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Partitioning

Partitioning. First Project. Synthesis by Design Compiler Physical design by SoC Encounter Use tutorials in \fileservercommonszamaniEDA (MS)Tools. Fiduccia-Mattheyses Algorithm. “ A linear time heuristic for improving network partitions ” , Fiduccia and Mattheyses, DAC 82. Basic Ideas.

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Partitioning

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  1. Partitioning مرتضي صاحب الزماني

  2. First Project • Synthesis by Design Compiler • Physical design by SoC Encounter • Use tutorials in \\fileserver\common\szamani\EDA (MS)\Tools\ مرتضي صاحب الزماني

  3. Fiduccia-Mattheyses Algorithm • “A linear time heuristic for improving network partitions”, Fiduccia and Mattheyses, DAC 82. مرتضي صاحب الزماني

  4. Basic Ideas • Differences from KL: • Move only one cell each time. • Cells can have different sizes. • Nets can be multi-terminal. • Maintain a balanced partition after every move. • O(#Pins) • #Pins = ∑eedge degrees مرتضي صاحب الزماني

  5. FM Algorithm • Start with a balanced partition P = {X, Y}. • Repeat • For i = 1 to n: • Choose a free cell b  XY s.t. moving b to the other side gives the highest gain, gain(b), and moving b preserves balance in P. • Move and lock b. • Let gi = gain(b). • Find k s.t. G = g1 + g2 + ….. + gk is maximized and shuffle the cells up to this kth step. • Until G = 0. مرتضي صاحب الزماني

  6. An Example g2 d e f a c a b c d e f g1 locked b a c g4 a c d f g3 f d b e b e مرتضي صاحب الزماني

  7. An Example c a a g5 a d d c g6 f d f c f e b e e b b If G = g1 + g2 + g3 + g4 is the largest partial sum, the final partition after this pass is: c d e a f b مرتضي صاحب الزماني

  8. Balanced Partition • A partition P = (X, Y) is balanced iff: • r ( 1): Ratio Factor • S(X): total size of cells in X. • Balancing condition: cell b is moved if, after moving, • Stot = S(XY) • Smax =maximum cell size مرتضي صاحب الزماني

  9. Digital Design Contest مرتضي صاحب الزماني

  10. i i نت بد: نتي که متصل به سلول i است و cut را قطع نکرده است ( cutsize را يک واحد افزايش مي دهد). نت خوب: نتي که متصل به سلول i است و به هيچ سلول ديگري در پارتيشن مبدأ وصل نيست (cutsize را يک واحد کاهش مي دهد). Cell Gain gain(i) = تعداد نتهاي خوب - تعداد نتهاي بد مرتضي صاحب الزماني

  11. i نت خنثي: نتي که متصل به سلول i است و به سلولهايي در هر دو پارتيشن وصل است. Cell Gain مرتضي صاحب الزماني

  12. Cell Gain • Note: -p(i)  gain(i)  p(i) where p(i) is the no. of pins in cell i. gain(i) = -p(i) i i gain(i) = p(i) مرتضي صاحب الزماني

  13. Gain Bucket +pmax Max Gain Cell # Cell # -pmax 2 n 1 Two “bucket lists”, one for X and one for Y مرتضي صاحب الزماني

  14. +pmax +pmax Max Gain Max Gain Cell # Cell # Cell # Cell # -pmax -pmax 2 2 n n 1 1 Gain Bucket مرتضي صاحب الزماني

  15. Gain Update • Let cell b move from X to Y. • Let N(b) be the set of nets connecting b. • This move will only affect those cells connected by the nets in N(b). مرتضي صاحب الزماني

  16. 2.4.3 Fiduccia-Mattheyses (FM) Algorithm – Terminology Gain g(c) for cell c g(c) = FS(c) – TE(c) FS(c): (moving force): No. of good nets connected to c TE(c): (retention force): No. bad nets connected to c 3 2 b e a 4 c d 1 5 Cell 2: FS(2) = 0 TE(2) = 1 g(2) = -1

  17. 2.4.3 Fiduccia-Mattheyses (FM) Algorithm – Terminology Gain g(c) for cell c g(c) = FS(c) – TE(c) , 3 2 b e a 4 c d 1 5 3 2 Cell 1: FS(1) = 2 TE(1) = 1 g(1) = 1 Cell 2: FS(2) = 0 TE(2) = 1 g(2) = -1 Cell 3: FS(3) = 1 TE(3) = 1 g(3) = 0 Cell 4: FS(4) = 1 TE(4) = 1 g(4) = 0 Cell 5: FS(5) = 1 TE(5) = 0 g(5) = 1 b e a 4 c d 1 5

  18. 2.4.3 Fiduccia-Mattheyses (FM) Algorithm – Terminology Maximum positive gain Gmof a pass

  19. 2.4.3 Fiduccia-Mattheyses (FM) Algorithm – Example 3 2 A B Given:Ratio factor r = 0,375 area(Cell_1) = 2 area(Cell_2) = 4 area(Cell_3) = 1 area(Cell_4) = 4 area(Cell_5) = 5. b e a 4 c d 1 5 Step 0: Compute the balance criterion [ r ∙ area(V) – Smax(V) ] ≤ area(A) ≤ [ r ∙ area(V) + areamax(V) ] 0,375 * 16 – 5 = 1 area(A)  11 = 0,375 * 16 +5.

  20. 2.4.3 Fiduccia-Mattheyses (FM) Algorithm – Example 3 2 A B b e a 4 c d 1 5 Step 1:Compute the gains of each cell Cell 1: FS(Cell_1) = 2 TE(Cell_1) = 1 g(Cell_1) = 1 Cell 2: FS(Cell_2) = 0 TE(Cell_2) = 1 g(Cell_2) = -1 Cell 3: FS(Cell_3) = 1 TE(Cell_3) = 1 g(Cell_3) = 0 Cell 4: FS(Cell_4) = 1 TE(Cell_4) = 1 g(Cell_4) = 0 Cell 5: FS(Cell_5) = 1 TE(Cell_5) = 0 g(Cell_5) = 1

  21. 2.4.3 Fiduccia-Mattheyses (FM) Algorithm – Example 3 2 A B Cell1: FS(Cell_1) = 2 TE(Cell_1) = 1 g(Cell_1) = 1 Cell 2: FS(Cell_2) = 0 TE(Cell_2) = 1 g(Cell_2) = -1 Cell 3: FS(Cell_3) = 1 TE(Cell_3) = 1 g(Cell_3) = 0 Cell 4: FS(Cell_4) = 1 TE(Cell_4) = 1 g(Cell_4) = 0 Cell 5: FS(Cell_5) = 1 TE(Cell_5) = 0 g(Cell_5) = 1 b e a 4 c d 1 5 Step 2:Select the base cell Possible base cells are Cell 1 and Cell 5 Balance criterion after moving Cell 1: area(A) = area(Cell_2) = 4 Balance criterion after moving Cell 5: area(A) = area(Cell_1) + area(Cell_2) + area(Cell_5) = 11 Both moves respect the balance criterion, but Cell 1 is selected, moved, and fixed as a result of the tie-breaking criterion.

  22. 2.4.3 Fiduccia-Mattheyses (FM) Algorithm – Example 3 2 A B b e a 4 c d 1 5 Step 3: Fix base cell, update g values Cell 2: FS(Cell_2) = 2 TE(Cell_2) = 0 g(Cell_2) = 2 Cell 3: FS(Cell_3) = 0 TE(Cell_3) = 1 g(Cell_3) = -1 Cell 4: FS(Cell_4) = 0 TE(Cell_4) = 2 g(Cell_4) = -2 Cell 5: FS(Cell_5) = 0 TE(Cell_5) = 1 g(Cell_5) = -1 After Iteration i = 1:Partition A1 = 2, Partition B1 = 1,3,4,5, with fixed cell 1.

  23. 2.4.3 Fiduccia-Mattheyses (FM) Algorithm – Example Iteration i = 1 3 2 A B Cell 2: FS(Cell_2) = 2 TE(Cell_2) = 0 g(Cell_2) = 2 Cell 3: FS(Cell_3) = 0 TE(Cell_3) = 1 g(Cell_3) = -1 Cell 4: FS(Cell_4) = 0 TE(Cell_4) = 2 g(Cell_4) = -2 Cell 5: FS(Cell_5) = 0 TE(Cell_5) = 1 g(Cell_5) = -1 b e a 4 c d 1 5 Iteration i = 2 Cell 2 has maximum gain g2 = 2, area(A) = 0, balance criterion is violated. Cell 3 has next maximum gain g2 = -1, area(A) = 5, balance criterion is met. Cell 5 has next maximum gain g2= -1, area(A) = 9, balance criterion is met. Move cell 3, updated partitions: A2 = {2,3}, B2 = {1,4,5}, with fixed cells {1,3}

  24. 2.4.3 Fiduccia-Mattheyses (FM) Algorithm – Example Iteration i = 2 3 2 A Cell 2: g(Cell_2) = 1 Cell 4: g(Cell_4) = 0 Cell 5: g(Cell_5) = -1 e B b a 4 c d 1 5 Iteration i = 3 Cell 2 has maximum gain g3 = 1, area(A) = 1, balance criterion is met.Move cell 2, updated partitions: A3 = {3}, B3 = {1,2,4,5}, with fixed cells {1,2,3}

  25. 2.4.3 Fiduccia-Mattheyses (FM) Algorithm – Example Iteration i = 3 3 2 B A e Cell 4: g(Cell_4) = 0 Cell 5: g(Cell_5) = -1 b a 4 c d 1 5 Iteration i = 4 Cell 4 has maximum gain g4 = 0, area(A) = 5, balance criterion is met.Move cell 4, updated partitions: A4 = {3,4}, B3 = {1,2,5}, with fixed cells {1,2,3,4}

  26. 2.4.3 Fiduccia-Mattheyses (FM) Algorithm – Example Iteration i = 4 3 2 B A e Cell 5: g(Cell_5) = -1 b a 4 c d 1 5 Iteration i = 5 Cell 5 has maximum gain g5 = -1, area(A) = 10, balance criterion is met.Move cell 5, updated partitions: A4 = {3,4,5}, B3 = {1,2}, all cells {1,2,3,4,5} fixed.

  27. 2.4.3 Fiduccia-Mattheyses (FM) Algorithm – Example 3 2 B A e b a 4 c d 1 5 Step 5: Find best move sequence c1 … cm G1 = g1 = 1G2 = g1 + g2 = 0G3 = g1 + g2 + g3 = 1G4 = g1 + g2 + g3 + g4 = 1G5 = g1 + g2 + g3 + g4 + g5 = 0. Maximum positive cumulative gain found in iterations 1, 3 and 4. The move prefix m = 4 is selected due to the better balance ratio (area(A) = 5); the four cells 1, 2, 3and 4 are then moved. Result of Pass 1: Current partitions: A = {3,4}, B = {1,2,5}, cut cost reduced from 3 to 2.

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