Joitsa 1500 A.D., presented by Andreas
White to move and win.
1Q3K2/6P1/4q3/P2b4/B7/8/k7/8 w - - 0 1
Thursday, May 08, 2008
An ancient puzzle
Joitsa 1500 A.D., presented by Andreas
White to move and win.
1Q3K2/6P1/4q3/P2b4/B7/8/k7/8 w - - 0 1
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21 comments:
1.Qb3+
cute
Does 1.Bb3+ cook the study? (I realize that study conventions hadn't evolved 500 years ago....)
Fritz tells me I missed something in the "cook"
Really nice!
I like the variations if black tries do postpone the capture:
1. Qb3+ Ka1 2. Qc3+ Kb1 3. Bc2+ Ka2 4. Qb3+ (4. Bb3+ would do the trick here too) B:b3 5. B:b3+ followed by g8=Q.
bill
If Bb3 then Ka1 draws
Artic Knight
1.Qb3+ Bxb3
2.Bxb3+ K or Q takes
3.g8=Q and the exchange of queen is force. White than promote his a pawn...
The most interesting Variation is
1. Qb3+ Lxb3
2. Lxb3+ Qxb3
and now if 3.d8Q Ka1! with a diffcult endgame, because 4. Qxb3 is stalemate.
Therefore 3. d8B ist the solution.
Thanks ano for showing the pointe with g8B.
I found the right starting moves but did not consider the finesse Ka1 after g8Q.
Hmmm...
Ah! This one really fooled me!
1. Bb3+?? is just a draw after 1. - Ka1 2. Bxd5 Qf7+!
and 1. Qb3+ Bxb3 2. Bxb3 Qxb3 3. g8=Q?? Ka1! is nothing but a draw according to the tablebases.
My mistake: I didn't expect that level of sofistication in a puzzle this old.
Nice.
/L
and 1. Qb3+ Bxb3 2. Bxb3 Qxb3 3. g8=Q?? Ka1! is nothing but a draw according to the tablebases.
Hm, according to my tablebases installation this is mate in 19 max.
Re: 3.g8=Q and tablebases -- I get that it's drawn. The FEN at that point is
5KQ1/8/8/P7/8/1q6/k7/8 b - - 0 1
Qb3+
white to move and win my ass;
I played Fritz 10 against Fritz 11,
starting with Qb3+ and it was a Draw in 24 moves after threefold repetition
since black has the Ka1 trick, this problem is either cooked, or the tablebase (showing the extra pawn) can not win. Also, I dont think she would print a puzzle where then win was actually a Q+P v Q 50 move grind.
right?
Actually white has to promote to a Bishop instead of a queen.
White wins, but obviously it's not so simple:
1. Qb3+ Ka1 2. Qc3+ Kb1 3. Bc2+ Ka2 4. Bb3+ Bxb3 5. Qxb3+ Kxb3 6. g8=Q Qxg8+ 7. Kxg8 -/+
The win is forced.
White wins, but obviously it's not so simple:
1. Qb3+ Ka1 2. Qc3+ Kb1 3. Bc2+ Ka2 4. Bb3+ Bxb3 5. Qxb3+ Kxb3 6. g8=Q Qxg8+ 7. Kxg8 -/+
The win is forced.
1. Qb3+ Ka1 2. Qc3+ Kb1 3. Bc2+ Ka2 4. Bb3+ Bxb3 5. Qxb3+ Kxb3 6. g8=Q Qxg8+ 7. Kxg8
Yes, this wins. If Black instead tries 5...Qxb3 White counters with 6.g8=B! (6.g8=Q? only draws).
Incidentally, I'm thoroughly skeptical of this "1500 A.D." business. Somebody named Joitsa was composing problems as recently as the 1980s. Here's a composition of his published by Tim Krabbé:
k7/2P5/b4B4/K6q/8/4Q3/8/8 w - - 0 5
White to move and win.
k7/2P5/b4B4/K6q/8/4Q3/8/8 w - - 0 5
Note to self: proofread more carefully before posting FEN strings. That ought to be:
k7/2P5/b3B3/K6q/8/4Q3/8/8 w - - 0 5
1.Qb3+ Bxb3
2.Bxb3+ if Q takes
3.g8=B and the exchange of queen is forced, promotion follows, White wins.
However if
2...K takes
3.g8=Q White wins.
(Not 3.g8=B?? Kb4 draw)
This position may have very well composed in 1500, having been a good illustration of the expanded powers of the queen and the bishop. However there is no way the underpromotion could have been found at that time. The first known study/problem with an underpromotion was by Stamma around 1750, and in the 16th century rules regarding pawn promotion and stalemate were not yet established. Thus I suppose the original solution was simply 1.Qb3+ Bxb3 2.Bxb3+ and 3.g8Q. The accidental finesse with stalemate and the bishop promotion is a highly unlikely coincidence, comparable to the Saavedra position.
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