How to draw rook endgames

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In the game, HIARCS played 52. Kxf3, and it then becomes too difficult to bring the King over to the queenside and not lose one of the kingside pawns (due to the forking power of the Queen). Even after this, it's still not completely obvious to me, as Black's h-pawn won't be passed, and White can aim for the same set-up as before. But HIARCS moved Nb3 rather than Nc2, and then let the a-pawn thru via snatching the c-pawn, and now I do think it is over. [I'm not completely convinced I saw a -5 position that could be drawn (regarding the resignation rules), but there might be some here -- IvanHoe was around -4 at worst, but those with higher relative scores might slip the trigger].

To reduce it to a simpler statement (though I admit the kingside pawns give Black more chances), can Black win this?
8/k7/8/2p5/p7/K1R5/2N4q/8 bI really don't see a way.

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Transitioning back into rook endgames in Wijk aan Zee, Giri has Rook+4 versus Rook+3 against Nakamura. Another lesson in How to Draw. r5k1/5ppp/8/8/8/6P1/4PP1P/1R4K1 w - - 0 38
From the Endgame Corner at the Chess Cafe: The Famous Rook Endgame with 4 vs. 3 Pawns http://www.chesscafe.com/text/mueller95.pdf

You can also try this yourself: http://www.chessvideos.tv/endgame-train ... ook-14.php

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Other notable 4 vs 3 instances:
Korchnoi-Karpov 1981: http://www.chessgames.com/perl/chessgame?gid=1068246
Piket-Kasparov 2000 (Internet G/60): http://www.chessgames.com/perl/chessgame?gid=1339092

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I might point out that, even in a reduced setting, 4 vs 3 pawns is likely too much for a TB solution now. However, 3 vs 2 (on the same side) might very well be "solvable". Here is a calculation, assuming White has fgh-pawns and Black has gh-pawns. By the assumptions, there are 15 configurations for pawns in each of the gh-files, and 6 spots for the f-pawn. There are 64^4 places for the Rooks/Kings. This leads to 22 billion total positions, which is well below the 496 billion needed for a generic 7-piece (and already done in practise a few years ago). Even 4 vs 3 is only 15x this, so only 340 billion. However, the captures and promotions will multiply the number by some nontrivial factor. This factor would be insurmountable if we demanded that QQRP vs QQR (among others) be completely solved first, but it is not always necessary to do this.

The best current practise is the complete solution of RPP vs RP assuming all promotions are to Queens (due to Bourzutschky-Konoval (March 2007, see. June 2007 issue of EG, reported by John Beasley). However, we are greatly restricting the problem by naming specific files for the pawns to reside, and so can push this further if enough assumptions are made.