The Little Master

Wednesday, August 22, 2012

Downloading a complete page from website

when working with css or js we often convenient to get the website page on our desktop. We generally endup saving as html, then saving each of the css and js files. I discovered an easier way to do the same.

This is supported in linux or cygwin (linux version on windows). Run this command

Command

wget -E -H -k -K -p http://www.something.com

Monday, July 23, 2012

Generating gradiants with out using images

This is written in JQuery. Below is the code.

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.7.2/jquery.min.js" > </script>

<script>
$(function(){
gradiant('div.newGradiant',[255,0,0],[0,0,0]);
});

function gradiant(selector,rgb1,rgb2){
var $src = $(selector);
var height = $src.height()*1;
$src.css('background','yellow');
var x = $src.offset().left + $src.width();
var y = $src.offset().top;
//var r=256,g=256;b=256;
var r=rgb1[0],g=rgb1[1];b=rgb1[2];
var rr=rgb2[0],gg=rgb2[1];bb=rgb2[2];
for(var i=0;i<height;i++){
var tr= Math.round(r+((rr-r)*i/height));
var tg=Math.round(g+((gg-g)*i/height));
var tb=Math.round(b+((bb-b)*i/height));
var newColor = hex(tr)+hex(tg)+hex(tb);
var div =('<DIV style="BORDER-BOTTOM: #ffffff 0px;'
+' BORDER-LEFT: #ffffff 0px solid; BACKGROUND-COLOR: '
+'#'+ newColor +'; MIN-HEIGHT: 1px; HEIGHT:'
+' 1px; FONT-SIZE: 1px; OVERFLOW: hidden; BORDER-TOP: '
+'#ffffff 0px; BORDER-RIGHT: #ffffff 0px solid"></DIV>');
$src.append($(div));
}
}

var hexDigits = new Array("0","1","2","3","4","5","6","7","8","9","a","b","c","d","e","f"); 

function hex(x) {
return isNaN(x) ? "00" : hexDigits[(x - x % 16) / 16] + hexDigits[x % 16];
}

</script>

<div class="newGradiant" style="width:100px; height:200px; background:yellow; border: 1px solid black">

This generates the gradiant as follows. The algo can be modified to take the edge rgb values and then do the magic. This is just a naive example ;)

Tested this on IE-7, IE-8, IE-9. Chrome 20.0, firefox 13.0 version.

The only issue i see is when we combine this with jQuery corner apis. The results are working in IE but not in chrome!!!

eclipse: deploying classpath jars into tomcat container

Aaaah! this is a big pain.. u will have all jars in classpath, and when u say run as server they dont get copied to the server. the Fix is simpl ( eclipse should have taken care of it by default.. however.. ).

click on the poject,
press ctrl + enter.
Now u see a properties window of the project
search for deployment assemble and click on it.
Now in web-deployment assembly plane click on add button
select java build path entities
click on next. now select all jars (ctrl + a); then OK
now click finish.

Deploy the stuff on tomcat and u will see classpath entities are being copied to the server

Sunday, July 8, 2012

Near Random number… 2

In continuation to my previous post.. i made a few more refinements and made the program more simple..

The advantage of this new one is it allows the total digits we want to alter w.r.t the previous one.
we can dictate the number of variants between consecutive integers

it reads as follows

import java.util.Random;

public class NearRandomIncrements2 {
    private static final Random    rand                = new Random(99L);
    private static final int    maxVariants            = 10;
    private static final int    maxVariableDigits    = 10000;
    private static int[]        variantArr            = new int[maxVariants];
    private static int[]        cummulativeArr        = new int[maxVariants];
    private static int            sum;

    static {
        for (int i = 0; i < maxVariants; i++) {
            int nextVariant = Math.abs(rand.nextInt()) % maxVariableDigits;
            variantArr[i] = nextVariant;
            sum += nextVariant;
            cummulativeArr[i] = sum;
            System.out.println(">>" + nextVariant + " cumm:" + cummulativeArr[i]);
        }
    }

    public static void main(String[] s) {
        long prev = 0;
        for (long i = 0L; i < 40; i++) {
            int index = (int) (i % (maxVariants));
            long cycles = (i / (maxVariants)) + 1;
            long newVal = cycles + cummulativeArr[index] + index;
            if (newVal < 0) {
                newVal = Long.MAX_VALUE + newVal + 1;
            }
            // if (newVal >= 0 && newVal < 100) {
            System.out.println(i + " new no:" + newVal + " cycle:" + cycles + " variant:" + (newVal - prev));
            // }
            prev = newVal;
        }
    }
}

For the given maxVariable digits the random numbers generated are as follows

>>5722 cumm:5722
>>6916 cumm:12638
>>4859 cumm:17497
>>1273 cumm:18770
>>696 cumm:19466
>>3525 cumm:22991
>>5948 cumm:28939
>>9283 cumm:38222
>>5856 cumm:44078
>>9198 cumm:53276

The numbers remain the same on multiple runs also.

0 new no:5723 cycle:1 variant:5723
1 new no:12640 cycle:1 variant:6917
2 new no:17500 cycle:1 variant:4860
3 new no:18774 cycle:1 variant:1274
4 new no:19471 cycle:1 variant:697
5 new no:22997 cycle:1 variant:3526
6 new no:28946 cycle:1 variant:5949
7 new no:38230 cycle:1 variant:9284
8 new no:44087 cycle:1 variant:5857
9 new no:53286 cycle:1 variant:9199
10 new no:5724 cycle:2 variant:-47562
11 new no:12641 cycle:2 variant:6917
12 new no:17501 cycle:2 variant:4860
13 new no:18775 cycle:2 variant:1274
14 new no:19472 cycle:2 variant:697
15 new no:22998 cycle:2 variant:3526
16 new no:28947 cycle:2 variant:5949
17 new no:38231 cycle:2 variant:9284
18 new no:44088 cycle:2 variant:5857
19 new no:53287 cycle:2 variant:9199
20 new no:5725 cycle:3 variant:-47562
21 new no:12642 cycle:3 variant:6917
22 new no:17502 cycle:3 variant:4860
23 new no:18776 cycle:3 variant:1274
24 new no:19473 cycle:3 variant:697
25 new no:22999 cycle:3 variant:3526
26 new no:28948 cycle:3 variant:5949
27 new no:38232 cycle:3 variant:9284
28 new no:44089 cycle:3 variant:5857
29 new no:53288 cycle:3 variant:9199
30 new no:5726 cycle:4 variant:-47562
31 new no:12643 cycle:4 variant:6917
32 new no:17503 cycle:4 variant:4860
33 new no:18777 cycle:4 variant:1274
34 new no:19474 cycle:4 variant:697
35 new no:23000 cycle:4 variant:3526
36 new no:28949 cycle:4 variant:5949
37 new no:38233 cycle:4 variant:9284
38 new no:44090 cycle:4 variant:5857
39 new no:53289 cycle:4 variant:9199

if u observe no two numbers are related to each other; and to know the sequence we need to see atleast 30 consecutive numbers (say we have maxVariant by 10 numbers) So by putting the maxVariants a high number and giving the gerated numbers to a trusted client we may use this algo to generate proxies not loosing even one number in the seq bts 0-> Long.Max_Value

any better soln?

Thursday, July 5, 2012

Near random–Converting a sequence no to random no

Q: so the statement of the problem is as follows. I want to generate unique numbers from 1 to Max (say max is Long.MAX_LONG). The generated sequence should not be predictable. SO it means we need to generate uniq random nubers and ensuer they dont repeate. Remember we cannot use current time stamp + some increment number as we have limited range and we need to use it as affectively as possible.

So i wrote a func that transforms a function to (seemingly) random number. Its as follows.

public class NearRandomIncrements {
    private static final long    salt    = 0xABCDEF12345628ACL;
    private static final long    salt2    = 0xeE72CC072837198DL;

    public static void main(String[] s) {
        long prev = 0;
        for (long i = 1L; i > 0 && i < Long.MAX_VALUE; i++) {
            long withSalt = Long.reverseBytes(Long.rotateLeft(i, 31) ^ salt);
            long withSalt2 = Long.rotateLeft(withSalt, 17) ^ salt2;
            long variation = withSalt2 - prev;
            prev = withSalt2;
            System.out.println(i + " " + withSalt2 + " variation:" + Math.abs(variation));
        }
    }
}

This genates the output as follows

1 4835434266867286493 variation:4835434266867286493
2 4763378871852614109 variation:72055395014672384
3 4835436465890542045 variation:72057594037927936
4 4763381070875869661 variation:72055395014672384
5 4835438664913797597 variation:72057594037927936
6 4763383269899125213 variation:72055395014672384
7 4835440863937053149 variation:72057594037927936
8 4763367876736336349 variation:72072987200716800
9 4835425470774264285 variation:72057594037927936
10 4763370075759591901 variation:72055395014672384

So if u closely observe based on variation we see a pattern; and we can predict the next number. But if u see the i=8 part; we see a new variation creeping in. So even if a person gets hold of the sequence he may predict only limited number of numbers only. The solution may not be perfect, but i think its a gud start. What do u think?