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 L02 CPG: Complex sentences - participle 'te_hue'


         Participle 'te_hue'

o  Ex.:  shikArI ne    bhAgate hue shera ko dekhA
         hunter  -erg  running     lion acc. saw
         (The hunter saw a running lion.)
                     OR
         (The hunter while running saw a lion.)

   -  Sentence is ambiguous:
       + 'running' modifies 'lion',  OR
       + 'running' modifies 'saw'

   -  Two different TAMs:
       +  'tA_huA' modifies a noun
       +  'te_hue' modifies a verb

       However, oblique form of former TAM becomes 'te_hue', which
       becomes ambiguous. Discussed in detail below.

+  Ex.1:
         shikArI ne    bhAgatA huA shera dekhA
         hunter  -erg  runn   -ing lion  saw
         (The hunter saw a running lion.)

            dekha[tA_hE]       
                  |            
                  |            
            .----------.       
            |k1        |k2     
            |          |       
         shikArI     shera     
                       |       
                       |       
                       |k1-INV 
                       |       
                 bhaga[tA_huA] 
                      

-  Demand chart for tA_huA:

   *  This together with karaka chart etc. produces m-m tree above.

+  Ex. 2:
     shikArI ne    bhAgate hue  shera ko   dekhA
     hunter -erg   run    -ing  lion acc.  saw
     (The hunter while running saw a lion.)

             dekha[tA_hE]           
                   |                
                   |                
          .----------------.        
          |k1    |k2       |simult.
          |      |         |        
       shikArI shera bhaga[tA_huA]  
                             
    -  Demand chart for te_hue

         [tA_huA]

       arc-lable  manda.  vibh.  lex-type  agreement-holds
        simult.     m       -      (v)     agr

       *  This together with karaka chart etc. produces m-m tree above.

    -  Karaka sharing rule for v[te_hue]

         If k1(v) = 0
         then k1(v) = k1(parent (v))
         *  Applying above rule
                 k1(bhAga)  = k1(dekha)
                            = shikArI

    *  Note: 'shikArI' is karta of 'bhaga' not directly but indirectly
       through sharing rule. ('bhAga' does not modify 'shikArI' directly.)  
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