Assassin 74
Posted: Fri Oct 26, 2007 1:52 am
Hi guys! I'm a newbie, so please be merciful. Please forgive any notation errors (turns out it's harder to explain one's thinking than to just solve these).
Editted thrice for corrections and clarifications (thanks to Mike and Andrew).
Thank you for your patience with the "New Kid." My apologies to everyone for using the wrong brackets throughout. I was a math major in a former life, and my brain has yet to adjust. I also apologize for mixed case in labeling nonets, rows and columns. I typed everything lower case, but my computer insists on capitalizing the first letter of every sentence.
Prelims:
1) 6(3) n1 = [1,2,3] locked for n1.
2) 20(3) n12: r1c4 => no 1,2
3) 24(3) n7 = [7,8,9] locked for n7
4) 7(3) n7 = [1,2,4] locked for n7
a) R7c123 = [3,5,6] locked for n7, r7.
5) Both 11(3) n2 no 9
6) 19(3) n23 no 1
7) 20(3) n8 no 1,2
8) 21(3) n3 no 1,2,3
9) 22(3) n6 no 1,2,3,4. 9 is locked into 22(3) cage for n6.
10) 7(3) n6 = [1,2,4] locked for n6
11) 20(3) n69 no 1,2
Solving:
Draw in Hidden cages with 45 and Outies
12) R123c3 is 21(3)
13) R123c4 is 12(3)
14) R123c7 is 8(3)
a) R123c7 no 6,7,8,9 and 1 is locked to 8(3) cage and r12c7 for n3, c7.
15) R123c6 is 22(3)
a) R1c6 no 1
b) R123c6 no 2,3,4
c) 9 is locked r23c6 for n2, 19(3) and c6.
i) 19(3) n23 no 5.
16) R6c123 is 12(3)
17) R7c123 is 14(3)
18) R6c456 is 17(3)
19) R7c456 is 13(3)
20) R6c789 is 16(3)
21) R7c789 is 18(3)
Using Venn Diagram logic and algebra, eliminate some choices with “paired” up hidden cages:
(Note: Venn Diagrams are the math major in me leaking out again. Think of the MasterCard symbol. One circle contains all left-handed persons. The other circle contains all blue-eyed people. The overlap area is left-handed blue-eyed people. The area on the card outside the circles is everybody else. I'm told that normal people call this a variant of innies and outies. Para can attest that my brain is having a hard time with that concept.)
22) R7c3 – 1 = r6c1 = (2,4,5)
a) If r6c1 is 5, conflict with n47 13(3), so r6c1 <> 5 and r7c3 <> 6.
b) R6c23 = (1,2,4,6,7,8,9)
23) R7c6 +2 = r6c4 = (3,4,6,9); r7c6 has no 8.
24) R6c7 + 4 = r7c9 = (7,9); r 6c7 is (3,5)
25) R3c7 + 3 = r1c6 = (5,6,7)
Innies & Outies:
26) Innies c89: r579c8 hidden 11(3)
a) No 9 in r579c8
b) See step 9), 9 is now locked to r45c7 for 22(3), n6 and c7.
27) Outies c6789: r468c5 hidden 20(3)
a) R468c5 no 1,2
28) Outies c1234: r579c5 hidden 14(3)
29) Innies c12: r468c2 hidden 13(3)
Picking off choices:
30) N6 hidden 16(3) r6c789 has locked 3, since n/e in r6.
a) No 3 in r6c456 => r7c6 no 1
31) N9 12(3) r89c7 min 5 => r9c8 max 7, so no 8.
32) Step 14a) locks 1 to n23 11(3) => no 2,5 in 11(3)
33) Hidden 8(3) from step 14a) => can’t have 2; r123c7 = (1,3,4) => Naked Triple (1,3,4) for hidden 8(3), n3 and c7.
a) Naked Single 5 at r6c7
i) Cleanup 5s from r7, c7 and n6.
ii) N69 14(3) r7c8 no 4,8.
34) N6 hidden 16(3): since r6c7 = 5, r6c89 = {3,8} locked to N6 hidden 16(3), n69 20(3) and r6.
a) Cleanup: n69 20(3): r7c9 = 9.
i) Cleanup 9s from r7, n9,c9.
35) Step 15c) has 9 locked in n2 19(3). Since r3c7 = (3,4), r23c6 <> 8.
a) N2 hidden 22(3) r123c6 is now a locked triple (6,7,9).
i) Cleanup 6s and 7s at n2, c6:
ii) R6c4 <> 9.
36) N58 15(3) at r6c56 + r7c6 must be r6c5 = 9 and r67c6 = {2,4} NP locked for N4.
a) NP {2,4} at r6c1 and r6c6 locked for R6.
b) Cleanup 9s r6, c5, n5:
c) Cleanup 2s and 4s r6, c6: NS r6c4 = 6.
d) Cleanup 6s r6, c4, n5.
e) N47 13(3) has r6c23 = {1,7} NP locked for N4, thus r7c3 = 5.
f) Cleanup 1s,7s n4, 5s r7, c3: r6c1 = 4 => r6c6 =2 => r7c6 = 4.
37) N58 15(3) at r67c4 + r7c5 has locked 2s in r7c45 => r7c45 = (2,7).
a) Cleanup of 2s and 7s => r7c7 = 8 and r7c8 = 1.
38) N9 12(3) r89c7 = (2,6,7) => r9c8 = (3,4)
39) N58 split hidden 20(3) r6c5 = 9 => r4c5 = (3,5,8) and r8c5 = (3,6,8)
40) N8 12(3) has locked 1 in r89c6 since n/e.
a) Cleanup c5: no 1 in r45c6 => NT at n5 16(3) = (3,5,8)
b) Cleanup 3s,5s,8s from N5: NT at n5 12(3) = (1,4,7)
41) N2 r3c45 locked 1 since n/e in r3.
42) N3 16(3) has locked 2 => no 7 in 16(3)
43) N1 hidden 21(3) must have an 8, so 8 locked in n1 r123c3 and c3.
44) Cannot be a 3 in r4c2.
45) R1c4 + 1 = r3c3 = (4,6,9) and r1c4 = (3,5,8)
46) N1 hidden 21(3) now has 8 locked in r12c3 => must be an 8 in n12 20(3)=> no 8 in r1c4 and no 4 or 6 in r12c3.
47) Repeat of step 45 => now no 9 in r3c3.
48) Check combinations in n12 13(3) => no 8 in r3c4.
49) N2 11(3) r12c5 min 5 => max 6 for r3c5 => no 8.
50) 8 now locked in n3 16(3) in r3c89 since n/e in row.
51) N3 check combos in 21(3) => no 6.=> r1c8 = 9 and r12c9 = (5,7) locked.
52) N1 6(3) has 1 locked in r12c1 since n/e in col.
53) N9 HS r8c8 = 5.
54) N2 HS r1c6 = 6.
a) Thus n23 11(3) r12c7 = (1,4) and r3c7 = 3.
55) N4 17(3) has 3 locked in r45c3 since n/e in col. => no 2 in 17(3).
56) N7 7(3) has 2 locked in r89c3 since n/e in col.
57) Check combos on n47 split hidden 13(3) r468c2 => no 6 or 9 in r4c2.
58) Naked Triple r4c256 = (3,5,8).
59) Cleanup results in HS at r5c3 = 3.
60) N9 12(3) has 7 locked in r89c7 since n/e in n9.
a) R89c7 = (2,7) and r9c8 = 3
b) Cleanup n6 22(3) r45c6 = (6,9) and r5c8 = 7.
61) Cleaning up now getting wild:
62) R4c4 = 7 Hidden Single.
63) NP (6,9) at r4c37 => r4c1 = 2
The rest just seems to fall out as hidden singles and naked singles with iterative cleanups.
Susan
Editted thrice for corrections and clarifications (thanks to Mike and Andrew).
Thank you for your patience with the "New Kid." My apologies to everyone for using the wrong brackets throughout. I was a math major in a former life, and my brain has yet to adjust. I also apologize for mixed case in labeling nonets, rows and columns. I typed everything lower case, but my computer insists on capitalizing the first letter of every sentence.
Prelims:
1) 6(3) n1 = [1,2,3] locked for n1.
2) 20(3) n12: r1c4 => no 1,2
3) 24(3) n7 = [7,8,9] locked for n7
4) 7(3) n7 = [1,2,4] locked for n7
a) R7c123 = [3,5,6] locked for n7, r7.
5) Both 11(3) n2 no 9
6) 19(3) n23 no 1
7) 20(3) n8 no 1,2
8) 21(3) n3 no 1,2,3
9) 22(3) n6 no 1,2,3,4. 9 is locked into 22(3) cage for n6.
10) 7(3) n6 = [1,2,4] locked for n6
11) 20(3) n69 no 1,2
Solving:
Draw in Hidden cages with 45 and Outies
12) R123c3 is 21(3)
13) R123c4 is 12(3)
14) R123c7 is 8(3)
a) R123c7 no 6,7,8,9 and 1 is locked to 8(3) cage and r12c7 for n3, c7.
15) R123c6 is 22(3)
a) R1c6 no 1
b) R123c6 no 2,3,4
c) 9 is locked r23c6 for n2, 19(3) and c6.
i) 19(3) n23 no 5.
16) R6c123 is 12(3)
17) R7c123 is 14(3)
18) R6c456 is 17(3)
19) R7c456 is 13(3)
20) R6c789 is 16(3)
21) R7c789 is 18(3)
Using Venn Diagram logic and algebra, eliminate some choices with “paired” up hidden cages:
(Note: Venn Diagrams are the math major in me leaking out again. Think of the MasterCard symbol. One circle contains all left-handed persons. The other circle contains all blue-eyed people. The overlap area is left-handed blue-eyed people. The area on the card outside the circles is everybody else. I'm told that normal people call this a variant of innies and outies. Para can attest that my brain is having a hard time with that concept.)
22) R7c3 – 1 = r6c1 = (2,4,5)
a) If r6c1 is 5, conflict with n47 13(3), so r6c1 <> 5 and r7c3 <> 6.
b) R6c23 = (1,2,4,6,7,8,9)
23) R7c6 +2 = r6c4 = (3,4,6,9); r7c6 has no 8.
24) R6c7 + 4 = r7c9 = (7,9); r 6c7 is (3,5)
25) R3c7 + 3 = r1c6 = (5,6,7)
Innies & Outies:
26) Innies c89: r579c8 hidden 11(3)
a) No 9 in r579c8
b) See step 9), 9 is now locked to r45c7 for 22(3), n6 and c7.
27) Outies c6789: r468c5 hidden 20(3)
a) R468c5 no 1,2
28) Outies c1234: r579c5 hidden 14(3)
29) Innies c12: r468c2 hidden 13(3)
Picking off choices:
30) N6 hidden 16(3) r6c789 has locked 3, since n/e in r6.
a) No 3 in r6c456 => r7c6 no 1
31) N9 12(3) r89c7 min 5 => r9c8 max 7, so no 8.
32) Step 14a) locks 1 to n23 11(3) => no 2,5 in 11(3)
33) Hidden 8(3) from step 14a) => can’t have 2; r123c7 = (1,3,4) => Naked Triple (1,3,4) for hidden 8(3), n3 and c7.
a) Naked Single 5 at r6c7
i) Cleanup 5s from r7, c7 and n6.
ii) N69 14(3) r7c8 no 4,8.
34) N6 hidden 16(3): since r6c7 = 5, r6c89 = {3,8} locked to N6 hidden 16(3), n69 20(3) and r6.
a) Cleanup: n69 20(3): r7c9 = 9.
i) Cleanup 9s from r7, n9,c9.
35) Step 15c) has 9 locked in n2 19(3). Since r3c7 = (3,4), r23c6 <> 8.
a) N2 hidden 22(3) r123c6 is now a locked triple (6,7,9).
i) Cleanup 6s and 7s at n2, c6:
ii) R6c4 <> 9.
36) N58 15(3) at r6c56 + r7c6 must be r6c5 = 9 and r67c6 = {2,4} NP locked for N4.
a) NP {2,4} at r6c1 and r6c6 locked for R6.
b) Cleanup 9s r6, c5, n5:
c) Cleanup 2s and 4s r6, c6: NS r6c4 = 6.
d) Cleanup 6s r6, c4, n5.
e) N47 13(3) has r6c23 = {1,7} NP locked for N4, thus r7c3 = 5.
f) Cleanup 1s,7s n4, 5s r7, c3: r6c1 = 4 => r6c6 =2 => r7c6 = 4.
37) N58 15(3) at r67c4 + r7c5 has locked 2s in r7c45 => r7c45 = (2,7).
a) Cleanup of 2s and 7s => r7c7 = 8 and r7c8 = 1.
38) N9 12(3) r89c7 = (2,6,7) => r9c8 = (3,4)
39) N58 split hidden 20(3) r6c5 = 9 => r4c5 = (3,5,8) and r8c5 = (3,6,8)
40) N8 12(3) has locked 1 in r89c6 since n/e.
a) Cleanup c5: no 1 in r45c6 => NT at n5 16(3) = (3,5,8)
b) Cleanup 3s,5s,8s from N5: NT at n5 12(3) = (1,4,7)
41) N2 r3c45 locked 1 since n/e in r3.
42) N3 16(3) has locked 2 => no 7 in 16(3)
43) N1 hidden 21(3) must have an 8, so 8 locked in n1 r123c3 and c3.
44) Cannot be a 3 in r4c2.
45) R1c4 + 1 = r3c3 = (4,6,9) and r1c4 = (3,5,8)
46) N1 hidden 21(3) now has 8 locked in r12c3 => must be an 8 in n12 20(3)=> no 8 in r1c4 and no 4 or 6 in r12c3.
47) Repeat of step 45 => now no 9 in r3c3.
48) Check combinations in n12 13(3) => no 8 in r3c4.
49) N2 11(3) r12c5 min 5 => max 6 for r3c5 => no 8.
50) 8 now locked in n3 16(3) in r3c89 since n/e in row.
51) N3 check combos in 21(3) => no 6.=> r1c8 = 9 and r12c9 = (5,7) locked.
52) N1 6(3) has 1 locked in r12c1 since n/e in col.
53) N9 HS r8c8 = 5.
54) N2 HS r1c6 = 6.
a) Thus n23 11(3) r12c7 = (1,4) and r3c7 = 3.
55) N4 17(3) has 3 locked in r45c3 since n/e in col. => no 2 in 17(3).
56) N7 7(3) has 2 locked in r89c3 since n/e in col.
57) Check combos on n47 split hidden 13(3) r468c2 => no 6 or 9 in r4c2.
58) Naked Triple r4c256 = (3,5,8).
59) Cleanup results in HS at r5c3 = 3.
60) N9 12(3) has 7 locked in r89c7 since n/e in n9.
a) R89c7 = (2,7) and r9c8 = 3
b) Cleanup n6 22(3) r45c6 = (6,9) and r5c8 = 7.
61) Cleaning up now getting wild:
62) R4c4 = 7 Hidden Single.
63) NP (6,9) at r4c37 => r4c1 = 2
The rest just seems to fall out as hidden singles and naked singles with iterative cleanups.
Susan