Banker's algorithm
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The Banker's algorithm is a resource allocation and deadlock avoidance algorithm developed by Edsger Dijkstra that tests for safety by simulating the allocation of predetermined maximum possible amounts of all resources, and then makes a "s-state" check to test for possible deadlock conditions for all other pending activities, before deciding whether allocation should be allowed to continue.
Example input
Assuming that the system distinguishes between four types of resources, (A, B, C and D), the following is an example of how those resources could be distributed. Note that this example shows the system at an instant before a new request for resources arrives. Also, the types and number of resources are abstracted. Real systems, for example, would deal with much larger quantities of each resource.
Total resources in system: A B C D 6 5 7 6
Available system resources are: A B C D 3 1 1 2
Processes (currently allocated resources): A B C D P1 1 2 2 1 P2 1 0 3 3 P3 1 2 1 0
Processes (maximum resources): A B C D P1 3 3 2 2 P2 1 2 3 4 P3 1 3 5 0
Need= maximum resources - currently allocated resources Processes (need resources): A B C D P1 2 1 0 1 P2 0 2 0 1 P3 0 1 4 0
C
Standard binary heap-as-priority queue affair. Only that each node links back to its heap position for easier update.
There are two main()
functions to choose from (look for #define BIG_EXAMPLE
), one is for task example, the other is a much heavier duty test case.
<lang c>#include <stdio.h>
- include <stdbool.h>
int main() {
int curr[5][5]; int max_claim[5][5]; int avl[5]; int alloc[5] = {0, 0, 0, 0, 0}; int max_res[5]; int running[5];
int i, j, exec, r, p; int count = 0; bool safe = false;
printf("\nEnter the number of resources: "); scanf("%d", &r);
printf("\nEnter the number of processes: "); scanf("%d", &p); for (i = 0; i < p; i++) { running[i] = 1; count++; }
printf("\nEnter Claim Vector: "); for (i = 0; i < r; i++) scanf("%d", &max_res[i]);
printf("\nEnter Allocated Resource Table: "); for (i = 0; i < p; i++) { for (j = 0; j < r; j++) scanf("%d", &curr[i][j]); }
printf("\nEnter Maximum Claim table: "); for (i = 0; i < p; i++) { for (j = 0; j < r; j++) scanf("%d", &max_claim[i][j]); }
printf("\nThe Claim Vector is: "); for (i = 0; i < r; i++) printf("%d ", max_res[i]);
printf("\nThe Allocated Resource Table:\n"); for (i = 0; i < p; i++) { for (j = 0; j < r; j++) printf("\t%d", curr[i][j]); printf("\n"); }
printf("\nThe Maximum Claim Table:\n"); for (i = 0; i < p; i++) { for (j = 0; j < r; j++) printf("\t%d", max_claim[i][j]); printf("\n"); }
for (i = 0; i < p; i++) for (j = 0; j < r; j++) alloc[j] += curr[i][j];
printf("\nAllocated resources: "); for (i = 0; i < r; i++) printf("%d ", alloc[i]); for (i = 0; i < r; i++) avl[i] = max_res[i] - alloc[i];
printf("\nAvailable resources: "); for (i = 0; i < r; i++) printf("%d ", avl[i]); printf("\n");
while (count != 0) { safe = false; for (i = 0; i < p; i++) { if (running[i]) { exec = 1; for (j = 0; j < r; j++) { if (max_claim[i][j] - curr[i][j] > avl[j]) { exec = 0; break; } }
if (exec) { printf("\nProcess%d is executing.\n", i + 1); running[i] = 0; count--; safe = true; for (j = 0; j < r; j++) avl[j] += curr[i][j]; break; } } }
if (!safe) { printf("\nThe processes are in unsafe state."); break; }
if (safe) printf("\nThe process is in safe state.");
printf("\nAvailable vector: "); for (i = 0; i < r; i++) printf("%d ", avl[i]); }
return 0;
}</lang>
- Input and Output:
Enter the number of resources: 4 Enter the number of processes: 5 Enter Claim Vector: 8 5 9 7 Enter Allocated Resource Table: 2 0 1 1 0 1 2 1 4 0 0 3 0 2 1 0 1 0 3 0 Enter Maximum Claim table: 3 2 1 4 0 2 5 2 5 1 0 5 1 5 3 0 3 0 3 3 The Claim Vector is: 8 5 9 7 The Allocated Resource Table: 2 0 1 1 0 1 2 1 4 0 0 3 0 2 1 0 1 0 3 0 The Maximum Claim Table: 3 2 1 4 0 2 5 2 5 1 0 5 1 5 3 0 3 0 3 3 Allocated resources: 7 3 7 5 Available resources: 1 2 2 2 Process3 is executing. The process is in safe state. Available vector: 5 2 2 5 Process1 is executing. The process is in safe state. Available vector: 7 2 3 6 Process2 is executing. The process is in safe state. Available vector: 7 3 5 7 Process4 is executing. The process is in safe state. Available vector: 7 5 6 7 Process5 is executing. The process is in safe state. Available vector: 8 5 9 7
Go
WP cites EWD-108 and has EWD-623 as further reading. In using the analogy of a money, EWD-108 considers only a single type of resource. EWD-623 seems mostly in terms of a single resource but a couple of times says "all resources." This subtly hints that the algorithm can be adapted for multiple resource types. WP then gives an array-based presentation that works for multiple resources.
This solution is more inspired by EWD-623 than WP. EWD-623, while it talks of finding a permutation of processes, notes that the "ordering effort" can be stopped as soon as the process requesting resources happens to be found satisfiable. The solution here attempts to make this finding as soon as possible by moving the process to the front of a list of unsatisfied processes. Also since the solved permutation of satisfied processes has no use, it is not kept which simplifies the algorithm a bit. <lang go>package bank
import (
"bytes" "errors" "fmt" "log" "sort" "sync"
)
type PID string type RID string type RMap map[RID]int
// format RIDs in order func (m RMap) String() string {
rs := make([]string, len(m)) i := 0 for r := range m { rs[i] = string(r) i++ } sort.Strings(rs) var b bytes.Buffer b.WriteString("{") for _, r := range rs { fmt.Fprintf(&b, "%q: %d, ", r, m[RID(r)]) } bb := b.Bytes() if len(bb) > 1 { bb[len(bb)-2] = '}' } return string(bb)
}
type Bank struct {
available RMap max map[PID]RMap allocation map[PID]RMap sync.Mutex
}
func (b *Bank) need(p PID, r RID) int {
return b.max[p][r] - b.allocation[p][r]
}
func New(available RMap) (b *Bank, err error) {
for r, a := range available { if a < 0 { return nil, fmt.Errorf("negative resource %s: %d", r, a) } } return &Bank{ available: available, max: map[PID]RMap{}, allocation: map[PID]RMap{}, }, nil
}
func (b *Bank) NewProcess(p PID, max RMap) (err error) {
b.Lock() defer b.Unlock() if _, ok := b.max[p]; ok { return fmt.Errorf("process %s already registered", p) } for r, m := range max { switch a, ok := b.available[r]; { case !ok: return fmt.Errorf("resource %s unknown", r) case m > a: return fmt.Errorf("resource %s: process %s max %d > available %d", r, p, m, a) } } b.max[p] = max b.allocation[p] = RMap{} return
}
func (b *Bank) Request(pid PID, change RMap) (err error) {
b.Lock() defer b.Unlock() if _, ok := b.max[pid]; !ok { return fmt.Errorf("process %s unknown", pid) } for r, c := range change { if c < 0 { return errors.New("decrease not allowed") } if _, ok := b.available[r]; !ok { return fmt.Errorf("resource %s unknown", r) } if c > b.need(pid, r) { return errors.New("increase exceeds declared max") } } // allocation is non-exported data so we can change it in place // then change it back if the request cannot be granted. for r, c := range change { b.allocation[pid][r] += c // change in place } defer func() { if err != nil { // if request not granted, for r, c := range change { b.allocation[pid][r] -= c // change it back } } }() // Collect list of process IDs, also compute cash // First in the list is always the requesting PID. cash := RMap{} for r, a := range b.available { cash[r] = a } perm := make([]PID, len(b.allocation)) i := 1 for pr, a := range b.allocation { if pr == pid { perm[0] = pr } else { perm[i] = pr i++ } for r, a := range a { cash[r] -= a } } ret := RMap{} // sum of loans m := len(perm) // number of processes still candidates for termination for { // find a process h that can terminate h := 0 h: for ; ; h++ { if h == m { // no process could terminate return errors.New("request would make deadlock possible") } for r := range b.available { if b.need(perm[h], r) > cash[r]+ret[r] { // h cannot terminate if any resource need cannot be met. continue h } } // log possible terimation, consistent with WP example program. log.Println(" ", perm[h], "could terminate") // all resource needs can be met. h can terminate. break } if h == 0 { // Zwanenburg condition: // if requesting process can terminate, pattern is safe and // remaining terminations do not need to be demonstrated. return nil } for r, a := range b.allocation[perm[h]] { ret[r] += a } m-- perm[h] = perm[m] }
}</lang> <lang go>package main
import (
"fmt" "bank"
)
func main() {
// Task example data: // create "bank" with available resources b, _ := bank.New(bank.RMap{"A": 6, "B": 5, "C": 7, "D": 6})
// add processes with their maximum allocation limits b.NewProcess("P1", bank.RMap{"A": 3, "B": 3, "C": 2, "D": 2}) b.NewProcess("P2", bank.RMap{"A": 1, "B": 2, "C": 3, "D": 4}) b.NewProcess("P3", bank.RMap{"A": 1, "B": 3, "C": 5})
// processes request resources. Each request is checked for safety. // <nil> returned error value means request was safe and was granted. fmt.Println("P1 request:") fmt.Println(b.Request("P1", bank.RMap{"A": 1, "B": 2, "C": 2, "D": 1}))
fmt.Println("\nP2 request:") fmt.Println(b.Request("P2", bank.RMap{"A": 1, "C": 3, "D": 3}))
fmt.Println("\nP3 request:") fmt.Println(b.Request("P3", bank.RMap{"A": 1, "B": 2, "C": 1}))
}</lang>
- Output:
P1 request: 2017/08/29 16:44:15 P1 could terminate <nil> P2 request: 2017/08/29 16:44:15 P2 could terminate <nil> P3 request: 2017/08/29 16:44:15 P1 could terminate 2017/08/29 16:44:15 P2 could terminate 2017/08/29 16:44:15 P3 could terminate <nil>
J
The task description currently does not define the process being run. So we follow the example set by other implementations and have each process free all resources after successfully being run. Also, since this is a demo, we give a blow-by-blow description of what's happening as it runs.
<lang j>bankrun=:1 :0
'MAX ALLOC TOTAL'=. y todo=.(#ALLOC)#1 whilst. (+./todo)*-. prior-:todo do. prior=. todo for_p.I.todo do. avail=. TOTAL-+/ALLOC echo 'currently available: ',":avail pALLOC=. p{ALLOC pMAX=. p{MAX request=. pMAX-pALLOC if.(0>request)+.&(+./)request>avail do. echo 'unsafe request ',(":request),', skipping ',":p continue. else. echo 'running process ',(":p),', allocating ',":request end. free=.request u pALLOC echo 'process ',(":p),' freeing ',":free assert (0<:free) *&(*/) free <: pMAX ALLOC=. (pALLOC-free) p} ALLOC todo=. 0 p} todo end. end. if.+./todo do. echo 'deadlocked processes: ',":I.todo end. echo 'DONE'
)</lang>
Definitions for task example:
<lang j>max=: 3 3 2 2,1 2 3 4,:1 3 5 0 alloc=: 1 2 2 1,1 0 3 3,:1 2 1 0 total=:6 5 7 6
NB. simulate running process NB. left arg: newly available resources, right: previously available NB. result: resources freed run=: +</lang>
Example run:
<lang J> run bankrun max;alloc;total currently available: 3 1 1 2 running process 0, allocating 2 1 0 1 process 0 freeing 3 3 2 2 currently available: 6 4 3 4 running process 1, allocating 0 2 0 1 process 1 freeing 1 2 3 4 currently available: 7 6 6 8 running process 2, allocating 0 1 4 0 process 2 freeing 1 3 5 0 DONE</lang>
Julia
<lang julia>function queryprompt(query, typ)
print(query, ": ") entry = uppercase(strip(readline(stdin))) return (typ <: Integer) ? parse(Int, entry) : (typ <: Vector) ? map(x -> parse(Int, x), split(entry, r"\s+")) : entry
end
function testbankers()
r = queryprompt("Enter the number of resources", Int) p = queryprompt("\nEnter the number of processes", Int) maxres = queryprompt("\nEnter Claim Vector", Vector{Int}) curr, maxclaim = zeros(Int, p, r), zeros(Int, p, r) for i in 1:p curr[i, :] .= queryprompt("\nEnter Allocated Resource Table, Row $i", Vector{Int}) end for i in 1:p maxclaim[i, :] .= queryprompt("\nEnter Maximum Claim Table, Row $i", Vector{Int}) end alloc = [sum(curr[:, j]) for j in 1:r] println("\nAllocated Resources: $alloc") avl = map(i -> maxres[i] - alloc[i], 1:r) println("\nAvailable Resources: $avl") running = trues(p) count = p while count != 0 safe = false for i in 1:p if running[i] exec = true for j in 1:r if maxclaim[i, j] - curr[i, j] > avl[j] exec = false break end end if exec println("\nProcess $i is executing.") running[i] = false count -= 1 safe = true for j in 1:r avl[j] += curr[i, j] end break end end end if !safe println("The processes are in an unsafe state.") break end println("\nThe process is in a safe state.") println("\nAvailable Vector: $avl") end
end
testbankers()
</lang>
- Output:
Enter the number of resources: 4 Enter the number of processes: 5 Enter Claim Vector: 8 5 9 7 Enter Allocated Resource Table, Row 1: 2 0 1 1 Enter Allocated Resource Table, Row 2: 0 1 2 1 Enter Allocated Resource Table, Row 3: 4 0 0 3 Enter Allocated Resource Table, Row 4: 0 2 1 0 Enter Allocated Resource Table, Row 5: 1 0 3 0 Enter Maximum Claim Table, Row 1: 3 2 1 4 Enter Maximum Claim Table, Row 2: 0 2 5 2 Enter Maximum Claim Table, Row 3: 5 1 0 5 Enter Maximum Claim Table, Row 4: 1 5 3 0 Enter Maximum Claim Table, Row 5: 3 0 3 3 Allocated Resources: [7, 3, 7, 5] Available Resources: [1, 2, 2, 2] Process 3 is executing. The process is in a safe state. Available Vector: [5, 2, 2, 5] Process 1 is executing. The process is in a safe state. Available Vector: [7, 2, 3, 6] Process 2 is executing. The process is in a safe state. Available Vector: [7, 3, 5, 7] Process 4 is executing. The process is in a safe state. Available Vector: [7, 5, 6, 7] Process 5 is executing. The process is in a safe state. Available Vector: [8, 5, 9, 7]
Kotlin
For simplicity, input checking is ignored: <lang scala>// version 1.1.4-3
fun main(args: Array<String>) {
print("Enter the number of resources: ") val r = readLine()!!.toInt()
print("\nEnter the number of processes: ") val p = readLine()!!.toInt()
print("\nEnter Claim Vector: ") val maxRes = readLine()!!.split(' ').map { it.toInt() } .toIntArray()
println("\nEnter Allocated Resource Table:") val curr = Array(p) { IntArray(r) } for (i in 0 until p) { print("Row ${i + 1}: ") curr[i] = readLine()!!.split(' ').map { it.toInt() }.toIntArray() }
println("\nEnter Maximum Claim Table: ") val maxClaim = Array(p) { IntArray(r) } for (i in 0 until p) { print("Row ${i + 1}: ") maxClaim[i] = readLine()!!.split(' ').map { it.toInt() }.toIntArray() }
val alloc = IntArray(r) for (i in 0 until p) { for (j in 0 until r) alloc[j] += curr[i][j] } println("\nAllocated Resources: ${alloc.joinToString(" ")}")
val avl = IntArray(r) { maxRes[it] - alloc[it] } println("\nAvailable Resources: ${avl.joinToString(" ")}")
val running = BooleanArray(p) { true } var count = p while (count != 0) { var safe = false for (i in 0 until p) { if (running[i]) { var exec = true for (j in 0 until r) { if (maxClaim[i][j] - curr[i][j] > avl[j]) { exec = false break } }
if (exec) { print("\nProcess ${i + 1} is executing.\n") running[i] = false count-- safe = true for (j in 0 until r) avl[j] += curr[i][j] break } } }
if (!safe) { print("The processes are in an unsafe state.") break }
print("\nThe process is in a safe state.") println("\nAvailable Vector: ${avl.joinToString(" ")}") }
}</lang>
Sample input/output:
Enter the number of resources: 4 Enter the number of processes: 5 Enter Claim Vector: 8 5 9 7 Enter Allocated Resource Table: Row 1: 2 0 1 1 Row 2: 0 1 2 1 Row 3: 4 0 0 3 Row 4: 0 2 1 0 Row 5: 1 0 3 0 Enter Maximum Claim Table: Row 1: 3 2 1 4 Row 2: 0 2 5 2 Row 3: 5 1 0 5 Row 4: 1 5 3 0 Row 5: 3 0 3 3 Allocated Resources: 7 3 7 5 Available Resources: 1 2 2 2 Process 3 is executing. The process is in a safe state. Available Vector: 5 2 2 5 Process 1 is executing. The process is in a safe state. Available Vector: 7 2 3 6 Process 2 is executing. The process is in a safe state. Available Vector: 7 3 5 7 Process 4 is executing. The process is in a safe state. Available Vector: 7 5 6 7 Process 5 is executing. The process is in a safe state. Available Vector: 8 5 9 7
M2000 Interpreter
<lang M2000 Interpreter> Module BankerAlgo {
Form 80, 44 Cls 5 Pen 14 Function Request(FromWhere as Inventory, What$, Many as long) { =FromWhere(What$)-FromWhere(What$+"_Request")-Many>=0 } Function RequestPreset(FromWhere as Inventory, What$, Many as long) { =FromWhere(What$+"_Request")-Many>=0 } Function Need(FromWhere as Inventory, What$, Many) { =FromWhere(What$ + "_max")-FromWhere(What$)-Many>=0 } Function AllProcessEnd(Where as Inventory) { Def Ok As Boolean=True M=Each(Where) While M { If Instr(Eval$(M, M^),"_")=0 Then Ok = Ok And Where(Eval$(M, M^)+"_max")=Where(Eval$(M, M^)+"_Request") } =Ok } Inventory System \\ Recource, Max, Available ApplyResources(System, "A", 6, 3,"B", 5,1,"C", 7, 1, "D", 6, 2) Function NewProcess { Inventory Process ApplyResources(Process) =Process } \\ Recource, Max, Available Inventory Processes = "P1":=NewProcess("A", 3, 1, "B", 3, 2, "C", 2, 2, "D", 2,1) Append Processes, "P2":=NewProcess("A", 1, 1, "B", 2, 0, "C", 3, 3, "D", 4,3) Append Processes, "P3":=NewProcess("A", 1, 1, "B", 3, 2, "C", 5, 1, "D", 0,0) Status(True) ' show all process, available resource and max SafeState=True Print "Current Status" RequestResource() ' display Safe State RequestResource("P2", "D", 1) ' display Safe State RequestResource("P1", "A", 1, "D", 1) ' display Safe State RequestResource("P1", "C", 1, "D", 1) ' display Too many resources ... RequestResource("P2", "B", 1) ' display Unsafe State RequestResource("P3", "C", 1) ' display Safe State Status() Sub Respond() If SafeState Then { Pen 15 {Print "Safe State"} } Else Pen 13 {Print "Unsafe State"} End Sub Sub RequestResource(ProcessName$="" ) SafeState=True If ProcessName$="" Then CheckNewState(&SafeState) : Respond() : Print :Exit Sub Local pro=Processes(ProcessName$), ResourceName$, many as long ClearAllRequest(pro) Local skip=False While Match("SN") { Read ResourceName$, many Print Format$("Claim {1} for type {0} resource ",ResourceName$, many) If skip Then Continue If Request(System, ResourceName$, many) Then { If Need(pro, ResourceName$, many) Then { Return pro, ResourceName$+"_Request":=many Return System, ResourceName$+"_Request":=-many } Else { Print "Too many Recources "+ResourceName$+" for Process "+ProcessName$ : Skip=True } } Else Print "Too many Recources for System" : Skip=True } If skip Else CheckNewState(&SafeState) : Respond() Print ' just a new line End Sub Sub ApplyResources(Where as Inventory, What$, MaxValue, InitialValue) Repeat { If Not Exist(Where, What$) Then { Append Where, What$:=InitialValue, What$+"_max":=MaxValue, What$+"_Request":=0 } If not Match("SNN") Then Exit Read What$, MaxValue, InitialValue } Always End Sub Sub ClearAllRequest(Where as Inventory) Local M=Each(Where) While M { If Instr(Eval$(M, M^),"_")=0 Then { Return Where, Eval$(M,M^)+"_Request":=0 } } End Sub Sub SetAllRequest(Where as Inventory) Local M=Each(Where) While M { If Instr(Eval$(M, M^),"_")=0 Then { Return Where, Eval$(M,M^)+"_Request":=Where(Eval$(M,M^))+Where(Eval$(M,M^)+"_Request") } } End Sub Sub PrintResources(Where as Inventory) Local M=Each(Where) While M { If Instr(Eval$(M, M^),"_")=0 Then Print Eval$(M, M^)+"="+Eval$(M), } Print Exit Sub Sub PrintEstimatedResources(Where as Inventory) Local M=Each(Where) While M { If Instr(Eval$(M, M^),"_")=0 Then Print Eval$(M, M^)+"="+Str$(Where(Eval$(M, M^)+"_Request"),""), } Print Exit Sub Sub PrintMax(Where as Inventory) Local M=Each(Where) While M { If Instr(Eval$(M, M^),"_max")>0 Then Print LeftPart$(Eval$(M, M^), "_")+"="+Eval$(M), } Print Exit Sub Sub Status(Ok as boolean=False) Print "Total System Resources" PrintMax(System) Print "Available Resources in System" PrintResources(System) If Not Ok Then Exit Sub Local M=Each(Processes) While M { Print "Process "+Eval$(M, M^) PrintResources(Processes(M^!)) ' give index M^ as Key index number (using !) Print "Maximum Resources for "+Eval$(M, M^) PrintMax(Processes(M^!)) } End Sub Sub CheckNewState(&Ok) local M=Each(Processes), M1 Local Z, Recource$ SetAllRequest(System) While M { Z=Processes(M^!) M1=Each(Z) Ok=True While M1 { Recource$=Eval$(M1, M1^) If Instr(Recource$,"_")=0 Then { Ok=Ok And RequestPreset(System, Recource$, Z(Recource$ + "_max") - Z(Recource$)-Z(Recource$ + "_Request")) } } if Ok then { M1=Each(Z) While M1 { Recource$=Eval$(M1, M1^) If Instr(Recource$,"_")=0 Then { Return System, Recource$+"_Request":= System(Recource$+"_Request") + Z(Recource$) + Z(Recource$+"_Request") Return Z, Recource$+"_Request":=0 } } } Else ClearAllRequest(Z) } If Ok Then Ok=AllProcessEnd(System) ClearAllRequest(System) End Sub
} BankerAlgo </lang>
Phix
<lang Phix>sequence max_res = {6, 5, 7, 6},
curr = {{1, 2, 2, 1}, {1, 0, 3, 3}, {1, 2, 1, 0}}, running = repeat(true,length(curr)), max_claim = {{3, 3, 2, 2}, {1, 2, 3, 4}, {1, 3, 5, 0}}, alloc = repeat(0,length(max_res))
integer count = length(curr) for i=1 to count do alloc = sq_add(alloc,curr[i]) end for sequence avl = sq_sub(max_res,alloc)
printf(1,"Available system resources: ") ?max_res printf(1,"Process allocated: ") ?curr printf(1,"Maximum resources: ") ?max_claim printf(1,"Allocated resources: ") ?alloc printf(1,"Available resources: ") ?avl
while count!=0 do bool safe = false for i=1 to length(curr) do if running[i] then bool execute = true for j=1 to length(max_res) do if max_claim[i][j]-curr[i][j] > avl[j] then execute = false exit end if end for if execute then printf(1,"Process%d is executing. ", i) running[i] = false count -= 1 safe = true for j=1 to length(max_res) do avl[j] += curr[i][j] end for exit end if end if end for if not safe then printf(1,"The processes are in an unsafe state.\n"); exit end if printf(1, "Safe state. Available resources: ") ?avl end while</lang>
- Output:
Available system resources: {6,5,7,6} Process allocated: {{1,2,2,1},{1,0,3,3},{1,2,1,0}} Maximum resources: {{3,3,2,2},{1,2,3,4},{1,3,5,0}} Allocated resources: {3,4,6,4} Available resources: {3,1,1,2} Process1 is executing. Safe state. Available resources: {4,3,3,3} Process2 is executing. Safe state. Available resources: {5,3,6,6} Process3 is executing. Safe state. Available resources: {6,5,7,6}
Changing the initial curr[2] to {1, 1, 3, 3}:
Available system resources: {6,5,7,6} Process allocated: {{1,2,2,1},{1,1,3,3},{1,2,1,0}} Maximum resources: {{3,3,2,2},{1,2,3,4},{1,3,5,0}} Allocated resources: {3,5,6,4} Available resources: {3,0,1,2} The processes are in an unsafe state.
Racket
<lang racket>#lang racket/base (require racket/block racket/pretty racket/port racket/vector)
(pretty-print-columns 20) ; make the matrices look a bit more matrixey
(define (bankers-algorithm p r maxres curr maxclaim)
(define running? (make-vector p #t)) (define alloc (for/vector #:length r ((j (in-range r))) (for/sum ((cu_i (in-vector curr))) (vector-ref cu_i j)))) (printf "Allocated resources:~%~a~%" (pretty-format alloc)) (define avl (for/vector #:length r ((m (in-vector maxres)) (a (in-vector alloc))) (- m a))) (printf "Available resources:~%~a~%~%" (pretty-format avl)) (define (safe-exec i mc_i cu_i) (define exec? (for/and ((a (in-vector avl)) (m (in-vector mc_i)) (c (in-vector cu_i))) (<= (- m c) a))) (cond [exec? (printf "Process ~a is executing~%" (add1 i)) (vector-set! running? i #f) (for ((j (in-range r)) (a (in-vector avl)) (c (in-vector cu_i))) (vector-set! avl j (+ a c))) #t] [else #f])) (let loop () (unless (zero? (vector-count values running?)) (define safe? (for/first ((i (in-range p)) (r? (in-vector running?)) (mc_i (in-vector maxclaim)) (cu_i (in-vector curr)) ;; the break condition for this is identical to safe?, so we have no ;; separate break? flag #:when r? #:when (safe-exec i mc_i cu_i)) #t)) (cond [safe? (printf "The process is in a safe state~%~%Available vector: ~a~%" (pretty-format avl)) (loop)] [else (displayln "The processes are in an unsafe state")]))))
(define (bankers-input)
(define ((n-vector? type? dims) x) ;; not the world's most efficient implementation! (cond [(null? dims) (type? x)] [(not (vector? x)) #f] [(not (= (car dims) (vector-length x))) #f] [else (for/and ((e (in-vector x))) (n-vector? type? (cdr dims)) e)])) (define-syntax-rule (prompted-input prompt valid?) (block (printf "Enter ~a:~%" prompt) (define rv (read)) (pretty-print rv) (unless (valid? rv) (raise-argument-error 'prompted-input (format "~a" 'valid?) rv)) rv)) (define p (prompted-input "the number of processes" exact-positive-integer?)) (define r (prompted-input "the number of resources" exact-positive-integer?)) (define maxres (prompted-input "Claim Vector" (n-vector? exact-positive-integer? (list r)))) (define curr (prompted-input "Allocated Resource Table" (n-vector? exact-positive-integer? (list p r)))) (define maxclaim (prompted-input "Maximum Claim Table" (n-vector? exact-positive-integer? (list p r)))) (values p r maxres curr maxclaim))
(module+ main
(with-input-from-string #<<EOS
5 4
- (8 5 9 7)
- (#(2 0 1 1)
#(0 1 2 1) #(4 0 0 3) #(0 2 1 0) #(1 0 3 0))
- (#(3 2 1 4)
#(0 2 5 2) #(5 1 0 5) #(1 5 3 0) #(3 0 3 3))
EOS
(λ () (call-with-values bankers-input bankers-algorithm))))</lang>
- Output:
Enter the number of processes: 5 Enter the number of resources: 4 Enter Claim Vector: '#(8 5 9 7) Enter Allocated Resource Table: '#(#(2 0 1 1) #(0 1 2 1) #(4 0 0 3) #(0 2 1 0) #(1 0 3 0)) Enter Maximum Claim Table: '#(#(3 2 1 4) #(0 2 5 2) #(5 1 0 5) #(1 5 3 0) #(3 0 3 3)) Allocated resources: '#(7 3 7 5) Available resources: '#(1 2 2 2) Process 3 is executing The process is in a safe state Available vector: '#(5 2 2 5) Process 1 is executing The process is in a safe state Available vector: '#(7 2 3 6) Process 2 is executing The process is in a safe state Available vector: '#(7 3 5 7) Process 4 is executing The process is in a safe state Available vector: '#(7 5 6 7) Process 5 is executing The process is in a safe state Available vector: '#(8 5 9 7)
Rust
Adapted from the C language version. It crashes for invalid input.
<lang rust> fn read_numbers<T>() -> Vec<T> where T: std::str::FromStr {
use std::io::Write; std::io::stdout().flush().unwrap();
let mut line = String::new(); std::io::stdin().read_line(&mut line).unwrap(); line.split(" ").map(|word| word.trim().parse::<T>().ok().unwrap()).collect()
}
fn main() {
print!("Enter the number of resources: "); let r = read_numbers()[0]; print!("Enter the number of processes: "); let p = read_numbers()[0]; let mut running = vec![true; p]; let mut count = p; print!("Enter the {}-item claim vector: ", r); let max_res = read_numbers::<u32>();
println!("Enter the {}-line {}-column allocated-resource table:", p, r); let mut curr = vec![vec![0; 0]; p]; for i in 0..p { curr[i] = read_numbers::<u32>(); } println!("Enter the {}-line {}-column maximum-claim table:", p, r); let mut max_claim = vec![vec![0; 0]; p]; for i in 0..p { max_claim[i] = read_numbers::<u32>(); } print!("The claim vector is: "); for i in 0..r { print!("{} ", max_res[i]); } println!();
println!("The allocated resources table is:"); for i in 0..p { for j in 0..r { print!("\t{}", curr[i][j]); } println!(); }
println!("The maximum claims table is:"); for i in 0..p { for j in 0..r { print!("\t{}", max_claim[i][j]); } println!(); } let mut alloc = vec![0; r]; for i in 0..p { for j in 0..r { alloc[j] += curr[i][j]; } } print!("The allocated resources are: "); for i in 0..r { print!("{} ", alloc[i]); } println!(); let mut avl = vec![0; r]; for i in 0..r { avl[i] = max_res[i] - alloc[i]; }
print!("The available resources are: "); for i in 0..r { print!("{} ", avl[i]); } println!();
while count != 0 { let mut safe = false; for i in 0..p { if running[i] { let mut exec = true; for j in 0..r { if max_claim[i][j] - curr[i][j] > avl[j] { exec = false; break; } }
if exec { println!("Process {} is executing.", i + 1); running[i] = false; count -= 1; safe = true; for j in 0..r { avl[j] += curr[i][j]; } break; } } }
if safe { println!("The process is in safe state."); } else { println!("The processes are in unsafe state."); break; }
print!("The available vector is: "); for i in 0..r { print!("{} ", avl[i]); } println!(); }
} </lang>
- Input and Output:
Enter the number of resources: 4 Enter the number of processes: 5 Enter the 4-item claim vector: 8 5 9 7 Enter the 5-line 4-column allocated-resource table: 2 0 1 1 0 1 2 1 4 0 0 3 0 2 1 0 1 0 3 0 Enter the 5-line 4-column maximum-claim table: 3 2 1 4 0 2 5 2 5 1 0 5 1 5 3 0 3 0 3 3 The claim vector is: 8 5 9 7 The allocated resources table is: 2 0 1 1 0 1 2 1 4 0 0 3 0 2 1 0 1 0 3 0 The maximum claims table is: 3 2 1 4 0 2 5 2 5 1 0 5 1 5 3 0 3 0 3 3 The allocated resources are: 7 3 7 5 The available resources are: 1 2 2 2 Process 3 is executing. The process is in safe state. The available vector is: 5 2 2 5 Process 1 is executing. The process is in safe state. The available vector is: 7 2 3 6 Process 2 is executing. The process is in safe state. The available vector is: 7 3 5 7 Process 4 is executing. The process is in safe state. The available vector is: 7 5 6 7 Process 5 is executing. The process is in safe state. The available vector is: 8 5 9 7