Mux(8-bits input and 4-bits output) in HDL - hdl

If sel = false, then out should contain the lower 4-bits of in (i.e. in[0], in[1], in[2], in[3])
If sel = true, then out should contain the upper 4-bits of in (i.e. in[4], in[5], in[6], in[7])
Here is my code and how to modify it....?
CHIP NewMux
{
IN in[8], sel;
OUT out[4];
PARTS:
Mux4(sel=sel, a=in[0], b=false, out=out[0]);
Mux4(sel=sel, a=in[1], b=false, out=out[1]);
Mux4(sel=sel, a=in[2], b=false, out=out[2]);
Mux4(sel=sel, a=in[3], b=false, out=out[3]);
Mux4(sel=sel, a=in[4], b=true, out=out[0]);
Mux4(sel=sel, a=in[5], b=true, out=out[1]);
Mux4(sel=sel, a=in[6], b=true, out=out[2]);
Mux4(sel=sel, a=in[7], b=true, out=out[3]);
}

You cannot have multiple components generating the same outputs. You have 4 output lines, so you need to use 4 simple 1 bit muxes. Each will take as inputs the sel value and two of your in lines and generate a single out line.

Related

Linear equation of two vaariable

I want to find the solutions of linear equation of two variable(i know that there are infinitly many solution but i have to print some of them!)
so i tried in C language
what i have did is simply taken the input of coefficient of X,Y and c. Then calucalted the values as you can see it in the code Then calculated that if the result is satisfying the equation(means it is =0 or not) then checked and print. After that i have incremented the value of x,y so that it can be calculaed with different numbers(in the first i have put it by default 0)..
here is the code please see the code and tell me what is problem and am i doing mistake to choose the type of variable etc..
if possible please give me the correct code(request).
code in C:-
#include <stdio.h>
int main()
{
float a, b, c, Rx, Ry;
int sol;
float x, y;
printf("The Linear equation in two variable\n");
printf("Formate is aX+bY+c=0\n");
printf("Enter the coefficient of 'X'\n");
scanf("%f", &a);
printf("Enter the coefficient of 'Y'\n");
scanf("%f", &b);
printf("Enter the coefficient of constant 'c'\n");
scanf("%f", &c);
printf("Your equation looks like\n");
printf("%0.0fX+%0.0fY+%0.0f=0\n", a, b, c);
x = 0;
y = 0;
for (int i = 0; i < 10; i++)
{
// for x
x = ((-c) + (-b * y)) / b;
printf("The value of x is %0.2f\n", x);
// for y
y = ((-c) + (-a * x)) / a;
printf("The value of y is %0.2f\n", y);
// checking wheter the value of x and y satisfy the equation (which is equal to 0)
sol = (a * x) + (b * y) + (c);
printf("The value of sol is %0.2f\n", sol);
x++;
y++;
if (sol == 0)
{
printf("The value of x at %d time is %0.2f\n", i, x);
printf("The value of y at %d time is %0.2f\n", i, y);
}
else
continue;
}
return 0;
}
am i doing mistake to choose the type of variable
int sol;
…
printf("The value of sol is %0.2f\n", sol);
Whether the choice of type int for sol is a mistake depends on which precision you want. In any case, the conversion specifier f and type int don't fit together; you have to bring those in line. Be aware that if you choose float for sol, the comparison sol == 0 will not very likely succeed.

How is the matrix created using Isomorphic transform fucntion and Isomorphic inverse transform function? [closed]

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Trying to implementation AES Sbox and InSbox in combination circuit. Here for Sbox two operation is done i.e. Multiplicative Inverse and Affine Transform. For Affine Transform finite field is converted into a composite field using isomorphic transform, of which I have no idea how is that done. Need help in getting matrix delta shown in the image(attached with the question) from the irreducible polynomial mentioned p(x).
The matrices in the question are used for inversion (1/x) step. Affine transformation is a separate step and normally involves a matrix multiply followed by a column xor as specified by AES algorithm. Link to wiki article, note that the wiki article has least significant bits at the top, while the article you reference and other articles have the most significant bit at the top.
https://en.wikipedia.org/wiki/Rijndael_S-box
Getting back to how those matrices are created, I found a few articles, but they not only don't explain how those matrices are created, they were also missing key information, such as the primitive element chosen for GF(2^8) based on
polynomial x^8 + x^4 + x^3 + x + 1 (0x11b) with 1 bit coefficients, which is irreducible, but not primitive, since its primitive element is not x (0x02).
GF(2^8) is mapped to GF(((2^2)^2)^2). From the questions information, GF(2^2) uses x^2 + x + 1 (hex 7) with 1 bit coefficients to produce a 2 bit field with primitive element x = 0x2. GF((2^2)^2) uses x^2 + x + 2 (hex 16) with 2 bit coefficients from GF(2^2) to produce a 4 bit field with primitive element x = 0x4. GF(((2^2)^2)^2) uses x^2 + x + c (hex 11c) with 4 bit coefficients from GF((2^2)^2) to produce an 8 bit field with primitive x = 0x10.
For GF(2^8) there are 128 possible primitive elements: {0x03, 0x05, 0x06, ... , 0xff}. The matrix δ can be used to identify which primitive element was chosen for GF(2^8), in this case x^4 + x^3 + x^2 + x + 1 (hex 1f).
The columns of the matrix δ correspond to the mapping from GF(2^8) to GF(((2^2)^2)^2) by bit: 1st column maps 0x80, 2nd column maps 0x40, ..., 7th column maps 0x02, 8th column maps 0x01. The columns are powers of 0x10 in GF(((2^2)^2)^2). For example the 7th column is 0x5f, which is 0x10^0xa0 in GF(((2^2)^2)^2). Since the 7th column is used to map 0x02 in GF(2^8), this means GF(2^8)log??(0x02) = 0xa0, and that the chosen primitive element is 0x1f, since GF(2^8)log1f(0x02) = 0xa0. The 6th column is 0x7c, which is 0x10^0x41 in GF(((2^2)^2)^2), and GF(2^8)log1f(0x04) = 0x41.
The table below shows the data for all 8 colums.
GF(2^8) log1f(0x80) = 0x64, GF(((2^2)^2)^2) 0x10^0x64 = 0xfc (1st column of matrix)
GF(2^8) log1f(0x40) = 0xc3, GF(((2^2)^2)^2) 0x10^0xc3 = 0x4b (2nd column of matrix)
GF(2^8) log1f(0x20) = 0x23, GF(((2^2)^2)^2) 0x10^0x23 = 0xb0 (3rd column of matrix)
GF(2^8) log1f(0x10) = 0x82, GF(((2^2)^2)^2) 0x10^0x82 = 0x46 (4th column of matrix)
GF(2^8) log1f(0x08) = 0xe1, GF(((2^2)^2)^2) 0x10^0xe1 = 0x74 (5th column of matrix)
GF(2^8) log1f(0x04) = 0x41, GF(((2^2)^2)^2) 0x10^0x41 = 0x7c (6th column of matrix)
GF(2^8) log1f(0x02) = 0xa0, GF(((2^2)^2)^2) 0x10^0xa0 = 0x5f (7th column of matrix)
GF(2^8) log1f(0x01) = 0x00, GF(((2^2)^2)^2) 0x10^0x00 = 0x01 (8th column of matrix)
The inverse mapping matrix can use the same logic:
GF(((2^2)^2)^2) log10(0x80) = 0x67, GF(2^8) 0x1f^0x67 = 0x84 (1st column of matrix)
GF(((2^2)^2)^2) log10(0x40) = 0xbc, GF(2^8) 0x1f^0xbc = 0xf1 (2nd column of matrix)
GF(((2^2)^2)^2) log10(0x20) = 0xab, GF(2^8) 0x1f^0xab = 0xbb (3rd column of matrix)
GF(((2^2)^2)^2) log10(0x10) = 0x01, GF(2^8) 0x1f^0x01 = 0x1f (4th column of matrix)
GF(((2^2)^2)^2) log10(0x08) = 0x66, GF(2^8) 0x1f^0x66 = 0x0c (5th column of matrix)
GF(((2^2)^2)^2) log10(0x04) = 0xbb, GF(2^8) 0x1f^0xbb = 0x5d (6th column of matrix)
GF(((2^2)^2)^2) log10(0x02) = 0xaa, GF(2^8) 0x1f^0xaa = 0xbc (7th column of matrix)
GF(((2^2)^2)^2) log10(0x01) = 0x00, GF(2^8) 0x1f^0x00 = 0x01 (8th column of matrix)
Note that in the questions image, the inverse matrix 1st and 6th columns have the least significant bit flipped. The pdf file linked to below has the proper matrices.
https://github.com/bpdegnan/aes/blob/master/aes-sbox/documentation/aessbox.pdf
I created a small pdf file that explains how the mapping matrices seen on page 4, matrix (8) and page 5, matrix (10) are generated and the logic behind them.
https://github.com/jeffareid/finite-field/blob/master/Composite%20Field%20Mapping%20Example.pdf
In order for sub-field aka composite field math to work, there are two main requirements. Using map() to represent the mapping from GF(2^8) to GF(((2^2)^2)^2), then while operating in GF(((2^2)^2)^2)
map(a + b) = map(a) + map(b) // addition (xor) is isomorphic
map(a · b) = map(a) · map(b) // multiplication is isomorphic
This can also be restated as: using α to represent the primitive element for GF(2^8) and β to represent the primitive element for GF(((2^2)^2)^2).
if α^i + α^j = α^k, then β^i + β^j = β^k // addition (xor) is isomporhic
if α^i · α^j = α^k, then β^i · β^j = β^k // multiplication is isomorphic
Normally β = 100002, and a brute force search is done for 3 constants α, φ, δ, that result in compatible mapping and minimizing gate count, where α is the primitive element for GF(2^8), φ is the constant term for GF((2^2)^2) = x^2 + x + φ, and δ is the constant term for GF(((2^2)^2)^2) = x^2 + x + δ. In this case, α = 111112, φ = 102, δ = 11002.

Error code not working

Hello I am trying to implement the gate MiniALU but the howard simulator give me this error: "has no source pin". I would be happy if you can help me solve this.
my code-
CHIP MiniALU {
IN
x[16], y[16], // 16-bit inputs
zx, // zero the x input?
zy, // zero the y input?
f; // compute out = x + y (if f == 1) or out = x & y (if == 0)
OUT
out[16]; // 16-bit output
PARTS:
// Zero the x input and y input
Mux16(a[0..15]=x, b[0..15]=false, sel=zx, out[0..15]=x1);
Mux16(a[0..15]=y, b[0..15]=false, sel=zy, out[0..15]=y1);
// Perform f
And16(a[0..15]=x2, b[0..15]=y2, out[0..15]=xandy);
Add16(a[0..15]=x2, b[0..15]=y2, out[0..15]=xaddy);
Mux16(a[0..15]=xandy, b[0..15]=xaddy, sel=f, out[0..15]=out);
}
You are connecting x2 and y2 to the inputs of And16 and Add16 but x2 and y2 are not defined anywhere.
You need to replace x2 and y2 with x1 and y1 in the connections to And16 and Add16.
Correct code:
Mux16(a=x, b=false, sel=zx, out=x1);
Mux16(a=y, b=false, sel=zy, out=y1);
And16(a=x1, b=y1, out=xandy);
Add16(a=x1, b=y1, out=xaddy);
Mux16(a=xandy, b=xaddy, sel=f, out=out);
Your problem was that you wrote y2 and x2 instead of y1 and x2.
Also, there is not need in [0..15]
You wrote:
And16(a[0..15]=x2, b[0..15]=y2, out[0..15]=xandy);
Add16(a[0..15]=x2, b[0..15]=y2, out[0..15]=xaddy);
instead of:
And16(a[0..15]=x1, b[0..15]=y1, out[0..15]=xandy);
Add16(a[0..15]=x1, b[0..15]=y1, out[0..15]=xaddy);

Looking for an efficient integer square root algorithm for ARM Thumb2

I am looking for a fast, integer only algorithm to find the square root (integer part thereof) of an unsigned integer.
The code must have excellent performance on ARM Thumb 2 processors. It could be assembly language or C code.
Any hints welcome.
Integer Square Roots by Jack W. Crenshaw could be useful as another reference.
The C Snippets Archive also has an integer square root implementation. This one goes beyond just the integer result, and calculates extra fractional (fixed-point) bits of the answer. (Update: unfortunately, the C snippets archive is now defunct. The link points to the web archive of the page.) Here is the code from the C Snippets Archive:
#define BITSPERLONG 32
#define TOP2BITS(x) ((x & (3L << (BITSPERLONG-2))) >> (BITSPERLONG-2))
struct int_sqrt {
unsigned sqrt, frac;
};
/* usqrt:
ENTRY x: unsigned long
EXIT returns floor(sqrt(x) * pow(2, BITSPERLONG/2))
Since the square root never uses more than half the bits
of the input, we use the other half of the bits to contain
extra bits of precision after the binary point.
EXAMPLE
suppose BITSPERLONG = 32
then usqrt(144) = 786432 = 12 * 65536
usqrt(32) = 370727 = 5.66 * 65536
NOTES
(1) change BITSPERLONG to BITSPERLONG/2 if you do not want
the answer scaled. Indeed, if you want n bits of
precision after the binary point, use BITSPERLONG/2+n.
The code assumes that BITSPERLONG is even.
(2) This is really better off being written in assembly.
The line marked below is really a "arithmetic shift left"
on the double-long value with r in the upper half
and x in the lower half. This operation is typically
expressible in only one or two assembly instructions.
(3) Unrolling this loop is probably not a bad idea.
ALGORITHM
The calculations are the base-two analogue of the square
root algorithm we all learned in grammar school. Since we're
in base 2, there is only one nontrivial trial multiplier.
Notice that absolutely no multiplications or divisions are performed.
This means it'll be fast on a wide range of processors.
*/
void usqrt(unsigned long x, struct int_sqrt *q)
{
unsigned long a = 0L; /* accumulator */
unsigned long r = 0L; /* remainder */
unsigned long e = 0L; /* trial product */
int i;
for (i = 0; i < BITSPERLONG; i++) /* NOTE 1 */
{
r = (r << 2) + TOP2BITS(x); x <<= 2; /* NOTE 2 */
a <<= 1;
e = (a << 1) + 1;
if (r >= e)
{
r -= e;
a++;
}
}
memcpy(q, &a, sizeof(long));
}
I settled on the following code. It's essentially from the Wikipedia article on square-root computing methods. But it has been changed to use stdint.h types uint32_t etc. Strictly speaking, the return type could be changed to uint16_t.
/**
* \brief Fast Square root algorithm
*
* Fractional parts of the answer are discarded. That is:
* - SquareRoot(3) --> 1
* - SquareRoot(4) --> 2
* - SquareRoot(5) --> 2
* - SquareRoot(8) --> 2
* - SquareRoot(9) --> 3
*
* \param[in] a_nInput - unsigned integer for which to find the square root
*
* \return Integer square root of the input value.
*/
uint32_t SquareRoot(uint32_t a_nInput)
{
uint32_t op = a_nInput;
uint32_t res = 0;
uint32_t one = 1uL << 30; // The second-to-top bit is set: use 1u << 14 for uint16_t type; use 1uL<<30 for uint32_t type
// "one" starts at the highest power of four <= than the argument.
while (one > op)
{
one >>= 2;
}
while (one != 0)
{
if (op >= res + one)
{
op = op - (res + one);
res = res + 2 * one;
}
res >>= 1;
one >>= 2;
}
return res;
}
The nice thing, I discovered, is that a fairly simple modification can return the "rounded" answer. I found this useful in a certain application for greater accuracy. Note that in this case, the return type must be uint32_t because the rounded square root of 232 - 1 is 216.
/**
* \brief Fast Square root algorithm, with rounding
*
* This does arithmetic rounding of the result. That is, if the real answer
* would have a fractional part of 0.5 or greater, the result is rounded up to
* the next integer.
* - SquareRootRounded(2) --> 1
* - SquareRootRounded(3) --> 2
* - SquareRootRounded(4) --> 2
* - SquareRootRounded(6) --> 2
* - SquareRootRounded(7) --> 3
* - SquareRootRounded(8) --> 3
* - SquareRootRounded(9) --> 3
*
* \param[in] a_nInput - unsigned integer for which to find the square root
*
* \return Integer square root of the input value.
*/
uint32_t SquareRootRounded(uint32_t a_nInput)
{
uint32_t op = a_nInput;
uint32_t res = 0;
uint32_t one = 1uL << 30; // The second-to-top bit is set: use 1u << 14 for uint16_t type; use 1uL<<30 for uint32_t type
// "one" starts at the highest power of four <= than the argument.
while (one > op)
{
one >>= 2;
}
while (one != 0)
{
if (op >= res + one)
{
op = op - (res + one);
res = res + 2 * one;
}
res >>= 1;
one >>= 2;
}
/* Do arithmetic rounding to nearest integer */
if (op > res)
{
res++;
}
return res;
}
If exact accuracy isn't required, I have a fast approximation for you, that uses 260 bytes of RAM (you could halve that, but don't).
int ftbl[33]={0,1,1,2,2,4,5,8,11,16,22,32,45,64,90,128,181,256,362,512,724,1024,1448,2048,2896,4096,5792,8192,11585,16384,23170,32768,46340};
int ftbl2[32]={ 32768,33276,33776,34269,34755,35235,35708,36174,36635,37090,37540,37984,38423,38858,39287,39712,40132,40548,40960,41367,41771,42170,42566,42959,43347,43733,44115,44493,44869,45241,45611,45977};
int fisqrt(int val)
{
int cnt=0;
int t=val;
while (t) {cnt++;t>>=1;}
if (6>=cnt) t=(val<<(6-cnt));
else t=(val>>(cnt-6));
return (ftbl[cnt]*ftbl2[t&31])>>15;
}
Here's the code to generate the tables:
ftbl[0]=0;
for (int i=0;i<32;i++) ftbl[i+1]=sqrt(pow(2.0,i));
printf("int ftbl[33]={0");
for (int i=0;i<32;i++) printf(",%d",ftbl[i+1]);
printf("};\n");
for (int i=0;i<32;i++) ftbl2[i]=sqrt(1.0+i/32.0)*32768;
printf("int ftbl2[32]={");
for (int i=0;i<32;i++) printf("%c%d",(i)?',':' ',ftbl2[i]);
printf("};\n");
Over the range 1 → 220, the maximum error is 11, and over the range 1 → 230, it's about 256. You could use larger tables and minimise this. It's worth mentioning that the error will always be negative - i.e. when it's wrong, the value will be LESS than the correct value.
You might do well to follow this with a refining stage.
The idea is simple enough: (ab)0.5 = a0.b × b0.5.
So, we take the input X = A×B where A = 2N and 1 ≤ B < 2
Then we have a lookup table for sqrt(2N), and a lookup table for sqrt(1 ≤ B < 2). We store the lookup table for sqrt(2N) as integer, which might be a mistake (testing shows no ill effects), and we store the lookup table for sqrt(1 ≤ B < 2) as 15-bit fixed-point.
We know that 1 ≤ sqrt(2N) < 65536, so that's 16-bit, and we know that we can really only multiply 16-bit × 15-bit on an ARM, without fear of reprisal, so that's what we do.
In terms of implementation, the while(t) {cnt++;t>>=1;} is effectively a count-leading-bits instruction (CLB), so if your version of the chipset has that, you're winning! Also, the shift instruction would be easy to implement with a bidirectional shifter, if you have one?
There's a Lg[N] algorithm for counting the highest set bit here.
In terms of magic numbers, for changing table sizes, THE magic number for ftbl2 is 32, though note that 6 (Lg[32]+1) is used for the shifting.
One common approach is bisection.
hi = number
lo = 0
mid = ( hi + lo ) / 2
mid2 = mid*mid
while( lo < hi-1 and mid2 != number ) {
if( mid2 < number ) {
lo = mid
else
hi = mid
mid = ( hi + lo ) / 2
mid2 = mid*mid
Something like that should work reasonably well. It makes log2(number) tests, doing
log2(number) multiplies and divides. Since the divide is a divide by 2, you can replace it with a >>.
The terminating condition may not be spot on, so be sure to test a variety of integers to be sure that the division by 2 doesn't incorrectly oscillate between two even values; they would differ by more than 1.
I find that most algorithms are based on simple ideas, but are implemented in a way more complicated manner than necessary. I've taken the idea from here: http://ww1.microchip.com/downloads/en/AppNotes/91040a.pdf (by Ross M. Fosler) and made it into a very short C-function:
uint16_t int_sqrt32(uint32_t x)
{
uint16_t res= 0;
uint16_t add= 0x8000;
int i;
for(i=0;i<16;i++)
{
uint16_t temp=res | add;
uint32_t g2= (uint32_t)temp * temp;
if (x>=g2)
{
res=temp;
}
add>>=1;
}
return res;
}
This compiles to 5 cycles/bit on my blackfin. I believe your compiled code will in general be faster if you use for loops instead of while loops, and you get the added benefit of deterministic time (although that to some extent depends on how your compiler optimizes the if statement.)
It depends about the usage of the sqrt function. I often use some approx to make fast versions. For example, when I need to compute the module of vector :
Module = SQRT( x^2 + y^2)
I use :
Module = MAX( x,y) + Min(x,y)/2
Which can be coded in 3 or 4 instructions as:
If (x > y )
Module = x + y >> 1;
Else
Module = y + x >> 1;
It's not fast but it's small and simple:
int isqrt(int n)
{
int b = 0;
while(n >= 0)
{
n = n - b;
b = b + 1;
n = n - b;
}
return b - 1;
}
I have settled to something similar to the binary digit-by-digit algorithm described in this Wikipedia article.
I recently encountered the same task on ARM Cortex-M3 (STM32F103CBT6) and after searching the Internet came up with the following solution. It's not the fastest comparing with solutions offered here, but it has good accuracy (maximum error is 1, i.e. LSB on the entire UI32 input range) and relatively good speed (about 1.3M square roots per second on a 72-MHz ARM Cortex-M3 or about 55 cycles per single root including the function call).
// FastIntSqrt is based on Wikipedia article:
// https://en.wikipedia.org/wiki/Methods_of_computing_square_roots
// Which involves Newton's method which gives the following iterative formula:
//
// X(n+1) = (X(n) + S/X(n))/2
//
// Thanks to ARM CLZ instruction (which counts how many bits in a number are
// zeros starting from the most significant one) we can very successfully
// choose the starting value, so just three iterations are enough to achieve
// maximum possible error of 1. The algorithm uses division, but fortunately
// it is fast enough here, so square root computation takes only about 50-55
// cycles with maximum compiler optimization.
uint32_t FastIntSqrt (uint32_t value)
{
if (!value)
return 0;
uint32_t xn = 1 << ((32 - __CLZ (value))/2);
xn = (xn + value/xn)/2;
xn = (xn + value/xn)/2;
xn = (xn + value/xn)/2;
return xn;
}
I'm using IAR and it produces the following assembler code:
SECTION `.text`:CODE:NOROOT(1)
THUMB
_Z11FastIntSqrtj:
MOVS R1,R0
BNE.N ??FastIntSqrt_0
MOVS R0,#+0
BX LR
??FastIntSqrt_0:
CLZ R0,R1
RSB R0,R0,#+32
MOVS R2,#+1
LSRS R0,R0,#+1
LSL R0,R2,R0
UDIV R3,R1,R0
ADDS R0,R3,R0
LSRS R0,R0,#+1
UDIV R2,R1,R0
ADDS R0,R2,R0
LSRS R0,R0,#+1
UDIV R1,R1,R0
ADDS R0,R1,R0
LSRS R0,R0,#+1
BX LR ;; return
The most cleverly coded bit-wise integer square root implementations for ARM achieve 3 cycles per result bit, which comes out to a lower bound of 50 cycles for the square root of a 32-bit unsigned integer. An example is shown in Andrew N. Sloss, Dominic Symes, Chris Wright, "ARM System Developer's Guide", Morgan Kaufman 2004.
Since most ARM processors also have very fast integer multipliers, and most even provide a very fast implementation of the wide multiply instruction UMULL, an alternative approach that can achieve execution times on the order of 35 to 45 cycles is computation via the reciprocal square root 1/√x using fixed-point computation. For this it is necessary to normalize the input with the help of a count-leading-zeros instruction, which on most ARM processors is available as an instruction CLZ.
The computation starts with an initial low-accuracy reciprocal square root approximation from a lookup table indexed by some most significant bit of the normalized argument. The Newton-Raphson iteration to refine the reciprocal square root r of a number a with quadratic convergence is rn+1 = rn + rn* (1 - a * rn2) / 2. This can be re-arranged into algebraically equivalent forms as is convenient. In the exemplary C99 code below, an 8-bit approximation r0 is read from a 96-entry lookup table. This approximation is accurate to about 7 bits. The first Newton-Raphson iteration computes r1 = (3 * r0 - a * r03) / 2 to potentially take advantage of small operand multiplication instructions. The second Newton-Raphson iteration computes r2 = (r1 * (3 - r1 * (r1 * a))) / 2.
The normalized square root is then computed by a back multiplication s2 = a * r2 and the final approximation is achieved by denormalizing based on the count of leading zeros of the original argument a. It is important that the desired result ⌊√a⌋ is approximated with underestimation. This simplifies the checking whether the desired result has been achieved by guaranteeing that the remainder ⌊√a⌋ - s2 * s2 is positive. If the final approximation is found to be too small, the result is increased by one. Correct operation of this algorithm can easily be demonstrated by exhaustive test of all possible 232 inputs against a "golden" reference, which takes only a few minutes.
One can speed up this implementation at the expense of additional storage for the lookup table, by pre-computing 3 * r0 and r03 to simplify the first Newton-Raphson iteration. The former requires 10 bit of storage and the latter 24 bits. In order to combine each pair into a 32-bit data item, the cube is rounded to 22 bits, which introduces negligible error into the computation. This results in a lookup table of 96 * 4 = 384 bytes.
An alternative approach uses the observation that all starting approximations have the same most significant bit set, which therefore can be assumed implicitly and does not have to be stored. This allows to squeeze a 9-bit approximation r0 into an 8-bit data item, with the leading bit restored after table lookup. With a lookup table of 384 entries, all underestimates, one can achieve an accuracy of about 7.5 bits. Combining the back multiply with the Newton-Raphson iteration for the reciprocal square root, one computes s0 = a * r0, s1 = s0 + r0 * (a - s0 * s0) / 2. Because the accuracy of the starting approximation is not high enough for a very accurate final square root approximation, it can be off by up to three, and an appropriate adjustment must be made based on the magnitude of the remainder floor (sqrt (a)) - s1 * s1.
One advantage of the alternative approach is that is halves the number of multiplies required, and in particular requires only a single wide multiplication UMULL. Especially processors where wide multiplies are fairly slow, this is an alternative worth trying.
#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
#include <string.h>
#include <math.h>
#if defined(_MSC_VER) && defined(_WIN64)
#include <intrin.h>
#endif // defined(_MSC_VER) && defined(_WIN64)
#define CLZ_BUILTIN (1) // use compiler's built-in count-leading-zeros
#define CLZ_FPU (2) // emulate count-leading-zeros via FPU
#define CLZ_CPU (3) // emulate count-leading-zeros via CPU
#define ALT_IMPL (0) // alternative implementation with fewer multiplies
#define LARGE_TABLE (0) // ALT_IMPL=0: incorporate 1st NR-iter into table
#define CLZ_IMPL (CLZ_CPU)// choose count-leading-zeros implementation
#define GEN_TAB (0) // generate tables
uint32_t umul32_hi (uint32_t a, uint32_t b);
uint32_t float_as_uint32 (float a);
int clz32 (uint32_t a);
#if ALT_IMPL
uint8_t rsqrt_tab [384] =
{
0xfe, 0xfc, 0xfa, 0xf8, 0xf6, 0xf4, 0xf2, 0xf0, 0xee, 0xed, 0xeb, 0xe9,
0xe7, 0xe6, 0xe4, 0xe2, 0xe1, 0xdf, 0xdd, 0xdc, 0xda, 0xd8, 0xd7, 0xd5,
0xd4, 0xd2, 0xd1, 0xcf, 0xce, 0xcc, 0xcb, 0xc9, 0xc8, 0xc7, 0xc5, 0xc4,
0xc2, 0xc1, 0xc0, 0xbe, 0xbd, 0xbc, 0xba, 0xb9, 0xb8, 0xb7, 0xb5, 0xb4,
0xb3, 0xb2, 0xb0, 0xaf, 0xae, 0xad, 0xac, 0xab, 0xa9, 0xa8, 0xa7, 0xa6,
0xa5, 0xa4, 0xa3, 0xa2, 0xa0, 0x9f, 0x9e, 0x9d, 0x9c, 0x9b, 0x9a, 0x99,
0x98, 0x97, 0x96, 0x95, 0x94, 0x93, 0x92, 0x91, 0x90, 0x8f, 0x8e, 0x8d,
0x8c, 0x8b, 0x8b, 0x8a, 0x89, 0x88, 0x87, 0x86, 0x85, 0x84, 0x83, 0x83,
0x82, 0x81, 0x80, 0x7f, 0x7e, 0x7d, 0x7d, 0x7c, 0x7b, 0x7a, 0x79, 0x79,
0x78, 0x77, 0x76, 0x75, 0x75, 0x74, 0x73, 0x72, 0x72, 0x71, 0x70, 0x6f,
0x6f, 0x6e, 0x6d, 0x6c, 0x6c, 0x6b, 0x6a, 0x6a, 0x69, 0x68, 0x67, 0x67,
0x66, 0x65, 0x65, 0x64, 0x63, 0x63, 0x62, 0x61, 0x61, 0x60, 0x5f, 0x5f,
0x5e, 0x5d, 0x5d, 0x5c, 0x5c, 0x5b, 0x5a, 0x5a, 0x59, 0x58, 0x58, 0x57,
0x57, 0x56, 0x55, 0x55, 0x54, 0x54, 0x53, 0x52, 0x52, 0x51, 0x51, 0x50,
0x50, 0x4f, 0x4e, 0x4e, 0x4d, 0x4d, 0x4c, 0x4c, 0x4b, 0x4b, 0x4a, 0x4a,
0x49, 0x48, 0x48, 0x47, 0x47, 0x46, 0x46, 0x45, 0x45, 0x44, 0x44, 0x43,
0x43, 0x42, 0x42, 0x41, 0x41, 0x40, 0x40, 0x3f, 0x3f, 0x3e, 0x3e, 0x3d,
0x3d, 0x3c, 0x3c, 0x3c, 0x3b, 0x3b, 0x3a, 0x3a, 0x39, 0x39, 0x38, 0x38,
0x37, 0x37, 0x36, 0x36, 0x36, 0x35, 0x35, 0x34, 0x34, 0x33, 0x33, 0x33,
0x32, 0x32, 0x31, 0x31, 0x30, 0x30, 0x30, 0x2f, 0x2f, 0x2e, 0x2e, 0x2d,
0x2d, 0x2d, 0x2c, 0x2c, 0x2b, 0x2b, 0x2b, 0x2a, 0x2a, 0x29, 0x29, 0x29,
0x28, 0x28, 0x27, 0x27, 0x27, 0x26, 0x26, 0x26, 0x25, 0x25, 0x24, 0x24,
0x24, 0x23, 0x23, 0x23, 0x22, 0x22, 0x21, 0x21, 0x21, 0x20, 0x20, 0x20,
0x1f, 0x1f, 0x1f, 0x1e, 0x1e, 0x1e, 0x1d, 0x1d, 0x1d, 0x1c, 0x1c, 0x1c,
0x1b, 0x1b, 0x1a, 0x1a, 0x1a, 0x19, 0x19, 0x19, 0x18, 0x18, 0x18, 0x17,
0x17, 0x17, 0x17, 0x16, 0x16, 0x16, 0x15, 0x15, 0x15, 0x14, 0x14, 0x14,
0x13, 0x13, 0x13, 0x12, 0x12, 0x12, 0x11, 0x11, 0x11, 0x11, 0x10, 0x10,
0x10, 0x0f, 0x0f, 0x0f, 0x0e, 0x0e, 0x0e, 0x0e, 0x0d, 0x0d, 0x0d, 0x0c,
0x0c, 0x0c, 0x0c, 0x0b, 0x0b, 0x0b, 0x0a, 0x0a, 0x0a, 0x0a, 0x09, 0x09,
0x09, 0x08, 0x08, 0x08, 0x08, 0x07, 0x07, 0x07, 0x07, 0x06, 0x06, 0x06,
0x05, 0x05, 0x05, 0x05, 0x04, 0x04, 0x04, 0x04, 0x03, 0x03, 0x03, 0x03,
0x02, 0x02, 0x02, 0x02, 0x01, 0x01, 0x01, 0x01, 0x00, 0x00, 0x00, 0x00,
};
/* compute floor (sqrt (a)) */
uint32_t my_isqrt32 (uint32_t a)
{
uint32_t b, r, s, scal, rem;
if (a == 0) return a;
/* Normalize argument */
scal = clz32 (a) & ~1;
b = a << scal;
/* Compute initial approximation to 1/sqrt(a) */
r = rsqrt_tab [(b >> 23) - 128] | 0x100;
/* Compute initial approximation to sqrt(a) */
s = umul32_hi (b, r << 8);
/* Refine sqrt approximation */
b = b - s * s;
s = s + ((r * (b >> 2)) >> 23);
/* Denormalize result*/
s = s >> (scal >> 1);
/* Ensure we got the floor correct */
rem = a - s * s;
if (rem < (2 * s + 1)) s = s + 0;
else if (rem < (4 * s + 4)) s = s + 1;
else if (rem < (6 * s + 9)) s = s + 2;
else s = s + 3;
return s;
}
#else // ALT_IMPL
#if LARGE_TABLE
uint32_t rsqrt_tab [96] =
{
0xfa0bfafa, 0xee6b2aee, 0xe5f02ae5, 0xdaf26ed9, 0xd2f002d0, 0xc890c2c4,
0xc1037abb, 0xb9a75ab2, 0xb4da42ac, 0xadcea2a3, 0xa6f27a9a, 0xa279c294,
0x9beb4a8b, 0x97a5ca85, 0x9163427c, 0x8d4fca76, 0x89500270, 0x8563ba6a,
0x818ac264, 0x7dc4ea5e, 0x7a120258, 0x7671da52, 0x72e4424c, 0x6f690a46,
0x6db24243, 0x6a52423d, 0x67042637, 0x6563c234, 0x62302a2e, 0x609cea2b,
0x5d836a25, 0x5bfd1a22, 0x58fd421c, 0x5783ae19, 0x560e4a16, 0x53300210,
0x51c7120d, 0x50623a0a, 0x4da4c204, 0x4c4c1601, 0x4af769fe, 0x49a6b9fb,
0x485a01f8, 0x471139f5, 0x45cc59f2, 0x448b5def, 0x4214fde9, 0x40df89e6,
0x3fade1e3, 0x3e8001e0, 0x3d55e1dd, 0x3c2f79da, 0x3c2f79da, 0x3b0cc5d7,
0x39edc1d4, 0x38d265d1, 0x37baa9ce, 0x36a689cb, 0x359601c8, 0x348909c5,
0x348909c5, 0x337f99c2, 0x3279adbf, 0x317741bc, 0x30784db9, 0x30784db9,
0x2f7cc9b6, 0x2e84b1b3, 0x2d9001b0, 0x2d9001b0, 0x2c9eb1ad, 0x2bb0b9aa,
0x2bb0b9aa, 0x2ac615a7, 0x29dec1a4, 0x29dec1a4, 0x28fab5a1, 0x2819e99e,
0x2819e99e, 0x273c599b, 0x273c599b, 0x26620198, 0x258ad995, 0x258ad995,
0x24b6d992, 0x24b6d992, 0x23e5fd8f, 0x2318418c, 0x2318418c, 0x224d9d89,
0x224d9d89, 0x21860986, 0x21860986, 0x20c18183, 0x20c18183, 0x20000180,
};
#else // LARGE_TABLE
uint8_t rsqrt_tab [96] =
{
0xfe, 0xfa, 0xf7, 0xf3, 0xf0, 0xec, 0xe9, 0xe6, 0xe4, 0xe1, 0xde, 0xdc,
0xd9, 0xd7, 0xd4, 0xd2, 0xd0, 0xce, 0xcc, 0xca, 0xc8, 0xc6, 0xc4, 0xc2,
0xc1, 0xbf, 0xbd, 0xbc, 0xba, 0xb9, 0xb7, 0xb6, 0xb4, 0xb3, 0xb2, 0xb0,
0xaf, 0xae, 0xac, 0xab, 0xaa, 0xa9, 0xa8, 0xa7, 0xa6, 0xa5, 0xa3, 0xa2,
0xa1, 0xa0, 0x9f, 0x9e, 0x9e, 0x9d, 0x9c, 0x9b, 0x9a, 0x99, 0x98, 0x97,
0x97, 0x96, 0x95, 0x94, 0x93, 0x93, 0x92, 0x91, 0x90, 0x90, 0x8f, 0x8e,
0x8e, 0x8d, 0x8c, 0x8c, 0x8b, 0x8a, 0x8a, 0x89, 0x89, 0x88, 0x87, 0x87,
0x86, 0x86, 0x85, 0x84, 0x84, 0x83, 0x83, 0x82, 0x82, 0x81, 0x81, 0x80,
};
#endif //LARGE_TABLE
/* compute floor (sqrt (a)) */
uint32_t my_isqrt32 (uint32_t a)
{
uint32_t b, r, s, t, scal, rem;
if (a == 0) return a;
/* Normalize argument */
scal = clz32 (a) & ~1;
b = a << scal;
/* Initial approximation to 1/sqrt(a)*/
t = rsqrt_tab [(b >> 25) - 32];
/* First NR iteration */
#if LARGE_TABLE
r = (t << 22) - umul32_hi (b, t);
#else // LARGE_TABLE
r = ((3 * t) << 22) - umul32_hi (b, (t * t * t) << 8);
#endif // LARGE_TABLE
/* Second NR iteration */
s = umul32_hi (r, b);
s = 0x30000000 - umul32_hi (r, s);
r = umul32_hi (r, s);
/* Compute sqrt(a) = a * 1/sqrt(a). Adjust to ensure it's an underestimate*/
r = umul32_hi (r, b) - 1;
/* Denormalize result */
r = r >> ((scal >> 1) + 11);
/* Make sure we got the floor correct */
rem = a - r * r;
if (rem >= (2 * r + 1)) r++;
return r;
}
#endif // ALT_IMPL
uint32_t umul32_hi (uint32_t a, uint32_t b)
{
return (uint32_t)(((uint64_t)a * b) >> 32);
}
uint32_t float_as_uint32 (float a)
{
uint32_t r;
memcpy (&r, &a, sizeof r);
return r;
}
int clz32 (uint32_t a)
{
#if (CLZ_IMPL == CLZ_FPU)
// Henry S. Warren, Jr, " Hacker's Delight 2nd ed", p. 105
int n = 158 - (float_as_uint32 ((float)(int32_t)(a & ~(a >> 1))+.5f) >> 23);
return (n < 0) ? 0 : n;
#elif (CLZ_IMPL == CLZ_CPU)
static const uint8_t clz_tab[32] = {
31, 22, 30, 21, 18, 10, 29, 2, 20, 17, 15, 13, 9, 6, 28, 1,
23, 19, 11, 3, 16, 14, 7, 24, 12, 4, 8, 25, 5, 26, 27, 0
};
a |= a >> 16;
a |= a >> 8;
a |= a >> 4;
a |= a >> 2;
a |= a >> 1;
return clz_tab [0x07c4acddu * a >> 27] + (!a);
#else // CLZ_IMPL == CLZ_BUILTIN
#if defined(_MSC_VER) && defined(_WIN64)
return __lzcnt (a);
#else // defined(_MSC_VER) && defined(_WIN64)
return __builtin_clz (a);
#endif // defined(_MSC_VER) && defined(_WIN64)
#endif // CLZ_IMPL
}
/* Henry S. Warren, Jr., "Hacker's Delight, 2nd e.d", p. 286 */
uint32_t ref_isqrt32 (uint32_t x)
{
uint32_t m, y, b;
m = 0x40000000U;
y = 0U;
while (m != 0) {
b = y | m;
y = y >> 1;
if (x >= b) {
x = x - b;
y = y | m;
}
m = m >> 2;
}
return y;
}
#if defined(_WIN32)
#if !defined(WIN32_LEAN_AND_MEAN)
#define WIN32_LEAN_AND_MEAN
#endif
#include <windows.h>
double second (void)
{
LARGE_INTEGER t;
static double oofreq;
static int checkedForHighResTimer;
static BOOL hasHighResTimer;
if (!checkedForHighResTimer) {
hasHighResTimer = QueryPerformanceFrequency (&t);
oofreq = 1.0 / (double)t.QuadPart;
checkedForHighResTimer = 1;
}
if (hasHighResTimer) {
QueryPerformanceCounter (&t);
return (double)t.QuadPart * oofreq;
} else {
return (double)GetTickCount() * 1.0e-3;
}
}
#elif defined(__linux__) || defined(__APPLE__)
#include <stddef.h>
#include <sys/time.h>
double second (void)
{
struct timeval tv;
gettimeofday(&tv, NULL);
return (double)tv.tv_sec + (double)tv.tv_usec * 1.0e-6;
}
#else
#error unsupported platform
#endif
int main (void)
{
#if ALT_IMPL
printf ("Alternate integer square root implementation\n");
#else // ALT_IMPL
#if LARGE_TABLE
printf ("Integer square root implementation w/ large table\n");
#else // LARGE_TAB
printf ("Integer square root implementation w/ small table\n");
#endif
#endif // ALT_IMPL
#if GEN_TAB
printf ("Generating lookup table ...\n");
#if ALT_IMPL
for (int i = 0; i < 384; i++) {
double x = 1.0 + (i + 1) * 1.0 / 128;
double y = 1.0 / sqrt (x);
uint8_t val = (uint8_t)((y * 512) - 256);
rsqrt_tab[i] = val;
printf ("0x%02x, ", rsqrt_tab[i]);
if (i % 12 == 11) printf("\n");
}
#else // ALT_IMPL
for (int i = 0; i < 96; i++ ) {
double x1 = 1.0 + i * 1.0 / 32;
double x2 = 1.0 + (i + 1) * 1.0 / 32;
double y = (1.0 / sqrt(x1) + 1.0 / sqrt(x2)) * 0.5;
uint32_t val = (uint32_t)(y * 256 + 0.5);
#if LARGE_TABLE
uint32_t cube = val * val * val;
rsqrt_tab[i] = (((cube + 1) / 4) << 10) + (3 * val);
printf ("0x%08x, ", rsqrt_tab[i]);
if (i % 6 == 5) printf ("\n");
#else // LARGE_TABLE
rsqrt_tab[i] = val;
printf ("0x%02x, ", rsqrt_tab[i]);
if (i % 12 == 11) printf ("\n");
#endif // LARGE_TABLE
}
#endif // ALT_IMPL
#endif // GEN_TAB
printf ("Running exhaustive test ... ");
uint32_t i = 0;
do {
uint32_t ref = ref_isqrt32 (i);
uint32_t res = my_isqrt32 (i);
if (res != ref) {
printf ("error: arg=%08x res=%08x ref=%08x\n", i, res, ref);
return EXIT_FAILURE;
}
i++;
} while (i);
printf ("PASSED\n");
printf ("Running benchmark ...\n");
i = 0;
uint32_t sum[8] = {0, 0, 0, 0, 0, 0, 0, 0};
double start = second();
do {
sum [0] += my_isqrt32 (i + 0);
sum [1] += my_isqrt32 (i + 1);
sum [2] += my_isqrt32 (i + 2);
sum [3] += my_isqrt32 (i + 3);
sum [4] += my_isqrt32 (i + 4);
sum [5] += my_isqrt32 (i + 5);
sum [6] += my_isqrt32 (i + 6);
sum [7] += my_isqrt32 (i + 7);
i += 8;
} while (i);
double stop = second();
printf ("%08x \relapsed=%.5f sec\n",
sum[0]+sum[1]+sum[2]+sum[3]+sum[4]+sum[5]+sum[6]+sum[7],
stop - start);
return EXIT_SUCCESS;
}
Here is a solution in Java that combines integer log_2 and Newton's method to create a loop free algorithm. As a downside, it needs division. The commented lines are required to upconvert to a 64-bit algorithm.
private static final int debruijn= 0x07C4ACDD;
//private static final long debruijn= ( ~0x0218A392CD3D5DBFL)>>>6;
static
{
for(int x= 0; x<32; ++x)
{
final long v= ~( -2L<<(x));
DeBruijnArray[(int)((v*debruijn)>>>27)]= x; //>>>58
}
for(int x= 0; x<32; ++x)
SQRT[x]= (int) (Math.sqrt((1L<<DeBruijnArray[x])*Math.sqrt(2)));
}
public static int sqrt(final int num)
{
int y;
if(num==0)
return num;
{
int v= num;
v|= v>>>1; // first round up to one less than a power of 2
v|= v>>>2;
v|= v>>>4;
v|= v>>>8;
v|= v>>>16;
//v|= v>>>32;
y= SQRT[(v*debruijn)>>>27]; //>>>58
}
//y= (y+num/y)>>>1;
y= (y+num/y)>>>1;
y= (y+num/y)>>>1;
y= (y+num/y)>>>1;
return y*y>num?y-1:y;
}
How this works: The first part produces a square root accurate to about three bits. The line y= (y+num/y)>>1; doubles the accuracy in bits. The last line eliminates the roof roots that can be generated.
If you need it just for ARM Thumb 2 processors, CMSIS DSP library by ARM is the best shot for you. It's made by people who designed Thumb 2 processors. Who else can beat it?
Actually you don't even need an algorithm but specialized square root hardware instructions such as VSQRT. The ARM company maintains math and DSP algorithm implementations highly optimized for Thumb 2 supported processors by trying to use its hardware like VSQRT. You can get the source code:
arm_sqrt_f32()
arm_sqrt_q15.c / arm_sqrt_q31.c (q15 and q31 are the fixed-point data types specialized for ARM DSP core, which is often comes with Thum 2 compatible processors.)
Note that ARM also maintains compiled binaries of CMSIS DSP that guarantees the best possible performance for ARM Thumb architecture-specific instructions. So you should consider statically link them when you use the library. You can get the binaries here.
This method is similar to long division: you construct a guess for the next digit of the root, do a subtraction, and enter the digit if the difference meets certain criteria. With a the binary version, your only choice for the next digit is 0 or 1, so you always guess 1, do the subtraction, and enter a 1 unless the difference is negative.
http://www.realitypixels.com/turk/opensource/index.html#FractSqrt
I implemented Warren's suggestion and the Newton method in C# for 64-bit integers. Isqrt uses the Newton method, while Isqrt uses Warren's method. Here is the source code:
using System;
namespace Cluster
{
public static class IntegerMath
{
/// <summary>
/// Compute the integer square root, the largest whole number less than or equal to the true square root of N.
///
/// This uses the integer version of Newton's method.
/// </summary>
public static long Isqrt(this long n)
{
if (n < 0) throw new ArgumentOutOfRangeException("n", "Square root of negative numbers is not defined.");
if (n <= 1) return n;
var xPrev2 = -2L;
var xPrev1 = -1L;
var x = 2L;
// From Wikipedia: if N + 1 is a perfect square, then the algorithm enters a two-value cycle, so we have to compare
// to two previous values to test for convergence.
while (x != xPrev2 && x != xPrev1)
{
xPrev2 = xPrev1;
xPrev1 = x;
x = (x + n/x)/2;
}
// The two values x and xPrev1 will be above and below the true square root. Choose the lower one.
return x < xPrev1 ? x : xPrev1;
}
#region Sqrt using Bit-shifting and magic numbers.
// From http://stackoverflow.com/questions/1100090/looking-for-an-efficient-integer-square-root-algorithm-for-arm-thumb2
// Converted to C#.
private static readonly ulong debruijn= ( ~0x0218A392CD3D5DBFUL)>>6;
private static readonly ulong[] SQRT = new ulong[64];
private static readonly int[] DeBruijnArray = new int[64];
static IntegerMath()
{
for(int x= 0; x<64; ++x)
{
ulong v= (ulong) ~( -2L<<(x));
DeBruijnArray[(v*debruijn)>>58]= x;
}
for(int x= 0; x<64; ++x)
SQRT[x]= (ulong) (Math.Sqrt((1L<<DeBruijnArray[x])*Math.Sqrt(2)));
}
public static long Isqrt2(this long n)
{
ulong num = (ulong) n;
ulong y;
if(num==0)
return (long)num;
{
ulong v= num;
v|= v>>1; // first round up to one less than a power of 2
v|= v>>2;
v|= v>>4;
v|= v>>8;
v|= v>>16;
v|= v>>32;
y= SQRT[(v*debruijn)>>58];
}
y= (y+num/y)>>1;
y= (y+num/y)>>1;
y= (y+num/y)>>1;
y= (y+num/y)>>1;
// Make sure that our answer is rounded down, not up.
return (long) (y*y>num?y-1:y);
}
#endregion
}
}
I used the following to benchmark the code:
using System;
using System.Diagnostics;
using Cluster;
using Microsoft.VisualStudio.TestTools.UnitTesting;
namespace ClusterTests
{
[TestClass]
public class IntegerMathTests
{
[TestMethod]
public void Isqrt_Accuracy()
{
for (var n = 0L; n <= 100000L; n++)
{
var expectedRoot = (long) Math.Sqrt(n);
var actualRoot = n.Isqrt();
Assert.AreEqual(expectedRoot, actualRoot, String.Format("Square root is wrong for N = {0}.", n));
}
}
[TestMethod]
public void Isqrt2_Accuracy()
{
for (var n = 0L; n <= 100000L; n++)
{
var expectedRoot = (long)Math.Sqrt(n);
var actualRoot = n.Isqrt2();
Assert.AreEqual(expectedRoot, actualRoot, String.Format("Square root is wrong for N = {0}.", n));
}
}
[TestMethod]
public void Isqrt_Speed()
{
var integerTimer = new Stopwatch();
var libraryTimer = new Stopwatch();
integerTimer.Start();
var total = 0L;
for (var n = 0L; n <= 300000L; n++)
{
var root = n.Isqrt();
total += root;
}
integerTimer.Stop();
libraryTimer.Start();
total = 0L;
for (var n = 0L; n <= 300000L; n++)
{
var root = (long)Math.Sqrt(n);
total += root;
}
libraryTimer.Stop();
var isqrtMilliseconds = integerTimer.ElapsedMilliseconds;
var libraryMilliseconds = libraryTimer.ElapsedMilliseconds;
var msg = String.Format("Isqrt: {0} ms versus library: {1} ms", isqrtMilliseconds, libraryMilliseconds);
Debug.WriteLine(msg);
Assert.IsTrue(libraryMilliseconds > isqrtMilliseconds, "Isqrt2 should be faster than Math.Sqrt! " + msg);
}
[TestMethod]
public void Isqrt2_Speed()
{
var integerTimer = new Stopwatch();
var libraryTimer = new Stopwatch();
var warmup = (10L).Isqrt2();
integerTimer.Start();
var total = 0L;
for (var n = 0L; n <= 300000L; n++)
{
var root = n.Isqrt2();
total += root;
}
integerTimer.Stop();
libraryTimer.Start();
total = 0L;
for (var n = 0L; n <= 300000L; n++)
{
var root = (long)Math.Sqrt(n);
total += root;
}
libraryTimer.Stop();
var isqrtMilliseconds = integerTimer.ElapsedMilliseconds;
var libraryMilliseconds = libraryTimer.ElapsedMilliseconds;
var msg = String.Format("isqrt2: {0} ms versus library: {1} ms", isqrtMilliseconds, libraryMilliseconds);
Debug.WriteLine(msg);
Assert.IsTrue(libraryMilliseconds > isqrtMilliseconds, "Isqrt2 should be faster than Math.Sqrt! " + msg);
}
}
}
My results on a Dell Latitude E6540 in Release mode, Visual Studio 2012 were
that the Library call Math.Sqrt is faster.
59 ms - Newton (Isqrt)
12 ms - Bit shifting (Isqrt2)
5 ms - Math.Sqrt
I am not clever with compiler directives, so it may be possible to tune the compiler to get the integer math faster. Clearly, the bit-shifting approach is very close to the library. On a system with no math coprocessor, it would be very fast.
I've designed a 16-bit sqrt for RGB gamma compression. It dispatches into 3 different tables, based on the higher 8 bits. Disadvantages: it uses about a kilobyte for the tables, rounds unpredictable, if exact sqrt is impossible, and, in the worst case, uses single multiplication (but only for a very few input values).
static uint8_t sqrt_50_256[] = {
114,115,116,117,118,119,120,121,122,123,124,125,126,127,128,129,130,131,132,
133,134,135,136,137,138,139,140,141,142,143,143,144,145,146,147,148,149,150,
150,151,152,153,154,155,155,156,157,158,159,159,160,161,162,163,163,164,165,
166,167,167,168,169,170,170,171,172,173,173,174,175,175,176,177,178,178,179,
180,181,181,182,183,183,184,185,185,186,187,187,188,189,189,190,191,191,192,
193,193,194,195,195,196,197,197,198,199,199,200,201,201,202,203,203,204,204,
205,206,206,207,207,208,209,209,210,211,211,212,212,213,214,214,215,215,216,
217,217,218,218,219,219,220,221,221,222,222,223,223,224,225,225,226,226,227,
227,228,229,229,230,230,231,231,232,232,233,234,234,235,235,236,236,237,237,
238,238,239,239,240,241,241,242,242,243,243,244,244,245,245,246,246,247,247,
248,248,249,249,250,250,251,251,252,252,253,253,254,254,255,255
};
static uint8_t sqrt_0_10[] = {
1,2,3,3,4,4,5,5,5,6,6,6,7,7,7,7,8,8,8,8,9,9,9,9,9,10,10,10,10,10,11,11,11,
11,11,11,12,12,12,12,12,12,13,13,13,13,13,13,13,14,14,14,14,14,14,14,15,15,
15,15,15,15,15,15,16,16,16,16,16,16,16,16,17,17,17,17,17,17,17,17,17,18,18,
18,18,18,18,18,18,18,19,19,19,19,19,19,19,19,19,19,20,20,20,20,20,20,20,20,
20,20,21,21,21,21,21,21,21,21,21,21,21,22,22,22,22,22,22,22,22,22,22,22,23,
23,23,23,23,23,23,23,23,23,23,23,24,24,24,24,24,24,24,24,24,24,24,24,25,25,
25,25,25,25,25,25,25,25,25,25,25,26,26,26,26,26,26,26,26,26,26,26,26,26,27,
27,27,27,27,27,27,27,27,27,27,27,27,27,28,28,28,28,28,28,28,28,28,28,28,28,
28,28,29,29,29,29,29,29,29,29,29,29,29,29,29,29,29,30,30,30,30,30,30,30,30,
30,30,30,30,30,30,30,31,31,31,31,31,31,31,31,31,31,31,31,31,31,31,31,32,32,
32,32,32,32,32,32,32,32,32,32,32,32,32,32,33,33,33,33,33,33,33,33,33,33,33,
33,33,33,33,33,33,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,34,35,35,
35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,35,36,36,36,36,36,36,36,36,36,
36,36,36,36,36,36,36,36,36,37,37,37,37,37,37,37,37,37,37,37,37,37,37,37,37,
37,37,37,38,38,38,38,38,38,38,38,38,38,38,38,38,38,38,38,38,38,38,39,39,39,
39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,39,40,40,40,40,40,40,40,40,
40,40,40,40,40,40,40,40,40,40,40,40,41,41,41,41,41,41,41,41,41,41,41,41,41,
41,41,41,41,41,41,41,41,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,42,
42,42,42,42,43,43,43,43,43,43,43,43,43,43,43,43,43,43,43,43,43,43,43,43,43,
43,44,44,44,44,44,44,44,44,44,44,44,44,44,44,44,44,44,44,44,44,44,44,45,45,
45,45,45,45,45,45,45,45,45,45,45,45,45,45,45,45,45,45,45,45,45,46,46,46,46,
46,46,46,46,46,46,46,46,46,46,46,46,46,46,46,46,46,46,46,47,47,47,47,47,47,
47,47,47,47,47,47,47,47,47,47,47,47,47,47,47,47,47,47,48,48,48,48,48,48,48,
48,48,48,48,48,48,48,48,48,48,48,48,48,48,48,48,48,49,49,49,49,49,49,49,49,
49,49,49,49,49,49,49,49,49,49,49,49,49,49,49,49,49,50,50,50,50,50,50,50,50,
50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,50,51,51,51,51,51,51,51,51,
51,51,51,51,51,51,51,51,51,51,51,51,51,51,51,51,51,51,52,52,52,52,52,52,52,
52,52,52,52,52,52,52,52,52,52,52,52,52,52,52,52,52,52,52,53,53
};
static uint8_t sqrt_11_49[] = {
54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,0,76,77,78,
0,79,80,81,82,83,0,84,85,86,0,87,88,89,0,90,0,91,92,93,0,94,0,95,96,97,0,98,0,
99,100,101,0,102,0,103,0,104,105,106,0,107,0,108,0,109,0,110,0,111,112,113
};
uint16_t isqrt16(uint16_t v) {
uint16_t a, b;
uint16_t h = v>>8;
if (h <= 10) return v ? sqrt_0_10[v>>2] : 0;
if (h >= 50) return sqrt_50_256[h-50];
h = (h-11)<<1;
a = sqrt_11_49[h];
b = sqrt_11_49[h+1];
if (!a) return b;
return b*b > v ? a : b;
}
I've compared it against the log2 based sqrt, using clang's __builtin_clz (which should expand to a single assembly opcode), and the library's sqrtf, called using (int)sqrtf((float)i). And got rather strange results:
$ gcc -O3 test.c -o test && ./test
isqrt16: 6.498955
sqrtf: 6.981861
log2_sqrt: 61.755873
Clang compiled the call to sqrtf to a sqrtss instruction, which is nearly as fast as that table sqrt. Lesson learned: on x86 the compiler can provide fast enough sqrt, which is less than 10% slower than what you yourself can come up with, wasting a lot of time, or can be 10 times faster, if you use some ugly bitwise hacks. And still sqrtss is a bit slower than custom function, so if you really need these 5%, you can get them, and ARM for example has no sqrtss, so log2_sqrt shouldn't lag that bad.
On x86, where FPU is available, the old Quake hack appears to be the fastest way to calculate integer sqrt. It is 2 times faster than this table or the FPU's sqrtss.

How do I set output flags for ALU in "Nand to Tetris" course?

Although I tagged this homework, it is actually for a course which I am doing on my own for free. Anyway, the course is called "From Nand to Tetris" and I'm hoping someone here has seen or taken the course so I can get some help. I am at the stage where I am building the ALU with the supplied hdl language. My problem is that I can't get my chip to compile properly. I am getting errors when I try to set the output flags for the ALU. I believe the problem is that I can't subscript any intermediate variable, since when I just try setting the flags to true or false based on some random variable (say an input flag), I do not get the errors. I know the problem is not with the chips I am trying to use since I am using all builtin chips.
Here is my ALU chip so far:
/**
* The ALU. Computes a pre-defined set of functions out = f(x,y)
* where x and y are two 16-bit inputs. The function f is selected
* by a set of 6 control bits denoted zx, nx, zy, ny, f, no.
* The ALU operation can be described using the following pseudocode:
* if zx=1 set x = 0 // 16-bit zero constant
* if nx=1 set x = !x // Bit-wise negation
* if zy=1 set y = 0 // 16-bit zero constant
* if ny=1 set y = !y // Bit-wise negation
* if f=1 set out = x + y // Integer 2's complement addition
* else set out = x & y // Bit-wise And
* if no=1 set out = !out // Bit-wise negation
*
* In addition to computing out, the ALU computes two 1-bit outputs:
* if out=0 set zr = 1 else zr = 0 // 16-bit equality comparison
* if out<0 set ng = 1 else ng = 0 // 2's complement comparison
*/
CHIP ALU {
IN // 16-bit inputs:
x[16], y[16],
// Control bits:
zx, // Zero the x input
nx, // Negate the x input
zy, // Zero the y input
ny, // Negate the y input
f, // Function code: 1 for add, 0 for and
no; // Negate the out output
OUT // 16-bit output
out[16],
// ALU output flags
zr, // 1 if out=0, 0 otherwise
ng; // 1 if out<0, 0 otherwise
PARTS:
// Zero the x input
Mux16( a=x, b=false, sel=zx, out=x2 );
// Zero the y input
Mux16( a=y, b=false, sel=zy, out=y2 );
// Negate the x input
Not16( in=x, out=notx );
Mux16( a=x, b=notx, sel=nx, out=x3 );
// Negate the y input
Not16( in=y, out=noty );
Mux16( a=y, b=noty, sel=ny, out=y3 );
// Perform f
Add16( a=x3, b=y3, out=addout );
And16( a=x3, b=y3, out=andout );
Mux16( a=andout, b=addout, sel=f, out=preout );
// Negate the output
Not16( in=preout, out=notpreout );
Mux16( a=preout, b=notpreout, sel=no, out=out );
// zr flag
Or8way( in=out[0..7], out=zr1 ); // PROBLEM SHOWS UP HERE
Or8way( in=out[8..15], out=zr2 );
Or( a=zr1, b=zr2, out=zr );
// ng flag
Not( in=out[15], out=ng );
}
So the problem shows up when I am trying to send a subscripted version of 'out' to the Or8Way chip. I've tried using a different variable than 'out', but with the same problem. Then I read that you are not able to subscript intermediate variables. I thought maybe if I sent the intermediate variable to some other chip, and that chip subscripted it, it would solve the problem, but it has the same error. Unfortunately I just can't think of a way to set the zr and ng flags without subscripting some intermediate variable, so I'm really stuck!
Just so you know, if I replace the problematic lines with the following, it will compile (but not give the right results since I'm just using some random input):
// zr flag
Not( in=zx, out=zr );
// ng flag
Not( in=zx, out=ng );
Anyone have any ideas?
Edit: Here is the appendix of the book for the course which specifies how the hdl works. Specifically look at section 5 which talks about buses and says: "An internal pin (like v above) may not be subscripted".
Edit: Here is the exact error I get: "Line 68, Can't connect gate's output pin to part". The error message is sort of confusing though, since that does not seem to be the actual problem. If I just replace "Or8way( in=out[0..7], out=zr1 );" with "Or8way( in=false, out=zr1 );" it will not generate this error, which is what lead me to look up in the appendix and find that the out variable, since it was derived as intermediate, could not be subscripted.
For anyone else interested, the solution the emulator supports is to use multiple outputs
Something like:
Mux16( a=preout, b=notpreout, sel=no, out=out,out=preout2,out[15]=ng);
This is how I did the ALU:
CHIP ALU {
IN // 16-bit inputs:
x[16], y[16],
// Control bits:
zx, // Zero the x input
nx, // Negate the x input
zy, // Zero the y input
ny, // Negate the y input
f, // Function code: 1 for add, 0 for and
no; // Negate the out output
OUT // 16-bit output
out[16],
// ALU output flags
zr, // 1 if out=0, 0 otherwise
ng; // 1 if out<0, 0 otherwise
PARTS:
Mux16(a=x, b=false, sel=zx, out=M16x);
Not16(in=M16x, out=Nx);
Mux16(a=M16x, b=Nx, sel=nx, out=M16M16x);
Mux16(a=y, b=false, sel=zy, out=M16y);
Not16(in=M16y, out=Ny);
Mux16(a=M16y, b=Ny, sel=ny, out=M16M16y);
And16(a=M16M16x, b=M16M16y, out=And16);
Add16(a=M16M16x, b=M16M16y, out=Add16);
Mux16(a=And16, b=Add16, sel=f, out=F16);
Not16(in=F16, out=NF16);
Mux16(a=F16, b=NF16, sel=no, out=out, out[15]=ng, out[0..7]=zout1, out[8..15]=zout2);
Or8Way(in=zout1, out=zr1);
Or8Way(in=zout2, out=zr2);
Or(a=zr1, b=zr2, out=zr3);
Not(in=zr3, out=zr);
}
The solution as Pax suggested was to use an intermediate variable as input to another chip, such as Or16Way. Here is the code after I fixed the problem and debugged:
CHIP ALU {
IN // 16-bit inputs:
x[16], y[16],
// Control bits:
zx, // Zero the x input
nx, // Negate the x input
zy, // Zero the y input
ny, // Negate the y input
f, // Function code: 1 for add, 0 for and
no; // Negate the out output
OUT // 16-bit output
out[16],
// ALU output flags
zr, // 1 if out=0, 0 otherwise
ng; // 1 if out<0, 0 otherwise
PARTS:
// Zero the x input
Mux16( a=x, b=false, sel=zx, out=x2 );
// Zero the y input
Mux16( a=y, b=false, sel=zy, out=y2 );
// Negate the x input
Not16( in=x2, out=notx );
Mux16( a=x2, b=notx, sel=nx, out=x3 );
// Negate the y input
Not16( in=y2, out=noty );
Mux16( a=y2, b=noty, sel=ny, out=y3 );
// Perform f
Add16( a=x3, b=y3, out=addout );
And16( a=x3, b=y3, out=andout );
Mux16( a=andout, b=addout, sel=f, out=preout );
// Negate the output
Not16( in=preout, out=notpreout );
Mux16( a=preout, b=notpreout, sel=no, out=preout2 );
// zr flag
Or16Way( in=preout2, out=notzr );
Not( in=notzr, out=zr );
// ng flag
And16( a=preout2, b=true, out[15]=ng );
// Get final output
And16( a=preout2, b=preout2, out=out );
}
Have you tried:
// zr flag
Or8way(
in[0]=out[ 0], in[1]=out[ 1], in[2]=out[ 2], in[3]=out[ 3],
in[4]=out[ 4], in[5]=out[ 5], in[6]=out[ 6], in[7]=out[ 7],
out=zr1);
Or8way(
in[0]=out[ 8], in[1]=out[ 9], in[2]=out[10], in[3]=out[11],
in[4]=out[12], in[5]=out[13], in[6]=out[14], in[7]=out[15],
out=zr2);
Or( a=zr1, b=zr2, out=zr );
I don't know if this will work but it seems to make sense from looking at this document here.
I'd also think twice about using out as a variable name since it's confusing trying to figure out the difference between that and the keyword out (as in "out=...").
Following your edit, if you cannot subscript intermediate values, then it appears you will have to implement a separate "chip" such as IsZero16 which will take a 16-bit value as input (your intermediate out) and return one bit indicating its zero-ness that you can load into zr. Or you could make an IsZero8 chip but you'd have to then call it it two stages as you're currently doing with Or8Way.
This seems like a valid solution since you can subscript the input values to a chip.
And, just looking at the error, this may be a different problem to the one you suggest. The phrase "Can't connect gate's output pin to part" would mean to me that you're unable to connect signals from the output parameter back into the chips processing area. That makes sense from an electrical point of view.
You may find you have to store the output into a temporary variable and use that to both set zr and out (since once the signals have been "sent" to the chips output pins, they may no longer be available).
Can we try:
CHIP SetFlags16 {
IN inpval[16];
OUT zflag,nflag;
PARTS:
Or8way(in=inpval[0.. 7],out=zr0);
Or8way(in=inpval[8..15],out=zr1);
Or(a=zr0,b=zr1,out=zflag);
Not(in=inpval[15],out=nflag);
}
and then, in your ALU chip, use this at the end:
// Negate the output
Not16( in=preout, out=notpreout );
Mux16( a=preout, b=notpreout, sel=no, out=tempout );
// flags
SetFlags16(inpval=tempout,zflag=zr,nflag=ng);
// Transfer tempout to out (may be a better way).
Or16(a=tempout,b=tempout,out=out);
Here's one also with a new chip but it feels cleaner
/**
* Negator16 - negates the input 16-bit value if the selection flag is lit
*/
CHIP Negator16 {
IN sel,in[16];
OUT out[16];
PARTS:
Not16(in=in, out=negateIn);
Mux16(a=in, b=negateIn, sel=sel, out=out);
}
CHIP ALU {
// IN and OUT go here...
PARTS:
//Zero x and y if needed
Mux16(a=x, b[0..15]=false, sel=zx, out=x1);
Mux16(a=y, b[0..15]=false, sel=zy, out=y1);
//Create x1 and y1 negations if needed
Negator16(in=x1, sel=nx, out=x2);
Negator16(in=y1, sel=ny, out=y2);
//Create x&y and x+y
And16(a=x2, b=y2, out=andXY);
Add16(a=x2, b=y2, out=addXY);
//Choose between And/Add according to selection
Mux16(a=andXY, b=addXY, sel=f, out=res);
// negate if needed and also set negative flag
Negator16(in=res, sel=no, out=res1, out=out, out[15]=ng);
// set zero flag (or all bits and negate)
Or16Way(in=res1, out=nzr);
Not(in=nzr, out=zr);
}