Mixing
Given starting number in sequence, find position of second number
$begingroup$
Given a sequence of numbers $1 - 12$, with starting number $X$ at position $1$, how would you go about finding the position $P$ of number $Y$?
For example, given $X$ = 5 and $Y$ = 4:
The sequence of numbers would be: $[5 6 7 8 9 10 11 12 1 2 3 4]$
So $P = 12$
Or if $X =12$ and $Y = 3$
$[12 1 2 3 4 5 6 7 8 9 10 11]$
$P = 4$
Hopefully this is clear. This is my first SO post, so feedback on formatting etc is welcome.
linear-algebra matrices
asked Jan 24 at 23:15
FaquarlFaquarl
11
$endgroup$
add a comment |
$begingroup$
Given a sequence of numbers $1 - 12$, with starting number $X$ at position $1$, how would you go about finding the position $P$ of number $Y$?
For example, given $X$ = 5 and $Y$ = 4:
The sequence of numbers would be: $[5 6 7 8 9 10 11 12 1 2 3 4]$
So $P = 12$
Or if $X =12$ and $Y = 3$
$[12 1 2 3 4 5 6 7 8 9 10 11]$
$P = 4$
Hopefully this is clear. This is my first SO post, so feedback on formatting etc is welcome.
linear-algebra matrices
asked Jan 24 at 23:15
FaquarlFaquarl
11
$endgroup$
$begingroup$
Did you leave "3" out on purpose?
$endgroup$
– Lee Mosher
Jan 24 at 23:18
$begingroup$
"Hopefully this is clear." Not in the slightest.
$endgroup$
– fleablood
Jan 24 at 23:18
$begingroup$
The are the numbers in consecutive order? With $1$ following $12$? If so then that's just $(Y+1) - X$ if $(Y+1)-X > 0$ and $(Y+1) -X+12$ if $(Y+1)-X le 0$.
$endgroup$
– fleablood
Jan 24 at 23:22
$begingroup$
Thanks @fleablood , and apologies for any confusion. I should have mentioned that I have come up with the two separate functions based on the value of Y < > X. I was mostly curious if there is a way to solve for P with one formula.
$endgroup$
– Faquarl
Jan 24 at 23:26
$begingroup$
$P = [(Y+1) - X]% 12$ where $%$ is the remainder function. i.e. if $a = qb +r$ were $r$ is the non-negative remainder then $a% b = r$.
$endgroup$
– fleablood
Jan 24 at 23:30
add a comment |
$begingroup$
Given a sequence of numbers $1 - 12$, with starting number $X$ at position $1$, how would you go about finding the position $P$ of number $Y$?
For example, given $X$ = 5 and $Y$ = 4:
The sequence of numbers would be: $[5 6 7 8 9 10 11 12 1 2 3 4]$
So $P = 12$
Or if $X =12$ and $Y = 3$
$[12 1 2 3 4 5 6 7 8 9 10 11]$
$P = 4$
Hopefully this is clear. This is my first SO post, so feedback on formatting etc is welcome.
linear-algebra matrices
asked Jan 24 at 23:15
FaquarlFaquarl
11
$endgroup$
Given a sequence of numbers $1 - 12$, with starting number $X$ at position $1$, how would you go about finding the position $P$ of number $Y$?
For example, given $X$ = 5 and $Y$ = 4:
The sequence of numbers would be: $[5 6 7 8 9 10 11 12 1 2 3 4]$
So $P = 12$
Or if $X =12$ and $Y = 3$
$[12 1 2 3 4 5 6 7 8 9 10 11]$
$P = 4$
Hopefully this is clear. This is my first SO post, so feedback on formatting etc is welcome.
linear-algebra matrices
linear-algebra matrices
asked Jan 24 at 23:15
FaquarlFaquarl
11
asked Jan 24 at 23:15
FaquarlFaquarl
11
edited Jan 24 at 23:19
Faquarl
asked Jan 24 at 23:15
FaquarlFaquarl
11
asked Jan 24 at 23:15
FaquarlFaquarl
11
asked Jan 24 at 23:15
FaquarlFaquarl
11
11
$begingroup$
Did you leave "3" out on purpose?
$endgroup$
– Lee Mosher
Jan 24 at 23:18
$begingroup$
"Hopefully this is clear." Not in the slightest.
$endgroup$
– fleablood
Jan 24 at 23:18
$begingroup$
The are the numbers in consecutive order? With $1$ following $12$? If so then that's just $(Y+1) - X$ if $(Y+1)-X > 0$ and $(Y+1) -X+12$ if $(Y+1)-X le 0$.
$endgroup$
– fleablood
Jan 24 at 23:22
$begingroup$
Thanks @fleablood , and apologies for any confusion. I should have mentioned that I have come up with the two separate functions based on the value of Y < > X. I was mostly curious if there is a way to solve for P with one formula.
$endgroup$
– Faquarl
Jan 24 at 23:26
$begingroup$
$P = [(Y+1) - X]% 12$ where $%$ is the remainder function. i.e. if $a = qb +r$ were $r$ is the non-negative remainder then $a% b = r$.
$endgroup$
– fleablood
Jan 24 at 23:30
add a comment |
$begingroup$
Did you leave "3" out on purpose?
$endgroup$
– Lee Mosher
Jan 24 at 23:18
$begingroup$
"Hopefully this is clear." Not in the slightest.
$endgroup$
– fleablood
Jan 24 at 23:18
$begingroup$
The are the numbers in consecutive order? With $1$ following $12$? If so then that's just $(Y+1) - X$ if $(Y+1)-X > 0$ and $(Y+1) -X+12$ if $(Y+1)-X le 0$.
$endgroup$
– fleablood
Jan 24 at 23:22
$begingroup$
Thanks @fleablood , and apologies for any confusion. I should have mentioned that I have come up with the two separate functions based on the value of Y < > X. I was mostly curious if there is a way to solve for P with one formula.
$endgroup$
– Faquarl
Jan 24 at 23:26
$begingroup$
$P = [(Y+1) - X]% 12$ where $%$ is the remainder function. i.e. if $a = qb +r$ were $r$ is the non-negative remainder then $a% b = r$.
$endgroup$
– fleablood
Jan 24 at 23:30
$begingroup$
Did you leave "3" out on purpose?
$endgroup$
– Lee Mosher
Jan 24 at 23:18
$begingroup$
Did you leave "3" out on purpose?
$endgroup$
– Lee Mosher
Jan 24 at 23:18
$begingroup$
"Hopefully this is clear." Not in the slightest.
$endgroup$
– fleablood
Jan 24 at 23:18
$begingroup$
"Hopefully this is clear." Not in the slightest.
$endgroup$
– fleablood
Jan 24 at 23:18
$begingroup$
The are the numbers in consecutive order? With $1$ following $12$? If so then that's just $(Y+1) - X$ if $(Y+1)-X > 0$ and $(Y+1) -X+12$ if $(Y+1)-X le 0$.
$endgroup$
– fleablood
Jan 24 at 23:22
$begingroup$
The are the numbers in consecutive order? With $1$ following $12$? If so then that's just $(Y+1) - X$ if $(Y+1)-X > 0$ and $(Y+1) -X+12$ if $(Y+1)-X le 0$.
$endgroup$
– fleablood
Jan 24 at 23:22
$begingroup$
Thanks @fleablood , and apologies for any confusion. I should have mentioned that I have come up with the two separate functions based on the value of Y < > X. I was mostly curious if there is a way to solve for P with one formula.
$endgroup$
– Faquarl
Jan 24 at 23:26
$begingroup$
Thanks @fleablood , and apologies for any confusion. I should have mentioned that I have come up with the two separate functions based on the value of Y < > X. I was mostly curious if there is a way to solve for P with one formula.
$endgroup$
– Faquarl
Jan 24 at 23:26
$begingroup$
$P = [(Y+1) - X]% 12$ where $%$ is the remainder function. i.e. if $a = qb +r$ were $r$ is the non-negative remainder then $a% b = r$.
$endgroup$
– fleablood
Jan 24 at 23:30
$begingroup$
$P = [(Y+1) - X]% 12$ where $%$ is the remainder function. i.e. if $a = qb +r$ were $r$ is the non-negative remainder then $a% b = r$.
$endgroup$
– fleablood
Jan 24 at 23:30
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
When $Xle Y$:
Let’s find out where 12 is first.
12 will be at (13-X)th place.
That means, Y will be at (13-X+Y)th place.
When $X>Y$:
Then it will just be Y-X+1.
answered Jan 24 at 23:27
Math LoverMath Lover
16010
$endgroup$
add a comment |
$begingroup$
If $Y ge X$ then $Y$ is in the $Y-X$ position past $1$. In other words $P = (Y-X) + 1$.
If $Y < X$ then we must go up to $12$ That is $12-X +1=13-X$ and then $Y$ more places $P= 13 - X +Y$
Or, $P = (Y-X) + 1$ if $Y ge X$ or $P = [(Y-X) + 1] + 12$ if $Y < X$.
edited Jan 24 at 23:52
Math Lover
16010
answered Jan 24 at 23:28
fleabloodfleablood
72.7k22788
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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oldest
votes
$begingroup$
When $Xle Y$:
Let’s find out where 12 is first.
12 will be at (13-X)th place.
That means, Y will be at (13-X+Y)th place.
When $X>Y$:
Then it will just be Y-X+1.
answered Jan 24 at 23:27
Math LoverMath Lover
16010
$endgroup$
add a comment |
$begingroup$
When $Xle Y$:
Let’s find out where 12 is first.
12 will be at (13-X)th place.
That means, Y will be at (13-X+Y)th place.
When $X>Y$:
Then it will just be Y-X+1.
answered Jan 24 at 23:27
Math LoverMath Lover
16010
$endgroup$
add a comment |
$begingroup$
When $Xle Y$:
Let’s find out where 12 is first.
12 will be at (13-X)th place.
That means, Y will be at (13-X+Y)th place.
When $X>Y$:
Then it will just be Y-X+1.
answered Jan 24 at 23:27
Math LoverMath Lover
16010
$endgroup$
When $Xle Y$:
Let’s find out where 12 is first.
12 will be at (13-X)th place.
That means, Y will be at (13-X+Y)th place.
When $X>Y$:
Then it will just be Y-X+1.
answered Jan 24 at 23:27
Math LoverMath Lover
16010
answered Jan 24 at 23:27
Math LoverMath Lover
16010
answered Jan 24 at 23:27
Math LoverMath Lover
16010
answered Jan 24 at 23:27
Math LoverMath Lover
16010
16010
add a comment |
add a comment |
$begingroup$
If $Y ge X$ then $Y$ is in the $Y-X$ position past $1$. In other words $P = (Y-X) + 1$.
If $Y < X$ then we must go up to $12$ That is $12-X +1=13-X$ and then $Y$ more places $P= 13 - X +Y$
Or, $P = (Y-X) + 1$ if $Y ge X$ or $P = [(Y-X) + 1] + 12$ if $Y < X$.
edited Jan 24 at 23:52
Math Lover
16010
answered Jan 24 at 23:28
fleabloodfleablood
72.7k22788
$endgroup$
add a comment |
$begingroup$
If $Y ge X$ then $Y$ is in the $Y-X$ position past $1$. In other words $P = (Y-X) + 1$.
If $Y < X$ then we must go up to $12$ That is $12-X +1=13-X$ and then $Y$ more places $P= 13 - X +Y$
Or, $P = (Y-X) + 1$ if $Y ge X$ or $P = [(Y-X) + 1] + 12$ if $Y < X$.
edited Jan 24 at 23:52
Math Lover
16010
answered Jan 24 at 23:28
fleabloodfleablood
72.7k22788
$endgroup$
add a comment |
$begingroup$
If $Y ge X$ then $Y$ is in the $Y-X$ position past $1$. In other words $P = (Y-X) + 1$.
If $Y < X$ then we must go up to $12$ That is $12-X +1=13-X$ and then $Y$ more places $P= 13 - X +Y$
Or, $P = (Y-X) + 1$ if $Y ge X$ or $P = [(Y-X) + 1] + 12$ if $Y < X$.
edited Jan 24 at 23:52
Math Lover
16010
answered Jan 24 at 23:28
fleabloodfleablood
72.7k22788
$endgroup$
If $Y ge X$ then $Y$ is in the $Y-X$ position past $1$. In other words $P = (Y-X) + 1$.
If $Y < X$ then we must go up to $12$ That is $12-X +1=13-X$ and then $Y$ more places $P= 13 - X +Y$
Or, $P = (Y-X) + 1$ if $Y ge X$ or $P = [(Y-X) + 1] + 12$ if $Y < X$.
edited Jan 24 at 23:52
Math Lover
16010
answered Jan 24 at 23:28
fleabloodfleablood
72.7k22788
edited Jan 24 at 23:52
Math Lover
16010
edited Jan 24 at 23:52
Math Lover
16010
edited Jan 24 at 23:52
Math Lover
16010
16010
answered Jan 24 at 23:28
fleabloodfleablood
72.7k22788
answered Jan 24 at 23:28
fleabloodfleablood
72.7k22788
answered Jan 24 at 23:28
fleabloodfleablood
72.7k22788
72.7k22788
add a comment |
add a comment |
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$begingroup$
Did you leave "3" out on purpose?
$endgroup$
– Lee Mosher
Jan 24 at 23:18
$begingroup$
"Hopefully this is clear." Not in the slightest.
$endgroup$
– fleablood
Jan 24 at 23:18
$begingroup$
The are the numbers in consecutive order? With $1$ following $12$? If so then that's just $(Y+1) - X$ if $(Y+1)-X > 0$ and $(Y+1) -X+12$ if $(Y+1)-X le 0$.
$endgroup$
– fleablood
Jan 24 at 23:22
$begingroup$
Thanks @fleablood , and apologies for any confusion. I should have mentioned that I have come up with the two separate functions based on the value of Y < > X. I was mostly curious if there is a way to solve for P with one formula.
$endgroup$
– Faquarl
Jan 24 at 23:26
$begingroup$
$P = [(Y+1) - X]% 12$ where $%$ is the remainder function. i.e. if $a = qb +r$ were $r$ is the non-negative remainder then $a% b = r$.
$endgroup$
– fleablood
Jan 24 at 23:30