fiduccia mattheyses

FIDUCCIA MATTHEYSES

FIDUCCIA MATTHEYSES

• Starts with possibly some random solution and changes the solution by sequence of moves. At the beginning of pass all vertices are free to move (unlocked) and each possible move is labeled with immediate changes in total cost called gain. Iteratively a move with highest gain is selected and locked.

• Selection of best gain and update are repeated until all nodes are locked.

•This method had three operations, first computation of initial gain value, second retrieval of best gain, and third update of all affected gain values.

FEATURES

• Aims at reducing the net-cut cost.

• Works well with weighted graphs.

• Only a single vertex is moved across the cut each time.

• Can handle unbalanced partitions .

• A special data structure is used to select vertices to be moved across.

STEPS…

• Compute gains of all cells, Gi = FS(i) – TE(i).

• Select base cell.

• Lock base cell.

• Check for more free cells.

• Choose for best sequence of moves for k-base cells.

• Make k-moves permanent and free all cells.

SOLVED EXAMPLE Random initial partition

Gain(c)=FS(c)-TS(C)=1

c is contained in net n1, n2 and n3. n3 contains cell c as its only cell from left partition, so FS(c)=1. Non of the nets are entirely located in left partition, so TE(c)=0.

Gain(c)=FS(c)-TS(C)=1

F(c) = 1 T(c) = 0; G(c) = 1

F(d) = 1 T(d) = 0; G(d) = 1

F(g) = 2 T(c) = 0; G(g) = 2

F(a) = 0 T(a) = 0; G(a) = 0

F(b) = 1 T(b) = 0; G(b) = 1

F(e) = 2 T(e) = 0; G(e) = 2

F(f) = 1 T(f) = 0; G(f) = 1

F(h) = 0 T(h) = 0; G(h) = 0

Move 1: e

F(a) = 0 T(a) = 1; G(a) = -1

F(c) = 0 T(c) = 1; G(c) = -1

F(g) = 1 T(g) = 1; G(g) = 0

F(d) = 1 T(d) = 0; G(d) = 1

F(b) = 1 T(b) = 0; G(b) = 1

F(f) = 2 T(f) = 0; G(f) = 2

F(h) = 0 T(h) = 0; G(h) = 0

MOVE 2: f has the max gain, but moving f will violate the area constraint. So we move d.

F(a) = 0 T(a) = 1; G(a) = -1 F(c) = 1 T(c) = 1; G(c) = 0 F(g) = 1 T(g) = 1; G(g) = 0 F(b) = 0 T(b) = 0; G(b) = 0 F(f) = 1 T(f) = 1; G(f) = 0 F(h) = 0 T(h) = 0; G(h) = 0

MOVE 3: consider b.

F(a) = 0 T(a) = 1; G(a) = -1

F(c) = 0 T(c) = 1; G(c) = -1

F(g) = 1 T(g) = 1; G(h) = 0

F(f) = 1 T(f) = 1; G(f) = 0

F(h) = 0 T(h) = 0; G(h) = 0

MOVE 4: choose g.

F(a) = 0 T(a) = 1; G(a) = -1

F(c) = 0 T(c) = 1; G(c) = -1

F(f) = 1 T(f) = 2; G(f) = -1

F(h) = 0 T(h) = 1; G(h) = -1

MOVE 5: We choose a.

F(c) = 0 T(c) = 0; G(c) = 0

F(f) = 1 T(f) = 2; G(f) = -1

F(h) = 0 T(h) = 1; G(h) = -1

MOVE 6: We consider f.

F(c) = 0 T(c) = 1; G(c) = -1

F(h) = 0 T(h) = 0; G(h) = 0

MOVE 7:

F(c) = 0 T(c) = 1; G(c) = -1

MOVE 8 (LAST MOVE):