining options for complex 20(3): r1c4 = 5, r4c2 = 6 ...
The puzzle falls out from here.

A good, challenging puzzle - helping me improve combination analysis! :)

Edit: Just had a scan of Richard's walkthrough - looks like mostly the same moves though some in a different order. Para seems to have taken a more varied route.
Last edited by CathyW on Sun May 13, 2007 3:25 pm, edited 4 times in total.
Para
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Post by Para »

CathyW wrote:Just had a scan of Richard's walkthrough - looks like mostly the same moves though some in a different order. Para seems to have taken a more varied route.
My routes tend to vary from Richards.
rcbroughton
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Post by rcbroughton »

Para wrote:My routes tend to vary from Richards.
Variety being the spice of life. :D
mhparker
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Post by mhparker »

All quiet on the Western Front...

Maybe everybody's still struggling with JC's V2?

Therefore I thought I'd generate a bit of interest by posting a walkthrough for it. Here it is:

Edit: Modifications to steps 20,23,29,31,38b,40c,43c
Edit: Simplified logic for steps 34 & 37, removed unnecessary step 36


Assassin 49V2 Walkthrough


1. 9/2 at R1C1: no 9

2. 8/2 at R1C5: no 4,8,9

3. 5/2 at R1C9 = {14|23}

4. 11/2 at R2C3: no 1

5. 12/2 at R2C7: no 1,2,6

6. 14/4 at R4C3: no 9

7. 26/4 at R4C7: no 1

8. 10/3 at R6C2: no 8,9

9. 14/2 at R7C3 = {59|68}

10. 5/2 at R7C7 = {14|23}

11. 10/2 at R8C1: no 5

12. 10/2 at R8C5: no 5

13. 9/2 at R8C9: no 9

14. Innies C5: R37C5 = 10/2: no 5

15. Innies R12: R2C37 = 14/2 = {59}|[68]
15a. Cleanup: R3C3 = {2356}, R3C7 = {347}

16. Innies R89: R8C37 = 9/2: [54|63|81]
16a. Cleanup R7C3: no 5, R7C7: no 3

17. 11/2 at R2C3 cannot be {56} (blocked by 14/2 at R7C3) -> {29|38}
17a. 11/2 at R2C3 and 14/2 at R7C3 form killer pair on {89} -> no 8,9 elsewhere in C3

18. Innies R12 (step 15): R2C37 = 14/2 = [95] -> R3C37 = [27]
18a. Cleanup: no 7 in 9/2 at R1C1
-> 7 in N1 now locked in 27/5 at R1C2 = {7...}, no 7 in R12C4
18b. Cleanup: no 3 in R1C5, no 4 in R1C1, no 3,8 in R7C5 (step 14)

19. 14/2 at R7C3 = {68}, locked for C3 and N7
19a. Cleanup: no 2,4 in 10/2 at R8C1 = {19|37}
19b. Cleanup: no 4 in R8C7 (step 16) -> no 1 in R7C7

20. {268} unavailable for in R456C3 for 14/4 at R4C3 -> 14/4 = {1247} (with 2 in R5C4),
{1346} (with 6 in R5C4) or {2345} (with 2 in R5C4) -> R5C4 = {26},
4 locked in R456C3 for C3 and N4

21. 26/4 at R4C7: cannot have both of {57} -> {9(278|368|458|467)} -> no 9 in R5C89
21a. 5,7 only available in R5C6 -> no 2,4 in R5C6

22. Innies C89: R1289C8 = 27/4 = {9(378|468|567)} (no 1,2), 9 locked for C8

23. <deleted>

24. 14/3 at R3C8 = {149|239|158|248|167|347|356} ({257} unplaceable)
24b. 2,5,7 only available in R4C8 -> no 3,6,8 in R4C8

25. 18/3 at R6C8: 9 only available in R7C9 -> no 1,2 in R7C9

26. Innie/outie difference, C9: R5C8 = R37C9 + 14
26a. Max. of R37C9 = 17 -> R5C8 = {123} -> R3C9 = {689}, R7C9 = {6789}
26b. 14/3 at R3C8 can only have one of {689} (step 24), which must now come from R3C9
-> no 6,8 in R3C8

27. Innie/outie difference, C9: R345C8 = R7C9 = 6,7,8 or 9 -> no 7 in R4C8

28. Common Peer Elimination (CPE): R7C9 can see all candidate positions for 7 in C8
-> no 7 in R7C9
28a. R37C9 cannot now sum to 16 (7 unavailable) -> no 2 in R5C8 (step 26)
28b. 9 now locked in R37C9 for C9
28c. 9 in N6 now locked in R456C7 -> not elsewhere in C7, no 9 in R5C6

29. 17/4 at R4C9 = {1268|1358|2348|1367|1457|2357}
29a. Of these, {2348} is blocked by 5/2 at R1C9
29b. Therefore 17/4 at r4C9 must contain 1 of {56} -> 14/3 at R3C8 cannot contain both of {56}
-> no 6 in R3C9
29c. 6 in N3 now locked in 27/5 at R1C6 = {6...} -> no 6 in R12C6

30. Important observation: 9/2 at R8C9 must contain 1 of {1234}
30a. Therefore, whichever combination it contains, it directly determines the combinations
{14|23} in the two 5/2 cages at R1C9 and R7C7
30b. These two 5/2 cages are therefore synchronized (i.e., must contain the same combination)
30c. Thus, they also lock the same 2 digits into R456C8 (i.e., 2 of these 3 cells must sum to 5)
(no eliminations yet)

31. 17/4 at R4C9 = {1268|1358|1367|1457|2357}
31a. AIC: (2)r456c9-(2=14)r12c9-(14=3)r3c8-(3=1)r5c8
-> if 17/4(R4C9) contains a 2, it must also contain a 1
31b. {2357} can thus be rejected as possible combination
31c. Therefore, 1 is locked for N6 in 17/4 at R4C9 -> not elsewhere in N6

32. Innie/outie difference, N6: R5C6+R7C89 = R4C8 + 16 -> no 2 in R7C8
(Reason: 2 in R7C8 forces R4C8 to at least 4, requiring R5C6+R7C9 to be at least 18 - unreachable)

33. 2 in C8 now locked in N6 -> not elsewhere in N6

34. 2 in C9 locked in 5/2 at R1C9 or 9/2 at R8C9
34a. Therefore, either 5/2 at R1C9 = {23}, or 9/2 at R8C9 = {27}
34b. In either case, 9/2 at R8C9 cannot be {36} -> no 3,6 in 9/2 at R8C9

35. 2 in C9 already locked in 5/2 at R1C9 and 9/2 at R8C9
35a. Due to synchronization of 5/2 cages (step 30b), 2 must therefore also be locked in N9 in
5/2 at R7C7 and 9/2 at R8C9 -> no 2 elsewhere in N9

36. <deleted>

37. R46C8 = {2..} (step 33), i.e., one of these 2 cells is a 2
37a. Innie/outie difference(N6) (R46C6 - R5C6 = 2) -> other cell in R46C6 = R5C6
37b. Thus, R46C6 cannot contain any candidate (apart from 2) not in R5C6
-> no 4 in R46C6

38. Hidden 5/2 pair in R456C8 (step 30c) must now be {23} (4 unavailable)
38a. Therefore (steps 30b, 30c) 5/2 at R7C7 = [23] and 5/2 at R1C9 = {23}, locked for C9 and N3
38b. Cleanup: no 7 in R89C9, no 7 in R9C5, no 7 in R9C1

39. 14/2 at R7C3 = [86] (step 16)
39a. Cleanup: no 4 in R9C5

40. 14/3 at R3C8 = {(15|24)8} -> R3C9 = 8 -> R7C9 = 9 (step 28b) -> R5C8 = 3 (step 26)
40a. Cleanup: no 1 in 9/2 at R8C9 = {45}, locked for C9 and N9
40b. R456C9 = {167}, locked for N6
40c. R456C7 = {489} -> R5C6 = 5; {489} locked for C7 and N6
40d. Split 9/2 at R67C8 = [27] (only remaining combination/permutation)

41. Naked singles at R48C8 = [58] -> R39C8 = [16] -> R9C7 = 1

42. NS at R2C8 = 4 -> R1C78 = [69]

43. Hidden single (HS) in C5 at R1C5 = 5 -> R2C5 = 3
43a. R12C9 = [32]
43b. Cleanup: 9/2 at R1C1 = {18}, locked for C1 and N1
43c. Cleanup: 10/2 at R8C1 = [73], locked for C1 and N7

44. NS at R19C3 = [75]

The rest is all naked and hidden singles.


P.S. I went in from the opposite side to the one that JSudoku took, just to show that there's more than one way to peel an onion (as Richard would say) and also to make it a bit more interesting.
Last edited by mhparker on Thu May 10, 2007 8:13 am, edited 4 times in total.
Cheers,
Mike
Glyn
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Post by Glyn »

Here is my walkthrough of V2, what a one to pick for my first try.

Redone following earlier booboo


1 ) R12 Innies=14 R2C37={59}|[68] (R2C7=6 not possible in 12(2))
Cleanup R3C3={256},R3C7={347} (Combo sum must=9 Outties R12)

2) In N1 11(2)R23C3=[92]|{56} must contain 5|9 leaves only {68} combo for 14(2) in N7.

3) Naked pair {68} in R3 and N7.

4) In N1 11(2)=[92]. R2C2=9 & R3C2=2.

5) Innies R12 = 14. => R2C7=5 & R3C7=7.

6) Innies R89 = 9. R8C37=[63]|[81] => R7C7=2|4.

7) Combinations for 14(4) cage in N45 (R456C3+R5C4)={134}6,{147}2,{345}2 other combos prohibited by repeated digits or exceeding cage sum.
=> R5C4=2|6.

8) 4 locked in C3 for 14(2) cage of N4.

9) Cage 26(4) in N56 = {2789},{3689},{4589},{4679},{5678}
Possible Combos for Cage 26 R5C6+R456C7= 3{689}|5{489}|6{389}|7{469}|7{289}|8{369}|9{368}.
R5C6<>124

10)Clean up in N1 9(2)={18}{36}[54], in N2 8(2)={17}(26}[53], in N7 10(2)={19}{37}

11) 7's in N1 locked in 27(5) cage. Not elsewhere in cage R12C4<>7.

12) N14 16(3) R3C12+R4C2={16}9|{18}7|{34}9|{35}8|{36}7|{38}5|{45}7|{58}3|{68}2 => R4C2<>1,6.

13) N36 14(3) R3C89+R4C8={14}9|{16}7|{18}5|{19}4|{34}7|{36}5|{39}2|{48}2|{49}1 => R4C8<>3,6,8

14) Outties N1=18 R12C4+R4C2={18}9|[91]8|{28}8|[92]7|{36}9|{38}7|[54]9|{46}8|[94]5|[56]7|[58]5|[96]3 => R4C2<>2.

15) R456C3 sum to either 8 or 12. Cage(14)-(2|6).Therefore R4C2+R6C2=9|13. Possible combos are [36][72][81][76][85] => R4C2<>5,9. R6C2<>3,7

16) Revisit the Outties of N1 R12C4+R4C2=[91]8|{28}8|[92]7|{38}7|{46}8|[56]7|[96]3 =>R1C4<>1.

17) Trying all combination with R46C2 in N4.
a) R46C2=[36] R456C3={147} Cage 24(4)={2589}
b) R46C2=[72] R456C3={345} Cage 24(4)={1689}
c) R46C2=[81] R456C3={345} Cage 24(4)={2679}
d) R46C2=[76] R456C3={134} Cage 24(4)={2589}
e) R46C2=[85] R456C3={134}.Cage 24(4)={2679}.

18) Thanks to Ed
a)2 in R6C2 blocks all 2's in N7
b)6 in R6C2 forces R7C12={13} blocks all combos of cage 10(2) in N7.

19) Now I can do it only better. All combos for R46C2 from step 17) require R4C2=8.
Cage 24(4) in N4={2679}.

20) 16(3) cage N14={358} => R3C12={35} locked for N1 and R3.

21) 9(2) cage N1 R12C1={18} locked for C1 and N1.

22) Naked single R1C3=7.

23) Hidden single R7C1=4.

24) R89C1={37} locked for C1 and N7.

25) Naked singles R3C1=[53].

26) Hidden single R5C2=7.

27) Naked singles R78C7=[23], R89C1=[73].

28) Innies R89=9. R8C3=6=> R7C3=8.

29) Hidden pair R89C2={29} locked for Cage 24(5) in N78. Remaining combos {12489}|{12579}. No 6.

30) Naked pair R12C2={46} locked for 27(2) cage in N12. Remaining combos {14679}|{24678}. No 3,5. => R12C4=[91]|{28}.

31) From Step 9. Remaining combos for 26(4) cage in N56 are 3{689}|5{489}. Others blocked by 12(2) and 5(2) cages in C7 or by R5C2.

32) 8 and 9 locked in R456C7 for N6 and C7.

33) Mandatory inclusion of a 1 in 24(5) cage Nonets 7 and 8 at either R9C3 or R89C4. Elimate from common peers R9C56<>1.

34) Innies C5=10 R37C5={19}|[46].

35) a) If R37C5={19} sole remaining combos for 10(2) cage in N8 are {28}|[46] leaves only {35} combo for 8(2) in N2.
b) If R37C5=[46] eliminates {26} combo from cage 8(2) in N2.
Cleanup 8(2) cage R12C5= [17]|[53]

36) Cleanup 10(2) cage in N8=[19]|{28}|[46].

37) N69 18(2) cannot contain 1.

38) N9 9(2) cannot contain 6 or 7.

39) 3's on N8 locked in 26(5). R6C46<>3.

40) outties of R123=18 R4C468 sum to 10. Maximimum value R4C468=7.

41) Innies and Outties C9 R37C9=R5C8+14
Possible LHS exceeding 14 are 15,16,17 => R5C8=1|2|3
R37C9={69}|[87]|[97]|[89]

42) N36 14(2) remaining combos {149}{158}{167}{248} Combo {257} blocked as only one cell is available containing 2 and 7.
Arrangements only formed from R3C8=1|4 R3C9=6|8|9 R4C8=1|2|4|5|7.

43) Cell R3C8 forces cage 5(2) of N3 to {23}. Naked pair {23} in C9 and N3.

44) Lets do something different an AIC
3[R5C8]=3[R6C8]-{69}[R7C89]=1[R7c5]=5[R7C2]=1[R6C2]-{345}[R456C3]=2[R5C4]-
which implies that if R5C8<>3 then R5C8<>2. (Obviously if R5C8=3 then R5C8<>2).

45) Remaining combos for 17(4) in N6 are {1367}|{1457} must contain 1 and 7. The 7 is locked in C9 of N6.

46) Unplaceable candidate R3C9<>6.

47) R3 6's locked in N2 and cage 24(5)

48) Cage 18(3) in N69 R6C8<>6.

49) The remaining combo for from Step 41) is R37C9=[89] & R5C8=3.

50) Naked single R5C6=5.

51) Hidden single R1C5=5 => R2C5=3.

52) Hidden single R2C6=7.

53) Naked single R2C9=2 => R1C9=3.

54) Hidden single R6C3=3.

55) Hidden single R1C8=9

56) Cage 27(5) in N12 remaining combo {24678} => R12C4=[28]

57) Naked singles R12C1=[81].

58) 9(2) cage R89C9={45} locked for C9 and N9.

59) Naked single R5C4=6 => 14(4) in N4 ={1346} => R45C3={14} => R9C3=5.

60) Just singles from here on.



Hope it makes sense.

All the best,

Glyn
Last edited by Glyn on Sun May 20, 2007 8:42 pm, edited 7 times in total.
I have 81 brain cells left, I think.
Jean-Christophe
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Post by Jean-Christophe »

mhparker wrote: Assassin 49V2 Walkthrough

Edit: Simplified complex step 31
Are you sure it's simpler ? :shock:

Thanks for the walkthrough. It's maybe a bit too much of combinations crushing as JSudoku does :roll:

Some comments:

Step 20 should read 14/4, not 14/3

Step 23 how about R4C2 = 9 & R3C12 = {16|34} ? Anyhow, you do not use it later on.

Should maybe move step 31a before step 29 to prove 17/4 at r4C9 must contain 1 of {56}


Now, I'll compare with Glyn's walkthrough
mhparker
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Post by mhparker »

Jean-Christophe wrote:Are you sure it's simpler ?
Elementary, my dear Watson! :)
Jean-Christophe wrote:It's maybe a bit too much of combinations crushing as JSudoku does
JSudoku is pretty good at that as well!
Last edited by mhparker on Wed May 09, 2007 7:48 am, edited 1 time in total.
Cheers,
Mike
Andrew
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Post by Andrew »

I only managed to finish Assassin 49 yesterday and then worked through the 3 posted walkthroughs which all contained some interesting moves and were quite a bit different from the way that I eventually solved it.

Most of my difficulties were my own fault. First I had a couple of flawed moves based on incorrect mental arithmetic. Then after fixing that I got stuck because I'd forgotten that 9 was locked in R12C4 for the 26(5) cage. After starting again to reach that position it came out fairly easily. Each time I restarted I spotted some moves earlier than I'd previously seen them, and moved them to the earlier position, so I no longer have a couple of "crossover" moves that were originally in the walkthrough. My final logic flaw is mentioned in the walkthrough.
CathyW wrote:Just had a scan of Richard's walkthrough - looks like mostly the same moves though some in a different order.
One thing Cathy had, which I don't remember in Richard's walkthrough, was the use of a UR step to eliminate {34} from R12C5.

Here is my walkthrough, modified as a result of several restarts.

Thanks Para for the comments and typo correction.

1. R12C1 = {19/28/37/46}, no 5

2. R12C5 = {16/25/34}, no 7,8,9

3. R12C9 = {16/25/34}, no 7,8,9

4. R23C3 = {59/68}

5. R23C7 = {59/68}

6. R78C3 = {29/38/47/56}, no 1

7. R78C7 = {39/48/57}, no 1,2,6

8. R89C1 = {29/38/47/56}, no 1

9. R89C5 = {29/38/47/56}, no 1

10. R89C9 = {17/26/35}, no 4,8,9

11. 11(3) cage in N36 = {128/137/146/236/245}, no 9

12. 23(3) cage in N69 = {689}, no 6,8,9 in R89C8

13. 13(4) cage in N45 = 1{237/246/345}, no 8,9, no 1 in R5C12

14. 45 rule on R12 2 innies R2C37 = 13 = {58}, locked for R2, clean-up: no 2 in R1C1, no 2 in R1C5, no 2 in R1C9, no 5,8 in R3C37
14a. Naked pair {69} in R3C37, locked for R3

15. 45 rule on R89 2 innies R8C37 = 15 = [69]/{78}, clean-up: R7C3 = {345}, R7C7 = {345}

16. 45 rule on C5 2 innies R37C5 = 11 = [29/56]/{38/47}, no 1, no 2,5 in R7C5

17. 45 rule on R89 2 outies R7C37 = 8 = {35}, locked for R7, clean-up: no 8 in R3C5, no 7 in R8C3, no 8 in R8C7

18. R23C3 = [59] ([86] clashes with R8C3) -> R78C3 = [38], R23C7 = [86], R78C7 = [57], clean-up: no 1 in R12C1, no 1 in R12C9, no 1,3 in R89C9, no 4 in R9C1, no 3,4 in R9C5

19. Naked pair {26} in R89C9, locked for C9 and N9, clean-up: no 5 in R1C9

20. Naked pair {34} in R12C9, locked for C9 and N3

21. Naked pair {89} in R7C89, locked for R7, N9 and 23(3) cage -> R6C8 = 6, clean-up: no 2,3 in R3C5 (step 16)

22. 22(5) cage in N89 naked triple {134} in R8C8 + R9C78 -> R89C6 = 14 = {59}/[68]

23. 13(4) cage in N45 (step 13) = 1{237/246} (cannot be {1345} because 3,5 only in R5C4) = 12{37/46}, no 5, no 2 in R5C12
23a. 3 only in R5C4 -> no 7 in R5C4

24. 45 rule on C9 2 innies R37C9 – 9 = 1 outie R5C8, max R37C9 = 16 -> max R5C8 = 7
Para wrote “You can also eliminate 2,3 from R5C8 with this move.
Doesn't complicate the solving process though.”

Yes, I must admit that I only looked at the max and min cases with the latter not helping at this time. With early multiple innies/outies there are usually too many candidates for anything more to be helpful. In this case it is already down to 3 candidates in R3C1 and 2 candidates in R7C1. Fortunately missing the elimination of 2,3 didn’t matter in this case; the 3 goes in step 29 and the 2 in step 31. I must look more carefully at cases like this.


25. 45 rule on C6789 4 innies R3467C6 = 14, no 9

26. 21(4) cage in N56 max R456C7 = 16 -> min R5C6 = 5
26a. 45 rule on N6 1 outie R5C6 – 3 = 1 remaining innie R4C8 -> R4C8 = {2345}, no 9 in R5C6
26b. Max R5C6 = 8 -> min R456C7 = 13 -> must contain 9, locked for C7 and N6

27. R7C8 = 8, R7C9 = 9 (hidden singles in C8 and C9)

28. 11(3) cage in N36 = {137/245}
28a. 4 only in R4C8 -> no 2,5 in R4C8
28b. 2 only in R3C8 -> no 5 in R3C8
28c. R4C8 = {34} -> R5C6 = {67} (step 26a)

29. Naked triple {134} in R489C8, locked for C8
29a. 1 in C8 locked in R89C8, locked for N9

30. 11(3) cage in N36 = {137/245}
30a. 1 only in R3C9 -> no 7 in R3C9

31. 7,8 in C9 locked in R456C9 -> 21(4) cage in N6 = {1578}, locked for N6 -> R5C8 = 5
31a.1 locked in R456C9 for C9 -> R3C9 = 5, R34C8 = [24] (step 30) , R5C6 = 7 (step 26a), clean-up: no 6 in R7C5

32. R1C7 = 1 (naked single), clean-up: no 6 in R2C5

33. R9C7 = 4 (hidden single in N9)

34. Naked pair {47} in R37C5, locked for C5, clean-up: no 3 in R12C5

35. R89C5 = {29}/[38] (cannot be {56} which clashes with R1C5), no 5,6

36. Naked pair {79} in R12C8, locked for 27(5) cage in N23
36a.R12C6 = 10 = {46}/[82], no 3,5, no 2 in R1C6

37. 9 in C6 locked in R89C6 = {59}, locked for C6 and N8, clean-up: no 2 in R89C5 = [38] -> R89C8 = [13]
37a. R3467C6 = 13{28/46}

38. 9 in C5 locked in R456C5, locked for N5
38a. R456C5 = 9{16/25}

39. 14(3) cage in N47 = {149/167/248/257} (cannot be {158/239/356} because 3,5,8,9 only in R6C2, cannot be {347} because {47} in R7C12 clashes with R7C5), no 3
39a. 5,8,9 only in R6C2 -> no 2,4 in R6C2

40. 15(3) cage in N14 = {168/348/357} (cannot be {159/249/258/267/456} because 2,5,6,9 only in R4C2), no 2,9
40a. 5 only in R4C2 -> no 7 in R4C2
Para wrote “Also 6 only in R4C2 -> no 1 in R4C2
You get this in step 46 with the 45-test on N1.”

Thanks. Don’t know why I missed that one.


41. 23(5) cage in N25 ={12578/13478/14567/23468/23567} (cannot be {13568} which doesn’t contain 4,7 for R3C5)

42. 22(5) cage in N58 = {12478/13468/13567/23467} (cannot be {12568} which doesn’t contain 4,7 in R7C5, cannot be {23458} because 3,5,8 only in R6C46)

43. 45 rule on R123 3 remaining outies R4C246 = 15 = {168/258}, no 3 = 8{16/25}, 8 locked for R4

44. 45 rule on R789 3 remaining outies R6C246 = 16 = {178/259/349/358/457}
44a. 7 only in R6C2 -> no 1 in R6C2
44b. 9 only in R6C2 and only other 5 in R6C4 -> no 2 in R6C4

45. 23(5) cage in N78 with 1 locked in R9C234 = 1{2479/2569/4567}

46. 45 rule on N1 3 outies R12C4 + R4C2 = 20 (doubles possible), no 1
46a. Max R4C2 = 8 -> min R12C4 = 12, no 2

47. 45 rule on N7 3 outies R6C2 + R89C4 = 14
47a. no 6 in R9C4 (no {17/26/35/44} in R6C2 + R8C4)
Para wrote “You should add no {44} to 47a, that could also be possible. ”

Good point! It’s in there now. I usually remember to look for “doubles possible” when I’m doing outies from a nonet.


48. 5 in R1 only in R1C4 or R1C5, 1 in R2 only in R2C2 or R2C5
48a. If R12C5 = [52], R2C2 = 1, if R12C5 = [61], R1C4 = 5 -> 26(5) cage in N12 must contain 1 or 5 but not both
48b. 9 locked in R12C4
48c. 26(5) cage = {12689/13679/23579/24569}
48d. 5,9 only in R12C4 -> no 4 in R12C4

49. 45 rule on N4 2 innies R46C2 – 9 = 1 outie R5C4
49a. Min R46C2 = 11 -> no 1 in R5C4
49b. 1 in 13(4) cage locked in R456C3, locked for C3 and N4

50. 1 in C1 locked in R37C1
50a. 45 rule on C1 2 innies R37C1 – 1 = 1 outie R5C2 -> R5C2 = R3C1 or R7C1 with the other being 1
50b. No 2,7 in R5C2 -> no 2,7 in R37C1
50c. No 9 in R37C1 -> no 9 in R5C2

51. 3 in N4 locked in 23(4) cage
51a.23(4) cage = 3{479/569/578}, no 2
51b. 5,7 only in R46C1 -> no 8 in R6C1
51c. 2 in N4 locked in R456C3, locked for C3 -> no 2 in R5C4
Para wrote “This step created a hidden single 2 in R1C2 for R1. And that breaks open the puzzle to a few naked pairs and singles.
I noticed this because the moved that broke the puzzle for me was that the 2 in R1 was locked in R1C234 in 26(5) in R1C2.”

That would be correct if I hadn’t missed 2 locked in R1C23 at step 48, which would have eliminated {13579} but probably not any candidates. With 2 locked in R1C23 then step 51c would have created the hidden single. In that sense I didn’t miss a hidden single at this stage.


52. R5C4 cannot be 3
52a. If R5C4 = 3 => R456C3 = {127} (step 23) clashes with R46C2 = 12 = [57] (step 49) -> no 3 in R5C4 [Typo corrected. Thanks Para.]
52b. 13(4) cage in N45 (step 23) = {1246}, no 7, no 4,6 in R5C12
52c. 4 in R5 locked in R5C34 -> no 4 in R6C3
52d. No 4,6 in R5C2 -> no 4,6 in R37C1 (step 50a)

53. R7C1 = 1 (naked single), R67C2 = 13 = [76/94]

54. 2,7 in R7 locked in R7C456, locked for N8 -> R9C4 = 1
[Alternatively R9C4 = 1 (hidden single) after step 53. I saw the locked 2,7 first.]
54a. 2,7 locked in R7C456 for N8 and 22(5) cage in N58 (step 42) = 27{148/346}, no 5 = 247{18/36}, no 2 in R6C6
54b. 1 only in R6C6 -> no 8 in R6C6
[My final logic flaw was to assume from step 54a that 4 was also locked in R7C456 rather than just in the 22(5) cage. Fortunately that was not correct or I might have ended with a "solution" based on false logic which Ed would have spotted pretty quickly.]

55. Naked pair {46} in R58C4, locked for C4

56. 22(5) cage in N58 (step 54a) = 247{18/36}
56a. 4 only in R6C6 for {23467} combination -> no 3 in R6C6
Para wrote “This also created a Hidden single 3 in R6C4 for N5”

Thanks. Missed that one.


57. 23(4) cage (step 51a) = 3{479/569/578}
57a. R5C12 = {389} -> no 3,9 in R46C1
57b. 3 in N4 locked in R5C12, locked for R5

58. 45 rule on C1234 4 innies R3467C4 = 20 = {2378} (only remaining combination), no 5, 3,7,8 locked for C4

59. R12C4 = [59] (naked singles) -> R12C5 = [61], R12C8 = [97], clean-up: no 3 in R1C1, no 4 in R2C1, no 4 in R12C6 = [82]

60. Naked pair {47} in R1C13, locked for R1 and N1 -> R12C9 = [34], R1C2 = 2
60a. Naked pair {36} in R2C12, locked for N1 -> R3C12 = [81], R4C2 = 6 (cage sum), R5C2 = 8 (step 50a)

and the rest is naked singles

Para wrote “That hidden single 2 could have saved you some work. I don't know if you regularly scan the grid for hidden singles? I usually do this after every few steps just to make sure. Even if i don't expect any.”

Good point. I should do it more often. They aren’t always easy to see on the Excel spreadsheet that I use for solving Sudokus. Maybe they are easier to see for people using software such as SumoCue but I’ve no current plans to download it. I like the way that I can save positions in Excel, make copies of positions on multiple worksheets and add notes below the diagram.
Last edited by Andrew on Fri May 11, 2007 9:44 pm, edited 1 time in total.
Para
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Post by Para »

Hi all

Here is my walk-through for JC's V2 with some interesting and also a bunch of redundant moves but those are always going to be in my walk-throughs. This took a bit more searching for the breakthrough move.

Walkthrough Assassin 49 V2

1. R1C12 + R89C9 = {18/27/36/45}: no 9

2. R12C5 = {17/26/35}: no 4, 8, 9

3. R12C9 + R78C7 = {14/23}: no 5, 6, 7, 8, 9

4. R23C3 = {29/38/47/56}: no 1

5. R23C7 = {39/48/57}: no 1, 2, 6

6. 26(4) in R4C7 = {2789/3689/4589/4679/5678}: no 1

7. 10(3) in R6C2 = {127/136/145/235}: no 8, 9

8. 14(2) in R78C3 = {59/68}: no 1, 2, 3, 4, 7

9. R89C1 + R89C5 = {19/28/37/46}: no 5

10. 45 on R12: 2 innies: R2C37 = 14 = {59}/[68]
10a. Clean up: R3C3: no 3, 4, 7, 8, 9; R3C7: no 5, 8, 9

11. R23C3: no {56}: clashes with R78C3 -->> R23C3 = [92]
11a. R23C7 = [57]
11b. R78C3 = {68}: locked for C3 and N7
11c. Clean up: R1C1: no 4,7; R2C1: no 7; R1C5: no 3; R89C1: no 2, 4

12. 14(4) in R4C3 needs 3 of {13457} in R456C3 -->> 14(4) = {1247/1346/2345}: R5C4 = {26}
12a. 4 locked in 14(4) cage in R456C3 -->> locked for C3 and N4
12b. 24(4) in R4C1 can’t have both {37}(one needed in 14(4) cage) -->> 24(4) = {1689/2589/2679}: no 3; 9 locked in 24(4) for N4

13. 45 on R89: 2 innies: R8C37 = 9 = [63/81]
13a. Clean up: R7C7: no 1, 3

14. 45 on C5: 2 innies: R37C5 = 10 = {19/46}/[37]/[82]: R3C5: no 5; R7C5: no 3, 5, 8

15. 10(3) in R6C2: no {136}: only possible with R7C12 = {13} which clashes with R89C1 -->> R6C2: no 6

16. 45 on N4: 2 innies – 1 outie: R46C2 – R5C4 = 7
16a. R5C4 = 6 -->> R46C2 = [67/85]
16b. R5C4 = 2 -->> R46C2 = [27/63/72/81]
16c. Conclusion: R4C2: no 1, 3, 5

17. 16(3) in R3C1 = {18}[7]/{68}[2]/{35}[8]/{36}[7]/{45}[7]
17a. R4C2: no 6
17b. R6C2: no 3(step 16 b)

18. 3 in N4 locked for C3
18a. 3 in N4 locked for 14(4) in R4C3 -->> 14(4) = {1346/2345}: no 7

19. 6 in N4 locked in 24(4) -->> 24(4) = {1689/2679}: no 5

20. 7 in N1 locked in 27(5) cage in R1C2: R12C4: no 7 -->> 27(5) = {12789/14679/15678/23679/24579/24678/34578}

21. When 8(2) in R1C5 = {17/26}, 5 in C5 locked in 17(3) in R4C5
21a. When 8{2} = [53], no 7 in R37C5 + R89C5 (both 10(2)) so 7 in C5 locked in 17(3)
21b. 17(3) in R4C5 needs one of {57} -->> 17(3) = {179/278/359/458/467}

22. 14(3) in R3C8 = {149}/{18}[5]/{16}[7]/{39}[2]/{48}[2]/{36}[5]: no {34}[7] clashes with R12C9 -->> R4C8: no 3, 6, 8

23. 10(3) in R6C2 = {127/145}-->> R89C2: no 1
23a. 10(3) = {235} (3 in R7C12) -->> R89C1 = {19} -->> R89C2: no 1
23b. Conclusion R89C2: no 1

24. 45 on C9: 5 outies : R34567C8 = 18 = {12348/12357/12456}: no 9; 1,2 locked for C8

25. 26(4) in R4C7 needs 3 of {234689} in R456C7 -->> 26(4) = [7]{289}/{3689}/[5]{489}/[7]{469} -->> R5C6: no 2,4; 9 locked in 26(4) cage: R5C9: no 9

26. Only place for {45} in C1 are R1237C1: R12C1 can only contain both {45} or neither so R37C1 can only contain both {45} or neither.

27. 45 on C1: 2 innies - 1 outie: R37C1 - R5C2 = 2
27a. R5C2 = 1 -->> R37C1 = [12] -->> R12C1 = [54]
27b. R5C2 = 2 -->> R37C1 = {13}: clash with R89C1
27c. R5C2 = 6 -->> R37C1 = [62]({17} clashes with R89C1} -->> R12C1 = [54]
27d. R5C2 = 7 -->> R456C1 = {269} -->> R37C1 = {45}/[81] -->> R12C1 = {18}/[54]
27e. R5C2 = 8 -->> R37C1 = [82/37] -->> R12C1 = [54]
27f. R5C2 = 9 -->> R37C1 = [83] -->> R12C1 = [54]
27g. Conclusions: R12C1: no {36}; R5C2: no 2
27h. R37C1 = [12/37/45/54/62/81/82/83]

28. R6C2: no 2: sees all 2’s in C1
28a. Clean up: R4C2: no 7(step 16b)
28b. 16(3) in R3C1 = {68}[2]/{35}[8]: no 1, 4
28c. Clean up: R7C1: no 5; R5C2: no 1 (step 27)

29. Killer Pair {58} in R12C1 + R3C12 locked for N1

30. Killer Pair {24} in R7C12 + R7C7 locked for R7
30a. Clean up: R3C5: no 6, 8 (step 14)

31. 5 in C1 locked in R13C1 for N1 -->> R3C2: no 5

32. Clean up: R3C1: no 3; R7C1: no 7 (step 27)

33. 27(5) in R1C2 needs one of {36} in R12C2 (only place left in N1) and 2 of {147} in R1C12 + R2C2 -->> 27(5) = {14769/15678/24679/34579}: R12C4: no 3, 6
33a. Only place for 9 is R1C4 -->> R1C4: no 1, 4

34. 3 in N1 locked for C2

35. 3’s in N1
35a. R12C2 = 3 -->> R12C4 = [58] -->> R3C46: no 5
35b. R3C2 = 3 -->> R3C1 = 5 -->> R3C46: no 5
35c. Conclusion R3C46: no 5

36. Hidden single: R3C1 = 5
36a. R34C2 = [38]; R7C1 = 4; R78C7 = [23]

37. R12C1 = {18} locked for C1 and N1
37a. R1C3 = 7; R89C1 = [73]
37b. Naked Triple {269} locked for N4
37c. R5C2 = 7
37d. Clean up: R89C9: no 6, 7; R9C5: no 7; R7C5: no 7

38. Hidden pair {29} in R89C2 locked for 24(5) in R8C2 -->> 24(5) = {12489/12579}: no 6;
38a. Naked Pair {46} in R12C2 locked for 27(5) in R1C2 -->> 27(5) = 7{1469/2468}: no 5; R12C4 = [28/82/91]

39. R12C9 = {23}: {14} clashes with R89C9 -->> {23} locked for C9 and N3

40. 3 in C8 locked in R56C8: R34567C8 = {12348/12357}(step 24): no 6

41. 26(4) in R4C7 needs 3 of {4689} in R456C7 -->> 26(4) = [3]{689}/[5]{489}: R5C6 = {35}; {89} locked in R456C7 for C7 and N6

42. 17(4) in R4C9 = {1367/1457/2357}: {2456} clashes with R456C7 -->> 7 locked in 17(4) in R26C9: locked for C9 and N6

43. 14(3) in R3C8 = {1[9]4}/{18}[5]/{48}[2]: no 6

44. 6 in R3 locked for N2 and 24(5) in R3C4
44a. Clean up: R12C5: no 2

45. 5’s in N2
45a. R1C5 = 5 -->> R2C5 = 3: R1C6: no 3
45b. R1C6 = 5: R1C6: no 3
45c. Conclusion: R1C6: no 3
45d. Hidden Single: R1C9 = 3; R2C9 = 2
45e. Clean up : R1C4: no 8

46. Naked Pair {18} in R2C14 locked for R2
46a. Naked Pair {46} in R2C28 locked for R2

Ugh completely forgot this
47. 45 R89: 1 innie: R8C3 = 6; R7C3 = 8

48. 18(3) in R6C8 = [279/459/576] -->> R6C8 = {245}; R7C8 = {57}; R7C9 = {69}
49. R5C8 = 3(hidden); R5C6 = 5; R1C5 = 5(hidden); R2C5 = 3; R2C6 = 7

50. 26(4) in R4C7 = 5{489} -->> R456C7 = {489}: locked for C7 and N6
50a. Clean up: R7C8: no 5(step 49)
50b. R7C8 = 7; R9C4 = 7(hidden)

51. 17(4) in R4C9 = 3{167}: R456C9 = {167} locked for C9 and N6
Now it is all singles and here is how:
51a. R7C9 = 9; R6C8 = 2; R4C8 = 5; R1C8 = 9(hidden); R1C4 = 2; R2C4 = 8(step 38)
51b. R12C1 = [81]; R5C4 = 6; R5C9 = 1; R5C3 = 4
51c. Hidden singles: R3C6 = 6; R7C5 = 6; R3C8 = 1; R9C7 = 1; R1C6 = 1
51d. R1C27 = [46]; R2C28 = [64]; R3C9 = 8; R89C8 = [86]; R9C3 = 5; R7C2 = 1; R7C46 = [53]; R6C2 = 1; R89C9 = [54]
51e. Hidden singles: R9C5 = 8; R6C6 = 8; R5C7 = 8
51f. R8C5 = 2; R5C5 = 9; R3C45 = [94]; R89C6 = [49]; R8C4 = 1; R89C2 = [92]
51g. R4C46 = [32]; R46C3 = [13]; R46C5 = [71]; R6C4 = 4; R5C1 = 2; R46C9 = [67]; R46C7 = [49]; R46C1 = [96]

greetings

Para
Last edited by Para on Thu May 31, 2007 1:30 pm, edited 1 time in total.
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Post by Jean-Christophe »

Glyn wrote:Here is my walkthrough of V2, what a one to pick for my first try.
I can't follow you logic, unless I missed something:
Step 14
I don't see why R5C6<>3 since R19C7 could add up to 5. These two could work:
Cage 5(2) in N9 = {14}, R19C7 = {23}, R5C6 = 3
Cage 5(2) in N9 = {23}, R19C7 = {14}, R5C6 = 3
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Post by Glyn »

Walkthrough now corrected thanks to Ed and JC for input.

All the best,

Glyn
I have 81 brain cells left, I think.
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