Here is a simple session with maxima, in which im trying to make the simplification (r-r0)=h
(%i1) ax: G*M*m*(r-r0)/r0^2 - G*M*m/r0 ;
G M m (r - r0) G M m
(%o1) -------------- - -----
2 r0
r0
(%i2) let(r-r0,h);
(%o2) r - r0 --> h
(%i3) expand(scanmap(letsimp,ax));
G M m r 2 G M m
(%o3) ------- - -------
2 r0
r0
I was expecting this in the last part:
G M m h 2 G M m
------- - -------
2 r0
r0
Why did maxima replace (r-r0) with r rather than with h?
Iv attempted letsimp and letrat as instructed in this other question: common subexpressions
r - r0 is not a form supported by let. From the documentation:
-- Function: let
let (, , , , ..., )
let ([, , , , ..., ],
)
Defines a substitution rule for 'letsimp' such that <prod> is
replaced by <repl>. <prod> is a product of positive or negative
powers of the following terms:
* Atoms which 'letsimp' will search for literally unless
previous to calling 'letsimp' the 'matchdeclare' function is
used to associate a predicate with the atom. In this case
'letsimp' will match the atom to any term of a product
satisfying the predicate.
* Kernels such as 'sin(x)', 'n!', 'f(x,y)', etc. As with atoms
above 'letsimp' will look for a literal match unless
'matchdeclare' is used to associate a predicate with the
argument of the kernel.
In this case one can start with a syntactic substitution:
(%i1) ax: G*M*m*(r-r0)/r0^2 - G*M*m/r0 $
(%i2) subst(h, r - r0, ax);
G M h m G M m
(%o2) ------- - -----
2 r0
r0
Related
Consider the following gretl script (hansl):
open bjg.gdt
arima 1 1 0 ; 2 1 0 ; g
series fitted = $yhat
g1 <- gnuplot g fitted --with-lines --time-series --output=display
What I want to do next is to make a forecast for, lets say, 24 steps ahead, that is from Jan 1961 to Dec 1962. I believe the fifth line should be something like
fcast [options] --plot=display
What options to use here? I have tried several combinations but none is successful.
After further experimentation, here is the solution:
open bjg.gdt
arima 1 1 0 ; 2 1 0 ; g
series fitted = $yhat
g1 <- gnuplot g fitted --with-lines --time-series --output=display
dataset addobs 24
g2 <- fcast --dynamic --out-of-sample --plot=display
I want to get a character from string but the path is like going downstairs in Google Sheets
The string is like:
y C I 6 8 V 5
~5 Z n I L w f
V ~s i w J d _
o R ~4 2 i v f
9 ^ j ~h r u #
Z y Q 7 ~1 u a
T t z u _ ~! Q
G Y n r * t ~^
J A l v F j d
a 2 l - y O B
h B w % n a 4
M t _ P D W a
And expect the output is:
5 s 4 h 1 ! ^
I put ~ in front of the character that demo the path I want to get.
The logic is I will find the first column which cell is 5 and then get start from that cell.
Example sheet link
https://docs.google.com/spreadsheets/d/1UQEGEl_rqAMFePDAGoueTI47xF8T_DyrvJ3de5pkLRA/edit#gid=306981885
I tried auto-fill but it could only incrementally either row or column. And I hope to incremental both row and column.
={INDIRECT("A"&MATCH(5, A1:A14, 0)),
INDIRECT("B"&MATCH(5, A1:A14, 0)+1),
INDIRECT("C"&MATCH(5, A1:A14, 0)+2),
INDIRECT("D"&MATCH(5, A1:A14, 0)+3),
INDIRECT("E"&MATCH(5, A1:A14, 0)+4),
INDIRECT("F"&MATCH(5, A1:A14, 0)+5),
INDIRECT("G"&MATCH(5, A1:A14, 0)+6)}
for a large scale scenario you can use this one and drag it to the right:
=INDIRECT(ADDRESS(MATCH(5, $A1:$A14, 0)+COLUMN()-1, COLUMN(), 4))
You can also do it with an array formula if you want to:
=ArrayFormula(hlookup(column(A2:G13),{column(A2:G13);A2:G13},match(5,A2:A13,0)+column(A2:G13)))
I have a array/ named vector that looks like this:
d f g
1 2 3
I want to fill up the empty slots, meaning I want this:
a b c d e f g
0 0 0 1 0 2 3
Is there an elegant way of doing this, without having to write loops and conditionals? In my actual problem, instead of abcd as my array names, it's numbers. Not sure if that makes a difference. Figured alphabet is easier to understand for a reproducible example.
Create a vector of the final names, nms and then create a named vector of zeros from it using sapply and replace the elements corresponding to input names with the input values.
v <- c(d = 1, f = 2, g = 3) # input
nms <- letters[letters <= max(names(v))] # names on output vector, i.e. letters[1:7]
replace(sapply(nms, function(x) 0), names(v), v) ##
giving:
a b c d e f g
0 0 0 1 0 2 3
If in your actual vector the names are not letters then just set nms yourself. For example, nms <- c("dogs", "cats", "d", "elephants", "f", "g") would work with the same line marked ## above.
2) An alternative is to replace the line marked ## above with:
unlist(modifyList(as.list(setNames(numeric(length(nms)), nms)), as.list(v)))
Data
x <- c(d=1L,f=2L,g=3L);
x;
## d f g
## 1 2 3
Solution 1: First match new names into x and extract values, then replace NAs with zero.
x <- setNames(x[match(letters[1:7],names(x))],letters[1:7]);
x[is.na(x)] <- 0L;
x;
## a b c d e f g
## 0 0 0 1 0 2 3
Solution 2: One-liner, using nomatch argument of match().
setNames(c(x,0L)[match(letters[1:7],names(x),nomatch=length(x)+1L)],letters[1:7]);
## a b c d e f g
## 0 0 0 1 0 2 3
I'm trying to find a efficient algorithm in C, which provides me all options of a given equation.
I have equation AX + BY = M, where A, B and M i got on input (scanf).
For example lets have: 5X + 10Y = 45
1st option: 5 * 9 + 10 * 0
2nd option: 5 * 7 + 10 * 1
n-th option: 5 * 1 +
10 * 4
And also I need to count how many possible options exist?
Some tips, hints?
I forgot to say that X and Y are in Z and >= 0, so there is no infinite options.
The question makes sense if you restrict to non-negative unknowns.
Rewrite the equation as
AX = M - BY.
There can be positive solutions as long as the RHS is positive, i.e.
BY ≤ M,
or
Y ≤ M/B.
Then for a given Y, there is a solution iff
A|(M - BY)
You can code this in Python as
for Y in range(M / B + 1):
if (M - B * Y) % A == 0:
X= (M - B * Y) / A
The solutions are
9 0
7 1
5 2
3 3
1 4
The number of iterations equals M / B. If A > B, it is better to swap X and Y.
you can calcule every solution if you put some limit in your input value, for example: use X and Y in a value included from 0 to 9... in this way you can use for to calculate every solution.
The number of solution is infinite:
find a first solution like: X=9, Y=0.
you can create another solution by using:
X' = X+2*p
Y' = Y-p
For any p in Z.
This proves your program will never terminate.
I am trying to generate a password with certain requirements.
When I enter the while loop to generate a random character from the array it is fine until I add a count for my index "$i"
With the following code:
#!/bin/bash
#SET ARRAY VALUES
all=( 0 1 2 3 4 5 6 7 8 9 a b c d e f g h i j k l m n o p q r s t u v w x y z A B C D E F G H I J K L M N O P Q R S T U V W X Y Z )
echo
#SET COUNT VALUES TO 0
numc=0
lowc=0
upc=0
i=0
while true;
do
#GENERATE PASSWORD
phrase[$i]=${all[$RANDOM%62]}
#CHECK IF PASSWORD MEETS REQUIREMENTS
for ((n=0; n<10; n++))
do
if [ ${phrase[$i]} == ${all[$n]} ]
then
echo num ${all[$n]}
let "numc++"
let "i++"
fi
done
I get the error "line 39: [: ==: unary operator expected"
but if I remove the let "i++" line then there is no error. But I need to increase my index in order to exit the loop and check the minimum length of the password
If ${phrase[$i]} (by the way you don't need $ in [$i] context ${phrase[i]} works too) is ever the empty string that if test will become [ == value-of-all-n ] which isn't a valid test.
Either quote the variables (which is almost always the right thing to do) or prevent that from ever being the empty string. (Was that i++ supposed to happen outside the inner loop?)