number pluzle solve try

[bluatigro]

fount the next puzle in a magazin :
acute = 30
blut = 25
boel = 17
cel = 18
clou = 26
club = 29
lat = 17
leut = 23
loep = 22
plat = 27
pol = 19
pub = 24
valt = 18
vla = 14
the char's are numbers added
what is the char's value

[bluatigro]

this wil take to long :


for a = 0 to 9
for b = 0 to 9
 for c = 0 to 9
  for e = 0 to 9
   for l = 0 to 9
    for o = 0 to 9
     for p = 0 to 9
      for t = 0 to 9
       for u = 0 to 9
        for v = 0 to 9
         if a+c+u+t+e=30 _
         and b+l+u+t=25 _
         and b+o+e+l=17 _
         and c+e+l=18 _
         and c+l+o+u= 26 _
         and c+l+u+b=29 _
         and l+a+t=17 _
         and l+e+u+t=23 _
         and l+o+e+p=22 _
         and p+l+a+t=27 _
         and p+o+l=19 _
         and p+u+b=24 _
         and v+a+l+t=18 _
         and v+l+a=14 then
           print "a = " ; a
           print "b = " ; b
           print "c = " ; c
           print "e = " ; e
           print "l = " ; l
           print "o = " ; o
           print "p = " ; p
           print "t = " ; t
           print "u = " ; u
           print "v = " ; v
           end
         end if
        next v
       next u
      next t
     next p
    next o
   next l
  next e
 next c
next b
next a


this can be done faster whit permutation's
but i do not know how

[Rod]

Answer is a=6 b=5 c=8 e=3 l=7 0=2 p=10 t=4 u=9 v=1

Like most folks a few minutes algebra brings the solution bubbling out.

How to code it? Interesting. My first thoughts are that you need two strategies. The first looks for single character differences between words and assigns the numeric difference. The second strategy would be a solve matrix. Here you would check repeatedly for single missing values as you store solutions. Again a single missing solution is the known sum deducted from the total.

e.g. VALT=18 ALT=17 so V=1
PLAT=27 ALT=17 so P=10

[Rod]

Having thought a little more I would sort each “word” into alpha order. So reorder the letters of each word. Then sort the words, that will help find single character differences. When I do find a single character and obtain its value I would then remove it from every word and deduct its value from the total. That way the words get shorter and shorter and single character differences easier to find. Probably a single iteritive solution.

[B+]

OK here is a solution finder in code. 

I used an anaCode$(letters$) function to arrange the letter counts of a word in alpha order into a 26 character string st that matching anaCodes are identical words or anagrams with same letter set.

I used a cycle of finding solutions then removing those solution letters and their values from the problem set until no new changes occurred.

'find letter values puzzle.txt for JB v2 B+ 2019-01-15

'first create a data$ from problem 14 items
data$ = "acute = 30, blut = 25, boel = 17, cel = 18, clou = 26, club = 29, lat = 17, leut = 23, loep = 22, "
data$ = data$ + "plat = 27, pol = 19, pub = 24, valt = 18, vla = 14"

' state problem and give choice of viewing intermediate calculations
print "Problem: Find the value for each letter such that:"
i = 1
while word$(data$, i, ",") <> ""
   print trim$(word$(data$, i, ","))
   i = i + 1
wend
print
print "Do you want to see the interim calculations"
print " as this program arrives at a solution set?"
input " Enter y for yes, or n for no ";a$
if a$ = "y" then debug = 1
cls

'now separate words from values
maxIndex = 14
dim words$(100), values(100), codes$(100), solutionValues(26)  '< leave room for more words
dim letters(26) 'this is for the anaCode$ funtion
if debug then print : print "Original problem:"
for i = 1 to maxIndex
   words$(i) = trim$(leftOf$(word$(data$, i, ","), "="))
   values(i) = val(trim$(rightOf$(word$(data$, i, ","), "=")))
   codes$(i) = anaCode$(words$(i))
   if debug then print words$(i), values(i), codes$(i)  'debug check data load
next
do
   scan
   'look for 1 letter difference between one word and another, if so we can isolte a letter and it's value
   if debug then print : print "Solutions found:"
   for i = 1 to maxIndex - 1
       for j = i + 1 to maxIndex
           d$ = oneLetterDifference$(codes$(i), codes$(j))
           if  d$ <> "" then
               v = abs(values(i) - values(j))
               if debug then print d$; " = "; v
               solutionValues(asc(d$) - 96) = v
           end if
       next
   next

   'now take these solutions and apply them back into the arrays for shorter words and smnaller values
   changes = 0
   for i = 1 to 26
       if solutionValues(i) then
           letter$ = chr$(i + 96)
           for w = 1 to maxIndex
               pos = instr(words$(w), letter$)
               if pos then 'rebuild word without letter, value without the letter's value and find new anaCode$
                   words$(w) = mid$(words$(w), 1, pos - 1) + mid$(words$(w), pos + 1)
                   values(w) = values(w) - solutionValues(i)
                   codes$(w) = anaCode$(words$(w))
                   changes = changes + 1
               end if
           next
       end if
   next
   if debug and changes <> 0 then
       print
       print "After ";changes; " changes, letters and their values removed."
       print "The current state of problem:"
       for i = 1 to maxIndex
           print words$(i), values(i), codes$(i)
       next
   end if
loop until changes = 0
print : print "The Solution Set is:"
for i = 1 to 26
   if solutionValues(i) then print chr$(i + 96); " = "; solutionValues(i)
next

function oneLetterDifference$(ana1$, ana2$)
   for i = 1 to 26 'letters
       if mid$(ana1$, i, 1) <> mid$(ana2$, i, 1) then
           diff = diff + 1
           saveLetter = i
       end if
   next
   if diff = 1 then oneLetterDifference$ = chr$(saveLetter + 96)
end function

'create a function that counts a set of letters in word
'such that words with same counts are the same or anagrams of each other
function anaCode$(s$)
   'since letters is global, zero out letters
   s$ = lower$(s$)
   for i = 1 to 26
       letters(i) = 0
   next
   'assuming s$ is just letters not digits or other stuff like punctuation
   for i = 1 to len(s$)
       letters(asc(mid$(s$, i, 1)) - 96) = letters(asc(mid$(s$, i, 1)) - 96) + 1
   next
   for i = 1 to 26
       b$ = b$ + str$(letters(i))
   next
   anaCode$ = b$
end function

'some handy word tool functions
FUNCTION leftOf$ (source$, of$)
   posOf = INSTR(source$, of$)
   IF posOf > 0 THEN leftOf$ = MID$(source$, 1, posOf - 1)
END FUNCTION

FUNCTION rightOf$ (source$, of$)
   posOf = INSTR(source$, of$)
   IF posOf > 0 THEN rightOf$ = MID$(source$, posOf + LEN(of$))
END FUNCTION


[tsh73]

Nice solution, B+

[Rod]

Beaten by B+ but I could not resist having a go. Here is where I got to, it finds the same solution more than once but that could be fixed.

dim d$(14,2)
dim found$(50)
dim found(50)
d$(1,1)="blut"
d$(1,2)="25"
d$(2,1)="boel"
d$(2,2)="17"
d$(3,1)="cel"
d$(3,2)="18"
d$(4,1)="clou"
d$(4,2)="26"
d$(5,1)="club"
d$(5,2)="29"
d$(6,1)="lat"
d$(6,2)="17"
d$(7,1)="leut"
d$(7,2)="23"
d$(8,1)="loep"
d$(8,2)="22"
d$(9,1)="plat"
d$(9,2)="27"
d$(10,1)="pol"
d$(10,2)="19"
d$(11,1)="pub"
d$(11,2)="24"
d$(12,1)="valt"
d$(12,2)="18"
d$(13,1)="vla"
d$(13,2)="14"
oldfound=1

[start]
'first put letters within words in alpha order
for n= 1 to 13
   l=len(d$(n,1))
   for m= 1 to l
       l$(m)=mid$(d$(n,1),m,1)
   next
   sort l$(,1,l
   w$=""
   for m= 1 to l
       w$=w$+l$(m)
   next
   d$(n,1)=w$
next
sort d$(,1,13,1

'now match each word against every other word looking for a single character difference
for m= 1 to 13
   m$=d$(m,1)
   lm=len(m$)
   'get the word we are matching to
   for n= 1 to m
       mm$=m$
       n$=d$(n,1)
       ln=len(n$)
       'get the word we are checking
       for o=1 to lm
           c$=mid$(m$,o,1)
           if instr(n$,c$,0)>0 then 'remove all matching characters
               n$=remchar$(n$,c$)
               mm$=remchar$(mm$,c$)
           end if
           'print m$ , mm$,n$
       next
       'now see if we have a single character left in either word
       'if so store it and deduce its value
       if mm$="" and len(n$)=1 then
            print "Found ";n$,abs(val(d$(n,2))-val(d$(m,2)))
            found=found+1
            found$(found)=n$
            found(found)=abs(val(d$(n,2))-val(d$(m,2)))
       end if
       if n$="" and len(mm$)=1 then
           print "Found ";mm$,abs(val(d$(n,2))-val(d$(m,2)))
           found=found+1
           found$(found)=mm$
           found(found)=abs(val(d$(n,2))-val(d$(m,2)))
       end if
   next
next

'now clear out all found characters from all words and deduct its value from the total
for f = oldfound to found
   for n= 1 to 13
       if instr(d$(n,1),found$(f),1)>0 then
           d$(n,1)=remchar$(d$(n,1),found$(f))
           d$(n,2)=str$(val(d$(n,2))-found(f))
       end if
   next
next
'remember where we ended last time
oldfound=found

scan

'now loop back and do it all again
if found<10 then goto [start]

'end




[B+]

Hi Rod,

I think your code is missing a remchar$ function after 'end.

And those sort lines like:
sort d$(,1,13,1

look wrong without the ) paired with the (, but probably right. 
I forgot that JB v2 has sort now.

[Rod]

Away from home and B+ is correct. I used Liberty BASIC to code the routine and forgot that remchar$() is a new Liberty function. Will roll my own for Just BASIC when I get home. Or perhaps someone had one in their code bank?

[B+]

OK Rod, your code works with this addition (now that I understand it stands for remove character):


function remchar$(s$, c$)
   pos = instr(s$, c$)
   if pos then remchar$ = mid$(s$, 1, pos -1) + mid$(s$, pos + 1) else remchar$ = s$
end function


And those sort lines like:
sort d$(,1,13,1


look wrong without the ) paired with the (, but probably right.
I forgot that JB v2 has sort now.

Oh! does work as:
sort d$(),1,13,1

So all JB v2 needs is the first ( to be able to tell it's an array type.

[Rod]

Actually I didn't need all the sorting. The remchar$() function did not care. So it simplifies to:



dim d$(14,2)
dim found$(50)
dim found(50)
d$(1,1)="blut"
d$(1,2)="25"
d$(2,1)="boel"
d$(2,2)="17"
d$(3,1)="cel"
d$(3,2)="18"
d$(4,1)="clou"
d$(4,2)="26"
d$(5,1)="club"
d$(5,2)="29"
d$(6,1)="lat"
d$(6,2)="17"
d$(7,1)="leut"
d$(7,2)="23"
d$(8,1)="loep"
d$(8,2)="22"
d$(9,1)="plat"
d$(9,2)="27"
d$(10,1)="pol"
d$(10,2)="19"
d$(11,1)="pub"
d$(11,2)="24"
d$(12,1)="valt"
d$(12,2)="18"
d$(13,1)="vla"
d$(13,2)="14"
oldfound=1
foundletters$=""

[start]
scan

'match each word against every other word looking for a single character difference
for m= 1 to 13
   m$=d$(m,1)
   lm=len(m$)
   for n= 1 to m
       mm$=m$
       n$=d$(n,1)
       ln=len(n$)
       for o=1 to lm
           c$=mid$(m$,o,1)
           if instr(n$,c$,0)>0 then
               n$=remchar$(n$,c$)
               mm$=remchar$(mm$,c$)
           end if
       next
       if mm$="" and len(n$)=1 and instr(foundletters$,n$,1)=0 then
           print "Found ";n$,abs(val(d$(n,2))-val(d$(m,2)))
           found=found+1
           found$(found)=n$
           found(found)=abs(val(d$(n,2))-val(d$(m,2)))
           foundletters$=foundletters$+n$
       end if
       if n$="" and len(mm$)=1 and instr(foundletters$,mm$,1)=0 then
           print "Found ";mm$,abs(val(d$(n,2))-val(d$(m,2)))
           found=found+1
           found$(found)=mm$
           found(found)=abs(val(d$(n,2))-val(d$(m,2)))
           foundletters$=foundletters$+mm$
       end if
   next
next

' eliminate the found letters and their values from the array
for f = oldfound to found
   for n= 1 to 13
       if instr(d$(n,1),found$(f),1)>0 then
           d$(n,1)=remchar$(d$(n,1),found$(f))
           d$(n,2)=str$(val(d$(n,2))-found(f))
       end if
   next
next
oldfound=found

if found<10 then goto [start]
end

function remchar$(s$, c$)
   pos = instr(s$, c$)
   if pos then remchar$ = mid$(s$, 1, pos -1) + mid$(s$, pos + 1) else remchar$ = s$
end function


[B+]

Oh I see you compare letter sets by comparing letters and removing like letters. Yes, then no need for sorting, just look for one letter left in one word and none left in the other.

I think bluatigro gave us a particularly easy puzzle. What if only 1 letter value could be discovered on first round, ie better = 28 and beer = 18 the difference is 2 t's = 10 st t = 5...

So instead of looking for 1 letter left, you'd see if all the letters left were the same. If so, divide the value difference of the 2 words by the number of like letters to solve for the letter.