As far as the game itself, I'm generally pretty critical of players who don't safeguard their king early on. In White's case, I suspect time pressure forced errors stemming from the fact that he had to play very accurately in an uncomfortable and complicated position leading up to time control.

The decision to not castle carries a heavy burden in that you must play very accurately throughout the game. And, very accurate play usually takes a great deal of time to implement over the board. To lighten this burden, a GM usually plans ahead to not castle and prepares accordingly.

In this game, by the end of move 25, the position slightly favors White, but Black probably has the more comfortable game. His knight's are about as poorly placed as knights can be, but other than that, he is fine. Meanwhile, White's pieces are hemmed in by his own pawns, and his king must use his knight to hold back Black's pieces.

And, while White correctly chooses to open up the position by exchanging with b5, calculated attacks against reasonably well defended positions, as in this case, usually take a great deal of time. White's attack is not the most accurate, but its hard to argue against it given the position after move 37. However, 30. axb7 allows White to play Nh7, Nf6, Rb1, e5 which also has promise.

White correctly turns down the exchange of queens (twice!), but I suspect under time pressure, allows Black a way out with 38. Qd3? More accurate would have been 38. Qe2 d5 39. Qf2+ Ka8 40. exd5 and White has a winning game despite the material disadvantage. 40. Qc3? which does little to help White's cause is also probably a time pressure error. Better would have been Qd5.

D Flores vs Radjabov, 2005 (49…?)

Black to play and win.

Material: R for B+3P. The White Kg2 has 2 legal moves. It’s the old replace-the-piece-with-something-more-expensive-- and-then-fork-with-a-N trick.

Candidates (49…): Qxf2+

49…Qxf2+ 50.Kh1

[50...Kxf2 Nxe4+ forks Qd6, winning a clear R in a placid position]

50…Qe1+ 51.Kh2 [or Kg2] Qg3+ 52.Qxg3 [else, drop Qd6]

52…hxg3 53.Kxg3 Nxe4+

The kibitzing about White's squandered chances led me to replay the entire game to see what they were. Indeed White missed great chances, starting at move 30 where the surprising 30 axb7 (intending to follow with Rb3) would have won according to Hiarcs analysis. After 30 axb7:

click for larger view

If Black takes on b7 then 31 Rb3 cannot be answered with b4 on account of 32 Bxb4 and White breaks through with overwhelming force. If black plays 30...b4 then 31 Rb3 still wins.

<49....?>

White: a6, e4, f5, h3, Ne6, Bg4, Rf2, Qd6, Kg2

Black: h4, Nf6, Re7, Rg8, Qe1, Ka8

Material: /+\ B+3P/R

Candidates: <[Qxf2†]>

-ML-
This position is insanely dynamic and White seems to have his forces, albeit the useless Bg4, well placed with three pawns(!) compensation for the exchange and a quite overwhelming position. Radjabov, however, bluntly throws a tactic in here that involves a simple fork to decide the game... or so it seems.

<49....Qxf2!> and suddenly the White combinations fall apart like a house of cards. Accepting this sacrifice tends to lead to an unwillingly useless position for White.

/A\
<50.Kxf2 Nxe4† 51.Kf3 Nxd6 > and Black's forces, in the end, work better than the White forces. So would not accepting the sacrifice, and thus giving away the Rook without the 'Queen trade' fair better for White?

/B\
<50.Kh1?! Rd7!> clearing both the Rook being en prise and threatening another nice combination. In general, that would happen if the Queen moves out of the Rook's fire without Qc5† and a trade, and aside giving away the full Queen with Qxd7 or Qc7 as Black's Queen is still on f2.

/BA\
<51.Qc6> Qe5 accomplishes the same for Black without a mate threat from White, so c6 is better.

<51....Rd1†! 52.Bxd1 Rg1‡ 0-1> which is another nifty combination. So White should, with a hopeless position to follow, still trade Queens.

/BC\
<51.Qc5† Qxc5 52.Nxc5 Rd1† > and Nxg4 with all the pawns dropping next. A tad bit complex for Monday.

<TheaN: How well does Rd7 fair? I believe it's pretty forced to an easy win for Black, and White can go horribly wrong with Qc6 or Qe5.>

Yes I think if white declines the queen, Rd7 is simple enough. At this point black is technically winning so even after 51.Qc6 if black didn't have 51...Rd1+! black is still winning. If nothing else, he needs to make sure white has no counterplay (like a draw by perpetual) and his material force should win. I really like other's ideas to force a queen trade and simplify.

But another way to win even more brilliant is

49.....Rxg4+!
50 hxg4 (no option the game will be finished much earlier if not taken)h3+!

Options are

51 Kf3
51 Kg3
51 Kxh3
51 Kh2 (mate after two moves)

Lets take a look Kf3 first

51 Kf3 Qxe4+
52 Kg3 Qe3+! (not Qxg4+ its a losing move)
53 Rf3 (not Kh4 or Kh2 or Kxh3 otherwise it will be mated soon) Ne4+ 54 Kh2 Qe2+ and mate after 1 or 2 moves.

What if Kg3...
51 Kg3 Qe3+!! (Ne4+ is also a clear win of white queen but will take a little longer to finish the game, remember, dont hurry to eat the white queen, its always there for free 52 Rf3 (no option else will get mated soon) Nxe4+ (now the timing of Ne4+ is better) 53 Kh2 (other move will get mated one move earlier) Qe2+ and mate in 1 or 2 moves.

Now Kxh3 a bit longer but still winning!
51. Kxh3 Rh7+
52. Kg2 (not Kg3 at it will be mated soon by 53 Qe3+,Kg2, 54 Kg1,Qh3+ then 55 Qh1 mate or if Rf3, 53 Nxe4+,54 Kg2 Qe2+ 55 Kg1 Qe1+ 56 Kg2 Qh1 mate or 56 Rf1 Qe3+Kg2 57 Qh3+ Kg1 58 Qh1 mate) Qh1+ 53 Kg3 Qh4+!(♘xe4+ also wins the queen but a still a long way to win)<<Once> And here Fritz announces mate in 17>. Mr.Once, there is a faster way than Fritz analysis for grabbing the queen and mate in 17. You don't need so many moves to mate...

Now we have 2 options Kf3 or Kf4

54 Kf3 (not Kg2 else it will be mated in two moves by Qh3 followed by Qh1 mate. Qxg4+ 55 Kf3 Rh3+
56 Kd4 (if Kd2 then Ne4+ then wins the queen and will get mated in less than 7 moves) Qd1+ (not Qxe4+ there is no mate in that move) 57 Kc5 (if Ke5, Ng4+ then followed by Qxd6+,Kxg4 Qg3 mate) Nxe4+ 58 Kb5 Rb3+
59 Ka5 Qa1+ and mate soon (if 59 Ka4 Nc3+ then 60... Qa1+ and mate soon, if Kc4 Rc3+, Kb5 (if Kb4 Qxd6 and mate soon) NXd6+ and mate soon

If 54 Kf4...

54 Kf4 Qxf2+
55 Ke4 Nxg4+
56 Kd5 Qa2+
56 Kd4 (if Kc6 Qa6+ 57 Kc5 Qa5+ 58 Kc6 Ne5+ and white queen is gone and will be mated soon or 58 Kd4 Qd2+ If 59 Kc4 Ne3+ and mate soon or 59 Kc5 Qc3+ 60 Kb5 (if Kd5 Ne3 mate) Rb7+ and mate in 2)

after 50.hxg4 Qxf2!
if here 51.Kh1 then 51... Qf1+ followed by ...Nxg4+

For 49... Qxf2 50.Kh1 , white has drawing chances due to black's exposed king

after 50... Qxf2
Black has several options, n all leads to mate or leads to a won ending of R+N vs N

For 51.Kxf2 then 51... Nxe4+
wins the queen n a won endgame

For 51.Kh3 then 51... Qf3+
here if 52.Kxh4 then 52... Rh7 followed by Nxe4
and if 52.Kh2 then 52... Nxg4 followed by mate

Then I looked at the game and saw that it had a much simpler solution than the one I found.

I zeroed in on 49...Rxg4+! 50.hxg3 h3+! and white is lost, even though our rook is threated to be captured with check by white's queen.

If, and if 51.Kxh3, then 51...Rh7+ (saving the rook with check) 52.Kg2 Qh1+ 53.Kg3 Nxe3 forking K+Q.

Of course, if 51.Kg3, then 50...Nxe4+ forks K+Q right away.

49... ♖xg4+ 50. hxg4 h3+ 51. ♔xh3 ♖h7+ 52. ♔g2 ♕h1+ 53. ♔g3 ♘xe4+ 54. ♔f4 ♘xd6

which leaves white in this position:

click for larger view

As a couple of others have commented this is a little stronger than the position after 49...Qxf2+, although obviously both win.>

<after 50.hxg4 Qxf2!
if here 51.Kh1 then 51... Qf1+ followed by ...Nxg4+

For 49... Qxf2 50.Kh1 , white has drawing chances due to black's exposed king>

after you move 50. hxg4 Qxf2?? should be h3+! White has no drawing chances at all due to white will never have a chance to attack.

<correct me if im wrong.. > You are now corrected :)

after 50... Qxf2
Black has several options, n all leads to mate or leads to a won ending of R+N vs N>

It is the very least material advantage and Rxg4+ sacrifice doesn't make sense if you will just continue with Qxf2+. You just move it on 49. Qxf2+ as the solution says then you have your two rooks and and knight vs. N+B then bigger advantage. But if you take 49. Rxg4 then it should be followed by a3+! that is the proper continuation.

And here Fritz announces mate in 17. >

<< spreadsanity:> I found this line as the solution, Shredder agreed with me: 49... xg4+ 50. hxg4 h3+ 51. xh3 h7+ 52. g2 h1+ 53. g3 xe4+ 54. f4 xd6 >

So let' s see. I noticed the fork against the Rf2 and the Qd6. Also, I saw that the Rf2 was already under attack, and only protected by the King. From there, it was easy to see the royal fork at e4.

But of course, White is not absolutely obliged to take back on f2. He could play 50. Kh1. At least, Black can get a perpitual, though, in a worse case scenario. But then, I saw 49. ...Qxf2+ 50. Kh1 Rd7(!!), pinning the white Queen, and supported by the possibility of an other pin a little later, against the white Bishop this time. For example: 51. Qc6 Rd1+ 52. Bxd1 Rg1+ mate.