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		<title>Problem 242: Alexandr Pankratiev - Helpmate</title>
		<description>Discuss Problem 242: Alexandr Pankratiev - Helpmate</description>
		<link>http://kobulchess.com/en/problems/originals2013/351-chess-problem-alexandr-pankratiev.html</link>
		<lastBuildDate>Tue, 20 Oct 2026 13:01:29 +0000</lastBuildDate>
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			<title>Diyan Kostadinov says:</title>
			<link>http://kobulchess.com/en/problems/originals2013/351-chess-problem-alexandr-pankratiev.html#comment-588</link>
			<description><![CDATA[The diagram and solutions of the suggested version by Nikola is on the following link: http://prntscr.com/1pvudm]]></description>
			<dc:creator>Diyan Kostadinov</dc:creator>
			<pubDate>Sat, 07 Sep 2013 16:16:57 +0000</pubDate>
			<guid>http://kobulchess.com/en/problems/originals2013/351-chess-problem-alexandr-pankratiev.html#comment-588</guid>
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			<title>Nikola Predrag says:</title>
			<link>http://kobulchess.com/en/problems/originals2013/351-chess-problem-alexandr-pankratiev.html#comment-587</link>
			<description><![CDATA[In the mate Qd5# wPg2 is superflouos, that might be a weakness in such type of light problems. With the intended mates it is unavoidable in this scheme. But the main question is about the originality and then the economy. At least, all black pieces should be fully and equally used, if that is possible. An example: W:Pg7,Kc4,Pg2;B:Rh5,Ke4,Q f4;h#2; 3 sol.]]></description>
			<dc:creator>Nikola Predrag</dc:creator>
			<pubDate>Sat, 07 Sep 2013 15:16:35 +0000</pubDate>
			<guid>http://kobulchess.com/en/problems/originals2013/351-chess-problem-alexandr-pankratiev.html#comment-587</guid>
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			<title>Kenneth Solja says:</title>
			<link>http://kobulchess.com/en/problems/originals2013/351-chess-problem-alexandr-pankratiev.html#comment-586</link>
			<description><![CDATA[Nice problem, but somehow I don't like black e3-pawn which is actually not needed in two of the mates at all. I made a following version: W: Kc4, pg2, pg7 = 3. B: Ke4, Qg5, Bh6, Sg6 = 4. Solutions: 1.Qg5-e3 g7-g8=Q 2.Bh6-f4 Qg8-d5 # 1.Sg6-h8 g7*h8=Q 2.Qg5-f5 Qh8-d4 # 1.Sg6-f8 g7*f8=Q 2.Qg5-e5 Qf8-f3 # But still also in here I don't like Qd5 mate, because f3 is taken twice. I think also this can be improved.]]></description>
			<dc:creator>Kenneth Solja</dc:creator>
			<pubDate>Sat, 07 Sep 2013 07:30:47 +0000</pubDate>
			<guid>http://kobulchess.com/en/problems/originals2013/351-chess-problem-alexandr-pankratiev.html#comment-586</guid>
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