-------------
| a | b | -> 0.ab
--------------
| c | d | -> 0.cd
-------------
| |
v v
0.ac 0.bdGiven the above grid find an assignment of the digits $0-9$ to $a,b,c,d$ so that no digit is used twice and the sum is $1.0$ or the closest that that the assignments can get to it
Note $a=0$, $b=1$, $c=2$, $d=3$ is different than $a=1$, $b=0$, $c=2$, $d=3$. In other word the same digits assigned in different locations is a unique "permutation" that must be checked but remember no digit can be repeated.
We can figure out how many different assignments that we have to check. Since there at $10$ choice for one of the locations then $9$ remain for the second, $8$ for the third and $7$ for the last one. So $10 \times 9 \times 8 \times 7=5040$ is the total number of assignments.
Formally this is called the permutations of $4$ digits from a group of $10$ and this is the formula for calculating the number permutations = $10!/(10-4)!$ Where $!$ is called factorial and is defined as multiplying a number by all the numbers less than it until you get to $1$. Eg. $5! = 5 \times 4 \times 3 \times 2 \times 1$
See Permutations on Khan Academy for more information about Permutations.
Write a program that check all possible assignments.
The following program searches all of the permutations and checks to see if any produce a sum to $1.0$ and prints the values for a,b,c,d
# Brute Force solution written in Python
# If you want to see all the permutaition tested and their sum set the
# following variable to True
seeAll=False
print("Results:")
# The following code searches all of them
i=1 # counter for each permutation values for a, b, c and d that we try
for a in range(10): # loop a over 0-9
for b in range(10): # loop b over 0-9
if (b==a): # if b is equal to a then
continue # skip this b and go on to next one
for c in range(10): # loop c over 0-9
if (c==a or c==b): # if c is equal to a or b then
continue # skip this c and go on to next one
for d in range(10): # loop d over 0-9
if (d==a or d==b or d==c): # if d is equal to a, b, or
continue # c skip and move on
# We have an a, b, c, d to check
# a/10 = .a and b/100 = 0.0b so a/10 + b/100 = .ab
# value = .ab
value=a/10 + b/100
# a/10 = .a and c/100 = 0.0c so a/10 + c/100 = .ac
# value = .ab + .ac
value+=a/10 + c/100
# b/10 = .b and d/100 = 0.0d so b/10 + d/100 = .bd
# value = .ab + .ac + .bd
value+=b/10 + d/100
# c/10 = .c and d/100 = 0.0d so b/10 + d/100 = .cd
# value = .ab + .ac + .bd + .cd
value+=c/10 + d/100
# check if .ab + .ac + .bd + .cd is equal to 1.0
if (value==1.0):
# yes then we found a solution print it out
print("* ", str(i) + ": " + str(a) + " " + str(b) + " "
+ str(c) + " " + str(d) + ": " + str(value))
if (seeAll and value != 1.0):
# if we want to see all the others print them out too
print(str(i) + ": " + str(a) + " " + str(b) + " "
+ str(c) + " "+ str(d) + ": " + str(value))
i=i+1 # increase count of the values we
# have tried
Results: * 40: 0 1 7 6: 1.0 * 137: 0 3 5 6: 1.0 * 242: 0 5 3 6: 1.0 * 341: 0 7 1 6: 1.0 * 537: 1 0 6 7: 1.0 * 579: 1 2 4 7: 1.0 * 684: 1 4 2 7: 1.0 * 789: 1 6 0 7: 1.0 * 1028: 2 0 4 8: 1.0 * 1077: 2 1 3 8: 1.0 * 1133: 2 3 1 8: 1.0 * 1526: 3 0 2 9: 1.0 * 1631: 3 2 0 9: 1.0