Windows에서 제공하는 GDI+를 이용한 이미지 파일 변환을 생각난 김에 함 만들어 봤다.

개인적으로 OS에 들어 있는 것을 사용하는 걸 좋아하기 땜시.... ㅋㅋ

쓸모가 있을지는 모르겠으나.... 함 맹글어 본다.....


// zImageCvt.cpp : 콘솔 응용 프로그램에 대한 진입점을 정의합니다.
//

#include "stdafx.h"
#include <windows.h>
#include <gdiplus.h>
#pragma comment (lib, "Gdiplus.lib")

// 시스템의 encoder 정보 load
INT nCntEncoder = 0;
Gdiplus::ImageCodecInfo* g_pImageCodecInfo = NULL;

int fnLoadEncoderInfo(void)
{

 UINT  num = 0;          // number of image encoders
 UINT  size = 0;         // size of the image encoder array in bytes

 Gdiplus::GetImageEncodersSize(&num, &size);
 if(size == 0)
  return -1;  // Failure

 g_pImageCodecInfo = (Gdiplus::ImageCodecInfo*)(malloc(size));
 if(g_pImageCodecInfo == NULL)
  return -1;  // Failure

 GetImageEncoders(num, size, g_pImageCodecInfo);
 nCntEncoder = num;

 return num;

}

int fnPrintEncoderInfo(BOOL bOnlyDesc)
{

 if( bOnlyDesc ){

  for (int i=0; i<nCntEncoder; i++)
  {
   wprintf(_T("/%s\n"), g_pImageCodecInfo[i].FormatDescription);   

  }
 }else{

  for (int i=0; i<nCntEncoder; i++)
  {
   wprintf(_T("Format Desc. = %s, MimeType = %s, Ext = %s\n"),
    g_pImageCodecInfo[i].FormatDescription,
    g_pImageCodecInfo[i].MimeType,
    g_pImageCodecInfo[i].FilenameExtension
    );  
  }

 }

 return nCntEncoder;

}

int GetEncoderClsidByMimeType(const WCHAR* format, CLSID* pClsid)
{

 for(UINT j = 0; j < (UINT)nCntEncoder; ++j)
 {

  if( wcscmp(g_pImageCodecInfo[j].MimeType, format) == 0 )
  {

   *pClsid = g_pImageCodecInfo[j].Clsid;
   return j;  // Success

  }    

 }
 return -1;  // Failure

}

int GetEncoderClsidByFormatDesc(const WCHAR* format, CLSID* pClsid)
{

 WCHAR *szFormat = (TCHAR *)::malloc( ::wcslen(format) + 1 );
 if( szFormat == NULL ) return -1;
 ::ZeroMemory( szFormat, ::wcslen(format) + 1 );
 ::wcscpy( szFormat, format );

 WCHAR *szReq = ::_wcsupr(szFormat);
 for(INT j = 0; j < nCntEncoder; ++j)
 {

  if( wcscmp(g_pImageCodecInfo[j].FormatDescription, szReq) == 0 )
  {
   *pClsid = g_pImageCodecInfo[j].Clsid;
   return j;  // Success
  }    

 }
 return -1;  // Failure

}

BOOL fnImageCvt(LPCTSTR lpszSRC, LPCTSTR lpszDst, const WCHAR* format)
{

 BOOL bretval = FALSE;
 do
 {

  // Create an Image object based on a PNG file.
  Gdiplus::Image  image(lpszSRC);
  if( image.GetLastStatus() != Gdiplus::Ok )
   break;

  // Save the image.
  CLSID jpgClsid;
   if( GetEncoderClsidByFormatDesc(format/*L"image/jpeg"*/, &jpgClsid) < 0 )
    break;

 image.Save(lpszDst, &jpgClsid, NULL);

  bretval = TRUE;

 } while (FALSE);

 return bretval;

}

void fnPrintSwitch(void)
{

 ::printf("이 명령에 대한 구문:\n\n");
 ::printf("zImageCvt\n");

 ::printf("zImageCvt\n");
 ::printf("[원본파일] [옵션] \n");
 ::printf("\t[원본파일] [결과파일] [옵션] \n");
 ::printf("옵션은 다음과 같음 \n");
 ::printf("옵션 \n");
 ::printf("/format : 입력한 format으로 변환합니다 \n");

 ::printf("----------------------------------------------------\n");
 ::printf("현재 변환가능 Descrions \n");

 fnPrintEncoderInfo(TRUE);

}

int _tmain(int argc, _TCHAR* argv[])
{

 Gdiplus::GdiplusStartupInput gdiplusStartupInput;
 ULONG_PTR gdiplusToken;
 Gdiplus::GdiplusStartup(&gdiplusToken, &gdiplusStartupInput, NULL);

 do
 {

 // image encoder load
if( fnLoadEncoderInfo() < 0 ){
   ::printf("Image Encoder load에 실패했습니다.!");
   break;
}

// 옵션맞는 함수 호출
TCHAR szSrc[MAX_PATH] = _T("");
TCHAR szDst[MAX_PATH] = _T("");
TCHAR szSwitch[MAX_PATH] = _T("");

TCHAR szTmp[MAX_PATH] = _T("");

switch( argc )
{
  default: break;
  case 2:

   // 옵션 보기
   if( wcscmp(argv[1], L"/h") == 0 ||
    wcscmp(argv[1], L"-h") == 0 )
   {
    fnPrintSwitch();
   }

   break;
  case 3: 

   ::_tcscpy( szSrc, argv[1] );
   ::_tcscpy( szSwitch, argv[2]+1 );

   break;
  case 4:

   ::_tcscpy( szSrc, argv[1] );
   ::_tcscpy( szDst, argv[2] );
   ::_tcscpy( szSwitch, argv[3]+1 );

   break;

  }

if( ::_tcslen(szSrc) < 3  )
   break;

//-------------------------------------------------------------------------------------
// 입력인자 보정
//-------------------------------------------------------------------------------------
if( ::_tcslen(szDst) < 3 )
{
   ::_tcsncpy( szDst, szSrc, ::_tcslen(szSrc) - 4 );
   ::_tcscat(szDst, _T("."));
   ::_tcscat(szDst, szSwitch);
}

// 실행
if( fnImageCvt( szSrc, szDst, szSwitch ) == FALSE )
   printf("변환 실패!!"); 

} while (FALSE);

 // 메모리 해제
if( g_pImageCodecInfo )
::free(g_pImageCodecInfo);

 Gdiplus::GdiplusShutdown(gdiplusToken);

 return 0;

}

소스코드 및 실행파일 : 

zImageCvt.exe

zImageCvt.cpp

ImageCodecInfo has the following types of members:

Data Members

The following table lists the members exposed by the ImageCodecInfo object.

Requirements

The GetImageEncoders function gets an array of ImageCodecInfo objects that contain information about the available image encoders.

Parameters

numEncoders [in]

Type: UINT

Integer that specifies the number of available image encoders. Call GetImageEncodersSize to determine this number.

size [in]

Type: UINT

Integer that specifies the size, in bytes, of the array of ImageCodecInfo objects. Call GetImageEncodersSize to determine this number.

encoders [out]

Type: ImageCodecInfo*

Pointer to a buffer that receives the array of ImageCodecInfo objects. You must allocate memory for this buffer. Call GetImageEncodersSize to determine the size of the required buffer.

Return Value

Type: Status

If the function succeeds, it returns Ok, which is an element of the Status enumeration.

If the function fails, it returns one of the other elements of the Status enumeration.

Examples

The following console application lists the available image encoders:

#include <windows.h>
#include <gdiplus.h>
#include <stdio.h>
using namespace Gdiplus;

INT main()
{
   // Initialize GDI+.
   GdiplusStartupInput gdiplusStartupInput;
   ULONG_PTR           gdiplusToken;
   GdiplusStartup(&gdiplusToken, &gdiplusStartupInput, NULL);
 
   UINT  num;        // number of image encoders
   UINT  size;       // size, in bytes, of the image encoder array

   ImageCodecInfo* pImageCodecInfo;

   // How many encoders are there?
   // How big (in bytes) is the array of all ImageCodecInfo objects?
   GetImageEncodersSize(&num, &size);

   // Create a buffer large enough to hold the array of ImageCodecInfo
   // objects that will be returned by GetImageEncoders.
   pImageCodecInfo = (ImageCodecInfo*)(malloc(size));

   // GetImageEncoders creates an array of ImageCodecInfo objects
   // and copies that array into a previously allocated buffer.
   // The third argument, imageCodecInfos, is a pointer to that buffer.
   GetImageEncoders(num, size, pImageCodecInfo);

   // Display the graphics file format (MimeType)
   // for each ImageCodecInfo object.
   for(UINT j = 0; j < num; ++j)
   {
      wprintf(L"%s\n", pImageCodecInfo[j].MimeType);  
   }

   free(pImageCodecInfo);
   GdiplusShutdown(gdiplusToken);
   return 0;
}

The preceding code produces the following output:

image/bmp
image/jpeg
image/gif
image/tiff
image/png

Discussed here are a number of interpolation methods, this is by no means an exhaustive list but the methods shown tend to be those in common use in computer graphics. The main attributes is that they are easy to compute and are stable. Interpolation as used here is different to "smoothing", the techniques discussed here have the characteristic that the estimated curve passes through all the given points. The idea is that the points are in some sense correct and lie on an underlying but unknown curve, the problem is to be able to estimate the values of the curve at any position between the known points.

double LinearInterpolate(
   double y1,double y2,
   double mu)
{
   return(y1*(1-mu)+y2*mu);
}

double CosineInterpolate(
   double y1,double y2,
   double mu)
{
   double mu2;

   mu2 = (1-cos(mu*PI))/2;
   return(y1*(1-mu2)+y2*mu2);
}

double CubicInterpolate(
   double y0,double y1,
   double y2,double y3,
   double mu)
{
   double a0,a1,a2,a3,mu2;

   mu2 = mu*mu;
   a0 = y3 - y2 - y0 + y1;
   a1 = y0 - y1 - a0;
   a2 = y2 - y0;
   a3 = y1;

   return(a0*mu*mu2+a1*mu2+a2*mu+a3);
}

   a0 = -0.5*y0 + 1.5*y1 - 1.5*y2 + 0.5*y3;
   a1 = y0 - 2.5*y1 + 2*y2 - 0.5*y3;
   a2 = -0.5*y0 + 0.5*y2;
   a3 = y1;

/*
   Tension: 1 is high, 0 normal, -1 is low
   Bias: 0 is even,
         positive is towards first segment,
         negative towards the other
*/
double HermiteInterpolate(
   double y0,double y1,
   double y2,double y3,
   double mu,
   double tension,
   double bias)
{
   double m0,m1,mu2,mu3;
   double a0,a1,a2,a3;

	mu2 = mu * mu;
	mu3 = mu2 * mu;
   m0  = (y1-y0)*(1+bias)*(1-tension)/2;
   m0 += (y2-y1)*(1-bias)*(1-tension)/2;
   m1  = (y2-y1)*(1+bias)*(1-tension)/2;
   m1 += (y3-y2)*(1-bias)*(1-tension)/2;
   a0 =  2*mu3 - 3*mu2 + 1;
   a1 =    mu3 - 2*mu2 + mu;
   a2 =    mu3 -   mu2;
   a3 = -2*mu3 + 3*mu2;

   return(a0*y1+a1*m0+a2*m1+a3*y2);
}

Trilinear interpolation is the name given to the process of linearly interpolating points within a box (3D) given values at the vertices of the box. Perhaps its most common application is interpolating within cells of a volumetric dataset.

Consider a unit cube with the lower/left/base vertex at the origin as shown here on the right.  The values at each vertex will be denoted V000, V100, V010, ....etc....V111

The value at position (x,y,z) within the cube will be denoted V_xyz and is given by

In general the box will not be of unit size nor will it be aligned at the origin. Simple translation and scaling (possibly of each axis independently) can be used to transform into and then out of this simplified situation.

Linear regression is a method to best fit a linear equation (straight line) of the form y(x) = a + b x to a collection of N points (x_i,y_i). Where b is the slope and a the intercept on the y axis.

The result will be stated below without derivation, that requires minimisation of the sum of the squared distance from the data points and the proposed line. This function is minimised by calculating the derivative with respect to a and b and setting these to zero. For a more complete derivation see the "Numerical Recipes in C".

The solution is clearer if we define the following

Then

and

And finally the regression coefficient is

This is 0 if there is no linear trend, 1 for perfect linear fit.

Note

• This discussion assumes there is no known variance for the x and y values. There are solutions which can take this into account, this is particularly important if some values are known with less error than others.

• The solution above requires that the slope is not infinite, S_xx is not zero.

Example

Source

C++ contributed by Charles Brown
RegressionLine.cpp
RegressionLine.hpp

The following introduces a method of immediately deriving a polynomial that passes through an arbitrary number of points. That is, find a polynomial f(x) that passes through the N points

The key to this solution is that we want an exact fit at the points given and we don't care what happens in between those points. The general solution is

To see how this works, consider the product term. When x = x_i the product term has the same denominator and numerator and thus equals 1 and therefore contributes y_i to the sum. All other terms in the summation contribute 0 since there exists a (x_j - x_j) in the numerator. Thanks to Simon Stegmaier for pointing out that this is known as a Lagrange Polynomial.

For a numerical example consider the polynomial that passes through the following points

(0,2) (1,1) (3,3) (4,0) (6,5)

The function using the above formula is

   f(x) = 2 * (x-1) * (x-3) * (x-4) * (x-6) / [ (0-1) * (0-3) * (0-4) * (0-6) ]
        + 1 * (x-0) * (x-3) * (x-4) * (x-6) / [ (1-0) * (1-3) * (1-4) * (1-6) ]
        + 3 * (x-0) * (x-1) * (x-4) * (x-6) / [ (3-0) * (3-1) * (3-4) * (3-6) ]
        + 0 * (x-0) * (x-1) * (x-3) * (x-6) / [ (4-0) * (4-1) * (4-3) * (4-6) ]
        + 5 * (x-0) * (x-1) * (x-3) * (x-4) / [ (6-0) * (6-1) * (6-3) * (6-4) ]

   f(x) = (x-1) * (x-3) * (x-4) * (x-6) / 36
        - (x-0) * (x-3) * (x-4) * (x-6) / 30
        + (x-0) * (x-1) * (x-4) * (x-6) / 6 
        + (x-0) * (x-1) * (x-3) * (x-4) / 36

   f(x) = 17*x4/90 - 181*x3/90 + 563*x2/90 - 163*x/30 + 2

By substituting the values of x for the points the function must pass through (x=0,1,3,4,6) it is easy to see that the expression above achieves the result, namely y=2,1,3,0,5 respectively.

All bets are off regarding the behavior between the fixed points. The polynomial is of degree N and could violently fly off anywhere. The continuous curve for the numerical example above is shown below.

A competition in the Mathematics news groups in October 1998

From: "John Santos" <santos_john@hotmail.com>
Newsgroups: alt.sci.math.probability,alt.tv.mathnet,aus.mathematics
Subject: $100.00 prize for the solution
Date: Tue, 15 Sep 1998 20:56:50 -0700
X-Newsreader: Microsoft Outlook Express 4.72.3110.1
X-MimeOLE: Produced By Microsoft MimeOLE V4.72.3110.3
NNTP-Posting-Host: 209.239.197.111
X-NNTP-Posting-Host: 209.239.197.111
Message-ID: <35ff36d1.0@blushng.jps.net>
X-NNTP-Posting-Host: 209.63.114.134

Hi everyone,
My name is John Santos and I am willing to pay anyone $100.00 for the first
person to solve this problem. I am providing some hints. This works as follows:
you will see nine digits - the tenth digit or letter is the answer. 
What I need is the formula or mathematical equation to get there...

   Number          Answer 
---------------------------
   749736637         1     
   713491024         8     
   523342792         D     
   749236871         P     
   727310078         E     
   746261832         4     
   733237527         L     
   743510589         9     
   715240338         K     
   722592910         1     
   739627071         R     

The first one with the answer and emails it to me wins the $100.00
Email address: santos_john@hotmail.com
Good Luck !!!!

The following is the solution to the posted problem, although it probably 
doesn't offer the insight you are seeking, it certainly falls within the 
scope of competition. To illustrate my solution I will use the following 
symbols instead of the numbers you used simply to save space. For your 
second column with letters and numbers use their ASCII values instead, 
this has no loss of generality as it is a simple 1 to 1 mapping.

x1 y1
x2 y2
x3 y3
x4 y4
etc

The following is a general method of making a function that passes 
through any pair of values (xi,yi).

        y1 (x-x2) (x-x3) (x-x4)
f(x) = --------------------------- 
         (x1-x2) (x1-x3) (x1-x4)

        y2 (x-x1) (x-x3) (x-x4)
     + --------------------------- 
         (x2-x1) (x2-x3) (x2-x4)

        y3 (x-x1) (x-x2) (x-x4)
     + --------------------------- 
         (x3-x1) (x3-x2) (x3-x4)

        y4 (x-x1) (x-x2) (x-x3)
     + --------------------------- 
         (x4-x1) (x4-x2) (x4-x3)

etc etc. As you can see, at x=x1 all the terms disappear except the 
first which equals y1, at x=x2 all the terms disappear except the
second which equals y2, etc etc.

   X                           Y
   Number           Answer     ASCII
---------------------------------------
   749736637        1          49
   713491024        8          56
   523342792        D          68
   749236871        P          80
   727310078        E          69
   746261832        4          52
   733237527        L          76
   743510589        9          57
   715240338        K          75
   722592910        1          49
   739627071        R          82

f(x) =
+ 49 ((x -  713491024) /   36245613) ((x -  523342792) /  226393845) ((x -  749236871) /     499766) ((x -  727310078) /   22426559) ((x -  746261832) /    3474805) ((x -  733237527) /   16499110) ((x -  743510589) /    6226048) ((x -  715240338) /   34496299) ((x -  722592910) /   27143727) ((x -  739627071) /   10109566) 
+ 56 ((x -  749736637) /  -36245613) ((x -  523342792) /  190148232) ((x -  749236871) /  -35745847) ((x -  727310078) /  -13819054) ((x -  746261832) /  -32770808) ((x -  733237527) /  -19746503) ((x -  743510589) /  -30019565) ((x -  715240338) /   -1749314) ((x -  722592910) /   -9101886) ((x -  739627071) /  -26136047) 
+ 68 ((x -  749736637) / -226393845) ((x -  713491024) / -190148232) ((x -  749236871) / -225894079) ((x -  727310078) / -203967286) ((x -  746261832) / -222919040) ((x -  733237527) / -209894735) ((x -  743510589) / -220167797) ((x -  715240338) / -191897546) ((x -  722592910) / -199250118) ((x -  739627071) / -216284279) 
+ 80 ((x -  749736637) /    -499766) ((x -  713491024) /   35745847) ((x -  523342792) /  225894079) ((x -  727310078) /   21926793) ((x -  746261832) /    2975039) ((x -  733237527) /   15999344) ((x -  743510589) /    5726282) ((x -  715240338) /   33996533) ((x -  722592910) /   26643961) ((x -  739627071) /    9609800) 
+ 69 ((x -  749736637) /  -22426559) ((x -  713491024) /   13819054) ((x -  523342792) /  203967286) ((x -  749236871) /  -21926793) ((x -  746261832) /  -18951754) ((x -  733237527) /   -5927449) ((x -  743510589) /  -16200511) ((x -  715240338) /   12069740) ((x -  722592910) /    4717168) ((x -  739627071) /  -12316993) 
+ 52 ((x -  749736637) /   -3474805) ((x -  713491024) /   32770808) ((x -  523342792) /  222919040) ((x -  749236871) /   -2975039) ((x -  727310078) /   18951754) ((x -  733237527) /   13024305) ((x -  743510589) /    2751243) ((x -  715240338) /   31021494) ((x -  722592910) /   23668922) ((x -  739627071) /    6634761) 
+ 76 ((x -  749736637) /  -16499110) ((x -  713491024) /   19746503) ((x -  523342792) /  209894735) ((x -  749236871) /  -15999344) ((x -  727310078) /    5927449) ((x -  746261832) /  -13024305) ((x -  743510589) /  -10273062) ((x -  715240338) /   17997189) ((x -  722592910) /   10644617) ((x -  739627071) /   -6389544) 
+ 57 ((x -  749736637) /   -6226048) ((x -  713491024) /   30019565) ((x -  523342792) /  220167797) ((x -  749236871) /   -5726282) ((x -  727310078) /   16200511) ((x -  746261832) /   -2751243) ((x -  733237527) /   10273062) ((x -  715240338) /   28270251) ((x -  722592910) /   20917679) ((x -  739627071) /    3883518) 
+ 75 ((x -  749736637) /  -34496299) ((x -  713491024) /    1749314) ((x -  523342792) /  191897546) ((x -  749236871) /  -33996533) ((x -  727310078) /  -12069740) ((x -  746261832) /  -31021494) ((x -  733237527) /  -17997189) ((x -  743510589) /  -28270251) ((x -  722592910) /   -7352572) ((x -  739627071) /  -24386733) 
+ 49 ((x -  749736637) /  -27143727) ((x -  713491024) /    9101886) ((x -  523342792) /  199250118) ((x -  749236871) /  -26643961) ((x -  727310078) /   -4717168) ((x -  746261832) /  -23668922) ((x -  733237527) /  -10644617) ((x -  743510589) /  -20917679) ((x -  715240338) /    7352572) ((x -  739627071) /  -17034161) 
+ 82 ((x -  749736637) /  -10109566) ((x -  713491024) /   26136047) ((x -  523342792) /  216284279) ((x -  749236871) /   -9609800) ((x -  727310078) /   12316993) ((x -  746261832) /   -6634761) ((x -  733237527) /    6389544) ((x -  743510589) /   -3883518) ((x -  715240338) /   24386733) ((x -  722592910) /   17034161) 

f( 749736637) = 49 (1)
f( 713491024) = 56 (8)
f( 523342792) = 68 (D)
f( 749236871) = 80 (P)
f( 727310078) = 69 (E)
f( 746261832) = 52 (4)
f( 733237527) = 76 (L)
f( 743510589) = 57 (9)
f( 715240338) = 75 (K)
f( 722592910) = 49 (1)
f( 739627071) = 82 (R)