Title: Challenge: Binary Search.Post by: Moneo on July 09, 2003, 08:14:35 PM
BINARY SEARCH:
1) Declare your definition of what a binary search is. 2) Define a table with 10 sorted entries (elements) which your program will do the search on. 3) Write and post the code to do a binary search on the above table, searching for a user provided argument, and issuing a found or not found message at the end. ***** Title: Challenge: Binary Search.Post by: whitetiger0990 on July 09, 2003, 09:31:20 PM
(http://www.geocities.com/edymnionii/KkatBlink.txt) Huh?
Title: Challenge: Binary Search.Post by: Ninkazu on July 09, 2003, 10:16:40 PM
a binary search goes through an array by first taking its size divided by 2, then only asking if the number is higher or lower than the number at the current entry. After knowing if it's higher or lower, it halves itself in the according direction.
I couldn't figure out how to code this (I'm stupid) and I'm not sure how accurate my description is. Title: Challenge: Binary Search.Post by: na_th_an on July 10, 2003, 12:16:33 AM
Code: DECLARE FUNCTION binarySearchRec% (Array%(), i1%, i2%, n%) DECLARE FUNCTION binarySearch% (Array%(), n%) ' Binary search example by Na Than CLS : RANDOMIZE TIMER ' First make a sorted array DIM Array%(15) PRINT "POSTN:" PRINT "CONT:" FOR i% = 0 TO 15 Array%(i%) = a% LOCATE 1, 8 + i% * 3: PRINT i% LOCATE 2, 8 + i% * 3: PRINT a% a% = a% + INT(RND * 5) NEXT i% ' Test loop. Enter sumthin <0 to exit DO PRINT "Find what"; INPUT n% IF n% < 0 THEN EXIT DO Res% = binarySearch%(Array%(), n%) IF Res% <> -1 THEN PRINT "Found at "; Res% ELSE PRINT "Not Found" END IF LOOP ' ' This function searches for n% in Array%() returning its position, ' or "-1" if n% is not in Array%(). ' FUNCTION binarySearch% (Array%(), n%) binarySearch% = binarySearchRec%(Array%(), LBOUND(Array%), UBOUND(Array%), n%) END FUNCTION ' ' binarySearchRec% is called by binarySearch% takes four parameters: ' ' Array%() -> The array to look inside of. ' i1% -> lower limit. ' i2% -> upper limit. ' n% -> Number that is searched. ' ' We basicly take the array, and look the element in the mid place. ' if it equals n%, we have found it! ' ' If the element in the array is SMALLER than n%, we should look in ' the right half of the array, so we call ourselves recursively with ' new i1%, i2% values. ' ' If the element in the array is BIGGER than n%, we should look in the ' left half of the array. ' ' This (of course) only works if the array is sorted. ' FUNCTION binarySearchRec% (Array%(), i1%, i2%, n%) Res% = -1 IF i1% > i2% THEN binarySearchRec% = Res%: EXIT FUNCTION midPoint% = i1% + (i2% - i1%) \ 2 IF Array%(midPoint%) = n% THEN Res% = midPoint% ELSEIF Array%(midPoint%) < n% THEN Res% = binarySearchRec%(Array%(), midPoint% + 1, i2%, n%) ELSE Res% = binarySearchRec%(Array%(), i1%, midPoint% - 1, n%) END IF binarySearchRec% = Res% END FUNCTION Title: Challenge: Binary Search.Post by: relsoft on July 10, 2003, 02:16:32 AM
Can anyone point to a DL of Address.bas? It has a binary search algo in there. ;*)
*curses CT.... Title: Challenge: Binary Search.Post by: Moneo on July 10, 2003, 02:11:57 PM
WHITETIGER:
--------------- See Ninkazu's definition which is pretty good. An example of a binary search that you may have seen is: You ask a person to think of a number from 1 to 100. You can find his number in 7 questions. Why? Because 2 to the 7th power is 128 which is greater than 100. Each of your questions is going to eliminate half of the possibiities. You start your questions at the mid-point of 50. You first ask if the answer is less than 50. If yes, your next question will be against 25. If no, your next question will be against 75, and so on. NINKAZU: ----------- I think you have the idea. Why don't you try it. NATHAN: ---------- I certainly did not expect a recursive solution. I honestly don't think recursion is required for this. It only takes about 16 lines of code to do it straightforward. I have not checked it out yet because the PC that I'm using right now doesn't have my compiler. Later. ***** Title: Challenge: Binary Search.Post by: Moneo on July 10, 2003, 09:54:13 PM
NATHAN,
Wonderful! Your program works vey well --- a very complete solution. It never would have occurred to me to use recursion. You should post this program in Antoni's recursion challenge. It's a very strong candidate to be the winner. ***** Title: Challenge: Binary Search.Post by: na_th_an on July 11, 2003, 12:27:52 AM
Thanks man. In fact, I did it very naturally. Sometimes, the recursive sollution is the easy one for me, as I tend to think recursively in most cases.
And, sinceramente, no se me ocurre cómo hacerlo de forma iterativa :P Title: Challenge: Binary Search.Post by: Moneo on July 11, 2003, 12:46:24 AM
Nathan,
Don't forget to post this under Antoni's topic. ***** Title: Challenge: Binary Search.Post by: Sterling Christensen on July 11, 2003, 01:15:02 AM
A binary search checks the middle item of those that have not yet been eliminated, and (depending on whether the item is too high or too low) eliminates either the top half or the bottom half from consideration. It does this over and over until it has either found the item or exhausted all possibilites.
Code: DECLARE FUNCTION binarySearch% (item%, first%, last%, array%(), where%) DEFINT A-Z '--------------------------------------- ' Define a random sorted array: RANDOMIZE TIMER DIM array(1 TO 10) n = 0 FOR i = 1 TO 10 n = n + INT(RND * 100) array(i) = n NEXT i '--------------------------------------- '--------------------------------------- ' Display the elements of the array: CLS PRINT "Items in the array:" FOR i = 1 TO 10 PRINT array(i); NEXT i '--------------------------------------- '--------------------------------------- ' Ask the user for an item to search for: INPUT "Item to search for: ", item '--------------------------------------- '--------------------------------------- ' Attempt to find it, then display the result: IF binarySearch(item, 1, 10, array(), position) THEN PRINT item; "was found at position"; position ELSE PRINT item; "was not found in the list." END IF '--------------------------------------- END ' ' Returns false if the item isn't found, or true if it was in which ' case where% is the item's index. ' FUNCTION binarySearch (item, first, last, array(), where) DO WHILE first <= last middle = (first + last) \ 2 i = array(middle) IF i = item THEN binarySearch = -1: where = middle: EXIT FUNCTION IF i < item THEN first = middle + 1 ELSE last = middle - 1 LOOP binarySearch = 0 END FUNCTION Title: Challenge: Binary Search.Post by: relsoft on July 11, 2003, 03:28:32 AM
Only problem I see is you have to have a soted list. ;) But that's just me. ;*)
Title: Challenge: Binary Search.Post by: Sterling Christensen on July 11, 2003, 04:29:46 AM
Quote from: "relsoft" Only problem I see is you have to have a soted list. ;) But that's just me. ;*) Yeah, any binary search algorithm requires a sorted list. Title: Challenge: Binary Search.Post by: Moneo on July 11, 2003, 01:56:04 PM
STERLING C,
--------------- Very nice, straightforward job. It works fine. I even modified it for only 1 table element, and it also works. RELSOFT, ----------- You're right, the list must be sorted. When the binary search function doesn't know where the list comes from, that is who made it, it would be adviseable for the function to do a quick sequence check on the list up front. If it's out-of-sequence, you should abort the program with a fatal error. TO ALL, -------- When you write this kind of general purpose routine for sorting or searching a table, you have to take into consideration that the routine may be given an empty table (null) or one with only 1 table element in it. ***** Title: Challenge: Binary Search.Post by: Moneo on August 18, 2003, 09:11:05 PM
I guess it's time I posted my old code for doing a binary search.
Code: REM "BINSER" BINARY SEARCH SUBROUTINE. Edward F. Moneo 02-Jun-89. REM REM CALLING SEQUENCE: REM BSTAB() = Name of the table to be searched. REM Z = Number of entries in table. REM BSARG = Search argument. REM REM ON OUTPUT: Z = Found table index, (Z=0=Not found) BINSER: IF Z=0 THEN RETURN ZTOP=0 'Pointer to the top -1 ZBOT=Z+1 'Pointer to the bottom+1 IF Z=1 THEN 'Set increment (table offset) ZINC=1 ELSE ZINC=Z/2 END IF DO Z=ZTOP+ZINC IF BSTAB(Z)=BSARG THEN RETURN IF BSTAB(Z) < BSARG THEN ZTOP=ZTOP+ZINC ELSE ZBOT=ZTOP+ZINC END IF ZINC=ZINC/2 IF ZINC=0 THEN ZINC=1 LOOP WHILE ZTOP+ZINC < ZBOT Z=0 RETURN ***** |