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Assassin 43 V0 - The Original Concept

 
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Ruud
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PostPosted: Thu Mar 22, 2007 10:26 pm    Post subject: Assassin 43 V0 - The Original Concept Reply with quote

To give you some insight into the way I make Assassins, here is a Version 0 of Assassin #43.

My first step is to draw an interesting cage pattern, avoiding single innies and outies and including at least a few large cages in the pattern. I know I have to break them up or rearrange them at some point, but it's worth trying to include several of them.

So this was my original cage pattern:



As you can see, it does not have any cages of size 2. Hardly any of my initial designs do...

After dozens of attempts, it turned out impossible to create a Killer with a single solution from this pattern, so I started to split up some of the cages. Initially, I did not want to break up the 4 size 6 cages in the corners, so I broke the 4 square cages instead. Splitting them vertically would create 2 pairs of outies - too easy - so I splitted them horizontally. After several attempts, the following puzzle came out:



3x3::k:5120:8193:8193:4611:4611:4611:9222:9222:2824:5120:8193:2315:2315:4611:3086:3086:9222:2824:5120:8193:2068:2068:5398:1559:1559:9222:2824:8193:8193:5661:5661:5398:7968:7968:9222:9222:3876:3876:3876:5661:5398:7968:4394:4394:4394:8493:8493:5661:5661:5398:7968:7968:7732:7732:1590:8493:4152:4152:5398:2619:2619:7732:3390:1590:8493:2369:2369:6723:1092:1092:7732:3390:1590:8493:8493:6723:6723:6723:7732:7732:3390:

SumoCue managed to place a single digit, and the you-know-who solver produced a solving path with more than 200 steps, using 25 different techniques. Too much for the intended audience.

More cage splitting was needed. I broke the size 6 cages in half, but now the pattern could only produce puzzles way below the Assassin standard. After I glued back 2 opposite square cages (preserving 180 degrees symmetry), the current assassin came out. It is tough, but very solvable for an advanced player.

Meanwhile, I'll leave this 43V0 for you to play with. There will certainly be some people who enjoy being tortured.

Ruud
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Para
Yokozuna
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Joined: 08 Nov 2006
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PostPosted: Thu Mar 22, 2007 11:04 pm    Post subject: Reply with quote

Hi

That is about the same i created my killer-X but i did give away a single innie mostly because otherwise the solving path seemed to be too clear and this innie didn't do much to lower the puzzle difficulty.
I usually set up a few 45-tests size 2 or 3 with my cages and try to keep those not to reveal too much. Try and keep the 2's between 7 and 13 and the 3's between 9 and 21.
But then usually my killers aren't assassin worthy. Mostly between the difficulty of the daily and extremes on sudoku.org.uk. But i used to always check my uniqueness by hand, avoiding obvious deadly patterns, but sometimes ran into a few curious ones(Bug-Lite variations mostly). This killer-X was easier to create because i checked uniqueness using Sumocue, and then examined the solving difficulty.

greetings

Para
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rcbroughton
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PostPosted: Sun Mar 25, 2007 1:17 pm    Post subject: Re: Assassin 43 V0 - The Original Concept Reply with quote

Ruud wrote:
Meanwhile, I'll leave this 43V0 for you to play with. There will certainly be some people who enjoy being tortured.


An interesting challenge - not in the same league as last week's V2, though.

Letting SumoCue run through we get to the following position:

Code:

.-------.--------------------.-----------------------------.-------------------.--------.
|456789 |123456789 12346789  |123467    123456789 123467   |123456789 123456789|12345678|
|       |         .----------'---------.         .---------'---------.         |        |
|456789 |12345679 |1234678    1235678  |123456789|345789    345789   |123456789|12345678|
|       |         :--------------------+---------+-------------------:         |        |
|456789 |123456789|12367      12567    |123456789|1245      1245     |123456789|12345678|
:-------'         :--------------------:         :-------------------:         '--------:
|4678    123456789|123456789  123456789|123456789|123456789 123456789|123456789 4789    |
:-----------------'----------.         |         |         .---------'------------------:
|46789   123456789 123456789 |123456789|123456789|123456789|123456789 123456789 46789   |
:-----------------.----------'         |         |         '---------.------------------:
|457     123      |123456789  123456789|123456789|123456789 123456789|13        68      |
:-------.         :--------------------:         :-------------------:         .--------:
|13     |4568     |79         79       |12345    |24        68       |134568   |1345    |
|       |         :--------------------+---------+-------------------:         |        |
|2      |6789     |45         45       |6789     |13        13       |6789     |6789    |
|       |         '----------.---------'         '---------.---------'         |        |
|13     |456789    456789    |346789    3456789   346789   |123456789 123456789|12345   |
'-------'--------------------'-----------------------------'-------------------'--------'


From there:

1. 13(3)n9 - combo {139} blocked by r8c7
1a. no other combinations with a 9 - so no 9 in r8c9

2. 9 now locked in n6 for c9

3. 26(4)n8 - combos {4589}/{2789}/{4679}/{3689}/{5678}
3a. {4589} - blocked by r8c4
3b. {2789} - not possible as no 2
3c. {4679} - blocked by r7c4
3d. leaves only {3689}/{5678} - no 4
3e. 5 only at r9c5 so no 7 at r9c5

4. from 45 rule on c1234 r19c4=10 - cleanup from 3d. no 6 at r1c4

5. from 45 rule on c6789 r19c6=10 - cleanup from 3d. no 6 at r1c6

6. 11(3) and 13(3) in c9 between them must use 1,2,3 & 6
6a. 11(3)={128}/{137}/{146}/{236}/{245}
6b. 13(3)={148}/{157}/{238}/{247}/{256}/{346}
6c. combined 11(3)&13(3)={128}{346}/{137}{256}/{146}{238}/{236}{148}/{236}{157}
6d. no 6 in r56c9 -> r6c9=8

7. Hidden single 8 at r7c7 for n9
7a. 10(2)n89=[28]
7b. cleanup from r7c6=2 -> 6(2)n23={45}/[12]
7c. cleanup from r7c7=8 -> 12(2)n23={39}/{57}/[84]
7d. cleanup from r7c6=2 -> (from step 5) no 8 at r9c6

8. Revisit step 6c. to remove combos with an 8 - leaves {127}{256}/{236}{157} - so must also use 7
8a. no 7 in r45c9
8b. leaves a naked pair {49} at r45c9 for n6 and c9

9. Cleanup from {49} naked pair
9a. 11(3)n3={137}/{236} - no 5
9b. 13(3)n9={157}/{256} - no 3

10. 3 locked in n3 for c9
10a. cleanup 12(2)n23=[39]/{57}/[84]

11. 9 now locked in c5 of n2

12. 5 now locked in n9 for c9

13. Cleanup from 12 - no 5 in 30(6)n69
13a. 30(6)n69=183{279} 381{279} 18{2469} 38{2467}
13b. no 1,3 at r9c78

14. 8(2) & 6(2) in r3 must use 2 between them
14a. 8(2)={17}/{26}/[35]
14b. 6(2)={15}/[24]
14c. combined 8(2) & 6(2) = {17}[24] {26}{15} [35][24] - must use 2 locked for r3

15. 31(5)n56 must use 9 - locked in r456c6 for c6 and n5
15a. cleanup from r9c6<>9 -> r1c6 no 1

16. following from step 13 - 30(6)n69 must use 2 - locked in r9c78 for r9
16a. cleanup r9c9<>2 -> 13(3)n9=7{15} - r8c9=7

17. cleanup from r8c9=7
17a. from 9a 11(3)n3={236} locked for n3
17b. from 13a. 30(6)n69=18{2469} - r6c8=1 no 1,3 in r7c8
17c. from 3d. 26(4)n8={3689}/{5678}

18. 36(6)n36={156789}/{246789}/{345789}
18a. {246789} - not possible since 2,6 only occur in cell r4c8
18b. {156789} - 6 must be at r4c8
18c. {345789} - 3 must be at r4c8
18d. r4c8=3,6 - no 2,5,7

19. cleanup from 17c. - hidden single 3 at r8c7 for n9
19a 4(2)n89=[13]

20. 6(2)n23=[51]
20a. from 14c. 8(2)n12={26} - locked for r3 -> r3c9=3
20b. from 10a. 12(2)n23=[39]/[75]/[84]
20c. 9(2)n12 - no 7 at r2c3

21. r3c7=1 -> no 1 in 36(6)n36 - from 18c. r4c8=3
21a. cleanup 17(3)n6 no 3 -> {269}/{467} -> no 5 either

22. 5 now locked in c7 of n6
22a. from 20b. [75] not possible in 12(2)n23 -> no 7 at r2c6

23. from 21a. 17(3)={26}9 or {76}4 - must use 6 - locked for r5 and n6

24. 31(5)n56 - combos are {16789}/{25789}/{34789}/{35689}/{45679}
24a. {16789} - only have 2,5,7 at r46c7 so can't place
24b. {34789} - ditto
24c. {35689} - ditto
24d. only combos {25789} and {45679} - no 3

25. cleanup from 23 - 15(3)n4 combos without 6 are {159}, {249}, {258}, {348}, {357}
25a. can't have {249} - blocked by r5c9
25b. {159} - must have the 9 at r5c1
25c. {357} - must have the 7 at r5c1
25d. no 7,9 at r5c23

26. 26(4)n8 - can only place {3689} or {5678}
26a. {3689} - {368} must be in r8c5, r9c56 -> no 3 at r9c4
26b. cleanup -> r1c4<>7

27. killer pair {26} in n2 - r3c4 & 18(4) must use 2 and 6. - no 2,6 at r2c4
27a. cleanup - no 7, 3 at r2c3

28. 45 rule on n1 outies minus innies is 7. minimum innies is 3, so min outies is 10 -> no 1 at r4c2
28a. innies = 3=[12], 7=[16], 8={26}, 10=[82], 14=[86] so outies total 10,14,15,17 or 21. Last is not possible in 2 cells!
28b. outies = 10=[82]/{64}
28c. outies = 14={68}
28d. outies = 15=[69], [87] - can't have 7 at r4c1 as this would block 2,6,7 in 20(3)n1
28c. outies = 17=[89]
28d. conclusion - no 5 used at r4c2, no 7 used at r4c1

29. 45 on n2 - innies total 17(4) (already know r3c6=5)
29a. {1259}, {1349}, {1358}, {1457}, {2456} not possible because r2c6&r3c4 only have 3/8 and 2/6
29b. {1268} - r2c6=8, r3c4=2 no 6 in either of other cells so can't be placed
29c. {1367} - r2c6=3, r3c4=6, r2c4=1, r3c5=7
29d. {2348} - r3c4=2, r3c5=4, r2c46={38}
29e. no 7 at r2c4, no 4 or 8 at r3c5
29f. cleanup r2c3<>2

30. 21(5)c5 must use 7 or 9 at r3c5 -> no 7 at r456c5 because:
30a. r3c5=7 - nowhere else
30b. r3c5=9 - combos are 9{1236} 9{1245}

31. 7 now locked for c5 in n2

32. 18(4)n2 - no 5 and must use 3 or 4 at r1c6
32a. r1c6=3 - combos are {1368} - blocked by r2c4, {2349} or {2367} - blocked by r3c4
32b. r1c6=4 - combos are {2349} or {1467} - {67} must be in r12c5
32c. deduction no 1,8 in r12c5

33. 1 now locked in c4 of n2

34. 8 locked in r2 of n2
34a. 9(2)n12=[18]/[63]

35. hidden single 1 at r1c4 for c4
35a. from 32b. r1c6=4, r12c5={67} locked for c5 and n2
35b. r9c4=9

36. cleanup from 35a/b.
36a. 8(2)n12=[62]
36b. r3c5=9
36c. r8c5=8
36d. 9(2)n12=[18]
36e. 12(2)n23=[39]
36f. 16(2)n78=[97]
36g. (from 17b.) r8c8=9

37. 33(6)n47 - no 9, - so combo is {345678} -> r6c2=3

... and the rest falls out very quickly

Richard


Last edited by rcbroughton on Sat Apr 14, 2007 12:39 pm; edited 1 time in total
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Para
Yokozuna
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Joined: 08 Nov 2006
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PostPosted: Tue Mar 27, 2007 12:47 pm    Post subject: Reply with quote

Hi

This is basically how i solved it from your marks pic on. I thought your route was a bit long, so thought i'd give my view on it.

1. 13(3) in R7C9 can't be {139}: clashes with R8C7 : no 9.

2. 45 on C9: 3 innies = R456C9 = 21 = {489/579} : {678} clashes with R8C9: no 6.
2a. R6C9 = 8

3. 45 on N9: 4 outies = R6C89 + R78C6 = 12 = 8{112}: R6C8 = 1; R7C6 = 2; R8C6 = 1
3a. R7C7 =8; R8C7 = 3

4. 33(6) = {245679/345678}: can't be {126789/1356789/1456789}: no 1; can't be {234789/235689}: can only have one of {23}
4a. Can't have both {45} in R789C2+R9C3 because of R8C3: R6C1 = {45}
4b. R6C1 = R8C3

5. 45 on N7: 4 outies = R6C12 + R78C4 = 19
5a. R6C1 + R8C4 = 9(step 4b); R6C2 + R7C4 = 10 = [37]
5b. R7C3 = 9

6. 26(4) = {3689} locked for N8: can't be {4589}, clashes with R8C4.
6a. Hidden singles: R7C1 = 3; R7C9 = 1
6b. R9C1 = 1
6c. R89C9 = 12 = [75]

7. Naked Pair: R45C9 = {49}: locked for C9 and N6
7a. 11(3) in R1C9 = {236}: locked for N3.
7b. 6(2) in R3C6 = [51]
7c. 8(2) in R3C3 = {62}: locked for R3
7d. R3C9 = 3

8. 36(6) in R1C7 = {345789}: can't be {246789}, only room for {236} in one cell.
8a. R4C8 = 3(only place in 36(6))
8b. {578} locked in 36(6) in N3
8c. R2C6 = {38}

9. 45 on c6789: 2 innies: R19C6 = 10: R1C6 = {47}; R9C6 = {36}
9a. 9 locked in C6 for N5.
9b. Hidden: R9C4 = 9; R8C8 = 9; R6C6 = 9
9c. 45 on C1234: 1 innie: R1C4 = 1

10. 8 locked in N8 for C5
10a. 18(4) in R1C4 = 1{467}, locked for N2 : 1{269} not possible: needs 4 or 7 in R1C6
10b. R4C34 = [62]; R4C5 = 9; R1C3 = 3(hidden)
10c. 9(2) in R2C3 = [18]; R2C67 = [39]
10d. R9C6 = 6; R1C6 = 4(step 9); R89C5 = [83]; R8C2 = 6; R7C8 = 6(hidden)

11. R4C9 = 9(only place for 9 in 36(6) in R1C7; R4C5 = 4
11a. 32(6) in R1C2 needs a 9: R1C2 = 9(only option); R5C1 = 9(hidden)
11b. 20(3) in R1C1 = {578}: locked for N1 and C1

Now just basics to the end.

greetings

Para


Last edited by Para on Sat Aug 11, 2007 12:17 pm; edited 1 time in total
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rcbroughton
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PostPosted: Wed Mar 28, 2007 7:32 am    Post subject: Reply with quote

Para wrote:
... so thought i'd give my view on it.

Very nicely done Para - a very compact walkthrough. Some similar steps but cutting out the unnecessary bits.

Good job!

Richard
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PsyMar
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PostPosted: Thu Apr 12, 2007 12:57 am    Post subject: Reply with quote

Here's my walkthrough. Move 14 is somewhere between conflicting combinations and trial-and-error; I'm not sure if it's kosher. Aside from that I didn't think this was that hard.

The entire thing's about 50-60 steps, maybe more depending on how you count them. I'd say well under 100 though.

0a. 4/2 in r3 = {13} naked pair -> elim from rest of r8
0b. 6/2 in r3 = {15|24} -> no 3|6..9
0c. 8/2 in r3 = {17|26|35} -> no 4|8|9
0d. 9/2 in r2 = {18|27|36|45} -> no 9
0e. 12/2 in r2 = {39|48|57} -> no 1|2|6
0f. 16/2 in r7 = {79} naked pair -> elim from rest of r7
0g. 6/3 in n7 = {123} naked triple -> elim from rest of n7 and c1; r8c1 = 2
0h. 11/3 in n3 = {128|137|146|236|245} -> no 9
0i. 20/3 in n1 = {479|569|578} (no 3 due to 6/3 in c1) -> no 1|2|3 (but those already eliminated)
0j. 26/4 in n8 = {2789|3689|4589|4679|5678} -> no 1
0k. 9/2 in r8 = {45} naked pair -> elim from rest of r8 (not 18 or 36 as 1|3 eliminated by 4/2 in r8 and not 27 as 2 eliminated by r8c1)
0l. 10/2 in r7 = {28|46} (19 and 37 elimed by 16/2 in r7)
1. innies of c9 = r456c9 = 21/3 = {489|579|678} -> no 1|2|3
2. 33/6 in n47 cannot have more than one of 123 -> is {145689|245679|345678}
3. 33/6 in n47 must have one of 123 -> only possibilities in r6c2 -> r6c2 = {123}
4. Outies of n9 = r78c6+r6c89 = 12/(2+2); r6c89+r8c6 >= 6 -> r7c6 <= 6 -> 10/2 in r7 = [28|46|64]
5. Outies of n7 = r6c12+r78c4 = 19/(2+2); r6c2+r78c4 >= 12 -> r6c1 <= 7
6. 8 of n7 locked in 33/6 in n47 -> 33/6 in n47 = {145689|345678}
7. 33/6 in n47 must have both 4 and 5; cannot have both 4 and 5 in n7 due to r8c3; thus r6c1 = {45}
8. 4|5: r6c1 != r789c2+r9c3 != r8c3 != r8c4 -- thus r6c1 != r8c4, forming a naked pair {45} eliminating from r6c4 (and in theory r8c1) -- this move may not be useful but I like it
9. 26/4 in n8 = {3689|5678} -> no 2 or 4; elim 6 and 8 from rest of n8 (cannot be {2789} or {4679} due to r7c4 and cannot be {4589} due to r8c4)
10. combinations for 10/2 in r7 = [28|46]
11. 13/3 in n9 has one and only one of {6789}; this must be in r8c9; elim 6..9 from r79c9
12. combinations for 33/6 in n47 -> no 2's
13. 2 of n8 locked in r7 -> elim from rest of r7
14. combinations for 13/3 in n9: {157|256} -> elim 5 from rest of n9/c9 (cannot be 139 due to r8c7, cannot be 148 as it can see all but one square of 30/6 in n69 which must either have two of 148 or be 234579, which would then leave no possibilities for r8c7; cannot be 238 as then r8c7 = 1 and then there are again no possibilities for 30/6 in n69; cannot be 247 as this conflicts with 11/3 in c9; cannot be 346 due to more r8c7 and 30/6 in n69 hijinx.)
15. combinations for 11/3 in n3: {137|236} -> elim 3 from rest of n3/c9 (128 and 146 both conflict with 13/3 in c9)
16. r456c9 = {489} hidden triple -> elim from rest of n6
17. 9 of n9 locked in 30/6 -> elim from rest of 30/6, specifically r6c9
18. combinations for 31/5 in c67 -> {9...}; 9 locked in n5; elim from rest of n5 and c6
19. 20/3 in c1 forms killer pair {45} with r6c1; elim from r45c1
20. 15/3 in r5 != {456} (conflicts with r6c1)
21. innies of c1234 = r19c4 = 10/2 = [19|28|37|46|73]
22. 9 of c4 locked in n8 -> elim from rest of n8
23. innies of c6789 = r19c6 = 10/2 = [28|37|46|73]
24. 4 and 9 of n9 locked in 30/6 in n69; thus r6c9 = 8, and since 13459 cannot be the other 5 (5 forced to r6c8 by 13/3 in n9, but then either 1 or 3 forced to r6c8 by r8c7), so 30/6 in n69 = {124689}
25. to avoid conflict between 30/6 in n69 and 13/3 in n9, r6c8 = {12}; thus 6 of 30/6 locked in n9; elim from rest of n9; 13/3 in n9 = [175|571] -> r6c8 = 2 and 4/2 in r8 = [13] -> r6c2=3
26. r7c7 = hidden single 8 -> r7c6 = 2
27. outies of n7 = r78c4+r6c1 = 19 = r7c4+4+5 = 7+4+5 -> 16/2 in r7 = [97]
28. r9c4 = hidden single 9
29. r8c8 = hidden single 9
30. r6c6 = hidden single 9
31. combinations for 11/3 in n3 = {236} triple -> elim from rest of n3
32. combinations for 6/2 in r3 = [51]
33. combinations for 26/4 in n8 = {3689} -> elim 3 from r7c5
34. r7c1 = hidden single 3 -> r9c1 = 1 -> r79c9 = [15]
35. innies of r1234 = r1c4 = 1
36. 21/5 in c5 = {12459|12567|23457} (all other possibilities conflict with r89c5) -> elim 2 from rest of c5
37. combinations for 36/6 in n36 = {345789} -> r4c8 = 3
38. combinations for 17/3 in r5 = {269|467} -> elim 6 from rest of r5/n6
39. combinations for 15/3 in r5 = {159|258} (249 would conflict with r5c9) -> elim 5 from rest of n4/r5 -> r6c1 = 4
40. r7c2 = 5 (hidden single in 33/6) -> 9/2 in r8 = [45] -> r7c5 = 4 -> r7c8 = 6
41. r5c3 = hidden single 5
42. r5c7 = hidden single 6
43. 5 of n6 locked in c7 -> elim from rest of c7
44. combinations for 18/4 in n2 = {1467} (1368 would conflict with r89c5) -> r1c6 = 4
45. r12c5 = {67} naked pair -> elim from rest of n2/c5 -> 26/4 in n8 = [8936] -> r8c2 = 6
46. 6 of c4 locked in 22/5 of c34 -> elim from rest of cage
47. r4c1 = hidden single 6
48. r6c4 = hidden single 6
49. r3c5 = hidden single 9
50. innies of c1 = r5c1 = 9 -> naked singles and last-digit-in-cage moves solve it
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