Longest string challenge: Difference between revisions
m (→read file until not ready: added a colon to a template text.) |
m (→read file until not ready: elided the need for the variable named ONE.) |
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<br>no arrays or lists are used, and |
<br>no arrays or lists are used, and |
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<br>no additions or subtractions are used. |
<br>no additions or subtractions are used. |
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<lang rexx>/*REXX program reads a file and displays the longest [widest] record(s) [line(s)]. |
<lang rexx>/*REXX program reads a file and displays the longest [widest] record(s) [line(s)].*/ |
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parse arg iFID . /*obtain optional argument from the CL.*/ |
parse arg iFID . /*obtain optional argument from the CL.*/ |
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if iFID=='' | iFID=="," then iFID= 'LONGEST.TXT' /*Not specified? Then use the default.*/ |
if iFID=='' | iFID=="," then iFID= 'LONGEST.TXT' /*Not specified? Then use the default.*/ |
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signal on notReady /*when E-O-F is reached, jump/branch. */ |
signal on notReady /*when E-O-F is reached, jump/branch. */ |
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one=1 /*used to cause 1st time initialization*/ |
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m=0 /*the maximum width (so far). */ |
m=0 /*the maximum width (so far). */ |
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do |
do rec=1; _=linein(iFID); w=length(_) /*read a line from the input file. */ |
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do y=rec to 1; m=w; w=0; $=y || _; end /*first initialization.*/ |
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do |
do n=m for 1; end /*for smaller records. */ |
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do |
do z=n to w for 1; $=; end /*found longest record.*/ |
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do |
do x=m to w for 1; $=$'a0d'x || _; end /*append record to $. */ |
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do x=m to w for 1; $=$'a0d'x || _; end /*append record to $. */ |
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m=max(m, w) /*M: is the maximum length record. */ |
m=max(m, w) /*M: is the maximum length record. */ |
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end /*forever*/ |
end /*forever*/ |
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Revision as of 09:38, 11 October 2017
You are encouraged to solve this task according to the task description, using any language you may know.
- Background
This "longest string challenge" is inspired by a problem that used to be given to students learning Icon. Students were expected to try to solve the problem in Icon and another language with which the student was already familiar. The basic problem is quite simple; the challenge and fun part came through the introduction of restrictions. Experience has shown that the original restrictions required some adjustment to bring out the intent of the challenge and make it suitable for Rosetta Code.
The original programming challenge and some solutions can be found at Unicon Programming TWiki / Longest Strings Puzzle. (See notes on the talk page if you have trouble with the site).
- Basic problem statement
Write a program that reads lines from standard input and, upon end of file, writes the longest line to standard output. If there are ties for the longest line, the program writes out all the lines that tie. If there is no input, the program should produce no output.
- Task
Implement a solution to the basic problem that adheres to the spirit of the restrictions (see below).
Describe how you circumvented or got around these 'restrictions' and met the 'spirit' of the challenge. Your supporting description may need to describe any challenges to interpreting the restrictions and how you made this interpretation. You should state any assumptions, warnings, or other relevant points. The central idea here is to make the task a bit more interesting by thinking outside of the box and perhaps by showing off the capabilities of your language in a creative way. Because there is potential for considerable variation between solutions, the description is key to helping others see what you've done.
This task is likely to encourage a variety of different types of solutions. They should be substantially different approaches.
Given the input:
a bb ccc ddd ee f ggg
the output should be (possibly rearranged):
ccc ddd ggg
- Original list of restrictions
- No comparison operators may be used.
- No arithmetic operations, such as addition and subtraction, may be used.
- The only datatypes you may use are integer and string. In particular, you may not use lists.
- Do not re-read the input file. Avoid using files as a replacement for lists (this restriction became apparent in the discussion).
- Intent of restrictions
Because of the variety of languages on Rosetta Code and the wide variety of concepts used in them, there needs to be a bit of clarification and guidance here to get to the spirit of the challenge and the intent of the restrictions.
The basic problem can be solved very conventionally, but that's boring and pedestrian. The original intent here wasn't to unduly frustrate people with interpreting the restrictions, it was to get people to think outside of their particular box and have a bit of fun doing it.
The guiding principle here should be to be creative in demonstrating some of the capabilities of the programming language being used. If you need to bend the restrictions a bit, explain why and try to follow the intent. If you think you've implemented a 'cheat', call out the fragment yourself and ask readers if they can spot why. If you absolutely can't get around one of the restrictions, explain why in your description.
Now having said that, the restrictions require some elaboration.
- In general, the restrictions are meant to avoid the explicit use of these features.
- "No comparison operators may be used" - At some level there must be some test that allows the solution to get at the length and determine if one string is longer. Comparison operators, in particular any less/greater comparison should be avoided. Representing the length of any string as a number should also be avoided. Various approaches allow for detecting the end of a string. Some of these involve implicitly using equal/not-equal; however, explicitly using equal/not-equal should be acceptable.
- "No arithmetic operations" - Again, at some level something may have to advance through the string. Often there are ways a language can do this implicitly advance a cursor or pointer without explicitly using a +, - , ++, --, add, subtract, etc.
- The datatype restrictions are amongst the most difficult to reinterpret. In the language of the original challenge strings are atomic datatypes and structured datatypes like lists are quite distinct and have many different operations that apply to them. This becomes a bit fuzzier with languages with a different programming paradigm. The intent would be to avoid using an easy structure to accumulate the longest strings and spit them out. There will be some natural reinterpretation here.
To make this a bit more concrete, here are a couple of specific examples:
In C, a string is an array of chars, so using a couple of arrays as strings is in the spirit while using a second array in a non-string like fashion would violate the intent.
In APL or J, arrays are the core of the language so ruling them out is unfair. Meeting the spirit will come down to how they are used.
Please keep in mind these are just examples and you may hit new territory finding a solution. There will be other cases like these. Explain your reasoning. You may want to open a discussion on the talk page as well.
- The added "No rereading" restriction is for practical reasons, re-reading stdin should be broken. I haven't outright banned the use of other files but I've discouraged them as it is basically another form of a list. Somewhere there may be a language that just sings when doing file manipulation and where that makes sense; however, for most there should be a way to accomplish without resorting to an externality.
At the end of the day for the implementer this should be a bit of fun. As an implementer you represent the expertise in your language, the reader may have no knowledge of your language. For the reader it should give them insight into how people think outside the box in other languages. Comments, especially for non-obvious (to the reader) bits will be extremely helpful. While the implementations may be a bit artificial in the context of this task, the general techniques may be useful elsewhere.
Ada
This first solution is prepended to the following earlier solutions that did not fully comply with the restrictions, namely the use of arithmatic operators and language features that return numbers used with them.
In order to comply with the avoidance of greater-than or less-than comparisons and iterations with operators, only Constraint_Error exception handling is used to obtain comparison of less-than, equal-to, or greater-than. With these, only the strings that are longer than previous strings and the succeeding strings, and equal in length to each other are printed. They are printed in reverse order, but this is specifically allowed by the instructions. In order to be clear to the reader, all cases of less-than, equal to, and greater-than are manually iterated through. Since equal-to/not-equal-to testing is allowed, there should be no question that a "case" statement is also allowed.
<lang Ada>with Ada.Text_IO;
procedure Longest_Strings is
use Ada.Text_IO;
-- first, in order to strictly use integer, I use integer in
-- place of an enumeration type: -1 => not-equal
-- 0 => shorter - ignore, no print current string
-- 1 => equal - print current and up-stream
-- 2 => longer - no print upstream, only current and equal subsequent
-- others => null; -- must never happen.
--
-- Anything else that is tested or used that is not a string or integer
-- is not used explicitly by me, but is a standard part of the language
-- as provided in the standard libraries (like boolean "End_Of_File").
function Measure_And_Print_N (O : String := ""; -- original/old string
N : String := "" -- next/new string
) return Integer is
T1 : String := O;
T2 : String := N;
L : Integer := 1; -- Length defaults to the same;
function Test_Length (O : in out String; -- original/old string
N : in out String) -- new/test-subject string
return Integer is
function Test_Equal (O : in out String; N : in out String)
return Integer is
begin
O := N;
return 1;
exception
when Constraint_Error =>
return -1;
end;
begin
case Test_Equal (O, N) is
when -1 =>
O (N'Range) := N;
return 0;
when 1 =>
return 1;
when others =>
return -1;
end case;
exception
when Constraint_Error =>
return 2;
end;
begin
case Test_Length (T1, T2) is
when 0 =>
-- N < O, so return "shorter" do not print N
if End_Of_File
then
return 0;
else
case Measure_And_Print_N (O, Get_Line) is
when 0 =>
return 0;
when 1 =>
return 0;
when 2 =>
return 2; -- carry up any subsequent canceling of print.
when others =>
raise Numeric_Error;
end case;
end if;
when 1 =>
-- O = N, so return "equal" print N if all subsequent values are
-- less than or equal to N
if End_Of_File
then
Put_Line (N);
return 1;
else
case Measure_And_Print_N (O, Get_Line) is
when 0 =>
Put_Line (N);
return 1;
when 1 =>
Put_Line (N);
return 1;
when 2 => -- carry up the subsequent canceling of print.
null;
return 2;
when others =>
raise Numeric_Error;
end case;
end if;
when 2 =>
-- N > O, so return "longer" to cancel printing all previous values
-- and print N if it is also equal to or greater than descendant
-- values.
if End_Of_File
then
Put_Line (N);
return 2;
else
case Measure_And_Print_N (N, Get_Line) is
when 0 =>
Put_Line (N);
return 2;
when 1 =>
Put_Line (N);
return 2;
when 2 => -- printing N cancelled by subsequent input.
null;
return 2;
when others =>
raise Numeric_Error;
end case;
end if;
when others =>
-- This should never happen - raise exception
raise Numeric_Error;
end case;
end;
begin
if End_Of_File
then
null;
else
case Measure_And_Print_N ("", Get_Line) is
when 0 =>
Put_Line (Current_Error,
"Error, Somehow the input line is calculated as less than zero!");
when 1 =>
Put_Line (Current_Error,
"All input lines appear to be blank.");
when 2 =>
null;
when others =>
raise Numeric_Error;
end case;
end if;
end; </lang>
The output, given the above example input, is:
ggg ddd ccc
Keeping the input in order means either appending to a long string with varying segments and a separater character, like below, or possible using a list feature that has been specifically disallowed.
Next follows previous attempts:
How to bypass the restrictions:
- All lines of potential output are stored in an (unbounded) string, named Buffer. On special character (Latin_1.Nul) is used to separate between different lines.
- We can't directly compare the lengths of two strings. So instead, we assign the difference of the lengths to a variable of type Natural (*). If the result is outside of Natural, this raises an exception. If there is no exception, we assign the result to a variable of type Positive (*), which raises an exception if the result is outside of Positive.
(*) Technically, Natural and Positive are not types but subtypes of Integer: Natural ranges from 0 to Integer'Last, Positive from 1 to Integer'Last.
So this is the first solution.
<lang Ada>with Ada.Strings.Unbounded; use Ada.Strings.Unbounded; with Ada.Text_IO, Ada.Characters.Latin_1;
procedure Longest_String_Challenge is
function "+"(S: String) return Unbounded_String renames To_Unbounded_String; Separator: constant Character := Ada.Characters.Latin_1.NUL;
procedure Funny_Stuff(B, L: in out Unbounded_String; N: Unbounded_String) is
-- B holds a list of all longest strings, separated by Separator
-- L holds longest string so far
-- N is the next string to be considered
Nat: Natural;
begin
Nat := Length(N) - Length(L);
-- (1) this raises exception if L longer then N
declare
Pos: Positive;
begin
Pos := Nat; -- (2) this raises exception if L at least as long as N
-- at this point, we know N is longer then L
B := N;
L := N;
exception
when Constraint_Error -- come from (2)
-- at this point, we know L and N are of the same length
=> B := B & Separator & N; -- add N to the set of solutions
end;
exception
when Constraint_Error => null; -- come from (1)
-- at this point, we know L is longer then N
end Funny_Stuff;
Buffer: Unbounded_String := +""; Longest: Unbounded_String := +""; Next: Unbounded_String;
begin
while True loop
Next := + Ada.Text_IO.Get_Line;
-- (3) raises exception when trying to read beyond the end of file
Funny_Stuff(Buffer, Longest, Next);
end loop;
exception
when others => -- come from (3)
for I in To_String(Buffer)'Range loop
if To_String(Buffer)(I) = Separator then
Ada.Text_IO.New_Line;
else
Ada.Text_IO.Put(To_String(Buffer)(I));
end if;
end loop;
end Longest_String_Challenge;</lang>
Output, when run with its own source code as the input:
function "+"(S: String) return Unbounded_String renames To_Unbounded_String; procedure Funny_Stuff(B, L: in out Unbounded_String; N: Unbounded_String) is
Here is the second solution. It also makes heavy use of exceptions, but it does not require to compute the difference (which is an arithmetic operation, i.e., a bit of a cheat). Instead, the procedure Funny_Stuff carries some auxiliary strings S, T. If they are unequal and neither is empty, it recursively calls itself with the same strings shortened by 1. At some point of time, either S is empty, or T is empty, or both are empty.
<lang Ada>with Ada.Strings.Unbounded; use Ada.Strings.Unbounded; with Ada.Text_IO, Ada.Characters.Latin_1;
procedure Longest_String_Challenge is
function "+"(S: String) return Unbounded_String renames To_Unbounded_String;
function "-"(U: Unbounded_String) return String renames To_String;
Separator: constant Character := Ada.Characters.Latin_1.NUL;
procedure Funny_Stuff(B, L: in out Unbounded_String;
N: Unbounded_String;
S, T: String) is
C: Character;
begin
C:= T(T'First); -- (1) raises Constraint_Error if T is empty
begin
C := S(S'First); -- (2) raises Constraint_Error if S is empty -- at this point, we know that neither S nor T are empty
Funny_Stuff(B,L,N,S(S'First+1 .. S'Last), T(T'First+1..T'Last));
exception
when Constraint_Error => -- come from (2), S is empty, T is not empty!
B := N; L := N;
end;
exception
when Constraint_Error => -- come from (1), T is empty
begin C := S(S'First); -- (3) raises Constraint_Error if S is empty -- at this point, we know that T is empty and S isn't null; exception when Constraint_Error => -- come from (3); both S and T are empty B := B & Separator & N; end;
end Funny_Stuff; Buffer: Unbounded_String := +""; Longest: Unbounded_String := +""; Next: Unbounded_String;
begin
while True loop
Next := + Ada.Text_IO.Get_Line;
-- (4) raises exception when trying to read beyond end of file
Funny_Stuff(Buffer, Longest, Next, -Longest, -Next);
end loop;
exception
when others => -- come from (4)
for I in To_String(Buffer)'Range loop
if To_String(Buffer)(I) = Separator then
Ada.Text_IO.New_Line;
else
Ada.Text_IO.Put(To_String(Buffer)(I));
end if;
end loop;
end Longest_String_Challenge;</lang>
The output, when given its own source code as the input:
function "+"(S: String) return Unbounded_String renames To_Unbounded_String;
when Constraint_Error => -- come from (2), S is empty, T is not empty!
ALGOL 68
Empty loops are used for comparison for equality; STRING slicing is used to implement subtraction; slicing and concatenation implements addition.
The STRING 'buffer' holds all the input strings and separating newline characters. A dummy zero-length string is tacked on the end so that the final loop terminates cleanly. The start of the STRING 'mask' is filled in with "1" characters at the position of the terminal character in each input string, overwriting any "0" that may have been there. A final scan of 'mask' identifies the maximum string length seen on the input.
The 'char in string' function, here used to slice up the buffer into the original inputs, is an Algol 68 Genie extension. The 'Pedantry' comment draws attention to the explicit voiding of the function result; not strictly necessary but it avoids a complaint if the --pedantic option is given to a68g.
Warts are that all input strings must be shorter than (bound -1) characters and it is assumed that ABS "1" > ABS "0"; this true for every known implementation of Algol 68. <lang algol68> BEGIN
INT bound = 1000000; CO Arbitrary upper limit on string lengths CO INT max; CO Length of longest string CO INT len; CO Length of string under examination CO STRING buffer := ""; CO All characters read from stand in CO STRING mask := bound * "0"; CO High water mark of string length seen so far CO
CO Standard boiler plate CO
on file end (stand in, (REF FILE f) BOOL: (close (f); GOTO finished));
DO
STRING line;
read ((line, newline));
buffer PLUSAB line + REPR 10; CO Concatenate string and newline CO
mask[UPB line] := "1" CO And set mask where character exists in line CO
OD;
finished:
buffer PLUSAB REPR 10; CO Guarantee there's a zero-length string at the end CO
CO
Scan backwards through mask looking for highest index used which is equal to the length of the longest string with its terminating newline.
CO
FOR i FROM bound BY -1 TO 1
DO
FROM ABS mask[i] TO ABS "0" DO max := i OD CO Exploit ABS "1" > ABS "0" CO
OD;
FROM 1 TO UPB buffer
DO CO Null loop if buffer is empty CO
VOID (char in string (REPR 10, len, buffer)); CO Pedantry and Algol68 Genie extension CO
FROM max TO len
DO CO Null loop if len < max CO
FOR i FROM 1 TO max DO printf (($a$, buffer[i])) CO Print string and newline CO OD
OD;
buffer := buffer[len : UPB buffer]; CO Step over string CO
buffer := buffer[2 : UPB buffer] CO Step over newline CO
OD
END</lang>
- Output:
printf "a\nbb\nccc\nddd\nee\nf\nggg\n" | a68g Longest_String.a68 ccc ddd ggg
Alternative recursive solution - the only standard operators used are string concatenation and array upper bound and SIGN (the signum function). Operators ATLEASTASLONGAS and LONGERTHAN are defined without the use of any comparison or arithmetic operators. The input file is processed recursively so this would run out of stack space for a large input file. Exploits some features of Algol 68 STRING slicing. <lang algol68># The standard SIGN operator returns -1 if its operand is < 0 #
- , 0 if its operand is 0 #
- , 1 if its operand is > 0 #
- This array maps he results of SIGN to FALSE or TRUE for the #
- ATLEASTASLONGAS operator defined below #
[ -1 : 1 ]BOOL not shorter; not shorter[ -1 ] := FALSE; not shorter[ 0 ] := FALSE; not shorter[ 1 ] := TRUE;
- Set the priorities for the dyadic operators defined below #
- 9 is the highest priority, so a LOMGERTHAN b AND ... #
- is parsed correctly #
PRIO ATLEASTASLONGAS = 9
, LONGERTHAN = 9 ;
OP NONEMPTYSTRING = ( STRING a )STRING: " " + a[ AT 1 ];
- STRING x is at least as long as STRING y if the substring #
- of x from the upper bound of y to the end of x is at least #
- one character long #
- Note that Algol 68 doesn't raise an error if the substring #
- start position is after the upper bound of the string, but #
- does object if the start position is before the lower bound #
- - hence the need for the NONEMPTYSTRING operator to ensure #
- we don't try executing a[ 0 : ] when b is "" #
OP ATLEASTASLONGAS = ( STRING x, STRING y )BOOL:
BEGIN
STRING a = NONEMPTYSTRING x;
STRING b = NONEMPTYSTRING y;
not shorter[ SIGN UPB a[ UPB b : ] ]
END # ATLEASTASLONGAS # ;
- x is longer than y if x is at least as long as y and #
- y is not at least as long as x #
OP LONGERTHAN = ( STRING x, STRING y )BOOL: x ATLEASTASLONGAS y AND NOT ( y ATLEASTASLONGAS x );
- additional LONGERTHAN operators to handle single chatracter #
- STRINGs which are actually CHAR values in Algol 68 #
- Not needed for the task, but useful for testing LONGERTHAN #
OP LONGERTHAN = ( CHAR x, CHAR y )BOOL: FALSE; OP LONGERTHAN = ( CHAR x, STRING y )BOOL: STRING( x ) LONGERTHAN y; OP LONGERTHAN = ( STRING x, CHAR y )BOOL: x LONGERTHAN STRING( y );
COMMENT # basic test of LONGERTHAN: # C-MMENT print( ( "abc" LONGERTHAN "bbcd", "ABC" LONGERTHAN "", "" LONGERTHAN "abc", "DEF" LONGERTHAN "DEF", "abcd" LONGERTHAN "a", newline ) ); C-MMENT COMMENT
PROC read line = ( REF FILE f )STRING:
BEGIN
STRING line;
get( f, ( line, newline ) );
IF at eof THEN "" ELSE line FI
END # read line # ;
- EOF handler for standard input #
BOOL at eof := FALSE; on logical file end( stand in, ( REF FILE f )BOOL:
BEGIN
at eof := TRUE;
TRUE
END
);
- recursively find the longest line(s) in the specified file #
- and print them #
PROC print longest lines = ( REF FILE f, STRING longest so far )STRING:
BEGIN
IF at eof THEN
longest so far
ELSE
STRING s = read line( f );
STRING t = IF s LONGERTHAN longest so far
THEN
print longest lines( f, s )
ELSE
print longest lines( f, longest so far )
FI;
IF s ATLEASTASLONGAS t AND t ATLEASTASLONGAS s
THEN
# this line is as long as the longest #
print( ( s, newline ) );
s
ELSE
# shorter line - return the longest #
t
FI
FI
END # print longest lines # ;
- find the logest lines from standard inoout #
VOID( print longest lines( stand in, read line( stand in ) ) ) </lang>
AutoHotkey
This was fun to implement. How I bypassed the restrictions: SubStr() returns part of a string starting from somewhere. If it goes past the end of a string, it returns "" which can be treated as false. StrLen() returns the length of a string. Thus: If this line contains a character at position "longestLength" then append it to the output. If the top line of the output does not contain a character at the position of the last character in this line of the input, then reset the output to be this line *and* set "LongestLength" to be the length of this line. did I break any rules? <lang AutoHotkey>input = ( a bb ccc ddd ee f ggg ) longestLen := 0, buffer := "" Loop Parse, input, `n {
top := SubStr(buffer, 1, InStr(buffer, "`n"))
StringReplace, top, top, `n
If SubStr(A_LoopField, LongestLen) ; at least as long
buffer .= A_LoopField "`n"
If !SubStr(top, StrLen(A_LoopField)) ; longer
buffer := A_LoopField "`n", LongestLen := StrLen(A_LoopField)
} MsgBox % buffer</lang>
AWK
<lang awk>#!/usr/bin/awk -f BEGIN {
maxlen = 0; lenList = 0;
}
{
if (length($0)>maxlen) {
lenList = 1; List[lenList] = $0; maxlen = length($0);
} else if (length($0)==maxlen)
List[++lenList]=$0; }
END {
for (k=1; k <= lenList; k++) print List[k];
}</lang>
Output:
ccc ddd ggg
BBC BASIC
Key to this solution are the functions FNcmp, which compares the lengths of two strings without using comparison operators, and FNinc, which increments an integer without using arithmetic operators. It also strictly adheres to the requirement to use only integer and string data types (no arrays or pointers) and avoids the use of LEN. <lang bbcbasic> DIM buffer% 65535
bufptr% = buffer%
longest$ = " "
ON ERROR PRINT $$buffer%; : END
REPEAT
READ A$
IF FNcmp(A$, longest$) THEN
IF FNcmp(longest$, A$) ELSE bufptr% = buffer%
longest$ = A$
$bufptr% = A$
WHILE ?bufptr%
bufptr% = FNinc(bufptr%)
ENDWHILE
?bufptr% = 10
bufptr% = FNinc(bufptr%)
ENDIF
UNTIL FALSE : REM Loops until 'Out of data' error
END
DATA a, bb, ccc, ddd, ee, f, ggg
DEF FNcmp(a$, b$) : REM Returns LEN(a$)>=LEN(b$) [if b$<>""]
LEFT$(a$, 65535) = b$
= INSTR(a$, b$)
DEF FNinc(i%) : REM Returns i%+1
FOR i% = i% TO i% : NEXT
= i%</lang>
Output:
ccc ddd ggg
C
<lang c>#include <stdio.h>
- include <string.h>
int cmp(const char *p, const char *q) { while (*p && *q) p = &p[1], q = &q[1]; return *p; }
int main() { char line[65536]; char buf[1000000] = {0}; char *last = buf; char *next = buf;
while (gets(line)) { strcat(line, "\n"); if (cmp(last, line)) continue; if (cmp(line, last)) next = buf; last = next; strcpy(next, line); while (*next) next = &next[1]; }
printf("%s", buf); return 0; }</lang>Running it:<lang>% printf "a\nbb\nccc\nddd\nee\nf\nggg" | ./a.out ccc ddd ggg</lang> Note that the above code never checked for memory bounds and long input can overrun the buffers. It's intentionally made this way to keep it simple, please don't complicate it by adding safety features: if you are really concerned with that, below is a second method that can handle arbitrary length input.<lang c>#include <stdio.h>
- include <stdio.h>
- include <stdlib.h>
- include <string.h>
int inc(int x) { return (int)&((char *)x)[1]; } int dec(int x) { return (int)&((char *)x)[-1]; } int gt(int x, int y) { while (y && x) y = dec(y), x = dec(x); return x; }
int eq(int x, int y) { return !gt(x, y) && !gt(y, x); }
int add(int x, int y) { while(y) x = inc(x), y = dec(y); return x; }
/* strlen(a) + 1 */ int length(const char *a) { char *x = 0; // assuming (int)(char*)0 == 0 if (!a) return 0; while (*a) a++, x++; return (int)x; }
char *str_cat(char *a, const char *b) { int len = add(1, add(length(a), length(b))); if (!(a = realloc(a, len))) abort(); return strcat(a, b); }
char *get_line(char *l, FILE *fp) { int c, len = 0; char tmp[2] = {0};
*l = 0; while ((c = fgetc(fp)) != EOF) { *tmp = c; len = inc(len);
l = str_cat(l, tmp); if (eq(*tmp, '\n')) return l; }
*tmp = '\n'; return len ? str_cat(l, tmp) : l; }
int main() { int l1, l2; char *line = malloc(1), *buf = malloc(1), *longest = malloc(1); while (1) { line = get_line(line, stdin);
if (!(l1 = length(line))) break; l2 = length(longest);
if (gt(l1, l2)) { *buf = *longest = 0; longest = str_cat(longest, line); } else if (gt(l2, l1)) continue;
buf = str_cat(buf, line); } printf("%s", buf);
free(buf); free(longest); free(line);
return 0; }</lang>
Here is a more concise variation which exits (with a non-zero return code) if it encounters a buffer overflow:
<lang c>#include <stdio.h>
- include <stdlib.h>
- include <string.h>
int longer(const char *p, const char *q) {
while (*p && *q) p = &p[1], q = &q[1];
return *p;
}
int main() {
char line[100000];
char buf[1100001];
char *linend= &line[99999];
char *bufend= &buf[1000000];
char *last = buf;
char *next = buf;
memset(line, 1, 100000);
memset(buf, 1, 1100001);
buf[0]= buf[1100000]= 0;
while (fgets(line, 100000, stdin)) {
if (!*linend) exit(1);
if (longer(last, line)) continue;
if (!longer(bufend, line)) exit(1);
if (longer(line, last)) next = buf;
last = next;
strcpy(next, line);
while (*next) next = &next[1];
}
printf("%s", buf);
exit(0);
}</lang>
Clojure
<lang lisp>ns longest-string
(:gen-class))
(defn longer [a b]
" if a is longer, it returns the characters in a after length b characters have been removed otherwise it returns nil " (if (or (empty? a) (empty? b)) (not-empty a) (recur (rest a) (rest b))))
(defn get-input []
" Gets the data from standard input as a lazy-sequence of lines (i.e. reads lines as needed by caller
Input is terminated by a zero length line (i.e. line with just <CR> "
(let [line (read-line)]
(if (> (count line) 0)
(lazy-seq (cons line (get-input)))
nil)))
(defn process []
" Returns list of longest lines "
(first ; takes lines from [lines longest]
(reduce (fn [[lines longest] x]
(cond
(longer x longest) [x x] ; new longer line
(not (longer longest x)) [(str lines "\n" x) longest] ; append x to previous longest
:else [lines longest])) ; keep previous lines & longest
["" ""] (get-input))))
(println "Input text:") (println "Output:\n" (process)) </lang>
- Output:
Input text: a bb ccc ddd ee f ggg Output: ccc ddd ggg
D
<lang d>import std.stdio, std.array;
/// Return a.length - b.length if positive, 0 otherwise. int longer(string a, string b) {
while (!a.empty && !b.empty)
a.popFront(), b.popFront();
return a.length;
}
void main() {
string longest, lines;
foreach (string line; stdin.lines())
if (longer(line, longest))
lines = longest = line;
else if (!longer(longest, line))
lines ~= line;
writeln(lines);
}</lang>
- Output:
ccc ddd ggg
Go
<lang go>package main
import (
"bufio" "os"
)
func main() {
in := bufio.NewReader(os.Stdin)
var blankLine = "\n"
var printLongest func(string) string
printLongest = func(candidate string) (longest string) {
longest = candidate
s, err := in.ReadString('\n')
defer func() {
recover()
defer func() {
recover()
}()
_ = blankLine[0]
func() {
defer func() {
recover()
}()
_ = s[len(longest)]
longest = s
}()
longest = printLongest(longest)
func() {
defer func() {
recover()
os.Stdout.WriteString(s)
}()
_ = longest[len(s)]
s = ""
}()
}()
_ = err.(error)
os.Stdout.WriteString(blankLine)
blankLine = ""
return
}
printLongest("")
}</lang> Description: It's basically the recursion+exceptions solution used by others, but staying close to the restrictions.
Restriction 1. The program actually has no operators at all, much less comparison operators. By the Go language specification, assignment is a statement, for example, and an index into a string is simply an "index." For this program, comparisons that control program flow are ones that happen during expression evaluation and then either do or do not trigger a run-time panic, the rough equivalent of throwing an exception in other languages.
Restriction 2. No arithmetic is done on numeric types, and in fact there are no variables of any numeric type in the program. While numeric values do appear at points during expression evaluation (the len function, for example) no arithmetic is explicitly done with them. The compiler certainly generates arithmetic instructions for address calculations underlying an index expression, for example; but the source code here simply supplies numbers as indexes, and relies on the compiler to figure out what arithmetic is appropriate.
Restriction 3. Other than integer and string, data types used are:
- Function: Main is a function, and there is extensive use of function literals in the program.
- os.File: os.Stdin and os.Stdout are predefined in package os.
- bufio.Reader: Used to wrap os.Stdin so the convenient ReadString function can be used to get individual input strings.
- error: A predefined type in Go, returned by many library functions (such as bufio.ReadString.)
- byte: While there are no variables of type byte in the program, single byte values appear at various points in expression evaluation. A byte is the result of indexing into a string, for example.
The spirit of the challenge seems to be prohibiting easy ways of doing things until the only ways left are considered novel. I don't consider recursion a particularly novel way of implementing a list, but it's obviously allowed as a solution so I used it. Avoiding arithmetic was fairly easy using the fact that the Go len function returns the length of a string, but that strings are zero based. Thus,
if a[len(b)] panics, it means that len(a) <= len(b) if a[len(b)] does not panic, it means that len(a) > len(b)
The above expressions avoid arithmetic, but not all comparisons, because error values are typically tested and branched on. Eliminating all comparisons leaves no boolean values in the program and no way to use if statements, which in Go require a boolean condition. Conditional flow control is implemented with the following device: <lang go>func() {
// 1. statements executed in either case
// 2. func below is a closure that captures free variables
// now, although the defer statement keeps the function
// from running until later
defer func() {
// 5. function runs either when panic happens, or
// at the time of a normal function return.
recover() // this stops panic mode
// 6. statements executed in either case, just
// before function returns
}()
// 3. more statements executed in either case
// 4. an expression that may or may not panic
// 4a. conditional code. executed only if no panic happens
return // 7. function return happens in either case
}()</lang> A complication of course is that sometimes you want to conditionally execute code if the expression panics. Without a boolean value to invert, this case requires introducing an extra layer of func..defer..recover with a different expression contrived that will panic with the opposite effect.
Groovy
Solution: recursive <lang groovy>def longer = { a, b ->
def aa = a, bb = b
while (bb && aa) {
bb = bb.substring(1)
aa = aa.substring(1)
}
aa ? a : b
}
def longestStrings longestStrings = { BufferedReader source, String longest = ->
String current = source.readLine()
def finalLongest = current == null \
? longest \
: longestStrings(source,longer(current,longest))
if (longer(finalLongest, current) == current) {
println current
}
return finalLongest
}</lang>
Test: <lang groovy>def source = new BufferedReader(new StringReader(a bb ccc ddd ee f ggg))
longestStrings(source)</lang>
Output:
ggg ddd ccc
Icon and Unicon
String Scanning / Pattern Matching Solution
<lang Icon>procedure main(arglist)
local b # the current buffer (string)
local l # the last string
local L # a \n delimited accumulation of all the longest strings
while b := read() do {
/l := b # primes l on first pass
b ? ( move(*l), if move(1) then L := (l := b) || "\n" else if move(0) then L := (\L|"") || b || "\n")
# move(*l) - fails if b is not l characters long
# move(1) - succeeds/fails if the string is longer and triggers a reset of L
}
write(\L)
end</lang>
Sample Output:
ccc ddd ggg
Recursive Solution
Here is a recursive solution using only single character substring-ing (no string scanning/pattern matching). <lang Icon>procedure main()
longest(".") # needs a single character seed to throw away
end
procedure longest(Longest)
Line := read() | return Longest # read until we can return the longest strings
if Line[*Longest] then Longest := Line # prime/reset Longest
Longest := longest(Longest) # use stack to hold multiples
if Line[*Longest] then write(Line) # write only longest strings,
# Line must be at least as long as Longest
return Longest # feed back longest for length
end</lang>
Sample Output:
ggg ddd ccc
J
<lang j> isempty =. (0 [ 0&{) :: 1: NB. 0=#
compare =. ($:&}.)`((0 1,:_1 0) {~ <@,&isempty)@.(+.&isempty) NB. *@-&#
add =. ,`(,:@[)`] @. (compare {:)
> add&.>/ (}: , ,:&.>@{:) ;: 'a bb ccc ddd ee f ggg'
ccc ddd ggg</lang>
Description:
isempty fetches the first element from an array, and traps the error. The result is false (0) if there was a first element, true (1) otherwise.
compare uses isempty recursively (dropping an element from the array each time) until one (or both) are empty. Once one of the lists is empty a result the pair of numbers from isempty on the two lists is used to pull a value out of an table - the sign of this value indicates which of the two lists was greater. In other words, you get a result of _1 0 or 1, depending on whether one is larger, or neither is larger or the other is larger.
add uses the result from compare to pick from a list of functions which do the right thing when combining lists (add to the current list of results or replace it).
The final expression sets things up for add to function properly and then extracts the result when add is done. (You might think of this as being somewhat like a Haskell monad - though perhaps that analogy should be avoided since a J monad is something very different. J's use of the word "monad" came about decades ago and is related to the use of the word in music theory.)
Java
Translation of Python via D <lang java>import java.io.File; import java.util.Scanner;
public class LongestStringChallenge {
public static void main(String[] args) throws Exception {
String lines = "", longest = "";
try (Scanner sc = new Scanner(new File("lines.txt"))) {
while(sc.hasNext()) {
String line = sc.nextLine();
if (longer(longest, line))
lines = longest = line;
else if (!longer(line, longest))
lines = lines.concat("\n").concat(line);
}
}
System.out.println(lines);
}
static boolean longer(String a, String b) {
try {
String dummy = a.substring(b.length());
} catch (StringIndexOutOfBoundsException e) {
return true;
}
return false;
}
}</lang>
ccc ddd ggg
Julia
<lang julia>function longer(a, b)
try b[endof(a)] catch return true end return false
end
function printlongest(io::IO)
lines = longest = ""
while !eof(io)
line = readline(io)
if longer(line, longest)
longest = lines = line
elseif !longer(longest, line)
lines *= "\n" * line
end
end
println(String(lines))
end
const str = IOBuffer("a\nbb\nccc\nddd\nee\nf\nggg") printlongest(str)</lang>
- Output:
ccc ddd ggg
Kotlin
<lang scala>// version 1.1.0
import java.io.File import java.util.*
fun longer(a: String, b: String): Boolean =
try {
a.substring(b.length)
false
}
catch (e: StringIndexOutOfBoundsException) {
true
}
fun main(args: Array<String>) {
var lines = ""
var longest = ""
val sc = Scanner(File("lines.txt"))
while(sc.hasNext()) {
val line = sc.nextLine()
if (longer(longest, line)) {
longest = line
lines = longest
}
else if (!longer(line, longest))
lines = lines.plus("\n").plus(line) // using 'plus' to avoid using '+'
}
sc.close()
println(lines);
println()
// alternatively (but cheating as library functions will use comparisons and lists under the hood)
println(File("lines.txt").readLines().groupBy { it.length }.maxBy { it.key }!!.value.joinToString("\n"))
}</lang>
- Output:
ccc ddd ggg ccc ddd ggg
Lua
<lang lua>function longer(s1, s2)
while true do
s1 = s1:sub(1, -2)
s2 = s2:sub(1, -2)
if s1:find('^$') and not s2:find('^$') then
return false
elseif s2:find('^$') then
return true
end
end
end
local output = local longest =
for line in io.lines() do
local islonger = longer(line, longest)
if islonger and longer(longest, line) then
output = output .. line .. '\n'
elseif islonger then
longest = line
output = line .. '\n'
end
end
print(output)</lang>
This solution is ispired by the Python one, but since in Lua even an empty string has a boolean true value, it had to be slightly altered. Testing whether a string is empty is done by searching for the Lua string pattern '^$'. If it is found, i.e. the examined string is empty, string.find returns the position of the match, or nil> if it didn't match. Since in Lua any number is true, we just test for the boolean value of the result.
Note that the longer function returns true even if both strings have the same length, so the return value can either be true or false and we can avoid using a comparison or equality operator in interpreting this return value.
The longer function also avoids using the length operator (#), because in the comments on the restrictions of this task it is stated that "Representing the length of any string as a number should also be avoided.". Otherwise it could have been written shorter and faster like this:
<lang lua>function longer(s1, s2)
if s1:sub(#s2):find('^$') then
return false
else
return true
end
end
</lang>
Mathematica
<lang mathematica>FixedPoint[
StringReplace[#,
x : "\n" | StartOfString ~~ a : Except["\n"] ... ~~ "\n" ~~
b : Except["\n"] ... ~~ y : "\n" | EndOfString :>
x <> Switch[((#1 + #2) + Abs[#1 - #2])/2 &[StringLength@a,
StringLength@b], Except[StringLength@a], b,
Except[StringLength@b], a, _, a <> "\n" <> b] <> y] &, "a
bb
ccc
ddd
ee
f
ggg"]</lang>
- Output:
ccc
ddd
ggg
MATLAB / Octave
<lang MATLAB>function longestString(file);
fid = fopen(file);
maxlen = 0; L = {};
while ~feof(fid)
line = fgetl(fid);
if (length(line)>maxlen)
maxlen = length(line);
L = {line};
elseif (length(line)==maxlen)
L{end+1} = line;
end;
end;
fclose(fid);
disp(L);
end;
</lang>
Output:
L = {
[1,1] = ccc
[1,2] = ddd
[1,3] = ggg
} OCaml
Without the restrictions, the standard way to solve the task would be to iterate the lines, compare their length, and accumulate the longest lines in a list. Here we bypass the restriction of not using comparison operators (in the function cmp) by taking advantage that characters of strings are indexed from 0 to [length - 1] and trying to access to a character which index is the length of the other string (if it is beyond the end of the string an exception is raised). The other restriction of not using lists is easily bypassed by concatenating the results instead of accumulating it in a list.
<lang ocaml>let input_line_opt ic =
try Some (input_line ic)
with End_of_file -> None
let cmp s1 s2 =
try ignore(s1.[String.length s2]); 1 (* s1 is longer *)
with _ ->
try ignore(s2.[String.length s1]); -1 (* s2 is longer *)
with _ -> 0 (* both same length *)
let () =
let ic = open_in Sys.argv.(1) in
let rec loop longest acc =
match input_line_opt ic with
| Some line ->
( match cmp line longest with
| 1 -> loop line (line ^ "\n")
| 0 -> loop line (acc ^ line ^ "\n")
| _ -> loop longest acc )
| None ->
close_in ic;
print_string acc
in
loop "" ""</lang>
Pascal
In this version, inc() is used instead of additions. It still has a comparison.
<lang pascal>program LongestStringChallenge_1(input, output);
var
Line: string;
Lines: array of string;
position, len: integer;
begin
if not eoln(input) then
begin
len := 1;
position := 0;
readln (line);
setlength(lines, len);
lines[position] := line;
while not eoln(input) do
begin
readln (line);
if length(line) = length(lines[0]) then
begin
inc(position);
inc(len);
setlength(lines, len);
lines[position] := line;
end;
if length(line) > length(lines[0]) then
begin
position := 0;
len := 1;
setlength(lines, 1);
lines[0] := line;
end;
end;
for position := low(lines) to high(lines) do
writeln (lines[position]);
end;
end.</lang>
Output:
% ./LongestStringChallenge_1
a
b
ccc
ddd
ee
f
ggg
ccc
ddd
ggg
This version uses range check exceptions for the comparisons of string lengths. The range checks are compiler specific. With FreePascal its requires the use of the type ANSIstring instead of "classic" type string.
<lang pascal>program LongestStringChallenge_2(input, output);
{$mode ObjFPC}
{$rangechecks on}
uses
SysUtils;
var
Line: ANSIstring;
Lines: array of ANSIstring;
position: integer;
tester: char;
begin
if not eoln(input) then
begin
readln (line);
position := 0;
setlength(lines, 1);
lines[0] := line;
while not eoln(input) do
begin
readln (line);
try
tester := lines[0][length(line)];
try
tester := line[length(lines[0])];
inc(position);
setlength(lines, succ(position));
lines[position] := line;
except
end;
except
position := 0;
setlength(lines, 1);
lines[0] := line;
end;
end;
for position := low(lines) to high(lines) do
writeln (lines[position]);
end;
end.</lang>
Output:
% ./LongestStringChallenge_2
a
b
ccc
ddd
ee
f
ggg
ccc
ddd
ggg
Perl
<lang perl>#!/usr/bin/perl -n
END{ print $all }
substr($_, length($l)) and $all = $l = $_
or substr($l, length) or $all .= $_;</lang>
Perl 6
<lang perl6>my $l = ; # Sample longest string seen.
my $a = ; # Accumulator to save longest strings.
while get() -> $s {
my $n = "$s\n";
if $n.substr($l.chars) { # Is new string longer?
$a = $l = $n; # Reset accumulator.
}
elsif !$l.substr($n.chars) { # Same length?
$a ~= $n; # Accumulate it.
}
}
print $a;</lang>
Given the example input, returns:
ccc
ddd
ggg
Phix
A recursive/substring based approach, I can't see any explicit maths/comparisons anyway:
<lang Phix>integer fn = open(command_line()[2],"r") -- (reading the source file)
function allx(string line)
line[1..-1] = 'x'
return line
end function
function longest(string mask)
object line = gets(fn)
string newmask
if atom(line) then return mask end if
newmask = allx(line)
if not match(mask,newmask) then return longest(mask) end if
mask = longest(newmask)
if match(mask,newmask) then
puts(1,line)
end if
return mask
end function
?longest("x")
close(fn)
</lang>
Of course it is just a thinly disguised version of:
<lang Phix>function longest(integer l)
object line = gets(fn)
if line=-1 then return l end if
if l>length(line) then return longest(l) end if
l = longest(length(line))
if l=length(line) then
puts(1,line)
end if
return l
end function
?longest(0)</lang>
PicoLisp
Not sure if this meets the spirit. I would implement it the same way if there were no "restrictions":
<lang PicoLisp>(mapc prinl
(maxi '((L) (length (car L)))
(by length group
(in NIL
(make (until (eof) (link (line)))) ) ) ) )</lang>
Another solution avoids 'group', and builds an associative buffer of lines instead:
<lang PicoLisp>(let Buf NIL
(in NIL
(until (eof)
(let (Line (line) Len (length Line))
(if (assoc Len Buf)
(conc @ (cons Line))
(push 'Buf (cons Len (cons Line))) ) ) ) )
(mapc prinl (cdr (maxi car Buf))) )</lang>
Pike
things of note:
the comparison of strings is done by converting the string into an array of indices: ("abc" becomes ({ 1,2,3 }))
the - operation is the set operation and not a numerical subtraction. it removes all the elements in the second array from the first.
if there are any left, we know that the string is longer.
now, once a longer string is found we call write() to print it.
however we don't write it out directly, but instead we store the call in queue of pikes backend. the backend is used to handle callbacks for non-blocking I/O, and it provides a facility to call functions after a delay of time. (call_out(write, 5, "foo"); calls write after 5 seconds with the argument foo)
before we add the new call to write, we remove all older calls to write since we don't want them anymore.
return -1; starts the backend, which allows pike to execute the remaining call_outs and exit.
<lang Pike>int main(int argc, array argv)
{
string longest = "";
foreach(Stdio.stdin.line_iterator();; string line)
{
if( sizeof(indices(line) - indices(longest)))
{
while(!zero_type(remove_call_out(write)));
longest = line;
call_out(write, 0, line+"\n");
}
else if( !sizeof(indices(longest) - indices(line)))
{
call_out(write, 0, line+"\n");
}
}
call_out(exit, 0.01, 0);
return -1;
}</lang>
PL/I
<lang PL/I>
read: procedure options (main); /* 18 January 2012. */
declare line character (100) varying controlled;
declare text character (100) varying;
declare max_length fixed binary;
declare in file input;
on endfile (in) go to done;
open file (in) title ('/readline.pli,type(text),recsize(100)');
max_length = 0;
do forever;
get file (in) edit (text) (L);
put skip list (text);
if length (text) > max_length then
do;
max_length = length(text);
/* empty the stack */
do while (allocation(line) > 0);
free line;
end;
allocate line;
line = text;
end;
else if length(text) = max_length then
do;
allocate line;
line = text;
end;
end;
done:
put skip list (max_length || ' is the length of the longest line(s)' );
do while (allocation(line) > 0);
put skip list (line);
free line;
end;
end read;
</lang>
output (the above file plus the following 3 lines):
74 is the length of the longest line(s)
put skip list (max_length || ' is the length of the longest line(s)' );
/* Read lines of a file, and print the longest. (If there is more than */
PowerShell
<lang PowerShell>
- Get-Content strips out any type of line break and creates an array of strings
- We'll join them back together and put a specific type of line break back in
$File = ( Get-Content C:\Test\File.txt ) -join "`n"
$LongestString = $LongestStrings =
- While the file string still still exists
While ( $File )
{
# Set the String to the first string and File to any remaining strings
$String, $File = $File.Split( "`n", 2 )
# Strip off characters until one or both strings are zero length
$A = $LongestString
$B = $String
While ( $A -and $B )
{
$A = $A.Substring( 1 )
$B = $B.Substring( 1 )
}
# If A is zero length...
If ( -not $A )
{
# If $B is not zero length (and therefore String is longer than LongestString)...
If ( $B )
{
$LongestString = $String
$LongestStrings = $String
}
# Else ($B is also zero length, and therefore String is the same length as LongestString)...
Else
{
$LongestStrings = $LongestStrings, $String -join "`n"
}
}
}
- Output longest strings
$LongestStrings.Split( "`n" )
</lang>
PowerShell Alternate Version
The list restrictions should not apply here because this is essentially one line of code using only the input and no variables.
<lang PowerShell>
@'
a
bb
ccc
ddd
ee
f
ggg
'@ -split "`r`n" |
Group-Object -Property Length |
Sort-Object -Property Name -Descending |
Select-Object -Property Count, @{Name="Length"; Expression={[int]$_.Name}}, Group -First 1
</lang>
- Output:
Count Length Group
----- ------ -----
3 3 {ccc, ddd, ggg}
PureBasic
<lang purebasic>
Procedure.i ConsoleWrite(t.s) ; compile using /CONSOLE option
OpenConsole()
PrintN (t.s)
CloseConsole()
ProcedureReturn 1
EndProcedure
Procedure.i StdOut(t.s) ; compile using /CONSOLE option
OpenConsole()
Print(t.s)
CloseConsole()
ProcedureReturn 1
EndProcedure
DataSection
s:
Data.s "a"
Data.s "bb"
Data.s "ccc"
Data.s "ddd"
Data.s "ee"
Data.s "f"
Data.s "ggg"
Data.s "~" ; the tilda is only to keep the code compact
e: ; and easy to understand
EndDataSection
l$="" ; memory allocation for strings is automatic
a$="" ; in fact these two lines are unnecessary
Restore s
Repeat
Read.s s$
If s$="~":Break:EndIf
s$+#CRLF$
s=Len(s$):l=Len(l$) ; using s$ allows the use of s as an integer type
If s>l :l$=s$:a$=l$
ElseIf s=l :a$+s$
EndIf
Forever
StdOut(a$)
</lang>
- Output
; Directory of C:\_sys\temp
; 07/14/2012 03:04 PM 4,608 LongestStringChallenge.exe
; 1 File(s) 4,608 bytes
; 0 Dir(s) 434,768,625,664 bytes free
; C:\_sys\temp>LongestStringChallenge.exe
; ccc
; ddd
; ggg
Python
<lang python>import fileinput
- This returns True if the second string has a value on the
- same index as the last index of the first string. It runs
- faster than trimming the strings because it runs len once
- and is a single index lookup versus slicing both strings
- one character at a time.
def longer(a, b):
try:
b[len(a)-1]
return False
except:
return True
longest, lines = ,
for x in fileinput.input():
if longer(x, longest):
lines, longest = x, x
elif not longer(longest, x):
lines += x
print(lines, end=)</lang>
- Sample runs
paddy@paddy-VirtualBox:~$ cat <<! | python3.2 longlines.py
a
bb
ccc
ddd
ee
f
ggg
!
ccc
ddd
ggg
paddy@paddy-VirtualBox:~$ touch nothing.txt
paddy@paddy-VirtualBox:~$ cat nothing.txt | python3.2 longlines.py
paddy@paddy-VirtualBox:~$
Racket
This is an attempt to follow the problem restrictions: use just one list of the complete output (so it's not used as a container of strings), and the work is done by manipulating this list vs the input instead of direct comparisons.
<lang Racket>
- lang racket
(define (newline? c) (equal? c #\newline))
(define eof? eof-object?)
(let loop ([O '()] [C '(#\newline)] [rI '()] [rO '()] [rC '()])
(let* ([i (read-char)] [o (car C)] [i:rI (cons i rI)] [i:rC (cons i rC)])
(cond [(eof? i) (for-each write-char O)]
[(and (newline? o) (newline? i))
(let ([O (reverse i:rC)]) (loop O O '() i:rC i:rC))]
[(newline? i) (loop O O '() rO rO)]
[(newline? o) (loop O C i:rI rO i:rI)]
[else (loop O (cdr C) i:rI rO i:rC)])))
</lang>
REXX
In the REXX language, everything is a string (characters).
read file until not ready
No comparators are used (such as if or when),
no output is produced when the file is empty (or contains all null strings),
no arrays or lists are used, and
no additions or subtractions are used.
<lang rexx>/*REXX program reads a file and displays the longest [widest] record(s) [line(s)].*/
parse arg iFID . /*obtain optional argument from the CL.*/
if iFID== | iFID=="," then iFID= 'LONGEST.TXT' /*Not specified? Then use the default.*/
signal on notReady /*when E-O-F is reached, jump/branch. */
m=0 /*the maximum width (so far). */
do rec=1; _=linein(iFID); w=length(_) /*read a line from the input file. */
do y=rec to 1; m=w; w=0; $=y || _; end /*first initialization.*/
do n=m for 1; end /*for smaller records. */
do z=n to w for 1; $=; end /*found longest record.*/
do x=m to w for 1; $=$'a0d'x || _; end /*append record to $. */
m=max(m, w) /*M: is the maximum length record. */
end /*forever*/
notReady: do j=m for m /*handle the case of no input. */
say substr($, 2) /*display (all) the longest records. */
exit /*stick a fork in it, we're all done. */
end /*j*/</lang>
- input file LONGEST.TXT:
a
bb
ccc
ddd
ee
f
ggg
- output when using the default input:
ccc
ddd
ggg
Dual code (works on TSO and PC)
<lang rexx>/* REXX ***************************************************************
- 27.10.2010 Walter Pachl
- /
Parse Arg fid
If fid= Then Do
"ALLOC FI(IN) DA('N561985.PRIV.V100(LL)') SHR REUSE"
'EXECIO * DISKR IN (STEM L. FINIS' /* read all lines */
'FREE FI(IN)'
End
Else Do
Do i=1 By 1 While lines(fid)>0
l.i=linein(fid)
End
l.0=i-1
End
maxl = 0 /* initialize maximum length */
Do i=1 To l.0 /* loop through all lines */
linl=length(l.i) /* length of current line */
Select
When linl>maxl Then Do /* line longer than preceding */
maxl=linl /* initialize maximum length */
mem.0=1 /* memory has one entry */
mem.1=l.i /* the current line */
lin.1=i /* its line number */
End
When linl=maxl Then Do /* line as long as maximum */
z=mem.0+1 /* new memory index */
mem.z=l.i /* the current line */
lin.z=i /* its line number */
mem.0=z /* memory size */
End
Otherwise /* line is shorter than max. */
Nop /* ignore */
End
End
If mem.0>0 Then Do
Say 'Maximum line length='maxl
Say ' Line Contents'
Do i=1 To mem.0
Say right(lin.i,5) mem.i
End
End
Else
Say 'No lines in input file or file does not exist' </lang>
Maximum line length=5
Line Contents
1 99999
3 +++++
Ring
<lang ring>
- Project : Longest string challenge
- Date : 2017/10/11
- Author : Gal Zsolt (~ CalmoSoft ~)
- Email : <calmosoft@gmail.com>
load "stdlib.ring"
test = ["a", "bb", "ccc", "ddd", "ee", "f", "ggg"]
test1 = []
test2 = []
for n = 1 to len(test)
add(test1, [test[n], len(test[n])])
next
sortFirstSecond(test1, 2)
for n = len(test1) to 2 step -1
if test1[n][2] = test1[n-1][2]
add(test2, test1[n][1])
else
add(test2, test1[n][1])
exit
ok
next
test2 = sort(test2)
see test2 + nl
</lang>
Output:
ccc
ddd
ggg
Ruby
<lang ruby># Without restrictions
BEGIN {
v = [ ]
m = 0
}
n = $_.length
if n == m then
v <<= $_
elsif n > m then
v = [$_]
m = n
end
END {
v.each { |s| puts s }
}</lang>
Then ruby -n longest.rb < file.txt
<lang ruby>h = $stdin.group_by(&:size)
puts h.max.last unless h.empty?</lang>
This uses a hash with arrays as values - but not explicit.
Run BASIC
Uses in memory database
<lang runbasic>sqliteconnect #mem, ":memory:" ' Create in memory DB
- mem execute("CREATE TABLE data(str)") ' And fields to hold the string data
strings$ = "a bb ccc ddd ee f ggg" ' The given string data
while word$(strings$,i + 1," ") <> ""
i = i + 1
#mem execute("INSERT INTO data VALUES('";word$(strings$,i," ");"')") ' insert the strings in to the DB
wend
- mem execute("SELECT length(str) as leng, str FROM data ORDER BY leng desc,str") ' pull data in reverse lenght sequence
WHILE #mem hasanswer()
#row = #mem #nextrow()
leng = #row leng()
str$ = #row str$()
print leng;" ";str$ ' print the data
WEND</lang>
Using a simple sort method
<lang runbasic>strings$ = "a bb ccc ddd ee f ggg" ' The given string data
while word$(strings$,numWords + 1," ") <> "" ' Count the words
numWords = numWords + 1
wend
dim string$(numWords) ' Dimension the string with the word cound
for j = 1 to numWords
string$(j) = word$(strings$,j," ") ' put the words from the string into the string array
next j
h$ = "1"
while h$ <> "" ' The good old simple bubble sort
h$ = ""
for i = 1 to numWords -1
if len(string$(i)) < len(string$(i+1)) then ' sort by length descending
h$ = string$(i)
string$(i) = string$(i+1)
string$(i+1) = h$
end if
next i
wend
for i = 1 to numWords
print len(string$(i));" ";string$(i) ' print out the words in length descending sequence
next i</lang>
3 ccc
3 ddd
3 ggg
2 bb
2 ee
1 a
1 f
Scala
<lang Scala>val longest = scala.io.Source.fromFile(args.head).getLines.toIterable.groupBy(_.length).max._2
println(longest mkString "\n")</lang>
Sidef
<lang ruby>var l = ; # Sample longest string seen.
var a = ; # Accumulator to save longest strings.
STDIN.each { |n|
n.substr(l.len) ? (a = n; l = n)
: (!l.substr(n.len) && a.concat!(n));
}
print a;</lang>
Tcl
Uses only string comparisons for equality and glob-style matching
<lang tcl>#!/usr/bin/env tclsh
set longest z
set output ""
while {[gets stdin line] != -1} {
set comparison [string repeat z [string length $line]]
if {$longest eq $comparison} {
# this line is equally long
append output $line \n
} elseif {[string match ${longest}z* $comparison]} {
# this line is longer
set longest $comparison
set output "$line\n"
}
}
puts -nonewline $output</lang>
Test:
$ ./longest.tcl <<END
> a
> bb
> ccc
> ddd
> ee
> f
> ggg
> END
ccc
ddd
ggg
$ ./longest.tcl </dev/null
$
VBScript
The solution uses the Mid function to compare string lengths.
<lang vb>
'Read the input file. This assumes that the file is in the same
'directory as the script.
Set objfso = CreateObject("Scripting.FileSystemObject")
Set objfile = objfso.OpenTextFile(objfso.GetParentFolderName(WScript.ScriptFullName) &_
"\input.txt",1)
list = ""
previous_line = ""
l = Len(previous_line)
Do Until objfile.AtEndOfStream
current_line = objfile.ReadLine
If Mid(current_line,l+1,1) <> "" Then
list = current_line & vbCrLf
previous_line = current_line
l = Len(previous_line)
ElseIf Mid(current_line,l,1) <> "" And Mid(current_line,(l+1),1) = "" Then
list = list & current_line & vbCrLf
End If
Loop
WScript.Echo list
objfile.Close
Set objfso = Nothing
</lang>
- Output:
ccc
ddd
ggg
XSLT 2.0
This XSLT 2.0 style-sheet...
<lang><xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output indent="yes" encoding="UTF-8" omit-xml-declaration="yes" />
<xsl:template match="/*">
<t><xsl:copy-of select="for $l in max( for $s in s return string-length($s))
return s[string-length(.) eq $l]" /></t>
</xsl:template>
</xsl:stylesheet></lang>
...when applied to this input...
<lang><t>
a
bb
ccc
ddd
ee
f
ggg
</t></lang>
...yields...
<lang><t>
ccc
ddd
ggg
</t></lang>
zkl
The ops used are: remove a character from the beginning of a string, testing if a string is "" or not, , string concatenation, integer equality.
To decide which of two strings is longer, a character is removed from each until one is empty. If one still has text, it is longer.
<lang zkl>fcn longer(a,b){ //-->0,1,2 (same, a longer, b longer)
while(a and b){a=a.del(0); b=b.del(0);}
if (not a and not b) return(0); // a & b same length
if(a) return(1); // a is longer
2 // b is longer
}
text:=a:=ask("text: ").strip();
while(b:=ask("text: ").strip()){
switch(longer(a,b)){
case(0){ text=String(text,"\n",b) } // a.len()==b.len()
case(1){ } // a.len>b.len()
case(2){ text=a=b } // a.len>b.len()
} //switch
}
println(text);</lang>
- Output:
text: a
text: bb
text: ccc
text: ddd
text: ee
text: f
text: ggg
text:
The longest line(s) are:
ccc
ddd
ggg
- Output:
text:
text:
The longest line(s) are:
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