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Jim Randell 11:18 am on 16 December 2021 Permalink | Reply
Tags: by: Angela Newing ( 39 ), flawed ( 54 )
Teaser 2844: Children’s children
From The Sunday Times, 26th March 2017 [link] [link]

The head teacher’s retirement celebration was attended by many former pupils. These included Adam, Brian, Colin and David, whose sons and grandsons had also attended the school – actually different numbers of grandsons for each of them, with Adam having the most. Their sons were Eric, Fred, George, Harry and Ivan, and the sons of the sons were John, Keith, Lawrence, Michael, Norman and Oliver.

Altogether Adam and Brian had sons Eric, Fred and one other; altogether Adam and Colin had sons George and one other; altogether Eric, George and Harry had sons Keith, Michael, Norman and one other; and altogether Harry and Ivan had sons Lawrence, Norman and one other.

The retirement gift was presented by the fathers of John’s and Oliver’s fathers.

Who were they?

As stated there are multiple solutions to this puzzle.

This puzzle was not included in the published collection of puzzles The Sunday Times Brainteasers Book 1 (2019).

There are now 600 puzzles available on S2T2.

[teaser2844]
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Hugh+Casement and Jim Randell are discussing. Toggle Comments
- Jim Randell 11:19 am on 16 December 2021 Permalink | Reply
  As stated the puzzle has multiple solutions.
  
  The following Python program builds the 8 possible family trees that fit the described situation.
  
  It runs in 85ms.
  
  Run: [ @replit ]
```
from enigma import (subsets, peek, product, map2str, join, printf)

# map ks to vs
def make_map(ks, vs):
  for ss in subsets(ks, size=len(vs), select='M'):
    d = dict((k, list()) for k in ks)
    for (k, v) in zip(ss, vs):
      d[k].append(v)
    yield d

# find viable father -> sons mappings
def check_fs(fs):

  # "A, B have sons E, F and one other"
  s = fs['A'] + fs['B']
  if not (len(s) == 3 and 'E' in s and 'F' in s): return False

  # "A, C have sons G and one other"
  s = fs['A'] + fs['C']
  if not (len(s) == 2 and 'G' in s): return False

  # looks OK
  return True

# collect viable maps
ffs = list(fs for fs in make_map('ABCD', 'EFGHI') if check_fs(fs))

# find viable son -> grandsons maps
def check_sg(sg):
    
  # "E, G, H have sons K, M, N and one other"
  s = sg['E'] + sg['G'] + sg['H']
  if not (len(s) == 4 and 'K' in s and 'M' in s and 'N' in s): return False

  # "H, I have sons L, N and one other"
  s = sg['H'] + sg['I']
  if not (len(s) == 3 and 'L' in s and 'N' in s): return False

  # looks OK
  return True

# collect viable maps
sgs = list(sg for sg in make_map('EFGHI', 'JKLMNO') if check_sg(sg))

# find the father of x in map d
def father(x, d):
  return peek(k for (k, v) in d.items() if x in v)

# choose the maps
for (fs, sg) in product(ffs, sgs):

  # A, B, C, D have different numbers of grandsons
  s = list(sum(len(sg[s]) for s in fs[f]) for f in 'ABCD')
  if len(set(s)) != 4: continue
  # A has the most
  if s[0] != max(s): continue

  # grandfather of J and O
  gJ = father(father('J', sg), fs)
  gO = father(father('O', sg), fs)
  # they must be different people
  if gJ == gO: continue

  # output solution
  f = lambda s: join(s, sep="+")
  fmt = lambda m: map2str((k, f(v)) for (k, v) in m.items())
  printf("{gs} [grandfather(J) = {gJ}; grandfather(O) = {gO}]", gs=f(sorted([gJ, gO])))
  printf("  father -> sons: {fs}", fs=fmt(fs))
  printf("  son -> grandsons: {sg}", sg=fmt(sg))
  printf()
```
  Solution: One of the presenters is Adam, but the other can be any of: Brian, Colin, or David.
  
  The originally published solution is: “Adam and David”, but a later correction was made (with Teaser 2846) noting there are three valid solutions to the puzzle (given above).
  
  However, there is a unique solution if the end of the puzzle is changed to:
  
  The retirement gift was presented by the paternal grandfather of John and Oliver.
  
  Who was he?
  
  We can change the sense of the test at line 68 to ensure that the grandfathers of John and Oliver are the same person, and we find there are 2 possible family trees that lead to this situation, but in both cases the grandfather is Brian.
  
  LikeLike
- Hugh+Casement 8:34 am on 17 December 2021 Permalink | Reply
  
  Without trying Jim’s program, I think it’s like this:
  A’s son is H whose sons are L, N, and either K or M. D’s son is I who has no son.
  C’s son is G whose son is M or K (whichever is not H’s son).
  Brian has sons E (who has no sons) and F who is the father of J and O.
  
  LikeLike
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Jim Randell 4:35 pm on 14 December 2021 Permalink | Reply
Tags: by: R Postill ( 20 )
Brain-Teaser 890: Round pond boat race
From The Sunday Times, 27th August 1978 [link]

In our park is a circular pond exactly 50 metres in diameter, which affords delight to small boys who go down with ships.

Two such youngsters, Arthur and Boris, took their model motor-cruisers there the other morning, and decided on a race. Each was to start his boat from the extreme North of the pond at the same moment, after which each was to run round the pond to await his boat’s arrival. The moment it touched shore its owner was to grab it by the bows and run with it directly to the South gate of the park, situated exactly 27 metres due South of the Southern edge of the pond. Both boats travelled at the same speed (1 metre in 3 seconds) but both owners, burdened with their respective craft, could manage only 3 metres in 4 seconds over the rough grass.

When the race started, Arthur’s boat headed due South, but that of Boris headed somewhat East of South and eventually touched shore after travelling 40 metres in a straight line.

Who arrived first at the gate, and by what time margin (to the nearest second)?

This puzzle is included in the book The Sunday Times Book of Brain-Teasers: Book 1 (1980). The puzzle text above is taken from the book.

[teaser890]
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Jim Randell and galoisest are discussing. Toggle Comments
- Jim Randell 4:36 pm on 14 December 2021 Permalink | Reply
  The boats start at S and travel to A and B, and are then carried to the gate at G. O is the centre of the pond, GX is tangent to the pond.
  
  For A, the distance travelled by the boat is SA = 50m, and the distance on foot is AG = 27m.
  
  So the total time taken is: 50 × 3 + 27 × 4/3 = 186 seconds.
  
  For B, we see that the triangle ABS is inscribed in a semicircle, so has a right angle at B, and the sides SA = 50m, SB = 40m (so AB = 30m), so if θ is the angle ASB we have:
  
  cos(θ) = 40/50 = 4/5
  
  We can then calculate the straight line distance BG using the cosine rule on triangle GSB:
  
  (BG)² = (GS)² + (BS)² − 2(GS)(BS)cos(θ)
  (BG)² = 77² + 40² − 2×77×40×4/5
  (BG)² = 2601
  BG = 51
  
  For B, the distance travelled by the boat is SB = 40m, and the straight line distance on foot is BG = 51m.
  
  So the total time taken is: 40 × 3 + 51 × 4/3 = 188 seconds.
  
  So it looks like A beats B by 2 seconds.
  
  However looking at the diagram, we note that the line BG actually passes through the pond, and so B would have to run around the circumference of the pool (along the arc BX), until he can make a straight line to G (along XG, a tangent to the circle).
  
  The straight line distance along the tangent, XG, is:
  
  (XG)² + 25² = (25 + 27)²
  (XG)² = 2079
  XG = 3√(231)
  
  We can calculate the angle SOB using the cosine rule:
  
  (SB)² = (OS)² + (OB)² − 2(OS)(OB)cos(𝛂)
  cos(𝛂) = (2×25² − 40²) / 2×25²
  cos(𝛂) = −7/25
  
  The amount of arc travelled, φ (the angle BOX), is then given by:
  
  φ = ψ − (𝛂 − 𝛑/2)
  
  Where ψ is the angle OGX, which we can calculate using the right angled triangle OXG:
  
  cos(ψ) = XG / (OB + BG)
  cos(ψ) = 3√(231)/52
  
  As expected, the difference made by following the arc instead of the straight line is not enough to change the result (which is given to the nearest second).
```
from math import acos
from enigma import (fdiv, sq, sqrt, pi, printf)

# boat time
boat = lambda d: d * 3

# foot time
foot = lambda d: fdiv(d, 0.75)

# given distances
R = 25
SA = 2 * R
SB = 40
AG = 27

# exact calculation?
exact = 1

# calculate time for A
A = boat(SA) + foot(AG)
printf("A = {A:.6f} sec")

# calculate time for B
SG = SA + AG
cos_theta = fdiv(SB, SA)
BG = sqrt(sq(SG) + sq(SB) - 2 * SG * SB * cos_theta)

B = boat(40) + foot(BG)
printf("B = {B:.6f} sec (approx)")

if exact:
  # calculate XG
  XG = sqrt(sq(R + AG) - sq(R))

  # alpha is the angle SOB
  cos_alpha = fdiv(2 * sq(R) - sq(SB), 2 * sq(R))
  
  # psi is the angle OGX
  cos_psi = fdiv(XG, R + AG)

  # calculate phi (amount of arc)
  psi = acos(cos_psi)
  alpha = acos(cos_alpha)
  phi = psi - (alpha - 0.5 * pi)

  # calculate the exact distance for B
  BX = R * phi
  B = boat(40) + foot(BX + XG)
  printf("B = {B:.6f} sec (exact) [extra distance = {xB:.6f} m]", xB=BX + XG - BG)
  
# calculate the difference
d = abs(A - B)
printf("diff = {d:.6f} sec")
```
  Solution: Arthur beats Boris by approximately 2 seconds.
  
  Following the arc is 3.95cm longer than the straight line distance, and adds 53ms to B’s time.
  
  The actual shortest distance from B to G can be expressed as:
  
  $3\sqrt{231}\; +\; 25\cos ^{-1}\left( \frac{7}{52}\; +\; \frac{18\sqrt{231}}{325} \right)$
  
  which is approximately 51.039494 m, compared to the straight line distance BG of 51 m.
  
  LikeLike
  - galoisest 4:28 pm on 15 December 2021 Permalink | Reply
    
    Nice analysis Jim, I completely missed that the straight line goes through the pond.
    
    The simplest expression I can get for the difference including the arc is:
    
    100/3*(pi-acos(25/52)-2*acos(3/5)) + 4sqrt(231) – 66
    
    LikeLike
    - Jim Randell 4:37 pm on 15 December 2021 Permalink | Reply
      
      It was more obvious in my original diagram (which had the line BG, instead of XG).
      
      But in this case it doesn’t matter if you spot this or not – the answer is unaffected. (I wonder if the setter intended us to calculate the exact distance, or just work out the length of BG).
      
      LikeLike
Jim Randell 10:33 am on 12 December 2021 Permalink | Reply
Tags: by: Mr. W. Blackman
Brain Teaser: Tea table talk
From The Sunday Times, 26th December 1954 [link]

The Sales Manager of the Kupper Tea Company came into the Chief Accountant’s office. “I’ve got to report to the Chairman on the results of our new policy of selling pound packets only in four qualities only”, he said, “Which is the most popular brand — the 3s. 11d., 4s. 11d., 5s. 11d., or 6s. 11d.?”.

“When reduced to pounds, shillings and pence”, replied the man of figures, the sales of each brand for last month differ from each other only in the pence column, and there only to the extent that the figures are 1d., 2d., 3d., and 4d., though not respectively”.

“Very Interesting,” said the Sales Manager, but the Chairman will want a plain “yes” or “no” to whether we have reached our sales target. What is the answer to that?”.

“The answer to that, my boy”, replied the Chief Accountant, “is all the further information you need to calculate the complete figures for yourself”.

What are the figures?

[Note: £1 = 20s, and 1s = 12d]

This is one of the occasional Holiday Brain Teasers published in The Sunday Times prior to the start of numbered Teasers in 1961. Prizes of £3, £2 and £1 were offered for the first 3 correct solutions.

[teaser-1954-12-26] [teaser-unnumbered]
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GeoffR, Hugh+Casement, and Jim Randell are discussing. Toggle Comments
- Jim Randell 10:35 am on 12 December 2021 Permalink | Reply
  The following program steps through multiples of the largest price, and looks for multiples of the other prices which come in close to the same price, but with differing “pence” values.
  
  We find that there are many sets of 4 sales figures that differ only in the pence column, with values of (1, 2, 3, 4).
  
  If the sales target had been met (or exceeded) we would not know which of these to choose (if we find one set of figures that works, any larger set of figures would also work).
  
  However, if the sales target had not been met, and that is enough information to work out what the figures are, then we must be interested in the lowest set of possible sales figures, and the target must be higher than this, but less than the second lowest possible set of figures.
  
  The program runs in 166ms, so is fast enough, but this isn’t a suitable approach for a manual solution.
  
  Run: [ @replit ]
```
from enigma import irange, inf, div, singleton, unpack, printf

# pence -> (pounds, shillings, pence)
def d2psd(n):
  (n, d) = divmod(n, 12)
  (p, s) = divmod(n, 20)
  return (p, s, d)

# (pounds, shillings, pence) -> pence
def psd2d(p, s, d):
  return 240 * p + 12 * s + d

# solve for the specified prices, (A, B, C, D)
def solve(A, B, C, D):
  # look for multiples of the largest price
  for n in irange(0, inf, step=D):
    (p, s, d) = d2psd(n)
    if 0 < d < 5:
      # so the other amounts were are interested in are...
      r = dict()
      for d_ in (1, 2, 3, 4):
        if d_ == d: continue
        k = singleton(x for x in (A, B, C) if div(psd2d(p, s, d_), x))
        if k:
          r[k] = (p, s, d_)
        else:
          break
      # have we found a value for each d?
      if len(r.keys()) == 3:
        r[D] = (p, s, d)
        yield r

# prices (in ascending order)
prices = list(psd2d(0, s, 11) for s in (3, 4, 5, 6))

# look for the smallest solution
for r in solve(*prices):
  for (k, (p, s, d)) in sorted(r.items(), key=unpack(lambda k, v: v)):
    n = div(psd2d(p, s, d), k)
    (_, ks, kd) = d2psd(k)    
    printf("({p}, {s}s, {d}d) = {n} pkts @ ({ks}s, {kd}d) each")
  printf()
  break
```
  Solution: The sales figures are:
  
  £53878, 1s, 1d = 182123 pkts @ 5s, 11d each
  £53878, 1s, 2d = 275122 pkts @ 3s, 11d each
  £53878, 1s, 3d = 219165 pkts @ 4s, 11d each
  £53878, 1s, 4d = 155792 pkts @ 6s, 11d each
  
  And the target must be above this.
  
  The next set of figures starts at £77098, 0s, 1d, so the target must be below this (otherwise there would be a choice of two sets of figures).
  
  LikeLike
- Hugh+Casement 2:10 pm on 12 December 2021 Permalink | Reply
  
  I wouldn’t like to have had to work that out in 1954, with no computer to hand.
  It would have been easier if the sales had been all the same: £68088 14s. 1d. is the LCM of the four packet prices, each a prime number of pence.
  
  In the days before the ubiquitous teabags, tea was normally sold in quarter-pound packets.
  I seem to remember a price of about 1s. 6d., but probably less in the ’50s.
  
  LikeLike
- GeoffR 9:00 am on 14 December 2021 Permalink | Reply
```
% A Solution in MiniZinc
include "globals.mzn";

% Four standard sale prices (3s, 11d, 4s, 11d, 5s, 11d, 6s, 11d)
int: P1 == 47; int: P2 == 59; int: P3 == 71; int: P4 == 83;

% Amounts sold for each packet price
var 100000..300000:pkt1; var 100000..300000:pkt2;
var 100000..300000:pkt3; var 100000..300000:pkt4;

constraint all_different([pkt1,pkt2,pkt3,pkt4]);

% testing range for sales totals is £50,000 to £60,000
% range values in old pence (240d = £1)
var 12000000..14400000: T1;  
var 12000000..14400000: T2;
var 12000000..14400000: T3;
var 12000000..14400000: T4;

constraint T1 == pkt1 * P1 /\ T2 == pkt2 * P2
/\ T3 == pkt3 * P3 /\ T4 == pkt4 * P4;

constraint all_different([T1, T2, T3, T4]);

% the four total sales values are 1d, 2d, 3d and 4d different 
constraint max([T1, T2, T3, T4]) - min([T1, T2, T3, T4]) == 3;

solve satisfy;

output ["Totals(old pence) [T1, T2, T3, T4] = " ++ show([T1, T2, T3, T4]) 
++ "\nPackets (No.) = " ++ show([pkt1, pkt2, pkt3, pkt4])];

% Totals(old pence) [T1, T2, T3, T4] = [12126799, 12126801, 12126800, 12126798]
% Packets (No.) = [258017, 205539, 170800, 146106]
% ----------
% Totals(old pence) [T1, T2, T3, T4] = [12193116, 12193117, 12193114, 12193115]
% Packets (No.) = [259428, 206663, 171734, 146905]
% ----------
% Totals(old pence) [T1, T2, T3, T4] = [12930734, 12930735, 12930733, 12930736]
% Packets (No.) = [275122, 219165, 182123, 155792]  *** Published answer ***
% ----------
% Totals(old pence) [T1, T2, T3, T4] = [13664874, 13664872, 13664873, 13664871]
% Packets (No.) = [290742, 231608, 192463, 164637]
% ----------
% ==========
```
  I worked in old decimal pence (240p = £1), looking for 1d differences in total sales values between 12,000,000 and 14,400,000 sales totals, in old pence. This search range is £50,000 to £60,000 in current money terms and was used to reduce search time. The Chuffed solver was used.
  
  Using Python’ divmod function for converting old money totals to current monetary values:
  
  T1 = 12930734 in old pence is:
  divmod(12930734,240) = (53878, 14) = £53878 1s 2d
  
  T2 = divmod(12930735,240)
  = (53878, 15) = £53878 1s 3d
  
  T3 = divmod(12930733,240)
  = (53878, 13) = £53878 1s 1d
  
  T4 = divmod(12930736,240)
  = (53878, 16) = £53878 1s 4d
  
  There were several solutions, including the published solution.
  
  LikeLike
Jim Randell 1:17 pm on 10 December 2021 Permalink | Reply
Tags: by: Andrew Skidmore ( 87 )
Teaser 3090: Main line
From The Sunday Times, 12th December 2021 [link] [link]

Anton and Boris live next to a railway line. One morning a goods train passed Anton’s house travelling south just as a slower train passed Boris’s house travelling north. The goods train passed Boris’s house at the same time as a passenger train, heading north at a speed that was half as fast again as the goods train. Similarly, as the slower train passed Anton’s house it passed a passenger train; this was heading south at a speed that was three times as great as that of the slower train.

The passenger trains then passed each other at a point 25 kilometres from Anton’s house before simultaneously passing the two houses.

All four trains travelled along the same route and kept to their own constant speeds.

How far apart do Anton and Boris live?

[teaser3090]
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- Jim Randell 1:38 pm on 10 December 2021 Permalink | Reply
  
  This is an exercise is generating and solving simultaneous equations. No programming necessary.
  
  If we suppose B lives a distance d from A.
  
  Initially (at time 0) if the goods train passes A travelling south at speed 2v, then it reaches B at a time of (d / 2v).
  
  At this time, the passenger train, with a speed of 3v passes B, heading north.
  
  And the slow train, travelling at speed 2fv (i.e. some fraction of v), reaches A at a time of (d / 2fv).
  
  And at this time a train travelling at 6fv passes A heading south.
  
  These trains pass at time t1 at a point 25 km south of A:
  
  t1 = (d / 2fv) + (25 / 6fv) = (3d + 25) / 6fv
  t1 = (d / 2v) + ((d − 25) / 3v) = (5d − 50) / 6v
  ⇒ f = (3d + 25) / (5d − 50)
  
  And then at time (t1 + t2) the two trains pass A and B:
  
  t2 = 25 / 3v
  t2 = (d − 25) / 6fv
  ⇒ f = (d − 25) / 50
  
  Equating these:
  
  (3d + 25) / (5d − 50) = (d − 25) / 50
  10(3d + 25) = (d − 25)(d − 10)
  30d + 250 = d² − 35d + 250
  d² − 65d = 0
  ⇒ d = 65
  
  So A and B are 65 km apart, and f = 4/5.
  
  We are not given any times, so we cannot determine the actual speeds of the trains.
  
  Solution: Anton and Boris live 65 km apart.
  
  LikeLike

Jim Randell 11:27 am on 9 December 2021 Permalink | Reply
Tags: by: Andrew Skidmore ( 87 )

Teaser 2843: Child’s play

From The Sunday Times, 19th March 2017 [link] [link]

Liam has a set of cards numbered from one to twelve. He can lay some or all of these in a row to form various numbers. For example the four cards:

6, 8, 11, 2

give the five-figure number 68112. Also, that particular number is exactly divisible by the number on each of the cards used.

In this way Liam used his set of cards to find another much larger number that was divisible by each of the numbers on the cards used — in fact he found the largest such possible number.

What was that number?

[teaser2843]

Jim Randell 11:29 am on 9 December 2021 Permalink | Reply

To reduce the search space we can use some shortcuts.

Considering the cards that are included in the number, if the “10” card is included, then it would have to appear at the end, and would preclude the use of cards “4”, “8”, “12”.

Similarly if “5” is included, and any even card, then the number must end in “10” and the cards “4”, “8”, “12” are excluded.

If any of the cards “3”, “6”, “9”, “12” are included then the overall digit sum must be a multiple of 3.

Similarly if “9” is included, then the overall digit sum must be a multiple of 9.

If any of the even cards (“2”, “4”, “6”, “8”, “10”) are used, the resulting number must also be even.

This Python program uses a number of shortcuts. It considers the total number of digits in the number, as once a number is found we don’t need to consider any number with fewer digits. It runs in 263ms.

Run: [ @replit ]

from enigma import (enigma, irange, group, subsets, Accumulator, join, printf)

# the cards
cards = list(irange(1, 12))

# map cards to their digit sum, and number of digits
dsum = dict((x, enigma.dsum(x)) for x in cards)
ndigits = dict((x, enigma.ndigits(x)) for x in cards)

# find maximum arrangement of cards in ss
def find_max(ss):
  # look for impossible combinations
  if (10 in ss) and (4 in ss or 8 in ss or 12 in ss): return
  even = any(x % 2 == 0 for x in ss)
  if (5 in ss and even) and (4 in ss or 8 in ss or 12 in ss): return
  # calculate the sum of the digits
  ds = sum(dsum[x] for x in ss)
  # check for divisibility by 9
  if (9 in ss) and ds % 9 != 0: return
  # check for divisibility by 3
  if (3 in ss or 6 in ss or 12 in ss) and ds % 3 != 0: return
  # look for max rearrangement
  r = Accumulator(fn=max)
  fn = (lambda x: (lambda x: (-len(x), x))(str(x)))
  for s in subsets(sorted(ss, key=fn, reverse=1), size=len, select="P"):
    if even and s[-1] % 2 != 0: continue
    # done, if leading digits are less than current max
    if r.value and s[0] != r.data[0] and str(s[0]) < str(r.value): break
    # form the number
    n = int(join(s))
    if all(n % x == 0 for x in ss): r.accumulate_data(n, s)
  return (r.value, r.data)

# sort subsets of cards into the total number of digits
d = group(subsets(cards, min_size=1), by=(lambda ss: sum(ndigits[x] for x in ss)))

r = Accumulator(fn=max)
for k in sorted(d.keys(), reverse=1):
  printf("[considering {k} digits ...]")

  for ss in d[k]:
    rs = find_max(ss)
    if rs is None: continue
    r.accumulate_data(*rs)

  if r.value: break

printf("max = {r.value} [{s}]", s=join(r.data, sep=","))

Solution: The number is 987362411112.

The cards are: (1, 2, 3, 4, 6, 7, 8, 9, 11, 12).

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Updates from December, 2021 Toggle Comment Threads | Keyboard Shortcuts

Jim Randell 11:18 am on 16 December 2021 Permalink | Reply Tags: by: Angela Newing ( 39 ), flawed ( 54 )

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