plz help me find about oscillating sort and its woking?
410833
An electronic oscillator is an electronic circuit that produces a repetitive electronic signal, often a sine wave or a square wave.\n\nA low-frequency oscillator (or LFO) is an electronic oscillator that generates an AC waveform between 0.1 Hz and 10 Hz. This term is typically used in the field of audio synthesizers, to distinguish it from an audio frequency oscillator.\n\nThere are two main types of electronic oscillator: the harmonic oscillator and the relaxation oscillator.\n\nThe harmonic oscillator produces a sinusoidal output. The basic form of an harmonic oscillator is an electronic amplifier with the output attached to a narrow-band electronic filter, and the output of the filter attached to the input of the amplifier. When the power supply to the amplifier is first switched on, the amplifier's output consists only of noise. The noise travels around the loop, being filtered and re-amplified until it increasingly resembles the desired signal.\n\nA piezoelectric crystal (commonly quartz) may be coupled to the filter to stabilise the frequency of oscillation, resulting in a crystal oscillator.\n\nThere are many ways to implement harmonic oscillators, because there are different ways to amplify and filter. For example oscillator variations include: Hartley oscillator, Colpitts oscillator, Clapp oscillator, Pierce crystal oscillator, Phase-shift oscillator, RC oscillator (Wien Bridge and "Twin-T") \n\nThe relaxation oscillator is often used to produce a non-sinusoidal output, such as a square wave or sawtooth. The oscillator contains a nonlinear component such as a transistor that periodically discharges the energy stored in a capacitor or inductor, causing abrupt changes in the output waveform.\n\nSquare-wave relaxation oscillators can be used to provide the clock signal for sequential logic circuits such as timers and counters, although crystal oscillators are often preferred for their greater stability.\n\nTriangle-wave or sawtooth oscillators are used in the timebase circuits that generate the horizontal deflection signals for cathode ray tubes in analogue oscilloscopes and television sets. In function generators, this triangle wave may then be further shaped into a close approximation of a sine wave.\n\nThe multivibrator and the rotary traveling wave oscillator are another types of relaxation oscillators.
115650
--5 2 0 1 -----  F L O W\n--4 2 0 1 -----  B L O W\n--1 0 2 5 -----  W O L F\n-------------------------\n1 0 4 2 7 ----- N O B L E\n\nBut still N and W value are coming same...does this problem have solution. If you have then can you please send the answer to my mail id hssiddhu@yahoo.com
410291
http://en.wikipedia.org/wiki/Permutation
235678
you can do that by sorting the data source from where you are populating the list. Or while you are populating the list.\nIf you can tell me how you are populating, I can give you more pointers. I am not sure if there is any way to sort a list directly in java script.\n\njainnys\nMind Experts Inc.\nhttp://www.mindexp.com
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Yes, when a list is already sorted, bubblesort will pass through the list once, and find that it does not need to swap any elements. \n\nIt will also use considerably less time if the elements in the list are not too far from their sorted places.
657904
<!-- bubble_sort.js by detour@metalshell.com -->\n\n<!-- Generate 100 random numbers the use bubble sort to -->\n<!-- put them in ascending order. -->\n\n<!-- http://www.metalshell.com/ -->\n\n<html>\n<head>\n<title>JavaScript Random Number Generator.</title>\n</head>\n<body>\n\n<script type="text/javascript">\n\nvar ranarray = new Array(100);\n\n// Generate random numbers to fill ranarray.\nfunction genNumbers(listbox) {\n  var i;\n\n  for(i = 0; i < ranarray.length; i++) {\n    ranarray[i] = Math.random()*100;\n    // Round to nearest integer.\n    ranarray[i] = Math.round(ranarray[i]);\n  }\n\n  // Update the select box list.\n  updateList(listbox);\n}\n\n\nfunction sortNumbers(listbox) {\n  var x, y, holder;\n  // The Bubble Sort method.\n  for(x = 0; x < ranarray.length; x++) {\n    for(y = 0; y < (ranarray.length-1); y++) {\n      if(ranarray[y] > ranarray[y+1]) {\n        holder = ranarray[y+1];\n        ranarray[y+1] = ranarray[y];\n        ranarray[y] = holder;\n      }\n    }\n  }\n\n  // Update the select box list.\n  updateList(listbox);\n}\n\n// Assign values in array to values in the select box.\nfunction updateList(listbox) {\n  var i;\n  for(i = 0; i < ranarray.length; i++) {\n    if(listbox.options[i] == null) {\n      listbox.options[i] = new Option(ranarray[i]);\n    } else {\n      listbox.options[i].text = ranarray[i];\n    }\n  }\n\n}\n\n</script>\n\n<form>\n  <select name="ranlist" size="20" style="width:200px">\n  </select><br><br>\n  <input type="button" value="Generate Numbers"\n         onclick="genNumbers(this.form.ranlist);">\n  <input type="button" value="Bubble Sort Numbers"\n         onclick="sortNumbers(this.form.ranlist);">\n</form>\n\n\n</body>\n</html>
982925
in this order:\n\nprominent\nquaint\nsnare\neradicate\nvengeance\nmakeshift
697965
every number from 0 to 10\n\nplus placed in alphabetical order
428865
order of operation - parenthesis, power, multi/divide, and add/subtract
663163
go to run-regedit\nand then open HKEY_CLASSES_ROOT&#92;CLSID&#92;{450D8... key\nthen see the SortOrderIndex and make its value 54.\n\n:-)
457195
the simple program given above if fine.. in terms of the conncept.. but there's a bug... \nthe problem is with looping.. coz when u enter a number such as 80301  or 55302 ( a no. having 0 in between), the loop breaks at the point when zero is encountered.\n\n so the output in the first eg. is  8 0 ,sum= 8 (instead of correct 8 0 3 0 1 , sum=12\n\n\nso the logic of handling the number is absolutely correct but the looping is a comparatively difficult job in this case!
350177
Sometimes adding some indents will help figure out what's going on.  In your algorithm, you have a lot of nested if and while statements, so space them out like this first to see how each is related to each other (I'm using periods because Yahoo won't show spaces):\n\ni <- 0\nwhile (i < n) do\n...j <- i+1\n...m <- a[i]\n...while (j < n) do\n......if (m > a[j]) then\n.........m <- a[j]\n......end if\n......j <- j + 1\n...end while\n...b[i] <- m\n...i <- i + 1\nend while\n\nYour i is a counter that starts with 0, and your j is a counter that starts with 1.  j will always be more than i.\n\nThe "while (i < n) do" line limits this algorithm to 3 loops, since n is 3.\n\nDirectly after this statement, j is established as a counter, and m is defined to be the i-th element of your matrix.  The next while statement then compares m (the i-th element in the matrix) to the element after this (using j to read the next element).\n\nIf m is larger, then the if statement makes m take on the smaller value.  If m is smaller, nothing happens and j is indexed to the next value.  Then the process repeats itself.  If we get to the end of the matrix and m isn't replaced anymore, it becomes the i-th value in matrix B.  After this, i is indexed to the next element in matrix A and the entire process is repeated once again, until all of matrix A has been checked.\n\nThe result then is that through m, matrix A is sorted least to greatest and is written into matrix B, which is answer (d).
473239
Several problems:\n\n>  bubble_sort(names[3]);\n\nthis is accessing the 4th element of the names array (first 3 elements are at indices 0, 1, and 2). this is why you're getting your output, it's not sorting anything! i'm surprised this program didn't crash. instead you simply want to pass 'names' as the parameter - this means you are passing the entire array as a parameter, not a single element of the array. proper:\n   bubble_sort(names);\n\n>           if (array[j+1] > array[j])\n\nthis would sort the array opposite the way you want it. simple fix is using less-than instead of greater-than. BUT, if i'm not mistaken, you can't compare strings like that. you have to use strcmp, like this:\n            if (strcmp(array[j+1], array[j]) < 0)\notherwise, you are just comparing pointers.
367272
Suppose the number of combinations of k objects out of n objects without respect to order chosen is C(n, k). If we wanted to count order, we would multiply each combination by the number of ways it could be permuted. Thus P(k)*C(n, k) is the number of ways one could choose k people out of n people with regard to order. \nBut we can easily count this number another way. The number of possible choices for the first person is n, for the second person n-1, and so on until we reach n - k + 1. Thus, we get the equation P(k)*C(n, k) = P(n, k) = n*(n - 1)*...*(n - k + 1) = n!/(n - k)!. Solving for C(n, k), we get C(n, k) = n!/[P(k)(n - k)!] = n!/[k!(n - k)!]. \nNow you should be able to derive it whenever you get confused.
713772
ok, create 3 methods, one called printHours(int time), printMinutes(int time) adn another called printSeconds(int time).\n\nin those methods just print to the screen(time / 24) for hours.\ntime  60 for seconds adn time devided by 60 for minutes, adn remeber to return any remainders the the calling function.(pass the remainder to the time unit below). thats it.\n\n\nP.S. welldone for giving it a try rather than just giveing the question and expecting the answers.
2509
Here's a java solution that uses recursion.  The permute function takes two arguments: a done portion and a todo portion.   The basic idea is that each of the characters in the todo portion could be the next character.  So, at each step, n recursive calls are made, where n is the length of todo.  Each call chooses a different character as the next character to go into the done portion.  So, the first time the number of recursive calls will equal the number of characters in the initial string.  Each of these calls will make further recursive calls equal to the number of characters in the initial string minus one.  And so on.  Thus, the number of solutions will be n! = n*n-1*n-2*...*1, where n is the number of characters in the initial string.\n\nThe program below was tested with java's sun compiler 1.4.2.\npublic class perm\n{\n  public perm(String S)\n  {     \n    permute("", S);\n  }     \n  public void permute(String done, String todo) \n  {     \n    if ((todo.length()==0)||\n        (todo.equals("")))\n      System.out.println(done);\n    else        \n    {           \n      for (int i=0; i<todo.length(); i++)\n      {                 \n        String c = todo.substring(0,1);\n        todo = todo.substring(1);\n        permute(done+c, todo);  \n        todo = todo + c;        \n      }                 \n    }           \n  }     \n  public static void main(String args[])\n  {     \n    if ((args.length==0)||\n        (args[0].length()==0)||\n        (args[0].equals("")))\n      return;           \n    perm p = new perm(args[0]);\n  }     \n}
634160
The fasting is Quicksort, but also the most complicated. If you are looking at sorting hundreds of numbers or more, Quicksort with its recursive elements are the way to go.