in Digital Logic edited by
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5 votes
5 votes

In the given network of AND and OR gates $f$ can be written as

  1. $\text{X}_0\text{X}_1\text{X}_2 \dots \text{X}_n + \text{X}_1\text{X}_2 \dots \text{X}_n + \text{X}_2\text{X}_3 \dots \text{X}_n + \dots + \text{X}_n$
  2. $\text{X}_0\text{X}_1 + \text{X}_2\text{X}_3+ \dots \text{X}_{n-1}\text{X}_n$
  3. $\text{X}_0+\text{X}_1 + \text{X}_2+ \dots +\text{X}_n $
  4. $\text{X}_0\text{X}_1 + \text{X}_3  \dots \text{X}_{n-1}+ \text{X}_2\text{X}_3 + \text{X}_5 \dots \text{X}_{n-1} + \dots +\text{X}_{n-2} \text{X}_{n-1} +\text{X}_n$
in Digital Logic edited by
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2 Answers

11 votes
11 votes
Best answer
$(X_0X_1+X_2)X_3+X_4)X_5+\cdots+X_N$

    $=(X_0X_1X_3+X_2X_3+X_4)X_5+\cdots+X_N$

    $=X_0X_1X_3X_5+X_2X_3X_5+X_4X_5+\cdots+X_N$

    $=X_0X_1X_3X_5\cdots X_{N-1}+X_2X_3X_5\cdots X_{N-1}+X_4X_5X_7\cdots X_{N-1}+\cdots + X_N$
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4 Comments

No option matching
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Agree. No option matching !
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its option D , there's typo  --> "X0X1+X3…Xn−1" should be "X0X1+X2…Xn−1" , but general term says correct.
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0 votes
0 votes
Option A is answer I think

1 comment

how
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