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digitalmars.D.learn - [BEGINNER] reccurence! and sequence!

reply helxi <brucewayneshit gmail.com> writes:
Can someone give me a very watered-down explanation of what 
std.range's recurrence! and sequence! do?

 auto tri = sequence!((a,n) => n*(n+1)/2)();
/** okay, it's a triangular number array * I understand n is the index number, the nth term * However where does this 'a' go? */
 auto odds = sequence!("a[0] + n * a[1]")(1, 2);
/** okay, this is a range of odd numbers / where and how do I plug (1, 2) into ""a[0] + n * a[1]"
 sequence!("n+2")(1).take(10).writeln;
//[2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
 recurrence!("n+2")(1).take(10).writeln;
//[1, 3, 4, 5, 6, 7, 8, 9, 10, 11]
Jun 26
parent reply ag0aep6g <anonymous example.com> writes:
On 06/26/2017 11:51 AM, helxi wrote:
 auto tri = sequence!((a,n) => n*(n+1)/2)();
/** okay, it's a triangular number array * I understand n is the index number, the nth term * However where does this 'a' go? */
`a` is a tuple of the run-time arguments you pass to `sequence`. In this example, no arguments are passed (empty parens at the end of the call), so `a` is empty.
 auto odds = sequence!("a[0] + n * a[1]")(1, 2);
/** okay, this is a range of odd numbers / where and how do I plug (1, 2) into ""a[0] + n * a[1]"
a[0] = 1 a[1] = 2
 sequence!("n+2")(1).take(10).writeln;
//[2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
`a` isn't used in the string lambda, and it's not considered an element of the range. So n starts at 0 and this just prints 0+2, 1+2, 2+2, etc.
 recurrence!("n+2")(1).take(10).writeln;
//[1, 3, 4, 5, 6, 7, 8, 9, 10, 11]
`a` is still not used in the string lambda, but `recurrence` uses the values in `a` as the first elements of the range. `n` is incremented accordingly (to 1), so this prints: 1 = a[0], 3 = (n = 1) + 2, 4 = (n = 2) + 2, etc. Another difference between `sequence` and `recurrence` is that `a` always refers to the same initial value(s) in `sequence`, while in `recurrence` it gets updated and refers to the previous element(s) of the range: ---- sequence!((a, n) => a[0] + 1)(1).take(10).writeln; // [2, 2, 2, 2, 2, 2, 2, 2, 2, 2] // because a[0] is always 1 recurrence!((a, n) => a[0] + 1)(1).take(10).writeln; // [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] // because a[0] refers to the previous value ----
Jun 26
parent reply helxi <brucewayneshit gmail.com> writes:
On Monday, 26 June 2017 at 10:34:22 UTC, ag0aep6g wrote:
 On 06/26/2017 11:51 AM, helxi wrote:
 [...]
`a` is a tuple of the run-time arguments you pass to `sequence`. In this example, no arguments are passed (empty parens at the end of the call), so `a` is empty.
 [...]
a[0] = 1 a[1] = 2
 [...]
`a` isn't used in the string lambda, and it's not considered an element of the range. So n starts at 0 and this just prints 0+2, 1+2, 2+2, etc.
 [...]
`a` is still not used in the string lambda, but `recurrence` uses the values in `a` as the first elements of the range. `n` is incremented accordingly (to 1), so this prints: 1 = a[0], 3 = (n = 1) + 2, 4 = (n = 2) + 2, etc. Another difference between `sequence` and `recurrence` is that `a` always refers to the same initial value(s) in `sequence`, while in `recurrence` it gets updated and refers to the previous element(s) of the range: ---- sequence!((a, n) => a[0] + 1)(1).take(10).writeln; // [2, 2, 2, 2, 2, 2, 2, 2, 2, 2] // because a[0] is always 1 recurrence!((a, n) => a[0] + 1)(1).take(10).writeln; // [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] // because a[0] refers to the previous value ----
Oh thank you. Just 2 follow-up questions:
 recurrence!((a, n) => a[0] + 1)(1).take(10).writeln;
1. In the last example of reccurence, what does n in (a,n) refer to? 2. How would you chain until! with reccurence? For example I want to compute 1, 10, 100, ..., (until the value remains smaller than 1000_000)?
Jul 05
parent reply =?UTF-8?Q?Ali_=c3=87ehreli?= <acehreli yahoo.com> writes:
On 07/05/2017 04:38 PM, helxi wrote:

 ----
 sequence!((a, n) => a[0] + 1)(1).take(10).writeln;
     // [2, 2, 2, 2, 2, 2, 2, 2, 2, 2]
     // because a[0] is always 1

 recurrence!((a, n) => a[0] + 1)(1).take(10).writeln;
     // [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
     // because a[0] refers to the previous value
 ----
Oh thank you. Just 2 follow-up questions:
 recurrence!((a, n) => a[0] + 1)(1).take(10).writeln;
1. In the last example of reccurence, what does n in (a,n) refer to?
n is "the index of the current value". Each time the lambda is called, a[n] is what is being generated a[n-1] is the previous value a[0] is the same as a[n-1]? (I find this confusing)
 2. How would you chain until! with reccurence? For example I want to
 compute 1, 10, 100, ..., (until the value remains smaller than 1000_000)?
import std.stdio; import std.algorithm; import std.range; void main() { auto r = recurrence!((a, n) => a[n-1] * 10)(1); auto u = r.until!(a => a >= 1_000_000); writeln(u); } [1, 10, 100, 1000, 10000, 100000] Ali
Jul 05
next sibling parent helxi <brucewayneshit gmail.com> writes:
On Thursday, 6 July 2017 at 00:21:44 UTC, Ali Çehreli wrote:
 On 07/05/2017 04:38 PM, helxi wrote:

 [...]
Oh thank you. Just 2 follow-up questions:
 [...]
1. In the last example of reccurence, what does n in (a,n)
refer to? n is "the index of the current value". Each time the lambda is called, a[n] is what is being generated a[n-1] is the previous value a[0] is the same as a[n-1]? (I find this confusing)
 2. How would you chain until! with reccurence? For example I
want to
 compute 1, 10, 100, ..., (until the value remains smaller
than 1000_000)? import std.stdio; import std.algorithm; import std.range; void main() { auto r = recurrence!((a, n) => a[n-1] * 10)(1); auto u = r.until!(a => a >= 1_000_000); writeln(u); } [1, 10, 100, 1000, 10000, 100000] Ali
Hmm, I get it now. Thank you very much.
Jul 06
prev sibling parent ag0aep6g <anonymous example.com> writes:
On 07/06/2017 02:21 AM, Ali Çehreli wrote:
 On 07/05/2017 04:38 PM, helxi wrote:
[...]
  >> recurrence!((a, n) => a[0] + 1)(1).take(10).writeln;
  > 1. In the last example of reccurence, what does n in (a,n) refer to?
 
 n is "the index of the current value". Each time the lambda is called,
 
    a[n] is what is being generated
    a[n-1] is the previous value
    a[0] is the same as a[n-1]? (I find this confusing)
Looks like I was wrong when I stated that "a[0] refers to the previous value". `a[0]` is actually invalid for n > 1. The documentation for `recurrence` says you have to index relative to n, and you can only go back as many values as you gave initially. I should have used `a[n - 1]`.
Jul 06