## KILLER No 74 - 12/12/05

For fans of Killer Sudoku, Samurai Sudoku and other variants

### KILLER No 74 - 12/12/05

Is it just me, or was this a lot harder than posted? It got done, but more like 45 minutes than 12.
Bigtone53

Posts: 413
Joined: 19 September 2005

### Agreed

Same here, it took about 45 minutes.

It was also interesting because it's the first killer puzzle where the rule "within each dotted-line shape, a digit cannot be repeated" was actually necessary to solve the puzzle rather than merely helpful.
carlosr

Posts: 7
Joined: 12 December 2005

It took me a lot longer than 12 minutes too.
afjt

Posts: 82
Joined: 07 September 2005

### Re: Agreed

carlosr wrote:It was also interesting because it's the first killer puzzle where the rule "within each dotted-line shape, a digit cannot be repeated" was actually necessary to solve the puzzle rather than merely helpful.

That rule is always neccesary if you are working on the principle that an area of 2 cells adding up to a 4 muist be 1&3

And the higher inverse

A tactic which i use in almost every puzzle
Pi

Posts: 389
Joined: 27 May 2005

### Re: Agreed

Pi wrote:That rule is always neccesary if you are working on the principle that an area of 2 cells adding up to a 4 muist be 1&3

No it isn't. Any two cell box must lie in a single row or column, so it can't contain two 2's anyway.
carlosr

Posts: 7
Joined: 12 December 2005

I've just replied to a post about this puzzle on DJApe's website (www.djape.net) - took me about half an hour. Definitely harder than 'Moderate' though I'm not complaining!
CathyW

Posts: 316
Joined: 20 June 2005

I'm sorry about my innacuracy, i was wrong.

The rule does still apply if you are working on the principle that an area with three cells with a value of 6 must by 1,2,3 or the high inverse.

sorry
Pi

Posts: 389
Joined: 27 May 2005

Pi wrote:The rule does still apply if you are working on the principle that an area with three cells with a value of 6 must by 1,2,3 or the high inverse.

Not necessarily. If the three cells are in a straight line, or all lie within a single nonet, then the sum must be 1+2+3. But if the three cells form a corner shape covering two or more nonets then the sum could be 1+4+1. Unless you apply the "no repeats" rule.

It so happens that this is the first puzzle that needs the "no repeats" rule.
carlosr

Posts: 7
Joined: 12 December 2005

Have you got proof of that, by keeping a record or something?

Anyway that puzzle was hard. It took me 20 mins when the time in the paper was 12
Pi

Posts: 389
Joined: 27 May 2005

Pi wrote:Have you got proof of that, by keeping a record or something?

I was interested because The Times kept changing the rule back and forth for the first few weeks. "A digit CANNOT be repeated" one day and "a digit CAN be repeated" the next.

When I thought I needed the rule I fed the puzzle into my sudoku-solving computer program, and it could do it without using the rule. Until puzzle 74, when the rule is obviously required.

So I'm 99% sure. But I wouldn't swear I haven't made a mistake.
carlosr

Posts: 7
Joined: 12 December 2005

well i appologise for questioning your statement
Pi

Posts: 389
Joined: 27 May 2005

Well, without just guessing (which you don't have to do with Times Sudoku) I couldn't even get the thing started.
mtr

Posts: 6
Joined: 23 September 2005

mtr - perhaps you haven't tried Killer Sudokus before?

Starting tips - you can enter candidates in the 2-cell cages that have only one pair as an option. Thus: Row 1, cage of 3 must have 1 and 2 as candidates, likewise row 7, cage of 16 = 7+9, row 9 cage of 4 = 1+3 though you don't know which way round they go at this stage.

After that, look at Box (nonet) 1. 13+9+9=31. Thus the remaining 3 cells must add to 14. Thus in box 2, the balance of the 8-cell 44 cage must add to 30. 30+3+13 = 46, therefore r4c6 = 1.

You can do the same for box 3 and 6 to calculate r6c6; box 9 and 8 to calculate r6c4; and box 7 and 4 to calculate r4c4. These 4 digits in the corners of box 5 should add up to 14!

Hope that helps.
CathyW

Posts: 316
Joined: 20 June 2005

Tried them, done them, never had a problem with them. Yesterday's took me about 10 minutes. Normally, there are nonets with either a single cell that's not part of any groups, so you can just add the groups up and subtract from 45, or one cell overlapping the next nonet, allowing you to fill that in on similar principles. Then you've got the two-cell 3s, 4s, 16s and 17s, and so on, where you can eliminate the candidates along the rest of the row or column. But number 74, frankly, especially with the 12 minute estimate, I assumed must be some kind of mistake, so I didn't really bother.
mtr

Posts: 6
Joined: 23 September 2005

It was tough, but in a good way. I liked this one. I did 78 straight after, and that was a stinker...
Lardarse

Posts: 106
Joined: 01 July 2005