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If I want to type the string aaa, the least keystrokes I can type it in is 3: a a a. But if I want to type the string aaaaaa, I can do it in 5: a a a ctrl-c ctrl-v, where the ctrl-c refers to copying aaa and the ctrl-v refers to pasting it.

Specifically, starting with an empty "buffer" and an empty "clipboard":

  • The keystroke a appends an a to the buffer.
  • ctrl-c takes some substring of the buffer and stores it into the clipboard. I'll notate it as ctrl-c(5) or similar to refer to 5 characters being stored. Only one string can be stored into the clipboard, and storing overwrites previous content.
  • ctrl-v appends the clipboard to the buffer.

Each of these counts as one keystroke.

With a larger example, the least keystrokes 17 as can be typed in is 8:

a a a ctrl-c(3) ctrl-v ctrl-v ctrl-c(8) ctrl-v

Your challenge is to, given a number n, return the number of keystrokes required to type n as. This is , shortest wins!

Testcases

These are done by hand, so tell me if any of these are wrong. Also, this doesn't appear to be on OEIS. I've written some not-quite-functional python code to find all possible outputs for a given length.

The first 30 terms of the sequence are:

1,2,3,4,5,5,6,6,6,7,7,7,8,8,8,8,8,8,9,9,9,9,9,9,10,9,10,10,10

And some more specific ones, with examples:

11 -> 7 (a a a ctrl-c(3) ctrl-v ctrl-c(5) ctrl-v)
17 -> 8 (a a a ctrl-c(3) ctrl-v ctrl-v ctrl-c(8) ctrl-v)
25 -> 9 (a a a ctrl-c(3) ctrl-v ctrl-v ctrl-c(8) ctrl-v ctrl-v, the python code doesn't find this one)
75 -> 12 (a a a ctrl-c(3) ctrl-v ctrl-v ctrl-c(9) ctrl-v ctrl-v ctrl-c(24) ctrl-v ctrl-v, python code also misses this one)
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5
  • 7
    \$\begingroup\$ Your exemple seem to imply that the mouse selection and clicking [at the end, to both unselect and choose a spot] do not count as keystrokes? My Vim blood boils ^^ \$\endgroup\$ Mar 14 at 12:26
  • 1
    \$\begingroup\$ I think this can be formulated as dynamic programming, where the state is the current length of string as well as the length in the buffer. \$\endgroup\$
    – qwr
    Mar 14 at 18:12
  • 2
    \$\begingroup\$ @OlivierDulac: yeah, seriously. If an arbitrary amount of mouse work is allowed, just middle-click to paste, or bring up a right-click context menu to copy and paste. I assumed when reading at first that the entire textbox was the "selection" copied by control-c, since that's the only thing plausible with just a keyboard without having hit any shift-arrow keys or even ctrl-a. Being able to select precisely 24 characters somehow, for the same time cost as hitting a or ctrl-v is total nonsense. This needs a new title and framing to make any sense as an application of this math problem. \$\endgroup\$ Mar 15 at 7:03
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    \$\begingroup\$ @PeterCordes this a simplified problem and not intended to be seen as any "application of [a] math problem." This is not any real world problem. \$\endgroup\$
    – Seggan
    Mar 15 at 12:49
  • 1
    \$\begingroup\$ @PeterCordes They probably don't allow using the autorepeat on the keyboard either. \$\endgroup\$ Mar 16 at 14:28

7 Answers 7

8
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JavaScript (Node.js), 47 bytes

  • -1 by Arnauld.
f=x=>x<6?x:Math.min(f(x+1>>1)+2,f(x-=x/3<<1)+3)

Try it online!


Python 2, 49 bytes

f=lambda x:x*(x<6)or min(f(x-x/2)+2,f(x-x/3*2)+3)

Try it online!

I don't really ensure this function is correct. But it at least pass all testcases given.

After I ask this solution on Math SE, caduk had provided a nice prove to it. You may read the prove by caduk on Math SE.

Anyway, the formula is

$$ f(x)= \begin{cases} n & n<6 \\ \min \left\{f\left(n-\left\lfloor\frac{n}{2}\right\rfloor\right)+2, f\left(n-2\left\lfloor\frac{n}{3}\right\rfloor\right)+3 \right\} & \text{otherwise} \end{cases} $$

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6
  • \$\begingroup\$ Is it possible that you type more than n-floor(n/2)c and copy, as the function isn't monotone? \$\endgroup\$
    – l4m2
    Mar 14 at 5:43
  • \$\begingroup\$ @l4m2 reasonable argue, but i cannot find a counterexample with n<2000 compared with the referenced implementation. So i have no idea how to either prove it or reject it. \$\endgroup\$
    – tsh
    Mar 14 at 6:00
  • 1
    \$\begingroup\$ I tested to 116M without counterexample \$\endgroup\$
    – l4m2
    Mar 14 at 7:23
  • \$\begingroup\$ If I understand the description correctly for 81 a's you could do: a a a c(3) v v c(9) v v c(27) v v for a total of 12 key strokes, way less then how I understand your algorithm which starts with typing either 41 or 27 a's first? \$\endgroup\$
    – quarague
    Mar 15 at 9:46
  • \$\begingroup\$ @quarague how can it be different from f(27)+3 in second part of min? \$\endgroup\$
    – tsh
    Mar 15 at 10:29
3
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JavaScript (Node.js), 68 bytes

f=(n,m,i=n>>1)=>i?Math.min(f(n-i,i)-~(i!=m),f(n,m,i-1),f(n-1,m)+1):n

Try it online!

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0
3
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Charcoal, 51 38 bytes

⊞υ⟦⁰⟧FN⊞υE⁺²ι⌊E⮌υ⌊Eμ⁺ξ⁺∨⁼π⊕ν⊕ν¬⁼πκI⌊⊟υ

Try it online! Link is to verbose version of code. Explanation: Uses dynamic programming on the basic recurrence relation, which may be equivalent to @l4m2's approach, but I can't read it.

⊞υ⟦⁰⟧

Start with 0 keystrokes to type 0 as.

FN

Loop over all numbers up to n.

⊞υE⁺²ι⌊E⮌υ⌊Eμ⁺ξ⁺∨⁼π⊕ν⊕ν¬⁼πκ

Loop over all clipboard sizes up to n and calculate the minimum number of keystrokes needed for the current number while finishing with that clipboard size. All previous entries are considered; 1 is added to the entry if it has a different clipboard size, as that means another ^C is needed, plus the number of extra as is added, unless the clipboard is of exactly the right size, in which case 1 is added, as in that case only a ^V is needed. (The values where more than one a is needed are unnecessary but it's golfier to calculate and discard them.)

I⌊⊟υ

Output the minimum number of keystrokes needed to output n in unary.

A port of @tsh's formula using dynamic programming takes 33 bytes:

F⊕N⊞υ⎇‹ι⁶ι⌊E²⁺⁺²κ§υ⁻ι×⊕κ÷ι⁺²κI⌊⊟υ

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

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2
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Swift 5.9, 48 bytes

let f={x in x<6 ?x:2+min(f(x-x/2),1+f(x-x/3*2))}

A direct port of @tsh's Python 2 answer, because I don't know how else to do this concisely.

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2
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Python3, 327 bytes

R=range
def f(n):
 q,s,r=[(0,0)],[],[]
 for a,c in q:
  if a==n:r+=[c];continue
  if r and c>=min(r):continue
  if a<3:q+=[(a+1,c+1)]
  else:
   U=[(a+1,c+1)]
   for i in R(3,min(a,n-a)+1):
    for j in R(1,(n-a)//i+1):U+=[(a+i*j,c+1+j)]
   for A,B in U:
    if all(k<A or K>B for k,K in s):q+=[(A,B)];s+=[(A,B)]
 return min(r)

Try it online!

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2
  • \$\begingroup\$ -6 if you change 0==any(k>=A and K<=B into all(k<A or K>B \$\endgroup\$
    – Vélimir
    Mar 14 at 23:38
  • \$\begingroup\$ @Vélimir Thank you, updated \$\endgroup\$
    – Ajax1234
    Mar 18 at 17:06
1
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Ruby, 42 bytes

f=->x{x<6?x:2-[-f[x-x/2],~f[x-x/3*2]].max}

Try it online!

Not very original, basically a port of tsh's answer.

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1
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Scala 3, 64 bytes

A direct port of @tsh's Python 2 answer.


def f(x:Int):Int=if(x<6)x else 2+Math.min(f(x-x/2),1+f(x-x/3*2))

Attempt This Online!

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