Computing With Limited Memory
Small RAM
e.g., chimpanzee
Large (but not infinite) ROM
e.g., stack of instruction cards
Computing With Limited Memory
m states
(log2 m bits memory)
n boolean inputs, n >> m
x1 ,, xn
Each instruction:
Examine a specific input bit xi .
Given xi and current state, lookup next state.
• Processes instructions sequentially.
• Result is final state.
S1 ,, Sm
Computing With Limited Memory
x2=0, current = S2
 next is S7
x2 current
next
0
0
S1
S2
S3
S7



1
1
S1
S2

S4
S1


Computing With Limited Memory
?
Compute
MAJ ( x1 ,, xn )
Fixed-Width BDD
S
S1
S2
S3
S4
x1
x1
x1
x1
x2
x2
x2
x2
x3
x3
x3
x3
x1
x1
x1
x1
0
1
Fixed-Width BDD
S
x1
x2
x2
x3
x3
x3
x3
x1
x1
x1
x1
0
1
Fixed-Width BDD for Any Fn.
S
x1
Suppose
f (0,0,,0)  1
x2

xn
1
c1
Fixed-Width BDD for Any Fn.
ci
x1
v1
i-th stage
f (v1 , v2 , , vn )  d
x2
v2
Chimpanzee computable
but depth n  2n

vn
d
xn
ci 1
Fixed-Width BDD
S
1
2
3
4
x1
x1
x1
x1
x2
x2
x2
x2
x3
x3
x3
x3
x1
x1
x1
x1
0
1
Fixed-Width BDD
1
2
3
4
xi  0
1 2 3 4
View each row as a
pair of permutations.
xi
xi 1
xi
xi 1
xi
xi 1
xi
xi 1
1 2 3 4
xi  1
1 2 3 4
4 3 2 1
Fixed-Width BDD
1
2
3
4
xi  0
1 2 3 4
Restrict ourselves to
cyclic permutations.
xi
xi 1
xi
xi 1
xi
xi 1
xi
xi 1
1 2 3 4
xi  1
1 2 3 4
4 3 2 1
Fixed-Width BDD
1
2
3
4
xi  0
1 2 3 4
Restrict ourselves to
cyclic permutations.
xi
xi 1
xi
xi 1
xi
xi 1
xi
xi 1
1 2 3 4
xi  1
1 2 3 4
4 3 1 2
Fixed-Width BDD
1
2
3
4
xi  0
1 2 3 4
Restrict ourselves to
cyclic permutations.
xi
xi 1
xi
xi 1
xi
xi 1
xi
xi 1
1 2 3 4
xi  1
1 2 3 4
4 3 1 2
1 4 2 3
Fixed-Width BDD
1
2
3
4
xi  0
1 2 3 4
Restrict ourselves to
cyclic permutations.
xi
xi 1
xi
xi 1
xi
xi 1
xi
xi 1
1 2 3 4
xi  1
1 2 3 4
4 3 1 2
(1 4 2 3)
Fixed-Width BDD
S
1
2
3
4
x1
x1
x1
x1
x2
x2
x2
x2
x3
x3
x3
x3
x4
x4
x4
x4
1 then (1 2 3 4)
x1  
0 then 
1 then (1 3 2 4)
x2  
0 then 
1 then (1 4 2 3)
x3  
0 then 
 is identity permutation
0
1
Fixed-Width BDD
S
1
2
3
4
x1
x1
x1
x1
 x1, , (1,2,3,4) 
x2
x2
x2
x2
 x2 ,  , (1,3,2,4) 
x3
x3
x3
x3
 x3 ,  , (1,4,2,3) 
x4
x4
x4
x4
 x4 ,  , (1,4,2,3) 
0
1
Fixed-Width BDD
S
1
2
3
4
5
x1
x1
x1
x1
x1
x2
x2
x2
x2
x2





1 if (1 3 2 5 4)
F (X )  
0 if 
xd
xd
xd
xd
xd
Say BDD “yields” (1 3 2 5 4).
0
1
Cyclic Permutations
Composition:
a b = (1 2 3 4 5)(1 4 2 3 5) = (1 3 2 5 4)
Identity:
a =  a = a
Inverse:
a = (1 2 3 4), a-1=(4 3 2 1), a a-1 = 
Cyclic Permutations
Transformation:
For cyclic permutations a and b, there exists a permutation
 such that:
a   b1
(Why?)
Cyclic Permutations
Commutation: aba-1b-1
Consider a = (1 2 3 4 5), b = (1 3 5 4 2)
Let S be the set of permutations obtained by taking
commutations of pairs, starting with a and b.
Cyclic Permutations
Example:
a = (1 2 3 4 5), b = (1 3 5 4 2)
aba-1b-1
= (1 2 3 4 5)(1 3 5 4 2)(5 4 3 2 1)(2 4 5 3 1)
= (1 3 2 5 4)
=c
Cyclic Permutations
Example:
a = (1 2 3 4 5), c = (1 3 2 5 4)
aca-1c-1
= (1 2 3 4 5)(1 3 2 5 4)(5 4 3 2 1)(4 5 2 3 1)
= (1 5 3 2 4)
=d
Cyclic Permutations
Commutation: aba-1b-1
Consider a = (1 2 3 4 5), b = (1 3 5 4 2)
Let S be the set of permutations obtained by taking
commutations of all pairs, starting with a and b.
S is a closed set under commutation (always inlcudes  ).
Such a set does not exist for cyclic permutations of length < 5.
(How many elements does S contain?)
Computing with Width-5 BDDs
Lemma 1 (negation):
If F is a width-5 BDD that yields  , then there exists a
1
width-5 BDD G of the same length that yields  .
(Why?)
Computing with Width-5 BDDs
Lemma 1 (negation):
G is identical to F except in last instruction.
If the last instruction of F is  xi ,  ,  ,
then the last instruction of G is  xi ,  1 , 1  .
Computing with Width-5 BDDs
Lemma 2 (transformation):
For  and  in S, a width-5 BDD F that yields  , can be
transformed into a width-5 BDD G of the same length that
yields  .
(How?)
Computing with Width-5 BDDs
Lemma 2 (transformation):
Find  such that    1 .
Change each instruction in F from  xi ,  , 
to
 xi , 1 , 1  .
Computing with Width-5 BDDs
Circuits: depth d, fan-in 2, AON gates
Computing with Width-5 BDDs
Theorem:
A function that is computed by a depth-d circuit of fan-in 2
AON gates can be computed by a width-5 BDD of length at
most 4d.
(How?)
Proof: Width-5 Result
By induction on d.
Base Case: d=0
Circuit is an input gate, we have a one instruction BDD.
Induction Hypothesis:
Any function of a circuit of depth d  1 can be computed with a
BDD of length 4d 1.
Proof: Width-5 Result
Inductive Step:
WLOG (by Lemma 1) assume that the output of the circuit is an
AND gate:
f  AND( f1 , f 2 )
f1 , f 2 have circuits of depth d  1 and thus width-5 BDDs
B1 , B2 of length 4d 1 .
Say B1 , B2 yield  1 and  2 . Construct BDDs B1, B2 that yield
 11 and  21 (by Lemma 2).
Proof: Width-5 Result
Inductive Step (cont.):
The concatenation
B  B1 , B2, B1, B2
computes f and has depth 4 d .
1 1  
B2  2   2 
B1  11  11  
B1
B2  21   21 
B




   1  2 11 21
Example: Width-5 Result
AND( x1 , x2 )
S
1
2
3
4
5
x1
x1
x1
x1
x1
 x1 ,  , (12345) 
x2
x2
x2
x2
x2
x1
x1
x1
x1
x1
x2
x2
x2
x2
x2
0
1
 x2 , , (13542) 
 x1 ,  , (54321) 
 x2 , , (24531) 
Majority
MAJ ( x1 ,, xn )
Can be computed by a circuit with depth O (log n) .
 Can be computed by a width-5 BDD of length polynomial in n.
MAJORITY of
10,000 inputs?
No problem!!!!!