This page was spun off from Solving cryptarithms, numeric puzzles in which digits have been replaced by letters
TFW (Feb, 23 2004) - OK, Here is another one, we must solve the equation below where ABCDE*4=EDCBA, nothing fancy but the brute force methods solved it before your finger leaves the keyboard.
#----------------------------------------------------------------------------
# Solve the problem
# ABCDE
# X4
#------
# EDCBA
#----------------------------------------------------------------------------
proc ABCDE {args} {
set nums {1 2 3 4 5 6 7 8 9}
set counter 0
foreach a $nums {
foreach b $nums {
foreach c $nums {
foreach d $nums {
foreach e $nums {
set n1 [expr {"$a$b$c$d$e"*4}]
set n2 "$e$d$c$b$a"
incr counter
if {$n1==$n2} {
puts "We solved it! $a$b$c$d$e * 4 = $n2 at $counter tries"
return
}
}
}
}
}
}
puts "Not Solved"
}Here is another one, we have $1 in coins but only one can be a nickel. So we know that 19 coins must add to 95 cents using only pennies, dimes, quarters and half-dollars. Again, the simple brute force method yields an answer before your finger leaves the keyboard
#----------------------------------------------------------------------------
# Solve the problem
# 20 coins = $1.00, only one is a nickel
# so we have 19 coins =95 cents
#----------------------------------------------------------------------------
proc solve2 {args} {
set nums {0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19}
set counter 0
foreach penny $nums {
foreach dime $nums {
foreach quarter $nums {
foreach halfdollar $nums {
if {($penny+$dime+$quarter+$halfdollar)==19} {
set value [expr {$penny*1+$dime*10+$quarter*25+$halfdollar*50}]
incr counter
if {$value==95} {
puts "We solved it! penny=$penny dime=$dime quarter=$quarter halfdollar=$halfdollar at $counter tries"
return
}
}
}
}
}
}
puts "Not Solved"
}RS experiments with the following "General Problem Solver" (for small values of General), which, with heavy metaprogramming, builds up a nest of foreachs suiting the problem, quick kills (with continue) to force unique values for the variables, and returns the first solution found, or else an empty string:
proc solve {problem {domain0 {0 1 2 3 4 5 6 7 8 9}}} {
set vars [lsort -u [split [regsub -all {[^A-Z]} $problem ""] ""]]
set map {= ==}
set outers {}
set initials [regexp -all -inline {[^A-Z]([A-Z])} /$problem]
set pos [lsearch $domain0 0]
set domain1 [lreplace $domain0 $pos $pos]
foreach var $vars {
append body "foreach $var \$domain[expr [lsearch $initials $var]>=0] \{\n"
lappend map $var $$var
foreach outer $outers {
append body "if {$$var eq $$outer} continue\n"
}
lappend outers $var
append epilog \}
}
set test [string map $map $problem]
append body "if {\[expr $test\]} {return \[subst $test\]}" $epilog
if 1 $body
}This passes the tests from earlier in this page:
% solve SEND+MORE=MONEY 9567+1085==10652 % solve SAVE+MORE=MONEY 9386+1076==10462 % solve YELLOW+YELLOW+RED=ORANGE 143329+143329+846==287504
So this routine is not blindingly fast, but can process a number of problems from earlier in this page, without other configuration than specifying the problem.
Another kind of cryptarithm I found in Martin Gardner's Mathematical Circus:
EVE/DID=.TALKTALKTALK...
requires epsilon comparison with a periodic fraction... Any takers? - Took it myself, by replacing == equality with abs(delta)<epsilon:
proc solve {problem {domain0 {0 1 2 3 4 5 6 7 8 9}}} {
set vars [lsort -u [split [regsub -all {[^A-Z]} $problem ""] ""]]
set map {= )-( ... ""}
set outers {}
set initials [regexp -all -inline {[^A-Z]([A-Z])} /$problem]
set pos [lsearch $domain0 0]
set domain1 [lreplace $domain0 $pos $pos]
foreach var $vars {
append body "foreach $var \$domain[expr [lsearch $initials $var]>=0] \{\n"
lappend map $var $$var
foreach outer $outers {
append body "if {$$var eq $$outer} continue\n"
}
lappend outers $var
append epilog \}
}
set test abs(([string map $map $problem]))<=.00000001
append body "if {\[expr $test\]} {return \[subst $test\]}" $epilog
if 1 $body
}The other tests still pass, but the output is a bit harder to read:
% solve EVE/DID.=.TALKTALK... abs((212/606.)-(.34983498))<=.00000001 % solve SEND+MORE=MONEY abs((9567+1085)-(10652))<=.00000001 % solve ABCDE*4=EDCBA abs((21978*4)-(87912))<=.00000001 % solve 7AB*CD=EFGHJ abs((713*59)-(42067))<=.00000001