I need to calculate correlations across columns.
The code below works when calculating correlations across rows.
What is needed to modify to calculate across columns?
Input file:
Name C1 C2 C3 C4 C5 C6
R1 1 2 3 4 5 6
R2 2 1 1 0 1 0
R3 1 3 1 1 2 1
R4 1 1 0 2 0 1
R5 1 2 2 2 0 2
R6 1 1 0 1 2 0
Desired Output:
C1 C1 1.00
C1 C2 -0.4
C1 C3 -0.069
C1 C4 -0.597
C1 C5 -0.175
C1 C5 -0.362
C2 C2 1.00
C2 C3 0.4889
etc.
Code:
awk '{
a = 0; for (i = 2; i <= NF; ++i) a += $i; a /= NF-1
b = 0; for (i = 2; i <= NF; ++i) b += ($i - a) ^ 2; b = sqrt(b)
if (b <= 0) next
for (i = 2; i <= NF; ++i) x[NR, i] = ($i - a) / b
n[NR] = $1
for (i = 2; i <= NR; ++i) {
if (!(i in n)) continue
a = 0
for (k = 2; k <= NF; ++k)
a += x[NR, k] * x[i, k]
print n[NR], n[i], a
}}'
Don't know if looking for this kind solution, but how about to transpose first with following awk:
awk '
{ for (i=1;i<=NF;i++) arr[i","NR]=$i; }
END {
for (i=1;i<=NF;i++) {
for (j=1;j<=NR;j++) printf("%s%s",arr[i","j],FS);
printf("%s",RS);
}
}
'
Output:
Name R1 R2 R3 R4 R5 R6
C1 1 2 1 1 1 1
C2 2 1 3 1 2 1
C3 3 1 1 0 2 0
C4 4 0 1 2 2 1
C5 5 1 2 0 0 2
C6 6 0 1 1 2 0
Then just combine with Your script to calculate column-column correlations:
awk '
{ for (i=1;i<=NF;i++) arr[i","NR]=$i; }
END {
for (i=1;i<=NF;i++) {
for (j=1;j<=NR;j++) printf("%s%s",arr[i","j],FS);
printf("%s",RS);
}
}
' roddy.txt | awk '{
a = 0; for (i = 2; i <= NF; ++i) a += $i; a /= NF-1
b = 0; for (i = 2; i <= NF; ++i) b += ($i - a) ^ 2; b = sqrt(b)
if (b <= 0) next
for (i = 2; i <= NF; ++i) x[NR, i] = ($i - a) / b
n[NR] = $1
for (i = 2; i <= NR; ++i) {
if (!(i in n)) continue
a = 0
for (k = 2; k <= NF; ++k)
a += x[NR, k] * x[i, k]
print n[NR], n[i], a
}}'
Output:
C1 C1 1
C2 C1 -0.4
C2 C2 1
C3 C1 -0.069843
C3 C2 0.488901
C3 C3 1
C4 C1 -0.597614
C4 C2 0.239046
C4 C3 0.667827
C4 C4 1
C5 C1 -0.175412
C5 C2 0.30697
C5 C3 0.581936
C5 C4 0.576557
C5 C5 1
C6 C1 -0.362738
C6 C2 0.362738
C6 C3 0.861381
C6 C4 0.932143
C6 C5 0.731727
C6 C6 1
Related
I am new to Verilog, and trying to write a traffic light code where the LED light changes after certain time. I'm keep getting on different errors while compiling. I tried to fix them by changing the arrangement, or variables in the code, but it still fails.
This is the code I wrote,
module traffic_light(clk, reset, G1, Y1, R1, G2, Y2, R2);
input clk, reset;
output reg G1, Y1, R1, G2, Y2, R2;
// parameters for each light control
parameter GREEN = 3'b001,
YELLOW = 3'b010,
RED = 3'b100,
LEFT_GREEN = 3'b101, // assume both red and green will be turned on
LEFT_YELLOW = 3'b110; // assume both red and yellow will be turned on
// finite-state definition (Moore Type):
// ---------------------------
// NSlight EWlight
// ---------------------------
parameter S0 = 3'd0, // GREEN RED
S1 = 3'd1, // YELLOW RED
S2 = 3'd2, // RED, GREEN RED
S3 = 3'd3, // RED, YELLOW RED
S4 = 3'd4, // RED GREEN
S5 = 3'd5, // RED YELLOW
S6 = 3'd6, // RED RED, GREEN
S7 = 3'd7; // RED RED, YELLOW
// internal state variables
reg [2:0] state, next_state;
integer t1 = 19, t2 = 4;
integer count;
// buttons are appropriate for use as clock or reset inputs in a circuit
always #(posedge clk, negedge reset)
if(reset == 'b0) // button pressed, when reset is active low
begin
next_state = S0;
count = t1;
end
else
state <= next_state;
always #(next_state)
begin
next_state = S0;
count = t1;
case(state)
S0:
if (count < 0) // load: if the count reaches below 0, reset
begin
count <= t2;
next_state <= S1;
end
else // enable
begin
// down count
count <= count - 1;
// assign LEDs
G1 <= 1;
Y1 <= 0;
R1 <= 0;
G2 <= 0;
Y2 <= 0;
R2 <= 1;
end
S1:
if (count < 0)
begin
count <= t1;
next_state <= S2;
end
else
begin
G1 <= 0;
Y1 <= 1;
R1 <= 0;
G2 <= 0;
Y2 <= 0;
R2 <= 1;
count <= count - 1;
end
S2:
if (count < 0)
begin
count <= t2;
next_state <= S3;
end
else
begin
G1 <= 1;
Y1 <= 0;
R1 <= 1;
G2 <= 0;
Y2 <= 0;
R2 <= 1;
count <= count - 1;
end
S3:
if (count < 0)
begin
count <= t1;
next_state <= S4;
end
else
begin
G1 <= 0;
Y1 <= 1;
R1 <= 1;
G2 <= 0;
Y2 <= 0;
R2 <= 1;
count <= count - 1;
end
S4:
if (count < 0)
begin
count <= t1;
next_state <= S5;
end
else
begin
G1 <= 0;
Y1 <= 0;
R1 <= 1;
G2 <= 1;
Y2 <= 0;
R2 <= 0;
count <= count - 1;
end
S5:
if (count < 0)
begin
count <= t1;
next_state <= S6;
end
else
begin
G1 <= 0;
Y1 <= 0;
R1 <= 1;
G2 <= 0;
Y2 <= 1;
R2 <= 0;
count <= count - 1;
end
S6:
if (count < 0)
begin
count <= t2;
next_state <= S7;
end
else
begin
G1 <= 0;
Y1 <= 0;
R1 <= 1;
G2 <= 1;
Y2 <= 0;
R2 <= 1;
count <= count - 1;
end
S7:
if (count < 0)
begin
count <= t1;
next_state <= S0;
end
else
begin
G1 <= 0;
Y1 <= 0;
R1 <= 1;
G2 <= 0;
Y2 <= 1;
R2 <= 1;
count <= count - 1;
end
endcase
end
endmodule
and these are the errors generated from the above code:
Error (10028): Can't resolve multiple constant drivers for net "next_state.S0" at traffic_light.v(41)
Error (10029): Constant driver at traffic_light.v(32)
Error (10028): Can't resolve multiple constant drivers for net "next_state.S1" at traffic_light.v(41)
Error (10028): Can't resolve multiple constant drivers for net "next_state.S2" at traffic_light.v(41)
Error (10028): Can't resolve multiple constant drivers for net "next_state.S3" at traffic_light.v(41)
Error (10028): Can't resolve multiple constant drivers for net "next_state.S6" at traffic_light.v(41)
Error (10028): Can't resolve multiple constant drivers for net "next_state.S7" at traffic_light.v(41)
Error (12152): Can't elaborate user hierarchy "traffic_light:inst"
Hope I can get any suggestions or solutions to this problem. Thank you in advance.
I solved the problem after few more searching.
Error (10028): Can't resolve multiple constant drivers for net... VHDL ERROR
"Multiple Constant Drivers" Error Verilog with Quartus Prime
These two links helped solving, the problem is that you cannot assign one variable in two different always block.
The below code is the fixed one.
module traffic_light(clk, reset, G1, Y1, R1, G2, Y2, R2, time_);
input clk, reset;
output reg G1, Y1, R1, G2, Y2, R2;
output reg [4:0] time_; // added, it displays the remaining time
// parameters for each light control
parameter GREEN = 3'b001,
YELLOW = 3'b010,
RED = 3'b100,
LEFT_GREEN = 3'b101, // assume both red and green will be turned on
LEFT_YELLOW = 3'b110; // assume both red and yellow will be turned on
// finite-state definition (Moore Type):
// ---------------------------
// NSlight EWlight
// ---------------------------
parameter S0 = 3'd0, // GREEN RED
S1 = 3'd1, // YELLOW RED
S2 = 3'd2, // RED, GREEN RED
S3 = 3'd3, // RED, YELLOW RED
S4 = 3'd4, // RED GREEN
S5 = 3'd5, // RED YELLOW
S6 = 3'd6, // RED RED, GREEN
S7 = 3'd7; // RED RED, YELLOW
// internal state variables and time settings
reg [2:0] state, next_state;
integer t1 = 19, t2 = 4;
reg count; // changed to count only
always #(posedge clk)
if(reset == 0) // button pressed, when reset is active low
begin
state <= S0;
time_ <= t1;
end
else
begin
state <= next_state;
time_ <= count;
end
always #(*) // changed according to advice in the comment
case(state)
S0: if (count < 0) // load: if the count reaches below 0, reset
begin
count <= t2;
next_state <= S1;
end
else // enable
begin
// down count
count <= count - 1;
// assign LEDs
G1 <= 1;
Y1 <= 0;
R1 <= 0;
G2 <= 0;
Y2 <= 0;
R2 <= 1;
end
S1: if (count < 0)
begin
count <= t1;
next_state <= S2;
end
else
begin
G1 <= 0;
Y1 <= 1;
R1 <= 0;
G2 <= 0;
Y2 <= 0;
R2 <= 1;
count <= count - 1;
end
S2: if (count < 0)
begin
count <= t2;
next_state <= S3;
end
else
begin
G1 <= 1;
Y1 <= 0;
R1 <= 1;
G2 <= 0;
Y2 <= 0;
R2 <= 1;
count <= count - 1;
end
S3: if (count < 0)
begin
count <= t1;
next_state <= S4;
end
else
begin
G1 <= 0;
Y1 <= 1;
R1 <= 1;
G2 <= 0;
Y2 <= 0;
R2 <= 1;
count <= count - 1;
end
S4: if (count < 0)
begin
count <= t2;
next_state <= S5;
end
else
begin
G1 <= 0;
Y1 <= 0;
R1 <= 1;
G2 <= 1;
Y2 <= 0;
R2 <= 0;
count <= count - 1;
end
S5: if (count < 0)
begin
count <= t1;
next_state <= S6;
end
else
begin
G1 <= 0;
Y1 <= 0;
R1 <= 1;
G2 <= 0;
Y2 <= 1;
R2 <= 0;
count <= count - 1;
end
S6: if (count < 0)
begin
count <= t2;
next_state <= S7;
end
else
begin
G1 <= 0;
Y1 <= 0;
R1 <= 1;
G2 <= 1;
Y2 <= 0;
R2 <= 1;
count <= count - 1;
end
S7: if (count < 0)
begin
count <= t1;
next_state <= S0;
end
else
begin
G1 <= 0;
Y1 <= 0;
R1 <= 1;
G2 <= 0;
Y2 <= 1;
R2 <= 1;
count <= count - 1;
end
default:
begin
next_state <= S0;
count <= t1;
end
endcase
endmodule
You have next_state assigned in both processes, and that is causing troubles. We can find it actually described in the Verilog Standard. section 14.5 Driving wired logic
Module path output nets shall not have more than one driver within the module. Therefore, wired logic is not allowed at module path outputs
And it also provides an example:
I have a data set: (file.txt)
X Y
1 a
2 b
3 c
10 d
11 e
12 f
15 g
20 h
25 i
30 j
35 k
40 l
41 m
42 n
43 o
46 p
I have two Up10 and Down10 columns,
Up10: From (X) to (X-10) count of row.
Down10 : From (X) to (X+10)
count of row
For example:
X Y Up10 Down10
35 k 3 5
For Up10; 35-10 X=35 X=30 X=25 Total = 3 row
For Down10; 35+10 X=35 X=40 X=41 X=42 X=42 Total = 5 row
I have tried, but i cant show 3rd and 4rth column:
awk 'BEGIN{ FS=OFS="\t" }
NR==FNR{
a[$1]+=$3
next
}
{ $(NF+10)=a[$3] }
{ $(NF-10)=a[$4] }
1
' file.txt file.txt > file-2.txt
Desired Output:
X Y Up10 Down10
1 a 1 5
2 b 2 5
3 c 3 4
10 d 4 5
11 e 5 4
12 f 5 3
15 g 4 3
20 h 5 3
25 i 3 3
30 j 3 3
35 k 3 5
40 l 3 5
41 m 3 4
42 n 4 3
43 o 5 2
46 p 5 1
This is the Pierre François' solution: Thanks again #Pierre François
awk '
BEGIN{OFS="\t"; print "X\tY\tUp10\tDown10"}
(NR == FNR) && (FNR > 1){a[$1] = $1 + 0}
(NR > FNR) && (FNR > 1){
up = 0; upl = $1 - 10
down = 0; downl = $1 + 10
for (i in a) { i += 0 # tricky: convert i to integer
if ((i >= upl) && (i <= $1)) {up++}
if ((i >= $1) && (i <= downl)) {down++}
}
print $1, $2, up, down;
}
' file.txt file.txt > file-2.txt
This is the Pierre François' solution: Thanks again #Pierre François
awk '
BEGIN{OFS="\t"; print "X\tY\tUp10\tDown10"}
(NR == FNR) && (FNR > 1){a[$1] = $1 + 0}
(NR > FNR) && (FNR > 1){
up = 0; upl = $1 - 10
down = 0; downl = $1 + 10
for (i in a) { i += 0 # tricky: convert i to integer
if ((i >= upl) && (i <= $1)) {up++}
if ((i >= $1) && (i <= downl)) {down++}
}
print $1, $2, up, down;
}
' file.txt file.txt > file-2.txt
I have a column that looks like this:
A 1
B 3
C 5
D 4
E 7
F 1
G 1
H 3
For every filed in column#2, I want to calculate the range (max-min) of 3 field radius up and down.
A range(1 3 5 4)
B range(1 3 5 4 7)
C range(1 3 5 4 7 1)
D range(1 3 5 4 7 1 1)
E range( 3 5 4 7 1 1 3)
F range( 5 4 7 1 1 3)
G range( 4 7 1 1 3)
H range( 7 1 1 3)
How can do this in awk?
I could do the same in perl using:
my $set_size = #values;
for ( my $i = 0 ; $i < $set_size ; $i++ ) {
my $min = $i - 4;
if ( $min < 0 ) { $min = 0; }
my $max = $i + 4;
if ( $max > ( $set_size - 1 ) ) { $max = $set_size - 1; }
my $min_val = $values[$min];
my $max_val = $values[$min];
for ( my $j = $min ; $j <= $max ; $j++ ) {
if ( $values[$j] <= $min_val ) { $min_val = $values[$j]; }
if ( $values[$j] >= $max_val ) { $max_val = $values[$j]; }
}
my $range = $max_val - $min_val;
printf "$points[$i] %.15f\n", $range;
}
idk exactly what I want to calculate the range (max-min) of 3 field radius up and down. means but to get the output you posted from the input you posted would be:
$ cat tst.awk
{
keys[NR] = $1
values[NR] = $2
}
END {
range = 3
for (i=1; i<=NR; i++) {
min = ( (i - range) >= 1 ? i - range : 1 )
max = ( (i + range) <= NR ? i + range : NR )
printf "%s range(", keys[i]
for (j=min; j<=max; j++) {
printf "%s%s", values[j], (j<max ? " " : ")\n")
}
}
}
$
$ awk -f tst.awk file
A range(1 3 5 4)
B range(1 3 5 4 7)
C range(1 3 5 4 7 1)
D range(1 3 5 4 7 1 1)
E range(3 5 4 7 1 1 3)
F range(5 4 7 1 1 3)
G range(4 7 1 1 3)
H range(7 1 1 3)
Your sample perl doesn't print out your sample output, it seems to do something different... so here's how I'd do it in perl:
#!/usr/bin/perl
use warnings;
use strict;
use feature qw/say/;
use List::Util qw/min max/;
my (#col1, #col2);
while (<>) {
chomp;
my ($v1, $v2) = split;
push #col1, $v1;
push #col2, $v2;
}
my #prefix;
for my $i (0 .. $#col1) {
my #range = #col2[max($i - 3, 0) .. min($i + 3, $#col2)];
push #prefix, ' ' if $i > 3;
unshift #range, #prefix;
say "$col1[$i] range(#range)"
}
running it:
$ perl range.pl input.txt
A range(1 3 5 4)
B range(1 3 5 4 7)
C range(1 3 5 4 7 1)
D range(1 3 5 4 7 1 1)
E range( 3 5 4 7 1 1 3)
F range( 5 4 7 1 1 3)
G range( 4 7 1 1 3)
H range( 7 1 1 3)
The formatting will break if any of the numbers are greater than 9, though.
Since you tagged tcl:
#!/usr/bin/env tclsh
lassign $argv file size
set fh [open $file r]
# assume exactly 2 space-separated words per line
set data [dict create {*}[split [read -nonewline $fh]]]
close $fh
set len [dict size $data]
set values [dict values $data]
set i 0
dict for {key _} $data {
set first [expr {max($i - $size, 0)}]
set last [expr {min($i + $size, $len)}]
puts [format "%s range(%s%s)" \
$key \
[string repeat " " $first] \
[lrange $values $first $last] \
]
incr i
}
outputs
A range(1 3 5 4)
B range(1 3 5 4 7)
C range(1 3 5 4 7 1)
D range(1 3 5 4 7 1 1)
E range( 3 5 4 7 1 1 3)
F range( 5 4 7 1 1 3)
G range( 4 7 1 1 3)
H range( 7 1 1 3)
TXR at the shell prompt:
bash$ txr -c '#(collect)
#label #num
#(end)
#(bind rng #[window-map 3 nil (opip list (remq nil) (mapcar toint)) num])
#(output)
# (repeat)
#label range(#rng) -> #(find-min rng)..#(find-max rng)
# (end)
#(end)'
A 1
B 3
C 5
D 4
E 7
F 1
G 1
H 3
[Ctrl-D][Enter]
A range(1 3 5 4) -> 1..5
B range(1 3 5 4 7) -> 1..7
C range(1 3 5 4 7 1) -> 1..7
D range(1 3 5 4 7 1 1) -> 1..7
E range(3 5 4 7 1 1 3) -> 1..7
F range(5 4 7 1 1 3) -> 1..7
G range(4 7 1 1 3) -> 1..7
H range(7 1 1 3) -> 1..7
I have data in the following format:
ID Date X1 X2 X3
1 01/01/00 1 2 3
1 01/02/00 7 8 5
2 01/03/00 9 7 1
2 01/04/00 1 4 5
I would like to group measurements into new rows according to ID, so I end up with:
ID Date X1 X2 X3 Date X1_2 X2_2 X3_2
1 01/01/00 1 2 3 01/02/00 7 8 5
2 01/03/00 9 7 1 01/04/00 1 4 5
etc.
I have as many as 20 observations for a given ID.
So far I have tried the technique given by http://gadgetsytecnologia.com/da622c17d34e6f13e/awk-transpose-childids-column-into-row.html
The code I have tried so far is:
awk -F, OFS = '\t' 'NR >1 {a[$1] = a[$1]; a[$2] = a[$2]; a[$3] = a[$3];a[$4] = a[$4]; a[$5] = a[$5] OFS $5} END {print "ID,Date,X1,X2,X3,Date_2,X1_2, X2_2 X3_2'\t' for (ID in a) print a[$1:$5] }' file.txt
The file is a tab delimited file. I don't know how to manipulate the data, or to account for the fact that there will be more than two observations per person.
Just keep track of what was the previous first field. If it changes, print the stored line:
awk 'NR==1 {print; next} # print header
prev && $1!=prev {print prev, line; line=""} # print on different $1
{prev=$1; $1=""; line=line $0} # store data and remove $1
END {print prev, line}' file # print trailing line
If you have tab-separated fields, just add -F"\t".
Test
$ awk 'NR==1 {print; next} prev && $1!=prev {print prev, line; line=""} {prev=$1; $1=""; line=line $0} END {print prev, line}' a
ID Date X1 X2 X3
1 01/01/00 1 2 3 01/02/00 7 8 5
2 01/03/00 9 7 1 01/04/00 1 4 5
you can try this (gnu-awk solution)
gawk '
NR == 1 {
N = NF;
MAX = NF-1;
for(i=1; i<=NF; i++){ #store columns names
names[i]=$i;
}
next;
}
{
for(i=2; i<=N; i++){
a[$1][length(a[$1])+1] = $i; #store records for each id
}
if(length(a[$1])>MAX){
MAX = length(a[$1]);
}
}
END{
firstline = names[1];
for(i=1; i<=MAX; i++){ #print first line
column = int((i-1)%(N-1))+2
count = int((i-1)/(N-1));
firstline=firstline OFS names[column];
if(count>0){
firstline=firstline"_"count
}
}
print firstline
for(id in a){ #print each record in store
line = id;
for(i=1; i<=length(a[id]); i++){
line=line OFS a[id][i];
}
print line;
}
}
' input
input
ID Date X1 X2 X3
1 01/01/00 1 2 3
1 01/02/00 7 8 5
2 01/03/00 9 7 1
2 01/04/00 1 4 5
1 01/03/00 72 28 25
you get
ID Date X1 X2 X3 Date_1 X1_1 X2_1 X3_1 Date_2 X1_2 X2_2 X3_2
1 01/01/00 1 2 3 01/02/00 7 8 5 01/03/00 72 28 25
2 01/03/00 9 7 1 01/04/00 1 4 5
I've been doing past paper questions and keep coming up against these questions that deal with 3 valued logic. My notes mention it but don't give examples that relate to those asked in exams.
I understand the basis that True = 1, False = 0 & Unknown = 1/2 as well as And = Min, Or = Max and Not(x) = 1-x. However I do not know how to apply it to questions such as those below:
In SQL, discuss the possible truth values of the following expression:
R.a > R.b OR R.a <= 0 OR R.b >= 0
Justify your answer.
And:
The phone and age fields of the Owner table might have null values in
them. Considering all possible combinations, show which of the three
truth values might be returned by the expression:
phone = ’141-3304913’ OR age <50 OR age >= 50
Any help in clarifying these for me would be really appreciated :)
I will focus on the concrete example, which is more proper for clarifying things.
Put simply, your logical expression is made of a conjunction of three clauses
C1: phone = '141-3304913'
C2: age < 50
C3: age >= 50
for which tri-boolean logic states that the result is
True, if any clause is true
False, if all clauses are false
Unknown, in all the other cases
Consequently, if the value associated with True is the largest, with False is the smallest, and with Unknown is any intermediate value, then taking the MAX for a conjunction proves correct. Similarly, a disjunction works with the MIN function. Negation works as long as we interpret any value between 0 and 1 (excluded) as Unknown; clearly, if we take 1/2 then the negation function is "stable", but that does not really matter in mathematical terms.
More operatively, the clauses clearly react to the following values (instances) of your phone variable P and your age variable A:
P1 such that P1 = '141-3304913'
P2 such that P2 <> '141-3304913'
P3 such that P3 = NULL
A1 such that A1 < 50
A2 such that A2 >= 50
A3 such that A3 = NULL
In terms of satisfaction of the clauses, we have
P1 -> C1 = 1
P2 -> C1 = 0
P3 -> C1 = 1/2
A1 -> C2 = 1, C3 = 0
A2 -> C2 = 0, C3 = 1
A3 -> C2 = C3 = 1/2
In general there exist 3*3 possible combinations, since each of your two variables takes three possible values:
P1 A1: C1 = 1, C2 = 1, C3 = 0 -> MAX(1,1,0) = 1 -> true
P1 A2: C1 = 1, C2 = 0, C3 = 1 -> MAX(1,0,1) = 1 -> true
P1 A3: C1 = 1, C2 = 1/2, C3 = 1/2 -> MAX(1,1/2,1/2) = 1 -> true
P2 A1: C1 = 0, C2 = 1, C3 = 0 -> MAX(0,1,0) = 1 -> true
P2 A2: C1 = 0, C2 = 0, C3 = 1 -> MAX(0,0,1) = 1 -> true
P2 A3: C1 = 0, C2 = 1/2, C3 = 1/2 -> MAX(0,1/2,1/2) = 1/2 -> unknown
P3 A1: C1 = 1/2, C2 = 1, C3 = 0 -> MAX(1/2,1,0) = 1 -> true
P3 A2: C1 = 1/2, C2 = 0, C3 = 1 -> MAX(1/2,0,1) = 1 -> true
P3 A3: C1 = 1/2, C2 = 1/2, C3 = 1/2 -> MAX(1/2,1/2,1/2) = 1/2 -> unknown
In particular, since C2 and C3 are mutually exclusive, you never get False as a result of the conjunction.
The expression R.a > R.b OR R.a <= 0 OR R.b >= 0 instead presents these cases:
R.a <= 0, R.a > 0, R.a = unknown
R.b >= 0, R.b < 0, R.b = unknown
R.a - R.b > 0, R.a - R.b <= 0, R.a - R.b = unknown
Apparently we have three variables and 27 possible cases, but several related to R.a - R.b can be trivially ruled out.