The code below works but I am wondering if there is a better way that maybe uses some of the features of coffeescript that I am unfamiliar with.
The problem is this, I need to page items but the paging increases each time.
If I take the number 20 for example, it would create the following pages:
1 - 3
4 - 7
8 - 15
16 - 20
I have the following test and code which does pass:
module 'Remainder',
setup: ->
#remainder = 20
test 'splits remainder incrementally', ->
parts = #remainder.increasingSplit()
equal parts[0], '1 - 3', ''
equal parts[1], '4 - 7', ''
equal parts[2], '8 - 15', ''
equal parts[3], '16 - 20', ''
Number.prototype.increasingSplit = ->
start = 1
nextSplit = 3
parts = []
finished = false
while !finished
if nextSplit > #
parts.push "#{start} - #{#}"
break
parts.push "#{start} - #{nextSplit}"
start = nextSplit + 1
nextSplit = nextSplit * 2 + 1
parts
Without changing the algorithm too much, you can try this:
Number::increasingSplit = ->
start = 1
nextSplit = 3
parts = []
while start <= #
parts.push "#{start} - #{Math.min nextSplit, #}"
start = nextSplit + 1
nextSplit = nextSplit * 2 + 1
parts
The changes were:
replacing .prototype with ::,
removing of the finished variable (which was not being used effectively because the break anyway) and the break altogether and changing the condition to start <= #,
using only one parts.push <part>, with the minimum between nextSplit and # as the top.
Also, i'd advice against extending the Number prototype in this case. Extending the prototype of primitive types can sometimes cause weird problems, like:
Number::isFour = -> # is 4
console.log 4.isFour() # -> false
That happens because inside that function # will be a Number object instead of a primitive number, thus making the === 4 comparison always fail. That would not happen if you define isFour as a standalone function:
isFour = (n) -> n is 4
console.log isFour 4 # -> true
So, i'd prefer this version of incrasingSplit:
increasingSplit = (n) ->
start = 1
nextSplit = 3
parts = []
while start <= n
parts.push "#{start} - #{Math.min nextSplit, n}"
start = nextSplit + 1
nextSplit = nextSplit * 2 + 1
parts
Finally, if you don't mind recursion, you can go with a more FP-style algorithm :)
increasingSplit = (n, start = 1, nextSplit = 3) ->
if start > n
[]
else
part = "#{start} - #{Math.min nextSplit, n}"
rest = increasingSplit n, nextSplit + 1, nextSplit * 2 + 1
[part, rest...]
Related
In Raku, given a list of pairs (2 => 3, 3 => 2, 5 => 1, 7 => 4) ( representing the prime factorization of n = 2 3 · 3 2 · 5 1 · 7 4 ), how does construct a Raku expression for σ(n) = ( 2 0 + 2 1 + 2 2 + 2 3 ) · ( 3 0 + 3 1 + 3 2 ) · ( 5 0 + 5 1 ) · ( 7 0 + 7 1 + 7 2 + 7 3 + 7 4 ) ?
sub MAIN()
{
my $pairList = (2 => 3, 3 => 2, 5 => 1, 7 => 4) ;
say '$pairList' ;
say $pairList ;
say $pairList.WHAT ;
# Goal:
# from $pairList,
# the product (1 + 2 + 4 + 8) * (1 + 3 + 9) * (1 + 5) * (1 + 7 + 49 + 343 + 2401)
# = sigma ( 2^3 * 3^2 * 5^1 * 7^4 )
} # end sub MAIN
Update 1
Based upon the answer of #raiph, the following program breaks the overall process into stages for the newcomer to Raku (such as me) …
sub MAIN()
{
my $pairList = (2 => 3, 3 => 2, 5 => 1, 7 => 4) ;
say '$pairList' ;
say $pairList ;
say $pairList.WHAT ;
# Goal:
# from $pairList,
# the product (1 + 2 + 4 + 8) * (1 + 3 + 9) * (1 + 5) * (1 + 7 + 49 + 343 + 2401)
# the product (15) * (13) * (6) * (2801)
# sigma ( 2^3 * 3^2 * 5^1 * 7^4 )
# 3277170
# Stage 1 : ((1 2 4 8) (1 3 9) (1 5) (1 7 49 343 2401))
my $stage1 = $pairList.map: { (.key ** (my $++)) xx (.value + 1) } ;
say '$stage1 : lists of powers' ;
say $stage1 ;
say $stage1.WHAT ;
# Stage 2 : ((1 + 2 + 4 + 8) (1 + 3 + 9) (1 + 5) (1 + 7 + 49 + 343 + 2401))
my $stage2 = $stage1.map: { sum $_ } ;
say '$stage2 : sum each list' ;
say $stage2 ;
say $stage2.WHAT ;
# Stage 3 : (1 + 2 + 4 + 8) * (1 + 3 + 9) * (1 + 5) * (1 + 7 + 49 + 343 + 2401)
my $stage3 = $stage2.reduce( &infix:<*> ) ;
say '$stage3 : product of list elements' ;
say $stage3 ;
say $stage3.WHAT ;
} # end sub MAIN
A related post appears on Mathematics Stack Exchange.
Update 2
My original motivation had been to calculate aliquot sum s(n) = σ(n) - n. I found that prime factorization of each n is not necessary and seems inefficient. Raku and C++ programs calculating s(n) for n = 0 … 10 6 follow …
Raku
sub MAIN()
{
constant $limit = 1_000_000 ;
my #s of Int = ( 1 xx ($limit + 1) ) ;
#s[0] = 0 ;
#s[1] = 0 ;
loop ( my $column = 2; $column <= ($limit + 1) div 2; $column++ )
{
loop ( my $row = (2 * $column); $row <= $limit; $row += $column )
{
#s[$row] += $column ;
} # end loop $row
} # end loop $column
say "s(", $limit, ") = ", #s[$limit] ; # s(1000000) = 1480437
} # end sub MAIN
C++
(Observed to execute significantly faster than Raku)
#include <iostream>
#include <vector>
using namespace std ;
int main ( void )
{
const int LIMIT = 1000000 ;
vector<int> s ( (LIMIT + 1), 1 ) ;
s[0] = 0 ;
s[1] = 0 ;
for ( int col = 2 ; col <= (LIMIT + 1) / 2 ; col++ )
for ( int row = (2 * col) ; row <= LIMIT ; row += col )
s[row] += col ;
cout << "s(" << LIMIT << ") = " << s[LIMIT] << endl ; // s(1000000) = 1480437
} // end function main
There'll be bazillions of ways. I've ignored algorithmic efficiency. The first thing I wrote:
say [*] (2 => 3, 3 => 2, 5 => 1, 7 => 4) .map: { sum .key ** my $++ xx .value + 1 }
displays:
3277170
Explanation
1 say
2 [*] # `[op]` is a reduction. `[*] 6, 8, 9` is `432`.
3 (2 => 3, 3 => 2, 5 => 1, 7 => 4)
4 .map:
5 {
6 sum
7 .key # `.key` of `2 => 3` is `2`.
8 **
9 my # `my` resets `$` for each call of enclosing `{...}`
10 $++ # `$++` integer increments from `0` per thunk evaluation.
11 xx # `L xx R` forms list from `L` thunk evaluated `R` times
12 .value + 1
13 }
It is unlikely that Raku is ever going to be faster than C++ for that kind of operation. It is still early in its life and there are lots of optimizations to be gained, but raw processing speed is not where it shines.
If you are trying to find the aliquot sum for all of the numbers in a continuous range then prime factorization is certainly less efficient than the method you arrived at. Sort of an inverse Sieve of Eratosthenes. There are a few things you could change to make it faster, though still probably much slower than C++
About twice as fast on my system:
constant $limit = 1_000_000;
my #s = 0,0;
#s.append: 1 xx $limit;
(2 .. $limit/2).race.map: -> $column {
loop ( my $row = (2 * $column); $row <= $limit; $row += $column ) {
#s[$row] += $column ;
}
}
say "s(", $limit, ") = ", #s[$limit] ; # s(1000000) = 1480437
Where the prime factorization method really shines is for finding arbitrary aliquot sums.
This produces an answer in fractions of a second when the inverse sieve would likely take hours. Using the Prime::Factor module: from the Raku ecosystem
use Prime::Factor;
say "s(", 2**97-1, ") = ", (2**97-1).&proper-divisors.sum;
# s(158456325028528675187087900671) = 13842607235828485645777841
function fnum = fib(n)
if (n == 1) || (n == 2)
fnum = 1;
else
fnum = fib(n-1) + fib(n-2);
end
Can you explain how does each step outputs for the given input. For example inputting 7 gives me 13, 5 gives me 5, but I am not able to track how. I would highly appreciate your reply.
Recursion basically means that the function calls itself.
If we follow your function for fib(3), you will see that what it does is call fib(2)+fib(1). The values of these are defined, and are 1, so it will return 2.
If you call it with fib(4), it will go and compute fib(3)+fib(2). You already know what fib(3) does (see previous paragraph), and we already mentioned that fib(2) returns 1.
If you call it with fib(5) it will go and compute fib(4)+fib(3). See previous paragraph.
This is a very useful way of programming as it is a very simple function to compute something that is arguably more complicated. The most important thing is that you make sure that any recursive function has strong stopping criteria, else it can go forever!
Do you know how Fibonacci series is defined? This function implements that recursively.
Longer answer
Fibonacci series is defined as
n(1) = 1
n(2) = 1
n(k+1) = n(k) + n(k-1)
So when you put 5 as argument, the expansion becomes
n(4+1) = n(4)+n(3)
= n(3)+n(2)+n(2)+n(1)
= n(2)+n(1)+1+1+1
= 1+1+1+1+1
= 5
A much easier back of envelop method is to start from first index and add last two terms to arrive at the next.
1, 1, 2 <- (1+1), 3 <- (2+1), 5 <- (3+2), ...
The Fibonnacci series is defined as f(1) = 1, f(2) = 1 and for all n > 2, f(n) = f(n-1) + f(n-2)
So when you call fib(1) it returns 1 same for fib(2). But when you call fib(3) it returns fib(3-1) + fib(3-2) which is fib(2) + fib(1) = 2. And then when you call fib(4)it returns fib(3) + fib(2) = (fib(2) + fib(1)) + fib(1) = 3. And recursively the fibonnaci series is equal to 1, 1, 3, 5, 8, 13, 21, ...
For the code when n is different than 1 or 2 it call the function fib recursively. And when is equals to 1 or 2 it returns 1.
I am trying to write a macro which enables me to transform
(a, b, c, d) to (a, a + b, a + b + c, a + b + c + d), etc. Here is what I have got so far:
macro_rules! pascal_next {
($x: expr) => ($x);
($x: expr, $y: expr) => (
($x, $x + $y)
);
($x: expr, $y: expr, $($rest: expr),+) => (
($x, pascal_next!(
$x + $y, $($rest),+
)
)
);
}
However, there is a problem that it would actually output (a, (a + b, (a + b + c, a + b + c +d))). The origin is that the second matching rule ($x: expr, $y: expr) => (($x, $x + $y));, produces an extra bracket, so that there would be nested brackets. If I don't put a bracket outside, I would get the error error:
unexpected token: ,
So is it possible to output a comma , in Rust macros?
No; the result of a macro must be a complete grammar construct like an expression or an item. You absolutely cannot have random bits of syntax like a comma or a closing brace.
You can get around this by simply not outputting anything until you have a complete, final expression. Behold!
#![feature(trace_macros)]
macro_rules! pascal_impl {
/*
The input to this macro takes the following form:
```ignore
(
// The current output accumulator.
($($out:tt)*);
// The current additive prefix.
$prefix:expr;
// The remaining, comma-terminated elements.
...
)
```
*/
/*
Termination condition: there is no input left. As
such, dump the output.
*/
(
$out:expr;
$_prefix:expr;
) => {
$out
};
/*
Otherwise, we have more to scrape!
*/
(
($($out:tt)*);
$prefix:expr;
$e:expr, $($rest:tt)*
) => {
pascal_impl!(
($($out)* $prefix+$e,);
$prefix+$e;
$($rest)*
)
};
}
macro_rules! pascal {
($($es:expr),+) => { pascal_impl!((); 0; $($es),+,) };
}
trace_macros!(true);
fn main() {
println!("{:?}", pascal!(1, 2, 3, 4));
}
Note: To use this on a stable compiler, you will need to delete the #![feature(trace_macros)] and trace_macros!(true); lines. Everything else should be fine.
What this does is it recursively munches away at the input, passing the partial (and potentially semantically invalid) output as input to the next level of recursion. This lets us build up an "open list", which we couldn't otherwise do.
Then, once we're out of input, we just re-interpret our partial output as a complete expression and... done.
The reason I including the tracing stuff is so I could show you what it looks like as it runs:
pascal! { 1 , 2 , 3 , 4 }
pascal_impl! { ( ) ; 0 ; 1 , 2 , 3 , 4 , }
pascal_impl! { ( 0 + 1 , ) ; 0 + 1 ; 2 , 3 , 4 , }
pascal_impl! { ( 0 + 1 , 0 + 1 + 2 , ) ; 0 + 1 + 2 ; 3 , 4 , }
pascal_impl! { ( 0 + 1 , 0 + 1 + 2 , 0 + 1 + 2 + 3 , ) ; 0 + 1 + 2 + 3 ; 4 , }
pascal_impl! { ( 0 + 1 , 0 + 1 + 2 , 0 + 1 + 2 + 3 , 0 + 1 + 2 + 3 + 4 , ) ; 0 + 1 + 2 + 3 + 4 ; }
And the output is:
(1, 3, 6, 10)
One thing to be aware of: large numbers of un-annotated integer literals can cause a dramatic increase in compile times. If this happens, you can solve it by simply annotating all of your integer literals (like 1i32).
I have a CSV file in a format similar to the following one:
title1
index columnA1 columnA2 columnA3
1 2 3 6
2 23 23 1
3 2 3 45
4 2 2 101
title2
index columnB1 columnB2 columnB3
1 23 53 6
2 22 13 1
3 5 4 43
4 8 6 102
I want to build a function readCustomCSV which receives a CSV file in the bellow illustrated format and a row index i and returns an output file with (for let's say i = 3) the following content:
title1
index columnA1 columnA2 columnA3
3 2 3 45
title2
index columnB1 columnB2 columnB3
3 5 4 43
Do you know how to use the csvread function in order to obtain this type of functionality?
It confuses me that there are 2 types sections. I was thinking at using the whole thing as a string and then split it into 2 .csv files and then read the corresponding line line.
try using this function :
I assumed that all tables have equal number of columns/rows. The code can definitely be shortened / improved / extended ;)
function multi_table_csvread (row_index)
filename_INPUT = 'multi_table.csv' ;
filename_OUTPUT = 'selected_row.csv' ;
fIN = fopen(filename_INPUT,'r');
nextLine = fgetl(fIN);
tableIndex = 0;
tableLine = 0;
csvTable = [];
% start reading the csv file, line by line
while nextLine ~= -1
lineStr = strtrim(strsplit(nextLine,',')) ;
% remove empty cells
lineStr(cellfun('isempty',lineStr)) = [] ;
tableLine = tableLine + 1 ;
% if 1 element start new table
if numel(lineStr) == 1
tableIndex = tableIndex + 1;
tableLine = 1;
csvTable{tableIndex,tableLine} = lineStr ;
else
lineStr = add_comas(lineStr) ;
csvTable{tableIndex,tableLine} = lineStr ;
end
nextLine = fgetl(fIN);
end
fclose(fIN);
fOUT = fopen(filename_OUTPUT,'w');
if row_index > size(csvTable,2) -2
error('The row index exceeds the maximum number of rows!')
end
for k = 1 : size(csvTable,1)
title = csvTable{k,1};
columnHeaders = csvTable{k,2};
selected_row = csvTable{k,row_index+2};
fprintf(fOUT,'%s\n',title{:});
fprintf(fOUT,'%s',columnHeaders{:});
fprintf(fOUT,'\n');
fprintf(fOUT,'%s',selected_row{:});
fprintf(fOUT,'\n');
end
fclose(fOUT);
function line_with_comas = add_comas(this_line)
for ii = 1 : length(this_line)-1
this_line{ii} = strcat(this_line{ii},',') ;
end
line_with_comas = this_line ;
I noticed that Verilog rounds my real number results into integer results. For example when I look at simulator, it shows the result of 17/2 as 9. What should I do? Is there anyway to define something like a: output real reg [11:0] output_value ? Or is it something that has to be done by simulator settings?
Simulation only (no synthesis). Example:
x defined as a signed input and output_value defined as output reg.
output_value = ((x >>> 1) + x) + 5;
If x=+1 then output value has to be: 13/2=6.5.
However when I simulate I see output_value = 6.
Code would help, but I suspect your not dividing reals at all. 17 and 2 are integers, and so a simple statement like that will do integer division.
17 / 2 = 8 (not 9, always rounds towards 0)
17.0 / 2.0 = 8.5
In your second case
output_value = ((x >>> 1) + x) + 5
If x is 1, x >>> 1 is 0, not 0.5 because you've just gone off the bottom of the word.
output_value = ((1 >>> 1) + 1) + 5 = 0 + 1 + 5 = 6
There's nothing special about verilog here. This is true for the majority of languages.