# Die Schwalbe

No. 237, June 2009

14123 -

Die Schwalbe 237, June 2009

(This problem was later found to be cooked)

11+10. Shortest proof game in 10.0 moves

[3qkbnr/3ppppp/8/8/8/8/PPPPPP2/RNBQK3]

Solution

14124 -

Die Schwalbe 237, June 2009

(In the printed issue of Die Schwalbe the diagram was misprinted; the bPe7 was missing. This problem was later found to be cooked.)

13+13. Shortest proof game in 19.0 moves

[4b2r/pQp1ppb1/n2qppkp/5n2/8/3P3N/PPPP3P/RNB2RK1]

Solution

14125 -

Die Schwalbe 237, June 2009

13+15. Shortest proof game in 8.5 moves (Madrasi)

[rnbq1bnr/p1p1pppp/3k4/3p4/8/QP6/2PPPPPP/KN3BNR]

Solution

14126 -

Die Schwalbe 237, June 2009

(This problem was later found to be cooked)

13+11. Shortest proof game in 9.5 moves (Einstein)

[1nbqkb1r/p1pppp2/8/8/8/8/P1PPP1PP/RN1QKBNR]

Solution

14127 -

Die Schwalbe 237, June 2009

9+7. -8 & #1 (Proca retractor, Marscirce)

[1R6/P4P2/2K4P/5P2/8/1N4P1/2P5/nb1rkqbn]

Solution

14128 -

Die Schwalbe 237, June 2009

4+2. ELM?

[8/8/8/8/8/8/5PPk/4KR1r]

Solution

14129 -

Die Schwalbe 237, June 2009

In an illegal cluster with (a) 2 white pieces and 2 black pieces (b) 3 white pieces and 1 black piece should the sum of the smallest distances between the diagram position and the origin square (either original or promotion) be as large as possible (example: Distance from origin square for a wBf4 is not sqrt(18) (distance to c1) but 4 (distance to f8).

Original text: In einem Illegal Cluster mit (a) zwei wei?#376;en Steinen und zwei schwarzen, (b) drei wei?#376;en Steinen und einem schwarze soll die Summe der kleinsten Abstände der Standfelder der (vier) Steine zu inhren Partieanfangs- oder Umwandlungsfeldern möglichst gro?#376; sein. (Beispiel: Für den wLf4 beträgt der kleinste Abstand 4, nämlich vron f4 bis f8, und nicht ca. 4,2 von f4 bis c1).

Solution