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help with Architecture homework

Name: Grif 2006-11-06 0:42

PROBLEM
Braille is a system which allows a blind person to "read" alphanumerics by feeling a pattern of six raised dots. (Click on the above image for a Braille chart.) For the digits 0-9, only four dots are used, placed essentially in the four corners of a square (the top four dots out of six). Design a ciruit that converts BCD to Braille. The table shows the correspondence between BCD and Braille. NOTE; A,B,C and D are the inputs and UL, UR, LL and LR are the outputs (UL represents the upper left corner, UR the upper right corner, LR the lower right corner, and LL the lower left corner). A 0 indicates no raised dot and a 1 indicates a raised dot in that position.

   inputs        outputs
 A  B  C  D |  UL  UR  LR  LL
-----------------------------
 0  0  0  0 |  0    1   1   1  
 0  0  0  1 |  1    0   0   0
 0  0  1  0 |  1    0   0   1
 0  0  1  1 |  1    1   0   0
 0  1  0  0 |  1    1   1   0
 0  1  0  1 |  1    0   1   0
 0  1  1  0 |  1    1   0   1
 0  1  1  1 |  1    1   1   1
 1  0  0  0 |  1    0   1   1
 1  0  0  1 |  0    1   0   1


PROGRESS
I took each output and assigned its values to four 4x4 k-maps, and came out with simplified boolean expressions.  Then I took the complement of each output and did the same.  Now I have to choose which combination of functions and complements to use that would require the smallest number of logic gates.

I'm not asking for anyone to do my homework for me.  I just need advice on what to do next...that is, do I figure out what products are shared between expressions?  Or do I lump them all together in some strange method I don't remember?  As a last resort, I can take all 16 possible combinations and compare the number of gates required, then decide on a circuit.

Name: Anonymous 2006-11-06 0:44

You add 3.

Name: Anonymous 2006-11-06 0:45

3 bits that is.

Name: Grif 2006-11-06 0:47

Add 3 bits??  To what?  And how would that help?

Name: Anonymous 2006-11-06 0:48

Actually that will only work for a few cases. Oh well.

Name: Anonymous 2006-11-06 0:50

Just manipulate the boolean expressions until you get phrases that are shared by different outputs.

Name: Anonymous 2006-11-06 0:55

>>4
I looked at the first couple of cases, and then one further down. Unfortunately, those were the only ones it worked for :D

input         output            difference
 0  0  0  0 |  0    1   1   1  = 111 *
 0  0  0  1 |  1    0   0   0  = 111 *
 0  0  1  0 |  1    0   0   1  = 111 *
 0  0  1  1 |  1    1   0   0  = 1001
 0  1  0  0 |  1    1   1   0  = 1010
 0  1  0  1 |  1    0   1   0  = 0101
 0  1  1  0 |  1    1   0   1  = 111 *
 0  1  1  1 |  1    1   1   1  = 1000
 1  0  0  0 |  1    0   1   1  = 0011
 1  0  0  1 |  0    1   0   1  = -0100

Name: Anonymous 2006-11-06 1:07

Karnaugh maps are your friend!

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