Thread: ordered or unordered selections

Consider the Boolean functions f(x, y, z) in three variables such that the table of values of f contains exactly four 1’s.

(i) Calculate the total number of such functions.

I was trying to think of this in a mentality of an injective functions, say A= (1,1,1,1) and B = (1,2,3,...,24)
So I thought 24P4 (ordered and with without repetition)




(ii) We apply the Karnaugh map method to such a function f. Suppose that the map does not contain any blocks of four 1’s, and all four 1’s are covered by three blocks of two 1’s. Moreover, we find that it is not possible to cover all 1’s by fewer than three blocks. Calculate the number of the functions with this property.

Here, I tried drawing it, I got 12 different ways.. But still trying to figure out how to put that in equation terms.

Thanks

Hey cellae.

Can you clarify with the boolean functions are x,y,z all just bits and what space does this function map to (i.e. x,y,z contains 2^3 = 8 possibilities for 3 independent bits, but f can map these either to a lower or higher dimensional space and this info is not given).

umm. Well, that is exactly the question I was given - but I have made a bit of progress:

(i) So the are the 2^3 possible functions.. And the from here I ask myself; how can I pick precisely 4 1's out of the 8.. which I did (8C4).. so 70

Then, for (ii)
I'm still trying to figure out how to calculate the total possibilities of the 4 grouped 1's - the 4 possible square grouped 1's.. this is painful.

Try treating a group as one outcome and everything else as the opposite outcome.

Ok, what about this..

There is 8 places to put the first 1... Then, because the second 1 needs to sit next to the other 1 then there is 3 possible spots.. Then again, the third 1 has to sit next to a 1, and again the fourth 1 has to be next to a 1.. and there are 4 possible squares of 4 ones..

sooo.. (8x3x2x1) - 4..so 44 possibilities?

Seems like a lot though ..

a

Can you clarify your twelve mine look like the following;

	YZ	YZ'	Y'Z'	Y'Z
x	0	1	1	0
x'	1	1	0	0
	YZ	YZ'	Y'Z'	Y'Z
x	1	1	0	0
x'	1	0	0	1
	YZ	YZ'	Y'Z'	Y'Z
x	1	1	0	0
x'	0	1	1	0
	YZ	YZ'	Y'Z'	Y'Z
x	1	0	0	1
x'	1	1	0	0
	YZ	YZ'	Y'Z'	Y'Z
x	0	0	1	1
x'	0	1	1	0
	YZ	YZ'	Y'Z'	Y'Z
x	1	0	0	1
x'	0	0	1	1
	YZ	YZ'	Y'Z'	Y'Z
x	0	1	1	0
x'	0	0	1	1
	YZ	YZ'	Y'Z'	Y'Z
x	0	0	1	1
x'	1	0	0	1

Giving me 8 for 1.b and not 12

well, there is definitely more that 8!

I got 16, and that was without overlapping the walls (if that makes sense, like in your last example)

hmmm... can you give me some hints on what I am doing wrong? for example

Like to have three overlapping of 2 1's mean's to me that the z-shape formed is rotated along....

Wait did you swap up the columns at all?

As i think I might be able to get another 8 by doing that?

I too am trying to figure this out ..

But how is the rest of your 'HW' going ? figure out the propositions?

not yet - going to revise logic quickly before attempting it

Well, I know a definite one is (q<->q)..
and I'm tossing whether (qv~q) is non equivalent or not..

hope that saves a bit of time !

[QUOTE=cellae;744469]Well, I know a definite one is (q<->q)..
and I'm tossing whether (qv~q) is non equivalent or not..

hope that saves a bit of time ![/QUOTE

Ignore this - not sure how to delete