Convert Prod of Sums to Sum of Prod

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The discussion focuses on converting the expression F = (p+q’+r+s’)(p’+q+r+s’)(p’+q’+r+s’) into Sum of Products (SOP) form. Participants detail two methods: simplification through algebraic manipulation and using a Karnaugh map (K-map) for minimization. The final simplified expression achieved is F = s' + (p'q') + r, confirming both methods yield the same result. The use of K-maps is highlighted as an efficient approach for longer expressions.

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  • De Morgan's Theorems
  • Inverse Law and Absorption Law in Boolean algebra
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Question: Convert to Sum of Products

(p+q’+r+s’). (p’+q+r+s’). (p’+q’+r+s’)

= ((p+q’+r+s’)’ + (p’+q+r+s’)’ + (p’+q’+r+s’)’)’

(Demorgans each Sum-Term)

= ((p’.q.r’.s) + (p.q’.r’.s) + (p.q.r’.s))’ (Factor p.r’.s)

= ((p’.q.r’.s) + (p.r’.s).(q’ + q))’ (Inverse Law)

= ((p’.q.r’.s) + (p.r’.s))’ (Factor r’.s)

= ((r’.s).(p’.q + p))’ (absorption Variant)

= ((r’.s). (q + p))’ (Distributive Law)

= (q.r’.s + p.r’.s)’ (Demorgans outer expression)

= ((q.r’.s)’. (p.r’.s)’) (Demorgans inner expression)

= .(q’+r+s’) . (p’+r+s’)

Still not in SOP
 
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There are many ways to solve this.
The BASIC WAY is to simplify the expression. Let's name the Expression F
F = (p+q’+r+s’). (p’+q+r+s’). (p’+q’+r+s’)
F = (p + q')(p'+q)(p'+q') + (r+s') ---By factoring out (r+s')
F = (pp' + pq + p'q' + qq')(p'+q') + (r+s') ---Multiplying (p + q')(p'+q)
(pq+p'q')(p'+q') + (r+s') --- Some terms are canceled pp' = 0, qq' = 0
(pp'q + pqq' + p'q' + p'q') + (r+s') --Multiplying (pq+p'q')(p'+q')
(p'q') + r + s' --- Some terms are canceled pp'q = 0, pqq' = 0 and p'q' + p'q' = p'q'

F = (p'q') + r + s'THE OTHER METHOD
I think the easiest and most efficient way to solve this is to use a K-map especially in really long expressions.
lets name your function F
Computing for complement of F
F' = ((p+q’+r+s’) + (p’+q+r+s’)’ + (p’+q’+r+s’))'= (p'qr's)(pq'r's)(pqr's)
Then plot it on the K-MAP
(SEE BELOW)Then group the remaining ones (1s)
Which will yield:

F = s'+(p'q') + r.

Which is the same as the above expression.
 

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Thanks!
 
kukumaluboy said:
Question: Convert to Sum of Products

(p+q’+r+s’). (p’+q+r+s’). (p’+q’+r+s’)

= ((p+q’+r+s’)’ + (p’+q+r+s’)’ + (p’+q’+r+s’)’)’

(Demorgans each Sum-Term)

= ((p’.q.r’.s) + (p.q’.r’.s) + (p.q.r’.s))’ (Factor p.r’.s)

= ((p’.q.r’.s) + (p.r’.s).(q’ + q))’ (Inverse Law)

= ((p’.q.r’.s) + (p.r’.s))’ (Factor r’.s)

= ((r’.s).(p’.q + p))’ (absorption Variant)

= ((r’.s). (q + p))’ (Distributive Law)

= (q.r’.s + p.r’.s)’ (Demorgans outer expression)

= ((q.r’.s)’. (p.r’.s)’) (Demorgans inner expression)

= .(q’+r+s’) . (p’+r+s’)

Still not in SOP
In order to get sum of minterms you should multiply what you have gotten so far, then simplify. Also, you could've done that at the very beginning. What electronicsguy suggested is preferred; when you complement the maxterms given in the original function, you get positions of zeros in your K-table, then minimize the ones as to get sum of minterms.
 

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