Jagannatha Svami Nayanapathagami Bhavatu Me
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  Jagannatha, Lord of the universe, in His Form in the great 10th century temple at Puri, Orissa, India

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  A Grammar of the Tihai
Acharya Dr Chintamani Rath
(Essay written in August 1989, Appendices in March 2008)
 



The object of this essay is to establish certain basic relationships connecting different time variables that naturally occur in the course of constructing a Tihai. Although there is some literature already published on Tihai, no attempt has so far been made, to my knowledge, to arrive at a perfectly general theory of the Tihai, such as a theory that can explain, in perfectly abstract algebraic terms, the construction of any kind of Tihai within the framework of any Tala. This essay seeks to fill this gap. It is submitted that a practical student working diligently through the method outlined in this "Grammar" will, in a short time, gain a confident command over improvising Tihai in any Tala. Of course, as in the case of most other areas of practical music making, once the student masters the craft of Tihai construction, he will transcend a mere "Grammar" of this nature.

From the general principles herein outlined, the student may easily extend the numerical examples given in this essay to cover a wide variety of particular cases. He should also note that this essay limits itself to the severely rhythmic analysis of the Tihai. No attempt has been made to translate, either generally or through specific examples, the principles herein discussed into melodic lines. The task of creating the melodic phrase to be used in the Tihai, depending upon the Raga being presented at the moment, is left to the student. This is the reason why the musical examples given in this essay use the Bol of the Tabla, although this essay is intended for all students of music in general and for the instrumentalist in particular.

This essay is divided into the following sections (click on section heading to jump to the section):-

  1. Nature of a Tihai
  2. Aesthetic Foundation of a Tihai
  3. Components of a Tihai
  4. Arithmetic Relationship Linking the Components of a Tihai:
    1. General Relationship
    2. Particular Relationship
      1. Case I – Remainder 2
      2. Case II – Remainder 1
      3. Case III – Remainder 0
    3. Cross Rhythm
  5. Relation Between Tala and Tihai
  6. Tihai Ending on the Sum of any Tala
  7. Conclusion
  8. Appendix I: Selected Tihai for use in Vilambit Teentala
  9. Appendix II: Selected Tihai for use in Maddhya Laya and Drut Teentala, Jhaptala, Roopakatala, Ektala and Adachoutala
For the sake of convenience, I have used the following symbols in this essay to denote the following Bol of the Tabla, used in the musical examples in this essay:-

SymbolBol
DDha
D!Dhin
DtDhit
dDi
TTi
NNa
tTa
T!Thum
t!Tin
TtktTirakita or Titakita
GGhi
gga
KKat or Kit
kkra

Further, a group of underlined Bol indicates that those Bol must be executed exactly in the space of one Matra or beat. And finally, the symbol " - " denotes the continuation of the preceding Bol and the symbol " x " denotes a Viram or rest.


A. Nature of a Tihai



A Tihai (literally, "three times") is a set of three successively executed (played, sung or danced) musical phrases, usually of similar rhythmic pattern and equal duration in time, designed to end on a predetermined point in the Tala cycle and having the effect of concluding the passage which immediately precedes it. The most usual point of such ending isn the Sum itself. Sometimes, the Tihai is made to end on a point just preceding the Matra from which the Mukhda starts.

As an example, consider the following composition for the Tabla, set to Teentala:-

Ex. 1:-

+
Dt  Dt  TtKt  Dt   |  TtKt  Dt  Dg  Tt  |  Tg  tt  kD  Tt  |  Dg  Tt  kD  -N  |
Dt   -t   Dt t   T!   |  NK    -t   D    -   | T!  NK  -t   D   |    -   T!  NK  -t  |  D


The phrases in yellow, green and yellow are identical in form and content. The set of these three phrases ends on the Sum. The effect of playing these phrases at the end of the composition is to round it off satisfactorily. This set of three phrases is a "Tihai".

The rendering of Tihai is an important skill in the craft of the performing musician, whether vocalist (in particular, a Dhrupad and Dhamar singer), instrumentalist, percussionist or dancer. Indeed, "Tihai ability" is even one indicator of both virtuosity and musicianship (although, like any other artistic tool, it can be overdone to the extent of detracting from the overall aesthetic appeal of the performance). A highly skilled performer is so developed in his sense of Laya (i.e., tempo) and Tala that he is able to fashion and execute perfect Tihai on the spur of the moment in a performance. Needless to say, such ability is largely the result of long periods of diligent, methodical and painstaking practice. For the student, of course, it is essential that a systametic "theory" or "grammar" of Tihai is available, which must be so diligently mastered by him that he in fact makes it his own and ultimately, with the flowering of the "artiste" in him, is able to effectively transcend the mundane level of grammatical calculations.

The purpose of a Tihai is, of course, to round off a passage of music. The passage itself may be in the form of a Vistara or a Tana / Toda. The ways in which a Tihai may be fashioned are limitless, subject only to the skill and musical sensitivity of the performer. For instance, one way might be to repeat the final phrase of a passage twice, so that the phrase is executed three times in all. Such a length of the phrase is chosen for repetition as will enable the conclusion of the Tihai at the Matra desired, A typical example of such a manoeuvre, to conclude on the Sum, in Teentala, might be as follows:

Ex. 2:-

+
D!N  D!D!  Nt!  Nt!  |  t!N  Tt  Kt  Kt  |   D!N  D!D!  Nt!  Nt!  |  t!N  Tt  Kt  Kt  |  
D!N  D!D!  Nt!  Nt!  |  t!N  Tt  Kt  Kt  |   D!N  D!D!  Nt!  Nt!  |  t!N  Tt  Kt  Kt  |  
D!N  D!D!  Nt!  Nt!  |  t!N  Tt  Kt  Kt  |   D       Tt     Kt    Kt  |  D     Tt  Kt  Kt  |  D


In Ex. 2, the passage is a syncopated one, being a Jhaptala phrase (10 beats, thus: D! N | D! D! N | T N | D! D! N played in the example in double time) used in conjunction with the phrase Tt Kt Kt (in blue). It is this last phrase of three Matra (the first yellow coloured phrase in line 3) that is the concluding phrase of the preceding passage. This phrase is repeated twice – once (the first repitition) typed in green and the second time (the second repition) typed in yellow again. The Tihai is thus complete.

Sometimes, however, the Tihai may be in the Alap, especially in the Jod portion or at the conclusion of the Jhala at the end of the Alap. Since such Tihai have no relation to any particular Tala, it is not necessary to think of them with the same rigour (rhythmically speaking) as is necessary in the case of Tihai that must be executed within a Tala framework. This is obvious, since the Jod and Jhala portions of an instrumental Alap (or in the Nibaddha portion of a vocal Alap), beats proceed along a "linear", and not a "cyclic", pattern. That is to say, the concept of a Sum is not incorporated into the Alap. Hence, it will not be improper to assert that the "grammar" of the Tihai, as outlined in this essay, need not apply to Tihai used in the Sum-less Alap. In those occasional moments when a Sum is implied in the Alap (usually in its concluding moments), the principles discussed herein may, of course, be applied with advantage.

[For a fuller discussion of the Alap and its structure, click here and here]


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B. Aesthetic Foundation of a Tihai



There can be no doubt that, in its proper context, a well-constructed and accurately rendered Tihai is extremely effective in enhancing the musical experience of both performer and listener. Such a Tihai rarely fails to draw appreciation and applause from the audience. A concert in which a Dhrupadiya, instrumentalist, percussionist or dancer does not present a single Tihai would tend to be rather monotonous and even insipid, unless the other contents of the concert and their presentation are so arresting as to compensate for the lack of Tihai. Why should this be so? What is it in a Tihai that makes it so effective as a musical event?

First, a Tihai is sequential. That is, the three phrases in it are constructed in the same rhythmic pattern. What is more, in the majority of cases, the three phrases are in fact identical to one another in every respect. Such repetition of phrases prepares the listener to the fact that the passage to which he has been listening is drawing to a close. That a repetition heralds a conclusion of a passage is an accepted practice in India, and has been so for millennia. In fact, this custom is not restricted to music alone. For example, in chanting Sanskrt Texts, especially religious ones, the last stanza, couplet, line or phrase often repeated to indicate than the end has been reached.

An interesting question in this regard is: "speaking of Tihai, why is the repetition always twice, and not more nor less?" The answer is that we are conditioned so since Vedic times (approx. 6000 BCE); however, it is beyond the scope of this essay to explore this area. Suffice it to recall that very often, the initial "Aum Vishnu" or "Namo Vishnu" at the start of almost all religious rites is chanted thrice and so is the concluding "Aum Shantihi"...

Thus, since the listener knows from the Tihai that the passage or episode is closing, such closing does not appear abrupt, and the conclusion is an expected, identified and on that account a happy one.

Second, a Tihai first creates, then resolves, rhythmic tension. Much of the beauty in music is the creation of musical tension followed by its skilful resolution. This includes melodic, harmonic and rhythmic tension. Since the Tihai has to do primarily with rhythm, it is sufficient to consider here only rhythmic tension.

It is commonly accepted that in triple time, the first beat is a strong beat, the other two being weak. Likewise, in quadruple time, the first and third beats are strong, though the third is not as much so as the first. In executing many Tihai, the weak beat often gets accented. This creates a rhythmic tension, because the accent shifts in this case from its natural place (i.e., a strong beat) to an "artificial" place (i.e., a weak beat). The tension is finally resolved in, say, the return to the regular or natural accents of the Tala and the final culmination of the Sum, which is a very strong beat indeed. For example, a standard Tihai in Teentala goes like this:

Ex. 3:-

+
DgTt  TgTt  T!NKt  TtKt  |  DgTt  D-Dg  TtTg  TtT!N  |  KtTt  KtDg  TtD-  DgTt  |  TgTt  T!NKt  TtKt  DgTt  |  D-


Here, the strong stroke D- (6th beat from Sum), which ends the first phrase of the Tihai is on the second beat (a weak beat) of the bar. The next strong stroke of the Tihai, i.e. the D- (in yellow), which ends the second phrase (shown in yellow) of the Tihai falls after the third beat but before the fourth beat of the third bar, owing to the syncopated nature of the second phrase. Here is the rhythmic tension. The resolution is effected in the third phrase of the Tihai (shown once again in orange, after the yellow). The third phrase happily coincides its accents with the basis beats of the Tala and also ends on the Sum.

Third, a Tihai changes the periodicity of the foregoing passage. If the preceding passage is in the form of a Vistara, without a fixed periodicity of beats, the Tihai introduces a periodicity or at least an approximation of periodicity by intimately linking the melodic development with the Matra of the Tala. If, however, the preceding passage is in the form of a Tana/Toda, where there is an identifiable periodicity of beats, the Tihai varies the measure of the periodicity. In either case, the monotony of a single rhythmic pattern is broken, thereby creating a new interest in the music.

For example, in Ex. 2 above, the passage before the Tihai has a regular periodicity of beats, thus — five beats (divided into two half-beats followed by three half-beats followed again by two and three further half-beats respectively), then three beats, then five beats again, then three beats again, and so on (assuming the first two lines are repeated in an episodic Tana/Toda improvisation for several Avartan (cycles of the Tala). It is in the Tihai that a sudden change occurs in the periodicity of the passage — it now becomes thus: three beats, one stroke [a 'stroke' is defined in section C of this essay] (lasting one beat), again three beats followed by a stroke and finally three beats leading to the conclusion of the Tihai on the Sum. Such variation in a previously established periodicity keeps the music "moving", thus sustaining its interest.

Finally, a Tihai is one indicator of the personal skill of the musician. There is surely an element of virtuositic display in a Tihai. While too many Tihai will undoubtedly smack of rank exhibitionism, there is no doubt that some amount of aplomb and showmanship is expected from a concert musician. To bring off a difficult Tihai perfectly is in itself a sign of considerable skill, and there is no reason why a musician sitting (or moving, in the case of a dancer) on the concert stage should not show off this ability, should he possess it and be in a mood for it at the time. Indeed, today, many audiences actually expect this from the performer. A well-fashioned Tihai can involve the audience to the point of nail-biting, edge-of-the-seat tension, and its final resolution on the Sum can bring the recital to a point of tremendous climax, heightening the musical experience for performer and listener alike.


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C. Components of a Tihai



Consider the following Tihai :—

Ex. 4(a):-
Tt   Kt   D   X   Tt   Kt   D   X   Tt   Kt   |   D


Ex. 4(b):-
DT!   -N   Tt   D   -D   -D   D   X   X   DT!   -N   Tt   D   -D   -D   D   X   X   DT!   -N   Tt   D   -D   -D   |   D


Ex. 4(c):-
gd   GN   Dg   dG   N D   gd   GN   |   D


In each example, suppose that the Tihai ends on the Sum of an arbitrary Tala. The Tala itself is left undefined, because this section will consider a Tihai per se, without reference to a Tala framework. How a Tihai may be fit into the cyclic pattern of a Tala will be discussed in section E.

In Ex. 4(a), the total numner of beats, not counting the concluding beat of the Tihai which coincides with the Sum, needed for the Tihai is 10. In Ex. 4(b), it is 24 and in Ex. 4(c) it is 7. In other words, it may be said that the three Tihai are of 10, 24 and 7 Matra duration (or beats long) respectively.

To generalise, Tihai is said to be of M Matra if it spans M beats of the Tala to complete all its three phrases, without taking into account its last beat or Matra which will coincide with the point in the Tala cycle where it (i.e., the Tihai) is meant to end. If the Tihai starts on the first beat or Matra, and is M Matra long, this ending beat sounds on the (M + 1)th Matra

In Ex. 4(a), the total of 10 Matra may be subdivided thus:—

(i) the phrase Tt   Kt comprising 2 Matra
(ii) an accented beat D comprising 1 Matra
(iii) a pause or rest X comprising 1 Matra
(iv) the phrase Tt   Kt comprising 2 Matra
(v) an accented beat D comprising 1 Matra
(vi) a pause or rest X comprising 1 Matra
(vii) the phrase Tt   Kt comprising 2 Matra
Total 10 Matra


At this juncture, it is necessary to introduce three defining words to signify three components of every Tihai. I introduce the following three Sanskrt words with the following three English equivalents. Henceforth in this essay (which is after all written in English), I will use the English equivalent of each term, leaving the Sanskrt words to be used in Indian language essays. The words I propose to use are:—


Sanskrt English
Shareera Body
Aghata Stroke
Virama Gap


In examples Ex. 4(a), Ex. 4(b) and Ex. 4(c), the phrases in blue, yellow and green may each be called a Shareera or "Body" of the Tihai. Thus in example Ex. 4(a), each "Tt  Kt" is a body of the Tihai. Every Tihai has 3 bodies.

In examples Ex. 4(a), Ex. 4(b) and Ex. 4(c), the accented beats D and D may each be called an Aghata or "Stroke" of the Tihai. The third accent which ends the Tihai is actually outside the length of the Tihai, coinciding as it does with the Matra (which may be the Sum or any other predetermined Matra of the Tala) to which the Tihai is intended to lead. In this sense, every Tihai has 2 strokes.

In examples Ex. 4(a), Ex. 4(b) and Ex. 4(c), the pauses X and X may each be called a Virama or "Gap" of the Tihai. Every Tihai has 2 gaps; the value of a gap may be any number of Matra, including 0 (zero).

Definitions:

The "stroke" of a Tihai is the accent at the end of each of the first two phrases of the Tihai. The accent or stroke at the end of the third phrase, coinciding with the Sum or other predetermined point, is outside the Matra count of the Tihai and so is not relevant for the present.

The "body" of a Tihai is the musical event (melodic/rhythmic figure or phrase) occurring from the start of the Tihai to the first stroke. The body occurs thrice in the course of a Tihai.

The "gap" in a Tihai is the rest, silence or pause occurring after a stroke and before the commencement of the subsequent body. A Tihai may or may not have a gap (i.e., the value of a gap may be zero); however, if there is are gaps, the duration of the gaps must be identical.

Thus, in Ex. 4(b):—

The bodies
DT!   -N   Tt   D   -D   -D,
DT!   -N   Tt   D   -D   -D and
DT!   -N   Tt   D   -D   -D
each being of 6 Matra, equal 6 x 3 =
18 Matra
The strokes D and D are each of 1 Matra, together making 1 x 2 = 2 Matra
The gaps X   X and X   X are each of 2 Matra, totalling 2 x 2 = 4 Matra
Total 24 Matra


Similarly, in Ex. 4(c):—

The bodies
gd  GN, gd  GN and gd  GN
each being of 2 Matra, equal 2 x 3 =
6 Matra
The strokes D and D are each of ½ Matra, together making ½ x 2 = 1 Matra
There are no gaps in this Tihai = 0 Matra
Total 7 Matra



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D. Arithmetic Relationship Linking the Components of a Tihai:

1. General Relationship



Since, by definition, in a Tihai of M Matra:—
  1. the body occurs thrice,
  2. the stroke and the gap occur twice each,
  3. the three bodies are of equal duration of, say, B Matra length each,
  4. the two strokes are of equal duration of, say, S Matra length each and
  5. the two gaps are of equal duration of, say, G Matra length each,
the following relation always holds —

M = 3B + 2S + 2G


This relationship may be re-written as —

M = 3B + 2(S + G)


This is the basic Tihai equation – the relationship of the component parts of the Tihai to be found in every Tihai (except in certain special forms of the Tihai that are not considered here for the moment. An indication of some of these special forms may be found in the concluding section of this essay). This form of the equation shows that S and G can be made inversely proportional to each other, without altering the numerical value of M: if S is made small, G must be made large and if S is made large, i.e., if the note comprising the stroke is held for a long duration, G must be made that much shorter. To illustrate, the Tihai of 10 Matra in Ex. 4(a) may be executed in any of the following ways:—

Ex. 5(a). S = ½, G = 1½ :—    Tt   Kt   DX   X   Tt   Kt   DX   X   Tt   Kt   |   D

Ex. 5(b). S = 1, G = 1 :—        Tt   Kt   D   X   Tt   Kt   D   X   Tt   Kt   |   D

Ex. 5(c). S = 1½, G = ½ :—    Tt   Kt   D   -X   Tt   Kt   D   -X   Tt   Kt   |   D

Ex. 5(d) S = 2, G = 0 :—         Tt   Kt   D   -   Tt   Kt   D   -   Tt   Kt   |   D

If the smallest rhythmic unit is taken to be a quarter-beat instead of a half-beat as implied in Ex. 5, several more combinations of S and G can be worked out. If G is relatively large, it is a common practice to play a short melodic phrase or figure of a contrasting nature to fill up the silent time covered by G.

It is obvious that for practical purposes, units of the duration of less than one beat should be taken as the basic unit for the purpose of counting time only if the tempo of the music is sufficiently slow. For fast tempi, the basic unit for counting time may well be several beats grouped into one unit. For the present purpose, the smallest unit of counting time will always be taken to be half a beat. This will lend itself to be applied to medium and slow (but not too slow!) speeds. For very slow speeds, and also for very fast speeds, other units may be chosen, but the principles on which this analysis of Tihai rests will obviously remain the same. It is left to the reader to adapt these principles to his requirements.

Since the speed of the music has been assumed to be so slow (or so fast) that half a Matra can be conveniently counted (this is medium tempo or Madhya Laya), the value of S may be taken to be ½. Now, the basic relation in a Tihai is, as stated earlier, given by the equation

M = 3B + 2S + 2G

Putting S = ½, the equation becomes

M = 3B + (2 x 1) + 2G

or,
M = 3B + 2G + 1

or,
(M − 1) = 3B + 2G

If G is taken to be the dependant variable, the equation takes the form

2G = (M − 1) − 3B

These equations may be applied in practice in the following manner :—

First, determine the value of M. This may be any arbitrary number that the performer chooses at will. Thus, he may wish to use a Tihai of 14 Matra, in which case M takes the value 14. Or he may wish to conclude a passage with a Tihai of 17½ Matra; here, M = 17½, and so on. In every case, however, as the smallest unit of countable time has been assumed to be half a Matra, the value of M will be in the form (P + Q), where P is an integer and Q = ½, or 0.

The second step is to choose, arbitrarily, any value of B from the arithmetic series ½, 1, 1½, 2, 2½, ..., so however that 3B always remnains less than (M − 1).

Next, the value of 2G must be found from the equation

2G = (M − 1) − 3B

for the value of B so chosen. If 2G is a whole number, even or odd, G will be expressible in terms of half a beat, i.e., the value of G will be in the form (P' + Q'), where P' is an integer and Q' = ½, or 0. It is clearly seen from the equation that if M is chosen to be (P1 + Q1), where P1 is an integer and Q1 is ½, or 0, and if B is chosen to be (P2 + Q2), where P2 is an integer and Q2 is ½, or 0, 2G will be a whole number if and only if Q1 and Q2 are at once both ½ or both 0.

If 2G is not a whole number, the value of G will be such as to make a conscious counting ot the Tihai impractical, since then G cannot be expressed in terms of half a beat, which is the beat previously assumed to be the smallest unit of time that can be conveniently counted consciously. In such a case, the previously chosen value of B must be discarded, a fresh value taken and the whole exercise undergone again. However, if care is taken in complying with the condition stated in the previous paragraph, viz., that Q1 and Q2 must be both ½ or both 0, such a necessity will not arise.

A practical example will make this process clearer. Suppose that the performer wishes to deliver a Tihai of 11½ Matra. Here, M = 11½. Hence, (M − 1) − (11½ − 1) = 10½. Now, choose a B from the series ½, 1, 1½, 2, 2½, ..., such that 3B < (M − 1). Let B be chosen to be 2½. Then, 3B = 3 x 2½ = 7½. This gives 2G =(M − 1) − 3B = 10½ − 7½ = 3. Hence, G = 3 ÷ 2 = 1½, and the calculation is complete. A Tihai of 11½ Matra may be obtained by rendering a body of 2½ Matra. sounding a stroke of ½ Matra and observing a gap of 1½ Matra.

Now suppose that instead of choosing B = 2½, the value B = 3 is taken. This gives 3B = 9. So 2G becomes (M − 1) − 3B = 10½ − 9 = 1½. In this case, G = ¾, i.e., it is not expressed in terms of half a Matra, and so cannot be counted in terms of the assumption that ½ beat is the minimum countable unit of time. Thus, B = 3 is not a suitable choice and must be discarded. Instead, if B is taken to be 3½, 3B becomes 10½, so that 2G = (M − 1) − 3B = 10½ − 10½ = 0, giving G = 0. Hence, a Tihai of 11½ Matra may also be executed with a body of 3½ Matra and a stroke of ½ Matra. No gap must be allowed.

As shown above, the relation

M = 3B + 2S + 2G

becomes

M = 3B + 2G + 1

Two variants of this equation, useful in practice, have already been stated :—

(M − 1) = 3B + 2G    ..   ..   ..   .. (1)

and
2G = (M − 1) − 3B    ..    ..   ..   .. (2)


Other useful variants are also possible. Thus :—

M = 3B + 2S + 2G

or,

2G = M − 3B − 1

or,

G = ½{M − (3B + 1)}    ..    ..   ..   .. (3)


Alternatively,

M = 3B + 2S + 2G

or,

3B = (M − 1) − 2G

or,

B = ⅓{(M − 1) − 2G}    ..    ..   ..   .. (4)


The performer may determine any variable (B, G or M) first and thereafter proceed to determine the other variables. For example, if a particular body of, say, 5 Matra comes to the mind of the performer and if he wishes to keep a gap of ½ Matra in his Tihai, the problem before him will be to find the total numnet of Matra that such a Tihai will span. He may find this by using the relation

M = 3B + 2S + 2G

or,

M = (3 x 5) + (2 x ½) + 1

or,

M = 15 + 1 + 1

or,

M = 17 Matra


Again, in some cases, the performer may desire to cover 10 Matra by using bodies of such a length that there is no gap. Here, M = 10 and G = 0. B must be determined. This can be done from variant (4) of the basic equation, stated above —

B = ⅓{(M − 1) − 2G}

or,

B = ⅓{(10 − 1) − 2 x 0}

or,

B = ⅓ x 9 = 3 Matra

Hence a body of 3 Matra, with a stroke of ½ Matra and no gap will result in the desired Tihai of 10 Matra.

If the tempo of the music is so fast that half a Matra cannot be conveniently counted and the basic unit of countable time becomes 1 Matra, the analysis given above will remain true, with the difference that S must now be made equal to 1. In this case, the basic relation

M = 3B + 2S + 2G

now takes the form of the following particular variants :—

M = 3B + 2S + 2G    ..    ..   ..   .. (i)

(M − 2) = 3B + 2G    ..    ..   ..   .. (ii)

2G = (M − 2) − 3B    ..    ..   ..   .. (iii)

G = ½{M − (3B +2)}    ..    ..   ..   .. (iv)

B = ⅓{(M − 2) − 2G}    ..    ..   ..   .. (v)



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D. Arithmetic Relationship Linking the Components of a Tihai (continued):

2. Particular Relationship



For the particular case where an integral M (i.e., where M is a whole number and not a mixed number comprising a number of whole Matra and a fraction of another Matra) is determined first, the following process will assist in determining B, G and S quickly.

For any such M, the quantity ⅓M will have an integer U as quotient and another value V as remainder. Obviously, in each case, V can have only one of three possible values — 2, 1 or 0. Thus, any integral M may be written in the form

M = 3U + V, where V = either 2 or 1 or 0


Case I ⇒ where V = 2.

Here, M = 3U + 2.

Comparing this equation with the basic equation of a Tihai, viz.,

M = 3B + 2S + 2G

or,

M = 3B + 2(S + G),

it is possible to set

3U = 3B

and

2(S + G) = 2.

[M = 3U + 2 — the starting point
M = 3B + 2(S + G) — the basic Tihai equation
3U + 2 = 3B + 2(S + G)
i.e., 3U = 3B and 2 = 2(S + G) are both at once true.]


It is clear from this that B = U and S + G = 1. That is to say, if S = 1, G = 0 and if S = ½, G = ½.

Thus, if the total number of desired Matra M of a Tihai, when divided by 3, leaves U as quotient and V as remainder, the Tihai may be formed by a body of of U Matra each, a stroke of 1 Matra and no gap. Some examples are given in the following table:—


for a desired M of quotient in (M ÷ 3) is B must be of S must be of G must be of
5 Matra 1 1 Matra 1 Matra 0 Matra
8 Matra 2 2 Matra 1 Matra 0 Matra
11 Matra 3 3 Matra 1 Matra 0 Matra
14 Matra 4 4 Matra 1 Matra 0 Matra
17 Matra 5 5 Matra 1 Matra 0 Matra
20 Matra 6 6 Matra 1 Matra 0 Matra
23 Matra 7 7 Matra 1 Matra 0 Matra
etc. etc. etc. etc. etc.



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Case II ⇒ where V = 1.

Here, M = 3U + 1.

Two sub-cases may be considered:—

Sub-case A: Where the speed of the music is sufficiently slow to make it possible for the performer to conveniently count half a Matra.

In this sub-case, the relation

M = 3U + 1

when compared with the basic Tihai equation

M = 3B + 2S + 2G

gives

3B = 3U

whence,

B = U and 2(S +G) = 1

i.e., S+G = ½, so that S = ½ and G = 0 (because a Tihai cannot not have a stroke!)

Thus, for a desired value of M such that U becomes the quotient in ⅓M and 1 the remainder, each body must comprise U Matra, followed by a stroke of ½ Matra and no gap. The following table gives some illustrations:—


for a desired M of quotient in (M ÷ 3) is B must be of S must be of G must be of
4 Matra 1 1 Matra ½ Matra 0 Matra
7 Matra 2 2 Matra ½ Matra 0 Matra
10 Matra 3 3 Matra ½ Matra 0 Matra
13 Matra 4 4 Matra ½ Matra 0 Matra
16 Matra 5 5 Matra ½ Matra 0 Matra
19 Matra 6 6 Matra ½ Matra 0 Matra
22 Matra 7 7 Matra ½ Matra 0 Matra
etc. etc. etc. etc. etc.



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Sub-case B: Where the speed of the music is such that the minimum time unit that can be conveniently counted is 1 Matra.

In this sub-case, B cannot equal U, for that (i.e., to be equal to U) would go against the basic assumption of this sub-case.

[Proof of B ≠ U:—
Assume B = U (if possible).
Then, substituting the value of B in the basic Tihai equation
M = 3B + 2(S + G),
we get:
M = 3U + 2(S + G) ... ... ... (A)
But since V = 1 for this sub-case,
M = 3U + 1
∴ eqn (A) now becomes
3U + 1 = 3U + 2(S + G)
or,
2(S + G) = 1
or,
S + G = ½ Matra
But, in this sub-case, ½ Matra cannot be counted.
Hence B ≠ U. (proved)]

Since B ≠ U, B < U. Now, the closest that B can approach U is up to (B − 1), for 1 Matra is the minimum conveniently countable unit of time in this sub-case. Taking B = (U − 1), S and G can be determined from the basic Tihai equation

M = 3B + 2(S + G)

Since B = (U − 1) and since M = 3U + 1, the equation becomes

3U + 1 = 3(U − 1) + 2(S + G)

or,
3U + 1 =3U − 3 + 2(S + G)

or,
2(S + G) = 4

or,
S + G = 2

It is convenient to take S = 1 and G = 1.

Thus, B = (U − 1), S = 1 and G = 1 are the parameters for constructing Tihai under this sub-case.

A few examples are given in the following table:—


for a desired M of quotient in (M ÷ 3) is B must be of S must be of G must be of
4 Matra 1 0 Matra 1 Matra 1 Matra
7 Matra 2 1 Matra 1 Matra 1 Matra
10 Matra 3 2 Matra 1 Matra 1 Matra
13 Matra 3 4 Matra 1 Matra 1 Matra
16 Matra 5 4 Matra 1 Matra 1 Matra
19 Matra 6 6 Matra 1 Matra 1 Matra
22 Matra 6 7 Matra 1 Matra 1 Matra
etc. etc. etc. etc. etc.



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Case III ⇒ where V = 0.

Here, M = 3U.

As in Case II above, so too here: two sub-cases may be considered :—
Sub-case A: Where the speed of the music is sufficiently slow to enable the performer to conveniently count half a Matra.

Since
M = 3B + 2(S + G),

and
S + G > 0,

obviously,
B > U, (as M = 3U in this sub-case).

This is obvious, for if m = U, there will be no time duration left, after 3B has been performed, to perform S, even assuming G = 0.

Since the minimum time unit that can be conveniently counted (in this sub-case) is ½ Matra, B may conceivably be taken to equal (U − ½). But this leads to difficulties, for, following the logic outlined in sub-case B of Case II above,

M = 3B + 2(S + G),

or,
3U = 3(U − ½) + 2(S + G),

or,
3U = 3U − (3 x ½) + 2(S + G),

or,
2(S + G) = 1½,

or,
S + G = ¾ Matra,

which involves the counting of ¼ Matra, an impossibility according to the assumption in this sub-case. Hence, B = (U − ½) cannot be a solution.

The next alternative is to consider B = U − 1). This leads to a fruitful conclusion, as follows:—

M = 3B + 2(S + G),

or,
3U = 3(U − 1) + 2(S + G),

or,
3U = 3U − 3 + 2(S + G),

or,
2(S + G) = 3,
i.e.,
S + G = 1½ Matra.

One easy possibility is to take S = 1, G = ½. Then the conditions for constructing a Tihai where ⅓M does not leave any remainder and ½ Matra can be conveniently counted are —

B = (U − 1), S = 1, G = ½.

The following table gives some examples:—


for a desired M of quotient in (M ÷ 3) is B must be of S must be of G must be of
3 Matra 1 0 Matra 1 Matra ½ Matra
6 Matra 2 1 Matra 1 Matra ½ Matra
9 Matra 3 2 Matra 1 Matra ½ Matra
12 Matra 4 3 Matra 1 Matra ½ Matra
15 Matra 5 4 Matra 1 Matra ½ Matra
18 Matra 6 5 Matra 1 Matra ½ Matra
21 Matra 7 6 Matra 1 Matra ½ Matra
etc. etc. etc. etc. etc.



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Sub-case B: Where the speed of the music is such that the minimum time unit that can be conveniently counted is 1 Matra.

For the same reason as stated in sub-case A of this Case, above, here too B < U. If B is taken to be equal to (U − 1) Matra, difficulties arise, as follows:—

M = 3B + 2(S + G),

or,
3U = 3(U − 1) + 2(S + G),

or,
3U = 3U − 3 + 2(S + G),

or,
2(S + G) = 3,

or,
S + G = 1½ Matra,

needing the ability to conveniently count in terms if ½ Matra; but this is denied by the assumption made in this sub-case. So B = (U − 1) cannot be a solution.

The next best alternative is to take B = (U − 2). Then the solution works out as follows:—

M = 3B + 2(S + G),

or,
3U = 3(U − 2) + 2(S + G),

or,
3U = 3U − 6 + 2(S + G),

or,
2(S + G) = 6,
i.e.,
S + G = 3. Matra.

One easy allocation is S = 1, G = 2.

The solution then becomes B = (U − 2), S = 1, G = 2. The following table gives some examples:—


for a desired M of quotient in (M ÷ 3) is B must be of S must be of G must be of
3 Matra 1 n o   s o l u t i o n   p o s s i b l e
6 Matra 2 0 Matra 1 Matra 2 Matra
9 Matra 3 1 Matra 1 Matra 2 Matra
12 Matra 4 2 Matra 1 Matra 2 Matra
15 Matra 5 3 Matra 1 Matra 2 Matra
18 Matra 6 4 Matra 1 Matra 2 Matra
21 Matra 7 5 Matra 1 Matra 2 Matra
etc. etc. etc. etc. etc.



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D. Arithmetic Relationship Linking the Components of a Tihai (continued):

3. Cross Rhythm



Sometimes, special and interesting rhythmic effects may be obtained by executing Tihai in a rhythm that has a basic pulse periodicity (i.e., duration between two consecutive pulses) different from that of the original rhythm (i.e., the pulse periodocity of the tala being used). For instance, in a Tihai having M = 12, it has been shown in the tables given above that S + G may be either 1½ or 3. However, if it is desired that the minimum conveniently countable time duration be set for S and at the same time that there should be no G [a Tihai where G = 0 is called a Bedum Tihai – pronounce the "u" in Bedum as in "but", "under", "umbrella" etc.)], one solution would be to divide the time span of 12 Matra into 16 equal pulses and consider thenceforth the total length of the desired Tihai to be 16 such Matra.This is known as cross rhythm.

Simply put, in this example of cross rhythm 16 Matra will be executed in the space of 12 Matra. To generalise, in cross rhythm, what happens is that a given number of Matra following a basic pulse periodicity is superimposed upon another given number of Matra following a different pulse periodicity.

In the above example, the original basic pulse patters (denoted by the symbol Mb) consists of 12 Matra. That is, Mb = 12. Exactly in the time (i.e., neither over nor under the time) it takes to perform these 12 Matra, the performer performs 16 Matra. This means that each of the new, or "superimposed", 16 Matra will be exactly ¾ the time of the original, or "basic", 12 Matra. Denote the superimposed Matra by the symbol Ms. So, Ms = 16.

Here, 12 Mb = 16 Ms. Therefore, 3 Mb = 4 Ms.

Setting up a cross rhythm of this sort can be acquired by correct practice and ought to be part of the standard rhythmic abiloty of any competent performer. Now that 12 Mb is converted into 16 Ms. it is easy to get the required Bedum Tihai that we set out to get by using B = 5 Ms, s = ½ Ms and G = 0.

The reader can use this principle to get any sort of cross-rhythmic superimposition he wishes.


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3. Relation Between Tala and Tihai



So far, the discussion has been limited to Tihai qua Tihai, without any attempt to see them within the Tala cycle into which they must be unltimately integrated. It is now necessary to relate the Tihai to the Tala that has been adopted for the recital. A formal statement of the problem is:—

"Given a Tihai of M Matra to be fit into a Tala of T Matra, from which Matra Ti of the Tala must the Tihai start so that it may end on [i.e., the final of third stroke on the Matra coincide with] a predetermined Matra Tf of the Tala?"

A numerical example should clarify the problem further. Suppost that the performer is performing to Teentala (= Tritala), and suppost that the Mukhda starts on the 12th Matra of the Tala i.e., it is a Mukhda consisting of 5 Matra (such a 5 beat Mukhda is in fact very common in Teentala). Now suppose that the performer decides to end a passage with a Tihai that ends on the 11th Matra, so that immediately after the Tihai, he may pick up the Mukhda from its natural place of the 12th matra. Also suppose that the performer has worked out that the Tihai should be of 8 Matra, with the properties B = 2, S = 1 and G = 0. The question now before the preformer is, and this is the question earlier enunciated: from which point in the Teentala cycle must the Tihai be started? The answer in this case is clearly seen to be the 3rd Matra. Only if the Tihai is started on the 3rd Matra of the Tala will M = 8 Matra be covered by the 10th Matra and the final stroke on the (M+1)th matra will fall on the 11th Matra of the Tala. The performer may now start on the Mukhda from the 12th Matra.

Here, I introduce two new expressions: the "initial Matra" and the "final Matra".

Definitions:

The initial Matra is the Matra in the Tala cycle, counting from the Sum, on which the Tihai should start to fulfil the condition enunciated in the statement of the problem, above. Denote the initial Matra by Ti.

The final Matra is the Matra after the Matra in the Tala cycle, counting from the Sum, on which the Tihai ends, i.e., the final Matra is the Matra on which the (M + 1)th Matra falls. Denote the final Matra by Tf.

Thus, in the example above, Ti = 3 and Tf = 11.

The problem, then, is to find Ti,, given M and Tf.

In this context, three possibilities present themselves:—
  • Case I: Where M < Tf
  • Case II: Where T > M ≥ Tf (as stated earlier, T is the number of Matra in the chosen Tala)
  • Case III: Where M > T.

Each of these cases may now be considered separately, thus:—


Case I ⇒ Where M < Tf

Where the Tihai consists of a fewer number of Matra than the number of Matra from the Sum to the Matra on which it should conclude, it is evident that the Tihai must begin either on or after the Sum, and never before it (the Sum). In this case, therefore, Ti is determined by the equation

Ti = Tf − M

This relation is too obvious to require proof. If a Tihai of 7 Matra is desired to be concluded on the 12th Matra, it must obviously start on the (12 − 7) = 5th Matra. Clearly, this case applies to short Tihai which begin and end within one Avartana (cycle) of a given Tala.


Case II ⇒ Where T > M ≥ Tf

In this case, because the Tihai has a greater number of Matra than the number of Matra lying between the Sum and Tf, it is clear that the Tihai must be started from a Matra lying in the Avartana or cycle preceding the Avartana in which Tf lies. In other words, in this case, the Sum of the Tala lies between Ti and Tf. Moreover, since T > M, no more than a single Sum will lie between Ti and Tf.

Here, counting backwards from Tf, the quantity (M − Tf) gives the number of Matra needed to be used from the Avartana preceding the one in which Tf exists, so that the Tihai on M Matra may end on Tf. This means that, starting from the Sum of the previous Avartana, the Tihai must start from the {T − (M − Tf)}th  Matra of that Avartana. It is only then that the Tihai will end on the Tfth  Matra of the desired Avartana. Hence, where T > M ≥ Tf, Ti is obtained from the equation

Ti = T − (M − Tf).

This relation may be easily verified thus:—

Starting from the {T − (M − Tf)}th  Matra of an Avartana, after counting (M − Tf)  Matra, the Sum is reached, since by definition the Sum occurs in the tala cycle after every T Matra. From this Sum, there now remain {M − (M − Tf)} Matra of the Tihai. After executing these, the Tfth  Matra is reached, and the relation

Ti = T − (M − Tf).

stands proved.

This relation may also be written as:

Ti = T + (T + Tf) − M.

The performer may use either form of the relation according to convenience.

A numerical example may be taken:— Suppose that a Tihai of 12 Matra must be made to end on the 9th Matra of a Dhamar cycle (which has 14 beats to the Avartana). Here, M = 12, Tf = 9 and T = 14. Therefore,

Ti = T − (M − Tf)

or,
Ti = 14 − (12 − 9) = 11.

Thus the Tihai must start on the 11th Matra of the Dhamar cycle so that it ends, i.e., the (M + 1)th Matra falls on the 9th Matra and the Mukhda, if any, can be started on the 10th Matra of the Tala.


Case III ⇒ Where M > T

Here, since the Tihai has a greater number of Matra than that of the Tala, there will obviously be more than one Avartana of the Tala in the course of the Tihai, counting from the Tith  Matra. For example, if a Tihai of 17 Matra is performed in the framework of Jhaptala (this has 10 Matra to the Avartana), and the Tihai starts on the 2nd Matra of the Tala, the Sum will fall on the 10th Matra of the Tihai, the Tihai continuing until it ends on the 9th Matra of the Tala. That is to say, the (M + 1)th or (17 + 1) = 18th Matra – the final or "third stroke" – falls on the 9th Matra of the Tala. Thus, counting one Avartana of Jhaptala, in this example, from the 2nd Matra, more than one such Avartana has elapsed before the Tihai has ended.

Indeed, if the Tihai is very long, it may span not merely one or two Avartana but many Avartana of the Tala. If a Tihai of, say, 66 Matra is rendered in the Jhaptala structure, there will be 6 full Avartana and the seventh will have started, before the Tihai comes to an end.

So, the problem in this case is to find Ti , when the Tihai spans more than one Avartana of the Taka.

To do this, it is first necessary to take away from M as many T as possible. That is to say, take awat as many Avartan of the Tala as possible from the Tihai. This may be done by dividing M by T —

M/T = A + R/T ,

where A is the quotient (number of completed Avartana) and R is the remainder (niumber of Matra covered in the last Avartana started but not finished, as M has run out).

Once R is obtained, there are two possibilities —

  1. R < Tf , and
  2. R ≥ Tf
Obviously, by T in the above two inequations is meant the number of Matra from the Sum to the Tf th  Matra.

a. If R < Tf  , the problem reduces itself to the same one as the one dealt with in Case I above, in this section. Following the reasoning there stated, Ti is found from the equation

Ti = Tf  − R.

b. If R ≥ Tf , the problem becomes similar to Case II of this section, above. Following similar lines of reasoning, T is revealed by the equation

Ti = T − (R − Tf).

The following numerical examples will serve to illustrate a. and b. above.

(i) Suppose that a Tihai of 68 Matra is to be executed in Roopaka Tala (this has 7 beats to the Avartana), so that it must end on the 6th Matra of the Tala. Here, M = 68, Tf = 6 and T = 7. Dividing 68 by 7 gives 9 as quotient and 5 as remainder. So, here A = 9 and R = 5. Thus R ≤ Tf. Therefore, the equation to be used is

Ti = Tf  − R

or,

Ti = 6 − 5 = 1.

Thus the desired Tihai must begin on the first matra, i.e., the Sum of the Tala. It will span 9 Avartana.

Suppose that a Tihai of 40 Matra is to end on the 4th Matra of a Roopaka Tala cycle. Here, M = 40, T f = 4 and T = 7. Dividing 40 by 7 gives 5 as the quotient and 5 as the remainder. So, A = 5 and R = 5. Thus, here R > Tf, so that the equation is

Ti = T − (R − Tf).

or,

Ti = 7 − (5 − 4) = 6.

So the Tihai must begin on the 6th Matra, and it will cover 5 Avartana.

For the particular case R = Tf, the equation

Ti = T − (R − Tf).

becomes

Ti = T,

and so the Tihai in this case must always begin on the Tth or the last Matra of the Tala.


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E. Tihai Ending on the Sum of any Tala



As the majority of Tihai ending in practice end on the Sum of the Tala, it is proper to consider such Tihai separately. Actually, of course, a Tihai ending on the Sum is only a particular case of the general relationship between Tala and Tihai discussed in the previous section.

Since the Sum is the Matra from where the Tala starts, and since the Tihai is to end on the Sum ( i.e., the (M + 1)th   Matra must fall on the Sum), Tf will always be a number in the arithmetic series (T + 1), (2T + 1), (3T + 1), . . . , (AT + 1), where A is the number of Avartana the Tihai will take (i.e., span) before completion. It is important to note that the value of A must be so chosen that, for a Tihai of M Matra, either of the following conditions is satisfied, viz.—

a. (AT − M) = 0, or
b. (AT − M) > 0 < T

The object of so choosing A is to eliminate all completed Avartana that are spanned in the course of executing M Matra.

Having chosen A in this fashion, Ti may be found quite simply from the equation

Ti = (AT + 1) − M.

To illustrate, suppose that a Tihai of 16 Matra is to end on the Sum of Teentala. here, M = 16 and T = 16. To find such a value of A as will satisfy condition a. or condition b. above, divide M by T. This gives 1 as quotient, without leaving any remainder. So the value of A in this case is 1. Then,

Ti = (AT + 1) − M

or,

Ti = (1 × 16 + 1) − 17 - 16 = 1.

Thus the Tihai must begin on the first Matra, i.e., the Sum itself, of the Tala.

As a second illustration, suppose that a Tihai of 23 Matra must be made to end on the Sum of a Roopaka Tala cycle. Then, M ÷ 7 gives 3 as the quotient, leaving over 2 as remainder. Since there is a remainder left over, A may be chosen to be (3 + 1) = 4. Then the condition

(AT − M) > 0 < T

is satisfied.

The required value of Ti in this case will therefore be as follows:—

Ti = (AT + 1) − M

or,

Ti = (4 × 7 + 1) − 23 = (28 + 1) − 23 = 6.

Hence the Tihai must begin on the 6th Matra of Roopaka Tala.


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F. Conclusion



In this essay, I have established a few basic numerical relationships for assisting the student of Indian art music in creating Tihai of his own choice. These relationships can be used, with modifications, by any reader who has taken the pains to work through the foregoing sections, to create all the different kinds of Tihai in use on the concert platform today. Thus:—

  • a Bedum Tihai is merely one where the stroke lasts the smallest conveniently countable unit of time chosen and there is no gap;
  • an Anaghata Tihai is one which ends without the "third stroke" – the (M + 1)th  Matra – sounding; instead, in its place the Mukhda starts;
  • an Ateet Sum Tihai is one which ends so that the "third stroke" – the (M + 1)th  Matra – sounds just before the Sum of the Tala;
  • a Chakradhara Tihai spans more than one Avartana and often consists of three sub-Tihai, i.e., each body consists of a small Tihai in itself. Ehen the last stroke of each such sub-Tihai coincides with the Sum of the Tala, the Tihai is called Farmaishi Chakradhara Tihai. It becomes a Kamali Chakradhara Tihai when the first stroke of the first sub-Tihai and the second stroke of the second subTihai coincide with the Sum of the Tala, and the entire Tihai also ends on the Sum of the Tala.
Many other variations are possible, such as the body consisting of several short phrases, the length of the phrases making up an arithmetical progression. This kind of vatiation is common among Katthak dancers, who call it Gintee. Again, the first body may end with one stroke, the second body with two strokes and the third with three strokes, the third stroke being the (M + 1)th  Matra or the Tf th Matra. I propose to expand this essay to include formulae for computing Gintee at some time in the future.

I end this essay with two appendices: Appendix I gives examples of Tihai that may be used in Vilambit Teentala, where the smallest conveniently countable unit of time is ⅛th of a Matra. Appendix II gives examples of Tihai for Maddhya Laya and Drut Laya Teentala, Jhaptala, Roopakatala, Ekatala and Adachoutala.




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Appendix I: Selected Tihai for use in Vilambit Teentala




In the following list, the number on the left is the number of Matra in the Tihai, i.e., the length of the Tihai measured in Matra. The letters represent Bol of the Tabla, as follows:

t = "Ti" or "Ta"; therefore, tt = "Ti Ta"
d = "Dha"
kt = "Ki Ta"; therefore, ttkt = "Ti Ta Ki Ta"
gdgn = "Ga Di Ghi Na"

Each Matra is divided into eight sub-Matra. Each dash ( - ) denotes an eighth of a Matra. A capital D denotes a stroke or Aghat of the Tihai. The vertical lines separate the bars of the Tala.

In Teentala, a Tihai of 16 Matra starts from the 1st Matra of the Tala, i.e., from the first Dha. A Tihai of 15 Matra starts from the 2nd Matra of the Tala, i.e., from the first Dhin or the second beat of the first bar. And so on.... That is to say:-

A Tihai of Starts from i.e., From Bol
16 Matra the 1st Matra of the Tala 1st beat of 1st bar Dha
15 Matra the 2nd Matra of the Tala 2nd beat of 1st barDhin
14 Matra the 3rd Matra of the Tala 3rd beat of 1st barDhin
13 Matra the 4th Matra of the Tala 4th beat of 1st barDha
12 Matra the 5th Matra of the Tala 1st beat of 2nd barDha
11 Matra the 6th Matra of the Tala 2nd beat of 2nd barDhin
10 Matra the 7th Matra of the Tala 3rd beat of 2nd barDhin
9 Matra the 8th Matra of the Tala 4th beat of 2nd barDha
8 Matra the 9th Matra of the Tala 1st beat of 3rd barDha
7 Matra the 10th Matra of the Tala 2nd beat of 3rd barTin
6 Matra the 11th Matra of the Tala 3rd beat of 3rd barTin
5 Matra the 12th Matra of the Tala 4th beat of 3rd barTa
4 Matra the 13th Matra of the Tala 1st beat of 4th barTiRaKiTa
3 Matra the 14th Matra of the Tala 2nd beat of 4th barDhin
2 Matra the 15th Matra of the Tala 3rd beat of 4th barDhin
1 Matra the 16th Matra of the Tala 4th beat of 4th barDha

The Tihai from the list below can be used while presenting any Raga where a Vilambit Gat is being presented. Simply substitute the Bol for notes. Most of the Tihai below are Navadha Tihai, that is to say, they comprise three sets of Tihai making a large Tihai. The second of the three smaller Tihai comprising the larger Navadha Tihai (or, where there is not a Navadha Tihai, the second or middle phrase of the Tihai) is typed in a different colour for easy reference.

Because the Tihai from Sum to Sum is so frequent and important, three possible Tihai of 16 Matra duration are shown below.








16 ttktgdgn  ttktd-tt  ktgdgntt  ktd-ttkt  |  gdgnttkt  D---ttkt  gdgnttkt  d-ttktgd  |  gnttktd-  ttktgdgn  ttktD---  ttktgdgn  |  ttktd-tt  ktgdgntt  ktd-ttkt  gdgnttkt  |  D


Or, an alternative Tihai of 16 Matra as under:

16 ttd-ttd-  ttd-d-tt  d-ttd-tt  d-d-ttd-  |  ttd-ttd-  D---ttd-  ttd-ttd-  d-ttd-tt  |  d-ttd-d-  ttd-ttd-  ttd-D---  ttd-ttd-  |  ttd-d-tt  d-ttd-tt  d-d-ttd-  ttd-ttd-  |  D

Or, yet another alternative Tihai of 16 Matra as under:

16 ttktgdgn  d---ttkt  gdgnd---  ttktd-tt  |  ktd-ttkt  D---ttkt  gdgnd---  ttktgdgn  |  d---ttkt  d-ttktd-  ttktD---  ttktgdgn  |  d---ttkt  gdgnd---  ttktd-tt  ktd-ttkt  |  D

15 ttktgdgn  ttd--ttk  tgdgnttd  |  --ttktgd  gnttD---  --ttktgd  gnttd--t  |  tktgdgnt  td--ttkt  gdgnttD-  ----ttkt  |  gdgnttd-  -ttktgdg  nttd--tt  ktgdgntt  |  D

14 ttktd-tt  ktd-ttkt  |  d-ttktd-  ttktd-tt  ktd-D-tt  ktd-ttkt  |  d-ttktd-  ttktd-tt  ktd-ttkt  d-D-ttkt  |  d-ttktd-  ttktd-tt  ktd-ttkt  d-ttked-  |  D

13 ttktgdgn  |  d---ttkt  gdgnd---  ttktgdgn  D---ttkt  |  gdgnd---  ttktgdgn  d---ttkt  gdgnD---  |  ttktgdgn  d---ttkt  gdgnd---  ttktgdgn  |  D

12 ttktgdgn  d--ttktg  dgnd--tt  ktgdgnD-  |  -ttktgdg  nd--ttkt  gdgnd--t  tktgdgnD  |  --ttktgd  gnd--ttk  tgdgnd--  ttktgdgn  |  D

11 ttktgdgn  d-ttktgd  gnd-ttkt  |  gdgnD-tt  ktgdgnd-  ttktgdgn  d-ttktgd  |  gnD-ttkt  gdgnd-tt  ktgdgnd-  ttktgdgn  |  D

10 ttktttd-  -ttktttd  |  --ttkttt  D---ttkt  ttd--ttk  tttd--tt  |  ktttD---  ttktttd-  -ttktttd  --ttkttt  |  D

 9  ttktttd-  |  ttktttd-  ttktttD-  -ttktttd  -ttktttd  |  -ttktttD  --ttkttt  d-ttkttt  d-ttkttt  |  D

 8  ttktd--t  tktd--tt  ktd-D-tt  ktd--ttk  |  td--ttkt  d-D-ttkt  d--ttktd  --ttktd-  |  D

 7  ttktd-tt  ktd-ttkt  D---ttkt  |  d-ttktd-  ttktD---  ttktd-tt  ktd-ttkt  |  D

 6  ttktdttk  tdttktD-  |  -ttktdtt  ktdttktD  --ttktdt  tktdttkt  |  D

 5  ttd-ttd-  |  ttd-D-tt  d-ttd-tt  d-D-ttd-  ttd-ttd-  |  D

 4  ttktgdgn  D---ttkt  gdgnD---  ttktgdgn  |  D

 3  ttktttD-  -ttktttD  --ttkttt  |  D

 2  ttktD-tt  ktD-ttkt  |  D

 1  ttDttDtt  |  D




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Appendix II: Selected Tihai for use in Selected Tala in Maddhya and Drut Laya



In the following table, I have listed a selection of Tihai that may be used in commonly used Tala played in medium or fast speeds. For example, if the Tihai length is 4 Matra and the Tala is Teentala (also called Tritala), the Tihai must start on the 13th Matra of the Tala. If it is Jhaptala that is being played, the Tihai must start from the 7th beat of the Tala, snd so on. The Tihai will gave B = 1, S = ½ and G = 0. The Tihai are as follows :—


Tihai attributes   Matra from which Tihai should start
Length Body Stroke Gap   Tritala Jhaptala Roopakatala Ekatala Adachoutala
4 1 ½ 0   13 7 4 9 11
5 1 1 0   12 6 3 8 10
6 0 1 2   11 5 2 7 9
7 1 1 1   10 4 1 6 8
7 2 ½ 0   10 4 1 6 8
8 2 1 0   9 3 7 5 7
9 1 1 2   8 2 6 4 6
10 2 1 1   7 1 5 3 5
10 3 ½ 0   7 1 5 3 5
11 3 1 0   6 10 4 2 4
12 2 1 2   5 9 3 1 3
13 3 1 1   4 8 2 12 2
13 4 ½ 0   4 8 2 12 2
14 4 1 0   3 7 1 11 1
15 3 1 2   2 6 7 10 12
16 4 1 1   1 5 6 9 11
16 5 ½ 0   1 5 6 9 11
17 5 1 0   16 4 5 8 10
18 4 1 2   15 3 4 7 9
19 5 1 1   14 2 3 6 8
19 6 ½ 0   14 2 3 6 8
20 6 1 0   13 1 2 5 7
21 5 1 2   12 10 1 4 6
22 6 1 1   11 9 7 3 5
22 7 ½ 0   11 9 7 3 5
23 7 1 0   10 8 6 2 4
24 6 1 2   9 7 5 1 3
25 7 1 1   8 6 4 12 2
25 8 ½ 0   8 6 4 12 2
26 8 1 0   7 5 3 11 1
27 7 1 2   6 4 2 10 12
28 8 1 1   5 3 1 9 11
28 9 ½ 0   5 3 1 9 11
29 9 1 0   4 2 7 8 10
30 8 1 2   3 1 6 7 9
31 9 1 1   2 10 5 6 8
31 10 ½ 0   2 10 5 6 8
32 10 1 0   1 9 4 5 7



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