how to plot function in scilab - function

I have 2 functions here:
function y=f(x)
y=(x*%e^x)-cos(x);
endfunction
function c=regulafalsi(f, x0, x1, tol)
i=1;
n(1)=100;
if f(x0)*f(x1) < 0
a(1)=x0;
b(1)=x1;
c(1)=a(1)-f(a(1))*(b(1)-a(1))/(f(b(1))-f(a(1)));
printf('i.\t\t a\t\t c\t\t b\t\t f(c)\t Galat \n');
printf('%2d \t %11.4f \t %11.4f \t %11.4f \t %11.4f \n',i,a(i),c(i),b(i),f(c(i)));
while abs(n(i))>=tol,
if f(a(i))*f(c(i))< 0
a(i+1)=a(i);
b(i+1)=c(i);
end
if f(a(i))*f(c(i))> 0
a(i+1)=c(i);
b(i+1)=b(i);
end
c(i+1)=a(i+1)-f(a(i+1))*(b(i+1)-a(i+1))/(f(b(i+1))-f(a(i+1)));
n(i+1)=abs((c(i+1)-c(i))/(c(i+1)));
printf('%2d \t %11.4f \t %11.4f \t %11.4f \t %11.4f \t %7.4f \n',i+1,a(i+1),c(i+1),b(i+1),f(c(i+1)),n(i+1));
i=i+1;
end
else
printf(' ');
end
endfunction
How to plot those function?
I tried plot(y,c) not working.


You have two Scilab functions but only the first of them represents a mathematical function that could be plotted. The second one is a root-finding procedure which outputs some numerical data; it's not meant to be fed into plot command.
To plot function f, you can do the following (e.g., at the end of your script, or in command window)
x = linspace(0,1)
plot(x,f(x))
Here 0, 1 are the endpoints of the interval on which to plot. The linspace command creates an array of numbers distributed over that interval, which is fed into f and then the result into plot.
In order for this to work, you will need to adjust your formula for f slightly:
function y=f(x)
y=(x.*exp(x))-cos(x);
endfunction
The dot before * indicates the multiplication is to be done component wise. This allows this function to process arrays of numbers at once.
Your root-finding function (the method of false position, apparently) works fine. For example, executed with
regulafalsi(f,0,1,0.01)
it returns
i. a c b f(c) Galat
1 0.0000 0.3147 1.0000 -0.5199
2 0.3147 0.4467 1.0000 -0.2035 0.2956
3 0.4467 0.4940 1.0000 -0.0708 0.0957
4 0.4940 0.5099 1.0000 -0.0236 0.0312
5 0.5099 0.5152 1.0000 -0.0078 0.0102
6 0.5152 0.5169 1.0000 -0.0025 0.0033
The root around 0.516 has been found.

Related

Finding the location of ones in a bit mask - Julia

I have a series of values that are each being stored as UInt16. Each of these numbers represents a bitmask - these numbers are commands that have been sent to a microprocessor telling it which pins to set high or low. I would like to parse this arrow of commands to find out which pins were being set high each time in such a way that is easier to analyse later.
Consider the example value 0x3c00, which in decimal is 15360 and in binary is 0011110000000000. Currently I have the following function
function read_message(hex_rep)
return findall.(x -> x .== '1',bitstring(hex_rep))
end
Which gets called on every element of the array of UInt16. Is there a better/more efficient way of doing this?

The best approach probably depends on how you want to handle vectors of hex-values. But here's an approach for processing a single hex which is much faster than the one in the OP:
function readmsg(x::UInt16)
N = count_ones(x)
inds = Vector{Int}(undef, N)
if N == 0
return inds
end
k = trailing_zeros(x)
x >>= k + 1
i = N - 1
inds[N] = n = 16 - k
while i >= 1
(x, r) = divrem(x, 0x2)
n -= 1
if r == 1
inds[i] = n
i -= 1
end
end
return inds
end

I can suggest padding your vector into a Vector{UInt64} and use that to manually construct a BitVector. The following should mostly work (even for input element types other than UInt16), but I haven't taken into account specific endianness you might want to respect:
julia> function read_messages(msgs)
bytes = reinterpret(UInt8, msgs)
N = length(bytes)
nchunks, remaining = divrem(N, sizeof(UInt64))
padded_bytes = zeros(UInt8, sizeof(UInt64) * cld(N, sizeof(UInt64)))
copyto!(padded_bytes, bytes)
b = BitVector(undef, N * 8)
b.chunks = reinterpret(UInt64, padded_bytes)
return b
end
read_messages (generic function with 1 method)
julia> msgs
2-element Vector{UInt16}:
0x3c00
0x8000
julia> read_messages(msgs)
32-element BitVector:
0
0
0
0
0
0
0
0
0
⋮
0
0
0
0
0
0
0
1
julia> read_messages(msgs) |> findall
5-element Vector{Int64}:
11
12
13
14
32
julia> bitstring.(msgs)
2-element Vector{String}:
"0011110000000000"
"1000000000000000"
(Getting rid of the unnecessary allocation of the undef bit vector would require some black magic, I belive.)

Want to be able to save cities that are large into a vector in Clojure

So i want to take cities from a csv file that passes through my large function(which determines if they have a population of over 500000) and if true the city gets saved into a vector. But when I try to save it with my function all the letters are double spaced from each other?
Here is what I have so far:
(defn closest-city-pairs []
(with-open [rdr (reader)]
(vec
(for [line (drop 1 (line-seq rdr))
:let [y (string/split line #",")]
:when (= true (large(y 0)))]
(let [newVec (into [] (y 0))]
(print newVec))))))
Here's the output I'm getting:
[T o r o n t o][M o n t r Θ a l][V a n c o u v e r]...
How do I go about making it single space so it outputs are [Toronto][Montreal]...?

There's a subtle detail: Strings are collections in Clojure (collections of the individual characters), which, mixed with into can cause these issues.
into takes 2 collections and "pours" the second into the first:
user=> (into [1 2 3] (range 4 7))
[1 2 3 4 5 6]
user=> (into [1 2 3] "Hello")
[1 2 3 \H \e \l \l \o]
In the second example, all 5 characters of the string Hello are "added" as 5 new elements of the vector.
In your case, what you probably want is conj:
user=> (conj [1 2 3] "Hello")
[1 2 3 "Hello"]
...now, since you are creating a new vector with just one element, you can use the vector's syntax directly, with something like (let [newVec [(y 0)]] ...), which means that newVec will be a 1-element vector containing the value returned by (y 0), which is the city name in your case.
PS: I see some weird characters in your city names in your example (Q u Θ b e c instead of Q u é b e c), so you might need to check the encoding of your CSV file (eg. save it as UTF-8 for better results).

Octave: "'ncx2cdf' undefined" error

I'm trying to run a M-file from octave but I get this error:
Octave evaluation error: 'ncx2cdf' undefined

Apparently the non-central chi-square distribution is simply defined as 1 minus the Marcum Q function. The signal package at octave-forge provides an implementation for this function (seemingly compatible with matlab).
Therefore you could presumably write your own ncx2cdf function simply as follows:
function Out = myncx2cdf (X, V, Delta)
Out = 1 - marcumq (sqrt (Delta), sqrt (X), V/2);
end
Confirmed in matlab:
>> X = randi(100, [1,20]); V = 4; Delta = 10;
>> ncx2cdf(X, V, Delta)
ans =
1.0000 0.9410 0.9999 1.0000 1.0000 1.0000 1.0000 0.5549 0.6093 0.9410 1.0000 0.9410 1.0000 0.9279 1.0000 0.9920 0.8183 0.9410 1.0000 0.9997
>> 1 - marcumq(sqrt(Delta), sqrt(X), V/2)
ans =
1.0000 0.9410 0.9999 1.0000 1.0000 1.0000 1.0000 0.5549 0.6093 0.9410 1.0000 0.9410 1.0000 0.9279 1.0000 0.9920 0.8183 0.9410 1.0000 0.9997
Octave session for the same X, V, and Delta:
octave:34> pkg load signal
octave:35> 1 - marcumq(sqrt(Delta), sqrt(X), V/2)
ans =
1.00000 0.94105 0.99988 1.00000 1.00000 1.00000 0.99996 0.55492 0.60929 0.94105 1.00000 0.94105 1.00000 0.92793 1.00000 0.99203 0.81831 0.94105 1.00000 0.99972
Note that the degrees of freedom parameter V is restricted to even values with this implementation; if you'd like to use odd degrees of freedom too, e.g. 5, this could be interpolated from the result for V=4 and V=6 (this seems to work well in practice).

here is a easy implementation of non-central chi square distribution:
function f = ncx2pdf(x, n, lambda, term = 32)
f = exp(-lambda/2) * arrayfun(#(x) sum_expression([0:term],x,n,lambda), x);
function t = sum_expression(j,v,n,l)
# j is vector, v is scalar.
numerator = (l/2).^j .* v.^(n/2+j-1) * exp(-v/2);
denominator = factorial(j) .* 2.^(n/2+j) .* gamma(n/2+j);
t = sum(numerator ./ denominator);
end
end
here is the function file , put it in your octave path.

Function does not return the list correctly

I have written a code for adding the numbers from two different text files. For a very big data 2-3 GB, I get the MemoryError. So, I am writing a new code using some functions to avoid loading the whole data into memory.
This code opens an input file 'd.txt' an reads the numbers after some lines from a bigger data as following:
SCALAR
ND 3
ST 0
TS 1000
1.0
1.0
1.0
SCALAR
ND 3
ST 0
TS 2000
3.3
3.4
3.5
SCALAR
ND 3
ST 0
TS 3000
1.7
1.8
1.9
and adds to the number have read from a smaller text file 'e.txt' as following:
SCALAR
ND 3
ST 0
TS 0
10.0
10.0
10.0
The result is written in a text file 'output.txt' like this:
SCALAR
ND 3
ST 0
TS 1000
11.0
11.0
11.0
SCALAR
ND 3
ST 0
TS 2000
13.3
13.4
13.5
SCALAR
ND 3
ST 0
TS 3000
11.7
11.8
11.9
The code which I prepared:
def add_list_same(list1, list2):
"""
list2 has the same size as list1
"""
c = [a+b for a, b in zip(list1, list2)]
print(c)
return c
def list_numbers_after_ts(n, f):
result = []
for line in f:
if line.startswith('TS'):
for node in range(n):
result.append(float(next(f)))
return result
def writing_TS(f1):
TS = []
ND = []
for line1 in f1:
if line1.startswith('ND'):
ND = float(line1.split()[-1])
if line1.startswith('TS'):
x = float(line1.split()[-1])
TS.append(x)
return TS, ND
with open('d.txt') as depth_dat_file, \
open('e.txt') as elev_file, \
open('output.txt', 'w') as out:
m = writing_TS(depth_dat_file)
print('number of TS', m[1])
for j in range(0,int(m[1])-1):
i = m[1]*j
out.write('SCALAR\nND {0:2f}\nST 0\nTS {0:2f}\n'.format(m[1], m[0][j]))
list1 = list_numbers_after_ts(int(m[1]), depth_dat_file)
list2 = list_numbers_after_ts(int(m[1]), elev_file)
Eh = add_list_same(list1, list2)
out.writelines(["%.2f\n" % item for item in Eh])
the output.txt is like this:
SCALAR
ND 3.000000
ST 0
TS 3.000000
SCALAR
ND 3.000000
ST 0
TS 3.000000
SCALAR
ND 3.000000
ST 0
TS 3.000000
The addition of lists does not work, besides I checked separately the functions, they work. I don't find the error. I changed it a lot, but it does not work. Any suggustion? I really appreciate any help you can provide!

You can use grouper to read files by fixed count of lines. Next code should works if order of lines in groups is unchanged.
from itertools import zip_longest
#Split by group iterator
#See http://stackoverflow.com/questions/434287/what-is-the-most-pythonic-way-to-iterate-over-a-list-in-chunks
def grouper(iterable, n, padvalue=None):
return zip_longest(*[iter(iterable)]*n, fillvalue=padvalue)
add_numbers = []
with open("e.txt") as f:
# Read data by 7 lines
for lines in grouper(f, 7):
# Suppress first SCALAR line
for line in lines[1:]:
# add last number in every line to array (6 elements)
add_numbers.append(float(line.split()[-1].strip()))
#template for every group
template = 'SCALAR\nND {:.2f}\nST {:.2f}\nTS {:.2f}\n{:.2f}\n{:.2f}\n{:.2f}\n'
with open("d.txt") as f, open('output.txt', 'w') as out:
# As before
for lines in grouper(f, 7):
data_numbers = []
for line in lines[1:]:
data_numbers.append(float(line.split()[-1].strip()))
# in result_numbers sum elements of two arrays by pair (6 elements)
result_numbers = [x + y for x, y in zip(data_numbers, add_numbers)]
# * unpack result_numbers as 6 arguments of function format
out.write(template.format(*result_numbers))

I had to change some small things in the code and now it works but just for small input files, because many variables are loaded into memory. Can you please tell me how can I work with yield.
from itertools import zip_longest
def grouper(iterable, n, padvalue=None):
return zip_longest(*[iter(iterable)]*n, fillvalue=padvalue)
def writing_ND(f1):
for line1 in f1:
if line1.startswith('ND'):
ND = float(line1.split()[-1])
return ND
def writing_TS(f):
for line2 in f:
if line2.startswith('TS'):
x = float(line2.split()[-1])
TS.append(x)
return TS
TS = []
ND = []
x = 0.0
n = 0
add_numbers = []
with open("e.txt") as f, open("d.txt") as f1,\
open('output.txt', 'w') as out:
ND = writing_ND(f)
TS = writing_TS(f1)
n = int(ND)+4
f.seek(0)
for lines in grouper(f, int(n)):
for item in lines[4:]:
add_numbers.append(float(item))
i = 0
for l in grouper(f1, n):
data_numbers = []
for line in l[4:]:
data_numbers.append(float(line.split()[-1].strip()))
result_numbers = [x + y for x, y in zip(data_numbers, add_numbers)]
del data_numbers
out.write('SCALAR\nND %d\nST 0\nTS %0.2f\n' % (ND, TS[i]))
i += 1
for item in result_numbers:
out.write('%s\n' % item)

Code Golf: Connecting the dots

Locked. This question and its answers are locked because the question is off-topic but has historical significance. It is not currently accepting new answers or interactions.
You may remember these drawings from when you were a child, but now it's time to let the computer draw them (in full ascii splendour). Have fun!
Description:
The input are multiple lines (terminated by a newline) which describe a 'field'. There are 'numbers' scattered across this field (seperated by whitespace). All lines can be considered to be the same length (you can pad spaces to the end).
the numbers always start at 1
they follow the ordering of the natural numbers: every 'next number' is incremented with 1
every number is surrounded by (at least) one whitespace on its left and right
Task:
Draw lines between these numbers in their natural order
(1 -> 2 -> 3 -> ...N) (assume N <= 99) with the following characteristics:
replace a number with a '+' character
for horizontal lines: use '-'
for vertical lines: use '|'
going left and down or right and up: /
going left and up or right and down: \
Important notes:
When drawing lines of type 4 and 5 you can assume (given the points to connect with coordinates x1, y1 and x2, y2) that distance(x1,x2) == distance(y1,y2). Or in other words (as user jball commented): "consecutive elements that are not horizontally or vertically aligned always align to the slope of the slash or backslash".
It is important to follow the order in which the dots are connected (newer lines can strike out older lines).
-- Sample input 1 --
8
7 6
10 9
5
3 4
11
12 13
1 2
-- Sample output 1 --
+
/|
/ +--+
+--------+ \
/ \
/ +
/ |
/ +--+
+ |
\ |
+------------------------+
+--------------------------+
-- Sample input 2 --
64
63
62 61
1 65
66 57 58
2 56 59 45
67 55 46
3 44
54 60 47
53 52 49 48
4 51 50 43
5 42
41
6 23
22 25 26 40
20 21 24 34
7 13 12 33
19 27 32
14 35
8 15
16
39
17 18 28 31 36
9 38
10 11 29 30 37
-- Sample output 2 -- (unicorn reference)
+
/+
//
//
//
/+--+
+ + \
| + +-\+
+ \ + \ +
/ + + \ +\
+ \ \ | +
| + + +/
| +--+ +-------+/
+ +--+ +
/ \
+ +
| +
+ + /
\ +\ +---+ +
\ +--+ + \ /+
+ +--+ / \ /+|
/ | |+ + /+ |
/ + || / // +
+ + || / // /
\ + || / // /
\ | || / +/ /
\ +---+ + +\ +
+ | | | +|
+--+ +---+ +
Winner:
Shortest solution (by code character count). Input can be read via standard input.

Commodore 64 BASIC - 313 chars
EDIT: See below for the golfed version
A little trip down the memory lane with PET graphics, POKEs and PEEKs and everything :)
The program operates directly in the screen memory, so you just go ahead, clear the screen, place your dots, and type RUN:
You have to wait a minute or so while it finds the dots and then it starts to draw. It isn't fast - you can actually see the lines being drawn, but that's the coolest part :)
Golfed version:
Commodore BASIC seems like a great language for golfing, because it doesn't require whitespace :) You can also shorten most of the commands by entering an unshifted first letter followed by a shifted second letter. For example, POKE can be typed as P[SHIFT+O], which appears as P┌ on the screen:

Perl, 222 char (211)
Perl, 384 365 276 273 253 225 222 218 211 chars (222 when contest ended). Newlines are for "readability" only and are not included in the character count.
Last edit: no longer overwriting $", and printing #S directly
$_=join'',#S=map{$n=s/$/$"x97/e;(/./g)[0..95],$/}<>;
while(/\b$n /){$S[$q=$-[0]]='+';($P,$Q)=sort{$a-$b}$q,$p||$q;
for(qw'\98 |97 /96 -1'){/\D/;$S[$P]=$&until($Q-$P)%$'||$Q<=($P+=$')}
$n++;$p=$q}s/\d/ /,print for#S
Explanation:
$_=join'',#S=map{$n=s/$/$"x97/e;(/./g)[0..95],$/}<>;
This task will be easier if all the lines are the same length (say, 97 characters).
This statement takes each line of input, replaces the end-of-line character with
96 spaces, then pushes the first 96 characters plus a newline into the array #S.
Note we are also setting $n=1, as 1 is the first number we'll look for in
the input.
The join statement creates a single string from the array #S.
It is more convenient to use the scalar variable $_ for pattern matching, and more convenient to use the array #S for making updates to the picture.
while(/\b$n /){
Search for the number $n in the variable $_. Evaluating regular expressions in Perl
has several side-effects. One is to set the special variable $-[0] with the position of the start of the matched pattern within the matched string. This gives us the position of the number $n in the string $_ and also the array #S.
Of course, the loop will end when $n is high enough that we can't find it in the input.
$S[$q=$-[0]]='+';
Let $q be the position of the number $n in the string $_ and the array #S,
and assign the character '+' at that position.
$P=($p||=$q)+$q-($Q=$q>$p?$q:$p)
($P,$Q)=sort{$a-$b}$p||$q,$q;
The first time through the loop, set $p to $q. After the
first time, $p will hold the previous value of $q (which
will refer to the position in the input of the previous number).
Assign $P and $Q such that $P=min($p,$q),
$Q=max($p,$q)
for(qw'\98 |97 /96 -1'){
By construction, consecutive numbers are either
connected by a vertical line. Since the input is constructed
to have 97 characters on each line, this case means that
$p-$q is divisible by 97.
"aligned to the slope of a backslash", which would make
$p-$q divisible by 98
"aligned to the slope of a forward slash", which would make
$p-$q divisible by 96
on the same horizontal line
The elements of this list encode the possible number of positions
between line segments, and the character to encode that segment.
/\D/;
Another trivial regex evaluation. As a side-effect, it sets the
special variable $& (the MATCH variable) to the line segment
character (\ | / or -) and $' (the POSTMATCH variable) to
the number (98 97 96 or 1) encoded in the list element.
$S[$P]=$&until($Q-$P)%$'||$Q<=($P+=$')
This statement draws the line segment between two numbers.
If $Q-$P is divisible by $', then keep incrementing $P by $'
and assigning the character $& to $S[$P] until $P reaches $Q.
More concretely, for example if $Q-$P is divisible by 97, then
increment $P by 97 and set $S[$P]='|'. Repeat until $P>=$Q.
$n++;$p=$q
Prepare for the next iteration of the loop. Increment $n to the
next number to search for in the input, and let $p hold the
position of the previous number.
s/\d/ /,print for#S
Output the array, converting any leftover digits (from double
digit identifiers in the input where we only overwrote the first
digit with a '+') to spaces as we go.

MS-DOS Batch (yes, you read right!)
I often hear (or read) people say batch isn't very powerful and you can't do much with them, well to them I say, behold, the power of BATCH!
The actual script (script.bat):
set file=%~1
call :FindNextNum 1
for /F "tokens=2 delims=:" %%i IN ('find /c /V "" "%file%"') DO set /a totalLines=%%i
set maxLen=0
for /F "delims=" %%i IN (%file%) DO (
call :CountChars "%%i"
if /i !charCount! gtr !maxLen! set maxLen=!charCount!
)
for /L %%i IN (0,1,%totalLines%) DO set "final_%%i=" & for /L %%j IN (0,1,%maxLen%) DO set "final_%%i=!final_%%i! "
:MainLoop
set currLineNum=%lineNum%
set currCol=%linePos%
set currNum=%nextNum%
set /a targetNum=%currNum%+1
call :FindNextNum %targetNum%
if "%nextNum%"=="" goto MainEnd
REM echo %currNum% -^> %nextNum%
if /I %currLineNum% lss %lineNum% (
call :DrawLine %currCol% %currLineNum% %linePos% %lineNum%
) else (
call :DrawLine %linePos% %lineNum% %currCol% %currLineNum%
)
goto MainLoop
:MainEnd
for /L %%i IN (0,1,%totalLines%) DO echo.!final_%%i!
goto:eof
:DrawLine
if /I %2 equ %4 goto:DrawHoriz
set "char=" & set "pos=%1" & set "inc=0"
if /I %1 LSS %3 set "char=\" & set "pos=%1" & set "inc=1"
if /I %1 GTR %3 set "char=/" & set "pos=%1" & set "inc=-1"
for /L %%i IN (%2,1,%4) DO call :DrawChar %%i !pos! %char% & set /a "pos+=%inc%"
goto:DrawEnds
:DrawHoriz
set "start=%1+1" & set "end=%3"
if /I %start% gtr %end% set "start=%3+1" & set "end=%1"
set /a lineEnd=%end%+1
set lineEnd=!final_%2:~%lineEnd%!
for /L %%i IN (%start%,1,%end%) DO set final_%2=!final_%2:~0,%%i!-
set final_%2=!final_%2!!lineEnd!
:DrawEnds
call :DrawChar %2 %1 +
call :DrawChar %4 %3 +
goto:eof
:DrawChar
set /a skip2=%2+1
if "%3"=="" (
set final_%1=!final_%1:~0,%2!^|!final_%1:~%skip2%!
) else (
set final_%1=!final_%1:~0,%2!%3!final_%1:~%skip2%!
)
goto:eof
:CountChars
set charCount=0
set val=%~1
:CountChars_loop
if not "%val:~1%"=="" (
set /a charCount+=1
set val=!val:~1!
goto CountChars_loop
)
goto:eof
:FindNextNum
for /F "delims=" %%i IN ('type "%file%" ^| find /V /N ""') DO (
for /F "tokens=1,2 delims=[]" %%j IN ("%%i") DO (
set /a lineNum=%%j-1
call :FindNext_internal "%%k" %1
if /I !nextNum! equ %1 goto :eof
)
)
goto:eof
:FindNext_internal
set currLine=%~1
set linePos=0
:FindNext_internal_loop
call :NextNumInLine "%currLine%"
set /a linePos+=%spaceInterval%
if "%nextNum%"=="" goto :EOF
if /I %nextNum% equ %2 goto :EOF
set /a spaceInterval+=1
set /a linePos+=1
if /I %nextNum% GTR 9 set /a "spaceInterval+=1" & set /a linePos+=1
set currLine=!currLine:~%spaceInterval%!
goto FindNext_internal_loop
:NextNumInLine
set nextNum=
for /F %%i IN (%1) DO set /a nextNum=%%i
if "%nextNum%"=="" goto :eof
set /a spaceInterval=0
set val=%~1
:NextNumInLine_loop
if "%val:~0,1%"==" " (
set /a spaceInterval+=1
set val=!val:~1!
goto NextNumInLine_loop
)
goto :eof
And this is how you call it
echo off
setlocal ENABLEDELAYEDEXPANSION
call script.bat input.txt
where "input.txt" is a file that contains the input for the "program".
P.S. This isn't actually optimized for line length yet, I've already spent a couple of hours getting to this point and now I need to sleep... I'll see if I can improve it tomorrow (currently 'script.bat' sits at 2755 bytes)

Rebmu: 218 chars
Ma L{-|\/}Qb|[sg?SBaB]Da|[feSm[TfiSrj[spAsp]iT[++Tbr]]t]Xa|[i?A]Ya|[i?FImHDa]Ca|[skPCmSCaBKfsA]wh[Jd++N][roG[xJyJ]]Bf+GwhB[JcB Ff+GiF[KcF HqXkXj VqYkYju[chCbPClEZv1[ezH2[eeHv3 4]]e?A+bRE[hV]f]]chJeFIlSCj{+}{+ }Jk Bf]wM
I'm getting pretty good at reading and editing it natively in its pig-latin form. (Though I do use line breaks!!) :)
But here's how the dialect is transformed by the interpreter when the case-insensitive "mushing" trick is boiled away, and one gets accustomed to it. I'll add some comments. (Tips: fi is find, fe is foreach, sp is a space character, i? is index, hd is head, ch is change, sk is skip, pc is pick, bk is break, i is if, e is either, ee is either equal, ad nauseum)
; copy program argument into variable (m)atrix
m: a
; string containing the (l)etters used for walls
l: {-|\/}
; q is a "b|function" (function that takes two parameters, a and b)
; it gives you the sign of subtracting b from a (+1, -1, or 0)
q: b| [sg? sb a b]
; d finds you the iterator position of the first digit of a two digit
; number in the matrix
d: a| [fe s m [t: fi s rj [sp a sp] i t [++ t br]] t]
; given an iterator position, this tells you the x coordinate of the cell
x: a| [i? a]
; given an iterator position, this tells you the y coordinate of the cell
y: a| [i? fi m hd a]
; pass in a coordinate pair to c and it will give you the iterator position
; of that cell
c: a| [sk pc m sc a bk fr a]
; n defaults to 1 in Rebmu. we loop through all the numbers up front and
; gather their coordinate pairs into a list called g
wh [j: d ++ n] [ro g [x j y j]]
; b is the (b)eginning coordinate pair for our stroke. f+ returns the
; element at G's current position and advances G (f+ = "first+")
; advance g's iteration position
b: f+ g
wh b [
; j is the iterator position of the beginning stroke
j: c b
; f is the (f)inishing coordinate pair for our stroke
f: f+ g
; if there is a finishing pair, we need to draw a line
i f [
; k is the iterator position of the end of the stroke
k: c f
; the (h)orizontal and (v)ertical offsets we'll step by (-1,0,1)
h: q x k x j
v: q y k y j
u [
; change the character at iterator location for b (now our
; current location) based on an index into the letters list
; that we figure out based on whether v is zero, h is zero,
; v equals h, or v doesn't equal h.
ch c b pc l ez v 1 [ez h 2 [ee h v 3 4]]
; if we update the coordinate pair by the offset and it
; equals finish, then we're done with the stroke
e? a+ b re [h v] f
]
]
; whether we overwrite the number with a + or a plus and space
; depends on whether we detect one of our wall "letters" already
; one step to the right of the iterator position
ch j e fi l sc j {+} {+ }
; update from finish pair to be new begin pair for next loop iteration
j: k
b: f
]
; write out m
w m
Both the language and sample are new and in an experimental stage. For instance, ad couldn't be used to add together vectors and matrices before I changed it to help with this sample. But I think that's just the sort of thing that a language designed specifically for code golf has to have anyway. It's a subtle line between "language" and "library".
Latest source with comments available on GitHub

Haskell, 424 chars
Current char count: 424 430 451 466 511 515 516 518 525 532 541 545 550 556 569 571 577 582 586 592.
import List
x%c=[(i,c)|i<-x]
l k p q|p>q=l k q p|True=head[[p,p+j..q]%c|m<-zip[k-1,k,k+1,1]"/|\\-",let (j,c)=m,mod(q-p)j==0]
w=map snd
q(k,m,x)z=w$sort$nubBy((==)&fst)$x%'+'++(concat$zipWith(l k)x$tail x)++z%'\n'++[1..m]%' '
r(z,m,x)=q(last z,m-1,w$sort x)z
u[(m,_)]n x=(-m::Int,n):x;u _ _ x=x
t(z,n,x)s|s=="\n"=(n:z,n+1,x)|True=(z,n+length s,u(reads s)n x)
y&x=(.x).y.x
main=interact$r.foldl t([],1,[]).groupBy((&&)&(>' '))
This version takes a lot of inspiration from the original Haskell entry below, but makes some significant changes. Most importantly, it represents image locations with a single index, not a pair of coordinates.
There are some changes:
The input must now have all lines padded to the same length (allowed by the rules.)
No longer needs either language extension
Original version:
(Needs -XTupleSections, and maybe -XNoMonomorphismRestriction)
import List
b=length
f=map
g=reverse
a(x,y)" "=(x,y+1)
a(x,y)z=([y,read z]:x,y+b z)
x%y=[min x y+1..max x y-1]
j([x,y],[w,z])|y==z=f(,'-')$f(y,)$x%w|x==w=f(,'|')$f(,x)$y%z|(y<z)==(x<w)=f(,'\\')$zip(y%z)$x%w|True=f(,'/')$zip(y%z)$g$x%w
k 0='\n'
k _=' '
y&x=(.x).y.x
y?x=f y.sort.x.concat
r z=snd?(nubBy((==)&fst).g)$[((y,x),k x)|x<-[0..maximum$f b d],y<-[1..b d]]:[((y,x),'+')|[x,y]<-e]:(f j$zip e$tail e)where d=f(groupBy$(&&)&(>' '))$lines z;e=tail?f g$zipWith(f.(:))[1..]$f(fst.foldl a([],1))d
main=interact r
Explanation:
(1) d=...: Splits the input into spaces and numbers, e.g.
z = " 6 5\n\n1 2\n\n 4 3\n\n 7"
=> d = [[" ","6"," "," ","5"],[],["1"," "," "," "," "," "," "," ","2"],[],[" "," "," "," ","4"," "," "," ","3"],[],[" ","7"]]
(2) e=...: Converts d into a list of (y, x) coordinates for each number.
e = [[1,3],[9,3],[9,5],[5,5],[5,1],[2,1],[2,7]]
--- // 1 2 3 4 5 6 7
(3)
[((y,x),k x)|...] is an empty board. (k returns a space or a \n depending on the x-coordinate.)
[((y,x),'+'))|...] are the plus signs at the numbers.
(f j$zip e$tail e) are the lines connecting the numbers. (j maps a pair of coordinates into a list of (coordinate, character) which represents a line.)
These 3 components are concatenated and filtered to form the actual output. Note that the order is important, so that nubBy(...).g can only keep the last character in the same location.

AWK - 296 317 321 324 334 340
Not a prize winner (yet), but I am pleased with the effort (line breaks for display). This new version uses VT-100 escape sequences. The '^[' is just one character, Escape!!! Cut and paste will not work with this version, since the sequence "^[" has to be replaced with the real ESC character. To make it forum friendly, ESC could be specified as "\0x1b", but it takes too much space...
BEGIN{FS="[ ]"}{for(j=i=0;i<NF;j+=length(g)){if(g=$++i){x[g]=k=i+j;y[g]=NR;
m=m>k?m:k}}}END{printf"^[[2J[%d;%dH+",Y=y[i=1],X=x[1];while(a=x[++i])
{a-=X;b=y[i]-Y;t=a?b?a*b>0?92:47:45:124;A=a?a>0?1:-1:0;B=b?b>0?1:-1:0;
for(r=a?a*A:b*B;--r;){printf"^[[%d;%dH%c",Y+=B,X+=A,t}
printf"^[[%d;%dH+",Y+=B,X+=A}}
The older standard version
BEGIN{FS="[ ]"}{for(j=i=0;i<NF;j+=length(g)){if(g=$++i){x[g]=k=i+j;y[g]=NR;
m=m>k?m:k}}}END{q[X=x[1],Y=y[i=1]]=43;while(a=x[++i]){a-=X;b=y[i]-Y;
t=a?b?a*b>0?92:47:45:124;A=a?a>0?1:-1:0;B=b?b>0?1:-1:0;for(r=a?a*A:b*B;--r;
q[X+=A,Y+=B]=t);q[X+=A,Y+=B]=43}for(j=0;++j<NR;){for(i=0;i<m;){t=q[i++,j];
printf"%c",t?t:32}print}}
Now a little explanation
# This will break the input in fields separated by exactly 1 space,
# i.e. the fields will be null or a number.
BEGIN{FS="[ ]"}
# For each line we loop over all fields, if the field is not null
# it is a number, hence store it.
# Also account for the fact the numbers use space.
# Also, find the maximum width of the line.
{
for(j=i=0;i<NF;j+=length(g)){
if(g=$++i){
k=j+i;x[g]=k;y[g]=NR;m=m>k?m:k
}
}
}
# Once we have all the data, let start cooking.
END{
# First, create a matrix with the drawing.
# first point is a +
q[X=x[1],Y=y[i=1]]=43;
# loop over all points
while(a=x[++i]){
# Check next point and select character
# If a == 0 -> -
# If b == 0 -> |
# If a and b have same sign -> \ else /
a-=X;b=y[i]-Y;t=a?b?a*b>0?92:47:45:124;
# there is no sgn() function
A=a?a>0?1:-1:0;B=b?b>0?1:-1:0;
# Draw the line between the points
for(k=0;++k<(a?a*A:b*B);){
q[X+=A,Y+=B]=t
}
# store + and move to next point
q[X+=A,Y+=B]=43
}
# Now output all lines. If value in point x,y is 0, emit space
for(j=0;++j<NR;){
for(i=0;i<m;){
t=q[i++,j];printf("%c",t?t:32)
}
print
}
}

C, 386
402 386 character in C. Newlines after the first are only for readability.
#include <stdio.h>
int x[101],y[101],c=1,r,w,h,b,i,j,k,m,n;
int main(){
while((b=getchar())-EOF)
b-' '?b-'\n'?ungetc(b,stdin),scanf("%d",&b),x[b]=c++,y[b]=h,c+=b>9:(w=c>w?c:w,++h,c=1):++c;
for(r=0;r<h&&putchar('\n');++r)
for(c=0;c<w;++c){
for(b=' ',i=2,m=x[1]-c,n=y[1]-r;j=m,k=n,m=x[i]-c,n=y[i]-r,x[i++];)
b=j|k&&m|n?j*m>0|k|n?k*n<0?(j-k|m-n?j+k|m+n?j|m?b:'|':'/':'\\'):b:'-':'+';
putchar(b);
}
}

Intel Assembler
Assembled size: 506 bytes
Source: 2252 bytes (hey, it's not a trivial problem this one)
To Assemble: Use A86
To Run: Tested with a WinXP DOS box. Invocation jtd.com < input > output
mov ax,3
int 10h
mov ax,0b800h
mov es,ax
mov ah,0bh
int 21h
mov bx,255
cmp al,bl
mov dh,bh
mov si,offset a12
push offset a24
je a1
mov si,offset a14
a1: inc bl
a2: mov dl,255
call si
cmp al,10
jb a4
a3: cmp al,10-48
jne a1
inc bh
mov bl,dh
jmp a2
a4: mov dl,al
call si
cmp al,10
jae a5
mov ah,dl
aad
mov dl,al
a5: mov di,dx
mov ch,al
shl di,2
mov [di+a32],bx
cmp bl,[offset a30]
jb a6
mov [offset a30],bl
a6: cmp bh,[offset a31]
jb a7
mov [offset a31],bh
a7: push offset a19
mov al,80
mul bh
add al,bl
adc ah,0
add ax,ax
lea di,[di+2+a32]
mov [di],ax
add di,2
cmp di,[a22-3]
jbe a8
mov [a22-3],di
mov [a25-3],di
a8: mov di,ax
mov al,dl
aam
cmp ah,0
je a10
a9: add ah,48
mov es:[di],ah
add di,2
a10:add al,48
mov es:[di],al
mov al,ch
inc bl
jmp a3
a11:jmp si
a12:mov ah,0bh
int 21h
cmp al,255
jne a15
mov ah,8
int 21h
a13:cmp al,13
je a11
sub al,48
ret
a14:mov ah,1
int 21h
cmp al,26
jne a13
mov si,offset a15
ret
a15:cmp dl,255
je a16
mov al,32
ret
a16:mov si,offset a32 + 4
lodsw
mov cx,ax
mov dx,ax
lodsw
mov di,ax
mov b es:[di],1
mov bp,0f000h
call a26
add sp,6
mov bx,[a22-3]
mov ax,[offset a31]
inc ax
a17:mov bp,[offset a30]
a18:mov b[bx],32
inc bx
dec bp
jnz a18
mov w[bx],0a0dh
add bx,2
dec ax
jnz a17
mov b[bx],'$'
add w[a30],2
a19:lodsw
xchg ax,dx
cmp ah,dh
lahf
mov bl,ah
cmp al,dl
lahf
shr bl,6
shr ah,4
and ah,12
or bl,ah
mov bh,0
shl bx,3
a20:mov b es:[di],43
a21:mov al,b[a30]
mul ch
add al,cl
adc ah,0
mov bp,ax
mov b[bp+100h],43
a22:add di,[bx + a29]
add cl,[bx + a29 + 4]
add ch,[bx + a29 + 6]
mov b es:[di],1
mov al,[bx + a29 + 2]
mov [a21-1],al
mov [a22-1],al
mov bp,01000h
call a26
cmp di,[si]
jne a20
mov al,es:[di+2]
sub al,48
cmp al,10
jae a23
mov b es:[di+2],0
a23:mov b[a21-1],43
mov b[a22-1],43
mov b es:[di],43
lodsw
ret
a24:mov al,b[a30]
mul ch
add al,cl
adc ah,0
mov bp,ax
mov b[bp+100h],43
a25:mov dx,[a22-3]
mov ah,9
int 21h
ret
a26:pusha
a27:mov cx,0ffffh
a28:loop a28
dec bp
jnz a27
popa
ret
a29:dw -162,92,-1,-1,-2,45,-1,0,158,47,-1,1,0,0,0,0,-160,124,0,-1
a30:dw 0
a31:dw 0,0,0,160,124,0,1,0,0,0,0,-158,47,1,-1,2,45,1,0,162,92,1,1
a32:
Interesting features: self modifying code, animated output (the second example works, but is too big to display), abuse of 'ret' to implement a loop counter, interesting way of determining line/movement direction.

F#, 725 chars
open System
let mutable h,s,l=0,Set.empty,Console.ReadLine()
while l<>null do
l.Split([|' '|],StringSplitOptions.RemoveEmptyEntries)
|>Seq.iter(fun t->s<-s.Add(int t,h,(" "+l+" ").IndexOf(" "+t+" ")))
h<-h+1;l<-Console.ReadLine()
let w=Seq.map(fun(k,h,x)->x)s|>Seq.max
let o=Array2D.create h (w+1)' '
Seq.sort s|>Seq.pairwise|>Seq.iter(fun((_,b,a),(_,y,x))->
let a,b,x,y=if b>y then x,y,a,b else a,b,x,y
o.[b,a]<-'+'
o.[y,x]<-'+'
if b=y then for x in(min a x)+1..(max a x)-1 do o.[y,x]<-'-'
elif a=x then for h in b+1..y-1 do o.[h,x]<-'|'
elif a<x then for i in 1..y-b-1 do o.[b+i,a+i]<-'\\'
else for i in 1..y-b-1 do o.[b+i,a-i]<-'/')
for h in 0..h-1 do
for x in 0..w do printf"%c"o.[h,x]
printfn""
Legend:
h = height
s = set
l = curLine
w = (one less than) width
o = output array of chars
Lines 1-6: I keep a set of (number, lineNum, xCoord) tuples; as I read in each line of input I find all the numbers and add them to the set.
Line 7-8: Then I create an array of output chars, initialized to all spaces.
Line 9: Sort the set (by 'number'), then take each adjacent pair and ...
Lines 10-16: ... sort so (a,b) is the 'highest' of the two points and (x,y) is the other. Put the '+' signs, and then if horizontal, draw that, else if vertical, draw that, else draw the correct diagonal. If the input is not 'valid', then who knows what happens (this code was littered with 'asserts' before I golf-ized it).
Lines 17-19: Print the result

Powershell, 328 304 characters
$i=$l=0;$k=#{}
$s=#($input|%{[regex]::matches($_,"\d+")|%{$k[1*$_.Value]=#{y=$l
x=$_.Index}};$l++;""})
while($a=$k[++$i]){
if($i-eq1){$x=$a.x;$y=$a.y}
do{$d=$a.x.CompareTo($x);$e=$a.y.CompareTo($y)
$s[$y]=$s[($y+=$e)].PadRight($x+1).Remove($x,1).Insert(($x+=$d),
"\-/|+|/-\"[4+$d*3+$e])}while($d-or$e)}$s
and here's a pretty-printed version with comments:
# Usage: gc testfile.txt | dots.ps1
$l=$i=0 # line, dot index (used below)
$k=#{} # hashtable that maps dot index to coordinates
# Apply regular expression to each line of the input
$s=#( $input | foreach{
[regex]::matches($_,"\d+") | foreach{
# Store each match in the hashtable
$k[ 1*$_.Value ] = #{ y = $l; x = $_.Index }
}
$l++; # Next line
"" # For each line return an empty string.
# The strings are added to the array $s which
# is used to produce the final output
}
)
# Connect the dots!
while( $a = $k[ ++$i ] )
{
if( $i -eq 1 ) # First dot?
{
# Current position is ($x, $y)
$x = $a.x;
$y = $a.y
}
do
{
$d = $a.x.CompareTo( $x ) # sign( $a.x - $x )
$e = $a.y.CompareTo( $y ) # sign( $a.y - $y )
$c = '\-/|+|/-\'[ 4 + $d * 3 + $e ] # character '
# Move
$x += $d
$y += $e
# "Replace" the charcter at the current position
# PadRight() ensures the string is long enough
$s[ $y ]=$s[ $y ].PadRight( $x+1 ).Remove( $x, 1 ).Insert( $x, $c )
} while( $d -or $e ) # Until the next dot is reached
}
# Print the resulting string array
$s

Python - 381
import re
b=list(iter(raw_input,''))
c=sum((zip([i]*999,re.finditer('\\d+',x))for i,x in enumerate(b)),[])
d=sorted((int(m.group()),i,m.start())for i,m in c)
e=[[' ']*max(map(len,b))for x in b]
for(t,u,v),(x,y,z)in zip(d,d[1:]+d[-1:]):
e[u][v]='+'
while u!=y or v!=z:i,j=(u<y)-(u>y),(v<z)-(v>z);u+=i;v+=j;e[u][v]=['|','/\\-'[(i==j)+2*(i==0)]][j!=0]
print'\n'.join(map(''.join,e))

C#, 422 chars
758 754 641 627 584 546 532 486 457 454 443 440 422 chars (next time maybe I won't submit so soon.)
using A=System.Console;class B{static int C,o,d,e,G,O=1,f,F,u,n;static
void Main(){var s=A.In.ReadToEnd();A.Clear();while(++u<s.Length){f++;if
(s[u]<32){u++;F++;f= 0;}if(s[u]>32){if(int.Parse(s[u]+""+s[++u])==O){o=
e>f?1:f>e?-1:0;C=d>F?1:F>d?-1:0 ;G=e+o;n=d+C;if(O++>1)while(n!=F||G!=f)
{A.SetCursorPosition(G-=o,n-=C);A.Write( "+/-|\\"[n==d&&G==e?0:n==F&&G
==f?0:C+o==0?1:C==0?2:o==0?3:4]);}e=f;d=F;F=0;f=u=-1 ;}f++;}}A.Read();}}
Usage: run, paste (or type) the input, ensure the last line is terminated, press CTRL-Z or F6, press Enter.
Formatted but still basically unintelligable version:
using A = System.Console;
class B
{
// code golf fun!
static int C, o, d, e, G, O = 1, f, F, u, n;
static void Main()
{
// read the input into a string char by char until EOF
var s = A.In.ReadToEnd();
A.Clear(); // clear console, ready to draw picture
// O is the "dot" number we're looking for
// f is current column
// F is current row
// loop over the field looking for numbers sequentially
// until no more are found
while (++u < s.Length)
{
f++;
// any char <32 is expected to be a CR/LF
// increment the current row and reset the current column
if (s[u] < 32)
{
u++; // skip the other half of the CR/LF pair
F++; // next row
f = 0; // column reset
}
// any char >32 is expected to be a number
if (s[u] > 32)
{
// parse the current + next char and see if it's
// the number we want
if (int.Parse(s[u] + "" + s[++u]) == O)
{
// set up coordinates, compare X1 with X2
// and Y1 with Y2 to figure out line direction
// horizontal direction (same as o=e.CompareTo(f))
o = e > f ? 1 : f > e ? - 1 : 0;
// vertical direction (same as C=d.CompareTo(F))
C = d > F ? 1 : F > d ? - 1 : 0;
// initial offsets compensate for off-by-one
G = e + o;
n = d + C;
// draw the line (except for the very first dot)
if (O++ > 1)
while (n != F || G != f)
{
// update coords and write desired char
A.SetCursorPosition(G -= o, n -= C);
// this lovely line decides which char to
// print, and prints it
A.Write(
"+/-|\\"[n == d && G == e ? 0 : n == F && G
== f ? 0 : C + o == 0 ? 1 : C == 0 ? 2 : o
== 0 ? 3 : 4]);
}
// remember end point of this line, to use as start point
// of next line
e = f;
d = F;
// reset current row (F), column (f), field position (u)
F = 0;
f = u = -1;
}
// bump current column because we parse 2 chars when we
// find a dot
f++;
}
}
A.Read(); // prevent command prompt from overwriting picture
}
}

Here goes!
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int sign(int x) {
if (x < 0)
return -1;
if (x > 0)
return +1;
return 0;
}
#define MAX_ROWS 100
#define MAX_COLS 100
#define MAX_DIGITS 100
int main(void)
{
// Read in the digits
int number[MAX_DIGITS][2];
int rows = 0;
int cols = 0;
char row[MAX_COLS];
int maxvalue = 0;
int i, j, value, x;
for (i = 0; i < MAX_ROWS; i++) {
if (row != fgets(row, MAX_COLS, stdin))
break;
value = 0;
for (j=0; row[j] != 0; j++) {
if (row[j] >= '0' && row[j] <= '9') {
x = j;
value = 0;
do {
value = 10*value + (row[j]-'0');
j++;
} while (row[j] >= '0' && row[j] <= '9');
number[value][0] = i;
number[value][1] = x;
if (maxvalue < value) maxvalue = value;
if (rows < i+1) rows = i+1;
if (cols < x+1) cols = x+1;
}
}
}
// Create an empty field
char field[rows][cols];
memset(field, ' ', rows*cols);
char lines[] = "\\|/-+-/|\\";
int dr,dc;
// Draw the numbers and lines
field[number[1][0]][number[1][1]] = '+';
for (i = 2; i <= maxvalue; ++i) {
int r = number[i-1][0];
int c = number[i-1][1];
int rt = number[i][0];
int ct = number[i][1];
dr = sign(rt-r);
dc = sign(ct-c);
char line = lines[(dr+1)*3+dc+1];
while (r != rt || c != ct) {
r += dr;
c += dc;
field[r][c] = line;
}
field[r][c] = '+';
}
for (i = 0; i < rows; ++i) {
for (j = 0; j < cols; ++j)
putchar(field[i][j]);
putchar('\n');
}
return 0;
}

C#, 638 chars
using System;
using System.Linq;
using System.Text.RegularExpressions;
class C
{
static void Main()
{
int i=0,j;
var p = Console.In.ReadToEnd()
.Split('\n')
.SelectMany(
r =>
{
i++; j =0;
return Regex.Matches(r, "\\s+(\\d+)").Cast<Match>()
.Select(m => { j += m.Length; return new { X = j, Y = i-1, N = int.Parse(m.Groups[1].Value) }; });
}
).OrderBy(a=>a.N).ToList();
var W = p.Max(a => a.X)+1;
var k = new char[W*i+W];
i = 0;
while (i < p.Count)
{
var b = p[i > 0 ? i - 1 : 0]; var a = p[i];
int h = a.Y - b.Y, w = a.X - b.X;
var s = "|-/\\"[h == 0 ? 1 : w == 0 ? 0 : h / w > 0 ? 3 : 2];
while ((h | w) != 0) { k[b.X + w + W * (b.Y + h)] = s; h -= h.CompareTo(0); w -= w.CompareTo(0); }
k[a.X + a.Y * W] = '+';
k[W * ++i] = '\n';
}
Console.Write(k);
}
}

I cannot do multi-line in a comment, so I will demonstrate here.
In the following examples, distance(x1,x2) == distance(y1,y2):
+
|\
+-+
+
|\
| \
+--+
+
|\
| \
| \
+---+
With the rules as explained, distance(x1,x2) == distance(y1,y2)+2:
+\
| \
+--\+
+\
| \
| \
+---\+
+\
| \
| \
| \
+----\+

C++ 637
#include <iostream>
#include <string>
#include <vector>
#define S(x)((x)<0?-1:x>0?1:0)
using namespace std;enum{R=100,C=100,D=100};int main(){string s;
int N[D][2],M=0,q=0,p=0,i,j,V,L,a,b;for(i=0;j=0,(i<R)&&getline(cin,s);i++)
while((j=s.find_first_not_of(" ",j))<=s.size()){L=sscanf(&s[j],"%d",&V);
N[V][0]=i;N[V][1]=j;if(M<V)M=V;if(q<=i)q=i+1;if(p<=j)p=j+1;j+=L+1;}
string F(q*p,' '),l="\\|/-+-/|\\";F[p*N[1][0]+N[1][1]]='+';for(i=2;i<=M;++i){
int r=N[i-1][0],c=N[i-1][1],d=N[i][0],e=N[i][1];for(a=S(d-r),b=S(e-c);r!=d||c!=e;)
r+=a,c+=b,F[p*r+c]=l[(a+1)*3+b+1];F[p*r+c]='+';}for(i=0;i<q;i++)
cout<<string(&F[i*p],p)+"\n";}
Indented, and with a few slightly more meaningful names, that looks like:
#include <iostream>
#include <string>
#include <vector>
#define S(x)((x)<0?-1:x>0?1:0)
using namespace std;
enum{R=100,C=100,D=100};
int main(){
string s;
int N[D][2],M=0,rs=0,cs=0,i,j,V,L,dr,dc;
for(i=0;j=0,(i<R)&&getline(cin,s);i++)
while((j=s.find_first_not_of(" ",j))<=s.size()){
L=sscanf(&s[j],"%d",&V);
N[V][0]=i;
N[V][1]=j;
if(M<V)M=V;
if(rs<=i)rs=i+1;
if(cs<=j)cs=j+1;
j+=L+1;
}
string F(rs*cs,' '),lines="\\|/-+-/|\\";
F[cs*N[1][0]+N[1][1]]='+';
for(i=2;i<=M;++i){
int r=N[i-1][0],c=N[i-1][1],rt=N[i][0],ct=N[i][1];
for(dr=S(rt-r),dc=S(ct-c);r!=rt||c!=ct;)
r+=dr,c+=dc,F[cs*r+c]=lines[(dr+1)*3+dc+1];
F[cs*r+c]='+';
}
for(i=0;i<rs;i++)
cout<<string(&F[i*cs],cs)+"\n";
}
Despite superficial differences, it's a blatant theft of morotspaj's code.