Tokitozuka Tokarinku

Computer Science Level 3

Using the above image, one would can Ninja-encode a phrase by replacing each letter by the corresponding phrase.

For example, my name (Lokesh) came out to be 'ta~mo~me~ku~ari~ri' . Notice, I have placed '~' among the sub-words to make the reverse translation easier. I will follow the same notation while converting any other word to Ninja name.

Now, write a program that converts any name in English to Ninja name. Solve this puzzle and enter the answer below

chi~ri~ku
ka~to~ari~mei~ku~shi
chi~mo
chi~ri~ki~ari
no~shi~mo~zu~ta~ku~rin
ki~ari
ari~ki~na
te~ki~ru~ki~te~ku~te
zu~fu
ri~ka~ta~lu

The answer is 12.

6 solutions

Discussions for this problem are now closed

Lokesh Sharma
Apr 11, 2014

Python Solution:

import string

# Makes a dictionary to convert English letters to Japense letters
EngtoJap = {'a':'ka', 'b':'zu', 'c':'mi', 'd':'te', 'e':'ku',
            'f':'lu', 'g':'ji', 'h':'ri', 'i':'ki', 'j':'zu',
            'k':'me', 'l':'ta', 'm':'rin', 'n':'to', 'o':'mo',
            'p':'no', 'q':'ke', 'r':'shi', 's':'ari', 't':'chi',
            'u':'do', 'v':'ru', 'w':'mei', 'x':'na', 'y':'fu',
            'z':'zi'}


# Makes a dictionary to convert Japenese letters to English letters
JaptoEng = {}

# Makes use of previous dictionary to create this one
for i in EngtoJap:
    JaptoEng[EngtoJap[i]] = i



def loadWords(fileName):
    '''
    Returns list of words from file fileName
    '''
    fileData = open(fileName, 'r')
    words = []
    for line in fileData:
        words.append(line.rstrip())
    return words

def convertToJap(words):
    '''
    Takes list of words in English and
    returns list of words in Japense
    '''
    convertedWords = []
    for word in words:
        convertedWord = ''
        for letter in word:
            convertedWord += EngtoJap[letter] + '~'
        convertedWords.append(convertedWord[:-1])
    return convertedWords

def convertToEng(words):
    '''
    Takes list of words in Japenese and
    returns list of words in English
    '''
    convertedWords = []
    for word in words:
        convertedWord = ''
        for subword in word.split('~'):
            convertedWord += JaptoEng[subword]
        convertedWords.append(convertedWord)
    return convertedWords


words = loadWords('nameProblem.txt')
toeng = convertToEng(words)
print ' '.join(toeng)

-By ta~mo~me~ku~ari~ri

Cool problem,but the mapping is not one to one. $B$ and $J$ are both $'ku'$ ,thus making the inverse function have to choose. I got "THE ANSWER TO THIS PROJLEM IS SIX DIVIDED JY HALF".

Also If I were a Ninja I would be killed before I finished saying my name 'chirikatetekudoari ' :)

Thaddeus Abiy - 7 years, 2 months ago

Yeah, I once got concerned about this regard of mapping being not one to one but later I thought that people would figure it out and hence left it out.

Nice name by the way :P

Lokesh Sharma - 7 years, 2 months ago

I know right :) -Zikashitamodorinkajitoku (zikashi for short XD)

Zarloumagne Paragas - 7 years, 1 month ago

C Plus Plus solution

#include <iostream>
#include <string.h>

/* run this program using the console pauser or add your own getch, system("pause") or input loop */

using namespace std;

int main(int argc, char** argv) 
{
    char njName[1024]="", name[1024]="";
    char ninja_dic[26][4]={"ka","zu","mi","te","ku","lu","ji","ri","ki","zu","me","ta","rin","to","mo","no","ke","shi","ari","chi","do","ru","mei","na","fu","zi"};
    char dic[]={'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'};

    cout<<" Your name: ";
    cin>>name;
    int namelen = strlen(name) ;
    strupr(name);

    for(int i=0;i<namelen;i++)
    {
        for(int j=0;j<26;j++)
        {
            if(name[i] == dic[j] )
            {
                strcat(njName,ninja_dic[j]);
                strcat(njName,"~");
                break;
            }
        }
    }

    cout<<" Ninja Name : "<<njName;
    cin>>name;

    return 0;
}

Shubham Gaikwad - 7 years, 1 month ago

Mekarukichirika - _ _-

Kavitha mahan - 7 years, 1 month ago

Cameron Bernhardt
Apr 21, 2014

I threw together a crude website to "ninja-ify" and "de-ninja-ify" phrases: tinyurl.com/ninjaify

As for the code, I used JavaScript:

var cypher = { //a JS object that holds the translation info
  "a": "ka",
  "b": "zu",
  "c": "mi",
  "d": "te",
  "e": "ku",
  "f": "lu",
  "g": "ji",
  "h": "ri",
  "i": "ki",
  "j": "zu",
  "k": "me",
  "l": "ta",
  "m": "rin",
  "n": "to",
  "o": "mo",
  "p": "no",
  "q": "ke",
  "r": "shi",
  "s": "ari",
  "t": "chi",
  "u": "do",
  "v": "ru",
  "w": "mei",
  "x": "na",
  "y": "fu",
  "z": "zi"
};

var alpha = []; //all this does is get the keys from the object

for(var k in cypher){//for cypher, that's just the English alphabet
  alpha.push(k);
}

Now, the decryption function:

function decode(code){//decodes ninja-ified string
  var str = code.toLowerCase(); //downcases input for compatability
  var newStr = ""; //will hold decrypted string

  str = str.split("~"); //splits input into an array of ninja 'letters'

  for(var i = 0; i < str.length; i++){//goes through encoded array and checks for matches
    for(var j = 0; j < alpha.length; j++){
      var b = alpha[j];
      if(cypher[b] === str[i]){
        newStr += b;
      }
    }
  }

  return newStr; //returns decrypted string
}
alert("Decoded: " + decode(prompt("Enter ninja-ified stuff.")));

And, just for fun, an encryption function:

function encode(notcode){//function for ninja-ifying
  var str = notcode.toLowerCase(); //downcases input for compatability
  var newStr = ""; //will hold decrypted string

  for(var i = 0; i < str.length; i++){//goes through string and encrypts each letter
    var b = str[i];//letter
    if(i != str.length - 1){//checks to see whether or not it's the last letter
      newStr += cypher[b] + "~"; //if not, adds ~ to separate words
    }else{
      newStr += cypher[b]; //if so, don't add it
    }
  }

  return newStr; //returns encrypted string
}
alert("Encoded: " + encode(prompt("Enter stuff to be ninja-ified.")));

Sudipta Kar
Apr 17, 2014

import java.util.HashMap;
import java.util.StringTokenizer;

public class NinjaName {

    public static void main(String[] args) {
        HashMap map = new HashMap();
        map.put("ka","A");
        map.put("zu","B");
        map.put("mi","C");
        map.put( "te", "D");
        map.put( "ku", "E");
        map.put( "lu", "F");
        map.put( "ji", "G");
        map.put( "ri", "H");
        map.put( "ki", "I");
        map.put( "zu", "J");
        map.put( "me", "K");
        map.put( "ta", "L");
        map.put( "rin", "M");
        map.put( "to", "N");
        map.put( "mo","O" );
        map.put( "no", "P");
        map.put( "ke", "Q");
        map.put( "shi", "R");
        map.put( "ari", "S");
        map.put( "chi", "T");
        map.put( "do", "U");
        map.put( "ru", "V");
        map.put( "mei", "W");
        map.put( "na", "X");
        map.put( "fu", "Y");
        map.put( "zi", "Z");

        String s = "chi~ri~ku~ka~to~ari~mei~ku~shi~chi~mo~chi~ri~ki~ari~no~shi~mo~zu~ta~ku~rin~ki~ari~ari~ki~na~te~ki~ru~ki~te~ku~te~zu~fu~ri~ka~ta~lu";
        StringTokenizer st = new StringTokenizer(s, "~");

        while(st.hasMoreTokens()){
            String ss = st.nextToken();
            System.out.print(map.get(ss));
        }
    }
}

Nicholasi Flamel
Apr 15, 2014

Lang used: java

import java.util.*;

public class NameNinja{

    static String[] ninjaAlphabet = {"ka", "zu", "mi", "te", "ku", "lu", "ji", "ri", "ki", "zu", "me", "ta", "rin", "to", "mo", "no", "ke", "shi", "ari", "chi", "do", "ru", "mei", "na", "fu", "zi"};

    static String[] alphabet = {"a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z"};

    public static void main(String[] args){
        getName();
    }
    public static void getName(){
        Scanner s = new Scanner(System.in);
        String nameString = new String();
        int size;
        try{
            while(true){
                nameString = s.nextLine();
                String[] ninjaName = nameString.split("~");
                System.out.println(getName(ninjaName));
            }
        }catch(Exception e){}
    }
    public static String getName(String[] ninjaName){
        String name = new String();
        for(int i = 0; i < ninjaName.length; i++){
            for(int j = 0; j < 26; j++)  
                if(ninjaName[i].equals(ninjaAlphabet[j])){
                    name += alphabet[j];
                    break;
                }
        }
        return name;
    }
}

Connor Kenway
Apr 14, 2014

include<iostream>

#include<stdio.h>
#include<string.h>

```c++ using namespace std; char store[26][6]={"ka","zu","mi","te","ku","lu","ji","ri","ki","zu","me","ta","rin","to","mo","no","ke","shi","ari","chi","do","ru","mei","na","fu","zi"}; int main() { int test=10; while(test--) { char ch[50]; gets(ch); int length=strlen(ch); for(int i=0;i<=length;i++) { if(ch[i]=='~'||ch[i]=='\0') { int j=i-1; for(;(j-1)!=-1&&(ch[j-1]!='\0');j--); ch[i]='\0'; for(int k=0;k<26;k++) { char *c=ch; c=c+j; if(strcmp(c,store[k])==0 ) {cout<<(char)(97+k);break;}

            }
        }
    }
    cout<<endl;
       }
    return 0;
}

``` btw the code for b and j is same that is 'zu' so my program shows b for a 'zu' .MY name - ri~ ki~ rin~ ka~ to~ ari~ ri~ do

Check the link .

Connor Kenway - 7 years, 2 months ago

Sanjay Kamath
Apr 14, 2014

Here is my solution in C#

         ArrayList ar = new ArrayList();

        Hashtable ht = new Hashtable();

        ht.Add("ka","A");
        ht.Add("zu", "B");
        ht.Add("mi", "C");
        ht.Add("te", "D");
        ht.Add("ku", "E");
        ht.Add("lu", "F");
        ht.Add("ji", "G");
        ht.Add("ri", "H");
        ht.Add("ki", "I");
        ht.Add("zu2", "J");
        ht.Add("me", "K");
        ht.Add("ta", "L");
        ht.Add("rin", "M");
        ht.Add("to", "N");
        ht.Add("mo", "O");
        ht.Add("no", "P");
        ht.Add("ke", "Q");
        ht.Add("shi", "R");
        ht.Add("ari", "S");
        ht.Add("chi", "T");
        ht.Add("do", "U");
        ht.Add("ru", "V");
        ht.Add("mei", "W");
        ht.Add("na", "X");
        ht.Add("fu", "Y");
        ht.Add("zi", "Z");

        string s1 = "chi~ri~ku";
        string s2 = "ka~to~ari~mei~ku~shi";
        string s3 = "chi~mo";
        string s4 = "chi~ri~ki~ari";
        string s5 = "no~shi~mo~zu~ta~ku~rin";
        string s6 = "ki~ari";
        string s7 = "ari~ki~na";
        string s8 = "te~ki~ru~ki~te~ku~te";
        string s9 = "zu~fu";
        string s10 = "ri~ka~ta~lu";

        string[] sarr = new string[10];
        ArrayList are = new ArrayList();

        are.Add(s1);
        are.Add(s2);
        are.Add(s3);
        are.Add(s4);
        are.Add(s5);
        are.Add(s6);
        are.Add(s7);
        are.Add(s8);
        are.Add(s9);
        are.Add(s10);


        string sg = String.Empty;

        for (int j = 0; j < 10; j++)
        {
            string[] res = are[j].ToString().Split('~');


            for (int y = 0; y < res.Length; y++)
            {
                sg += ht[res[y]].ToString();
            }

            ar.Add(sg);

            sg = "";
        }

        for (int h = 0; h < ar.Count; h++)
        {
            MessageBox.Show(ar[h].ToString());
        }

       Output is "THE ANSWER TO THIS PROBLEM IS SIX DIVIDED BY HALF" - equals 12.

Tokitozuka Tokarinku

The answer is 12.

6 solutions

Discussions for this problem are now closed

include<iostream>

0 pending reports