Rebuild docs
This commit is contained in:
Simon Forman
@@ -1,17 +1,17 @@
|
||||
`Project Euler, first problem: "Multiples of 3 and 5" <https://projecteuler.net/problem=1>`__
|
||||
`Project Euler, first problem: “Multiples of 3 and 5” <https://projecteuler.net/problem=1>`__
|
||||
=============================================================================================
|
||||
|
||||
::
|
||||
|
||||
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
|
||||
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
|
||||
|
||||
Find the sum of all the multiples of 3 or 5 below 1000.
|
||||
Find the sum of all the multiples of 3 or 5 below 1000.
|
||||
|
||||
.. code:: ipython2
|
||||
|
||||
from notebook_preamble import J, V, define
|
||||
|
||||
Let's create a predicate that returns ``True`` if a number is a multiple
|
||||
Let’s create a predicate that returns ``True`` if a number is a multiple
|
||||
of 3 or 5 and ``False`` otherwise.
|
||||
|
||||
.. code:: ipython2
|
||||
@@ -44,7 +44,7 @@ Given the predicate function ``P`` a suitable program is:
|
||||
|
||||
::
|
||||
|
||||
PE1 == 1000 range [P] filter sum
|
||||
PE1 == 1000 range [P] filter sum
|
||||
|
||||
This function generates a list of the integers from 0 to 999, filters
|
||||
that list by ``P``, and then sums the result.
|
||||
@@ -68,22 +68,22 @@ Consider the first few terms in the series:
|
||||
|
||||
::
|
||||
|
||||
3 5 6 9 10 12 15 18 20 21 ...
|
||||
3 5 6 9 10 12 15 18 20 21 ...
|
||||
|
||||
Subtract each number from the one after it (subtracting 0 from 3):
|
||||
|
||||
::
|
||||
|
||||
3 5 6 9 10 12 15 18 20 21 24 25 27 30 ...
|
||||
0 3 5 6 9 10 12 15 18 20 21 24 25 27 ...
|
||||
-------------------------------------------
|
||||
3 2 1 3 1 2 3 3 2 1 3 1 2 3 ...
|
||||
3 5 6 9 10 12 15 18 20 21 24 25 27 30 ...
|
||||
0 3 5 6 9 10 12 15 18 20 21 24 25 27 ...
|
||||
-------------------------------------------
|
||||
3 2 1 3 1 2 3 3 2 1 3 1 2 3 ...
|
||||
|
||||
You get this lovely repeating palindromic sequence:
|
||||
|
||||
::
|
||||
|
||||
3 2 1 3 1 2 3
|
||||
3 2 1 3 1 2 3
|
||||
|
||||
To make a counter that increments by factors of 3 and 5 you just add
|
||||
these differences to the counter one-by-one in a loop.
|
||||
@@ -95,7 +95,7 @@ the counter to the running sum. This function will do that:
|
||||
|
||||
::
|
||||
|
||||
PE1.1 == + [+] dupdip
|
||||
PE1.1 == + [+] dupdip
|
||||
|
||||
.. code:: ipython2
|
||||
|
||||
@@ -276,8 +276,8 @@ get to 990 and then the first four numbers 3 2 1 3 to get to 999.
|
||||
233168
|
||||
|
||||
|
||||
This form uses no extra storage and produces no unused summands. It's
|
||||
good but there's one more trick we can apply. The list of seven terms
|
||||
This form uses no extra storage and produces no unused summands. It’s
|
||||
good but there’s one more trick we can apply. The list of seven terms
|
||||
takes up at least seven bytes. But notice that all of the terms are less
|
||||
than four, and so each can fit in just two bits. We could store all
|
||||
seven terms in just fourteen bits and use masking and shifts to pick out
|
||||
@@ -286,8 +286,8 @@ integer terms from the list.
|
||||
|
||||
::
|
||||
|
||||
3 2 1 3 1 2 3
|
||||
0b 11 10 01 11 01 10 11 == 14811
|
||||
3 2 1 3 1 2 3
|
||||
0b 11 10 01 11 01 10 11 == 14811
|
||||
|
||||
.. code:: ipython2
|
||||
|
||||
@@ -516,14 +516,14 @@ And so we have at last:
|
||||
233168
|
||||
|
||||
|
||||
Let's refactor.
|
||||
Let’s refactor.
|
||||
|
||||
::
|
||||
|
||||
14811 7 [PE1.2] times pop
|
||||
14811 4 [PE1.2] times pop
|
||||
14811 n [PE1.2] times pop
|
||||
n 14811 swap [PE1.2] times pop
|
||||
14811 7 [PE1.2] times pop
|
||||
14811 4 [PE1.2] times pop
|
||||
14811 n [PE1.2] times pop
|
||||
n 14811 swap [PE1.2] times pop
|
||||
|
||||
.. code:: ipython2
|
||||
|
||||
@@ -545,21 +545,21 @@ Now we can simplify the definition above:
|
||||
233168
|
||||
|
||||
|
||||
Here's our joy program all in one place. It doesn't make so much sense,
|
||||
Here’s our joy program all in one place. It doesn’t make so much sense,
|
||||
but if you have read through the above description of how it was derived
|
||||
I hope it's clear.
|
||||
I hope it’s clear.
|
||||
|
||||
::
|
||||
|
||||
PE1.1 == + [+] dupdip
|
||||
PE1.2 == [3 & PE1.1] dupdip 2 >>
|
||||
PE1.3 == 14811 swap [PE1.2] times pop
|
||||
PE1 == 0 0 66 [7 PE1.3] times 4 PE1.3 pop
|
||||
PE1.1 == + [+] dupdip
|
||||
PE1.2 == [3 & PE1.1] dupdip 2 >>
|
||||
PE1.3 == 14811 swap [PE1.2] times pop
|
||||
PE1 == 0 0 66 [7 PE1.3] times 4 PE1.3 pop
|
||||
|
||||
Generator Version
|
||||
=================
|
||||
|
||||
It's a little clunky iterating sixty-six times though the seven numbers
|
||||
It’s a little clunky iterating sixty-six times though the seven numbers
|
||||
then four more. In the *Generator Programs* notebook we derive a
|
||||
generator that can be repeatedly driven by the ``x`` combinator to
|
||||
produce a stream of the seven numbers repeating over and over again.
|
||||
@@ -591,8 +591,8 @@ terms to reach up to but not over one thousand.
|
||||
466
|
||||
|
||||
|
||||
Here they are...
|
||||
~~~~~~~~~~~~~~~~
|
||||
Here they are…
|
||||
~~~~~~~~~~~~~~
|
||||
|
||||
.. code:: ipython2
|
||||
|
||||
@@ -604,8 +604,8 @@ Here they are...
|
||||
3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3
|
||||
|
||||
|
||||
...and they do sum to 999.
|
||||
~~~~~~~~~~~~~~~~~~~~~~~~~~
|
||||
…and they do sum to 999.
|
||||
~~~~~~~~~~~~~~~~~~~~~~~~
|
||||
|
||||
.. code:: ipython2
|
||||
|
||||
@@ -618,7 +618,7 @@ Here they are...
|
||||
|
||||
|
||||
Now we can use ``PE1.1`` to accumulate the terms as we go, and then
|
||||
``pop`` the generator and the counter from the stack when we're done,
|
||||
``pop`` the generator and the counter from the stack when we’re done,
|
||||
leaving just the sum.
|
||||
|
||||
.. code:: ipython2
|
||||
@@ -652,21 +652,21 @@ Instead of summing them,
|
||||
|
||||
::
|
||||
|
||||
10 9 8 7 6
|
||||
+ 1 2 3 4 5
|
||||
---- -- -- -- --
|
||||
11 11 11 11 11
|
||||
|
||||
11 * 5 = 55
|
||||
10 9 8 7 6
|
||||
+ 1 2 3 4 5
|
||||
---- -- -- -- --
|
||||
11 11 11 11 11
|
||||
|
||||
11 * 5 = 55
|
||||
|
||||
From the above example we can deduce that the sum of the first N
|
||||
positive integers is:
|
||||
|
||||
::
|
||||
|
||||
(N + 1) * N / 2
|
||||
(N + 1) * N / 2
|
||||
|
||||
(The formula also works for odd values of N, I'll leave that to you if
|
||||
(The formula also works for odd values of N, I’ll leave that to you if
|
||||
you want to work it out or you can take my word for it.)
|
||||
|
||||
.. code:: ipython2
|
||||
@@ -695,20 +695,20 @@ Generalizing to Blocks of Terms
|
||||
|
||||
We can apply the same reasoning to the PE1 problem.
|
||||
|
||||
Between 0 and 990 inclusive there are sixty-six "blocks" of seven terms
|
||||
Between 0 and 990 inclusive there are sixty-six “blocks” of seven terms
|
||||
each, starting with:
|
||||
|
||||
::
|
||||
|
||||
[3 5 6 9 10 12 15]
|
||||
[3 5 6 9 10 12 15]
|
||||
|
||||
And ending with:
|
||||
|
||||
::
|
||||
|
||||
[978 980 981 984 985 987 990]
|
||||
[978 980 981 984 985 987 990]
|
||||
|
||||
If we reverse one of these two blocks and sum pairs...
|
||||
If we reverse one of these two blocks and sum pairs…
|
||||
|
||||
.. code:: ipython2
|
||||
|
||||
@@ -749,9 +749,9 @@ additional unpaired terms between 990 and 1000:
|
||||
|
||||
::
|
||||
|
||||
993 995 996 999
|
||||
993 995 996 999
|
||||
|
||||
So we can give the "sum of all the multiples of 3 or 5 below 1000" like
|
||||
So we can give the “sum of all the multiples of 3 or 5 below 1000” like
|
||||
so:
|
||||
|
||||
.. code:: ipython2
|
||||
@@ -764,7 +764,7 @@ so:
|
||||
233168
|
||||
|
||||
|
||||
It's worth noting, I think, that this same reasoning holds for any two
|
||||
It’s worth noting, I think, that this same reasoning holds for any two
|
||||
numbers :math:`n` and :math:`m` the multiples of which we hope to sum.
|
||||
The multiples would have a cycle of differences of length :math:`k` and
|
||||
so we could compute the sum of :math:`Nk` multiples as above.
|
||||
@@ -774,14 +774,14 @@ interval spanning the least common multiple of :math:`n` and :math:`m`:
|
||||
|
||||
::
|
||||
|
||||
| | | | | | | |
|
||||
| | | | |
|
||||
| | | | | | | |
|
||||
| | | | |
|
||||
|
||||
Here we have 4 and 7, and you can read off the sequence of differences
|
||||
directly from the diagram: 4 3 1 4 2 2 4 1 3 4.
|
||||
|
||||
Geometrically, the actual values of :math:`n` and :math:`m` and their
|
||||
*lcm* don't matter, the pattern they make will always be symmetrical
|
||||
*lcm* don’t matter, the pattern they make will always be symmetrical
|
||||
around its midpoint. The same reasoning holds for multiples of more than
|
||||
two numbers.
|
||||
|
||||
@@ -793,6 +793,6 @@ is just:
|
||||
|
||||
::
|
||||
|
||||
PE1 == 233168
|
||||
PE1 == 233168
|
||||
|
||||
Fin.
|
||||
|
||||
Reference in New Issue
Block a user