How to generate a multiplication table sequence? - f#

Question

I want to generate a sequence like a multiplication table. So for a start of 1 and a stop of 10 I am looking for a sequence like
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, // 1*1 - 1*10
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, // 2*1 - 2*10
3, 6, 9, 12, ... // 3*1 - 3*10
Here is my lame start at it, however I can't seem to figure out how to cleanly increment j when the stop is reached, or how to reset i back to the start.
let multable (start,stop) =
(start,start)
|> Seq.unfold(
fun (i,j) ->
Some(i*j, (i+1, j)))
let its = multable(1, 1)
let first10 = Seq.take 10 its
printf "%A" (Seq.to_list first10)
Which of course gives me 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

Answer 1

I can't really think of many cases where I'd prefer Seq.unfold over an equivalent list comprehension:
> let multiplication n m = [for a in 1 .. n -> [for b in 1 .. m -> a * b ] ];;
val multiplication : int -> int -> int list list
> multiplication 5 5;;
val it : int list list =
[[1; 2; 3; 4; 5]; [2; 4; 6; 8; 10]; [3; 6; 9; 12; 15]; [4; 8; 12; 16; 20];
[5; 10; 15; 20; 25]]
Occasionally the Array.init methods are useful too:
> let multiplication n m = Array2D.init n m (fun n m -> n * m);;
val multiplication : int -> int -> int [,]
> multiplication 5 5;;
val it : int [,] = [[0; 0; 0; 0; 0]
[0; 1; 2; 3; 4]
[0; 2; 4; 6; 8]
[0; 3; 6; 9; 12]
[0; 4; 8; 12; 16]]

Answer 2

Use a sequence expression:
let multable start stop = seq{
for a = start to stop do
for b = start to stop do
yield a*b
}
Output:
> multable 1 3 |> Seq.to_list;;
val it : int list = [1; 2; 3; 2; 4; 6; 3; 6; 9]
It's weird to represent a fundamentally 2d structure in this fashion. Why not a sequence of sequences:
let multable2 start stop = seq {
for a = start to stop do yield seq {
for b = start to stop do
yield a*b
}
}
Output:
val multable2 : int -> int -> seq<seq<int>>
> multable2 1 3 |> Seq.to_list;;
val it : seq<int> list = [seq [1; 2; 3]; seq [2; 4; 6]; seq [3; 6; 9]]
If you want to be "clever" and avoid multiplication:
let multable4 start stop = seq {
for a = start to stop do yield seq {
let s = ref 0 in
for b = start to stop do
s:=!s+a
yield !s
}
}
I don't actually see any nice prepackaged "sequence from a to b" outside of comprehensions/sequence expressions, although there's obviously [a..b] (list) and [|a..b|] (array) which you can project through Seq.unfold, Seq.map, etc. to make a Seq.

Answer 3

let Table r c =
[for row in 1..r do
yield [for col in 1..c do
yield row * col]]
printfn "%A" (Table 5 4)
// [[1; 2; 3; 4]; [2; 4; 6; 8]; [3; 6; 9; 12]; [4; 8; 12; 16]; [5; 10; 15; 20]]

Answer 4

Here's another way to use sequences:
let rec fromXToY x y =
seq {
if x <= y
then yield x; yield! fromXToY (x + 1) y;
else ()
}
let scaledSequence factor items =
Seq.map (fun x -> factor * x) items
let mulTable x y =
let sequenceOfIntegersMultipliedByValue = (fun n -> scaledSequence n (fromXToY x y))
let seqOfSeqOfValues = Seq.map sequenceOfIntegersMultipliedByValue (fromXToY x y)
// Convert the sequence of sequences to a simple sequence of values
Seq.fold Seq.append Seq.empty seqOfSeqOfValues