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Posers and Puzzles

Posers and Puzzles

  1. 27 Sep '07 17:01


    Release the position
  2. 27 Sep '07 17:14 / 2 edits
    Originally posted by David113
    [fen]2b5/1p1p1p2/1p6/b6N/1P3p2/RPP4P/RBpPPPkp/nRKQ4[/fen]

    Release the position
    Who's to play?

    If I was white, I'd play 1. Qh1+ (1. ... Kxf2 2. Rxa1 +-) Kxh1 2. Rxa1 Kg2 3. Kxc2 h1=Q 3. Rxh1 Kxh1 4. bxa4 and white seems better with a great material advantage and no immediate threats (as much as I see).
  3. Standard member SwissGambit
    Caninus Interruptus
    27 Sep '07 18:25 / 1 edit
    Originally posted by kbaumen
    Who's to play?

    If I was white, I'd play 1. Qh1+ (1. ... Kxf2 2. Rxa1 +-) Kxh1 2. Rxa1 Kg2 3. Kxc2 h1=Q 3. Rxh1 Kxh1 4. bxa4 and white seems better with a great material advantage and no immediate threats (as much as I see).
    The stipulation "Release the position" is for retrograde problems. It means that you're supposed to take back moves and reach a position that is obviously legal.

    Edit: "It's always better to sacrifice your opponent's men." -Tartakower.
    In forward play, 1.Nxf4+ Kxf2 2.Qh1 wins trivially.
  4. Standard member SwissGambit
    Caninus Interruptus
    28 Sep '07 07:37
    Originally posted by David113
    [fen]2b5/1p1p1p2/1p6/b6N/1P3p2/RPP4P/RBpPPPkp/nRKQ4 w[/fen]

    Release the position
    Finally, a retro to test my mettle! And what a masterpiece of a problem - has to be Volet, Ceriani, or Fabel; all masters of the lengthy retro.

    Analysis:

    Black\'s two captures were made by the two doubled pawns. c2 and h2 have never left their file. To accommodate them, White needs the capture b2xc3, followed by two more caps to get the other 2P off the b-file, and the capture g2xh3.

    The 3rd wR must have been the promoted h2 pawn. To avoid collision with bPh, a 5th capture, hxg, is needed. This accounts for all White captures.

    So, the key question: How did the bK enter White\'s fortress? h2 never captured, so that route is out. As clogged as it looks, the only route left is the Queenside.

    As if that wasn\'t intimidating enough, White cannot retract a2xb3 anytime soon. The long-term problem is the collision between the bK and wBb2. If wBb2 stays behind the pawn chain, bK can\'t cross b2, because the wB gives an impossible check. If it goes outside the pawn chain, White has to retract b4-b3, and the bK goes to a3, but then wB can\'t get back to c1 without once again administering an impossible check.

    The only solution is to hide bK on a2 (!), move wB to c1, and provide a screen on b2, making ...Ka3 possible. OK, great, but what the hell to do with all of those White Rooks?!!

    The answer is astounding.

    Retract: 1.Ng3 Kg1 2.Nf1. Black now shuffles Kg1-g2 for a looong time, while White goes thru the sequence Ra4 Ba3 Qe1 Kb2 Rd1 Kc1 Rb2-b1 Kb2-a2 Bc1 Kb2 Ra2 and 18.Kb5.

    Next, wQ is extracted: Ra4 Ba3 Rb1-a2 Rb1 Qc1-b2 Rd1 Qb1 Bc1 Rb2 Qa2 Rb1 Qb2 Ra2 and 35.Qa4.

    Next, wN is released from the 1st rank. Kc4 Qb5 Ra4 Ba3 Rb1-a2 Rb1 43.Bc1.

    wN redeploys: Ng3 Nf4 Nd4 Nf3 48.Ne1! Dangerously close to the 50 move rule, but precisely calculated. 48...Kf1 49.Ba3 Kg2 50.Rh1 f5! Just in time. And the point of the wN maneuver is revealed: Black only had to burn one P tempo while waiting for wR to cross over. He needs to save the other two P tempos for the other two White Rooks! Incredible.

    51.Rh1 Kf1 52.Ng1 Ke1 bK heads west. wK will come in, Ng1 will leave, followed by Rh1 escaping via g1. All Rooks will be brought out using this method.

    53.Rb2 allows bK to shuffle from c1 to d1. wK makes it to f1 on move 58. 59.Nf3 Rg1-g8-d8 Kg2 64.Kg3 Ke1 65.Ne5 Kf1 66.Kf4 Kg2 67.Nf3 Kf1 68.Ne1 69.Rb1 Kg2 70.Nf3 fxe6(+B) 71.Rh1 Kf1 72.Ng1. Lather, rinse repeat. Again, bK goes west. This time, wB saves time by shielding on f1, allowing Ng1 and Rh1 to leave, and bK doesn\'t need c1. This R might as well unpromote on g8, on move 65.

    Ra4 is brought to c1 and out through g1 with the same technique. 89...Kg2 90.Nf3 e7 91.Rh1 Kf1 92.Ng1 and now the final act.

    92...Ke1 93.Bg2 Kd1 94.Bf1 Ke1 95.Qg5 [there\'s confidence for ya - White clears the square a4 (!) in advance] 96.Nf3 97.Ne5 98.Nd3 Kc1 99.Nb2 [obviously the only piece that can shield] 99...Kb1 100.Nc4 Ka2 101.Ne3 Kb1 102.Nd1 Ka2 103.Bg2 104.Re1 105.Bf1 106.hxg2(+R) Finally, Black has a free piece to move. 106...Ka2 107.Ne3 108.Rb1 109.Bc1 Ka3 110.Rb2 Ka4 111.bxa2(+R) and now, with the bK freed, the rest is [comparatively] easy.

    Whew! That was a lot of typing for something very few people will read. [But I can\'t help it; I\'m addicted to these damn things.]
  5. Standard member Peakite
    Sais
    28 Sep '07 16:21
    Big problem with these is that it's very hard to visualise the solution, nice to see someone can get it. In saying that I do like some of the retro probelms that have been posted here (got a few too, though this one is beyond me).
  6. 29 Sep '07 20:01
    SwissGambit, excellent job!

    here is a proof game:

    1. h4 c5 2. h5 Qb6 3. h6 Kd8 4. hxg7 Kc7
    5. Nc3 Nf6 6. Na4 Qb3 7. Nb6 axb6 8.cxb3 Ra3
    9. g8=R Bg7 10. Qc2 h5 11. Qf5 Ne4 12. Kd1 Bc3
    13. Kc2 Ba5 14. Rh4 c4 15. Rf4 c3 16. Kd3 c2
    17. Nf3 Nc3 18. bxc3 h4 19. b4 h3 20. Bb2 h2
    21. Ke3 Rh3 22. gxh3 Kc6 23. Bg2 Na6 24. Rh1 Nc5
    25. Bf1 Kb5 26. Bc1 Ka4 27. Ne5 Nb3 28. Nd3 Na1
    29. Rh8 Rb3 30. Nb2+ Ka3 31. axb3 Ka2 32. Qb5 Kb1
    33. Nd3 Ka2 34. Ba3 Kb1 35. Nb2 Kc1 36. Nc4+ Kd1
    37. Ne5 Ke1 38. Nf3+ Kd1 39. Ng1 Ke1 40. Bg2 Kd1
    41. Bd5 Ke1 42. Qa4 Kf1 43. Nf3+ Kg2 44. Rb1 e6
    45. Rb2 Kf1 46. Nd4 Kg1 47. Ra2 Kf1 48. Bg2+ Ke1
    49. Bf1 Kd1 50. Rg4 Ke1 51. Rg1 Kd1 52. Rh1 Ke1
    53. Nf3+ Kd1 54. Ng1 Ke1 55. Bg2 Kd1 56. Be4 Ke1
    57. Bf5 Kf1 58. Nf3+ Kg2 59. Rb1 exf5 60. Rbb2 Kf1
    61. Nd4 Ke1 62. Kf3 Kd1 63. Kg2 Kc1 64. Kf1 Kd1
    65. Rg8 Kc1 66. Rg1 Kd1 67. Rh1 Kc1 68. Nf3 Kd1
    69. Ng1 Kc1 70. Kg2 Kd1 71. Kf3 Ke1 72. Ke3 Kf1
    73. Kd3 Ke1 74. Kc4 Kf1 75. Rb1+ Kg2 76. Bc1 Kf1
    77. Qa3 Kg2 78. Qb2 Kf1 79. Ra4 Kg2 80. Qa3 Kf1
    81. Rb2 Kg2 82. Ra2 Kf1 83. Bb2 Kg2 84. Kb5 Kf1
    85. Nf3+ Kg2 86. Rb1 f4 87. Bc1 Kf1 88. Ne5 Kg1
    89. Ng4 Kf1 90. Ne3+ Kg1 91. Nf1 Kg2 92. Bb2 Kg1
    93. Rd1 Kg2 94. Bc1 Kg1 95. Qb2 Kg2 96. Qb1 Kg1
    97. Ba3 Kg2 98. Qb2 Kg1 99. Rb1 Kg2 100. Qc1 Kg1
    101. Qe1 Kg2 102. Rd1 Kg1 103. Rb2 Kg2 104. Rbb1 Kg1
    105. Bc1 Kg2 106. Ra2 Kg1 107. Ka4 Kg2 108. Ka3 Kg1
    109. Kb2 Kg2 110. Ra4 Kg1 111. Ka2 Kg2 112. Ba3 Kg1
    113. Kb2 Kg2 114. Kc1 Kg1 115. Rb2 Kg2 116. Ra2 Kg1
    117. Kb2 Kg2 118. Rb1 Kg1 119. Qd1 Kg2 120. Kc1 Kg1
    121. Bb2 Kg2 122. R4a3 Kg1 123. Ng3+ Kg2 124. Nh5

    The following problem may be even worse.



    Release the position...
  7. Standard member SwissGambit
    Caninus Interruptus
    30 Sep '07 04:10
    Originally posted by David113

    The following problem may be even worse.

    [fen][/fen]

    Release the position...
    Maybe if I\'m really, really bored one day, I\'ll tackle this...
  8. Standard member SwissGambit
    Caninus Interruptus
    02 Oct '07 05:13
    Originally posted by David113
    The following problem may be even worse.

    [fen]Kr6/8/1r6/8/8/1PPPPpPP/kpbrrP1R/qbrnrrbn[/fen]

    Release the position...
    OK, I think I\'ve solved this monster.

    Black obviously promoted 5 R and 1 B. White never captured. All Black\'s pawns are on the board, in the form of either pawns or promotees. Minimally, Black\'s a-pawn and f-pawn do not need to capture; every other pawn needs to capture at least once to get behind a wP. The h7 pawn needs 2 caps, because Bg1 isn\'t promoted, and the only way to get it out is by retracting g2-g3, then h2-h3. From the forward perspective, ...hxg2 was never possible. That\'s 7 minimum P captures, which equals the number of missing W units.

    Now, the mess on the first 3 ranks!

    Given the capturing constraints, the only scheme that works is ...bxa, ...cxb2, ...dxc2, ...exd2 and ...h7xgxh2 with ...g7-g3xh2. ...bxa was followed by White\'s b2-b3, then ...cxb2, then c2-c3, then d3-c2, etc.

    Black must have promoted twice on a1, once on c1 and d1, and twice on h1. A Bishop couldn\'t have come from h1 [N must retract, but this requires g2-g3 to be retracted first, which precludes any Bh1], so it must have come from d1, the only remaining light square.

    Of the W Qside pawns, e3 must retract first, and continue in the order right-to-left. However, if e3 comes back right away, Bf1 has to get home via g2, which doesn\'t allow enough room for the requisite 2 bR to be stuffed into the SE corner [they have to go there to unpromote] before the gate closes for good with -g2-g3. This means that e2-e3 cannot be retracted before g2-g3.

    This realization complicates matters. It is very appealing to pull back e2-e3 at the outset and free Nd1, followed by unpromoting the Qside, but the Kside tangle kills the idea.

    The Nd1 is going to be stuck there for a very very long time.
  9. Standard member SwissGambit
    Caninus Interruptus
    02 Oct '07 05:22
    (continued)
    So, how to make progress? A simple shift left, or shift right, leads nowhere. Instead, we have something that resembles a parting of the Red Sea.

    0... Rb7-b8 -1. Rg2-h2 Bh2-g1 -2. Ka7-a8 Rb8-b7 -3. Ka6-a7 Rb7-b6 -4. Ka5-a6
    Rg1-f1 -5. Ka6-a5 Rf1-e1 -6. Ka5-a6 Re1-e2 -7. Ka6-a5 Re2-d2 -8. Ka5-a6
    Ka3-a2 -9. Ka6-a5 Ba2-b1 -10. Ka5-a6 Bb1-c2 -11. Ka6-a5 Rc2-c1 -12. Ka5-a6
    Rd2-c2 -13. Ka6-a5 Bc2-b1 -14. Ka5-a6 Qc1-a1

    Every little bit of room helps. The most nimble piece, the Q, is the first to evacuate. Next, the Qside is cleared out as much as possible, to prepare bK\'s evacuation.

    -15. Ka6-a5 Bb1-c2 -16. Ka5-a6
    Rc2-d2 -17. Ka6-a5 Qd2-c1 -18. Ka5-a6 Rc1-c2 -19. Ka6-a5 Bc2-b1 -20. Ka5-a6
    Ra1-c1 -21. Ka6-a5 Bb1-c2 -22. Ka5-a6 Qc1-d2 -23. Ka6-a5 Rc2-e2 -24. Ka5-a6
    Re2-e1 -25. Ka6-a5 Qd2-c1 -26. Ka5-a6 Rc1-c2 -27. Ka6-a5 Qe1-d2 -28. Ka5-a6
    Rc2-e2 -29. Ka6-a5 Qe2-e1 -30. Ka5-a6 Re1-f1 -31. Ka6-a5 Qf1-e2 -32. Ka5-a6
    Re2-e1 -33. Ka6-a5 Rd2-e2 -34. Ka5-a6 Qe2-f1 -35. Ka6-a5 Re1-g1 -36. Rg1-g2
    Qf1-e2 -37. Ka5-a6 Qg2-f1 -38. Ka6-a5 f4-f3 -39. Ka5-a6 Qc6-g2 -40. Ka6-a5
    Qe8-c6 -41. Rg2-g1 Rg1-e1 -42. Ka5-a6 Re2-d2 -43. Ka6-a5 Re1-e2 -44. Ka5-a6
    Rf1-e1 -45. Ka6-a5 Re2-c2 -46. Ka5-a6 Re1-e2 -47. Ka6-a5 Rc2-c1 -48. Ka5-a6
    Re2-c2 -49. Ka6-a5 Bc2-b1 -50. Ka5-a6

  10. Standard member SwissGambit
    Caninus Interruptus
    02 Oct '07 05:29
    (continued) (numbering resets)
    0... Rc1-a1 -1. Ka6-a5 Bb1-c2 -2. Ka5-a6 Rc2-c1 -3. Ka6-a5 Rd2-c2 -4. Ka5-a6
    Bc2-b1 -5. Ka6-a5 Bb1-a2 -6. Ka5-a6 Ka2-a3 -7. Ka6-a5 Ka1-a2 -8. Ka5-a6
    Ba2-b1 -9. Ka6-a5 Kb1-a1 -10. Ka5-a6 Kc1-b1 -11. Ka6-a5 Bb1-c2 -12. Ka5-a6
    Rc2-d2 -13. Ka6-a5 Kd2-c1 -14. Ka5-a6 Rc1-c2 -15. Ka6-a5 Bc2-b1 -16. Ka5-a6
    Ra1-c1 -17. Ka6-a5 Bb1-c2 -18. Ka5-a6 Kc1-d2 -19. Ka6-a5 Rc2-e2 -20. Ka5-a6
    Kd2-c1 -21. Ka6-a5 Ke2-d2 -22. Ka5-a6 Kf3-e2

    Now that his majesty is out of the \"Rook corridor\", we can free up yet more room by unpromoting the two Rooks on the a-file.

    -23. Ka6-a5 Rd7-b7 -24. Kb6-a6
    Rd8-b8 -25. Kb5-b6 Rc1-c2 -26. Kb6-b5 Bc2-b1 -27. Kb5-b6 Bb1-a2 -28. Kb6-b5
    a2-a1=R -29. Kb5-b6 a3-a2 -30. Kb6-b5 a4-a3 -31. Kb5-b6 a5-a4 -32. Kb6-b5
    Ba2-b1 -33. Kb5-b6 Ra1-c1 -34. Kb6-b5 Bb1-a2 -35. Kb5-b6 a2-a1=R

    And now to clear the pawns.

    -36. Kb6-b5
    a3-a2 -37. Kb5-b6 Ba2-b1 -38. Kb6-b5 Re2-e1 -39. Kb5-b6 Bb1-c2 -40. Kb6-b5
    Rc2-e2 -41. Kb5-b6 Rc1-c2 -42. Kb6-b5 Bc2-b1 -43. Kb5-b6 Ra1-c1 -44. Kb6-b5
    a4-a3 -45. Kb5-b6 Bb1-a2 -46. Kb6-b5 a6-a5 -47. Kc5-b6 a5-a4

    Now the wR needs to be freed, so that 2 bR can get into the SE corner. This means all bR in the way must exit via the a-file and even the 4th rank to make way.

    -48. Kb6-c5
    Ra4-a1 -49. Kc5-b6 Re4-a4 -50. Kb6-c5 Ba2-b1 -51. Kc5-b6 Re1-f1 -52. Kb6-c5
    Re2-e1 -53. Kc5-b6 Bb1-c2 -54. Kb6-c5 Rc2-e2 -55. Kc5-b6 Rc1-c2 -56. Kb6-c5
    Bc2-b1

  11. Standard member SwissGambit
    Caninus Interruptus
    02 Oct '07 05:36
    (continued)
    It\'s really not necessary to give actual moves anymore.

    The two bB\'s on the Qside work like turnstiles. All bR\'s exit through them, followed by wR. Two bR\'s go back to f1/g1, and Black retracts ...Ke2, then White retracts g2-g3, letting bB and bN leave the Qside. The two bR unpromote on h1. Black retracts h2-h1=R, g3xh2 and h5xBg4 to restore Bf1, which then goes home, which allows e2-e3 to come back, which allows Nd1 to leave, which allows an unpromotion of Bd1, which allows -...e3xd2, etc. etc.

    I didn\'t worry about the 50-move rule in my retractions. Black can simply make a timely retraction with the f-pawn to avoid it.

    Despite the longer solution, this problem actually didn\'t seem harder than the original problem in the thread. Thanks David113 for posting it.