mysql – Condition filtering estimate calculating double dipped extra value without performing range reduction

In a query like below

select  
e.emp_no,  
concat(e.first_name, ' ', e.last_name),  
d.dept_name,  
t.title_name,  
es.salary,  
es.insurance,  
es.pf,  
ea.city,  
ea.state,  
ea.phone  
from employee e  
join emp_salary es on e.emp_no = es.emp_no  
join emp_title et on et.emp_no = e.emp_no  
join title t on t.title_no = et.title_no  
join emp_address_phone ea on ea.emp_no = e.emp_no  
join emp_dept ed on e.emp_no = ed.emp_no  
join department d on d.dept_no = ed.dept_no  
where  
(e.hire_date > '2020-01-01' or e.hire_date < '1972-01-01' or e.hire_date < '1971-01-01') 
and (t.title_created > '2006-01-01') 
and (ea.country = 'Spain' or ea.country = 'Samoa' or ea.country = 'India');

Access happens through the criteria on ea.country, and when e is added to join order, condition filtering is applied for the criteria on e.hire_date.

When there is histogram:

"filtering_effect": (
  {
    "condition": "(`e`.`hire_date` > DATE'2020-01-01')",
    "histogram_selectivity": 0.0085
  },
  {
    "condition": "(`e`.`hire_date` < DATE'1972-01-01')",
    "histogram_selectivity": 0.0426
  },
  {
    "condition": "(`e`.`hire_date` < DATE'1971-01-01')",
    "histogram_selectivity": 0.0204
  }
),
"final_filtering_effect": 0.0701

The value (0.0701) of final_filtering_effect comes from the probability formula, i.e.

final_filtering_effect = 0.0085 + 0.0426 + 0.0204 - (0.0085 * 0.0426) - (0.0426 * 0.0204) - (0.0085 * 0.0204) + (0.0085 * 0.0426 * 0.0204)
final_filtering_effect = 0.0701

In collates the probability of all the ranges, without considering to reduce range. This causes double dipping of number of rows for the values that are less than ‘1971-01-01’ as they are included twice.

If the final_filtering_effect was calculated only using the reduced range, the value would have been

final_filtering_effect = 0.0085 + 0.0426 - (0.0085 * 0.0426)
final_filtering_effect = 0.0507

There are total of 10000 rows in the relation, hence the condition seems to filter around 507 rows, and if we see the real count, that seems to align right. It is 506 and not 701.

mysql > select count(*) from employee where hire_date > '2020-01-01' or hire_date < '1972-01-01' or hire_date < '1971-01-01';
+----------+
| count(*) |
+----------+
|      506 |
+----------+

When there is no histogram, just going on with heuristics of 33.33 percent for col (non eq op) val criteria, the value is double dipped,

"filtering_effect": (
),
"final_filtering_effect": 0.7037

Is there a reason why there is no range reduction performed in this stage?

Log space reduction from STCONN to CYCLE

I read this post: Showing Cycle is NL-complete?, but I am not sure why the reduction is log space, as it requires keeping track of the new graph, which has $n^n$ nodes.

dnd 5e – Does reduction of maximum hit points stick to the form it is applied to?

Following up on How does Max-HP reduction affect wild-shaped/polymorphed creatures?, which states:

Damage taken in animal form doesn’t affect your original form’s HP unless you’re dropped to 0 HP in animal form and there’s excess damage. Nowhere is it suggested that max-HP reduction would work any differently. Because Wild Shape/Polymorph gives you a new pool of HP, only that pool is affected by the reduction.

Context

A druid gets seduced by a succubus. They kiss while the druid is in bear form – this is not hypothetical as yesterday exactly this had happened. The druid gets lowered Maximum Hit Points because of this forced romance. So according to the linked Q&A, the reduction would only apply to the bear form.

Question

If the druid reverts back to normal, the HP reduction is not active anymore. What if the druid wild shapes another time, back into a bear: does it get a fresh “pool of HP”, or does the Reduced Max HP stay with its bear form until it gets “cured”?

In other words: are shapeshifters actually really resilient against abilities that reduce maximum hit points?

Example

For clarity, let’s use these numbers:

  1. Druid: 45 HP
  2. Wild Shapes into Brown Bear: 34 HP, but reduced to 10 Maximum HP (after two kisses).
  3. Druid reverts back to normal: 45 HP
  4. Wild Shapes back into Brown Bear: 34 HP, or still at 10 HP?

turing machines – Reduction from HALT to MPCP via TM Configuration Transitions

I have book with a very simple reduction from HALT to MPCP and I suspect that it is wrong.

Add the following dominos $(top, bottom$):

  • Start: $(langle varepsilon rangle ,langle varepsilonrangle langlekappa_0rangle)$ for initial configuration $kappa_0$.

  • Step: $(langlekappa_irangle,
    langlekappa_jrangle)$
    for all configuration transitions $kappa_i rightarrow kappa_j$

  • Finish: $(langlekappa_erangle langlevarepsilonrangle, langlevarepsilonrangle)$ for all end final configurations $kappa_e$

This is much easier than what I see e.g. in Sipser. Is there an error in this proof?


You may need the definition of configuration transitions:

Let $T = (S, Sigma, Pi, delta, s_0, _, E)$ be a TM. Every triple
$(nu,s,omega)$ with $nu, omega in Pi^{+}$ and $s
in S$
is called configuration of $T$. The configuration transitions
$rightarrow_{T}$ is defined as

movement to the right: $delta(s,sigma) = (s’, sigma’, rightarrow), varrho, sigma in Pi$

begin{align*}
(nu varrho, s, sigma omega) rightarrow_{T} left{begin{array}{ll}%
(nu varrho sigma’ ,s’, omega) & mbox{if $omega neq epsilon$} \
(nu varrho sigma’ ,s’, _) & mbox{if $omega = epsilon$}
end{array}right.
end{align*}

movement to the left: $delta(s, sigma) = (s’, sigma’, leftarrow), varrho, sigma in Pi$
begin{align*}
(nu varrho, s, sigma omega) rightarrow_{T} left{begin{array}{ll}%
(nu, {s’}, varrho {sigma’} omega) & mbox{if $nu neq epsilon$} \
(_, {s’}, varrho {sigma’} omega) & mbox{if $nu = epsilon$}
end{array}right.
end{align*}


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block – Half reduction formula – Bitcoin Stack Exchange

The formula is correct – halving occurs every 210,000 blocks. If the average time between each block is exactly 10 minutes, you end up with a total time of around 3,995 years.

However, if the blocks are extracted a little more slowly, it will take a little more than 4 years – if they are extracted faster than 10 minutes, it will take less than 4 years to move to the next half.

It is practically impossible to predict the date of the next halving at this time, as it would vary greatly depending on the changes in hashrate and difficulty over the next few years.

turing machines – Reduction from $ HALT $ to $ A_ {TM} $

I know the reduction of $ A_ {TM} $ at $ HALT $. But is
the next reduction of $ HALT $ at $ A_ {TM} $ correct?

We are looking for a total calculable function $ f $ mapping from $ HALT $ at $ A_ {TM} $.
The following TM $ F $ calculate reduction $ f $.

F = on input 
    create the following TM T':
    T' = on input v:
       start T on v
       if T accepts or rejects, *accept*
       if T loops, *reject*
    return 

I think the line if T accepts or rejects, *accept* is correct, but it would be nice if someone could verify this.

Will this reduction in exact subtotal coverage fail because of a potential false positive?

After deleting multiple sets and sets that contain items that do not exist in $ S $.

$ S $ = (9,6,7,4,5,1,8) $

$ C $ =$ ((9,6,7), (4,5), (1,8)) $

Turn values ​​into $ C $ shared index values ​​with $ S $. (This must be done before touching $ S $)

$ C $ = $ ((1,2,3), (4,5), (6,7)) $

And do the same for $ S $

$ S $ = $ (1,2,3,4,5,6,7) $

Square each $ x $ integer in both $ S $ and $ C $

$ f (x) $ = $ x ^ 2 $, $ x ∈ S $ then $ C $

$ S $ = $ (1, 4, 9, 16, 25, 36, 49) $

$ C $ = $ ((1,4,9), (16,25), (36,49)) $

Then delete all sets with repetitive sums to avoid false positives. That means no $ (1), (1) s … $ that could be used, to sum up the total sum of $ S $, It also means $ (1,4,9) $ or $ (4,9,1) $. (leave one!)

  1. Once the transformation is finished, use the subset solver and define the total sum as $ 140 (total sum of $ s $)
  2. Define the list of integers as a summary set of $ C $ = $ ((14), (41), (85)) $
  3. Run the algorithm and get the solution

Question

Will this reduction of the exact coverage as a sub sum give a false positive?

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