Participant number

Question

A math Olympiad will be held, and participants are being registered. The highest number of participants is 100. Each participant is given an ID number. It is given in a sequence like \$100, 97, 94, 91, 88, ...., 1\$, and when the first sequence is over, then \$99, 96, 93, 90, 87, ...., 3\$ sequence and so on.

Let's assume one of the participant's ID number is \$k\$ and he/she is the \$n^{th}\$ participant. Given the value of \$k\$, return the value of \$n\$.

Test Cases

59 -> 81
89 -> 71
16 -> 29
26 -> 92
63 -> 47
45 -> 53
91 -> 4
18 -> 62
19 -> 28

There will be no leading zero in input. Standard loopholes apply, shortest code wins. In output there can be trailing whitespace.

Can you specify how this sequence works? Also this would work better under standard sequence rules in my opinion. — Rydwolf Programs, Apr 21, 2021 at 2:32
These test cases are awful. They are extremely inconvenient for testing and there is a duplicate. Did you generate them randomly? — Rydwolf Programs, Apr 21, 2021 at 2:45
Just so we're absolutely clear here, from what I can infer the third sequence id the final one and goes 98,95,...,2? Because if not then you need to specify it in the description. — user100690, Apr 21, 2021 at 5:18
2 -> 100 should definitely be a test case (since it catches an edge-case if using a mod 100). — Jonathan Allan, Apr 21, 2021 at 16:27

xnor · Accepted Answer · 2021-04-21 05:25:43Z

8

Python 2, 23 bytes

lambda n:1-199*~n/3%~99

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I kind-of just tried stuff until I got something that worked.

answered Apr 21, 2021 at 5:25

xnor

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dingledooper · Accepted Answer · 2021-04-21 03:21:56Z

6

Python 2, 27 bytes

Some magic formula after a bit of trial and error.

lambda n:~-~n%3*33+34-~-n/3

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answered Apr 21, 2021 at 3:21

dingledooper

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Razetime · Accepted Answer · 2021-04-21 04:04:42Z

6

Husk, ¹¹ 9 bytes

€ΣTC3ṫ100

Try it online! or Verify all tests

-2 bytes, borrowing Unrelated String's idea.

Explanation

€ΣTC3ṫ100
     ṫ100 [100..1]
   C3     slices of 3
  T       transpose
 Σ        join
€         index of input

edited Apr 21, 2021 at 4:04

answered Apr 21, 2021 at 2:57

Razetime

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Unrelated String · Accepted Answer · 2021-04-21 03:54:39Z

5

Jelly, 9 bytes

ȷ2RUs3ZFi

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   U         Reverse
  R          the inclusive range from 1 to
ȷ2           100.
    s3       Split it into slices of length 3,
      Z      zip the slices,
       F     flatten the columns,
        i    and find the index of the input in the resulting list.

edited Apr 21, 2021 at 3:54

answered Apr 21, 2021 at 3:23

Unrelated String

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1

\$\begingroup\$ Right, idk how I didn't think of using zip. I was thinking it'd be convenient if there were a way to take each block of three and then go through them column-wise... \$\endgroup\$
– hyper-neutrino ♦
Apr 21, 2021 at 3:25

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Rydwolf Programs · Accepted Answer · 2021-04-21 12:26:40Z

5

JavaScript (V8), 33 20 bytes

Thanks to @tsh for -13

n=>101+~n%3*33-n/3|0

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Old:

n=>(n=100-n)%3*33+n/3+1+!!(n%3)|0

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I don't know why this works. Don't question it.

edited Apr 21, 2021 at 12:26

answered Apr 21, 2021 at 2:57

Rydwolf Programs

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\$\begingroup\$ Do you need |0 at the end? \$\endgroup\$
– radrow
Apr 21, 2021 at 6:06
1

\$\begingroup\$ After some math simplify: Maybe n=>101+~n%3*33-n/3|0 \$\endgroup\$
– tsh
Apr 21, 2021 at 9:44

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Jonah · Accepted Answer · 2021-04-21 04:12:16Z

4

J, ^{28 27 26} 24 bytes

(_,;(</.~3|])i._100)i.<:

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^{-2 bytes thanks to Bubbler}

edited Apr 21, 2021 at 4:12

answered Apr 21, 2021 at 2:39

Jonah

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1

\$\begingroup\$ 24 bytes using parenthesized noun expression. \$\endgroup\$
– Bubbler
Apr 21, 2021 at 4:08

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hakr14 · Accepted Answer · 2021-04-21 20:46:05Z

4

Desmos, 30 32 31 bytes

ceil(mod(67.1-101\frac n3,100))

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+2 bytes to account for edge case at 2
-1 byte thanks to Jonathan Allan!

edited Apr 21, 2021 at 20:46

answered Apr 21, 2021 at 5:14

hakr14

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\$\begingroup\$ @JonathanAllan well, darn. I've always thought there should be an equivalent to \$\mod n\$ where multiples of \$n\$ go to \$n\$ instead of \$0\$, but no such operation exists... this workaround will have to do. \$\endgroup\$
– hakr14
Apr 21, 2021 at 17:56
1

\$\begingroup\$ @JonathanAllan brilliant, thanks! \$\endgroup\$
– hakr14
Apr 21, 2021 at 18:36

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Joe Slater · Accepted Answer · 2021-04-22 08:52:18Z

3

Pari/GP, 23 bytes

p(n)=-n\3+(1-n)%3*33+35

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2 more bytes saved thanks to Dominic van Essen

edited Apr 22, 2021 at 8:52

answered Apr 21, 2021 at 4:42

Joe Slater

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\$\begingroup\$ Nice! Looks like you can swap -n+1 for n-1 to save 1 byte... \$\endgroup\$
– Dominic van Essen
Apr 21, 2021 at 16:20
\$\begingroup\$ No, I want a function f: f=(0,1,2) for n=(100,99,98). To do that I need to reverse the results given by n%3, making the results (2,0,1) instead of (1,0,2) and cycle them to the right, giving (0,1,2). But I clearly should have done (1-n) instead of (-n+1), is it too late to fix it? \$\endgroup\$
– Joe Slater
Apr 22, 2021 at 2:15
\$\begingroup\$ You're right: that was a typo by me. Of course not too late to fix it - many of us 'fix' our answers continuously (look at all the crossed-out previous byte scores)! Just click 'edit' below the post. If you like (there's no requirement), you can leave the old score of 26 crossed out by surrounding it with <s> & </s>... \$\endgroup\$
– Dominic van Essen
Apr 22, 2021 at 5:07
\$\begingroup\$ Done, thanks for the tip! \$\endgroup\$
– Joe Slater
Apr 22, 2021 at 6:09
\$\begingroup\$ Also: if you rearrange by pre-dividing 99 and 105 by 3, you'll save 2 more bytes (and end-up with the same formula that I used, but arriving by a different route!)... \$\endgroup\$
– Dominic van Essen
Apr 22, 2021 at 7:00

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hyper-neutrino · Accepted Answer · 2021-04-21 02:43:56Z

2

Jelly, 12 bytes

101ḶUẋ3m3ḟ0i

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answered Apr 21, 2021 at 2:43

hyper-neutrino♦

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M Virts · Accepted Answer · 2021-04-22 19:53:13Z

2

Python 3.8 (pre-release), 34 bytes

-8 bytes thanks to ovs!

lambda a:101-round(-~a%3*33.3+a/3)

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edited Apr 22, 2021 at 19:53

answered Apr 21, 2021 at 3:53

M Virts

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1

\$\begingroup\$ [1,2,0][a%3] can be golfed to -~a%3* and int(...+.5) is the same as round(...). 34 bytes \$\endgroup\$
– ovs
Apr 21, 2021 at 6:46

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Aaroneous Miller · Accepted Answer · 2021-04-28 04:56:26Z

2

Vyxal, ¹¹ 10 bytes

Thanks to @Razetime for -1 byte by writing 199 as ⁺b.

ꜝ⁺b*3ḭ₁N%⌐

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Vyxal port of @xnor's Python answer, with a bit of extra golfing.

Explanation:

            # Note: 'X' denotes current value.

            # Implicit input
ꜝ           # ( X + 1 ) * -1
 ⁺b*        # X * 199
    3ḭ      # X // 3
      ₁N%   # X % -100
         ⌐  # 1 - X
            # Implicit output
```

edited Apr 28, 2021 at 4:56

answered Apr 21, 2021 at 15:37

Aaroneous Miller

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\$\begingroup\$ 10 bytes using ⁺b for 199 \$\endgroup\$
– lyxal
Apr 28, 2021 at 4:40

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Neil · Accepted Answer · 2021-04-21 12:29:57Z

1

Charcoal, 18 bytes

Ｉ⊕÷﹪⁻²⁰¹×¹⁰¹Ｎ³⁰⁰¦³

Try it online! Link is to verbose version of code. Explanation: Implements the formula n=1+(201-101*k)%300/3, which I found through trial and error.

answered Apr 21, 2021 at 12:29

Neil

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Etheryte · Accepted Answer · 2021-04-21 13:31:49Z

1

Japt, 13 bytes

Lõ Ôó3 c aU Ä
Lõ            // Create the range [1..100],
   Ô          // reverse it and
    ó3        // group by every third item.
       c      // Flatten the result, then
         aU Ä // return the index of input plus one.

Try it here.

answered Apr 21, 2021 at 13:31

Etheryte

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Dominic van Essen · Accepted Answer · 2021-04-21 15:49:21Z

1

R, 30 bytes

k=scan();35+-k%/%3+33*(1-k)%%3

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There are already 12 other answers in various languages before I managed to post this, and - amazingly for what seems such a simple task - they mostly seem to use different formulas/strategies to each other, and all appear to be different to this one.

I suspect this means that if I go-through all the other approaches, I'll find that at least some of them will turn-out to be golfier than this one, though...

answered Apr 21, 2021 at 15:49

Dominic van Essen

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ZippyMagician · Accepted Answer · 2021-04-22 01:47:47Z

Arn `-IF`, 12 bytes

Found a few bugs while writing this, lol. This is 0-indexed (e.g. 59 -> 80), I'm not sure if that's okay so I can change the answer if it isn't (somebody asked in the comments but there was no response).

&»¨)¢►†3K1v&

Explained

Unpacked: e2.~::%3-1& .@

I see some answers explain like this:

  e2.~    Descending range from 100 to 1
  ::%3-1  Sort in chunks of 3
  & .@    Bind the transpose operator to the output

And others like this:

    e2   1 * 10 ^ 2 = 100
  .~     Descending range
::       Sort adjacently while the right is truthy
      _    The currently value
    %      Modulo
      3    Three
  -        Minus
    1      One
&        Bind
  .@     The transpose operation

So I might as well include both. Either way, there is an implicit :_ and :i surrounding the program. The space in & .@ is required because &. is a symbol in the language already. I admit the flags are a bit cheaty, so for anybody curious this would be 15 bytes without the two flags.

Kevin Cruijssen · Accepted Answer · 2021-05-27 08:30:39Z

05AB1E, 11 bytes

тLR3ôζ˜þIk>

Port of @UnrelatedString's Jelly answer.

Try it online or verify the entire sequence.

Alternatively, a port of @xnor's Python answer (which only works in the legacy version of 05AB1E due to the negative modulo) would be 11 bytes as well:

$±Ƶ$*3÷т(%-

Try it online or verify the entire sequence.

Explanation:

тLR          # Push a list in the range [100,1]
   3ô        # Split it into parts of size 3
     ζ       # Zip/transpose the list, using a space as filler
      ˜      # Flatten this list of triplets
       þ     # Remove the spaces by only leaving digits
        Ik   # Get the (0-based) index of the input in this list
          >  # Increase it by 1 to make it a 1-based index
             # (after which the result is output implicitly)

$            # Push 1 and the input
 ±           # Get the Bitwise-NOT of that: -input-1
  Ƶ$*        # Multiply it by 199
     3÷      # Integer-divide it by 3
       т(%   # Modulo -100
$         -  # Subtract it from the 1 we pushed at the start
             # (after which the result is output implicitly)

See this 05AB1E tip of mine (section How to compress large integers?) to understand why Ƶ$ is 199.

Stack Exchange Network

Participant number

16 Answers 16

Python 2, 23 bytes

Python 2, 27 bytes

Husk, ¹¹ 9 bytes

Explanation

Jelly, 9 bytes

JavaScript (V8), 33 20 bytes

J, ^{28 27 26} 24 bytes

Desmos, 30 32 31 bytes

Pari/GP, 23 bytes

Jelly, 12 bytes

Python 3.8 (pre-release), 34 bytes

Vyxal, ¹¹ 10 bytes

Charcoal, 18 bytes

Japt, 13 bytes

R, 30 bytes

Arn `-IF`, 12 bytes

Explained

05AB1E, 11 bytes

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Participant number

16 Answers 16

Python 2, 23 bytes

Python 2, 27 bytes

Husk, 11 9 bytes

Explanation

Jelly, 9 bytes

JavaScript (V8), 33 20 bytes

J, 28 27 26 24 bytes

Desmos, 30 32 31 bytes

Pari/GP, 23 bytes

Jelly, 12 bytes

Python 3.8 (pre-release), 34 bytes

Vyxal, 11 10 bytes

Charcoal, 18 bytes

Japt, 13 bytes

R, 30 bytes

Arn -IF, 12 bytes

Explained

05AB1E, 11 bytes

Your Answer

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Hot Network Questions

Husk, ¹¹ 9 bytes

J, ^{28 27 26} 24 bytes

Vyxal, ¹¹ 10 bytes

Arn `-IF`, 12 bytes

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math
or ask your own question.