Brain teaser 1005: Words and numbers
From The Sunday Times, 1st November 1981 [link]
A familiar letters-for-digits problem is to take a correct addition sum such as:
1 + 1 = 2
and write it in words so:
ONE + ONE = TWO
The problem is to assign numbers from 0 to 9 to the letters so that we get a correct addition sum, e.g. E = 6, N = 3, O = 2, T = 4, W = 7 gives:
236 + 236 = 472
Two rules to be kept are that the left hand digit of any number is not zero, and different letters are assigned different numbers.
Some addition sums clearly have no solution. e.g.:
ONE + THREE = FOUR
In the Kudali Language the words for 1, 2, 3, 4 are AA, BA, EB, DE, although not necessarily in that order. Now all the addition sums:
1 + 1 = 2
1 + 2 = 3
1 + 3 = 4
2 + 2 = 4
1 + 4 = 2 + 3when written in Kudali, give letters-for-digits problems each of which has at least one solution.
What are the Kudali words for 1, 2, 3, 4, in that order?
This puzzle is included in the books Microteasers (1986) and The Sunday Times Book of Brainteasers (1994).
[teaser1005]
Frits and 
Hugo, and 
John Crabtree, 
GeoffR, and 
S2T2 
Jim Randell 11:57 am on 23 April 2023 Permalink |
This Python program uses the [[
SubstitutedExpression()]] solver from the enigma.py library to see if there are solutions to the alphametic expressions.It runs in 85ms.
Run: [ @replit ]
from enigma import (SubstitutedExpression, subsets, sprintf, printf) # does an alphametic sum have answers? solve = lambda expr: any(SubstitutedExpression([expr]).solve(verbose=0)) # expressions involving words for 1, 2, 3, 4 exprs = [ "{w1} + {w1} = {w2}", "{w1} + {w2} = {w3}", "{w1} + {w3} = {w4}", "{w2} + {w2} = {w4}", "{w1} + {w4} == {w2} + {w3}", ] # allocate the words for (w1, w2, w3, w4) in subsets(["AA", "BA", "EB", "DE"], size=4, select='P'): # are all the expressions solvable? if all(solve(sprintf(x)) for x in exprs): # output solution printf("1 = {w1}; 2 = {w2}; 3 = {w3}; 4 = {w4}")Solution: 1 = DE; 2 = BA; 3 = AA; 4 = EB.
So we have:
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Frits 5:10 pm on 23 April 2023 Permalink |
from enigma import subsets ''' 0: ab + ab = ef reject if: b in {a, f}, e = f, 1: ab + cd = ef reject if: b = f and d in {a, e}, d = f and b in {c, e} e = f and (b = c or a = d) 2: ab + cd = ef + gh reject if not: b = d or f = h ''' # reject words depending on the type of equation def rej(t, w, x, y, z=0): (a, b) = (w[0], w[1]) (c, d) = (x[0], x[1]) (e, f) = (y[0], y[1]) if t == 0 and (e == f or b in {a, f}): return True if t == 1 and any([b == f and d in {a, e}, d == f and b in {c, e}, e == f and (b == c or a == d)]): return True if t == 2 and not (b == d or f == z[1]): return True return False # allocate the words for (w1, w2, w3, w4) in subsets(["AA", "BA", "EB", "DE"], size=4, select='P'): if rej(0, w1, w1, w2): continue if rej(1, w1, w2, w3): continue if rej(1, w1, w3, w4): continue if rej(0, w2, w2, w4): continue if rej(2, w1, w4, w2, w3): continue print(*zip(range(1, 5), (w1, w2, w3, w4)))LikeLike
Frits 5:23 pm on 23 April 2023 Permalink |
This is more of a program saying: “if there is a solution it must be one of these: ….”
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