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O problema pe zi (30.10.2010)

Posted by dan bujor on October 30, 2010


Problema 1         

M. B. Newman – Chess Amateur – 1913

Albul muta si castiga


Problema 2

S. M. Birnov – 64 – 1935 – Premiul II-III

Albul muta si castiga


 

 

Problema 3

A. Kovalenko – Schachmatny Listok – 1927

Albul muta si remizeaza


 

Problema 4

 

B. A. Bron – Chess in USSR – 1939

Albul muta si castiga


 

Problema 5

B. F. Didrichson –
Schachmaty – 1936

Albul muta si remizeaza



 

7 Responses to “O problema pe zi (30.10.2010)”

  1. ghita said

    P4. 1.c5+! D:c5(…R:c5 2.Ce4+! +-) 2.Ca4+! b:a4 3.Ne3! D:e3 4.Cc4+! +-

  2. dan bujor said

    Si la 3…Ra5! ?

  3. ghita said

    Urmeaza 4.b4+! D:b4(4…R:b4 5.N:c5+! +-)5.Nd2! D:d2 6.Cc4+! +-

  4. ghita said

    P1. 1.Nb5+! C:b5(…D:b5 2.Cc3+! +-) 2.Cc5+! R:a3 3.Cc2# sau 2…Ca5 3.Cc4#

  5. dan bujor said

    P4 e completa acum.

    Si P1 e aproape terminata, dar cum se finalizeaza lucrurile in varianta din paranteza dupa… 2.Cc3+ Rxa3 3.Cxb5+ Cxb5?

  6. ghita said

    P1.Probabil 4.Cc4+! Rb4 5.h6!! cu transformare imparabila!

  7. dan bujor said

    Corect. Mergea si 4.Cf5, si negrul nu avea intrare.

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