[Retros] N2A: 14 variation

Noam Elkies elkies at math.harvard.edu
Tue Mar 15 13:33:48 EDT 2011


I wrote:

` " I was actually hoping for an 8Q+0 or if necessary 9Q+0 position
` " where *both* Kings' positions are uniquely determined.

` While we don't have that, here's a 14-man realization of N2A:

` Q4/R7/8/R7/4Q3/5n2/1PPPPPP1/2BN1B2

Sorry, FEN botch: should be 3Q4/R7/8/R7/4Q3/5n2/1PPPPPP1/2BN1B2

` . . . Q . . . .
` R . . . . . . .
` . . . . . . . .
` R . . . . . . .
` . . . . Q . . .
` . . . . . n . .
` . P P P P P P .
` . . B N . B . .

[solution: +wKe1, +bkh6]

Possibly a bit more amusing:

2Q5/1R6/5Q2/R7/8/3n4/1PPPPPP1/2BN1B2 [hope I got the FEN right this time]

. . Q . . . . .
. R . . . . . .
. . . . . Q . .
R . . . . . . .
. . . . . . . .
. . . n . . . .
. P P P P P P .
. . B N . B . .

Add 2 missing units. What result?

NDE



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