It can be reduced further by changing LEFT(String,1) to LEFT(String), which appears to give the first letter by default. Also, you do not need to specify the formula to give 0 if the test is false, so the ,0 part can also be removed: Code: =IF(SUM(IF(MID(A1,ROW(A:A),1)=LEFT(RIGHT(A1,ROW(A:A))),1))=LEN(A1),,"No ")&"Palindrome" which is 86 characters. I feel we're just refining it now so...back to study!
This may be considered cheating , but if you change the cell formatting to Code: "Palindrome";;"No palindrome" then the following solution only requires 66 characters Code: =--(SUM(IF(MID(A1,ROW(A:A),1)=LEFT(RIGHT(A1,ROW(A:A))),1))=LEN(A1))
CA2 examiners would go mad.....and I like it! edit: by the same logic, we could use: Code: (SUM((MID(A1,ROW(A:A),1)=LEFT(RIGHT(A1,ROW(A:A))))*1)=LEN(A1))*1 for 64 characters!
This attempt has a character count of 56. Code: =--AND(RIGHT(LEFT(A1,ROW(A:A)))=LEFT(RIGHT(A1,ROW(A:A)))) The MID function had to be abandoned because MID(string,x,1) returns "" when x>len(string), whereas RIGHT(LEFT(string,x)) just gives RIGHT(string) for large values of x.
Nice one! I tried to change the approach by using the SEARCH() function to reduce the character count but to no avail. Although only one "-" sign is needed, with a result of -1 verifying it is a palindrome
There haven't been any new suggestions for a couple of days, so it looks as though maz1987 is this month's winner - well done! Here's a summary of the final submission; Code: =-AND(RIGHT(LEFT(A1,ROW(A:A)))=LEFT(RIGHT(A1,ROW(A:A)))) with the number formatting set to Code: ;"Palindrome";"No palindrome" That puts the character count at 55. Watch out for Excel Challenge #3 - launching Friday 10th April 2015.
I couldn't possibly accept that after I simply removed one character from your attempt! Look forward to the third..
