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Talk:Seven-sided dice from five-sided dice: Difference between revisions
Talk:Seven-sided dice from five-sided dice (view source)
Revision as of 23:31, 28 December 2018
, 5 years ago→numbers on a die: added comments about binary dice.
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== dead link ==
The Stack Overflow link in the task description is dead --Thu Apr 10 20:57:05 PDT 2014
== numbers on a die ==
It's more common for computer random number generators to generate a random number from 0 to n-1, than from 1 to n. So I propose changing the definitions of dice5() and dice7() to generate integers from 0..4 and 0..6, respectively. It will make the math a little simpler. --[[Special:Contributions/96.238.211.175|96.238.211.175]] 08:26, 9 August 2009 (UTC)
:Hi, please don't change this as it is more common for dice to count from 1. It is better to make the program adapt to the problem in this case. --[[User:Paddy3118|Paddy3118]] 08:56, 9 August 2009 (UTC)
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::As noted in the Tcl explanatory text, this is explicitly about making a primitive D5 and creating a D7 from it. (That's also why I use the terms D5 and D7; what programmer hasn't played at least ''some'' D&D? :-)) In any case, no conventional die (the correct singular form of “dice”) numbers from 0. —[[User:Dkf|Donal Fellows]] 10:04, 9 August 2009 (UTC)
::I'm probably the exception that proves the rule about D&D. (My great time waster was PacMan)! --[[User:Paddy3118|Paddy3118]] 11:10, 9 August 2009 (UTC)
::: I have quite a collection of dice and none of which have a zero (or blank) on them. --- Well, all except one set. They are ''binary dice'' (and pretty hard to find one in the wild), and are six sized, with just three sets of a '''one''' and a '''zero'''. But other than that anomaly, I have no dice with zero (or no) pips. Note that some binary die have the pips numbered (one through six) in binary, but no zero. -- [[User:Gerard Schildberger|Gerard Schildberger]] ([[User talk:Gerard Schildberger|talk]]) 23:30, 28 December 2018 (UTC)
== J solution seems ugly ==
I'd really like someone knowledgeable to look and make J solution more elegant. This straightforward solution doesn't look very good. [[User:Avmich|Avmich]] 21:03, 13 September 2009 (UTC)
:You could use this <code>dice5</code> and either <code>dice7a</code> or <code>dice7b</code> for the main bit:
<lang j>dice5=: >:@:?@$&5
dice7a=: 0 8 -.~ 3 >.@%~ 5 #. [: <:@dice5 2 ,~ */
dice7b=: [: (#~ 7&>:) 3 >.@%~ [: 5&#.&.:<:@dice5 */ , 2: </lang>
:Then it's just a question on ensuring that you've got enough rolls. You could use the following explicit:
<lang j>
dice7=: monad define
res=. 0$0
while. (*/y) > #res do.
res=. res, dice7a >. 0.75 * y
end.
y $ res
)
</lang>
:...or you could create a tacit equivalent using the <code>^:</code> conjunction.--[[User:Tikkanz|Tikkanz]] 00:15, 14 September 2009 (UTC)
:Here is a more instructive version of the main bit of code:
<lang j>
dice5=: [: >: ] ?@$ 5: NB. makes a y shape array of 5s, "rolls" the array and increments.
rolltwice=: [: dice5 2 ,~ */ NB. rolls dice5 twice for each desired dice7 roll (*/y rows, 2 cols)
base5to10=: 5 #. <: NB. decrements and converts rows from base 5 to 10
keepgood=: #~ 21&> NB. compress out values not less than 21
groupsof3=: [: >. >: % 3: NB. increments, divides by 3 and takes ceiling
dice7c=: groupsof3@keepgood@base5to10@rolltwice
</lang>--[[User:Tikkanz|Tikkanz]] 01:20, 14 September 2009 (UTC)
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