The New Sudoku Players' Forum

[scuba_chris]

Hi,
>
> I'm having a go at my second "killer". I managed to complete
> the one in yesterday's Times, however the one published today
> appears to be illogical ( although I accept it's more likely
> to be me who is missing something.)
>
> Let me explain the problem : -
>
> If I notate the 9 * 9 grid, the numbers and letters, giving
> the rows numbers 1 - 9, and the columns letters A - I, then
> each little square can be addressed uniquely ( e.g. A1 is
> the top LH little square, and I9 is bottom RH one).
>
> Now, the sum of the values in each 3 * 3 box should be 45 (
> the sum of digits 1 - 9).
> As G1+ G2 + H3 = 15, and H1 + I1 + I2 = 13.
> then it follows that G3 + H3 + I3 = 45 - (13 + 15) = 17.
>
> Now for 3 unique digits to sum to 17, only the digits 4, 5
> and 8 can be used , so G3 = 4,5 or 8 ( as is H3 and I3).
>
> We also now that I3 + I4 + I5 = 18.
> For 3 unique digits to sum to 18, only the digits 4, 5 and 9
> can be used , so I3 = 4,5 or 9 ( as is I4 and I5).
>
> But we already know that I3 can only be 4,5 or 8, so for both
> these conditions to be satisfied, I3 must be 4 or 5 (only)
> and I4 and I5 contain
> 4,5 or 9.
>
> As G3 + G4 = 9, and G3 is 4,5 or 8, then G4 must be 4,5 or 1.
> As H3 + H4 = 9, and H3 is 4,5 or 8, then H4 must be 4,5 or 1.
>
> What we are seeking to show now, is that the square G5 must
> have the value 8. We can conclude this, because none of the
> other squares (G4, H4, I4,H5, I5, G6, H6 or I6) can have the
> value 8. ( So far we have shown that G4 and
> H4 have values 4,5 or 1 , and also that I4 and I5 have the
> value 4,5 or 9).
> We can also see that as H5 + H6 = 8, neither H5 or H6 can
> have the value 8.
> This leaves squares G5, G6 or I6 as each possibly having the values 8.
>
> We will now show that G6 and I6 cannot contain 8, which
> leaves only G5 ( so that must be 8).
> Neither G6 or I6 can have the value 8, as D6 has this value.
>
> We can see that D6 = 8 because A4 + B4 + C4 + A5 + B5 + C5 +
> A6 + B6 + C6 = 45; however A4 + B4 + C4 + A5 + B5 + C5 + A6 =
> 36. Therefore B6 + C6 = 45
> - 36 = 9.
> Now B6 + C6 + D6 = 17, so D6 = 17 -9 = 8.
>
> So we can now see that G5 must = 8 (above).
> As F5 + G5 = 12, F5 = 4.
>
> Now F6 + G6 = 6, (which means that F6 and G6 must have a
> value 5,4,2 or 1).
> As we already know F5 = 4, this means F6 can only be 5,2 or 1
> - and consequently G6 = 5,4 or 1.
>
> This leaves three squares within a 3 * 3 grid with 5,4 or 1
> as the only possible values (G6, G4 and H4). As each of these
> square must contain one of these values, none of the other
> squares within this 3 * 3 grid can contain the values.
>
> Before we saw that I4 and I5 both had possible values 4,5 or
> 9. ( So if we discount the values 4 or 5), this means that
> I4 and I5 can only contain 9, which is not permissible.
>
> Can anyone tell me what I've done wrong please ?

Kind regards

Chris

[possum]

You state that 'for 3 unique digits to sum to 17, only the digits 4, 5
and 8 can be used... '

1, 7 and 9
2, 6 and 9
2, 7 and 8
3, 5 and 9
3, 6 and 8
4, 5 and 8
4, 6 and 7

These are all the possible options for 3 digits to equal 17.

[kevkemp]

Hi
As just stated above, 3 unique digits to sum to 17 could be 4 5 8, 4 6 7, 3 5 9, 3 6 8, 2 6 9, 2 7 8 or 1 7 9

Also, 3 unique digits to sum to 18 could be 4 5 9, 3 6 9, 2 7 9, 1 8 9, 3 7 8, 4 6 8 or 5 6 7

The rest of the logic is based on these two flawed assumptions!

I found the middle "box" was the only place I could get started initially - try there.

[possum]

Similarly, when you say that 'For 3 unique digits to sum to 18, only the digits 4, 5 and 9 can be used ...'

there are several combinations of 3 digits that will produce 18:

1, 8 and 9
2, 7 and 9
3, 7 and 8
3, 6 and 9
4, 5 and 9
4, 6 and 8
5, 6 and 7

[possum]

I'm still stuck with this killer myself. So far I've only got a definite 8 in the middle box.

[scuba_chris]

Thanks. for that. I don't know what I was thinking about.

[jf27]

Chris, see http://www7a.biglobe.ne.jp/~sumnumberplace for very useful lists of possible number combinations

[dalek]

This was probably the hardest killer for weeks. It took me about half an hour to do. Nice to see that the standard is improving (or maybe I am just having an off day....)

[Spa]

With an 8 in the middle box, and a little calculation using another box you should be able to identify the two possibles numbers to go into r4c4 and r5c4. Which should in turn help with the possibles in r4c5 and r4c6.

[Karyobin]

Well, this is as far as I've got:

Code: Select all

. . . | . . . | . . .
. . . | . . . | . . .
. . . | . . . | . . .
------+-------+------
. . . | 9 . . | . . .
. . . | 5 . 3 | 9 . .
9 . . | 8 . . | . . .
------+-------+------
. . . | . . . | . . .
. . . | . . . | . . .
. . . | . . . | . . .

(Sorry for the mess, first time I've tried to use the 'Code' command.)

It seems to me that, in the absence of anyone at The Times knowing what they're doing, they're just grading everything as 'Tricky' by default.
There have been quite a few variations in degree of difficulty lately, even before they got the rules sorted out.

To the point however, if anyone wants to give me a pointer on where to go next, I'd be grateful. I'm in the second day of a heavy cold and it's hard to write, much less concentrate, when the world is falling out of your nose wrapped in white blood cells (which are showing a decidedly green hue).

[Edit: Yaaay! Made the code look nice. What a bl**dy hassle.]

[Spa]

Karyobin:

See my previous comment, and try to fill in the middle box

[Karyobin]

Sorry, already know that they're {4,7}, some way round. I also know a few more candidates in the central area, I'm just having trouble turning them into definites.

[Spa]

I think it tiik me a while to decide which way around the 7 & 4 were but just knowing they were there helped. If memory serves me right I worked on the groups at and around the bottom of the central box, and to its right

[Spa]

You should be able to work r6c6 out quickly and then r7c5 followed by r1c4

[Bigtone53]

Yes, Filling in the middle box, even with the unceratinty over the 4/7, enables you to work to the right and then up/down. I agree that this was a surprisingly challenging one, taking me over the 28 minutes.

There is a thread in the general group involving an Australian who was upset that when he asked for help, he was given hints not the 'straight bloody answer'. I hope that the suggestions above are what you are looking for.