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In a standard dice (die) the numbers are arranged so that opposite faces add to seven. Write the shortest possible program in your preferred language which outputs a random throw followed by 9 random tippings. A tipping is a quarter turn of the dice, e.g. if the dice is facing 5, all possible tippings are 1,3,4 and 6.
Example of desired output:
1532131356
c=0;10.times{$><<c=([*1..6]-[c,7-c]).sample}
I found the [*1..6] trick by lucky experimenting.
0{(.6,5@--\-.,rand=).}10*;
A slight more compressed version of Joey's, basically working around the issue with zero-indexing.
You may need to replace print with alert or something else, depending on your JavaScript environment.
for(C=L=T=0;C++<10;print(L=T))while(!(T-L&&T+L-7))T=Math.random()*6+1|0
0:|;{7,[0|7|-]-.,rand=:|}10*
Bash
#/!bin/bash
f=`expr $RANDOM % 6`
f=`expr $f + 1`
printf "$f"
for ((i=0; i<9; i++))
do
((bad=7-$f))
next=`expr $RANDOM % 6`
next=`expr $next + 1`
while [ $next -eq $bad ] || [ $next -eq $f ]
do
next=`expr $RANDOM % 6`
next=`expr $next + 1`
done
printf "$next"
f=$next
done
sample code: http://ideone.com/CCfro
((var=expression)) is very nice - I thought that the shortest way was var=$((expression)) But why do you use that once only and waste tonnes of characters on expr in backticks?
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Feb 16 '11 at 10:00
for(i in 0:9)cat(F<-sample(setdiff(1:6,c(F,7-F)),1))
for((i=10,f=RANDOM%6+1;i--;))do printf $f ((n=RANDOM%4+1,m=f<4?f:7-f,f=n<m||++n<7-m?n:n+1)) done
The key idea is that to pick a random number in [1,x] which isn't equal to y you can pick a random number in [1,x-1] and then increment if it's >= y. For this problem we want a random number in [1,6] which isn't equal to f or 7-f. We have to do the two tests in order min(f,7-f), max(f,7-f).
Assuming an initially empty environment could save 2 chars by not initialising i and changing the loop condition to i++<10
Heavily golfed from Aman ZeeK Verma's answer:
for((i=10,f=0;i--;))do for((n=f;n==f||n+f==7;f=RANDOM%6+1))do : done printf $f done
NB arguably it can be shrunk by 5 chars by changing the first line to for((;i++<10;)) but that makes assumptions which aren't always valid. It would work ok in ideone but someone running it from a shell could have i or f exported to something non-zero.
man bash, though, I recommend it. I once read it cover to cover, and just having a vague idea of what's possible and worth looking up has served me well.
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Feb 16 '11 at 11:36
-join(0..9|%{($d=1..6-ne(7-$d)-ne$d|random)})
Pretty trivial, actually. I generate a list of possible dice rolls 1..6 and then select only those not equal to seven minus the last roll and then only those not equal to the last roll. From the remaining list I then select a random item and assign it to $d. Since $d is initially treated as 0 it rolls a normal die the first time.
Test script:
for($i=0;$i-lt20;$i++){
$o=@(./tipping.ps1)
if ($i-gt0-and$o-eq$o2) { throw "Must have random output" }
if ($o.count-ne1) { throw "Must only have one line of output" }
if ($o[0]-match'[^1-6]'){ throw "Invalid characters" }
if($o[0].length-ne10){ throw "Wrong length: $($o[0].length)" }
$r=[char[]]($o[0])|%{$_-48}
for ($x=1;$x-lt$r.count;$x++){
if ($r[$x-1]+$r[$x]-eq7) { throw "Not a tipping: $($r[$x-1]) and $($r[$x])" }
}
$o2=$o
}
History:
The term 'random' is not recognized as a cmdlet, function, operable program, or script file. Verify the term and try again.
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Feb 18 '11 at 12:17
Get-Random cmdlet didn't exist in v1.
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This should work but unfortunately J's random generator gets stuck after the 3rd iteration:
a=:>:i.6
f=:a#~1-(+&(a=])7&-)
((,(?4)&{@f@(_1&{))^:9)>:?6
6 4 5 4 5 4 5 4 5 4
(?4) tends to be rolled once and treated as a constant for subsequent iterations if you're not careful around it. I worked around it using a (?@4:)-like construct.
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(0..9).reduce([]){|m|m<<((1..6).to_a-[d=m[-1]||0,7-d]).shuffle[0]}
>:(?@4:{(i.6)-.],5&-)^:(<10)?6
6 2 3 5 4 2 4 1 3 6
Explanations (read from right to left):
?6 returns a random number between 0 and 5^:(<10) applies a function 9 times, accumulating the results along the way. The function is:?@4:{(i.6)-.],5&-
] , 5&- returns an array of the input number and its complement to 5 (we're handling 0-based numbers currently, so the sum of opposite faces is 5)(i. 6) -. removes them from the full set of integers 0 to 5. We're left with all valid positions after a single tipping operation from the input position.?@4: { picks one of those at random.>: increments the whole sequence to bring the figures back to the 1 to 6 interval.
16 2f 25 08 41 20 17 30 16 2f 31 31 25 09 19 32
Here's how it works
16 2f 25 # make range from 1 to 6 and push random element
08 # start block
41 # duplicate top of stack twice
20 17 30 # negate top of stack and add 7
16 2f # push range from 1 to 6
31 31 # do set-wise difference with each of the two previous numbers
25 # push a random element from the list
09 # end block
19 32 # repeat block 9 times
The two newlines are necessary and included in the character count as one character each.
RANDOMIZE:FOR I=0TO 9
1N=INT(RND*6)+1:IF L=N OR L+N=7THEN 1
?N:L=N:NEXT
For(I,1,9
Ans→X
Repeat Ans≠X and Ans≠7-X
randInt(1,6
End
Disp Ans
End
Boring solution, but it's shorter than the previous revision. I take advantage of the fact that on a fresh calculator, Ans is initialized to zero.
v->{int d=(int)(Math.random()*6+1),i=10,p;String r=""+d;for(;i-->0;r+=d)for(p=d;p==d|p+d==7;d=(int)(Math.random()*6+1));return r;}
As full program with verbose main-method this would be 178 bytes instead:
interface M{static void main(String[]a){int d=(int)(Math.random()*6+1),i=10,p;String r=""+d;for(;i-->0;r+=d)for(p=d;p==d|p+d==7;d=(int)(Math.random()*6+1));System.out.print(r);}}
Semi-port of @AmanZeeKVerma's Bash answer.
Explanation:
v->{ // Method with empty unused parameter and String return-type
int d=(int)(Math.random()*6+1),
// Random dice-roll 1-6
i=10, // Counter-integer, starting at 10
p; // Temp integer to store new side
String r=""+d; // Result-String, starting at the first dice-roll
for(;i-->0; // Loop (1) 10 times:
r+=d) // After every iteration, append the result with a random side
for(p=d; // Set the new side to the current side
p==d // Loop (2) as long as the new side and current side are the same
|p+d==7; // or as long as both combined are exactly 7:
d=(int)(Math.random()*6+1)
// Set the new side to a random side 1-6
); // End of loop (2)
// End of loop (1) (implicit / single-line body)
return r; // Return the result-String
} // End of method
a=randi(6)
for i=1:9;b=1:6;b([a,7-a])=[];a=b(randi(4))
end
I'm glad I could showcase ><>'s x code-pointer randomization as I don't recall seeing it here.
a&0 v
/2v
1x3v
>x< <<
6x4v
\5v ~
:{:/ ^?=}
:{:/ ^?=7+}
:~$<^&;!?:-1&n
You can try it on this online interpreter (paste code, submit, start).
c(3,5,1,4,2,6)[(sample(1:6,1)+cumsum(sample((-2:2)[-3],9,T)))%%6+1]
There is a golfier R answer but this I think is a different approach from the answers submitted so far.
c(3,5,1,4,2,6) #A dice and its facets
(sample(1:6,1) #Initial dice roll
+cumsum(sample((-2:2)[-3],9,T))) #9 tippings in c(-2,-1,1,2)
%%6+1 #converts to values in [0,6]
[ ]#
6LΩUTFX?6LʒDXÊsX+7Ê*}ΩU
Can definitely be golfed, but I'm currently not seeing it..
Explanation:
6LΩ # Pick a random value from the range [1,6]
# i.e. [1,2,3,4,5,6] → 3
U # Save this random value in variable `X`
TF # Loop 10 times:
X? # Print `X` without newline to STDOUT
6Lʒ } # Create a range [1,6] again, and filter it by:
DXÊ # Check if the current value is not equal to `X`
# i.e. 1 and 3 → 1 (truthy)
# i.e. 3 and 3 → 0 (falsey)
sX+ # Sum the current value with `X`
# i.e. 1 and 3 → 4
# i.e. 3 and 3 → 6
7Ê # And check if it's not equal to 7
# i.e. 4 and 7 → 1 (truthy)
# i.e. 6 and 7 → 1 (truthy)
* # If both checks are truthy, keep it in the filtered list
# i.e. 1 and 1 → 1 (truthy)
# i.e. 0 and 1 → 0 (falsey)
Ω # Pick a random value from the filtered list
# i.e. [1,2,5,6] → 1
U # And save it in variable `X` for the next iteration of the loop