एरियर बचना - वहाँ से निकल जाना

Question

एक दिन आप केवल अपने आप को एक सरणी में पकड़े जाने के लिए जागते हैं। आप उस समय केवल एक इंडेक्स लेते हुए वहां से बाहर जाने की कोशिश करते हैं, लेकिन ऐसा लगता है कि अन्य नियम भी हैं:

सरणी पूरी तरह से प्राकृतिक संख्याओं से भरी हुई है।

• यदि आप एक सूचकांक पर खुद को पाते हैं n , तो आप सूचकांक array[n]को छोड़कर, को छोड़कर:
• यदि आप अपने आप को एक इंडेक्स पर पाते हैं nजो एक प्रमुख संख्या है, तो आप array[n]कदम वापस लेते हैं

उदाहरण: आप 4इस सरणी में सूचकांक पर शुरू करते हैं (प्रारंभ सूचकांक 0 है):

array = [1,4,5,6,8,10,14,15,2,2,4,5,7];
-----------------^ you are here

आप जिस क्षेत्र पर हैं 8, उसके मूल्य के रूप में , आप 8पहले चरण के रूप में सूचकांक में जाते हैं। आप जिस क्षेत्र पर उतरते हैं, उसमें मान होता है 2। फिर आप 2अपने दूसरे चरण के रूप में इंडेक्स पर जाते हैं । जैसा 2कि एक प्रमुख संख्या है, आप 5 कदम पीछे ले जाते हैं, जो कि आपका तीसरा चरण है। जैसा कि कोई सूचकांक नहीं है -3, आप कुल 3 चरणों में सरणी से सफलतापूर्वक बच गए।

आपका कार्य है:

एक प्रोग्राम या फ़ंक्शन लिखने के लिए, जो एक सरणी और स्टार्ट इंडेक्स को पैरामीटर के रूप में स्वीकार करता है, और सरणी से बचने के लिए चरणों की मात्रा को आउटपुट करता है। यदि आप सरणी से नहीं बच सकते (उदाहरण [2,0,2]के लिए स्टार्ट-इंडेक्स 2=> आप लगातार इंडेक्स से जाते हैं2 से जाते हैं 0), तो एक फाल्सी वैल्यू आउटपुट। आप एक-आधारित अनुक्रमण या शून्य-आधारित अनुक्रमण का उपयोग कर सकते हैं, लेकिन कृपया निर्दिष्ट करें कि आप किसका उपयोग करते हैं।

परीक्षण के मामलों

इनपुट: [2,5,6,8,1,2,3], 3

आउटपुट: 1

इनपुट: [2, 0, 2], 2

आउटपुट: false

इनपुट: [14,1,2,5,1,3,51,5,12,3,4,41,15,4,12,243,51,2,14,51,12,11], 5 ;

आउटपुट: 6

सबसे छोटा जवाब जीत जाता है।

Comments

Welcome to PPCG! This is a decent first challenge. :) Can we use 1-based indexing as well? Also it might be good to have a few more test cases. For future challenges you can also consider using the sandbox where you can get feedback from the community before a challenge goes live.

— Martin Ender

Very closely related

— Peter Taylor

@Martin Ender it's not related to the question... but me as a mobile user find it impossible to use the sandbox. What should I do to get a feedback on my questions before actually posting them?

— user6245072

@JerryJeremiah why can't you take 3 steps back? you'll land on index 2 if you start at 5 and take 3 steps back

— Michael Kunst

@user902383 going to index 2, which is prime, so we do 2 steps back and go to index 0, which is not prime. The value at index 0 is 2, so we go to index 2, which is prime ... repeat

— edc65

Answer 1

Pyth, 31 Bytes

KlQ%tl-.u?}NUK?P_N-N@QN@QNKQEKK

The test cases

It uses zero to indicate a false value, the number of hops otherwise.

Answer 2

Python, 161 138 bytes

Credits for factorial.

g=lambda x:0**x or x*g(x-1)
f=lambda a,i,n=0,l=[]:(i<0)+(i>=len(a))and n or(0 if i in l else f(a,[a[i],i-a[i]][i and-g(i-1)%i],n+1,l+[i]))

Ideone it!

How it works

Wilson's theorem is used for prime checking.

Loop detection by storing seen indices to an array (l) and checking whether current index is in l.

Answer 3

Python, 107 bytes

import sympy
f=lambda a,i,n=0:0if n>len(a)else f(a,[a[i],i-a[i]][sympy.isprime(i)],n+1)if 0<=i<len(a)else n

Usage: f(list, start) ex: f([2,5,6,8,1,2,3], 3)

Returns 0 for loops (detected when n > len(a))

Answer 4

Matlab, 138 bytes

This a straighforward approach, using 1-based indices because Matlab uses 1-based indices by default. To count the number of steps we use a for loop counting from 1 to infinity(!). For the case were we cannot escape the array, we use a vector v to keep track of which entries we already visited. If we visit an entry twice, we know we are stuck in an unescapeable cycle. To see check whether we are outside of an array, we use the try/catch structure, which also catches out of bounds exceptions.

function r=f(a,i);v=a*0;v(i)=1;for k=1:Inf;if isprime(i);i=i-a(i);else;i=a(i);end;try;if v(i);r=0;break;end;v(i)=1;catch;r=k;break;end;end

Answer 5

05AB1E, 32 bytes

ï[U¯Xåi0,q}²gL<Xå_#X²XèXDˆpi-]¯g

Explanation

ï                                 # explicitly convert input to int
 [                            ]   # infinite loop
  U                               # store current index in X
   ¯Xåi0,q}                       # if we've already been at this index, print 0 and exit
           ²gL<Xå_#               # if we've escaped, break out of infinite loop
                   X²XèXDˆpi-     # else calculate new index
                               ¯g # print nr of indices traversed

Try it online

Answer 6

JavaScript (ES6), 100

Index base 0. Note: this function modifies the input array

(a,p)=>eval("for(s=0;1/(q=a[p]);++s,p=p>1&&p%i||p==2?p-q:q)for(a[p]=NaN,i=1;p%++i&&i*i<p;);q==q&&s")

Less golfed

(a,p)=>
{
  for(s = 0; 
      1/ (q = a[p]); 
      ++s)
  {
    a[p] = NaN; // mark visited position with NaN to detect loops
    for(i = 1; p % ++i && i*i < p;); // prime check
    p = p > 1 && p % i || p == 2 ? p-q : q;
  }
  return q==q && s // return false if landed on NaN as NaN != NaN
}

Test

F=
(a,p)=>eval("for(s=0;1/(q=a[p]);++s,p=p>1&&p%i||p==2?p-q:q)for(a[p]=NaN,i=1;p%++i&&i*i<p;);q==q&&s")

;[
 [[2,5,6,8,1,2,3], 3, 1]
,[[2, 0, 2], 2, false]
,[[14,1,2,5,1,3,51,5,12,3,4,41,15,4,12,243,51,2,14,51,12,11], 5, 6]
].forEach(t=>{
  var [a,b,k]=t, i=a+' '+b,r=F(a,b)
  console.log(r==k?'OK':'KO',i+' -> '+r)
  
})  

Answer 7

JAVA, 229 218 Bytes

Object e(int[]a,int b){Stack i=new Stack();int s=0;for(;!(a.length<b|b<0);s++){if(i.contains(b))return 1>2;i.add(b);b=p(b)>0?b-a[b]:a[b];}return s;}int p(int i){for(int j=2;j<i/2;j++)if(i%j<1)return 0;return i<2?0:1;}

Thanks to Kevin, 11 bytes bites the dust.

Comments

A few things to golf it down some more: Stack<Integer>i=new Stack<>(); can be changed to Stack i=new Stack(); and return 1==2; can be changed to return 0>1;. Also, you might want to mention it's Java 7 instead of Java in general.

— Kevin Cruijssen

@KevinCruijssen I'm not sure is it point to mention that it is java 7, as especially now this solution is compatible with most java versions.

— user902383

Well, in Java 8 you can use a lambdas which is shorter: a,b->{...} instead of Object e(int[]a,int b){...}, which is why I personally mention Java 7 to let people know I purposely haven't used Java 8 lambdas, but it's up to you.

— Kevin Cruijssen

@KevinCruijssen fair enough, when I'm using lamda, I'm specifying java version, but when solution works with java 7, it's usually works with java 8 as well, so it was pointless to add version. But you might be right, i should specify minimum version.

— user902383

Answer 8

CJam, 44 bytes

Expects index array on the stack.

:G\{_G,,&{G=_L#)0{_L+:L;_mp3T?-F}?}L,?}:F~o@

Try it online!

My first CJam answer, hence why it's so terrible and imperative...

:G\{_G,,&{G=_L#)0{_L+:L;_mp3T?-F}?}L,?}:F~o@
:G                                              Store the array as G
  \                                             Put the index first
   {                                  }:F~      The recursive F function
     G,,                                        Generate a 0..length(G) sequence
    _   &                            ?          Check that the index is contained
         {                        }             If so, then...
          G=                                    Get the value at the index
            _L#)                 ?              If the value is in L (`-1)` gives `0` which is falsy)
                0                               Return 0 (infinite loop)
                 {              }               Otherwise...
                  _L+:L;                        Store the value we're accessing in L (infinite loop check)
                        _mp3T?-                 Remove 3 if the number is prime
                               F                Then recursively call F
                                   L,           We escaped! Return the size of "L" (number of steps)
                                          o     Print the top value of the stack
                                           @    Tries to swap 3 elements, which will error out

(it is considered okay to crash after the correct output as printed, which is what the program here does)

Answer 9

C, 121 bytes

Function f accepts array, starting index (0-based) and number of elements in the array, since there is no way how to test the end of an array in C (at least I don't know any).

p(n,i,z){return--i?p(n,i,z*i*i%n):z%n;}c;f(a,i,n)int*a;{return i<0||i/n?c:c++>n?0:i&&p(i,i,1)?f(a,i-a[i],n):f(a,a[i],n);}

Try it on ideone!

Note: function p(n) tests if n is prime or not. Credit for this goes to @Lynn and his answer for Is this number a prime?

Comments

@raznagul nonsense, you can't determine the length of an input parameter array. See answer 2 on the same question

— edc65

@edc65: Sorry, I should have read beyond the first answer.

— raznagul

@Jasmes - In code golf, a function should be able to be called multiple times to get the same output. Your code requires resetting c to call the function again.

— owacoder

Answer 10

JavaScript, 121 132 bytes

p=n=>t=i=>n%i&&n>i?t(i+1):(0<n&&n<=i?1:0),c=-1,a=>r=s=>(++c,0<=s&&s<a.length?(p(s)(2)?r(s-a[s]):0||([a[s],s]=[0,a[s]])[1]?r(s):0):c)

f=(p=n=>t=i=>n%i&&n>i?t(i+1):(0<n&&n<=i?1:0),c=-1,a=>r=s=>(++c,0<=s&&s<a.length?(p(s)(2)?r(s-a[s]):0||([a[s],s]=[0,a[s]])[1]?r(s):0):c));

let test_data = [[[1,4,5,6,8,10,14,15,2,2,4,5,7],4],
                 [[2,5,6,8,1,2,3],3],
                 [[2,0,2],2],
                 [[14,1,2,5,1,3,51,5,12,3,4,41,15,4,12,243,51,2,14,51,12,11],5]];
for (test of test_data) {
    c = -1;
    console.log(f(test[0])(test[1]));
}

edit 1: oops, missed the bit about returning number of steps. fix coming soonish.

edit 2: fixed

Answer 11

Racket, 183 156 bytes

Probably more bytes savable with further golfing, but that's it for me. :)

(require math)(define(e l i[v'()][g length])(cond[(memq i v)#f][(not(< -1 i(g l)))(g v)][else(e l((λ(a)(if(prime? i)(- i a)a))(list-ref l i))(cons i v))]))

Complete module with test suite with cleaner function:

#lang racket

(require math)

(define (e l i [v'()] [g length])
  (cond
    [(memq i v) #f]
    [(not (< -1 i (g l))) (g v)]
    [else (e l
             ((λ (a) (if (prime? i)
                         (- i a)
                         a))
              (list-ref l i))
             (cons i v))]))

(module+ test
  (require rackunit)
  (define escape-tests
    '((((2 5 6 8 1 2 3) 3) . 1)
      (((2 0 2) 2) . #f)
      (((14 1 2 5 1 3 51 5 12 3 4 41 15 4 12 243 51 2 14 51 12 11) 5) . 6)))
  (for ([t escape-tests])
    (check-equal? (apply e (car t)) (cdr t) (~a t))))

Run it like raco test e.rkt

Major kudos for @cat discovering the undocumented prime? function.

Answer 12

Java, 163 160 bytes

boolean p(int n){for(int i=2;i<n;)if(n%i++==0)return 0>1;return 1>0;}
int f(int[]a,int n){return n<0||n>=a.length?1:p(n)?n<a[n]?1:1+f(a,a[n-a[n]]):1+f(a,a[n]);}

p(n) is for prime testing, f(a,n) is for the escape function. Usage:

public static void main(String[] args) {
    int[] array = {14,1,2,5,1,3,51,5,12,3,4,41,15,4,12,243,51,2,14,51,12,11};
    System.out.println(f(array, 5));
}

Ungolfed version:

static boolean isPrime(int n) {
    for (int i = 2; i < n; i++) {
        if (n % i == 0) {
            return false;
        }
    }
    return true;
}

static int escape(int[] array, int n) {
    if (n < 0 || n >= array.length) {
        return 1;
    } else if (isPrime(n)) {
        if (n < array[n]) {
            return 1;
        } else {
            return 1 + escape(array, array[n - array[n]]);
        }
    } else {
        return 1 + escape(array, array[n]);
    }
}

Answer 13

Perl 6, 85 bytes

->\n,\a{{.[+a].defined??0!!+$_}(lazy n,{.is-prime??$_- a[$_]!!a[$_]}...^!(0 <=* <a))}

Explanation:

lazy n, { .is-prime ?? $_ - a[$_] !! a[$_] } ...^ !(0 <= * < a)

This is a lazy sequence of the indices traversed according to the rule. If the index eventually exceeds the input array bounds (the !(0 <= * < a) condition), the sequence is finite; otherwise, the indices cycle infinitely.

That sequence is fed to the internal anonymous function:

{ .[+a].defined ?? 0 !! +$_ }

If the sequence is defined at the index given by the size of the input array, it must have entered an infinite cycle, so 0 is returned. Otherwise, the size of the sequence +$_ is returned.

Answer 14

Perl 5, 107 + 1 (-a) = 108 bytes

for($i=<>;!$k{$i}++&&$i>=0&&$i<@F;$s++){$f=0|sqrt$i||2;1while$i%$f--;$i=$f?$F[$i]:$i-$F[$i]}say$k{$i}<2&&$s

Try it online!

0-based list. Returns false (blank) if the list is unescapable.