Welcome to PR3 The Art of Optimization. Lecture 3 May 22nd, IGAD - Hopmanstraat, Breda

Transcription

1 Lecture 3 May 22nd, IGAD - Hopmanstraat, Breda

2 Lecture 3 May 22nd, IGAD - Hopmanstraat, Breda > Recap > Demo Time > SIMD part 1/3 > Fixed Point part 2/2 > Homework part 3/4 > Coding Time

3 Lecture 3 May 22nd, IGAD - Hopmanstraat, Breda > Recap > Demo Time > SIMD part 1/3 > Fixed Point part 2/2 > Homework part 3/4 > Coding Time

4 Lecture 3 May 22nd, IGAD - Hopmanstraat, Breda > Recap >> Demo Time > SIMD part 1/3 > Fixed Point part 2/2 > Homework part 3/4 > Coding Time

5 Lecture 3 May 22nd, IGAD - Hopmanstraat, Breda Coding time: Homework in 15 minutes.

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10 7. DO THINGS SIMULTANEOUSLY.

11 Lecture 3 May 22nd, IGAD - Hopmanstraat, Breda > Recap > Demo Time >> SIMD part 1/3 > Fixed Point part 2/2 > Homework part 3/4 > Coding Time

12 Welcome to PR3 The Art of Optimization SIMD: S ingle I nstruction M ultiple D ata

13 SIMD: S ingle I nstruction M ultiple D ata int columntotal[16]; for ( int x = 0; x < 16; x++ ) { columntotal[x] = 0; for ( int y = 0; y < 16; y++ ) { if (pixelset( x, y )) { columntotal[x]++; } } }

14 SIMD: S ingle I nstruction M ultiple D ata int columntotal[16]; for ( int x = 0; x < 16; x++ ) { columntotal[x] = 0; for ( int y = 0; y < 16; y++ ) if (pixelset( x, y )) columntotal[x]++; }

15 SIMD: S ingle I nstruction M ultiple D ata int x, y, ct[16]; for ( x = 0; x < 16; x++ ) for ( ct[x] = 0, y = 0; y < 16;; ) if (pixelset( x, y++ )) ct[x]++;

16 SIMD: S ingle I nstruction M ultiple D ata for(int ct[16],y,x=0;x<16;x++) for(ct[x]=0,y=0;y<16;) if(pixelset(x,y++))ct[x]++;

17 Welcome to PR3 The Art of Optimization SIMD: S ingle I nstruction M ultiple D ata

18 SIMD: S ingle I nstruction M ultiple D ata

19 SIMD: S ingle I nstruction M ultiple D ata char a[4] = { 192, 168, 1, 1 }; int a4 = (a[0] << 24) + (a[1] << 16) + (a[2] << 8) + (a[3]);

20 SIMD: S ingle I nstruction M ultiple D ata char a[4] = { 192, 168, 1, 1 }; int a4 = (a[0] << 24) + (a[1] << 16) + (a[2] << 8) + (a[3]); = (173<<24) + (194<<16) + (65<<8) + (113) =

21 ?

22 { 2, 3, 1, 6 } 0x { 2, 3, 1, 6 } 0x { 4, 6, 2, 12 } 0x C

23 { 2, 3, 1, 6 } { 2, 3, 1, 6 } * { 4, 9, 1, 36 }

24 { 2, 3, 1, 6 } * { 96, 144, 48, 288 } { 96, 144, 49, 32 }

25 red = (color >> 16) & 255; green = (color >> 8) & 255; blue = color & 255; scaled_red = (red * scale) >> 8; scaled_green = (green * scale) >> 8; scaled_blue = (blue * scale) >> 8; final_color = (scaled_red << 16) + (scaled_green << 8) + scaled_blue;

26 red_blue = color & 0xff00ff; green = color & 0x00ff00; red_blue_scaled = ((red_blue * scale) >> 8) & 0xff00ff; green_scaled = ((green * scale) >> 8) & 0x00ff00; final_color = red_blue_scaled + green_scaled;

27 char a[] = optimization skills rule ; char b[] = optimization kicks ass!! ; bool equal = true; for ( int l = strlen(a), i = 0; i < l; i++ ) { if (a[i]!= b[i]) { equal = false; break; } }

28 Lecture 3 May 22nd, IGAD - Hopmanstraat, Breda > Recap > Demo Time > SIMD part 1/3 >> Fixed Point part 2/2 > Homework part 3/4 > Coding Time

29 The Trick with the Imaginary Dot =? =?

30 The Trick with the Imaginary Dot =? =?

31 The Trick with the Imaginary Dot =? =?

32 The Trick with the Imaginary Dot =? =?

33 The Trick with the Imaginary Dot =? =?

34 The Trick with the Imaginary Dot =? =? INTEGER PART FRACTIONAL PART

35 The Trick with the Imaginary Dot =? =? INTEGER PART FRACTIONAL PART

36 The Trick with the Imaginary Dot =? =? INTEGER PART FRACTIONAL PART float pi = ; int fp_pi = (int)(pi * f);

37 And so the suffering starts * 2 =?

38 And so the suffering starts * 2 =? * =?

39 And so the suffering starts * 2 =? * =? Multiplication affects your end result!

40 And so the suffering starts * 2 =? * =? Multiplication affects your end result! int a = f * 1024; // 10-bit fractional, approx. 3 decimal digits int b = f * 1024; int c = a * b; // warning: c is now 1024 times too large!

41 And so the suffering starts * 2 =? * =? Multiplication affects your end result! int a = f * 1024; // 10-bit fractional, approx. 3 decimal digits int b = f * 1024; int c = a * b; // warning: c is now 1024 times too large! Solve: Option 1: int c = (a * b) >> 10; c = ~1.047 * 1024 Option 2: int c = a * (b >> 10); c = 0 Option 3: int c = (a >> 10) * b; c = 1 Option 4: int c = (a >> 5) * (b >> 5); Option 5: int c = (a >> 2) * (b >> 8); Option 6: int c = ((a >> 2) * (b >> 2)) >> 6;

42 Lecture 3 May 22nd, IGAD - Hopmanstraat, Breda > Recap > Demo Time > SIMD part 1/3 > Fixed Point part 1/2 >> Homework part 1/4 > Coding Time

43 ftp://chiptune.untergrund.net/users/in4kadmin/files/the_neglected_art_of_fixed_point_arithmetic_ pdf

44 Lecture 3 May 22nd, IGAD - Hopmanstraat, Breda > Recap > Demo Time > SIMD part 1/3 > Fixed Point part 1/2 > Homework part 1/4 >> Coding Time