I'm learning smalltalk right now and am trying to make a very simple program that creates an array of numbers and then finds the largest number. My code looks like this:
| list max |
list := #(1 8 4 5 3).
1 to: list size do: [:i |
max < (list at: i)
ifTrue: [max := (list at: i)].
ifFalse: [max := max].
].
When I run this code, I get "stdin:7: parse error, expected ']'". I'm a bit confused as to what is causing this. It looks to me like all of my square brackets correspond. Help?
Alexandre told you already that is likely that collection provides a max method. You might be interested in a few ways how to do it.
Using collection max (maximum of all elements)
#(1 8 4 5 3) max
Using number max: (which of two numbers is bigger)
#(1 8 4 5 3) inject: 0 into: [:max :elem|
max max: elem ]
Or using just using the internal iterator
#(1 8 4 5 3) inject: 0 into: [:max :elem|
max < elem
ifTrue: [ elem ]
ifFalse:[ max ] ]
Together with your solution to use an external iteration you can see there are a lot of possibilities.
Hope it adds something
The line number seems off but looking at your code it seems the error is probably caused by the period after ifTrue: [max := (list at: i)]. #ifTrue:ifFalse is a single method selector and it doesn't make sense to break it into two statements.
Actually you could remove the ifFalse part of the code entirely. Assigning max to itself has no effect. Also looping on the indices is not needed here. You can work on the values directly with list do: […].
The max variable should also be initialized. Zero seems like a good choice to compare against the positive numbers in your array.
But instead of doing all that look into the Collection class. Your Smalltalk dialect may already offer a max method for this task.
Your immediate problem is that you terminate the ifTrue section with a period. You could normally just remove the period but, since your ifFalse section is effectively a non-operation, it's probably better to remove that bit.
But even when you fix that, you still need to initialise max so that you don't get a < message being sent to the nil object. You can initialise it to the first element, if there are any, then you can change the loop to start at the second.
Of course, initialising to the first element is problematic when the list is empty so you should handle that as well. In the code below, I've set it to a suitable small value then initialised it from the first element of the list only if it's available.
Corrected code is:
| list max |
list := #(1 8 4 5 3).
max := -99999.
(list size) > 0 ifTrue: [max := (list at: 1)].
2 to: list size do: [:i |
max < (list at: i) ifTrue: [max := (list at: i)].
]
(max) displayNl
This outputs 8 as expected and also works fine on the edge cases (list size of zero and one).
Related
I'm working on this weeks PerlWChallenge.
You are given an array of integers #A. Write a script to create an
array that represents the smaller element to the left of each
corresponding index. If none found then use 0.
Here's my approach:
my #A = (7, 8, 3, 12, 10);
my $L = #A.elems - 1;
say gather for 1 .. $L -> $i { take #A[ 0..$i-1 ].grep( * < #A[$i] ).min };
Which kinda works and outputs:
(7 Inf 3 3)
The Infinity obviously comes from the empty grep. Checking:
> raku -e "().min.say"
Inf
But why is the minimum of an empty Seq Infinity? If anything it should be -Infinity. Or zero?
It's probably a good idea to test for the empty sequence anyway.
I ended up using
take .min with #A[ 0..$i-1 ].grep( * < #A[$i] ) or 0
or
take ( #A[ 0..$i-1 ].grep( * < #A[$i] ) or 0 ).min
Generally, Inf works out quite well in the face of further operations. For example, consider a case where we have a list of lists, and we want to find the minimum across all of them. We can do this:
my #a = [3,1,3], [], [-5,10];
say #a>>.min.min
And it will just work, since (1, Inf, -5).min comes out as -5. Were min to instead have -Inf as its value, then it'd get this wrong. It will also behave reasonably in comparisons, e.g. if #a.min > #b.min { }; by contrast, an undefined value will warn.
TL;DR say min displays Inf.
min is, or at least behaves like, a reduction.
Per the doc for reduction of a List:
When the list contains no elements, an exception is thrown, unless &with is an operator with a known identity value (e.g., the identity value of infix:<+> is 0).
Per the doc for min:
a comparison Callable can be specified with the named argument :by
by is min's spelling of with.
To easily see the "identity value" of an operator/function, call it without any arguments:
say min # Inf
Imo the underlying issue here is one of many unsolved wide challenges of documenting Raku. Perhaps comments here in this SO about doc would best focus on the narrow topic of solving the problem just for min (and maybe max and minmax).
I think, there is inspiration from
infimum
(the greatest lower bound). Let we have the set of integers (or real
numbers) and add there the greatest element Inf and the lowest -Inf.
Then infimum of the empty set (as the subset of the previous set) is the
greatest element Inf. (Every element satisfies that is smaller than
any element of the empty set and Inf is the greatest element that
satisfies this.) Minimum and infimum of any nonempty finite set of real
numbers are equal.
Similarly, min in Raku works as infimum for some Range.
1 ^.. 10
andthen .min; #1
but 1 is not from 1 ^.. 10, so 1 is not minimum, but it is infimum
of the range.
It is useful for some algorithm, see the answer by Jonathan
Worthington or
q{3 1 3
-2
--
-5 10
}.lines
andthen .map: *.comb( /'-'?\d+/ )».Int # (3, 1, 3), (-2,), (), (-5, 10)
andthen .map: *.min # 1,-2,Inf,-5
andthen .produce: &[min]
andthen .fmt: '%2d',',' # 1,-2,-2,-5
this (from the docs) makes sense to me
method min(Range:D:)
Returns the start point of the range.
say (1..5).min; # OUTPUT: «1»
say (1^..^5).min; # OUTPUT: «1»
and I think the infinimum idea is quite a good mnemonic for the excludes case which also could be 5.1^.. , 5.0001^.. etc.
I am trying to rank these functions — 2n, n100, (n + 1)2, n·lg(n), 100n, n!, lg(n), and n99 + n98 — so that each function is the big-O of the next function, but I do not know a method of determining if one function is the big-O of another. I'd really appreciate if someone could explain how I would go about doing this.
Assuming you have some programming background. Say you have below code:
void SomeMethod(int x)
{
for(int i = 0; i< x; i++)
{
// Do Some Work
}
}
Notice that the loop runs for x iterations. Generalizing, we say that you will get the solution after N iterations (where N will be the value of x ex: number of items in array/input etc).
so This type of implementation/algorithm is said to have Time Complexity of Order of N written as O(n)
Similarly, a Nested For (2 Loops) is O(n-squared) => O(n^2)
If you have Binary decisions made and you reduce possibilities into halves and pick only one half for solution. Then complexity is O(log n)
Found this link to be interesting.
For: Himanshu
While the Link explains how log(base2)N complexity comes into picture very well, Lets me put the same in my words.
Suppose you have a Pre-Sorted List like:
1,2,3,4,5,6,7,8,9,10
Now, you have been asked to Find whether 10 exists in the list. The first solution that comes to mind is Loop through the list and Find it. Which means O(n). Can it be made better?
Approach 1:
As we know that List of already sorted in ascending order So:
Break list at center (say at 5).
Compare the value of Center (5) with the Search Value (10).
If Center Value == Search Value => Item Found
If Center < Search Value => Do above steps for Right Half of the List
If Center > Search Value => Do above steps for Left Half of the List
For this simple example we will find 10 after doing 3 or 4 breaks (at: 5 then 8 then 9) (depending on how you implement)
That means For N = 10 Items - Search time was 3 (or 4). Putting some mathematics over here;
2^3 + 2 = 10 for simplicity sake lets say
2^3 = 10 (nearly equals --- this is just to do simple Logarithms base 2)
This can be re-written as:
Log-Base-2 10 = 3 (again nearly)
We know 10 was number of items & 3 was the number of breaks/lookup we had to do to find item. It Becomes
log N = K
That is the Complexity of the alogorithm above. O(log N)
Generally when a loop is nested we multiply the values as O(outerloop max value * innerloop max value) n so on. egfor (i to n){ for(j to k){}} here meaning if youll say for i=1 j=1 to k i.e. 1 * k next i=2,j=1 to k so i.e. the O(max(i)*max(j)) implies O(n*k).. Further, if you want to find order you need to recall basic operations with logarithmic usage like O(n+n(addition)) <O(n*n(multiplication)) for log it minimizes the value in it saying O(log n) <O(n) <O(n+n(addition)) <O(n*n(multiplication)) and so on. By this way you can acheive with other functions as well.
Approach should be better first generalised the equation for calculating time complexity. liken! =n*(n-1)*(n-2)*..n-(n-1)so somewhere O(nk) would be generalised formated worst case complexity like this way you can compare if k=2 then O(nk) =O(n*n)
I am currently building my first stuff on Smalltalk and I have hit an issue. I have to deal with a user-entered number, and I need to div it by 2 and still be an integer. If an user inputs 10, I will work with 5, if they input 11, I have to work with 6, but I will obviously get 5.5.
If I could get the mod of a number I could simply make sure mod = 0 else add 0.5 and it would do just as good, but I just can't find how to make a mod operation in SmallTalk, all my searches end up in unrelated stuff about actual social smalltalk, which is extremely frustrating.
So if you could tell me how to get the mod of a number it would be great, if you could tell me how to round up with a separate function, even better. Thanks for your help and time beforehand.
UPDATE: After some research, I tried to do it this way:
mod := par rem: 2.
mod = 0 ifFalse: [ par := par + 0.5 ].
where as "mod" is mod of the variable "par", and if it isn't 0, it should add up 0.5 to par.
My issue now is that trying to use par in a timesRepeat brings up a "BoxedFloat64 did not understand #timesRepeat" error. So I am still in the same issue, or just need a way to make a float into an integer.
There are a lot of ways. For example
Add 1 to entered number before div by 2 if entered number is odd
temp := enteredNumber.
temp odd ifTrue: [temp := temp + 1 ].
^temp / 2
Using ceiling method
^(enteredNumber / 2) ceiling
In Smalltalk, we have an operator for integer division (and even two operators):
11 / 2
would answer a Fraction, not a whole Integer
But:
11 // 2
would answer the quotient of division, rounded toward negative infinity
And the corresponding remainder will be:
11 \\ 2
The second operator quo: for quotient and rem: for remainder
The difference is only with negative receiver/operand: the later ones are truncating the quotient toward zero.
-11 // 4 = -3. "floored toward negative infinity"
-11 \\ 4 = 1.
(-11 quo: 4) = -2. "truncated toward zero"
(-11 rem: 4) = -3.
If you want to round the quotient upper (toward positive infinity), then you can write:
(anInteger + 1) // 2.
Or same without parenthesis if you are confident enough in binary operator precedence:
anInteger + 1 // 2.
I've been trying to exercise my Perl 6 chops by looking at some golfing problems. One of them involved extracting the bits of an integer. I haven't been able to come up with a succinct way to write such an expression.
My "best" tries so far follow, using 2000 as the number. I don't care whether the most or least significant bit comes first.
A numeric expression:
map { $_ % 2 }, (2000, * div 2 ... * == 0)
A recursive anonymous subroutine:
{ $_ ?? ($_ % 2, |&?BLOCK($_ div 2)) !! () }(2000)
Converting to a string:
2000.fmt('%b') ~~ m:g/./
Of these, the first feels cleanest to me, but it would be really nice to be able to generate the bits in a single step, rather than mapping over an intermediate list.
Is there a cleaner, shorter, and/or more idiomatic way to get the bits, using a single expression? (That is, without writing a named function.)
The easiest way would be:
2000.base(2).comb
The .base method returns a string representation, and .comb splits it into characters - similar to your third method.
An imperative solution, least to most significant bit:
my $i = 2000; say (loop (; $i; $i +>= 1) { $i +& 1 })
The same thing rewritten using hyperoperators on a sequence:
say (2000, * +> 1 ...^ !*) >>+&>> 1
An alternative that is more useful when you need to change the base to anything above 36, is to use polymod with an infinite list of that base.
Most of the time you will have to reverse the order though.
say 2000.polymod(2 xx *);
# (0 0 0 0 1 0 1 1 1 1 1)
say 2000.polymod(2 xx *).reverse;
say [R,] 2000.polymod(2 xx*);
# (1 1 1 1 1 0 1 0 0 0 0)
As title says, I don't understand why f^:proposition^:_ y is a while loop. I have actually used it a couple times, but I don't understand how it works. I get that ^: repeats functions, but I'm confused by its double use in that statement.
I also can't understand why f^:proposition^:a: y works. This is the same as the previous one but returns the values from all the iterations, instead of only the last one as did the one above.
a: is an empty box and I get that has a special meaning used with ^: but even after having looked into the dictionary I couldn't understand it.
Thanks.
Excerpted and adapted from a longer writeup I posted to the J forums in 2009:
while =: ^:break_clause^:_
Here's an adverb you can apply to any code (which would equivalent of the
loop body) to create a while loop. In case you haven't seen it before, ^: is the power conjunction. More specifically, the phrase f^:n y applies the function f to the argument y exactly n times. The count n maybe be an integer or a function which applied to y produces an integer¹.
In the adverb above, we see the power conjunction twice, once in ^:break_clause and again in ^:_ . Let's first discuss the latter. That _ is J's notation for infinity. So, read literally, ^:_ is "apply the function an infinite number of times" or "keep reapplying forever". This is related to a while-loop's function, but it's not very useful if applied literally.
So, instead, ^:_ and its kin were defined to mean "apply a function to its limit", that is, "keep applying the function until its output matches its input". In that case, applying the function again would have no effect, because the next iteration would have the same input as the previous (remember that J is a functional language). So there's
no point in applying the function even once more: it has reached its limit.
For example:
cos=: 2&o. NB. Cosine function
pi =: 1p1 NB. J's notation for 1*pi^1 analogous to scientific notation 1e1
cos pi
_1
cos cos cos pi
0.857553
cos^:3 pi
0.857553
cos^:10 pi
0.731404
cos^:_ pi NB. Fixed point of cosine
0.739085
Here, we keep applying cosine until the answer stops changing: cosine has reached its fixed point, and more applications are superfluous. We can visualize this by showing the
intermediate steps:
cos^:a: pi
3.1415926535897 _1 0.54030230586813 ...73 more... 0.73908513321512 0.73908513321
So ^:_ applies a function to its limit. OK, what about ^:break_condition? Again, it's the same concept: apply the function on the left the number of times specified by the function on the right. In the case of _ (or its function-equivalent, _: ) the output is "infinity", in the case of break_condition the output will be 0 or 1 depending on the input (a break condition is boolean).
So if the input is "right" (i.e. processing is done), then the break_condition will be 0, whence loop_body^:break_condition^:_ will become loop_body^:0^:_ . Obviously, loop_body^:0 applies the loop_body zero times, which has no effect.
To "have no effect" is to leave the input untouched; put another way, it copies the input to the output ... but if the input matches the output, then the function has reached its limit! Obviously ^:_: detects this fact and terminates. Voila, a while loop!
¹ Yes, including zero and negative integers, and "an integer" should be more properly read as "an arbitrary array of integers" (so the function can be applied at more than one power simultaneously).
f^:proposition^:_ is not a while loop. It's (almost) a while loop when proposition returns 1 or 0. It's some strange kind of while loop when proposition returns other results.
Let's take a simple monadic case.
f =: +: NB. Double
v =: 20 > ] NB. y less than 20
(f^:v^:_) 0 NB. steady case
0
(f^:v^:_) 1 NB. (f^:1) y, until (v y) = 0
32
(f^:v^:_) 2
32
(f^:v^:_) 5
20
(f^:v^:_) 21 NB. (f^:0) y
21
This is what's happening: every time that v y is 1, (f^:1) y is executed. The result of (f^:1) y is the new y and so on.
If y stays the same for two times in a row → output y and stop.
If v y is 0→ output y and stop.
So f^:v^:_ here, works like double while less than 20 (or until the result doesn't change)
Let's see what happens when v returns 2/0 instead of 1/0.
v =: 2 * 20 > ]
(f^:v^:_) 0 NB. steady state
0
(f^:v^:_) 1 NB. (f^:2) 1 = 4 -> (f^:2) 4 = 16 -> (f^:2) 16 = 64 [ -> (f^:0) 64 ]
64
(f^:v^:_) 2 NB. (f^:2) 2 = 8 -> (f^:2) 8 = 32 [ -> (f^:0) 32 ]
32
(f^:v^:_) 5 NB. (f^:2) 5 = 20 [ -> (f^:0) 20 ]
20
(f^:v^:_) 21 NB. [ (f^:0) 21 ]
21
You can have many kinds of "strange" loops by playing with v. (It can even return negative integers, to use the inverse of f).