Lesson 3. Negative numbers and subtraction: two's complement

After this lesson the same adder learns to subtract: the SUB switch turns A + B into A − B. Four lamps show the flags — what the processor knows about every answer: was there a carry, is the answer zero, is it negative and did it fit in 4 bits.

3.1. The idea: numbers on a circle

A 4-bit adder counts in a circle: 15 + 1 gives 0 — the fifth digit goes into the carry. So 1111 behaves like −1: add 1 to it and you get 0. In the same way 1110 is −2 and 1000 is −8.

Let us agree: if the top bit is 1, the number is negative. Then 4 bits hold the numbers from −8 to 7. This way of writing is called two's complement. Nothing in the adder has to change: it adds bits, and what they mean is up to us.

Four bits are a circle of 16 numbers: +1 is one step clockwise, after 1111 comes 0000 again. If numbers with the top bit 1 count as negative, 1111 is −1 and 1000 is −8. The adder is the same — only the way we read the bits changes.

How to get −B: flip all the bits of B and add 1. For example, 3 = 0011 → 1100 → 1101 = −3. Check: 3 + (−3) = 0011 + 1101 = 1 0000 — zero in 4 bits.

And subtracting means adding −B:

XOR with the SUB switch flips the bits of B: with SUB = 0 a bit passes as it is, with SUB = 1 it is flipped;
the "plus 1" comes from the same SUB fed to the carry input CI of the adder.

The ADDSUB4 subtractor for 3 − 5. SUB = 1 flips the bits of B in the XOR gates (0101 → 1010) and adds one through CI: 3 + 1010 + 1 = 1110, that is −2. With SUB = 0 the XOR gates pass B as it is and CI = 0 — plain addition.

3.2. Flags

After an addition or a subtraction every processor keeps four bits about the answer:

C (carry) — the carry out of the top digit. When subtracting, C = 1 means "no borrow": A ≥ B unsigned;
Z (zero) — the answer is zero: all digits are 0, that is a NOR of all the bits;
N (negative) — the answer is negative: simply the top bit of the sum;
V (overflow) — the answer did not fit in the signed numbers: both operands have the same sign and the sum the other one. For example, 7 + 1 gives 1000 — "−8".

The processor decides by these flags: "if zero — jump", "if less — repeat".

3.3. Building

The same project. Delete the lesson 2 setup (8 switches, ADDER4 on the canvas, three displays); keep Power/Ground.

The idea. Subtracting B means adding −B. In two's complement −B = all the bits of B flipped, plus 1. XOR with the SUB switch does the flipping (with SUB = 0 a bit passes as it is, with SUB = 1 it is flipped), and the "plus 1" comes from the same SUB on the carry input CI of the adder.

Step 1. ADDSUB4 — addition and subtraction

Part	Pin	Label
Input port	out	A0
Input port	out	A1
Input port	out	A2
Input port	out	A3
Input port	out	B0
Input port	out	B1
Input port	out	B2
Input port	out	B3
Input port	out	SUB
Input port	out	CI
XOR (bit 0)	I0 I1 O0	B0 SUB bx0
XOR (bit 1)	I0 I1 O0	B1 SUB bx1
XOR (bit 2)	I0 I1 O0	B2 SUB bx2
XOR (bit 3)	I0 I1 O0	B3 SUB bx3
ADDER4	A0 A1 A2 A3 B0 B1 B2 B3 CI S0 S1 S2 S3 CO	A0 A1 A2 A3 bx0 bx1 bx2 bx3 CI S0 S1 S2 S3 CO
XOR (sign A ≠ sign S)	I0 I1 O0	A3 S3 va
XOR (sign B ≠ sign S)	I0 I1 O0	bx3 S3 vb
AND (overflow)	I0 I1 O0	va vb V
Output port	in	S0
Output port	in	S1
Output port	in	S2
Output port	in	S3
Output port	in	CO
Output port	in	V

Ctrl+G → name ADDSUB4. Pins: inputs A0–A3, B0–B3, SUB, CI; outputs S0–S3, CO, V.

V is overflow: both operands have the same sign and the sum has the other one (so 7 + 1 gives "−8").

Step 2. Canvas: switches, subtractor, flags

Part	Pin	Label
Input	out	A0
Input	out	A1
Input	out	A2
Input	out	A3
Input	out	B0
Input	out	B1
Input	out	B2
Input	out	B3
Input	out	SUB
ADDSUB4	A0 A1 A2 A3 B0 B1 B2 B3 SUB CI S0 S1 S2 S3 CO V	A0 A1 A2 A3 B0 B1 B2 B3 SUB SUB S0 S1 S2 S3 C V
OR	I0 I1 O0	S0 S1 z01
OR	I0 I1 O0	S2 S3 z23
NOR (zero)	I0 I1 O0	z01 z23 Z
LED (C — carry)	in	C
LED (Z — zero)	in	Z
LED (N — sign)	in	S3
LED (V — overflow)	in	V
Display	⚙	width 4, DEC, bus A
Display	⚙	width 4, DEC, bus B
Display	⚙	width 4, BIN, bus S

The subtractor's CI goes to the same label SUB: that is the "plus 1". Lamp N is simply the top bit of the sum S3.

3.4. Check

Negative numbers in 4 bits: 1111 = −1, 1110 = −2 … 1000 = −8. The top bit is the sign.

A	Switches A3…A0	B	Switches B3…B0	SUB	Display S	Lamps	Meaning
3	0011	4	0100	0	0111	—	3 + 4 = 7
3	0011	5	0101	1	1110	N	3 − 5 = −2
5	0101	5	0101	1	0000	Z, C	5 − 5 = 0
2	0010	3	0011	1	1111	N	2 − 3 = −1 (C = 0: a borrow)
7	0111	1	0001	0	1000	N, V	7 + 1 does not fit in 4 signed bits
9	1001	9	1001	0	0010	C, V	−7 + −7 = −14 — also an overflow

In the "Switches" columns the top switch (A3) is on the left and A0 on the right; 1 — on, 0 — off.

When subtracting, C is lit when there was no borrow (A ≥ B unsigned) — as in most processors, the 6502 for one.

If with SUB = 1 the answer is one less, check that the subtractor's CI is on the label SUB, not on GND.

Lesson 3 paste (`Cmd+I`)

Needs XOR, AND, OR, NOR and ADDER4. Delete the lesson 2 setup, keep the power.

# Lesson 3 — two's complement subtraction and the flags C, Z, N, V.chip ADDSUB4 {  portIn(A0)  portIn(A1)  portIn(A2)  portIn(A3)  portIn(B0)  portIn(B1)  portIn(B2)  portIn(B3)  portIn(SUB)  portIn(CI)  portOut(S0)  portOut(S1)  portOut(S2)  portOut(S3)  portOut(CO)  portOut(V)  XOR(B0, SUB, bx0)  XOR(B1, SUB, bx1)  XOR(B2, SUB, bx2)  XOR(B3, SUB, bx3)  ADDER4(A0=A0, A1=A1, A2=A2, A3=A3, B0=bx0, B1=bx1, B2=bx2, B3=bx3, CI=CI, S0=S0, S1=S1, S2=S2, S3=S3, CO=CO)  XOR(A3, S3, va)  XOR(bx3, S3, vb)  AND(va, vb, V)} input(A0)input(A1)input(A2)input(A3)input(B0)input(B1)input(B2)input(B3)input(SUB)ADDSUB4(A0=A0, A1=A1, A2=A2, A3=A3, B0=B0, B1=B1, B2=B2, B3=B3, SUB=SUB, CI=SUB, S0=S0, S1=S1, S2=S2, S3=S3, CO=C, V=V)OR(S0, S1, z01)OR(S2, S3, z23)NOR(z01, z23, Z)led(C)led(Z)led(S3)led(V)

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