A minifloat is a binary representation of a floating-point number that has very few bits.
The minifloat in this question will be defined as an 6-bit number m, which has the following representation:
1 bit to repesent the sign of the number. This bit will be
0if the number is positive, and1if the number is negative.3 bits to represent the exponent of the number, offset by
3(i.e. an exponent of110actually represents a factor of 23, not 26).- An exponent of
000refers to a subnormal number. The mantissa refers to the fractional part of a number with an integer part of0multiplied by a factor of the lowest possible exponent (in this case, 2-2).
- An exponent of
2 bits to represent the mantissa of the number. If the exponent is anything other than
000or111, the 2 bits represent the fractional part after a1.- An exponent of
111representsinfinityif the mantissa is0, andNaN(not a number) otherwise.
- An exponent of
In the Wikipedia article, this would be referred to as a (1.3.2.3) minifloat.
Some examples of this minifloat's representation:
000000 = 0.00 = 0
000110 = 1.10 ×ばつ 2^(1-3) = 0.375
001100 = 1.00 ×ばつ 2^(3-3) = 1
011001 = 1.01 ×ばつ 2^(6-3) = 10
011100 = infinity
011101 = NaN
100000 = -0.00 = -0
100011 = -0.11 ×ばつ 2^(1-3) = -0.1875 (subnormal)
101011 = -1.11 ×ばつ 2^(2-3) = -0.875
110100 = -1.00 ×ばつ 2^(5-3) = -4
111100 = -infinity
111111 = NaN
Your task is to build a network of two-input NAND gates that takes 6 inputs representing a minifloat a and 6 inputs representing a minifloat b, and returns 6 outputs representing the minifloat a + b.
Your network must properly add subnormals. For example,
000001+000010must equal000011, and001001+000010=001010.Your network must add and subtract infinities properly. Anything finite added to an infinity is the same infinity. Positive infinity plus negative infinity is
NaN.A
NaNplus anything must equal aNaN, although whichNaNit equals is up to you.How you handle adding positive zero and negative zero to each other is up to you, although zero plus zero must equal zero.
Your network can implement any of the following rounding rules depending on convenience:
- Round down (toward negative infinity)
- Round up (toward positive infinity)
- Round toward zero
- Round away from zero
- Round to nearest, with halves rounded according to any of the above rules
To simplify things, you may use AND, OR, NOT, and XOR gates in your diagram, with the following corresponding scores:
NOT: 1AND: 2OR: 3XOR: 4
Each of these scores corresponds to the number of NAND gates that it takes to construct the corresponding gate.
The logic circuit that uses the fewest NAND gates to correctly implement all the above requirements wins.
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3\$\begingroup\$ Nice challenge - I'd have to seriously think about this to implement it in code, let alone NAND gates. \$\endgroup\$Digital Trauma– Digital Trauma2014年04月22日 20:26:28 +00:00Commented Apr 22, 2014 at 20:26
1 Answer 1
830 NANDs
It uses 24 NOTs, 145 ANDs, 128 ORs, 33 XORs. It always rounds towards zero, it may return either -0 or +0 for zero values, and I believe it treats Infinities and NaNs correctly:
- ± INF ± INF = ± INF
- ± INF + NaN = ± INF
- ± INF ∓ INF = NaN
- ± INF + number = ± INF
- NaN + NaN = NaN
- NaN + number = NaN
Below I have a coded representation of the circuit. I have little experience annotating these types of things, so I don't really know what the typical way to do this is, but every variable is a Boolean so it's clear to see that it describes a circuit. Another thing, I have neither the know-how nor probably the tenacity to try and make a diagram of this, but if there is any easy to use software out there anyone wants to point out I would be interested in taking a look.
a0,a1,a2,a3,a4,a5 = mini0
b0,b1,b2,b3,b4,b5 = mini1
neg = XOR(a0,b0)
nneg = NOT(neg)
na1 = NOT(a1)
na2 = NOT(a2)
na3 = NOT(a3)
a2_a3 = AND(a2,a3)
a2_na3 = AND(a2,na3)
na2_a3 = AND(na2,a3)
na2_na3 = AND(na2,na3)
a123 = AND(a1,a2_a3)
l0 = AND(a1,a2_na3)
l1 = AND(a1,na2_a3)
l2 = AND(a1,na2_na3)
l3 = AND(na1,a2_a3)
l4 = AND(na1,a2_na3)
l5 = AND(na1,na2_a3)
l6 = AND(na1,na2_na3)
a45 = OR(a4,a5)
a_nan = AND(a123,a45)
a_inf = AND(a123,NOT(a45))
m0 = l0
m1 = OR(l1,AND(l0,a4))
m2 = OR(l2,OR(AND(l1,a4),AND(l0,a5)))
m3 = OR(l3,OR(AND(l2,a4),AND(l1,a5)))
m4 = OR(l4,OR(AND(l3,a4),AND(l2,a5)))
m5 = OR(l5,OR(AND(l4,a4),AND(l3,a5)))
l5_l6 = OR(l5,l6)
m6 = OR(AND(l4,a5),AND(l5_l6,a4))
m7 = AND(l5_l6,a5)
nb1 = NOT(b1)
nb2 = NOT(b2)
nb3 = NOT(b3)
b2_b3 = AND(b2,b3)
b2_nb3 = AND(b2,nb3)
nb2_b3 = AND(nb2,b3)
nb2_nb3 = AND(nb2,nb3)
b123 = AND(b1,b2_b3)
k0 = AND(b1,b2_nb3)
k1 = AND(b1,nb2_b3)
k2 = AND(b1,nb2_nb3)
k3 = AND(nb1,b2_b3)
k4 = AND(nb1,b2_nb3)
k5 = AND(nb1,nb2_b3)
k6 = AND(nb1,nb2_nb3)
b45 = OR(b4,b5)
b_nan = AND(b123,b45)
b_inf = AND(b123,NOT(b45))
n0 = k0
n1 = OR(k1,AND(k0,b4))
n2 = OR(k2,OR(AND(k1,b4),AND(k0,b5)))
n3 = OR(k3,OR(AND(k2,b4),AND(k1,b5)))
n4 = OR(k4,OR(AND(k3,b4),AND(k2,b5)))
n5 = OR(k5,OR(AND(k4,b4),AND(k3,b5)))
k5_k6 = OR(k5,k6)
n6 = OR(AND(k4,b5),AND(k5_k6,b4))
n7 = AND(k5_k6,b5)
first = n0,n1,n2,n3,n4,n5,n6,n7
i7 = n7
i6 = XOR(n6,n7)
carry_6 = OR(n6,n7)
i5 = XOR(n5,carry_6)
carry_5 = OR(n5,carry_6)
i4 = XOR(n4,carry_5)
carry_4 = OR(n4,carry_5)
i3 = XOR(n3,carry_4)
carry_3 = OR(n3,carry_4)
i2 = XOR(n2,carry_3)
carry_2 = OR(n2,carry_3)
i1 = XOR(n1,carry_2)
carry_1 = OR(n1,carry_2)
i0 = XOR(n0,carry_1)
i_sign = OR(n0,carry_1)
n7 = OR(AND(nneg,n7),AND(neg,i7))
n6 = OR(AND(nneg,n6),AND(neg,i6))
n5 = OR(AND(nneg,n5),AND(neg,i5))
n4 = OR(AND(nneg,n4),AND(neg,i4))
n3 = OR(AND(nneg,n3),AND(neg,i3))
n2 = OR(AND(nneg,n2),AND(neg,i2))
n1 = OR(AND(nneg,n1),AND(neg,i1))
n0 = OR(AND(nneg,n0),AND(neg,i0))
n_sign = AND(neg,i_sign)
r7 = XOR(m7,n7)
carry_7 = AND(m7,n7)
hr6 = XOR(m6,n6)
hcarry_6 = AND(m6,n6)
r6 = XOR(hr6,carry_7)
carry_6 = OR(hcarry_6,AND(hr6,carry_7))
hr5 = XOR(m5,n5)
hcarry_5 = AND(m5,n5)
r5 = XOR(hr5,carry_6)
carry_5 = OR(hcarry_5,AND(hr5,carry_6))
hr4 = XOR(m4,n4)
hcarry_4 = AND(m4,n4)
r4 = XOR(hr4,carry_5)
carry_4 = OR(hcarry_4,AND(hr4,carry_5))
hr3 = XOR(m3,n3)
hcarry_3 = AND(m3,n3)
r3 = XOR(hr3,carry_4)
carry_3 = OR(hcarry_3,AND(hr3,carry_4))
hr2 = XOR(m2,n2)
hcarry_2 = AND(m2,n2)
r2 = XOR(hr2,carry_3)
carry_2 = OR(hcarry_2,AND(hr2,carry_3))
hr1 = XOR(m1,n1)
hcarry_1 = AND(m1,n1)
r1 = XOR(hr1,carry_2)
carry_1 = OR(hcarry_1,AND(hr1,carry_2))
hr0 = XOR(m0,n0)
hcarry_0 = AND(m0,n0)
r0 = XOR(hr0,carry_1)
carry_0 = OR(hcarry_0,AND(hr0,carry_1))
r_sign = XOR(n_sign,carry_0)
s7 = r7
s6 = XOR(r6,r7)
carry_6 = OR(r6,r7)
s5 = XOR(r5,carry_6)
carry_5 = OR(r5,carry_6)
s4 = XOR(r4,carry_5)
carry_4 = OR(r4,carry_5)
s3 = XOR(r3,carry_4)
carry_3 = OR(r3,carry_4)
s2 = XOR(r2,carry_3)
carry_2 = OR(r2,carry_3)
s1 = XOR(r1,carry_2)
carry_1 = OR(r1,carry_2)
s0 = XOR(r0,carry_1)
n_r_sign = NOT(r_sign)
r0 = OR(AND(n_r_sign,r0),AND(r_sign,s0))
r1 = OR(AND(n_r_sign,r1),AND(r_sign,s1))
r2 = OR(AND(n_r_sign,r2),AND(r_sign,s2))
r3 = OR(AND(n_r_sign,r3),AND(r_sign,s3))
r4 = OR(AND(n_r_sign,r4),AND(r_sign,s4))
r5 = OR(AND(n_r_sign,r5),AND(r_sign,s5))
r6 = OR(AND(n_r_sign,r6),AND(r_sign,s6))
r7 = OR(AND(n_r_sign,r7),AND(r_sign,s7))
h0 = r0
rest = h0
h1 = AND(r1,NOT(rest))
rest = OR(rest,h1)
h2 = AND(r2,NOT(rest))
rest = OR(rest,h2)
h3 = AND(r3,NOT(rest))
rest = OR(rest,h3)
h4 = AND(r4,NOT(rest))
rest = OR(rest,h4)
h5 = AND(r5,NOT(rest))
rest = OR(rest,h5)
h6 = AND(r6,NOT(rest))
rest = OR(rest,h6)
h7 = AND(r7,NOT(rest))
e0 = OR(h0,OR(h1,h2))
e1 = OR(h0,OR(h3,h4))
e2 = OR(h1,OR(h3,h5))
ne0 = NOT(e0)
ne1 = NOT(e1)
ne2 = NOT(e2)
e0e1 = AND(e0,e1)
e0ne1 = AND(e0,ne1)
ne0e1 = AND(ne0,e1)
ne0ne1 = AND(ne0,ne1)
x0 = AND(e0e1, ne2)
x1 = AND(e0ne1, e2 )
x2 = AND(e0ne1, ne2)
x3 = AND(ne0e1, e2 )
x4 = AND(ne0e1, ne2)
x5 = AND(ne0ne1,e2 )
x6 = AND(ne0ne1,ne2)
u0 = AND(x0,r1)
u1 = AND(x1,r2)
u2 = AND(x2,r3)
u3 = AND(x3,r4)
u4 = AND(x4,r5)
u5 = AND(x5,r6)
u6 = AND(x6,r6)
v0 = AND(x0,r2)
v1 = AND(x1,r3)
v2 = AND(x2,r4)
v3 = AND(x3,r5)
v4 = AND(x4,r6)
v5 = AND(x5,r7)
v6 = AND(x6,r7)
f0 = OR(u0,OR(u1,OR(u2,OR(u3,OR(u4,OR(u5,u6))))))
f1 = OR(v0,OR(v1,OR(v2,OR(v3,OR(v4,OR(v5,v6))))))
sign = XOR(a0,r_sign)
either_nan = OR(a_nan,b_nan)
either_inf = OR(a_inf,b_inf)
ans_nan = OR(AND(AND(a_inf,b_inf),XOR(a0,b0)),AND(NOT(either_inf),either_nan))
nans_nan = NOT(ans_nan)
ans_inf = AND(nans_nan,OR(either_nan,either_inf))
ans_none = AND(nans_nan,NOT(ans_inf))
nans_none = NOT(ans_none)
result0 = OR(OR(AND(a_inf,a0),AND(b_inf,b0)),AND(ans_none,sign))
result1 = OR( nans_none, AND(ans_none,e0) )
result2 = OR( nans_none, AND(ans_none,e1) )
result3 = OR( nans_none, AND(ans_none,e2) )
result4 = OR( ans_nan, AND(ans_none,f0) )
result5 = OR( ans_nan, AND(ans_none,f1) )
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1\$\begingroup\$ When it's finished, will it round "down" towards zero or towards negative infinity? Just curious. \$\endgroup\$Joe Z.– Joe Z.2014年04月24日 04:24:40 +00:00Commented Apr 24, 2014 at 4:24
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1\$\begingroup\$ @JoeZ. I will definitely try to make it round towards zero, and I think that doing so shouldn't be a problem, though I can't be sure without writing it out. It is obviously trivial to make this add two negative numbers (having them round towards zero) so I think it will probably be easier to stick with that. \$\endgroup\$KSab– KSab2014年04月24日 04:33:58 +00:00Commented Apr 24, 2014 at 4:33
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2\$\begingroup\$ Well done for coming up with a complete solution. There are some easy optimisations.
OR(AND(w,x),AND(y,z))isNAND(NAND(w,x),NAND(y,z))saving 4, and you've used the first construction a few times; and your NaN treatment is slightly wrong becauseInf + NaNshould beNaN. \$\endgroup\$Peter Taylor– Peter Taylor2014年04月26日 19:32:28 +00:00Commented Apr 26, 2014 at 19:32