Answer / kiran

Record split into 2 halfs

compare last value of 1st half and 1st value of 2nd half

ex
we have 100 numbers
if we give 58

1-50 51-100
1st half second half

it ill check the last value of 1st half it is 50 and 1st
value of 2nd half it is 51

so it ill omit 1st half contains 1-50

again 51-100 split in 2 halfs

51-75 75-100
1st half 2nd half


it ill check the last value of 1st half it is 75 and 1st
value of 2nd half it is 100

so it ill omit 2nd half contains 75-100

like that process ill continue upto we get desired element

plz correct if i m worng

Answer / rohit

hi kiran ... on the last you are comparing wrong values ...
that is last value of first half 75 and first value of 2nd
half 100 (it should be 76)..... please correct ....

Answer / srikanth doki

Binary search is also known as search all.
For large tables with more than 10 records this search is
used. The records must be either in ascending or descending
order. First it will divide it into two parts and then it
will search for the center one if it not match then it will
search for the left or right half most ones.

Answer / sumankumar

IT MEANS DATA SPLIT INTO 2 HALFS AND COMPARE FIRST HALF LAST
VALUE AND SECOND HALF FIRST VALUE WITH GIVEN DATA THEN
ELIMINATE HALF OF THE DATA AND AGAIN SPLIT INTO NEXT 2 HALFS
THIS PROCCESS WILL BE CONTINUED STILL VALUE IS IDENTIFIED

LET ME KNOW ONE EX:
PROGRAM: CONSIDER 1,2,3,4,5,6,7,8,9,10.
SET I TO 2
SEARCH ALL DATA
AT END DISPLY " EOT"
WHEN DATA(I)= 4
END-SEARCH ALL

SOL: THE DATA DIVIDE INTO 2 HALFS 1 TO 4, 5 TO 10 U R
REQUIREMENT IS SERCH AT POSITION OF 4 SO IT COMPARE FIRST
HALFS OF LAST VALUE i'e 4 AND SECOND HALFS OF FIRST VALUE
i'e 5 AND ELIMINATE THE HALF OF DATA AND AGAIN SPLIT IT INTO
NEXET 2 HALFS THIS PROCCESS WILL BE CONTINUED STILL VALUE IS
IDENTIFIED

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