sequence.h

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/*   This program is free software: you can redistribute it and/or modify
 *   it under the terms of the GNU General Public License as published by
 *   the Free Software Foundation, either version 3 of the License, or
 *   (at your option) any later version.
 *
 *   This program is distributed in the hope that it will be useful,
 *   but WITHOUT ANY WARRANTY; without even the implied warranty of
 *   MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
 *   GNU General Public License for more details.
 *
 *   You should have received a copy of the GNU General Public License
 *   along with this program.  If not, see <http://www.gnu.org/licenses/>.
 *
 *   Copyright (C) 2009 - 2012 Jun Liu and Jieping Ye 
 */

#ifndef  SEQUENCE_SLEP
#define  SEQUENCE_SLEP

#include <shogun/lib/config.h>

#include <stdlib.h>
#include <stdio.h>
#include <time.h>
#include <math.h>


/*
 * In this file, we propose the algorithms for solving the problem:
 *
 * min   1/2 \|x - u\|^2
 * s.t.  x1 \ge x2 \ge x3 \ge ... \ge xn \ge 0
 *
 *
 */

/*
 *
 * x= sequence_bottomup(u,n)
 *
 * we compute using a bottom up order
 * 
 */

void sequence_bottomup(double *x, double *u, int n){
    int i, j;
    int *location=(int *)malloc(sizeof(int)*n);
    int num;

    if(!location){
        printf("\n Allocation of array failure!");
        exit(1);
    }


    /*
     * compute the maximal mean from the root to the i-th point:
     *
     * x[i]: the maximal mean
     * location[i]: the ending index of the mean
     *
     */


    /* process the last element*/
    if (n<1){
        printf("\n n=%d should be an integer over 1!",n);
        exit(1);
    }
    else{
        i=n-1;        
        x[i]=u[i];
        location[i]=i; 
    }

    /*process the remaining elements in a bottom-up recursive manner*/
    for(i=n-2;i>=0;i--){


        if (u[i]>x[i+1]){
            x[i]=u[i];
            location[i]=i;            
        }
        else{
            /*make use of x[i: (n-1)] and location[i: (n-1)] for update*/

            /*merge with the first group*/
            num=location[i+1]-i;
            x[i]=(u[i] + x[i+1]*num)/(num+1);
            location[i]=location[i+1];
            j=location[i+1]+1;

            /*If necessary, we need to further merge with the remainig groups */
            for(;j<n;){
                if(x[i] <= x[j]){

                    num=location[j]-j +1;
                    x[i]=( x[i]* (j-i) + x[j]* num ) / (location[j] -i +1);
                    location[i]=location[j];

                    j=location[j]+1;
                }
                else
                    break;
            }                
        }
    }

    /*
       for(i=0;i<30;i++)
       printf("\n x[%d]=%2.5f, location[%d]=%d",i+1, x[i], i+1, location[i]+1);
       */

    /*
     * compute the solution x with the mean and location
     *
     */

    for(i=0;i<n;){

        if (x[i]>0){
            for(j=i+1;j<=location[i];j++){
                x[j]=x[i];
            }

            i=location[i]+1;
        }
        else{
            for(j=i;j<n;j++)
                x[j]=0;
            break;
        }
    }

    free(location);
}


/*
 *
 * x= sequence_topdown(u,n)
 *
 * we compute using a top to down order
 * 
 */

void sequence_topdown(double *x, double *u, int n){
    int i, j;
    double sum, max, mean;
    int num;
    int *location=(int *)malloc(sizeof(int)*n);


    if(!location){
        printf("\n Allocation of array failure!");
        exit(1);
    }

    for(i=0;i<n;){

        /*
         * From each root node i, we compute the maximal mean from.
         *
         */

        max=0;
        location[i]=i;

        sum=0;
        num=1;        
        for(j=i;j<n;j++){
            sum+=u[j];
            mean=sum/num;            
            num++;

            /* record the most largest mean and the location*/
            if (mean >= max){                
                max=mean;
                location[i]=j;
            }
        }

        if (max>0){
            x[i]=max; /*record the maximal mean*/
            i=location[i]+1; /*the next i*/
        }
        else{
            x[i]=-1; /* any value less or equal to 0
                      *
                      * This shows that the remaining elements should be zero
                      *
                      */
            break;
        }
    }


    /*
     * compute the solution x with the mean and location
     *
     */

    for(i=0;i<n;){

        if (x[i]>0){
            for(j=i+1;j<=location[i];j++){
                x[j]=x[i];
            }

            i=location[i]+1;
        }
        else{
            for(j=i;j<n;j++)
                x[j]=0;
            break;
        }
    }

    free(location);
}
#endif   /* ----- #ifndef SEQUENCE_SLEP  ----- */