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:

Triple click to see the walkthrough I wrote: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.