Importantly, Anand has won one of his personal encounters with Aronian and Mamedaryov. He is also the only unbeaten player. So most tie-breakers will favour him. Aronian would need to win at least one game against Karjakin and Dmitry Andreikin, while Anand loses both his remaining encounters, with Karjakin and Peter Svidler. Karjakin could also have a chance if he beats Anand and then Aronian in the last-round. In that sense, Karjakin is the pivot.Or else, Aronian would need to win both games, while Anand loses at least one.
The key point here is that Anand would need to lose at least one game for any of those scenarios and that would require a sudden attack of nerves, or reversal of form. So far, the Indian GM has played solid, pragmatic chess in his best tournament performance in several years. In round 12, Anand took a pragmatic decision, opting not to play out a winning game against Andreikin, because it was an extremely high-risk situation. If he does fail to qualify, that decision would haunt him.
Aronian and Vladimir Kramnik started as favourites. Kramnik is completely out of it after blundering into two losses against Svidler and Karjakin, and suffering an earlier loss in a grudge match with Veselin Topalov. Kramnik was so disheartened he let Aronian off the hook by refusing to play out a winning position. The Armenian GM has also lost a crazy encounter with Mamedaryov.
The diagram, WHITE TO PLAY (White: Anand Vs Black: Andreikin Candidates 2014) is the point at which fireworks started. White played 26.Nh5! Re6 27.Nxf6+?! Rxf6 28.d7 Qc7 29.Qg4 c4
Anand missed the direct 27.d7! Nbxd7 28. Nxf6+ winning a piece or 27...Qc7 28.f5 Rxe5 29.Nxf6+ gxf6 30.Bxh6 with a mating attack. But this is a matter of taste since white has a big attack anyhow.
Play continued 30.Rg3! g6 31.h5! cxb3 32.Rxb3! Na4! Black's dead if he defends with 32.-- Bf8 33. hxg6 fxg6 34. Nxg6 so the piece sac is justified. Play continued 33.hxg6 fxg6 34.Rxb4! Nc3+ 35.Kc2! b5! Black would lose simply after 35...Nxd1+ 36.Rc4! Qd6 37.Qxd1 36.Kb3 Na4 37.Qf3?! Instead 37.Bd2! wins more convincingly after 37...Nc5+ 38.Kxa3 Ra6+ 39.Kb2 Qa5 40.a4! Nxa4+ 41.Kc1! Qc7+ 42.Kb1 Nc3+ 43. Bxc3 Qxc3 44. Rb2.
The game ended 37...Nc5+ 38.Kc2 Na4+ 39.Kb3 Nc5+ 40.Kc2 Na4+ 41.Kb3?! (½-½).
White can win by 41.Kd2 Qd6+ 42.Nd3 or 41.Rc4!! bxc4 42.Bxa3 but Anand did not want to calculate through the high-risk variations that ensue.
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