% CHANGES TO VOLUME 4B OF THE ART OF COMPUTER PROGRAMMING
%
% Copyright (C) 2022, 2023 by Donald E. Knuth
% This file may be freely copied provided that no modifications are made.
% All other rights are reserved.
%
% Three levels of changes to the books are distinguished here:
%
% "\bugonpage" introduces the correction of an error;
% "\amendpage" introduces new material for future editions;
% "\improvepage" introduces ameliorations of lesser importance.
%
% (Changes introduced by \improvepage do not appear in the hardcopy listing.)
%
% Also, "\planforpage" introduces some of the author's half-baked intentions.
%

% NOTE: TO PUT THE INDEX ON A SEPARATE PAGE, RUN THIS WITH THE COMMAND LINE
%   tex "\let\indexeject+ \input err4b"

\newif\ifall % \alltrue means show the trivial items too
\relax % hook

\def\vertical{|}
\def\inref#1 #{\expandafter\def\csname\vertical#1\endcsname}
\catcode`|=\active
\let|\inref \input \jobname.ref
\catcode`|=12

\input taocpmac % use the format for TAOCP, with modifications below

\def\becomes{\ifmmode\ \hbox\fi{\manfnt y}\ } % wiggly arrow indicates a change

\def\bugonpage#1.#2 #3 (#4) {
  \medbreak\defaultpointsize
  \line{\kern-5pt\llap{\manfnt x}% print a black triangle in left margin
   {\bf Page #2}\enspace #3
   \leaders\hrule\hfill\ \eightrm\date#4.}
  \nobreak\smallskip\iftrue\noindent}
\def\amendpage#1.#2 #3 (#4) {
  \medbreak\defaultpointsize
  \line{\kern-5pt{\bf Page #2}\enspace #3
   \leaders\hrule\hfill\ \eightrm\date#4.}
  \nobreak\smallskip\iftrue\noindent}
\def\improvepage#1.#2 #3 (#4) {\ifall
  \medbreak\ninepoint
  \line{\kern-6pt{\sl Page #2\enspace #3\/}
   \leaders\hrule\hfill\ \eightrm\date#4.}
  \nobreak\smallskip\noindent}
\def\planforpage#1.#2 #3 (#4) {
  \medbreak\defaultpointsize
  \line{\kern-5pt{\bf Page #2}\enspace #3
   \leaders\hbox to 5pt{\hss.\hss}\hfill\ \eightrm\date#4.}
  \nobreak\smallskip\begingroup\let\endchange=\endgroup\sl\noindent}
\let\endchange=\fi
\def\nl{\par\noindent}
\def\nlh{\par\noindent\hangit}
\def\hangit{\hangindent2em}
\def\cutpar{{\parfillskip=0pt\par}}

\def\date#1.#2.#3.{% convert "yy.mm.dd." to "dd Mon 19yy"
  #3
  \ifcase#2\or Jan\or Feb\or Mar\or Apr\or May\or Jun\or
               Jul\or Aug\or Sep\or Oct\or Nov\or Dec\fi
  \ \ifnum #1<97 \hundred#1\else19#1\fi}
\def\hundred{20} % the "century" for dates before '97

\def\ex #1. [#2]{\ninepoint
  \textindent{\bf#1.}[{\it#2\/}]\kern6pt}
\def\EX #1. [#2]{\ninepoint
  \textindent{\llap{\manfnt x}\bf#1.}[{\it#2\/}]\kern6pt}

\def\foottext#1{\medskip
        \hrule height\ruleht width5pc \kern-\ruleht \kern3pt \eightpoint
        \smallskip\textindent{#1}}
\def\volheadline#1{\line{\cleaders\hbox{\raise3pt\hbox{\manfnt\char'36}}\hfill
       \titlefont\ #1\ \cleaders\hbox{\raise3pt\hbox{\manfnt\char'36}}\hfill}}
\def\refin#1 {\let|\inref \input #1.ref \let|\crossref}

\let\defaultpointsize=\tenpoint

%%%%%%%%%%%%%% opening remarks %%%%%%%%%%%%%%%%%%%%%%%%

\def\lhead{INTRODUCTION}
\let\rhead=\lhead
\titlepage
\volheadline{THE ART OF COMPUTER PROGRAMMING}
\bigskip
\volheadline{ERRATA TO VOLUME 4B}
\bigskip

\noindent This document is a transcript of the notes that I have been making
in my personal copy of {\sl The Art of Computer Programming}, Volume~4B
since it was first printed in 2022.

\ifall Four levels of updates\dash---``errors,'' ``amendments,'' ``plans,''
   and ``improvements''\dash---appear, indicated by four
\else Three levels of updates\dash---``errors,'' ``amendments,'' and
   ``plans''\dash---appear, indicated by three \fi
different typographic conventions:
\begingroup\def\hundred{17}

\bugonpage 0.666 line 1 (76.07.04)
Technical or typographical errors (aka bugs) are the most critical items, so
they are flagged with a `\thinspace{\manfnt x}\thinspace' preceding the page
number. The date on which I first was told about the bug is shown; this is the
effective date on which I paid the finder's fee. The necessary
corrections are indicated in
a straightforward way. If,~for example, the book says
`$n$' where it should have said `$n+1$', the change is shown thus:
\smallskip
$n$ \becomes $n+1$
\endchange

\amendpage 0.666 line 2 (89.07.14)
Amendments to the text appear in the same format as bugs, but without
the~`\thinspace{\manfnt x}\thinspace'. These are things I wish I had known
about or thought of when I wrote the original text, so I added them later.
The date is the date I drafted the new text.
\endchange

\def\hundred{19}

\planforpage 0.666 line 3 (17.11.20)
Plans for the future represent a third kind of item. In such notes I~sketched
my intentions about things that I wasn't ready to flesh out further when
I~wrote them down. You can identify these items because they're written in
slanted type, and preceded by a bunch of dots
`\hbox to 6em{\leaders\hbox to 5pt{\hss.\hss}\hfill}' leading to the date on
which I recorded the plan in my files.
\endchange

\improvepage 0.666 line 4 (38.01.10)
The fourth and final category\dash---indicated by page and line number in
smaller, slanted type\dash---consists of minor corrections or improvements
that most readers don't want to know about, because they are so trivial.
You wouldn't even
be seeing these items if you hadn't specifically chosen to print the complete
errata list in all its gory details.
Are you sure you wanted to do that?
\endchange

\endgroup
\ifall\else\medskip\ninepoint
My personal file of updates also includes a fourth category of items, not
shown in this list. They are miscellaneous minor corrections or improvements
that most readers don't want to know about, because they are so trivial.
If you really want to see all of the gory details,
you can download the full list from Internet webpage
$$\.{http://www-cs-faculty.stanford.edu/\char`\~knuth/taocp.html}$$
by selecting the ``long form'' of the errata.
\fi

\medskip
\tenpoint

\beginconstruction
The ultimate, glorious, future editions of {\sl The Art of Computer
Programming\/} are works in progress.
Please let me know of any improvements that you think I ought to make.
Send your comments either by snail mail to D.~E. Knuth, Computer
Science, Gates Building 1B, Stanford University, Stanford CA~94305-9015,
or by email to
{\tt taocp{\char`\@}cs.stanford.edu}. (Use email for book suggestions
only, please\dash---all
other correspondence is returned unread to the sender, or discarded,
because I have no time to
read ordinary email.) Although I'm working full time on
Volume~4C these days, I~will try to reply to all such messages
within a year of receipt. Current news is posted on
$$\.{http://www-cs-faculty.stanford.edu/\char`~knuth/taocp.html}$$
and updated regularly.
\par\endconstruction
\rightline{\dash---Don Knuth, July 2022}

\bigskip
\bigskip
{\quoteformat
Next to the promulgation of new truth,
the best thing, I conceive, that a man can do,
is the recantation of published error.
\author JOSEPH LISTER (1878)
% Trans. Pathol. Soc. Lond. 29 (1878), 425--467;
% I saw it in Notes&Records 67(2013)228 in note 23 by R Richardson

\bigskip\bigskip
\vfill\eject
}

\def\today{\number\day\space\ifcase\month\or
  January\or February\or March\or April\or May\or June\or
  July\or August\or September\or October\or November\or December\fi
  \space\number\year}

%%%%%%%%%%%%%%% CHANGES FOR VOLUME 4B %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\def\lhead{CHANGES TO VOLUME 4B: COMBINATORIAL ALGORITHMS, PART 2}
\let\rhead=\lhead
\titlepage
\volheadline{COMBINATORIAL ALGORITHMS, PART 2}
\bigskip
\rightline{Copyright \copyright\ 2022, 2023
  Addison\with Wesley}
\rightline{Last updated \today}
\bigskip
\rightline{\sl Most of these corrections have already been made in
                  recent printings.}
\smallskip
\let\defaultpointsize=\tenpoint
\def\bland{@\land@} % \land as wide as \xor
\def\blor{@\lor@}
\def\nbool#1{\mathbin{\mkern1.5mu\overline{\mkern-1.5mu#1\mkern-1.5mu}\mkern1.5mu}}
\def\nimp{\?\nbool\supset\?}

\bugonpage 4b.ix lines 10 and 11 (23.01.06)
Mar-ijn \becomes Ma-rijn
\endchange

\improvepage 4b.xiv running head (23.04.09)
[delete this page number and running head; put `xiv' at bottom, centered]
\endchange

\amendpage 4b.xv replacement for lines 11--13 (22.11.26)
\ninepoint\smallskip
\def\0 #1 |#2|{\line{{#1}\leaders\hbox to
1em{\hfil.\hfil}\hfil\hbox to 5.7em{\hfil#2}}}%
\def\1 #1 #2 |#3|{\line{\hbox to 2.25em{#1\hfil}{#2}\leaders\hbox to
1em{\hfil.\hfil}\hfil\hbox to 5.7em{\hfil#3}}}%
\def\2 #1 #2 |#3|{\line{\hskip2.25em \hbox to
2.95em{#1\hfil}{#2}\leaders \hbox to 1em{\hfil.\hfil}\hfil\hbox to 5.7em{\hfil#3}}}%
\0 {\tenssbx Chapter 7\dash---Combinatorial Searching} |[4A.1\rlap]|
\smallskip
\1 7.2. Generating All Possibilities |[4A.281\rlap]|
\2 7.2.1. Generating Basic Combinatorial Patterns |[4A.281\rlap]|
\endchange

\amendpage 4b.2 in {\eq(9)} (22.12.25)
$(\E XY)$ \becomes $\E(XY)$
\endchange

\amendpage 4b.27 lines 3 and 4 of exercise 135 (22.12.05)
\ninepoint
either ($p_k<l$ and $p_{k+1}>l$ and $q_l>k$ and $q_{l+1}<k$) or
       ($q_k<l$ and $q_{k+1}>l$ and $p_l>k$ and $p_{l+1}<k$). \becomes
either ($q_k<l$ and $p_l>k$ and $p_{l+1}<k$ and $q_{k+1}>l$) or
($q_{k+1}<l$ and $p_l<k$ and $p_{l+1}>k$ and $q_k>l$).
\endchange

\improvepage 4b.33 in step W4 (22.11.27)
[Append `(Otherwise stop.)', to match steps B5 and B5*.]
\endchange

\bugonpage 4b.37 line 17 (23.02.15)
to run though \becomes to run through
\endchange

\bugonpage 4b.42 in the caption to Table 1 (22.11.27)
\ninepoint \.{0000} \becomes \.{000}\nl
\.{150f} \becomes \.{15f}\nl
\.{MEM[40d]} \becomes \.{MEM[4d]}
\endchange

\bugonpage 4b.43 line 2 after Table 2 (22.11.27)
\.{200} through \.{204} \becomes \.{20} through \.{24}
\endchange

\bugonpage 4b.58 exercises 41 and 42 (22.11.27)
\ninepoint
\.{MEM[40d]} \becomes \.{MEM[4d]}\nl
\.{MEM[904]} \becomes \.{MEM[94]}\nl
\.{MEM[a0d]} \becomes \.{MEM[ad]}
\endchange

\improvepage a4b.68 line 2 after Table 1 (22.12.16)
to cover a given item~$i$: \becomes
to cover the item whose number is~$i$:
\endchange

\improvepage 4b.90 near the top (22.09.30)
line 2: like cover$(i)$ \becomes like cover$(i)$ in \eq(12)\nl
line 3: like hide$(p)$ \becomes like hide$(p)$ in \eq(13)
\endchange

\bugonpage 4b.91 line $-6$ (22.12.17)
only 5.2 G$\mu$ \becomes only 4.5 G$\mu$
\endchange

\bugonpage 4b.102 in {\eq(86)} (23.01.19)
$60741+$ \becomes $60742-$
\endchange

\bugonpage 4b.110 line 1 of step P5 (23.01.02)
of its option, go \becomes
of its option, or if $\.{COLOR($x$)}\ne0$, go
\endchange

\bugonpage 4b.111 line 11 (22.12.14)
30,286,432 solutions \becomes 30,258,432 solutions
\endchange

\improvepage 4b.135 lines 8 and 9 of exercise 103 (23.01.03)
\ninepoint twelve-tone row \becomes {\it twelve-tone row}
\endchange

\bugonpage 4b.138 line 8 of exercise 121 (23.01.07)
\ninepoint (2, 3, 3, 57) \becomes (2, 2, 3, 57)
\endchange

\improvepage 4b.142 line 7 (23.02.13)
\ninepoint the boundary of \becomes the boundary path of
\endchange

\amendpage 4b.145 and 146 in exercise 172 (23.01.03)
\ninepoint [change `snake-in-the-box' to simply `snake', in five places]
\endchange

\amendpage 4b.160 new rating for exercise 305 (23.02.03)
\ninepoint {\it 25\/} \becomes {\it 28}
\endchange

\amendpage 4b.160 new rating for exercise 306 (23.02.04)
\ninepoint {\it 30\/} \becomes {\it 32}
\endchange

\amendpage 4b.161 in exercise 306 (23.01.03)
\ninepoint [change `snake-in-the-box' to simply `snake', in two places]
\endchange

\amendpage 4b.172 new first paragraph for exercise 372{(b)} (22.12.06)
\begingroup\advance\parindent by 4pt\advance\thickmuskip-1mu
\ninepoint\item{b)} A {\it twintree\/} is a data structure whose nodes~$x$
have four pointer fields, \.{L0($x$)}, \.{R0($x$)},
\.{L1($x$)}, \.{R1($x$)}. It defines
two binary trees $T_0$ and $T_1$ on the nodes, where
$T_\theta$ is rooted at \.{ROOT$\theta$} and has child links
$(\.{L$\theta$},\.{R$\theta$})$.
These two trees satisfy inorder$(T_0)={\rm inorder}(T_1)=
x_1\ldots x_n$;
$\.{L0($x_k$)}=\Lambda$ $\iff$ $\.{L1($x_k$)}\ne\Lambda$,
for $1<k\le n$;
$\.{R0($x_k$)}=\Lambda$ $\iff$ $\.{R1($x_k$)}\ne\Lambda$,
for $1\le k<n$.
\par\endgroup
\endchange

\amendpage 4b.172 last line of exercise 372 (22.11.30)
\ninepoint twin tree \becomes twintree
\endchange

\bugonpage 4b.175 in exercise 398 (23.04.07)
\ninepoint entries in each of \becomes entries in some or all of\nl
simply states its contents, without any operation; hence its solution
is a no-brainer. \becomes without any operation, is a no-brainer, unless no
clue has been specified for it.
\endchange

\amendpage 4b.176 in exercise 399 (23.04.07)
\ninepoint [{\it22\/}] \becomes [{\it23}]\quad and\quad
Algorithm C \becomes Algorithm X
\endchange

\improvepage 4b.177 in exercise 408 (23.02.06)
\ninepoint only five clues \becomes at most five clues
\endchange

% ANSWERS BEGIN ON PAGE 370
\let\defaultpointsize=\ninepoint % get ready for answer pages

\bugonpage 4b.394 last line of answer 133 (22.11.10)
143--145 \becomes 145--147
\endchange

\bugonpage 4b.394 in answer 135 (22.11.10)
line 19: the interesting recurrence \becomes
an interesting recurrence for nonnegative $i$, $j$, $k$, $n$:\nl
line 20: $B_1(i,j,k)$ \becomes $B_0(i,j,k)$\nl
line 21: the solution \becomes the solution for $i+j\le n$
\endchange

\bugonpage 4b.395 line $-2$ of answer 135 (22.11.10)
$b_1=1$ \becomes $b_0=1$
\endchange

\bugonpage 4b.396 line 2 of answer 140{(e)} (23.01.20)
($**$) ($\epsilon$ \becomes ($**$) $\epsilon$
\endchange

\bugonpage 4b.397 line 2 of answer 144{(d)} (23.01.21)
$X_1^+\cdots+X_n^+$ \becomes $X_1^++\cdots+X_n^+$
\endchange

\amendpage 4b.406 append new line to answer 48 (23.02.18)
See {\sl Journal of Automated Reasoning\/ \bf67} (2023), 12:1--12:10.
\endchange

\bugonpage 4b.410 line 10 (23.02.18)
pages 28 \becomes page 28
\endchange

\amendpage 4b.411 replacement for answer 69 (23.03.14)
\ans69. (The main point is that the 33rd digit of $\pi$ is zero.)
On 14 March 2023,
Germ\'an ^{Gonz\'alez-Morris} rose to this challenge by
promulgating the puzzle shown here, which actually exhibits
35 of the requested digits(!).
(The answer can be found in Appendix E.)\hfil\break
\indent Is it possible that there might also be a puzzle
of this kind whose empty boxes are symmetrically located
about the center, as they are in exercise 68?
\hangindent-2.68in\hangafter0
{\parfillskip=0pt\hfil
 \rlap{\def\epsfsize#1#2{.54#1}\hskip6em\smash{\raise0pt\vbox{
     \epsfbox{\figdir/v4b.3344}}}}\tighten\par}
\endchange

\amendpage 4b.431 and 432, replacement for answer 73 (22.12.12)
\ans73. \def\\{\mkern.25mu/\mkern.25mu}From
$06110144\\66611514\\56132143\\55326224\\53032022\\36400525\\33640054$
we get 764,128 solutions, which is the current record. This array, found by
Filip Stappers in 2022 based on previous work by Michael Keller,
has the surprising property that {\it every\/}
candidate placement occurs in at least 33,204 solutions!
\endchange

\amendpage 4b.442 line 2 (22.12.19)
found in 20 G$\mu$ \becomes
found in 23 G$\mu$
\endchange

\bugonpage 4b.443 in the rightmost Torto of {(c)} (23.04.08)
\.{ETV} \becomes \.{YTE}
\endchange

\amendpage 4b.444 line 4 (23.01.07)
225 G$\mu$  \becomes 202 G$\mu$
\endchange

\bugonpage 4b.444 line 5 (23.01.07)
$\{\.C,\.E,\.T,\.R,\.U,\.Y\}$ \becomes
$\{\.C,\.E,\.I,\.R,\.U,\.Y\}$
\endchange

\amendpage 4b.444 replacement for answer 114 (23.01.02)
\ans114. (Solution improved by P. Weigel.)
Besides the primary items $p_{ij}$, $r_{ik}$, $c_{jk}$, $b_{xk}$
of~\eq(30), introduce $u_k$ and $v_l$ to define a permutation. Also introduce
secondary items $\pi_k$ to record the permutation and $ij$ to record the
value at cell $(i,j)$. The permutation is defined by 81 options
`$u_k$ $v_l$ $\pi_k{:}l$' for $1\le k,l\le 9$. And there are $9^4=6561$
other options, one for each cell $(i,j)$ of the board and each pair $(k,l)$
of values before and after $\alpha$ is applied. If $(ij)\alpha=i'\?\?j'$,
let $x'=3\lfloor i'\!/3\rfloor+\lfloor j'\!/3\rfloor$. Then option
$(i,j,k,l)$ is normally `$p_{i'j'}$ $r_{i'l}$ $c_{j'l}$ $b_{x'l}$
$ij{:}k$ $i'\?\?j'\?{:}l$ $\pi_k{:}l$'. However,
if $i'=i$ and $j'=j$, that option is shortened to
`$p_{ij}$ $r_{ik}$ $c_{jk}$ $b_{xk}$ $ij{:}k$ $\pi_k{:}k$', and
omitted when $k\ne l$.
The options $(i,j,k,l)$ are also omitted when $i=0$ and $k\ne j+1$,
or $i'=0$ and $l\ne j'+1$, or $i>0$ and $k=j+1$, or $i'>0$ and $l=j'+1$,
in order to force `\.{123456789}' on the top row.\par
With that top row and with $\alpha={}$transposition, Algorithm C produces
30,258,432 solutions in 171~gigamems. (These solutions were first enumerated
in 2005 by E.~Russell; see {\tt
www.afjarvis.org.uk/sudoku/sudgroup.html}.)
\endchange

\amendpage 4b.444 line 1 of answer 115 (23.01.02)
$b'_{yk}$ and $B'_{y'l}$ \becomes $b'_{yk}$
\endchange

\bugonpage 4b.446 new solution for answer {121(c)} (23.01.08)
\indent(c) With $\delta RU$ in the middle, another solution has
$C_{k-1}$, $D_{k-1}$, $A_{k-1}$, $B_{k-1}$ at the corners. With $\delta LD$,
another solution has corners $B_{k-1}$, $A_{k-1}$, $D_{k-1}$, $C_{k-1}$.
Both of those solutions work with $\delta LU$ and $\delta SU\?$;
and $\delta SU$ also has 54 additional solutions, with $C_{k-2}$ in the
upper left corner. They use
$\delta\{\{L,P,S,T\}\{J,U\},RU\}$ in the middle of row $2^{k-2}$,
and independently $\delta\{L,K,P,R,S,T\}U$ in the middle of row
$3\cdot2^{k-2}$.
\endchange

\bugonpage 4b.450 lines 4 and 5 of answer 133 (23.01.16)
We save a factor \dots\ another factor \becomes
Remove options with non-\.a on the border; also
limit \.{aaaa} to four border positions; and save another factor
\endchange

\bugonpage 4b.455 line 11 (22.12.20)
$q\equiv x$ (modulo~2) \becomes $q\equiv x/q$ (modulo~2)
\endchange

\amendpage 4b.459 line 13 of answer 151 (22.11.05)
with Algorithm X \becomes with Algorithm X and exercise 413
\endchange

\amendpage 4b.464 lines 5 and 6 of answer 172 (23.02.02)
${}=2$. For the path \dots\ $=1$ \becomes
${}=2$. We can safely omit options where
$v_i\adj v_j$ and $a_i=a_j=1$; they force a triangle.
For the path problem, the starting vertex should have $d$ options,
with $a_1+\cdots+a_d=1$
\endchange

\amendpage 4b.464 line 4 of answer 172{(a)} (23.01.03)
snake-in-the-box \becomes snake
\endchange

\amendpage 4b.464 in answer 172{(a)} (23.02.02)
line 4: 6 T$\mu$ \becomes 270 G$\mu$\nl
line 8: 13 M$\mu$ \becomes 54 M$\mu$\nl
line 12: (Algorithm M \dots\ 625 G$\mu$ \becomes
(If we force the first step downward,
Algorithm~M finds the $7!^2=25401600$ solutions in 365~G$\mu$\nl
line 17: in 17~G$\mu$, despite having 16788 options of total
length 454380(!). \becomes
in 9.4~G$\mu$, despite having 10422 options of total
length 281934(!).
\endchange

\bugonpage 4b.464 replacement for the last paragraph (23.02.02)
\indent
Now let's consider paths that start in cell $(0,1)$ and do not end in a corner.
(i)~No such solutions have 32 kings (proved in 166 G$\mu$).
(ii)~Knights, however, yield a big surprise:
There's a unique path of length~33, doubly counted! (Found
in 43~G$\mu$.) (iii)~Bishop paths can't have length~12 unless they start or
end in a corner. (iv)~There are $N=(n-1)!^2-2(n-2)!^2$ solutions where the
rook first moves down, and $N$ where it first moves sideways. Of these,
$2(n-2)!^2$ end at $(1,n-1)$ and are equivalent to those ending at $(n-2,0)$;
$(n-2)!^2$ end at $(n-1,n-2)$ and are double-counted by central symmetry;
$(n-2)!^2-(n-2)!$ end at $(1,0)$ and are double-counted by transposition;
$(n-2)!^2-(n-2)!$ end at $(n-2,n-1)$ and are double-counted by dual
transposition. 
\endchange

\bugonpage 4b.465 line 1 (23.02.02)
So there are \dots\ 47691936 \becomes
So there are $2(n-1)!^2-9(n-2)!^2+2(n-2)!=46139040$
\endchange

\amendpage 4b.465 line 6 (23.02.02)
fast, except that the kings need 6.3 T$\mu$. \becomes
fast; kings need the max time (170~G$\mu$).
\endchange

\bugonpage 4b.465 line 5 of answer 172{(b)} (23.3.16)
bishop has 36 \becomes bishop has 72
\endchange

\bugonpage 4b.465 lines 7 and 8 of answer 172{(b)} (23.01.03)
under both horizontal and vertical reflection. Every rook snake-in-the-box
\becomes
under reflection about either diagonal. Every rook snake
\endchange

\amendpage 4b.465 and 466, replacement for final paragraphs of answer 172 (23.01.03)
Nikolai \dots\ [To appear.] \becomes\nl
\indent Nikolai Beluhov has shown that the maximum length of an $n\times n$
snake king path is $\lceil n^2\!/2\rceil-1$. Indeed, those longest paths
can be characterized completely; they're related to spirals when $n$ is even,
and to stamp-folding when $n$ is odd(!). When $n\ge6$ is even, 
exactly $2n+(n\mod4)/2$ such paths are distinct
under symmetry, and they all start in a corner.
Furthermore there are exactly six distinct
snake king {\it cycles\/} of length $n^2\!/2-1$, when $n\ge8$
is a multiple of~4.\par
With arguments of a different kind, Beluhov has also proved that the longest
snake paths and cycles of a {\it knight}, on an $m\times n$ board,
have length $mn/2-O(m+n)$. [See \arXiv:2301.01152 [math.CO] (2023), 36~pages.]
\endchange

\amendpage 4b.482 line $-4$ (22.11.25)
44 pages \becomes 46 pages
\endchange

\bugonpage 4b.497 replacement for answer 306{(a)} (23.02.04)
\indent (a) Two cases: We can use a $5\times5$ box, and require that
the small squares of each option are either
$\{\.{34},\.{45}\}$,
$\{\.{47},\.{56}\}$,
$\{\.{76},\.{65}\}$, or
$\{\.{63},\.{54}\}$; or a $6\times6$ box, with those small-square coordinates
shifted by~\.{11}.
For example, if we call the leftmost piece `\.0', its $5\times5$ options are
`\.0 \.{35} \.{37} \.{34} \.{45}',
`\.0 \.{57} \.{77} \.{47} \.{56}',
`\.0 \.{53} \.{33} \.{63} \.{54}',
`\.0 \.{75} \.{73} \.{76} \.{65}'.
The $5\times5$ problem has $4\cdot183$ solutions, in groups of four that are
related by $90^\circ$ rotation; the $6\times6$ problem, similarly, has
$4\cdot209$ solutions. Reflection gives $4\cdot183+4\cdot209$ more.
Here are some of the $8+5$ classes of
equivalent solutions whose large squares form a symmetric shape; the
last two of these look the same, but use different pieces(!):
$$\def\\#1{\vcenter{\epsfbox{\figdir/v4b.331#1}}}
  \def\|#1{\vcenter{\epsfbox{\figdir/v4b.334#1}}}
\def\epsfsize#1#2{.7#1}
\kern-20pt\\0\quad\\1\!\!\quad\\2\quad\\4\quad\\5\quad\|6\ \|7\ \|8\kern-20pt$$
\endchange

\amendpage 4b.498 and 499 in answer 306 (23.01.03)
[change `snake-in-the-box' to simply `snake', in two places]
\endchange

\bugonpage 4b.498 and 499 in answer 306 (23.01.22)
line 2: both even, and $0<x<2n$, $0<y<2m$. \becomes
both even. But now we extend the ranges to $0\le x\le 2n$ and
$0\le y\le 2m$, allowing small squares to cross the boundary.\nl
in 3rd paragraph: Algorithm M finds \dots\ single 20-cycle. \becomes
Algorithm M finds respectively (0, 0, $4\cdot10$, $8\cdot10$) solutions in those
four cases. However, (0, 0, $4\cdot1$, $8\cdot8$) of those solutions are
spurious, because their small squares form disjoint cycles instead of a single
20-cycle.
\endchange

\bugonpage 4b.499 last line of answer 308 (23.02.13)
3, 1, and 4 distinct \becomes 23, 7, and~8 distinct
\endchange

\bugonpage 4b.500 line 6 (22.12.21)
168~G$\mu$ to 135~G$\mu$ \becomes 16.8~G$\mu$ to 13.5~G$\mu$
\endchange

\bugonpage 4b.502 line 1 of answer 315 (23.04.09)
$(x,y)\mapsto(x+y,x_{\max}-x)$ \becomes
$(x,y)\mapsto(x+y,-x)$
\endchange

\bugonpage 4b.503 in answer 319 (22.12.21)
[The H-grid representation of the L tetrabolo should be
$$\catcode`\*=\active \def*{$\cdot$}
\def\\#1/{\vcenter{\seventt\baselineskip6pt
              \halign{&\hbox to\baselineskip{\hss##\hss}\cr#1}}}
\\*&*&*&*&L\cr L&L&*&*&L\cr *&*&L&L&*\cr *&*&L&L&*\cr/$$
and the rightmost illustration in the display of line $-4$
should be flipped vertically.]
\endchange

\amendpage 4b.506 lines 3 and 4 (23.01.23)
{\tt userpages.monmouth.com/\char`\~colonel/polycur.html} \becomes\nl
{\tt sicherman.net/polycur.html}
\endchange

\amendpage 4b.509 and 510, replacement for answer 336 (22.12.31)
\begingroup\font\tiny=cmr5 at 4pt
\def\|{\vrule height3.2pt depth0pt width0pt}%
\def\3#1#2#3{$\vcenter{\tiny\offinterlineskip
    \hbox to.5em{\|\hss#1\hss}%
    \hbox to.5em{\|\hss#2\hss}%
    \hbox to.5em{\|\hss#3\hss}}$}%
\ans336. (Solution by P. Weigel.) We can model such a piece by three cubies
and one dowel, where the dowel is not allowed to have the same coordinates
as the solid cubie: Let there be 27 primary items $C(x,y,z)$ and
27 secondary items $D(x,y,z)$, for $0\le x,y,z<3$, representing cubies and
dowels. There are $2\cdot2\cdot2\cdot24=192$ options, obtained by
translation and rotation of the prototype piece
`$C(0,1,0)$ $C(1,1,0)$ $C(0,0,0)$ $D(0,0,0)$ $D(0,0,1)$'.
(In other words, we consider the solid cubie to be essentially the same
as a drilled cubie that's occupied by a dowel.)\par
Algorithm X needs only 500 kilomems to discover 24 solutions, all of which
are equivalent under rotation. One solution has the cubies in positions
\3199\3117\3277 \3339\3288\3286 \3355\3445\3466 and their dowels in positions
\3190\3090\3070 \3150\3228\3470 \3350\3368\3460$@$.\par
\endgroup % end of \def\3 and \font\tiny
Our model has a surprising corollary that was noted by
Ronald Kint-Bruynseels, the designer of this remarkable puzzle:
It's possible to drill two holes in each solid cubie, perpendicular to the
dowel, without destroying the uniqueness of the solution.
[{\sl Cubism For Fun\/ \bf75} (2008), 16--19; {\bf 77} (2008), 13--18.]
\endchange

\bugonpage 4b.512 in answer 342 (22.12.15)
line 6: 2.5 T$\mu$ \becomes 240 G$\mu$\nl
line 8: 2075 sets \becomes 2074 sets
\endchange

\bugonpage 4b.512 line 8 of answer 343 (22.09.12)
`y 212 311 312 412 512' \becomes
`y 122 113 123 124 125'
\endchange

\bugonpage 4b.513 in answer 345's solution (22.12.17)
{\tt nnnn} \becomes {\tt bbbb}
\endchange

\amendpage 4b.522 and 523 in answer 372{(b)} (22.11.30)
twin tree \becomes twintree\quad (6 times)
\endchange

\bugonpage a4b.532 in answer 398{(b)} (23.03.16)
in the leftmost column, second row from bottom, \.3 \becomes \.2
\endchange

\amendpage 4b.532 replacement for the first paragraph of answer 399 (23.04.07)
\ans399. (Solution by P. Weigel.)
Let $r_{ik}$ and $c_{jk}$ be primary items for $0\le i,j<n$ and $1\le k\le n$.
Also, if there are $w$ cages, introduce primary items $g_t$ for
$1\le t\le w$. Furthermore, introduce primary items $p_{ij}$ for all
cells $ij$ that belong to clueless cages. Every such $p_{ij}$ has
$n$ options `$p_{ij}$ $r_{ik}$ $c_{jk}$', for $1\le k\le n$. Finally,
let $C_t$ be the cells of the $t$th cage, and let there
be an option
$$\hbox{`$g_t$ $\bigcup\bigl\{r_{il(i_t,j_t)},\;
                              c_{jl(i_t,j_t)}\bigr\}$'}$$
for every feasible way to assign labels $l(i_t,j_t)$ to the cells of $C_t$.
For example, there are two labelings that satisfy the clue `$15\times\!@$' in
the third cage in puzzle 398(a), namely
either $l(0,3)=3$ and $l(0,4)=5$ or
       $l(0,3)=5$ and $l(0,4)=3$; the two options for $g_3$ are therefore
`$g_3$ $r_{03}$ $c_{33}$ $r_{05}$ $c_{45}$' and
`$g_3$ $r_{05}$ $c_{35}$ $r_{03}$ $c_{43}$'.
The cage of that puzzle whose clue is `$9\times\!@$' has just one option:
`$g_4$ $r_{13}$ $c_{03}$ $r_{11}$ $c_{11}$ $r_{23}$ $c_{13}$'.\par
\endchange

\amendpage 4b.533 at the beginning of answer 400 (23.04.07)
The formulation \dots\ problems, and \becomes
Algorithm X breezes through those 2048 problems almost instantaneously, and
\endchange

\amendpage 4b.533 in answer 401 (23.04.07)
within the leftmost puzzle on line $-3$:\quad $2\div$ \becomes $2-$\quad and\quad
$2-$ \becomes $3\div$\nl
line $-2$: the most \becomes among the most\nl
line $-1$: (134) \becomes (12)
\endchange

\amendpage 4b.533 replacement for answer 402 (23.04.07)
\ans402. (The solution appears below, following answer 403.)
The author constructed this puzzle by first designing the
cages, then generating a dozen or so random latin squares via exercise~86
until finding one that had a unique solution. Then the domino clues were
permuted at random, ten times; the most difficult of those ten puzzles
was selected.\tighten\par
The construction of answer 399 gives an {\mc XC} problem with
9186 options, 342 items, and 107388 total entries. There
are respectively (720, 684, 744, 1310, 990, 360, 792, 708, 568, 1200, 606, 30)
options for the pentominoes (O, P, \dots,~Z); preprocessing with
Algorithm~P reduces those counts to
(486, 231, 60, 957, 852, 186, 360, 572, 175, 765, 477, 22).
Overall, the reduced problem has 5528 options, 324
items, and 62914 total entries. The total time to find the solution and
prove its uniqueness was 2.5~G$\mu$ for Algorithm~P and 11~G$\mu$ for
Algorithm~X, with a search tree of 829 kilonodes.
(Without preprocessing, Algorithm~X would have taken
27.4~G$\mu$, and its search tree would have had 1.8 meganodes.
Could a human being solve this puzzle by hand?)
\endchange

\amendpage 4b.533 replacement for answer 403 (23.04.07)
\ans403. Filip Stappers has devised the stunning puzzle shown below,
which matches $\pi$'s first 42 digits perfectly(!).
The construction of answer 399
fails spectacularly on this particular instance, because the monster cage
for `$64+$' has 354,896,640 options. We can, however, work around that problem
by simply treating row~4 as an additional cage without a clue, and
subtracting $1+2+\cdots+9=45$ from the original
cage total. (A latin square is determined by any $n-1$ of its rows.)
Then Algorithm~X solves the problem handily, with a cost of only 292~M$\mu$
(from 173875 options, 200 items, 2766956 total entries),
and with a search tree of only 194 nodes.
(See Appendix~E\null.)
$$\vcenter{\def\epsfsize#1#2{.7#1}\epsfbox{\figdir/v4b.5875}}\qquad\qquad
  \vcenter{\def\epsfsize#1#2{.9#1}\epsfbox{\figdir/v4b.4876}}$$
\endchange

\bugonpage 4b.534 in answer 404 (23.02.13)
line 16: `$s_3$ $+10$ $p_{10}{:}4$ $q_4{:}10$' \becomes
         `$s_3$ $+11$ $p_{11}{:}4$ $q_4{:}11$'\nl
line 17: `$s_4$ $-11$ $p_{11}{:}4$ $q_4{:}11$' \becomes
         `$s_4$ $-10$ $p_{10}{:}4$ $q_4{:}10$'
\endchange

\amendpage 4b.535 replacement for answer 408 (23.02.04)
\ans408. \vrule height9pt width0pt \vtop{\null\vskip-8pt
\hbox{\def\\#1{\vcenter{\offinterlineskip\everycr{\noalign{\hrule}}
  \halign{\strut\vrule##&&\hbox to11pt{\hss##\hss}&\vrule##\cr#1}}}%
$\llap{(a)\quad}\ \\{
    &  &&  &&  &&  &&  && 6&\cr
    &28&&  &&  &&  &&  &&  &\cr
    &  &&  &&  &&  &&  &&  &\cr
    &  &&  &&  &&  &&  &&  &\cr
    &  &&  &&  &&  &&  &&  &\cr
    &11&&  &&  &&24&&  &&33&\cr
}\;;\quad
\hbox{(b)}\quad\\{
    &  && 2&&  && 6&&  &&10&\cr
    & 1&&  && 5&&  && 9&&  &\cr
    &23&&21&&  &&  &&13&&  &\cr
    &24&&  &&19&&17&&  &&14&\cr
    &  &&25&&  &&30&&  &&36&\cr
    &26&&  &&29&&  &&35&&  &\cr
}\;.\ \quad \vcenter{\hsize=10pc\raggedright\noindent
            This 19-clue puzzle is due\break
            to~Germ\'an ^{Gonz\'alez-Morris}.\break
            (See Appendix E for solu\-tions. Are
            the numbers 5\break and 19
            best possible for\break
            $6\times6$ hidato?)}$}\medskip}
\endchange

\improvepage 4b.536 line $-5$ of answer 413 (23.01.21)
we unclose the loop (and set $\.E\gets0$) if $i=\.E$;
otherwise we zero the mates and set \becomes
we set $\.E\gets0$ and unclose the loop, if $i=\.E$;
otherwise we set $\.{MATE($u$)}\gets\.{MATE($v$)}\gets0$ and
\endchange

\bugonpage 4b.539 in answer 422 (22.12.31)
line 1: $2m-2$ \becomes $2m-1$ \quad and\quad $2n-2$ \becomes $2n-1$\nl
lines 9 and 10: $N(i,j)=C(1,j)-10$, \dots\ Edges \becomes
$N(i,j)=C(i,j)-21$,
$NN(i,j)=C(i,j)-41$,
$W(i,j)=C(i,j)-12$,
$WW(i,j)=C(i,j)-14$; $N(i,j)$ is the edge between cells $(i,j)$ and
$(i-1,j)$. Edges
\endchange

\amendpage 4b.540 last lines of answer 425 (22.11.10)
solutions for $k=5$ \dots\ 9, 10. \becomes
solutions for $5\le k\le 10$, due to G.~J.~H. Goh, are also known.
(See (xviii)--(xxiv) in Fig.~\fig A--5/;
{\tt https://boonsuan.github.io/masyu.pdf}.)
\endchange

\amendpage 4b.548 in answer 449 (23.02.20)
The current record for largest \becomes An even larger\nl
{\sl Chuzzlewit}.] \becomes
{\sl Chuzzlewit}.]
The current record for longevity, $12\times6=72$, was found by
Filip Stappers in February 2023:
``$@$`Well, help him practice for this one, and I'll be very surprised
if he doesn't win,' said Moody.''
[J.~K. Rowling, {\sl Harry Potter and the Goblet of Fire\/}
(Scholastic Press, 2000), 571.]
\endchange

\bugonpage 4b.624 last line of answer 379 (22.09.13)
property. The erp rule \dots\ sufficient. \becomes
property, and no erp rule is needed.
\endchange

\bugonpage 4b.632 in answer 410 (22.10.08)
line 4: $(c_0z_0)_2\ge x_0+x_1$ \becomes
        $(c_0z_0)_2\ge x_0+x_2$\nl
line 8: $z_i\ge v\xor s$, $q_i\ge v\land s$ \becomes
        $z_i\ge v\xor s_i$, $q_i\ge v\land s_i$\nl
lines 11--13: ternary clauses; \dots But the order encoding \becomes
 ternary clauses, together with $(\bar c_7)\land(\bar z_7)$.
 Unit propagation, and the removal
of clauses with the pure literals $z_0$, $z_2$, $z_4$, $z_6$,
now reduce the result. But the order encoding
\endchange

\amendpage 4b.653 line 2 of ansert 521 (23.01.19)
\arXiv:1402.1956 [cs.AI] (2014), 13~pages. \becomes
{\sl Int.\ J. Artificial Intelligence Tools\/ \bf27},8 (2018), 1850033:1--19.
\endchange

\bugonpage 4b.662 line $-4$ (22.08.15)
saturated subtraction \becomes saturating subtraction
\endchange

\bugonpage 4b.668 in the Clique entry (23.01.19)
${\sim}G$ \becomes $\overline G$
\endchange

\amendpage 4b.671 line 3 (23.01.03)
in Section 7.2.2.1.  \becomes  in Section 7.2.2.1, unless stated otherwise.
\endchange

\amendpage 4b.671 new answer inserted at left of line 4 (23.03.14)
$$\vcenter{\def\epsfsize#1#2{.38#1}\epsfbox{\figdir/v4b.3345}}\quad
\vcenter{\halign{&\hfil#\hfil\cr\llap(see\cr answer\cr 7.2.2--69\rlap)\cr}}$$
\endchange

\amendpage 4b.672 solution to the puzzle in the new answer 403 (23.02.18)
$$\def\epsfsize#1#2{.5#1}
\vcenter{\epsfbox{\figdir/v4b.5876}}\quad\hbox{(see answer 403)}$$
\endchange

\amendpage 4b.672 rightmost solution for answer 408 (23.02.04)
\def\\#1{\vcenter{\offinterlineskip\everycr{\noalign{\hrule}}%
  \halign{\strut\vrule##&&\hbox to11pt{\hss##\hss}&\vrule##\cr#1}}}$\\{
    & 3&& 2&& 7&& 6&&11&&10&\cr
    & 1&& 4&& 5&& 8&& 9&&12&\cr
    &23&&21&&20&&18&&13&&15&\cr
    &24&&22&&19&&17&&16&&14&\cr
    &27&&25&&31&&30&&33&&36&\cr
    &26&&28&&29&&32&&35&&34&\cr}$
\endchange

\expandafter\ifx\csname indexeject\endcsname\relax\else\vfill\eject\fi
\amendpage 4b.674 and following (22.07.23)
Miscellaneous changes to the existing index of Volume~4B are collected here,
including corrections and amendments to the old entries as well as new entries
that are occasioned by the new material. Thus, the lines of the full index
that have changed serve also as an index to the present document. However,
when a correction or amendment has caused an old index entry to be deleted,
the deletion is usually not indicated.

\input exotic
\begindoublecolumns
\indexformat
\gdef\Uni1.08:{\bitmap24:1.08:}

\hangindent2em
Ahlswede, Rudolf Friedrich, 379. % 3rd
{\mc AND} operation, bitwise ($x\band y$), 17, 128, 212, 213, 215, 221, 222, 250, 252, 260, 265, 344, 400, 410, 419, 542, 560, 573--575, 584, 605. % 3rd
Bellman, Richard Ernest, xi. % 3rd
Bent trominoes, 79, 82. % 3rd
Burnside, William, lemma, 450, 465. % 3rd
Carlsen, Ingwer Curt, 482. % 3rd
Cartier, Pierre \'Emile, 267, 270, 347. % 3rd
Chordless path, \see Snake path. % 3rd
Coil: A snake cycle, 146, 161, 465. % 3rd
Factoring an exact cover problem, 83--86, 102, 109, 124, 137, 164, 167, 459--460, 482, 491, 492, 499, 523, 527; \also Relaxation of constraints. % 3rd
Factorization of problems, 52--53, 59, 60, 62, 192--194, 320, 368, 415, 556. % 3rd
Factorization of traces, 270, 346, 614. % 3rd
Generating functions, 15, 22, 24, 28, 58, 148, 149, 160, 269, 273--274, 335, 342--344, 348, 376, 378, 381, 384, 385, 392, 393, 406, 408, 415, 418, 430, 481, 552, 558, 561, 583, 586, 594, 617, 648. % 3rd
Gonz\'alez-Morris, Germ\'an Antonio, 411, 535. % 3rd
Gr{\=\i}nbergs (= Grinberg), Emanuels Donats Fr{\=\i}drihs J\=anis, 582. % 3rd
Guibert, Olivier Patrick Serge, 395, 523. % 3rd
Induced subgraphs, 63, 114, 145--146, 153, 183, 265, 436, 626. % 3rd
Keller, Michael, 132, 432, 494, 505. % 3rd
Kint-Bruynseels, Ronald Odilon Bondewijn, 509. % 3rd
Knuth, Donald Ervin (\GC75:75:-3:62% G2463
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), ii, iv, vi, viii, ix, xvii, 46, 54, 55, 63, 73, 77--79, 100, 118, 123, 185, 198, 200, 203, 235--236, 256, 258, 277, 278, 302, 309--311, 384, 389, 397, 401, 406, 412, 413, 419, 420, 424--425, 427, 429, 431--432, 440, 446, 459, 460, 463, 466, 473, 475--478, 480, 481, 484, 485, 501--503, 508, 511, 514, 523, 527, 532, 533, 538, 540, 541, 544, 548, 556, 557, 559, 561, 566, 574, 576, 577, 580, 583, 584, 591, 599--601, 604, 606, 613, 624, 628--631, 638, 639, 642, 643, 646, 650, 654, 80, 686, 714. % 3rd
Model RB, 333--334. % 2nd
Moody, Alastor, 548. % 3rd
{\mc OR} operation, bitwise ($x\bor y$), 17, 128, 227, 410, 560, 605, 622--623.
Pi day puzzle, 60, 411, 439. % 3rd
Potter, Harry James, 548. % 3rd
Proof logging, \see Certificates of unsatisfiability. % 3rd
Propagation algorithm, 412. % 3rd
RB random instances, 333--334. % 2nd
Rowling, Joanne (= J\period\ K\period), 548.
Search trees, 31, 32, 35, 37--39, 46--48, 52, 54, 73, 98--100, 104--107, 126, 147, 212--213, 216--218, 239, 308, 336, 399, 406, 407, 412, 434. % 3rd
Self-equivalent sudoku solutions, 111, 137. % 3rd
Slack options, 71, 478. % 3rd
Slack variables, 71. % 3rd
Snake cycles, 146, 161, 465. % 3rd
Snake paths, 145. % 3rd
Stamp folding, 465. % 3rd
Stappers, Filip Jan Jos, ix, 426, 431, 533, 548. % 3rd
Sudoku, symmetries of, 74, 111, 137. % 3rd
Tagged vertices, 414. % 3rd
Tetraboloes, 163. % 3rd
Twintree structure, 172. % 3rd
Utility fields in SGB format, 414. % 3rd
Weigel, Peter Heinrich, 444, 509, 532. % 3rd

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\bugonpage 4b.714 line $-6$ (22.11.14)
\eightpoint on an Dell Precision \becomes on a Dell Precision
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\bye

[The next printing will be the 3rd.]
not listed above:
page 54, line -18, add "in"
page 59, add a period following display in exercise 64
page 176, line 1, I adjusted the spacing in `9\times'
page 533, similar spacing adjustments as on page 176
page 674, adjusted spacing slightly in the epsilon entries
L-cube puzzle leaves the index
Solid bent trominoes leaves the index
The change to page 411 affects also pages 412, 413, 414, 415
The change to page 497 affects also page 496
Apge 540, line -3 when -> when we have

ARTICLES "TO APPEAR" THAT ARE STILL PENDING:
Simkin arXiv 2107.13460 [ans 7.2.2--15]
Nobel,Agrawal,Boyd arXiv 2112.03336 about Simkin's constant [ans 7.2.2--15]