Find the binary input function given the outputs (part 2)
$begingroup$
Here we have three binary variables $x_1$, $x_2$, $x_3$ $in {0,1}$.
I want to find the form of the function $f(x_1, x_2, x_3)$ such that the following are satisfied:
if $ x_1 = 0, x_2 = 0, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 6$
if $ x_1 = 0, x_2 = 0, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 4$
if $ x_1 = 0, x_2 = 1, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 5$
if $ x_1 = 0, x_2 = 1, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 2$
if $ x_1 = 1, x_2 = 0, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 5$
if $ x_1 = 1, x_2 = 0, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 3$
if $ x_1 = 1, x_2 = 1, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 4$
if $ x_1 = 1, x_2 = 1, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 0$
I imagine it like having a "virtual sum" which starts at 1 and it is increased by 1 at each step in the sequence. Every time I see a zero, this virtual sum "becomes real" and is reset. For instance:
$ x_1 = 1, x_2 = 1, x_3 = 0 $ After seeing the first one the virtual sum is 2. After the second one the virtual sum is 3. Finally there is a zero, so the virtual sum becomes 4, "becomes real" and is reset. The final result is 4.
$ x_1 = 0, x_2 = 1, x_3 = 0 $ After seeing the first zero the sum is 2 and the virtual sum is reset to 1. After the second one the virtual sum is 2. Finally there is a zero, so the virtual sum is incremented at 3 and "becomes real". The final result is 2 + 3 = 5.
$ x_1 = 1, x_2 = 1, x_3 = 1 $ We have three ones, so the virtual sum is 4 at the end of the sequence. However, since there are no zeroes it does not "become real" and the result is 0.
$ x_1 = 0, x_2 = 0, x_3 = 0 $ We have three zeroes, so we get 2 at each step and the sum is 6.
I had already asked a similar question (Find the binary input function given the outputs), which was (brilliantly) solved. This is a generalisation in which the "virtual sum" starts from one rather then zero and that, with respect to the previous question, inverts the binary value of the variables $x_1$, $x_2$, $x_3$.
combinatorics binary binary-operations
asked Jan 16 at 15:58
aprosperoaprospero
103
$endgroup$
add a comment |
$begingroup$
Here we have three binary variables $x_1$, $x_2$, $x_3$ $in {0,1}$.
I want to find the form of the function $f(x_1, x_2, x_3)$ such that the following are satisfied:
if $ x_1 = 0, x_2 = 0, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 6$
if $ x_1 = 0, x_2 = 0, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 4$
if $ x_1 = 0, x_2 = 1, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 5$
if $ x_1 = 0, x_2 = 1, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 2$
if $ x_1 = 1, x_2 = 0, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 5$
if $ x_1 = 1, x_2 = 0, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 3$
if $ x_1 = 1, x_2 = 1, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 4$
if $ x_1 = 1, x_2 = 1, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 0$
I imagine it like having a "virtual sum" which starts at 1 and it is increased by 1 at each step in the sequence. Every time I see a zero, this virtual sum "becomes real" and is reset. For instance:
$ x_1 = 1, x_2 = 1, x_3 = 0 $ After seeing the first one the virtual sum is 2. After the second one the virtual sum is 3. Finally there is a zero, so the virtual sum becomes 4, "becomes real" and is reset. The final result is 4.
$ x_1 = 0, x_2 = 1, x_3 = 0 $ After seeing the first zero the sum is 2 and the virtual sum is reset to 1. After the second one the virtual sum is 2. Finally there is a zero, so the virtual sum is incremented at 3 and "becomes real". The final result is 2 + 3 = 5.
$ x_1 = 1, x_2 = 1, x_3 = 1 $ We have three ones, so the virtual sum is 4 at the end of the sequence. However, since there are no zeroes it does not "become real" and the result is 0.
$ x_1 = 0, x_2 = 0, x_3 = 0 $ We have three zeroes, so we get 2 at each step and the sum is 6.
I had already asked a similar question (Find the binary input function given the outputs), which was (brilliantly) solved. This is a generalisation in which the "virtual sum" starts from one rather then zero and that, with respect to the previous question, inverts the binary value of the variables $x_1$, $x_2$, $x_3$.
combinatorics binary binary-operations
asked Jan 16 at 15:58
aprosperoaprospero
103
$endgroup$
add a comment |
$begingroup$
Here we have three binary variables $x_1$, $x_2$, $x_3$ $in {0,1}$.
I want to find the form of the function $f(x_1, x_2, x_3)$ such that the following are satisfied:
if $ x_1 = 0, x_2 = 0, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 6$
if $ x_1 = 0, x_2 = 0, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 4$
if $ x_1 = 0, x_2 = 1, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 5$
if $ x_1 = 0, x_2 = 1, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 2$
if $ x_1 = 1, x_2 = 0, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 5$
if $ x_1 = 1, x_2 = 0, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 3$
if $ x_1 = 1, x_2 = 1, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 4$
if $ x_1 = 1, x_2 = 1, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 0$
I imagine it like having a "virtual sum" which starts at 1 and it is increased by 1 at each step in the sequence. Every time I see a zero, this virtual sum "becomes real" and is reset. For instance:
$ x_1 = 1, x_2 = 1, x_3 = 0 $ After seeing the first one the virtual sum is 2. After the second one the virtual sum is 3. Finally there is a zero, so the virtual sum becomes 4, "becomes real" and is reset. The final result is 4.
$ x_1 = 0, x_2 = 1, x_3 = 0 $ After seeing the first zero the sum is 2 and the virtual sum is reset to 1. After the second one the virtual sum is 2. Finally there is a zero, so the virtual sum is incremented at 3 and "becomes real". The final result is 2 + 3 = 5.
$ x_1 = 1, x_2 = 1, x_3 = 1 $ We have three ones, so the virtual sum is 4 at the end of the sequence. However, since there are no zeroes it does not "become real" and the result is 0.
$ x_1 = 0, x_2 = 0, x_3 = 0 $ We have three zeroes, so we get 2 at each step and the sum is 6.
I had already asked a similar question (Find the binary input function given the outputs), which was (brilliantly) solved. This is a generalisation in which the "virtual sum" starts from one rather then zero and that, with respect to the previous question, inverts the binary value of the variables $x_1$, $x_2$, $x_3$.
combinatorics binary binary-operations
asked Jan 16 at 15:58
aprosperoaprospero
103
$endgroup$
Here we have three binary variables $x_1$, $x_2$, $x_3$ $in {0,1}$.
I want to find the form of the function $f(x_1, x_2, x_3)$ such that the following are satisfied:
if $ x_1 = 0, x_2 = 0, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 6$
if $ x_1 = 0, x_2 = 0, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 4$
if $ x_1 = 0, x_2 = 1, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 5$
if $ x_1 = 0, x_2 = 1, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 2$
if $ x_1 = 1, x_2 = 0, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 5$
if $ x_1 = 1, x_2 = 0, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 3$
if $ x_1 = 1, x_2 = 1, x_3 = 0 $ then $ f(x_1, x_2, x_3) = 4$
if $ x_1 = 1, x_2 = 1, x_3 = 1 $ then $ f(x_1, x_2, x_3) = 0$
I imagine it like having a "virtual sum" which starts at 1 and it is increased by 1 at each step in the sequence. Every time I see a zero, this virtual sum "becomes real" and is reset. For instance:
$ x_1 = 1, x_2 = 1, x_3 = 0 $ After seeing the first one the virtual sum is 2. After the second one the virtual sum is 3. Finally there is a zero, so the virtual sum becomes 4, "becomes real" and is reset. The final result is 4.
$ x_1 = 0, x_2 = 1, x_3 = 0 $ After seeing the first zero the sum is 2 and the virtual sum is reset to 1. After the second one the virtual sum is 2. Finally there is a zero, so the virtual sum is incremented at 3 and "becomes real". The final result is 2 + 3 = 5.
$ x_1 = 1, x_2 = 1, x_3 = 1 $ We have three ones, so the virtual sum is 4 at the end of the sequence. However, since there are no zeroes it does not "become real" and the result is 0.
$ x_1 = 0, x_2 = 0, x_3 = 0 $ We have three zeroes, so we get 2 at each step and the sum is 6.
I had already asked a similar question (Find the binary input function given the outputs), which was (brilliantly) solved. This is a generalisation in which the "virtual sum" starts from one rather then zero and that, with respect to the previous question, inverts the binary value of the variables $x_1$, $x_2$, $x_3$.
combinatorics binary binary-operations
combinatorics binary binary-operations
asked Jan 16 at 15:58
aprosperoaprospero
103
asked Jan 16 at 15:58
aprosperoaprospero
103
asked Jan 16 at 15:58
aprosperoaprospero
103
asked Jan 16 at 15:58
aprosperoaprospero
103
asked Jan 16 at 15:58
aprosperoaprospero
103
103
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
How about (completed) $$f(x_1,x_2,x_3) = 6cdot (1-x_1)(1-x_2)(1-x_3) + 4cdot (1-x_1)(1-x_2)x_3 \+ 5cdot (1-x_1)x_2(1-x_3) + 2cdot (1-x_1)x_2x_3
+5cdot x_1(1-x_2)(1-x_3) + 3cdot x_1(1-x_2)x_3 + 4cdot x_1x_2(1-x_3) + 0cdot x_1x_2x_3.$$
answered Jan 16 at 16:01
WuestenfuxWuestenfux
4,5141413
$endgroup$
$begingroup$
Is the complete formula $f(x_1,x_2,x_3) = 6(1-x_1)(1-x_2)(1-x_3) + 4(1-x_1)(1-x_2)x_3 + 2(1-x_1)x_2x_3$? Because it doesn't work for many inputs. Try (0,1,0) for example.
$endgroup$
– aprospero
Jan 16 at 16:26
$begingroup$
No I guess you got the idea. You need to complete it.
$endgroup$
– Wuestenfux
Jan 16 at 16:30
$begingroup$
Oooh! I got the point. So it is enough to hardcode the output sequence I want and then simplify the expression. Wow, did not come into my mind! Also I can easily generalise at 4, 5 variables and search for a pattern. Thank you @Wuestenfux, really helpful!
$endgroup$
– aprospero
Jan 16 at 17:06
add a comment |
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1 Answer
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1 Answer
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$begingroup$
How about (completed) $$f(x_1,x_2,x_3) = 6cdot (1-x_1)(1-x_2)(1-x_3) + 4cdot (1-x_1)(1-x_2)x_3 \+ 5cdot (1-x_1)x_2(1-x_3) + 2cdot (1-x_1)x_2x_3
+5cdot x_1(1-x_2)(1-x_3) + 3cdot x_1(1-x_2)x_3 + 4cdot x_1x_2(1-x_3) + 0cdot x_1x_2x_3.$$
answered Jan 16 at 16:01
WuestenfuxWuestenfux
4,5141413
$endgroup$
$begingroup$
Is the complete formula $f(x_1,x_2,x_3) = 6(1-x_1)(1-x_2)(1-x_3) + 4(1-x_1)(1-x_2)x_3 + 2(1-x_1)x_2x_3$? Because it doesn't work for many inputs. Try (0,1,0) for example.
$endgroup$
– aprospero
Jan 16 at 16:26
$begingroup$
No I guess you got the idea. You need to complete it.
$endgroup$
– Wuestenfux
Jan 16 at 16:30
$begingroup$
Oooh! I got the point. So it is enough to hardcode the output sequence I want and then simplify the expression. Wow, did not come into my mind! Also I can easily generalise at 4, 5 variables and search for a pattern. Thank you @Wuestenfux, really helpful!
$endgroup$
– aprospero
Jan 16 at 17:06
add a comment |
$begingroup$
How about (completed) $$f(x_1,x_2,x_3) = 6cdot (1-x_1)(1-x_2)(1-x_3) + 4cdot (1-x_1)(1-x_2)x_3 \+ 5cdot (1-x_1)x_2(1-x_3) + 2cdot (1-x_1)x_2x_3
+5cdot x_1(1-x_2)(1-x_3) + 3cdot x_1(1-x_2)x_3 + 4cdot x_1x_2(1-x_3) + 0cdot x_1x_2x_3.$$
answered Jan 16 at 16:01
WuestenfuxWuestenfux
4,5141413
$endgroup$
$begingroup$
Is the complete formula $f(x_1,x_2,x_3) = 6(1-x_1)(1-x_2)(1-x_3) + 4(1-x_1)(1-x_2)x_3 + 2(1-x_1)x_2x_3$? Because it doesn't work for many inputs. Try (0,1,0) for example.
$endgroup$
– aprospero
Jan 16 at 16:26
$begingroup$
No I guess you got the idea. You need to complete it.
$endgroup$
– Wuestenfux
Jan 16 at 16:30
$begingroup$
Oooh! I got the point. So it is enough to hardcode the output sequence I want and then simplify the expression. Wow, did not come into my mind! Also I can easily generalise at 4, 5 variables and search for a pattern. Thank you @Wuestenfux, really helpful!
$endgroup$
– aprospero
Jan 16 at 17:06
add a comment |
$begingroup$
How about (completed) $$f(x_1,x_2,x_3) = 6cdot (1-x_1)(1-x_2)(1-x_3) + 4cdot (1-x_1)(1-x_2)x_3 \+ 5cdot (1-x_1)x_2(1-x_3) + 2cdot (1-x_1)x_2x_3
+5cdot x_1(1-x_2)(1-x_3) + 3cdot x_1(1-x_2)x_3 + 4cdot x_1x_2(1-x_3) + 0cdot x_1x_2x_3.$$
answered Jan 16 at 16:01
WuestenfuxWuestenfux
4,5141413
$endgroup$
How about (completed) $$f(x_1,x_2,x_3) = 6cdot (1-x_1)(1-x_2)(1-x_3) + 4cdot (1-x_1)(1-x_2)x_3 \+ 5cdot (1-x_1)x_2(1-x_3) + 2cdot (1-x_1)x_2x_3
+5cdot x_1(1-x_2)(1-x_3) + 3cdot x_1(1-x_2)x_3 + 4cdot x_1x_2(1-x_3) + 0cdot x_1x_2x_3.$$
answered Jan 16 at 16:01
WuestenfuxWuestenfux
4,5141413
edited Jan 16 at 16:34
answered Jan 16 at 16:01
WuestenfuxWuestenfux
4,5141413
answered Jan 16 at 16:01
WuestenfuxWuestenfux
4,5141413
answered Jan 16 at 16:01
WuestenfuxWuestenfux
4,5141413
4,5141413
$begingroup$
Is the complete formula $f(x_1,x_2,x_3) = 6(1-x_1)(1-x_2)(1-x_3) + 4(1-x_1)(1-x_2)x_3 + 2(1-x_1)x_2x_3$? Because it doesn't work for many inputs. Try (0,1,0) for example.
$endgroup$
– aprospero
Jan 16 at 16:26
$begingroup$
No I guess you got the idea. You need to complete it.
$endgroup$
– Wuestenfux
Jan 16 at 16:30
$begingroup$
Oooh! I got the point. So it is enough to hardcode the output sequence I want and then simplify the expression. Wow, did not come into my mind! Also I can easily generalise at 4, 5 variables and search for a pattern. Thank you @Wuestenfux, really helpful!
$endgroup$
– aprospero
Jan 16 at 17:06
add a comment |
$begingroup$
Is the complete formula $f(x_1,x_2,x_3) = 6(1-x_1)(1-x_2)(1-x_3) + 4(1-x_1)(1-x_2)x_3 + 2(1-x_1)x_2x_3$? Because it doesn't work for many inputs. Try (0,1,0) for example.
$endgroup$
– aprospero
Jan 16 at 16:26
$begingroup$
No I guess you got the idea. You need to complete it.
$endgroup$
– Wuestenfux
Jan 16 at 16:30
$begingroup$
Oooh! I got the point. So it is enough to hardcode the output sequence I want and then simplify the expression. Wow, did not come into my mind! Also I can easily generalise at 4, 5 variables and search for a pattern. Thank you @Wuestenfux, really helpful!
$endgroup$
– aprospero
Jan 16 at 17:06
$begingroup$
Is the complete formula $f(x_1,x_2,x_3) = 6(1-x_1)(1-x_2)(1-x_3) + 4(1-x_1)(1-x_2)x_3 + 2(1-x_1)x_2x_3$? Because it doesn't work for many inputs. Try (0,1,0) for example.
$endgroup$
– aprospero
Jan 16 at 16:26
$begingroup$
Is the complete formula $f(x_1,x_2,x_3) = 6(1-x_1)(1-x_2)(1-x_3) + 4(1-x_1)(1-x_2)x_3 + 2(1-x_1)x_2x_3$? Because it doesn't work for many inputs. Try (0,1,0) for example.
$endgroup$
– aprospero
Jan 16 at 16:26
$begingroup$
No I guess you got the idea. You need to complete it.
$endgroup$
– Wuestenfux
Jan 16 at 16:30
$begingroup$
No I guess you got the idea. You need to complete it.
$endgroup$
– Wuestenfux
Jan 16 at 16:30
$begingroup$
Oooh! I got the point. So it is enough to hardcode the output sequence I want and then simplify the expression. Wow, did not come into my mind! Also I can easily generalise at 4, 5 variables and search for a pattern. Thank you @Wuestenfux, really helpful!
$endgroup$
– aprospero
Jan 16 at 17:06
$begingroup$
Oooh! I got the point. So it is enough to hardcode the output sequence I want and then simplify the expression. Wow, did not come into my mind! Also I can easily generalise at 4, 5 variables and search for a pattern. Thank you @Wuestenfux, really helpful!
$endgroup$
– aprospero
Jan 16 at 17:06
add a comment |
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