- Updated solutions by Ulrich Rieke.

3 files changed, 52 insertions, 16 deletions

diff --git a/challenge-157/ulrich-rieke/cpp/ch-2.cpp b/challenge-157/ulrich-rieke/cpp/ch-2.cpp
index 5f6a402345..f2c7158cff 100644
--- a/challenge-157/ulrich-rieke/cpp/ch-2.cpp
+++ b/challenge-157/ulrich-rieke/cpp/ch-2.cpp
@@ -1,12 +1,20 @@
 #include <iostream>
 #include <vector>
-#include <algorithm>
 #include <cstdlib>
+#include <map>
+#include <string>
+#include <numeric>
+#include <algorithm>
 
-std::vector<int> toBaseDigits( int n , int base ) {
-  std::vector<int> baseDigits ;
+std::vector<std::string> toBaseDigits( int n , int base ,
+      const std::map<int,std::string> & extendedBases ) {
+  std::vector<std::string> baseDigits ;
   while ( n != 0 ) {
-      baseDigits.push_back( n % base ) ;
+      int remainder = n % base ;
+      if ( remainder > 10 )
+    baseDigits.push_back( extendedBases.find(remainder)->second ) ;
+      else
+    baseDigits.push_back( std::to_string( remainder )) ;
       n /= base ;
   }
   return baseDigits ;
@@ -14,17 +22,27 @@ std::vector<int> toBaseDigits( int n , int base ) {
 
 int main( int argc, char * argv[] ) {
   int n = std::atoi( argv[1] ) ;
-  while ( n <= 3 ) {
-      std::cout << "number must be greater than 3! Re-enter!\n" ;
+  //we allow bases up to 36 there numbers from 3 to 36 + 2 are admitted
+  while ( n <= 3 || n > 38 ) {
+      std::cout << "number must be greater than 3 and smaller than 39! Re-enter!\n" ;
       std::cin >> n ;
   }
+  //every number from 11 to 36 corresponds to a letter from A to Z
+  //this way remainders greater than 10 can be denoted
+  std::vector<std::string> alphabet( 26 ) ;
+  std::iota( alphabet.begin( ) , alphabet.end( ) , 'A' ) ;
+  std::vector<int> greaterBases( 26 ) ;
+  std::iota( greaterBases.begin( ) , greaterBases.end( ) , 11 ) ;
+  std::map<int, std::string> extendedBases ;
+  for ( int i = 0 ; i < 26 ; i++ )  //26 numbers from 11 to 36
+      extendedBases[ *(greaterBases.begin( ) + i)] = *(alphabet.begin( ) + i) ;
   bool isBrazilian = false ;
   for ( int i = 2 ; i < n - 1 ; i++ ) {
-      std::vector<int> digits { toBaseDigits( n , i ) } ;
+      std::vector<std::string> digits { toBaseDigits( n , i , extendedBases) } ;
       if ( std::count( digits.begin( ) , digits.end( ) , *(digits.begin( ) )) ==
-        digits.size( ) ) {
-    isBrazilian = true ;
-    break ;
+            digits.size( ) ) {
+        isBrazilian = true ;
+        break ;
       }
   }
   if ( isBrazilian )
diff --git a/challenge-157/ulrich-rieke/haskell/ch-2.hs b/challenge-157/ulrich-rieke/haskell/ch-2.hs
index be824168b8..5f0e977ad5 100644
--- a/challenge-157/ulrich-rieke/haskell/ch-2.hs
+++ b/challenge-157/ulrich-rieke/haskell/ch-2.hs
@@ -28,10 +28,10 @@ where
 
 askForCorrectNumber ::IO Int
 askForCorrectNumber = do
-  putStrLn "Please enter an integer greater than 3!"
+  putStrLn "Please enter an integer greater than 3 and smaller than 39!"
   num <- getLine
   let n = read num
-  if n > 3 then return n
+  if (n > 3 && n < 39) then return n
   else do
       askForCorrectNumber
 
diff --git a/challenge-157/ulrich-rieke/perl/ch-2.pl b/challenge-157/ulrich-rieke/perl/ch-2.pl
index 48a932e8fb..7f445eca57 100644
--- a/challenge-157/ulrich-rieke/perl/ch-2.pl
+++ b/challenge-157/ulrich-rieke/perl/ch-2.pl
@@ -15,24 +15,42 @@ sub allSame {
 sub toBase {
   my $number = shift ;
   my $base = shift ;
+  my $extended = shift ;
   my @digits ;
   while ( $number != 0 ) {
-      push @digits , $number % $base ;
+      my $remainder = $number % $base ;
+      if ( $remainder > 10 ) {
+    push @digits , $extended->{$remainder} ;
+      }
+      else {
+    push @digits, $remainder ;
+      }
       $number = int( $number / $base ) ;
   }
   return @digits ;
 }
 
 my $n = $ARGV[0] ;
-while ( $n <= 3 ) {
-  say "number should be greater than 3! Re-enter!" ;
+#the algorithm allows bases from 2 to 36 and therefore numbers from 3 + 1 to
+#36 + 2
+while ( $n <= 3 || $n > 38 ) {
+  say "number should be greater than 3 and smaller than 39! Re-enter!" ;
   my $line = <STDIN> ;
   chomp $line ;
   $n = $line ;
 }
+#if the number base is greater than 10 we introduce letters to denote the rest
+#of dividing a number by the base
+my @letters = ('A' .. 'Z') ;
+my @numbers = (11 .. 36 ) ;
+my %extendedBases ;
+#each number from 11 to 36 corresponds to a letter
+for my $num (0 .. 25 ) {
+  $extendedBases{ $numbers[$num] } = $letters[$num] ;
+}
 my $isBrazilian = 0 ;
 for my $i ( 2 .. $n - 2 ) {
-  my @digits = toBase( $n , $i ) ;
+  my @digits = toBase( $n , $i , \%extendedBases ) ;
   if ( allSame( \@digits ) ) {
       $isBrazilian = 1 ;
       last ;