Computer Design Fun Party!

1. To compute expression A = (B-C)*(D-E)

Using accumulator:

Load B ; acc = B; opcode=1 B, oper=2B; mem.traf = 2 B

Sub C ; acc = B – C; 1 byte opcode, 2-bytes operand = 3 bytes

Store A ; A = B -C; 1 byte opcode, 2-bytes operand = 3 bytes

Load D ; acc = D; 1 byte opcode, 2-bytes operand = 3 bytes

Sub E ; acc = D -E; 1 byte opcode, 2-bytes operand = 3 bytes

Mul A ; acc = (B-C)* (D-E) ; 1 byte opcode, 2-bytes operand = 3 bytes

Store A ; A = (B-C) * (D-E) ; 1 byte opcode, 2-bytes operand = 3 bytes

Assume each opcode takes 8-bit, operands are 16-bit and data is move in 16-bit chunks:

Total bytes: 7 * 3 byte = 21 bytes

Memory traffic: 21 + 7 * 2 = 35 bytes

Using load-store:

Ld B, r1 ; r1 = B; opcode=1 B, op = 2 B; mem. traff = 2 bytes

Ld C, r2 ; r2 = C; memory traffic = 2 bytes

Sub r1, r2,r3 ; r3 = r1 * r2 = B – C; size=2 bytes, memory traffic = 0

Ld D, r1 ; r1 = D; memory traffic = 2 bytes

Ld E, r2 ; r2 = E; memory traffic = 2 bytes

Sub r1,r2,r4 ; r4 = r1 * r2 = D-E; memory traffic = 0 bytes

Mul r3, r4, r1 ; r1 = r3 * r4 = (B-C) * (D-E); memory traffic = 0 bytes

St r1,A ; A = r1 = (B-C) * (D-E); memory traffic = 2 bytes

Number of bytes for instructions: 3*7 = 21 bytes

Memory traffic: 21 + 5 *2 = 31 bytes

Using stack:

Push B ; mem. Traf = 2 bytes

Push C ; mem. Traf = 2 bytes

Sub ; B – C; mem. Traf. = 2 bytes

Push D ; mem traf = 2 bytes

Push E ; mem.traf = 2 bytes

Sub ; top of the stack contains D-E, below it is B-C

Mul ; top of the stack now contains (B-C) * (D-E)

Each instruction occupies 3 bytes (1 byte opcode and 2-byte operand), so total instructions use 3*7 = 21 bytes.

Memory traffic: 21 + 2*7 = 35 bytes

2. Register 0, 1, 2 (R0, R1, R2) are GPRs. R3 is initialized to 1 and make sure it is unchanged.
   1. Control sequence that forms the 2’s complement difference (or, R0 ← R0 + (-R1) = R0 + 2^’s complement of R1) of the contents of R0 and R1:

Write Enable				A-bus enables				B-bus enables				F0­F1		Time
0	1	2	3	0	1	2	3	0	1	2	3	F0	F1	T
0	1	0	0	0	1	0	0	0	0	0	0	1	1	0
0	1	0	0	0	1	0	0	0	0	0	1	0	0	1
1	0	0	0	1	1	0	0	1	0	0	0	0	0	2

Explanation:

At T=0:

F₀F₁ = 11 (A’), A-bus enable = 1, B-bus enable = all zeros. Data (say, Y) is placed in A-bus. C-bus contains Y’, Write-enable = pin 1. So Y is written to register 1 (R1 = Y’).

At T = 1:

F₀F₁ = 00 (add), A-bus enable = pin 1 (content of R1 which is X, is in A-bus), B-bus enable= pin 3 (content of R3 which is 1 is in B-bus), write-enable = pin 1 (C-bus contains Y’ + 1). So R1 = Y’ + 1 = 2’s compl. Of X.(or R1 = -Y)

At T = 2

B-bus = 0, hence data from R0, say X, is placed in B-bus. F₀F₁ = 00 (add), A-bus enable = 1 (which contains –Y). C-bus contains R0 + R1 = X - Y . Write-enable = pin 0, so result is stored in R0.

2. R0 XOR R1 (R0’ & R1 + R0 & R1’) = (R0 + R1)(R0’ + R1’), leaving the result in R0.

Write Enable				A-bus enables				B-bus enables				F0­F1		Time	Note
0	1	2	3	0	1	2	3	0	1	2	3	F0	F1	T	(Movements etc.)
0	0	1	0	1	0	0	0	0	1	0	0	0	0	0	R2 ← A + B
1	0	0	0	1	0	0	0	0	0	0	0	1	1	1	R0 ← A’
0	1	0	0	0	1	0	0	0	0	0	0	1	1	2	R1 ← B’
0	1	0	0	1	0	0	0	0	1	0	0	0	0	3	R1 ← A’ + B’
1	0	0	0	0	1	0	0	0	0	1		0	1	4	R0 ← (A+B)&(A’+B’)

3. Booth Multiplication Algorithm:

/***********************************************************
Title: Booth Algorithm sample
Desc: To simulate booth multiplication algorithm
Author: Buhadram
Notes:

Ref:
- Computer Architecture: A Quantitative Approach
*************************************************************/
#include 
#include 
#include 
#include 

#define NUMTYPE	   	short int
#define NUMBITS		(8*sizeof(NUMTYPE))
#define MAX(x,y) 		((x) > (y) ? x : y)
#define MIN(x,y) 		((x) < (y) ? x : y)
#define BITMASK(x,mask) 	((x) & mask)
/*********************************
proc: bitmask
desc: return 1 if bit[pos] is set
      pos=0 for LSB, and p=sizeof(x)-1
      for MSB
*********************************/
#define BIT(x,pos)		( (BITMASK(x,1<<(pos))) >> (pos) )
#define ASCII_BIT(x)	((x) + '0')

char *get_input(const char *prompt, char *buf, int length)
{
	char *p;
	int i;
    printf("%s",prompt);
	// fgets is safer to avoid buffer overflow
	p = fgets(buf, length, stdin);
	// remove newline character
	buf[strlen(buf)-1] = 0;
	return p;
}

/********************************************
* Swapbits
* Desc: To swap bit orders, e.g LSB becomes MSB
        Note: b[0] is LSB, b[n-1] is MSB
********************************************/
void swapbits(NUMTYPE *b)
{
    int i;
    int tmp;

	tmp = 0;
	for(i=0; i base)
       base = 10;                    /* can only use 0-9, A-Z        */
    tail = &buf[BUFSIZE - 1];           /* last character position      */
    *tail-- = '\0';

    if (10 == base && N < 0L) {
	   *head++ = '-';
       uarg = -N;
    }
    else uarg = N;

    if (uarg) {
  	   	for (i = 1; uarg; ++i) {
       	   register ldiv_t r;
		   r = ldiv(uarg, base);
           *tail-- = (char)(r.rem + ((9L < r.rem) ? ('A' - 10L) : '0'));
           uarg    = r.quot;
    	}
    }
    else  *tail-- = '0';
	memcpy(head, ++tail, i);
    return str;
}

#endif


/******************************
proc: btol
desc: converts binary string to
	  long integer
******************************/
long btol(const char *bin)
{
    int n,i;
    long d;

	n = strlen(bin);
	d = 0;
	for(i=n-1; i>=0; i--) {
		d += (bin[i]-'0') * (1 << i);
    }
    return d;
}

/**************************************
Proc: Ashift_bits_right
desc: perform arithmetic shift right on
    bits in x, but preserving the sign bit.
***************************************/
void ashift_bits_right(long *x, int numbits)
{
    int neg;

    neg = (*x<0) ? 1 : 0;
    *x >>= numbits;

    if (neg)
       //*x |= (1<>= 1;
		qm = q0;
		q >>= 1;
		if (a0)
		   r |= b1;
		b1 <<= 1;
    }
    return r;
}


int main(int argc, char *argv[])
{
	#define BUF_SIZE	(1024+1)
	char *buf;
	NUMTYPE A, B;
	long res;
	char str[256];

	buf = (char *)malloc(BUF_SIZE+1);
	assert(buf);
    get_input("A = ", buf, BUF_SIZE);
    A = atol(buf);
    printf("A (in Binary) = %s\n", ltoa(A, str, 2));
    get_input("B = ", buf, BUF_SIZE);
    B = atol(buf);
    printf("B (in Binary) = %s\n", ltoa(B, str, 2));
    res = A+B;
    printf("A + B = %ld = 0x%0X\n", res, res);
    res = A - B;
	printf("A - B = %ld = 0x%0X\n", res, res);
    puts("\n\nBOOTH MULTIPLICATION\n");
    res = booth_mult(A,B);
	printf("A * B = %ld = 0x%0X = %sb\n", res, res, ltoa(B,str,2));
    free(buf);
	return 0;
}

Counting the number of "1" bits

Another "ridiculous" interview I had was when an interviewer asked me how to count the number of set bits in a 32-bit or 8-bit. I told him that there are various algorithms to do this, but the easiest way but yet good enough in performance is a function like this:

uint32_t count_bits(uint32_t b)
{
    int i;
    int n = 0;
    
    for(i=0; i<sizeof(b)*8; i++)
    {
        n += (b & 1);
        b >>= 1;
    }   
    return n;
}

I also told him that there are many ways to do this (referred him to look at "Beautiful Code" to see some of the beautiful algorithms).

But then the interviewer argued.  FIrst, he said the upper border of the for-loop should be to "sizeof(b)*8" (meaning, instead of using "<", I should have used "<="). To bad, I said yes to him without thinking further (Probably got nervous.  Later at home, I checked that he was wrong).  Second, he said that the fastest is to use look-up table (n[0] = 0, n[1]=1, n[2]=1,...,n[255]=8,...). Although I agreed with his argument, but then he become inconsistent with his original question and there is a big downside of his argument. I said that this wouldn't be a choice in embedded programming where memory resource is limited. Secondly, his original question was to compute ones in a 32-bit number. That means we have to 2^32 (4,294,967,296 or about 4 GBytes) items set manually in an array. That's so ridiculous. Who wants to waste that much space just to calculate the number of bits in an integer? For a byte, although it seems acceptable ("only" 256 bytes of table), but compare that to the above algorithm:

movl $32,  %edx
 xorl %ecx, %ecx

.L2:
 movl %ecx, %ebx
 shrl %ecx
 andl $1,   %ebx
 addl %ebx, %eax
 decl %edx
 jne .L2

It's only 7 x86 mnemonic instructions (about 12-18 bytes) and no memory involvement in the loop!. Yes, it is O(32) for 32-bit, but it's a constant speed.  Even further, if the function is called only once, the fully optimized version of the instructions would be inserted inline in the place (without calling the function), so we can CPU time from doing push-pop stack frames. That's another tiny speedup.  It's probably fine for table lookup (yes, it's the fastest, O(1)), but it is limited to 8-bit.  For 16 or higher, seems the space-time trade-off is too big not to consider.  Besides, the above algorithm extendable into polymorphism in Object-Oriented code.

When I searched Google, I found the following (on Stanford's web):

v = v - ((v >> 1) & 0x55555555);                    // reuse input as temporary
v = (v & 0x33333333) + ((v >> 2) & 0x33333333);     // temp
c = ((v + (v >> 4) & 0xF0F0F0F) * 0x1010101) >> 24; // count

Which renders to instruction codes like below:

shrl %edx
 andl $1431655765, %edx
 subl %edx, %eax
 movl %eax, %edx
 shrl $2, %eax
 andl $858993459, %edx
 andl $858993459, %eax
 addl %edx, %eax
 movl %eax, %edx
 shrl $4, %edx
 leal (%edx,%eax), %eax
 andl $252645135, %eax
 imull $16843009, %eax, %eax
 shrl $24, %eax

(14 mnemonics).  This seems the fastest and most optimized.  It took some time for me to understand it.

Seems the interviewer doesn't have enough experience in embedded or microcontroller  or doesn't think much about optimization. Nevertheless, I failed an interview again.   I start to think that many interview questions are really not to dig the thinking ability of a candidate, but rather to get instant answers meet their mind.  I just hope big software companies like IBM, Microsoft, Google or Apple shouldn't fall to this kind of trap, but rather ask logical questions to know if the candidate can think systematically.

What a waste!

Parallelizable integer multiplication

At a recent interview with an Amazon, the hiring manager asked me about an algorithm to do integer multiplication without using any multiplication/division symbol in parallel computing.  The following algorithm was suggested to do integer multiplication without using any "MULT" instruction:

int mult(int x, int y)
{
 int z;

 if ((x==0) || (y==0))
  return 0;
 if (x==1)
  return y;

 if (y==1)
  return x;

 z = mult(x, y>>1);
 if (y % 2 == 0)
  return z<<1;
 else
  return x + (z<<1);
}

My question came up, is this parallelizable?  My argument to the interviewer was that this is not a good approach if he is asking from parallel computation perspective (See Robertazzi's papers on ACM journals to know more detail about parallel computation)

When a recursive call is made, the parameters are pushed into the stack before calling each subsequence recursive call.  Involving stack in this manner is hardly parallelizable, because there is coupling of current result based on previous results.  One of the principle of parallel computing is to decouple two or more computation as much as we can.

A dig into highly-optimized assembly language of the above code:

mult:
 pushl %ebp
 movl %esp, %ebp
 subl $24, %esp
 movl %ebx, -8(%ebp)
 movl %esi, -4(%ebp)
 movl 12(%ebp), %ebx
 movl 8(%ebp), %esi
 testl %ebx, %ebx
 je .L4
 testl %esi, %esi
 je .L4
 cmpl $1, %esi
 je .L2
 cmpl $1, %ebx
 .p2align 4,,3
 je .L5
 movl %ebx, %eax
 movl %esi, (%esp)
 sarl %eax
 movl %eax, 4(%esp)
 call mult
 andl $1, %ebx
 je .L7
 leal (%esi,%eax,2), %ebx
.L2:
 movl %ebx, %eax
 movl -4(%ebp), %esi
 movl -8(%ebp), %ebx
 leave
 ret
 .p2align 4,,10
 .p2align 3
.L4:
 xorl %ebx, %ebx
 movl -4(%ebp), %esi
 movl %ebx, %eax
 movl -8(%ebp), %ebx
 leave
 ret
 .p2align 4,,10
 .p2align 3
.L7:
 leal (%eax,%eax), %ebx
 movl -4(%ebp), %esi
 movl %ebx, %eax
 movl -8(%ebp), %ebx
 leave
 ret
 .p2align 4,,10
 .p2align 3
.L5:
 movl %esi, %ebx
 movl -4(%ebp), %esi
 movl %ebx, %eax
 movl -8(%ebp), %ebx
 leave
 ret

As can be seen, "SARL" (Shift Arithmetic Left) for the shift-left operation.  Some issues after analyzing this code.  First, there are multiple register-to-memory flow while doing the recursive:

movl %ebx, %eax
 movl %esi, (%esp)
 sarl %eax
 movl %eax, 4(%esp)

Memory access is expensive and the instructions might cause cache miss thus forcing CPU to reread data from RAM.
Another issue, how do we parallelize this, if there is stack push-pop coupling for the result?


My argument was to decompose the computation into two (or more) portions.



For example:

int pivot = b/2;

for(i=0; i<pivot; i++)
    sum1 += a;

for(i=pivot+1; i<b; i++)
    sum2 += a;
sum = sum1 + sum2;

No, let's check what we can see in the x86 assembly (gcc-generated):

mult:
 pushl %ebp
 movl %esp, %ebp
 subl $8, %esp
 movl 12(%ebp), %eax
 movl %ebx, (%esp)
 movl %esi, 4(%esp)
 movl 8(%ebp), %edx
 testl %eax, %eax
 je .L5
 testl %edx, %edx
 je .L5
 cmpl $1, %edx
 je .L2
 cmpl $1, %eax
 .p2align 4,,3
 je .L6
 movl %eax, %ecx
 xorl %esi, %esi
 shrl $31, %ecx
 xorl %ebx, %ebx
 addl %eax, %ecx
 sarl %ecx
 testl %ecx, %ecx
 jle .L3
 movl %ecx, %esi
 movl %ecx, %ebx
 imull %edx, %esi
.L3:
 xorl %ecx, %ecx
 cmpl %ebx, %eax
 jle .L4
 movl %eax, %ecx
 subl %ebx, %ecx
 imull %edx, %ecx
.L4:
 leal (%ecx,%esi), %eax
.L2:
 movl (%esp), %ebx
 movl 4(%esp), %esi
 leave
 ret
 .p2align 4,,10
 .p2align 3
.L5:
 xorl %eax, %eax
 movl (%esp), %ebx
 movl 4(%esp), %esi
 leave
 ret
 .p2align 4,,10
 .p2align 3
.L6:
 movl %edx, %eax
 jmp .L2

As can be seen, most of operations are in registers, also there is decoupling between 2 partitions (e.g, sum1 does not rely on sum2). The more we partition the calculation (e.g, pivot1 = b/4, and do 4 for-next loops for each partition), the more divisible partition can be accomplished by parallel computer. I was arguing that to optimize computation, hence lowering O(n), we should use divisible loops ("divide-et-empera") instead of recursive approach (seems the interviewer disagreed)

PS: My argument above caused me not to land job at Amazon.

Simple UDP Client

<pre>

/* UDP client in the internet domain */
#include <sys/types.h>
#include <sys/socket.h>
#include <netinet/in.h>
#include <arpa/inet.h>
#include <netdb.h>
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <string.h>

#define MIN(x, y)  ((x) < (y) ? (x) : (y))
#define MAX(x, y)  ((x) > (y) ? (x) : (y))

void error(const char *);

int main(int argc, char *argv[])
{
   int sock, n;
   unsigned int length;
   struct sockaddr_in server, from;
   struct hostent *hp;
   char buffer[256];

   if (argc < 3) {
        printf("Usage: %s <port>\n", argv[0]);
        exit(1);
   }
   sock= socket(AF_INET, SOCK_DGRAM, 0);
   if (sock < 0) error("socket");

   server.sin_family = AF_INET;
   hp = gethostbyname(argv[1]);
   if (hp==0) error("Unknown host");

   bcopy((char *)hp->h_addr,
        (char *)&server.sin_addr,
         hp->h_length);
   server.sin_port = htons(atoi(argv[2]));
   length=sizeof(struct sockaddr_in);
   printf("Please enter the message: ");
   bzero(buffer, sizeof(buffer));
   fgets(buffer, sizeof(buffer)-1, stdin);
   n=sendto(sock,buffer,
            strlen(buffer),0,(const struct sockaddr *)&server,length);
   if (n < 0)
       error("Sendto");
   n = recvfrom(sock, buffer, sizeof(buffer), 0, (struct sockaddr *)&from, &length);
   if (n < 0)
       error("recvfrom");
   printf("Got an ack: %d bytes\n", n);
   write(1, buffer, MIN(n, strlen(buffer)));
   close(sock);
   return 0;
}

void error(const char *msg)
{
    perror(msg);
    exit(0);
}

</pre>

Simple UDP Server

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <netdb.h>
#include <unistd.h>
#include <errno.h>
#include <sys/socket.h>
#include <sys/types.h>
#include <arpa/inet.h>

extern int h_errno;

void error(const char *msg);

int UdpSocket(int protocol)
{
 int sockfd;

    sockfd = socket(AF_INET, SOCK_DGRAM, 0);
 if (-1 == sockfd)
 {
  error(__func__);
 }
 return sockfd;
}


void error(const char *msg)
{
 printf("Error Code: %d", errno);
 perror(msg);
}


int BindSocketToAddr(int sock, struct sockaddr_in *addr, int port)
{
 int rc;

    if (!addr)
    {
        printf("%s: NULL pointer in addr\n", __func__);
        return -1;
    }
 bzero(addr, sizeof(struct sockaddr_in));
 addr->sin_family = AF_INET;
 addr->sin_addr.s_addr = INADDR_ANY;
 addr->sin_port = htons(port);
 rc = bind(sock, (struct sockaddr *)addr, sizeof(struct sockaddr));
 if (rc < 0)
 {
  perror(__func__);
 }
    return rc;
}



int main(int argc, char **argv)
{
 int sockfd;
 struct sockaddr_in fromSockAddr;
 struct sockaddr_in serverSockAddr;
 socklen_t fromlen;
 char buf[1024];
 char fromAddrStr[512];
 int n;

    if (argc < 2)
    {
        printf("\n%s <PORT NUMBER>\n\n", argv[0]);
        exit(0);
    }
    setuid(0);

 printf("UDP Server\n");

 sockfd = UdpSocket(0);
 if (sockfd > 0)
 {
  BindSocketToAddr(sockfd, &serverSockAddr, atoi(argv[1]));
  fromlen = sizeof(fromSockAddr);
  while (1) {
   /* wait for client to send a message */
   n = recvfrom(sockfd, buf, sizeof(buf), 0, (struct sockaddr *)&fromSockAddr, &fromlen);
   if (n < 0)
   {
    error("recvfrom");
    break;
   }
   else {
    inet_ntop(fromSockAddr.sin_family, &fromSockAddr.sin_addr, fromAddrStr, sizeof(fromAddrStr));
    printf("GOT msg from: %s\n", fromAddrStr);
    /* echo back */
    n = sendto(sockfd, buf, n, 0, (struct sockaddr *)&fromSockAddr, sizeof(fromSockAddr));
    if (n < 0)
    {
     error("sendto");
     break;
    }
   }
  }
 }
 return 0;
}

Intro to Ruby

Just got my hands dirty with Ruby today. It's an interesting language, yet simple to learn. Unlike some other dynamic programming languages that too cryptic to learn, Ruby is easy to learn, similar to Tcl.

Here's an example of a program I tested:

#!/usr/bin/ruby

presidents = ["Ford", "Carter", "Reagan", "Bush1", "Clinton", "Bush2", "Obama"]

for fwd in [-1,1]
    if fwd == 1 
        print "Forward/ascending\n"
        dec = 0
    else
        print "Backward/descending\n"
        dec = -1
    end
    for ss in 0...presidents.length
        print ss+1, ": ", presidents[ss*fwd+dec], "\n";
    end
end

The output:


Backward/descending
1: Obama
2: Bush2
3: Clinton
4: Bush1
5: Reagan
6: Carter
7: Ford
Forward/ascending
1: Ford
2: Carter
3: Reagan
4: Bush1
5: Clinton
6: Bush2
7: Obama