Walkthru for Assassin June 23

[sudokuEd]

Trying to learn to write non-cryptic walkthru's. How did I go? Any feedback welcomed.

This puzzle seemed a lot easier than last week.

The walkthru is in tiny text - copy and paste to a text doc.

Assassin Killer 23 June 06 Walkthru by SudokuEd

Step 1.
24(3) cage in N4 is {789} only -> no 789 elsewhere in N4.

Every combination for 31(6) in N1 uses at least two out of 7, 8 or 9. Since no 7, 8 or 9 in r4c3 -> 7, 8 or 9 used in 31(6) must be in N1. Also, since 12(2) in N1 must use a 7, 8 or 9 -> no 7, 8 or 9 can be in r2c3 or r3c1.

Step 2.
“45” on c1 -> innies = (hidden) h28(4) -> no 1, 2 or 3 in those 4 cells ->1, 2 and 3 locked in 11(3) or 6(2) in c1.
Since 6(2) uses 1 or 2 -> 11(3) must have two out of 1, 2 or 3 -> only combination in 11(3) is {236}
->6(2) in c1 is {15}
->r6c1= 4,

Step 3a.
17(2) cage in r9 is {89} only -> no 8 or 9 elsewhere in r9 or N9.
Since the 29(6) cage in N9 must use 8 or 9 -> r6c7 can only be {89}

Can also now see that r9c1= 7
->{89} in r12c1
->{57} in 12(2) in N1

Step 3b.
3(2) cage in r6 is {12} only -> no 1 or 2 elsewhere in r6. Also, now have an 8(2) cage in r67c2 ->no 3 in r6c2 since no 5 in r7c2.

In N7, 31(6) + 8(2) + 6(2) + r9c3 – 2 outies = 45 ie. 45 + r9c3 – 2 outies = 45
-> r6c2 + r7c4 = r9c3.
Since maximum digit in r9c3 = 6 and since minimum digit in r6c2 = 5 -> r7c4 =1, r6c2=5, r9c3=6.
->r7c2=3, r6c34=[12], r78c1=[51], r1c23=[75]

->{89} naked pair in r45c2 -> r5c3=7
->{24} naked pair in r89c2
->{16} naked pair in r23c2

Step 4.
In N1, 31(6) + 12(2) – 1 outie + 2 innies = 45 ie. 43 – r4c3 + 2 innies = 45
-> r3c1 + r2c3 – r4c3 = 2
Since r4c3={23} -> 2 innies = 4 or 5. But only option is {23} which = 5 -> [3] in r4c3.
-> r2c3 [2], r3c1 [3], r3c3 [4]

Step 5.
“45” on r9 -> innies = h7(3) ie {124} only in r9c269
->{24} naked pair in r9c26 -> r9c9=1

Step 6.
“45” on N8 -> 2 outies = 6 -> r6c6 =3, r8c7 = 3, r7c6 = 8, r78c3 = [98], r89c6 = [64], r89c2 = [42]

Step 7.
“45” on N5 -> 2 innies = 15 -> r4c46 = [87], r3c4 = 5, r4c7 = [2], r45c2 = [98], etc, all singles

[nd]

Haven't tried out the solving path yet but you will want to renumber the steps -- two step 3s!