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Cheryl's birthday: Difference between revisions
→{{header|Lua}}
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Cheryl's birthday is July 16
</pre>
=={{header|Lua}}==
<lang lua>-- Cheryl's Birthday in Lua 6/15/2020 db
local function Date(mon,day)
return { mon=mon, day=day, valid=true }
end
local choices = {
Date(5,15), Date(5,16), Date(5,19),
Date(6,17), Date(6,18),
Date(7,14), Date(7,16),
Date(8,14), Date(8,15), Date(8,17)
}
local function apply(t,f)
for k,v in pairs(t) do
f(k,v)
end
end
local function filter(t,f)
local result = {}
for k,v in pairs(t) do
if f(k,v) then
result[#result+1] = v
end
end
return result
end
local function count(t) return #t end
local function isvalid(k,v) return v.valid end
local function invalidate(k,v) v.valid = false end
local function remaining()
io.write(" ")
apply(filter(choices, isvalid), function(k,v) io.write(v.mon.."/"..v.day..", ") end)
print("\n")
end
print("Cheryl offers these ten choices:")
remaining()
print("1a) Albert knows month but not day, so month cannot be unique, obviously/redundantly, leaving still:")
apply(filter(choices, isvalid), function(k,v)
if count(filter(choices, function(k2,v2) return v2.valid and v.mon==v2.mon end)) == 1 then
apply(filter(choices, function(k2,v2) return v2.valid and v.mon==v2.mon end), invalidate)
end
end)
remaining()
print("1b) Albert also knows that Bernard cannot yet know, so cannot be a month with a unique day, leaving:")
apply(filter(choices, isvalid), function(k,v)
if count(filter(choices, function(k2,v2) return v.day==v2.day end)) == 1 then
apply(filter(choices, function(k2,v2) return v2.valid and v.mon==v2.mon end), invalidate)
end
end)
remaining()
print("2) After Albert's revelation, Bernard now knows, so day must be unique, leaving:")
apply(filter(choices, isvalid), function(k,v)
if count(filter(choices, function(k2,v2) return v2.valid and v.day==v2.day end)) > 1 then
apply(filter(choices, function(k2,v2) return v2.valid and v.day==v2.day end), invalidate)
end
end)
remaining()
print("3) After Bernard's revelation, Albert now knows, so month must be unique, leaving only:")
apply(filter(choices, isvalid), function(k,v)
if count(filter(choices, function(k2,v2) return v2.valid and v.mon==v2.mon end)) > 1 then
apply(filter(choices, function(k2,v2) return v2.valid and v.mon==v2.mon end), invalidate)
end
end)
remaining()</lang>
{{out}}
<pre>Cheryl offers these ten choices:
5/15, 5/16, 5/19, 6/17, 6/18, 7/14, 7/16, 8/14, 8/15, 8/17,
1a) Albert knows month but not day, so month cannot be unique, obviously/redundantly, leaving still:
5/15, 5/16, 5/19, 6/17, 6/18, 7/14, 7/16, 8/14, 8/15, 8/17,
1b) Albert also knows that Bernard cannot yet know, so cannot be a month with a unique day, leaving:
7/14, 7/16, 8/14, 8/15, 8/17,
2) After Albert's revelation, Bernard now knows, so day must be unique, leaving:
7/16, 8/15, 8/17,
3) After Bernard's revelation, Albert now knows, so month must be unique, leaving only:
7/16,</pre>
=={{header|Perl}}==
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