Multiplicative undelta

Question

Let us define the "multiplicative deltas" of values \$[\; a_0, \cdots a_N, \;]\$ as:

$$ [\; a_1 / a_0, \cdots, a_{i+1} / a_i, \cdots, a_N / a_{N-1} \;] $$

The reverse operation - namely "multiplicative undelta" - returns values such that the above operation results in the given values.

Example

Given values \$[\; 1, 5, 3, 2 \;]\$ a general solution to the "multiplicative undelta" operation is:

$$ [\; a_0, \quad \underset{a_1}{\underbrace{a_0 \cdot 1}}, \quad \underset{a_2}{\underbrace{a_1 \cdot 5}}, \quad \underset{a_3}{\underbrace{a_2 \cdot 3}}, \quad \underset{a_4}{\underbrace{a_3 \cdot 2}} \;] $$

A particular solution can be obtained by setting \$a_0\$ to any value other than zero, for example by setting \$a_0 := 1 \$ we would get:

$$ [\; 1, 1, 5, 15, 30 \;] $$

Challenge

Your task for this challenge is to implement the operation "multiplicative undelta" as defined above.

Rules

Inputs are:

a non-zero value \$a_0\$
a non-empty list/array/vector/... of non-zero "multiplicative deltas"

Output is a list/array/vector/... of values such that the first element is \$a_0\$ and for which the "multiplicative deltas" are the input.

Note: If your language has no support of negative integers you may replace non-zero by positive.

Test cases

2 [21] -> [2,42]
1 [1,5,3,2] -> [1,1,5,15,30]
-1 [1,5,3,2] -> [-1,-1,-5,-15,-30]
7 [1,-5,3,2] -> [7,7,-35,-105,-210]
2 [-12,3,-17,1311] -> [2,-24,-72,1224,1604664]
-12 [7,-1,-12,4] -> [-12,-84,84,-1008,-4032]
1 [2,2,2,2,2,2,2,2] -> [1,2,4,8,16,32,64,128,256]

May we take a single list of length 1+number of deltas, where the first item is a₀? — Adám, Oct 9, 2018 at 8:08
@Adám: I'm gonna say no to that because it would change things for existing answers. — ბიმო, Oct 9, 2018 at 9:15

nimi · Accepted Answer · 2018-10-06 16:04:48Z

18

Haskell, 8 bytes

scanl(*)

Try it online!

answered Oct 6, 2018 at 16:04

nimi

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\$\begingroup\$ I was checking if anyone wrote this, and I thought noone wrote it and I said "cool, then I'll write" and then saw your answer at the bottom of the page. upvoted. \$\endgroup\$
– Windmill Cookies
Oct 7, 2018 at 7:06

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Quintec · Accepted Answer · 2018-10-06 18:41:23Z

9

APL(Dyalog), 3 bytes

×\∊

Try it online!

If I have to take the number on the left and the array on the right:

-2 thanks to @H.PWiz

7 5 3 bytes

×\,

edited Oct 6, 2018 at 18:41

answered Oct 6, 2018 at 16:20

Quintec

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2

\$\begingroup\$ ⊣,⊢ can just be , \$\endgroup\$
– H.PWiz
Oct 6, 2018 at 17:52
\$\begingroup\$ @H.PWiz thanks, my tacit is extremely weak \$\endgroup\$
– Quintec
Oct 6, 2018 at 17:54
\$\begingroup\$ You don't need the parens; ×\, evaluates to a function. \$\endgroup\$
– Dennis
Oct 6, 2018 at 17:54
\$\begingroup\$ @Dennis Doesn't seem to work unfortunately \$\endgroup\$
– Quintec
Oct 6, 2018 at 17:56
\$\begingroup\$ You don't need to count the parens, because they're not needed in some situations. f←×\ works, for example. Btw, the (-12 3 -17 1311) in your permalink should be (¯12 3 ¯17 1311). \$\endgroup\$
– Dennis
Oct 6, 2018 at 18:27

| Show 4 more comments

Kirill L. · Accepted Answer · 2018-10-06 16:55:13Z

8

R, 15 bytes

cumprod(scan())

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Full program. Function is longer (unless we were allowed to "glue" the inputs together, so that built-in cumprod would suffice as a complete answer):

R, 28 bytes

function(i,x)cumprod(c(i,x))

Try it online!

edited Oct 6, 2018 at 16:55

answered Oct 6, 2018 at 16:50

Kirill L.

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Giuseppe · Accepted Answer · 2018-10-10 15:43:04Z

6

MATL, 3 bytes

hYp

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            #implicit input, x_0 and A
h           #horizontally concatenate
Yp          #cumulative product
            #implicit output

The hYpe about MATL is real.

edited Oct 10, 2018 at 15:43

answered Oct 6, 2018 at 16:48

Giuseppe

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Quintec · Accepted Answer · 2018-10-07 14:49:12Z

4

Python 3, 39 bytes

def f(a,b):
    for x in[1]+b:a*=x;yield a

Alternative approach. Returns a generator.

Try it online!

answered Oct 7, 2018 at 14:49

Quintec

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Misha Lavrov · Accepted Answer · 2018-10-06 16:14:52Z

3

Wolfram Language (Mathematica), 14 bytes

FoldList@Times

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FoldList[Times, a0, {x1, x2, ..., xn}] produces the desired output. FoldList[Times] is the curried form that produces a pure function still waiting for its a0 and {x1, x2, ..., xn}.

answered Oct 6, 2018 at 16:14

Misha Lavrov

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Community · Accepted Answer · 2020-06-17 09:04:33Z

3

JavaScript (ES6), 27 bytes

Takes input as (m)(a).

m=>a=>[m,...a.map(x=>m*=x)]

Try it online!

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Oct 6, 2018 at 16:16

Arnauld

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Community · Accepted Answer · 2020-06-17 09:04:33Z

3

Japt, 3 bytes

å*V

Try it

Explanation

        :Implicit input of array U and integer V
å       :Cumulatively reduce U
 *      :By multiplication
  V     :With an initial value of V

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Oct 6, 2018 at 19:20

Shaggy

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Erik the Outgolfer · Accepted Answer · 2018-10-06 16:05:13Z

2

Jelly, 3 bytes

;×\

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answered Oct 6, 2018 at 16:05

Erik the Outgolfer

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nwellnhof · Accepted Answer · 2018-10-06 16:35:46Z

2

Perl 6, 9 bytes

{[\*] @_}

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answered Oct 6, 2018 at 16:35

nwellnhof

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Laikoni · Accepted Answer · 2018-10-06 22:28:52Z

2

Standard ML, 32 bytes

fun f(x::r)a=a::f r(a*x)|f&a=[a]

Try it online!

Ungolfed:

fun f (x::xr) a = a :: f xr (a*x)
  | f    _    a = [a]

answered Oct 6, 2018 at 22:28

Laikoni

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Galen Ivanov · Accepted Answer · 2018-10-12 08:56:10Z

2

J, 6 5 bytes

-1 byte thanks to Bubbler

*/\@,

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Explanation:

A diadic verb, the left argument is a0, the rigth one - the vector

     @, - append the vector to a0 and 
  */\   - find the running product

edited Oct 12, 2018 at 8:56

answered Oct 7, 2018 at 8:05

Galen Ivanov

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1

\$\begingroup\$ */\@, works exactly the same. \$\endgroup\$
– Bubbler
Oct 11, 2018 at 23:02
\$\begingroup\$ @Bubbler Yes, of course! Thank you! \$\endgroup\$
– Galen Ivanov
Oct 12, 2018 at 3:19

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chunes · Accepted Answer · 2022-05-07 06:40:41Z

2

Factor, 22 bytes

[ prefix cum-product ]

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answered May 7, 2022 at 6:40

chunes

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Neil · Accepted Answer · 2018-10-07 00:14:18Z

1

Batch, 69 bytes

@set/pa=
@echo %a%
@for %%d in (%*) do @set/aa*=%%d&call echo %%a%%

Takes input of \$a_0\$ on STDIN and the deltas as command-line arguments.

answered Oct 7, 2018 at 0:14

Neil

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Renzo · Accepted Answer · 2018-10-07 08:32:23Z

1

Common Lisp, 67 bytes

(lambda(a l &aux(y 1))(mapcar(lambda(x)(setf y(* y x)))(cons a l)))

Try it online!

answered Oct 7, 2018 at 8:32

Renzo

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Leo · Accepted Answer · 2018-10-08 03:50:34Z

1

Husk, 2 bytes

G*

Try it online!

This is equivalent to nimi's answer in Haskell: scanl(*), which means reduce from the left using multiplication, and return all partial results.

answered Oct 8, 2018 at 3:50

Leo

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

05AB1E, 5 3 bytes

šηP

-2 bytes thanks to @BMO.

Try it online or verify all test cases.

Explanation:

š      # Prepend the (implicit) input-integer at the start of the (implicit) input-list
       #  i.e. -12 and [7,-1,-12,4] → ["-12",7,-1,-12,4]
 η     # Prefixes of this new list
       #  i.e. ["-12",7,-1,-12,4]
       #   → [["-12"],["-12",7],["-12",7,-1],["-12",7,-1,-12],["-12",7,-1,-12,4]]
  P    # Take the product of each inner list (and output implicitly)
       #  i.e. [["-12"],["-12",7],["-12",7,-1],["-12",7,-1,-12],["-12",7,-1,-12,4]]
       #   → [-12,-84,84,-1008,-4032]

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Oct 8, 2018 at 6:56

Kevin Cruijssen

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\$\begingroup\$ No, the s was there to get the other input. :) s is swap, and š is prepending at the start of the list. Either way, thanks for the -2 \$\endgroup\$
– Kevin Cruijssen
Oct 8, 2018 at 14:42

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Pyth, 6 bytes

*FR._s

Test that one here!

Alternatively, 7 bytes:

.u*NYEQ

Test it here!

The first takes input as a tuple, the second takes input as two separate lines.

Thanks to @Sok for helping me with getting good at mapping and saving 1 byte.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Oct 7, 2018 at 7:59

Steven H.

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\$\begingroup\$ You can save a byte by implementing the map in the second solution as R, as in *FR._s - demonstration \$\endgroup\$
– Sok
Oct 8, 2018 at 10:27
\$\begingroup\$ @Sok Nice catch! I had tried M assuming it'd work, then got a sugar parse failure - mostly because I didn't remember how the parsing for F <pf2> worked. \$\endgroup\$
– Steven H.
Oct 9, 2018 at 1:51
\$\begingroup\$ I'll be honest, I don't really understand how nested M, F, L, and R work, I just tried a bunch until one worked :o) \$\endgroup\$
– Sok
Oct 9, 2018 at 7:23

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Steven H. · Accepted Answer · 2018-10-10 14:20:25Z

1

Python 2, 40 bytes

f=lambda a,b:[a]+(b and f(a*b[0],b[1:]))

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Surprisingly, the move to Python 3 and use of generators there only saves 1 byte over the recursive solution.

answered Oct 10, 2018 at 14:20

Steven H.

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Veskah · Accepted Answer · 2018-10-11 23:21:17Z

1

PowerShell, 29 bytes

param($a,$b)$a;$b|%{($a*=$_)}

Try it online!

This assumes just outputting the values is fine.

> .\scratch.ps1 1 (1,5,3,2)
1
1
5
15
30

If that's not fine, this actually builds the list and then pushes it to toString which prints the same way.

param($a,$b)$c=,$a;$b|%{$c+=$_*$c[-1]};$c #41 bytes

edited Oct 11, 2018 at 23:21

answered Oct 11, 2018 at 22:58

Veskah

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\$\begingroup\$ Rules are: Output is a list/array/vector/. So, the first is fine. \$\endgroup\$
– mazzy
Oct 12, 2018 at 16:00

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Jo King · Accepted Answer · 2018-11-28 01:08:17Z

1

MathGolf, 6 5 bytes

\{\o*

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~~I think this could be 5 bytes (\{\o*) but \ instruction seems a little off when dealing with input.~~ This is now fixed in the latest version.

Explanation:

\       Swap arguments, pushing both to stack
 {      Foreach loop over second argument
  \o    Output counter with newline
    *   Multiply counter by current element
        Implicitly output the last element

edited Nov 28, 2018 at 1:08

answered Oct 7, 2018 at 0:33

Jo King♦

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\$\begingroup\$ I have redesigned the handling of `` with implicit input. It still handles items on the stack the same way, but now it pops two elements from the input onto the stack instead of just one. I think it could be useful for this challenge. \$\endgroup\$
– maxb
Nov 27, 2018 at 11:42

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Aiden Chow · Accepted Answer · 2022-05-07 06:15:04Z

1

Desmos, 47 bytes

f(k,l)=∏_{n=1}^{[1...l.length+1]}join(k,l)[n]

Try It On Desmos!

Try It On Desmos! - Prettified

answered May 7, 2022 at 6:15

Aiden Chow

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oeuf · Accepted Answer · 2022-05-07 10:06:11Z

1

K (ngn/k), 8 bytes

{x,x*\y}

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Another easier-than-I-thought problem. \ is the scan operator.

answered May 7, 2022 at 10:06

oeuf

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1

\$\begingroup\$ Shuffling thing around a bit, I think this can just be *\, \$\endgroup\$
– coltim
May 8, 2022 at 12:51

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Chas Brown · Accepted Answer · 2018-10-06 23:25:12Z

0

Python 2, 47 bytes

lambda n,A:reduce(lambda a,c:a+[a[-1]*c],A,[n])

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edited Oct 6, 2018 at 23:25

answered Oct 6, 2018 at 23:17

Chas Brown

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\$\begingroup\$ It may be of interest to you that a python solution is shorter in python 3 using generators \$\endgroup\$
– Quintec
Oct 7, 2018 at 23:34

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Neil · Accepted Answer · 2018-10-07 00:11:45Z

Charcoal, 12 bytes

ＩＥ⁺⟦Ｎ⟧ＡΠ⊞Ｏυι

Try it online! Link is to verbose version of code. Explanation:

    Ｎ           Input a₀
   ⟦ ⟧          Wrap in a list
      Ａ         Input deltas as a list
  ⁺              Concatenate lists
 Ｅ              Map over elements
           ι    Current element
          υ     Predefined empty list variable
        ⊞Ｏ      Push and return updated list
       Π        Product
Ｉ               Cast to string
                Implicitly print each value on its own line

Thaufeki · Accepted Answer · 2018-10-10 14:42:52Z

0

K (oK), 9 bytes

{(*\)x,y}

Try it online!

Joins the first number to the second input as a list, and then returns successive results of multiplication

Test Cases

Enter your input after the function like below and then run, as I'm unsure how to use input properly for this language in TiO

{(*\)x,y}[1;1 5 3 2]

answered Oct 10, 2018 at 14:42

Thaufeki

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eush77 · Accepted Answer · 2018-10-12 07:16:40Z

0

dc, 13 bytes

p[*pz1<A]dsAx

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p[*pz1<A]dsAx
p               # Print the first element
  *p            # Multiply top elements on the stack
 [  z1<A]dsAx   # until only one element is left

edited Oct 12, 2018 at 7:16

answered Oct 12, 2018 at 7:11

eush77

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Multiplicative undelta

Example

Challenge

Rules

Test cases

27 Answers 27

Haskell, 8 bytes

APL(Dyalog), 3 bytes

R, 15 bytes

R, 28 bytes

MATL, 3 bytes

Python 3, 39 bytes

Wolfram Language (Mathematica), 14 bytes

JavaScript (ES6), 27 bytes

Japt, 3 bytes

Explanation

Jelly, 3 bytes

Perl 6, 9 bytes

Standard ML, 32 bytes

J, 6 5 bytes

Explanation:

Factor, 22 bytes

Batch, 69 bytes

Common Lisp, 67 bytes

Husk, 2 bytes

05AB1E, 5 3 bytes

Pyth, 6 bytes

Python 2, 40 bytes

PowerShell, 29 bytes

MathGolf, 6 5 bytes

Explanation:

Desmos, 47 bytes

K (ngn/k), 8 bytes

Python 2, 47 bytes

Charcoal, 12 bytes

K (oK), 9 bytes

dc, 13 bytes

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