Teaser 3169: Cancel culture — amp it up!
From The Sunday Times, 18th June 2023 [link] [link]
A heckled lecturer used her megaphone. The power reaching its amplifier’s input was multiplied by a two-figure whole number (the “intrinsic gain”). In each of three stages a fraction of the power was lost but, by good design, each of the three fractions was under one-twentieth. Curiously, each happened to have a different two-figure prime denominator. It turned out that the effective gain in the process (the ratio of the output power to the input power) was a whole number. In addition, the intrinsic gain and the effective gain both started with the same digit.
In ascending order, what were the three fractions?
[teaser3169]
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Jim Randell 4:58 pm on 16 June 2023 Permalink |
If I understand this correctly the mechanics of it are:
The input signal is initially multiplied by a 2-digit value (= I, the “intrinsic gain”), and then goes through 3 stages where it is reduced by fractions a/p, b/q, c/r, where each of the fractions is less than 1/20, and p, q and r are prime numbers. This gives the “effective gain” E, which is a whole number, and which has the same first digit as I.
So:
Here’s my initial stab at this.
It runs in 137ms. (Internal runtime is 38ms).
from enigma import ( primes, irange, first, subsets, cproduct, div, nsplit, fdiv, unpack, seq2str, sprintf, printf ) # map prime denominators to numerators that give fractions < 1/20 nums = lambda p: first(irange(1, p), count=(lambda n: 20 * n < p)) d = dict((p, nums(p)) for p in primes.between(20, 99)) # choose the prime denominators (p, q, r) for (p, q, r) in subsets(d.keys(), size=3): Rd = p * q * r # and choose the numerators (a, b, c) for (a, b, c) in cproduct(d[k] for k in (p, q, r)): # calculate the ratio E/I (= Rn / Rd) Rn = (p - a) * (q - b) * (r - c) # choose a 2-digit number for I (= intrinsic gain) for I in irange(10, 99): # calculate E (= effective gain) (a whole number) E = div(I * Rn, Rd) if E is None: continue # check E and I have the same first digit if nsplit(E)[0] != nsplit(I)[0]: continue # order the fractions fs = sorted([(a, p), (b, q), (c, r)], key=unpack(fdiv)) # output solution printf("{fs} [I={I} E={E}]", fs=seq2str(sprintf("{x}/{y}") for (x, y) in fs))Solution: The three fractions are: 1/41, 3/89, 2/43.
The intrinsic gain (I) is 89, and so the effective gain (E) is:
Without the requirement that the first digits of E and I are the same we find there are 5 potential solutions:
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Hugo 6:03 pm on 16 June 2023 Permalink |
A more realistic way of looking at it is that each stage is designed to have a certain gain, the product of the three gains being I. In practice the circuitry is imperfect and each gain is reduced by a certain fraction, so that the overall gain E is somewhat lower. The overall effect is the same as in the situation described by Jim, but the losses occur during amplification, not after it.
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Jim Randell 7:56 am on 17 June 2023 Permalink |
Here is a faster version. Once the ratio E:I is determined we can look for equivalent fractions with a 2-digit denominator to give values for E and I.
It runs in 77ms. (Internal runtime is 17ms).
from enigma import ( primes, irange, divf, subsets, cproduct, fraction, nsplit, fdiv, unpack, seq2str, format_fraction, printf ) # map denominators to numerators that give fractions < 1/20 nums = lambda n: irange(1, divf(n - 1, 20)) # choose the prime denominators (p, q, r) for (p, q, r) in subsets(primes.between(20, 99), size=3): Rd = p * q * r # and choose the numerators (a, b, c) for (a, b, c) in cproduct(nums(k) for k in (p, q, r)): # calculate the ratio E/I (= Rn / Rd) Rn = (p - a) * (q - b) * (r - c) # consider fractions E/I = Rn/Rd where I is 2-digits (n, d) = fraction(Rn, Rd) for k in irange(1, divf(99, d)): (E, I) = (k * n, k * d) if I < 10: continue # check E and I have the same first digit if nsplit(E)[0] != nsplit(I)[0]: continue # order the fractions fs = sorted([(a, p), (b, q), (c, r)], key=unpack(fdiv)) # output solution printf("{fs} [I={I} E={E}]", fs=seq2str(format_fraction(x, y) for (x, y) in fs))LikeLike
GeoffR 11:15 pm on 16 June 2023 Permalink |
A slow solution – about 8.5 sec.
It finds the three fractions, but not in ascending order.
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Jim Randell 11:31 pm on 16 June 2023 Permalink |
You can add the following to get the fractions in order:
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GeoffR 11:08 am on 17 June 2023 Permalink |
@Jim:
Yes, that extra line of code works well.
it also reduces the run-time for me to just over 2sec.
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Frits 12:31 am on 17 June 2023 Permalink |
from enigma import SubstitutedExpression from fractions import Fraction as rf # invalid digit / symbol assignments d2i = dict() for d in range(0, 10): vs = set() if d == 0: vs.update('ACEPXYZ') if d == 1: vs.update('ACE') if d > 4: vs.update('XYZ') if d % 2 == 0 or d == 5: vs.update('BDF') d2i[d] = vs # remaining checks for prime check = lambda n: all(n % x for x in [3, 7]) # PQ intrinsic gain, PR = effective gain # primes AB, CD and EF # the alphametic puzzle p = SubstitutedExpression( [ "20 * (AB - X) > 19 * AB", # prime check "check(AB)", "CD > AB", # prime check "check(CD)", "20 * (CD - Y) > 19 * CD", "EF > CD", # prime check "check(EF)", # if (AB - X) * (CD - Y) is not divisible by either prime # then (EF - Z) * PQ must be a multiple of AB * CD "any(((AB - X) * (CD - Y)) % x == 0 for x in [AB, CD]) or \ ((AB - X) * (CD - Y)) % EF == 0", "20 * (EF - Z) > 19 * EF", "PQ * (AB - X) * (CD - Y) * (EF - Z) == PR * (AB * CD * EF)" ], answer="ordered(rf(X, AB), rf(Y, CD), rf(Z, EF))", d2i=d2i, env=dict(check=check, rf=rf), distinct="", verbose=0, # use 256 to see the generated code ) # print answers for (_, ans) in p.solve(): print(f"{ans[0]}, {ans[1]} and {ans[2]}")LikeLike
GeoffR 12:31 pm on 17 June 2023 Permalink |
from enigma import is_prime from itertools import permutations PR = [ x for x in range(11, 98) if is_prime(x)] for A, B, C in permutations(range(1, 5), 3): for P in PR: if not(20 * A < P):continue for Q in PR: if Q == P:continue if not (20 * B < Q):continue if not(A * Q < B * P):continue for R in PR: if R in (P, Q):continue if not (20 * C < R):continue if not(B * R < C * Q):continue for E in range(10, 99): for I in range(10, 99): if E // 10 != I // 10:continue # Same equation as the MiniZinc solution if E * P * Q * R == I * (P - A) * (Q - B) * (R - C): print(f"Three fractions: {A}/{P}, {B}/{Q}, {C}/{R}.")LikeLike