DPM(Defomable Parts Model) 源码分析-训练（三）

申明：本文非笔者原创，原文转载自：http://blog.csdn.net/ttransposition/article/details/12954631

DPM(Defomable Parts Model)原理

首先调用格式：

example:
pascal('person', 2);   % train and evaluate a 2 component person model

pascal_train.m

 function model = pascal_train(cls, n) % n=2

 % model = pascal_train(cls)
 % Train a model using the PASCAL dataset.

 globals;
 %----------读取正负样本-----------------------
 % pos.im,neg.im存储了图像路径，pos.x1..pos.y2为box,负样本无box
 [pos, neg] = pascal_data(cls);

 % 按照长宽比，分成等量的两部分? 即将 component label  固定，phase2时，该值为latent variable。  spos为索引
 spos = split(pos, n);

 % -----------phase 1 : train root filters using warped positives & random negatives-----------
 try
   load([cachedir cls '_random']);
 catch
 % -----------------------------phas 1--------------------------------
 % 初始化 rootfilters
   for i=1:n
     models{i} = initmodel(spos{i});
     %---------train-------------
     % model.rootfilters{i}.w
     % model.offsets{i}.w
     models{i} = train(cls, models{i}, spos{i}, neg, 1, 1, 1, 1, 2^28);

   end
   save([cachedir cls '_random'], 'models');
 end

 % -----------------phase2-------------------------------------------
 % :merge models and train using latent detections & hard negatives
 try
   load([cachedir cls '_hard']);
 catch
   model = mergemodels(models);
   model = train(cls, model, pos, neg(1:200), 0, 0, 2, 2, 2^28, true, 0.7);
   save([cachedir cls '_hard'], 'model');
 end
 %----------------phase 3----------------------------------------------
 % add parts and update models using latent detections & hard negatives.
 try
   load([cachedir cls '_parts']);
 catch
   for i=1:n
     model = addparts(model, i, 6);
   end
   % use more data mining iterations in the beginning
   model = train(cls, model, pos, neg(1:200), 0, 0, 1, 4, 2^30, true, 0.7);
   model = train(cls, model, pos, neg(1:200), 0, 0, 6, 2, 2^30, true, 0.7, true);
   save([cachedir cls '_parts'], 'model');
 end

 % update models using full set of negatives.
 try
   load([cachedir cls '_mine']);
 catch
   model = train(cls, model, pos, neg, 0, 0, 1, 3, 2^30, true, 0.7, true, ...
                 0.003*model.numcomponents, 2);
   save([cachedir cls '_mine'], 'model');
 end

 % train bounding box prediction
 try
   load([cachedir cls '_final']);
 catch
  % 论文中说用最小二乘，怎么直接相除了，都不考虑矩阵的奇异性
   model = trainbox(cls, model, pos, 0.7);
   save([cachedir cls '_final'], 'model');
 end

initmodel.m

 function model = initmodel(pos, sbin, size)

 % model = initmodel(pos, sbin, size)
 % Initialize model structure.
 %
 % If not supplied the dimensions of the model template are computed
 % from statistics in the postive examples.
 %
 % This should be documented! :-)
 % model.sbin         8
 % model.interval     10
 % model.numblocks     phase 1 ：单独训练rootfilter时为2，offset,rootfilter；phase 2，为 4
 % model.numcomponents  1
 % model.blocksizes     （1）=1，（2）= root.h*root.w/2*31
 % model.regmult        0,1
 % model.learnmult      20,1
 % model.maxsize        root 的size
 % model.minsize
 % model.rootfilters{i}
 %   .size               以sbin为单位，尺寸为综合各样本的h/w，area计算出来的
 %   .w
 %   .blocklabel        blocklabel是为编号，offset（2）,rootfilter（2）,partfilter（12 or less）,def （12 same as part）虽然意义不同但是放在一起统一编号
 % model.partfilters{i}
 %   .w
 %   .blocklabel
 % model.defs{i}
 %   .anchor
 %   .w
 %   .blocklabel
 % model.offsets{i}
 %   .w               0
 %   .blocklabel       1
 % model.components{i}
 %   .rootindex    1
 %   .parts{j}
 %     .partindex
 %     .defindex
 %   .offsetindex    1
 %   .dim             2 + model.blocksizes(1) + model.blocksizes(2)
 %   .numblocks       2

 % pick mode of aspect ratios
 h = [pos(:).y2]' - [pos(:).y1]' + 1;
 w = [pos(:).x2]' - [pos(:).x1]' + 1;
 xx = -2:.02:2;
 filter = exp(-[-100:100].^2/400); % e^-25,e^25
 aspects = hist(log(h./w), xx); %
 aspects = convn(aspects, filter, 'same');
 [peak, I] = max(aspects);
 aspect = exp(xx(I)); %滤波后最大的h/w，作为最典型的h/w

 % pick 20 percentile area
 areas = sort(h.*w);
 area = areas(floor(length(areas) * 0.2)); % 比它大的，可以缩放，比该尺寸小的呢？
 area = max(min(area, 5000), 3000); %限制在 3000-5000

 % pick dimensions
 w = sqrt(area/aspect);
 h = w*aspect;

 % size of HOG features
 if nargin < 4
   model.sbin = 8;
 else
   model.sbin = sbin;
 end

 % size of root filter
 if nargin < 5
   model.rootfilters{1}.size = [round(h/model.sbin) round(w/model.sbin)];
 else
   model.rootfilters{1}.size = size;
 end

 % set up offset
 model.offsets{1}.w = 0;
 model.offsets{1}.blocklabel = 1;
 model.blocksizes(1) = 1;
 model.regmult(1) = 0;
 model.learnmult(1) = 20;
 model.lowerbounds{1} = -100;

 % set up root filter
 model.rootfilters{1}.w = zeros([model.rootfilters{1}.size 31]);
 height = model.rootfilters{1}.size(1);
 % root filter is symmetricf
 width = ceil(model.rootfilters{1}.size(2)/2);  % ？？？ /2
 model.rootfilters{1}.blocklabel = 2;
 model.blocksizes(2) = width * height * 31;
 model.regmult(2) = 1;
 model.learnmult(2) = 1;
 model.lowerbounds{2} = -100*ones(model.blocksizes(2),1);

 % set up one component model
 model.components{1}.rootindex = 1;
 model.components{1}.offsetindex = 1;
 model.components{1}.parts = {};
 model.components{1}.dim = 2 + model.blocksizes(1) + model.blocksizes(2);
 model.components{1}.numblocks = 2;

 % initialize the rest of the model structure
 model.interval = 10;
 model.numcomponents = 1;
 model.numblocks = 2;
 model.partfilters = {};
 model.defs = {};
 model.maxsize = model.rootfilters{1}.size;
 model.minsize = model.rootfilters{1}.size;

 

learn.cc

 #include <stdio.h>
 #include <stdlib.h>
 #include <string.h>
 #include <math.h>
 #include <sys/time.h>
 #include <errno.h>

 /*
  * Optimize LSVM objective function via gradient descent.
  *
  * We use an adaptive cache mechanism.  After a negative example
  * scores beyond the margin multiple times it is removed from the
  * training set for a fixed number of iterations.
  */

 // Data File Format
 // EXAMPLE*
 //
 // EXAMPLE:
 //  long label          ints
 //  blocks              int
 //  dim                 int
 //  DATA{blocks}
 //
 // DATA:
 //  block label         float
 //  block data          floats
 //
 // Internal Binary Format
 //  len           int (byte length of EXAMPLE)
 //  EXAMPLE       <see above>
 //  unique flag   byte

 // number of iterations
 #define ITER 5000000

 // small cache parameters
 #define INCACHE 3
 #define WAIT 10

 // error checking
 #define check(e) \
 (e ? (void)0 : (printf("%s:%u error: %s\n%s\n", __FILE__, __LINE__, #e, strerror(errno)), exit(1)))

 // number of non-zero blocks in example ex
 #define NUM_NONZERO(ex) (((int *)ex)[labelsize+1])

 // float pointer to data segment of example ex
 #define EX_DATA(ex) ((float *)(ex + sizeof(int)*(labelsize+3)))

 // class label (+1 or -1) for the example
 #define LABEL(ex) (((int *)ex)[1])

 // block label (converted to 0-based index)
 #define BLOCK_IDX(data) (((int)data[0])-1)

 int labelsize;
 int dim;

 // comparison function for sorting examples
 // 参见 http://blog.sina.com.cn/s/blog_5155e8d401009145.html
 int comp(const void *a, const void *b) {
   // sort by extended label first, and whole example second...

   //逐字节比较的，当buf1<buf2时，返回值<0，当buf1=buf2时，返回值=0，当buf1>buf2时，返回值>0
   // 先比较这五个量 [label id level x y]，也就是说按照 样本类别->id->level->x->y排序样本
   int c = memcmp(*((char **)a) + sizeof(int),
          *((char **)b) + sizeof(int),
          labelsize*sizeof(int));// 5
   if (c) //label 不相等
     return c;

   // labels are the same ，怎么可能会一样呢 id在正负样本集内从1开始是递增的啊  phase 2 阶段同一张图片产生的样本，id都是一样的
   int alen = **((int **)a);
   int blen = **((int **)b);
   if (alen == blen) //长度一样
     return memcmp(*((char **)a) + sizeof(int),
           *((char **)b) + sizeof(int),
           alen); //真霸气，所有字节都比较……
   return ((alen < blen) ? -1 : 1);//按长度排序
 }

 // a collapsed example is a sequence of examples
 struct collapsed {
   char **seq;
   int num;
 };

 // set of collapsed examples
 struct data {
   collapsed *x;
   int num;
   int numblocks;
   int *blocksizes;
   float *regmult;
   float *learnmult;
 };

 // seed the random number generator with the current time
 void seed_time() {
  struct timeval tp;
  check(gettimeofday(&tp, NULL) == 0);
  srand48((long)tp.tv_usec);
 }

 static inline double min(double x, double y) { return (x <= y ? x : y); }
 static inline double max(double x, double y) { return (x <= y ? y : x); }

 // gradient descent
 //---------------参照论文公式17 后的步骤---------------------------------------
 void gd(double C, double J, data X, double **w, double **lb) {
 //  C=0.0002, J=1, X, w==0, lb==-100);
 //
   int num = X.num; //组数

   // state for random permutations
   int *perm = (int *)malloc(sizeof(int)*X.num);
   check(perm != NULL);

   // state for small cache
   int *W = (int *)malloc(sizeof(int)*num);
   check(W != NULL);
   for (int j = 0; j < num; j++)
     W[j] = 0;

   int t = 0;
   while (t < ITER) {  // 5000000 ，霸气……
     // pick random permutation
     for (int i = 0; i < num; i++) //组数
       perm[i] = i;
     //-------打乱顺序-----
     // 论文中是随机选择一个样本，这里是随机排好序，再顺序取。
     // 类似于随机取，但是这里能保证取到全部样本，避免单个样本重复被抽到，重复作用
     for (int swapi = 0; swapi < num; swapi++) {
       int swapj = (int)(drand48()*(num-swapi)) + swapi; //drand48 产生 0-1之间的均匀分布
       int tmp = perm[swapi];
       perm[swapi] = perm[swapj];
       perm[swapj] = tmp;
     }

     // count number of examples in the small cache
     int cnum = 0; //下面的循环部分的实际循环次数
     for (int i = 0; i < num; i++) {
       if (W[i] <= INCACHE) // 3
         cnum++;
     }
     //-------------------------------------------------------
     for (int swapi = 0; swapi < num; swapi++) {
       // select example
       int i = perm[swapi];
       collapsed x = X.x[i];

       // skip if example is not in small cache
       //负样本分对一次+1，分错一次清为0
       //连续三次都分对了，那么这个样本很有可能是 easy 样本
       //直接让他罚停四次迭代
       if (W[i] > INCACHE) { //3
             W[i]--;
             continue;
       }

       // learning rate
       double T = t + 1000.0; //学习率，直接1/t太大了
       double rateX = cnum * C / T;
       double rateR = 1.0 / T;

       if (t % 10000 == 0) {
         printf(".");
         fflush(stdout); //清除文件缓冲区，文件以写方式打开时将缓冲区内容写入文件
       }
       t++;

       // compute max over latent placements
       //  -----step 3----
       int M = -1;
       double V = 0;
       // 组内循环，选择 Zi=argmax β*f 即文中的第3部
       // 训练rootfiter时，x.num=1,因为随机产生的负样本其id不同
       for (int m = 0; m < x.num; m++) {
         double val = 0;
         char *ptr = x.seq[m];
         float *data = EX_DATA(ptr); //特征数据的地址 第9个数据开始，
         //后面跟着是 block1 label | block2 data|block2 lable | block2 data
         //                 1      |       1    |     2       |  h*w/2*31个float
         int blocks = NUM_NONZERO(ptr); // phase 1，phase 2 : 2 个,offset,rootfilter
         for (int j = 0; j < blocks; j++) {
           int b = BLOCK_IDX(data); //
           data++;
           for (int k = 0; k < X.blocksizes[b]; k++)//（1）=1，（2）= root.h*root.w/2*31
             val += w[b][k] * data[k]; //第一次循环是0
           data += X.blocksizes[b];
         }
         if (M < 0 || val > V) {
           M = m;
           V = val;
         }
       }

       // update model
       //-----step.4 也算了step.5 的一半 ---------------
       // 梯度下降，减小 w
       for (int j = 0; j < X.numblocks; j++) {// 2
         double mult = rateR * X.regmult[j] * X.learnmult[j]; // 0,1  20,1,1/T，对于block2,学习率at就是 1/t,block 1 为0
         for (int k = 0; k < X.blocksizes[j]; k++) {
           w[j][k] -= mult * w[j][k]; //不管是分对了，还是分错了，都要减掉 at*β,见公式17下的4,5
         }
       }
       char *ptr = x.seq[M];
       int label = LABEL(ptr);
       //----step.5----------分错了，往梯度的负方向移动
       if (label * V < 1.0)
       {
         W[i] = 0;
         float *data = EX_DATA(ptr);
         int blocks = NUM_NONZERO(ptr);
         for (int j = 0; j < blocks; j++) {
             int b = BLOCK_IDX(data);
             //  yi*cnum * C / T*1,见论文中 公式16,17
             double mult = (label > 0 ? J : -1) * rateX * X.learnmult[b];
             data++;
             for (int k = 0; k < X.blocksizes[b]; k++)
                 w[b][k] += mult * data[k];
             data += X.blocksizes[b];
         }
       } else if (label == -1)
       {
             if (W[i] == INCACHE) //3
                 W[i] = WAIT; //10
             else
                 W[i]++;
       }
     }

     // apply lowerbounds
     for (int j = 0; j < X.numblocks; j++) {
       for (int k = 0; k < X.blocksizes[j]; k++) {
         w[j][k] = max(w[j][k], lb[j][k]);
       }
     }

   }

   free(perm);
   free(W);
 }

 // score examples
 double *score(data X, char **examples, int num, double **w) {
   double *s = (double *)malloc(sizeof(double)*num);
   check(s != NULL);
   for (int i = 0; i < num; i++) {
     s[i] = 0.0;
     float *data = EX_DATA(examples[i]);
     int blocks = NUM_NONZERO(examples[i]);
     for (int j = 0; j < blocks; j++) {
       int b = BLOCK_IDX(data);
       data++;
       for (int k = 0; k < X.blocksizes[b]; k++)
         s[i] += w[b][k] * data[k];
       data += X.blocksizes[b];
     }
   }
   return s;
 }

 // merge examples with identical labels
 void collapse(data *X, char **examples, int num) {
 //&X, sorted, num_unique
   collapsed *x = (collapsed *)malloc(sizeof(collapsed)*num);
   check(x != NULL);
   int i = 0;
   x[0].seq = examples;
   x[0].num = 1;
   for (int j = 1; j < num; j++) {
     if (!memcmp(x[i].seq[0]+sizeof(int), examples[j]+sizeof(int),
         labelsize*sizeof(int))) {
       x[i].num++; //如果label 五个量相同
     } else {
       i++;
       x[i].seq = &(examples[j]);
       x[i].num = 1;
     }
   }
   X->x = x;
   X->num = i+1;
 }

 //调用参数 C=0.0002, J=1, hdrfile, datfile, modfile, inffile, lobfile
 int main(int argc, char **argv) {
   seed_time();
   int count;
   data X;

   // command line arguments
   check(argc == 8);
   double C = atof(argv[1]);
   double J = atof(argv[2]);
   char *hdrfile = argv[3];
   char *datfile = argv[4];
   char *modfile = argv[5];
   char *inffile = argv[6];
   char *lobfile = argv[7];

   // read header file
   FILE *f = fopen(hdrfile, "rb");
   check(f != NULL);
   int header[3];
   count = fread(header, sizeof(int), 3, f);
   check(count == 3);
   int num = header[0]; //正负样本总数
   labelsize = header[1]; // labelsize = 5;  [label id level x y]
   X.numblocks = header[2]; // 2
   X.blocksizes = (int *)malloc(X.numblocks*sizeof(int)); //（1）=1，（2）= root.h*root.w/2*31
   count = fread(X.blocksizes, sizeof(int), X.numblocks, f);
   check(count == X.numblocks);
   X.regmult = (float *)malloc(sizeof(float)*X.numblocks); //0 ，1
   check(X.regmult != NULL);
   count = fread(X.regmult, sizeof(float), X.numblocks, f);
   check(count == X.numblocks);
   X.learnmult = (float *)malloc(sizeof(float)*X.numblocks);//20， 1
   check(X.learnmult != NULL);
   count = fread(X.learnmult, sizeof(float), X.numblocks, f);
   check(count == X.numblocks);
   check(num != 0);
   fclose(f);
   printf("%d examples with label size %d and %d blocks\n",
      num, labelsize, X.numblocks);
   printf("block size, regularization multiplier, learning rate multiplier\n");
   dim = 0;
   for (int i = 0; i < X.numblocks; i++) {
     dim += X.blocksizes[i];
     printf("%d, %.2f, %.2f\n", X.blocksizes[i], X.regmult[i], X.learnmult[i]);
   }

   // ---------------从 datfile 读取  正负 examples----------------
   // examples [i] 存储了第i个样本的信息 长度为 1 int + 7 int +dim 个float + 1 byte
   // 1 int      legth 样本包括信息头在内的总字节长度
   // 7 int      [1/-1 id 0 0 0 2 dim] ,id为样本编号，[label id level centry_x centry_y]，2是block个数
   // dim float  feature,dim=2+1+root.h*root.w/2*31,意义如下
   //         block1 label | block2 data|block2 lable | block2 data
   //               1      |       1    |     2       |  h*w/2*31个float
   // 1 byte     unique=0
   f = fopen(datfile, "rb");
   check(f != NULL);
   printf("Reading examples\n");

   //+,-example数据
   char **examples = (char **)malloc(num*sizeof(char *));

   check(examples != NULL);
     for (int i = 0; i < num; i++) {
     // we use an extra byte in the end of each example to mark unique
     // we use an extra int at the start of each example to store the
     // example's byte length (excluding unique flag and this int)
     //[legth label id level x y  unique] unique=0
     int buf[labelsize+2];
     //写入时的值为[1/-1 i 0 0 0 2 dim]
     count = fread(buf, sizeof(int), labelsize+2, f);
     check(count == labelsize+2);
     // byte length of an example's data segment

     //---前面七个是头，后面dim个float是样本特征数据，dim=2+1+root.h*root.w/2*31
     int len = sizeof(int)*(labelsize+2) + sizeof(float)*buf[labelsize+1];
     // memory for data, an initial integer, and a final byte
     examples[i] = (char *)malloc(sizeof(int)+len+1);

     check(examples[i] != NULL);
     // set data segment's byte length
     ((int *)examples[i])[0] = len;
     // set the unique flag to zero
     examples[i][sizeof(int)+len] = 0;
     // copy label data into example
     for (int j = 0; j < labelsize+2; j++)
       ((int *)examples[i])[j+1] = buf[j];
     // read the rest of the data segment into the example
     count = fread(examples[i]+sizeof(int)*(labelsize+3), 1,
           len-sizeof(int)*(labelsize+2), f);
     check(count == len-sizeof(int)*(labelsize+2));
   }
   fclose(f);
   printf("done\n");

   // sort
   printf("Sorting examples\n");
   char **sorted = (char **)malloc(num*sizeof(char *));
   check(sorted != NULL);
   memcpy(sorted, examples, num*sizeof(char *));

   //qsort 库函数，真正的比较函数为 comp
   //从小到大，快速排序
   //依次按照 样本类别->id->level->cx->cy  排序样本
   //如果前面五个量都一样……
   //1.等长度，比较所有字节；
   //2.谁长谁小，长度不同是因为不同的component的 尺寸不一致

   qsort(sorted, num, sizeof(char *), comp);
   printf("done\n");

   // find unique examples
   // 唯一的样本，unique flag=1,
   // 相同的样本第一个样本的unique flag为1，其余为0 ，有的样本的位置被，unique替代了，但是并没有完全删除掉
   int i = 0;
   int len = *((int *)sorted[0]); //负样本的第一个
   sorted[0][sizeof(int)+len] = 1; // unique flag 置 1
   for (int j = 1; j < num; j++) {
     int alen = *((int *)sorted[i]);
     int blen = *((int *)sorted[j]);
     if (alen != blen || memcmp(sorted[i] + sizeof(int), sorted[j] + sizeof(int), alen)) //component不同 || 不同样本
     {
       i++;
       sorted[i] = sorted[j];
       sorted[i][sizeof(int)+blen] = 1; //标记为 unique
     }
   }
   int num_unique = i+1;
   printf("%d unique examples\n", num_unique);

   // -------------------collapse examples----------------
   // 前面是找完全不一样的样本，这里是分组
   // label 的五个量 [label id level centry_x centry_y] 相同的分为一组，在detect时，写入了datfile
   // 负样本的 cx,cy都是相对于整张图片的，正样本是相对于剪切后的图像
   // 前面五个全相同，
   // 对于phase1 不可能，因为正负样本的id都不相同
   // 对于phase2 正样本只保留了最有可能是正样本的样本，只有一种情况,
   // rootfilter1,rootfilter2在同一张图片(id相同)，检测出来的 Hard负样本 的cx,cy相同，因此一组最多应该只能出现2个 （待验证）
   // 原因是此时的latent variable 为（cx,cy,component），上述情况相下，我们只能保留component1或者component2
   // 后续训练时，这两个量是连续使用的，为什么呢？？
   // collapse.seq(char **) 记录了每一组的第一个样本
   // collapse.num 每组的个数
   // X.num 组数
   // X.x=&collapse[0]，也就是第一个 collapse的地址
   collapse(&X, sorted, num_unique);
   printf("%d collapsed examples\n", X.num);

   // initial model
   // 读modfile文件，得到w的初始值。phase 1 初始化为全 0，phase 2 为上一次训练的结果……
   double **w = (double **)malloc(sizeof(double *)*X.numblocks);//2
   check(w != NULL);
   f = fopen(modfile, "rb");
   for (int i = 0; i < X.numblocks; i++) {
     w[i] = (double *)malloc(sizeof(double)*X.blocksizes[i]); //（1）=1，（2）= root.h*root.w/2*31
     check(w[i] != NULL);
     count = fread(w[i], sizeof(double), X.blocksizes[i], f);
     check(count == X.blocksizes[i]);
   }
   fclose(f);

   // lower bounds
   // 读lobfile文件，初始化为全 滤波器参数下线-100 ……
   double **lb = (double **)malloc(sizeof(double *)*X.numblocks);
   check(lb != NULL);
   f = fopen(lobfile, "rb");
   for (int i = 0; i < X.numblocks; i++) {
     lb[i] = (double *)malloc(sizeof(double)*X.blocksizes[i]);
     check(lb[i] != NULL);
     count = fread(lb[i], sizeof(double), X.blocksizes[i], f);
     check(count == X.blocksizes[i]);
   }
   fclose(f);


   printf("Training");
   //-------------------------------- train -------------------------------
   //-----梯度下降发训练参数 w，参见论文 公式17 后面的步骤
   gd(C, J, X, w, lb);
   printf("done\n");

   // save model
   printf("Saving model\n");
   f = fopen(modfile, "wb");
   check(f != NULL);
   //   存储 block1,block2的训练结果，w
   for (int i = 0; i < X.numblocks; i++) {
     count = fwrite(w[i], sizeof(double), X.blocksizes[i], f);
     check(count == X.blocksizes[i]);
   }
   fclose(f);

   // score examples
   // ---所有的样本都的得分,没有乘以 label y
   printf("Scoring\n");
   double *s = score(X, examples, num, w);

   // ---------Write info file-------------
   printf("Writing info file\n");
   f = fopen(inffile, "w");
   check(f != NULL);
   for (int i = 0; i < num; i++) {
     int len = ((int *)examples[i])[0];
     // label, score, unique flag
     count = fprintf(f, "%d\t%f\t%d\n", ((int *)examples[i])[1], s[i],
                     (int)examples[i][sizeof(int)+len]);
     check(count > 0);
   }
   fclose(f);

   printf("Freeing memory\n");
   for (int i = 0; i < X.numblocks; i++) {
     free(w[i]);
     free(lb[i]);
   }
   free(w);
   free(lb);
   free(s);
   for (int i = 0; i < num; i++)
     free(examples[i]);
   free(examples);
   free(sorted);
   free(X.x);
   free(X.blocksizes);
   free(X.regmult);
   free(X.learnmult);

   return 0;
 }