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Thread: Binary search
 11202011, 06:09 PM #1Member
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Binary search
Hello :)
I'm trying to make an interactive program (guess the number)
the idea is that i put a number in my head and the program tries to guess it.
the program should ask whether the number is less or greater than the middle of the search interval.
the number should be between 1 & 1000.
so what should the psuedo code be ?
Thanks!!
thirdage.

Re: Binary search
 11202011, 06:58 PM #3Member
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Re: Binary search
I'am just not getting it !!
Java Code:public class guessNumber { public static void main(String[] args) { System.out.print("Is your number between 1 and 1000 ?"); int start=1; int end=1000; int middle =(((endstart)/2)+start); int oldmiddle = 0; boolean numfound = false; while (!numfound) { boolean usersAnswer = StdIn.readBoolean(); if (usersAnswer ==true) { ifTrue(middle,start); oldmiddle = (((end middle)/2)+middle); middle = (((middlestart)/2)+start); } else { ifFalse(oldmiddle,start); } } } public static void ifTrue(int mid,int start) { System.out.print("is your number less than "+mid+" ?"); } public static void ifFalse(int mid,int start) { System.out.print("is your number less than "+mid+" ?"); } }
 11202011, 07:08 PM #4
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Re: Binary search
In a range lo ... hi, you pick an element in the middle: lo ... mid ... hi; if the guess is correct, you're done; otherwise if your guess is too high, pick the interval lo ... mid1, otherwise pick the interval mid+1 ... hi.
kind regards,
JosThe only person who got everything done by Friday was Robinson Crusoe.
 11202011, 07:44 PM #5Member
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Re: Binary search
example :
am picking 40
the program asks if it's between 1 & 1000
my answer: true
between 1 & 500
: true
1&250
:true
//
//
//
1&32
: false
32 & 63
:true
then 47 & 63
:false
I mean each time the interval gets small ... and we ask whether it's between the midpoint of interval and end if false
or start and midpoint if true ... hope you get what I mean.
Thanks again!
 11202011, 07:47 PM #6
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 11202011, 07:51 PM #7Member
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Re: Binary search
If true:
11000 becomes [1500] where we got the midpoint of 1 & 1000.
if false after [1500] was true:
we check on [5001000]
then true
we check on [7501000] .... (where 750 = ((1000500)/2) +500).
 11202011, 07:57 PM #8
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