Euler_Problem-036.b93

v      
0             v                      \<     v            *2\<     >:.55+,\   v
>093194*+**>::>55+*10p:55+%10g+\55+/:#^_$::0>10p:2%10g+\2/:#^_$-!#^_$        v
           |:-1                                                              <
v          <     v                      \<     v            *2\<     >:.55+,\v
>93194*+**>::55+/>55+*10p:55+%10g+\55+/:#^_$::0>10p:2%10g+\2/:#^_$-!#^_$     v
          |:-1                                                               <
          >"= ",,>+\:#<_+.@
       
       

[10] temp




93194*+** => 999
       
// Test number for binary palindrome (stack -> inversed stack )

  v            *2\<
:0>10p:2%10g+\2/:#^_$-  @

// enumerate Palindromes



             v                      \<
>93194*+**>::>55+*10p:55+%10g+\55+/:#^_$                v
          |-1                                           <
v         <      v                      \<
>93194*+**>::55+/>55+*10p:55+%10g+\55+/:#^_$            v
          |-1                                           <




// generate palindrome not-odd

 v                      \<
:>55+*10p:55+%10g+\55+/:#^_$


// generate palindrome odd

     v                      \<
:55+/>55+*10p:55+%10g+\55+/:#^_$


---------------------------------------

The *trick* here is that we only need to test 1998 numbers. Because there are only so much base10 palindromes:

- The numbers from `1` to `999` mirrored result into to palindromes `11` to `999999`
- The numbers from `1` to `999` mirrored at the last digit result into to palindromes `1` to `99999`

Then we only need to test these numbers whether they are also binary palindromes.