Signed binary numbers multiplication - chip HDL code

Question

As a part of Nand2Tetris course I'm taking this term, I have to build a binary number multiplication chip.

I have built some chip which probably treats the most cases well enough.

But it has some bugs when I multiply number likes -5 and -2.

It gives me -32758.

Here is the HDL code:

// This file is part of www.nand2tetris.org
// and the book "The Elements of Computing Systems"
// by Nisan and Schocken, MIT Press.
// File name: projects/02/Mul.hdl

/**
 * The chip will multiply 2 numbers.
 * Handling overflows: any number larger than 16 bits
 * can be truncated to include only the 16 least significant bits.
 */

CHIP Mul{
    IN a[16], b[16]; // Two 16-bit numbers to multiply
    OUT out[16]; // 16-bit output number

    PARTS:
    Mux16(a=false, b=a, sel=b[0], out=out0); // If the current bit of b is 1 then output a, else 0
    ShiftLeft(in=a, out=shift1); // Shift a left as we grow to ten's position
    Mux16(a=false, b= shift1, sel=b[1], out=out1);
    ShiftLeft(in=shift1, out=shift2);
    Mux16(a=false, b= shift2, sel=b[2], out=out2);
    ShiftLeft(in=shift2, out=shift3);
    Mux16(a=false, b= shift3, sel=b[3], out=out3);
    ShiftLeft(in=shift3, out=shift4);
    Mux16(a=false, b= shift4, sel=b[4], out=out4);
    ShiftLeft(in=shift4, out=shift5);
    Mux16(a=false, b= shift5, sel=b[5], out=out5);
    ShiftLeft(in=shift5, out=shift6);
    Mux16(a=false, b= shift6, sel=b[6], out=out6);
    ShiftLeft(in=shift6, out=shift7);
    Mux16(a=false, b= shift7, sel=b[7], out=out7);
    ShiftLeft(in=shift7, out=shift8);
    Mux16(a=false, b= shift8, sel=b[8], out=out8);
    ShiftLeft(in=shift8, out=shift9);
    Mux16(a=false, b= shift9, sel=b[9], out=out9);
    ShiftLeft(in=shift9, out=shift10);
    Mux16(a=false, b= shift10, sel=b[10], out=out10);
    ShiftLeft(in=shift10, out=shift11);
    Mux16(a=false, b= shift11, sel=b[11], out=out11);
    ShiftLeft(in=shift11, out=shift12);
    Mux16(a=false, b= shift12, sel=b[12], out=out12);
    ShiftLeft(in=shift12, out=shift13);
    Mux16(a=false, b= shift13, sel=b[13], out=out13);
    ShiftLeft(in=shift13, out=shift14);
    Mux16(a=false, b= shift14, sel=b[14], out=out14);
    ShiftLeft(in=shift14, out=shift15);
    Mux16(a=false, b= shift15, sel=b[15], out=out15);

    //add all options
    Add16(a=out0, b=out1, out=firstAdd0);
    Add16(a=out2, b=out3, out=firstAdd1);
    Add16(a=out4, b=out5, out=firstAdd2);
    Add16(a=out6, b=out7, out=firstAdd3);
    Add16(a=out8, b=out9, out=firstAdd4);
    Add16(a=out10, b=out11, out=firstAdd5);
    Add16(a=out12, b=out13, out=firstAdd6);
    Add16(a=out14, b=out15, out=firstAdd7);
    Add16(a=firstAdd0, b=firstAdd1, out=secondAdd0);
    Add16(a=firstAdd2, b=firstAdd3, out=secondAdd1);
    Add16(a=firstAdd4, b=firstAdd5, out=secondAdd2);
    Add16(a=firstAdd6, b=firstAdd7, out=secondAdd3);
    Add16(a=secondAdd0, b=secondAdd1, out=thirdAdd0);
    Add16(a=secondAdd2, b=secondAdd3, out=thirdAdd1);
    Add16(a=thirdAdd0, b=thirdAdd1, out=out);
}

Does someone know what's the issue?

Thanks!

score 0 · Answer 1 · answered Nov 07 '19 at 13:19

0

In signed 2's complement representation the most significant bit is negative. As a result, the last partial product (out15) and must be subtracted rather than added from the sum.

Have a look at http://www-inst.eecs.berkeley.edu/~eecs151/sp18/files/Lecture21.pdf for more information on signed 2's complement multiplies.

answered Nov 07 '19 at 13:19

gatecat

1,156
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I've seen this suggestion several so far... And I still don't really get that. Why do you say: "the MSB is negative"? To my understanding, it's not only the MSB which is negative but the entire number is negative! The MSB only indicates that this is a negative number but I really don't understand why only the MSB is considered as a negative partial product? – HelpMe Nov 07 '19 at 21:44
In two's complement; the MSB tells you to subtract 2^i from the current value; whereas the other bits still add 2^i Consider in two's complement 1000 = -(2^3) = -8 whereas 1111 = -(2^3)+2^2+2^1+2^0= -1 – gatecat Nov 08 '19 at 09:56
I don't understand why is only the MSB negative and the others are positive... What is the rational? – HelpMe Nov 08 '19 at 16:08
This is the purpose of two's complement. The ultimate effect is that addition and subtraction with two's complement require no special logic – gatecat Nov 08 '19 at 16:56

Signed binary numbers multiplication - chip HDL code

1 Answers1