The New Sudoku Players' Forum

[udosuk]

Has anyone tried #16 of the following page:

http://entertainment.timesonline.co.uk/tol/arts_and_entertainment/games_and_puzzles/article3599073.ece

I've tried it multiple times and needed very advanced moves to crack it every time. I don't think puzzles of this difficulty are suitable to be published on a newspaper (i.e. requiring people to solve it using pen & paper). I don't recall EVER seeing any number placement puzzle on the Times newspaper to be nearly as tough as this one. (Then again, I don't solve many of their puzzles these days).

Here is my own complete walkthrough of this puzzle:

11/2 @ r2c2={56}+ (NP @ r2)
11/2 @ r4c1={56}+ (NP @ c1)
30/2 @ r3c4={56}x (NP @ c4)
=> r4c14={56} (NP @ r4)
=> 36/3 @ r3c3=[6{23}|334]x
=> r3c3 from {36}, r4c23=[{23}|34] (3 @ r4 locked)

3/2 @ r5c4 from {1234}={12}+|{14}-|{13}x (1 @ c4 locked)
=> 2/2 @ r1c4 from {234}={24}- (or {24}/) (NP @ c4)
=> 3/2 @ r5c4={13}x

2/2 @ r1c2={13|24|35|46}-|{12}x|{36}/ ({12|24}/ covered)
1/2 @ r1c5={12|23|34|45|56}-

r1c4 from {24}
=> 2/2 @ r1c2 can't be {24}-

6 @ r1 locked @ r1c2356
=> Either r1c23={36|46} or r1c56={56}
=> r1c23 can't have 5
=> 2/2 @ r1c2 can't be {35}-

5 @ r1 locked @ r1c56
=> 1/2 @ r1c5={45|56}- has 4|6
=> 2/2 @ r1c2 can't be {46}-

c23 must have a total of four of {36}
11/2 @ r2c2={56} has one
36/3 @ r3c3=[6{23}|334]x has two
=> 2/2 @ r1c2 can have at most one of {36}
=> 2/2 @ r1c2 can't be {36}/
=> 2/2 @ r1c2={13}-|{12}x (1 @ r1 locked)

HP @ r1: 1/2 @ r1c5={56}-

3/2 @ r1c1 from {1234}=[21]+|[41]-|[31]x => r2c1=1

3 @ r2 locked @ r2c56 from {234}
=> r2c56={23|34}
=> 12/3 @ r2c5=[{34}5]+|[322|{34}1]x
=> r3c5 from {125}

36/3 @ r3c3=[6{23}|334]x
=> Either r3c3=3 or r4c23={23}
=> r34c6 can't be [32]
=> 5/2 @ r3c6={14}+|[61]-|[51]x (1 @ c6 locked)

Swordfish: 3 @ r134 locked @ r134c123
=> r56c123 can't have 3

9/2 @ r6c5={36|45}+ has 3|4
=> r6c124 can't be [413]
=> r6c12 can't be [41]
=> 3/2 @ r6c1=[21]+|[25]-|[26]/
=> r6c1=2
=> r3c1 from {34}

4 @ r6 locked @ r6c356
=> Either r6c3=4 or r6c56={45}
=> r6c3 can't be 5

r3c34 from {356} has 3|5 & 3|6
=> 2/2 @ r3c1 can't be {35}-|{36}/
=> 2/2 @ r3c1=[31|42|46]-
=> r3c2 from {126}

HS @ c2: r5c2=4
=> 10/3 @ r5c2=[451|424]+
=> r5c3 from {25}

r5c4 from {13}, r5c13 from {256} has 2|6 & 5|6
=> r5c56 can't be {13|26|56}

Also r5c2=4 => r5c56 can't be {34}

c56 must have a total of four of {35}
1/2 @ r1c5={56}- has one
r2c56={23|34} has one
9/2 @ r6c5={36|45}+ has one
=> r5c56 can have at most one of {35}
=> r5c56 can't be {35}

Hence r5c56 can't be {13|26|34|35|56}
=> 12/3 @ r4c5 can't be [1{56}]+|[1{26|34}]x
=> it also can't be [4{26|35}]+|[4{13}]x
=> r4c5 can't be 1 or 4
=> r4c5=2
=> 36/3 @ r3c3=[334]x

All naked singles from here.

I welcome anyone to suggest a easier way to solve this puzzle, or inform me of a tougher Times puzzle.

[Bigtone53]

I did it like this (as did my colleague, as I have now found out) - about 20 minutes but not very elegant

1. By inspection, column 4 is 2/4 in some order, 5/6 iso and 1/3 iso

2. The two 11s are 5/6 iso

3. The 9 on the bottom right is either 4/5 iso or 3/6 iso

4. If 3/6, the 3 on the bottom left can only be 2/5. Working on upwards from the bottom leads to a contradiction as you run out of numbers for the 36

5. If 4/5, the 3 on the bottom left is 2/6 iso or 3/6 iso

6. If 2/6 you get the same contradiction trying to fill in the 36

7. 3/6 works though, to give the answer

312465
156243
423516
534621
645132
261354

[udosuk]

Thanks Bigtone53 for the reply!

So you guys did it the T&E way. Perhaps that's what the Times people feel about these puzzles, that the readers should not try to solve them with pure logic?

[Bigtone53]

To be honest, I doubt if anyone at The Times even understands these things. Remember the confusion when Killers first appeared in The Times about whether a number could be repeated in a cage or not. They just flip-fopped on this one until it was brought up in the weekly Complaints column.

I don't like T&E but needs must ..... . Perhaps your first Magic Hexagon involved a bit of T&E in orientating the magix hexagon figures before starting to put in numbers. As you said at the time, one way leads to a quick contradiction.

[Glyn]

Matt I have copied this one out. Just a thought if the puzzle is assumed to be unique then the operators should be assumed to be unique as well, it was posed as a NOP KenKen. If so then that may reduce the choices. Of course that won't work on a standard KenKen where we ignore the operators by choice. I haven't looked at the two walkthroughs so maybe you have applied that rule.

My first dumb move was thinking KenKen for Adults(13-18) was the age range.

EDIT: And the answer to my question about Unique Operators is no in this case. It fails immediately in column 4 and checking Bigtone53 for the solution the cage at r1c4 has two valid operators. The puzzle is not unique in the complete sense. Slapped wrists for the Times.

[udosuk]

Bigtone53, after studying your "walkthrough" I think you (and your colleague) have skipped quite a lot of scenarios, thus your solving path is not even complete in the T&E sense. (And there is a minor mistake with the 2/6 & 3/6 incorrectly swapped.)

Perhaps your first Magic Hexagon involved a bit of T&E in orientating the magix hexagon figures before starting to put in numbers. As you said at the time, one way leads to a quick contradiction.

Actually the "contradiction chain" was so short that I should be able to present it in a very "non T&E" manner. Will follow that up in the Magic Hexagon Killer thread later.

My first dumb move was thinking KenKen for Adults(13-18) was the age range.

Exactly same here! Now the word "Adults" with an age range can elicit some very unsavoury thoughts. (Associating those aged 13-18 with the word "Adults"? Did I smell something fishy like Bigtone53 did in this thread? )

Matt I have copied this one out. Just a thought if the puzzle is assumed to be unique then the operators should be assumed to be unique as well, it was posed as a NOP KenKen. If so then that may reduce the choices. Of course that won't work on a standard KenKen where we ignore the operators by choice. I haven't looked at the two walkthroughs so maybe you have applied that rule.

EDIT: And the answer to my question about Unique Operators is no in this case. It fails immediately in column 4 and checking Bigtone53 for the solution the cage at r1c4 has two valid operators. The puzzle is not unique in the complete sense. Slapped wrists for the Times.

Yes I think Maurice and I have agreed long ago that the uniqueness of operators is never required, just the cell values are important.

Note I never apply any uniqueness technique when solving any KenKen. They could often be helpful but I still prefer to try something else when there is an option. Call me stubborn.

[Bigtone53]

Bigtone53, after studying your "walkthrough" I think you (and your colleague) have skipped quite a lot of scenarios, thus your solving path is not even complete in the T&E sense. (And there is a minor mistake with the 2/6 & 3/6 incorrectly swapped.)

Once you accept T&E, you cannot reject a correct answer reached from it. I assumed that you would fill in the obvious missing steps that did not work (like why !/4 iso in the left will not work with 5/4 as the 9 in the botton right).

You are right though that I have muddled myself in my typing, as it is 26 in the bottom left that is the winner, as indicated before. I am happy to try and slog trough this again for you when I get the time but the fun has gone once there is an answer

[udosuk]

Once you accept T&E, you cannot reject a correct answer reached from it. I assumed that you would fill in the obvious missing steps that did not work (like why !/4 iso in the left will not work with 5/4 as the 9 in the botton right).

You are right though that I have muddled myself in my typing, as it is 26 in the bottom left that is the winner, as indicated before. I am happy to try and slog trough this again for you when I get the time but the fun has gone once there is an answer

Some of the cases you've skipped are not that obvious.

E.g. If r6c56={45}, r6c12={12|13|26|36}

{13} can be obviously eliminated by r6c4.

But how about {12}? Don't you think you should have elaborated more?

And the description that "Working on upwards from the bottom leads to a contradiction as you run out of numbers for the 36" is so vague. Even if I accept T&E I don't think this is acceptable. For example, did you make use of the 12(3) cage? If so how deep did you go in the branch?

[Bigtone53]

Okaaaaayyyyyy

Bearing in mind that I am supposed to be at work, I have scribbled this out. Likely to be both full of errors and typos

By inspection, C4 is 24 in some order, 56 iso, 13 iso
The two 11s are 5/6 iso
R9c89 is 36 iso or 45 iso

1.) If 36 iso, then c4r56 is 31 and r6c123 is 254
bottom row is 25416~3 and r2c23 is 65
R5c23 total 6 and cannot be 51 iso as 5s in c23 already used, so has to be 42
If the 12 cage is an addition, it has to be 156 iso with the 1 in r5 but this clashes with the 6/5 iso in c1r45
It is therefore a multiplication and filling in, I am forced to
000000
065000
000600
600520
54231~6
25416~3
The 36 cage cannot have any 6s or 4s in it. ERROR

2) It is therefore 45 iso, giving r6c12 the possibilities of 36 26 or 12 iso (13 is not an option)

2a) If 36, r5c23 is 15iso as I have used up the 6s in c23 (once56 is inserted in r2c23).

if 15iso, I have

000000
056000
000500
500600
61532~4
36215~4 and I cannot complete the 12 ERROR

2b) If r6c12 is 12 iso r6c3 is 6 and the 1 in r5c4 prevents me completing the 10 ERROR

2c) If r6c12 is 26, r6c3 is a 3 or a 1. If 3, I have the r5c23 as 25 iso or 16 iso

2c)i) If 25, I have
000000
056000
000600
500630
62~531~4
26315~4 and I cannot fit any of 623,334 or 661 into the 36 ERROR

2c)ii) If 16 iso, I have
000000
056000
000600
600540
51~632~4
26315~4 and I cannot fit any of 623,334 or 661 into the 36 ERROR

The bottom line is therefore

26135~4

The balance of the 10 cannot be 63 as I have used up the 6s in c23 (once56 is inserted in r2c23).

I now have
000000
056000
000500
500620
64~513~2
26135~4
The 36 is 334 and it is plain sailing from there.

[udosuk]

Not bad, but you still missed a case in the last step, while added some redundant branches:

2c) If r6c12 is 26, r6c3 is a 3 or a 1. If 3, I have the r5c23 as 25 iso or 16 iso

(udosuk: the 2 branches below are redundant, as you now have r6c23=[63] & r2c3=6, so the 36-cage can't have any 6, and it can't be [334]. No way to fill the 36-cage, ERROR.)

2c)i) If 25, I have
000000
056000
000500
500630
62~531~4
26315~4 and I cannot fit any of 623,334 or 661 into the 36 ERROR

2c)ii) If 16 iso, I have
000000
056000
000600
600540
51~632~4
26315~4 and I cannot fit any of 623,334 or 661 into the 36 ERROR

The bottom line is therefore

26135~4

The balance of the 10 cannot be 63 as I have used up the 6s in c23 (once56 is inserted in r2c23).

(udosuk: You should also consider the case that r5c23=[25] here as 10 could also be 2x5x1 besides 4+5+1.)

I now have
000000
056000
000500
500620
64~513~2
26135~4
The 36 is 334 and it is plain sailing from there.

[Glyn]

Here is my effort, apologies for lack of step numbering and irregularities in notation.

Initial cage possibilities
r12c1 +{12},x/{13},-{14},{25},/{26},-{36}
r1c23 x/{12},-{13},-/{24},-{35},/{36},-{46}
r12c4 x/{12},-{13}-/{24},-{35},/{36},-{46}
r1c56 -{12},-{23},-{34},-{45},-{56}
r2c23 +{56}
r2c56,r3c5 +{156},+{246},+{255},+{336},+{345},x{126},x{134},x{223}
r3c12 x/{12},-{13},-/{24},-{35},/{36},-{46}
r3c3,r4c23 x{166},x{236},x{334}
r34c4 x{56}
r34c6 +{14},x/{15},-{16},+{23}
r45c1 +{56}
r4c5,r5c56 +{156},+{246},+{255},+{336},+{345},x{126},x{134},x{223}
r5c23,r6c3 +{136},+{145},+{226},+{235},+{244},+{334},+{125}
r56c4 +{12},x/{13},-{14},-{25},/{26},-{36}
r6c12 +{12},x/{13},-{14},-{25},/{26},-{36}
r6c56 +{36},+{45}

Set r2c23,r34c4,r45c1={56} Clear all naked pairs
r12c1 remaining options +{12},x/{13},-{14} 1 locked in this cage
r12c4 remaining x/{12},-{13},-/{24}
r56c4 remaining options +{12},x/{13},-{14} 1 locked in this cage
r12c4 must be -/{24} naked pair
r56c4 must be x/{13}
r1c23 remaining options x/{12},-{13},-{35},/{36},-{46}
r6c12 remaining options +[21],-[41],-[25],/[26],-[36] => r6c2={156}
r3c3,r4c23 remaining options x6{23} x[334] => r3c3={36},r4c2={23},r4c3={234}
r2c56,r3c5 remaining options +{34}5,x{12}6,x{134},x[322]
r4c5,r5c56 remaining options {+246}{+336}{+345}{x126}{x134}{x223}
If r6c56=+{36} => r6c4=1 => r6c12=-[25]
If r6c56=+{45} => r6c12=> +[21],/[26],-[36]
r6c12 remaining options +[21],-[25],/[26],-[36] => r6c1={23},r6c2={156}

In row 1 one of these cages must contain 6
r1c23 {x/12}{-13}{-35}{/36}{-46}
r1c56 {-12}{-23}{-34}{-45}{-56}
Combos {/36} and {-12}{-45}
{-46} and {-12}{-23}
{x/12}{-13} and {-56} =>
r1c23 {x/12}{-13}{/36}{-46} r1c23<>5
5 locked in r1c56 remaining combos {-45}{-56} ie no 1,2,3
Combos again r1c23 remaining {x/12}{-13}{/36} ie no 4
r3c12 x/[21],-[31],-/{24},-[46]
not {-35} pushes 2 6's in row 3, {/36} blocked => r3c1={234},r3c2=[1246]

r5c23,r6c3 +{16}3,+[36]1,{+145},+[262],{+235},+[424],+[343],+{125}

Bilocation the 4's of c2
r3c2=4 => r3c1=2 => r6c1=3 => r6c3=2
r5c2=4 => r56c3={15}[24]
Conclusion r6c3<>3,6

Possibilities Row 6= {2,5,4,1,36,36} or {23,6,12,13,45,45}
=>r6c2<>1,r6c3<>5
r5c23,r6c3 remaining options are +[36]1,{+145},+[262],{+235},+[424],+{125}

Naked pair {56}r26c2

If r5c23,r6c3 contains 5 or 6 it must be in r5c3. The other option is [424] with r5c3=2. => r5c3<>1,3,4
Bivalue r2c3=5 => r6c2=5 => r6c3=4 => r5c3<>6 or r2c3=6 => r5c3<>6

Bilocation 4's of c3
If r6c3=4 => r1c3=1, r1c2={23}
If r4c3=4 => r3c3,r4c2=3 => r1c23<>3 r1c23={12}
Combined result r1c3<>3,6
Hidden pair r1c56={56}

(2)r2c4=(2)r1c4-(2=1)r1c3-(1)r1c1=(1)r2c1 => r2c1<>2
(5=6)r3c4-(6)r3c3=(6)r2c3-(5)r2c3=(5)r5c3-(5)r5c1=(5)r4c1 => r4c4<>5
Singles r3c4,r4c1=5 & r5c1,r4c4=6
r2c56,r3c5 remaining options x{12}6,x{134},x[322]

If r4c2=2 => r34c3=[63] => r6c3=4 =>r5c2={14}
If r4c2=3 => r5c2<>3
r5c23,r6c3 is +[154],+[451],+[424],+[152],+[251]

r1c23 = x{12},-{31} locked for 1 => r1c1<>1 Hidden single r1c2=1

Bilocation 3's of c1
If r1c1=3 => r6c1=2
If r1c2=3 => r1c3=1 => r6c3=1
Conclusion r6c3<>4

Branch on r1c2=1 and r1c3=1 => r6c3<>4
Hidden single r4c3=4. => r3c3,r4c2=3
String of singles leaving c56 unsolved.
r2c56,r3c5 only possible option is x{34}1 => r3c5=1
Singles to finish

Solution 312465156243423516534621645132261354

[Bigtone53]

udosuk,

I do these things for fun, and largely in my head. I am sorry that my one effort at spelling it out does not meet your high standards, but what I wrote is not how I actually think when I attack these things. It is more complicated than that.

The fun is leeching away

[udosuk]

I do these things for fun, and largely in my head. I am sorry that my one effort at spelling it out does not meet your high standards, but what I wrote is not how I actually think when I attack these things. It is more complicated than that.

Perhaps I should have been more tactful when I was discussing your solving path but I thought that was all for academic purposes and nothing personal. So I apologize if you somehow feel offended by my innocent comments.

[Bigtone53]

No, not offended. I just have some difficulty putting my thought process down on paper as I think quicker than I type and by the time I have written it down I have often forgotten the next step. That is what happened to the first Magic Hexagon. I did it but had real difficulties retracing my steps

[udosuk]

A bit of personal experience: when I need to type a walkthrough I usually use a solver program or Excel and then open a Notepad window, also the puzzle pic if necessary. Then I tile all the windows up on the screen and do it simultaneously. I figure that's the easiest way to write a walkthrough.

If I have to do a puzzle on paper I'll usually use an exercise book and hand draw the puzzle grid. Then I'll write down the steps on the opposite page of the grid. Also my "scripting font" is quite tiny so I can often pack in a lot of words in one page.