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The logic circuit has 3 inputs x2, x1, x0, where x2 is the most significant bit and x0 is the least significant one. The output is 1 if the decimal equivalent of the input (x2,x1,x0) is 1, 4, 5, or 6. For other inputs the output is zero.
I need to Simplify the output expression
I tried making a truth table for it, I get:
x1'(x1'x0+x2x0')+x2(x1'x0+x1x0') [Here 'x1 means not]
at the end all I get is:
x1'(x2'*x0+x2*x0')+x2(x1'*x0+x1*x0')
How can I simplify it beyond XOR expression? enter image description here

**EDIT:** Thanks to everyone! I got the answer, I didn't do the Boolean Algebra properly. Another way to get the answer is to simply make a K-Map make sure to use GreyCode and not BCD. The answer is option **b**
Mayur
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  • Remember that this community isn't for us to do your homework. It is important that you try to solve the problem yourself before asking. – Julien Aug 28 '23 at 17:15
  • Please post an attempt at a solution, if copying from a text or online, please cite a source. – Voltage Spike Aug 28 '23 at 17:35
  • @VoltageSpike I currently don't have the solution I'm trying to solve the question. – Mayur Aug 28 '23 at 17:37
  • @Julien I have added my the work that I tried, I'm stuck at a stage so I'm asking – Mayur Aug 28 '23 at 17:38
  • How did you get to that point, can you post the steps in between? – Voltage Spike Aug 28 '23 at 17:39
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    Make a Karnaugh map. Mine has x2x1 down one side, and x0 across. Fill it in with the True cases -- what's left is the False cases. – Scott Seidman Aug 28 '23 at 17:48
  • @ScottSeidman I got x2(x0'+x1')+x1'*x2'*x3' as the final expression from the Karnaugh map, is that correct? I wonder why I'm getting different answers from truth table and the K-map? – Mayur Aug 28 '23 at 18:57
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    Make sure you make the 2-element side change in a Grey code, one changed bit at a time, to make it easier to identify patterns. I did it for your problem some hours ago, and it's not too tough to see which multiple l e cho8ce answer works. – Scott Seidman Aug 28 '23 at 21:59
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    That's 00, 01 , 11, 10 – Scott Seidman Aug 28 '23 at 22:00
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    You should see that 4,6 can be handled by \$ x_2\bar{x_0}\$. This leaves 1,5 for you to handle – Scott Seidman Aug 28 '23 at 22:04
  • @ScottSeidman thanks for the reply! I did the K-Map and I'm getting \$x_2\bar{x_0} + \bar{x_0}\bar{x_1}\$ , is that correct. – Mayur Aug 30 '23 at 18:15
  • @ScottSeidman Thank you! I got the answer \$x_2\bar{x_0} + x_0\bar{x_1}\$. Can you tell me why you 1) used Gray code instead of normal BCD, shouldn't BCD also work? 2) Without using K-Map, can the expression I got in the question be simplified? Thank you once again – Mayur Aug 30 '23 at 18:26
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    I'll try to frame it up as an answer when I have a moment to LaTex up some K-maps -- but when you arrange them like that in Gray code (which is actually the guy's name, not a color), it's substantially easier to see patterns that match up to logical statements, and indeed the "wrap arounds" from bottom to top and from left to right are much easier to see, too. – Scott Seidman Aug 30 '23 at 19:44
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    I'll leave it for you to check if it's correct -- you need to evaluate the statement for only 8 cases (0 - 7 in binary) to see if it matches your requirement. It's a good exercise for you that should take much less than a half hour. – Scott Seidman Aug 30 '23 at 19:46
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    @ScottSeidman Thanks for the response! 1) It seems that I haven't fully understood K-maps, it seems that K-Map only works with Grey code and not BCD ( based on a similar [question here](https://electronics.stackexchange.com/questions/161888/why-does-the-karnaugh-map-use-gray-code-and-not-binary) – Mayur Aug 31 '23 at 13:15
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    @ScottSeidman I understand the solution now, thank you for clarifying my doubts and answering my questions! Thank you very much – Mayur Aug 31 '23 at 13:44

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