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2004 ‫ אביב‬- ‫מבנה המחשב‬
#3 ‫תרגול‬
Topics:
1. Finding a cycle in a graph
2. Propagation delay - example
3. Trees - properties
Checking if a Circuit is combinational
• Mapping the circuit into a directed graph DG(V,E).


IN
Each gate is a vertex.
Each net defines a star of edges emanating out of an
output terminal.
Single net
AND
OR
IN
NOT
XOR
OUT
Checking a Circuit (cont.)
• If an input terminal has two incoming edges,
then it is fed by two nets, which is illegal. In that
case we stop and return FALSE.
IN
AND
Two
inputs to
one
terminal
!
OR
IN
NOT
XOR
OUT
Checking a Circuit (cont.)
• Is this condition enough for a circuit to be
valid?
• No, we still need to check that the graph of
the circuit contains no cycles.
Testing for Cycles
• Two classical algorithms:
• Topological sort based:
– While “peeling off” sources, if there are no
sources but the set of non-sorted vertices is
not empty, then return FALSE (proof: next
slide).
• DFS based:
– If a backwards edge is encountered, return
FALSE (proof: auxiliary slides at the end).
Existence of cycles
• Claim: If a directed (sub-)graph G has no sources, then it
contains a cycle.
• Proof (algorithmic):
– Note: No sources  For all u, in(u) > 0  |E| > |V|-1.
– Delete repeatedly all sinks (this does not create
sources)
– Pick any vertex u and mark it.
– Pick an edge (u,v)
• If v is marked: a cycle was found!!!
• Else: pick an edge emanating from v…repeat
– Observation: The process must encounter a marked
vertex (can’t have an infinite sequence of unmarked
vertices). Hence, we close a cycle!
Propagation Delay

Max
{
t
}
pd
T pd
all paths
There is always at least one “critical path”.
What is the propagation delay of a circuit that is not
acyclic?
Finding the Propagation Delay
t (v ) 
inputs
stable
Max
{ t (v' )  t pd (v' )}
v '  predecessor ( v ) inputs
stable
acyclic graph  topological order.
Each time updating the “inputs stable” time of the next
vertex. This update is often called relaxation.
The total propagation delay is the maximal “inputs
stable” time (assuming that we used output nodes).
This is also the delay of a critical path in DG(V,E).
Finding the Maximum Delay
0
0
OR
AND
NOT
XOR
OUT
A circuit with 2n/2 paths
2 options
2 options
Note: n/2 stages with 2 options each, resulting in 2n/2
paths.
 There is no need to check all the paths, only the
longest. This takes a linear time.
A circuit with 2n paths
• …cannot be built!
• Why? A combinational circuit is a DAG, therefore
we cannot reorder the gates to create different
paths. Our only option is to include or exclude
gates to create different paths.
• But, having n gates, we only have 2n such paths.
Each gate can be included or excluded,
therefore 2n.
• We cannot build this circuit since we will require
an unbounded in-degree of the gates.
XOR is Associative
•Define mod(a,q) to be the remainder of a/q.
•Claim:
– mod(a+b,q) = mod( mod(a,q) + mod(b,q),q)
– mod(a · b,q) = mod( mod(a,q) · mod(b,q),q)
•Proof:
– Let a = kq+j and b = sq+t.  a+b = q(k+s) + j+t
mod(a+b,q) = mod(j+t,q) =
mod(mod(a,q)+mod(b,q),q)
–The same proof for multiplication.
XOR is Associative (cont.)
•Note that xor(x,y) = mod(x+y,2)
xor(xor(x,y),z) = mod(mod(x+y,2)+mod(z,2),2)

= mod((x+y)+z,2)

= mod(x+(y+z),2)

= mod(mod(x,2)+mod(y+z,2),2)

= xor(x,xor(y,z))
•Hence we may denote it by xor(x,y,z).
Tree: Basic properties
•An undirected graph is a tree if it is:
•Connected, i.e., there is a path between every
two vertices.
•Contains no cycles.
•One of these properties can be replaced with the
following:
•|E| = |V|-1
Directed Tree: Basic properties
•An directed graph is a directed tree if it is:
•In-degree = 1 for all v, besides the root.
•In degree of the root is 0.
•There is a directed path from the root to every
node.
•Tree  directed tree:
•Pick one of its vertices to be a root
•Direct all the edges out from the root
•Repeat it with the vertices as the roots.
Directed Tree (cont.)
•A leaf is a node with out-degree zero.
•We define on the vertices of a directed tree:
•The parent of a node v. Note there is only one
such vertices.
•A child of a node v (there may be many).
•Ancestor and Descendent of v.
The root (source): Ancestor for all x, but has no
parent.
Internal node: has a parent, may have children.
A leaf (sink): has no children.
A claim on Trees
•In a directed tree, any non-leaf is called an
internal node.
•For a tree: a leaf is a node whose degree is 1.
•Claim: If the degree of any vertex in the tree  3,
then the number of internal nodes  the number of
the leaves –2.
•Note: the underlying graph of a directed binary
tree is such a tree, but since the root does not
count as a leaf, we have in directed trees:
#internal nodes  #internal leaves -1
A claim on Trees (cont.)
•Proof (Induction on the number of nodes):
•Basis: trivial for n=1,2 , check for all trees with
3 nodes (there is only one such tree).
•Assumption: correct for all trees of size  n.
•Step: prove for any tree of size n+1. Given
such a tree, it must contain an internal node v.
•There exists two trees T1,T2 such that
connecting them yields T.
•Define |Ti|=ni = mi+ki, (m is #leaf, k is # internal.
A claim on Trees (cont.)
n = n1 + n2.
ki  mi - 2 (From the induction hypothesis).
Note that the connection may reduce the
number of leaves by at most two, the claim
follows.
The next slides are for the
curious students (home reading)
DFS Example
Stack
1
2
3
4
5
DFS
Stack
1
2
5
3
1
4
= Currently in the stack
DFS
Stack
1
2
5
3
2
1
4
DFS
Stack
1
2
5
3
3
2
1
4
DFS
Stack
1
2
5
4
3
3
2
Backwards Edge
4
This graph has a cycle!
1
Elements of the Proof
• The algorithm terminates regardless of the
structure of the input graph.
• If there is a cycle in the graph, the
algorithm will find it (return FALSE).
• If there is no cycle in the graph, the
algorithm will return TRUE.
The Algorithm Terminates
• The algorithm passes through every vertex
only once, therefore it will always
terminate after visiting all of the vertices
regardless of the edges.
The Proof
• Assume there is a cycle in the graph. At
some point a first vertex that belongs to
the cycle will be reached. All other vertices
of the cycle have not been reached yet.
• Before that first vertex is popped out of the
stack, the DFS procedure guarantees that
an edge closing the cycle and entering
that vertex will be tested.
• It is a backwards edge.
The Proof
• Immediate, but nevertheless:
• Assume there are no cycles in the graph.
• Backwards edges cannot exist since they
require a path from a successor to a
predecessor, which means there is a
cycle.