Thursday, June 30, 2011

नेपालीले बनाए देशकै पहिलो सर्च इन्जिन

स्वदेशी सर्च इन्जिनबाटै अब इन्टरनेटमा रहेका जानकारी र तस्बिर खोज्न सकिने भएको छ । ईबजएसिया प्रालिले अंग्रेजी भाषाको नेपालको पहिलो ईबजएसिया डटकम नामक सर्च इन्जिन सञ्चालनमा ल्याएपछि त्यस्तो सुविधा प्राप्त भएको हो।

सन् २००६ देखि अनुसन्धान तथा विकास सुरु भएको यो सर्च इन्जिन २०११ मा बजारमा आएको हो। अनडिजिटलाइज डाटालाई डिजिटलाइज गर्न यसले महत्त्वपूर्ण भूमिका निर्वाह गर्न सक्ने बताउँदै कम्पनीका अध्यक्ष ईश्वरचन्द्र प्रधान भन्छन्, "नेपालको सम्पूर्ण डाटा राख्न अझै ५० वर्ष लाग्नेछ।" प्रत्येक दिन करबि ३० हजार सामग्री राखिरहेको इबजएसियाले निकट भविष्यमै सोसियल नेटवर्क स्थापना गरी फ्रि इमेल सेवा सुरु गर्ने भएको छ।

नेपालका लागि सबैभन्दा बढी वेबसाइट, तस्िबर, सूचना सर्च गर्नका लागि ईबजएसिया नै निकट भविष्यमा भरपर्दो माध्यम बन्ने कम्पनीको दाबी छ। अन्य सर्च इन्जिनहरूले नेपाललाई त्यति प्राथमिकतामा नराखेको अवस्थामा यो सर्च इन्जिन महत्त्वपूर्ण हुने पनि कम्पनीको विश्व्ाास छ।

Source:http://www.ekantipur.com/nepal/article

Wednesday, June 22, 2011

आकृतिलाई तान्न झुम्मिए कलेजहरू

भक्तपुर,क्तपुर, अषाढ ८ - एसएलसीमा ९४.२५ प्रतिशत ल्याएकी कटुन्जेकी आकृति पाण्डेलाई भेट्न मंगलबार दिनभर राजधानीका नाम चलेका कलेजहरूको घुइँचो लाग्यो । पेन्टागन, नासा, पिरामिड, सीसीआरसी, युनिभर्सल, लिभरपुल, ह्वाइट गोल्डलगायत कलेजका प्रतिनिधिले उनी पढेको ओम सेकेन्डरी स्कुल कटुन्जे पुगेर आकृतिलाई आफ्नो कलेज भ्रमणको निम्तो दिएका थिए ।
'एक चोटि आएर हाम्रो कलेज हेरिदिनु भनेर कलेजका प्रतिनिधिहरूले निम्तो दिनुभएको छ, पढ्नै पर्छ भनेर दबाब दिएका छैनन्,' उनले भनिन्, 'कुन कलेज पढ्ने भन्ने निधो मैले गरेकी छैन ।'
उनले ८ सय पूर्णाङ्कमा ७ सय ५४ अंक ल्याएकी छन् । उनको जति अरू कसैले अंक ल्याएको हालसम्म नपाइएपछि उनलाई फोनमा र भेटेर बधाई दिनेको ओइरो लागेको छ ।
'फोन उठाएर साध्य छैन, कहिले मामुको मोबाइलमा कहिले दाइको मोबाइलमा फोन आउँछ,' उनले भनिन्, 'भेट्न आउने आफन्त र शुभचिन्तक पनि उत्तिकै छन् ।' छोरीले धेरै नम्बर ल्याएपछि आकृतिका आमा मुना र बुबा सानुभाइमा हर्षको सीमा छैन । पत्रकार र कलेजका प्रतिनिधिले बारम्बार भेट्न आए पनि आमा-छोरी दिनभर विद्यालयमा थिए । 'छिन-छिनमै छोरीलाई खोज्दै आउँछन्, त्यही भएर पायक पार्न विद्यालयमा बसेका हौं,' मुनाले भनिन् । समयतालिका मिलाएर पढेकै कारण राम्रो अंक ल्याउन सफल भएको आकृतिले बताइन् ।
'घोकेर राम्रो नम्बर आउँछ भन्ने होइन, समयतालिका मिलाएर पढ्नुपर्छ, कोर्सको विषयसँगै बाहिरी ज्ञान पनि हुनु जरुरी छ,' उनले भनिन्, 'सर्वोत्कृष्ट हुन्छु भन्ने त सोचेको थिइनँ, ९० प्रतिशत कट्नेमा ढुक्क थिएँ ।'
आठ कक्षाको जिल्लास्तरीय परीक्षामा पनि उनी जिल्ला प्रथम भएकी थिइन् । सेन्डअप परीक्षामा जिल्ला चौथो भएकी थिइन् । डाक्टर बन्ने लक्ष्य लिएकी उनी साइन्स पढ्न बि्रजकोर्स गरिरहेकी छन् ।
उनको सफलताले आकृतिको परिवार मात्रै नभई विद्यालय र सिंगो भक्तपुरवासीमा खुसियाली छाएको छ । खुसियालीमा विद्यालयमा मंगलबार दिनभर नाचगान भएको थियो । विद्यालय वरिपरिका बासिन्दा पनि नाचगान हेर्न झुम्मिएका थिए । 'आकृति सर्वोत्कृष्ट हुनु भनेको सिंगो भक्तपुरले शिक्षा क्षेत्रमा फड्को मारेको हामीले लिएका छौं,' प्याब्सन भक्तपुरका उपाध्यक्ष राजकुमार महर्जनले भने, 'भक्तपुरमा स्तरीय विद्यालय छैन भन्नेलाई यो नतिजाले चुनौती दिएको छ ।' आकृतिबारे समाचार आएपछि विद्यालयकी पि्रन्सिपल सानु घिमिरेलाई पनि बिहानैदेखि बधाई लिन भ्याईनभ्याई थियो । उनले भनिन्, 'आकृतिले मेरो र स्कुलको सपना पूरा गरिदिइन् ।' ३५ विद्यार्थीले परीक्षा दिएकामा यो विद्यालयबाट १२ विशिष्ट र २३ जना प्रथम श्रेणीमा उत्तीर्ण भए । उनी खेलकुदमा पनि रुचि राख्छिन् । गत वर्षको राष्ट्रपति कप भलिबलमा उनको टिम प्रथम भएको थियो । उनका पिता कृष िविकास बैंकमा कार्यरत छन् भने आमा गुन्डुको ज्योति निमाविकी शिक्षकिा हुन् । दाइ पुल्चोकमा इन्जिनियरिङ विषय पढ्दै छन् ।
आकृतिको एक दिन
बिहान सबेरै उठें । उत्कृष्ट नम्बर आएको हो कि होइन भन्ने कौतूहल रातभरि भयो । निद्रा लागेन । बिहानै आत्मा अधिकारी भन्ने साथीले फोन गरेर तिमी त नेपाल टप भएछौ पत्रिकामा छापिएको छ भनिन् । त्यसपछि खुसीको सीमा रहेन । केही खान मन लागेन । स्कुल गएर कहिले म्याडमलाई भेटांै जस्तो लाग्यो । घरको फोन र मोबाइलमा बधाई आउन थाल्यो । चिया मात्रै खाएर साढे सात बजे नै ममी र म स्कुलमा गयौं । घरबाट स्कुल पुग्न २० मिनेट लाग्छ । पुग्नासाथ म्याडमले अँगालोमा बाँध्नु भयो ।
बधाई दिन स्कुलका भाइबहिनी पनि आए । दिनभर स्कुलमा बित्यो । अन्तर्वार्ता लिन सञ्चारमाध्यमहरू पनि आए । खाना खान घर जानसमेत फुर्सद मिलेन । राजधानीका नाम चलेका कलेजका प्रतिनिधिहरू पनि आएका थिए । उहाँहरूले जहाँ पढे पनि आफ्नो कलेजको एकपल्ट अवलोकन गरिदिन आग्रह गर्नुभएको छ । बल्ल साँझमा विद्यालयबाट घर फर्किन पाएँ ।
 
प्रकाशित मिति: २०६८ अषाढ ८ ०८:३०
Source:http://www.ekantipur.com/np

पन्ध्र दिनभित्र मार्कसिट र प्रोभिजनल सर्टिफिकेटसँगै दिइने

भक्तपुर,असार ७/परीक्षा नियन्त्रण कार्यालय सानोठिमीले यसवर्षदेखि विद्यार्थीको 'मार्कसिट'सँगै 'प्रोभिजनल सर्टिफिकेट' पनि वितरण गर्ने भएको छ । विगतका वर्षहरुमा परीक्षाफल प्रकाशित भएको दुईसातामा 'मार्कसिट' र छ महिनापछि 'प्रोभिजनल सर्टिफिकेट' पठाइने गरेकामा यसवर्षदेखि विद्यार्थीलाई सहज बनाउन कार्यालयले सँगसँगै 'मार्कसिट' र 'प्रोभिजनल  सर्टिफिकेट' पठाउन लागेको हो ।

सवर्ष १५ दिनभित्र 'मार्कसिट' र 'प्रोभिजनल सर्टिफिकेट'सँगै जिल्ला-जिल्ला पठाउन सुरु गरिने कार्यालयका उपनियन्त्रक डिल्लीराम लुइँटेलले बताए । 

'मार्कसिट' र 'प्रोभिजनल सर्टिफिकेट' फरक-फरक समयमा पठाउँदा विद्यार्थीले 'मार्कसिट' लैजाने तर 'प्रोभिजनल सर्टिफिकेट' लिन नआउने गरेकाले 'प्रोभिजनल सर्टिफिकेट' जिल्लामै थन्किएर बसेको हुँदा यसवर्षदेखि सँगै पठाउन सुरु गर्न लागेको बताए ।

'प्रोभिजनल सर्टिफिकेट' अहिले हरेक क्षेत्रमा अनिवार्य भए पनि विद्यार्थीले लैजान वास्ता नगर्ने गरेकाले जिल्ला-जिल्लामा थन्किएर रहेको छ । 
विद्यार्थीले आवश्यक पर्दामात्र यो 'सर्टिफिकेट' लिन आउने भएकाले विद्यार्थीलाई पनि झन्झट हुने र कार्यालयले पनि तयार पारेको 'प्रोभिजनल सर्टिफिकेट' जिल्लामै थन्किएर बस्ने गरेकाले यस्तो नहोस् भनेर यसवर्षदेखि 'प्रोभिजनल सर्टिफिकेट' पठाउन लागेको हो ।

त्यसैगरी कार्यालयले यसवर्षको परीक्षामा दुई विषयमा अनत्तीर्ण भएकालाई मौका परीक्षाका लागि साउन १८ गतेसम्म परीक्षाको फाराम भर्ने समय तोकेको छ ।
त्यसका लागि कार्यालयले 'मार्कसिट' र 'प्रोभिजनल सर्टिफिकेट' जिल्ला पठाउँदा मौका परीक्षाको फाराम पनि सँगै पठाउने भएको छ ।

उपनियन्त्रक लुइँटेलले फाराम बुझाउन ढिला भए पनि परीक्षाको राजस्व भने विद्यार्थीले साउन १८ गतेभित्र बुझाइसक्नुपर्ने बताए । 

यसवर्षको मौका परीक्षा साउन २३ देखि ३० गतेसम्म सञ्चालन हुने भएको छ । - रासस

Tuesday, June 21, 2011

Notice from the Ministry of Home Affairs

बिज्ञप्ति २०६८।३।७ –२०६८ साल असार ८ गतेका दिनलाइ राष्ट्रिय जनगणना दिवसका रूपमा मनाउन मुलुकभर सार्वजिनक विदा दिने नेपाल सरकारबाट निर्णय भएकोले सर्वसाधारण सबैलाइ जानकारीका लागि यो विज्ञप्ति प्रकाशित गरिएको छ।

Source:
Government of Nepal
Ministry of Home Affairs
Singha Durbar,Kathmandu,Nepal
web:http://www.moha.gov.np/

Monday, June 20, 2011

Congratulation!!! Congratulation!!! Congratulation!!!

Congratulation!!! Congratulation!!! 

0300793Y       Distinction Marks.
0300794Z        Distinction Marks.
0300795A       Distinction Marks.
0300798D       FIRST Division.
0300799E        Distinction Marks.
0300800F        FIRST Division.
0300801G       Distinction Marks.
0300802H       SECOND Division.
0300803I         FIRST Division.
0300804J         FIRST Division.      
0300805K       SECOND Division.
0300806L        FIRST Division.
0300807M       FIRST Division.
 0300809O       SECOND Division.
0300810P        FIRST Division.
   

Friday, June 17, 2011

Qbasic Examples

lab1.
REM “My first QBASIC program”
REM “The sales tax on an article is 25% of its price.”
REM “This program finds the sales tax on an article”
LET price =200
LET tax =price*25/100
PRINT “The sales tax is::”; tax
END

Lab2
CLS
REM “This program finds the area of a circle”
PI=3.14159
LET R=8
Area =PI*(R^2)
PRINT “The area is:::”; Area
END

lab 3.
CLS
Rate = 50
Pas = 20
Total = Pas*rate
PRINT “The total amount is=”; Total
END

Lab4
REM “This program finds the area of a rectangle”
CLS
INPUT “Enter the length of a rectangle”; l
INPUT “Enter the width of a rectangle”; w
Area= l*b
PRINT “The area of the rectangle is:::”; Area
END

Lab5.
X = 1
Top :
Y = x*x
PRINT y
X=x+1
GOTO top
END

lab 6.
CLS
INPUT “Enter test score 1”; score1
INPUT “Enter test score 2”; Score 2
INPUT “Enter test score 3”; Score 3
Average = (score1+score2+score3)/3
PRINT “The score is=”; average
IF (average>40) THEN PRINT “congratulation”
END

Thursday, June 16, 2011

Computer Science

Computer science is the status of the operating principles of computers,computer programming language and algorithms for solving theoretical as well as practical problems.

The Binary System

The Binary System
Basic Concepts Behind the Binary SystemTo understand binary numbers, begin by recalling elementary school math. When we first learned about numbers, we were taught that, in the decimal system, things are organized into columns:
H | T | O
1 | 9 | 3
such that "H" is the hundreds column, "T" is the tens column, and "O" is the ones column. So the number "193" is 1-hundreds plus 9-tens plus 3-ones.
Years later, we learned that the ones column meant 10^0, the tens column meant 10^1, the hundreds column 10^2 and so on, such that
10^2|10^1|10^0
1 | 9 | 3
the number 193 is really {(1*10^2)+(9*10^1)+(3*10^0)}.
As you know, the decimal system uses the digits 0-9 to represent numbers. If we wanted to put a larger number in column 10^n (e.g., 10), we would have to multiply 10*10^n, which would give 10^(n+1), and be carried a column to the left. For example, putting ten in the 10^0 column is impossible, so we put a 1 in the 10^1 column, and a 0 in the 10^0 column, thus using two columns. Twelve would be 12*10^0, or 10^0(10+2), or 10^1+2*10^0, which also uses an additional column to the left (12).
The binary system works under the exact same principles as the decimal system, only it operates in base 2 rather than base 10. In other words, instead of columns being

10^2|10^1|10^0
they are
2^2|2^1|2^0Instead of using the digits 0-9, we only use 0-1 (again, if we used anything larger it would be like multiplying 2*2^n and getting 2^n+1, which would not fit in the 2^n column. Therefore, it would shift you one column to the left. For example, "3" in binary cannot be put into one column. The first column we fill is the right-most column, which is 2^0, or 1. Since 3>1, we need to use an extra column to the left, and indicate it as "11" in binary (1*2^1) + (1*2^0).
Examples: What would the binary number 1011 be in decimal notation?

Try converting these numbers from binary to decimal:
10
111
10101
11110
Remember:
2^4| 2^3| 2^2| 2^1| 2^0
| | | 1 | 0
| | 1 | 1 | 1
1 | 0 | 1 | 0 | 1
1 | 1 | 1 | 1 | 0

Binary AdditionConsider the addition of decimal numbers:
23
+48
___
We begin by adding 3+8=11. Since 11 is greater than 10, a one is put into the 10's column (carried), and a 1 is recorded in the one's column of the sum. Next, add {(2+4) +1} (the one is from the carry)=7, which is put in the 10's column of the sum. Thus, the answer is 71.
Binary addition works on the same principle, but the numerals are different. Begin with one-bit binary addition:
0 0 1
+0 +1 +0
___ ___ ___
0 1 1
1+1 carries us into the next column. In decimal form, 1+1=2. In binary, any digit higher than 1 puts us a column to the left (as would 10 in decimal notation). The decimal number "2" is written in binary notation as "10" (1*2^1)+(0*2^0). Record the 0 in the ones column, and carry the 1 to the twos column to get an answer of "10." In our vertical notation,
1
+1
___
10
The process is the same for multiple-bit binary numbers:
1010
+1111
______
Alternately:
11 (carry)
1010
+1111
______
11001
Always remember
0+0=0
1+0=1
1+1=10
Try a few examples of binary addition:
111 101 111
+110 +111 +111
______ _____ _____

Binary MultiplicationMultiplication in the binary system works the same way as in the decimal system:
1*1=1
1*0=0
0*1=0
101
* 11
____
101
1010
_____
1111
Note that multiplying by two is extremely easy. To multiply by two, just add a 0 on the end.
Return to Table of Contents
Binary DivisionFollow the same rules as in decimal division. For the sake of simplicity, throw away the remainder.
For Example: 111011/11
10011 r 10
_______
11)111011
-11
______
101
-11
______
101
11
______
10
Return to Table of Contents
Decimal to BinaryConverting from decimal to binary notation is slightly more difficult conceptually, but can easily be done once you know how through the use of algorithms. Begin by thinking of a few examples. We can easily see that the number 3= 2+1. and that this is equivalent to (1*2^1)+(1*2^0). This translates into putting a "1" in the 2^1 column and a "1" in the 2^0 column, to get "11". Almost as intuitive is the number 5: it is obviously 4+1, which is the same as saying [(2*2) +1], or 2^2+1. This can also be written as [(1*2^2)+(1*2^0)]. Looking at this in columns,
2^2 | 2^1 | 2^0
1 0 1
or 101.
What we're doing here is finding the largest power of two within the number (2^2=4 is the largest power of 2 in 5), subtracting that from the number (5-4=1), and finding the largest power of 2 in the remainder (2^0=1 is the largest power of 2 in 1). Then we just put this into columns. This process continues until we have a remainder of 0. Let's take a look at how it works. We know that:
2^0=1
2^1=2
2^2=4
2^3=8
2^4=16
2^5=32
2^6=64
2^7=128
and so on. To convert the decimal number 75 to binary, we would find the largest power of 2 less than 75, which is 64. Thus, we would put a 1 in the 2^6 column, and subtract 64 from 75, giving us 11. The largest power of 2 in 11 is 8, or 2^3. Put 1 in the 2^3 column, and 0 in 2^4 and 2^5. Subtract 8 from 11 to get 3. Put 1 in the 2^1 column, 0 in 2^2, and subtract 2 from 3. We're left with 1, which goes in 2^0, and we subtract one to get zero. Thus, our number is 1001011.
Making this algorithm a bit more formal gives us:
Let D=number we wish to convert from decimal to binary
Repeat until D=0
a. Find the largest power of two in D. Let this equal P.
b. Put a 1 in binary column P.
c. Subtract P from D.
Put zeros in all columns which don't have ones.
This algorithm is a bit awkward. Particularly step 3, "filling in the zeros." Therefore, we should rewrite it such that we ascertain the value of each column individually, putting in 0's and 1's as we go:
Let D= the number we wish to convert from decimal to binary
Find P, such that 2^P is the largest power of two smaller than D.
Repeat until P<0
If 2^P<=D then
put 1 into column P
subtract 2^P from D
Else
put 0 into column P End if
Subtract 1 from P
Now that we have an algorithm, we can use it to convert numbers from decimal to binary relatively painlessly. Let's try the number D=55.
Our first step is to find P. We know that 2^4=16, 2^5=32, and 2^6=64. Therefore, P=5.
2^5<=55, so we put a 1 in the 2^5 column: 1-----.
Subtracting 55-32 leaves us with 23. Subtracting 1 from P gives us 4.
Following step 3 again, 2^4<=23, so we put a 1 in the 2^4 column: 11----.
Next, subtract 16 from 23, to get 7. Subtract 1 from P gives us 3.
2^3>7, so we put a 0 in the 2^3 column: 110---
Next, subtract 1 from P, which gives us 2.
2^2<=7, so we put a 1 in the 2^2 column: 1101--
Subtract 4 from 7 to get 3. Subtract 1 from P to get 1.
2^1<=3, so we put a 1 in the 2^1 column: 11011-
Subtract 2 from 3 to get 1. Subtract 1 from P to get 0.
2^0<=1, so we put a 1 in the 2^0 column: 110111
Subtract 1 from 1 to get 0. Subtract 1 from P to get -1.
P is now less than zero, so we stop.
Another algorithm for converting decimal to binaryHowever, this is not the only approach possible. We can start at the right, rather than the left.
All binary numbers are in the form
a[n]*2^n + a[n-1]*2^(n-1)+...+a[1]*2^1 + a[0]*2^0where each a[i] is either a 1 or a 0 (the only possible digits for the binary system). The only way a number can be odd is if it has a 1 in the 2^0 column, because all powers of two greater than 0 are even numbers (2, 4, 8, 16...). This gives us the rightmost digit as a starting point.
Now we need to do the remaining digits. One idea is to "shift" them. It is also easy to see that multiplying and dividing by 2 shifts everything by one column: two in binary is 10, or (1*2^1). Dividing (1*2^1) by 2 gives us (1*2^0), or just a 1 in binary. Similarly, multiplying by 2 shifts in the other direction: (1*2^1)*2=(1*2^2) or 10 in binary. Therefore
{a[n]*2^n + a[n-1]*2^(n-1) + ... + a[1]*2^1 + a[0]*2^0}/2is equal to
a[n]*2^(n-1) + a[n-1]*2^(n-2) + ... + a[1]2^0 Let's look at how this can help us convert from decimal to binary. Take the number 163. We know that since it is odd, there must be a 1 in the 2^0 column (a[0]=1). We also know that it equals 162+1. If we put the 1 in the 2^0 column, we have 162 left, and have to decide how to translate the remaining digits.
Two's column: Dividing 162 by 2 gives 81. The number 81 in binary would also have a 1 in the 2^0 column. Since we divided the number by two, we "took out" one power of two. Similarly, the statement a[n-1]*2^(n-1) + a[n-2]*2^(n-2) + ... + a[1]*2^0 has a power of two removed. Our "new" 2^0 column now contains a1. We learned earlier that there is a 1 in the 2^0 column if the number is odd. Since 81 is odd, a[1]=1. Practically, we can simply keep a "running total", which now stands at 11 (a[1]=1 and a[0]=1). Also note that a1 is essentially "remultiplied" by two just by putting it in front of a[0], so it is automatically fit into the correct column.
Four's column: Now we can subtract 1 from 81 to see what remainder we still must place (80). Dividing 80 by 2 gives 40. Therefore, there must be a 0 in the 4's column, (because what we are actually placing is a 2^0 column, and the number is not odd).
Eight's column: We can divide by two again to get 20. This is even, so we put a 0 in the 8's column. Our running total now stands at a[3]=0, a[2]=0, a[1]=1, and a[0]=1.
We can continue in this manner until there is no remainder to place.
Let's formalize this algorithm:
1. Let D= the number we wish to convert from decimal to binary.
2. Repeat until D=0:
a) If D is odd, put "1" in the leftmost open column, and subtract 1 from D.
b) If D is even, put "0" in the leftmost open column.
c) Divide D by 2.
End Repeat
For the number 163, this works as follows:
1. Let D=163
2. b) D is odd, put a 1 in the 2^0 column.
Subtract 1 from D to get 162.
c) Divide D=162 by 2.
Temporary Result: 01 New D=81
D does not equal 0, so we repeat step 2.
2. b) D is odd, put a 1 in the 2^1 column.
Subtract 1 from D to get 80.
c) Divide D=80 by 2.
Temporary Result: 11 New D=40
D does not equal 0, so we repeat step 2.
2. b) D is even, put a 0 in the 2^2 column.
c) Divide D by 2.
Temporary Result:011 New D=20
2. b) D is even, put a 0 in the 2^3 column.
c) Divide D by 2.
Temporary Result: 0011 New D=10
2. b) D is even, put a 0 in the 2^4 column.
c) Divide D by 2.
Temporary Result: 00011 New D=5
2. a) D is odd, put a 1 in the 2^5 column.
Subtract 1 from D to get 4.
c) Divide D by 2.
Temporary Result: 100011 New D=2
2. b) D is even, put a 0 in the 2^6 column.
c) Divide D by 2.
Temporary Result: 0100011 New D=1
2. a) D is odd, put a 1 in the 27 column.
Subtract 1 from D to get D=0.
c) Divide D by 2.
Temporary Result: 10100011 New D=0
D=0, so we are done, and the decimal number 163 is equivalent to the binary number 10100011.
Since we already knew how to convert from binary to decimal, we can easily verify our result. 10100011=(1*2^0)+(1*2^1)+(1*2^5)+(1*2^7)=1+2+32+128= 163.
Return to Table of Contents
Negation in the Binary System
These techniques work well for non-negative integers, but how do we indicate negative numbers in the binary system?
Before we investigate negative numbers, we note that the computer uses a fixed number of "bits" or binary digits. An 8-bit number is 8 digits long. For this section, we will work with 8 bits.
Signed Magnitude:
The simplest way to indicate negation is signed magnitude. In signed magnitude, the left-most bit is not actually part of the number, but is just the equivalent of a +/- sign. "0" indicates that the number is positive, "1" indicates negative. In 8 bits, 00001100 would be 12 (break this down into (1*2^3) + (1*2^2) ). To indicate -12, we would simply put a "1" rather than a "0" as the first bit: 10001100.
One's Complement:
In one's complement, positive numbers are represented as usual in regular binary. However, negative numbers are represented differently. To negate a number, replace all zeros with ones, and ones with zeros - flip the bits. Thus, 12 would be 00001100, and -12 would be 11110011. As in signed magnitude, the leftmost bit indicates the sign (1 is negative, 0 is positive). To compute the value of a negative number, flip the bits and translate as before.
Two's Complement:
Begin with the number in one's complement. Add 1 if the number is negative. Twelve would be represented as 00001100, and -12 as 11110100. To verify this, let's subtract 1 from 11110100, to get 11110011. If we flip the bits, we get 00001100, or 12 in decimal.

In this notation, "m" indicates the total number of bits. For us (working with 8 bits), it would be excess 2^7. To represent a number (positive or negative) in excess 2^7, begin by taking the number in regular binary representation. Then add 2^7 (=128) to that number. For example, 7 would be 128 + 7=135, or 2^7+2^2+2^1+2^0, and, in binary,10000111. We would represent -7 as 128-7=121, and, in binary, 01111001.
Note:
Unless you know which representation has been used, you cannot figure out the value of a number.
A number in excess 2^(m-1) is the same as that number in two's complement with the leftmost bit flipped.
To see the advantages and disadvantages of each method, let's try working with them.
Using the regular algorithm for binary adition, add (5+12), (-5+12), (-12+-5), and (12+-12) in each system. Then convert back to decimal numbers.



Answers
What would the binary number 1011 be in decimal notation?1011=(1*2^3)+(0*2^2)+(1*2^1)+(1*2^0)
= (1*8) + (0*4) + (1*2) + (1*1)
= 11 (in decimal notation)

Try converting these numbers from binary to decimal:10=(1*2^1) + (0*2^0) = 2+0 = 2
111 = (1*2^2) + (1*2^1) + (1*2^0) = 4+2+1=7
10101= (1*2^4) + (0*2^3) + (1*2^2) + (0*2^1) + (1*2^0)=16+0+4+0+1=21
11110= (1*2^4) + (1*2^3) + (1*2^2) + (1*2^1) + (0*2^0)=16+8+4+2+0=30

Try a few examples of binary addition:1 1
111 111 111
+110 +110 +110
______ ______ _____
1 01 1101
1 11 1
101 101 101
+111 +111 +111
_____ ____ _____
0 00 1100
1 1 1
111 111 111
+111 +111 +111
_____ _____ _____
0 10 1110
Using the regular algorithm for binary adition, add (5+12), (-5+12), (-12+-5), and (12+-12) in each system. Then convert back to decimal numbers.
Signed Magnitude:
5+12 -5+12 -12+-5 12+-12
00000101 10000101 10001100 00001100
00001100 00001100 10000101 10001100
__________ ________ ________ _________
00010001 10010001 00010000 10011000
17 -17 16 -24
One' Complement:
00000101 11111010 11110011 00001100
00001100 00001100 11111010 11110011
_________ ________ ________ ________
00010001 00000110 11101101 11111111
17 6 -18 0
Two's Complement:
00000101 11111011 11110100 00001100
00001100 00001100 11111011 11110100
________ ________ ________ ________
00010001 00000111 11101111 00000000
17 7 -17 0
Signed Magnitude:
10000101 01111011 01110100 00001100
10001100 10001100 01111011 01110100
________ ________ ________ ________
00010001 00000111 11101111 01111100
109 119 111 124
Signed Magnitude
One's Complement
Two's Complement
Excess 2^(m-1)
Step one:
Column 2^0: 0+1=1.
Record the 1.
Temporary Result: 1; Carry: 0
Step two:
Column 2^1: 1+1=10.
Record the 0, carry the 1.
Temporary Result: 01; Carry: 1
Step three:
Column 2^2: 1+0=1 Add 1 from carry: 1+1=10.
Record the 0, carry the 1.
Temporary Result: 001; Carry: 1
Step four:
Column 2^3: 1+1=10. Add 1 from carry: 10+1=11.
Record the 11.
Final result: 11001

Sunday, June 12, 2011

एसएलसीमा किन बढी विद्यार्थी अनुत्तीर्ण हुन्छन् ?

सरकारी विद्यालयका विद्यार्थीहरूमा अभ्यासको कमी, घोकन्ते र परीक्षामुखी प्रवृत्ति भएकाले पनि उत्तीर्ण प्रतिशत बढ्न नसकेको हो । उनीहरू परीक्षाका बेलाबाहेक मेहनत गर्दैनन् ।

एसएलसी परीक्षामा उत्तीर्ण प्रतिशत नबढ्नुमा कुनै एउटा कारण नभै थुप्रै कारण छन् । नेपालमा बढी संख्यामा विद्यार्थी अनुत्तीर्ण हुनुमा पाठयक्रम, पाठयपुस्तक एवं परीक्षा प्रणाली प्रमुख रूपमा दोषी पाइन्छ । एकथरि, शिक्षाविद्का अनुसार पाठ्यपुस्तकमा रहेका कमी-कमजोरीले यस्तो स्थिति सिर्जना भएको छ भने अर्काथरी शिक्षाविद्को ठम्याइचाहिँ परीक्षामा सोधिने प्रश्नावलीका कारण यस्तो समस्या भोग्नुपरेको हो । एक विशेषज्ञका अनुसार एसएलसीमा सोधिने प्रश्न धेरै भएकाले समय अभावका कारण सबै प्रश्न हल गर्न नपाउँदा धेरै विद्यार्थी अनुत्तीर्ण हुन्छन् । अहिले एलएलसी तहको पाठयपुस्तकमा विषय बढी भएको, क्षेत्र एवं क्रम नमिलेको, अभ्यास कम भएको जस्ता त्रुटि पाइन्छन् ।

एसएलसी परीक्षाको उत्तर पुस्तिका ठीकसँग परीक्षण नहुनाले पनि अनुत्तीर्ण प्रतिशत बढेको हो । एक शिक्षकले एक सयभन्दा बढी उत्तरपुस्तिका जाँच्न नपाइने नियम भए पनि उक्त नियम व्यवहारमा लागू हुनसकेको छैन । शिक्षकले जति बढी उत्तरपुस्तिका जाँच गर्‍यो उति धेरै फाइदा हुने हुँदा हतारमा परीक्षण गर्ने भएकाले पनि यस्तो अवस्था सिर्जना भएको हुनसक्छ । प्रत्येक वर्षको अनुत्तीर्ण प्रतिशत देखेर पनि परीक्षार्थीहरूले निराश भएर आफ्नो पढाइ र परीक्षामा राम्रो प्रदर्शन गर्न सकिरहेका हुँदैनन् ।

सरकारी विद्यालयका विद्यार्थीहरूमा अभ्यासको कमी, घोकन्ते र परीक्षामुखी प्रवृत्ति भएकाले पनि उत्तीर्ण प्रतिशत बढ्न नसकेको हो । उनीहरू परीक्षाका बेलाबाहेक मेहनत गर्दैनन् । अहिलेका छात्रछात्रामा कोचिङ वा ट्युसन गए पास भइन्छ भन्ने प्रवृत्ति देखा परेको छ । शिक्षकले पनि विद्यार्थीलाई मिहिनेत गराएको देखिँदैन । शिक्षक विद्यार्थीप्रति उत्तरदायी हुनुपर्छ । सहयोगी सामग्रीहरू अध्ययन नगरेकाले पनि एसएलसीमा धेरै विद्यार्थी अनुत्तीर्ण भैरहेका हुन्छन् । उनीहरूलाई परीक्षाका बारेमा अभ्यास गराउन बराबर परीक्षा लिइरहनुपर्छ । अभ्यासको अभावले धेरैले एसएलसीमा राम्रो प्रदर्शन गर्न सक्दैनन् । अहिले सरकारी विद्यालयहरूमा तालिम योग्य एवं विषयगत शिक्षकहरूको अभावका कारण पनि परीक्षामा बढी विद्यार्थी अनुत्तीर्ण हुने गरेका छन् । कमजोर विद्यार्थीलाई कक्षा चढाउने चलनका कारण पनि एसएलसीको स्तर बढ्न नसकेको हो ।

विद्यार्थीहरूको उमेर र स्तर विद्यालय र अभिभावकहरूमा समयमै विद्यार्थीहरूको शैक्षिक सुधारप्रति विशेष चनाखो रहने प्रवृत्तिमा कमी, विद्यार्थीहरूको ज्ञान, सीप र अवधारणामा परिस्कार ल्याउने व्यावहारिक शिक्षाभन्दा सैद्धान्तिक शिक्षाप्रति बढी जोड, अघिपछि पनि पढ्ने कार्यमा उचित ध्यान नदिई परीक्षा नजिकिएपछि मात्र पढ्न तत्पर हुने अधिकांश विद्यार्थीको आनीवानी आदिले पनि एसएलसी परीक्षाको परिणाम कमजोर देखापर्ने गरेको छ । यस्तो हुनुमा मूलरूपमा शैक्षिक व्यवस्थापन, गलत शिक्षा नीति एवं असमान शैक्षिक वातावरण पनि केही हदसम्म जिम्मेवार छ ।

यर्थात् जे भए पनि प्रत्येक परीक्षार्थीले आफ्नो सफल जीवनका लागि आत्मसंयमी भै परीक्षामा सम्मिलित हुनुको विकल्प हुँदैन । आफ्नो भविष्य निर्माण केही हदसम्म आफैमा नीहित हुने यथार्थलाई परीक्षामा सहभागी हुने परीक्षार्थीहरूले बिर्सनु हुँदैन । जीवनका कुनै पनि क्षणमा हतोत्साहित नभै अध्ययनप्रतिको आफ्नो कर्तव्यमा लीन हुनसके निश्चित रूपमा सफलता हासिल गर्न सकिन्छ । परीक्षा हुनुअघि नै समयको सदुपयोग गरी हरेक विषयमा होसियार भएर आफूले पढेका सबै कुरा पटक-पटक पढ्ने, अध्ययनका क्रममा सबै विषयलाई समान महत्त्व दिने, नबुझेको कुरा बुझ्न प्रयत्न गरे सफलता प्राप्त गर्न सकिन्छ ।
Source:http://www.ekantipur.com/saptahik/article/?id=4513 Friday/10/June/2011

Friday, June 10, 2011

School's News


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source :http\\www.annapurnapost.com

Tuesday, June 7, 2011

Tips

Tips: 1
Tips: John Von Neumann was a Hungarian-born US mathematician. He was a leader in the design and development of high-speed electronic computers .He designed the first flowchart in 1945.
Tips: 2
The word algorithm oriented from the name of a famous Arab mathematician, “Abu Jafar Muhammad Ibn Musa Al-Khowarizmi. He wrote a famous book called “Rules of Restoration and Reduction”, where he explained the concept of algorithm.


Wednesday, June 1, 2011

Chinese Provert

The beginning of wisdom is to call things by their right names.
                                                 Chinese Provert