Wang Xiaotong on right
triangles: Forthcoming in East Asian
Science, Technology, and Medicine. Tina Su-lyn Lim 林淑铃 Donald B. Wagner 华道安 4 May 2012
Tina Su-lyn Lim is a project manager with NNIT (Novo Nordisk Information Technology) in Copenhagen, Denmark. This article formed part of her M.Sc. thesis in Mathematics at the University of Copenhagen, 2006. Donald
B. Wagner has taught
at the University of Copenhagen, the University of
Victoria (British Columbia), and the Technical
University of Berlin. He has written widely on the
history of science and technology in China; his most
recent book is the volume on Ferrous Metallurgy in
Joseph Needham’s Science and Civilisation in China. Abstract: Wang Xiaotong’s Jigu suanjing 緝古算經 is primarily concerned with problems in
solid and plane geometry leading to polynomial
equations which are to be solved numerically using a
procedure similar to Horner’s Method. We translate and
analyze here six problems in plane geometry. In each
case the solution is derived using a dissection of a
3-dimensional object. We suggest an interpretation of
one fragmentary comment which at first sight appears
to refer to a dissection of a 4-dimensional object. • Wang Xiaotong (late 6th–7th century
AD, exact dates unknown) served the Sui and Tang
dynasties in posts concerned with calendrical
calculations, and presented his book, Jigu suanshu 緝 古算術, ‘Continuation of ancient mathematics’, to the Imperial
court at some time after AD 626. In 656 it was made
one of ten official ‘canons’ (jing 經) for mathematical education, and was
retitled Jigu
suanjing.[2] The book contains 20 problems: one
astronomical problem, then 13 on solid geometry, then
six on right triangles. All but the first provide
extensions of the methods in the mathematical classic
Jiuzhang
suanshu 九章算術 (‘Arithmetic in nine
chapters’, perhaps
1st century AD): the solid-geometry methods in Chapter
5 and the right-triangle methods in Chapter 9,[3] thus ‘continuing’ ancient
mathematics. All but the first require the extraction
of a root of a cubic or (in two cases) quadratic
equation. The present article is concerned with
problems 15–20, on right triangles. Each problem describes a geometric
situation, states some given quantities, and asks for
one or more other quantities. Then the answer is
given, and finally an algorithm for computing the
answer. Printed in smaller characters in the text are
comments which generally give explanations of the
algorithms of the main text. All commentators appear
to agree that the comments are by Wang Xiaotong
himself, but we have noticed some differences in
terminology between the comments and the main text,[4] and feel therefore that the question
of the authorship of the comments should be left open.
Perhaps most but not all of the comments are by him. The text
The history of the text is discussed
by the modern editors, Qian Baocong (1963: 490–491)
and Guo Shuchun and Liu Dun (1998: 1: 21–22); see also
He Shaogeng (1989). All extant editions of Jigu suanjing
go back to an edition printed in the early 13th
century AD. One hand-copy of this edition survived to
the 20th century; Qian Baocong appears to have seen
it.[5] It is
now lost, but in 1684 a copy was included in a
collectaneum, Jiguge
congshu 汲古閣叢書. This version,
the oldest surviving text of the Jigu suanjing,
is now available in facsimile on the World Wide Web.[6] The best Qing
critical edition is that of Li Huang (1832);[7] others are by
Dai Zhen (1777), Bao Tingbo (1780), Zhang Dunren
(1803), and the Korean mathematician Nam Pyŏng-Gil (1820–1869).[8] All of these
are available on the Web. The standard modern edition has long been
Qian Baocong’s (1963, 2: 487–527; important
corrections, 1966), but that of Guo Shuchun and Liu
Dun (1998) has much to recommend it. In the parts of
the text treated in the present article there is no
important difference between the two. The 1684 edition is marred by numerous
obvious scribal errors which must be corrected by
reference to the mathematical context. Guo and Liu
(1998, 1: 22) note that Li Huang introduced ca. 700
emendations to the text, and that Qian Baocong
followed most of these but introduced 20 new
emendations. The part of the text considered here
gives special difficulties, for it appears that the
last few pages of the Southern Song hand copy were
damaged, and a great many characters are missing. The
problems involved in reconstructing the text are
considered further below. • The Jigu suanjing
has not been much studied in modern times. The two
critical editions, already mentioned, do not
explicitly comment on the mathematical content. Lin
Yanquan (2001) translates the text into modern
Chinese, expresses the calculations in modern
notation, and gives derivations of some of the
formulas. Deeper studies of individual parts of the
text are by Shen Kangshen (1964), Qian Baocong (1966),
He Shaogeng (1989), Wang Rongbin (1990), Guo Shiying
(1994), and Andrea Bréard (1999: 95–99, 333–336,
353–356). The derivations of Problems 15 and 17
(Figures 2 and 3–4 below) have been reconstructed by
Lin Yanquan, by He Shaogeng, and by Wang Rongbin; our
reconstruction of the derivation of Problem 19 (Figure
5) is new. The comments in the text give special
difficulties, for two reasons. Being written in
smaller characters, they are more subject to banal
scribal errors, and their content is more abstract and
complex than the main text. They have hardly been
studied at all by modern scholars: the most recent
serious study we have found is that of Luo Tengfeng
(1770–1841) [1993].[9] Solving cubic equations by ‘Horner’s
method’ In the problem solutions given by Wang Xiaotong the coefficients of a cubic equation are calculated, after which the reader is instructed to ‘extract the cube root’. Wang Xiaotong gives no indication of how this was done, but it was obviously a well-known procedure, for algorithms are described in detail in the Jiuzhang suanshu and the Zhang Qiujian suanjing 張 邱建算經 (5th century AD) for the special case of extracting a cube root (the cubic equation x3 = A), and extensions of the latter algorithm to general polynomials of all orders are given in several Chinese books from after Wang Xiaotong’s time.[10] So much has been written about Chinese methods for extracting the roots of polynomial equations that it is not necessary to go into detail here. Let it suffice here to say that the algorithm is equivalent to ‘Horner’s method’ as sometimes taught in modern schools and colleges.[11] The first digit of the root is determined, the roots of the equation are reduced by the value of that digit (an operation which amounts to a change of variable), the next digit is determined, and this step is repeated until the desired precision is reached. The operation is carried out on paper in the Western version, in the Chinese version with ‘calculating rods’ laid out on a table. Wang Xiaotong’s text describes how each of the elements of the cubic equation is calculated, using the following terminology: shi 實, the constant term fangfa 方 法, the linear coefficient lianfa 廉 法, the quadratic coefficient The cubic coefficient is always 1,
and is never mentioned.[12] The triangle problems The Chinese text of problems 15–20 and our translation are given in the final section of this article. Here we give an overview using modern terminology. In pre-modern Chinese mathematics the sides of a right triangle are referred to as gou 句, ‘shorter leg’, gu 股, ‘longer leg’, and xian 弦, ‘hypotenuse’. We translate these terms as ‘base’, ‘leg’, and ‘hypotenuse’, and in equations denote them as a, b, and c respectively; see Figure 1. A peculiarity of the six triangle
problems is that no units are given for the quantities
involved. This is virtually unique in all of
pre-modern Chinese mathematics (including Wang
Xiaotong’s Problems 1–14), in which problems are
normally stated with reference to practical
situations. In Problem 15
the given quantities are: ab = 7061/50
c–a = 369/10 and the values of a, b, and c are
required. In the solution the coefficients of a cubic
equation are calculated:
This has one real root, a = 147/20 whereafter it is straightforward to calculate the remaining quantities: b = c = a + (c–a) = 511/4 A comment, printed in smaller
characters, explains (1) in a mixture of algebraic and
geometric reasoning. It first notes that (ab)2 = a2b2
The comment refers to a fang 方 (rectangular
parallelipiped) and to quantities
‘lined up’; this suggests a geometric rather than
algebraic derivation of the method. We reconstruct
this derivation as in Figure 2. We note that
(which Wang Xiaotong does not state explicitly). Then using Figure 2,
It is not known whether Wang Xiaotong’s book
originally included illustrations like Figure 2. We
think it probably did not; he seems simply to give
verbal descriptions of the geometric constructions
which he uses. Problem 16 is equivalent to Problem 15. The given quantities are ab = 40361/5 c–b = 61/5 and c is required. The cubic equation here is
This has one real root, b = 1081/2 whence c = b + (c–b) = 1147/10 Problem 17 gives the quantities ac = 13371/20 c–b = 11/10 and b is required. The cubic equation arrived at is
This has one real root, b = 922/5 A comment explains (3) geometrically, but here our textual problems start, for a large number of characters are missing in the extant editions. In spite of the lacunae in the comment text it is clear that the geometrical construction is similar to that shown in Figures 3 and 4. The solid in Figure 3 has volume which is the right side of (3). The sum of the volumes of the blocks into which the solid is divided (Figure 4) is which is the left
side of (3). • Note that the dimensions of the triangles in Problems 15–17 are based on to Pythagorean triples: 15.
[147/20, 491/5,
511/4] = [7, 24, 25] × 41/20 16. [371/5, 1081/2, 1147/10] = [12, 35, 37] × 31/10 (4) 17. [143/10, 922/5, 931/2] = [13, 84, 85] × 11/10 In Problems 18–20
so much is missing from the text that any
reconstruction will to some extent be speculative. The
Chinese mathematician Zhang Dunren 張敦
仁 (1754-1834) and the Korean mathematician Nam
Pyŏng-Gil 南秉吉
(1820–1869) have given two reconstructions which agree
in principle but differ in detail. Neither explains
how he arrived at his reconstruction, but some matters
seem clear. Any
reconstruction should satisfy the approximate number
of characters seen to be missing from the text.[13] In
addition, certain assumptions can safely be made: (1)
The problems all concern right triangles; (2) the
dimensions of the triangles are derived from
Pythagorean triples, as in Problems 15–17; (3) the
problem statements all follow the same rigid pattern
as in Problems 15–17; and (4) the problems come in
pairs, so that, just as 15 and 16 are equivalent, so
are the pairs 17–18 and 19–20. Assumption
(4) implies that the fragmentary problem statements
are 17.
Given ac
and c–b, determine
b. 18.
Given bc
and c–a, [determine
either a or
b]. 19.
Given bc
[and a],
determine b. 20.
Given [ac
and] b,
[determine a]. What
remains of the ‘method’ for Problem 18 indicates that
the quantity to be determined is b. Zhang Dunren and Nam Pyŏng-Gil
agree on the form of the problem statements as given
here, but they differ on the numerical values of the
given quantities in Problems 18 and 19. Their
reconstructions are as follows:[14]
• The Pythagorean triples on which these
reconstructions are based are: 18. [121/10, 66, 671/10] = [11, 60, 61] × 11/10 [153/10, 68, 697/10] = [9, 40, 41] × 17/10 19 [7/100, 6/25, 1/4] = [7, 24, 25] × 1/100 [77/10, 262/5, 271/2] = [7, 24, 25] × 11/10 20. [84/5, 161/2, 187/10] = [8, 15, 17] × 11/10 • In the following we have arbitrarily chosen to follow Zhang Dunren’s reconstructions of the given quantities. Zhang Dunren also attempts – with much less success – to reconstruct large parts of the ‘method’ sections, and we have not followed him in the translation. Nam Pyŏng-Gil’s edition does not include Wang Xiaotong’s methods, but instead gives his own, which appear to follow the methods of Li Ye 李冶 (1192–1279) in Ceyuan haijing 測圓海 鏡.[15] Problem 18 is, by assumption (4) above, equivalent to Problem 17. It results in the cubic equation
(cf. (3) above). In Zhang Dunren’s reconstruction,
This has one real root, a = 153/10 And
Problems 19 and 20 are extremely fragmentary, but the assumption that they are equivalent makes Zhang Dunren’s and Nam Pyŏng-Gil’s reconstructions of the problems and their methods plausible. The method for Problem 19 leads in each case to the quadratic equation
which in Zhang Dunren’s version gives x2 + 5929/100 x = 52706 where x = b2. This has one positive root. b = 262/5 It is easy to derive (5) algebraically, for but what little remains of the comment appears to imply a geometric explanation. Commentators have had difficulty here, for a geometric explanation would seem to involve four-dimensional quantities. The lack of units in the quantities involved, however, makes the construction shown in Figure 5 plausible, in which both b and b2 are used as linear measures. The volume of the solid is b2c2 = (bc)2 which is the right side of (5), and the sum of the two blocks into which the solid is divided is
which is the left side of (5). If our reconstruction is correct, the ‘length’ mentioned in one of the remaining fragments of the comment is probably x, the length of the solid. Problem 20, assumed to be equivalent to Problem 19, results in the quadratic equation where x = a2. In Zhang Dunren’s reconstrruction, x2 + 2721/4 x = 27079621/625 This has one positive root, a = 84/5 Concluding remarksHow did Wang Xiaotong arrive at the results in his book? The question is probably unanswerable, but some guesses may be permissible. We presume that he started with one or more general methods and derived a number of concrete geometric constructions which could solved by their use. Our study of several of his problems in solid geometry[16] suggests that his most important method was dissections of 3-dimensional objects leading to (in modern terminology) cubic equations. (We hope that further research will lead to a more specific description.) Our guess is then that Wang Xiaotong’s study of the possibilities of this method led, in addition to the problems in solid geometry, to the three dissections related to right triangles reconstructed here. Having found a relation between the dimensions of a right triangle and a cubic equation, Wang Xiaotong then needed only to fit a specific right triangle to it. In each case he used a Pythagorean triple, making the student’s work more difficult by multiplying by a fraction. A general method for generating Pythagorean triples seems to have been known to Chinese mathematicians since the Jiuzhang suanshu,[17] but the three triples known to have been used by Wang Xiaotong (see (4) above) all have c–b equal to 1 or 2. They could therefore be generated using the simpler special cases ascribed by Proclus to Pythagoras and Plato respectively:[18]
We confidently expect that careful study of the solid-geometry problems in the book will reveal further examples of disguised Pythagorean triples, and that these will provide useful clues to Wang Xiaotong’s general methods. • A further question is the function of this book in mathematical education. Though we know that it was adopted as one of the textbooks of Imperial mathematical education, we do not know whether it was originally written for this purpose, or perhaps was intended as a tour de force in pure mathematics. The main governmental uses of advanced mathematics – beyond accounting and the like – were in astronomy and in public works. It is difficult to see any use for these geometrical methods in astronomy, but in public works a thorough understanding of dissection methods (not necessarily the specific problems given in the book) might have been very useful to an official calculating the volumes of earthworks and their labour requirements. Chinese text and translation In the Chinese text we indicate text variants with the notation {a/b/c}, where a is the 1684 version, b is Qian Baocong’s version, which is usually the same as Zhang Dunren’s, and c, if present, is from some other version, indicated in a footnote. We use Qian Baocong’s punctuation. Our translations of the smaller-character comments in the Chinese text are printed in a sans-serif font and with a sideline. Our mathematical comments are given indented in the text, while a few philological comments are given in footnotes. Problem 15 假令有句股相乘冪七百六 、五十分之一,弦多於句三十六、十分之九。問三事各多少? 答曰: 句十四、二十分之七, 股四十九、五分之一, 弦五十一、四分之一。 術曰:冪自乘,倍多數而一,為實。半多{ /
數為}廉法,從。開立方除之,即句。以弦多{ /數 加之},即弦。以句
除冪,即股。句股相乘冪自{ /乘,即}句冪乘股冪{ /之積。故}以倍句弦差而一,得一句與半差{ /相連,乘/再乘得[19]}句冪為方。故半差為廉{/
法},從,開立方除之。
[In a right triangle] the area obtained by multiplying the base [a] by the leg [b] is [ab =] 7061/50 and the hypotenuse [c] is greater than the base [a] by [c–a =] 369/10. How large are the three quantities? Answer: base, [a =] 147/20 leg, [b =] 491/5 hypotenuse, [c =] 511/4 Method: Multiply the area [ab] by itself and divide by twice the difference [2(c–a)] to make the shi [the constant term]. Halve the difference [c–a] to make the lianfa [the quadratic coefficient]. Extract the cube root; this is the base [a].
Add the hypotenuse difference [c–a]; this is the hypotenuse [c]. Divide the product [ab] by the base [a]; this is the leg [b]. c = a + (c – a)
[Comment:] The product of the base and the leg, multiplied by itself [(ab)2], is the ‘volume’ [ji 積] obtained by multiplying the area of [the square on] the base by the area of [the square on] the leg [a2b2]. (ab)2 = a2b2 In the following see Figure 2. Therefore dividing by the doubled difference between the base and the hypotenuse [2(c – a)] gives the base and the halved difference lined up [a + (c–a)/2] multiplied by the area of [the square on] the base [a2] to make a box [fang 方, rectangular parallelipiped].
This is why the difference is halved to make the lianfa, and the extraction of the cube root is carried out. Problem 16 假令有句股相乘冪四千三十六、五分之{ /一,股}少
於弦六、五分之 一。問弦多少? 答曰:弦一百一十四、十分之七。 術曰:冪自乘,倍少數而
一,為實。半少為廉法,從。開立方除之,即股。加差,即弦。
[In a right triangle], the area obtained by multiplying the base [a] by the leg [b] is [ab=] 40361/5. The leg is less than the hypotenuse [c] by [c–b =] 61/5. How large is the hypotenuse? Answer: The hypotenuse is [c=] 1147/10. This problem is equivalent to Problem 15. Method: Multiply the area [ab] by itself and divide by twice the difference [c–b] to make the shi [the constant term]. Halve the difference [c–b] to make the lianfa [the quadratic coefficient]. Extract the cube root; this is the leg [b].
Add the difference [c–b] to obtain the hypotenuse [c]. c = b + (c–b) = 1147/10 Problem 17 假令有句弦相乘冪一千三百三十七、二十分之 一,弦多於股一、 十分之一。問股多少? 答曰:九十二、五分之二。 術曰:冪自乘,倍多而一,為立冪。又多再{ /自}乘,半之,減立冪,餘為實。又多數自乘,{ /倍之, }為方法。又置多數,五之,二而一,為廉{ /法,從}。開立方除之,即股。句弦相乘冪自{ /乘,即句}冪乘弦冪之{ /積。故以倍}股弦差而一,得一股與{半差 /□□□□□半差}為方今多再自乘半之為隅{ /□□□□□}橫虛二立廉{ /□□□□□□□□□□□}倍之為從隅{ /□□□□□□□□□□□}多為上廉即二多{ //□□□□□□□□□[20]}法故五之二而一{ /□□□} [In a right
triangle] the area obtained by multiplying the base [a] by the
hypotenuse [c]
is [ac =]
13371/20, and the hypotenuse is
greater than the leg by [c–b =] 11/10.
How large is the leg [b]? Answer: [b=] 922/5. Method: Multiply
the area [ac]
by itself and divide by twice the difference [c–b] to make a
volume Further multiply the difference [c–b] by itself and double it to make the fangfa [the linear coefficient]. Further lay out the difference [c–b], multiply by 5, and divide by 2 to make the lianfa [the quadratic coefficient]. Extract the cube root. This is the leg [b].
[Comment:] Multiplying the area [ac] obtained by multiplying the base by the hypotenuse by itself gives the ‘volume’ [ji 積] obtained by multiplying the area of [the square on] the base [a2] by the area of [the square on] the hypotenuse [c2]. (ac)2 = a2c2 From here on the text is very fragmentary. Therefore
dividing by twice the difference [c-b] gives
one leg [b]
and . . . Problem 18 假令有股弦相乘冪{ /四千七百三十九、五分之/四千四百二十八五分之}三,句少於弦五十{ /四、五分之二。問股多少?/五問股多少} 答曰:六{ /十八。/十六} 術曰:冪自乘,{ /倍少數而一,為立冪。又少數}再自乘,半之,以{ /減立冪,餘為實。又少數自}乘,倍之,為方法。{ /又置少數,五之,二而一,為}廉法,從。開立方{ /除之,即句。加差即弦。弦除}冪,即股。 [In a right
triangle] the area obtained by multiplying the leg [b] by the
hypotenuse [c]
is [bc =]
. . . Problem 19 假令有股弦相乘冪{ /七百二十六,句七、十分之/五十分之三句一百分之}七。
問股多少? 答曰:股二十{ /六、五分之二。/五分之
六} 術曰:冪自{ /
乘,為實。句自乘,為方法,從。開方}除 之,所得,{ /又開方,即股}{/□□□□□□□□□□□□□□}數亦是股{ /□□□□□□□□□□□□}為長以股{ /□□□□□□□□□□□□}得股冪又開{ /。。。}股北分母常{ /。。。} [In a right
triangle] the area obtained by multiplying the leg [b] by the
hypotenuse [c]
is [bc =]
. . . [Comment:]
[characters missing] . . . the number is
also the leg [b]
. . . [characters missing] . . . to obtain the area of [the square on] the leg [b]. Extract the . . . root . . . [characters
missing] . . . gu bei fen mu
chang 股
北分母常
[?] Problem 20 假令有股十六、二分{ /
之一,句弦相乘冪 一百六}十四、二十五分{ /之十四。問句多少?} 答曰:{ /
句八、五分之四。/八五分之四} 術曰:冪自乘{ /
為實。股自乘,為 方法,從。開方}除之,所得,又開方{ /即句。}
[In a right
triangle] the leg [b] is 16 and
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[1] We are grateful to Karine Chemla for
drawing our attention to Wang Xiaotong’s book, and
for detailed comments on this article; to Jesper
Lützen and Ivan Tafteberg for comments on an
earlier version; and to Mark Elvin, Jia-ming
Ying 英家 銘, and
four anonymous reviewers for additional comments. [2] Jiu Tang
shu 1975, 32: 1168, 47: 2039, 4: 76,
44: 1892; Qian Baocong 1963: 487. [3] Chemla and
Guo 2004: 387–457, 661–745; see also Cullen 1993;
Shen Kangshen et al. 1999: 250–306, 439–517. [4] E.g. in the
commentary to problem 3, Qian Baocong 1963: 506,
lines 5–8. Here bienao 鼈
臑 is written in full, while it is always
abbreviated as bie in
the main text; yanchu 羡除 is once written in full and twice
abbreviated as chu,
while in the main text it is always abbreviated as
yan. [5] Qian
Baocong describes the physical condition of the
manuscript, and refers to it throughout his
critical edition. He Shaogeng (1989: 37) states
that the manuscript is ‘now’ in the Palace Museum
in Beijing, but he does not appear to have seen it
himself. Guo Shuchun and Liu Dun state that it is
now lost. [6] This is the
Tianlu
Linlang congshu 天祿琳瑯叢書 edition, a facsimile printed in 1931,
available at http://www.scribd.com/doc/78515037. [7] On Li
Huang’s edition see Huang Juncai 2008. [8] On Nam Pyŏng-Gil see Jia-ming Ying 2011;
Horng Wann-sheng 2002; Zhang Fukai 2005. [9] The only
exceptions to this statement appear to be the
highly speculative reconstructions of the
fragmentary commentary to Problem 17 by He
Shaogeng (1989) and Wang Rongbin (1990). [10] Chemla
& Guo 2004: 322–335, 363–379; Shen et al.
1999: 175–195, 204–226; Wang & Needham 1955;
Lam Lay Yong 1970; 1977: 195–196, 251–285; 1986;
Libbrecht 1973: 175–191; Chemla 1994; Martzloff
1997: 221–249. [11] See e.g.
Rees & Sparks 1967: 294–297 as well as
numerous pages on the World Wide Web. Horner
(1819) presented a procedure for approximating
roots of any infinitely differentiable function,
but modern descriptions of ‘Horner’s method’
consider only the special case of polynomial
functions. [12] Commentators are divided on the function of the word cong / zong 從in the phrase
從開立方除. Qian Baocong (1966: 46–47) reviews the
various attempts to explain it, and comes to the
conclusion that in each case it belongs at the end
of the previous sentence, and means ‘follow,
accompany’ (gensui
跟随). This is difficult for us to understand,
and we omit the word in our translation. Another
interpretation is given by Chemla and Guo (2004:
65, 912). [13] Assuming
that the Tianlu
Linlang congshu edition used the same page
and line lengths as the Southern Song edition.
This assumption is not certain, but the counts of
missing characters may be taken as approximations. [14] In all but one case the two editors
insert the same number of characters counted by
the Zhibuzuzhai
edition. In giving the value of bc in
Problem 18, both insert 10 characters where the Zhibuzhai
edition indicates only 9 missing. [15] See e.g. Mei Rongzhao 1966; Chemla
1982; Martzloff 1997: 143–149; Zhang Fukai 2005. [16] These are
Problems 2, 3, 4, 5, and 7, each of which consists
of several related but distinct problems. Our
interpretation of the four problems of Problem 2
is given in a forthcoming article, ‘The Grand
Astrologer’s platform and ramp’, at present under
review by another journal. [17] Chemla and
Guo (2004: 727), under the interpretation of van
der Waerden (1983: 5–7). [18] Heath 1921,
1: 80–81. [19] Zhang Dunren 1803. [20] Guo Shuchun & Liu Dun 1998, 2: 18. |