**Hazil shaxmat muammolari** shaxmatda hazilni asosiy yoki ikkilamchi element sifatida ishlatadigan jumboqlardir. Garchi koʻpchilik shaxmat muammolari, boshqa ijodiy shakllar kabi, jiddiy badiiy mavzular (Grimshaw, Novotny va Lacny) uchun qadrlangan boʻlsa-da, hazil shaxmat muammolari baʼzi burilishlar uchun zavqlanadi. Baʼzi hollarda bastakor hal qiluvchining odatiy tahlil bilan muvaffaqiyat qozonishiga yoʻl qoʻymaslik uchun hiyla oʻynaydi. Boshqa hollarda, hazil gʻayrioddiy yakuniy pozitsiyadan kelib chiqadi. Oddiy shaxmat jumboqlaridan farqli oʻlaroq, hazil muammolari ichki mantiq yoki oʻyin qoidalarini buzadigan echimni oʻz ichiga olishi mumkin.

Oʻz-oʻzini hal qilish muammolari
--------------------------------

V. Ropke, Skakbladet 1942

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| White to move, mate in 6. Can it be stopped? |

Baʼzi shaxmat jumboqlari umuman boshqotirma emas. Diagrammada Oqdan oltita harakatda Qora bilan mat qilish soʻraladi. Bu holatda hazil, shaxmat qoidalariga koʻra, Uaytning bu masalada tanlovi yoʻq; yagona qonuniy harakatlar toʻgʻridan-toʻgʻri „yechim" ga olib keladi:

**1. d4 b5 2. d5 b4 3. axb4 a3 4. b5 a2 5. b6 a1=har qanday 6. b7 #**

Tim Krabbé oʻzining shaxmat veb-saytida boshqa misollarni taqdim etadi

Shaxmat qoidalarining gʻayrioddiy talqinlari
--------------------------------------------



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| Mate in 1 |

Shaxmat qoidalari juda oddiy va aniq belgilangan. Shunga qaramay, bu har doim ham shunday boʻlmagan va baʼzi bastakorlar kulgili jumboqlarni yaratishda qoidalardagi noaniqliklarga tayangan. Oʻngdagi diagrammada shunday bir misol koʻrsatilgan. Shaxmat afsonasiga koʻra, bastakor „Bir harakatda oq juftlar" deb shart qoʻygan. Bu imkonsiz boʻlib tuyuladi va hozir ham shunday, lekin u taqdim etilganda, koʻtarilish qoidasida piyon koʻtarilishi mumkin boʻlgan buyumning rangi koʻrsatilmagan; Shunday qilib, „yechim" Oq uchun b8 da *qora* ritsarga koʻtarilib, qora qirolni potentsial parvoz maydonidan mahrum qiladi. Shunga oʻxshash (boʻshliqqa asoslangan boʻlmasa-da) muammolar qirollar yoki piyonlar darajasiga koʻtarilish bilan bogʻliq.

Tim Krabbé, 1972

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| Mate in 3 |

Keyinchalik murakkab misol Tim Krabbé tomonidan tuzilgan va goʻyoki castling taʼrifida mavjud boʻlgan boʻshliqqa tayangan. Diagrammada Oq uchta harakatda juftlashishi kerak. Yechim **1. e7** dan boshlanadi, keyin asosiy oʻzgarishlar:

* **1... Kd3 2. e8=Q gxf3** (other moves allow Qe2#) **3. 0-0-0#**
* **1… Kxf3 2. e8=RAndoza:Chesspunc** (an underpromotion), and now:
* **2… d4 3. 0-0#**
* **2… Kg2 3. 0-0-0-0#!**

Josef Klinger, 1980s

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| Mate in 2 |

Oxirgi variatsiyada Oq qal’alar oʻzlarining yangi koʻtarilgan qal’asi bilan qirolni e3 ga, qal’ani e2 ga oʻtkazadilar. Krabbening taʼkidlashicha, oʻsha paytdagi shaxmat qoidalariga koʻra, bu harakat shaxmatchi hali qimirlamaganligi sababli qonuniy boʻlgan va keyinroq FIDE qoidalarga oʻzgartirish kiritib, kasting qalqonsi qirol bilan bir xil *darajani* egallashini talab qilgan. Biroq, aslida, FIDE ning 1930-yildagi asl qonunlarida rok va qirol bir xil darajaga ega boʻlishlari kerak edi.

Uchinchi diagrammada yechim **1 dan boshlanadi. e8=R Kb1 2. 0-0-0-0#** qirolni e3 ga va rookni e2 ga koʻchirish.

Qismning gʻayrioddiy joylashishi yoki harakati
----------------------------------------------

Baʼzi muammolar qismlarni joylashtirishning gʻayrioddiy naqshlari bilan ajralib turadi. Misol uchun, toʻgʻridan-toʻgʻri turmush oʻrtoqlar va ayniqsa yordamchilar harf yoki raqam yoki hatto daraxt shaklida boʻlaklar bilan tuzilgan.

= "Uyga qaytish vazifasi"
-------------------------

M. Kirtley, *The Problemist* 1986

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| Selfmate in 8 |

M. Kirtlining ushbu muammosida yakuniy pozitsiya tanish naqshni aks ettiradi. Krabbé bu muammoni „uyga qaytish vazifasi" deb ataydi, chunki barcha sakkizta oq boʻlak oʻzlarining dastlabki pozitsiyalariga chekinadi. U shunday yozadi: „Strategiya va chuqur mavzular yoʻq, Qora faqat majburiy harakatlarga ega, ammo bu men koʻrgan eng kulgili shaxmat muammolaridan biri". Oq sakkizta harakatda oʻz- oʻzini juftlashi kerak; yaʼni, ular Qoraning irodasiga qarshi Qorani Oq bilan mat qilishga majburlashlari kerak. Yechim quyidagicha:

 **1. Nb1+ Kb3 2. Qd1+ Rc2 3. Bc1 axb6 4. Ra1 b5 5. Rh1 bxc4 6. Ke1 c3 7. Ng1 f3 8. Bf1 f2#**

= "Tırtıl mavzusi"
------------------

G. Bridgewater, *Chess* 1936

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| Mate in 6 |

Krabbé ikki yoki uch marta piyonlar birin-ketin harakatlanadigan muammolar va tadqiqotlar uchun „tırtıl mavzusi" deb nom berdi. Oʻngdagi diagrammada, ayniqsa, ahmoqona misol koʻrsatilgan, Oq oltita harakatda juftlikni majburlaydi. Yechim quyidagicha:

**1.** **Bb1 b2 2.** **Ra2 b3 3.** **Ra3 b4 4.** **Ra4 b5 5.** **Ra5 b6 6.** **Be4#**

Krabbé oʻnga yaqin misollar keltirgan holda, tırtıllar mavzusida butun maqola yozdi.

W. A. Shinkman, 1887

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| Mate in 8 |

Amerikalik bastakor Uilyam A. Shinkman (1847-1933) diagrammadagi masalani a-faylda sekstupli piyonlar bilan tuzganligi bilan mashhur. Krabbé oʻz veb-saytida yozganidek, "Hazilda boʻlishi kerak boʻlgan yechim qiyin emas: **1. 0-0-0 Kxa7 2.** **Rd8 Kxa6 3.** **Rd7 Kxa5 4.** **Rd6 Kxa4 5.** **Rd5 Kxa3 6.** **Rd4 Kxa2 7.** **Rd3 Ka1 8.** **Ra3#** ".

Muammo "cooked "(shaxmat kompozitsiyasining lingosida buzilgan), ammo, chunki 1. Kd2 shuningdek, sakkizta harakatda juftlikni majburlaydi. Ammo bu muammo sakkizta toʻgʻri yoʻldosh boʻlish uchun moʻljallanmagan va buning oʻrniga Blek tomonidan yetti marta ketma-ket ushlash bilan 34 ta harakatda isbotlash uchun moʻljallangan edi.

Koʻproq anʼanaviy shaxmat muammolarida hazil
--------------------------------------------

Hazil baʼzi anʼanaviy mavzularning tarkibiy qismidir, masalan, grotesk va Excelsior .

2004-yilda Hans Böhm hazil-mutoyiba yakuniy tadqiqotlar uchun shaxmat kompozitsiyasi turniriga homiylik qildi. Eng yaxshi ikkita yozuv Krabbé veb-saytida echimlar bilan koʻrinadi.

T. R. Dawson

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| White to play |

Bunday masalalarda, masalan, standart shaxmat muammosi sifatida qoʻyilgan boʻlsa-da, yechimning shaxmat harakatlariga hech qanday aloqasi yoʻq. Masalan, ingliz bastakori Tomas Rayner Douson tomonidan yaratilgan ushbu muammo:

> In the position in the diagram, Black had decided to resign (although it wasn't his turn). White advised his opponent not to give up so quickly. "But I am bound to lose, and there is nothing I—or you, for that matter—can do," said Black. But White insisted: "I'll bet you 1700 that I can lose!" So the two made the bet and White actually lost. He did not resign, lose on time, or anything like that. How did he lose?
> 
> 

Yechim soʻzlarda (yaʼni bu topishmoqdir): Oq shaxmat oʻyinida yutqazmadi (diagramma holatida bu mumkin emas, chunki yagona qonuniy harakatlar 1.g7+ Qxg7+ 2.hxg7#), aksincha tikish.

Yana qarang
-----------

* Sem Loyd

Manbalar
--------

1. ↑  Examples – see items 276, 267, and 265
2. ↑
3. ↑  Open chess diary – see item 289
4. ↑  The article is not available via direct link, but it can be downloaded on the archives at  (krabbe08.pdf).
5. 1 2  „The Kuwait Immortal".  Xs4all.nl. Qaraldi: 7-dekabr 2011-yil.
6. ↑  Open chess diary – see item 281
7. ↑  Open chess diary – see item 276
8. ↑  „ChessBase Chrismas Puzzles 2014 – solutions" (25-yanvar 2015-yil). Qaraldi: 7-sentabr 2015-yil.

1. ↑  When this problem was presented, running out of time always resulted in a loss, regardless of whether or not it would have been possible to be checkmated had the game continued.