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(This article was first published on <strong><a href="https://xianblog.wordpress.com/2017/11/02/le-monde-last-puzzle-1026/"> R – Xi'an's Og</a></strong>, and kindly contributed to <a rel="nofollow" href="https://www.r-bloggers.com/">R-bloggers)</a> </div>
<p><strong><img src="https://encrypted-tbn3.gstatic.com/images?q=tbn:ANd9GcRsQd3v7efgSlmsNPKddzl-oQS9fZOF8a68V6KPmf5dnmI_EJ66rw" alt="" width="273" height="185"></strong></p>
<p><strong>T</strong>he last and final <a rel="nofollow" href="http://www.affairedelogique.com/index.php" target="_blank">Le Monde</a> puzzle is a bit of a disappointment, to wit:</p>
<blockquote>
<p><em>A 4×4 table is filled with positive and different integers. A 3×3 table is then deduced by adding four adjacent [i.e. sharing a common corner] entries of the original table. Similarly with a 2×2 table, summing up to a unique integer. What is the minimal value of this integer? And by how much does it increase if all 29 integers in the tables are different?</em></p>
</blockquote>
<p><strong>F</strong>or the first question, the resulting integer writes down as the sum of the corner values, plus 3 times the sum of the side values, plus 9 times the sum of the 4 inner values [of the 4×4 table]. Hence, minimising the overall sum means taking the inner values as 1,2,3,4, the side values as 5,…,12, and the corner values as 13,…,16. Resulting in a total sum of 352. As checked in this computer code in APL by Jean-Louis:</p>
<p><a rel="nofollow" href="https://xianblog.files.wordpress.com/2017/10/lemonde1026.png" target="_blank"><img data-orig-file="https://xianblog.files.wordpress.com/2017/10/lemonde1026.png" data-orig-size="804,699" data-medium-file="https://xianblog.files.wordpress.com/2017/10/lemonde1026.png?w=300&amp;h=261&amp;fit=300%2C261&amp;resize=300%2C261" data-attachment-id="34626" data-image-meta="{" data-image-description="" src="https://xianblog.files.wordpress.com/2017/10/lemonde1026.png?w=300&h=261&fit=300%2C261&resize=300%2C261" data-permalink="https://xianblog.wordpress.com/2017/11/02/le-monde-last-puzzle-1026/lemonde1026/" data-large-file="https://xianblog.files.wordpress.com/2017/10/lemonde1026.png?w=450" alt="" data-image-title="lemonde1026" title="Jean-Foulley's resolution of the first question, in APL" sizes="(max-width: 300px) 100vw, 300px" data-comments-opened="1" class="wp-image-34626" data-recalc-dims="1"></a>This configuration does not produce 29 distinct values, but moving one value higher in one corner does: I experimented with different upper bounds on the numbers and 17 always provided with the smallest overall sum, 365.</p>
<pre>firz=matrix(0,3,3)#second level
thirz=matrix(0,2,2)#third level
for (t in 1:1e8){
flor=matrix(sample(1:17,16),4,4)
for (i in 1:3) for (j in 1:3)
firz[i,j]=sum(flor[i:(i+1),j:(j+1)])
for (i in 1:2) for (j in 1:2)
thirz[i,j]=sum(firz[i:(i+1),j:(j+1)])
#last
if (length(unique(c(flor,firz,thirz)))==29)
solz=min(solz,sum(thirz))}
</pre>
<p>and a further simulated annealing attempt did not get me anywhere close to this solution.</p>
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