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retiming
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Retiming – 3Solving System of in-equalities by
Floyd-Warshall algorithm
Arunachalam VAP SG, SENSE
Definition• Re-timing is a architectural transformation technique used to change the
locations of delay elements in a circuit without affecting the input – outputcharacteristics of the circuit.
Solving system of in-equalities• There are “M” in-equalities and “n” variables.
each inequality is represented as where “k” is an integer value.• Step 1: Draw a constraint graph.
a) Draw the node ‘i’ for each of the n variables, ri, i=1,2,3,…,nb) Draw the (n+1)th node.c) For each draw an edge with length k.
d) For each node i, i=1,2,3,…,n , draw the edge with length 0.
;krr ji
;krr ji
ij
e
in
e1
An ExampleSystem of in-equalities Constraint graph “Gc”
21
450
23
34
14
13
21
rrrrrrrrrr
Step 2: Solve the constraint graph “Gc”
• Solving the constraint graph using shortest path algorithm (Bellman-Ford algorithm / Floyd-Warshall algorithm).a) A system of in-equalities has a solution if and only if the “Gc” contains no
negative cycles.b) If a solution exists, one solution is where “ri” is the maximum length path
from (n+1) to the node “i” .
FLOYD-WARSHALL ALGORITHMDETERMINING SHORTEST PATH IN A GIVEN GRAPHALL-POINTS SHORTEST PATH ALGORITHM
Part – I Initialize (steps 1 to 3)
1;1.1 nNNtoVFor
Gc
NtoUFor 1.2
VUrelse
exsistspathifVUwVUre
,
,.3
1
1
0000
120
45
1R
Part – II Updating the r(k)(U,V)(steps 4 to 9)
NtokFor 1.4 NtoVFor 1.5
VUrVUr kk ,,.7 1
VkrkUrVUrIf kkk ,,,.8 1
NtoUFor 1.6
VkrkUrVUrthen kkk ,,,.9 1
0000
120
45
1R
Application of the algorithm• k = 1,
1,11,1 111 rr 1,11,1 12 rr 1,11,11,1 112 rrr
0
01,1 UR
false 1,12r
1,21,2 111 rr 01,21,2 12 rr 1,11,21,2 112 rrr false 00 01,22 r
1,31,3 111 rr 1,31,3 12 rr 1,11,31,3 112 rrr false 1,32r 1,41,4 111 rr 1,41,4 12 rr 1,11,41,4 112 rrr false 1,42r 1,51,5 111 rr 01,51,5 12 rr 1,11,51,5 112 rrr false 00 01,52 r
0
01,2 UR
V= 1, 2, 3, 4, 5 and U = 1, 2, 3, 4, 5
Application of the algorithm• k = 1, V=2 and U = 1, 2, 3, 4, 5
2,12,1 111 rr 2,12,1 12 rr 2,11,12,1 112 rrr false 2,12r
2,22,2 111 rr 2,22,2 12 rr 2,11,22,2 112 rrr false 0 2,22r
2,32,3 111 rr 2,32,3 12 rr 2,11,32,3 112 rrr false 2,32r 2,42,4 111 rr 2,42,4 12 rr 2,11,42,4 112 rrr false 2,42r 2,52,5 111 rr 02,52,5 12 rr 2,11,52,5 112 rrr false 00 02,52 r
0
2,2 UR
00
01R
Step 1
0000
120
45
1R
0000
142045
2R
k – select kth row and kth columnV – select Vth columnU – Uth row in Vth column
211
,1RR
kfor
Step 2
0000
142045
2R
k – select kth row and kth columnV – select Vth columnU – Uth row in Vth column
312
,2RR
kfor
0000
142045
3R
Step 3
0000
142045
3R
k – select kth row and kth columnV – select Vth columnU – Uth row in Vth column
413
,3RR
kfor
1000
112045
4R
Step 4
k – select kth row and kth columnV – select Vth columnU – Uth row in Vth column
514
,4RR
kfor
1000
112045
5R
1000
112045
4R
Step 5k – select kth row and kth column
615
,5RR
kfor
1000
112045
6R
1000
112045
5RV – select Vth column
U – Uth row in Vth column
Part – III Check for the negative cycles(steps 10 to 14)
10. For k =1 to N11. For U = 1 to N12. If13. Return false and exit14. Return true and exit
0, UUr k
Check for negative cycle and solution
k = 4
4
1
-1
∞
-1
U = 1
U = 2
U = 3
U = 4
U = 5
k = 1
∞
0
∞
∞
0
k = 2
∞
∞
∞
∞
0
k = 3
5
2
∞
∞
0
k = 5
∞
∞
∞
∞
∞
• If “true” is returned because all diagonal entries are non negative.
• The solution is : r(1) = 0; r(2) = 0; r(3) = 0;r(4) = -1
Try these problems
121 rr
313 rr214 rr115 rr
123 rr134 rr
254 rr153 rr
313 rr214 rr215 rr
123 rr134 rr
754 rr653 rr
121 rr
Problem 1 Problem 2
SPECIAL CASES OF RETIMINGNext Class
