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Probabilistic Testability Measures

Probabilistic Testability Measures. Student: Shih-Chieh Wu Advisor: Chun-Yao Wang 2005.12.26 Department of Computer Science National Tsing Hua University, Taiwan. Outline. Introduction Controllability Calculation with Aliasing-Free Assignments Conclusion & Future Work. Introduction.

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Probabilistic Testability Measures

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  1. Probabilistic Testability Measures Student: Shih-Chieh Wu Advisor: Chun-Yao Wang 2005.12.26 Department of Computer Science National Tsing Hua University, Taiwan

  2. Outline • Introduction • Controllability Calculation with Aliasing-Free Assignments • Conclusion & Future Work

  3. Introduction • As circuits become larger, the testing complexity ↑ • Every testing technique needs a basis to analyze the circuits • Testability measures • Testability of a circuit is defined by controllability and observability of it

  4. Definitions • Controllability • The difficulty of setting a particular logic signal to 0 or 1 ( 0’s controllability or 1’ controllability ) • Observability • The difficulty of observing the state of a logic signal

  5. COP Controllability Calculation(1/2) • COP controllability: • a = pr(A=1), b = pr(B=1) are 1'controllability 1 - a a a a b b a 1 - (1 - a)  (1 - b) = a + b - a  b b

  6. COP Controllability Calculation(2/2) • Example: • Problem: • Signal correlation → inaccurate result 0.5 0.25 A 0.4375 (COP) 0.5000 (exact) G2 B G1 0.5 G4 0.5 0.25 0.5 G3 C

  7. COP Observability Calculation • COP observatility: obv(G1) = pr(A=1) × obv(G2) = pr(A=1) × pr(G3=0) × obv(G4) = 0.5 × (1-0.25) × 1 = 0.375 0.5 0.25 A G2 B G1 0.5 0.4375 G4 0.5 0.25 0.5 G3 C

  8. Calculate Controllability with Accuracy • Exhausted simulation • Assume pr(A=1) = pr(B=1) = 0.5 • pr ( F=1) = 0.25 + 0.25 = 0.5 F B A controllability of minterm • 0 0 • 0 1 • 0 • 1 1 0 1 0 1 (1-0.5) × 0.5 = 0.25 0.5 × 0.5 = 0.25 A F B

  9. Outline • Introduction • Controllability Calculation with Aliasing-Free Assignments • Conclusion & Future Work

  10. Aliasing-Free Assignments (I) • For any 2-input circuits • There are = 16 functions • The denominator of output probability is 5 × 3 = 15 • The numerators of output probability distribute from 0~15 (at most = 4 bits) Assume x2 = , x1 = X2 X1 prob. of minterms (numerator) • 0 0 • 0 1 • 0 • 1 1 1 = (bit 0=2’b00) 2 = (bit 1=2’b01) 4 = (bit 2=2’b10) 8 = (bit 3=2’b11)

  11. Accurate Controllability Calculation • Associate aliasing-free probability with input controllability • Ex: pr(X2=1)=0.5, pr(X1=1)=0.5, x2= , x1= bit 0=2’b0 is 1, NOT gate: pr(X1=0) = 1-pr(X1=1) = 0.5 bit 3=2’b11 is 1, AND gate: pr(X2=1) × pr(X1=1) = 0.25 bit 1=2’b01, bit 2=2’b10, bit 3=2’b11is 1, OR gate: pr(X2=0) × pr(X1=1) + pr(X2=1) × pr(X1=0) + pr(X2=1) × pr(X1=1) = 0.75

  12. Controllability of Signal Correlation (1/2) • Assume c= , b= , a= , and controllability of C , B and A is 0.5 • bit 1 (2’b001): pr(C=0) × pr(B=0) × pr(A=1) = 0.25 • bit 5 (2’b101): pr(C=1) × pr(B=0) × pr(A=1) = 0.25 • bit 6 (2’b110): pr(C=1) × pr(B=1) × pr(A=0) = 0.25 • bit 7 (2’b111): pr(C=1) × pr(B=1) × pr(A=1) = 0.25 • Controllability of gate G4 is 0.5 + A G2 B G1 G4 Order: C, B, A G3 C

  13. Controllability of Signal Correlation (2/2) • How to calculate controllability of gate G3 • The output probability of gate G3 is • bit 6 (2’b110) is 1, denominator is 110 (order: C, B, A) • Whatever the corresponding bit of A ( bit 0 of 2’b110 ) is 0 or 1, it dose not matter • pr(C=1) × pr(B=1) × pr(A=0) = 0.125 is incorrect • pr(C=1) × pr(B=1) = 0.25 is correct

  14. Aliasing-Free Assignments (II) • For any 2-input circuits • There are = 16 functions • The denominator of output probability is 5 × 3 = 15 • The numerators of output probability distribute from 0~15 (at most = 4 bits) Assume x2 = , x1 = X2 X1 prob. of minterms (numerator) • 0 0 • 0 1 • 0 • 1 1 8 = = (bit 3=2’b11) 4 = = (bit 2=2’b10) 2 = = (bit 1=2’b01) 1 = = (bit 0=2’b00)

  15. Accurate Controllability Calculation • Associate aliasing-free probability with input controllability • Ex: pr(X2=1)=0.5, pr(X1=1)=0.5, x2= , x1= bit 0=2’b00 is 1, 2’b00=2’b11 AND gate: pr(X2=1) × pr(X1=1) = 0.25

  16. Summary • For any n-input circuits • There are functions • The denominator of output probability is an× an-1 × … × a1 • The numerators of output probability distribute from 0~ (at most bits) • bit i 's binary representation correspond to the minterm order Assume x2 = , x1 = X2 X1 prob. of minterms (numerator) • 0 0 • 0 1 • 0 • 1 1 1 = (bit 0=2’b00) 2 = (bit 1=2’b01) 4 = (bit 2=2’b10) 8 = (bit 3=2’b11)

  17. Internal Tree-Structure Replacement • Internal tree-structure gates can be re-assigned after its controllability is calculated • Assume pr(A=1)=pr(B=1)=pr(C=1)=0.5 : 0.5 0.5 order A G1 0.25 : 0.5 B G2 C C(G1)=0.25 C(G2)=0.5×0.25+0.5×(1-0.25) + (1-0.5) ×0.25=0.625

  18. Outline • Introduction • Controllability Calculation with Aliasing-Free Assignments • Conclusions & Future Work

  19. Conclusions • Aliasing-free assignments can be used to calculate accurate controllability with some rules • The required input assignments is still the limitation • Internal tree-structure replacement can be used to reduce the required input assignments

  20. Future Work • Improve the controllability calculation • 24 input assignments: 0.20 s • 25 input assignments: 0.37 s • 29 input assignments: 5.87 s • 30 input assignments: 11.77 s • 31 input assignments: 32.59 s • Combine other techniques to calculate controllability with little inaccuracy under the input assignment constrain

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