Brainteaser 1635: Double anagram
From The Sunday Times, 9th January 1994 [link]
In the woodwork class the ABLEST students made STABLE TABLES.
In the arithmetic class the cleverest students took those three six-letter words which were anagrams of each other, and they then assigned a different digit to each of the six letters involved. Substituting letters for digits then gave them three six-figure numbers.
They found that one of the numbers was the sum of the other two. Furthermore, no matter what alternative substitution of digits they had used, they could never have achieved this coincidence with a lower sum.
(a) Which word was the sum?
(b) What was its numerical value?
This puzzle is included in the book Brainteasers (2002). The text was changed slightly, but the puzzle remains the same.
[teaser1635]
Frits, 
John Crabtree, and 1 other are discussing. 
GeoffR and 
Robert Brown, and 
Lise Andreasen, and 
S2T2 
Jim Randell 8:37 am on 6 August 2019 Permalink |
This program uses the [[
SubstitutedExpression()]] solver from the enigma.py library to solve the alphametic problem, and an [[Accumulator()]] object to find the smallest solution. It runs in 299ms.Run: [ @replit ]
from enigma import (SubstitutedExpression, Accumulator, sprintf as f, join, printf) # the three words words = ("ABLEST", "STABLE", "TABLES") # the alphametic puzzle ws = join(words, sep=",", enc="()") p = SubstitutedExpression(f("sum({ws}) == 2 * max({ws})"), answer=ws) # look for a minimal solution r = Accumulator(fn=min) # solve the alphametic for ans in p.answers(verbose=0): # find the result of the sum s = max(ans) # and which word is the result i = ans.index(s) r.accumulate_data(s, i) # output the minimal solution printf("{word} = {r.value} [of {r.count} possible sums]", word=words[r.data])Solution: TABLES = 417582.
The corresponding alphametic sum is: ABLEST + STABLE = TABLES.
There are 18 possible solutions to this alphametic.
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GeoffR 1:29 pm on 6 August 2019 Permalink |
I tried the three possible addition sums, looking for the smallest total. Two of these sums proved unsatisfiable. The third addition sum gave two possible values for TABLES, the smallest of which agreed with Jim’s solution.
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John Crabtree 6:21 pm on 8 August 2019 Permalink |
With TABLES as the sum, ABLE = 999(T-S) – 100S – 10T. This enables the desired solution, as well as all of the possible 18 found by Jim, to be easily found.
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Frits 2:36 pm on 27 July 2020 Permalink |
Substituted expression which returns only one solution (minimized on first part of answer):
# The following function has been manually added to enigma.py # # def substituted_expression_minimum(*args, **kw): # if 'verbose' not in kw: kw['verbose'] = 0 # answs = [] # for r in SubstitutedExpression(*args, **kw).solve(): # answs.append(r) # answs.sort(key=lambda x: x[1]) # if len(answs) > 0: # yield answs[0] from enigma import substituted_expression_minimum, printf, irange # the alphametic puzzle p = substituted_expression_minimum(\ [\ # if one element in (X,Y,Z) is sum of 2 others then sum(X,Y,Z) = 2*max(X,Y,Z) # in this way we don't have to specify it is Z=X+Y, Y=X+Z or X=Y+Z "TABLES + ABLEST + STABLE == 2 * max(TABLES, ABLEST, STABLE)",\ "max(TABLES, ABLEST, STABLE) = HIGNUM",\ "not(is_duplicate(TABLES))",\ "not(is_duplicate(ABLEST))",\ "not(is_duplicate(STABLE))"],\ distinct="", # needed for HIGNUM \ answer="HIGNUM, ABLEST, STABLE, TABLES",\ verbose=0) for (_, ans) in p: if ans[1] == ans[0]: printf("ABLEST is the sum with value {ans[1]}") if ans[2] == ans[0]: printf("STABLE is the sum with value {ans[2]}") if ans[3] == ans[0]: printf("TABLES is the sum with value {ans[3]}")LikeLike
Frits 3:12 pm on 27 July 2020 Permalink |
@Jim:
Any idea why I needed the HIGNUM field?
p = substituted_expression_minimum(\ [\ "TABLES + ABLEST + STABLE == 2 * max(TABLES, ABLEST, STABLE)"],\ answer="TABLES + ABLEST + STABLE, ABLEST, STABLE, TABLES",\ verbose=0)Using this setup I got no results (even with distinct=””).
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Jim Randell 3:33 pm on 27 July 2020 Permalink |
@Frits: I think you get no answers because you are checking to see when one of the summands is equal to the sum total (which is never).
Try using:
BTW: You don’t need to modify the enigma.py library, you can just define the [[
substituted_expression_minimum]] function in your own program.LikeLike
Frits 2:41 pm on 29 July 2020 Permalink |
@Jim: Thanks for the internal accumulate function
from enigma import SubstitutedExpression, printf # the alphametic puzzle p = SubstitutedExpression( [ # if one element in (X,Y,Z) is sum of 2 others then sum(X,Y,Z) = 2*max(X,Y,Z) # in this way we don't have to specify it is Z=X+Y, Y=X+Z or X=Y+Z "TABLES + ABLEST + STABLE == 2 * max(TABLES, ABLEST, STABLE)", "not(is_duplicate(TABLES))", "not(is_duplicate(ABLEST))", "not(is_duplicate(STABLE))" ], answer="(TABLES + ABLEST + STABLE)/2, ABLEST, STABLE, TABLES", accumulate=min, verbose=0) # solve the puzzle r = p.run() # Print answer lst = ["ABLEST", "STABLE", "TABLES"] cnt = 0 for ans in r.accumulate: if cnt == 0: hghnum = ans else: if ans == hghnum: printf("{lst[cnt-1]} is the sum with value {ans}") cnt += 1LikeLike