Sunday Times Teaser 3278 – Page Weight


# the number of pages in chapter one
ch1 = 34
# consider the number of pages in chapter two
for ch2 in range(1, ch1):
  # the number of A4 sheets needed
  nsh = (ch1 + ch2 + 3) // 4
  
  # collect the page numbers on the A4 sheets
  sheets, ns, rev = [], 0, False
  
  # add page numbers from chapter one ...
  for p in range(1, ch1, 2):
    # ... on the first half of the booklet ...
    if not rev:
      if ns <= nsh:
        sheets += [(p, p + 1) if p < ch1 else (p,)]
        ns += 1
        rev = ns == nsh
    else:
      # ... and complete it on the second half
      ns -= 1
      sheets[ns] += (p, p + 1)
      
  # add page numbers from chapter two
  for p in range(1, ch2 + 1, 2):
    ns -= 1
    if ns >= 0:
      sheets[ns] += (p, p + 1) if p < ch2 else (p,)
    else:
      break
    
  # the page number sum for the complete booklet
  pgs_sum = (ch1 * (ch1 + 1) + ch2 * (ch2 + 1)) // 2
  # find the number of sheets with half this page number sum
  sm = 0
  for i, sh in enumerate(sheets, 1):
    sm += sum(sh)
    if 2 * sm == pgs_sum:
      print(f"{ch2} pages in chapter 2 (sheet page number sums: first "
            f"{i} == last {nsh - i} == {sm}).")
      break

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42

43

# the number of pages in chapter one

ch1 = 34

# consider the number of pages in chapter two

for ch2 in range(1, ch1):

# the number of A4 sheets needed

nsh = (ch1 + ch2 + 3) // 4

# collect the page numbers on the A4 sheets

sheets, ns, rev = [], 0, False

# add page numbers from chapter one ...

for p in range(1, ch1, 2):

# ... on the first half of the booklet ...

if not rev:

if ns <= nsh:

sheets += [(p, p + 1) if p < ch1 else (p,)]

ns += 1

rev = ns == nsh

else:

# ... and complete it on the second half

ns -= 1

sheets[ns] += (p, p + 1)

# add page numbers from chapter two

for p in range(1, ch2 + 1, 2):

ns -= 1

if ns >= 0:

sheets[ns] += (p, p + 1) if p < ch2 else (p,)

else:

break

# the page number sum for the complete booklet

pgs_sum = (ch1 * (ch1 + 1) + ch2 * (ch2 + 1)) // 2

# find the number of sheets with half this page number sum

sm = 0

for i, sh in enumerate(sheets, 1):

sm += sum(sh)

if 2 * sm == pgs_sum:

print(f"{ch2} pages in chapter 2 (sheet page number sums: first "

f"{i} == last {nsh - i} == {sm}).")

break

Reply

John Z permalink


ch1 = 34

# consider the number of pages in chapter 2
for ch2 in range(1, ch1):

  # the number of blank pages at the end of the booklet
  blanks = (4 - (ch1 + ch2) % 4) & 3
  
  # constitute booklet with ch1 + ch2 + up to 3 blank pages
  p = tuple(range(1, 35)) + tuple(range(1, ch2 + 1)) + (0,) * blanks
  half = (ch1 + ch2 + blanks) // 2

  # make a list of the sum of the pages on each sheet
  sheets = [p[i] + p[i + 1] + p[-2 - i] + p[-1 - i] for i in range(0, half, 2)]

  # get the total of all of the page numbers
  tot = sum(p) 

  # consider sums of the first n sheets
  for i in range(half // 2):
    if sum(sheets[:i + 1]) * 2 == tot: # (credit: Brian Gladman)
      print(ch2, "pages in chapter 2; ", tot // 2, "total of each pile")

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

ch1 = 34

# consider the number of pages in chapter 2

for ch2 in range(1, ch1):

# the number of blank pages at the end of the booklet

blanks = (4 - (ch1 + ch2) % 4) & 3

# constitute booklet with ch1 + ch2 + up to 3 blank pages

p = tuple(range(1, 35)) + tuple(range(1, ch2 + 1)) + (0,) * blanks

half = (ch1 + ch2 + blanks) // 2

# make a list of the sum of the pages on each sheet

sheets = [p[i] + p[i + 1] + p[-2 - i] + p[-1 - i] for i in range(0, half, 2)]

# get the total of all of the page numbers

tot = sum(p)

# consider sums of the first n sheets

for i in range(half // 2):

if sum(sheets[:i + 1]) * 2 == tot: # (credit: Brian Gladman)

print(ch2, "pages in chapter 2; ", tot // 2, "total of each pile")

Reply

BRG permalink

A particularly nice feature of this version (when the ’35’ in line 11 is changed to ‘ch1 + 1’) is that it then works correctly for other values of the number of pages in chapter one. Here is a version that provides more detailed output and also finds multiple solutions.


# given the lengths of chapters one and two and a requirement to use
# at least some pages in the second half of the booklet, generate the
# number of A4 sheets used, the sum of all page numbers and a list of
# the page numbers on each sheet for possible solutions. Minimise the
# number of A4 sheets used if min_sheets is True.
def page_numbers(ch1, ch2, min_sheets=True):
  # consider the number of A4 sheets needed with at least some 
  # use of second half of the booklet
  lo = (ch1 + ch2 + 3) // 4
  hi = lo + 1 if min_sheets else (ch1 + ch2 + 1) // 2
  for a4n in range(lo, hi, 1):
    p = (tuple(range(1, ch1 + 1)) + tuple(range(1, ch2 + 1))
         + (0,) * (4 * a4n - (ch1 + ch2)))
    # collect the page numbers on each A4 sheet
    yield a4n, sum(p), [(p[k], p[k + 1], p[-2 - k], p[-1 - k])
                        for k in range(0, 2 * a4n, 2)]

# list the page numbers in chapter one
ch1 = 19

# the desired split ratio
ratio = 2

# minimise paper sheets
min_sheets = True

# consider the number of pages in chapter two
for ch2 in range(1, ch1):
  
  # consider the number of A4 sheets needed with at least some 
  # use of second half of the booklet
  for a4n, total, sheets in page_numbers(ch1, ch2, min_sheets):
    # split sheets into two piles for with equal page number sums
    sh_sum = 0
    for i, s in enumerate(sheets, 1):
      sh_sum += sum(s)
      if ratio * sh_sum == total:
        print(f"Chapters 1/2: {ch1}/{ch2} pages on {a4n} sheets (page "
              f"number sums: first {i} == last {a4n - i} == {sh_sum}).\n")
        # list the page numbers on each A4 sheet
        for i, s in enumerate(sheets, 1):
          print(f"{i:2} ({', '.join(f'{str(p):>2}' for p in s)}) {sum(s):3}")
        print()

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

# given the lengths of chapters one and two and a requirement to use

# at least some pages in the second half of the booklet, generate the

# number of A4 sheets used, the sum of all page numbers and a list of

# the page numbers on each sheet for possible solutions. Minimise the

# number of A4 sheets used if min_sheets is True.

def page_numbers(ch1, ch2, min_sheets=True):

# consider the number of A4 sheets needed with at least some

# use of second half of the booklet

lo = (ch1 + ch2 + 3) // 4

hi = lo + 1 if min_sheets else (ch1 + ch2 + 1) // 2

for a4n in range(lo, hi, 1):

p = (tuple(range(1, ch1 + 1)) + tuple(range(1, ch2 + 1))

+ (0,) * (4 * a4n - (ch1 + ch2)))

# collect the page numbers on each A4 sheet

yield a4n, sum(p), [(p[k], p[k + 1], p[-2 - k], p[-1 - k])

for k in range(0, 2 * a4n, 2)]

# list the page numbers in chapter one

ch1 = 19

# the desired split ratio

ratio = 2

# minimise paper sheets

min_sheets = True

# consider the number of pages in chapter two

for ch2 in range(1, ch1):

# consider the number of A4 sheets needed with at least some

# use of second half of the booklet

for a4n, total, sheets in page_numbers(ch1, ch2, min_sheets):

# split sheets into two piles for with equal page number sums

sh_sum = 0

for i, s in enumerate(sheets, 1):

sh_sum += sum(s)

if ratio * sh_sum == total:

print(f"Chapters 1/2: {ch1}/{ch2} pages on {a4n} sheets (page "

f"number sums: first {i} == last {a4n - i} == {sh_sum}).\n")

# list the page numbers on each A4 sheet

for i, s in enumerate(sheets, 1):

print(f"{i:2} ({', '.join(f'{str(p):>2}' for p in s)}) {sum(s):3}")

print()

This shows that this teaser actually has two solutions because no constraint was placed on the number of A4 pages that could be used. The use of the minimum number of A4 sheets was intended

Reply

John Z permalink

Testing that the sum of the page numbers is even improves speed


# number of pages in chapter 1
ch1 = 34

# consider the number of pages in chapter 2
for ch2 in range(1, ch1):

  # the number of blank pages at the end of the booklet
  blanks = 4 - (ch1 + ch2) & 3
  
  # make a list of booklet page numbers with ch1 + ch2 + the blank pages
  p = tuple(range(1, ch1 + 1)) + tuple(range(1, ch2 + 1)) + (0,) * blanks

  # the total of all of the page numbers must be even for two equal piles
  if (tot := sum(p)) & 1:
    continue

  # half of the number of pages
  half = (ch1 + ch2 + blanks) // 2

  # make a list of the sum of the page numbers on each sheet
  sheets = [p[i] + p[i + 1] + p[-2 - i] + p[-1 - i] for i in range(0, half, 2)]   

  # consider sums of the first n sheets
  for n in range(half // 2):
    if sum(sheets[:n + 1]) * 2 == tot: # (credit: Brian Gladman)
      print(ch2, "pages in chapter 2; ", tot // 2, "total of each pile")
      break

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

# number of pages in chapter 1

ch1 = 34

# consider the number of pages in chapter 2

for ch2 in range(1, ch1):

# the number of blank pages at the end of the booklet

blanks = 4 - (ch1 + ch2) & 3

# make a list of booklet page numbers with ch1 + ch2 + the blank pages

p = tuple(range(1, ch1 + 1)) + tuple(range(1, ch2 + 1)) + (0,) * blanks

# the total of all of the page numbers must be even for two equal piles

if (tot := sum(p)) & 1:

continue

# half of the number of pages

half = (ch1 + ch2 + blanks) // 2

# make a list of the sum of the page numbers on each sheet

sheets = [p[i] + p[i + 1] + p[-2 - i] + p[-1 - i] for i in range(0, half, 2)]

# consider sums of the first n sheets

for n in range(half // 2):

if sum(sheets[:n + 1]) * 2 == tot: # (credit: Brian Gladman)

print(ch2, "pages in chapter 2; ", tot // 2, "total of each pile")

break

Reply

BRG permalink

Hi John, Your version is really nice because it is compact and your approach is easy to understand. Ad there is room to make it even shorter and faster since you don’t need to pre-compute the sheets list on line 22:


# list the page numbers in chapter one
ch1 = 34
p1 =  tuple(range(1, ch1 + 1))

# consider the number of pages in chapter two
for ch2 in range(1, ch1):
  
  # the number of A4 sheets needed
  a4n = (ch1 + ch2 + 3) // 4

  # list all page numbers (with zeros for blanks)
  p = p1 + tuple(range(1, ch2 + 1)) + (0,) * (4 * a4n - (ch1 + ch2))
  
  # the total of all of the page numbers must be even for two equal piles
  if not (tot := sum(p)) & 1:
 
    # consider page sums for increasing numers of sheets
    sh_sum = 0
    for k in range(0, 2 * a4n, 2):
      sh_sum += p[k] + p[k + 1] + p[-2 - k] + p[-1 - k]
      if 2 * sh_sum == tot:
        print(f"{ch2} pages in chapter 2 (sheet page number sums: first "
              f"{k // 2 + 1} == last {a4n - k // 2 - 1} == {sh_sum}).")
        break

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

# list the page numbers in chapter one

ch1 = 34

p1 = tuple(range(1, ch1 + 1))

# consider the number of pages in chapter two

for ch2 in range(1, ch1):

# the number of A4 sheets needed

a4n = (ch1 + ch2 + 3) // 4

# list all page numbers (with zeros for blanks)

p = p1 + tuple(range(1, ch2 + 1)) + (0,) * (4 * a4n - (ch1 + ch2))

# the total of all of the page numbers must be even for two equal piles

if not (tot := sum(p)) & 1:

# consider page sums for increasing numers of sheets

sh_sum = 0

for k in range(0, 2 * a4n, 2):

sh_sum += p[k] + p[k + 1] + p[-2 - k] + p[-1 - k]

if 2 * sh_sum == tot:

print(f"{ch2} pages in chapter 2 (sheet page number sums: first "

f"{k // 2 + 1} == last {a4n - k // 2 - 1} == {sh_sum}).")

break

Reply

Frits permalink

@John, the sum all page numbers is not only even but also a multiple of 4. This can be seen if you consider summing the page numbers from the most inner sheet towards the most outer sheet.

Reply

John Z permalink

Hi Brian,
shouldn’t your line 21 be:


      sh_sum += p[2 * n] + p[2 * n + 1] + p[-2 - 2 * n] + p[-1 - 2 * n]

1

2

3

sh_sum += p[2 * n] + p[2 * n + 1] + p[-2 - 2 * n] + p[-1 - 2 * n]

otherwise some page numbers are summed twice and the central page numbers not at all?
Curiously, this doesn’t seem to affect the result.

Reply

BRG permalink

Yes, I meant to copy your line 22 exactly but I missed copying the increment and, because it produced the correct result, I didn’t notice the error. I will correct it

Reply

John Z permalink

I should have thought of this earlier: It would have been easier and more computationally efficient to rewrite line 20:

Python

for n in range(0, 2 * a4n, 2):

1
2
3

for n in range(0, 2 * a4n, 2):

Reply
- BRG permalink
  
  Yes, another good idea. I’ll change it again (and correct other errors).
  
  Reply

Sunday Times Teaser 3278 – Page Weight

by Mark Valentine

Published Sunday July 20 2025 (link)

Leave a Reply Cancel reply

Recent Posts

Recent Comments

Archives

Categories

Meta