this may be silly question but yet i am unable to figure it out...
syntax of abs and abs_diff is
ugentype abs (gentype x)
ugentype abs_diff (gentype x,gentype y)
let's take x=-4 and y=3
is there any difference between abs(-4-3) and abs_diff(-4,3) the result
of both operation is same... if i can rewrite abs_diff as abs then
why khronos gave 2 abs function
thank you
According to abs, abs_diff man:
abs returns |x|.
abs_diff returns |x-y| without modulo overflow.
abs(-4-3) = abs(-7) = 7
abs(-4, 3) = abs(-4 - 3) = abs(-7) = 7
there are two functions for convenience depending on which you think will be more consice for what you are doing.
Related
To split a number into digits in a given base, Julia has the digits() function:
julia> digits(36, base = 4)
3-element Array{Int64,1}:
0
1
2
What's the reverse operation? If you have an array of digits and the base, is there a built-in way to convert that to a number? I could print the array to a string and use parse(), but that sounds inefficient, and also wouldn't work for bases > 10.
The previous answers are correct, but there is also the matter of efficiency:
sum([x[k]*base^(k-1) for k=1:length(x)])
collects the numbers into an array before summing, which causes unnecessary allocations. Skip the brackets to get better performance:
sum(x[k]*base^(k-1) for k in 1:length(x))
This also allocates an array before summing: sum(d.*4 .^(0:(length(d)-1)))
If you really want good performance, though, write a loop and avoid repeated exponentiation:
function undigit(d; base=10)
s = zero(eltype(d))
mult = one(eltype(d))
for val in d
s += val * mult
mult *= base
end
return s
end
This has one extra unnecessary multiplication, you could try to figure out some way of skipping that. But the performance is 10-15x better than the other approaches in my tests, and has zero allocations.
Edit: There's actually a slight risk to the type handling above. If the input vector and base have different integer types, you can get a type instability. This code should behave better:
function undigits(d; base=10)
(s, b) = promote(zero(eltype(d)), base)
mult = one(s)
for val in d
s += val * mult
mult *= b
end
return s
end
The answer seems to be written directly within the documentation of digits:
help?> digits
search: digits digits! ndigits isdigit isxdigit disable_sigint
digits([T<:Integer], n::Integer; base::T = 10, pad::Integer = 1)
Return an array with element type T (default Int) of the digits of n in the given base,
optionally padded with zeros to a specified size. More significant digits are at higher
indices, such that n == sum([digits[k]*base^(k-1) for k=1:length(digits)]).
So for your case this will work:
julia> d = digits(36, base = 4);
julia> sum([d[k]*4^(k-1) for k=1:length(d)])
36
And the above code can be shortened with the dot operator:
julia> sum(d.*4 .^(0:(length(d)-1)))
36
Using foldr and muladd for maximum conciseness and efficiency
undigits(d; base = 10) = foldr((a, b) -> muladd(base, b, a), d, init=0)
I'm struggling to amend the Julia-specific tutorial on NLopt to meet my needs and would be grateful if someone could explain what I'm doing wrong or failing to understand.
I wish to:
Minimise the value of some objective function myfunc(x); where
x must lie in the unit hypercube (just 2 dimensions in the example below); and
the sum of the elements of x must be one.
Below I make myfunc very simple - the square of the distance from x to [2.0, 0.0] so that the obvious correct solution to the problem is x = [1.0,0.0] for which myfunc(x) = 1.0. I have also added println statements so that I can see what the solver is doing.
testNLopt = function()
origin = [2.0,0.0]
n = length(origin)
#Returns square of the distance between x and "origin", and amends grad in-place
myfunc = function(x::Vector{Float64}, grad::Vector{Float64})
if length(grad) > 0
grad = 2 .* (x .- origin)
end
xOut = sum((x .- origin).^2)
println("myfunc: x = $x; myfunc(x) = $xOut; ∂myfunc/∂x = $grad")
return(xOut)
end
#Constrain the sums of the x's to be 1...
sumconstraint =function(x::Vector{Float64}, grad::Vector{Float64})
if length(grad) > 0
grad = ones(length(x))
end
xOut = sum(x) - 1
println("sumconstraint: x = $x; constraint = $xOut; ∂constraint/∂x = $grad")
return(xOut)
end
opt = Opt(:LD_SLSQP,n)
lower_bounds!(opt, zeros(n))
upper_bounds!(opt,ones(n))
equality_constraint!(opt,sumconstraint,0)
#xtol_rel!(opt,1e-4)
xtol_abs!(opt,1e-8)
min_objective!(opt, myfunc)
maxeval!(opt,20)#to ensure code always terminates, remove this line when code working correctly?
optimize(opt,ones(n)./n)
end
I have read this similar question and documentation here and here, but still can't figure out what's wrong. Worryingly, each time I run testNLopt I see different behaviour, as in this screenshot including occasions when the solver uselessly evaluates myfunc([NaN,NaN]) many times.
You aren't actually writing to the grad parameters in-place, as you write in the comments;
grad = 2 .* (x .- origin)
just overrides the local variable, not the array contents -- and I guess that's why you see these df/dx = [NaN, NaN] everywhere. The simplest way to fix that would be with broadcasting assignment (note the dot):
grad .= 2 .* (x .- origin)
and so on. You can read about that behaviour here and here.
I am a newbie to programming .here I have been solving a simple problem in functional programming (OZ) which is finding the sum of the Digits of a 6 digit positive integer.
Example:- if n = 123456 then
output = 1+2+3+4+5+6 which is 21.
here I found a solution like below
fun {SumDigits6 N}
{SumDigits (N Div 1000) + SumDigits (N mod 1000)}
end
and it says the argument (N Div 1000) gives first 3 digits and the argument (N mod 1000) gives us last 3 digits. and yes I getting the correct solution but my Doubt is how could they giving correct solutions. I mean in given example isn't (N Div 1000) of 123456 gives 123 right not 1+2+3 and similarly (N mod 1000) of123456 gives us 456 not 4+5+6 right ?. in that case, the answer should be 123+456 which is equals to 579 not 21 right ? what Iam missing here.I apologize for asking such simple question but any help would be appreciated.
Thank you :)
You are missing the most important thing here.
It is supposed to happen in a loop and each time the value of N changes.
For example
in the first iteration
the Div gives 1 and mod gives 6 so you add 1 and 6 and store the result and the number is also modified (it becomes 2345)
in the second iteration
the div gives 2 and mod gives 5 you add 2+5+previous result and the number is also modified..
This goes on till the number becomes zero
Your function is recursive, so every time the number get smaller, untill is just 0, then it goes back summing all the partial result. You can do it with an accumulator to store the result, in this simple way:
declare
fun {SumDigit N Accumulator}
if N==0 then Accumulator
else {SumDigit (N div 10) Accumulator+(N mod 10)}
end
end
{Browse {SumDigit 123456 0}}
i think the most elegant way is the function --
static int SumOfDigit(int n)
{
if (n < 10) return n;
return SumOfDigit(SumOfDigit(n/10)+n%10);
}
simple and true :-)
int main()
{
int n,m,d,s=0;
scanf("%d",&n);
m=n;
while(m!=0)
{
d=m%10;
s=s+d;
m=m/10;
}
printf("Sum of digits of %d is %d",n,s);
}
I want to find the key corresponding to the min or max value of a dictionary in julia. In Python I would to the following:
my_dict = {1:20, 2:10}
min(my_dict, my_dict.get)
Which would return the key 2.
How can I do the same in julia ?
my_dict = Dict(1=>20, 2=>10)
minimum(my_dict)
The latter returns 1=>20 instead of 2=>10 or 2.
You could use reduce like this, which will return the key of the first smallest value in d:
reduce((x, y) -> d[x] ≤ d[y] ? x : y, keys(d))
This only works for non-empty Dicts, though. (But the notion of the “key of the minimal value of no values” does not really make sense, so that case should usually be handled seperately anyway.)
Edit regarding efficiency.
Consider these definitions (none of which handle empty collections)...
m1(d) = reduce((x, y) -> d[x] ≤ d[y] ? x : y, keys(d))
m2(d) = collect(keys(d))[indmin(collect(values(d)))]
function m3(d)
minindex(x, y) = d[x] ≤ d[y] ? x : y
reduce(minindex, keys(d))
end
function m4(d)
minkey, minvalue = next(d, start(d))[1]
for (key, value) in d
if value < minvalue
minkey = key
minvalue = value
end
end
minkey
end
...along with this code:
function benchmark(n)
d = Dict{Int, Int}(1 => 1)
m1(d); m2(d); m3(d); m4(d); m5(d)
while length(d) < n
setindex!(d, rand(-n:n), rand(-n:n))
end
#time m1(d)
#time m2(d)
#time m3(d)
#time m4(d)
end
Calling benchmark(10000000) will print something like this:
1.455388 seconds (30.00 M allocations: 457.748 MB, 4.30% gc time)
0.380472 seconds (6 allocations: 152.588 MB, 0.21% gc time)
0.982006 seconds (10.00 M allocations: 152.581 MB, 0.49% gc time)
0.204604 seconds
From this we can see that m2 (from user3580870's answer) is indeed faster than my original solution m1 by a factor of around 3 to 4, and also uses less memory. This is appearently due to the function call overhead, but also the fact that the λ expression in m1 is not optimized very well. We can alleviate the second problem by defining a helper function like in m3, which is better than m1, but not as good as m2.
However, m2 still allocates O(n) memory, which can be avoided: If you really need the efficiency, you should use an explicit loop like in m4, which allocates almost no memory and is also faster.
another option is:
collect(keys(d))[indmin(collect(values(d)))]
it depends on properties of keys and values iterators which are not guaranteed, but in fact work for Dicts (and are guaranteed for OrderedDicts). like the reduce answer, d must be non-empty.
why mention this, when the reduce, pretty much nails it? it is 3 to 4 times faster (at least on my computer) !
Here is another way to find Min with Key and Value
my_dict = Dict(1 => 20, 2 =>10)
findmin(my_dict) gives the output as below
(10, 2)
to get only key use
findmin(my_dict)[2]
to get only value use
findmin(my_dict)[1]
Hope this helps.
If you only need the minimum value, you can use
minimum(values(my_dict))
If you need the key as well, I don't know a built-in function to do so, but you can easily write it yourself for numeric keys and values:
function find_min_key{K,V}(d::Dict{K,V})
minkey = typemax(K)
minval = typemax(V)
for key in keys(d)
if d[key] < minval
minkey = key
minval = d[key]
end
end
minkey => minval
end
my_dict = Dict(1=>20, 2=>10)
find_min_key(my_dict)
findmax(dict)[2]
findmin(dict)[2]
Should also return the key corresponding to the max and min value(s). Here [2] is the index of the key in the returned tuple.
I am new to Prolog and am having some difficulty fixing the errors of my first program.
The program requirement is that it divides the 2 inputs using recursion, returning 0 if the dividend is larger than the divisor, and ignores remainders.
%Author: Justin Taylor
testquotient :-
repeat,
var(Divident), var(Divisor), var(Answer), var(End),
write('Enter Divident: '),
read(Divident),
write('Enter Divisor: '),
read(Divisor),
quotient(Divident, Divisor, Answer),
nl,
write('Quotient is = '),
write(Answer),
nl,
write('Enter 0 to quit, 1 to continue: '),
read(End),
(End =:= 0),!.
quotient(_, 0, 'Undefined').
quotient(0, _, 0).
quotient(Divisor == Divident -> Answer = 1).
quotient(Divisor < Divident -> Answer = 0).
quotient(Divident, Divisor, Answer) :-
(Divisor > Divident -> Divisor = Divisor - Divident,
quotient(Divident, Divisor, Answer + 1);
Answer = Answer).
First, read up on is. Type help(is). at the SWI-Prolog's prompt. Read the whole section about "Arithmetic" carefully. Second, your first few clauses for quotient are completely off-base, invalid syntax. I'll show you how to rewrite one of them, you'll have to do the other yourself:
%% WRONG: quotient(Divisor == Divident -> Answer = 1).
quotient(Divisor, Divident, Answer) :-
Divisor =:= Divident -> Answer = 1.
%% WRONG: quotient(Divisor < Divident -> Answer = 0).
....
Note the use of =:= instead of ==.
Your last clause for quotient looks almost right at the first glance, save for the major faux pas: prolog's unification, =, is not, repeat not, an assignment operator! We don't change values assigned to logical variables (if X is 5, what's there to change about it? It is what it is). No, instead we define new logical variable, like this
( Divisor > Divident -> NewDivisor = Divisor - Divident,
and we use it in the recursive call,
%% WRONG: quotient(Divident, NewDivisor, Answer + 1) ;
but this is wrong too, w.r.t. the new Answer. If you add 1 on your way down (as you subtract Divident from your Divisor - btw shouldn't it be the other way around?? check your logic or at least swap your names, "divisor" is what you divide by ) that means you should've supplied the initial value. But you seem to supply the terminal value as 0, and that means that you should build your result on your way back up from the depths of recursion:
%%not quite right yet
quotient(Divident, NewDivisor, NewAnswer), Answer = NewAnswer + 1 ;
Next, Answer = Answer succeeds always. We just write true in such cases.
Lastly, you really supposed to use is on each recursion step, and not just in the very end:
( Divisor > Divident -> NewDivisor is Divisor - Divident, %% use "is"
quotient(Divident, NewDivisor, NewAnswer), Answer is NewAnswer+1 %% use "is"
; true ). %% is this really necessary?
Your 'Undefined' will cause an error on 0, but leave it at that, for now. Also, you don't need to "declare" your vars in Prolog. The line var(Divident), ..., var(End), serves no purpose.