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We are supposed to multiply two 8 bit numbers using shift and add operations of 8085 microprocessor. Answer should be a 16 bit number. Use of shift and add operation is compulsory
To understand the solution you must be familiar with Rotate instructions in 8085,particularly for this solution you need to understand two things
RRC instruction rotates bits to the right and LSB can be checked from carry flag
multiplying number by 2(10 in binary) results in left shif by one bit (verify yourself)
adding number with itself is equivaent to multiplying number by 2(10 in binary) and hence also shift bits by 1 bit
#ORG 8000
//initializing operands
LXI H,7000H //data is stored in 7000H
MOV E,M
MVI D,00H
INX H
MOV A,M
MVI C,08H
LXI H, 0000H
//multiplication algorithm starts here
LOOP : RRC
JNC SKIP
DAD D
//left shift is performed by adding number with itself
//three lines just below this comment is shifting DE pair to left by 1 bit
SKIP: XCHG //exchange HL and DE pair
DAD H //Add HL with itself and store in HL
XCHG //exchange HL and DE
DCR C
JNZ LOOP
SHLD 7050H
HLT
#ORG7000
#DB 25,2A
Suppose we want to multiply two integers 27 and 23. Since 23 (10111 in binary) can be written as
2*11 + 1 = 2*(2*5 + 1) + 1 = ... = 2*(2*(2*(2*(2*0 + 1) + 0) + 1) + 1) + 1. Thus, x * 23 can be expressed as: 2*(2*(2*(2*(2*0 + x) + 0) + x) + x) + x. Observe that the addend terms in each step follows the binary representation of 23 (1, 0, 1, 1, 1). With this observation, we can write the following pseudocode to perform the multiplication x * y using shift and add operations.
let x be the first operand and y be the second one
set sum = 0
repeat
set sum = sum * 2
left shift y by one place
if the overflow bit after the shift is set then
set sum = sum + x
until y ≠ 0
output the sum as the result of x*y
Let x=27 (0x1B) and y=23 (0x17) be two 8-bit intergers, the follwing microprogram performs the required multiplication. As the multiplication may require 16 bits to store the result we use the HL register pair for the calculation.
LXI D,001BH ; DE <- 27(x)
MVI A,17H ; A <- 23(y)
LXI H,0000H ; HL <- 0(sum)
LOOP: DAD H ; sum <- sum*2
STC
CMC ; clear the carry flag before rotate
RAL ; y <- y<<1
JNC SKIP ; if overflow bit from y was not set
DAD D ; sum <- sum + x
SKIP: ORA A ; to update the zero flag
JNZ LOOP
HLT
The result, 27*23 = 621 (0x026D), is available in the HL register pair.
Trying to get the KissFFT fixed point implementation to line up with the DSPIC. The issue is that the fixed point implementation in Kiss is a true fixed point but the dspic does the multiplies and sums in a 40 bit register then shifts down to 16 bits after rounding. The KissFFT can be 16/32 bit fixed point or float. So far the float is the closest matching but I need them to be exact. I'm not sure how many times each frequency bin is updated in the ASM code but as you can see the accumulator result is shifted and rounded each time the bin is updated. I don't have enough FFT knowledge to solve this. If anyone can point me in the correct direction it would be greatly appreciated.
Here is the ASM Code:
.global _FFT
_FFT:
push.d w8
push.d w10
push.d w12
push w14
push CORCON
mov #0x00f1, w7
mov w7, CORCON
push PSVPAG
push w1 ; save return value
mov #0xff00, w7 ; check if w3==COEFFS_IN_DATA
cp w7, w3
bra z, $+6
bset CORCON, #2
mov w3, PSVPAG
mov #1, w3
sl w3, w0, w3 ; w3 = N (1<<log2N)
mov #0x8000, w14
dec2 w0, w12 ; w12 is # of non-trivial stages
mov #4, w0 ; w0 = 4 * (1, 2, 4, ..., (N/2))
mov #0x0018, w9 ; w9->w12
mov #0x8000, w6
lsr w3, w3 ; start of outer loop, w3 = N/2, N/4, N/8, ..., 1
sl w3, #2, w1 ; w1 = 4 * (N/2, N/4, N/8, ..., 1)
mov [w15-2], w10 ; w10->start of butterfly
lsr w0, #2, w4 ; w4 = groups per stage
dec w4, w4
do w4, $+88 ; first butterfly in group has trivial multiplications
add w0, w2, w8
mov w10, w13
add w1, w10, w11
mov [w10++], w4
mpy.n w4*w6, a, [w11]+=2, w5 ; a = Ar/2
msc w5*w6, a, [w10]+=2, w7 ; a = (Ar+Br)/2
mpy.n w6*w7, b, [w11]+=2, w4 ; b = Ai/2
msc w4*w6, b, [w13]+=2 ; b = (Ai+Bi)/2
mac w5*w6, a, [w13]+=2
sub w11, #4, w13
mac w5*w6, a, [w11]+=2, w5 ; a = (Ar-Br)/2
mac w4*w6, b
mac w4*w6, b, [w8]+=2, w6, [w11]+=2, w7, [w13]+=2 ; b = (Ai-Bi)/2
sub w3, #3, w4
do w4, $+28 ; b = previous Bi, w5 = Br, w6 = Wr, w7 = Bi, w8-> Wi, w10-> Ar, w11-> next Br, w13-> previous Bi
lac [w10], a ; a=Ar
add w5, a ; a=Ar+Br
subr w5, [w10], w4 ; w4=Ar-Br
sac.r a, #1, [w10] ; *w10++=__real__(A+B)
lac [++w10], a ; a=Ai
add w7, a ; a=Ai+Bi
subr w7, [w10], w5 ; w5=Ai-Bi
sac.r a, #1, [w10++] ; *w10++=__imag__(A+B)
mpy w4*w6, a, [w8]-=2, w7 ; a=(Ar-Br)*Wr, w7=Wi
msc w5*w7, a, [w13]+=2 ; a=(Ar-Br)*Wr-(Ai-Bi)*Wi, *w13++ = previous Bi
add w0, w8, w8 ; w8->next Wr
mpy w5*w6, b, [w11]+=2, w5 ; b=(Ai-Bi)*Wr, w5=next Br, w11->next Bi
mac w4*w7, b, [w8]+=2, w6, [w11]+=2, w7, [w13]+=2 ; b=(Ai-Bi)*Wr+(Ar-Br)*Wi, w6=next Wr=*w8++, w7=next Bi=*w11++, *w13++=__real__(A-B)*W
lac [w10], a ; epilog
add w5, a
subr w5, [w10], w4
sac.r a, #1, [w10]
lac [++w10], a
add w7, a
subr w7, [w10], w5
sac.r a, #1, [w10++]
mpy w4*w6, a, [w8]+=2, w7
msc w5*w7, a, [w13]+=2
mpy w5*w6, b, [w9]+=4, w6
mac w4*w7, b, [w9]-=4, w6, [w13]+=2
clr a, [w13]+=2
mov w11, w10 ; last instruction in group
sl w0, w0 ; next stage, double twiddle factor offset
dec w12, w12
bra gt, $-104 ; if w12 > 0, do next stage
mov [w15-2], w10 ; last two stages are done simultaneously
mov [w15-2], w13
add w10, #8, w11
lsr w0, #2, w3
dec w3, w3
clr w8
mov #0x4000, w12
clr a, [w9]+=4, w6, [w10]+=2, w4 ; initialize Ar, w6=0x4000
mov [w10++], w5 ; initialize Ai, w10->Br
do w3, $+58
mov #12, w0 ; adjust DOSTART to run prolog only once
add DOSTARTL
bra NC, $+4
inc DOSTARTH
sub w4, [w11], w0 ; w0 = Ar-Cr
bra $+10 ; w4 = Ar, w5 = Ai, w6 = 0x4000, w8->w0, w9->w14, w10->Br, w11->Cr, w12= 0x4000, w13->last Di, w14= 0x8000
add #12, w11 ; start of 22-cycle do loop
msc w5*w7, b, [w10]+=2, w4, [w13]+=2 ; b = new Di
sub w4, [w11], w0 ; w0 = Ar-Cr
clr a, [w9]+=4, w6, [w10]+=2, w5, [w13]+=2
add w4, [w11], w4 ; w4 = Ar+Cr
sub w5, [++w11], w1 ; w1 = Ai-Ci
add w5, [w11++], w5 ; w5 = Ai+Ci, w11->Dr
mpy w4*w6, a, [w10]+=2, w4 ; a = Ar+Cr, w4 = Br, *w13++ = Di
mpy w5*w6, b, [w9]-=4, w7, [w10]+=6, w5 ; b = Ai+Ci, w5 = Bi
sub w4, [w10], w3 ; w3 = Br-Dr
add w4, [w10], w4 ; w4 = Br+Dr
sub w5, [++w10], w2 ; w2 = Bi-Di
add w5, [w10++], w5 ; w5 = Bi+Di, w10->next Ar
mac w4*w6, a ; a = new Ar
mac w5*w6, b, [w13]+=2 ; b = new Ai, *w13++ = Ar
mac w4*w7, a, [w8]+=2, w4, [w13]+=2 ; a = new Br, w4=Ar-Cr, *w13++ = Ai
mac w5*w7, b, [w8]+=2, w5, [w13]+=2 ; b = new Bi, w5=Ai-Ci, *w13++ = Br
mpy w4*w6, a, [w8]+=2, w4 ; a = Ar-Cr, w4 = Bi-Di
mac w4*w6, a, [w13]+=2 ; a = new Cr, *w13++ = Bi
mpy w5*w6, b, [w8]-=6, w5 ; b = Ai-Ci, w5 = Br-Dr
msc w5*w6, b, [w13]+=2 ; b = new Ci, w6 = *w10++, *w13++ = Cr
mac w4*w7, a, [w13]+=2 ; a = new Dr (last instruction of do loop)
msc w5*w7, b, [w13]+=2 ; epilog
sac.r b, [w13]
pop w0 ; cleanup
pop PSVPAG
pop CORCON
pop w14
pop.d w12
pop.d w10
pop.d w8
return
Kiss Code
http://sourceforge.net/projects/kissfft/
This is where I think I need to modify kiss to line up with dspic
# define S_MUL(a,b) ( (a)*(b) )
#define C_MUL(m,a,b) \
do{ (m).r = (a).r*(b).r - (a).i*(b).i;\
(m).i = (a).r*(b).i + (a).i*(b).r; }while(0)
# define C_FIXDIV(c,div) /* NOOP */
# define C_MULBYSCALAR( c, s ) \
do{ (c).r *= (s);\
(c).i *= (s); }while(0)
#define C_ADD( res, a,b)\
do { \
CHECK_OVERFLOW_OP((a).r,+,(b).r)\
CHECK_OVERFLOW_OP((a).i,+,(b).i)\
(res).r=(a).r+(b).r; (res).i=(a).i+(b).i; \
}while(0)
#define C_SUB( res, a,b)\
do { \
CHECK_OVERFLOW_OP((a).r,-,(b).r)\
CHECK_OVERFLOW_OP((a).i,-,(b).i)\
(res).r=(a).r-(b).r; (res).i=(a).i-(b).i; \
}while(0)
#define C_ADDTO( res , a)\
do { \
CHECK_OVERFLOW_OP((res).r,+,(a).r)\
CHECK_OVERFLOW_OP((res).i,+,(a).i)\
(res).r += (a).r; (res).i += (a).i;\
}while(0)
#define C_SUBFROM( res , a)\
do {\
CHECK_OVERFLOW_OP((res).r,-,(a).r)\
CHECK_OVERFLOW_OP((res).i,-,(a).i)\
(res).r -= (a).r; (res).i -= (a).i; \
}while(0)
The dsPIC DSP has a few settings that you can change, I would try disable super saturation in the CORCON register. bit ACCSAT.
You could also try find a Q15() implementation of the dspic fft and use the builtin functions to convert from Q15 to float. I think its _Q15ftoi() and _itofQ15()
Double check if your using the dsPICFJ series or the dsPICEP ? The CORCON register is different between them be careful that you check for the PSV bit.
I have a shader that looks like this:
void main( in float2 pos : TEXCOORD0,
in uniform sampler2D data : TEXUNIT0,
in uniform sampler2D palette : TEXUNIT1,
in uniform float c,
in uniform float th0,
in uniform float th1,
in uniform float th2,
in uniform float4 BackGroundColor,
out float4 color : COLOR
)
{
const float4 dataValue = tex2D( data, pos );
const float vValue = dataValue.x;
const float tValue = dataValue.y;
color = BackGroundColor;
if ( tValue <= th2 )
{
if ( tValue < th1 )
{
const float vRealValue = abs( vValue - 0.5 );
if ( vRealValue > th0 )
{
// determine value and color
const float power = ( c > 0.0 ) ? vValue : ( 1.0 - vValue );
color = tex2D( palette, float2( power, 0.0 ) );
}
}
else
{
color = float4( 0.0, tValue, 0.0, 1.0 );
}
}
}
and I am compiling it like this:
cgc -profile arbfp1 -strict -O3 -q sh.cg -o sh.asm
Now, different versions of Cg compiler creating different output.
cgc version 2.2.0006 is compiling the shader into an assembler code using 18 instructions:
!!ARBfp1.0
PARAM c[6] = { program.local[0..4],{ 0, 1, 0.5 } };
TEMP R0;
TEMP R1;
TEMP R2;
TEX R0.xy, fragment.texcoord[0], texture[0], 2D;
ADD R0.z, -R0.x, c[5].y;
CMP R0.z, -c[0].x, R0.x, R0;
MOV R0.w, c[5].x;
TEX R1, R0.zwzw, texture[1], 2D;
SLT R0.z, R0.y, c[2].x;
ADD R0.x, R0, -c[5].z;
ABS R0.w, R0.x;
SGE R0.x, c[3], R0.y;
MUL R2.x, R0, R0.z;
SLT R0.w, c[1].x, R0;
ABS R2.y, R0.z;
MUL R0.z, R2.x, R0.w;
CMP R0.w, -R2.y, c[5].x, c[5].y;
CMP R1, -R0.z, R1, c[4];
MUL R2.x, R0, R0.w;
MOV R0.xzw, c[5].xyxy;
CMP result.color, -R2.x, R0, R1;
END
# 18 instructions, 3 R-regs
cgc version 3.0.0016 is compiling the shader into an assembler code using 23 instructions:
!!ARBfp1.0
PARAM c[6] = { program.local[0..4], { 0, 1, 0.5 } };
TEMP R0;
TEMP R1;
TEMP R2;
TEX R0.xy, fragment.texcoord[0], texture[0], 2D;
ADD R1.y, R0.x, -c[5].z;
MOV R1.z, c[0].x;
ABS R1.y, R1;
SLT R1.z, c[5].x, R1;
SLT R1.x, R0.y, c[2];
SGE R0.z, c[3].x, R0.y;
MUL R0.w, R0.z, R1.x;
SLT R1.y, c[1].x, R1;
MUL R0.w, R0, R1.y;
ABS R1.z, R1;
CMP R1.y, -R1.z, c[5].x, c[5];
MUL R1.y, R0.w, R1;
ADD R1.z, -R0.x, c[5].y;
CMP R1.z, -R1.y, R1, R0.x;
ABS R0.x, R1;
CMP R0.x, -R0, c[5], c[5].y;
MOV R1.w, c[5].x;
TEX R1, R1.zwzw, texture[1], 2D;
CMP R1, -R0.w, R1, c[4];
MUL R2.x, R0.z, R0;
MOV R0.xzw, c[5].xyxy;
CMP result.color, -R2.x, R0, R1;
END
# 23 instructions, 3 R-regs
The strange thing is that the optimization level for the cg 3.0 doesn't seems to influence anything.
Can someone explain what is going on? Why is the optimization not working and why is the shader longer when I compiled with cg 3.0?
Take a note that I removed comments from the compiled shaders.
This might not be a real answer to the problem but maybe give some more insight. I inspected the generated assembly code a bit and converted it back to high-level code. I tried to compress it as much as possible and remove all copies and temporaries that follow implicitly from the high-level operations. I used b variables as temporary bools and fs as temporary floats. The first one (with the 2.2 version) is:
power = ( c > 0.0 ) ? vValue : ( 1.0 - vValue );
R1 = tex2D( palette, float2( power, 0.0 ) );
vRealValue = abs( vValue - 0.5 );
b1 = ( tValue < th1 );
b2 = ( tValue <= th2 );
b3 = b1;
b1 = b1 && b2 && ( vRealValue > th0 );
R1 = b1 ? R1 : BackGroundColor;
color = ( b2 && !b3 ) ? float4( 0.0, tValue, 0.0, 1.0 ) : R1;
and the second (with 3.0) is:
vRealValue = abs( vValue - 0.5 );
f0 = c;
b0 = ( 0 < f0 );
b1 = ( tValue < th1 );
b2 = ( tValue <= th2 );
b4 = b1 && b2 && ( vRealValue > th0 );
b0 = b0;
b3 = b1;
power = ( b4 && !b0 ) ? ( 1.0 - vValue ) : vValue;
R1 = tex2D( palette, float2( power, 0.0 ) );
R1 = b4 ? R1 : BackGroundColor;
color = ( b2 && !b3 ) ? float4( 0.0, tValue, 0.0, 1.0 ) : R1;
Most parts are essentially the same. The second program does some unneccessary operations. It copies the c variable into a temporary instead of using it directly. Moreover does it switch vValue and 1-vValue in the power computation, so it needs to negate b0 (resulting in one more CMP), whereas the first one does not use a temporary at all (it uses CMP directly instead of SLT and CMP). It also uses b4 in this computation, which is completely unneccessary, because when b4 is false, the result of the texture access is irrelevant, anyway. This results in one more && (implemented with MUL). There is also the unneccessary copy from b1 to b3 (in the first program it is neccessary, but not in the second). And the extremely useless copy from b0 into itself (which is disguised as an ABS, but as the value comes from an SLT, it can only be 0.0 or 1.0 and the ABS degenerates to a MOV).
So the second program is quite similar to the first one with just some additional, but IMHO completely useless instructions. The optimizer seems to have done a worse job compared to the previous(!) version. As the Cg compiler is an nVidia product (and not from some other not to be named graphics company) this behaviour is really strange.
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.
Inspired by http://xkcd.com/710/ here is a code golf for it.
The Challenge
Given a positive integer greater than 0, print out the hailstone sequence for that number.
The Hailstone Sequence
See Wikipedia for more detail..
If the number is even, divide it by two.
If the number is odd, triple it and add one.
Repeat this with the number produced until it reaches 1. (if it continues after 1, it will go in an infinite loop of 1 -> 4 -> 2 -> 1...)
Sometimes code is the best way to explain, so here is some from Wikipedia
function collatz(n)
show n
if n > 1
if n is odd
call collatz(3n + 1)
else
call collatz(n / 2)
This code works, but I am adding on an extra challenge. The program must not be vulnerable to stack overflows. So it must either use iteration or tail recursion.
Also, bonus points for if it can calculate big numbers and the language does not already have it implemented. (or if you reimplement big number support using fixed-length integers)
Test case
Number: 21
Results: 21 -> 64 -> 32 -> 16 -> 8 -> 4 -> 2 -> 1
Number: 3
Results: 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1
Also, the code golf must include full user input and output.
x86 assembly, 1337 characters
;
; To assemble and link this program, just run:
;
; >> $ nasm -f elf collatz.asm && gcc -o collatz collatz.o
;
; You can then enjoy its output by passing a number to it on the command line:
;
; >> $ ./collatz 123
; >> 123 --> 370 --> 185 --> 556 --> 278 --> 139 --> 418 --> 209 --> 628 --> 314
; >> --> 157 --> 472 --> 236 --> 118 --> 59 --> 178 --> 89 --> 268 --> 134 --> 67
; >> --> 202 --> 101 --> 304 --> 152 --> 76 --> 38 --> 19 --> 58 --> 29 --> 88
; >> --> 44 --> 22 --> 11 --> 34 --> 17 --> 52 --> 26 --> 13 --> 40 --> 20 --> 10
; >> --> 5 --> 16 --> 8 --> 4 --> 2 --> 1
;
; There's even some error checking involved:
; >> $ ./collatz
; >> Usage: ./collatz NUMBER
;
section .text
global main
extern printf
extern atoi
main:
cmp dword [esp+0x04], 2
jne .usage
mov ebx, [esp+0x08]
push dword [ebx+0x04]
call atoi
add esp, 4
cmp eax, 0
je .usage
mov ebx, eax
push eax
push msg
.loop:
mov [esp+0x04], ebx
call printf
test ebx, 0x01
jz .even
.odd:
lea ebx, [1+ebx*2+ebx]
jmp .loop
.even:
shr ebx, 1
cmp ebx, 1
jne .loop
push ebx
push end
call printf
add esp, 16
xor eax, eax
ret
.usage:
mov ebx, [esp+0x08]
push dword [ebx+0x00]
push usage
call printf
add esp, 8
mov eax, 1
ret
msg db "%d --> ", 0
end db "%d", 10, 0
usage db "Usage: %s NUMBER", 10, 0
Befunge
&>:.:1-|
>3*^ #
|%2: <
v>2/>+
LOLCODE: 406 CHARAKTERZ
HAI
BTW COLLATZ SOUNDZ JUS LULZ
CAN HAS STDIO?
I HAS A NUMBAR
BTW, I WANTS UR NUMBAR
GIMMEH NUMBAR
VISIBLE NUMBAR
IM IN YR SEQUENZ
MOD OF NUMBAR AN 2
BOTH SAEM IT AN 0, O RLY?
YA RLY, NUMBAR R QUOSHUNT OF NUMBAR AN 2
NO WAI, NUMBAR R SUM OF PRODUKT OF NUMBAR AN 3 AN 1
OIC
VISIBLE NUMBAR
DIFFRINT 2 AN SMALLR OF 2 AN NUMBAR, O RLY?
YA RLY, GTFO
OIC
IM OUTTA YR SEQUENZ
KTHXBYE
TESTD UNDR JUSTIN J. MEZA'S INTERPRETR. KTHXBYE!
Python - 95 64 51 46 char
Obviously does not produce a stack overflow.
n=input()
while n>1:n=(n/2,n*3+1)[n%2];print n
Perl
I decided to be a little anticompetitive, and show how you would normally code such problem in Perl.
There is also a 46 (total) char code-golf entry at the end.
These first three examples all start out with this header.
#! /usr/bin/env perl
use Modern::Perl;
# which is the same as these three lines:
# use 5.10.0;
# use strict;
# use warnings;
while( <> ){
chomp;
last unless $_;
Collatz( $_ );
}
Simple recursive version
use Sub::Call::Recur;
sub Collatz{
my( $n ) = #_;
$n += 0; # ensure that it is numeric
die 'invalid value' unless $n > 0;
die 'Integer values only' unless $n == int $n;
say $n;
given( $n ){
when( 1 ){}
when( $_ % 2 != 0 ){ # odd
recur( 3 * $n + 1 );
}
default{ # even
recur( $n / 2 );
}
}
}
Simple iterative version
sub Collatz{
my( $n ) = #_;
$n += 0; # ensure that it is numeric
die 'invalid value' unless $n > 0;
die 'Integer values only' unless $n == int $n;
say $n;
while( $n > 1 ){
if( $n % 2 ){ # odd
$n = 3 * $n + 1;
} else { #even
$n = $n / 2;
}
say $n;
}
}
Optimized iterative version
sub Collatz{
my( $n ) = #_;
$n += 0; # ensure that it is numeric
die 'invalid value' unless $n > 0;
die 'Integer values only' unless $n == int $n;
#
state #next;
$next[1] //= 0; # sets $next[1] to 0 if it is undefined
#
# fill out #next until we get to a value we've already worked on
until( defined $next[$n] ){
say $n;
#
if( $n % 2 ){ # odd
$next[$n] = 3 * $n + 1;
} else { # even
$next[$n] = $n / 2;
}
#
$n = $next[$n];
}
say $n;
# finish running until we get to 1
say $n while $n = $next[$n];
}
Now I'm going to show how you would do that last example with a version of Perl prior to v5.10.0
#! /usr/bin/env perl
use strict;
use warnings;
while( <> ){
chomp;
last unless $_;
Collatz( $_ );
}
{
my #next = (0,0); # essentially the same as a state variable
sub Collatz{
my( $n ) = #_;
$n += 0; # ensure that it is numeric
die 'invalid value' unless $n > 0;
# fill out #next until we get to a value we've already worked on
until( $n == 1 or defined $next[$n] ){
print $n, "\n";
if( $n % 2 ){ # odd
$next[$n] = 3 * $n + 1;
} else { # even
$next[$n] = $n / 2;
}
$n = $next[$n];
}
print $n, "\n";
# finish running until we get to 1
print $n, "\n" while $n = $next[$n];
}
}
Benchmark
First off the IO is always going to be the slow part. So if you actually benchmarked them as-is you should get about the same speed out of each one.
To test these then, I opened a file handle to /dev/null ($null), and edited every say $n to instead read say {$null} $n. This is to reduce the dependence on IO.
#! /usr/bin/env perl
use Modern::Perl;
use autodie;
open our $null, '>', '/dev/null';
use Benchmark qw':all';
cmpthese( -10,
{
Recursive => sub{ Collatz_r( 31 ) },
Iterative => sub{ Collatz_i( 31 ) },
Optimized => sub{ Collatz_o( 31 ) },
});
sub Collatz_r{
...
say {$null} $n;
...
}
sub Collatz_i{
...
say {$null} $n;
...
}
sub Collatz_o{
...
say {$null} $n;
...
}
After having run it 10 times, here is a representative sample output:
Rate Recursive Iterative Optimized
Recursive 1715/s -- -27% -46%
Iterative 2336/s 36% -- -27%
Optimized 3187/s 86% 36% --
Finally, a real code-golf entry:
perl -nlE'say;say$_=$_%2?3*$_+1:$_/2while$_>1'
46 chars total
If you don't need to print the starting value, you could remove 5 more characters.
perl -nE'say$_=$_%2?3*$_+1:$_/2while$_>1'
41 chars total
31 chars for the actual code portion, but the code won't work without the -n switch. So I include the entire example in my count.
Haskell, 62 chars 63 76 83, 86, 97, 137
c 1=[1]
c n=n:c(div(n`mod`2*(5*n+2)+n)2)
main=readLn>>=print.c
User input, printed output, uses constant memory and stack, works with arbitrarily big integers.
A sample run of this code, given an 80 digit number of all '1's (!) as input, is pretty fun to look at.
Original, function only version:
Haskell 51 chars
f n=n:[[],f([n`div`2,3*n+1]!!(n`mod`2))]!!(1`mod`n)
Who the #&^# needs conditionals, anyway?
(edit: I was being "clever" and used fix. Without it, the code dropped to 54 chars.
edit2: dropped to 51 by factoring out f())
Golfscript : 20 chars
~{(}{3*).1&5*)/}/1+`
#
# Usage: echo 21 | ruby golfscript.rb collatz.gs
This is equivalent to
stack<int> s;
s.push(21);
while (s.top() - 1) {
int x = s.top();
int numerator = x*3+1;
int denominator = (numerator&1) * 5 + 1;
s.push(numerator/denominator);
}
s.push(1);
return s;
bc 41 chars
I guess this kind of problems is what bc was invented for:
for(n=read();n>1;){if(n%2)n=n*6+2;n/=2;n}
Test:
bc1 -q collatz.bc
21
64
32
16
8
4
2
1
Proper code:
for(n=read();n>1;){if(n%2)n=n*3+1else n/=2;print n,"\n"}
bc handles numbers with up to INT_MAX digits
Edit: The Wikipedia article mentions this conjecture has been checked for all values up to 20x258 (aprox. 5.76e18). This program:
c=0;for(n=2^20000+1;n>1;){if(n%2)n=n*6+2;n/=2;c+=1};n;c
tests 220,000+1 (aprox. 3.98e6,020) in 68 seconds, 144,404 cycles.
Perl : 31 chars
perl -nE 'say$_=$_%2?$_*3+1:$_/2while$_>1'
# 123456789 123456789 123456789 1234567
Edited to remove 2 unnecessary spaces.
Edited to remove 1 unnecessary space.
MS Excel, 35 chars
=IF(A1/2=ROUND(A1/2,0),A1/2,A1*3+1)
Taken straight from Wikipedia:
In cell A1, place the starting number.
In cell A2 enter this formula =IF(A1/2=ROUND(A1/2,0),A1/2,A1*3+1)
Drag and copy the formula down until 4, 2, 1
It only took copy/pasting the formula 111 times to get the result for a starting number of 1000. ;)
C : 64 chars
main(x){for(scanf("%d",&x);x>=printf("%d,",x);x=x&1?3*x+1:x/2);}
With big integer support: 431 (necessary) chars
#include <stdlib.h>
#define B (w>=m?d=realloc(d,m=m+m):0)
#define S(a,b)t=a,a=b,b=t
main(m,w,i,t){char*d=malloc(m=9);for(w=0;(i=getchar()+2)/10==5;)
B,d[w++]=i%10;for(i=0;i<w/2;i++)S(d[i],d[w-i-1]);for(;;w++){
while(w&&!d[w-1])w--;for(i=w+1;i--;)putchar(i?d[i-1]+48:10);if(
w==1&&*d==1)break;if(*d&1){for(i=w;i--;)d[i]*=3;*d+=1;}else{
for(i=w;i-->1;)d[i-1]+=d[i]%2*10,d[i]/=2;*d/=2;}B,d[w]=0;for(i=0
;i<w;i++)d[i+1]+=d[i]/10,d[i]%=10;}}
Note: Do not remove #include <stdlib.h> without at least prototyping malloc/realloc, as doing so will not be safe on 64-bit platforms (64-bit void* will be converted to 32-bit int).
This one hasn't been tested vigorously yet. It could use some shortening as well.
Previous versions:
main(x){for(scanf("%d",&x);printf("%d,",x),x-1;x=x&1?3*x+1:x/2);} // 66
(removed 12 chars because no one follows the output format... :| )
Another assembler version. This one is not limited to 32 bit numbers, it can handle numbers up to 1065534 although the ".com" format MS-DOS uses is limited to 80 digit numbers. Written for A86 assembler and requires a Win-XP DOS box to run. Assembles to 180 bytes:
mov ax,cs
mov si,82h
add ah,10h
mov es,ax
mov bh,0
mov bl,byte ptr [80h]
cmp bl,1
jbe ret
dec bl
mov cx,bx
dec bl
xor di,di
p1:lodsb
sub al,'0'
cmp al,10
jae ret
stosb
loop p1
xor bp,bp
push es
pop ds
p2:cmp byte ptr ds:[bp],0
jne p3
inc bp
jmp p2
ret
p3:lea si,[bp-1]
cld
p4:inc si
mov dl,[si]
add dl,'0'
mov ah,2
int 21h
cmp si,bx
jne p4
cmp bx,bp
jne p5
cmp byte ptr [bx],1
je ret
p5:mov dl,'-'
mov ah,2
int 21h
mov dl,'>'
int 21h
test byte ptr [bx],1
jz p10
;odd
mov si,bx
mov di,si
mov dx,3
dec bp
std
p6:lodsb
mul dl
add al,dh
aam
mov dh,ah
stosb
cmp si,bp
jnz p6
or dh,dh
jz p7
mov al,dh
stosb
dec bp
p7:mov si,bx
mov di,si
p8:lodsb
inc al
xor ah,ah
aaa
stosb
or ah,ah
jz p9
cmp si,bp
jne p8
mov al,1
stosb
jmp p2
p9:inc bp
jmp p2
p10:mov si,bp
mov di,bp
xor ax,ax
p11:lodsb
test ah,1
jz p12
add al,10
p12:mov ah,al
shr al,1
cmp di,bx
stosb
jne p11
jmp p2
dc - 24 chars 25 28
dc is a good tool for this sequence:
?[d5*2+d2%*+2/pd1<L]dsLx
dc -f collatz.dc
21
64
32
16
8
4
2
1
Also 24 chars using the formula from the Golfscript entry:
?[3*1+d2%5*1+/pd1<L]dsLx
57 chars to meet the specs:
[Number: ]n?[Results: ]ndn[d5*2+d2%*+2/[ -> ]ndnd1<L]dsLx
dc -f collatz-spec.dc
Number: 3
Results: 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1
Scheme: 72
(define(c n)(if(= n 1)`(1)(cons n(if(odd? n)(c(+(* n 3)1))(c(/ n 2))))))
This uses recursion, but the calls are tail-recursive so I think they'll be optimized to iteration. In some quick testing, I haven't been able to find a number for which the stack overflows anyway. Just for example:
(c 9876543219999999999000011234567898888777766665555444433332222
7777777777777777777777777777777798797657657651234143375987342987
5398709812374982529830983743297432985230985739287023987532098579
058095873098753098370938753987)
...runs just fine. [that's all one number -- I've just broken it to fit on screen.]
Mathematica, 45 50 chars
c=NestWhileList[If[OddQ##,3#+1,#/2]&,#,#>1&]&
Ruby, 50 chars, no stack overflow
Basically a direct rip of makapuf's Python solution:
def c(n)while n>1;n=n.odd?? n*3+1: n/2;p n end end
Ruby, 45 chars, will overflow
Basically a direct rip of the code provided in the question:
def c(n)p n;n.odd?? c(3*n+1):c(n/2)if n>1 end
import java.math.BigInteger;
public class SortaJava {
static final BigInteger THREE = new BigInteger("3");
static final BigInteger TWO = new BigInteger("2");
interface BiFunc<R, A, B> {
R call(A a, B b);
}
interface Cons<A, B> {
<R> R apply(BiFunc<R, A, B> func);
}
static class Collatz implements Cons<BigInteger, Collatz> {
BigInteger value;
public Collatz(BigInteger value) { this.value = value; }
public <R> R apply(BiFunc<R, BigInteger, Collatz> func) {
if(BigInteger.ONE.equals(value))
return func.call(value, null);
if(value.testBit(0))
return func.call(value, new Collatz((value.multiply(THREE)).add(BigInteger.ONE)));
return func.call(value, new Collatz(value.divide(TWO)));
}
}
static class PrintAReturnB<A, B> implements BiFunc<B, A, B> {
boolean first = true;
public B call(A a, B b) {
if(first)
first = false;
else
System.out.print(" -> ");
System.out.print(a);
return b;
}
}
public static void main(String[] args) {
BiFunc<Collatz, BigInteger, Collatz> printer = new PrintAReturnB<BigInteger, Collatz>();
Collatz collatz = new Collatz(new BigInteger(args[0]));
while(collatz != null)
collatz = collatz.apply(printer);
}
}
Python 45 Char
Shaved a char off of makapuf's answer.
n=input()
while~-n:n=(n/2,n*3+1)[n%2];print n
TI-BASIC
Not the shortest, but a novel approach. Certain to slow down considerably with large sequences, but it shouldn't overflow.
PROGRAM:COLLATZ
:ClrHome
:Input X
:Lbl 1
:While X≠1
:If X/2=int(X/2)
:Then
:Disp X/2→X
:Else
:Disp X*3+1→X
:End
:Goto 1
:End
Haskell : 50
c 1=[1];c n=n:(c$if odd n then 3*n+1 else n`div`2)
not the shortest, but an elegant clojure solution
(defn collatz [n]
(print n "")
(if (> n 1)
(recur
(if (odd? n)
(inc (* 3 n))
(/ n 2)))))
C#: 216 Characters
using C=System.Console;class P{static void Main(){var p="start:";System.Action<object> o=C.Write;o(p);ulong i;while(ulong.TryParse(C.ReadLine(),out i)){o(i);while(i > 1){i=i%2==0?i/2:i*3+1;o(" -> "+i);}o("\n"+p);}}}
in long form:
using C = System.Console;
class P
{
static void Main()
{
var p = "start:";
System.Action<object> o = C.Write;
o(p);
ulong i;
while (ulong.TryParse(C.ReadLine(), out i))
{
o(i);
while (i > 1)
{
i = i % 2 == 0 ? i / 2 : i * 3 + 1;
o(" -> " + i);
}
o("\n" + p);
}
}
}
New Version, accepts one number as input provided through the command line, no input validation. 173 154 characters.
using System;class P{static void Main(string[]a){Action<object>o=Console.Write;var i=ulong.Parse(a[0]);o(i);while(i>1){i=i%2==0?i/2:i*3+1;o(" -> "+i);}}}
in long form:
using System;
class P
{
static void Main(string[]a)
{
Action<object>o=Console.Write;
var i=ulong.Parse(a[0]);
o(i);
while(i>1)
{
i=i%2==0?i/2:i*3+1;
o(" -> "+i);
}
}
}
I am able to shave a few characters by ripping off the idea in this answer to use a for loop rather than a while. 150 characters.
using System;class P{static void Main(string[]a){Action<object>o=Console.Write;for(var i=ulong.Parse(a[0]);i>1;i=i%2==0?i/2:i*3+1)o(i+" -> ");o(1);}}
Ruby, 43 characters
bignum supported, with stack overflow susceptibility:
def c(n)p n;n%2>0?c(3*n+1):c(n/2)if n>1 end
...and 50 characters, bignum supported, without stack overflow:
def d(n)while n>1 do p n;n=n%2>0?3*n+1:n/2 end end
Kudos to Jordan. I didn't know about 'p' as a replacement for puts.
nroff1
Run with nroff -U hail.g
.warn
.pl 1
.pso (printf "Enter a number: " 1>&2); read x; echo .nr x $x
.while \nx>1 \{\
. ie \nx%2 .nr x \nx*3+1
. el .nr x \nx/2
\nx
.\}
1. groff version
Scala + Scalaz
import scalaz._
import Scalaz._
val collatz =
(_:Int).iterate[Stream](a=>Seq(a/2,3*a+1)(a%2)).takeWhile(1<) // This line: 61 chars
And in action:
scala> collatz(7).toList
res15: List[Int] = List(7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2)
Scala 2.8
val collatz =
Stream.iterate(_:Int)(a=>Seq(a/2,3*a+1)(a%2)).takeWhile(1<) :+ 1
This also includes the trailing 1.
scala> collatz(7)
res12: scala.collection.immutable.Stream[Int] = Stream(7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1)
With the following implicit
implicit def intToEven(i:Int) = new {
def ~(even: Int=>Int, odd: Int=>Int) = {
if (i%2==0) { even(i) } else { odd(i) }
}
}
this can be shortened to
val collatz = Stream.iterate(_:Int)(_~(_/2,3*_+1)).takeWhile(1<) :+ 1
Edit - 58 characters (including input and output, but not including initial number)
var n=readInt;while(n>1){n=Seq(n/2,n*3+1)(n%2);println(n)}
Could be reduced by 2 if you don't need newlines...
F#, 90 characters
let c=Seq.unfold(function|n when n<=1->None|n when n%2=0->Some(n,n/2)|n->Some(n,(3*n)+1))
> c 21;;
val it : seq<int> = seq [21; 64; 32; 16; ...]
Or if you're not using F# interactive to display the result, 102 characters:
let c=Seq.unfold(function|n when n<=1->None|n when n%2=0->Some(n,n/2)|n->Some(n,(3*n)+1))>>printf"%A"
Common Lisp, 141 characters:
(defun c ()
(format t"Number: ")
(loop for n = (read) then (if(oddp n)(+ 1 n n n)(/ n 2))
until (= n 1)
do (format t"~d -> "n))
(format t"1~%"))
Test run:
Number: 171
171 -> 514 -> 257 -> 772 -> 386 -> 193 -> 580 -> 290 -> 145 -> 436 ->
218 -> 109 -> 328 -> 164 -> 82 -> 41 -> 124 -> 62 -> 31 -> 94 -> 47 ->
142 -> 71 -> 214 -> 107 -> 322 -> 161 -> 484 -> 242 -> 121 -> 364 ->
182 -> 91 -> 274 -> 137 -> 412 -> 206 -> 103 -> 310 -> 155 -> 466 ->
233 -> 700 -> 350 -> 175 -> 526 -> 263 -> 790 -> 395 -> 1186 -> 593 ->
1780 -> 890 -> 445 -> 1336 -> 668 -> 334 -> 167 -> 502 -> 251 -> 754 ->
377 -> 1132 -> 566 -> 283 -> 850 -> 425 -> 1276 -> 638 -> 319 ->
958 -> 479 -> 1438 -> 719 -> 2158 -> 1079 -> 3238 -> 1619 -> 4858 ->
2429 -> 7288 -> 3644 -> 1822 -> 911 -> 2734 -> 1367 -> 4102 -> 2051 ->
6154 -> 3077 -> 9232 -> 4616 -> 2308 -> 1154 -> 577 -> 1732 -> 866 ->
433 -> 1300 -> 650 -> 325 -> 976 -> 488 -> 244 -> 122 -> 61 -> 184 ->
92 -> 46 -> 23 -> 70 -> 35 -> 106 -> 53 -> 160 -> 80 -> 40 -> 20 ->
10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1
The program frm Jerry Coffin has integer over flow, try this one:
#include <iostream>
int main(unsigned long long i)
{
int j = 0;
for( std::cin>>i; i>1; i = i&1? i*3+1:i/2, ++j)
std::cout<<i<<" -> ";
std::cout<<"\n"<<j << " iterations\n";
}
tested with
The number less than 100 million with the longest total stopping time is 63,728,127, with 949 steps.
The number less than 1 billion with the longest total stopping time is 670,617,279, with 986 steps.
ruby, 43, possibly meeting the I/O requirement
Run with ruby -n hail
n=$_.to_i
(n=n%2>0?n*3+1: n/2
p n)while n>1
C# : 659 chars with BigInteger support
using System.Linq;using C=System.Console;class Program{static void Main(){var v=C.ReadLine();C.Write(v);while(v!="1"){C.Write("->");if(v[v.Length-1]%2==0){v=v.Aggregate(new{s="",o=0},(r,c)=>new{s=r.s+(char)((c-48)/2+r.o+48),o=(c%2)*5}).s.TrimStart('0');}else{var q=v.Reverse().Aggregate(new{s="",o=0},(r, c)=>new{s=(char)((c-48)*3+r.o+(c*3+r.o>153?c*3+r.o>163?28:38:48))+r.s,o=c*3+r.o>153?c*3+r.o>163?2:1:0});var t=(q.o+q.s).TrimStart('0').Reverse();var x=t.First();q=t.Skip(1).Aggregate(new{s=x>56?(x-57).ToString():(x-47).ToString(),o=x>56?1:0},(r,c)=>new{s=(char)(c-48+r.o+(c+r.o>57?38:48))+r.s,o=c+r.o>57?1:0});v=(q.o+q.s).TrimStart('0');}C.Write(v);}}}
Ungolfed
using System.Linq;
using C = System.Console;
class Program
{
static void Main()
{
var v = C.ReadLine();
C.Write(v);
while (v != "1")
{
C.Write("->");
if (v[v.Length - 1] % 2 == 0)
{
v = v
.Aggregate(
new { s = "", o = 0 },
(r, c) => new { s = r.s + (char)((c - 48) / 2 + r.o + 48), o = (c % 2) * 5 })
.s.TrimStart('0');
}
else
{
var q = v
.Reverse()
.Aggregate(
new { s = "", o = 0 },
(r, c) => new { s = (char)((c - 48) * 3 + r.o + (c * 3 + r.o > 153 ? c * 3 + r.o > 163 ? 28 : 38 : 48)) + r.s, o = c * 3 + r.o > 153 ? c * 3 + r.o > 163 ? 2 : 1 : 0 });
var t = (q.o + q.s)
.TrimStart('0')
.Reverse();
var x = t.First();
q = t
.Skip(1)
.Aggregate(
new { s = x > 56 ? (x - 57).ToString() : (x - 47).ToString(), o = x > 56 ? 1 : 0 },
(r, c) => new { s = (char)(c - 48 + r.o + (c + r.o > 57 ? 38 : 48)) + r.s, o = c + r.o > 57 ? 1 : 0 });
v = (q.o + q.s)
.TrimStart('0');
}
C.Write(v);
}
}
}
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.