User:Yeti: Difference between revisions

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n += 1</lang>
n += 1</lang>


{{out}}
<pre>
2
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29
</pre>
...the program has to be terminated by the user e.g. by typing ctrl-c.

=Python/Primes/FSOE3=
{{works with|Python|2.x}}
<lang python>
L = {}
n = 2

while 1:

if n in L:
P = L[n]
else:
P = n
print n

m = n+P
while m in L:
m += P
L[m] = P

n += 1</lang>
{{out}}
{{out}}
<pre>
<pre>

Revision as of 04:23, 1 March 2019

DC/Questions/DROP

TODO Drop top of stack without side effects

# assume there is a value in x[1337], e.g. "gold"
[gold] 1337 :x

# "trash" on the stack:
[trash]

# push x[1337] to x-stack
1337;x Sx
# deposit TOS in x[1337]
1337:
# top of x-stack to x[1337]
Lx 1337:

1337 ;x

[Stack: ]PAP f
Output:
Stack: 
gold

Sigh!

Is there really no simpler way to drop the top of stack that works for strings and numbers and in all versions of Dc?

Python/Primes/FSOE1

Works with: Python version 2.x

This is a sequentialised variant of the well known sieve method (ab)using a dictionary as sparse array.

Alternatively you can describe it as finding the wheels and moving them on a tape. The only knowledge is to start looking 2 instead of starting with 1, but the original sieve does skip 1 too. 2 is not hard coded to be prime. The algorithm detects it as 1st prime and then puts the 2-wheel on the tape.

A tape position can hold multiple wheels.

<lang python>L = {} n = 2

while 1:

       if n in L:
               P = L[n]
               del L[n] # optional - saves some memory.
       else:
               print n
               P = [n]
       for p in P:
               npp = n+p
               if npp in L:
                       L[npp].add(p)
               else:
                       L[npp] = set([p])
       n += 1</lang>
Output:
2
3
5
7
11
13
17
19
23
29

...the program has to be terminated by the user e.g. by typing ctrl-c.

Python/Primes/FSOE3

Works with: Python version 2.x

<lang python> L = {} n = 2

while 1:

       if n in L:
               P = L[n]
       else:
               P = n
               print n
       m = n+P
       while m in L:
               m += P
       L[m] = P
       n += 1</lang>
Output:
2
3
5
7
11
13
17
19
23
29

...the program has to be terminated by the user e.g. by typing ctrl-c.


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