Generate lazy values

Question

Related: Program my microwave oven. Inspired by Generate lazy microwave input.

The lazy value of the non-negative integer N is the smallest of the integers that are closest to N while all their digits are identical.

Return (by any means) the lazy value of a given (by any means) N.

N ≤ (削除) the largest integer that your language represents in non-exponent form by default. (削除ここまで) 1000000 (A lot of interesting solutions are lost because of this too-high requirement.)

Test cases:

 0 → 0
 8 → 8
 9 → 9
 10 → 9
 16 → 11
 17 → 22
 27 → 22
 28 → 33
 100 → 99
 105 → 99
 106 → 111
 610 → 555
 611 → 666
7221 → 6666
7222 → 7777 

---

The colleague in question proved that there will be no ties: Except for 9/11, 99/111, etc. for which one is shorter than the other, two consecutive valid answers are always an odd distance apart, so no integer can be exactly equidistant from them.

Answer 1

JavaScript (ES6), 31 bytes

n=>~-(n*9+4).toPrecision(1)/9|0

Directly computes the lazy value for each n.

Edit: Only works up to 277777778 due to the limitations of JavaScript's integer type. Alternative versions:

n=>((n*9+4).toPrecision(1)-1)/9>>>0

35 bytes, works up to 16666666667.

n=>((n=(n*9+4).toPrecision(1))-n[0])/9

38 bytes, works up to 944444444444443. But that's still some way short of 2⁵³ which is 9007199254740992.

• Neil: @user81655 I've added some alternative versions with their numeric limitations.
• user81655: I couldn't get this algorithm to work with Number.MAX_SAFE_INTEGER either because 8e16 - 1 is expressed as 8e16. Sadly, it looks like the only way would be hard-coding the maximum result. +1 nonetheless.
• Adám: @user81655 I lowered the upper bound to allow the solution.
• NiCk Newman: Got you to 10k @Neil, love the golfs!
• Neil: @NiCkNewman Woohoo! Thanks!

Answer 2

Jelly, 16 bytes

ḤRμDIASμÐḟμạ3ỤḢị

Try it online!

How it works

ḤRμDIASμÐḟμạ3ỤḢị Main link. Input: n
Ḥ Compute 2n.
 R Yield [1, ..., 2n] or [0].
 μ Begin a new, monadic chain. Argument: R (range)
 D Convert to base 10.
 I Compute all differences of consecutive decimal digits.
 A Take the absolute values of the differences.
 S Sum the absolute values.
 μÐḟ Filter-false by the chain to the left.
 μ Begin a new, monadic chain. Argument: L (lazy integers)
 ạ3 Take the absolute difference of each lazy integer and n (input).
 Ụ Grade up; sort the indices of L by the absolute differences.
 This is stable, so ties are broken by earlier occurrence and,
 therefore, lower value.
 Ḣ Head; retrieve the first index, corresponding to the lowest
 absolute difference.
 ị Retrieve the item of L at that index.

Answer 3

Oracle SQL 11.2, 200 bytes

WITH v(i)AS(SELECT 0 FROM DUAL UNION ALL SELECT DECODE(SIGN(i),0,-1,-1,-i,-i-1)FROM v WHERE LENGTH(REGEXP_REPLACE(:1+i,'([0-9])1円+','1円'))>1)SELECT:1+MIN(i)KEEP(DENSE_RANK LAST ORDER BY rownum)FROM v;

Un-golfed

WITH v(i) AS
(
 SELECT 0 FROM DUAL -- Starts with 0
 UNION ALL
 SELECT DECODE(SIGN(i),0,-1,-1,-i,-i-1) -- Increments i, alternating between negatives and positives
 FROM v 
 WHERE LENGTH(REGEXP_REPLACE(:1+i,'([0-9])1円+','1円'))>1 -- Stop when the numbers is composed of only one digit
)
SELECT :1+MIN(i)KEEP(DENSE_RANK LAST ORDER BY rownum) FROM v;

Answer 4

Pyth - 26 bytes

(削除) This answer doesn't always return the smallest value in a tie, but that isn't in the specs, so awaiting clarification (削除ここまで) fixed for 3 bytes.

hSh.g.a-kQsmsM*RdjkUTtBl`Q

Test Suite.

Answer 5

Pyth, 16 bytes

haDQsM*M*`MTSl`Q

Try it online: Demonstration or Test Suite

Explanation:

haDQsM*M*`MTSl`Q implicit: Q = input number
 `Q convert Q to a string
 l take the length
 S create the list [1, 2, ..., len(str(Q))]
 `MT create the list ["0", "1", "2", "3", ..., "9"]
 * create every combination of these two lists:
 [[1, "0"], [1, "1"], [1, "2"], ..., [len(str(Q)), "9"]]
 *M repeat the second char of each pair according to the number:
 ["0", "1", "2", ..., "9...9"]
 sM convert each string to a number [0, 1, 2, ..., 9...9]
 D order these numbers by:
 a Q their absolute difference with Q
h print the first one

Answer 6

MATL, 25 bytes

2*:"@Vt!=?@]]N$vtG-|4#X<)

Uses brute force, so it may take a while for large numbers.

Try it online!

2*: % range [1,2,...,2*N], where is input
" % for each number in that range
 @V % push that number, convert to string
 t!= % test all pair-wise combinations of digits for equality
 ? % if they are all equal
 @ % push number: it's a valid candidate
 ] % end if
] % end for each
N$v % column array of all stack contents, that is, all candidate numbers
t % duplicate
G-| % absolute difference of each candidate with respect to input
4#X< % arg min
) % index into candidate array to obtain the minimizer. Implicitly display

Answer 7

Perl, 32

Based on the beautiful JavaScript solution by Neil.

$_=0|1/9*~-sprintf"%.e",$_*9+4.1

Starts to fail at 5e15

Answer 8

Mathematica, 122 bytes

f@x_:=Last@Sort[Flatten@Table[y*z,{y,1,9},{z,{FromDigits@Table[1,10~Log~x+1-Log[10,1055555]~Mod~1]}}],Abs[x-#]>Abs[x-#2]&]

Function named x.

Answer 9

JavaScript (ES6), 59 bytes

n=>eval(`for(i=a=0;i<=n;a=i%10?a:++i)p=i,i+=a;n-p>i-n?i:p`)

Recursive Solution (56 bytes)

This is a bit shorter but does not work for n > 1111111110 because the maximum call stack size is exceeded, so it is technically invalid.

f=(n,p,a,i=0)=>n<i?n-p>i-n?i:p:f(n,i,(i-=~a)%10?a:i++,i)

Explanation

Iterates through every lazy number until it gets to the first which is greater than n, then compares n to this and the previous number to determine the result.

var solution =
n=>
 eval(` // eval enables for loop without {} or return
 for(
 i=a=0; // initialise i and a to 0
 i<=n; // loop until i > n, '<=' saves having to declare p above
 a=i%10?a:++i // a = amount to increment i each iteration, if i % 10 == 0 (eg.
 ) // 99 + 11 = 110), increment i and set a to i (both become 111)
 p=i, // set p before incrementing i
 i+=a; // add the increment amount to i
 n-p>i-n?i:p // return the closer value of i or p
 `)

N = <input type="number" oninput="R.textContent=solution(+this.value)"><pre id="R"></pre>

• Adám: I lowered the upper bound to allow your solution.

Answer 10

Japt, 18 bytes

9*U+4 rApUs l1/9|0

Try it online!

_{Based on Neil's technique}

Non-competing solution:

*9+4 h /9|0

• ETHproductions: And now you can do *9+4 h /9|0 :-)
• Oliver: @ETHproductions Thanks! I'm having a lot of fun with Japt :)

Answer 11

05AB1E, 20 bytes

×ばつ}) ̃ïD1-ÄWQÏ{¬

Try it online!

9Ý # Push 0..9
 ×ばつ}) ̃ # For each digit, 0-9, push 1-7 copies of that number.
 ïD # Convert to integers, dupe the list.
 1 # Push original input (n).
 -Ä # Push absolute differences.
 WQ # Get min, push 1 for min indices.
 Ï{¬ # Push indices from original array that are the min, sort, take first.

• Adám: 99 is surely more lazy than 111, as it only requires two button presses.
• Magic Octopus Urn: @Adám fair enough, added head command.

Answer 12

Mathematica, 56 bytes

Min@Nearest[##&@@@Table[d(10^n-1)/9,{n,0,6},{d,0,9}],#]&

Pure function with first argument #, works for inputs up to 10^6.

For a nonnegative integer n and a digit d, 10^n-1 = 99...9 (9 repeated n times), so d(10^n-1)/9 = dd...d (d repeated n times). Creates a Table of values for 0 <= n <= 6 and 0 <= d <= 9, then flattens the table, finds the list of elements Nearest to # and takes the Min.

I believe this version will work for arbitrarily large integers:

Min@Nearest[##&@@@Table[d(10^n-1)/9,{n,0,IntegerLength@#},{d,0,9}],#]&