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i know pi to 101 digits!
i can rattle it off extremely rapid.
You can break the problem into five parts, for which can be totaled up for the answer.
1.) How many 10-bit binary strings have 4 ones?
2.) How many 10-bit binary strings have 3 ones?
3.) How many 10-bit binary strings have 2 ones?
4.) How many 10-bit binary strings have 1 one? Answer = ten
5.) How many 10-bit binary strings only zeros? Answer = one
So this is a start... Sorry I don't know the rest off the bat.
Here is the link to the number of boolean "shapes" using n-variable cubes.
oeis.org/a000616/list
Okay I'm back.
It turns out the number of irreducible boolean functions of 4 variables is 402.
So the sequence I found on
the web is 3, 6, 22, 402, 1228158, 400507806843728,
527471432057653004017274030725792, (edit: fixed typo on seven variable number.--J.E.F.)
and for eight variables it is very long number according to the internet and is:
11218076601767519586965281984173341005925142853855481024470471657123840.
11218076601767519586965281984173341005925142853855481024470471657123840.
I typed it twice to hope I did it right. (Looks right).
Notice that for 2 variables, the number is 6 "shapes" or 6 irreducible boolean functions.
00
00
01
00
11
00
10
01
11
10
11
11
And that's the six shapes because you can
rotate them to get all sixteen.
(end of post)
...So for 3 variables, the 3 circles of a Venn diagram may end up as the front, top, and left of a cube.
The 8 corners of a cube designate the 8 areas in a Venn diagram. You must count the outside area
as the eighth area in a Venn diagram with 3 circles. Now if you have this cube with 8 corners, the
corners can have colored beads on the corners. Since this is boolean algebra, and not ternary, there
are 2 colors (boolean = 2 values, black and white, orange and pear, etc.) Now the input variables
are usually A, B, and C. The colors of the corners is the output of true or false (boolean, true or false),
such as the outputs on a "truth table". Now if you choose the left or right side of the cube, that could
be the variable A. (Any way you choose is okay, left/right, top/bottom, front/back) Now the B
variable could be the top and the negation of B would then be the bottom of the cube, 4 corners on
the bottom. Finally the C variable could the front 4 corners of the cube and the negation of C (called
"C not") could be the back 4 corners of the cube. Now the fun begins if you combine the different
halves of the cube together with boolean operators such as NAND, AND, OR, NOR, XOR, EQUALITYGATE,
IMPLICATIONGATE, etc.... When you combine two halves of the cube together, then you get a
different shape or pattern of black and white beads in the corners. What I did is write a short
computer program to iterate all possible combinations of operators and keep track of the complexity
of the equation, by counting the operators in the equation. Operators is called "gates" in digital
electronics. I ran the 3 variable case on a PC, and it runs in under a second. Then I changed
the program (written in C) to 4 variable case (hypercube) and that runs in 20 minutes. The
result is an equation for all 65536 four variable cases... Anyway, it was a lot of the fun.
That's all for now, maybe I'll write more later on this in a couple months... Bye.
0 2 6 4
1 3 7 5
2**0 = 1 2**2=4 2**6=64=40Hex 2**4=16=10Hex
2**1 = 2 2**3=8 2**7=128=80Hex 2**5=32=20Hex
01 04 40 10
02 08 80 20
Add these numbers up.
FF = tautology, always true
00 = falsehood, contradiciton, always false.
F0 = A
0F = not A = /A
55 = B
AA = not B = /B
CC = C
33 = not C = /C
...Note A, B, and C are interchangeable to your liking when starting the definitions...(it depends on the definition of your axes.)
The variables A, B, and C are inputs that create the output conditions when combined together with AND, OR, and XOR,
and also IMP and others...00 = 0 trivial case, FALSE
01 = 33 & 05, where 05 = 55 & 0F
03 = 33 & 0F
09 = 99 & 0F, where 99 = CC XOR 55
81 = C3 & A5, where C3 = F0 XOR 33, and where A5 = F0 XOR 55
07 = 0F & 77, where 77 = 55 OR 33
83 = F0 XOR 73, where 73 = 50 OR 33, where 50 = F0 & 55
49 = AA XOR E3, where E3 = C3 OR A0, and where C3 = F0 XOR 33, and A0 = F0 & AA
0F = trivial case, a boolean variable.
96 = CC XOR 5A, where 5A = F0 XOR AA
35 = C0 XOR F5, where C0 = F0 & CC and where F5 = F0 OR 55
39 = CC XOR F5, where F5 = F0 OR 55
17 = C0 XOR D7, where C0 = F0 & CC and where D7 = C3 OR 55, and where C3 = F0 XOR 33
99 = CC XOR 55
These are 3-variable cases above.
Use a cube by folding the 2 by 4 Karnaugh map in half.
There are 22 shapes for 3 variables.
Almost perfect. Just a tiny error in 66 + 660 + 6600...
I am quite happy with each and every three number set.
Thanks for being concerned though.
Push a button to see the answer.
Push buttons for answer.
Press button to get the answer.
51.) 1 + 3 + 5 =
52.) 2 + 4 + 6 =
53.) 3 + 5 + 7 =
54.) 4 + 6 + 8 =
55.) 5 + 7 + 9 =
56.) 45 + 65 + 85 =
57.) 32 + 40 + 66 =
58.) 2 + 12 + 22 =
59.) 1 + 14 + 28 =
60.) 3 + 115 + 117 =
Good Job, ganesh. From John.
Check out #49 for why you say 85?
Press buttons for answers.
1.) 1 + 2 + 3 =
2.) 4 + 5 + 13 =
3.) 0 + 1 + 2 =
4.) 5 + 7 + 10 =
5.) 6 + 7 + 9 =
6.) 4 + 5 + 7 =
7.) 1 + 3 + 5 =
8.) 2 + 4 + 6 =
9.) 4 + 5 + 7 =
10.) 2 + 2 + 3 =
11.) 13 + 14 + 15 =
12.) 12 + 13 + 15 =
13.) 9 + 13 + 19 =
14.) 3 + 3 + 3 =
15.) 1 + 4 + 7 =
16.) 1 + 3 + 6 =
17.) 11 + 18 + 25 =
18.) 3 + 6 + 9 =
19.) 2 + 6 + 10 =
20.) 1 + 1 + 4 =
21.) 1 + 6 + 7 =
22.) 4 + 6 + 6 =
23.) 5 + 7 + 8 =
24.) 7 + 9 + 9 =
25.) 2 + 2 + 4 =
26.) 3 + 8 + 9 =
27.) 2 + 3 + 4 =
28.) 3 + 4 + 9 =
29.) 9 + 12 + 14 =
30.) 4 + 6 + 7 =
31.) 5 + 8 + 11 =
32.) 4 + 8 + 8 =
33.) 6 + 6 + 12 =
34.) 4 + 17 + 56 =
35.) 22 + 33 + 44 =
36.) 40 + 41 + 44 =
37.) 2 + 3 + 6 =
38.) 3 + 3 + 4 =
39.) 1 + 1 + 1 =
40.) 1 + 1 + 5 =
41.) 3 + 5 + 6 =
42.) 4 + 9 + 19 =
43.) 1 + 2 + 4 =
44.) 4 + 6 + 8 =
45.) 5 + 8 + 9 =
46.) 6 + 8 + 9 =
47.) 8 + 12 + 25 =
48.) 3 + 5 + 9 =
49.) 11 + 26 + 88 =
50.) 0 + 0 + 2 =
Way to go!
When you start seeing beautiful things in math, your life will forever be more fun!!
Bonjour!
(Hello!)
Jouis le forum!
(Enjoy the forum!)
My how the years have past!
I'm John E. Franklin with (2) after it now.
I still live near 72W & 43N within a radius of 200 km.
Je parle un peu de francais a coup!
(I speak a little French to boot!)
;;le fin::
;;the end::
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