Matematikada (chiziqli algebra ), the Faddeev - LeVerrier algoritmi  a rekursiv  ning koeffitsientlarini hisoblash usuli xarakterli polinom                      p         (         λ         )         =         det         (         λ                   Men                       n           −         A         )       { displaystyle p ( lambda) =  det ( lambda I_ {n} -A)}     kvadrat matritsa , A nomi bilan nomlangan Dmitriy Konstantinovich Faddeev  va Urbain Le Verrier . Ushbu polinomni hisoblash natijasida hosil bo'ladi o'zgacha qiymatlar   ning A  uning ildizi sifatida; matritsada matritsali polinom sifatida A  o'zi, u fundamental tomonidan g'oyib bo'ladi Keyli-Gemilton teoremasi . Determinantlarni hisoblash, ammo hisoblashda noqulay, ammo bu samarali algoritm hisoblashda ancha samaraliroq ( NC murakkabligi sinfi ).
Algoritm bir necha marta mustaqil ravishda biron bir shaklda yoki boshqacha tarzda kashf etilgan. Birinchi marta 1840 yilda nashr etilgan Urbain Le Verrier , keyinchalik P. Xorst tomonidan qayta ishlab chiqilgan, Jan-Mari Souriau , hozirgi shaklda bu erda Faddeev va Sominskiy, shuningdek J. S. Frame va boshqalar.[1] [2] [3] [4] [5]   (Tarixiy fikrlar uchun "Uy egasi" ga qarang.[6]    Atrofga o'tish uchun oqlangan yorliq Nyuton polinomlari , Hou tomonidan kiritilgan.[7]   Taqdimotning asosiy qismi Gantmaxer, p. 88.[8]  )
Algoritm  
Maqsad koeffitsientlarni hisoblashdir vk    ning xarakterli polinomining n ×n   matritsa A ,
                    p         (         λ         )         ≡         det         (         λ                   Men                       n           −         A         )         =                   ∑                       k             =             0                        n                     v                       k                     λ                       k                     ,       { displaystyle p ( lambda)  equiv  det ( lambda I_ {n} -A) =  sum _ {k = 0} ^ {n} c_ {k}  lambda ^ {k} ~,}   qaerda, aniq, vn    = 1 va v 0  = (−1)n   det A .
Koeffitsientlar rekursiv tarzda yuqoridan pastga qarab, yordamchi matritsalarning zarbalari bilan aniqlanadi M , 
                                                                                          M                                       0                                                   ≡                 0                                                  v                                       n                                                   =                 1                                                (                 k                 =                 0                 )                                                               M                                       k                                                   ≡                 A                                   M                                       k                     −                     1                   +                                   v                                       n                     −                     k                     +                     1                   Men                                                                                   v                                       n                     −                     k                                                   =                 −                                                       1                     k                                     t                   r                  (                 A                                   M                                       k                   )                                                k                 =                 1                 ,                 …                 ,                 n                                   .           { displaystyle { begin {aligned} M_ {0} &  equiv 0 & c_ {n} & = 1  qquad & (k = 0)  M_ {k} &  equiv AM_ {k-1} + c_ {n -k + 1} I  qquad  qquad & c_ {nk} & = - { frac {1} {k}}  mathrm {tr} (AM_ {k})  qquad & k = 1,  ldots, n ~.  end {hizalangan}}}   Shunday qilib, 
                              M                       1           =         Men                   ,                            v                       n             −             1           =         −                   t           r          A         =         −                   v                       n                     t           r          A         ;       { displaystyle M_ {1} = I ~,  quad c_ {n-1} = -  mathrm {tr} A = -c_ {n}  mathrm {tr} A;}                                 M                       2           =         A         −         Men                   t           r          A         ,                            v                       n             −             2           =         −                               1             2                                 (                     t           r                    A                       2           −         (                   t           r          A                   )                       2                                 )           =         −                               1             2           (                   v                       n                     t           r                    A                       2           +                   v                       n             −             1                     t           r          A         )         ;       { displaystyle M_ {2} = AI  mathrm {tr} A,  quad c_ {n-2} = - { frac {1} {2}} { Bigl (}  mathrm {tr} A ^ {2 } - ( mathrm {tr} A) ^ {2} { Bigr)} = - { frac {1} {2}} (c_ {n}  mathrm {tr} A ^ {2} + c_ {n -1}  mathrm {tr} A);}                                 M                       3           =                   A                       2           −         A                   t           r          A         −                               1             2                                 (                     t           r                    A                       2           −         (                   t           r          A                   )                       2                                 )           Men         ,       { displaystyle M_ {3} = A ^ {2} -A  mathrm {tr} A - { frac {1} {2}} { Bigl (}  mathrm {tr} A ^ {2} - ( mathrm {tr} A) ^ {2} { Bigr)} I,}                                 v                       n             −             3           =         −                                             1               6                                  (           (         tr                  A                   )                       3           −         3         tr                  (                   A                       2           )         (         tr                  A         )         +         2         tr                  (                   A                       3           )                               )           =         −                               1             3           (                   v                       n                     t           r                    A                       3           +                   v                       n             −             1                     t           r                    A                       2           +                   v                       n             −             2                     t           r          A         )         ;       { displaystyle c_ {n-3} = - { tfrac {1} {6}} { Bigl (} ( operatorname {tr} A) ^ {3} -3  operatorname {tr} (A ^ {2) }) ( operator nomi {tr} A) +2  operator nomi {tr} (A ^ {3}) { Bigr)} = = - { frac {1} {3}} (c_ {n}  mathrm {tr } A ^ {3} + c_ {n-1}  mathrm {tr} A ^ {2} + c_ {n-2}  mathrm {tr} A);}   va boshqalar.,[9] [10]     ...; 
                              M                       m           =                   ∑                       k             =             1                        m                     v                       n             −             m             +             k                     A                       k             −             1                     ,       { displaystyle M_ {m} =  sum _ {k = 1} ^ {m} c_ {n-m + k} A ^ {k-1} ~,}                                 v                       n             −             m           =         −                               1             m           (                   v                       n                     t           r                    A                       m           +                   v                       n             −             1                     t           r                    A                       m             −             1           +         .         .         .         +                   v                       n             −             m             +             1                     t           r          A         )         =         −                               1             m                     ∑                       k             =             1                        m                     v                       n             −             m             +             k                     t           r                    A                       k                     ;         .         .         .       { displaystyle c_ {nm} = - { frac {1} {m}} (c_ {n}  mathrm {tr} A ^ {m} + c_ {n-1}  mathrm {tr} A ^ {m -1} + ... + c_ {n-m + 1}  mathrm {tr} A) = - { frac {1} {m}}  sum _ {k = 1} ^ {m} c_ {n -m + k}  mathrm {tr} A ^ {k} ~; ...}   Kuzatib boring A−1  = - Mn  / c0   = (−1)n −1Mn  / detA   da rekursiyani tugatadi  λ . Buning teskari yoki determinantini olish uchun foydalanish mumkin A .
Hosil qilish  
Dalil rejimlariga asoslanadi yordamchi matritsa , Bk  ≡ Mn − k   , duch kelgan yordamchi matritsalar. Ushbu matritsa tomonidan belgilanadi 
                    (         λ         Men         −         A         )         B         =         Men                   p         (         λ         )       { displaystyle ( lambda I-A) B = I ~ p ( lambda)}   va shu bilan mutanosib hal qiluvchi 
                    B         =         (         λ         Men         −         A                   )                       −             1           Men                   p         (         λ         )                   .       { displaystyle B = ( lambda I-A) ^ {- 1} I ~ p ( lambda) ~.}   Bu, shubhasiz, matritsa polinomidir λ   daraja n-1  . Shunday qilib,
                    B         ≡                   ∑                       k             =             0                        n             −             1                     λ                       k                               B                       k           =                   ∑                       k             =             0                        n                     λ                       k                               M                       n             −             k           ,       { displaystyle B  equiv  sum _ {k = 0} ^ {n-1}  lambda ^ {k} ~ B_ {k} =  sum _ {k = 0} ^ {n}  lambda ^ {k} ~ M_ {nk},}   bu erda zararsizni aniqlash mumkin M 0  ≡0.
Yuqoridagi yordamchi uchun aniq polinom shakllarini aniqlovchi tenglamaga kiritish, 
                              ∑                       k             =             0                        n                     λ                       k             +             1                     M                       n             −             k           −                   λ                       k           (         A                   M                       n             −             k           +                   v                       k           Men         )         =         0                   .       { displaystyle  sum _ {k = 0} ^ {n}  lambda ^ {k + 1} M_ {nk} -  lambda ^ {k} (AM_ {nk} + c_ {k} I) = 0 ~. }   Endi, eng yuqori tartibda, birinchi muddat yo'qoladi M 0  = 0; pastki tartibda esa (doimiy ichida λ , yordamchining aniqlovchi tenglamasidan, yuqorida),
                              M                       n           A         =                   B                       0           A         =                   v                       0                     ,       { displaystyle M_ {n} A = B_ {0} A = c_ {0} ~,}   shuning uchun birinchi davrning qo'pol indekslari o'zgarishi hosil beradi 
                              ∑                       k             =             1                        n                     λ                       k                                 (                     M                       1             +             n             −             k           −         A                   M                       n             −             k           +                   v                       k           Men                               )           =         0                   ,       { displaystyle  sum _ {k = 1} ^ {n}  lambda ^ {k} { Big (} M_ {1 + nk} -AM_ {nk} + c_ {k} I { Big)} = 0 ~,}   bu esa rekursiyani belgilaydi
                    ∴                            M                       m           =         A                   M                       m             −             1           +                   v                       n             −             m             +             1           Men                   ,       { displaystyle  Shuning uchun  qquad M_ {m} = AM_ {m-1} + c_ {n-m + 1} I ~,}   uchun m =1,...,n . Ko'tarilayotgan indeksning kuchi bo'yicha kamayib borishiga e'tibor bering λ , lekin polinom koeffitsientlari v  jihatidan hali aniqlanmagan M s va A .
Bunga quyidagi yordamchi tenglama orqali erishish mumkin (Hou, 1998),
                    λ                                             ∂               p               (               λ               )                            ∂               λ            −         n         p         =         tr                  A         B                   .       { displaystyle  lambda { frac { qismli p ( lambda)} { qismli  lambda}} - np =  operator nomi {tr} AB ~.}   Bu faqat uchun belgilovchi tenglamaning izidir B  tomonidan Jakobining formulasi ,
                                                        ∂               p               (               λ               )                            ∂               λ            =         p         (         λ         )                   ∑                       m             =             0                        ∞                     λ                       −             (             m             +             1             )           tr                            A                       m           =         p         (         λ         )                   tr                                        Men                           λ               Men               −               A            ≡         tr                  B                   .       { displaystyle { frac { qismli p ( lambda)} { qismli  lambda}} = p ( lambda)  sum _ {m = 0} ^ { infty}  lambda ^ {- (m + 1 )}  operator nomi {tr} A ^ {m} = p ( lambda) ~  operator nomi {tr} { frac {I} { lambda IA}}  equiv  operator nomi {tr} B ~.}   Ushbu yordamchi tenglamada polinom rejimini qo'shish hosil bo'ladi
                              ∑                       k             =             1                        n                     λ                       k                                 (           k                   v                       k           −         n                   v                       k           −         tr                  A                   M                       n             −             k                                 )           =         0                   ,       { displaystyle  sum _ {k = 1} ^ {n}  lambda ^ {k} { Big (} kc_ {k} -nc_ {k} -  operatorname {tr} AM_ {nk} { Big)} = 0 ~,}   Shuning uchun; ... uchun; ... natijasida 
                              ∑                       m             =             1                        n             −             1                     λ                       n             −             m                                 (           m                   v                       n             −             m           +         tr                  A                   M                       m                                 )           =         0                   ,       { displaystyle  sum _ {m = 1} ^ {n-1}  lambda ^ {nm} { Big (} mc_ {nm} +  operator nomi {tr} AM_ {m} { Big)} = 0 ~ ,}   va nihoyat 
                    ∴                            v                       n             −             m           =         −                               1             m           tr                  A                   M                       m                     .       { displaystyle  Shuning uchun  qquad c_ {n-m} = - { frac {1} {m}}  operatorname {tr} AM_ {m} ~.}   Bu avvalgi qismning kamayish kuchlari bo'yicha takrorlanishini yakunlaydi λ .
Algoritmda qo'shimcha ravishda to'g'ridan-to'g'ri, 
                              M                       m           =         A                   M                       m             −             1           −                               1                           m               −               1            (         tr                  A                   M                       m             −             1           )         Men                   ,       { displaystyle M_ {m} = AM_ {m-1} - { frac {1} {m-1}} ( operatorname {tr} AM_ {m-1}) I ~,}   va bilan kelishilgan holda Keyli-Gemilton teoremasi ,
                    adj                  (         A         )         =         (         −                   )                       n             −             1                     M                       n           =         (         −                   )                       n             −             1           (                   A                       n             −             1           +                   v                       n             −             1                     A                       n             −             2           +         .         .         .         +                   v                       2           A         +                   v                       1           Men         )         =         (         −                   )                       n             −             1                     ∑                       k             =             1                        n                     v                       k                     A                       k             −             1                     .       { displaystyle  operator nomi {adj} (A) = (-) ^ {n-1} M_ {n} = (-) ^ {n-1} (A ^ {n-1} + c_ {n-1} A ^ {n-2} + ... + c_ {2} A + c_ {1} I) = (-) ^ {n-1}  sum _ {k = 1} ^ {n} c_ {k} A ^ {k-1} ~.}   
Yakuniy echim to'liq eksponensial jihatdan qulayroq ifodalanishi mumkin Qo'ng'iroq polinomlari  kabi 
                              v                       n             −             k           =                                             (               −               1                               )                                   n                   −                   k                              k               !                                                B                         k                                 (           tr                  A         ,         −         1         !                   tr                            A                       2           ,         2         !                   tr                            A                       3           ,         …         ,         (         −         1                   )                       k             −             1           (         k         −         1         )         !                   tr                            A                       k                                 )           .       { displaystyle c_ {nk} = { frac {(-1) ^ {nk}} {k!}} { mathcal {B}} _ {k} { Bigl (}  operator nomi {tr} A, - 1! ~  Operator nomi {tr} A ^ {2}, 2! ~  Operator nomi {tr} A ^ {3},  ldots, (- 1) ^ {k-1} (k-1)! ~  Operator nomi {tr} A ^ {k} { Bigr)}.}   Misol  
                                          A             =                           [                                                                                           3                                            1                                            5                                                               3                                            3                                            1                                                               4                                            6                                            4                   ]          { displaystyle { displaystyle A =  chap [{ begin {array} {rrr} 3 & 1 & 5  3 & 3 & 1  4 & 6 & 4  end {array}}  o'ng]}}                                                                                                                                       M                                               0                                                               =                                           [                                                                                                                                   0                                                            0                                                            0                                                                                       0                                                            0                                                            0                                                                                       0                                                            0                                                            0                           ]                                                                                                                         v                                               3                                                                                                                                           =                                        1                                                                               M                                                                                                     1                                                                 =                                           [                                                                                                                                   1                                                            0                                                            0                                                                                       0                                                            1                                                            0                                                                                       0                                                            0                                                            1                           ]                                         A                                                                 M                                               1                                                               =                                           [                                                                                                                                                                                                     3                                                              1                                                            5                                                                                       3                                                                                                                              3                                                              1                                                                                       4                                                            6                                                                                                                              4                             ]                                                               v                                               2                                                                                                     =                     −                                                                   1                                                                               1                                                                       10                                                                                  =                                        −                     10                                                                               M                                                                                                     2                                                                 =                                           [                                                                                                                                   −                               7                                                            1                                                            5                                                                                       3                                                            −                               7                                                            1                                                                                       4                                                            6                                                            −                               6                           ]                                                             A                                                                 M                                               2                                                               =                                           [                                                                                                                                                                                                     2                                                              26                                                            −                               14                                                                                       −                               8                                                                                                                              −                                   12                                                              12                                                                                       6                                                            −                               14                                                                                                                              2                             ]                                                                                   v                                               1                                                                                                     =                     −                                                                   1                                                                               2                                                                       (                         −                         8                         )                                                                                  =                                        4                                                                               M                                                                                                     3                                                                 =                                           [                                                                                                                                   6                                                            26                                                            −                               14                                                                                       −                               8                                                            −                               8                                                            12                                                                                       6                                                            −                               14                                                            6                           ]                                                             A                                                                 M                                               3                                                               =                                           [                                                                                                                                                                                                     40                                                              0                                                            0                                                                                       0                                                                                                                              40                                                              0                                                                                       0                                                            0                                                                                                                              40                             ]                                                                                   v                                               0                                                                                                     =                     −                                                                   1                                                                               3                                                                       120                                                                                  =                                        −                     40             { displaystyle { displaystyle { begin {aligned} M_ {0} & =  left [{ begin {array} {rrr} 0 & 0 & 0  0 & 0 & 0  0 & 0 & 0  end {array}}  right]  quad &&& c_ { 3} &&&&& = & 1  M _ { mathbf { color {blue} 1}} & =  left [{ begin {array} {rrr} 1 & 0 & 0  0 & 1 & 0  0 & 0 & 1  end {array}}  right] & A ~ M_ {1} & =  chap [{ begin {array} {rrr}  mathbf { color {red} 3} & 1 & 5  3 &  mathbf { color {red} 3} & 1  4 & 6 &  mathbf { color {red} 4}  end {array}}  right] & c_ {2} &&& = - { frac {1} { mathbf { color {blue} 1}}}  mathbf { color {red } 10} && = & - 10  M _ { mathbf { color {blue} 2}} & =  left [{ begin {array} {rrr} -7 & 1 & 5  3 & -7 & 1  4 & 6 & -6  end {array}}  right]  qquad & A ~ M_ {2} & =  chap [{ begin {array} {rrr}  mathbf { color {red} 2} & 26 & -14  - 8 &  mathbf {  color {red} -12} & 12  6 & -14 &  mathbf { color {red} 2}  end {array}}  right]  qquad & c_ {1} &&& = - { frac {1} { mathbf { color {blue} 2}}}  mathbf { color {red} (- 8)} && = & 4  M _ { mathbf { color {blue} 3}} & =  left [{ begin {array} {rrr} 6 & 26 & -14  - 8 & -8 & 12  6 & -14 & 6  end {array}}  right]  qquad & A ~ M_ {3} & =  left [{ begin {array} {rrr }  mathbf { color {red} 40} & 0 & 0  0 &  mathbf { color {re d} 40} & 0  0 & 0 &  mathbf { color {red} 40}  end {array}}  right]  qquad & c_ {0} &&& = - { frac {1} { mathbf { color {blue } 3}}}  mathbf { color {red} 120} && = & - 40  end {hizalanmış}}}}   
Bundan tashqari,                                                         M                               4               =             A                                         M                               3               +                           v                               0                             Men             =             0         { displaystyle { displaystyle M_ {4} = A ~ M_ {3} + c_ {0} ~ I = 0}}    , bu yuqoridagi hisob-kitoblarni tasdiqlaydi.
Matritsaning xarakterli polinomiyasi A  shunday                                                         p                               A               (             λ             )             =                           λ                               3               −             10                           λ                               2               +             4             λ             −             40         { displaystyle { displaystyle p_ {A} ( lambda) =  lambda ^ {3} -10  lambda ^ {2} +4  lambda -40}}    ; ning determinanti A  bu                                           det             (             A             )             =             (             −             1                           )                               3                             v                               0               =             40         { displaystyle { displaystyle  det (A) = (- 1) ^ {3} c_ {0} = 40}}    ; iz 10 = -v 2 ; va teskari A  bu 
                                                        A                               −                 1               =             −                                           1                                   v                                       0                                             M                               3               =                                           1                 40                             [                                                                                           6                                            26                                            −                       14                                                               −                       8                                            −                       8                                            12                                                               6                                            −                       14                                            6                   ]              =                           [                                                                                           0                                               .                        15                                            0                                               .                        65                                            −                       0                                               .                        35                                                               −                       0                                               .                        20                                            −                       0                                               .                        20                                            0                                               .                        30                                                               0                                               .                        15                                            −                       0                                               .                        35                                            0                                               .                        15                   ]          { displaystyle { displaystyle A ^ {- 1} = - { frac {1} {c_ {0}}} ~ M_ {3} = { frac {1} {40}}  chap [{ begin { massiv}} {rrr} 6 & 26 & -14  - 8 & -8 & 12  6 & -14 & 6  end {array}}  right] =  left [{ begin {array} {rrr} 0 {.} 15 & 0 {.} 65 & -0 {.} 35  - 0 {.} 20 & -0 {.} 20 va 0 {.} 30  0 {.} 15 & -0 {.} 35 & 0 {.} 15  end {array}}  o'ng] }}    .Ekvivalent, ammo aniq ifoda  
An ning ixcham determinanti m ×m Yuqoridagi Jakobi formulasi uchun matritsali eritma koeffitsientlarni muqobil ravishda belgilashi mumkin v ,[11] [12] 
                              v                       n             −             m           =                                             (               −               1                               )                                   m                              m               !                                  |                                                             tr                                      A                                    m                   −                   1                                    0                                    ⋯                                                   tr                                                          A                                           2                                      tr                                      A                                    m                   −                   2                                    ⋯                                                   ⋮                                    ⋮                                                                      ⋮                                                   tr                                                          A                                           m                       −                       1                                      tr                                                          A                                           m                       −                       2                                      ⋯                                    ⋯                                    1                                                   tr                                                          A                                           m                                      tr                                                          A                                           m                       −                       1                                      ⋯                                    ⋯                                    tr                                      A                |                     .       { displaystyle c_ {nm} = { frac {(-1) ^ {m}} {m!}} { begin {vmatrix}  operatorname {tr} A & m-1 & 0 &  cdots  operatorname {tr} A ^ {2} &  operatorname {tr} A & m-2 &  cdots  vdots &  vdots &&&  vdots  operatorname {tr} A ^ {m-1} &  operatorname {tr} A ^ {m- 2} &  cdots &  cdots & 1  operatorname {tr} A ^ {m} &  operatorname {tr} A ^ {m-1} &  cdots &  cdots &  operatorname {tr} A  end { vmatrix}} ~.}   Shuningdek qarang  
Adabiyotlar  
^   Urbain Le Verrier : Sur les variations séculaires des éléments des orbites pour les sept planètes principales. , J. de Matematik.  (1) 5 , 230 (1840), Onlayn ^   Pol Xorst: Xarakteristik tenglamaning koeffitsientlarini aniqlash usuli . Ann. Matematika. Stat.  6  83-84 (1935), doi :10.1214 / aoms / 1177732612  ^   Jan-Mari Souriau , Une méthode pour la décomposition spectrale et l'inversion des matrices , Comptes Rend.   227 , 1010-1011 (1948).^   D. K. Faddeev va I. S. Sominskiy, Sbornik zadatch po vyshej algebra  (Oliy algebra masalalari , Mir nashriyotlari, 1972), Moskov-Leningrad (1949). Muammo 979 . ^   J. S. ramkasi: Matritsani teskari aylantirish uchun oddiy rekursiya formulasi (mavhum) , Buqa. Am. Matematika. Soc.  55  1045 (1949), doi :10.1090 / S0002-9904-1949-09310-2  ^   Uy egasi, Alston S.  (2006). Raqamli analizda matritsalar nazariyasi . Matematikadan Dover kitoblari. ISBN  0486449726  .CS1 maint: ref = harv (havola ) ^   Hou, S. H. (1998). "Sinf uchun eslatma: Leverrierning oddiy isboti - Faddeevning xarakterli polinom algoritmi"  SIAM sharhi   40(3)  706-709, doi :10.1137 / S003614459732076X  . ^   Gantmaxer, F.R. (1960). Matritsalar nazariyasi . Nyu-York: Chelsi nashriyoti. ISBN  0-8218-1376-5  . CS1 maint: ref = harv (havola ) ^   Zadeh, Lotfi A. va Desoer, Charlz A. (1963, 2008). Lineer tizim nazariyasi: Davlat kosmik yondashuvi  (Mc Graw-Hill; Dover Fuqarolik va mashinasozlik) ISBN  9780486466637  , 303-305 betlar;  ^   Abdelxouid, Jounaidi va Lombardi, Anri (2004). Méthodes matricielles - Kirish à la complexité algébrique , (Mathématiques et Applications, 42) Springer, ISBN  3540202471  . ^   Brown, Lowell S. (1994). Kvant maydoni nazariyasi , Kembrij universiteti matbuoti. ISBN  978-0-521-46946-3 , p. 54; Shuningdek qarang: Curtright, T. L., Fairlie, D. B. va Alshal, H. (2012). "Galiley astarlari", arXiv: 1212.6972, 3-qism. ^   Rid, M.; Simon, B. (1978). Zamonaviy matematik fizika metodikasi . Vol. 4 Operatorlar tahlili. AQSh: ACADEMIC PRESS, Inc. 323–333, 340, 343-betlar. ISBN  0-12-585004-2  .