Assassin July 7 walk-thru

[sudokuEd]

Am finally over the shock - ND doesn't use a combo chart!!! You've inspired me ND - so trying to relearn my bad habits. This weeks assassin forced me to use min/max to avoid the combo chart. Much better. Too many 'elbows' (Ruud is a tennis fan??) to use 'subtraction combo'.

I've included my walk-thru in tiny text below. (Copy/paste into a text doc). I had to do some combo hunting in Step 4 - any simpler way to have done that?

Thanks for a great puzzle Ruud. What about a cycling theme??

Assassin 7 July

Step 1.
“45” on N9->outies at r789c6=24-> {789} only -> no 789 elsewhere in N8 or c6
->20(3) cage in N2, r12c6 Maximum = 6+5=11 -> minimum r1c5 = 9 -> r1c5 = 9, r12c6 {56} only
-> no 56 elsewhere in N2 or c6
->8(3) in N2 {134} only ->not elsewhere in N2
->r123c4 {278} only -> not elsewhere in c4
->r6c45=[97]
also r5c5=8 (hidden single N5)

Step 2.
“45” on r12 -> 3 innies=10
->no 7 or 8 possible in r2c4 since r2c1=>4
->r2c4=2, r3c34=[68], r1c4=7

Step 3.
now 2 innies left from step 2 means that r2c15= 8 -> r2c5 {13}, r2c1 {57}
->4 in N2 only in r 3 -> no 4 elsewhere in r3
also 2 in r3 locked in 6(3) cage in N3 -> no 2 elsewhere in N3 or r4c9
also 9 in N1 only in r3 ->no 9 in r3c7
also 21(3) cage in N1 {579} only -> no 579 elsewhere in N1
->11(3) cage in N1 = {128} only -> no 128 elsewhere in N1
->r12c3 now 7(2) -> {34} only -> no 3 or 4 elsewhere in c3

Step 4.
“45” on N7 ->
a. innies r7c123 = h16(3)
b. outies r6c123 = h17(3)
c. r6c1 – r7c3 = -2 (or r7c3 – 2= r6c1) -> minimum r7c3 = 3, maximum r6c1 = 7
->r7c3 = {578} (no 9 since no 7 in r6c1); r6c1 = {356}
-> 19(3) cage at r6c23r7c3 = [487/658/685] ->r6c2 = {46}, r6c3={58}, r7c3={578}
->h17(3) cage in N4 = [368/548] only
-> r6c3 = 8

Step 5.
8 in N6 now locked in r4c78 in 21(3) cage-> {678} only
->r3c7=7, r2c1=7, r4c9 = 3 (outie on N3)
Also r4c78 {68} only -> no 6 or 8 elsewhere in N6 or r4
->r4c56 = [52], r5c4 = 6 (hidden single N5), r6c2 = 6 (hidden single N4), r7c3= 5, r6c1 =3, r5c6 = 3

Step 6.
r3c89 ={12} -> no 1 or2 elsewhere in N3 or r3
->r3c56 = [34], r2c5 and r6c6 = 1, r4c4 = 4, r2c2 = 8

Step 7.
r4c123 = triple on {179} -> no 179 elsewhere in N4
->r5c3=2, r4c3 = 9, r5c2 = 5, r45c1=[14], r4c2=7, r3c12=[59], r1c12=[21]

Step 8.
r7c12 now 11(2) -> [92]
r89c3 = {17} = 8 -> r89c2 = [34]

Step 9.
In N8, 5 is now only in 10(3) cage -> {235} only
->r89c4=[53], r9c5=2, r7c4=1

Step 10.
"45" on r89 -> innies = h16(3) at r8c569
Minimum r8c56 sum=11 -> max. R8c9 =5
->h16(3) can only be [682/691]
->r8c5=6, r7c5=4, r89c1=[86]

Step 11
"45" on N9 ->r7c7 - r9c6= -5.
->r7c7=3, r9c6=8
.....etc

[nd]

Thanks for the walkthrough--just did this one yesterday: a fairly tricky one! No combination charts needed, but it does take some finesse!

[nd]

Here's my walkthrough.

1. R3C89 + R4C9 = {123}. 45 on N8 -> R789C6 = {789}, R7C7 = {234}. R1C5 = 9, R12C6 = {56}, R2C5 + R3C56 = {134}, R123C4 = {278}, R6C45 = [97], R5C5 = 8.

2. Naked quad in R3 blocks {1234} from R4C1234; 4 in N2 is locked in R3C56, 2 in N3 is locked in R3C89. The 16(3) cage in N2 cannot have {78} in it (since this would make R3C3 = 1), so the only solution is R123 = [728], R3C3 = 6, R3C127 = {579}.

3. 45 on N3 -> R3C7 = R4C9 + 4 = {57}, so R4C9 = {13}, 9 in N1 is locked in R3C12, and R2C1 = {57}. R12C3 = {34}, R1C12 + R2C2 = {128} (with the 2 locked in R1).

4. 45 on R6 -> R6C6789 = 12 = {12(36|45)}. 45 on N7 -> R6C1 = {356}, R7C3 = {578}. But 8 is now locked in R6 in R6C23, so R6C1 = {35}, R7C3 = {57}. The 19(3) cage must be {478} or {568}, so R6C2 = {46}, R6C3 = 8, R4C78 = {68}, R3C7 = R2C1 = 7, R3C12 = {59}.

5. 45 rule on R12 -> R2C5 = 1 -> R3C89 = {12}, R4C9 = 3, R4C56 = [52]. R4C4 = {14} and we have a naked triplet in N5 so R5C4 = 6, R6C12 = [36], R7C3 = 5, R5C6 = 3, R3C56 = [34], R6C6 = 1, R4C4 = 4.

The puzzle from this point on is easy sailing.

[sudokuEd]

Nicely finessed ND - worthy of the Agassi of the Killer world!
Using the naked quad in 3. and the h12(4) in r6 in 4. made a big difference compared to me.

Just a couple of quesions -
step 1 - you say the 4 is locked in r3c56 - how?

I can see the 4 locked in r3c56 in step 2 with the naked quad - but did you lock it in a different way in step 1?

Step 4 - you say "45 on N7 -> R6C1 = {35}, R7C3 = {57}". How did you eliminate 6 from r6c1 and 8 from r7c3 using this rule?

Again, I can see how those can be eliminated after the next part with the 12(4) in r6 which means 8 in r6 must be in r6c23.

Thanks

[nd]

Yeah there were a couple of mixups there resulting from my cutting & pasting (& thus getting steps out of order). See if the revision is clearer now.

Nope, I'm definitely not Killer Sudoku's Agassi: udosuk wins hands down. He does them about 5 times faster than me & he often solves things much more elegantly than I do. Shai & jcbonsai are pretty amazing solvers too.

[sudokuEd]

but they don't have Agassi's humility and grace

I'll stop embarrassing you any more. The changes to the walk-thru look great.

I see you've already put up a walkthrough for this week's assassin - is it that time of the week already??